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http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #COBOL | COBOL | identification division.
program-id. arbitrary-precision-integers.
remarks. Uses opaque libgmp internals that are built into libcob.
data division.
working-storage section.
01 gmp-number.
05 mp-alloc usage binary-long.
05 mp-size usage binary-long.
05 mp-limb usage pointer.
01 gmp-build.
05 mp-alloc usage binary-long.
05 mp-size usage binary-long.
05 mp-limb usage pointer.
01 the-int usage binary-c-long unsigned.
01 the-exponent usage binary-c-long unsigned.
01 valid-exponent usage binary-long value 1.
88 cant-use value 0 when set to false 1.
01 number-string usage pointer.
01 number-length usage binary-long.
01 window-width constant as 20.
01 limit-width usage binary-long.
01 number-buffer pic x(window-width) based.
procedure division.
arbitrary-main.
*> calculate 10 ** 19
perform initialize-integers.
display "10 ** 19 : " with no advancing
move 10 to the-int
move 19 to the-exponent
perform raise-pow-accrete-exponent
perform show-all-or-portion
perform clean-up
*> calculate 12345 ** 9
perform initialize-integers.
display "12345 ** 9 : " with no advancing
move 12345 to the-int
move 9 to the-exponent
perform raise-pow-accrete-exponent
perform show-all-or-portion
perform clean-up
*> calculate 5 ** 4 ** 3 ** 2
perform initialize-integers.
display "5 ** 4 ** 3 ** 2: " with no advancing
move 3 to the-int
move 2 to the-exponent
perform raise-pow-accrete-exponent
move 4 to the-int
perform raise-pow-accrete-exponent
move 5 to the-int
perform raise-pow-accrete-exponent
perform show-all-or-portion
perform clean-up
goback.
*> **************************************************************
initialize-integers.
call "__gmpz_init" using gmp-number returning omitted
call "__gmpz_init" using gmp-build returning omitted
.
raise-pow-accrete-exponent.
*> check before using previously overflowed exponent intermediate
if cant-use then
display "Error: intermediate overflow occured at "
the-exponent upon syserr
goback
end-if
call "__gmpz_set_ui" using gmp-number by value 0
returning omitted
call "__gmpz_set_ui" using gmp-build by value the-int
returning omitted
call "__gmpz_pow_ui" using gmp-number gmp-build
by value the-exponent
returning omitted
call "__gmpz_set_ui" using gmp-build by value 0
returning omitted
call "__gmpz_get_ui" using gmp-number returning the-exponent
call "__gmpz_fits_ulong_p" using gmp-number
returning valid-exponent
.
*> get string representation, base 10
show-all-or-portion.
call "__gmpz_sizeinbase" using gmp-number
by value 10
returning number-length
display "GMP length: " number-length ", " with no advancing
call "__gmpz_get_str" using null by value 10
by reference gmp-number
returning number-string
call "strlen" using by value number-string
returning number-length
display "strlen: " number-length
*> slide based string across first and last of buffer
move window-width to limit-width
set address of number-buffer to number-string
if number-length <= window-width then
move number-length to limit-width
display number-buffer(1:limit-width)
else
display number-buffer with no advancing
subtract window-width from number-length
move function max(0, number-length) to number-length
if number-length <= window-width then
move number-length to limit-width
else
display "..." with no advancing
end-if
set address of number-buffer up by
function max(window-width, number-length)
display number-buffer(1:limit-width)
end-if
.
clean-up.
call "free" using by value number-string returning omitted
call "__gmpz_clear" using gmp-number returning omitted
call "__gmpz_clear" using gmp-build returning omitted
set address of number-buffer to null
set cant-use to false
.
end program arbitrary-precision-integers.
|
http://rosettacode.org/wiki/Zhang-Suen_thinning_algorithm | Zhang-Suen thinning algorithm | This is an algorithm used to thin a black and white i.e. one bit per pixel images.
For example, with an input image of:
################# #############
################## ################
################### ##################
######## ####### ###################
###### ####### ####### ######
###### ####### #######
################# #######
################ #######
################# #######
###### ####### #######
###### ####### #######
###### ####### ####### ######
######## ####### ###################
######## ####### ###### ################## ######
######## ####### ###### ################ ######
######## ####### ###### ############# ######
It produces the thinned output:
# ########## #######
## # #### #
# # ##
# # #
# # #
# # #
############ #
# # #
# # #
# # #
# # #
# ##
# ############
### ###
Algorithm
Assume black pixels are one and white pixels zero, and that the input image is a rectangular N by M array of ones and zeroes.
The algorithm operates on all black pixels P1 that can have eight neighbours.
The neighbours are, in order, arranged as:
P9 P2 P3
P8 P1 P4
P7 P6 P5
Obviously the boundary pixels of the image cannot have the full eight neighbours.
Define
A
(
P
1
)
{\displaystyle A(P1)}
= the number of transitions from white to black, (0 -> 1) in the sequence P2,P3,P4,P5,P6,P7,P8,P9,P2. (Note the extra P2 at the end - it is circular).
Define
B
(
P
1
)
{\displaystyle B(P1)}
= The number of black pixel neighbours of P1. ( = sum(P2 .. P9) )
Step 1
All pixels are tested and pixels satisfying all the following conditions (simultaneously) are just noted at this stage.
(0) The pixel is black and has eight neighbours
(1)
2
<=
B
(
P
1
)
<=
6
{\displaystyle 2<=B(P1)<=6}
(2) A(P1) = 1
(3) At least one of P2 and P4 and P6 is white
(4) At least one of P4 and P6 and P8 is white
After iterating over the image and collecting all the pixels satisfying all step 1 conditions, all these condition satisfying pixels are set to white.
Step 2
All pixels are again tested and pixels satisfying all the following conditions are just noted at this stage.
(0) The pixel is black and has eight neighbours
(1)
2
<=
B
(
P
1
)
<=
6
{\displaystyle 2<=B(P1)<=6}
(2) A(P1) = 1
(3) At least one of P2 and P4 and P8 is white
(4) At least one of P2 and P6 and P8 is white
After iterating over the image and collecting all the pixels satisfying all step 2 conditions, all these condition satisfying pixels are again set to white.
Iteration
If any pixels were set in this round of either step 1 or step 2 then all steps are repeated until no image pixels are so changed.
Task
Write a routine to perform Zhang-Suen thinning on an image matrix of ones and zeroes.
Use the routine to thin the following image and show the output here on this page as either a matrix of ones and zeroes, an image, or an ASCII-art image of space/non-space characters.
00000000000000000000000000000000
01111111110000000111111110000000
01110001111000001111001111000000
01110000111000001110000111000000
01110001111000001110000000000000
01111111110000001110000000000000
01110111100000001110000111000000
01110011110011101111001111011100
01110001111011100111111110011100
00000000000000000000000000000000
Reference
Zhang-Suen Thinning Algorithm, Java Implementation by Nayef Reza.
"Character Recognition Systems: A Guide for Students and Practitioners" By Mohamed Cheriet, Nawwaf Kharma, Cheng-Lin Liu, Ching Suen
| #JavaScript | JavaScript | function Point(x, y) {
this.x = x;
this.y = y;
}
var ZhangSuen = (function () {
function ZhangSuen() {
}
ZhangSuen.image =
[" ",
" ################# ############# ",
" ################## ################ ",
" ################### ################## ",
" ######## ####### ################### ",
" ###### ####### ####### ###### ",
" ###### ####### ####### ",
" ################# ####### ",
" ################ ####### ",
" ################# ####### ",
" ###### ####### ####### ",
" ###### ####### ####### ",
" ###### ####### ####### ###### ",
" ######## ####### ################### ",
" ######## ####### ###### ################## ###### ",
" ######## ####### ###### ################ ###### ",
" ######## ####### ###### ############# ###### ",
" "];
ZhangSuen.nbrs = [[0, -1], [1, -1], [1, 0], [1, 1], [0, 1], [-1, 1], [-1, 0], [-1, -1], [0, -1]];
ZhangSuen.nbrGroups = [[[0, 2, 4], [2, 4, 6]], [[0, 2, 6], [0, 4, 6]]];
ZhangSuen.toWhite = new Array();
;
ZhangSuen.main = function (args) {
ZhangSuen.grid = new Array(ZhangSuen.image.length);
for (var r = 0; r < ZhangSuen.image.length; r++)
ZhangSuen.grid[r] = (ZhangSuen.image[r]).split('');
ZhangSuen.thinImage();
};
ZhangSuen.thinImage = function () {
var firstStep = false;
var hasChanged;
do {
hasChanged = false;
firstStep = !firstStep;
for (var r = 1; r < ZhangSuen.grid.length - 1; r++) {
for (var c = 1; c < ZhangSuen.grid[0].length - 1; c++) {
if (ZhangSuen.grid[r][c] !== '#')
continue;
var nn = ZhangSuen.numNeighbors(r, c);
if (nn < 2 || nn > 6)
continue;
if (ZhangSuen.numTransitions(r, c) !== 1)
continue;
if (!ZhangSuen.atLeastOneIsWhite(r, c, firstStep ? 0 : 1))
continue;
ZhangSuen.toWhite.push(new Point(c, r));
hasChanged = true;
}
}
for (let i = 0; i < ZhangSuen.toWhite.length; i++) {
var p = ZhangSuen.toWhite[i];
ZhangSuen.grid[p.y][p.x] = ' ';
}
ZhangSuen.toWhite = new Array();
} while ((firstStep || hasChanged));
ZhangSuen.printResult();
};
ZhangSuen.numNeighbors = function (r, c) {
var count = 0;
for (var i = 0; i < ZhangSuen.nbrs.length - 1; i++)
if (ZhangSuen.grid[r + ZhangSuen.nbrs[i][1]][c + ZhangSuen.nbrs[i][0]] === '#')
count++;
return count;
};
ZhangSuen.numTransitions = function (r, c) {
var count = 0;
for (var i = 0; i < ZhangSuen.nbrs.length - 1; i++)
if (ZhangSuen.grid[r + ZhangSuen.nbrs[i][1]][c + ZhangSuen.nbrs[i][0]] === ' ') {
if (ZhangSuen.grid[r + ZhangSuen.nbrs[i + 1][1]][c + ZhangSuen.nbrs[i + 1][0]] === '#')
count++;
}
return count;
};
ZhangSuen.atLeastOneIsWhite = function (r, c, step) {
var count = 0;
var group = ZhangSuen.nbrGroups[step];
for (var i = 0; i < 2; i++)
for (var j = 0; j < group[i].length; j++) {
var nbr = ZhangSuen.nbrs[group[i][j]];
if (ZhangSuen.grid[r + nbr[1]][c + nbr[0]] === ' ') {
count++;
break;
}
}
return count > 1;
};
ZhangSuen.printResult = function () {
for (var i = 0; i < ZhangSuen.grid.length; i++) {
var row = ZhangSuen.grid[i];
console.log(row.join(''));
}
};
return ZhangSuen;
}());
ZhangSuen.main(null); |
http://rosettacode.org/wiki/Zeckendorf_number_representation | Zeckendorf number representation | Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.
Recall that the first six distinct Fibonacci numbers are: 1, 2, 3, 5, 8, 13.
The decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.
10100 is not the only way to make 11 from the Fibonacci numbers however; 0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that no two consecutive Fibonacci numbers can be used which leads to the former unique solution.
Task
Generate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.
The intention in this task to find the Zeckendorf form of an arbitrary integer. The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task.
Also see
OEIS A014417 for the the sequence of required results.
Brown's Criterion - Numberphile
Related task
Fibonacci sequence
| #EchoLisp | EchoLisp |
;; special fib's starting with 1 2 3 5 ...
(define (fibonacci n)
(+ (fibonacci (1- n)) (fibonacci (- n 2))))
(remember 'fibonacci #(1 2))
(define-constant Φ (// (1+ (sqrt 5)) 2))
(define-constant logΦ (log Φ))
;; find i : fib(i) >= n
(define (iFib n)
(floor (// (log (+ (* n Φ) 0.5)) logΦ)))
;; left trim zeroes
(string-delimiter "")
(define (zeck->string digits)
(if (!= 0 (first digits))
(string-join digits "")
(zeck->string (rest digits))))
(define (Zeck n)
(cond
(( < n 0) "no negative zeck")
((inexact? n) "no floating zeck")
((zero? n) "0")
(else (zeck->string
(for/list ((s (reverse (take fibonacci (iFib n)))))
(if ( > s n) 0
(begin (-= n s) 1 )))))))
|
http://rosettacode.org/wiki/100_doors | 100 doors | There are 100 doors in a row that are all initially closed.
You make 100 passes by the doors.
The first time through, visit every door and toggle the door (if the door is closed, open it; if it is open, close it).
The second time, only visit every 2nd door (door #2, #4, #6, ...), and toggle it.
The third time, visit every 3rd door (door #3, #6, #9, ...), etc, until you only visit the 100th door.
Task
Answer the question: what state are the doors in after the last pass? Which are open, which are closed?
Alternate:
As noted in this page's discussion page, the only doors that remain open are those whose numbers are perfect squares.
Opening only those doors is an optimization that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.
| #Burlesque | Burlesque |
blsq ) 10ro2?^
{1 4 9 16 25 36 49 64 81 100}
|
http://rosettacode.org/wiki/Arrays | Arrays | This task is about arrays.
For hashes or associative arrays, please see Creating an Associative Array.
For a definition and in-depth discussion of what an array is, see Array.
Task
Show basic array syntax in your language.
Basically, create an array, assign a value to it, and retrieve an element (if available, show both fixed-length arrays and
dynamic arrays, pushing a value into it).
Please discuss at Village Pump: Arrays.
Please merge code in from these obsolete tasks:
Creating an Array
Assigning Values to an Array
Retrieving an Element of an Array
Related tasks
Collections
Creating an Associative Array
Two-dimensional array (runtime)
| #Eiffel | Eiffel |
class
APPLICATION
inherit
ARGUMENTS
create
make
feature {NONE} -- Initialization
make
-- Run application.
do
-- initialize the array, index starts at 1 (not zero) and prefill everything with the letter z
create my_static_array.make_filled ("z", 1, 50)
my_static_array.put ("a", 1)
my_static_array.put ("b", 2)
my_static_array [3] := "c"
-- access to array fields
print (my_static_array.at(1) + "%N")
print (my_static_array.at(2) + "%N")
print (my_static_array [3] + "%N")
-- in Eiffel static arrays can be resized in three ways
my_static_array.force ("c", 51) -- forces 'c' in position 51 and resizes the array to that size (now 51 places)
my_static_array.automatic_grow -- adds 50% more indices (having now 76 places)
my_static_array.grow (100) -- resizes the array to 100 places
end
my_static_array: ARRAY [STRING]
end
|
http://rosettacode.org/wiki/Arithmetic/Complex | Arithmetic/Complex | A complex number is a number which can be written as:
a
+
b
×
i
{\displaystyle a+b\times i}
(sometimes shown as:
b
+
a
×
i
{\displaystyle b+a\times i}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are real numbers, and
i
{\displaystyle i}
is √ -1
Typically, complex numbers are represented as a pair of real numbers called the "imaginary part" and "real part", where the imaginary part is the number to be multiplied by
i
{\displaystyle i}
.
Task
Show addition, multiplication, negation, and inversion of complex numbers in separate functions. (Subtraction and division operations can be made with pairs of these operations.)
Print the results for each operation tested.
Optional: Show complex conjugation.
By definition, the complex conjugate of
a
+
b
i
{\displaystyle a+bi}
is
a
−
b
i
{\displaystyle a-bi}
Some languages have complex number libraries available. If your language does, show the operations. If your language does not, also show the definition of this type.
| #Pascal | Pascal | program complexDemo(output);
const
{ I experienced some hiccups with -1.0 using GPC (GNU Pascal Compiler) }
negativeOne = -1.0;
type
line = string(80);
{ as per task requirements wrap arithmetic operations into separate functions }
function sum(protected x, y: complex): complex;
begin
sum := x + y
end;
function product(protected x, y: complex): complex;
begin
product := x * y
end;
function negative(protected x: complex): complex;
begin
negative := -x
end;
function inverse(protected x: complex): complex;
begin
inverse := x ** negativeOne
end;
{ only this function is not covered by Extended Pascal, ISO 10206 }
function conjugation(protected x: complex): complex;
begin
conjugation := cmplx(re(x), im(x) * negativeOne)
end;
{ --- test suite ------------------------------------------------------------- }
function asString(protected x: complex): line;
const
totalWidth = 5;
fractionDigits = 2;
var
result: line;
begin
writeStr(result, '(', re(x):totalWidth:fractionDigits, ', ',
im(x):totalWidth:fractionDigits, ')');
asString := result
end;
{ === MAIN =================================================================== }
var
x: complex;
{ for demonstration purposes: how to initialize complex variables }
y: complex value cmplx(1.0, 4.0);
z: complex value polar(exp(1.0), 3.14159265358979);
begin
x := cmplx(-3, 2);
writeLn(asString(x), ' + ', asString(y), ' = ', asString(sum(x, y)));
writeLn(asString(x), ' * ', asString(z), ' = ', asString(product(x, z)));
writeLn;
writeLn(' −', asString(z), ' = ', asString(negative(z)));
writeLn(' inverse(', asString(z), ') = ', asString(inverse(z)));
writeLn(' conjugation(', asString(y), ') = ', asString(conjugation(y)));
end. |
http://rosettacode.org/wiki/Arithmetic/Rational | Arithmetic/Rational | Task
Create a reasonably complete implementation of rational arithmetic in the particular language using the idioms of the language.
Example
Define a new type called frac with binary operator "//" of two integers that returns a structure made up of the numerator and the denominator (as per a rational number).
Further define the appropriate rational unary operators abs and '-', with the binary operators for addition '+', subtraction '-', multiplication '×', division '/', integer division '÷', modulo division, the comparison operators (e.g. '<', '≤', '>', & '≥') and equality operators (e.g. '=' & '≠').
Define standard coercion operators for casting int to frac etc.
If space allows, define standard increment and decrement operators (e.g. '+:=' & '-:=' etc.).
Finally test the operators:
Use the new type frac to find all perfect numbers less than 219 by summing the reciprocal of the factors.
Related task
Perfect Numbers
| #Quackery | Quackery | [ $ "bigrat.qky" loadfile ] now!
[ -2 n->v rot
factors witheach
[ n->v 1/v v+ ]
v0= ] is perfect ( n -> b )
19 bit times [ i^ perfect if [ i^ echo cr ] ] |
http://rosettacode.org/wiki/Arithmetic-geometric_mean | Arithmetic-geometric mean |
This page uses content from Wikipedia. The original article was at Arithmetic-geometric mean. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Write a function to compute the arithmetic-geometric mean of two numbers.
The arithmetic-geometric mean of two numbers can be (usefully) denoted as
a
g
m
(
a
,
g
)
{\displaystyle \mathrm {agm} (a,g)}
, and is equal to the limit of the sequence:
a
0
=
a
;
g
0
=
g
{\displaystyle a_{0}=a;\qquad g_{0}=g}
a
n
+
1
=
1
2
(
a
n
+
g
n
)
;
g
n
+
1
=
a
n
g
n
.
{\displaystyle a_{n+1}={\tfrac {1}{2}}(a_{n}+g_{n});\quad g_{n+1}={\sqrt {a_{n}g_{n}}}.}
Since the limit of
a
n
−
g
n
{\displaystyle a_{n}-g_{n}}
tends (rapidly) to zero with iterations, this is an efficient method.
Demonstrate the function by calculating:
a
g
m
(
1
,
1
/
2
)
{\displaystyle \mathrm {agm} (1,1/{\sqrt {2}})}
Also see
mathworld.wolfram.com/Arithmetic-Geometric Mean
| #Smalltalk | Smalltalk | agm:y
"return the arithmetic-geometric mean agm(x, y)
of the receiver (x) and the argument, y.
See https://en.wikipedia.org/wiki/Arithmetic-geometric_mean"
|ai an gi gn epsilon delta|
ai := (self + y) / 2.
gi := (self * y) sqrt.
epsilon := self ulp.
[
an := (ai + gi) / 2.
gn := (ai * gi) sqrt.
delta := (an - ai) abs.
ai := an.
gi := gn.
] doUntil:[ delta < epsilon ].
^ ai |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Frink | Frink | println[0^0] |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #FutureBasic | FutureBasic | window 1
print 0^0
HandleEvents |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Gambas | Gambas | Public Sub Main()
Print 0 ^ 0
End |
http://rosettacode.org/wiki/Zig-zag_matrix | Zig-zag matrix | Task
Produce a zig-zag array.
A zig-zag array is a square arrangement of the first N2 natural numbers, where the
numbers increase sequentially as you zig-zag along the array's anti-diagonals.
For a graphical representation, see JPG zigzag (JPG uses such arrays to encode images).
For example, given 5, produce this array:
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24
Related tasks
Spiral matrix
Identity matrix
Ulam spiral (for primes)
See also
Wiktionary entry: anti-diagonals
| #Ada | Ada | with Ada.Text_IO; use Ada.Text_IO;
procedure Test_Zig_Zag is
type Matrix is array (Positive range <>, Positive range <>) of Natural;
function Zig_Zag (Size : Positive) return Matrix is
Data : Matrix (1..Size, 1..Size);
I, J : Integer := 1;
begin
Data (1, 1) := 0;
for Element in 1..Size**2 - 1 loop
if (I + J) mod 2 = 0 then
-- Even stripes
if J < Size then
J := J + 1;
else
I := I + 2;
end if;
if I > 1 then
I := I - 1;
end if;
else
-- Odd stripes
if I < Size then
I := I + 1;
else
J := J + 2;
end if;
if J > 1 then
J := J - 1;
end if;
end if;
Data (I, J) := Element;
end loop;
return Data;
end Zig_Zag;
procedure Put (Data : Matrix) is
begin
for I in Data'Range (1) loop
for J in Data'Range (2) loop
Put (Integer'Image (Data (I, J)));
end loop;
New_Line;
end loop;
end Put;
begin
Put (Zig_Zag (5));
end Test_Zig_Zag; |
http://rosettacode.org/wiki/Yellowstone_sequence | Yellowstone sequence | The Yellowstone sequence, also called the Yellowstone permutation, is defined as:
For n <= 3,
a(n) = n
For n >= 4,
a(n) = the smallest number not already in sequence such that a(n) is relatively prime to a(n-1) and
is not relatively prime to a(n-2).
The sequence is a permutation of the natural numbers, and gets its name from what its authors felt was a spiking, geyser like appearance of a plot of the sequence.
Example
a(4) is 4 because 4 is the smallest number following 1, 2, 3 in the sequence that is relatively prime to the entry before it (3), and is not relatively prime to the number two entries before it (2).
Task
Find and show as output the first 30 Yellowstone numbers.
Extra
Demonstrate how to plot, with x = n and y coordinate a(n), the first 100 Yellowstone numbers.
Related tasks
Greatest common divisor.
Plot coordinate pairs.
See also
The OEIS entry: A098550 The Yellowstone permutation.
Applegate et al, 2015: The Yellowstone Permutation [1].
| #Arturo | Arturo | yellowstone: function [n][
result: new [1 2 3]
present: new [1 2 3]
start: new 4
while [n > size result][
candidate: new start
while ø [
if all? @[
not? contains? present candidate
1 = gcd @[candidate last result]
1 <> gcd @[candidate get result (size result)-2]
][
'result ++ candidate
'present ++ candidate
while [contains? present start] -> inc 'start
break
]
inc 'candidate
]
]
return result
]
print yellowstone 30 |
http://rosettacode.org/wiki/Yellowstone_sequence | Yellowstone sequence | The Yellowstone sequence, also called the Yellowstone permutation, is defined as:
For n <= 3,
a(n) = n
For n >= 4,
a(n) = the smallest number not already in sequence such that a(n) is relatively prime to a(n-1) and
is not relatively prime to a(n-2).
The sequence is a permutation of the natural numbers, and gets its name from what its authors felt was a spiking, geyser like appearance of a plot of the sequence.
Example
a(4) is 4 because 4 is the smallest number following 1, 2, 3 in the sequence that is relatively prime to the entry before it (3), and is not relatively prime to the number two entries before it (2).
Task
Find and show as output the first 30 Yellowstone numbers.
Extra
Demonstrate how to plot, with x = n and y coordinate a(n), the first 100 Yellowstone numbers.
Related tasks
Greatest common divisor.
Plot coordinate pairs.
See also
The OEIS entry: A098550 The Yellowstone permutation.
Applegate et al, 2015: The Yellowstone Permutation [1].
| #AutoHotkey | AutoHotkey | A := [], in_seq := []
loop 30 {
n := A_Index
if n <=3
A[n] := n, in_seq[n] := true
else while true
{
s := A_Index
if !in_seq[s] && relatively_prime(s, A[n-1]) && !relatively_prime(s, A[n-2])
{
A[n] := s
in_seq[s] := true
break
}
}
}
for i, v in A
result .= v ","
MsgBox % result := "[" Trim(result, ",") "]"
return
;--------------------------------------
relatively_prime(a, b){
return (GCD(a, b) = 1)
}
;--------------------------------------
GCD(a, b) {
while b
b := Mod(a | 0x0, a := b)
return a
} |
http://rosettacode.org/wiki/Arithmetic_evaluation | Arithmetic evaluation | Create a program which parses and evaluates arithmetic expressions.
