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http://rosettacode.org/wiki/Doubly-linked_list/Definition | Doubly-linked list/Definition | Define the data structure for a complete Doubly Linked List.
The structure should support adding elements to the head, tail and middle of the list.
The structure should not allow circular loops
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Java | Java |
import java.util.LinkedList;
public class DoublyLinkedList {
public static void main(String[] args) {
LinkedList<String> list = new LinkedList<String>();
list.addFirst("Add First");
list.addLast("Add Last 1");
list.addLast("Add Last 2");
list.addLast("Add Last 1");
traverseList(list);
list.removeFirstOccurrence("Add Last 1");
traverseList(list);
}
private static void traverseList(LinkedList<String> list) {
System.out.println("Traverse List:");
for ( int i = 0 ; i < list.size() ; i++ ) {
System.out.printf("Element number %d - Element value = '%s'%n", i, list.get(i));
}
System.out.println();
}
}
|
http://rosettacode.org/wiki/Digital_root/Multiplicative_digital_root | Digital root/Multiplicative digital root | The multiplicative digital root (MDR) and multiplicative persistence (MP) of a number,
n
{\displaystyle n}
, is calculated rather like the Digital root except digits are multiplied instead of being added:
Set
m
{\displaystyle m}
to
n
{\displaystyle n}
and
i
{\displaystyle i}
to
0
{\displaystyle 0}
.
While
m
{\displaystyle m}
has more than one digit:
Find a replacement
m
{\displaystyle m}
as the multiplication of the digits of the current value of
m
{\displaystyle m}
.
Increment
i
{\displaystyle i}
.
Return
i
{\displaystyle i}
(= MP) and
m
{\displaystyle m}
(= MDR)
Task
Tabulate the MP and MDR of the numbers 123321, 7739, 893, 899998
Tabulate MDR versus the first five numbers having that MDR, something like:
MDR: [n0..n4]
=== ========
0: [0, 10, 20, 25, 30]
1: [1, 11, 111, 1111, 11111]
2: [2, 12, 21, 26, 34]
3: [3, 13, 31, 113, 131]
4: [4, 14, 22, 27, 39]
5: [5, 15, 35, 51, 53]
6: [6, 16, 23, 28, 32]
7: [7, 17, 71, 117, 171]
8: [8, 18, 24, 29, 36]
9: [9, 19, 33, 91, 119]
Show all output on this page.
Similar
The Product of decimal digits of n page was redirected here, and had the following description
Find the product of the decimal digits of a positive integer n, where n <= 100
The three existing entries for Phix, REXX, and Ring have been moved here, under ===Similar=== headings, feel free to match or ignore them.
References
Multiplicative Digital Root on Wolfram Mathworld.
Multiplicative digital root on The On-Line Encyclopedia of Integer Sequences.
What's special about 277777788888899? - Numberphile video
| #AWK | AWK | # Multiplicative Digital Roots
BEGIN {
printMdrAndMp( 123321 );
printMdrAndMp( 7739 );
printMdrAndMp( 893 );
printMdrAndMp( 899998 );
tabulateMdr( 5 );
} # BEGIN
function printMdrAndMp( n )
{
calculateMdrAndMp( n );
printf( "%6d: MDR: %d, MP: %2d\n", n, MDR, MP );
} # printMdrAndMp
function calculateMdrAndMp( n, mdrStr, digit )
{
MP = 0; # global Multiplicative Persistence
MDR = ( n < 0 ? -n : n ); # global Multiplicative Digital Root
while( MDR > 9 )
{
MP ++;
mdrStr = "" MDR;
MDR = 1;
for( digit = 1; digit <= length( mdrStr ); digit ++ )
{
MDR *= ( substr( mdrStr, digit, 1 ) * 1 );
} # for digit
} # while MDR > 9
} # calculateMdrAndMp
function tabulateMdr( n, rqdValues, valueCount, value, pos )
{
# generate a table of the first n numbers with each possible MDR
rqdValues = n * 10;
valueCount = 0;
for( value = 0; valueCount < rqdValues; value ++ )
{
calculateMdrAndMp( value );
if( mdrCount[ MDR ] < n )
{
# still need another value with this MDR
valueCount ++;
mdrCount[ MDR ] ++;
mdrValues[ MDR ":" mdrCount[ MDR ] ] = value;
} # if mdrCount[ MDR ] < n
} # for value
# print the table
printf( "MDR: [n0..n%d]\n", n - 1 );
printf( "=== ========\n" );
for( pos = 0; pos < 10; pos ++ )
{
printf( "%3d:", pos );
separator = " [";
for( value = 1; value <= n; value ++ )
{
printf( "%s%d", separator, mdrValues[ pos ":" value ] );
separator = ", "
} # for value
printf( "]\n" );
} # for pos
} # tabulateMdr |
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #Scala | Scala | import javax.swing.JFrame
import java.awt.Graphics
class DragonCurve(depth: Int) extends JFrame(s"Dragon Curve (depth $depth)") {
setBounds(100, 100, 800, 600);
setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
val len = 400 / Math.pow(2, depth / 2.0);
val startingAngle = -depth * (Math.PI / 4);
val steps = getSteps(depth).filterNot(c => c == 'X' || c == 'Y')
def getSteps(depth: Int): Stream[Char] = {
if (depth == 0) {
"FX".toStream
} else {
getSteps(depth - 1).flatMap{
case 'X' => "XRYFR"
case 'Y' => "LFXLY"
case c => c.toString
}
}
}
override def paint(g: Graphics): Unit = {
var (x, y) = (230, 350)
var (dx, dy) = ((Math.cos(startingAngle) * len).toInt, (Math.sin(startingAngle) * len).toInt)
for (c <- steps) c match {
case 'F' => {
g.drawLine(x, y, x + dx, y + dy)
x = x + dx
y = y + dy
}
case 'L' => {
val temp = dx
dx = dy
dy = -temp
}
case 'R' => {
val temp = dx
dx = -dy
dy = temp
}
}
}
}
object DragonCurve extends App {
new DragonCurve(14).setVisible(true);
} |
http://rosettacode.org/wiki/Dinesman%27s_multiple-dwelling_problem | Dinesman's multiple-dwelling problem | Task
Solve Dinesman's multiple dwelling problem but in a way that most naturally follows the problem statement given below.
Solutions are allowed (but not required) to parse and interpret the problem text, but should remain flexible and should state what changes to the problem text are allowed. Flexibility and ease of expression are valued.
Examples may be be split into "setup", "problem statement", and "output" sections where the ease and naturalness of stating the problem and getting an answer, as well as the ease and flexibility of modifying the problem are the primary concerns.
Example output should be shown here, as well as any comments on the examples flexibility.
The problem
Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house that contains only five floors.
Baker does not live on the top floor.
Cooper does not live on the bottom floor.
Fletcher does not live on either the top or the bottom floor.
Miller lives on a higher floor than does Cooper.
Smith does not live on a floor adjacent to Fletcher's.
Fletcher does not live on a floor adjacent to Cooper's.
Where does everyone live?
| #11l | 11l | -V
BAKER = 0
COOPER = 1
FLETCHER = 2
MILLER = 3
SMITH = 4
names = [‘Baker’, ‘Cooper’, ‘Fletcher’, ‘Miller’, ‘Smith’]
V floors = Array(1..5)
L
I floors[BAKER] != 5 &
floors[COOPER] != 1 &
floors[FLETCHER] !C (1, 5) &
floors[MILLER] > floors[COOPER] &
abs(floors[SMITH] - floors[FLETCHER]) != 1 &
abs(floors[FLETCHER] - floors[COOPER]) != 1
L(floor) floors
print(names[L.index]‘ lives on floor ’floor)
L.break
I !floors.next_permutation()
print(‘No solution found.’)
L.break |
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #11l | 11l | F sentenceType(s)
I s.empty
R ‘’
[Char] types
L(c) s
I c == ‘?’
types.append(Char(‘Q’))
E I c == ‘!’
types.append(Char(‘E’))
E I c == ‘.’
types.append(Char(‘S’))
I s.last !C ‘?!.’
types.append(Char(‘N’))
R types.join(‘|’)
V s = ‘hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it’
print(sentenceType(s)) |
http://rosettacode.org/wiki/Display_a_linear_combination | Display a linear combination | Task
Display a finite linear combination in an infinite vector basis
(
e
1
,
e
2
,
…
)
{\displaystyle (e_{1},e_{2},\ldots )}
.
Write a function that, when given a finite list of scalars
(
α
1
,
α
2
,
…
)
{\displaystyle (\alpha ^{1},\alpha ^{2},\ldots )}
,
creates a string representing the linear combination
∑
i
α
i
e
i
{\displaystyle \sum _{i}\alpha ^{i}e_{i}}
in an explicit format often used in mathematics, that is:
α
i
1
e
i
1
±
|
α
i
2
|
e
i
2
±
|
α
i
3
|
e
i
3
±
…
{\displaystyle \alpha ^{i_{1}}e_{i_{1}}\pm |\alpha ^{i_{2}}|e_{i_{2}}\pm |\alpha ^{i_{3}}|e_{i_{3}}\pm \ldots }
where
α
i
k
≠
0
{\displaystyle \alpha ^{i_{k}}\neq 0}
The output must comply to the following rules:
don't show null terms, unless the whole combination is null.
e(1) is fine, e(1) + 0*e(3) or e(1) + 0 is wrong.
don't show scalars when they are equal to one or minus one.
e(3) is fine, 1*e(3) is wrong.
don't prefix by a minus sign if it follows a preceding term. Instead you use subtraction.
e(4) - e(5) is fine, e(4) + -e(5) is wrong.
Show here output for the following lists of scalars:
1) 1, 2, 3
2) 0, 1, 2, 3
3) 1, 0, 3, 4
4) 1, 2, 0
5) 0, 0, 0
6) 0
7) 1, 1, 1
8) -1, -1, -1
9) -1, -2, 0, -3
10) -1
| #Vlang | Vlang | import strings
fn linear_combo(c []int) string {
mut sb := strings.new_builder(128)
for i, n in c {
if n == 0 {
continue
}
mut op := ''
match true {
n < 0 && sb.len == 0 {
op = "-"
}
n < 0{
op = " - "
}
n > 0 && sb.len == 0 {
op = ""
}
else{
op = " + "
}
}
mut av := n
if av < 0 {
av = -av
}
mut coeff := "$av*"
if av == 1 {
coeff = ""
}
sb.write_string("$op${coeff}e(${i+1})")
}
if sb.len == 0 {
return "0"
} else {
return sb.str()
}
}
fn main() {
combos := [
[1, 2, 3],
[0, 1, 2, 3],
[1, 0, 3, 4],
[1, 2, 0],
[0, 0, 0],
[0],
[1, 1, 1],
[-1, -1, -1],
[-1, -2, 0, -3],
[-1],
]
for c in combos {
println("${c:-15} -> ${linear_combo(c)}")
}
} |
http://rosettacode.org/wiki/Display_a_linear_combination | Display a linear combination | Task
Display a finite linear combination in an infinite vector basis
(
e
1
,
e
2
,
…
)
{\displaystyle (e_{1},e_{2},\ldots )}
.
Write a function that, when given a finite list of scalars
(
α
1
,
α
2
,
…
)
{\displaystyle (\alpha ^{1},\alpha ^{2},\ldots )}
,
creates a string representing the linear combination
∑
i
α
i
e
i
{\displaystyle \sum _{i}\alpha ^{i}e_{i}}
in an explicit format often used in mathematics, that is:
α
i
1
e
i
1
±
|
α
i
2
|
e
i
2
±
|
α
i
3
|
e
i
3
±
…
{\displaystyle \alpha ^{i_{1}}e_{i_{1}}\pm |\alpha ^{i_{2}}|e_{i_{2}}\pm |\alpha ^{i_{3}}|e_{i_{3}}\pm \ldots }
where
α
i
k
≠
0
{\displaystyle \alpha ^{i_{k}}\neq 0}
The output must comply to the following rules:
don't show null terms, unless the whole combination is null.
e(1) is fine, e(1) + 0*e(3) or e(1) + 0 is wrong.
don't show scalars when they are equal to one or minus one.
e(3) is fine, 1*e(3) is wrong.
don't prefix by a minus sign if it follows a preceding term. Instead you use subtraction.
e(4) - e(5) is fine, e(4) + -e(5) is wrong.
Show here output for the following lists of scalars:
1) 1, 2, 3
2) 0, 1, 2, 3
3) 1, 0, 3, 4
4) 1, 2, 0
5) 0, 0, 0
6) 0
7) 1, 1, 1
8) -1, -1, -1
9) -1, -2, 0, -3
10) -1
| #Wren | Wren | import "/fmt" for Fmt
var linearCombo = Fn.new { |c|
var sb = ""
var i = 0
for (n in c) {
if (n != 0) {
var op = (n < 0 && sb == "") ? "-" :
(n < 0) ? " - " :
(n > 0 && sb == "") ? "" : " + "
var av = n.abs
var coeff = (av == 1) ? "" : "%(av)*"
sb = sb + "%(op)%(coeff)e(%(i + 1))"
}
i = i + 1
}
return (sb == "") ? "0" : sb
}
var combos = [
[1, 2, 3],
[0, 1, 2, 3],
[1, 0, 3, 4],
[1, 2, 0],
[0, 0, 0],
[0],
[1, 1, 1],
[-1, -1, -1],
[-1, -2, 0, -3],
[-1]
]
for (c in combos) {
Fmt.print("$-15s -> $s", c.toString, linearCombo.call(c))
} |
http://rosettacode.org/wiki/Dot_product | Dot product | Task
Create a function/use an in-built function, to compute the dot product, also known as the scalar product of two vectors.
If possible, make the vectors of arbitrary length.
As an example, compute the dot product of the vectors:
[1, 3, -5] and
[4, -2, -1]
If implementing the dot product of two vectors directly:
each vector must be the same length
multiply corresponding terms from each vector
sum the products (to produce the answer)
Related task
Vector products
| #BASIC | BASIC |
100 :
110 REM DOT PRODUCT
120 :
130 REM INITIALIZE VECTORS OF LENGTH N
140 N = 3
150 DIM V1(N): DIM V2(N)
160 FOR I = 1 TO N
170 V1(I) = INT ( RND (1) * 20 - 9.5)
180 V2(I) = INT ( RND (1) * 20 - 9.5)
190 NEXT I
300 :
310 REM CALCULATE THE DOT PRODUCT
320 :
330 FOR I = 1 TO N:DP = DP + V1(I) * V2(I): NEXT I
400 :
410 REM DISPLAY RESULT
420 :
430 PRINT "[";: FOR I = 1 TO N: PRINT " ";V1(I);: NEXT I
440 PRINT "] . [";: FOR I = 1 TO N: PRINT " ";V2(I);: NEXT I
450 PRINT "] = ";DP
|
http://rosettacode.org/wiki/Doubly-linked_list/Definition | Doubly-linked list/Definition | Define the data structure for a complete Doubly Linked List.
The structure should support adding elements to the head, tail and middle of the list.
The structure should not allow circular loops
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #JavaScript | JavaScript | show(DLNode) |
http://rosettacode.org/wiki/Digital_root/Multiplicative_digital_root | Digital root/Multiplicative digital root | The multiplicative digital root (MDR) and multiplicative persistence (MP) of a number,
n
{\displaystyle n}
, is calculated rather like the Digital root except digits are multiplied instead of being added:
Set
m
{\displaystyle m}
to
n
{\displaystyle n}
and
i
{\displaystyle i}
to
0
{\displaystyle 0}
.
While
m
{\displaystyle m}
has more than one digit:
Find a replacement
m
{\displaystyle m}
as the multiplication of the digits of the current value of
m
{\displaystyle m}
.
Increment
i
{\displaystyle i}
.
Return
i
{\displaystyle i}
(= MP) and
m
{\displaystyle m}
(= MDR)
Task
Tabulate the MP and MDR of the numbers 123321, 7739, 893, 899998
Tabulate MDR versus the first five numbers having that MDR, something like:
MDR: [n0..n4]
=== ========
0: [0, 10, 20, 25, 30]
1: [1, 11, 111, 1111, 11111]
2: [2, 12, 21, 26, 34]
3: [3, 13, 31, 113, 131]
4: [4, 14, 22, 27, 39]
5: [5, 15, 35, 51, 53]
6: [6, 16, 23, 28, 32]
7: [7, 17, 71, 117, 171]
8: [8, 18, 24, 29, 36]
9: [9, 19, 33, 91, 119]
Show all output on this page.
Similar
The Product of decimal digits of n page was redirected here, and had the following description
Find the product of the decimal digits of a positive integer n, where n <= 100
The three existing entries for Phix, REXX, and Ring have been moved here, under ===Similar=== headings, feel free to match or ignore them.
References
Multiplicative Digital Root on Wolfram Mathworld.
Multiplicative digital root on The On-Line Encyclopedia of Integer Sequences.
What's special about 277777788888899? - Numberphile video
| #Bracmat | Bracmat | (
& ( MP/MDR
= prod L n
. ( prod
= d
. @(!arg:%@?d ?arg)&!d*prod$!arg
| 1
)
& !arg:?L
& whl
' ( @(!arg:? [>1)
& (prod$!arg:?arg) !L:?L
)
& !L:? [?n
& (!n+-1.!arg)
)
& ( test
= n
. !arg:%?n ?arg
& out$(!n "\t:" MP/MDR$!n)
& test$!arg
|
)
& test$(123321 7739 893 899998)
& 0:?i
& 1:?collecting:?done
& whl
' ( !i+1:?i
& MP/MDR$!i:(?MP.?MDR)
& ( !done:?*(!MDR.)^((?.)+?)*?
| (!MDR.)^(!i.)*!collecting:?collecting
& ( !collecting:?A*(!MDR.)^(?is+[5)*?Z
& !A*!Z:?collecting
& (!MDR.)^!is*!done:?done
|
)
)
& !collecting:~1
)
& whl
' ( !done:(?MDR.)^?is*?done
& put$(!MDR ":")
& whl'(!is:(?i.)+?is&put$(!i " "))
& put$\n
)
); |
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #Scilab | Scilab | n_folds=10
folds=[];
folds=[0 1];
old_folds=[];
vectors=[];
i=[];
for i=2:n_folds+1
curve_length=length(folds);
vectors=folds(1:curve_length-1)-folds(curve_length);
vectors=vectors.*exp(90/180*%i*%pi);
new_folds=folds(curve_length)+vectors;
j=curve_length;
while j>1
folds=[folds new_folds(j-1)]
j=j-1;
end
end
scf(0); clf();
xname("Dragon curve: "+string(n_folds)+" folds")
plot2d(real(folds),imag(folds),5);
set(gca(),"isoview","on");
set(gca(),"axes_visible",["off","off","off"]); |
http://rosettacode.org/wiki/Dinesman%27s_multiple-dwelling_problem | Dinesman's multiple-dwelling problem | Task
Solve Dinesman's multiple dwelling problem but in a way that most naturally follows the problem statement given below.
Solutions are allowed (but not required) to parse and interpret the problem text, but should remain flexible and should state what changes to the problem text are allowed. Flexibility and ease of expression are valued.
Examples may be be split into "setup", "problem statement", and "output" sections where the ease and naturalness of stating the problem and getting an answer, as well as the ease and flexibility of modifying the problem are the primary concerns.
Example output should be shown here, as well as any comments on the examples flexibility.
The problem
Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house that contains only five floors.
Baker does not live on the top floor.
Cooper does not live on the bottom floor.
Fletcher does not live on either the top or the bottom floor.
Miller lives on a higher floor than does Cooper.
Smith does not live on a floor adjacent to Fletcher's.
Fletcher does not live on a floor adjacent to Cooper's.
Where does everyone live?
| #Ada | Ada | with Ada.Text_IO; use Ada.Text_IO;
procedure Dinesman is
subtype Floor is Positive range 1 .. 5;
type People is (Baker, Cooper, Fletcher, Miller, Smith);
type Floors is array (People'Range) of Floor;
type PtFloors is access all Floors;
function Constrained (f : PtFloors) return Boolean is begin
if f (Baker) /= Floor'Last and
f (Cooper) /= Floor'First and
Floor'First < f (Fletcher) and f (Fletcher) < Floor'Last and
f (Miller) > f (Cooper) and
abs (f (Smith) - f (Fletcher)) /= 1 and
abs (f (Fletcher) - f (Cooper)) /= 1
then return True; end if;
return False;
end Constrained;
procedure Solve (list : PtFloors; n : Natural) is
procedure Swap (I : People; J : Natural) is
temp : constant Floor := list (People'Val (J));
begin list (People'Val (J)) := list (I); list (I) := temp;
end Swap;
begin
if n = 1 then
if Constrained (list) then
for p in People'Range loop
Put_Line (p'Img & " on floor " & list (p)'Img);
end loop;
end if;
return;
end if;
for i in People'First .. People'Val (n - 1) loop
Solve (list, n - 1);
if n mod 2 = 1 then Swap (People'First, n - 1);
else Swap (i, n - 1); end if;
end loop;
end Solve;
thefloors : aliased Floors;
begin
for person in People'Range loop
thefloors (person) := People'Pos (person) + Floor'First;
end loop;
Solve (thefloors'Access, Floors'Length);
end Dinesman; |
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #ALGOL_68 | ALGOL 68 | BEGIN # determuine the type of a sentence by looking at the final punctuation #
CHAR exclamation = "E"; # classification codes... #
CHAR question = "Q";
CHAR serious = "S";
CHAR neutral = "N";
# returns the type(s) of the sentence(s) in s - exclamation, question, #
# serious or neutral; if there are multiple sentences #
# the types are separated by | #
PROC classify = ( STRING s )STRING:
BEGIN
STRING result := "";
BOOL pending neutral := FALSE;
FOR s pos FROM LWB s TO UPB s DO
IF pending neutral := FALSE;
CHAR c = s[ s pos ];
c = "?"
THEN result +:= question + "|"
ELIF c = "!"
THEN result +:= exclamation + "|"
ELIF c = "."
THEN result +:= serious + "|"
ELSE pending neutral := TRUE
FI
OD;
IF pending neutral
THEN result +:= neutral + "|"
FI;
# if s was empty, then return an empty string, otherwise remove the final separator #
IF result = "" THEN "" ELSE result[ LWB result : UPB result - 1 ] FI
END # classify # ;
# task test case #
print( ( classify( "hi there, how are you today? I'd like to present to you the washing machine 9001. "
+ "You have been nominated to win one of these! Just make sure you don't break it"
)
, newline
)
)
END |
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #AutoHotkey | AutoHotkey | Sentence := "hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it"
Msgbox, % SentenceType(Sentence)
SentenceType(Sentence) {
Sentence := Trim(Sentence)
Loop, Parse, Sentence, .?!
{
N := (!E && !Q && !S)
, S := (InStr(SubStr(Sentence, InStr(Sentence, A_LoopField)+StrLen(A_LoopField), 3), "."))
, Q := (InStr(SubStr(Sentence, InStr(Sentence, A_LoopField)+StrLen(A_LoopField), 3), "?"))
, E := (InStr(SubStr(Sentence, InStr(Sentence, A_LoopField)+StrLen(A_LoopField), 3), "!"))
, type .= (E) ? ("E|") : ((Q) ? ("Q|") : ((S) ? ("S|") : "N|"))
, D := SubStr(Sentence, InStr(Sentence, A_LoopField)+StrLen(A_LoopField), 3)
}
return (D = SubStr(Sentence, 1, 3)) ? RTrim(RTrim(type, "|"), "N|") : RTrim(type, "|")
} |
http://rosettacode.org/wiki/Display_a_linear_combination | Display a linear combination | Task
Display a finite linear combination in an infinite vector basis
(
e
1
,
e
2
,
…
)
{\displaystyle (e_{1},e_{2},\ldots )}
.
Write a function that, when given a finite list of scalars
(
α
1
,
α
2
,
…
)
{\displaystyle (\alpha ^{1},\alpha ^{2},\ldots )}
,
creates a string representing the linear combination
∑
i
α
i
e
i
{\displaystyle \sum _{i}\alpha ^{i}e_{i}}
in an explicit format often used in mathematics, that is:
α
i
1
e
i
1
±
|
α
i
2
|
e
i
2
±
|
α
i
3
|
e
i
3
±
…
{\displaystyle \alpha ^{i_{1}}e_{i_{1}}\pm |\alpha ^{i_{2}}|e_{i_{2}}\pm |\alpha ^{i_{3}}|e_{i_{3}}\pm \ldots }
where
α
i
k
≠
0
{\displaystyle \alpha ^{i_{k}}\neq 0}
The output must comply to the following rules:
don't show null terms, unless the whole combination is null.
e(1) is fine, e(1) + 0*e(3) or e(1) + 0 is wrong.
don't show scalars when they are equal to one or minus one.
e(3) is fine, 1*e(3) is wrong.
don't prefix by a minus sign if it follows a preceding term. Instead you use subtraction.
e(4) - e(5) is fine, e(4) + -e(5) is wrong.
Show here output for the following lists of scalars:
1) 1, 2, 3
2) 0, 1, 2, 3
3) 1, 0, 3, 4
4) 1, 2, 0
5) 0, 0, 0
6) 0
7) 1, 1, 1
8) -1, -1, -1
9) -1, -2, 0, -3
10) -1
| #zkl | zkl | fcn linearCombination(coeffs){
[1..].zipWith(fcn(n,c){ if(c==0) "" else "%s*e(%s)".fmt(c,n) },coeffs)
.filter().concat("+").replace("+-","-").replace("1*","")
or 0
} |
http://rosettacode.org/wiki/Dot_product | Dot product | Task
Create a function/use an in-built function, to compute the dot product, also known as the scalar product of two vectors.
If possible, make the vectors of arbitrary length.
As an example, compute the dot product of the vectors:
[1, 3, -5] and
[4, -2, -1]
If implementing the dot product of two vectors directly:
each vector must be the same length
multiply corresponding terms from each vector
sum the products (to produce the answer)
Related task
Vector products
| #BASIC256 | BASIC256 | dim zero3d = {0.0, 0.0, 0.0}
dim zero5d = {0.0, 0.0, 0.0, 0.0, 0.0}
dim x = {1.0, 0.0, 0.0}
dim y = {0.0, 1.0, 0.0}
dim z = {0.0, 0.0, 1.0}
dim q = {1.0, 1.0, 3.14159}
dim r = {-1.0, 2.618033989, 3.0}
print " q dot r = "; dot(q, r)
print " zero3d dot zero5d = "; dot(zero3d, zero5d)
print " zero3d dot x = "; dot(zero3d, x)
print " z dot z = "; dot(z, z)
print " y dot z = "; dot(y, z)
end
function dot(a, b)
if a[?] <> b[?] then return "NaN"
dp = 0.0
for i = 0 to a[?]-1
dp += a[i] * b[i]
next i
return dp
end function |
http://rosettacode.org/wiki/Dot_product | Dot product | Task
Create a function/use an in-built function, to compute the dot product, also known as the scalar product of two vectors.
If possible, make the vectors of arbitrary length.
As an example, compute the dot product of the vectors:
[1, 3, -5] and
[4, -2, -1]
If implementing the dot product of two vectors directly:
each vector must be the same length
multiply corresponding terms from each vector
sum the products (to produce the answer)
Related task
Vector products
| #bc | bc | /* Calculate the dot product of two vectors a and b (represented as
* arrays) of size n.