Requirements
An abstract-syntax tree (AST) for the expression must be created from parsing the input.
The AST must be used in evaluation, also, so the input may not be directly evaluated (e.g. by calling eval or a similar language feature.)
The expression will be a string or list of symbols like "(1+3)*7".
The four symbols + - * / must be supported as binary operators with conventional precedence rules.
Precedence-control parentheses must also be supported.
Note
For those who don't remember, mathematical precedence is as follows:
Parentheses
Multiplication/Division (left to right)
Addition/Subtraction (left to right)
C.f
24 game Player.
Parsing/RPN calculator algorithm.
Parsing/RPN to infix conversion.
| #Racket | Racket | #lang racket
(require parser-tools/yacc
parser-tools/lex
(prefix-in ~ parser-tools/lex-sre))
(define-tokens value-tokens (NUM))
(define-empty-tokens op-tokens (OPEN CLOSE + - * / EOF NEG))
(define lex
(lexer [(eof) 'EOF]
[whitespace (lex input-port)]
[(~or "+" "-" "*" "/") (string->symbol lexeme)]
["(" 'OPEN]
[")" 'CLOSE]
[(~: (~+ numeric) (~? (~: #\. (~* numeric))))
(token-NUM (string->number lexeme))]))
(define parse
(parser [start E] [end EOF]
[tokens value-tokens op-tokens]
[error void]
[precs (left - +) (left * /) (left NEG)]
[grammar (E [(NUM) $1]
[(E + E) (+ $1 $3)]
[(E - E) (- $1 $3)]
[(E * E) (* $1 $3)]
[(E / E) (/ $1 $3)]
[(- E) (prec NEG) (- $2)]
[(OPEN E CLOSE) $2])]))
(define (calc str)
(define i (open-input-string str))
(displayln (parse (λ () (lex i)))))
(calc "(1 + 2 * 3) - (1+2)*-3") |
http://rosettacode.org/wiki/Arithmetic_evaluation | Arithmetic evaluation | Create a program which parses and evaluates arithmetic expressions.
Requirements
An abstract-syntax tree (AST) for the expression must be created from parsing the input.
The AST must be used in evaluation, also, so the input may not be directly evaluated (e.g. by calling eval or a similar language feature.)
The expression will be a string or list of symbols like "(1+3)*7".
The four symbols + - * / must be supported as binary operators with conventional precedence rules.
Precedence-control parentheses must also be supported.
Note
For those who don't remember, mathematical precedence is as follows:
Parentheses
Multiplication/Division (left to right)
Addition/Subtraction (left to right)
C.f
24 game Player.
Parsing/RPN calculator algorithm.
Parsing/RPN to infix conversion.
| #Raku | Raku | sub ev (Str $s --> Numeric) {
grammar expr {
token TOP { ^ <sum> $ }
token sum { <product> (('+' || '-') <product>)* }
token product { <factor> (('*' || '/') <factor>)* }
token factor { <unary_minus>? [ <parens> || <literal> ] }
token unary_minus { '-' }
token parens { '(' <sum> ')' }
token literal { \d+ ['.' \d+]? || '.' \d+ }
}
my sub minus ($b) { $b ?? -1 !! +1 }
my sub sum ($x) {
[+] flat product($x<product>), map
{ minus($^y[0] eq '-') * product $^y<product> },
|($x[0] or [])
}
my sub product ($x) {
[*] flat factor($x<factor>), map
{ factor($^y<factor>) ** minus($^y[0] eq '/') },
|($x[0] or [])
}
my sub factor ($x) {
minus($x<unary_minus>) * ($x<parens>
?? sum $x<parens><sum>
!! $x<literal>)
}
expr.parse([~] split /\s+/, $s);
$/ or fail 'No parse.';
sum $/<sum>;
}
# Testing:
say ev '5'; # 5
say ev '1 + 2 - 3 * 4 / 5'; # 0.6
say ev '1 + 5*3.4 - .5 -4 / -2 * (3+4) -6'; # 25.5
say ev '((11+15)*15)* 2 + (3) * -4 *1'; # 768 |
http://rosettacode.org/wiki/Zeckendorf_arithmetic | Zeckendorf arithmetic | This task is a total immersion zeckendorf task; using decimal numbers will attract serious disapprobation.
The task is to implement addition, subtraction, multiplication, and division using Zeckendorf number representation. Optionally provide decrement, increment and comparitive operation functions.
Addition
Like binary 1 + 1 = 10, note carry 1 left. There the similarity ends. 10 + 10 = 101, note carry 1 left and 1 right. 100 + 100 = 1001, note carry 1 left and 2 right, this is the general case.
Occurrences of 11 must be changed to 100. Occurrences of 111 may be changed from the right by replacing 11 with 100, or from the left converting 111 to 100 + 100;
Subtraction
10 - 1 = 1. The general rule is borrow 1 right carry 1 left. eg:
abcde
10100 -
1000
_____
100 borrow 1 from a leaves 100
+ 100 add the carry
_____
1001
A larger example:
abcdef
100100 -
1000
______
1*0100 borrow 1 from b
+ 100 add the carry
______
1*1001
Sadly we borrowed 1 from b which didn't have it to lend. So now b borrows from a:
1001
+ 1000 add the carry
____
10100
Multiplication
Here you teach your computer its zeckendorf tables. eg. 101 * 1001:
a = 1 * 101 = 101
b = 10 * 101 = a + a = 10000
c = 100 * 101 = b + a = 10101
d = 1000 * 101 = c + b = 101010
1001 = d + a therefore 101 * 1001 =
101010
+ 101
______
1000100
Division
Lets try 1000101 divided by 101, so we can use the same table used for multiplication.
1000101 -
101010 subtract d (1000 * 101)
_______
1000 -
101 b and c are too large to subtract, so subtract a
____
1 so 1000101 divided by 101 is d + a (1001) remainder 1
Efficient algorithms for Zeckendorf arithmetic is interesting. The sections on addition and subtraction are particularly relevant for this task.
| #Julia | Julia | import Base.*, Base.+, Base.-, Base./, Base.show, Base.!=, Base.==, Base.<=, Base.<, Base.>, Base.>=, Base.divrem
const z0 = "0"
const z1 = "1"
const flipordered = (z1 < z0)
mutable struct Z s::String end
Z() = Z(z0)
Z(z::Z) = Z(z.s)
pairlen(x::Z, y::Z) = max(length(x.s), length(y.s))
tolen(x::Z, n::Int) = (s = x.s; while length(s) < n s = z0 * s end; s)
<(x::Z, y::Z) = (l = pairlen(x, y); flipordered ? tolen(x, l) > tolen(y, l) : tolen(x, l) < tolen(y, l))
>(x::Z, y::Z) = (l = pairlen(x, y); flipordered ? tolen(x, l) < tolen(y, l) : tolen(x, l) > tolen(y, l))
==(x::Z, y::Z) = (l = pairlen(x, y); tolen(x, l) == tolen(y, l))
<=(x::Z, y::Z) = (l = pairlen(x, y); flipordered ? tolen(x, l) >= tolen(y, l) : tolen(x, l) <= tolen(y, l))
>=(x::Z, y::Z) = (l = pairlen(x, y); flipordered ? tolen(x, l) <= tolen(y, l) : tolen(x, l) >= tolen(y, l))
!=(x::Z, y::Z) = (l = pairlen(x, y); tolen(x, l) != tolen(y, l))
function tocanonical(z::Z)
while occursin(z0 * z1 * z1, z.s)
z.s = replace(z.s, z0 * z1 * z1 => z1 * z0 * z0)
end
len = length(z.s)
if len > 1 && z.s[1:2] == z1 * z1
z.s = z1 * z0 * z0 * ((len > 2) ? z.s[3:end] : "")
end
while (len = length(z.s)) > 1 && string(z.s[1]) == z0
if len == 2
if z.s == z0 * z0
z.s = z0
elseif z.s == z0 * z1
z.s = z1
end
else
z.s = z.s[2:end]
end
end
z
end
function inc(z)
if z.s[end] == z0[1]
z.s = z.s[1:end-1] * z1[1]
elseif z.s[end] == z1[1]
if length(z.s) > 1
if z.s[end-1:end] == z0 * z1
z.s = z.s[1:end-2] * z1 * z0
end
else
z.s = z1 * z0
end
end
tocanonical(z)
end
function dec(z)
if z.s[end] == z1[1]
z.s = z.s[1:end-1] * z0
else
if (m = match(Regex(z1 * z0 * '+' * '$'), z.s)) != nothing
len = length(m.match)
if iseven(len)
z.s = z.s[1:end-len] * (z0 * z1) ^ div(len, 2)
else
z.s = z.s[1:end-len] * (z0 * z1) ^ div(len, 2) * z0
end
end
end
tocanonical(z)
z
end
function +(x::Z, y::Z)
a = Z(x.s)
b = Z(y.s)
while b.s != z0
inc(a)
dec(b)
end
a
end
function -(x::Z, y::Z)
a = Z(x.s)
b = Z(y.s)
while b.s != z0
dec(a)
dec(b)
end
a
end
function *(x::Z, y::Z)
if (x.s == z0) || (y.s == z0)
return Z(z0)
elseif x.s == z1
return Z(y.s)
elseif y.s == z1
return Z(x.s)
end
a = Z(x.s)
b = Z(z1)
while b != y
c = Z(z0)
while c != x
inc(a)
inc(c)
end
inc(b)
end
a
end
function divrem(x::Z, y::Z)
if y.s == z0
throw("Zeckendorf division by 0")
elseif (y.s == z1) || (x.s == z0)
return Z(x.s)
end
a = Z(x.s)
b = Z(y.s)
c = Z(z0)
while a > b
a = a - b
inc(c)
end
tocanonical(c), tocanonical(a)
end
function /(x::Z, y::Z)
a, _ = divrem(x, y)
a
end
show(io::IO, z::Z) = show(io, parse(BigInt, tocanonical(z).s))
function zeckendorftest()
a = Z("10")
b = Z("1001")
c = Z("1000")
d = Z("10101")
println("Addition:")
x = a
println(x += a)
println(x += a)
println(x += b)
println(x += c)
println(x += d)
println("\nSubtraction:")
x = Z("1000")
println(x - Z("101"))
x = Z("10101010")
println(x - Z("1010101"))
println("\nMultiplication:")
x = Z("1001")
y = Z("101")
println(x * y)
println(Z("101010") * y)
println("\nDivision:")
x = Z("1000101")
y = Z("101")
println(x / y)
println(divrem(x, y))
end
zeckendorftest()
|
http://rosettacode.org/wiki/Zeckendorf_arithmetic | Zeckendorf arithmetic | This task is a total immersion zeckendorf task; using decimal numbers will attract serious disapprobation.
The task is to implement addition, subtraction, multiplication, and division using Zeckendorf number representation. Optionally provide decrement, increment and comparitive operation functions.
Addition
Like binary 1 + 1 = 10, note carry 1 left. There the similarity ends. 10 + 10 = 101, note carry 1 left and 1 right. 100 + 100 = 1001, note carry 1 left and 2 right, this is the general case.
Occurrences of 11 must be changed to 100. Occurrences of 111 may be changed from the right by replacing 11 with 100, or from the left converting 111 to 100 + 100;
Subtraction
10 - 1 = 1. The general rule is borrow 1 right carry 1 left. eg:
abcde
10100 -
1000
_____
100 borrow 1 from a leaves 100
+ 100 add the carry
_____
1001
A larger example:
abcdef
100100 -
1000
______
1*0100 borrow 1 from b
+ 100 add the carry
______
1*1001
Sadly we borrowed 1 from b which didn't have it to lend. So now b borrows from a:
1001
+ 1000 add the carry
____
10100
Multiplication
Here you teach your computer its zeckendorf tables. eg. 101 * 1001:
a = 1 * 101 = 101
b = 10 * 101 = a + a = 10000
c = 100 * 101 = b + a = 10101
d = 1000 * 101 = c + b = 101010
1001 = d + a therefore 101 * 1001 =
101010
+ 101
______
1000100
Division
Lets try 1000101 divided by 101, so we can use the same table used for multiplication.
1000101 -
101010 subtract d (1000 * 101)
_______
1000 -
101 b and c are too large to subtract, so subtract a
____
1 so 1000101 divided by 101 is d + a (1001) remainder 1
Efficient algorithms for Zeckendorf arithmetic is interesting. The sections on addition and subtraction are particularly relevant for this task.
| #Kotlin | Kotlin | // version 1.1.51
class Zeckendorf(x: String = "0") : Comparable<Zeckendorf> {
var dVal = 0
var dLen = 0
private fun a(n: Int) {
var i = n
while (true) {
if (dLen < i) dLen = i
val j = (dVal shr (i * 2)) and 3
when (j) {
0, 1 -> return
2 -> {
if (((dVal shr ((i + 1) * 2)) and 1) != 1) return
dVal += 1 shl (i * 2 + 1)
return
}
3 -> {
dVal = dVal and (3 shl (i * 2)).inv()
b((i + 1) * 2)
}
}
i++
}
}
private fun b(pos: Int) {
if (pos == 0) {
var thiz = this
++thiz
return
}
if (((dVal shr pos) and 1) == 0) {
dVal += 1 shl pos
a(pos / 2)
if (pos > 1) a(pos / 2 - 1)
}
else {
dVal = dVal and (1 shl pos).inv()
b(pos + 1)
b(pos - (if (pos > 1) 2 else 1))
}
}
private fun c(pos: Int) {
if (((dVal shr pos) and 1) == 1) {
dVal = dVal and (1 shl pos).inv()
return
}
c(pos + 1)
if (pos > 0) b(pos - 1) else { var thiz = this; ++thiz }
}
init {
var q = 1
var i = x.length - 1
dLen = i / 2
while (i >= 0) {
dVal += (x[i] - '0').toInt() * q
q *= 2
i--
}
}
operator fun inc(): Zeckendorf {
dVal += 1
a(0)
return this
}
operator fun plusAssign(other: Zeckendorf) {
for (gn in 0 until (other.dLen + 1) * 2) {
if (((other.dVal shr gn) and 1) == 1) b(gn)
}
}
operator fun minusAssign(other: Zeckendorf) {
for (gn in 0 until (other.dLen + 1) * 2) {
if (((other.dVal shr gn) and 1) == 1) c(gn)
}
while ((((dVal shr dLen * 2) and 3) == 0) || (dLen == 0)) dLen--
}
operator fun timesAssign(other: Zeckendorf) {
var na = other.copy()
var nb = other.copy()
var nt: Zeckendorf
var nr = "0".Z
for (i in 0..(dLen + 1) * 2) {
if (((dVal shr i) and 1) > 0) nr += nb
nt = nb.copy()
nb += na
na = nt.copy()
}
dVal = nr.dVal
dLen = nr.dLen
}
override operator fun compareTo(other: Zeckendorf) = dVal.compareTo(other.dVal)
override fun toString(): String {
if (dVal == 0) return "0"
val sb = StringBuilder(dig1[(dVal shr (dLen * 2)) and 3])
for (i in dLen - 1 downTo 0) {
sb.append(dig[(dVal shr (i * 2)) and 3])
}
return sb.toString()
}
fun copy(): Zeckendorf {
val z = "0".Z
z.dVal = dVal
z.dLen = dLen
return z
}
companion object {
val dig = listOf("00", "01", "10")
val dig1 = listOf("", "1", "10")
}
}
val String.Z get() = Zeckendorf(this)
fun main(args: Array<String>) {
println("Addition:")
var g = "10".Z
g += "10".Z
println(g)
g += "10".Z
println(g)
g += "1001".Z
println(g)
g += "1000".Z
println(g)
g += "10101".Z
println(g)
println("\nSubtraction:")
g = "1000".Z
g -= "101".Z
println(g)
g = "10101010".Z
g -= "1010101".Z
println(g)
println("\nMultiplication:")
g = "1001".Z
g *= "101".Z
println(g)
g = "101010".Z
g += "101".Z
println(g)
} |
http://rosettacode.org/wiki/Zumkeller_numbers | Zumkeller numbers | Zumkeller numbers are the set of numbers whose divisors can be partitioned into two disjoint sets that sum to the same value. Each sum must contain divisor values that are not in the other sum, and all of the divisors must be in one or the other. There are no restrictions on how the divisors are partitioned, only that the two partition sums are equal.
E.G.
6 is a Zumkeller number; The divisors {1 2 3 6} can be partitioned into two groups {1 2 3} and {6} that both sum to 6.
10 is not a Zumkeller number; The divisors {1 2 5 10} can not be partitioned into two groups in any way that will both sum to the same value.
12 is a Zumkeller number; The divisors {1 2 3 4 6 12} can be partitioned into two groups {1 3 4 6} and {2 12} that both sum to 14.
Even Zumkeller numbers are common; odd Zumkeller numbers are much less so. For values below 10^6, there is at least one Zumkeller number in every 12 consecutive integers, and the vast majority of them are even. The odd Zumkeller numbers are very similar to the list from the task Abundant odd numbers; they are nearly the same except for the further restriction that the abundance (A(n) = sigma(n) - 2n), must be even: A(n) mod 2 == 0
Task
Write a routine (function, procedure, whatever) to find Zumkeller numbers.
Use the routine to find and display here, on this page, the first 220 Zumkeller numbers.
Use the routine to find and display here, on this page, the first 40 odd Zumkeller numbers.
Optional, stretch goal: Use the routine to find and display here, on this page, the first 40 odd Zumkeller numbers that don't end with 5.
See Also
OEIS:A083207 - Zumkeller numbers to get an impression of different partitions OEIS:A083206 Zumkeller partitions
OEIS:A174865 - Odd Zumkeller numbers
Related Tasks
Abundant odd numbers
Abundant, deficient and perfect number classifications
Proper divisors , Factors of an integer | #jq | jq | # The factors, sorted
def factors:
. as $num
| reduce range(1; 1 + sqrt|floor) as $i
([];
if ($num % $i) == 0 then
($num / $i) as $r
| if $i == $r then . + [$i] else . + [$i, $r] end
else .
end
| sort) ;
# If the input is a sorted array of distinct non-negative integers,
# then the output will be a stream of [$x,$y] arrays,
# where $x and $y are non-empty arrays that partition the
# input, and where ($x|add) == $sum.
# If [$x,$y] is emitted, then [$y,$x] will not also be emitted.
# The items in $x appear in the same order as in the input, and similarly
# for $y.
#
def distinct_partitions($sum):
# input: [$array, $n, $lim] where $n>0
# output: a stream of arrays, $a, each with $n distinct items from $array,
# preserving the order in $array, and such that
# add == $lim
def p:
. as [$in, $n, $lim]
| if $n==1 # this condition is very common so it saves time to check early on
then ($in | bsearch($lim)) as $ix
| if $ix < 0 then empty
else [$lim]
end
else ($in|length) as $length
| if $length <= $n then empty
elif $length==$n then $in | select(add == $lim)
elif ($in[-$n:]|add) < $lim then empty
else ($in[:$n]|add) as $rsum
| if $rsum > $lim then empty
elif $rsum == $lim then "amazing" | debug | $in[:$n]
else range(0; 1 + $length - $n) as $i
| [$in[$i]] + ([$in[$i+1:], $n-1, $lim - $in[$i]]|p)
end
end
end;
range(1; (1+length)/2) as $i
| ([., $i, $sum]|p) as $pi
| [ $pi, (. - $pi)]
| select( if (.[0]|length) == (.[1]|length) then (.[0] < .[1]) else true end) #1
;
def zumkellerPartitions:
factors
| add as $sum
| if $sum % 2 == 1 then empty
else distinct_partitions($sum / 2)
end;
def is_zumkeller:
first(factors
| add as $sum
| if $sum % 2 == 1 then empty
else distinct_partitions($sum / 2)
| select( (.[0]|add) == (.[1]|add)) // ("internal error: \(.)" | debug | empty) #2
end
| true)
// false; |
http://rosettacode.org/wiki/Zumkeller_numbers | Zumkeller numbers | Zumkeller numbers are the set of numbers whose divisors can be partitioned into two disjoint sets that sum to the same value. Each sum must contain divisor values that are not in the other sum, and all of the divisors must be in one or the other. There are no restrictions on how the divisors are partitioned, only that the two partition sums are equal.
E.G.
6 is a Zumkeller number; The divisors {1 2 3 6} can be partitioned into two groups {1 2 3} and {6} that both sum to 6.
10 is not a Zumkeller number; The divisors {1 2 5 10} can not be partitioned into two groups in any way that will both sum to the same value.
12 is a Zumkeller number; The divisors {1 2 3 4 6 12} can be partitioned into two groups {1 3 4 6} and {2 12} that both sum to 14.
Even Zumkeller numbers are common; odd Zumkeller numbers are much less so. For values below 10^6, there is at least one Zumkeller number in every 12 consecutive integers, and the vast majority of them are even. The odd Zumkeller numbers are very similar to the list from the task Abundant odd numbers; they are nearly the same except for the further restriction that the abundance (A(n) = sigma(n) - 2n), must be even: A(n) mod 2 == 0
Task
Write a routine (function, procedure, whatever) to find Zumkeller numbers.
Use the routine to find and display here, on this page, the first 220 Zumkeller numbers.
Use the routine to find and display here, on this page, the first 40 odd Zumkeller numbers.
Optional, stretch goal: Use the routine to find and display here, on this page, the first 40 odd Zumkeller numbers that don't end with 5.
See Also
OEIS:A083207 - Zumkeller numbers to get an impression of different partitions OEIS:A083206 Zumkeller partitions
OEIS:A174865 - Odd Zumkeller numbers
Related Tasks
Abundant odd numbers
Abundant, deficient and perfect number classifications
Proper divisors , Factors of an integer | #Julia | Julia | using Primes
function factorize(n)
f = [one(n)]
for (p, x) in factor(n)
f = reduce(vcat, [f*p^i for i in 1:x], init=f)
end
f
end
function cansum(goal, list)
if goal == 0 || list[1] == goal
return true
elseif length(list) > 1
if list[1] > goal
return cansum(goal, list[2:end])
else
return cansum(goal - list[1], list[2:end]) || cansum(goal, list[2:end])
end
end
return false
end
function iszumkeller(n)
f = reverse(factorize(n))
fsum = sum(f)
return iseven(fsum) && cansum(div(fsum, 2) - f[1], f[2:end])
end
function printconditionalnum(condition, maxcount, numperline = 20)
count, spacing = 1, div(80, numperline)
for i in 1:typemax(Int)
if condition(i)
count += 1
print(rpad(i, spacing), (count - 1) % numperline == 0 ? "\n" : "")
if count > maxcount
return
end
end
end
end
println("First 220 Zumkeller numbers:")
printconditionalnum(iszumkeller, 220)
println("\n\nFirst 40 odd Zumkeller numbers:")
printconditionalnum((n) -> isodd(n) && iszumkeller(n), 40, 8)
println("\n\nFirst 40 odd Zumkeller numbers not ending with 5:")
printconditionalnum((n) -> isodd(n) && (string(n)[end] != '5') && iszumkeller(n), 40, 8)
|
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #Common_Lisp | Common Lisp | (let ((s (format () "~s" (expt 5 (expt 4 (expt 3 2))))))
(format t "~a...~a, length ~a" (subseq s 0 20)
(subseq s (- (length s) 20)) (length s))) |
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #D | D | void main() {
import std.stdio, std.bigint, std.conv;
auto s = text(5.BigInt ^^ 4 ^^ 3 ^^ 2);
writefln("5^4^3^2 = %s..%s (%d digits)", s[0..20], s[$-20..$], s.length);
} |
http://rosettacode.org/wiki/Zhang-Suen_thinning_algorithm | Zhang-Suen thinning algorithm | This is an algorithm used to thin a black and white i.e. one bit per pixel images.
For example, with an input image of:
################# #############
################## ################
################### ##################
######## ####### ###################
###### ####### ####### ######
###### ####### #######
################# #######
################ #######
################# #######
###### ####### #######
###### ####### #######
###### ####### ####### ######
######## ####### ###################
######## ####### ###### ################## ######
######## ####### ###### ################ ######
######## ####### ###### ############# ######
It produces the thinned output:
# ########## #######
## # #### #
# # ##
# # #
# # #
# # #
############ #
# # #
# # #
# # #
# # #
# ##
# ############
### ###
Algorithm
Assume black pixels are one and white pixels zero, and that the input image is a rectangular N by M array of ones and zeroes.
The algorithm operates on all black pixels P1 that can have eight neighbours.
The neighbours are, in order, arranged as:
P9 P2 P3
P8 P1 P4
P7 P6 P5
Obviously the boundary pixels of the image cannot have the full eight neighbours.
Define
A
(
P
1
)
{\displaystyle A(P1)}
= the number of transitions from white to black, (0 -> 1) in the sequence P2,P3,P4,P5,P6,P7,P8,P9,P2. (Note the extra P2 at the end - it is circular).
Define
B
(
P
1
)
{\displaystyle B(P1)}
= The number of black pixel neighbours of P1. ( = sum(P2 .. P9) )
Step 1
All pixels are tested and pixels satisfying all the following conditions (simultaneously) are just noted at this stage.
(0) The pixel is black and has eight neighbours
(1)
2
<=
B
(
P
1
)
<=
6
{\displaystyle 2<=B(P1)<=6}
(2) A(P1) = 1
(3) At least one of P2 and P4 and P6 is white
(4) At least one of P4 and P6 and P8 is white
After iterating over the image and collecting all the pixels satisfying all step 1 conditions, all these condition satisfying pixels are set to white.
Step 2
All pixels are again tested and pixels satisfying all the following conditions are just noted at this stage.
(0) The pixel is black and has eight neighbours
(1)
2
<=
B
(
P
1
)
<=
6
{\displaystyle 2<=B(P1)<=6}
(2) A(P1) = 1
(3) At least one of P2 and P4 and P8 is white
(4) At least one of P2 and P6 and P8 is white
After iterating over the image and collecting all the pixels satisfying all step 2 conditions, all these condition satisfying pixels are again set to white.
Iteration
If any pixels were set in this round of either step 1 or step 2 then all steps are repeated until no image pixels are so changed.
Task
Write a routine to perform Zhang-Suen thinning on an image matrix of ones and zeroes.
Use the routine to thin the following image and show the output here on this page as either a matrix of ones and zeroes, an image, or an ASCII-art image of space/non-space characters.
00000000000000000000000000000000
01111111110000000111111110000000
01110001111000001111001111000000
01110000111000001110000111000000
01110001111000001110000000000000
01111111110000001110000000000000
01110111100000001110000111000000
01110011110011101111001111011100
01110001111011100111111110011100
00000000000000000000000000000000
Reference
Zhang-Suen Thinning Algorithm, Java Implementation by Nayef Reza.
"Character Recognition Systems: A Guide for Students and Practitioners" By Mohamed Cheriet, Nawwaf Kharma, Cheng-Lin Liu, Ching Suen
| #Julia | Julia |
const pixelstring =
"00000000000000000000000000000000" *
"01111111110000000111111110000000" *
"01110001111000001111001111000000" *
"01110000111000001110000111000000" *
"01110001111000001110000000000000" *
"01111111110000001110000000000000" *
"01110111100000001110000111000000" *
"01110011110011101111001111011100" *
"01110001111011100111111110011100" *
"00000000000000000000000000000000"
const pixels = reshape([UInt8(c- 48) for c in pixelstring], (32,10))'
function surroundtesting(px, i, j, step)
if px[i,j] == 0
return false
end
isize, jsize = size(px)
if i < 1 || j < 1 || i == isize || j == jsize # criteria 0.both
return false
end
s = Array{Int,1}(9)
s[1] = s[9] = px[i-1,j]; s[2] = px[i-1,j+1]; s[3] = px[i,j+1]; s[4] = px[i+1,j+1]
s[5] = px[i+1,j]; s[6] = px[i+1,j-1]; s[7] = px[i,j-1]; s[8] = px[i-1,j-1]
b = sum(s[1:8])
if b < 2 || b > 6 # criteria 1.both
return false
end
if sum([(s[i] == 0 && s[i+1] == 1) for i in 1:length(s)-1]) != 1 # criteria 2.both
return false
end
if step == 1
rightwhite = s[1] == 0 || s[3] == 0 || s[5] == 0 # 1.3
downwhite = s[3] == 0 || s[5] == 0 || s[7] == 0 # 1.4
return rightwhite && downwhite
end
upwhite = s[1] == 0 || s[3] == 0 || s[7] == 0 # 2.3
leftwhite = s[1] == 0 || s[5] == 0 || s[7] == 0 # 2.4
return upwhite && leftwhite
end
function zsthinning(mat)
retmat = copy(mat)
testmat = zeros(Int, size(mat))
isize, jsize = size(testmat)
needredo = true
loops = 0
while(needredo)
loops += 1
println("loop number $loops")
needredo = false
for n in 1:2
for i in 1:isize, j in 1:jsize
testmat[i,j] = surroundtesting(retmat, i, j, n) ? 1 : 0
end
for i in 1:isize, j in 1:jsize
if testmat[i,j] == 1
retmat[i,j] = 0
needredo = true
end
end
end
end
retmat
end
function asciiprint(mat)
for i in 1:size(mat)[1]
println(join(map(i -> i == 1 ? '#' : ' ', mat[i,:])))
end
end
asciiprint(zsthinning(pixels)) |
http://rosettacode.org/wiki/Zeckendorf_number_representation | Zeckendorf number representation | Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.
Recall that the first six distinct Fibonacci numbers are: 1, 2, 3, 5, 8, 13.
The decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.
10100 is not the only way to make 11 from the Fibonacci numbers however; 0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that no two consecutive Fibonacci numbers can be used which leads to the former unique solution.
Task
Generate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.
The intention in this task to find the Zeckendorf form of an arbitrary integer. The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task.
Also see
OEIS A014417 for the the sequence of required results.
Brown's Criterion - Numberphile
Related task
Fibonacci sequence
| #Elena | Elena | import system'routines;
import system'collections;
import system'text;
import extensions;
extension op
{
fibonacci()
{
if (self < 2)
{
^ self
}
else
{
^ (self - 1).fibonacci() + (self - 2).fibonacci()
};
}
zeckendorf()
{
var fibonacciNumbers := new List<int>();
int num := self;
int fibPosition := 2;
int currentFibonaciNum := fibPosition.fibonacci();
while (currentFibonaciNum <= num)
{
fibonacciNumbers.append:currentFibonaciNum;
fibPosition := fibPosition + 1;
currentFibonaciNum := fibPosition.fibonacci()
};
auto output := new TextBuilder();
int temp := num;
fibonacciNumbers.sequenceReverse().forEach:(item)
{
if (item <= temp)
{
output.write("1");
temp := temp - item
}
else
{
output.write("0")
}
};
^ output.Value
}
}
public program()
{
for(int i := 1, i <= 20, i += 1)
{
console.printFormatted("{0} : {1}",i,i.zeckendorf()).writeLine()
};
console.readChar()
} |
http://rosettacode.org/wiki/100_doors | 100 doors | There are 100 doors in a row that are all initially closed.
You make 100 passes by the doors.
The first time through, visit every door and toggle the door (if the door is closed, open it; if it is open, close it).
The second time, only visit every 2nd door (door #2, #4, #6, ...), and toggle it.
The third time, visit every 3rd door (door #3, #6, #9, ...), etc, until you only visit the 100th door.
Task
Answer the question: what state are the doors in after the last pass? Which are open, which are closed?
Alternate:
As noted in this page's discussion page, the only doors that remain open are those whose numbers are perfect squares.
Opening only those doors is an optimization that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.
| #BQN | BQN | swch ← ≠´{100⥊1«𝕩⥊0}¨1+↕100
¯1↓∾{𝕩∾@+10}¨•Fmt¨⟨swch,/swch⟩ |
http://rosettacode.org/wiki/Arrays | Arrays | This task is about arrays.
For hashes or associative arrays, please see Creating an Associative Array.
For a definition and in-depth discussion of what an array is, see Array.
Task
Show basic array syntax in your language.
Basically, create an array, assign a value to it, and retrieve an element (if available, show both fixed-length arrays and
dynamic arrays, pushing a value into it).
Please discuss at Village Pump: Arrays.
Please merge code in from these obsolete tasks:
Creating an Array
Assigning Values to an Array
Retrieving an Element of an Array
Related tasks
Collections
Creating an Associative Array
Two-dimensional array (runtime)
| #Elena | Elena | var staticArray := new int[]{1, 2, 3}; |
http://rosettacode.org/wiki/Arithmetic/Complex | Arithmetic/Complex | A complex number is a number which can be written as:
a
+
b
×
i
{\displaystyle a+b\times i}
(sometimes shown as:
b
+
a
×
i
{\displaystyle b+a\times i}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are real numbers, and
i
{\displaystyle i}
is √ -1
Typically, complex numbers are represented as a pair of real numbers called the "imaginary part" and "real part", where the imaginary part is the number to be multiplied by
i
{\displaystyle i}
.
Task
Show addition, multiplication, negation, and inversion of complex numbers in separate functions. (Subtraction and division operations can be made with pairs of these operations.)
Print the results for each operation tested.
Optional: Show complex conjugation.
By definition, the complex conjugate of
a
+
b
i
{\displaystyle a+bi}
is
a
−
b
i
{\displaystyle a-bi}
Some languages have complex number libraries available. If your language does, show the operations. If your language does not, also show the definition of this type.
| #Perl | Perl | use Math::Complex;
my $a = 1 + 1*i;
my $b = 3.14159 + 1.25*i;
print "$_\n" foreach
$a + $b, # addition
$a * $b, # multiplication
-$a, # negation
1 / $a, # multiplicative inverse
~$a; # complex conjugate |
http://rosettacode.org/wiki/Arithmetic/Rational | Arithmetic/Rational | Task
Create a reasonably complete implementation of rational arithmetic in the particular language using the idioms of the language.
Example
Define a new type called frac with binary operator "//" of two integers that returns a structure made up of the numerator and the denominator (as per a rational number).
Further define the appropriate rational unary operators abs and '-', with the binary operators for addition '+', subtraction '-', multiplication '×', division '/', integer division '÷', modulo division, the comparison operators (e.g. '<', '≤', '>', & '≥') and equality operators (e.g. '=' & '≠').
Define standard coercion operators for casting int to frac etc.
If space allows, define standard increment and decrement operators (e.g. '+:=' & '-:=' etc.).
Finally test the operators:
Use the new type frac to find all perfect numbers less than 219 by summing the reciprocal of the factors.
Related task
Perfect Numbers
| #Racket | Racket |
-> (* 1/7 14)
2
|
http://rosettacode.org/wiki/Arithmetic-geometric_mean | Arithmetic-geometric mean |
This page uses content from Wikipedia. The original article was at Arithmetic-geometric mean. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Write a function to compute the arithmetic-geometric mean of two numbers.
The arithmetic-geometric mean of two numbers can be (usefully) denoted as
a
g
m
(
a
,
g
)
{\displaystyle \mathrm {agm} (a,g)}
, and is equal to the limit of the sequence:
a
0
=
a
;
g
0
=
g
{\displaystyle a_{0}=a;\qquad g_{0}=g}
a
n
+
1
=
1
2
(
a
n
+
g
n
)
;
g
n
+
1
=
a
n
g
n
.
{\displaystyle a_{n+1}={\tfrac {1}{2}}(a_{n}+g_{n});\quad g_{n+1}={\sqrt {a_{n}g_{n}}}.}
Since the limit of
a
n
−
g
n
{\displaystyle a_{n}-g_{n}}
tends (rapidly) to zero with iterations, this is an efficient method.
Demonstrate the function by calculating:
a
g
m
(
1
,
1
/
2
)
{\displaystyle \mathrm {agm} (1,1/{\sqrt {2}})}
Also see
mathworld.wolfram.com/Arithmetic-Geometric Mean
| #SQL | SQL | WITH
rec (rn, a, g, diff) AS (
SELECT 1, 1, 1/SQRT(2), 1 - 1/SQRT(2)
FROM dual
UNION ALL
SELECT rn + 1, (a + g)/2, SQRT(a * g), (a + g)/2 - SQRT(a * g)
FROM rec
WHERE diff > 1e-38
)
SELECT *
FROM rec
WHERE diff <= 1e-38
; |
http://rosettacode.org/wiki/Arithmetic-geometric_mean | Arithmetic-geometric mean |
This page uses content from Wikipedia. The original article was at Arithmetic-geometric mean. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Write a function to compute the arithmetic-geometric mean of two numbers.
The arithmetic-geometric mean of two numbers can be (usefully) denoted as
a
g
m
(
a
,
g
)
{\displaystyle \mathrm {agm} (a,g)}
, and is equal to the limit of the sequence:
a
0
=
a
;
g
0
=
g
{\displaystyle a_{0}=a;\qquad g_{0}=g}
a
n
+
1
=
1
2
(
a
n
+
g
n
)
;
g
n
+
1
=
a
n
g
n
.
{\displaystyle a_{n+1}={\tfrac {1}{2}}(a_{n}+g_{n});\quad g_{n+1}={\sqrt {a_{n}g_{n}}}.}
Since the limit of
a
n
−
g
n
{\displaystyle a_{n}-g_{n}}
tends (rapidly) to zero with iterations, this is an efficient method.
Demonstrate the function by calculating:
a
g
m
(
1
,
1
/
2
)
{\displaystyle \mathrm {agm} (1,1/{\sqrt {2}})}
Also see
mathworld.wolfram.com/Arithmetic-Geometric Mean
| #Standard_ML | Standard ML |
fun agm(a, g) = let
fun agm'(a, g, eps) =
if Real.abs(a-g) < eps then
a
else
agm'((a+g)/2.0, Math.sqrt(a*g), eps)
in agm'(a, g, 1e~15)
end;
|
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Go | Go | package main
import (
"fmt"
"math"
"math/big"
"math/cmplx"
)
func main() {
fmt.Println("float64: ", math.Pow(0, 0))
var b big.Int
fmt.Println("big integer:", b.Exp(&b, &b, nil))
fmt.Println("complex: ", cmplx.Pow(0, 0))
} |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Groovy | Groovy | println 0**0 |
http://rosettacode.org/wiki/Zig-zag_matrix | Zig-zag matrix | Task
Produce a zig-zag array.
A zig-zag array is a square arrangement of the first N2 natural numbers, where the
numbers increase sequentially as you zig-zag along the array's anti-diagonals.
For a graphical representation, see JPG zigzag (JPG uses such arrays to encode images).
For example, given 5, produce this array:
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24
Related tasks
Spiral matrix
Identity matrix
Ulam spiral (for primes)
See also
Wiktionary entry: anti-diagonals
| #Agena | Agena | # zig-zag matrix
makeZigZag := proc( n :: number ) :: table is
local move := proc( x :: number, y :: number, upRight :: boolean ) is
if y = n then
upRight := not upRight;
x := x + 1
elif x = 1 then
upRight := not upRight;
y := y + 1
else
x := x - 1;
y := y + 1
fi;
return x, y, upRight
end ;
# create empty table
local result := [];
for i to n do
result[ i ] := [];
for j to n do result[ i, j ] := 0 od
od;
# fill the table
local x, y, upRight := 1, 1, true;
for i to n * n do
result[ x, y ] := i - 1;
if upRight then
x, y, upRight := move( x, y, upRight )
else
y, x, upRight := move( y, x, upRight )
fi
od;
return result
end;
scope
local m := makeZigZag( 5 );
for i to size m do
for j to size m do
printf( " %3d", m[ i, j ] )
od;
print()
od
epocs |
http://rosettacode.org/wiki/Yellowstone_sequence | Yellowstone sequence | The Yellowstone sequence, also called the Yellowstone permutation, is defined as:
For n <= 3,
a(n) = n
For n >= 4,
a(n) = the smallest number not already in sequence such that a(n) is relatively prime to a(n-1) and
is not relatively prime to a(n-2).
The sequence is a permutation of the natural numbers, and gets its name from what its authors felt was a spiking, geyser like appearance of a plot of the sequence.
Example
a(4) is 4 because 4 is the smallest number following 1, 2, 3 in the sequence that is relatively prime to the entry before it (3), and is not relatively prime to the number two entries before it (2).
Task
Find and show as output the first 30 Yellowstone numbers.
Extra
Demonstrate how to plot, with x = n and y coordinate a(n), the first 100 Yellowstone numbers.
Related tasks
Greatest common divisor.
Plot coordinate pairs.
See also
The OEIS entry: A098550 The Yellowstone permutation.
Applegate et al, 2015: The Yellowstone Permutation [1].
| #C | C | #include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
typedef struct lnode_t {
struct lnode_t *prev;
struct lnode_t *next;
int v;
} Lnode;
Lnode *make_list_node(int v) {
Lnode *node = malloc(sizeof(Lnode));
if (node == NULL) {
return NULL;
}
node->v = v;
node->prev = NULL;
node->next = NULL;
return node;
}
void free_lnode(Lnode *node) {
if (node == NULL) {
return;
}
node->v = 0;
node->prev = NULL;
free_lnode(node->next);
node->next = NULL;
}
typedef struct list_t {
Lnode *front;
Lnode *back;
size_t len;
} List;
List *make_list() {
List *list = malloc(sizeof(List));
if (list == NULL) {
return NULL;
}
list->front = NULL;
list->back = NULL;
list->len = 0;
return list;
}
void free_list(List *list) {
if (list == NULL) {
return;
}
list->len = 0;
list->back = NULL;
free_lnode(list->front);
list->front = NULL;
}
void list_insert(List *list, int v) {
Lnode *node;
if (list == NULL) {
return;
}
node = make_list_node(v);
if (list->front == NULL) {
list->front = node;
list->back = node;
list->len = 1;
} else {
node->prev = list->back;
list->back->next = node;
list->back = node;
list->len++;
}
}
void list_print(List *list) {
Lnode *it;
if (list == NULL) {
return;
}
for (it = list->front; it != NULL; it = it->next) {
printf("%d ", it->v);
}
}
int list_get(List *list, int idx) {
Lnode *it = NULL;
if (list != NULL && list->front != NULL) {
int i;
if (idx < 0) {
it = list->back;
i = -1;
while (it != NULL && i > idx) {
it = it->prev;
i--;
}
} else {
it = list->front;
i = 0;
while (it != NULL && i < idx) {
it = it->next;
i++;
}
}
}
if (it == NULL) {
return INT_MIN;
}
return it->v;
}
///////////////////////////////////////
typedef struct mnode_t {
int k;
bool v;
struct mnode_t *next;
} Mnode;
Mnode *make_map_node(int k, bool v) {
Mnode *node = malloc(sizeof(Mnode));
if (node == NULL) {
return node;
}
node->k = k;
node->v = v;
node->next = NULL;
return node;
}
void free_mnode(Mnode *node) {
if (node == NULL) {
return;
}
node->k = 0;
node->v = false;
free_mnode(node->next);
node->next = NULL;
}
typedef struct map_t {
Mnode *front;
} Map;
Map *make_map() {
Map *map = malloc(sizeof(Map));
if (map == NULL) {
return NULL;
}
map->front = NULL;
return map;
}
void free_map(Map *map) {
if (map == NULL) {
return;
}
free_mnode(map->front);
map->front = NULL;
}
void map_insert(Map *map, int k, bool v) {
if (map == NULL) {
return;
}
if (map->front == NULL) {
map->front = make_map_node(k, v);
} else {
Mnode *it = map->front;
while (it->next != NULL) {
it = it->next;
}
it->next = make_map_node(k, v);
}
}
bool map_get(Map *map, int k) {
if (map != NULL) {
Mnode *it = map->front;
while (it != NULL && it->k != k) {
it = it->next;
}
if (it != NULL) {
return it->v;
}
}
return false;
}
///////////////////////////////////////
int gcd(int u, int v) {
if (u < 0) u = -u;
if (v < 0) v = -v;
if (v) {
while ((u %= v) && (v %= u));
}
return u + v;
}
List *yellow(size_t n) {
List *a;
Map *b;
int i;
a = make_list();
list_insert(a, 1);
list_insert(a, 2);
list_insert(a, 3);
b = make_map();
map_insert(b, 1, true);
map_insert(b, 2, true);
map_insert(b, 3, true);
i = 4;
while (n > a->len) {
if (!map_get(b, i) && gcd(i, list_get(a, -1)) == 1 && gcd(i, list_get(a, -2)) > 1) {
list_insert(a, i);
map_insert(b, i, true);
i = 4;
}
i++;
}
free_map(b);
return a;
}
int main() {
List *a = yellow(30);
list_print(a);
free_list(a);
putc('\n', stdout);
return 0;
} |
http://rosettacode.org/wiki/Arithmetic_evaluation | Arithmetic evaluation | Create a program which parses and evaluates arithmetic expressions.
Requirements
An abstract-syntax tree (AST) for the expression must be created from parsing the input.
The AST must be used in evaluation, also, so the input may not be directly evaluated (e.g. by calling eval or a similar language feature.)
The expression will be a string or list of symbols like "(1+3)*7".
The four symbols + - * / must be supported as binary operators with conventional precedence rules.
Precedence-control parentheses must also be supported.
Note
For those who don't remember, mathematical precedence is as follows:
Parentheses
Multiplication/Division (left to right)
Addition/Subtraction (left to right)
C.f
24 game Player.
Parsing/RPN calculator algorithm.
Parsing/RPN to infix conversion.
| #REXX | REXX | /*REXX program evaluates an infix─type arithmetic expression and displays the result.*/
nchars = '0123456789.eEdDqQ' /*possible parts of a number, sans ± */
e='***error***'; $=" "; doubleOps= '&|*/'; z= /*handy─dandy variables.*/
parse arg x 1 ox1; if x='' then call serr "no input was specified."
x=space(x); L=length(x); x=translate(x, '()()', "[]{}")
j=0
do forever; j=j+1; if j>L then leave; _=substr(x, j, 1); _2=getX()
newT=pos(_,' ()[]{}^÷')\==0; if newT then do; z=z _ $; iterate; end
possDouble=pos(_,doubleOps)\==0 /*is _ a possible double operator?*/
if possDouble then do /* " this " " " " */
if _2==_ then do /*yupper, it's one of a double operator*/
_=_ || _ /*create and use a double char operator*/
x=overlay($, x, Nj) /*blank out 2nd symbol.*/
end
z=z _ $; iterate
end
if _=='+' | _=="-" then do; p_=word(z, max(1,words(z))) /*last Z token. */
if p_=='(' then z=z 0 /*handle a unary ± */
z=z _ $; iterate
end
lets=0; sigs=0; #=_
do j=j+1 to L; _=substr(x,j,1) /*build a valid number.*/
if lets==1 & sigs==0 then if _=='+' | _=="-" then do; sigs=1
#=# || _
iterate
end
if pos(_,nchars)==0 then leave
lets=lets+datatype(_,'M') /*keep track of the number of exponents*/
#=# || translate(_,'EEEEE', "eDdQq") /*keep building the number. */
end /*j*/
j=j-1
if \datatype(#,'N') then call serr "invalid number: " #
z=z # $
end /*forever*/
_=word(z,1); if _=='+' | _=="-" then z=0 z /*handle the unary cases. */
x='(' space(z) ")"; tokens=words(x) /*force stacking for the expression. */
do i=1 for tokens; @.i=word(x,i); end /*i*/ /*assign input tokens. */
L=max(20,length(x)) /*use 20 for the minimum display width.*/
op= ')(-+/*^'; Rop=substr(op,3); p.=; s.=; n=length(op); epr=; stack=
do i=1 for n; _=substr(op,i,1); s._=(i+1)%2; p._=s._ + (i==n); end /*i*/
/* [↑] assign the operator priorities.*/
do #=1 for tokens; ?=@.# /*process each token from the @. list.*/
if ?=='**' then ?="^" /*convert to REXX-type exponentiation. */
select /*@.# is: ( operator ) operand*/
when ?=='(' then stack="(" stack
when isOp(?) then do /*is the token an operator ? */
!=word(stack,1) /*get token from stack.*/
do while !\==')' & s.!>=p.?; epr=epr ! /*addition.*/
stack=subword(stack, 2) /*del token from stack*/
!= word(stack, 1) /*get token from stack*/
end /*while*/
stack=? stack /*add token to stack*/
end
when ?==')' then do; !=word(stack, 1) /*get token from stack*/
do while !\=='('; epr=epr ! /*append to expression*/
stack=subword(stack, 2) /*del token from stack*/
!= word(stack, 1) /*get token from stack*/
end /*while*/
stack=subword(stack, 2) /*del token from stack*/
end
otherwise epr=epr ? /*add operand to epr.*/
end /*select*/
end /*#*/
epr=space(epr stack); tokens=words(epr); x=epr; z=; stack=
do i=1 for tokens; @.i=word(epr,i); end /*i*/ /*assign input tokens.*/
Dop='/ // % ÷'; Bop="& | &&" /*division operands; binary operands.*/
Aop='- + * ^ **' Dop Bop; Lop=Aop "||" /*arithmetic operands; legal operands.*/
do #=1 for tokens; ?=@.#; ??=? /*process each token from @. list. */
w=words(stack); b=word(stack, max(1, w ) ) /*stack count; the last entry. */
a=word(stack, max(1, w-1) ) /*stack's "first" operand. */
division =wordpos(?, Dop)\==0 /*flag: doing a division operation. */
arith =wordpos(?, Aop)\==0 /*flag: doing arithmetic operation. */
bitOp =wordpos(?, Bop)\==0 /*flag: doing binary mathematics. */
if datatype(?, 'N') then do; stack=stack ?; iterate; end
if wordpos(?,Lop)==0 then do; z=e "illegal operator:" ?; leave; end
if w<2 then do; z=e "illegal epr expression."; leave; end
if ?=='^' then ??="**" /*REXXify ^ ──► ** (make it legal).*/
if ?=='÷' then ??="/" /*REXXify ÷ ──► / (make it legal).*/
if division & b=0 then do; z=e "division by zero" b; leave; end
if bitOp & \isBit(a) then do; z=e "token isn't logical: " a; leave; end
if bitOp & \isBit(b) then do; z=e "token isn't logical: " b; leave; end
select /*perform an arithmetic operation. */
when ??=='+' then y = a + b
when ??=='-' then y = a - b
when ??=='*' then y = a * b
when ??=='/' | ??=="÷" then y = a / b
when ??=='//' then y = a // b
when ??=='%' then y = a % b
when ??=='^' | ??=="**" then y = a ** b
when ??=='||' then y = a || b
otherwise z=e 'invalid operator:' ?; leave
end /*select*/
if datatype(y, 'W') then y=y/1 /*normalize the number with ÷ by 1. */
_=subword(stack, 1, w-2); stack=_ y /*rebuild the stack with the answer. */
end /*#*/
if word(z, 1)==e then stack= /*handle the special case of errors. */
z=space(z stack) /*append any residual entries. */
say 'answer──►' z /*display the answer (result). */
parse source upper . how . /*invoked via C.L. or REXX program ? */
if how=='COMMAND' | \datatype(z, 'W') then exit /*stick a fork in it, we're all done. */
return z /*return Z ──► invoker (the RESULT). */
/*──────────────────────────────────────────────────────────────────────────────────────*/
isBit: return arg(1)==0 | arg(1) == 1 /*returns 1 if 1st argument is binary*/
isOp: return pos(arg(1), rOp) \== 0 /*is argument 1 a "real" operator? */
serr: say; say e arg(1); say; exit 13 /*issue an error message with some text*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
getX: do Nj=j+1 to length(x); _n=substr(x, Nj, 1); if _n==$ then iterate
return substr(x, Nj, 1) /* [↑] ignore any blanks in expression*/
end /*Nj*/
return $ /*reached end-of-tokens, return $. */ |
http://rosettacode.org/wiki/Zeckendorf_arithmetic | Zeckendorf arithmetic | This task is a total immersion zeckendorf task; using decimal numbers will attract serious disapprobation.