*/
define d(a[], b[], n) {
auto d, i
for (i = 0; i < n; i++) {
d += a[i] * b[i]
}
return(d)
}
a[0] = 1
a[1] = 3
a[2] = -5
b[0] = 4
b[1] = -2
b[2] = -1
d(a[], b[], 3) |
http://rosettacode.org/wiki/Doubly-linked_list/Definition | Doubly-linked list/Definition | Define the data structure for a complete Doubly Linked List.
The structure should support adding elements to the head, tail and middle of the list.
The structure should not allow circular loops
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Julia | Julia | show(DLNode) |
http://rosettacode.org/wiki/Digital_root/Multiplicative_digital_root | Digital root/Multiplicative digital root | The multiplicative digital root (MDR) and multiplicative persistence (MP) of a number,
n
{\displaystyle n}
, is calculated rather like the Digital root except digits are multiplied instead of being added:
Set
m
{\displaystyle m}
to
n
{\displaystyle n}
and
i
{\displaystyle i}
to
0
{\displaystyle 0}
.
While
m
{\displaystyle m}
has more than one digit:
Find a replacement
m
{\displaystyle m}
as the multiplication of the digits of the current value of
m
{\displaystyle m}
.
Increment
i
{\displaystyle i}
.
Return
i
{\displaystyle i}
(= MP) and
m
{\displaystyle m}
(= MDR)
Task
Tabulate the MP and MDR of the numbers 123321, 7739, 893, 899998
Tabulate MDR versus the first five numbers having that MDR, something like:
MDR: [n0..n4]
=== ========
0: [0, 10, 20, 25, 30]
1: [1, 11, 111, 1111, 11111]
2: [2, 12, 21, 26, 34]
3: [3, 13, 31, 113, 131]
4: [4, 14, 22, 27, 39]
5: [5, 15, 35, 51, 53]
6: [6, 16, 23, 28, 32]
7: [7, 17, 71, 117, 171]
8: [8, 18, 24, 29, 36]
9: [9, 19, 33, 91, 119]
Show all output on this page.
Similar
The Product of decimal digits of n page was redirected here, and had the following description
Find the product of the decimal digits of a positive integer n, where n <= 100
The three existing entries for Phix, REXX, and Ring have been moved here, under ===Similar=== headings, feel free to match or ignore them.
References
Multiplicative Digital Root on Wolfram Mathworld.
Multiplicative digital root on The On-Line Encyclopedia of Integer Sequences.
What's special about 277777788888899? - Numberphile video
| #C | C |
#include <stdio.h>
#define twidth 5
#define mdr(rmdr, rmp, n)\
do { *rmp = 0; _mdr(rmdr, rmp, n); } while (0)
void _mdr(int *rmdr, int *rmp, long long n)
{
/* Adjust r if 0 case, so we don't return 1 */
int r = n ? 1 : 0;
while (n) {
r *= (n % 10);
n /= 10;
}
(*rmp)++;
if (r >= 10)
_mdr(rmdr, rmp, r);
else
*rmdr = r;
}
int main(void)
{
int i, j, vmdr, vmp;
const int values[] = { 123321, 7739, 893, 899998 };
const int vsize = sizeof(values) / sizeof(values[0]);
/* Initial test values */
printf("Number MDR MP\n");
for (i = 0; i < vsize; ++i) {
mdr(&vmdr, &vmp, values[i]);
printf("%6d %3d %3d\n", values[i], vmdr, vmp);
}
/* Determine table values */
int table[10][twidth] = { 0 };
int tfill[10] = { 0 };
int total = 0;
for (i = 0; total < 10 * twidth; ++i) {
mdr(&vmdr, &vmp, i);
if (tfill[vmdr] < twidth) {
table[vmdr][tfill[vmdr]++] = i;
total++;
}
}
/* Print calculated table values */
printf("\nMDR: [n0..n4]\n");
for (i = 0; i < 10; ++i) {
printf("%3d: [", i);
for (j = 0; j < twidth; ++j)
printf("%d%s", table[i][j], j != twidth - 1 ? ", " : "");
printf("]\n");
}
return 0;
}
|
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #Seed7 | Seed7 | $ include "seed7_05.s7i";
include "float.s7i";
include "math.s7i";
include "draw.s7i";
include "keybd.s7i";
var float: angle is 0.0;
var integer: x is 220;
var integer: y is 220;
const proc: turn (in integer: degrees) is func
begin
angle +:= flt(degrees) * PI / 180.0
end func;
const proc: forward (in float: length) is func
local
var integer: x2 is 0;
var integer: y2 is 0;
begin
x2 := x + trunc(cos(angle) * length);
y2 := y + trunc(sin(angle) * length);
lineTo(x, y, x2, y2, black);
x := x2;
y := y2;
end func;
const proc: dragon (in float: length, in integer: split, in integer: direct) is func
begin
if split = 0 then
forward(length);
else
turn(direct * 45);
dragon(length/1.4142136, pred(split), 1);
turn(-direct * 90);
dragon(length/1.4142136, pred(split), -1);
turn(direct * 45);
end if;
end func;
const proc: main is func
begin
screen(976, 654);
clear(curr_win, white);
KEYBOARD := GRAPH_KEYBOARD;
dragon(768.0, 14, 1);
ignore(getc(KEYBOARD));
end func; |
http://rosettacode.org/wiki/Dinesman%27s_multiple-dwelling_problem | Dinesman's multiple-dwelling problem | Task
Solve Dinesman's multiple dwelling problem but in a way that most naturally follows the problem statement given below.
Solutions are allowed (but not required) to parse and interpret the problem text, but should remain flexible and should state what changes to the problem text are allowed. Flexibility and ease of expression are valued.
Examples may be be split into "setup", "problem statement", and "output" sections where the ease and naturalness of stating the problem and getting an answer, as well as the ease and flexibility of modifying the problem are the primary concerns.
Example output should be shown here, as well as any comments on the examples flexibility.
The problem
Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house that contains only five floors.
Baker does not live on the top floor.
Cooper does not live on the bottom floor.
Fletcher does not live on either the top or the bottom floor.
Miller lives on a higher floor than does Cooper.
Smith does not live on a floor adjacent to Fletcher's.
Fletcher does not live on a floor adjacent to Cooper's.
Where does everyone live?
| #ALGOL_68 | ALGOL 68 | # attempt to solve the dinesman Multiple Dwelling problem #
# SETUP #
# special floor values #
INT top floor = 4;
INT bottom floor = 0;
# mode to specify the persons floor constraint #
MODE PERSON = STRUCT( STRING name, REF INT floor, PROC( INT )BOOL ok );
# yields TRUE if the floor of the specified person is OK, FALSE otherwise #
OP OK = ( PERSON p )BOOL: ( ok OF p )( floor OF p );
# yields TRUE if floor is adjacent to other persons floor, FALSE otherwise #
PROC adjacent = ( INT floor, other persons floor )BOOL: floor >= ( other persons floor - 1 ) AND floor <= ( other persons floor + 1 );
# displays the floor of an occupant #
PROC print floor = ( PERSON occupant )VOID: print( ( whole( floor OF occupant, -1 ), " ", name OF occupant, newline ) );
# PROBLEM STATEMENT #
# the inhabitants with their floor and constraints #
PERSON baker = ( "Baker", LOC INT := 0, ( INT floor )BOOL: floor /= top floor );
PERSON cooper = ( "Cooper", LOC INT := 0, ( INT floor )BOOL: floor /= bottom floor );
PERSON fletcher = ( "Fletcher", LOC INT := 0, ( INT floor )BOOL: floor /= top floor AND floor /= bottom floor
AND NOT adjacent( floor, floor OF cooper ) );
PERSON miller = ( "Miller", LOC INT := 0, ( INT floor )BOOL: floor > floor OF cooper );
PERSON smith = ( "Smith", LOC INT := 0, ( INT floor )BOOL: NOT adjacent( floor, floor OF fletcher ) );
# SOLUTION #
# "brute force" solution - we run through the possible 5^5 configurations #
# we cold optimise this by e.g. restricting f to bottom floor + 1 TO top floor - 1 #
# at the cost of reducing the flexibility of the constraints #
# alternatively, we could add minimum and maximum allowed floors to the PERSON #
# STRUCT and loop through these instead of bottom floor TO top floor #
FOR b FROM bottom floor TO top floor DO
floor OF baker := b;
FOR c FROM bottom floor TO top floor DO
IF b /= c THEN
floor OF cooper := c;
FOR f FROM bottom floor TO top floor DO
IF b /= f AND c /= f THEN
floor OF fletcher := f;
FOR m FROM bottom floor TO top floor DO
IF b /= m AND c /= m AND f /= m THEN
floor OF miller := m;
FOR s FROM bottom floor TO top floor DO
IF b /= s AND c /= s AND f /= s AND m /= s THEN
floor OF smith := s;
IF OK baker AND OK cooper AND OK fletcher AND OK miller AND OK smith
THEN
# found a solution #
print floor( baker );
print floor( cooper );
print floor( fletcher );
print floor( miller );
print floor( smith )
FI
FI
OD
FI
OD
FI
OD
FI
OD
OD |
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #AWK | AWK |
# syntax: GAWK -f DETERMINE_SENTENCE_TYPE.AWK
BEGIN {
str = "hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it"
main(str)
main("Exclamation! Question? Serious. Neutral")
exit(0)
}
function main(str, c) {
while (length(str) > 0) {
c = substr(str,1,1)
sentence = sentence c
if (c == "!") {
prn("E")
}
else if (c == ".") {
prn("S")
}
else if (c == "?") {
prn("Q")
}
str = substr(str,2)
}
prn("N")
print("")
}
function prn(type) {
gsub(/^ +/,"",sentence)
printf("%s %s\n",type,sentence)
sentence = ""
}
|
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #CLU | CLU | % This iterator takes a string and yields one of 'E', 'Q',
% 'S' or 'N' for every sentence found.
% Because sentences are separated by punctuation, only the
% last one can be 'N'.
sentence_types = iter (s: string) yields (char)
own punct: string := "!?." % relevant character classes
own space: string := " \t\n"
own types: string := "EQS" % sentence type characters
prev_punct: bool := false % whether the previous character was punctuation
last_punct: int := 0 % index of last punctuation character encountered
sentence: bool := true % whether there are words since the last punctuation
for c: char in string$chars(s) do
pu: int := string$indexc(c, punct)
sp: int := string$indexc(c, space)
if pu ~= 0 then
prev_punct := true
last_punct := pu
elseif sp ~= 0 then
if prev_punct then
% a space after punctuation means a sentence has ended here
yield(types[last_punct])
sentence := false
end
prev_punct := false
sentence := false
else
sentence := true
end
end
% handle the last sentence
if prev_punct then yield(types[last_punct])
elseif sentence then yield('N')
end
end sentence_types
% Test
start_up = proc ()
po: stream := stream$primary_output()
test: string :=
"hi there, how are you today? I'd like to " ||
"present to you the washing machine 9001. You " ||
"have been nominated to win one of these! Just " ||
"make sure you don't break it"
% print the type of each sentence
for c: char in sentence_types(test) do
stream$putc(po, c)
end
end start_up |
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Epoxy | Epoxy | const SentenceTypes: {
["?"]:"Q",
["."]:"S",
["!"]:"E"
}
fn DetermineSentenceType(Char)
return SentenceTypes[Char]||"N"
cls
fn GetSentences(Text)
var Sentences: [],
Index: 0,
Length: #Text
loop i:0;i<Length;i+:1 do
var Char: string.subs(Text,i,1)
var Type: DetermineSentenceType(Char)
if Type != "N" || i==Length-1 then
log(string.sub(Text,Index,i+1)+" ("+Type+")")
Index:i+2;
cls
cls
cls
GetSentences("hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it") |
http://rosettacode.org/wiki/Dot_product | Dot product | Task
Create a function/use an in-built function, to compute the dot product, also known as the scalar product of two vectors.
If possible, make the vectors of arbitrary length.
As an example, compute the dot product of the vectors:
[1, 3, -5] and
[4, -2, -1]
If implementing the dot product of two vectors directly:
each vector must be the same length
multiply corresponding terms from each vector
sum the products (to produce the answer)
Related task
Vector products
| #BCPL | BCPL | get "libhdr"
let dotproduct(A, B, len) = valof
$( let acc = 0
for i=0 to len-1 do
acc := acc + A!i * B!i
resultis acc
$)
let start() be
$( let A = table 1, 3, -5
let B = table 4, -2, -1
writef("%N*N", dotproduct(A, B, 3))
$) |
http://rosettacode.org/wiki/Doubly-linked_list/Definition | Doubly-linked list/Definition | Define the data structure for a complete Doubly Linked List.
The structure should support adding elements to the head, tail and middle of the list.
The structure should not allow circular loops
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Kotlin | Kotlin | // version 1.1.2
class LinkedList<E> {
class Node<E>(var data: E, var prev: Node<E>? = null, var next: Node<E>? = null) {
override fun toString(): String {
val sb = StringBuilder(this.data.toString())
var node = this.next
while (node != null) {
sb.append(" -> ", node.data.toString())
node = node.next
}
return sb.toString()
}
}
var first: Node<E>? = null
var last: Node<E>? = null
fun addFirst(value: E) {
if (first == null) {
first = Node(value)
last = first
}
else {
val node = first!!
first = Node(value, null, node)
node.prev = first
}
}
fun addLast(value: E) {
if (last == null) {
last = Node(value)
first = last
}
else {
val node = last!!
last = Node(value, node, null)
node.next = last
}
}
fun insert(after: Node<E>?, value: E) {
if (after == null)
addFirst(value)
else if (after == last)
addLast(value)
else {
val next = after.next
val new = Node(value, after, next)
after.next = new
if (next != null) next.prev = new
}
}
override fun toString() = first.toString()
}
fun main(args: Array<String>) {
val ll = LinkedList<Int>()
ll.addFirst(1)
ll.addLast(4)
ll.insert(ll.first, 2)
ll.insert(ll.last!!.prev, 3)
println(ll)
} |
http://rosettacode.org/wiki/Digital_root/Multiplicative_digital_root | Digital root/Multiplicative digital root | The multiplicative digital root (MDR) and multiplicative persistence (MP) of a number,
n
{\displaystyle n}
, is calculated rather like the Digital root except digits are multiplied instead of being added:
Set
m
{\displaystyle m}
to
n
{\displaystyle n}
and
i
{\displaystyle i}
to
0
{\displaystyle 0}
.
While
m
{\displaystyle m}
has more than one digit:
Find a replacement
m
{\displaystyle m}
as the multiplication of the digits of the current value of
m
{\displaystyle m}
.
Increment
i
{\displaystyle i}
.
Return
i
{\displaystyle i}
(= MP) and
m
{\displaystyle m}
(= MDR)
Task
Tabulate the MP and MDR of the numbers 123321, 7739, 893, 899998
Tabulate MDR versus the first five numbers having that MDR, something like:
MDR: [n0..n4]
=== ========
0: [0, 10, 20, 25, 30]
1: [1, 11, 111, 1111, 11111]
2: [2, 12, 21, 26, 34]
3: [3, 13, 31, 113, 131]
4: [4, 14, 22, 27, 39]
5: [5, 15, 35, 51, 53]
6: [6, 16, 23, 28, 32]
7: [7, 17, 71, 117, 171]
8: [8, 18, 24, 29, 36]
9: [9, 19, 33, 91, 119]
Show all output on this page.
Similar
The Product of decimal digits of n page was redirected here, and had the following description
Find the product of the decimal digits of a positive integer n, where n <= 100
The three existing entries for Phix, REXX, and Ring have been moved here, under ===Similar=== headings, feel free to match or ignore them.
References
Multiplicative Digital Root on Wolfram Mathworld.
Multiplicative digital root on The On-Line Encyclopedia of Integer Sequences.
What's special about 277777788888899? - Numberphile video
| #C.23 | C# | using System;
using System.Collections.Generic;
using System.Linq;
class Program
{
static Tuple<int, int> DigitalRoot(long num)
{
int mp = 0;
while (num > 9)
{
num = num.ToString().ToCharArray().Select(x => x - '0').Aggregate((a, b) => a * b);
mp++;
}
return new Tuple<int, int>(mp, (int)num);
}
static void Main(string[] args)
{
foreach (long num in new long[] { 123321, 7739, 893, 899998 })
{
var t = DigitalRoot(num);
Console.WriteLine("{0} has multiplicative persistence {1} and multiplicative digital root {2}", num, t.Item1, t.Item2);
}
const int twidth = 5;
List<long>[] table = new List<long>[10];
for (int i = 0; i < 10; i++)
table[i] = new List<long>();
long number = -1;
while (table.Any(x => x.Count < twidth))
{
var t = DigitalRoot(++number);
if (table[t.Item2].Count < twidth)
table[t.Item2].Add(number);
}
for (int i = 0; i < 10; i++)
Console.WriteLine(" {0} : [{1}]", i, string.Join(", ", table[i]));
}
} |
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #SequenceL | SequenceL | import <Utilities/Math.sl>;
import <Utilities/Conversion.sl>;
initPoints := [[0,0],[1,0]];
f1(point(1)) :=
let
matrix := [[cos(45 * (pi/180)), -sin(45 * (pi/180))],
[sin(45 * (pi/180)), cos(45 * (pi/180))]];
in
head(transpose((1/sqrt(2)) * matmul(matrix, transpose([point]))));
f2(point(1)) :=
let
matrix := [[cos(135 * (pi/180)), -sin(135 * (pi/180))],
[sin(135 * (pi/180)), cos(135 * (pi/180))]];
in
head(transpose((1/sqrt(2)) * matmul(matrix, transpose([point])))) + initPoints[2];
matmul(X(2),Y(2))[i,j] := sum(X[i,all]*Y[all,j]);
entry(steps(0), maxX(0), maxY(0)) :=
let
scaleX := maxX / 1.5;
scaleY := maxY;
shiftX := maxX / 3.0 / 1.5;
shiftY := maxY / 3.0;
in
round(run(steps, initPoints) * [scaleX, scaleY] + [shiftX, shiftY]);
run(steps(0), result(2)) :=
let
next := f1(result) ++ f2(result);
in
result when steps <= 0
else
run(steps - 1, next); |
http://rosettacode.org/wiki/Dinesman%27s_multiple-dwelling_problem | Dinesman's multiple-dwelling problem | Task
Solve Dinesman's multiple dwelling problem but in a way that most naturally follows the problem statement given below.
Solutions are allowed (but not required) to parse and interpret the problem text, but should remain flexible and should state what changes to the problem text are allowed. Flexibility and ease of expression are valued.
Examples may be be split into "setup", "problem statement", and "output" sections where the ease and naturalness of stating the problem and getting an answer, as well as the ease and flexibility of modifying the problem are the primary concerns.
Example output should be shown here, as well as any comments on the examples flexibility.
The problem
Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house that contains only five floors.
Baker does not live on the top floor.
Cooper does not live on the bottom floor.
Fletcher does not live on either the top or the bottom floor.
Miller lives on a higher floor than does Cooper.
Smith does not live on a floor adjacent to Fletcher's.
Fletcher does not live on a floor adjacent to Cooper's.
Where does everyone live?
| #AutoHotkey | AutoHotkey |
# syntax: GAWK -f DINESMANS_MULTIPLE-DWELLING_PROBLEM.AWK
BEGIN {
for (Baker=1; Baker<=5; Baker++) {
for (Cooper=1; Cooper<=5; Cooper++) {
for (Fletcher=1; Fletcher<=5; Fletcher++) {
for (Miller=1; Miller<=5; Miller++) {
for (Smith=1; Smith<=5; Smith++) {
if (rules() ~ /^1+$/) {
printf("%d Baker\n",Baker)
printf("%d Cooper\n",Cooper)
printf("%d Fletcher\n",Fletcher)
printf("%d Miller\n",Miller)
printf("%d Smith\n",Smith)
}
}
}
}
}
}
exit(0)
}
function rules( stmt1,stmt2,stmt3,stmt4,stmt5,stmt6,stmt7) {
# The following problem statements may be changed:
#
# Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house
# that contains only five floors numbered 1 (ground) to 5 (top)
stmt1 = Baker!=Cooper && Baker!=Fletcher && Baker!=Miller && Baker!=Smith &&
Cooper!=Fletcher && Cooper!=Miller && Cooper!=Smith &&
Fletcher!=Miller && Fletcher!=Smith &&
Miller!=Smith
stmt2 = Baker != 5 # Baker does not live on the top floor
stmt3 = Cooper != 1 # Cooper does not live on the bottom floor
stmt4 = Fletcher != 5 && Fletcher != 1 # Fletcher does not live on either the top or the bottom floor
stmt5 = Miller > Cooper # Miller lives on a higher floor than does Cooper
stmt6 = abs(Smith-Fletcher) != 1 # Smith does not live on a floor adjacent to Fletcher's
stmt7 = abs(Fletcher-Cooper) != 1 # Fletcher does not live on a floor adjacent to Cooper's
return(stmt1 stmt2 stmt3 stmt4 stmt5 stmt6 stmt7)
}
function abs(x) { if (x >= 0) { return x } else { return -x } }
|
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Factor | Factor | USING: combinators io kernel regexp sequences sets splitting
wrap.strings ;
! courtesy of https://www.infoplease.com/common-abbreviations
CONSTANT: common-abbreviations {
"A.B." "abbr." "Acad." "A.D." "alt." "A.M." "Assn."
"at. no." "at. wt." "Aug." "Ave." "b." "B.A." "B.C." "b.p."
"B.S." "c." "Capt." "cent." "co." "Col." "Comdr." "Corp."
"Cpl." "d." "D.C." "Dec." "dept." "dist." "div." "Dr." "ed."
"est." "et al." "Feb." "fl." "gal." "Gen." "Gov." "grad."
"Hon." "i.e." "in." "inc." "Inst." "Jan." "Jr." "lat."
"Lib." "long." "Lt." "Ltd." "M.D." "Mr." "Mrs." "mt." "mts."
"Mus." "no." "Nov." "Oct." "Op." "pl." "pop." "pseud." "pt."
"pub." "Rev." "rev." "R.N." "Sept." "Ser." "Sgt." "Sr."
"St." "uninc." "Univ." "U.S." "vol." "vs." "wt."
}
: sentence-enders ( str -- newstr )
R/ \)/ "" re-replace
" " split harvest
unclip-last swap
[ common-abbreviations member? ] reject
[ last ".!?" member? ] filter
swap suffix ;
: serious? ( str -- ? ) last CHAR: . = ;
: neutral? ( str -- ? ) last ".!?" member? not ;
: mixed? ( str -- ? ) "?!" intersect length 2 = ;
: exclamation? ( str -- ? ) last CHAR: ! = ;
: question? ( str -- ? ) last CHAR: ? = ;
: type ( str -- newstr )
{
{ [ dup serious? ] [ drop "S" ] }
{ [ dup neutral? ] [ drop "N" ] }
{ [ dup mixed? ] [ drop "EQ" ] }
{ [ dup exclamation? ] [ drop "E" ] }
{ [ dup question? ] [ drop "Q" ] }
[ drop "UNKNOWN" ]
} cond ;
: sentences ( str -- newstr )
sentence-enders [ type ] map "|" join ;
: show ( str -- )
dup sentences " -> " glue 60 wrap-string print ;
"Hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it" show
nl
"(There was nary a mouse stirring.) But the cats were going
bonkers!" show
nl
"\"Why is the car so slow?\" she said." show
nl
"Hello, Mr. Anderson!" show
nl
"Are you sure?!?! How can you know?" show |
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #FreeBASIC | FreeBASIC | function sentype( byref s as string ) as string
'determines the sentence type of the first sentence in the string
'returns "E" for an exclamation, "Q" for a question, "S" for serious
'and "N" for neutral.
'modifies the string to remove the first sentence
for i as uinteger = 1 to len(s)
if mid(s, i, 1) = "!" then
s=right(s,len(s)-i)
return "E"
end if
if mid(s, i, 1) = "." then
s=right(s,len(s)-i)
return "S"
end if
if mid(s, i, 1) = "?" then
s=right(s,len(s)-i)
return "Q"
end if
next i
'if we get to the end without encountering punctuation, this
'must be a neutral sentence, which can only happen as the last one
s=""
return "N"
end function
dim as string spam = "hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it"
while len(spam)>0
print sentype(spam)
wend |
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Go | Go | package main
import (
"fmt"
"strings"
)
func sentenceType(s string) string {
if len(s) == 0 {
return ""
}
var types []string
for _, c := range s {
if c == '?' {
types = append(types, "Q")
} else if c == '!' {
types = append(types, "E")
} else if c == '.' {
types = append(types, "S")
}
}
if strings.IndexByte("?!.", s[len(s)-1]) == -1 {
types = append(types, "N")
}
return strings.Join(types, "|")
}
func main() {
s := "hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it"
fmt.Println(sentenceType(s))
} |
http://rosettacode.org/wiki/Dot_product | Dot product | Task
Create a function/use an in-built function, to compute the dot product, also known as the scalar product of two vectors.
If possible, make the vectors of arbitrary length.
As an example, compute the dot product of the vectors:
[1, 3, -5] and
[4, -2, -1]
If implementing the dot product of two vectors directly:
each vector must be the same length
multiply corresponding terms from each vector
sum the products (to produce the answer)
Related task
Vector products
| #Befunge_93 | Befunge 93 |
v Space for variables
v Space for vector1
v Space for vector2
v http://rosettacode.org/wiki/Dot_product
>00pv
>>55+":htgneL",,,,,,,,&:0` |
v,,,,,,,"Length can't be negative."+55<
>,,,,,,,,,,,,,,,,,,,@ |!`-10<
>0.@
v,")".g00,,,,,,,,,,,,,,"Vector a(size " <
0v01g00,")".g00,,,,,,,,,,,,,,"Vector b"<
0pvp2g01&p01-1g01< "
g>> 10g0`| @.g30<(
1 >03g:-03p>00g1-` |s
0 vp00-1g00p30+g30*g2-1g00g1-1g00<i
p > v # z
vp1g01&p01-1g01<> ^ e
> 10g0` | vp01-1g01.g1<
>00g1-10p>10g:01-` | "
> ^
|
http://rosettacode.org/wiki/Doubly-linked_list/Definition | Doubly-linked list/Definition | Define the data structure for a complete Doubly Linked List.
The structure should support adding elements to the head, tail and middle of the list.
The structure should not allow circular loops
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Lua | Lua | -- Doubly Linked List in Lua 6/15/2020 db
-------------------
-- IMPLEMENTATION:
-------------------
local function Node(data)
return { data=data } --implied: return { data=data, prev=nil, next=nil }
end
local List = {
head = nil,
tail = nil,
insertHead = function(self, data)
local node = Node(data)
if (self.head) then
self.head.prev = node
node.next = self.head
self.head = node
else
self.head = node
self.tail = node
end
return node
end,
insertTail = function(self, data)
local node = Node(data)
if self.tail then
self.tail.next = node
node.prev = self.tail
self.tail = node
else
self.head = node
self.tail = node
end
return node
end,
insertBefore = function(self,mark,data)
if (mark) then
local node = Node(data)
if (mark == self.head) then
self.head.next = node
node.next = self.head
self.head = node
else
mark.prev.next = node
node.prev = mark.prev
mark.prev = node
node.next = mark
end
return node
else
-- if no mark given, then insertBefore()==insertHead()
return self:insertHead(data)
end
end,
insertAfter = function(self,mark,data)
if (mark) then
local node = Node(data)
if (mark == self.tail) then
self.tail.next = node
node.prev = self.tail
self.tail = node
else
mark.next.prev = node
node.next = mark.next
mark.next = node
node.prev = mark
end
return node
else
-- if no mark given, then insertAfter()==insertTail()
return self:insertTail(data)
end
end,
values = function(self)
local result, node = {}, self.head
while (node) do result[#result+1], node = node.data, node.next end
return result
end,
}
List.__index = List
setmetatable(List, {__call=function(self) return setmetatable({},self) end })
---------
-- TEST:
---------
local function validate(list, expected)
local values = list:values()
local actual = table.concat(values, ",")
print(actual==expected, actual)
end
local list = List() validate(list, "")
local n1 = list:insertTail(1) validate(list, "1")
local n2 = list:insertTail(2) validate(list, "1,2")
local n3 = list:insertTail(3) validate(list, "1,2,3")
local n4 = list:insertTail(4) validate(list, "1,2,3,4")
local n33 = list:insertAfter(n3, 33) validate(list, "1,2,3,33,4")
local n22 = list:insertAfter(n2, 22) validate(list, "1,2,22,3,33,4")
local n11 = list:insertAfter(n1, 11) validate(list, "1,11,2,22,3,33,4")
local n44 = list:insertAfter(n4, 44) validate(list, "1,11,2,22,3,33,4,44")
local n5 = list:insertTail(5) validate(list, "1,11,2,22,3,33,4,44,5")
local n444 = list:insertBefore(n5, 444) validate(list, "1,11,2,22,3,33,4,44,444,5")
local n55 = list:insertAfter(nil, 55) validate(list, "1,11,2,22,3,33,4,44,444,5,55")
local n0 = list:insertHead(0) validate(list, "0,1,11,2,22,3,33,4,44,444,5,55")
local nm1 = list:insertBefore(nil, -1) validate(list, "-1,0,1,11,2,22,3,33,4,44,444,5,55")
local n111 = list:insertBefore(n2, 111) validate(list, "-1,0,1,11,111,2,22,3,33,4,44,444,5,55")
local n222 = list:insertAfter(n22, 222) validate(list, "-1,0,1,11,111,2,22,222,3,33,4,44,444,5,55")
local n333 = list:insertBefore(n4, 333) validate(list, "-1,0,1,11,111,2,22,222,3,33,333,4,44,444,5,55")
local n555 = list:insertAfter(n55, 555) validate(list, "-1,0,1,11,111,2,22,222,3,33,333,4,44,444,5,55,555") |
http://rosettacode.org/wiki/Discordian_date | Discordian date |
Task
Convert a given date from the Gregorian calendar to the Discordian calendar.
| #8080_Assembly | 8080 Assembly | ;; On the 44th day of Discord in the YOLD 3186, ddate
;; has finally come to CP/M.
bdos: equ 5 ; CP/M syscalls
puts: equ 9
putch: equ 2
fcb: equ 5ch
month: equ fcb + 1 ; use the FCB as the date argument
day: equ month + 2
year: equ day + 2
org 100h
;; CP/M will try to parse the command line arguments as if
;; they are filenames. As luck would have it, MMDDYYYY is
;; 8 characters. It would be a shame not to use this.
lxi h,month ; check that the 'filename' is
lxi b,0800h ; all digits (and zero out C)
argcheck: mov a,m
call isdigit
jnc argerror ; if not, give an error and exit
inx h
dcr b
jnz argcheck
;; Fix the year (add 1166 to it, digit by digit)
lxi d,year + 3
lxi h,yearoffset + 3
mvi b,4
ana a ; Clear carry
yeardgt: ldax d ; Get digit
adc m ; Add offset digit
cpi '9' + 1 ; Did we overshoot?
cmc ; Carry is opposite of what we need
jnc yearnextdgt ; No carry = no adjustment
sui 10 ; Compensate
stc ; Carry the one
yearnextdgt: stax d ; Save digit
dcx d ; Look at more significant digit
dcx h
dcr b ; Until we're out of digits
jnz yeardgt
lxi h,year + 4 ; Terminate the year with a $
mvi m,'$' ; for easy output.
;; Is it St. Tib's Day?
lxi d,month ; Check month and day
lxi h,leap ; Against '0229'
mvi b,4
tibcheck: ldax d ; Get date byte
cmp m ; Match against leap day
jnz notibs ; No match = not Tibs
inx h
inx d
dcr b
jnz tibcheck ; if they all match it _is_ Tibs
;; Print "St. Tib's Day in the YOLD NNNN."
lxi d,tibsday
call outs ; fall through into printyear
;; Print " in the YOLD NNNN."
printyear: lxi d,yold
call outs
lxi d,year
jmp outs ; if Tibs, this ends the program
;; It isn't St. Tib's Day. We'll need to do real work :(
notibs: lxi h,month ; Find days at beginning of month
call parsenum ; (not counting Tibs)
dcr a ; subtract one (table is 0-indexed)
cpi 12 ; check that month < 12
jnc argerror ; otherwise, argument error
ana a ; multiply month by 2
ral ; (entries are 2 bytes wide)
lxi h,monthdays ; get days table
mov d,c ; look up the entry (C is zero here)
mov e,a
dad d
mov e,m ; load the 16-bit entry into DE
inx h
mov d,m
lxi h,day ; Find day number
call parsenum
mov l,a ; Add it to the start of the month
mov h,c ; (C is still zero) - to get the day
dad d ; number (still not counting Tibs)
dcx h ; One less (so we have day numbering at 0)
push h ; Keep day number
lxi d,-365 ; Make sure it isn't >365
dad d
jc argerror ; Give an error otherwise
pop h ; Restore day number
push h ; Keep it around
;; Calculate Erisian weekday
lxi d,-5 ; It's not worth it being clever here
weekcalc: dad d
jc weekcalc
lxi d,5
dad d
mov b,l
lxi h,weekdays ; Print the day of the week
call strselect
lxi d,commaday ; Print ", day "
call outs
;; Calculate season and season day number
pop h ; Restore day number
mvi b,-1 ; B will be season
lxi d,-73 ; L will be day
seasoncalc: inr b ; One season further on
dad d ; Is 73 days less
jc seasoncalc
lxi d,73 ; Correct overshoot
dad d
mov h,b ; H:L = season:day
inr l ; One based for output
push h ; Push season and day
mov a,l ; Print day of season
mvi c,'0'-1 ; Tens digit in C
seasondgts: inr c
sui 10
jnc seasondgts
adi '0' + 10 ; Ones digit in A
mov e,c ; Tens digit
call oute
mov e,a ; Ones digit
call oute
lxi d,of ; Print " of "
call outs
pop b ; Retrieve season:day
push b
lxi h,seasons ; Print the season name
call strselect
call printyear ; "... in the YOLD NNNN ..."
;; Is there any reason to celebrate? (day=5 or day=50)
pop b ; Retrieve season:day
mov a,c ; Day.
cpi 5 ; is it 5?
jz party ; Then we party
cpi 50 ; otherwise, is it 50?
rnz ; If not, we're done
party: push b ; Keep day
lxi d,celebrate ; "... Celebrate ..."
call outs
pop b ; Retrieve day
push b
lxi h,holydays5 ; Get holyday from 5 or 50 table
mov a,c
cpi 50
jnz dayname
lxi h,holydays50
dayname: call strselect ; the season is still in B
lxi d,xday
pop b ; Retrieve day once more
mov a,c
cpi 50
jnz outs ; Print 'day' or 'flux' depending
lxi d,xflux
jmp outs
;; Parse the 2-digit number in HL. Return in B (and A)
parsenum: mvi b,0 ; zero accumulator
call parsedigit ; this runs it twice
parsedigit: mov a,b ; B *= 10
add a
mov b,a
add a
add a
add b
add m ; Add the digit
sui '0' ; Subtract '0'
mov b,a
inx h
ret
;; Print the B'th string from HL
strselect: mvi a,'$'
strsearcho: dcr b
jm strfound
strsearchi: cmp m
inx h
jnz strsearchi
jmp strsearcho
strfound: xchg
jmp outs
;; Print the argument error, and exit
argerror: lxi d,argfmt
;; Print the string in DE and exit
error: call outs
rst 0
;; Returns with carry flag set if A is a digit ('0'-'9').
isdigit: cpi '0'
cmc
rnc
cpi '9' + 1
ret
;; Print the string in D.
outs: mvi c,puts
jmp bdos
;; Print character in E, keeping registers.
oute: push psw
push b
push d
push h
mvi c,putch
call bdos
pop h
pop d
pop b
pop psw
ret
;; Accumulated days at start of Gregorian months
;; (in a non-leap year)
monthdays: dw 0,31,59,90,120,151,181,212,243,273,304,334
;; Difference between Gregorian and Erisian year count
;; (we don't need to bother with the year otherwise)
yearoffset: db 1,1,6,6
;; This is matched to MMDD to handle St. Tib's Day
leap: db '0229'
;; Strings
argfmt: db 'DDATE MMDDYYYY$'
weekdays: db 'Sweetmorn$Boomtime$Pungenday$Prickle-Prickle$'
db 'Setting Orange$'
commaday: db ', day $'
of: db ' of $'
seasons: db 'Chaos$Discord$Confusion$Bureaucracy$The Aftermath$'
celebrate: db ': celebrate $'
holydays5: db 'Mung$Mojo$Sya$Zara$Mala$'
xday: db 'day!$'
holydays50: db 'Chao$Disco$Confu$Bure$Af$'
xflux: db 'flux!$'
tibsday: db 'Saint Tib',39,'s Day$'
yold: db ' in the YOLD $'
|
http://rosettacode.org/wiki/Digital_root/Multiplicative_digital_root | Digital root/Multiplicative digital root | The multiplicative digital root (MDR) and multiplicative persistence (MP) of a number,
n
{\displaystyle n}
, is calculated rather like the Digital root except digits are multiplied instead of being added:
Set
m
{\displaystyle m}
to
n
{\displaystyle n}
and
i
{\displaystyle i}
to
0
{\displaystyle 0}
.
While
m
{\displaystyle m}
has more than one digit:
Find a replacement
m
{\displaystyle m}
as the multiplication of the digits of the current value of
m
{\displaystyle m}
.
Increment
i
{\displaystyle i}
.
Return
i
{\displaystyle i}
(= MP) and
m
{\displaystyle m}
(= MDR)
Task
Tabulate the MP and MDR of the numbers 123321, 7739, 893, 899998
Tabulate MDR versus the first five numbers having that MDR, something like:
MDR: [n0..n4]
=== ========
0: [0, 10, 20, 25, 30]
1: [1, 11, 111, 1111, 11111]
2: [2, 12, 21, 26, 34]
3: [3, 13, 31, 113, 131]
4: [4, 14, 22, 27, 39]
5: [5, 15, 35, 51, 53]
6: [6, 16, 23, 28, 32]
7: [7, 17, 71, 117, 171]
8: [8, 18, 24, 29, 36]
9: [9, 19, 33, 91, 119]
Show all output on this page.
Similar
The Product of decimal digits of n page was redirected here, and had the following description
Find the product of the decimal digits of a positive integer n, where n <= 100
The three existing entries for Phix, REXX, and Ring have been moved here, under ===Similar=== headings, feel free to match or ignore them.
References
Multiplicative Digital Root on Wolfram Mathworld.
Multiplicative digital root on The On-Line Encyclopedia of Integer Sequences.
What's special about 277777788888899? - Numberphile video
| #C.2B.2B | C++ |
#include <iomanip>
#include <map>
#include <vector>
#include <iostream>
using namespace std;
void calcMDR( int n, int c, int& a, int& b )
{
int m = n % 10; n /= 10;
while( n )
{
m *= ( n % 10 );
n /= 10;
}
if( m >= 10 ) calcMDR( m, ++c, a, b );
else { a = m; b = c; }
}
void table()
{
map<int, vector<int> > mp;
int n = 0, a, b;
bool f = true;
while( f )
{
f = false;
calcMDR( n, 1, a, b );
mp[a].push_back( n );
n++;
for( int x = 0; x < 10; x++ )
if( mp[x].size() < 5 )
{ f = true; break; }
}
cout << "| MDR | [n0..n4]\n+-------+------------------------------------+\n";
for( int x = 0; x < 10; x++ )
{
cout << right << "| " << setw( 6 ) << x << "| ";
for( vector<int>::iterator i = mp[x].begin(); i != mp[x].begin() + 5; i++ )
cout << setw( 6 ) << *i << " ";
cout << "|\n";
}
cout << "+-------+------------------------------------+\n\n";
}
int main( int argc, char* argv[] )
{
cout << "| NUMBER | MDR | MP |\n+----------+----------+----------+\n";
int numbers[] = { 123321, 7739, 893, 899998 }, a, b;
for( int x = 0; x < 4; x++ )
{
cout << right << "| " << setw( 9 ) << numbers[x] << "| ";
calcMDR( numbers[x], 1, a, b );
cout << setw( 9 ) << a << "| " << setw( 9 ) << b << "|\n";
}
cout << "+----------+----------+----------+\n\n";
table();
return system( "pause" );
}
|
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #Sidef | Sidef | var rules = Hash(
x => 'x+yF+',
y => '-Fx-y',
)
var lsys = LSystem(
width: 600,
height: 600,
xoff: -430,
yoff: -380,
len: 8,
angle: 90,
color: 'dark green',
)
lsys.execute('Fx', 11, "dragon_curve.png", rules) |
http://rosettacode.org/wiki/Dinesman%27s_multiple-dwelling_problem | Dinesman's multiple-dwelling problem | Task
Solve Dinesman's multiple dwelling problem but in a way that most naturally follows the problem statement given below.
Solutions are allowed (but not required) to parse and interpret the problem text, but should remain flexible and should state what changes to the problem text are allowed. Flexibility and ease of expression are valued.
Examples may be be split into "setup", "problem statement", and "output" sections where the ease and naturalness of stating the problem and getting an answer, as well as the ease and flexibility of modifying the problem are the primary concerns.
Example output should be shown here, as well as any comments on the examples flexibility.
The problem
Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house that contains only five floors.
Baker does not live on the top floor.
Cooper does not live on the bottom floor.
Fletcher does not live on either the top or the bottom floor.
Miller lives on a higher floor than does Cooper.
Smith does not live on a floor adjacent to Fletcher's.
Fletcher does not live on a floor adjacent to Cooper's.
Where does everyone live?
| #AWK | AWK |
# syntax: GAWK -f DINESMANS_MULTIPLE-DWELLING_PROBLEM.AWK
BEGIN {
for (Baker=1; Baker<=5; Baker++) {
for (Cooper=1; Cooper<=5; Cooper++) {
for (Fletcher=1; Fletcher<=5; Fletcher++) {
for (Miller=1; Miller<=5; Miller++) {
for (Smith=1; Smith<=5; Smith++) {
if (rules() ~ /^1+$/) {
printf("%d Baker\n",Baker)
printf("%d Cooper\n",Cooper)
printf("%d Fletcher\n",Fletcher)
printf("%d Miller\n",Miller)
printf("%d Smith\n",Smith)
}
}
}
}
}
}
exit(0)
}
function rules( stmt1,stmt2,stmt3,stmt4,stmt5,stmt6,stmt7) {
# The following problem statements may be changed:
#
# Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house
# that contains only five floors numbered 1 (ground) to 5 (top)
stmt1 = Baker!=Cooper && Baker!=Fletcher && Baker!=Miller && Baker!=Smith &&
Cooper!=Fletcher && Cooper!=Miller && Cooper!=Smith &&
Fletcher!=Miller && Fletcher!=Smith &&
Miller!=Smith
stmt2 = Baker != 5 # Baker does not live on the top floor
stmt3 = Cooper != 1 # Cooper does not live on the bottom floor
stmt4 = Fletcher != 5 && Fletcher != 1 # Fletcher does not live on either the top or the bottom floor
stmt5 = Miller > Cooper # Miller lives on a higher floor than does Cooper
stmt6 = abs(Smith-Fletcher) != 1 # Smith does not live on a floor adjacent to Fletcher's
stmt7 = abs(Fletcher-Cooper) != 1 # Fletcher does not live on a floor adjacent to Cooper's
return(stmt1 stmt2 stmt3 stmt4 stmt5 stmt6 stmt7)
}
function abs(x) { if (x >= 0) { return x } else { return -x } }
|
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #jq | jq |
# Input: a string
# Output: a stream of sentence type indicators
def sentenceTypes:
def trim: sub("^ +";"") | sub(" +$";"");
def parse:
capture("(?<s>[^?!.]*)(?<p>[?!.])(?<remainder>.*)" )
// {p:"", remainder:""};
def encode:
if . == "?" then "Q"
elif . == "!" then "E"
elif . == "." then "S"
else "N"
end;
trim
| select(length>0)
| parse
| (.p | encode), (.remainder | sentenceTypes);
def s: "hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it";
s | sentenceTypes |
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Julia | Julia | const text = """
Hi there, how are you today? I'd like to present to you the washing machine 9001.
You have been nominated to win one of these! Just make sure you don't break it"""
haspunctotype(s) = '.' in s ? "S" : '!' in s ? "E" : '?' in s ? "Q" : "N"
text = replace(text, "\n" => " ")
parsed = strip.(split(text, r"(?:(?:(?<=[\?\!\.])(?:))|(?:(?:)(?=[\?\!\.])))"))
isodd(length(parsed)) && push!(parsed, "") # if ends without pnctuation
for i in 1:2:length(parsed)-1
println(rpad(parsed[i] * parsed[i + 1], 52), " ==> ", haspunctotype(parsed[i + 1]))
end
|
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Lua | Lua | text = "hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it"
p2t = { [""]="N", ["."]="S", ["!"]="E", ["?"]="Q" }
for s, p in text:gmatch("%s*([^%!%?%.]+)([%!%?%.]?)") do
print(s..p..": "..p2t[p])
end |
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Perl | Perl | use strict;
use warnings;
use feature 'say';
use Lingua::Sentence;
my $para1 = <<'EOP';
hi there, how are you today? I'd like to present to you the washing machine
9001. You have been nominated to win one of these! Just make sure you don't
break it
EOP
my $para2 = <<'EOP';
Just because there are punctuation characters like "?", "!" or especially "."
present, it doesn't necessarily mean you have reached the end of a sentence,
does it Mr. Magoo? The syntax highlighting here for Perl isn't bad at all.
EOP
my $splitter = Lingua::Sentence->new("en");
for my $text ($para1, $para2) {
for my $s (split /\n/, $splitter->split( $text =~ s/\n//gr ) {
print "$s| ";
if ($s =~ /!$/) { say 'E' }
elsif ($s =~ /\?$/) { say 'Q' }
elsif ($s =~ /\.$/) { say 'S' }
else { say 'N' }
}
} |
http://rosettacode.org/wiki/Dot_product | Dot product | Task
Create a function/use an in-built function, to compute the dot product, also known as the scalar product of two vectors.
If possible, make the vectors of arbitrary length.
As an example, compute the dot product of the vectors:
[1, 3, -5] and
[4, -2, -1]
If implementing the dot product of two vectors directly:
each vector must be the same length
multiply corresponding terms from each vector
sum the products (to produce the answer)
Related task
Vector products
| #BQN | BQN | •Show 1‿3‿¯5 +´∘× 4‿¯2‿¯1
# as a tacit function
DotP ← +´×
•Show 1‿3‿¯5 DotP 4‿¯2‿¯1 |
http://rosettacode.org/wiki/Doubly-linked_list/Definition | Doubly-linked list/Definition | Define the data structure for a complete Doubly Linked List.
The structure should support adding elements to the head, tail and middle of the list.
The structure should not allow circular loops
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #M2000_Interpreter | M2000 Interpreter |
Module Checkit {
Form 80, 50
Class Null {}
Global Null->Null()
Class Node {
group pred, succ
dat=0
Remove {
Print "destroyed", .dat
}
class:
module Node {
.pred->Null
.succ->Null
if match("N") Then Read .dat
}
}
Class LList {
Group Head, Tail
Module PushTail(k as pointer) {
if .Tail is Null then {
.Head<=k
.Tail<=k
} else {
n=.Tail
.Tail<=k
k=>pred=n=>pred
n=>pred=k
k=>succ=n
}
}
Function RemoveTail {
n=.Tail
if n is .Head then {
.Head->Null
.Tail->Null
} Else {
.Tail<=n=>succ
.Tail=>pred=n=>pred
n=>pred->Null
}
for n {
.succ->Null
.pred->Null
}
=n
}
Module PushHead(k as pointer) {
if .head is Null then {
.Head<=k
.Tail<=k
} else {
n=.head
.head<=k
k=>succ=n=>succ
n=>succ=k
k=>pred=n
}
}
Function RemoveHead {
n=.Head
if n is .Tail then {
.Head->Null
.Tail->Null
} Else {
.Head<=n=>pred
.Head=>succ=n=>succ
n=>succ->Null
}
for n {
.succ->Null
.pred->Null
}
=n
}
Module RemoveNode(k as pointer) {
pred=k=>pred
succ=k=>succ
if pred is succ then {
if .head is k else Error "Can't remove this node"
k=.RemoveHead()
clear k
} else {
pred=>succ=succ
succ=>pred=pred
}
}
Module InsertAfter(k as pointer, n as pointer) {
pred=k=>pred
n=>pred=pred
n=>succ=k
pred=>succ=n
k=>pred=n
}
Function IsEmpty {
= .Head is null or .tail is null
}
class:
Module LList {
.Head->Null
.Tail->Null
}
}
m->Node(100)
L=LList()
L.PushTail m
If not L.Head is Null then Print L.Head=>dat=100
for i=101 to 103 {
m->Node(i)
L.PushTail m
Print "ok....", i
}
for i=104 to 106 {
m->Node(i)
L.PushHead m
Print "ok....", i
}
Print "Use Head to display from last to first"
m=L.Head
do {
Print m=>dat
m=m=>pred
} Until m is null
Print "ok, now find 3rd and remove it"
m1=L.Head
i=1
Index=3
While i<Index {
if m1 is null then exit
m1=m1=>pred
i++
}
If i<>Index then {
Print "List has less than "; Index;" Items"
} Else {
Print "First add one new node"
newNode->Node(1000)
L.InsertAfter m1, newNode
L.RemoveNode m1
clear m1 ' last time m1 used here
newNode=Null
Print "ok.............."
}
Print "Use Tail to display from first to last"
m=L.Tail
do {
Print m=>dat
m=m=>succ
} Until m is null
useother=True
While not L.IsEmpty(){
For This {
\\ we have to use a temporary variable name, here A
A=If(useother->L.RemoveTail(),L.RemoveHead())
? A=>dat
useother~
\\ now we can try to perform removing
clear A
}
}
Print "list is empty:"; L.IsEmpty()
}
Checkit
|
http://rosettacode.org/wiki/Discordian_date | Discordian date |
Task
Convert a given date from the Gregorian calendar to the Discordian calendar.
| #8086_Assembly | 8086 Assembly | ;; DDATE for MS-DOS (assembles using nasm)
bits 16
cpu 8086
tlength: equ 80h
cmdtail: equ 81h
putch: equ 2
puts: equ 9
getdate: equ 2Ah
org 100h
section .text
;; Check if a date was given
mov bl,[tlength] ; Length of command line
and bl,bl ; Is it zero?
jz gettoday ; Then get today's date
;; Parse the date given on the command line
cmp bl,10+1 ; MM/DD/YYYY is ten characters long
jne printusage ; If it doesn't match, print usage.
xor bh,bh
add bx,cmdtail
mov byte [bx],0 ; Zero terminate the date
mov si,cmdtail+1 ; Get month,
call atoi
mov dh,al ; store in DH,
inc si ; get day,
call atoi
mov dl,al ; store in DL,
inc si ; get year,
call atoi
mov cx,ax ; and store in CX.
jmp convertdate
;; Ask MS-DOS for the date
gettoday: mov ah,puts ; Prefix output with 'Today is '
mov dx,todayis
int 21h
mov ah,getdate ; And get the current date.
int 21h
;; Convert the date to the Discordian calendar
;; (DL=day, DH=month, CX=year)
convertdate: add cx,1166 ; Erisian year is 1166 years onwards.
cmp dx,21dh ; Is it St. Tib's Day?
jne notibs
mov dx,tibsday ; If yes, print "St. Tib's Day"
mov ah,puts
int 21h
;; Print "in the YOLD NNNN" and then end.
intheyold: mov dx,yold ; In the YOLD
mov ah,puts
int 21h
mov ax,cx ; print the year
call printn
ret ; If Tibs, this ends the program.
;; It isn't St. Tib's Day.
notibs: cmp dh,1 ; Month < 1 = error
jb printinval
cmp dl,1 ; Day < 1 = error
jb printinval
cmp dh,12 ; Month > 12 = error
ja printinval
mov bx,monthlengths-1
xor ah,ah ; Day higher than it should be
mov al,dh ; (according to month table)
add bx,ax ; = error
cmp dl,[bx]
ja printinval
mov bx,monthdays-2 ; Calculate day of year
mov al,dh ; Cumulative months
sal al,1 ; Multiply by 2 (2 bits per entry)
add bx,ax
mov ax,[bx] ; Days since start of month
xor dh,dh
add dx,ax ; Add day of month
dec dx ; Start at 0 (For array lookup)
mov ax,dx ; Get weekday
mov bl,5 ; Divide by 5
div bl ; AH = weekday (modulo)
mov bl,ah
mov di,weekdays
call strselect ; Print the weekday
mov bx,dx ; Keep day of year in bx for now
mov dx,commaday ; ... ', day '
call printstr
mov dl,73 ; Each season is 73 days long
mov ax,bx ; Divide day of year by 73
div dl ; DH=AH=day of season (modulo)
mov dx,ax ; DL=AL=season number
xor ah,ah ; Print day of season
mov al,dh
inc al ; One more (days do start at 1)
call printn
mov bx,dx ; Store day and season in bx
mov dx,of ; Print ... ' of '
call printstr
mov dx,bx ; Day and season in dx again
mov di,seasons ; Print season
call strselect ; BL is still season number
mov bx,dx ; Day and season in BX again.
call intheyold ; Print the year
cmp bh,5-1 ; Something to celebrate?
je party ; (Day 5 or 50?
cmp bh,50-1
je party
ret ; If not, stop.
party: mov al,bh ; Day in AL.
mov bx,holydays5 ; Day 5 holydays.
mov cx,xday
cmp al,50-1 ; Unless it's 50
jne holyday
mov bx,holydays50 ; Then we need day 50 holydays.
mov cx,xflux
holyday: mov dx,celebrate ; Print ... ', celebrate: '
call printstr
mov dx,bx
call printstr
mov dx,cx
jmp printstr
;; Print "invalid date".
printinval: mov dx,inval
jmp printstr
;; Print usage.
printusage: mov dx,usage
;; Print DX.
printstr: mov ah,puts
int 21h
ret
;; Subroutine: print the BL'th string from DI
strselect: inc bl ; Counter the 'dec bl' later
mov al,'$' ; String end
push cx ; Keep registers
push dx
.search dec bl ; Are we there yet?
jz .found
mov cx,-1
repne scasb ; Scan to end of string
jmp .search
.found mov dx,di ; Found the string, print it
mov ah,puts
int 21h
pop dx ; Restore registers
pop cx
ret
;; Subroutine: print number in AX.
printn: push cx ; Don't trample registers
push dx
mov si,numend ; End of number string.
mov cx,10 ; Divisor, ten.