The task is to implement addition, subtraction, multiplication, and division using Zeckendorf number representation. Optionally provide decrement, increment and comparitive operation functions.
Addition
Like binary 1 + 1 = 10, note carry 1 left. There the similarity ends. 10 + 10 = 101, note carry 1 left and 1 right. 100 + 100 = 1001, note carry 1 left and 2 right, this is the general case.
Occurrences of 11 must be changed to 100. Occurrences of 111 may be changed from the right by replacing 11 with 100, or from the left converting 111 to 100 + 100;
Subtraction
10 - 1 = 1. The general rule is borrow 1 right carry 1 left. eg:
abcde
10100 -
1000
_____
100 borrow 1 from a leaves 100
+ 100 add the carry
_____
1001
A larger example:
abcdef
100100 -
1000
______
1*0100 borrow 1 from b
+ 100 add the carry
______
1*1001
Sadly we borrowed 1 from b which didn't have it to lend. So now b borrows from a:
1001
+ 1000 add the carry
____
10100
Multiplication
Here you teach your computer its zeckendorf tables. eg. 101 * 1001:
a = 1 * 101 = 101
b = 10 * 101 = a + a = 10000
c = 100 * 101 = b + a = 10101
d = 1000 * 101 = c + b = 101010
1001 = d + a therefore 101 * 1001 =
101010
+ 101
______
1000100
Division
Lets try 1000101 divided by 101, so we can use the same table used for multiplication.
1000101 -
101010 subtract d (1000 * 101)
_______
1000 -
101 b and c are too large to subtract, so subtract a
____
1 so 1000101 divided by 101 is d + a (1001) remainder 1
Efficient algorithms for Zeckendorf arithmetic is interesting. The sections on addition and subtraction are particularly relevant for this task.
| #Nim | Nim | type Zeckendorf = object
dVal: Natural
dLen: Natural
const
Dig = ["00", "01", "10"]
Dig1 = ["", "1", "10"]
# Forward references.
func b(z: var Zeckendorf; pos: Natural)
func inc(z: var Zeckendorf)
func a(z: var Zeckendorf; n: Natural) =
var i = n
while true:
if z.dLen < i: z.dLen = i
let j = z.dVal shr (i * 2) and 3
case j
of 0, 1:
return
of 2:
if (z.dVal shr ((i + 1) * 2) and 1) != 1: return
z.dVal += 1 shl (i * 2 + 1)
return
of 3:
z.dVal = z.dVal and not (3 shl (i * 2))
z.b((i + 1) * 2)
else:
assert(false)
inc i
func b(z: var Zeckendorf; pos: Natural) =
if pos == 0:
inc z
return
if (z.dVal shr pos and 1) == 0:
z.dVal += 1 shl pos
z.a(pos div 2)
if pos > 1: z.a(pos div 2 - 1)
else:
z.dVal = z.dVal and not(1 shl pos)
z.b(pos + 1)
z.b(pos - (if pos > 1: 2 else: 1))
func c(z: var Zeckendorf; pos: Natural) =
if (z.dVal shr pos and 1) == 1:
z.dVal = z.dVal and not(1 shl pos)
return
z.c(pos + 1)
if pos > 0:
z.b(pos - 1)
else:
inc z
func initZeckendorf(s = "0"): Zeckendorf =
var q = 1
var i = s.high
result.dLen = i div 2
while i >= 0:
result.dVal += (ord(s[i]) - ord('0')) * q
q *= 2
dec i
func inc(z: var Zeckendorf) =
inc z.dVal
z.a(0)
func `+=`(z1: var Zeckendorf; z2: Zeckendorf) =
for gn in 0 .. (2 * z2.dLen + 1):
if (z2.dVal shr gn and 1) == 1:
z1.b(gn)
func `-=`(z1: var Zeckendorf; z2: Zeckendorf) =
for gn in 0 .. (2 * z2.dLen + 1):
if (z2.dVal shr gn and 1) == 1:
z1.c(gn)
while z1.dLen > 0 and (z1.dVal shr (z1.dLen * 2) and 3) == 0:
dec z1.dLen
func `*=`(z1: var Zeckendorf; z2: Zeckendorf) =
var na, nb = z2
var nr: Zeckendorf
for i in 0 .. (z1.dLen + 1) * 2:
if (z1.dVal shr i and 1) > 0: nr += nb
let nt = nb
nb += na
na = nt
z1 = nr
func`$`(z: var Zeckendorf): string =
if z.dVal == 0: return "0"
result.add Dig1[z.dVal shr (z.dLen * 2) and 3]
for i in countdown(z.dLen - 1, 0):
result.add Dig[z.dVal shr (i * 2) and 3]
when isMainModule:
var g: Zeckendorf
echo "Addition:"
g = initZeckendorf("10")
g += initZeckendorf("10")
echo g
g += initZeckendorf("10")
echo g
g += initZeckendorf("1001")
echo g
g += initZeckendorf("1000")
echo g
g += initZeckendorf("10101")
echo g
echo "\nSubtraction:"
g = initZeckendorf("1000")
g -= initZeckendorf("101")
echo g
g = initZeckendorf("10101010")
g -= initZeckendorf("1010101")
echo g
echo "\nMultiplication:"
g = initZeckendorf("1001")
g *= initZeckendorf("101")
echo g
g = initZeckendorf("101010")
g += initZeckendorf("101")
echo g |
http://rosettacode.org/wiki/Zumkeller_numbers | Zumkeller numbers | Zumkeller numbers are the set of numbers whose divisors can be partitioned into two disjoint sets that sum to the same value. Each sum must contain divisor values that are not in the other sum, and all of the divisors must be in one or the other. There are no restrictions on how the divisors are partitioned, only that the two partition sums are equal.
E.G.
6 is a Zumkeller number; The divisors {1 2 3 6} can be partitioned into two groups {1 2 3} and {6} that both sum to 6.
10 is not a Zumkeller number; The divisors {1 2 5 10} can not be partitioned into two groups in any way that will both sum to the same value.
12 is a Zumkeller number; The divisors {1 2 3 4 6 12} can be partitioned into two groups {1 3 4 6} and {2 12} that both sum to 14.
Even Zumkeller numbers are common; odd Zumkeller numbers are much less so. For values below 10^6, there is at least one Zumkeller number in every 12 consecutive integers, and the vast majority of them are even. The odd Zumkeller numbers are very similar to the list from the task Abundant odd numbers; they are nearly the same except for the further restriction that the abundance (A(n) = sigma(n) - 2n), must be even: A(n) mod 2 == 0
Task
Write a routine (function, procedure, whatever) to find Zumkeller numbers.
Use the routine to find and display here, on this page, the first 220 Zumkeller numbers.
Use the routine to find and display here, on this page, the first 40 odd Zumkeller numbers.
Optional, stretch goal: Use the routine to find and display here, on this page, the first 40 odd Zumkeller numbers that don't end with 5.
See Also
OEIS:A083207 - Zumkeller numbers to get an impression of different partitions OEIS:A083206 Zumkeller partitions
OEIS:A174865 - Odd Zumkeller numbers
Related Tasks
Abundant odd numbers
Abundant, deficient and perfect number classifications
Proper divisors , Factors of an integer | #Kotlin | Kotlin | import java.util.ArrayList
import kotlin.math.sqrt
object ZumkellerNumbers {
@JvmStatic
fun main(args: Array<String>) {
var n = 1
println("First 220 Zumkeller numbers:")
run {
var count = 1
while (count <= 220) {
if (isZumkeller(n)) {
print("%3d ".format(n))
if (count % 20 == 0) {
println()
}
count++
}
n += 1
}
}
n = 1
println("\nFirst 40 odd Zumkeller numbers:")
run {
var count = 1
while (count <= 40) {
if (isZumkeller(n)) {
print("%6d".format(n))
if (count % 10 == 0) {
println()
}
count++
}
n += 2
}
}
n = 1
println("\nFirst 40 odd Zumkeller numbers that do not end in a 5:")
var count = 1
while (count <= 40) {
if (n % 5 != 0 && isZumkeller(n)) {
print("%8d".format(n))
if (count % 10 == 0) {
println()
}
count++
}
n += 2
}
}
private fun isZumkeller(n: Int): Boolean { // numbers congruent to 6 or 12 modulo 18 are Zumkeller numbers
if (n % 18 == 6 || n % 18 == 12) {
return true
}
val divisors = getDivisors(n)
val divisorSum = divisors.stream().mapToInt { i: Int? -> i!! }.sum()
// divisor sum cannot be odd
if (divisorSum % 2 == 1) {
return false
}
// numbers where n is odd and the abundance is even are Zumkeller numbers
val abundance = divisorSum - 2 * n
if (n % 2 == 1 && abundance > 0 && abundance % 2 == 0) {
return true
}
divisors.sort()
val j = divisors.size - 1
val sum = divisorSum / 2
// Largest divisor larger than sum - then cannot partition and not Zumkeller number
return if (divisors[j] > sum) false else canPartition(j, divisors, sum, IntArray(2))
}
private fun canPartition(j: Int, divisors: List<Int>, sum: Int, buckets: IntArray): Boolean {
if (j < 0) {
return true
}
for (i in 0..1) {
if (buckets[i] + divisors[j] <= sum) {
buckets[i] += divisors[j]
if (canPartition(j - 1, divisors, sum, buckets)) {
return true
}
buckets[i] -= divisors[j]
}
if (buckets[i] == 0) {
break
}
}
return false
}
private fun getDivisors(number: Int): MutableList<Int> {
val divisors: MutableList<Int> = ArrayList()
val sqrt = sqrt(number.toDouble()).toLong()
for (i in 1..sqrt) {
if (number % i == 0L) {
divisors.add(i.toInt())
val div = (number / i).toInt()
if (div.toLong() != i) {
divisors.add(div)
}
}
}
return divisors
}
} |
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #Dart | Dart | import 'dart:math' show pow;
int fallingPowers(int base) =>
base == 1 ? 1 : pow(base, fallingPowers(base - 1));
void main() {
final exponent = fallingPowers(4),
s = BigInt.from(5).pow(exponent).toString();
print('First twenty: ${s.substring(0, 20)}');
print('Last twenty: ${s.substring(s.length - 20)}');
print('Number of digits: ${s.length}');
|
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #dc | dc | [5432.dc]sz
5 4 3 2 ^ ^ ^ sy [y = 5 ^ 4 ^ 3 ^ 2]sz
ly Z sc [c = length of y]sz
[ First 20 digits: ]P ly 10 lc 20 - ^ / p sz [y / (10 ^ (c - 20))]sz
[ Last 20 digits: ]P ly 10 20 ^ % p sz [y % (10 ^ 20)]sz
[Number of digits: ]P lc p sz |
http://rosettacode.org/wiki/Zhang-Suen_thinning_algorithm | Zhang-Suen thinning algorithm | This is an algorithm used to thin a black and white i.e. one bit per pixel images.
For example, with an input image of:
################# #############
################## ################
################### ##################
######## ####### ###################
###### ####### ####### ######
###### ####### #######
################# #######
################ #######
################# #######
###### ####### #######
###### ####### #######
###### ####### ####### ######
######## ####### ###################
######## ####### ###### ################## ######
######## ####### ###### ################ ######
######## ####### ###### ############# ######
It produces the thinned output:
# ########## #######
## # #### #
# # ##
# # #
# # #
# # #
############ #
# # #
# # #
# # #
# # #
# ##
# ############
### ###
Algorithm
Assume black pixels are one and white pixels zero, and that the input image is a rectangular N by M array of ones and zeroes.
The algorithm operates on all black pixels P1 that can have eight neighbours.
The neighbours are, in order, arranged as:
P9 P2 P3
P8 P1 P4
P7 P6 P5
Obviously the boundary pixels of the image cannot have the full eight neighbours.
Define
A
(
P
1
)
{\displaystyle A(P1)}
= the number of transitions from white to black, (0 -> 1) in the sequence P2,P3,P4,P5,P6,P7,P8,P9,P2. (Note the extra P2 at the end - it is circular).
Define
B
(
P
1
)
{\displaystyle B(P1)}
= The number of black pixel neighbours of P1. ( = sum(P2 .. P9) )
Step 1
All pixels are tested and pixels satisfying all the following conditions (simultaneously) are just noted at this stage.
(0) The pixel is black and has eight neighbours
(1)
2
<=
B
(
P
1
)
<=
6
{\displaystyle 2<=B(P1)<=6}
(2) A(P1) = 1
(3) At least one of P2 and P4 and P6 is white
(4) At least one of P4 and P6 and P8 is white
After iterating over the image and collecting all the pixels satisfying all step 1 conditions, all these condition satisfying pixels are set to white.
Step 2
All pixels are again tested and pixels satisfying all the following conditions are just noted at this stage.
(0) The pixel is black and has eight neighbours
(1)
2
<=
B
(
P
1
)
<=
6
{\displaystyle 2<=B(P1)<=6}
(2) A(P1) = 1
(3) At least one of P2 and P4 and P8 is white
(4) At least one of P2 and P6 and P8 is white
After iterating over the image and collecting all the pixels satisfying all step 2 conditions, all these condition satisfying pixels are again set to white.
Iteration
If any pixels were set in this round of either step 1 or step 2 then all steps are repeated until no image pixels are so changed.
Task
Write a routine to perform Zhang-Suen thinning on an image matrix of ones and zeroes.
Use the routine to thin the following image and show the output here on this page as either a matrix of ones and zeroes, an image, or an ASCII-art image of space/non-space characters.
00000000000000000000000000000000
01111111110000000111111110000000
01110001111000001111001111000000
01110000111000001110000111000000
01110001111000001110000000000000
01111111110000001110000000000000
01110111100000001110000111000000
01110011110011101111001111011100
01110001111011100111111110011100
00000000000000000000000000000000
Reference
Zhang-Suen Thinning Algorithm, Java Implementation by Nayef Reza.
"Character Recognition Systems: A Guide for Students and Practitioners" By Mohamed Cheriet, Nawwaf Kharma, Cheng-Lin Liu, Ching Suen
| #Kotlin | Kotlin | // version 1.1.2
class Point(val x: Int, val y: Int)
val image = arrayOf(
" ",
" ################# ############# ",
" ################## ################ ",
" ################### ################## ",
" ######## ####### ################### ",
" ###### ####### ####### ###### ",
" ###### ####### ####### ",
" ################# ####### ",
" ################ ####### ",
" ################# ####### ",
" ###### ####### ####### ",
" ###### ####### ####### ",
" ###### ####### ####### ###### ",
" ######## ####### ################### ",
" ######## ####### ###### ################## ###### ",
" ######## ####### ###### ################ ###### ",
" ######## ####### ###### ############# ###### ",
" "
)
val nbrs = arrayOf(
intArrayOf( 0, -1), intArrayOf( 1, -1), intArrayOf( 1, 0),
intArrayOf( 1, 1), intArrayOf( 0, 1), intArrayOf(-1, 1),
intArrayOf(-1, 0), intArrayOf(-1, -1), intArrayOf( 0, -1)
)
val nbrGroups = arrayOf(
arrayOf(intArrayOf(0, 2, 4), intArrayOf(2, 4, 6)),
arrayOf(intArrayOf(0, 2, 6), intArrayOf(0, 4, 6))
)
val toWhite = mutableListOf<Point>()
val grid = Array(image.size) { image[it].toCharArray() }
fun thinImage() {
var firstStep = false
var hasChanged: Boolean
do {
hasChanged = false
firstStep = !firstStep
for (r in 1 until grid.size - 1) {
for (c in 1 until grid[0].size - 1) {
if (grid[r][c] != '#') continue
val nn = numNeighbors(r, c)
if (nn !in 2..6) continue
if (numTransitions(r, c) != 1) continue
val step = if (firstStep) 0 else 1
if (!atLeastOneIsWhite(r, c, step)) continue
toWhite.add(Point(c, r))
hasChanged = true
}
}
for (p in toWhite) grid[p.y][p.x] = ' '
toWhite.clear()
}
while (firstStep || hasChanged)
for (row in grid) println(row)
}
fun numNeighbors(r: Int, c: Int): Int {
var count = 0
for (i in 0 until nbrs.size - 1) {
if (grid[r + nbrs[i][1]][c + nbrs[i][0]] == '#') count++
}
return count
}
fun numTransitions(r: Int, c: Int): Int {
var count = 0
for (i in 0 until nbrs.size - 1) {
if (grid[r + nbrs[i][1]][c + nbrs[i][0]] == ' ') {
if (grid[r + nbrs[i + 1][1]][c + nbrs[i + 1][0]] == '#') count++
}
}
return count
}
fun atLeastOneIsWhite(r: Int, c: Int, step: Int): Boolean {
var count = 0;
val group = nbrGroups[step]
for (i in 0..1) {
for (j in 0 until group[i].size) {
val nbr = nbrs[group[i][j]]
if (grid[r + nbr[1]][c + nbr[0]] == ' ') {
count++
break
}
}
}
return count > 1
}
fun main(args: Array<String>) {
thinImage()
} |
http://rosettacode.org/wiki/Zeckendorf_number_representation | Zeckendorf number representation | Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.
Recall that the first six distinct Fibonacci numbers are: 1, 2, 3, 5, 8, 13.
The decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.
10100 is not the only way to make 11 from the Fibonacci numbers however; 0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that no two consecutive Fibonacci numbers can be used which leads to the former unique solution.
Task
Generate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.
The intention in this task to find the Zeckendorf form of an arbitrary integer. The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task.
Also see
OEIS A014417 for the the sequence of required results.
Brown's Criterion - Numberphile
Related task
Fibonacci sequence
| #Elixir | Elixir | defmodule Zeckendorf do
def number do
Stream.unfold(0, fn n -> zn_loop(n) end)
end
defp zn_loop(n) do
bin = Integer.to_string(n, 2)
if String.match?(bin, ~r/11/), do: zn_loop(n+1), else: {bin, n+1}
end
end
Zeckendorf.number |> Enum.take(21) |> Enum.with_index
|> Enum.each(fn {zn, i} -> IO.puts "#{i}: #{zn}" end) |
http://rosettacode.org/wiki/100_doors | 100 doors | There are 100 doors in a row that are all initially closed.
You make 100 passes by the doors.
The first time through, visit every door and toggle the door (if the door is closed, open it; if it is open, close it).
The second time, only visit every 2nd door (door #2, #4, #6, ...), and toggle it.
The third time, visit every 3rd door (door #3, #6, #9, ...), etc, until you only visit the 100th door.
Task
Answer the question: what state are the doors in after the last pass? Which are open, which are closed?
Alternate:
As noted in this page's discussion page, the only doors that remain open are those whose numbers are perfect squares.
Opening only those doors is an optimization that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.
| #C | C | #include <stdio.h>
int main()
{
char is_open[100] = { 0 };
int pass, door;
/* do the 100 passes */
for (pass = 0; pass < 100; ++pass)
for (door = pass; door < 100; door += pass+1)
is_open[door] = !is_open[door];
/* output the result */
for (door = 0; door < 100; ++door)
printf("door #%d is %s.\n", door+1, (is_open[door]? "open" : "closed"));
return 0;
} |
http://rosettacode.org/wiki/Arrays | Arrays | This task is about arrays.
For hashes or associative arrays, please see Creating an Associative Array.
For a definition and in-depth discussion of what an array is, see Array.
Task
Show basic array syntax in your language.
Basically, create an array, assign a value to it, and retrieve an element (if available, show both fixed-length arrays and
dynamic arrays, pushing a value into it).
Please discuss at Village Pump: Arrays.
Please merge code in from these obsolete tasks:
Creating an Array
Assigning Values to an Array
Retrieving an Element of an Array
Related tasks
Collections
Creating an Associative Array
Two-dimensional array (runtime)
| #Elixir | Elixir | ret = {:ok, "fun", 3.1415} |
http://rosettacode.org/wiki/Arithmetic/Complex | Arithmetic/Complex | A complex number is a number which can be written as:
a
+
b
×
i
{\displaystyle a+b\times i}
(sometimes shown as:
b
+
a
×
i
{\displaystyle b+a\times i}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are real numbers, and
i
{\displaystyle i}
is √ -1
Typically, complex numbers are represented as a pair of real numbers called the "imaginary part" and "real part", where the imaginary part is the number to be multiplied by
i
{\displaystyle i}
.
Task
Show addition, multiplication, negation, and inversion of complex numbers in separate functions. (Subtraction and division operations can be made with pairs of these operations.)
Print the results for each operation tested.
Optional: Show complex conjugation.
By definition, the complex conjugate of
a
+
b
i
{\displaystyle a+bi}
is
a
−
b
i
{\displaystyle a-bi}
Some languages have complex number libraries available. If your language does, show the operations. If your language does not, also show the definition of this type.
| #Phix | Phix | -- demo\rosetta\ArithComplex.exw
with javascript_semantics
include complex.e
complex a = complex_new(1,1), -- (or just {1,1})
b = complex_new(3.14159,1.25),
c = complex_new(1,0),
d = complex_new(0,1)
printf(1,"a = %s\n",{complex_sprint(a)})
printf(1,"b = %s\n",{complex_sprint(b)})
printf(1,"c = %s\n",{complex_sprint(c)})
printf(1,"d = %s\n",{complex_sprint(d)})
printf(1,"a+b = %s\n",{complex_sprint(complex_add(a,b))})
printf(1,"a*b = %s\n",{complex_sprint(complex_mul(a,b))})
printf(1,"1/a = %s\n",{complex_sprint(complex_inv(a))})
printf(1,"c/a = %s\n",{complex_sprint(complex_div(c,a))})
printf(1,"c-a = %s\n",{complex_sprint(complex_sub(c,a))})
printf(1,"d-a = %s\n",{complex_sprint(complex_sub(d,a))})
printf(1,"-a = %s\n",{complex_sprint(complex_neg(a))})
printf(1,"conj a = %s\n",{complex_sprint(complex_conjugate(a))})
|
http://rosettacode.org/wiki/Arithmetic/Rational | Arithmetic/Rational | Task
Create a reasonably complete implementation of rational arithmetic in the particular language using the idioms of the language.
Example
Define a new type called frac with binary operator "//" of two integers that returns a structure made up of the numerator and the denominator (as per a rational number).
Further define the appropriate rational unary operators abs and '-', with the binary operators for addition '+', subtraction '-', multiplication '×', division '/', integer division '÷', modulo division, the comparison operators (e.g. '<', '≤', '>', & '≥') and equality operators (e.g. '=' & '≠').
Define standard coercion operators for casting int to frac etc.
If space allows, define standard increment and decrement operators (e.g. '+:=' & '-:=' etc.).
Finally test the operators:
Use the new type frac to find all perfect numbers less than 219 by summing the reciprocal of the factors.
Related task
Perfect Numbers
| #Raku | Raku | (2..2**19).hyper.map: -> $candidate {
my $sum = 1 / $candidate;
for 2 .. ceiling(sqrt($candidate)) -> $factor {
if $candidate %% $factor {
$sum += 1 / $factor + 1 / ($candidate / $factor);
}
}
if $sum.nude[1] == 1 {
say "Sum of reciprocal factors of $candidate = $sum exactly", ($sum == 1 ?? ", perfect!" !! ".");
}
} |
http://rosettacode.org/wiki/Arithmetic-geometric_mean | Arithmetic-geometric mean |
This page uses content from Wikipedia. The original article was at Arithmetic-geometric mean. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Write a function to compute the arithmetic-geometric mean of two numbers.
The arithmetic-geometric mean of two numbers can be (usefully) denoted as
a
g
m
(
a
,
g
)
{\displaystyle \mathrm {agm} (a,g)}
, and is equal to the limit of the sequence:
a
0
=
a
;
g
0
=
g
{\displaystyle a_{0}=a;\qquad g_{0}=g}
a
n
+
1
=
1
2
(
a
n
+
g
n
)
;
g
n
+
1
=
a
n
g
n
.