.loop xor dx,dx ; Zero DX.
dec si ; Back up one digit.
div cx ; Divide AX by ten.
add dl,'0' ; Remainder is now in DX; make ASCII
mov [si],dl ; and store.
and ax,ax ; Quotient is in AX, check if zero
jnz .loop ; loop for next digit if there is one.
mov dx,si ; Print SI - move it to DX,
mov ah,puts ; and then call the DOS print function
int 21h
pop dx ; Restore the registers
pop cx
ret
;; Subroutine: parse number at [SI], store in AX.
atoi: push bx ; Don't trample registers
push cx
push dx
xor ax,ax ; Zero AX.
xor bh,bh ; BH as well.
mov cx,10 ; Multiplier, ten.
.loop mov bl,[si] ; Current number.
sub bl,'0' ; Subtract '0'.
jc .done ; If <0, then done.
cmp bl,9 ; If it's higher than 9,
jnbe .done ; also done.
mul cx ; Multiply accumulator by 10
add ax,bx ; Add the digit in.
inc si ; Next digit
jmp .loop
.done pop dx ; Restore registers
pop cx
pop bx
ret
section .data
;; Accumulated days at start of Gregorian months
monthdays: dw 0,31,59,90,120,151,181,212,243,273,304,334
;; Days per month
monthlengths: db 31,28,31,30,31,30,31,31,30,31,30,31
;; Strings
inval: db 'Invalid date.$'
usage: db `DDATE [MM/DD/YYYY]\r\n`
db `\r\n\tPrint today's date or the given date in the`
db ` Discordian calendar.$`
number: db '00000'
numend: db '$'
todayis: db 'Today is $'
weekdays: db 'Sweetmorn$Boomtime$Pungenday$Prickle-Prickle$'
db 'Setting Orange$'
commaday: db ', day $'
of: db ' of $'
seasons: db 'Chaos$Discord$Confusion$Bureaucracy$The Aftermath$'
celebrate: db ': celebrate $'
holydays5: db 'Mung$Mojo$Sya$Zara$Mala$'
xday: db 'day!$'
holydays50: db 'Chao$Disco$Confu$Bure$Af$'
xflux: db 'flux!$'
tibsday: db "Saint Tib's Day$"
yold: db ' in the YOLD $' |
http://rosettacode.org/wiki/Dijkstra%27s_algorithm | Dijkstra's algorithm | This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Dijkstra's algorithm, conceived by Dutch computer scientist Edsger Dijkstra in 1956 and published in 1959, is a graph search algorithm that solves the single-source shortest path problem for a graph with non-negative edge path costs, producing a shortest path tree.
This algorithm is often used in routing and as a subroutine in other graph algorithms.
For a given source vertex (node) in the graph, the algorithm finds the path with lowest cost (i.e. the shortest path) between that vertex and every other vertex.
For instance
If the vertices of the graph represent cities and edge path costs represent driving distances between pairs of cities connected by a direct road, Dijkstra's algorithm can be used to find the shortest route between one city and all other cities.
As a result, the shortest path first is widely used in network routing protocols, most notably:
IS-IS (Intermediate System to Intermediate System) and
OSPF (Open Shortest Path First).
Important note
The inputs to Dijkstra's algorithm are a directed and weighted graph consisting of 2 or more nodes, generally represented by:
an adjacency matrix or list, and
a start node.
A destination node is not specified.
The output is a set of edges depicting the shortest path to each destination node.
An example, starting with
a──►b, cost=7, lastNode=a
a──►c, cost=9, lastNode=a
a──►d, cost=NA, lastNode=a
a──►e, cost=NA, lastNode=a
a──►f, cost=14, lastNode=a
The lowest cost is a──►b so a──►b is added to the output.
There is a connection from b──►d so the input is updated to:
a──►c, cost=9, lastNode=a
a──►d, cost=22, lastNode=b
a──►e, cost=NA, lastNode=a
a──►f, cost=14, lastNode=a
The lowest cost is a──►c so a──►c is added to the output.
Paths to d and f are cheaper via c so the input is updated to:
a──►d, cost=20, lastNode=c
a──►e, cost=NA, lastNode=a
a──►f, cost=11, lastNode=c
The lowest cost is a──►f so c──►f is added to the output.
The input is updated to:
a──►d, cost=20, lastNode=c
a──►e, cost=NA, lastNode=a
The lowest cost is a──►d so c──►d is added to the output.
There is a connection from d──►e so the input is updated to:
a──►e, cost=26, lastNode=d
Which just leaves adding d──►e to the output.
The output should now be:
[ d──►e
c──►d
c──►f
a──►c
a──►b ]
Task
Implement a version of Dijkstra's algorithm that outputs a set of edges depicting the shortest path to each reachable node from an origin.
Run your program with the following directed graph starting at node a.
Write a program which interprets the output from the above and use it to output the shortest path from node a to nodes e and f.
Vertices
Number
Name
1
a
2
b
3
c
4
d
5
e
6
f
Edges
Start
End
Cost
a
b
7
a
c
9
a
f
14
b
c
10
b
d
15
c
d
11
c
f
2
d
e
6
e
f
9
You can use numbers or names to identify vertices in your program.
See also
Dijkstra's Algorithm vs. A* Search vs. Concurrent Dijkstra's Algorithm (youtube)
| #11l | 11l | T Edge
String start
String end
Int cost
F (start, end, cost)
.start = start
.end = end
.cost = cost
T Graph
[Edge] edges
Set[String] vertices
F (edges)
.edges = edges.map((s, e, c) -> Edge(s, e, c))
.vertices = Set(.edges.map(e -> e.start)).union(Set(.edges.map(e -> e.end)))
F dijkstra(source, dest)
assert(source C .vertices)
V dist = Dict(.vertices, vertex -> (vertex, Float.infinity))
V previous = Dict(.vertices, vertex -> (vertex, ‘’))
dist[source] = 0
V q = copy(.vertices)
V neighbours = Dict(.vertices, vertex -> (vertex, [(String, Int)]()))
L(start, end, cost) .edges
neighbours[start].append((end, cost))
L !q.empty
V u = min(q, key' vertex -> @dist[vertex])
q.remove(u)
I dist[u] == Float.infinity | u == dest
L.break
L(v, cost) neighbours[u]
V alt = dist[u] + cost
I alt < dist[v]
dist[v] = alt
previous[v] = u
Deque[String] s
V u = dest
L previous[u] != ‘’
s.append_left(u)
u = previous[u]
s.append_left(u)
R s
V graph = Graph([(‘a’, ‘b’, 7), (‘a’, ‘c’, 9), (‘a’, ‘f’, 14), (‘b’, ‘c’, 10),
(‘b’, ‘d’, 15), (‘c’, ‘d’, 11), (‘c’, ‘f’, 2), (‘d’, ‘e’, 6),
(‘e’, ‘f’, 9)])
print(graph.dijkstra(‘a’, ‘e’)) |
http://rosettacode.org/wiki/Digital_root | Digital root | The digital root,
X
{\displaystyle X}
, of a number,
n
{\displaystyle n}
, is calculated:
find
X
{\displaystyle X}
as the sum of the digits of
n
{\displaystyle n}
find a new
X
{\displaystyle X}
by summing the digits of
X
{\displaystyle X}
, repeating until
X
{\displaystyle X}
has only one digit.
The additive persistence is the number of summations required to obtain the single digit.
The task is to calculate the additive persistence and the digital root of a number, e.g.:
627615
{\displaystyle 627615}
has additive persistence
2
{\displaystyle 2}
and digital root of
9
{\displaystyle 9}
;
39390
{\displaystyle 39390}
has additive persistence
2
{\displaystyle 2}
and digital root of
6
{\displaystyle 6}
;
588225
{\displaystyle 588225}
has additive persistence
2
{\displaystyle 2}
and digital root of
3
{\displaystyle 3}
;
393900588225
{\displaystyle 393900588225}
has additive persistence
2
{\displaystyle 2}
and digital root of
9
{\displaystyle 9}
;
The digital root may be calculated in bases other than 10.
See
Casting out nines for this wiki's use of this procedure.
Digital root/Multiplicative digital root
Sum digits of an integer
Digital root sequence on OEIS
Additive persistence sequence on OEIS
Iterated digits squaring
| #11l | 11l | F digital_root(=n)
V ap = 0
L n >= 10
n = sum(String(n).map(digit -> Int(digit)))
ap++
R (ap, n)
L(n) [Int64(627615), 39390, 588225, 393900588225, 55]
Int64 persistance, root
(persistance, root) = digital_root(n)
print(‘#12 has additive persistance #2 and digital root #..’.format(n, persistance, root)) |
http://rosettacode.org/wiki/Digital_root/Multiplicative_digital_root | Digital root/Multiplicative digital root | The multiplicative digital root (MDR) and multiplicative persistence (MP) of a number,
n
{\displaystyle n}
, is calculated rather like the Digital root except digits are multiplied instead of being added:
Set
m
{\displaystyle m}
to
n
{\displaystyle n}
and
i
{\displaystyle i}
to
0
{\displaystyle 0}
.
While
m
{\displaystyle m}
has more than one digit:
Find a replacement
m
{\displaystyle m}
as the multiplication of the digits of the current value of
m
{\displaystyle m}
.
Increment
i
{\displaystyle i}
.
Return
i
{\displaystyle i}
(= MP) and
m
{\displaystyle m}
(= MDR)
Task
Tabulate the MP and MDR of the numbers 123321, 7739, 893, 899998
Tabulate MDR versus the first five numbers having that MDR, something like:
MDR: [n0..n4]
=== ========
0: [0, 10, 20, 25, 30]
1: [1, 11, 111, 1111, 11111]
2: [2, 12, 21, 26, 34]
3: [3, 13, 31, 113, 131]
4: [4, 14, 22, 27, 39]
5: [5, 15, 35, 51, 53]
6: [6, 16, 23, 28, 32]
7: [7, 17, 71, 117, 171]
8: [8, 18, 24, 29, 36]
9: [9, 19, 33, 91, 119]
Show all output on this page.
Similar
The Product of decimal digits of n page was redirected here, and had the following description
Find the product of the decimal digits of a positive integer n, where n <= 100
The three existing entries for Phix, REXX, and Ring have been moved here, under ===Similar=== headings, feel free to match or ignore them.
References
Multiplicative Digital Root on Wolfram Mathworld.
Multiplicative digital root on The On-Line Encyclopedia of Integer Sequences.
What's special about 277777788888899? - Numberphile video
| #CLU | CLU | digits = iter (n: int) yields (int)
while n>0 do
yield(n//10)
n := n/10
end
end digits
mdr = proc (n: int) returns (int,int)
i: int := 0
while n>=10 do
m: int := 1
for d: int in digits(n) do
m := m * d
end
n := m
i := i+1
end
return (i,n)
end mdr
first_mdr = iter (target_mdr, n: int) yields (int)
i: int := 0
while n>0 do
x, m: int := mdr(i)
if m=target_mdr then
yield(i)
n := n -1
end
i := i+1
end
end first_mdr
start_up = proc ()
po: stream := stream$primary_output()
nums: sequence[int] := sequence[int]$[123321, 7739, 893, 899998]
stream$putl(po, " N MDR MP")
stream$putl(po, "====== === ==")
for num: int in sequence[int]$elements(nums) do
stream$putright(po, int$unparse(num), 6)
stream$puts(po, " ")
i, m: int := mdr(num)
stream$putright(po, int$unparse(m), 3)
stream$puts(po, " ")
stream$putright(po, int$unparse(i), 3)
stream$putl(po, "")
end
stream$putl(po, "\nMDR: [n0..n4]")
stream$putl(po, "=== ========")
for dgt: int in int$from_to(0,9) do
stream$putright(po, int$unparse(dgt), 3)
stream$puts(po, ": ")
for num: int in first_mdr(dgt, 5) do
stream$puts(po, int$unparse(num) || " ")
end
stream$putl(po, "")
end
end start_up |
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #Smalltalk | Smalltalk | levels = 16
level = 0
step = 1
>
draw(level)
level += step
? level>levels
step = -1
level += step*2
.
? level=0, step = 1
#.delay(1)
<
draw(level)=
mx,my = #.scrsize()
fs = #.min(mx,my)/2
r = fs/2^((level-1)/2)
x = mx/2+fs*#.sqrt(2)/2
y = my/2+fs/4
a = #.pi/4*(level-2)
#.scroff()
#.scrclear()
#.drawline(x,y,x,y)
ss = 2^level-1
> i, 0..ss
? #.and(#.and(i,-i)*2,i)
a += #.pi/2
!
a -= #.pi/2
.
x += r*#.cos(a)
y += r*#.sin(a)
#.drawcolor(#.hsv2rgb(i/(ss+1)*360,1,1):3)
#.drawline(x,y)
<
#.scr()
. |
http://rosettacode.org/wiki/Dinesman%27s_multiple-dwelling_problem | Dinesman's multiple-dwelling problem | Task
Solve Dinesman's multiple dwelling problem but in a way that most naturally follows the problem statement given below.
Solutions are allowed (but not required) to parse and interpret the problem text, but should remain flexible and should state what changes to the problem text are allowed. Flexibility and ease of expression are valued.
Examples may be be split into "setup", "problem statement", and "output" sections where the ease and naturalness of stating the problem and getting an answer, as well as the ease and flexibility of modifying the problem are the primary concerns.
Example output should be shown here, as well as any comments on the examples flexibility.
The problem
Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house that contains only five floors.
Baker does not live on the top floor.
Cooper does not live on the bottom floor.
Fletcher does not live on either the top or the bottom floor.
Miller lives on a higher floor than does Cooper.
Smith does not live on a floor adjacent to Fletcher's.
Fletcher does not live on a floor adjacent to Cooper's.
Where does everyone live?
| #BASIC | BASIC | print "Los apartamentos están numerados del 0 (bajo) al 4 (ático)."
print "Baker, Cooper, Fletcher, Miller y Smith viven en apartamentos diferentes."
print "- Baker no vive en el último apartamento (ático)."
print "- Cooper no vive en el piso inferior (bajo)."
print "- Fletcher no vive ni en el ático ni en el bajo."
print "- Miller vive en un apartamento más alto que Cooper."
print "- Smith no vive en un apartamento adyacente al de Fletcher."
print "- Fletcher no vive en un apartamento adyacente al de Cooper." & chr(10)
for Baker = 0 to 3
for Cooper = 1 to 4
for Fletcher = 1 to 3
for Miller = 0 to 4
for Smith = 0 to 4
if Baker<>Cooper and Baker<>Fletcher and Baker<>Miller and Baker<>Smith and Cooper<>Fletcher and Cooper<>Miller and Cooper<>Smith and Fletcher<>Miller and Fletcher<>Smith and Miller<>Smith and Miller>Cooper and abs(Smith-Fletcher)<>1 and abs(Fletcher-Cooper)<>1 then
print "Baker vive en el piso "; Baker
print "Cooper vive en el piso "; Cooper
print "Fletcher vive en el piso "; Fletcher
print "Miller vive en el piso "; Miller
print "Smith vive en el piso "; Smith
end if
next Smith
next Miller
next Fletcher
next Cooper
next Baker
end |
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Phix | Phix | with javascript_semantics
constant s = `hi there, how are you today? I'd like to present
to you the washing machine 9001. You have been nominated to win
one of these! Just make sure you don't break it`
sequence t = split_any(trim(s),"?!."),
u = substitute_all(s,t,repeat("|",length(t))),
v = substitute_all(u,{"|?","|!","|.","|"},"QESN"),
w = join(v,'|')
?w
|
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Python | Python | import re
txt = """
Hi there, how are you today? I'd like to present to you the washing machine 9001.
You have been nominated to win one of these! Just make sure you don't break it"""
def haspunctotype(s):
return 'S' if '.' in s else 'E' if '!' in s else 'Q' if '?' in s else 'N'
txt = re.sub('\n', '', txt)
pars = [s.strip() for s in re.split("(?:(?:(?<=[\?\!\.])(?:))|(?:(?:)(?=[\?\!\.])))", txt)]
if len(pars) % 2:
pars.append('') # if ends without punctuation
for i in range(0, len(pars)-1, 2):
print((pars[i] + pars[i + 1]).ljust(54), "==>", haspunctotype(pars[i + 1]))
|
http://rosettacode.org/wiki/Dot_product | Dot product | Task
Create a function/use an in-built function, to compute the dot product, also known as the scalar product of two vectors.
If possible, make the vectors of arbitrary length.
As an example, compute the dot product of the vectors:
[1, 3, -5] and
[4, -2, -1]
If implementing the dot product of two vectors directly:
each vector must be the same length
multiply corresponding terms from each vector
sum the products (to produce the answer)
Related task
Vector products
| #Bracmat | Bracmat | ( dot
= a A z Z
. !arg:(%?a ?z.%?A ?Z)
& !a*!A+dot$(!z.!Z)
| 0
)
& out$(dot$(1 3 -5.4 -2 -1)); |
http://rosettacode.org/wiki/Dot_product | Dot product | Task
Create a function/use an in-built function, to compute the dot product, also known as the scalar product of two vectors.
If possible, make the vectors of arbitrary length.
As an example, compute the dot product of the vectors:
[1, 3, -5] and
[4, -2, -1]
If implementing the dot product of two vectors directly:
each vector must be the same length
multiply corresponding terms from each vector
sum the products (to produce the answer)
Related task
Vector products
| #C | C | #include <stdio.h>
#include <stdlib.h>
int dot_product(int *, int *, size_t);
int
main(void)
{
int a[3] = {1, 3, -5};
int b[3] = {4, -2, -1};
printf("%d\n", dot_product(a, b, sizeof(a) / sizeof(a[0])));
return EXIT_SUCCESS;
}
int
dot_product(int *a, int *b, size_t n)
{
int sum = 0;
size_t i;
for (i = 0; i < n; i++) {
sum += a[i] * b[i];
}
return sum;
} |
http://rosettacode.org/wiki/Doubly-linked_list/Definition | Doubly-linked list/Definition | Define the data structure for a complete Doubly Linked List.
The structure should support adding elements to the head, tail and middle of the list.
The structure should not allow circular loops
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Mathematica_.2F_Wolfram_Language | Mathematica / Wolfram Language | ds = CreateDataStructure["DoublyLinkedList"];
ds["Append", #] & /@ {1, 5, 7, 0, 3, 2};
ds["Prepend", 9];
ds["Append", 4];
(* This is adding at the end and then swap to insert in to the middle: *)
ds["Append", 14]; ds["SwapPart", Round[ds["Length"]/2], ds["Length"]];
ds["Elements"] |
http://rosettacode.org/wiki/Dice_game_probabilities | Dice game probabilities | Two players have a set of dice each. The first player has nine dice with four faces each, with numbers one to four. The second player has six normal dice with six faces each, each face has the usual numbers from one to six.
They roll their dice and sum the totals of the faces. The player with the highest total wins (it's a draw if the totals are the same). What's the probability of the first player beating the second player?
Later the two players use a different set of dice each. Now the first player has five dice with ten faces each, and the second player has six dice with seven faces each. Now what's the probability of the first player beating the second player?
This task was adapted from the Project Euler Problem n.205:
https://projecteuler.net/problem=205
| #11l | 11l | F throw_die(n_sides, n_dice, s, [Int] &counts)
I n_dice == 0
counts[s]++
R
L(i) 1..n_sides
throw_die(n_sides, n_dice - 1, s + i, counts)
F beating_probability(n_sides1, n_dice1,
n_sides2, n_dice2)
V len1 = (n_sides1 + 1) * n_dice1
V C1 = [0] * len1
throw_die(n_sides1, n_dice1, 0, &C1)
V len2 = (n_sides2 + 1) * n_dice2
V C2 = [0] * len2
throw_die(n_sides2, n_dice2, 0, &C2)
Float p12 = (n_sides1 ^ n_dice1) * (n_sides2 ^ n_dice2)
V tot = 0.0
L(i) 0 .< len1
L(j) 0 .< min(i, len2)
tot += Float(C1[i]) * C2[j] / p12
R tot
print(‘#.16’.format(beating_probability(4, 9, 6, 6)))
print(‘#.16’.format(beating_probability(10, 5, 7, 6))) |
http://rosettacode.org/wiki/Discordian_date | Discordian date |
Task
Convert a given date from the Gregorian calendar to the Discordian calendar.
| #Ada | Ada | with Ada.Calendar.Arithmetic;
with Ada.Text_IO;
with Ada.Integer_Text_IO;
with Ada.Strings.Unbounded;
with Ada.Strings.Unbounded.Text_IO;
with Ada.Command_Line;
procedure Discordian is
use Ada.Calendar;
use Ada.Strings.Unbounded;
use Ada.Command_Line;
package UStr_IO renames Ada.Strings.Unbounded.Text_IO;
subtype Year_Number is Integer range 3067 .. 3565;
type Seasons is (Chaos, Discord, Confusion, Bureaucracy, The_Aftermath);
type Days_Of_Week is (Sweetmorn, Boomtime, Pungenday,
Prickle_Prickle, Setting_Orange);
subtype Day_Number is Integer range 1 .. 73;
type Discordian_Date is record
Year : Year_Number;
Season : Seasons;
Day : Day_Number;
Week_Day : Days_Of_Week;
Is_Tibs_Day : Boolean := False;
end record;
function Week_Day_To_Str(Day : Days_Of_Week) return Unbounded_String is
s : Unbounded_String;
begin
case Day is
when Sweetmorn => s := To_Unbounded_String("Sweetmorn");
when Boomtime => s := To_Unbounded_String("Boomtime");
when Pungenday => s := To_Unbounded_String("Pungenday");
when Prickle_Prickle => s := To_Unbounded_String("Prickle-Prickle");
when Setting_Orange => s := To_Unbounded_String("Setting Orange");
end case;
return s;
end Week_Day_To_Str;
function Holiday(Season: Seasons) return Unbounded_String is
s : Unbounded_String;
begin
case Season is
when Chaos => s := To_Unbounded_String("Chaoflux");
when Discord => s := To_Unbounded_String("Discoflux");
when Confusion => s := To_Unbounded_String("Confuflux");
when Bureaucracy => s := To_Unbounded_String("Bureflux");
when The_Aftermath => s := To_Unbounded_String("Afflux");
end case;
return s;
end Holiday;
function Apostle(Season: Seasons) return Unbounded_String is
s : Unbounded_String;
begin
case Season is
when Chaos => s := To_Unbounded_String("Mungday");
when Discord => s := To_Unbounded_String("Mojoday");
when Confusion => s := To_Unbounded_String("Syaday");
when Bureaucracy => s := To_Unbounded_String("Zaraday");
when The_Aftermath => s := To_Unbounded_String("Maladay");
end case;
return s;
end Apostle;
function Season_To_Str(Season: Seasons) return Unbounded_String is
s : Unbounded_String;
begin
case Season is
when Chaos => s := To_Unbounded_String("Chaos");
when Discord => s := To_Unbounded_String("Discord");
when Confusion => s := To_Unbounded_String("Confusion");
when Bureaucracy => s := To_Unbounded_String("Bureaucracy");
when The_Aftermath => s := To_Unbounded_String("The Aftermath");
end case;
return s;
end Season_To_Str;
procedure Convert (From : Time; To : out Discordian_Date) is
use Ada.Calendar.Arithmetic;
First_Day : Time;
Number_Days : Day_Count;
Leap_Year : boolean;
begin
First_Day := Time_Of (Year => Year (From), Month => 1, Day => 1);
Number_Days := From - First_Day;
To.Year := Year (Date => From) + 1166;
To.Is_Tibs_Day := False;
Leap_Year := False;
if Year (Date => From) mod 4 = 0 then
if Year (Date => From) mod 100 = 0 then
if Year (Date => From) mod 400 = 0 then
Leap_Year := True;
end if;
else
Leap_Year := True;
end if;
end if;
if Leap_Year then
if Number_Days > 59 then
Number_Days := Number_Days - 1;
elsif Number_Days = 59 then
To.Is_Tibs_Day := True;
end if;
end if;
To.Day := Day_Number (Number_Days mod 73 + 1);
case Number_Days / 73 is
when 0 => To.Season := Chaos;
when 1 => To.Season := Discord;
when 2 => To.Season := Confusion;
when 3 => To.Season := Bureaucracy;
when 4 => To.Season := The_Aftermath;
when others => raise Constraint_Error;
end case;
case Number_Days mod 5 is
when 0 => To.Week_Day := Sweetmorn;
when 1 => To.Week_Day := Boomtime;
when 2 => To.Week_Day := Pungenday;
when 3 => To.Week_Day := Prickle_Prickle;
when 4 => To.Week_Day := Setting_Orange;
when others => raise Constraint_Error;
end case;
end Convert;
procedure Put (Item : Discordian_Date) is
begin
if Item.Is_Tibs_Day then
Ada.Text_IO.Put ("St. Tib's Day");
else
UStr_IO.Put (Week_Day_To_Str(Item.Week_Day));
Ada.Text_IO.Put (", day" & Integer'Image (Item.Day));
Ada.Text_IO.Put (" of ");
UStr_IO.Put (Season_To_Str (Item.Season));
if Item.Day = 5 then
Ada.Text_IO.Put (", ");
UStr_IO.Put (Apostle(Item.Season));
elsif Item.Day = 50 then
Ada.Text_IO.Put (", ");
UStr_IO.Put (Holiday(Item.Season));
end if;
end if;
Ada.Text_IO.Put (" in the YOLD" & Integer'Image (Item.Year));
Ada.Text_IO.New_Line;
end Put;
Test_Day : Time;
Test_DDay : Discordian_Date;
Year : Integer;
Month : Integer;
Day : Integer;
YYYYMMDD : Integer;
begin
if Argument_Count = 0 then
Test_Day := Clock;
Convert (From => Test_Day, To => Test_DDay);
Put (Test_DDay);
end if;
for Arg in 1..Argument_Count loop
if Argument(Arg)'Length < 8 then
Ada.Text_IO.Put("ERROR: Invalid Argument : '" & Argument(Arg) & "'");
Ada.Text_IO.Put("Input format YYYYMMDD");
raise Constraint_Error;
end if;
begin
YYYYMMDD := Integer'Value(Argument(Arg));
exception
when Constraint_Error =>
Ada.Text_IO.Put("ERROR: Invalid Argument : '" & Argument(Arg) & "'");
raise;
end;
Day := YYYYMMDD mod 100;
if Day < Day_Number'First or Day > Day_Number'Last then
Ada.Text_IO.Put("ERROR: Invalid Day:" & Integer'Image(Day));
raise Constraint_Error;
end if;
Month := ((YYYYMMDD - Day) / 100) mod 100;
if Month < Month_Number'First or Month > Month_Number'Last then
Ada.Text_IO.Put("ERROR: Invalid Month:" & Integer'Image(Month));
raise Constraint_Error;
end if;
Year := ((YYYYMMDD - Day - Month * 100) / 10000);
if Year < 1901 or Year > 2399 then
Ada.Text_IO.Put("ERROR: Invalid Year:" & Integer'Image(Year));
raise Constraint_Error;
end if;
Test_Day := Time_Of (Year => Year, Month => Month, Day => Day);
Convert (From => Test_Day, To => Test_DDay);
Put (Test_DDay);
end loop;
end Discordian;
|
http://rosettacode.org/wiki/Dijkstra%27s_algorithm | Dijkstra's algorithm | This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Dijkstra's algorithm, conceived by Dutch computer scientist Edsger Dijkstra in 1956 and published in 1959, is a graph search algorithm that solves the single-source shortest path problem for a graph with non-negative edge path costs, producing a shortest path tree.