{\displaystyle a_{n+1}={\tfrac {1}{2}}(a_{n}+g_{n});\quad g_{n+1}={\sqrt {a_{n}g_{n}}}.}
Since the limit of
a
n
−
g
n
{\displaystyle a_{n}-g_{n}}
tends (rapidly) to zero with iterations, this is an efficient method.
Demonstrate the function by calculating:
a
g
m
(
1
,
1
/
2
)
{\displaystyle \mathrm {agm} (1,1/{\sqrt {2}})}
Also see
mathworld.wolfram.com/Arithmetic-Geometric Mean
| #Stata | Stata | mata
real scalar agm(real scalar a, real scalar b) {
real scalar c
do {
c=0.5*(a+b)
b=sqrt(a*b)
a=c
} while (a-b>1e-15*a)
return(0.5*(a+b))
}
agm(1,1/sqrt(2))
end |
http://rosettacode.org/wiki/Arithmetic-geometric_mean | Arithmetic-geometric mean |
This page uses content from Wikipedia. The original article was at Arithmetic-geometric mean. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Write a function to compute the arithmetic-geometric mean of two numbers.
The arithmetic-geometric mean of two numbers can be (usefully) denoted as
a
g
m
(
a
,
g
)
{\displaystyle \mathrm {agm} (a,g)}
, and is equal to the limit of the sequence:
a
0
=
a
;
g
0
=
g
{\displaystyle a_{0}=a;\qquad g_{0}=g}
a
n
+
1
=
1
2
(
a
n
+
g
n
)
;
g
n
+
1
=
a
n
g
n
.
{\displaystyle a_{n+1}={\tfrac {1}{2}}(a_{n}+g_{n});\quad g_{n+1}={\sqrt {a_{n}g_{n}}}.}
Since the limit of
a
n
−
g
n
{\displaystyle a_{n}-g_{n}}
tends (rapidly) to zero with iterations, this is an efficient method.
Demonstrate the function by calculating:
a
g
m
(
1
,
1
/
2
)
{\displaystyle \mathrm {agm} (1,1/{\sqrt {2}})}
Also see
mathworld.wolfram.com/Arithmetic-Geometric Mean
| #Swift | Swift | import Darwin
enum AGRError : Error {
case undefined
}
func agm(_ a: Double, _ g: Double, _ iota: Double = 1e-8) throws -> Double {
var a = a
var g = g
var a1: Double = 0
var g1: Double = 0
guard a * g >= 0 else {
throw AGRError.undefined
}
while abs(a - g) > iota {
a1 = (a + g) / 2
g1 = sqrt(a * g)
a = a1
g = g1
}
return a
}
do {
try print(agm(1, 1 / sqrt(2)))
} catch {
print("agr is undefined when a * g < 0")
} |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #GW-BASIC | GW-BASIC | PRINT 0^0 |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Haskell | Haskell | import Data.Complex
main = do
print $ 0 ^ 0
print $ 0.0 ^ 0
print $ 0 ^^ 0
print $ 0 ** 0
print $ (0 :+ 0) ^ 0
print $ (0 :+ 0) ** (0 :+ 0) |
http://rosettacode.org/wiki/Zig-zag_matrix | Zig-zag matrix | Task
Produce a zig-zag array.
A zig-zag array is a square arrangement of the first N2 natural numbers, where the
numbers increase sequentially as you zig-zag along the array's anti-diagonals.
For a graphical representation, see JPG zigzag (JPG uses such arrays to encode images).
For example, given 5, produce this array:
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24
Related tasks
Spiral matrix
Identity matrix
Ulam spiral (for primes)
See also
Wiktionary entry: anti-diagonals
| #ALGOL_68 | ALGOL 68 | PROC zig zag = (INT n)[,]INT: (
PROC move = (REF INT i, j)VOID: (
IF j < n THEN
i := ( i <= 1 | 1 | i-1 );
j +:= 1
ELSE
i +:= 1
FI
);
[n, n]INT a;
INT x:=LWB a, y:=LWB a;
FOR v FROM 0 TO n**2-1 DO
a[y, x] := v;
IF ODD (x + y) THEN
move(x, y)
ELSE
move(y, x)
FI
OD;
a
);
INT dim = 5;
#IF formatted transput possible THEN
FORMAT d = $z-d$;
FORMAT row = $"("n(dim-1)(f(d)",")f(d)")"$;
FORMAT block = $"("n(dim-1)(f(row)","lx)f(row)")"l$;
printf((block, zig zag(dim)))
ELSE#
[,]INT result = zig zag(dim);
FOR i TO dim DO
print((result[i,], new line))
OD
#FI# |
http://rosettacode.org/wiki/Yellowstone_sequence | Yellowstone sequence | The Yellowstone sequence, also called the Yellowstone permutation, is defined as:
For n <= 3,
a(n) = n
For n >= 4,
a(n) = the smallest number not already in sequence such that a(n) is relatively prime to a(n-1) and
is not relatively prime to a(n-2).
The sequence is a permutation of the natural numbers, and gets its name from what its authors felt was a spiking, geyser like appearance of a plot of the sequence.
Example
a(4) is 4 because 4 is the smallest number following 1, 2, 3 in the sequence that is relatively prime to the entry before it (3), and is not relatively prime to the number two entries before it (2).
Task
Find and show as output the first 30 Yellowstone numbers.
Extra
Demonstrate how to plot, with x = n and y coordinate a(n), the first 100 Yellowstone numbers.
Related tasks
Greatest common divisor.
Plot coordinate pairs.
See also
The OEIS entry: A098550 The Yellowstone permutation.
Applegate et al, 2015: The Yellowstone Permutation [1].
| #C.2B.2B | C++ | #include <iostream>
#include <numeric>
#include <set>
template <typename integer>
class yellowstone_generator {
public:
integer next() {
n2_ = n1_;
n1_ = n_;
if (n_ < 3) {
++n_;
} else {
for (n_ = min_; !(sequence_.count(n_) == 0
&& std::gcd(n1_, n_) == 1
&& std::gcd(n2_, n_) > 1); ++n_) {}
}
sequence_.insert(n_);
for (;;) {
auto it = sequence_.find(min_);
if (it == sequence_.end())
break;
sequence_.erase(it);
++min_;
}
return n_;
}
private:
std::set<integer> sequence_;
integer min_ = 1;
integer n_ = 0;
integer n1_ = 0;
integer n2_ = 0;
};
int main() {
std::cout << "First 30 Yellowstone numbers:\n";
yellowstone_generator<unsigned int> ygen;
std::cout << ygen.next();
for (int i = 1; i < 30; ++i)
std::cout << ' ' << ygen.next();
std::cout << '\n';
return 0;
} |
http://rosettacode.org/wiki/Yahoo!_search_interface | Yahoo! search interface | Create a class for searching Yahoo! results.
It must implement a Next Page method, and read URL, Title and Content from results.
| #AArch64_Assembly | AArch64 Assembly |
/* ARM assembly AARCH64 Raspberry PI 3B */
/* program yahoosearch64.s */
/* access RosettaCode.org and data extract */
/* use openssl for access to port 443 */
/* test openssl : package libssl-dev */
/*******************************************/
/* Constantes file */
/*******************************************/
/* for this file see task include a file in language AArch64 assembly*/
.include "../includeConstantesARM64.inc"
.equ TAILLEBUFFER, 500
.equ SSL_OP_NO_SSLv3, 0x02000000
.equ SSL_OP_NO_COMPRESSION, 0x00020000
.equ SSL_MODE_AUTO_RETRY, 0x00000004
.equ SSL_CTRL_MODE, 33
.equ BIO_C_SET_CONNECT, 100
.equ BIO_C_DO_STATE_MACHINE, 101
.equ BIO_C_SET_SSL, 109
.equ BIO_C_GET_SSL, 110
.equ LGBUFFERREQ, 512001
/*********************************/
/* Initialized data */
/*********************************/
.data
szMessDebutPgm: .asciz "Début du programme. \n"
szRetourLigne: .asciz "\n"
szMessFinOK: .asciz "Fin normale du programme. \n"
szMessErreur: .asciz "Erreur !!!"
szMessExtractArea: .asciz "Extraction = "
szNomSite1: .asciz "search.yahoo.com:443" // host name and port
szLibStart: .asciz ">Rosetta Code" // search string
szNomrepCertif: .asciz "/pi/certificats"
szRequete1: .asciz "GET /search?p=\"Rosettacode.org\"&b=1 HTTP/1.1 \r\nHost: search.yahoo.com\r\nConnection: keep-alive\r\nContent-Type: text/plain\r\n\r\n"
/*********************************/
/* UnInitialized data */
/*********************************/
.bss
.align 4
sBufferreq: .skip LGBUFFERREQ
szExtractArea: .skip TAILLEBUFFER
stNewSSL: .skip 200
/*********************************/
/* code section */
/*********************************/
.text
.global main
main:
ldr x0,qAdrszMessDebutPgm
bl affichageMess // start message
/* connexion host port 443 and send query */
bl envoiRequete
cmp x0,#-1
beq 99f // error ?
bl analyseReponse
ldr x0,qAdrszMessFinOK // end message
bl affichageMess
mov x0, #0 // return code ok
b 100f
99:
ldr x0,qAdrszMessErreur // error
bl affichageMess
mov x0, #1 // return code error
b 100f
100:
mov x8,EXIT // program end
svc 0 // system call
qAdrszMessDebutPgm: .quad szMessDebutPgm
qAdrszMessFinOK: .quad szMessFinOK
qAdrszMessErreur: .quad szMessErreur
/*********************************************************/
/* connexion host port 443 and send query */
/*********************************************************/
envoiRequete:
stp x1,lr,[sp,-16]! // save registers
stp x2,x3,[sp,-16]! // save registers
stp x4,x5,[sp,-16]! // save registers
//*************************************
// openSsl functions use *
//*************************************
//init ssl
bl OPENSSL_init_crypto
bl ERR_load_BIO_strings
mov x2, #0
mov x1, #0
mov x0, #2
bl OPENSSL_init_crypto
mov x2, #0
mov x1, #0
mov x0, #0
bl OPENSSL_init_ssl
cmp x0,#0
blt erreur
bl TLS_client_method
bl SSL_CTX_new
cmp x0,#0
ble erreur
mov x20,x0 // save contex
ldr x1,iFlag
bl SSL_CTX_set_options
mov x0,x20 // contex
mov x1,#0
ldr x2,qAdrszNomrepCertif
bl SSL_CTX_load_verify_locations
cmp x0,#0
ble erreur
mov x0,x20 // contex
bl BIO_new_ssl_connect
cmp x0,#0
ble erreur
mov x21,x0 // save bio
mov x1,#BIO_C_GET_SSL
mov x2,#0
ldr x3,qAdrstNewSSL
bl BIO_ctrl
ldr x0,qAdrstNewSSL
ldr x0,[x0]
mov x1,#SSL_CTRL_MODE
mov x2,#SSL_MODE_AUTO_RETRY
mov x3,#0
bl SSL_ctrl
mov x0,x21 // bio
mov x1,#BIO_C_SET_CONNECT
mov x2,#0
ldr x3,qAdrszNomSite1
bl BIO_ctrl
mov x0,x21 // bio
mov x1,#BIO_C_DO_STATE_MACHINE
mov x2,#0
mov x3,#0
bl BIO_ctrl
// compute query length
mov x2,#0
ldr x1,qAdrszRequete1 // query
1:
ldrb w0,[x1,x2]
cmp x0,#0
add x8,x2,1
csel x2,x8,x2,ne
bne 1b
// send query
mov x0,x21 // bio
// x1 = address query
// x2 = length query
mov x3,#0
bl BIO_write // send query
cmp x0,#0
blt erreur
ldr x22,qAdrsBufferreq // buffer address
2: // begin loop to read datas
mov x0,x21 // bio
mov x1,x22 // buffer address
ldr x2,qLgBuffer
mov x3,#0
bl BIO_read
cmp x0,#0
ble 4f // error ou pb server
add x22,x22,x0
sub x2,x22,#8
ldr x2,[x2]
ldr x3,qCharEnd
cmp x2,x3 // text end ?
beq 4f
mov x1,#0xFFFFFF // delay loop
3:
subs x1,x1,1
bgt 3b
b 2b // loop read other chunk
4: // read end
//ldr x0,qAdrsBufferreq // to display buffer response of the query
//bl affichageMess
mov x0, x21 // close bio
bl BIO_free_all
mov x0,#0
b 100f
erreur: // error display
ldr x1,qAdrszMessErreur
bl afficheErreur
mov x0,#-1 // error code
b 100f
100:
ldp x4,x5,[sp],16 // restaur 2 registers
ldp x2,x3,[sp],16 // restaur 2 registers
ldp x1,lr,[sp],16 // restaur 2 registers
ret // return to address lr x30
qAdrszRequete1: .quad szRequete1
qAdrsBufferreq: .quad sBufferreq
iFlag: .quad SSL_OP_NO_SSLv3 | SSL_OP_NO_COMPRESSION
qAdrstNewSSL: .quad stNewSSL
qAdrszNomSite1: .quad szNomSite1
qAdrszNomrepCertif: .quad szNomrepCertif
qCharEnd: .quad 0x0A0D0A0D300A0D0A
qLgBuffer: .quad LGBUFFERREQ - 1
/*********************************************************/
/* response analyze */
/*********************************************************/
analyseReponse:
stp x1,lr,[sp,-16]! // save registers
stp x2,x3,[sp,-16]! // save registers
ldr x0,qAdrsBufferreq // buffer address
ldr x1,qAdrszLibStart // key text address
mov x2,#2 // occurence key text
mov x3,#-11 // offset
ldr x4,qAdrszExtractArea // address result area
bl extChaine
cmp x0,#-1
beq 99f
ldr x0,qAdrszMessExtractArea
bl affichageMess
ldr x0,qAdrszExtractArea // résult display
bl affichageMess
ldr x0,qAdrszRetourLigne
bl affichageMess
b 100f
99:
ldr x0,qAdrszMessErreur // error
bl affichageMess
mov x0, #-1 // error return code
b 100f
100:
ldp x2,x3,[sp],16 // restaur 2 registers
ldp x1,lr,[sp],16 // restaur 2 registers
ret // return to address lr x30
qAdrszLibStart: .quad szLibStart
qAdrszExtractArea: .quad szExtractArea
qAdrszMessExtractArea: .quad szMessExtractArea
qAdrszRetourLigne: .quad szRetourLigne
/*********************************************************/
/* Text Extraction behind text key */
/*********************************************************/
/* x0 buffer address */
/* x1 key text to search */
/* x2 number occurences to key text */
/* x3 offset */
/* x4 result address */
extChaine:
stp x1,lr,[sp,-16]! // save registers
stp x2,x3,[sp,-16]! // save registers
stp x4,x5,[sp,-16]! // save registers
stp x6,x7,[sp,-16]! // save registers
mov x5,x0 // save buffer address
mov x6,x1 // save key text
// compute text length
mov x8,#0
1: // loop
ldrb w0,[x5,x8] // load a byte
cmp x0,#0 // end ?
add x9,x8,1
csel x8,x9,x8,ne
bne 1b // no -> loop
add x8,x8,x5 // compute text end
mov x7,#0
2: // compute length text key
ldrb w0,[x6,x7]
cmp x0,#0
add x9,x7,1
csel x7,x9,x7,ne
bne 2b
3: // loop to search niéme(x2) key text
mov x0,x5
mov x1,x6
bl rechercheSousChaine
cmp x0,#0
blt 100f
subs x2,x2,1
ble 31f
add x5,x5,x0
add x5,x5,x7
b 3b
31:
add x0,x0,x5 // add address text to index
add x3,x3,x0 // add offset
sub x3,x3,1
// and add length key text
add x3,x3,x7
cmp x3,x8 // > at text end
bge 98f
mov x0,0
4: // character loop copy
ldrb w2,[x3,x0]
strb w2,[x4,x0]
cbz x2,99f // text end ? return zero
cmp x0,48 // extraction length
beq 5f
add x0,x0,1
b 4b // and loop
5:
mov x2,0 // store final zéro
strb w2,[x4,x0]
add x0,x0,1
add x0,x0,x3 // x0 return the last position of extraction
// it is possible o search another text
b 100f
98:
mov x0,-1 // error
b 100f
99:
mov x0,0
100:
ldp x6,x7,[sp],16 // restaur 2 registers
ldp x4,x5,[sp],16 // restaur 2 registers
ldp x2,x3,[sp],16 // restaur 2 registers
ldp x1,lr,[sp],16 // restaur 2 registers
ret // return to address lr x30
/******************************************************************/
/* search substring in string */
/******************************************************************/
/* x0 contains address string */
/* x1 contains address substring */
/* x0 return start index substring or -1 if not find */
rechercheSousChaine:
stp x1,lr,[sp,-16]! // save registers
stp x2,x3,[sp,-16]! // save registers
stp x4,x5,[sp,-16]! // save registers
stp x6,x7,[sp,-16]! // save registers
mov x2,#0 // index position string
mov x3,#0 // index position substring
mov x6,#-1 // search index
ldrb w4,[x1,x3] // load first byte substring
cbz x4,99f // zero final ? error
1:
ldrb w5,[x0,x2] // load string byte
cbz x5,99f // zero final ? yes -> not found
cmp x5,x4 // compare character two strings
beq 2f
mov x6,-1 // not equal - > raz index
mov x3,0 // and raz byte counter
ldrb w4,[x1,x3] // and load byte
add x2,x2,1 // and increment byte counter
b 1b // and loop
2: // characters equal
cmp x6,-1 // first character equal ?
csel x6,x2,x6,eq // yes -> start index in x6
add x3,x3,1 // increment substring counter
ldrb w4,[x1,x3] // and load next byte
cbz x4,3f // zero final ? yes -> search end
add x2,x2,1 // else increment string index
b 1b // and loop
3:
mov x0,x6 // return start index substring in the string
b 100f
99:
mov x0,-1 // not found
100:
ldp x6,x7,[sp],16 // restaur 2 registers
ldp x4,x5,[sp],16 // restaur 2 registers
ldp x2,x3,[sp],16 // restaur 2 registers
ldp x1,lr,[sp],16 // restaur 2 registers
ret // return to address lr x30
/********************************************************/
/* File Include fonctions */
/********************************************************/
/* for this file see task include a file in language AArch64 assembly */
.include "../includeARM64.inc"
|
http://rosettacode.org/wiki/Arithmetic_evaluation | Arithmetic evaluation | Create a program which parses and evaluates arithmetic expressions.
Requirements
An abstract-syntax tree (AST) for the expression must be created from parsing the input.
The AST must be used in evaluation, also, so the input may not be directly evaluated (e.g. by calling eval or a similar language feature.)
The expression will be a string or list of symbols like "(1+3)*7".
The four symbols + - * / must be supported as binary operators with conventional precedence rules.
Precedence-control parentheses must also be supported.
Note
For those who don't remember, mathematical precedence is as follows:
Parentheses
Multiplication/Division (left to right)
Addition/Subtraction (left to right)
C.f
24 game Player.
Parsing/RPN calculator algorithm.
Parsing/RPN to infix conversion.
| #Ruby | Ruby | $op_priority = {"+" => 0, "-" => 0, "*" => 1, "/" => 1}
class TreeNode
OP_FUNCTION = {
"+" => lambda {|x, y| x + y},
"-" => lambda {|x, y| x - y},
"*" => lambda {|x, y| x * y},
"/" => lambda {|x, y| x / y}}
attr_accessor :info, :left, :right
def initialize(info)
@info = info
end
def leaf?
@left.nil? and @right.nil?
end
def to_s(order)
if leaf?
@info
else
left_s, right_s = @left.to_s(order), @right.to_s(order)
strs = case order
when :prefix then [@info, left_s, right_s]
when :infix then [left_s, @info, right_s]
when :postfix then [left_s, right_s, @info]
else []
end
"(" + strs.join(" ") + ")"
end
end
def eval
if !leaf? and operator?(@info)
OP_FUNCTION[@info].call(@left.eval, @right.eval)
else
@info.to_f
end
end
end
def tokenize(exp)
exp
.gsub('(', ' ( ')
.gsub(')', ' ) ')
.gsub('+', ' + ')
.gsub('-', ' - ')
.gsub('*', ' * ')
.gsub('/', ' / ')
.split(' ')
end
def operator?(token)
$op_priority.has_key?(token)
end
def pop_connect_push(op_stack, node_stack)
temp = op_stack.pop
temp.right = node_stack.pop
temp.left = node_stack.pop
node_stack.push(temp)
end
def infix_exp_to_tree(exp)
tokens = tokenize(exp)
op_stack, node_stack = [], []
tokens.each do |token|
if operator?(token)
# clear stack of higher priority operators
until (op_stack.empty? or
op_stack.last.info == "(" or
$op_priority[op_stack.last.info] < $op_priority[token])
pop_connect_push(op_stack, node_stack)
end
op_stack.push(TreeNode.new(token))
elsif token == "("
op_stack.push(TreeNode.new(token))
elsif token == ")"
while op_stack.last.info != "("
pop_connect_push(op_stack, node_stack)
end
# throw away the '('
op_stack.pop
else
node_stack.push(TreeNode.new(token))
end
end
until op_stack.empty?
pop_connect_push(op_stack, node_stack)
end
node_stack.last
end |
http://rosettacode.org/wiki/Zeckendorf_arithmetic | Zeckendorf arithmetic | This task is a total immersion zeckendorf task; using decimal numbers will attract serious disapprobation.
The task is to implement addition, subtraction, multiplication, and division using Zeckendorf number representation. Optionally provide decrement, increment and comparitive operation functions.
Addition
Like binary 1 + 1 = 10, note carry 1 left. There the similarity ends. 10 + 10 = 101, note carry 1 left and 1 right. 100 + 100 = 1001, note carry 1 left and 2 right, this is the general case.
Occurrences of 11 must be changed to 100. Occurrences of 111 may be changed from the right by replacing 11 with 100, or from the left converting 111 to 100 + 100;
Subtraction
10 - 1 = 1. The general rule is borrow 1 right carry 1 left. eg:
abcde
10100 -
1000
_____
100 borrow 1 from a leaves 100
+ 100 add the carry
_____
1001
A larger example:
abcdef
100100 -
1000
______
1*0100 borrow 1 from b
+ 100 add the carry
______
1*1001
Sadly we borrowed 1 from b which didn't have it to lend. So now b borrows from a:
1001
+ 1000 add the carry
____
10100
Multiplication
Here you teach your computer its zeckendorf tables. eg. 101 * 1001:
a = 1 * 101 = 101
b = 10 * 101 = a + a = 10000
c = 100 * 101 = b + a = 10101
d = 1000 * 101 = c + b = 101010
1001 = d + a therefore 101 * 1001 =
101010
+ 101
______
1000100
Division
Lets try 1000101 divided by 101, so we can use the same table used for multiplication.
1000101 -
101010 subtract d (1000 * 101)
_______
1000 -
101 b and c are too large to subtract, so subtract a
____
1 so 1000101 divided by 101 is d + a (1001) remainder 1
Efficient algorithms for Zeckendorf arithmetic is interesting. The sections on addition and subtraction are particularly relevant for this task.
| #Perl | Perl | #!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Zeckendorf_arithmetic
use warnings;
for ( split /\n/, <<END ) # test cases
1 + 1
10 + 10
10100 + 1010
10100 - 1010
10100 * 1010
100010 * 100101
10100 / 1010
101000 / 1000
100001000001 / 100010
100001000001 / 100101
END
{
my ($left, $op, $right) = split;
my ($x, $y) = map Zeckendorf->new($_), $left, $right;
my $answer =
$op eq '+' ? $x + $y :
$op eq '-' ? $x - $y :
$op eq '*' ? $x * $y :
$op eq '/' ? $x / $y :
die "bad op <$op>";
printf "%12s %s %-9s => %12s in Zeckendorf\n", $x, $op, $y, $answer;
printf "%12d %s %-9d => %12d in decimal\n\n",
$x->asdecimal, $op, $y->asdecimal, $answer->asdecimal;
}
package Zeckendorf;
use overload qw("" zstring + zadd - zsub ++ zinc -- zdec * zmul / zdiv ge zge);
sub new
{
my ($class, $value) = @_;
bless \$value, ref $class || $class;
}
sub zinc
{
my ($self, $other, $swap) = @_;
local $_ = $$self;
s/0$/1/ or s/(?:^|0)1$/10/;
1 while s/(?:^|0)11/100/;
$_[0] = $self->new( s/^0+\B//r );
}
sub zdec
{
my ($self, $other, $swap) = @_;
local $_ = $$self;
1 while s/100(?=0*$)/011/;
s/1$/0/ or s/10$/01/;
$_[0] = $self->new( s/^0+\B//r );
}
sub zstring { ${ shift() } }
sub zadd
{
my ($self, $other, $swap) = @_;
my ($x, $y) = map $self->new($$_), $self, $other; # copy
++$x, $y-- while $$y ne 0;
return $x;
}
sub zsub
{
my ($self, $other, $swap) = @_;
my ($x, $y) = map $self->new($$_), $self, $other; # copy
--$x, $y-- while $$y ne 0;
return $x;
}
sub zmul
{
my ($self, $other, $swap) = @_;
my ($x, $y) = map $self->new($$_), $self, $other; # copy
my $product = Zeckendorf->new(0);
$product = $product + $x, --$y while "$y" ne 0;
return $product;
}
sub zdiv
{
my ($self, $other, $swap) = @_;
my ($x, $y) = map $self->new($$_), $self, $other; # copy
my $quotient = Zeckendorf->new(0);
++$quotient, $x = $x - $y while $x ge $y;
return $quotient;
}
sub zge
{
my ($self, $other, $swap) = @_;
my $l = length( $$self | $$other );
0 x ($l - length $$self) . $$self ge 0 x ($l - length $$other) . $$other;
}
sub asdecimal
{
my ($self) = @_;
my $n = 0;
my $aa = my $bb = 1;
for ( reverse split //, $$self )
{
$n += $bb * $_;
($aa, $bb) = ($bb, $aa + $bb);
}
return $n;
}
sub fromdecimal
{
my ($self, $value) = @_;
my $z = $self->new(0);
++$z for 1 .. $value;
return $z;
} |
http://rosettacode.org/wiki/Zumkeller_numbers | Zumkeller numbers | Zumkeller numbers are the set of numbers whose divisors can be partitioned into two disjoint sets that sum to the same value. Each sum must contain divisor values that are not in the other sum, and all of the divisors must be in one or the other. There are no restrictions on how the divisors are partitioned, only that the two partition sums are equal.