This algorithm is often used in routing and as a subroutine in other graph algorithms.
For a given source vertex (node) in the graph, the algorithm finds the path with lowest cost (i.e. the shortest path) between that vertex and every other vertex.
For instance
If the vertices of the graph represent cities and edge path costs represent driving distances between pairs of cities connected by a direct road, Dijkstra's algorithm can be used to find the shortest route between one city and all other cities.
As a result, the shortest path first is widely used in network routing protocols, most notably:
IS-IS (Intermediate System to Intermediate System) and
OSPF (Open Shortest Path First).
Important note
The inputs to Dijkstra's algorithm are a directed and weighted graph consisting of 2 or more nodes, generally represented by:
an adjacency matrix or list, and
a start node.
A destination node is not specified.
The output is a set of edges depicting the shortest path to each destination node.
An example, starting with
a──►b, cost=7, lastNode=a
a──►c, cost=9, lastNode=a
a──►d, cost=NA, lastNode=a
a──►e, cost=NA, lastNode=a
a──►f, cost=14, lastNode=a
The lowest cost is a──►b so a──►b is added to the output.
There is a connection from b──►d so the input is updated to:
a──►c, cost=9, lastNode=a
a──►d, cost=22, lastNode=b
a──►e, cost=NA, lastNode=a
a──►f, cost=14, lastNode=a
The lowest cost is a──►c so a──►c is added to the output.
Paths to d and f are cheaper via c so the input is updated to:
a──►d, cost=20, lastNode=c
a──►e, cost=NA, lastNode=a
a──►f, cost=11, lastNode=c
The lowest cost is a──►f so c──►f is added to the output.
The input is updated to:
a──►d, cost=20, lastNode=c
a──►e, cost=NA, lastNode=a
The lowest cost is a──►d so c──►d is added to the output.
There is a connection from d──►e so the input is updated to:
a──►e, cost=26, lastNode=d
Which just leaves adding d──►e to the output.
The output should now be:
[ d──►e
c──►d
c──►f
a──►c
a──►b ]
Task
Implement a version of Dijkstra's algorithm that outputs a set of edges depicting the shortest path to each reachable node from an origin.
Run your program with the following directed graph starting at node a.
Write a program which interprets the output from the above and use it to output the shortest path from node a to nodes e and f.
Vertices
Number
Name
1
a
2
b
3
c
4
d
5
e
6
f
Edges
Start
End
Cost
a
b
7
a
c
9
a
f
14
b
c
10
b
d
15
c
d
11
c
f
2
d
e
6
e
f
9
You can use numbers or names to identify vertices in your program.
See also
Dijkstra's Algorithm vs. A* Search vs. Concurrent Dijkstra's Algorithm (youtube)
| #Ada | Ada | private with Ada.Containers.Ordered_Maps;
generic
type t_Vertex is (<>);
package Dijkstra is
type t_Graph is limited private;
-- Defining a graph (since limited private, only way to do this is to use the Build function)
type t_Edge is record
From, To : t_Vertex;
Weight : Positive;
end record;
type t_Edges is array (Integer range <>) of t_Edge;
function Build (Edges : in t_Edges; Oriented : in Boolean := True) return t_Graph;
-- Computing path and distance
type t_Path is array (Integer range <>) of t_Vertex;
function Shortest_Path (Graph : in out t_Graph;
From, To : in t_Vertex) return t_Path;
function Distance (Graph : in out t_Graph;
From, To : in t_Vertex) return Natural;
private
package Neighbor_Lists is new Ada.Containers.Ordered_Maps (Key_Type => t_Vertex, Element_Type => Positive);
type t_Vertex_Data is record
Neighbors : Neighbor_Lists.Map; -- won't be affected after build
-- Updated each time a function is called with a new source
Previous : t_Vertex;
Distance : Natural;
end record;
type t_Graph is array (t_Vertex) of t_Vertex_Data;
end Dijkstra; |
http://rosettacode.org/wiki/Digital_root | Digital root | The digital root,
X
{\displaystyle X}
, of a number,
n
{\displaystyle n}
, is calculated:
find
X
{\displaystyle X}
as the sum of the digits of
n
{\displaystyle n}
find a new
X
{\displaystyle X}
by summing the digits of
X
{\displaystyle X}
, repeating until
X
{\displaystyle X}
has only one digit.
The additive persistence is the number of summations required to obtain the single digit.
The task is to calculate the additive persistence and the digital root of a number, e.g.:
627615
{\displaystyle 627615}
has additive persistence
2
{\displaystyle 2}
and digital root of
9
{\displaystyle 9}
;
39390
{\displaystyle 39390}
has additive persistence
2
{\displaystyle 2}
and digital root of
6
{\displaystyle 6}
;
588225
{\displaystyle 588225}
has additive persistence
2
{\displaystyle 2}
and digital root of
3
{\displaystyle 3}
;
393900588225
{\displaystyle 393900588225}
has additive persistence
2
{\displaystyle 2}
and digital root of
9
{\displaystyle 9}
;
The digital root may be calculated in bases other than 10.
See
Casting out nines for this wiki's use of this procedure.
Digital root/Multiplicative digital root
Sum digits of an integer
Digital root sequence on OEIS
Additive persistence sequence on OEIS
Iterated digits squaring
| #360_Assembly | 360 Assembly | * Digital root 21/04/2017
DIGROOT CSECT
USING DIGROOT,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
STM R14,R12,12(R13) save previous context
ST R13,4(R15) link backward
ST R15,8(R13) link forward
LR R13,R15 set addressability
LA R6,1 i=1
DO WHILE=(C,R6,LE,=A((PG-T)/4)) do i=1 to hbound(t)
LR R1,R6 i
SLA R1,2 *4
L R10,T-4(R1) nn=t(i)
LR R7,R10 n=nn
SR R9,R9 ap=0
DO WHILE=(C,R7,GE,=A(10)) do while(n>=10)
SR R8,R8 x=0
DO WHILE=(C,R7,GE,=A(10)) do while(n>=10)
LR R4,R7 n
SRDA R4,32 >>r5
D R4,=A(10) m=n//10
LR R7,R5 n=n/10
AR R8,R4 x=x+m
ENDDO , end
AR R7,R8 n=x+n
LA R9,1(R9) ap=ap+1
ENDDO , end
XDECO R10,XDEC nn
MVC PG+7(10),XDEC+2
XDECO R9,XDEC ap
MVC PG+31(3),XDEC+9
XDECO R7,XDEC n
MVC PG+41(1),XDEC+11
XPRNT PG,L'PG print
LA R6,1(R6) i++
ENDDO , enddo i
L R13,4(0,R13) restore previous savearea pointer
LM R14,R12,12(R13) restore previous context
XR R15,R15 rc=0
BR R14 exit
T DC F'627615',F'39390',F'588225',F'2147483647'
PG DC CL80'number=xxxxxxxxxx persistence=xxx root=x'
XDEC DS CL12
YREGS
END DIGROOT |
http://rosettacode.org/wiki/Digital_root | Digital root | The digital root,
X
{\displaystyle X}
, of a number,
n
{\displaystyle n}
, is calculated:
find
X
{\displaystyle X}
as the sum of the digits of
n
{\displaystyle n}
find a new
X
{\displaystyle X}
by summing the digits of
X
{\displaystyle X}
, repeating until
X
{\displaystyle X}
has only one digit.
The additive persistence is the number of summations required to obtain the single digit.
The task is to calculate the additive persistence and the digital root of a number, e.g.:
627615
{\displaystyle 627615}
has additive persistence
2
{\displaystyle 2}
and digital root of
9
{\displaystyle 9}
;
39390
{\displaystyle 39390}
has additive persistence
2
{\displaystyle 2}
and digital root of
6
{\displaystyle 6}
;
588225
{\displaystyle 588225}
has additive persistence
2
{\displaystyle 2}
and digital root of
3
{\displaystyle 3}
;
393900588225
{\displaystyle 393900588225}
has additive persistence
2
{\displaystyle 2}
and digital root of
9
{\displaystyle 9}
;
The digital root may be calculated in bases other than 10.
See
Casting out nines for this wiki's use of this procedure.
Digital root/Multiplicative digital root
Sum digits of an integer
Digital root sequence on OEIS
Additive persistence sequence on OEIS
Iterated digits squaring
| #Ada | Ada | package Generic_Root is
type Number is range 0 .. 2**63-1;
type Number_Array is array(Positive range <>) of Number;
type Base_Type is range 2 .. 16; -- any reasonable base to write down numb
generic
with function "&"(X, Y: Number) return Number;
-- instantiate with "+" for additive digital roots
-- instantiate with "*" for multiplicative digital roots
procedure Compute_Root(N: Number;
Root, Persistence: out Number;
Base: Base_Type := 10);
-- computes Root and Persistence of N;
end Generic_Root; |
http://rosettacode.org/wiki/Digital_root/Multiplicative_digital_root | Digital root/Multiplicative digital root | The multiplicative digital root (MDR) and multiplicative persistence (MP) of a number,
n
{\displaystyle n}
, is calculated rather like the Digital root except digits are multiplied instead of being added:
Set
m
{\displaystyle m}
to
n
{\displaystyle n}
and
i
{\displaystyle i}
to
0
{\displaystyle 0}
.
While
m
{\displaystyle m}
has more than one digit:
Find a replacement
m
{\displaystyle m}
as the multiplication of the digits of the current value of
m
{\displaystyle m}
.
Increment
i
{\displaystyle i}
.
Return
i
{\displaystyle i}
(= MP) and
m
{\displaystyle m}
(= MDR)
Task
Tabulate the MP and MDR of the numbers 123321, 7739, 893, 899998
Tabulate MDR versus the first five numbers having that MDR, something like:
MDR: [n0..n4]
=== ========
0: [0, 10, 20, 25, 30]
1: [1, 11, 111, 1111, 11111]
2: [2, 12, 21, 26, 34]
3: [3, 13, 31, 113, 131]
4: [4, 14, 22, 27, 39]
5: [5, 15, 35, 51, 53]
6: [6, 16, 23, 28, 32]
7: [7, 17, 71, 117, 171]
8: [8, 18, 24, 29, 36]
9: [9, 19, 33, 91, 119]
Show all output on this page.
Similar
The Product of decimal digits of n page was redirected here, and had the following description
Find the product of the decimal digits of a positive integer n, where n <= 100
The three existing entries for Phix, REXX, and Ring have been moved here, under ===Similar=== headings, feel free to match or ignore them.
References
Multiplicative Digital Root on Wolfram Mathworld.
Multiplicative digital root on The On-Line Encyclopedia of Integer Sequences.
What's special about 277777788888899? - Numberphile video
| #Common_Lisp | Common Lisp |
(defun mdr/p (n)
"Return a list with MDR and MP of n"
(if (< n 10)
(list n 0)
(mdr/p-aux n 1 1)))
(defun mdr/p-aux (n a c)
(cond ((and (zerop n) (< a 10)) (list a c))
((zerop n) (mdr/p-aux a 1 (+ c 1)))
(t (mdr/p-aux (floor n 10) (* (rem n 10) a) c))))
(defun first-n-number-for-each-root (n &optional (r 0) (lst nil) (c 0))
"Return the first m number with MDR = 0 to 9"
(cond ((and (= (length lst) n) (= r 9)) (format t "~3@a: ~a~%" r (reverse lst)))
((= (length lst) n) (format t "~3@a: ~a~%" r (reverse lst))
(first-n-number-for-each-root n (+ r 1) nil 0))
((= (first (mdr/p c)) r) (first-n-number-for-each-root n r (cons c lst) (+ c 1)))
(t (first-n-number-for-each-root n r lst (+ c 1)))))
(defun start ()
(format t "Number: MDR MD~%")
(loop for el in '(123321 7739 893 899998)
do (format t "~6@a: ~{~3@a ~}~%" el (mdr/p el)))
(format t "~%MDR: [n0..n4]~%")
(first-n-number-for-each-root 5)) |
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #SPL | SPL | levels = 16
level = 0
step = 1
>
draw(level)
level += step
? level>levels
step = -1
level += step*2
.
? level=0, step = 1
#.delay(1)
<
draw(level)=
mx,my = #.scrsize()
fs = #.min(mx,my)/2
r = fs/2^((level-1)/2)
x = mx/2+fs*#.sqrt(2)/2
y = my/2+fs/4
a = #.pi/4*(level-2)
#.scroff()
#.scrclear()
#.drawline(x,y,x,y)
ss = 2^level-1
> i, 0..ss
? #.and(#.and(i,-i)*2,i)
a += #.pi/2
!
a -= #.pi/2
.
x += r*#.cos(a)
y += r*#.sin(a)
#.drawcolor(#.hsv2rgb(i/(ss+1)*360,1,1):3)
#.drawline(x,y)
<
#.scr()
. |
http://rosettacode.org/wiki/Dinesman%27s_multiple-dwelling_problem | Dinesman's multiple-dwelling problem | Task
Solve Dinesman's multiple dwelling problem but in a way that most naturally follows the problem statement given below.
Solutions are allowed (but not required) to parse and interpret the problem text, but should remain flexible and should state what changes to the problem text are allowed. Flexibility and ease of expression are valued.
Examples may be be split into "setup", "problem statement", and "output" sections where the ease and naturalness of stating the problem and getting an answer, as well as the ease and flexibility of modifying the problem are the primary concerns.
Example output should be shown here, as well as any comments on the examples flexibility.
The problem
Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house that contains only five floors.
Baker does not live on the top floor.
Cooper does not live on the bottom floor.
Fletcher does not live on either the top or the bottom floor.
Miller lives on a higher floor than does Cooper.
Smith does not live on a floor adjacent to Fletcher's.
Fletcher does not live on a floor adjacent to Cooper's.
Where does everyone live?
| #Bracmat | Bracmat | ( Baker Cooper Fletcher Miller Smith:?people
& ( constraints
=
. !arg
: ~(? Baker)
: ~(Cooper ?)
: ~(Fletcher ?|? Fletcher)
: ? Cooper ? Miller ?
: ~(? Smith Fletcher ?|? Fletcher Smith ?)
: ~(? Cooper Fletcher ?|? Fletcher Cooper ?)
)
& ( solution
= floors persons A Z person
. !arg:(?floors.?persons)
& ( !persons:
& constraints$!floors
& out$("Inhabitants, from bottom to top:" !floors)
& ~ { The ~ always fails on evaluation. Here, failure forces Bracmat to backtrack and find all solutions, not just the first one. }
| !persons
: ?A
%?`person
(?Z&solution$(!floors !person.!A !Z))
)
)
& solution$(.!people)
| { After outputting all solutions, the lhs of the | operator fails. The rhs of the | operator, here an empty string, is the final result. }
); |
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Raku | Raku | use Lingua::EN::Sentence;
my $paragraph = q:to/PARAGRAPH/;
hi there, how are you today? I'd like to present to you the washing machine
9001. You have been nominated to win one of these! Just make sure you don't
break it
Just because there are punctuation characters like "?", "!" or especially "."
present, it doesn't necessarily mean you have reached the end of a sentence,
does it Mr. Magoo? The syntax highlighting here for Raku isn't the best.
PARAGRAPH
say join "\n\n", $paragraph.&get_sentences.map: {
/(<:punct>)$/;
$_ ~ ' | ' ~ do
given $0 {
when '!' { 'E' };
when '?' { 'Q' };
when '.' { 'S' };
default { 'N' };
}
} |
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Vlang | Vlang | fn sentence_type(s string) string {
if s.len == 0 {
return ""
}
mut types := []string{}
for c in s.split('') {
if c == '?' {
types << "Q"
} else if c == '!' {
types << "E"
} else if c == '.' {
types << "S"
}
}
if s[s.len-1..s.len].index_any('?!.') == -1 {
types << "N"
}
return types.join("|")
}
fn main() {
s := "hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it"
println(sentence_type(s))
} |
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Wren | Wren | var sentenceType = Fn.new { |s|
if (s.count == 0) return ""
var types = []
for (c in s) {
if (c == "?") {
types.add("Q")
} else if (c == "!") {
types.add("E")
} else if (c == ".") {
types.add("S")
}
}
if (!"?!.".contains(s[-1])) types.add("N")
return types.join("|")
}
var s = "hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it"
System.print(sentenceType.call(s)) |
http://rosettacode.org/wiki/Dot_product | Dot product | Task
Create a function/use an in-built function, to compute the dot product, also known as the scalar product of two vectors.
If possible, make the vectors of arbitrary length.
As an example, compute the dot product of the vectors:
[1, 3, -5] and
[4, -2, -1]
If implementing the dot product of two vectors directly:
each vector must be the same length
multiply corresponding terms from each vector
sum the products (to produce the answer)
Related task
Vector products
| #C.23 | C# | static void Main(string[] args)
{
Console.WriteLine(DotProduct(new decimal[] { 1, 3, -5 }, new decimal[] { 4, -2, -1 }));
Console.Read();
}
private static decimal DotProduct(decimal[] vec1, decimal[] vec2)
{
if (vec1 == null)
return 0;
if (vec2 == null)
return 0;
if (vec1.Length != vec2.Length)
return 0;
decimal tVal = 0;
for (int x = 0; x < vec1.Length; x++)
{
tVal += vec1[x] * vec2[x];
}
return tVal;
} |
http://rosettacode.org/wiki/Doubly-linked_list/Definition | Doubly-linked list/Definition | Define the data structure for a complete Doubly Linked List.
The structure should support adding elements to the head, tail and middle of the list.
The structure should not allow circular loops
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Nim | Nim | type
List[T] = object
head, tail: Node[T]
Node[T] = ref TNode[T]
TNode[T] = object
next, prev: Node[T]
data: T
proc initList[T](): List[T] = discard
proc newNode[T](data: T): Node[T] =
new(result)
result.data = data
proc prepend[T](l: var List[T], n: Node[T]) =
n.next = l.head
if l.head != nil: l.head.prev = n
l.head = n
if l.tail == nil: l.tail = n
proc append[T](l: var List[T], n: Node[T]) =
n.next = nil
n.prev = l.tail
if l.tail != nil:
l.tail.next = n
l.tail = n
if l.head == nil:
l.head = n
proc insertAfter[T](l: var List[T], r, n: Node[T]) =
n.prev = r
n.next = r.next
n.next.prev = n
r.next = n
if r == l.tail: l.tail = n
proc remove[T](l: var List[T], n: Node[T]) =
if n == l.tail: l.tail = n.prev
if n == l.head: l.head = n.next
if n.next != nil: n.next.prev = n.prev
if n.prev != nil: n.prev.next = n.next
proc `$`[T](l: var List[T]): string =
result = ""
var n = l.head
while n != nil:
if result.len > 0: result.add(" -> ")
result.add($n.data)
n = n.next
var l = initList[int]()
var n = newNode(12)
var m = newNode(13)
var i = newNode(14)
var j = newNode(15)
l.append(n)
l.prepend(m)
l.insertAfter(m, i)
l.prepend(j)
l.remove(m)
echo l
var l2 = initList[string]()
l2.prepend newNode("hello")
l2.append newNode("world")
echo l2 |
http://rosettacode.org/wiki/Dice_game_probabilities | Dice game probabilities | Two players have a set of dice each. The first player has nine dice with four faces each, with numbers one to four. The second player has six normal dice with six faces each, each face has the usual numbers from one to six.
They roll their dice and sum the totals of the faces. The player with the highest total wins (it's a draw if the totals are the same). What's the probability of the first player beating the second player?
Later the two players use a different set of dice each. Now the first player has five dice with ten faces each, and the second player has six dice with seven faces each. Now what's the probability of the first player beating the second player?
This task was adapted from the Project Euler Problem n.205:
https://projecteuler.net/problem=205
| #Action.21 | Action! | INCLUDE "D2:REAL.ACT" ;from the Action! Tool Kit
BYTE FUNC Roll(BYTE sides,dices)
BYTE i,sum
sum=0
FOR i=1 TO dices
DO
sum==+Rand(sides)+1
OD
RETURN (sum)
PROC Test(BYTE sides1,dices1,sides2,dices2)
INT i,count=[10000],sum1,sum2,wins1,wins2,draws
REAL r1,r2,p
wins1=0 wins2=0 draws=0
FOR i=1 TO count
DO
sum1=Roll(sides1,dices1)
sum2=Roll(sides2,dices2)
IF sum1>sum2 THEN
wins1==+1
ELSEIF sum1<sum2 THEN
wins2==+1
ELSE
draws==+1
FI
OD
IntToReal(wins1,r1)
IntToReal(wins2,r2)
RealDiv(r2,r1,p)
PrintF("Tested %I times%E",count)
PrintF("Player 1 with %B dice of %B sides%E",dices1,sides1)
PrintF("Player 2 with %B dice of %B sides%E",dices2,sides2)
PrintF("Player 1 wins %I times%E",wins1)
PrintF("Player 2 wins %I times%E",wins2)
PrintF("Draw %I times%E",draws)
Print("Player 1 beating Player 2 probability:")
PrintRE(p)
PutE()
RETURN
PROC Main()
Put(125) PutE() ;clear the screen
Test(4,9,6,6)
Test(10,5,7,6)
RETURN |
http://rosettacode.org/wiki/Dice_game_probabilities | Dice game probabilities | Two players have a set of dice each. The first player has nine dice with four faces each, with numbers one to four. The second player has six normal dice with six faces each, each face has the usual numbers from one to six.
They roll their dice and sum the totals of the faces. The player with the highest total wins (it's a draw if the totals are the same). What's the probability of the first player beating the second player?
Later the two players use a different set of dice each. Now the first player has five dice with ten faces each, and the second player has six dice with seven faces each. Now what's the probability of the first player beating the second player?
This task was adapted from the Project Euler Problem n.205:
https://projecteuler.net/problem=205
| #Ada | Ada | with Ada.Text_IO; use Ada.Text_IO;
with Ada.Numerics.Discrete_Random;
procedure Main is
package real_io is new Float_IO (Long_Float);
use real_io;
type Dice is record
Faces : Positive;
Num_Dice : Positive;
end record;
procedure Roll_Dice (The_Dice : in Dice; Count : out Natural) is
subtype Faces is Integer range 1 .. The_Dice.Faces;
package Die_Random is new Ada.Numerics.Discrete_Random (Faces);
use Die_Random;
Seed : Generator;
begin
Reset (Seed);
Count := 0;
for I in 1 .. The_Dice.Num_Dice loop
Count := Count + Random (Seed);
end loop;
end Roll_Dice;
function Win_Prob
(Dice_1 : Dice; Dice_2 : Dice; Tries : Positive) return Long_Float
is
Count_1 : Natural := 0;
Count_2 : Natural := 0;
Count_1_Wins : Natural := 0;
begin
for I in 1 .. Tries loop
Roll_Dice (Dice_1, Count_1);
Roll_Dice (Dice_2, Count_2);
if Count_1 > Count_2 then
Count_1_Wins := Count_1_Wins + 1;
end if;
end loop;
return Long_Float (Count_1_Wins) / Long_Float (Tries);
end Win_Prob;
D1 : Dice := (Faces => 4, Num_Dice => 9);
D2 : Dice := (Faces => 6, Num_Dice => 6);
D3 : Dice := (Faces => 10, Num_Dice => 5);
D4 : Dice := (Faces => 7, Num_Dice => 6);
P1 : Long_Float := Win_Prob (D1, D2, 1_000_000);
P2 : Long_Float := Win_Prob (D3, D4, 1_000_000);
begin
Put ("Dice D1 wins = ");
Put (Item => P1, Fore => 1, Aft => 7, Exp => 0);
New_Line;
Put ("Dice D2 wins = ");
Put (Item => P2, Fore => 1, Aft => 7, Exp => 0);
New_Line;
end Main;
|
http://rosettacode.org/wiki/Dining_philosophers | Dining philosophers | The dining philosophers problem illustrates non-composability of low-level synchronization primitives like semaphores. It is a modification of a problem posed by Edsger Dijkstra.
Five philosophers, Aristotle, Kant, Spinoza, Marx, and Russell (the tasks) spend their time thinking and eating spaghetti. They eat at a round table with five individual seats. For eating each philosopher needs two forks (the resources). There are five forks on the table, one left and one right of each seat. When a philosopher cannot grab both forks it sits and waits. Eating takes random time, then the philosopher puts the forks down and leaves the dining room. After spending some random time thinking about the nature of the universe, he again becomes hungry, and the circle repeats itself.
It can be observed that a straightforward solution, when forks are implemented by semaphores, is exposed to deadlock. There exist two deadlock states when all five philosophers are sitting at the table holding one fork each. One deadlock state is when each philosopher has grabbed the fork left of him, and another is when each has the fork on his right.
There are many solutions of the problem, program at least one, and explain how the deadlock is prevented.
| #Ada | Ada | with Ada.Numerics.Float_Random; use Ada.Numerics.Float_Random;
with Ada.Text_IO; use Ada.Text_IO;
with Synchronization.Generic_Mutexes_Array;
procedure Test_Dining_Philosophers is
type Philosopher is (Aristotle, Kant, Spinoza, Marx, Russel);
package Fork_Arrays is new Synchronization.Generic_Mutexes_Array (Philosopher);
use Fork_Arrays;
Life_Span : constant := 20; -- In his life a philosopher eats 20 times
Forks : aliased Mutexes_Array; -- Forks for hungry philosophers
function Left_Of (Fork : Philosopher) return Philosopher is
begin
if Fork = Philosopher'First then
return Philosopher'Last;
else
return Philosopher'Pred (Fork);
end if;
end Left_Of;
task type Person (ID : Philosopher);
task body Person is
Cutlery : aliased Mutexes_Set := ID or Left_Of (ID);
Dice : Generator;
begin
Reset (Dice);
for Life_Cycle in 1..Life_Span loop
Put_Line (Philosopher'Image (ID) & " is thinking");
delay Duration (Random (Dice) * 0.100);
Put_Line (Philosopher'Image (ID) & " is hungry");
declare
Lock : Set_Holder (Forks'Access, Cutlery'Access);
begin
Put_Line (Philosopher'Image (ID) & " is eating");
delay Duration (Random (Dice) * 0.100);
end;
end loop;
Put_Line (Philosopher'Image (ID) & " is leaving");
end Person;
Ph_1 : Person (Aristotle); -- Start philosophers
Ph_2 : Person (Kant);
Ph_3 : Person (Spinoza);
Ph_4 : Person (Marx);
Ph_5 : Person (Russel);
begin
null; -- Nothing to do in the main task, just sit and behold
end Test_Dining_Philosophers; |
http://rosettacode.org/wiki/Discordian_date | Discordian date |
Task
Convert a given date from the Gregorian calendar to the Discordian calendar.
| #ALGOL_68 | ALGOL 68 | BEGIN # DISCORDIAN DATE CALCULATION - TRANSLATION OF MAD VIA ALGOL W #
INT greg, gmonth, gday, gyear;
[]STRING holys = []STRING( "MUNG", "MOJO", "SYA", "ZARA", "MALA" )[ AT 0 ];
[]STRING holys0 = []STRING( "CHAO", "DISCO", "CONFU", "BURE", "AF" )[ AT 0 ];
[]STRING disday = []STRING( "SWEETMORN", "BOOMTIME", "PUNGENday", "PRICKLE-PRICKLE", "SETTING ORANGE" )[ AT 0 ];
[]STRING disssn = []STRING( "CHAOS", "DISCORD", "CONFUSION", "BUREAUCRACY", "THE AFTERMATH" )[ AT 0 ];
[]INT mlengt = []INT( 0, 0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334 )[ AT 0 ];
CHAR slash1, slash2;
# input date should contain MM/DD/YYYY in the gregorian calendar #
read( ( gmonth, slash1, gday, slash2, gyear ) );
IF slash1 /= "/" OR slash2 /= "/" THEN print( ( "Invalid date format", newline ) ); stop FI;
IF gmonth = 2 AND gday = 29
THEN print( ( "SAINT TIB'S DAY IN THE Y.O.L.D. ", whole( gyear + 1166, -4 ), newline ) )
ELSE
INT yrday := mlengt[ gmonth ] + gday;
INT season := yrday OVER 73;
INT day := yrday - ( season * 73 );
INT wkday := ( yrday - 1 ) MOD 5;
print( ( disday[ wkday ], ", DAY ", whole( day, -2 ), " OF ", disssn[ season ]
, " IN THE Y.O.L.D ", whole( gyear + 1166, 0 ), newline
)
);
IF day = 5 THEN print( ( "CELEBRATE ", holys[ season ], "DAY" ) )
ELIF day = 50 THEN print( ( "CELEBRATE ", holys0[ season ], "FLUX" ) )
FI
FI
END |
http://rosettacode.org/wiki/Dijkstra%27s_algorithm | Dijkstra's algorithm | This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Dijkstra's algorithm, conceived by Dutch computer scientist Edsger Dijkstra in 1956 and published in 1959, is a graph search algorithm that solves the single-source shortest path problem for a graph with non-negative edge path costs, producing a shortest path tree.
This algorithm is often used in routing and as a subroutine in other graph algorithms.
For a given source vertex (node) in the graph, the algorithm finds the path with lowest cost (i.e. the shortest path) between that vertex and every other vertex.
For instance
If the vertices of the graph represent cities and edge path costs represent driving distances between pairs of cities connected by a direct road, Dijkstra's algorithm can be used to find the shortest route between one city and all other cities.
As a result, the shortest path first is widely used in network routing protocols, most notably:
IS-IS (Intermediate System to Intermediate System) and
OSPF (Open Shortest Path First).
Important note
The inputs to Dijkstra's algorithm are a directed and weighted graph consisting of 2 or more nodes, generally represented by:
an adjacency matrix or list, and
a start node.
A destination node is not specified.
The output is a set of edges depicting the shortest path to each destination node.
An example, starting with
a──►b, cost=7, lastNode=a
a──►c, cost=9, lastNode=a
a──►d, cost=NA, lastNode=a
a──►e, cost=NA, lastNode=a
a──►f, cost=14, lastNode=a
The lowest cost is a──►b so a──►b is added to the output.
There is a connection from b──►d so the input is updated to:
a──►c, cost=9, lastNode=a
a──►d, cost=22, lastNode=b
a──►e, cost=NA, lastNode=a
a──►f, cost=14, lastNode=a
The lowest cost is a──►c so a──►c is added to the output.
Paths to d and f are cheaper via c so the input is updated to:
a──►d, cost=20, lastNode=c
a──►e, cost=NA, lastNode=a
a──►f, cost=11, lastNode=c
The lowest cost is a──►f so c──►f is added to the output.
The input is updated to:
a──►d, cost=20, lastNode=c
a──►e, cost=NA, lastNode=a
The lowest cost is a──►d so c──►d is added to the output.
There is a connection from d──►e so the input is updated to:
a──►e, cost=26, lastNode=d
Which just leaves adding d──►e to the output.
The output should now be:
[ d──►e
c──►d
c──►f
a──►c
a──►b ]
Task
Implement a version of Dijkstra's algorithm that outputs a set of edges depicting the shortest path to each reachable node from an origin.
Run your program with the following directed graph starting at node a.
Write a program which interprets the output from the above and use it to output the shortest path from node a to nodes e and f.
Vertices
Number
Name
1
a
2
b
3
c
4
d
5
e
6
f
Edges
Start
End
Cost
a
b
7
a
c
9
a
f
14
b
c
10
b
d
15
c
d
11
c
f
2
d
e
6
e
f
9
You can use numbers or names to identify vertices in your program.
See also
Dijkstra's Algorithm vs. A* Search vs. Concurrent Dijkstra's Algorithm (youtube)
| #ALGOL_68 | ALGOL 68 | # -*- coding: utf-8 -*- #
COMMENT REQUIRED BY "prelude_dijkstras_algorithm.a68" CO
MODE ROUTELEN = ~;
ROUTELEN route len infinity = max ~;
PROC route len add = (VERTEX v, ROUTE r)ROUTELEN:
route len OF v + route len OF r; # or MAX(v,r) #
MODE VERTEXPAYLOAD = ~;
PROC dijkstra fix value error = (STRING msg)BOOL:
(put(stand error, (msg, new line)); FALSE);
#PROVIDES:#
# VERTEX*=~* #
# ROUTE*=~* #
# vertex route*=~* #
END COMMENT
MODE VALVERTEX = STRUCT(
ROUTELEN route len,
FLEX[0]ROUTE route,
ROUTE shortest route,
VERTEXPAYLOAD vertex data
);
MODE VERTEX = REF VALVERTEX;
MODE VERTEXYIELD = PROC(VERTEX)VOID; # used to "generate" VERTEX path #
PRIO INIT = 1; # The same PRIOrity as +:= etc #
OP INIT = (VERTEX self, VERTEXPAYLOAD vertex data)VERTEX:
self := (route len infinity, (), NIL, vertex data);
# It may be faster to preallocate "queue", rather then grow a FLEX #
OP +:= = (REF FLEX[]VERTEX in list, VERTEX rhs)REF FLEX[]VERTEX: (
[UPB in list+1]VERTEX out list;
out list[:UPB in list] := in list;
out list[UPB out list] := rhs;
in list := out list # EXIT #
);
MODE VALROUTE = STRUCT(VERTEX from, to, ROUTELEN route len#, ROUTEPAYLOAD#);
MODE ROUTE = REF VALROUTE;
OP +:= = (REF FLEX[]ROUTE in list, ROUTE rhs)REF FLEX[]ROUTE: (
[UPB in list+1]ROUTE out list;
out list[:UPB in list] := in list;
out list[UPB out list] := rhs;
in list := out list # EXIT #
);
MODE VERTEXROUTE = UNION(VERTEX, ROUTE);
MODE VERTEXROUTEYIELD = PROC(VERTEXROUTE)VOID;
################################################################
# Finally: now the strong typing is in place, the task code... #
################################################################
PROC vertex route gen dijkstra = (
VERTEX source, target,
REF[]VALROUTE route list,
VERTEXROUTEYIELD yield
)VOID:(
# initialise the route len for BOTH directions on each route #
FOR this TO UPB route list DO
ROUTE route = route list[this];
route OF from OF route +:= route;
# assume route lens is the same in both directions, this i.e. NO A-B gradient NOR 1-way streets #
route OF to OF route +:= (HEAP VALROUTE := (to OF route, from OF route, route len OF route))
OD;
COMMENT
Algorithium Performance "about" O(n**2)...
Optimisations:
a) bound index in [lwb queue:UPB queue] for search
b) delay adding vertices until they are actually encountered
It may be faster to preallocate "queue" vertex list, rather then grow a FLEX
END COMMENT
PROC vertex gen nearest = (REF FLEX[]VERTEX queue, VERTEXYIELD yield)VOID: (
INT vertices done := 0, lwb queue := 1;
ROUTELEN shortest route len done := -route len infinity;
WHILE vertices done <= UPB queue ANDF shortest route len done NE route len infinity DO
ROUTELEN shortest route len := route len infinity;
# skip done elements: #
FOR this FROM lwb queue TO UPB queue DO
VERTEX this vertex := queue[this];
IF NOT(shortest route len done < route len OF this vertex) THEN
lwb queue := this; # remember for next time #
break
FI
OD;
break:
# find vertex with shortest path attached #
FOR this FROM lwb queue TO UPB queue DO VERTEX this vertex := queue[this];
IF shortest route len done < route len OF this vertex ANDF
route len OF this vertex < shortest route len THEN
shortest route len := route len OF this vertex FI
OD;
# update the other vertices with shortest path found #
FOR this FROM lwb queue TO UPB queue DO VERTEX this vertex := queue[this];
IF route len OF this vertex = shortest route len THEN
vertices done +:= 1; yield(this vertex) FI
OD;
shortest route len done := shortest route len
OD
);
route len OF target := 0;
FLEX[0]VERTEX queue := target;
# FOR VERTEX this vertex IN # vertex gen nearest(queue#) DO (#,
## (VERTEX this vertex)VOID: (
FOR this TO UPB route OF this vertex DO ROUTE this route = (route OF this vertex)[this];
# If this vertex has not been encountered before, then add to queue #
IF route len OF to OF this route = route len infinity THEN queue +:= to OF this route FI;
ROUTELEN route len = route len add(this vertex, this route);
IF route len < route len OF to OF this route THEN
route len OF to OF this route := route len;
shortest route OF to OF this route := this route
FI
OD;
IF this vertex IS source THEN done FI
# OD#));
IF NOT dijkstra fix value error("no path found") THEN stop FI;
############################
# Now: generate the result #
############################
done: (
VERTEX this vertex := source;
WHILE
yield(this vertex);
ROUTE this route = shortest route OF this vertex;
# WHILE # this route ISNT ROUTE(NIL) DO
yield(this route);
this vertex := from OF this route
OD
)
);
SKIP |
http://rosettacode.org/wiki/Digital_root | Digital root | The digital root,
X
{\displaystyle X}
, of a number,
n
{\displaystyle n}
, is calculated:
find
X
{\displaystyle X}
as the sum of the digits of
n
{\displaystyle n}
find a new
X
{\displaystyle X}
by summing the digits of
X
{\displaystyle X}
, repeating until
X
{\displaystyle X}
has only one digit.
The additive persistence is the number of summations required to obtain the single digit.
The task is to calculate the additive persistence and the digital root of a number, e.g.:
627615
{\displaystyle 627615}
has additive persistence
2
{\displaystyle 2}
and digital root of
9
{\displaystyle 9}
;
39390
{\displaystyle 39390}
has additive persistence
2
{\displaystyle 2}
and digital root of
6
{\displaystyle 6}
;
588225
{\displaystyle 588225}
has additive persistence
2
{\displaystyle 2}
and digital root of
3
{\displaystyle 3}
;
393900588225
{\displaystyle 393900588225}
has additive persistence
2
{\displaystyle 2}
and digital root of
9
{\displaystyle 9}
;
The digital root may be calculated in bases other than 10.
See
Casting out nines for this wiki's use of this procedure.
Digital root/Multiplicative digital root
Sum digits of an integer
Digital root sequence on OEIS
Additive persistence sequence on OEIS
Iterated digits squaring
| #ALGOL_68 | ALGOL 68 | # calculates the digital root and persistance of n #
PROC digital root = ( LONG LONG INT n, REF INT root, persistance )VOID:
BEGIN
LONG LONG INT number := ABS n;
persistance := 0;
WHILE persistance PLUSAB 1;
LONG LONG INT digit sum := 0;
WHILE number > 0
DO
digit sum PLUSAB number MOD 10;
number OVERAB 10
OD;
number := digit sum;
number > 9
DO
SKIP
OD;
root := SHORTEN SHORTEN number
END; # digital root #
# calculates and prints the digital root and persistace of number #
PROC print digital root and persistance = ( LONG LONG INT number )VOID:
BEGIN
INT root, persistance;
digital root( number, root, persistance );
print( ( whole( number, -15 ), " root: ", whole( root, 0 ), " persistance: ", whole( persistance, -3 ), newline ) )
END; # print digital root and persistance #
# test the digital root proc #
BEGIN print digital root and persistance( 627615 )
; print digital root and persistance( 39390 )
; print digital root and persistance( 588225 )
; print digital root and persistance( 393900588225 )
END |
http://rosettacode.org/wiki/Digital_root/Multiplicative_digital_root | Digital root/Multiplicative digital root | The multiplicative digital root (MDR) and multiplicative persistence (MP) of a number,
n
{\displaystyle n}
, is calculated rather like the Digital root except digits are multiplied instead of being added:
Set
m
{\displaystyle m}
to
n
{\displaystyle n}
and
i
{\displaystyle i}
to
0
{\displaystyle 0}
.
While
m
{\displaystyle m}
has more than one digit:
Find a replacement
m
{\displaystyle m}
as the multiplication of the digits of the current value of
m
{\displaystyle m}
.
Increment
i
{\displaystyle i}
.
Return
i
{\displaystyle i}
(= MP) and
m
{\displaystyle m}
(= MDR)
Task
Tabulate the MP and MDR of the numbers 123321, 7739, 893, 899998
Tabulate MDR versus the first five numbers having that MDR, something like:
MDR: [n0..n4]
=== ========
0: [0, 10, 20, 25, 30]
1: [1, 11, 111, 1111, 11111]
2: [2, 12, 21, 26, 34]
3: [3, 13, 31, 113, 131]
4: [4, 14, 22, 27, 39]
5: [5, 15, 35, 51, 53]
6: [6, 16, 23, 28, 32]
7: [7, 17, 71, 117, 171]
8: [8, 18, 24, 29, 36]
9: [9, 19, 33, 91, 119]
Show all output on this page.
Similar
The Product of decimal digits of n page was redirected here, and had the following description
Find the product of the decimal digits of a positive integer n, where n <= 100
The three existing entries for Phix, REXX, and Ring have been moved here, under ===Similar=== headings, feel free to match or ignore them.
References
Multiplicative Digital Root on Wolfram Mathworld.
Multiplicative digital root on The On-Line Encyclopedia of Integer Sequences.
What's special about 277777788888899? - Numberphile video
| #Component_Pascal | Component Pascal |
MODULE MDR;
IMPORT StdLog, Strings, TextMappers, DevCommanders;
PROCEDURE CalcMDR(x: LONGINT; OUT mdr, mp: LONGINT);
VAR
str: ARRAY 64 OF CHAR;
i: INTEGER;
BEGIN
mdr := 1; mp := 0;
LOOP
Strings.IntToString(x,str);
IF LEN(str$) = 1 THEN mdr := x; EXIT END;
i := 0;mdr := 1;
WHILE i < LEN(str$) DO
mdr := mdr * (ORD(str[i]) - ORD('0'));
INC(i)
END;
INC(mp);
x := mdr
END
END CalcMDR;
PROCEDURE Do*;
VAR
mdr,mp: LONGINT;
s: TextMappers.Scanner;
BEGIN
s.ConnectTo(DevCommanders.par.text);
s.SetPos(DevCommanders.par.beg);
REPEAT
s.Scan;
IF (s.type = TextMappers.int) OR (s.type = TextMappers.lint) THEN
CalcMDR(s.int,mdr,mp);
StdLog.Int(s.int);
StdLog.String(" MDR: ");StdLog.Int(mdr);
StdLog.String(" MP: ");StdLog.Int(mp);StdLog.Ln
END
UNTIL s.rider.eot;
END Do;
PROCEDURE Show(i: INTEGER; x: ARRAY OF LONGINT);
VAR
k: INTEGER;
BEGIN
StdLog.Int(i);StdLog.String(": ");
FOR k := 0 TO LEN(x) - 1 DO
StdLog.Int(x[k])
END;
StdLog.Ln
END Show;
PROCEDURE FirstFive*;
VAR
i,j: INTEGER;
five: ARRAY 5 OF LONGINT;
x,mdr,mp: LONGINT;
BEGIN
FOR i := 0 TO 9 DO
j := 0;x := 0;
WHILE (j < LEN(five)) DO
CalcMDR(x,mdr,mp);
IF mdr = i THEN five[j] := x; INC(j) END;
INC(x)
END;
Show(i,five)
END
END FirstFive;
END MDR.
|
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #SVG | SVG | <?xml version="1.0" standalone="yes"?>
<!DOCTYPE svg PUBLIC "-//W3C//DTD SVG 20010904//EN"
"http://www.w3.org/TR/2001/REC-SVG-20010904/DTD/svg10.dtd">
<svg xmlns="http://www.w3.org/2000/svg"
xmlns:l="http://www.w3.org/1999/xlink"
width="400" height="400">
<style type="text/css"><![CDATA[
line { stroke: black; stroke-width: .05; }
circle { fill: black; }
]]></style>
<defs>
<g id="marks">
<circle cx="0" cy="0" r=".03"/>
<circle cx="1" cy="0" r=".03"/>
</g>
<g id="l0">
<line x1="0" y1="0" x2="1" y2="0"/>
<!-- useful for studying the transformation stages:
<line x1="0.1" y1="0" x2="0.9" y2="0.1"/> -->
</g>
<!-- These are identical except for the id and href. -->
<g id="l1"> <use l:href="#l0" transform="matrix( .5 .5 -.5 .5 0 0)"/>
<use l:href="#l0" transform="matrix(-.5 .5 -.5 -.5 1 0)"/>
<use l:href="#marks"/></g>
<g id="l2"> <use l:href="#l1" transform="matrix( .5 .5 -.5 .5 0 0)"/>
<use l:href="#l1" transform="matrix(-.5 .5 -.5 -.5 1 0)"/>
<use l:href="#marks"/></g>
<g id="l3"> <use l:href="#l2" transform="matrix( .5 .5 -.5 .5 0 0)"/>
<use l:href="#l2" transform="matrix(-.5 .5 -.5 -.5 1 0)"/>
<use l:href="#marks"/></g>
<g id="l4"> <use l:href="#l3" transform="matrix( .5 .5 -.5 .5 0 0)"/>
<use l:href="#l3" transform="matrix(-.5 .5 -.5 -.5 1 0)"/>
<use l:href="#marks"/></g>
<g id="l5"> <use l:href="#l4" transform="matrix( .5 .5 -.5 .5 0 0)"/>
<use l:href="#l4" transform="matrix(-.5 .5 -.5 -.5 1 0)"/>
<use l:href="#marks"/></g>
<g id="l6"> <use l:href="#l5" transform="matrix( .5 .5 -.5 .5 0 0)"/>
<use l:href="#l5" transform="matrix(-.5 .5 -.5 -.5 1 0)"/>
<use l:href="#marks"/></g>
<g id="l7"> <use l:href="#l6" transform="matrix( .5 .5 -.5 .5 0 0)"/>
<use l:href="#l6" transform="matrix(-.5 .5 -.5 -.5 1 0)"/>
<use l:href="#marks"/></g>
<g id="l8"> <use l:href="#l7" transform="matrix( .5 .5 -.5 .5 0 0)"/>
<use l:href="#l7" transform="matrix(-.5 .5 -.5 -.5 1 0)"/>
<use l:href="#marks"/></g>
<g id="l9"> <use l:href="#l8" transform="matrix( .5 .5 -.5 .5 0 0)"/>
<use l:href="#l8" transform="matrix(-.5 .5 -.5 -.5 1 0)"/>
<use l:href="#marks"/></g>
</defs>
<g transform="translate(100, 200) scale(200)">
<use l:href="#marks"/>
<use l:href="#l9"/>
</g>
</svg> |
http://rosettacode.org/wiki/Dinesman%27s_multiple-dwelling_problem | Dinesman's multiple-dwelling problem | Task
Solve Dinesman's multiple dwelling problem but in a way that most naturally follows the problem statement given below.
Solutions are allowed (but not required) to parse and interpret the problem text, but should remain flexible and should state what changes to the problem text are allowed. Flexibility and ease of expression are valued.
Examples may be be split into "setup", "problem statement", and "output" sections where the ease and naturalness of stating the problem and getting an answer, as well as the ease and flexibility of modifying the problem are the primary concerns.
Example output should be shown here, as well as any comments on the examples flexibility.
The problem
Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house that contains only five floors.
Baker does not live on the top floor.
Cooper does not live on the bottom floor.
Fletcher does not live on either the top or the bottom floor.
Miller lives on a higher floor than does Cooper.
Smith does not live on a floor adjacent to Fletcher's.
Fletcher does not live on a floor adjacent to Cooper's.
Where does everyone live?
| #C | C | #include <stdio.h>
#include <stdlib.h>
int verbose = 0;
#define COND(a, b) int a(int *s) { return (b); }
typedef int(*condition)(int *);
/* BEGIN problem specific setup */
#define N_FLOORS 5
#define TOP (N_FLOORS - 1)
int solution[N_FLOORS] = { 0 };
int occupied[N_FLOORS] = { 0 };
enum tenants {
baker = 0,
cooper,
fletcher,
miller,
smith,
phantom_of_the_opera,
};
const char *names[] = {
"baker",
"cooper",
"fletcher",
"miller",
"smith",
};
COND(c0, s[baker] != TOP);
COND(c1, s[cooper] != 0);
COND(c2, s[fletcher] != 0 && s[fletcher] != TOP);
COND(c3, s[miller] > s[cooper]);
COND(c4, abs(s[smith] - s[fletcher]) != 1);
COND(c5, abs(s[cooper] - s[fletcher]) != 1);
#define N_CONDITIONS 6
condition cond[] = { c0, c1, c2, c3, c4, c5 };
/* END of problem specific setup */
int solve(int person)
{
int i, j;
if (person == phantom_of_the_opera) {
/* check condition */
for (i = 0; i < N_CONDITIONS; i++) {
if (cond[i](solution)) continue;
if (verbose) {
for (j = 0; j < N_FLOORS; j++)
printf("%d %s\n", solution[j], names[j]);
printf("cond %d bad\n\n", i);
}
return 0;
}
printf("Found arrangement:\n");
for (i = 0; i < N_FLOORS; i++)
printf("%d %s\n", solution[i], names[i]);
return 1;
}
for (i = 0; i < N_FLOORS; i++) {
if (occupied[i]) continue;
solution[person] = i;
occupied[i] = 1;
if (solve(person + 1)) return 1;
occupied[i] = 0;
}
return 0;
}
int main()
{
verbose = 0;
if (!solve(0)) printf("Nobody lives anywhere\n");
return 0;
} |
http://rosettacode.org/wiki/Determine_sentence_type | Determine sentence type | Use these sentences:
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
Task
Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
Output one of these letters
"E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
Extra
Make your code able to determine multiple sentences.
Don't leave any errors!
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #XPL0 | XPL0 | include xpllib; \for StrLen
int Sentence, N, Len;
char Str;
[Sentence:= ["hi there, how are you today?",
"I'd like to present to you the washing machine 9001.",
"You have been nominated to win one of these!",
"Just make sure you don't break it"];
for N:= 0 to 3 do
[Str:= Sentence(N);
Len:= StrLen(Str);
case Str(Len-1) of
^!: ChOut(0, ^E);
^?: ChOut(0, ^Q);
^.: ChOut(0, ^S)
other ChOut(0, ^N);
if N < 3 then ChOut(0, ^|);
];
] |
http://rosettacode.org/wiki/Dot_product | Dot product | Task
Create a function/use an in-built function, to compute the dot product, also known as the scalar product of two vectors.
If possible, make the vectors of arbitrary length.
As an example, compute the dot product of the vectors:
[1, 3, -5] and
[4, -2, -1]
If implementing the dot product of two vectors directly:
each vector must be the same length
multiply corresponding terms from each vector
sum the products (to produce the answer)
Related task
Vector products
| #C.2B.2B | C++ | #include <iostream>
#include <numeric>
int main()
{
int a[] = { 1, 3, -5 };
int b[] = { 4, -2, -1 };
std::cout << std::inner_product(a, a + sizeof(a) / sizeof(a[0]), b, 0) << std::endl;
return 0;
} |
http://rosettacode.org/wiki/Doubly-linked_list/Definition | Doubly-linked list/Definition | Define the data structure for a complete Doubly Linked List.
The structure should support adding elements to the head, tail and middle of the list.
The structure should not allow circular loops
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Oberon-2 | Oberon-2 |
IMPORT Basic;
TYPE
Node* = POINTER TO NodeDesc;
NodeDesc* = (* ABSTRACT *) RECORD
prev-,next-: Node;
END;
DLList* = POINTER TO DLListDesc;
DLListDesc* = RECORD
first-,last-: Node;
size-: INTEGER;
END;
|
http://rosettacode.org/wiki/Dice_game_probabilities | Dice game probabilities | Two players have a set of dice each. The first player has nine dice with four faces each, with numbers one to four. The second player has six normal dice with six faces each, each face has the usual numbers from one to six.
They roll their dice and sum the totals of the faces. The player with the highest total wins (it's a draw if the totals are the same). What's the probability of the first player beating the second player?
Later the two players use a different set of dice each. Now the first player has five dice with ten faces each, and the second player has six dice with seven faces each. Now what's the probability of the first player beating the second player?
This task was adapted from the Project Euler Problem n.205:
https://projecteuler.net/problem=205
| #BASIC | BASIC | dado1 = 9: lado1 = 4
dado2 = 6: lado2 = 6
total1 = 0: total2 = 0
for i = 0 to 1
for cont = 1 to 100000
jugador1 = lanzamiento(dado1, lado1)
jugador2 = lanzamiento(dado2, lado2)
if jugador1 > jugador2 then
total1 = total1 + 1
else
if jugador1 <> jugador2 then total2 = total2 + 1
endif
next cont
print "Lanzado el dado "; (cont - 1); " veces"
print "jugador1 con "; dado1; " dados de "; lado1; " lados"
print "jugador2 con "; dado2; " dados de "; lado2; " lados"
print "Total victorias jugador1 = "; total1; " => "; left(string(total2 / total1), 9)
print "Total victorias jugador2 = "; total2
print (cont - 1) - (total1 + total2); " empates" + chr(10)
dado1 = 5: lado1 = 10
dado2 = 6: lado2 = 7
total1 = 0: total2 = 0
next i
end
function lanzamiento(dado, lado)
total = 0
for lanza = 1 to dado
total = total + int(rand * lado) + 1
next lanza
return total
end function |
http://rosettacode.org/wiki/Dice_game_probabilities | Dice game probabilities | Two players have a set of dice each. The first player has nine dice with four faces each, with numbers one to four. The second player has six normal dice with six faces each, each face has the usual numbers from one to six.
They roll their dice and sum the totals of the faces. The player with the highest total wins (it's a draw if the totals are the same). What's the probability of the first player beating the second player?
Later the two players use a different set of dice each. Now the first player has five dice with ten faces each, and the second player has six dice with seven faces each. Now what's the probability of the first player beating the second player?