E.G.
6 is a Zumkeller number; The divisors {1 2 3 6} can be partitioned into two groups {1 2 3} and {6} that both sum to 6.
10 is not a Zumkeller number; The divisors {1 2 5 10} can not be partitioned into two groups in any way that will both sum to the same value.
12 is a Zumkeller number; The divisors {1 2 3 4 6 12} can be partitioned into two groups {1 3 4 6} and {2 12} that both sum to 14.
Even Zumkeller numbers are common; odd Zumkeller numbers are much less so. For values below 10^6, there is at least one Zumkeller number in every 12 consecutive integers, and the vast majority of them are even. The odd Zumkeller numbers are very similar to the list from the task Abundant odd numbers; they are nearly the same except for the further restriction that the abundance (A(n) = sigma(n) - 2n), must be even: A(n) mod 2 == 0
Task
Write a routine (function, procedure, whatever) to find Zumkeller numbers.
Use the routine to find and display here, on this page, the first 220 Zumkeller numbers.
Use the routine to find and display here, on this page, the first 40 odd Zumkeller numbers.
Optional, stretch goal: Use the routine to find and display here, on this page, the first 40 odd Zumkeller numbers that don't end with 5.
See Also
OEIS:A083207 - Zumkeller numbers to get an impression of different partitions OEIS:A083206 Zumkeller partitions
OEIS:A174865 - Odd Zumkeller numbers
Related Tasks
Abundant odd numbers
Abundant, deficient and perfect number classifications
Proper divisors , Factors of an integer | #Lobster | Lobster | import std
// Derived from Julia and Python versions
def get_divisors(n: int) -> [int]:
var i = 2
let d = [1, n]
let limit = sqrt(n)
while i <= limit:
if n % i == 0:
let j = n / i
push(d,i)
if i != j:
push(d,j)
i += 1
return d
def isPartSum(divs: [int], sum: int) -> bool:
if sum == 0:
return true
let len = length(divs)
if len == 0:
return false
let last = pop(divs)
if last > sum:
return isPartSum(divs, sum)
return isPartSum(copy(divs), sum) or isPartSum(divs, sum-last)
def isZumkeller(n: int) -> bool:
let divs = get_divisors(n)
let sum = fold(divs, 0): _a+_b
if sum % 2 == 1:
// if sum is odd can't be split into two partitions with equal sums
return false
if n % 2 == 1:
// if n is odd use 'abundant odd number' optimization
let abundance = sum - 2 * n
return abundance > 0 and abundance % 2 == 0
return isPartSum(divs, sum/2)
def printZumkellers(q: int, oddonly: bool):
var nprinted = 0
var res = ""
for(100000) n:
if (!oddonly or n % 2 != 0):
if isZumkeller(n):
let s = string(n)
let z = length(s)
res = concat_string([res, repeat_string(" ",8-z), s], "")
nprinted += 1
if nprinted % 10 == 0 or nprinted >= q:
print res
res = ""
if nprinted >= q:
return
print "220 Zumkeller numbers:"
printZumkellers(220, false)
print "\n\n40 odd Zumkeller numbers:"
printZumkellers(40, true)
|
http://rosettacode.org/wiki/Zumkeller_numbers | Zumkeller numbers | Zumkeller numbers are the set of numbers whose divisors can be partitioned into two disjoint sets that sum to the same value. Each sum must contain divisor values that are not in the other sum, and all of the divisors must be in one or the other. There are no restrictions on how the divisors are partitioned, only that the two partition sums are equal.
E.G.
6 is a Zumkeller number; The divisors {1 2 3 6} can be partitioned into two groups {1 2 3} and {6} that both sum to 6.
10 is not a Zumkeller number; The divisors {1 2 5 10} can not be partitioned into two groups in any way that will both sum to the same value.
12 is a Zumkeller number; The divisors {1 2 3 4 6 12} can be partitioned into two groups {1 3 4 6} and {2 12} that both sum to 14.
Even Zumkeller numbers are common; odd Zumkeller numbers are much less so. For values below 10^6, there is at least one Zumkeller number in every 12 consecutive integers, and the vast majority of them are even. The odd Zumkeller numbers are very similar to the list from the task Abundant odd numbers; they are nearly the same except for the further restriction that the abundance (A(n) = sigma(n) - 2n), must be even: A(n) mod 2 == 0
Task
Write a routine (function, procedure, whatever) to find Zumkeller numbers.
Use the routine to find and display here, on this page, the first 220 Zumkeller numbers.
Use the routine to find and display here, on this page, the first 40 odd Zumkeller numbers.
Optional, stretch goal: Use the routine to find and display here, on this page, the first 40 odd Zumkeller numbers that don't end with 5.
See Also
OEIS:A083207 - Zumkeller numbers to get an impression of different partitions OEIS:A083206 Zumkeller partitions
OEIS:A174865 - Odd Zumkeller numbers
Related Tasks
Abundant odd numbers
Abundant, deficient and perfect number classifications
Proper divisors , Factors of an integer | #Mathematica_.2F_Wolfram_Language | Mathematica / Wolfram Language | ClearAll[ZumkellerQ]
ZumkellerQ[n_] := Module[{d = Divisors[n], t, ds, x},
ds = Total[d];
If[Mod[ds, 2] == 1,
False
,
t = CoefficientList[Product[1 + x^i, {i, d}], x];
t[[1 + ds/2]] > 0
]
];
i = 1;
res = {};
While[Length[res] < 220,
r = ZumkellerQ[i];
If[r, AppendTo[res, i]];
i++;
];
res
i = 1;
res = {};
While[Length[res] < 40,
r = ZumkellerQ[i];
If[r, AppendTo[res, i]];
i += 2;
];
res |
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #Delphi | Delphi |
program Arbitrary_precision_integers;
{$APPTYPE CONSOLE}
uses
System.SysUtils,
Velthuis.BigIntegers;
var
value: BigInteger;
result: string;
begin
value := BigInteger.pow(3, 2);
value := BigInteger.pow(4, value.AsInteger);
value := BigInteger.pow(5, value.AsInteger);
result := value.tostring;
Write('5^4^3^2 = ');
Write(result.substring(0, 20), '...');
Write(result.substring(result.length - 20, 20));
Writeln(' (', result.Length,' digits)');
readln;
end. |
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #E | E | ? def value := 5**(4**(3**2)); null
? def decimal := value.toString(10); null
? decimal(0, 20)
# value: "62060698786608744707"
? decimal(decimal.size() - 20)
# value: "92256259918212890625"
? decimal.size()
# value: 183231 |
http://rosettacode.org/wiki/Zhang-Suen_thinning_algorithm | Zhang-Suen thinning algorithm | This is an algorithm used to thin a black and white i.e. one bit per pixel images.
For example, with an input image of:
################# #############
################## ################
################### ##################
######## ####### ###################
###### ####### ####### ######
###### ####### #######
################# #######
################ #######
################# #######
###### ####### #######
###### ####### #######
###### ####### ####### ######
######## ####### ###################
######## ####### ###### ################## ######
######## ####### ###### ################ ######
######## ####### ###### ############# ######
It produces the thinned output:
# ########## #######
## # #### #
# # ##
# # #
# # #
# # #
############ #
# # #
# # #
# # #
# # #
# ##
# ############
### ###
Algorithm
Assume black pixels are one and white pixels zero, and that the input image is a rectangular N by M array of ones and zeroes.
The algorithm operates on all black pixels P1 that can have eight neighbours.
The neighbours are, in order, arranged as:
P9 P2 P3
P8 P1 P4
P7 P6 P5
Obviously the boundary pixels of the image cannot have the full eight neighbours.
Define
A
(
P
1
)
{\displaystyle A(P1)}
= the number of transitions from white to black, (0 -> 1) in the sequence P2,P3,P4,P5,P6,P7,P8,P9,P2. (Note the extra P2 at the end - it is circular).
Define
B
(
P
1
)
{\displaystyle B(P1)}
= The number of black pixel neighbours of P1. ( = sum(P2 .. P9) )
Step 1
All pixels are tested and pixels satisfying all the following conditions (simultaneously) are just noted at this stage.
(0) The pixel is black and has eight neighbours
(1)
2
<=
B
(
P
1
)
<=
6
{\displaystyle 2<=B(P1)<=6}
(2) A(P1) = 1
(3) At least one of P2 and P4 and P6 is white
(4) At least one of P4 and P6 and P8 is white
After iterating over the image and collecting all the pixels satisfying all step 1 conditions, all these condition satisfying pixels are set to white.
Step 2
All pixels are again tested and pixels satisfying all the following conditions are just noted at this stage.
(0) The pixel is black and has eight neighbours
(1)
2
<=
B
(
P
1
)
<=
6
{\displaystyle 2<=B(P1)<=6}
(2) A(P1) = 1
(3) At least one of P2 and P4 and P8 is white
(4) At least one of P2 and P6 and P8 is white
After iterating over the image and collecting all the pixels satisfying all step 2 conditions, all these condition satisfying pixels are again set to white.
Iteration
If any pixels were set in this round of either step 1 or step 2 then all steps are repeated until no image pixels are so changed.
Task
Write a routine to perform Zhang-Suen thinning on an image matrix of ones and zeroes.
Use the routine to thin the following image and show the output here on this page as either a matrix of ones and zeroes, an image, or an ASCII-art image of space/non-space characters.
00000000000000000000000000000000
01111111110000000111111110000000
01110001111000001111001111000000
01110000111000001110000111000000
01110001111000001110000000000000
01111111110000001110000000000000
01110111100000001110000111000000
01110011110011101111001111011100
01110001111011100111111110011100
00000000000000000000000000000000
Reference
Zhang-Suen Thinning Algorithm, Java Implementation by Nayef Reza.
"Character Recognition Systems: A Guide for Students and Practitioners" By Mohamed Cheriet, Nawwaf Kharma, Cheng-Lin Liu, Ching Suen
| #Lua | Lua | function zhangSuenThin(img)
local dirs={
{ 0,-1},
{ 1,-1},
{ 1, 0},
{ 1, 1},
{ 0, 1},
{-1, 1},
{-1, 0},
{-1,-1},
{ 0,-1},
}
local black=1
local white=0
function A(x, y)
local c=0
local current=img[y+dirs[1][2]][x+dirs[1][1]]
for i=2,#dirs do
local to_compare=img[y+dirs[i][2]][x+dirs[i][1]]
if current==white and to_compare==black then
c=c+1
end
current=to_compare
end
return c
end
function B(x, y)
local c=0
for i=2,#dirs do
local value=img[y+dirs[i][2]][x+dirs[i][1]]
if value==black then
c=c+1
end
end
return c
end
function common_step(x, y)
if img[y][x]~=black or x<=1 or x>=#img[y] or y<=1 or y>=#img then
return false
end
local b_value=B(x, y)
if b_value<2 or b_value>6 then
return false
end
local a_value=A(x, y)
if a_value~=1 then
return false
end
return true
end
function step_one(x, y)
if not common_step(x, y) then
return false
end
local p2=img[y+dirs[1][2]][x+dirs[1][1]]
local p4=img[y+dirs[3][2]][x+dirs[3][1]]
local p6=img[y+dirs[5][2]][x+dirs[5][1]]
local p8=img[y+dirs[7][2]][x+dirs[7][1]]
if p4==white or p6==white or p2==white and p8==white then
return true
end
return false
end
function step_two(x, y)
if not common_step(x, y) then
return false
end
local p2=img[y+dirs[1][2]][x+dirs[1][1]]
local p4=img[y+dirs[3][2]][x+dirs[3][1]]
local p6=img[y+dirs[5][2]][x+dirs[5][1]]
local p8=img[y+dirs[7][2]][x+dirs[7][1]]
if p2==white or p8==white or p4==white and p6==white then
return true
end
return false
end
function convert(to_do)
for k,v in pairs(to_do) do
img[v[2]][v[1]]=white
end
end
function do_step_on_all(step)
local to_convert={}
for y=1,#img do
for x=1,#img[y] do
if step(x, y) then
table.insert(to_convert, {x,y})
end
end
end
convert(to_convert)
return #to_convert>0
end
local continue=true
while continue do
continue=false
if do_step_on_all(step_one) then
continue=true
end
if do_step_on_all(step_two) then
continue=true
end
end
for y=1,#img do
for x=1,#img[y] do
io.write(img[y][x]==black and '#' or ' ')
end
io.write('\n')
end
end
local image = {
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0},
{0,1,1,1,0,0,0,1,1,1,1,0,0,0,0,0,1,1,1,1,0,0,1,1,1,1,0,0,0,0,0,0},
{0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,0,0},
{0,1,1,1,0,0,0,1,1,1,1,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,1,1,1,0,1,1,1,1,0,0,0,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,0,0},
{0,1,1,1,0,0,1,1,1,1,0,0,1,1,1,0,1,1,1,1,0,0,1,1,1,1,0,1,1,1,0,0},
{0,1,1,1,0,0,0,1,1,1,1,0,1,1,1,0,0,1,1,1,1,1,1,1,1,0,0,1,1,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
}
zhangSuenThin(image)
|
http://rosettacode.org/wiki/Zeckendorf_number_representation | Zeckendorf number representation | Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.
Recall that the first six distinct Fibonacci numbers are: 1, 2, 3, 5, 8, 13.
The decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.
10100 is not the only way to make 11 from the Fibonacci numbers however; 0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that no two consecutive Fibonacci numbers can be used which leads to the former unique solution.
Task
Generate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.
The intention in this task to find the Zeckendorf form of an arbitrary integer. The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task.
Also see
OEIS A014417 for the the sequence of required results.
Brown's Criterion - Numberphile
Related task
Fibonacci sequence
| #F.23 | F# | let fib = Seq.unfold (fun (x, y) -> Some(x, (y, x + y))) (1,2)
let zeckendorf n =
if n = 0 then ["0"]
else
let folder k state =
let (n, z) = (fst state), (snd state)
if n >= k then (n - k, "1" :: z)
else (n, "0" :: z)
let fb = fib |> Seq.takeWhile (fun i -> i<=n) |> Seq.toList
snd (List.foldBack folder fb (n, []))
|> List.rev
for i in 0 .. 20 do printfn "%2d: %8s" i (String.concat "" (zeckendorf i)) |
http://rosettacode.org/wiki/100_doors | 100 doors | There are 100 doors in a row that are all initially closed.
You make 100 passes by the doors.
The first time through, visit every door and toggle the door (if the door is closed, open it; if it is open, close it).
The second time, only visit every 2nd door (door #2, #4, #6, ...), and toggle it.
The third time, visit every 3rd door (door #3, #6, #9, ...), etc, until you only visit the 100th door.
Task
Answer the question: what state are the doors in after the last pass? Which are open, which are closed?
Alternate:
As noted in this page's discussion page, the only doors that remain open are those whose numbers are perfect squares.
Opening only those doors is an optimization that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.
| #C.23 | C# | namespace ConsoleApplication1
{
using System;
class Program
{
static void Main(string[] args)
{
bool[] doors = new bool[100];
//Close all doors to start.
for (int d = 0; d < 100; d++) doors[d] = false;
//For each pass...
for (int p = 0; p < 100; p++)//number of passes
{
//For each door to toggle...
for (int d = 0; d < 100; d++)//door number
{
if ((d + 1) % (p + 1) == 0)
{
doors[d] = !doors[d];
}
}
}
//Output the results.
Console.WriteLine("Passes Completed!!! Here are the results: \r\n");
for (int d = 0; d < 100; d++)
{
if (doors[d])
{
Console.WriteLine(String.Format("Door #{0}: Open", d + 1));
}
else
{
Console.WriteLine(String.Format("Door #{0}: Closed", d + 1));
}
}
Console.ReadKey(true);
}
}
} |
http://rosettacode.org/wiki/Arrays | Arrays | This task is about arrays.
For hashes or associative arrays, please see Creating an Associative Array.
For a definition and in-depth discussion of what an array is, see Array.
Task
Show basic array syntax in your language.
Basically, create an array, assign a value to it, and retrieve an element (if available, show both fixed-length arrays and
dynamic arrays, pushing a value into it).
Please discuss at Village Pump: Arrays.
Please merge code in from these obsolete tasks:
Creating an Array
Assigning Values to an Array
Retrieving an Element of an Array
Related tasks
Collections
Creating an Associative Array
Two-dimensional array (runtime)
| #Erlang | Erlang |
%% Create a fixed-size array with entries 0-9 set to 'undefined'
A0 = array:new(10).
10 = array:size(A0).
%% Create an extendible array and set entry 17 to 'true',
%% causing the array to grow automatically
A1 = array:set(17, true, array:new()).
18 = array:size(A1).
%% Read back a stored value
true = array:get(17, A1).
%% Accessing an unset entry returns the default value
undefined = array:get(3, A1).
%% Accessing an entry beyond the last set entry also returns the
%% default value, if the array does not have fixed size
undefined = array:get(18, A1).
%% "sparse" functions ignore default-valued entries
A2 = array:set(4, false, A1).
[{4, false}, {17, true}] = array:sparse_to_orddict(A2).
%% An extendible array can be made fixed-size later
A3 = array:fix(A2).
%% A fixed-size array does not grow automatically and does not
%% allow accesses beyond the last set entry
{'EXIT',{badarg,_}} = (catch array:set(18, true, A3)).
{'EXIT',{badarg,_}} = (catch array:get(18, A3)).
|
http://rosettacode.org/wiki/Arithmetic/Complex | Arithmetic/Complex | A complex number is a number which can be written as:
a
+
b
×
i
{\displaystyle a+b\times i}
(sometimes shown as:
b
+
a
×
i
{\displaystyle b+a\times i}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are real numbers, and
i
{\displaystyle i}
is √ -1
Typically, complex numbers are represented as a pair of real numbers called the "imaginary part" and "real part", where the imaginary part is the number to be multiplied by
i
{\displaystyle i}
.
Task
Show addition, multiplication, negation, and inversion of complex numbers in separate functions. (Subtraction and division operations can be made with pairs of these operations.)
Print the results for each operation tested.
Optional: Show complex conjugation.
By definition, the complex conjugate of
a
+
b
i
{\displaystyle a+bi}
is
a
−
b
i
{\displaystyle a-bi}
Some languages have complex number libraries available. If your language does, show the operations. If your language does not, also show the definition of this type.
| #PicoLisp | PicoLisp | (load "@lib/math.l")
(de addComplex (A B)
(cons
(+ (car A) (car B)) # Real
(+ (cdr A) (cdr B)) ) ) # Imag
(de mulComplex (A B)
(cons
(-
(*/ (car A) (car B) 1.0)
(*/ (cdr A) (cdr B) 1.0) )
(+
(*/ (car A) (cdr B) 1.0)
(*/ (cdr A) (car B) 1.0) ) ) )
(de invComplex (A)
(let Denom
(+
(*/ (car A) (car A) 1.0)
(*/ (cdr A) (cdr A) 1.0) )
(cons
(*/ (car A) 1.0 Denom)
(- (*/ (cdr A) 1.0 Denom)) ) ) )
(de negComplex (A)
(cons (- (car A)) (- (cdr A))) )
(de fmtComplex (A)
(pack
(round (car A) (dec *Scl))
(and (gt0 (cdr A)) "+")
(round (cdr A) (dec *Scl))
"i" ) )
(let (A (1.0 . 1.0) B (cons pi 1.2))
(prinl "A = " (fmtComplex A))
(prinl "B = " (fmtComplex B))
(prinl "A+B = " (fmtComplex (addComplex A B)))
(prinl "A*B = " (fmtComplex (mulComplex A B)))
(prinl "1/A = " (fmtComplex (invComplex A)))
(prinl "-A = " (fmtComplex (negComplex A))) ) |
http://rosettacode.org/wiki/Arithmetic/Complex | Arithmetic/Complex | A complex number is a number which can be written as:
a
+
b
×
i
{\displaystyle a+b\times i}
(sometimes shown as:
b
+
a
×
i
{\displaystyle b+a\times i}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are real numbers, and
i
{\displaystyle i}
is √ -1
Typically, complex numbers are represented as a pair of real numbers called the "imaginary part" and "real part", where the imaginary part is the number to be multiplied by
i
{\displaystyle i}
.
Task
Show addition, multiplication, negation, and inversion of complex numbers in separate functions. (Subtraction and division operations can be made with pairs of these operations.)
Print the results for each operation tested.
Optional: Show complex conjugation.
By definition, the complex conjugate of
a
+
b
i
{\displaystyle a+bi}
is
a
−
b
i
{\displaystyle a-bi}
Some languages have complex number libraries available. If your language does, show the operations. If your language does not, also show the definition of this type.
| #PL.2FI | PL/I | /* PL/I complex numbers may be integer or floating-point. */
/* In this example, the variables are floating-pint. */
/* For integer variables, change 'float' to 'fixed binary' */
declare (a, b) complex float;
a = 2+5i;
b = 7-6i;
put skip list (a+b);
put skip list (a - b);
put skip list (a*b);
put skip list (a/b);
put skip list (a**b);
put skip list (1/a);
put skip list (conjg(a)); /* gives the conjugate of 'a'. */
/* Functions exist for extracting the real and imaginary parts */
/* of a complex number. */
/* As well, trigonometric functions may be used with complex */
/* numbers, such as SIN, COS, TAN, ATAN, and so on. */ |
http://rosettacode.org/wiki/Arithmetic/Rational | Arithmetic/Rational | Task
Create a reasonably complete implementation of rational arithmetic in the particular language using the idioms of the language.
Example
Define a new type called frac with binary operator "//" of two integers that returns a structure made up of the numerator and the denominator (as per a rational number).
Further define the appropriate rational unary operators abs and '-', with the binary operators for addition '+', subtraction '-', multiplication '×', division '/', integer division '÷', modulo division, the comparison operators (e.g. '<', '≤', '>', & '≥') and equality operators (e.g. '=' & '≠').
Define standard coercion operators for casting int to frac etc.
If space allows, define standard increment and decrement operators (e.g. '+:=' & '-:=' etc.).
Finally test the operators:
Use the new type frac to find all perfect numbers less than 219 by summing the reciprocal of the factors.
Related task
Perfect Numbers
| #REXX | REXX | /*REXX program implements a reasonably complete rational arithmetic (using fractions).*/
L=length(2**19 - 1) /*saves time by checking even numbers. */
do j=2 by 2 to 2**19 - 1; s=0 /*ignore unity (which can't be perfect)*/
mostDivs=eDivs(j); @= /*obtain divisors>1; zero sum; null @. */
do k=1 for words(mostDivs) /*unity isn't return from eDivs here.*/
r='1/'word(mostDivs, k); @=@ r; s=$fun(r, , s)
end /*k*/
if s\==1 then iterate /*Is sum not equal to unity? Skip it.*/
say 'perfect number:' right(j, L) " fractions:" @
end /*j*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
$div: procedure; parse arg x; x=space(x,0); f= 'fractional division'
parse var x n '/' d; d=p(d 1)
if d=0 then call err 'division by zero:' x
if \datatype(n,'N') then call err 'a non─numeric numerator:' x
if \datatype(d,'N') then call err 'a non─numeric denominator:' x
return n/d
/*──────────────────────────────────────────────────────────────────────────────────────*/
$fun: procedure; parse arg z.1,,z.2 1 zz.2; arg ,op; op=p(op '+')
F= 'fractionalFunction'; do j=1 for 2; z.j=translate(z.j, '/', "_"); end /*j*/
if abbrev('ADD' , op) then op= "+"
if abbrev('DIVIDE' , op) then op= "/"
if abbrev('INTDIVIDE', op, 4) then op= "÷"
if abbrev('MODULUS' , op, 3) | abbrev('MODULO', op, 3) then op= "//"
if abbrev('MULTIPLY' , op) then op= "*"
if abbrev('POWER' , op) then op= "^"
if abbrev('SUBTRACT' , op) then op= "-"
if z.1=='' then z.1= (op\=="+" & op\=='-')
if z.2=='' then z.2= (op\=="+" & op\=='-')
z_=z.2
/* [↑] verification of both fractions.*/
do j=1 for 2
if pos('/', z.j)==0 then z.j=z.j"/1"; parse var z.j n.j '/' d.j
if \datatype(n.j,'N') then call err 'a non─numeric numerator:' n.j
if \datatype(d.j,'N') then call err 'a non─numeric denominator:' d.j
if d.j=0 then call err 'a denominator of zero:' d.j
n.j=n.j/1; d.j=d.j/1
do while \datatype(n.j,'W'); n.j=(n.j*10)/1; d.j=(d.j*10)/1
end /*while*/ /* [↑] {xxx/1} normalizes a number. */
g=gcd(n.j, d.j); if g=0 then iterate; n.j=n.j/g; d.j=d.j/g
end /*j*/
select
when op=='+' | op=='-' then do; l=lcm(d.1,d.2); do j=1 for 2; n.j=l*n.j/d.j; d.j=l
end /*j*/
if op=='-' then n.2= -n.2; t=n.1 + n.2; u=l
end
when op=='**' | op=='↑' |,
op=='^' then do; if \datatype(z_,'W') then call err 'a non─integer power:' z_
t=1; u=1; do j=1 for abs(z_); t=t*n.1; u=u*d.1
end /*j*/
if z_<0 then parse value t u with u t /*swap U and T */
end
when op=='/' then do; if n.2=0 then call err 'a zero divisor:' zz.2
t=n.1*d.2; u=n.2*d.1
end
when op=='÷' then do; if n.2=0 then call err 'a zero divisor:' zz.2
t=trunc($div(n.1 '/' d.1)); u=1
end /* [↑] this is integer division. */
when op=='//' then do; if n.2=0 then call err 'a zero divisor:' zz.2
_=trunc($div(n.1 '/' d.1)); t=_ - trunc(_) * d.1; u=1
end /* [↑] modulus division. */
when op=='ABS' then do; t=abs(n.1); u=abs(d.1); end
when op=='*' then do; t=n.1 * n.2; u=d.1 * d.2; end
when op=='EQ' | op=='=' then return $div(n.1 '/' d.1) = fDiv(n.2 '/' d.2)
when op=='NE' | op=='\=' | op=='╪' | ,
op=='¬=' then return $div(n.1 '/' d.1) \= fDiv(n.2 '/' d.2)
when op=='GT' | op=='>' then return $div(n.1 '/' d.1) > fDiv(n.2 '/' d.2)
when op=='LT' | op=='<' then return $div(n.1 '/' d.1) < fDiv(n.2 '/' d.2)
when op=='GE' | op=='≥' | op=='>=' then return $div(n.1 '/' d.1) >= fDiv(n.2 '/' d.2)
when op=='LE' | op=='≤' | op=='<=' then return $div(n.1 '/' d.1) <= fDiv(n.2 '/' d.2)
otherwise call err 'an illegal function:' op
end /*select*/
if t==0 then return 0; g=gcd(t, u); t=t/g; u=u/g
if u==1 then return t
return t'/'u
/*──────────────────────────────────────────────────────────────────────────────────────*/
eDivs: procedure; parse arg x 1 b,a
do j=2 while j*j<x; if x//j\==0 then iterate; a=a j; b=x%j b; end
if j*j==x then return a j b; return a b
/*───────────────────────────────────────────────────────────────────────────────────────────────────*/
err: say; say '***error*** ' f " detected" arg(1); say; exit 13
gcd: procedure; parse arg x,y; if x=0 then return y; do until _==0; _=x//y; x=y; y=_; end; return x
lcm: procedure; parse arg x,y; if y=0 then return 0; x=x*y/gcd(x, y); return x
p: return word( arg(1), 1) |
http://rosettacode.org/wiki/Arithmetic-geometric_mean | Arithmetic-geometric mean |
This page uses content from Wikipedia. The original article was at Arithmetic-geometric mean. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Write a function to compute the arithmetic-geometric mean of two numbers.
The arithmetic-geometric mean of two numbers can be (usefully) denoted as
a
g
m
(
a
,
g
)
{\displaystyle \mathrm {agm} (a,g)}
, and is equal to the limit of the sequence:
a
0
=
a
;
g
0
=
g
{\displaystyle a_{0}=a;\qquad g_{0}=g}
a
n
+
1
=
1
2
(
a
n
+
g
n
)
;
g
n
+
1
=
a
n
g
n
.
{\displaystyle a_{n+1}={\tfrac {1}{2}}(a_{n}+g_{n});\quad g_{n+1}={\sqrt {a_{n}g_{n}}}.}
Since the limit of
a
n
−
g
n
{\displaystyle a_{n}-g_{n}}
tends (rapidly) to zero with iterations, this is an efficient method.
Demonstrate the function by calculating:
a
g
m
(
1
,
1
/
2
)
{\displaystyle \mathrm {agm} (1,1/{\sqrt {2}})}
Also see
mathworld.wolfram.com/Arithmetic-Geometric Mean
| #Tcl | Tcl | proc agm {a b} {
set old_b [expr {$b<0?inf:-inf}]
while {$a != $b && $b != $old_b} {
set old_b $b
lassign [list [expr {0.5*($a+$b)}] [expr {sqrt($a*$b)}]] a b
}
return $a
}
puts [agm 1 [expr 1/sqrt(2)]] |
http://rosettacode.org/wiki/Arithmetic-geometric_mean | Arithmetic-geometric mean |
This page uses content from Wikipedia. The original article was at Arithmetic-geometric mean. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Write a function to compute the arithmetic-geometric mean of two numbers.
The arithmetic-geometric mean of two numbers can be (usefully) denoted as
a
g
m
(
a
,
g
)
{\displaystyle \mathrm {agm} (a,g)}
, and is equal to the limit of the sequence:
a
0
=
a
;
g
0
=
g
{\displaystyle a_{0}=a;\qquad g_{0}=g}
a
n
+
1
=
1
2
(
a
n
+
g
n
)
;
g
n
+
1
=
a
n
g
n
.
{\displaystyle a_{n+1}={\tfrac {1}{2}}(a_{n}+g_{n});\quad g_{n+1}={\sqrt {a_{n}g_{n}}}.}
Since the limit of
a
n
−
g
n
{\displaystyle a_{n}-g_{n}}
tends (rapidly) to zero with iterations, this is an efficient method.
Demonstrate the function by calculating:
a
g
m
(
1
,
1
/
2
)
{\displaystyle \mathrm {agm} (1,1/{\sqrt {2}})}
Also see
mathworld.wolfram.com/Arithmetic-Geometric Mean
| #TI-83_BASIC | TI-83 BASIC | 1→A:1/sqrt(2)→G
While abs(A-G)>e-15
(A+G)/2→B
sqrt(AG)→G:B→A
End
A |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #HolyC | HolyC | F64 a = 0 ` 0;
Print("0 ` 0 = %5.3f\n", a); |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Icon_and_Unicon | Icon and Unicon | procedure main()
write(0^0)
end |
http://rosettacode.org/wiki/Zig-zag_matrix | Zig-zag matrix | Task
Produce a zig-zag array.
A zig-zag array is a square arrangement of the first N2 natural numbers, where the
numbers increase sequentially as you zig-zag along the array's anti-diagonals.
For a graphical representation, see JPG zigzag (JPG uses such arrays to encode images).
For example, given 5, produce this array:
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24
Related tasks
Spiral matrix
Identity matrix
Ulam spiral (for primes)
See also
Wiktionary entry: anti-diagonals
| #ALGOL_W | ALGOL W | begin % zig-zag matrix %
% z is returned holding a zig-zag matrix of order n, z must be at least n x n %
procedure makeZigZag ( integer value n
; integer array z( *, * )
) ;
begin
procedure move ;
begin
if y = n then begin
upRight := not upRight;
x := x + 1
end
else if x = 1 then begin
upRight := not upRight;
y := y + 1
end
else begin
x := x - 1;
y := y + 1
end
end move ;
procedure swapXY ;
begin
integer swap;
swap := x;
x := y;
y := swap;
end swapXY ;
integer x, y;
logical upRight;
% initialise the n x n matrix in z %
for i := 1 until n do for j := 1 until n do z( i, j ) := 0;
% fill in the zig-zag matrix %
x := y := 1;
upRight := true;
for i := 1 until n * n do begin
z( x, y ) := i - 1;
if upRight then move
else begin
swapXY;
move;
swapXY
end;
end;
end makeZigZap ;
begin
integer array zigZag( 1 :: 10, 1 :: 10 );
for n := 5 do begin
makeZigZag( n, zigZag );
for i := 1 until n do begin
write( i_w := 4, s_w := 1, zigZag( i, 1 ) );
for j := 2 until n do writeon( i_w := 4, s_w := 1, zigZag( i, j ) );
end
end
end
end. |
http://rosettacode.org/wiki/Yellowstone_sequence | Yellowstone sequence | The Yellowstone sequence, also called the Yellowstone permutation, is defined as:
For n <= 3,
a(n) = n
For n >= 4,
a(n) = the smallest number not already in sequence such that a(n) is relatively prime to a(n-1) and
is not relatively prime to a(n-2).
The sequence is a permutation of the natural numbers, and gets its name from what its authors felt was a spiking, geyser like appearance of a plot of the sequence.
Example
a(4) is 4 because 4 is the smallest number following 1, 2, 3 in the sequence that is relatively prime to the entry before it (3), and is not relatively prime to the number two entries before it (2).
Task
Find and show as output the first 30 Yellowstone numbers.
Extra
Demonstrate how to plot, with x = n and y coordinate a(n), the first 100 Yellowstone numbers.
Related tasks
Greatest common divisor.
Plot coordinate pairs.
See also
The OEIS entry: A098550 The Yellowstone permutation.
Applegate et al, 2015: The Yellowstone Permutation [1].
| #D | D | import std.numeric;
import std.range;
import std.stdio;
class Yellowstone {
private bool[int] sequence_;
private int min_ = 1;
private int n_ = 0;
private int n1_ = 0;
private int n2_ = 0;
public this() {
popFront();
}
public bool empty() {
return false;
}
public int front() {
return n_;
}
public void popFront() {
n2_ = n1_;
n1_ = n_;
if (n_ < 3) {
++n_;
} else {
for (n_ = min_;
!(n_ !in sequence_ && gcd(n1_, n_) == 1 && gcd(n2_, n_) > 1);
++n_) {
// empty
}
}
sequence_[n_] = true;
while (true) {
if (min_ !in sequence_) {
break;
}
sequence_.remove(min_);
++min_;
}
}
}
void main() {
new Yellowstone().take(30).writeln();
} |
http://rosettacode.org/wiki/Yahoo!_search_interface | Yahoo! search interface | Create a class for searching Yahoo! results.
It must implement a Next Page method, and read URL, Title and Content from results.
| #ARM_Assembly | ARM Assembly |
/* ARM assembly Raspberry PI */
/* program yahoosearch.s */
/* access RosettaCode.org and data extract */
/* use openssl for access to port 443 */
/* test openssl : package libssl-dev */
/* REMARK 1 : this program use routines in a include file
see task Include a file language arm assembly
for the routine affichageMess conversion10
see at end of this program the instruction include */
/*******************************************/
/* Constantes */
/*******************************************/
.equ STDOUT, 1 @ Linux output console
.equ EXIT, 1 @ Linux syscall
.equ WRITE, 4 @ Linux syscall
.equ BRK, 0x2d @ Linux syscall
.equ CHARPOS, '@'
.equ EXIT, 1
.equ TAILLEBUFFER, 500
.equ SSL_OP_NO_SSLv3, 0x02000000
.equ SSL_OP_NO_COMPRESSION, 0x00020000
.equ SSL_MODE_AUTO_RETRY, 0x00000004
.equ SSL_CTRL_MODE, 33
.equ BIO_C_SET_CONNECT, 100
.equ BIO_C_DO_STATE_MACHINE, 101
.equ BIO_C_SET_SSL, 109
.equ BIO_C_GET_SSL, 110
.equ LGBUFFERREQ, 512001
.equ LGBUFFER2, 128001
/*********************************/
/* Initialized data */
/*********************************/
.data
szMessDebutPgm: .asciz "Début du programme. \n"
szRetourLigne: .asciz "\n"
szMessFinOK: .asciz "Fin normale du programme. \n"
szMessErreur: .asciz "Erreur !!!"
szMessExtractArea: .asciz "Extraction = "
szNomSite1: .asciz "search.yahoo.com:443" @ host name and port
szLibStart: .asciz ">Rosetta Code" @ search string
szNomrepCertif: .asciz "/pi/certificats"
szRequete1: .asciz "GET /search?p=\"Rosettacode.org\"&b=1 HTTP/1.1 \r\nHost: search.yahoo.com\r\nConnection: keep-alive\r\nContent-Type: text/plain\r\n\r\n"
/*********************************/
/* UnInitialized data */
/*********************************/
.bss
.align 4
sBufferreq: .skip LGBUFFERREQ
szExtractArea: .skip TAILLEBUFFER
stNewSSL: .skip 200
/*********************************/
/* code section */
/*********************************/
.text
.global main
main:
ldr r0,iAdrszMessDebutPgm
bl affichageMess @ start message
/* connexion host port 443 and send query */
bl envoiRequete
cmp r0,#-1
beq 99f @ error ?
bl analyseReponse
ldr r0,iAdrszMessFinOK @ end message
bl affichageMess
mov r0, #0 @ return code ok
b 100f
99:
ldr r0,iAdrszMessErreur @ error
bl affichageMess
mov r0, #1 @ return code error
b 100f
100:
mov r7,#EXIT @ program end
svc #0 @ system call
iAdrszMessDebutPgm: .int szMessDebutPgm
iAdrszMessFinOK: .int szMessFinOK
iAdrszMessErreur: .int szMessErreur
/*********************************************************/
/* connexion host port 443 and send query */
/*********************************************************/
envoiRequete:
push {r2-r8,lr} @ save registers
@*************************************
@ openSsl functions use *
@*************************************
@init ssl
bl OPENSSL_init_crypto
bl ERR_load_BIO_strings
mov r2, #0
mov r1, #0
mov r0, #2
bl OPENSSL_init_crypto
mov r2, #0
mov r1, #0
mov r0, #0
bl OPENSSL_init_ssl
cmp r0,#0
blt erreur
bl TLS_client_method
bl SSL_CTX_new
cmp r0,#0
ble erreur
mov r6,r0 @ save ctx
ldr r1,iFlag
bl SSL_CTX_set_options
mov r0,r6
mov r1,#0
ldr r2,iAdrszNomrepCertif
bl SSL_CTX_load_verify_locations
cmp r0,#0
ble erreur
mov r0,r6
bl BIO_new_ssl_connect
cmp r0,#0
ble erreur
mov r5,r0 @ save bio
mov r1,#BIO_C_GET_SSL
mov r2,#0
ldr r3,iAdrstNewSSL
bl BIO_ctrl
ldr r0,iAdrstNewSSL
ldr r0,[r0]
mov r1,#SSL_CTRL_MODE
mov r2,#SSL_MODE_AUTO_RETRY
mov r3,#0
bl SSL_ctrl
mov r0,r5 @ bio
mov r1,#BIO_C_SET_CONNECT
mov r2,#0
ldr r3,iAdrszNomSite1
bl BIO_ctrl
mov r0,r5 @ bio
mov r1,#BIO_C_DO_STATE_MACHINE
mov r2,#0
mov r3,#0
bl BIO_ctrl
@ compute query length
mov r2,#0
ldr r1,iAdrszRequete1 @ query
1:
ldrb r0,[r1,r2]
cmp r0,#0
addne r2,#1
bne 1b
@ send query
mov r0,r5 @ bio
@ r1 = address query
@ r2 = length query
mov r3,#0
bl BIO_write @ send query
cmp r0,#0
blt erreur
ldr r7,iAdrsBufferreq @ buffer address
2: @ begin loop to read datas
mov r0,r5 @ bio
mov r1,r7 @ buffer address
mov r2,#LGBUFFERREQ - 1
mov r3,#0
bl BIO_read
cmp r0,#0
ble 4f @ error ou pb server
add r7,r0
sub r2,r7,#6
ldr r2,[r2]
ldr r3,iCharEnd
cmp r2,r3 @ text end ?
beq 4f
mov r1,#0xFFFFFF @ delay loop
3:
subs r1,#1
bgt 3b
b 2b @ loop read other chunk
4: @ read end
//ldr r0,iAdrsBufferreq @ to display buffer response of the query
//bl affichageMess
mov r0, r5 @ close bio
bl BIO_free_all
mov r0,#0
b 100f
erreur: @ error display
ldr r1,iAdrszMessErreur
bl afficheerreur
mov r0,#-1 @ error code
b 100f
100:
pop {r2-r8,lr} @ restaur registers
bx lr
iAdrszRequete1: .int szRequete1
iAdrsBufferreq: .int sBufferreq
iFlag: .int SSL_OP_NO_SSLv3 | SSL_OP_NO_COMPRESSION
iAdrstNewSSL: .int stNewSSL
iAdrszNomSite1: .int szNomSite1
iAdrszNomrepCertif: .int szNomrepCertif
iCharEnd: .int 0x0A0D300A
/*********************************************************/
/* response analyze */
/*********************************************************/
analyseReponse:
push {r1-r4,lr} @ save registers
ldr r0,iAdrsBufferreq @ buffer address
ldr r1,iAdrszLibStart @ key text address
mov r2,#2 @ occurence key text
mov r3,#-11 @ offset
ldr r4,iAdrszExtractArea @ address result area
bl extChaine
cmp r0,#-1
beq 99f
ldr r0,iAdrszMessExtractArea
bl affichageMess
ldr r0,iAdrszExtractArea @ résult display
bl affichageMess
ldr r0,iAdrszRetourLigne
bl affichageMess
b 100f
99:
ldr r0,iAdrszMessErreur @ error
bl affichageMess
mov r0, #-1 @ error return code
b 100f
100:
pop {r1-r4,lr} @ restaur registers
bx lr
iAdrszLibStart: .int szLibStart
iAdrszExtractArea: .int szExtractArea
iAdrszMessExtractArea: .int szMessExtractArea
iAdrszRetourLigne: .int szRetourLigne
/*********************************************************/
/* Text Extraction behind text key */
/*********************************************************/
/* r0 buffer address */
/* r1 key text to search */
/* r2 number occurences to key text */
/* r3 offset */
/* r4 result address */
extChaine:
push {r2-r8,lr} @ save registers
mov r5,r0 @ save buffer address
mov r6,r1 @ save key text
@ compute text length
mov r7,#0
1: @ loop
ldrb r0,[r5,r7] @ load a byte
cmp r0,#0 @ end ?
addne r7,#1 @ no -> loop
bne 1b
add r7,r5 @ compute text end
mov r8,#0
2: @ compute length text key
ldrb r0,[r6,r8]
cmp r0,#0
addne r8,#1
bne 2b
3: @ loop to search nième(r2) key text
mov r0,r5
mov r1,r6
bl rechercheSousChaine
cmp r0,#0
blt 100f
subs r2,#1
addgt r5,r0
addgt r5,r8
bgt 3b
add r0,r5 @ add address text to index
add r3,r0 @ add offset
sub r3,#1
@ and add length key text
add r3,r8
cmp r3,r7 @ > at text end
movge r0,#-1 @ yes -> error
bge 100f
mov r0,#0
4: @ character loop copy
ldrb r2,[r3,r0]
strb r2,[r4,r0]
cmp r2,#0 @ text end ?
moveq r0,#0 @ return zero
beq 100f
cmp r0,#48 @ extraction length
beq 5f
add r0,#1
b 4b @ and loop
5:
mov r2,#0 @ store final zéro
strb r2,[r4,r0]
add r0,#1
add r0,r3 @ r0 return the last position of extraction
@ it is possible o search another text
100:
pop {r2-r8,lr} @ restaur registers
bx lr
/******************************************************************/
/* search substring in string */
/******************************************************************/
/* r0 contains address string */
/* r1 contains address substring */
/* r0 return start index substring or -1 if not find */
rechercheSousChaine:
push {r1-r6,lr} @ save registers
mov r2,#0 @ index position string
mov r3,#0 @ index position substring
mov r6,#-1 @ search index
ldrb r4,[r1,r3] @ load first byte substring
cmp r4,#0 @ zero final ?
moveq r0,#-1 @ error
beq 100f
1:
ldrb r5,[r0,r2] @ load string byte
cmp r5,#0 @ zero final ?
moveq r0,#-1 @ yes -> not find
beq 100f
cmp r5,r4 @ compare character two strings
beq 2f
mov r6,#-1 @ not equal - > raz index
mov r3,#0 @ and raz byte counter
ldrb r4,[r1,r3] @ and load byte
add r2,#1 @ and increment byte counter
b 1b @ and loop
2: @ characters equal
cmp r6,#-1 @ first character equal ?
moveq r6,r2 @ yes -> start index in r6
add r3,#1 @ increment substring counter
ldrb r4,[r1,r3] @ and load next byte
cmp r4,#0 @ zero final ?
beq 3f @ yes -> search end
add r2,#1 @ else increment string index
b 1b @ and loop
3:
mov r0,r6 @ return start index substring in the string
100:
pop {r1-r6,lr} @ restaur registres
bx lr
/***************************************************/
/* ROUTINES INCLUDE */
/***************************************************/
.include "../affichage.inc"
|
http://rosettacode.org/wiki/Arithmetic_evaluation | Arithmetic evaluation | Create a program which parses and evaluates arithmetic expressions.
Requirements
An abstract-syntax tree (AST) for the expression must be created from parsing the input.
The AST must be used in evaluation, also, so the input may not be directly evaluated (e.g. by calling eval or a similar language feature.)
The expression will be a string or list of symbols like "(1+3)*7".
The four symbols + - * / must be supported as binary operators with conventional precedence rules.
Precedence-control parentheses must also be supported.
Note
For those who don't remember, mathematical precedence is as follows:
Parentheses
Multiplication/Division (left to right)
Addition/Subtraction (left to right)
C.f
24 game Player.
Parsing/RPN calculator algorithm.
Parsing/RPN to infix conversion.
| #Rust | Rust | //! Simple calculator parser and evaluator
/// Binary operator
#[derive(Debug)]
pub enum Operator {
Add,
Substract,
Multiply,
Divide
}
/// A node in the tree
#[derive(Debug)]
pub enum Node {
Value(f64),
SubNode(Box<Node>),
Binary(Operator, Box<Node>,Box<Node>),
}
/// parse a string into a node
pub fn parse(txt :&str) -> Option<Node> {
let chars = txt.chars().filter(|c| *c != ' ').collect();
parse_expression(&chars, 0).map(|(_,n)| n)
}
/// parse an expression into a node, keeping track of the position in the character vector
fn parse_expression(chars: &Vec<char>, pos: usize) -> Option<(usize,Node)> {
match parse_start(chars, pos) {
Some((new_pos, first)) => {
match parse_operator(chars, new_pos) {
Some((new_pos2,op)) => {
if let Some((new_pos3, second)) = parse_expression(chars, new_pos2) {
Some((new_pos3, combine(op, first, second)))
} else {
None
}
},
None => Some((new_pos,first)),
}
},
None => None,
}
}
/// combine nodes to respect associativity rules
fn combine(op: Operator, first: Node, second: Node) -> Node {
match second {
Node::Binary(op2,v21,v22) => if precedence(&op)>=precedence(&op2) {
Node::Binary(op2,Box::new(combine(op,first,*v21)),v22)
} else {
Node::Binary(op,Box::new(first),Box::new(Node::Binary(op2,v21,v22)))
},
_ => Node::Binary(op,Box::new(first),Box::new(second)),
}
}
/// a precedence rank for operators
fn precedence(op: &Operator) -> usize {
match op{
Operator::Multiply | Operator::Divide => 2,
_ => 1
}
}
/// try to parse from the start of an expression (either a parenthesis or a value)
fn parse_start(chars: &Vec<char>, pos: usize) -> Option<(usize,Node)> {
match start_parenthesis(chars, pos){
Some (new_pos) => {
let r = parse_expression(chars, new_pos);
end_parenthesis(chars, r)
},
None => parse_value(chars, pos),
}
}
/// match a starting parentheseis
fn start_parenthesis(chars: &Vec<char>, pos: usize) -> Option<usize>{
if pos<chars.len() && chars[pos] == '(' {
Some(pos+1)
} else {
None
}
}
/// match an end parenthesis, if successful will create a sub node contained the wrapped expression
fn end_parenthesis(chars: &Vec<char>, wrapped :Option<(usize,Node)>) -> Option<(usize,Node)>{
match wrapped {
Some((pos, node)) => if pos<chars.len() && chars[pos] == ')' {
Some((pos+1,Node::SubNode(Box::new(node))))
} else {
None
},
None => None,
}
}
/// parse a value: an decimal with an optional minus sign
fn parse_value(chars: &Vec<char>, pos: usize) -> Option<(usize,Node)>{
let mut new_pos = pos;
if new_pos<chars.len() && chars[new_pos] == '-' {
new_pos = new_pos+1;
}
while new_pos<chars.len() && (chars[new_pos]=='.' || (chars[new_pos] >= '0' && chars[new_pos] <= '9')) {
new_pos = new_pos+1;
}
if new_pos>pos {
if let Ok(v) = dbg!(chars[pos..new_pos].iter().collect::<String>()).parse() {
Some((new_pos,Node::Value(v)))
} else {
None
}
} else {
None
}
}
/// parse an operator
fn parse_operator(chars: &Vec<char>, pos: usize) -> Option<(usize,Operator)> {
if pos<chars.len() {
let ops_with_char = vec!(('+',Operator::Add),('-',Operator::Substract),('*',Operator::Multiply),('/',Operator::Divide));
for (ch,op) in ops_with_char {
if chars[pos] == ch {
return Some((pos+1, op));
}
}
}
None
}
/// eval a string
pub fn eval(txt :&str) -> f64 {
match parse(txt) {
Some(t) => eval_term(&t),
None => panic!("Cannot parse {}",txt),
}
}
/// eval a term, recursively
fn eval_term(t: &Node) -> f64 {
match t {
Node::Value(v) => *v,
Node::SubNode(t) => eval_term(t),
Node::Binary(Operator::Add,t1,t2) => eval_term(t1) + eval_term(t2),
Node::Binary(Operator::Substract,t1,t2) => eval_term(t1) - eval_term(t2),
Node::Binary(Operator::Multiply,t1,t2) => eval_term(t1) * eval_term(t2),
Node::Binary(Operator::Divide,t1,t2) => eval_term(t1) / eval_term(t2),
}
}
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_eval(){
assert_eq!(2.0,eval("2"));
assert_eq!(4.0,eval("2+2"));
assert_eq!(11.0/4.0, eval("2+3/4"));
assert_eq!(2.0, eval("2*3-4"));
assert_eq!(3.0, eval("1+2*3-4"));
assert_eq!(89.0/6.0, eval("2*(3+4)+5/6"));
assert_eq!(14.0, eval("2 * (3 -1) + 2 * 5"));
assert_eq!(7000.0, eval("2 * (3 + (4 * 5 + (6 * 7) * 8) - 9) * 10"));
assert_eq!(-9.0/4.0, eval("2*-3--4+-.25"));
assert_eq!(1.5, eval("1 - 5 * 2 / 20 + 1"));
assert_eq!(3.5, eval("2 * (3 + ((5) / (7 - 11)))"));
}
}
|
http://rosettacode.org/wiki/Zeckendorf_arithmetic | Zeckendorf arithmetic | This task is a total immersion zeckendorf task; using decimal numbers will attract serious disapprobation.