This task was adapted from the Project Euler Problem n.205:
https://projecteuler.net/problem=205
| #C | C | #include <stdio.h>
#include <stdint.h>
typedef uint32_t uint;
typedef uint64_t ulong;
ulong ipow(const uint x, const uint y) {
ulong result = 1;
for (uint i = 1; i <= y; i++)
result *= x;
return result;
}
uint min(const uint x, const uint y) {
return (x < y) ? x : y;
}
void throw_die(const uint n_sides, const uint n_dice, const uint s, uint counts[]) {
if (n_dice == 0) {
counts[s]++;
return;
}
for (uint i = 1; i < n_sides + 1; i++)
throw_die(n_sides, n_dice - 1, s + i, counts);
}
double beating_probability(const uint n_sides1, const uint n_dice1,
const uint n_sides2, const uint n_dice2) {
const uint len1 = (n_sides1 + 1) * n_dice1;
uint C1[len1];
for (uint i = 0; i < len1; i++)
C1[i] = 0;
throw_die(n_sides1, n_dice1, 0, C1);
const uint len2 = (n_sides2 + 1) * n_dice2;
uint C2[len2];
for (uint j = 0; j < len2; j++)
C2[j] = 0;
throw_die(n_sides2, n_dice2, 0, C2);
const double p12 = (double)(ipow(n_sides1, n_dice1) * ipow(n_sides2, n_dice2));
double tot = 0;
for (uint i = 0; i < len1; i++)
for (uint j = 0; j < min(i, len2); j++)
tot += (double)C1[i] * C2[j] / p12;
return tot;
}
int main() {
printf("%1.16f\n", beating_probability(4, 9, 6, 6));
printf("%1.16f\n", beating_probability(10, 5, 7, 6));
return 0;
} |
http://rosettacode.org/wiki/Dining_philosophers | Dining philosophers | The dining philosophers problem illustrates non-composability of low-level synchronization primitives like semaphores. It is a modification of a problem posed by Edsger Dijkstra.
Five philosophers, Aristotle, Kant, Spinoza, Marx, and Russell (the tasks) spend their time thinking and eating spaghetti. They eat at a round table with five individual seats. For eating each philosopher needs two forks (the resources). There are five forks on the table, one left and one right of each seat. When a philosopher cannot grab both forks it sits and waits. Eating takes random time, then the philosopher puts the forks down and leaves the dining room. After spending some random time thinking about the nature of the universe, he again becomes hungry, and the circle repeats itself.
It can be observed that a straightforward solution, when forks are implemented by semaphores, is exposed to deadlock. There exist two deadlock states when all five philosophers are sitting at the table holding one fork each. One deadlock state is when each philosopher has grabbed the fork left of him, and another is when each has the fork on his right.
There are many solutions of the problem, program at least one, and explain how the deadlock is prevented.
| #AutoHotkey | AutoHotkey | #Persistent
SetWorkingDir, %A_ScriptDir%
FileDelete, output.txt
EnoughForks := 2 ; required forks to begin eating
Fork1 := Fork2 := Fork3 := Fork4 := Fork5 := 1 ; fork supply per philosopher
SetTimer, AristotleWaitForLeftFork
SetTimer, KantWaitForLeftFork
SetTimer, SpinozaWaitForLeftFork
SetTimer, MarxWaitForLeftFork
SetTimer, RussellWaitForLeftFork
Return ;---------------------------------------------------------------
AristotleWaitForLeftFork:
WaitForFork("Aristotle", "left", Fork1, Fork2, AristotleLeftForkCount, AristotleRightForkCount, AristotleWaitCount, EnoughForks)
Return
AristotleWaitForRightFork:
WaitForFork("Aristotle", "right", Fork2, Fork1, AristotleRightForkCount, AristotleLeftForkCount, AristotleWaitCount, EnoughForks)
Return
AristotleFinishEating:
ReturnForks("Aristotle", Fork1, Fork2, AristotleLeftForkCount, AristotleRightForkCount, EnoughForks)
Return
KantWaitForLeftFork:
WaitForFork("Kant", "left", Fork2, Fork3, KantLeftForkCount, KantRightForkCount, KantWaitCount, EnoughForks)
Return
KantWaitForRightFork:
WaitForFork("Kant", "right", Fork3, Fork2, KantRightForkCount, KantLeftForkCount, KantWaitCount, EnoughForks)
Return
KantFinishEating:
ReturnForks("Kant", Fork2, Fork3, KantLeftForkCount, KantRightForkCount, EnoughForks)
Return
SpinozaWaitForLeftFork:
WaitForFork("Spinoza", "left", Fork3, Fork4, SpinozaLeftForkCount, SpinozaRightForkCount, SpinozaWaitCount, EnoughForks)
Return
SpinozaWaitForRightFork:
WaitForFork("Spinoza", "right", Fork4, Fork3, SpinozaRightForkCount, SpinozaLeftForkCount, SpinozaWaitCount, EnoughForks)
Return
SpinozaFinishEating:
ReturnForks("Spinoza", Fork3, Fork4, SpinozaLeftForkCount, SpinozaRightForkCount, EnoughForks)
Return
MarxWaitForLeftFork:
WaitForFork("Marx", "left", Fork4, Fork5, MarxLeftForkCount, MarxRightForkCount, MarxWaitCount, EnoughForks)
Return
MarxWaitForRightFork:
WaitForFork("Marx", "right", Fork5, Fork4, MarxRightForkCount, MarxLeftForkCount, MarxWaitCount, EnoughForks)
Return
MarxFinishEating:
ReturnForks("Marx", Fork4, Fork5, MarxLeftForkCount, MarxRightForkCount, EnoughForks)
Return
RussellWaitForLeftFork:
WaitForFork("Russell", "left", Fork5, Fork1, RussellLeftForkCount, RussellRightForkCount, RussellWaitCount, EnoughForks)
Return
RussellWaitForRightFork:
WaitForFork("Russell", "right", Fork1, Fork5, RussellRightForkCount, RussellLeftForkCount, RussellWaitCount, EnoughForks)
Return
RussellFinishEating:
ReturnForks("Russell", Fork5, Fork1, RussellLeftForkCount, RussellRightForkCount, EnoughForks)
Return
ReturnForks(Philosopher, ByRef ThisFork, ByRef OtherFork, ByRef CurrentThisForkCount, ByRef CurrentOtherForkCount, EnoughForks) {
OutputDebug, %Philosopher% finishes eating.
FileAppend, %Philosopher% finishes eating.`n,output.txt
ThisFork += CurrentThisForkCount ; return this fork
OtherFork += CurrentOtherForkCount ; return other fork
CurrentThisForkCount := 0 ; release this fork
CurrentOtherForkCount := 0 ; release other fork
OutputDebug, %Philosopher% returns all forks.
FileAppend, %Philosopher% returns all forks.`n,output.txt
; do something while resting
Random, Rand, 0, 1
Rand := Rand ? "Left" : "Right"
SetTimer, %Philosopher%WaitFor%Rand%Fork
}
WaitForFork(Philosopher, This, ByRef ThisFork, ByRef OtherFork, ByRef CurrentThisForkCount, ByRef CurrentOtherForkCount, ByRef CurrentWaitCount, EnoughForks) {
If This not in Left,Right
Return Error
Other := (This="right") ? "left" : "right"
OutputDebug, %Philosopher% is hungry.
FileAppend, %Philosopher% is hungry.`n,output.txt
If (ThisFork) ; if this fork available
{
SetTimer, %Philosopher%WaitFor%This%Fork, Off
CurrentWaitCount := 0
ThisFork-- ; take this fork
CurrentThisForkCount++ ; receive this fork
OutputDebug, %Philosopher% grabs %This% fork.
FileAppend, %Philosopher% grabs %This% fork.`n,output.txt
If (CurrentThisForkCount + CurrentOtherForkCount = EnoughForks) ; if philosopher has enough forks
{
OutputDebug, %Philosopher% starts eating.
FileAppend, %Philosopher% starts eating.`n,output.txt
; do something while eating
SetTimer, %Philosopher%FinishEating, -250
}
Else If (EnoughForks=2)
{
SetTimer, %Philosopher%WaitFor%Other%Fork
}
Else
{
Random, Rand, 0, 1
Rand := Rand ? "Left" : "Right"
SetTimer, %Philosopher%WaitFor%Rand%Fork
}
}
Else If (CurrentOtherForkCount and CurrentWaitCount > 5) ; if we've been holding other fork too long
{
SetTimer, %Philosopher%WaitFor%This%Fork, Off
CurrentWaitCount := 0
OtherFork++ ; return other fork
CurrentOtherForkCount-- ; release other fork
OutputDebug, %Philosopher% drops %Other% fork.
FileAppend, %Philosopher% drops %Other% fork.`n,output.txt
Random, Rand, 0, 1
Rand := Rand ? "Left" : "Right"
SetTimer, %Philosopher%WaitFor%Rand%Fork
}
Else If (CurrentThisForkCount and CurrentWaitCount > 5) ; if we've been holding one of this fork too long
{
SetTimer, %Philosopher%WaitFor%This%Fork, Off
CurrentWaitCount := 0
ThisFork++ ; return other fork
CurrentThisForkCount-- ; release other fork
OutputDebug, %Philosopher% drops %This% fork.
FileAppend, %Philosopher% drops %This% fork.`n,output.txt
Random, Rand, 0, 1
Rand := Rand ? "Left" : "Right"
SetTimer, %Philosopher%WaitFor%Rand%Fork
}
Else
{
CurrentWaitCount++
}
} |
http://rosettacode.org/wiki/Discordian_date | Discordian date |
Task
Convert a given date from the Gregorian calendar to the Discordian calendar.
| #ALGOL_W | ALGOL W | BEGIN % DISCORDIAN DATE CALCULATION - TRANSLATION OF MAD %
INTEGER GREG, GMONTH, GDAY, GYEAR;
STRING(16) ARRAY HOLY5 ( 0 :: 4 );
STRING(16) ARRAY HOLY50 ( 0 :: 4 );
STRING(16) ARRAY DISDAY ( 0 :: 4 );
STRING(16) ARRAY DISSSN ( 0 :: 4 );
INTEGER ARRAY MLENGT ( 0 :: 12 );
INTEGER APOS;
STRING(1) SLASH1, SLASH2;
% WRITES A "$" TERMINATED STRING %
PROCEDURE WRITEONTEXT( STRING(16) VALUE TEXT ) ;
BEGIN
INTEGER TPOS;
TPOS := 0;
WHILE TPOS < 16 DO BEGIN
IF TEXT( TPOS // 1 ) = "$"
THEN TPOS := 32
ELSE WRITEON( TEXT( TPOS // 1 ) );
;
TPOS := TPOS + 1
END WHILE_TPOS_LT_16
END WRITEONTEXT;
APOS := 0;
FOR M := 0,0,31,59,90,120,151,181,212,243,273,304,334 DO BEGIN MLENGT(APOS) := M; APOS := APOS + 1 END;
HOLY5 (0) := "MUNG$";HOLY5 (1) := "MOJO$"; HOLY5 (2) := "SYA$"; HOLY5 (3) := "ZARA$"; HOLY5 (4) := "MALA$";
HOLY50(0) := "CHAO$";HOLY50(1) := "DISCO$";HOLY50(2) := "CONFU$";HOLY50(3) := "BURE$"; HOLY50(4) := "AF$";
DISDAY(0) := "SWEETMORN$"; DISDAY(1) := "BOOMTIME$"; DISDAY(2) := "PUNGENDAY$";
DISDAY(3) := "PRICKLE-PRICKLE$"; DISDAY(4) := "SETTING ORANGE$";
DISSSN(0) := "CHAOS$"; DISSSN(1) := "DISCORD$"; DISSSN(2) := "CONFUSION$";
DISSSN(3) := "BUREAUCRACY$"; DISSSN(4) := "THE AFTERMATH$";
% INPUT DATE SHOULD CONTAIN MM/DD/YYYY IN GREGORIAN CALENDAR %
READ( GMONTH, SLASH1, GDAY, SLASH2, GYEAR );
IF GMONTH = 2 AND GDAY = 29
THEN WRITE( I_W := 4, S_W := 0, "SAINT TIB'S DAY IN THE Y.O.L.D. ", GYEAR + 1166 )
ELSE BEGIN
INTEGER YRDAY, SEASON, DAY, WKDAY;
YRDAY := MLENGT(GMONTH)+GDAY;
SEASON := YRDAY DIV 73;
DAY := YRDAY-SEASON*73;
WKDAY := (YRDAY-1) REM 5;
WRITEONTEXT( DISDAY(WKDAY) );
WRITEON( S_W := 0, ", DAY ", I_W := 2, DAY, " OF " );
WRITEONTEXT( DISSSN(SEASON) );
WRITEON( S_W := 0, " IN THE Y.O.L.D ", I_W := 4, GYEAR + 1166 );
IF DAY = 5 THEN BEGIN
WRITE( "CELEBRATE " );WRITEONTEXT( HOLY5(SEASON) ); WRITEON( "DAY" )
END
ELSE IF DAY = 50 THEN BEGIN
WRITE( "CELEBRATE " );WRITEONTEXT( HOLY50(SEASON) ); WRITEON( "FLUX" )
END IF_FAY_EQ_5__DAY_EQ_50
END
END. |
http://rosettacode.org/wiki/Dijkstra%27s_algorithm | Dijkstra's algorithm | This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Dijkstra's algorithm, conceived by Dutch computer scientist Edsger Dijkstra in 1956 and published in 1959, is a graph search algorithm that solves the single-source shortest path problem for a graph with non-negative edge path costs, producing a shortest path tree.
This algorithm is often used in routing and as a subroutine in other graph algorithms.
For a given source vertex (node) in the graph, the algorithm finds the path with lowest cost (i.e. the shortest path) between that vertex and every other vertex.
For instance
If the vertices of the graph represent cities and edge path costs represent driving distances between pairs of cities connected by a direct road, Dijkstra's algorithm can be used to find the shortest route between one city and all other cities.
As a result, the shortest path first is widely used in network routing protocols, most notably:
IS-IS (Intermediate System to Intermediate System) and
OSPF (Open Shortest Path First).
Important note
The inputs to Dijkstra's algorithm are a directed and weighted graph consisting of 2 or more nodes, generally represented by:
an adjacency matrix or list, and
a start node.
A destination node is not specified.
The output is a set of edges depicting the shortest path to each destination node.
An example, starting with
a──►b, cost=7, lastNode=a
a──►c, cost=9, lastNode=a
a──►d, cost=NA, lastNode=a
a──►e, cost=NA, lastNode=a
a──►f, cost=14, lastNode=a
The lowest cost is a──►b so a──►b is added to the output.
There is a connection from b──►d so the input is updated to:
a──►c, cost=9, lastNode=a
a──►d, cost=22, lastNode=b
a──►e, cost=NA, lastNode=a
a──►f, cost=14, lastNode=a
The lowest cost is a──►c so a──►c is added to the output.
Paths to d and f are cheaper via c so the input is updated to:
a──►d, cost=20, lastNode=c
a──►e, cost=NA, lastNode=a
a──►f, cost=11, lastNode=c
The lowest cost is a──►f so c──►f is added to the output.
The input is updated to:
a──►d, cost=20, lastNode=c
a──►e, cost=NA, lastNode=a
The lowest cost is a──►d so c──►d is added to the output.
There is a connection from d──►e so the input is updated to:
a──►e, cost=26, lastNode=d
Which just leaves adding d──►e to the output.
The output should now be:
[ d──►e
c──►d
c──►f
a──►c
a──►b ]
Task
Implement a version of Dijkstra's algorithm that outputs a set of edges depicting the shortest path to each reachable node from an origin.
Run your program with the following directed graph starting at node a.
Write a program which interprets the output from the above and use it to output the shortest path from node a to nodes e and f.
Vertices
Number
Name
1
a
2
b
3
c
4
d
5
e
6
f
Edges
Start
End
Cost
a
b
7
a
c
9
a
f
14
b
c
10
b
d
15
c
d
11
c
f
2
d
e
6
e
f
9
You can use numbers or names to identify vertices in your program.
See also
Dijkstra's Algorithm vs. A* Search vs. Concurrent Dijkstra's Algorithm (youtube)
| #Applesoft_BASIC | Applesoft BASIC | 100 O$ = "A" : T$ = "EF"
110 DEF FN N(P) = ASC(MID$(N$,P+(P=0),1))-64
120 DIM D(26),UNVISITED(26)
130 DIM PREVIOUS(26) : TRUE = 1
140 LET M = -1 : INFINITY = M
150 FOR I = 0 TO 26
160 LET D(I) = INFINITY : NEXT
170 FOR NE = M TO 1E38 STEP .5
180 READ C$
190 IF LEN(C$) THEN NEXT
200 DIM C(NE),FROM(NE),T(NE)
210 DIM PC(NE) : RESTORE
220 FOR I = 0 TO NE
230 READ C(I), N$
240 LET FROM(I) = FN N(1)
250 LET UNVISITED(FR(I)) = TRUE
260 LET T(I) = FN N(3)
270 LET UNVISITED(T(I)) = TRUE
290 NEXT
300 N$ = O$ : O = FN N(0)
310 D(O) = 0
320 FOR CV = O TO EMPTY STEP 0
330 FOR I = 0 TO NE
340 IF FROM(I) = CV THEN N = T(I) : D = D(CV) + C(I) : IF (D(N) = INFINITY) OR (D < D(N)) THEN D(N) = D : PREVIOUS(N) = CV : PC(N) = C(I)
350 NEXT I
360 LET UNVISITED(CV) = FALSE
370 LET MV = EMPTY
380 FOR I = 1 TO 26
390 IF UNVISITED(I) THEN MD = D(MV) * (MV <> INFINITY) + INFINITY * (MV = INFINITY) : IF (D(I) <> INFINITY) AND ((MD = INFINITY) OR (D(I) < MD)) THEN MV = I
400 NEXT I
410 LET CV = MV * (MD <> INF)
420 NEXT CV : M$ = CHR$(13)
430 PRINT "SHORTEST PATH";
440 FOR I = 1 TO LEN(T$)
450 LET N$ = MID$(T$, I, 1)
460 PRINT M$ " FROM " O$;
470 PRINT " TO "; : N = FN N(0)
480 IF D(N) = INFINITY THEN PRINT N$" DOES NOT EXIST.";
490 IF D(N) <> INFINITY THEN FOR N = N TO FALSE STEP 0 : PRINT CHR$(N + 64); : IF N < > O THEN PRINT " <- "; : N = PREVIOUS(N): NEXT N
500 IF D(N) <> INFINITY THEN PRINT : PRINT " IS "; : N = FN N(0) : PRINT D(N); : H = 15 : FOR N = N TO O STEP 0: IF N < > O THEN P = PREVIOUS(N): PRINT TAB(H)CHR$(43+18*(h=15));TAB(H+2)PC(N); :N = P: H=H+5: NEXT N
510 NEXT I
600 DATA 7,A-B
610 DATA 9,A-C
620 DATA 14,A-F
630 DATA 10,B-C
640 DATA 15,B-D
650 DATA 11,C-D
660 DATA 2,C-F
670 DATA 6,D-E
680 DATA 9,E-F
690 DATA |
http://rosettacode.org/wiki/Digital_root | Digital root | The digital root,
X
{\displaystyle X}
, of a number,
n
{\displaystyle n}
, is calculated:
find
X
{\displaystyle X}
as the sum of the digits of
n
{\displaystyle n}
find a new
X
{\displaystyle X}
by summing the digits of
X
{\displaystyle X}
, repeating until
X
{\displaystyle X}
has only one digit.
The additive persistence is the number of summations required to obtain the single digit.
The task is to calculate the additive persistence and the digital root of a number, e.g.:
627615
{\displaystyle 627615}
has additive persistence
2
{\displaystyle 2}
and digital root of
9
{\displaystyle 9}
;
39390
{\displaystyle 39390}
has additive persistence
2
{\displaystyle 2}
and digital root of
6
{\displaystyle 6}
;
588225
{\displaystyle 588225}
has additive persistence
2
{\displaystyle 2}
and digital root of
3
{\displaystyle 3}
;
393900588225
{\displaystyle 393900588225}
has additive persistence
2
{\displaystyle 2}
and digital root of
9
{\displaystyle 9}
;
The digital root may be calculated in bases other than 10.
See
Casting out nines for this wiki's use of this procedure.
Digital root/Multiplicative digital root
Sum digits of an integer
Digital root sequence on OEIS
Additive persistence sequence on OEIS
Iterated digits squaring
| #ALGOL_W | ALGOL W | begin
% calculates the digital root and persistence of an integer in base 10 %
% in order to allow for numbers larger than 2^31, the number is passed %
% as the lower and upper digits e.g. 393900588225 can be processed by %
% specifying upper = 393900, lower = 58825 %
procedure findDigitalRoot( integer value upper, lower
; integer result digitalRoot, persistence
) ;
begin
integer procedure sumDigits( integer value n ) ;
begin
integer digits, sum;
digits := abs n;
sum := 0;
while digits > 0
do begin
sum := sum + ( digits rem 10 );
digits := digits div 10
end % while digits > 0 % ;
% result: % sum
end sumDigits;
digitalRoot := sumDigits( upper ) + sumDigits( lower );
persistence := 1;
while digitalRoot > 9
do begin
persistence := persistence + 1;
digitalRoot := sumDigits( digitalRoot );
end % while digitalRoot > 9 % ;
end findDigitalRoot ;
% calculates and prints the digital root and persistence %
procedure printDigitalRootAndPersistence( integer value upper, lower ) ;
begin
integer digitalRoot, persistence;
findDigitalRoot( upper, lower, digitalRoot, persistence );
write( s_w := 0 % set field saeparator width for this statement %
, i_w := 8 % set integer field width for this statement %
, upper
, ", "
, lower
, i_w := 2 % change integer field width %
, ": digital root: "
, digitalRoot
, ", persistence: "
, persistence
)
end printDigitalRootAndPersistence ;
% test the digital root and persistence procedures %
printDigitalRootAndPersistence( 0, 627615 );
printDigitalRootAndPersistence( 0, 39390 );
printDigitalRootAndPersistence( 0, 588225 );
printDigitalRootAndPersistence( 393900, 588225 )
end. |
http://rosettacode.org/wiki/Digital_root | Digital root | The digital root,
X
{\displaystyle X}
, of a number,
n
{\displaystyle n}
, is calculated:
find
X
{\displaystyle X}
as the sum of the digits of
n
{\displaystyle n}
find a new
X
{\displaystyle X}
by summing the digits of
X
{\displaystyle X}
, repeating until
X
{\displaystyle X}
has only one digit.
The additive persistence is the number of summations required to obtain the single digit.
The task is to calculate the additive persistence and the digital root of a number, e.g.:
627615
{\displaystyle 627615}
has additive persistence
2
{\displaystyle 2}
and digital root of
9
{\displaystyle 9}
;
39390
{\displaystyle 39390}
has additive persistence
2
{\displaystyle 2}
and digital root of
6
{\displaystyle 6}
;
588225
{\displaystyle 588225}
has additive persistence
2
{\displaystyle 2}
and digital root of
3
{\displaystyle 3}
;
393900588225
{\displaystyle 393900588225}
has additive persistence
2
{\displaystyle 2}
and digital root of
9
{\displaystyle 9}
;
The digital root may be calculated in bases other than 10.
See
Casting out nines for this wiki's use of this procedure.
Digital root/Multiplicative digital root
Sum digits of an integer
Digital root sequence on OEIS
Additive persistence sequence on OEIS
Iterated digits squaring
| #AppleScript | AppleScript | on digitalroot(N as integer)
script math
to sum(L)
if L = {} then return 0
(item 1 of L) + sum(rest of L)
end sum
end script
set i to 0
set M to N
repeat until M < 10
set digits to the characters of (M as text)
set M to math's sum(digits)
set i to i + 1
end repeat
{N:N, persistences:i, root:M}
end digitalroot
digitalroot(627615) |
http://rosettacode.org/wiki/Digital_root/Multiplicative_digital_root | Digital root/Multiplicative digital root | The multiplicative digital root (MDR) and multiplicative persistence (MP) of a number,
n
{\displaystyle n}
, is calculated rather like the Digital root except digits are multiplied instead of being added:
Set
m
{\displaystyle m}
to
n
{\displaystyle n}
and
i
{\displaystyle i}
to
0
{\displaystyle 0}
.
While
m
{\displaystyle m}
has more than one digit:
Find a replacement
m
{\displaystyle m}
as the multiplication of the digits of the current value of
m
{\displaystyle m}
.
Increment
i
{\displaystyle i}
.
Return
i
{\displaystyle i}
(= MP) and
m
{\displaystyle m}
(= MDR)
Task
Tabulate the MP and MDR of the numbers 123321, 7739, 893, 899998
Tabulate MDR versus the first five numbers having that MDR, something like:
MDR: [n0..n4]
=== ========
0: [0, 10, 20, 25, 30]
1: [1, 11, 111, 1111, 11111]
2: [2, 12, 21, 26, 34]
3: [3, 13, 31, 113, 131]
4: [4, 14, 22, 27, 39]
5: [5, 15, 35, 51, 53]
6: [6, 16, 23, 28, 32]
7: [7, 17, 71, 117, 171]
8: [8, 18, 24, 29, 36]
9: [9, 19, 33, 91, 119]
Show all output on this page.
Similar
The Product of decimal digits of n page was redirected here, and had the following description
Find the product of the decimal digits of a positive integer n, where n <= 100
The three existing entries for Phix, REXX, and Ring have been moved here, under ===Similar=== headings, feel free to match or ignore them.
References
Multiplicative Digital Root on Wolfram Mathworld.
Multiplicative digital root on The On-Line Encyclopedia of Integer Sequences.
What's special about 277777788888899? - Numberphile video
| #D | D | import std.stdio, std.algorithm, std.typecons, std.range, std.conv;
/// Multiplicative digital root.
auto mdRoot(in int n) pure /*nothrow*/ {
auto mdr = [n];
while (mdr.back > 9)
mdr ~= reduce!q{a * b}(1, mdr.back.text.map!(d => d - '0'));
//mdr ~= mdr.back.text.map!(d => d - '0').mul;
//mdr ~= mdr.back.reverseDigits.mul;
return tuple(mdr.length - 1, mdr.back);
}
void main() {
"Number: (MP, MDR)\n====== =========".writeln;
foreach (immutable n; [123321, 7739, 893, 899998])
writefln("%6d: (%s, %s)", n, n.mdRoot[]);
auto table = (int[]).init.repeat.enumerate!int.take(10).assocArray;
auto n = 0;
while (table.byValue.map!walkLength.reduce!min < 5) {
table[n.mdRoot[1]] ~= n;
n++;
}
"\nMP: [n0..n4]\n== ========".writeln;
foreach (const mp; table.byKey.array.sort())
writefln("%2d: %s", mp, table[mp].take(5));
} |
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #Tcl | Tcl | package require Tk
set pi [expr acos(-1)]
set r2 [expr sqrt(2)]
proc turn {degrees} {
global a pi
set a [expr {$a + $degrees*$pi/180}]
}
proc forward {len} {
global a coords
lassign [lrange $coords end-1 end] x y
lappend coords [expr {$x + cos($a)*$len}] [expr {$y + sin($a)*$len}]
}
proc dragon {len split {d 1}} {
global r2 coords
if {$split == 0} {
forward $len
return
}
# This next part is only necessary to allow the illustration of progress
if {$split == 10 && [llength $::coords]>2} {
.c coords dragon $::coords
update
}
incr split -1
set sublen [expr {$len/$r2}]
turn [expr {$d*45}]
dragon $sublen $split 1
turn [expr {$d*-90}]
dragon $sublen $split -1
turn [expr {$d*45}]
}
set coords {150 180}
set a 0.0
pack [canvas .c -width 700 -height 500]
.c create line {0 0 0 0} -tag dragon
dragon 400 17
.c coords dragon $coords |
http://rosettacode.org/wiki/Dinesman%27s_multiple-dwelling_problem | Dinesman's multiple-dwelling problem | Task
Solve Dinesman's multiple dwelling problem but in a way that most naturally follows the problem statement given below.
Solutions are allowed (but not required) to parse and interpret the problem text, but should remain flexible and should state what changes to the problem text are allowed. Flexibility and ease of expression are valued.
Examples may be be split into "setup", "problem statement", and "output" sections where the ease and naturalness of stating the problem and getting an answer, as well as the ease and flexibility of modifying the problem are the primary concerns.
Example output should be shown here, as well as any comments on the examples flexibility.
The problem
Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house that contains only five floors.
Baker does not live on the top floor.
Cooper does not live on the bottom floor.
Fletcher does not live on either the top or the bottom floor.
Miller lives on a higher floor than does Cooper.
Smith does not live on a floor adjacent to Fletcher's.
Fletcher does not live on a floor adjacent to Cooper's.
Where does everyone live?
| #C.2B.2B | C++ | #include <algorithm>
#include <array>
#include <cmath>
#include <functional>
#include <string>
#include <iostream>
#include <list>
int main() {
constexpr auto floors = 5u;
constexpr auto top = floors - 1u, bottom = 0u;
using namespace std;
array<string, floors> tenants = { "Baker", "Cooper", "Fletcher", "Miller", "Smith" };
const auto floor_of = [&tenants](string t) {
for (int i = bottom; i <= top; i++)
if (tenants[i] == t) return i;
throw "invalid tenant";
};
const list<function<bool()>> constraints = {
[&tenants]() { return tenants[top] != "Baker"; },
[&tenants]() { return tenants[bottom] != "Cooper"; },
[&tenants]() { return tenants[top] != "Fletcher"; },
[&tenants]() { return tenants[bottom] != "Fletcher"; },
[&floor_of]() { return floor_of("Miller") > floor_of("Cooper"); },
[&floor_of]() { return abs(floor_of("Fletcher") - floor_of("Smith")) != 1; },
[&floor_of]() { return abs(floor_of("Fletcher") - floor_of("Cooper")) != 1; }
};
sort(tenants.begin(), tenants.end());
do {
if (all_of(constraints.begin(), constraints.end(), [](auto f) { return f(); } )) {
for (const auto &t : tenants) cout << t << ' ';
cout << endl;
}
} while (next_permutation(tenants.begin(), tenants.end()));
return EXIT_SUCCESS;
} |
http://rosettacode.org/wiki/Dot_product | Dot product | Task
Create a function/use an in-built function, to compute the dot product, also known as the scalar product of two vectors.
If possible, make the vectors of arbitrary length.
As an example, compute the dot product of the vectors:
[1, 3, -5] and
[4, -2, -1]
If implementing the dot product of two vectors directly:
each vector must be the same length
multiply corresponding terms from each vector
sum the products (to produce the answer)
Related task
Vector products
| #Clojure | Clojure | (defn dot-product [& matrix]
{:pre [(apply == (map count matrix))]}
(apply + (apply map * matrix)))
(defn dot-product2 [x y]
(->> (interleave x y)
(partition 2 2)
(map #(apply * %))
(reduce +)))
(defn dot-product3
"Dot product of vectors. Tested on version 1.8.0."
[v1 v2]
{:pre [(= (count v1) (count v2))]}
(reduce + (map * v1 v2)))
;Example Usage
(println (dot-product [1 3 -5] [4 -2 -1]))
(println (dot-product2 [1 3 -5] [4 -2 -1]))
(println (dot-product3 [1 3 -5] [4 -2 -1]))
|
http://rosettacode.org/wiki/Doubly-linked_list/Definition | Doubly-linked list/Definition | Define the data structure for a complete Doubly Linked List.
The structure should support adding elements to the head, tail and middle of the list.
The structure should not allow circular loops
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Objeck | Objeck |
use Collection;
class Program {
function : Main(args : String[]) ~ Nil {
list := List->New();
list->AddFront("first");
list->AddBack("last");
list->Insert("middle");
list->Forward();
do {
list->Get()->As(String)->PrintLine();
list->Previous();
}
while(list->Get() <> Nil);
}
} |
http://rosettacode.org/wiki/Dice_game_probabilities | Dice game probabilities | Two players have a set of dice each. The first player has nine dice with four faces each, with numbers one to four. The second player has six normal dice with six faces each, each face has the usual numbers from one to six.
They roll their dice and sum the totals of the faces. The player with the highest total wins (it's a draw if the totals are the same). What's the probability of the first player beating the second player?
Later the two players use a different set of dice each. Now the first player has five dice with ten faces each, and the second player has six dice with seven faces each. Now what's the probability of the first player beating the second player?
This task was adapted from the Project Euler Problem n.205:
https://projecteuler.net/problem=205
| #C.2B.2B | C++ | #include <cmath>
#include <cstdint>
#include <iomanip>
#include <iostream>
#include <map>
// Returns map from sum of faces to number of ways it can happen
std::map<uint32_t, uint32_t> get_totals(uint32_t dice, uint32_t faces) {
std::map<uint32_t, uint32_t> result;
for (uint32_t i = 1; i <= faces; ++i)
result.emplace(i, 1);
for (uint32_t d = 2; d <= dice; ++d) {
std::map<uint32_t, uint32_t> tmp;
for (const auto& p : result) {
for (uint32_t i = 1; i <= faces; ++i)
tmp[p.first + i] += p.second;
}
tmp.swap(result);
}
return result;
}
double probability(uint32_t dice1, uint32_t faces1, uint32_t dice2, uint32_t faces2) {
auto totals1 = get_totals(dice1, faces1);
auto totals2 = get_totals(dice2, faces2);
double wins = 0;
for (const auto& p1 : totals1) {
for (const auto& p2 : totals2) {
if (p2.first >= p1.first)
break;
wins += p1.second * p2.second;
}
}
double total = std::pow(faces1, dice1) * std::pow(faces2, dice2);
return wins/total;
}
int main() {
std::cout << std::setprecision(10);
std::cout << probability(9, 4, 6, 6) << '\n';
std::cout << probability(5, 10, 6, 7) << '\n';
return 0;
} |
http://rosettacode.org/wiki/Dining_philosophers | Dining philosophers | The dining philosophers problem illustrates non-composability of low-level synchronization primitives like semaphores. It is a modification of a problem posed by Edsger Dijkstra.
Five philosophers, Aristotle, Kant, Spinoza, Marx, and Russell (the tasks) spend their time thinking and eating spaghetti. They eat at a round table with five individual seats. For eating each philosopher needs two forks (the resources). There are five forks on the table, one left and one right of each seat. When a philosopher cannot grab both forks it sits and waits. Eating takes random time, then the philosopher puts the forks down and leaves the dining room. After spending some random time thinking about the nature of the universe, he again becomes hungry, and the circle repeats itself.
It can be observed that a straightforward solution, when forks are implemented by semaphores, is exposed to deadlock. There exist two deadlock states when all five philosophers are sitting at the table holding one fork each. One deadlock state is when each philosopher has grabbed the fork left of him, and another is when each has the fork on his right.
There are many solutions of the problem, program at least one, and explain how the deadlock is prevented.
| #BBC_BASIC | BBC BASIC | INSTALL @lib$+"TIMERLIB"
nSeats% = 5
DIM Name$(nSeats%-1), Fork%(nSeats%-1), tID%(nSeats%-1), Leftie%(nSeats%-1)
Name$() = "Aristotle", "Kant", "Spinoza", "Marx", "Russell"
Fork%() = TRUE : REM All forks are initially on the table
Leftie%(RND(nSeats%)-1) = TRUE : REM One philosopher is lefthanded
tID%(0) = FN_ontimer(10, PROCphilosopher0, 1)
tID%(1) = FN_ontimer(10, PROCphilosopher1, 1)
tID%(2) = FN_ontimer(10, PROCphilosopher2, 1)
tID%(3) = FN_ontimer(10, PROCphilosopher3, 1)
tID%(4) = FN_ontimer(10, PROCphilosopher4, 1)
ON CLOSE PROCcleanup : QUIT
ON ERROR PRINT REPORT$ : PROCcleanup : END
DEF PROCphilosopher0 : PROCtask(0) : ENDPROC
DEF PROCphilosopher1 : PROCtask(1) : ENDPROC
DEF PROCphilosopher2 : PROCtask(2) : ENDPROC
DEF PROCphilosopher3 : PROCtask(3) : ENDPROC
DEF PROCphilosopher4 : PROCtask(4) : ENDPROC
REPEAT
WAIT 0
UNTIL FALSE
END
DEF PROCtask(n%)
PRIVATE state%(), lh%(), rh%()
DIM state%(nSeats%-1), lh%(nSeats%-1), rh%(nSeats%-1)
REM States: 0 = waiting for forks, > 0 = eating, < 0 = left the room
CASE TRUE OF
WHEN state%(n%) < 0:
state%(n%) += 1 : REM Waiting to get hungry again
IF state%(n%) = 0 PRINT Name$(n%) " is hungry again"
WHEN state%(n%) > 0:
state%(n%) -= 1 : REM Eating
IF state%(n%) = 0 THEN
SWAP Fork%((n%-1+nSeats%) MOD nSeats%), lh%(n%)
SWAP Fork%((n% + 1) MOD nSeats%), rh%(n%)
state%(n%) = -RND(100)
PRINT Name$(n%) " is leaving the room"
ENDIF
WHEN state%(n%) = 0:
IF Leftie%(n%) THEN
IF NOT lh%(n%) SWAP Fork%((n%-1+nSeats%) MOD nSeats%), lh%(n%)
IF lh%(n%) IF NOT rh%(n%) SWAP Fork%((n% + 1) MOD nSeats%), rh%(n%)
ELSE
IF NOT rh%(n%) SWAP Fork%((n% + 1) MOD nSeats%), rh%(n%)
IF rh%(n%) IF NOT lh%(n%) SWAP Fork%((n%-1+nSeats%) MOD nSeats%), lh%(n%)
ENDIF
IF lh%(n%) AND rh%(n%) THEN
state%(n%) = RND(100)
PRINT Name$(n%) " is eating (" ; state%(n%) " ticks)"
ENDIF
ENDCASE
ENDPROC
DEF PROCcleanup
LOCAL I%
FOR I% = 0 TO nSeats%-1
PROC_killtimer(tID%(I%))
NEXT
ENDPROC |
http://rosettacode.org/wiki/Discordian_date | Discordian date |
Task
Convert a given date from the Gregorian calendar to the Discordian calendar.
| #AppleScript | AppleScript | on gregorianToDiscordian(inputDate) -- Input: AppleScript date object.
(*
Discordian years are aligned with, and the same length as, Gregorian years.
Each has 73 5-day weeks and 5 73-day seasons. (73 * 5 = 365.)
The first day of a Discordian year is also that of its first week and first season.
In leap years, an extra day called "St. Tib's Day", is inserted between days 59 and 60.
It's considered to be outside the calendar, so the day after it is Setting Orange, Chaos 60,
not Sweetmorn, Chaos 61. Year 1 YOLD is 1166 BC, but this handler takes an AS date object
as its input and is only good for AD Gregorian dates. Since the Discordian calendar's an
American invention, the output here's in the US style: "Weekday, Season day, year".
*)
-- Calculate the input date's day-of-year number.
copy inputDate to startOfYear
tell startOfYear to set {its day, its month} to {1, January}
set dayOfYear to (inputDate - startOfYear) div days + 1
-- If it's a leap year, special-case St. Tib's Day, or adjust the day-of-year number if the day comes after that.
set y to inputDate's year
if ((y mod 4 is 0) and ((y mod 100 > 0) or (y mod 400 is 0))) then
if (dayOfYear is 60) then
set dayOfYear to "St. Tib's Day"
else if (dayOfYear > 60) then
set dayOfYear to dayOfYear - 1
end if
end if
-- Start the output text with either "St. Tib's Day" or the weekday, season, and day-of-season number.
if (dayOfYear is "St. Tib's Day") then
set outputText to dayOfYear
else
tell dayOfYear - 1
set dayOfWeek to it mod 5 + 1
set seasonNumber to it div 73 + 1
set dayOfSeason to it mod 73 + 1
end tell
set theWeekdays to {"Sweetmorn", "Boomtime", "Pungenday", "Prickle-Prickle", "Setting Orange"}
set theSeasons to {"Chaos ", "Discord ", "Confusion ", "Bureaucracy ", "The Aftermath "}
set outputText to (item dayOfWeek of theWeekdays) & ", " & (item seasonNumber of theSeasons) & dayOfSeason
end if
-- Append the Discordian year number and return the result.
return outputText & ", " & (y + 1166)
end gregorianToDiscordian
set ASDate to (current date)
set gregorianDate to ASDate's date string
set discordianDate to gregorianToDiscordian(ASDate)
return {Gregorian:gregorianDate, Discordian:discordianDate} |
http://rosettacode.org/wiki/Dijkstra%27s_algorithm | Dijkstra's algorithm | This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Dijkstra's algorithm, conceived by Dutch computer scientist Edsger Dijkstra in 1956 and published in 1959, is a graph search algorithm that solves the single-source shortest path problem for a graph with non-negative edge path costs, producing a shortest path tree.
This algorithm is often used in routing and as a subroutine in other graph algorithms.
For a given source vertex (node) in the graph, the algorithm finds the path with lowest cost (i.e. the shortest path) between that vertex and every other vertex.
For instance
If the vertices of the graph represent cities and edge path costs represent driving distances between pairs of cities connected by a direct road, Dijkstra's algorithm can be used to find the shortest route between one city and all other cities.
As a result, the shortest path first is widely used in network routing protocols, most notably:
IS-IS (Intermediate System to Intermediate System) and
OSPF (Open Shortest Path First).
Important note
The inputs to Dijkstra's algorithm are a directed and weighted graph consisting of 2 or more nodes, generally represented by:
an adjacency matrix or list, and
a start node.
A destination node is not specified.
The output is a set of edges depicting the shortest path to each destination node.
An example, starting with
a──►b, cost=7, lastNode=a
a──►c, cost=9, lastNode=a
a──►d, cost=NA, lastNode=a
a──►e, cost=NA, lastNode=a
a──►f, cost=14, lastNode=a
The lowest cost is a──►b so a──►b is added to the output.
There is a connection from b──►d so the input is updated to:
a──►c, cost=9, lastNode=a
a──►d, cost=22, lastNode=b
a──►e, cost=NA, lastNode=a
a──►f, cost=14, lastNode=a
The lowest cost is a──►c so a──►c is added to the output.
Paths to d and f are cheaper via c so the input is updated to:
a──►d, cost=20, lastNode=c
a──►e, cost=NA, lastNode=a
a──►f, cost=11, lastNode=c
The lowest cost is a──►f so c──►f is added to the output.
The input is updated to:
a──►d, cost=20, lastNode=c
a──►e, cost=NA, lastNode=a
The lowest cost is a──►d so c──►d is added to the output.
There is a connection from d──►e so the input is updated to:
a──►e, cost=26, lastNode=d
Which just leaves adding d──►e to the output.
The output should now be:
[ d──►e
c──►d
c──►f
a──►c
a──►b ]
Task
Implement a version of Dijkstra's algorithm that outputs a set of edges depicting the shortest path to each reachable node from an origin.
Run your program with the following directed graph starting at node a.
Write a program which interprets the output from the above and use it to output the shortest path from node a to nodes e and f.
Vertices
Number
Name
1
a
2
b
3
c
4
d
5
e
6
f
Edges
Start
End
Cost
a
b
7
a
c
9
a
f
14
b
c
10
b
d
15
c
d
11
c
f
2
d
e
6
e
f
9
You can use numbers or names to identify vertices in your program.
See also
Dijkstra's Algorithm vs. A* Search vs. Concurrent Dijkstra's Algorithm (youtube)
| #Arturo | Arturo | define :graph [vertices, neighbours][]
initGraph: function [edges][
vs: []
ns: #[]
loop edges 'e [
[src, dst, cost]: e
vs: sort unique append vs src
vs: sort unique append vs dst
if not? key? ns src -> ns\[src]: []
ns\[src]: ns\[src] ++ @[@[dst, cost]]
]
to :graph @[vs ns]
]
Inf: 1234567890
dijkstraPath: function [gr, fst, lst][
dist: #[]
prev: #[]
result: new []
notSeen: new gr\vertices
loop gr\vertices 'vertex ->
dist\[vertex]: Inf
dist\[fst]: 0
while [0 < size notSeen][
vertex1: ""
mindist: Inf
loop notSeen 'vertex [
if dist\[vertex] < mindist [
vertex1: vertex
mindist: dist\[vertex]
]
]
if vertex1 = lst -> break
'notSeen -- vertex1
if key? gr\neighbours vertex1 [
loop gr\neighbours\[vertex1] 'v [
[vertex2, cost]: v
if contains? notSeen vertex2 [
altdist: dist\[vertex1] + cost
if altdist < dist\[vertex2][
dist\[vertex2]: altdist
prev\[vertex2]: vertex1
]
]
]
]
]
vertex: lst
while [not? empty? vertex][
'result ++ vertex
vertex: (key? prev vertex)? -> prev\[vertex] -> null
]
reverse 'result
return result
]
graph: initGraph [
["a" "b" 7] ["a" "c" 9] ["a" "f" 14]
["b" "c" 10] ["b" "d" 15] ["c" "d" 11]
["c" "f" 2] ["d" "e" 6] ["e" "f" 9]
]
print ["Shortest path from 'a' to 'e': " join.with:" -> " dijkstraPath graph "a" "e"]
print ["Shortest path from 'a' to 'f': " join.with:" -> " dijkstraPath graph "a" "f"] |
http://rosettacode.org/wiki/Digital_root | Digital root | The digital root,
X
{\displaystyle X}
, of a number,
n
{\displaystyle n}
, is calculated:
find
X
{\displaystyle X}
as the sum of the digits of
n
{\displaystyle n}
find a new
X
{\displaystyle X}
by summing the digits of
X
{\displaystyle X}
, repeating until
X
{\displaystyle X}
has only one digit.
The additive persistence is the number of summations required to obtain the single digit.
The task is to calculate the additive persistence and the digital root of a number, e.g.:
627615
{\displaystyle 627615}
has additive persistence
2
{\displaystyle 2}
and digital root of
9
{\displaystyle 9}
;
39390
{\displaystyle 39390}
has additive persistence
2
{\displaystyle 2}
and digital root of
6
{\displaystyle 6}
;
588225
{\displaystyle 588225}
has additive persistence
2
{\displaystyle 2}
and digital root of
3
{\displaystyle 3}
;
393900588225
{\displaystyle 393900588225}
has additive persistence
2
{\displaystyle 2}
and digital root of
9
{\displaystyle 9}
;
The digital root may be calculated in bases other than 10.
See
Casting out nines for this wiki's use of this procedure.
Digital root/Multiplicative digital root
Sum digits of an integer
Digital root sequence on OEIS
Additive persistence sequence on OEIS
Iterated digits squaring
| #Applesoft_BASIC | Applesoft BASIC | 1 GOSUB 430"BASE SETUP
2 FOR E = 0 TO 1 STEP 0
3 GOSUB 7"READ
4 ON E + 1 GOSUB 50, 10
5 NEXT E
6 END
7 READ N$
8 E = N$ = ""
9 RETURN
10 GOSUB 7"READ BASE
20 IF E THEN RETURN
30 BASE = VAL(N$)
40 READ N$
50 GOSUB 100"DIGITAL ROOT
60 GOSUB 420: PRINT " HAS AD";
70 PRINT "DITIVE PERSISTENCE";
80 PRINT " "P" AND DIGITAL R";
90 PRINT "OOT "X$";" : RETURN
REM DIGITAL ROOT OF N$, RETURNS X$ AND P
100 P = 0 : L = LEN(N$)
110 X$ = MID$(N$, 2, L - 1)
120 N = LEFT$(X$, 1) = "-"
130 IF NOT N THEN X$ = N$
140 FOR P = 0 TO 1E38
150 L = LEN(X$)
160 IF L < 2 THEN RETURN
170 GOSUB 200"DIGIT SUM
180 X$ = S$
190 NEXT P : STOP
REM DIGIT SUM OF X$, RETURNS S$
200 S$ = "0"
210 R$ = X$
220 L = LEN(R$)
230 FOR L = L TO 1 STEP -1
240 E$ = "" : V$ = RIGHT$(R$, 1)
250 GOSUB 400 : S = LEN(S$)
260 ON R$ <> "0" GOSUB 300
270 R$ = MID$(R$, 1, L - 1)
280 NEXT L
290 RETURN
REM ADD V TO S$
300 FOR C = V TO 0 STEP 0
310 V$ = RIGHT$(S$, 1)
320 GOSUB 400 : S = S - 1
330 S$ = MID$(S$, 1, S)
340 V = V + C : C = V >= BASE
350 IF C THEN V = V - BASE
360 GOSUB 410 : E$ = V$ + E$
370 IF S THEN NEXT C
380 IF C THEN S$ = "1"
390 S$ = S$ + E$ : RETURN
REM BASE VAL
400 V = V(ASC(V$)) : RETURN
REM BASE STR$
410 V$ = V$(V) : RETURN
REM BASE DISPLAY
420 PRINT N$;
421 IF BASE = 10 THEN RETURN
422 PRINT "("BASE")";
423 RETURN
REM BASE SETUP
430 IF BASE = 0 THEN BASE = 10
440 DIM V(127), V$(35)
450 FOR I = 0 TO 35
460 V = 55 + I - (I < 10) * 7
470 V$(I) = CHR$(V)
480 V(V) = I
490 NEXT I : RETURN
500 DATA627615,39390,588225
510 DATA393900588225
1000 DATA,30
1010 DATADIGITALROOT
63999DATA, |
http://rosettacode.org/wiki/Digital_root | Digital root | The digital root,
X
{\displaystyle X}
, of a number,
n
{\displaystyle n}
, is calculated:
find
X
{\displaystyle X}
as the sum of the digits of
n
{\displaystyle n}
find a new
X
{\displaystyle X}
by summing the digits of
X
{\displaystyle X}
, repeating until
X
{\displaystyle X}
has only one digit.
The additive persistence is the number of summations required to obtain the single digit.
The task is to calculate the additive persistence and the digital root of a number, e.g.:
627615
{\displaystyle 627615}
has additive persistence
2
{\displaystyle 2}
and digital root of
9
{\displaystyle 9}
;
39390
{\displaystyle 39390}
has additive persistence
2
{\displaystyle 2}
and digital root of
6
{\displaystyle 6}
;
588225
{\displaystyle 588225}
has additive persistence
2
{\displaystyle 2}
and digital root of
3
{\displaystyle 3}
;
393900588225
{\displaystyle 393900588225}
has additive persistence
2
{\displaystyle 2}
and digital root of
9
{\displaystyle 9}
;
The digital root may be calculated in bases other than 10.
See
Casting out nines for this wiki's use of this procedure.
Digital root/Multiplicative digital root
Sum digits of an integer
Digital root sequence on OEIS
Additive persistence sequence on OEIS
Iterated digits squaring
| #Arturo | Arturo | droot: function [num][
persistence: 0
until [
num: sum to [:integer] split to :string num
persistence: persistence + 1
][ num < 10 ]
return @[num, persistence]
]
loop [627615, 39390, 588225, 393900588225] 'i [
a: droot i
print [i "has additive persistence" a\0 "and digital root of" a\1]
] |
http://rosettacode.org/wiki/Digital_root/Multiplicative_digital_root | Digital root/Multiplicative digital root | The multiplicative digital root (MDR) and multiplicative persistence (MP) of a number,
n
{\displaystyle n}
, is calculated rather like the Digital root except digits are multiplied instead of being added:
Set
m
{\displaystyle m}
to
n
{\displaystyle n}
and
i
{\displaystyle i}
to
0
{\displaystyle 0}
.
While
m
{\displaystyle m}
has more than one digit:
Find a replacement
m
{\displaystyle m}
as the multiplication of the digits of the current value of
m
{\displaystyle m}
.
Increment
i
{\displaystyle i}
.
Return
i
{\displaystyle i}
(= MP) and
m
{\displaystyle m}
(= MDR)
Task
Tabulate the MP and MDR of the numbers 123321, 7739, 893, 899998
Tabulate MDR versus the first five numbers having that MDR, something like:
MDR: [n0..n4]
=== ========
0: [0, 10, 20, 25, 30]
1: [1, 11, 111, 1111, 11111]
2: [2, 12, 21, 26, 34]
3: [3, 13, 31, 113, 131]
4: [4, 14, 22, 27, 39]
5: [5, 15, 35, 51, 53]
6: [6, 16, 23, 28, 32]
7: [7, 17, 71, 117, 171]
8: [8, 18, 24, 29, 36]
9: [9, 19, 33, 91, 119]
Show all output on this page.
Similar
The Product of decimal digits of n page was redirected here, and had the following description
Find the product of the decimal digits of a positive integer n, where n <= 100
The three existing entries for Phix, REXX, and Ring have been moved here, under ===Similar=== headings, feel free to match or ignore them.
References
Multiplicative Digital Root on Wolfram Mathworld.
Multiplicative digital root on The On-Line Encyclopedia of Integer Sequences.
What's special about 277777788888899? - Numberphile video
| #Elixir | Elixir | defmodule Digital do
def mdroot(n), do: mdroot(n, 0)
defp mdroot(n, persist) when n < 10, do: {n, persist}
defp mdroot(n, persist), do: mdroot(product(n, 1), persist+1)
defp product(0, prod), do: prod
defp product(n, prod), do: product(div(n, 10), prod*rem(n, 10))
def task1(data) do
IO.puts "Number: MDR MP\n====== === =="
Enum.each(data, fn n ->
{mdr, persist} = mdroot(n)
:io.format "~6w: ~w ~2w~n", [n, mdr, persist]
end)
end
def task2(m \\ 5) do
IO.puts "\nMDR: [n0..n#{m-1}]\n=== ========"
map = add_map(0, m, Map.new)
Enum.each(0..9, fn i ->
first = map[i] |> Enum.reverse |> Enum.take(m)
IO.puts " #{i}: #{inspect first}"
end)
end
defp add_map(n, m, map) do
{mdr, _persist} = mdroot(n)
new_map = Map.update(map, mdr, [n], fn vals -> [n | vals] end)
min_len = Map.values(new_map) |> Enum.map(&length(&1)) |> Enum.min
if min_len < m, do: add_map(n+1, m, new_map),
else: new_map
end
end
Digital.task1([123321, 7739, 893, 899998])
Digital.task2 |
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #Plain_TeX | Plain TeX | \input tikz.tex
\usetikzlibrary{lindenmayersystems}
\pgfdeclarelindenmayersystem{Dragon curve}{
\symbol{S}{\pgflsystemdrawforward}
\rule{F -> F+S}
\rule{S -> F-S}
}
\tikzpicture
\draw
[lindenmayer system={Dragon curve, step=10pt, axiom=F, order=8}]
lindenmayer system;
\endtikzpicture
\bye |
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