The task is to implement addition, subtraction, multiplication, and division using Zeckendorf number representation. Optionally provide decrement, increment and comparitive operation functions.
Addition
Like binary 1 + 1 = 10, note carry 1 left. There the similarity ends. 10 + 10 = 101, note carry 1 left and 1 right. 100 + 100 = 1001, note carry 1 left and 2 right, this is the general case.
Occurrences of 11 must be changed to 100. Occurrences of 111 may be changed from the right by replacing 11 with 100, or from the left converting 111 to 100 + 100;
Subtraction
10 - 1 = 1. The general rule is borrow 1 right carry 1 left. eg:
abcde
10100 -
1000
_____
100 borrow 1 from a leaves 100
+ 100 add the carry
_____
1001
A larger example:
abcdef
100100 -
1000
______
1*0100 borrow 1 from b
+ 100 add the carry
______
1*1001
Sadly we borrowed 1 from b which didn't have it to lend. So now b borrows from a:
1001
+ 1000 add the carry
____
10100
Multiplication
Here you teach your computer its zeckendorf tables. eg. 101 * 1001:
a = 1 * 101 = 101
b = 10 * 101 = a + a = 10000
c = 100 * 101 = b + a = 10101
d = 1000 * 101 = c + b = 101010
1001 = d + a therefore 101 * 1001 =
101010
+ 101
______
1000100
Division
Lets try 1000101 divided by 101, so we can use the same table used for multiplication.
1000101 -
101010 subtract d (1000 * 101)
_______
1000 -
101 b and c are too large to subtract, so subtract a
____
1 so 1000101 divided by 101 is d + a (1001) remainder 1
Efficient algorithms for Zeckendorf arithmetic is interesting. The sections on addition and subtraction are particularly relevant for this task.
| #Phix | Phix | with javascript_semantics
sequence fib = {1,1}
function zeckendorf(atom n)
-- Same as Zeckendorf_number_representation#Phix
atom r = 0
while fib[$]<n do
fib &= fib[$] + fib[$-1]
end while
integer k = length(fib)
while k>2 and n<fib[k] do
k -= 1
end while
for i=k to 2 by -1 do
integer c = n>=fib[i]
r += r+c
n -= c*fib[i]
end for
return r
end function
function decimal(object z)
-- Convert Zeckendorf number(s) to decimal
if sequence(z) then
sequence res = repeat(0,length(z))
for i=1 to length(z) do
res[i] = decimal(z[i])
end for
return res
end if
atom dec = 0, bit = 2
while z do
if and_bits(z,1) then
dec += fib[bit]
end if
bit += 1
if bit>length(fib) then
fib &= fib[$] + fib[$-1]
end if
z = floor(z/2)
end while
return dec
end function
function to_bits(integer x)
-- Simplified copy of int_to_bits(), but in reverse order,
-- and +ve only but (also only) as many bits as needed, and
-- ensures there are *two* trailing 0 (most significant)
if x<0 then ?9/0 end if -- sanity/avoid infinite loop
sequence bits = {}
while 1 do
bits &= remainder(x,2)
if x=0 then exit end if
x = floor(x/2)
end while
bits &= 0 -- (since eg 101+101 -> 10000)
return bits
end function
function to_bits2(integer a,b)
-- Apply to_bits() to a and b, and pad to the same length
sequence sa = to_bits(a),
sb = to_bits(b)
integer diff = length(sa)-length(sb)
if diff!=0 then
if diff<0 then sa &= repeat(0,-diff)
else sb &= repeat(0,+diff)
end if
end if
return {sa,sb}
end function
function to_int(sequence bits)
-- Copy of bits_to_int(), but in reverse order (lsb last)
atom val = 0, p = 1
for i=length(bits) to 1 by -1 do
if bits[i] then
val += p
end if
p += p
end for
return val
end function
function zstr(object z)
if sequence(z) then
sequence res = repeat(0,length(z))
for i=1 to length(z) do
res[i] = zstr(z[i])
end for
return res
end if
return sprintf("%b",z)
end function
function rep(sequence res, integer ds, sequence was, wth)
-- helper for cleanup, validates replacements
integer de = ds+length(was)-1
if res[ds..de]!=was then ?9/0 end if
if length(was)!=length(wth) then ?9/0 end if
res = deep_copy(res)
res[ds..de] = wth
return res
end function
function zcleanup(sequence res)
-- (shared by zadd and zsub)
integer l = length(res)
res = deep_copy(res)
-- first stage, left to right, {020x -> 100x', 030x -> 110x', 021x->110x, 012x->101x}
for i=1 to l-3 do
sequence s3 = res[i..i+2]
if s3={0,2,0} then res[i..i+2] = {1,0,0} res[i+3] += 1
elsif s3={0,3,0} then res[i..i+2] = {1,1,0} res[i+3] += 1
elsif s3={0,2,1} then res[i..i+2] = {1,1,0}
elsif s3={0,1,2} then res[i..i+2] = {1,0,1}
end if
end for
-- first stage cleanup
if l>1 then
if res[l-1]=3 then res = rep(res,l-2,{0,3,0},{1,1,1}) -- 030 -> 111
elsif res[l-1]=2 then
if res[l-2]=0 then res = rep(res,l-2,{0,2,0},{1,0,1}) -- 020 -> 101
else res = rep(res,l-3,{0,1,2,0},{1,0,1,0}) -- 0120 -> 1010
end if
end if
end if
if res[l]=3 then res = rep(res,l-1,{0,3},{1,1}) -- 03 -> 11
elsif res[l]=2 then
if res[l-1]=0 then res = rep(res,l-1,{0,2},{1,0}) -- 02 -> 10
else res = rep(res,l-2,{0,1,2},{1,0,1}) -- 012 -> 101
end if
end if
-- second stage, pass 1, right to left, 011 -> 100
for i=length(res)-2 to 1 by -1 do
if res[i..i+2]={0,1,1} then res[i..i+2] = {1,0,0} end if
end for
-- second stage, pass 2, left to right, 011 -> 100
for i=1 to length(res)-2 do
if res[i..i+2]={0,1,1} then res[i..i+2] = {1,0,0} end if
end for
return to_int(res)
end function
function zadd(integer a, b)
sequence {sa,sb} = to_bits2(a,b)
return zcleanup(reverse(sq_add(sa,sb)))
end function
function zinc(integer a)
return zadd(a,0b1)
end function
function zsub(integer a, b)
sequence {sa,sb} = to_bits2(a,b)
sequence res = reverse(sq_sub(sa,sb))
-- (/not/ combined with the first pass of the add routine!)
for i=1 to length(res)-2 do
sequence s3 = res[i..i+2]
if s3={1, 0, 0} then res[i..i+2] = {0,1,1}
elsif s3={1,-1, 0} then res[i..i+2] = {0,0,1}
elsif s3={1,-1, 1} then res[i..i+2] = {0,0,2}
elsif s3={1, 0,-1} then res[i..i+2] = {0,1,0}
elsif s3={2, 0, 0} then res[i..i+2] = {1,1,1}
elsif s3={2,-1, 0} then res[i..i+2] = {1,0,1}
elsif s3={2,-1, 1} then res[i..i+2] = {1,0,2}
elsif s3={2, 0,-1} then res[i..i+2] = {1,1,0}
end if
end for
-- copied from PicoLisp: {1,-1} -> {0,1} and {2,-1} -> {1,1}
for i=1 to length(res)-1 do
sequence s2 = res[i..i+1]
if s2={1,-1} then res[i..i+1] = {0,1}
elsif s2={2,-1} then res[i..i+1] = {1,1}
end if
end for
if find(-1,res) then ?9/0 end if -- sanity check
return zcleanup(res)
end function
function zdec(integer a)
return zsub(a,0b1)
end function
function zmul(integer a, b)
sequence mult = {a,zadd(a,a)} -- (as per task desc)
integer bits = 2
while bits<b do
mult = append(mult,zadd(mult[$],mult[$-1]))
bits *= 2
end while
integer res = 0,
bit = 1
while b do
if and_bits(b,1) then
res = zadd(res,mult[bit])
end if
b = floor(b/2)
bit += 1
end while
return res
end function
function zdiv(integer a, b)
sequence mult = {b,zadd(b,b)}
integer bits = 2
while mult[$]<a do
mult = append(mult,zadd(mult[$],mult[$-1]))
bits *= 2
end while
integer res = 0
for i=length(mult) to 1 by -1 do
integer mi = mult[i]
if mi<=a then
res = zadd(res,bits)
a = zsub(a,mi)
if a=0 then exit end if
end if
bits = floor(bits/2)
end for
return {res,a} -- (a is the remainder)
end function
for i=0 to 20 do
integer zi = zeckendorf(i)
atom d = decimal(zi)
printf(1,"%2d: %7b (%d)\n",{i,zi,d})
end for
procedure test(atom a, string op, atom b, object res, string expected)
string zres = iff(atom(res)?zstr(res):join(zstr(res)," rem ")),
dres = sprintf(iff(atom(res)?"%d":"%d rem %d"),decimal(res)),
aka = sprintf("aka %d %s %d = %s",{decimal(a),op,decimal(b),dres}),
ok = iff(zres=expected?"":" *** ERROR ***!!")
printf(1,"%s %s %s = %s, %s %s\n",{zstr(a),op,zstr(b),zres,aka,ok})
end procedure
test(0b0,"+",0b0,zadd(0b0,0b0),"0")
test(0b101,"+",0b101,zadd(0b101,0b101),"10000")
test(0b10100,"-",0b1000,zsub(0b10100,0b1000),"1001")
test(0b100100,"-",0b1000,zsub(0b100100,0b1000),"10100")
test(0b1001,"*",0b101,zmul(0b1001,0b101),"1000100")
test(0b1000101,"/",0b101,zdiv(0b1000101,0b101),"1001 rem 1")
test(0b10,"+",0b10,zadd(0b10,0b10),"101")
test(0b101,"+",0b10,zadd(0b101,0b10),"1001")
test(0b1001,"+",0b1001,zadd(0b1001,0b1001),"10101")
test(0b10101,"+",0b1000,zadd(0b10101,0b1000),"100101")
test(0b100101,"+",0b10101,zadd(0b100101,0b10101),"1010000")
test(0b1000,"-",0b101,zsub(0b1000,0b101),"1")
test(0b10101010,"-",0b1010101,zsub(0b10101010,0b1010101),"1000000")
test(0b1001,"*",0b101,zmul(0b1001,0b101),"1000100")
test(0b101010,"+",0b101,zadd(0b101010,0b101),"1000100")
test(0b10100,"+",0b1010,zadd(0b10100,0b1010),"101000")
test(0b101000,"-",0b1010,zsub(0b101000,0b1010),"10100")
test(0b100010,"*",0b100101,zmul(0b100010,0b100101),"100001000001")
test(0b100001000001,"/",0b100,zdiv(0b100001000001,0b100),"101010001 rem 0")
test(0b101000101,"*",0b101001,zmul(0b101000101,0b101001),"101010000010101")
test(0b101010000010101,"/",0b100,zdiv(0b101010000010101,0b100),"1001010001001 rem 10")
test(0b10100010010100,"+",0b1001000001,zadd(0b10100010010100,0b1001000001),"100000000010101")
test(0b10100010010100,"-",0b1001000001,zsub(0b10100010010100,0b1001000001),"10010001000010")
test(0b10000,"*",0b1001000001,zmul(0b10000,0b1001000001),"10100010010100")
test(0b1010001010000001001,"/",0b100000000100000,zdiv(0b1010001010000001001,0b100000000100000),"10001 rem 10100001010101")
test(0b10100,"+",0b1010,zadd(0b10100,0b1010),"101000")
test(0b10100,"-",0b1010,zsub(0b10100,0b1010),"101")
test(0b10100,"*",0b1010,zmul(0b10100,0b1010),"101000001")
test(0b10100,"/",0b1010,zdiv(0b10100,0b1010),"1 rem 101")
integer m = zmul(0b10100,0b1010)
test(m,"/",0b1010,zdiv(m,0b1010),"10100 rem 0")
|
http://rosettacode.org/wiki/Zumkeller_numbers | Zumkeller numbers | Zumkeller numbers are the set of numbers whose divisors can be partitioned into two disjoint sets that sum to the same value. Each sum must contain divisor values that are not in the other sum, and all of the divisors must be in one or the other. There are no restrictions on how the divisors are partitioned, only that the two partition sums are equal.
E.G.
6 is a Zumkeller number; The divisors {1 2 3 6} can be partitioned into two groups {1 2 3} and {6} that both sum to 6.
10 is not a Zumkeller number; The divisors {1 2 5 10} can not be partitioned into two groups in any way that will both sum to the same value.
12 is a Zumkeller number; The divisors {1 2 3 4 6 12} can be partitioned into two groups {1 3 4 6} and {2 12} that both sum to 14.
Even Zumkeller numbers are common; odd Zumkeller numbers are much less so. For values below 10^6, there is at least one Zumkeller number in every 12 consecutive integers, and the vast majority of them are even. The odd Zumkeller numbers are very similar to the list from the task Abundant odd numbers; they are nearly the same except for the further restriction that the abundance (A(n) = sigma(n) - 2n), must be even: A(n) mod 2 == 0
Task
Write a routine (function, procedure, whatever) to find Zumkeller numbers.
Use the routine to find and display here, on this page, the first 220 Zumkeller numbers.
Use the routine to find and display here, on this page, the first 40 odd Zumkeller numbers.
Optional, stretch goal: Use the routine to find and display here, on this page, the first 40 odd Zumkeller numbers that don't end with 5.
See Also
OEIS:A083207 - Zumkeller numbers to get an impression of different partitions OEIS:A083206 Zumkeller partitions
OEIS:A174865 - Odd Zumkeller numbers
Related Tasks
Abundant odd numbers
Abundant, deficient and perfect number classifications
Proper divisors , Factors of an integer | #Nim | Nim | import math, strutils
template isEven(n: int): bool = (n and 1) == 0
template isOdd(n: int): bool = (n and 1) != 0
func getDivisors(n: int): seq[int] =
result = @[1, n]
for i in 2..sqrt(n.toFloat).int:
if n mod i == 0:
let j = n div i
result.add i
if i != j: result.add j
func isPartSum(divs: seq[int]; sum: int): bool =
if sum == 0: return true
if divs.len == 0: return false
let last = divs[^1]
let divs = divs[0..^2]
result = isPartSum(divs, sum)
if not result and last <= sum:
result = isPartSum(divs, sum - last)
func isZumkeller(n: int): bool =
let divs = n.getDivisors()
let sum = sum(divs)
# If "sum" is odd, it can't be split into two partitions with equal sums.
if sum.isOdd: return false
# If "n" is odd use "abundant odd number" optimization.
if n.isOdd:
let abundance = sum - 2 * n
return abundance > 0 and abundance.isEven
# If "n" and "sum" are both even, check if there's a partition which totals "sum / 2".
result = isPartSum(divs, sum div 2)
when isMainModule:
echo "The first 220 Zumkeller numbers are:"
var n = 2
var count = 0
while count < 220:
if n.isZumkeller:
stdout.write align($n, 3)
inc count
stdout.write if count mod 20 == 0: '\n' else: ' '
inc n
echo()
echo "The first 40 odd Zumkeller numbers are:"
n = 3
count = 0
while count < 40:
if n.isZumkeller:
stdout.write align($n, 5)
inc count
stdout.write if count mod 10 == 0: '\n' else: ' '
inc n, 2
echo()
echo "The first 40 odd Zumkeller numbers which don't end in 5 are:"
n = 3
count = 0
while count < 40:
if n mod 10 != 5 and n.isZumkeller:
stdout.write align($n, 7)
inc count
stdout.write if count mod 8 == 0: '\n' else: ' '
inc n, 2 |
http://rosettacode.org/wiki/Zumkeller_numbers | Zumkeller numbers | Zumkeller numbers are the set of numbers whose divisors can be partitioned into two disjoint sets that sum to the same value. Each sum must contain divisor values that are not in the other sum, and all of the divisors must be in one or the other. There are no restrictions on how the divisors are partitioned, only that the two partition sums are equal.
E.G.
6 is a Zumkeller number; The divisors {1 2 3 6} can be partitioned into two groups {1 2 3} and {6} that both sum to 6.
10 is not a Zumkeller number; The divisors {1 2 5 10} can not be partitioned into two groups in any way that will both sum to the same value.
12 is a Zumkeller number; The divisors {1 2 3 4 6 12} can be partitioned into two groups {1 3 4 6} and {2 12} that both sum to 14.
Even Zumkeller numbers are common; odd Zumkeller numbers are much less so. For values below 10^6, there is at least one Zumkeller number in every 12 consecutive integers, and the vast majority of them are even. The odd Zumkeller numbers are very similar to the list from the task Abundant odd numbers; they are nearly the same except for the further restriction that the abundance (A(n) = sigma(n) - 2n), must be even: A(n) mod 2 == 0
Task
Write a routine (function, procedure, whatever) to find Zumkeller numbers.
Use the routine to find and display here, on this page, the first 220 Zumkeller numbers.
Use the routine to find and display here, on this page, the first 40 odd Zumkeller numbers.
Optional, stretch goal: Use the routine to find and display here, on this page, the first 40 odd Zumkeller numbers that don't end with 5.
See Also
OEIS:A083207 - Zumkeller numbers to get an impression of different partitions OEIS:A083206 Zumkeller partitions
OEIS:A174865 - Odd Zumkeller numbers
Related Tasks
Abundant odd numbers
Abundant, deficient and perfect number classifications
Proper divisors , Factors of an integer | #Pascal | Pascal | program zumkeller;
//https://oeis.org/A083206/a083206.txt
{$IFDEF FPC}
{$MODE DELPHI} {$OPTIMIZATION ON,ALL} {$COPERATORS ON}
// {$O+,I+}
{$ELSE}
{$APPTYPE CONSOLE}
{$ENDIF}
uses
sysutils
{$IFDEF WINDOWS},Windows{$ENDIF}
;
//######################################################################
//prime decomposition
const
//HCN(86) > 1.2E11 = 128,501,493,120 count of divs = 4096 7 3 1 1 1 1 1 1 1
HCN_DivCnt = 4096;
//stop never ending recursion
RECCOUNTMAX = 100*1000*1000;
DELTAMAX = 1000*1000;
type
tItem = Uint64;
tDivisors = array [0..HCN_DivCnt-1] of tItem;
tpDivisor = pUint64;
const
SizePrDeFe = 12697;//*72 <= 1 or 2 Mb ~ level 2 cache -32kB for DIVS
type
tdigits = packed record
dgtDgts : array [0..31] of Uint32;
end;
//the first number with 11 different divisors =
// 2*3*5*7*11*13*17*19*23*29*31 = 2E11
tprimeFac = packed record
pfSumOfDivs,
pfRemain : Uint64; //n div (p[0]^[pPot[0] *...) can handle primes <=821641^2 = 6.7e11
pfpotPrim : array[0..9] of UInt32;//+10*4 = 56 Byte
pfpotMax : array[0..9] of byte; //10 = 66
pfMaxIdx : Uint16; //68
pfDivCnt : Uint32; //72
end;
tPrimeDecompField = array[0..SizePrDeFe-1] of tprimeFac;
tPrimes = array[0..65535] of Uint32;
var
SmallPrimes: tPrimes;
//######################################################################
//prime decomposition
procedure InitSmallPrimes;
//only odd numbers
const
MAXLIMIT = (821641-1) shr 1;
var
pr : array[0..MAXLIMIT] of byte;
p,j,d,flipflop :NativeUInt;
Begin
SmallPrimes[0] := 2;
fillchar(pr[0],SizeOf(pr),#0);
p := 0;
repeat
repeat
p +=1
until pr[p]= 0;
j := (p+1)*p*2;
if j>MAXLIMIT then
BREAK;
d := 2*p+1;
repeat
pr[j] := 1;
j += d;
until j>MAXLIMIT;
until false;
SmallPrimes[1] := 3;
SmallPrimes[2] := 5;
j := 3;
d := 7;
flipflop := 3-1;
p := 3;
repeat
if pr[p] = 0 then
begin
SmallPrimes[j] := d;
inc(j);
end;
d += 2*flipflop;
p+=flipflop;
flipflop := 3-flipflop;
until (p > MAXLIMIT) OR (j>High(SmallPrimes));
end;
function OutPots(const pD:tprimeFac;n:NativeInt):Ansistring;
var
s: String[31];
Begin
str(n,s);
result := s+' :';
with pd do
begin
str(pfDivCnt:3,s);
result += s+' : ';
For n := 0 to pfMaxIdx-1 do
Begin
if n>0 then
result += '*';
str(pFpotPrim[n],s);
result += s;
if pfpotMax[n] >1 then
Begin
str(pfpotMax[n],s);
result += '^'+s;
end;
end;
If pfRemain >1 then
Begin
str(pfRemain,s);
result += '*'+s;
end;
str(pfSumOfDivs,s);
result += '_SoD_'+s+'<';
end;
end;
function CnvtoBASE(var dgt:tDigits;n:Uint64;base:NativeUint):NativeInt;
//n must be multiple of base
var
q,r: Uint64;
i : NativeInt;
Begin
with dgt do
Begin
fillchar(dgtDgts,SizeOf(dgtDgts),#0);
i := 0;
// dgtNum:= n;
n := n div base;
result := 0;
repeat
r := n;
q := n div base;
r -= q*base;
n := q;
dgtDgts[i] := r;
inc(i);
until (q = 0);
result := 0;
while (result<i) AND (dgtDgts[result] = 0) do
inc(result);
inc(result);
end;
end;
function IncByBaseInBase(var dgt:tDigits;base:NativeInt):NativeInt;
var
q :NativeInt;
Begin
with dgt do
Begin
result := 0;
q := dgtDgts[result]+1;
// inc(dgtNum,base);
if q = base then
begin
repeat
dgtDgts[result] := 0;
inc(result);
q := dgtDgts[result]+1;
until q <> base;
end;
dgtDgts[result] := q;
result +=1;
end;
end;
procedure SieveOneSieve(var pdf:tPrimeDecompField;n:nativeUInt);
var
dgt:tDigits;
i, j, k,pr,fac : NativeUInt;
begin
//init
for i := 0 to High(pdf) do
with pdf[i] do
Begin
pfDivCnt := 1;
pfSumOfDivs := 1;
pfRemain := n+i;
pfMaxIdx := 0;
end;
//first 2 make n+i even
i := n AND 1;
repeat
with pdf[i] do
if n+i > 0 then
begin
j := BsfQWord(n+i);
pfMaxIdx := 1;
pfpotPrim[0] := 2;
pfpotMax[0] := j;
pfRemain := (n+i) shr j;
pfSumOfDivs := (1 shl (j+1))-1;
pfDivCnt := j+1;
end;
i += 2;
until i >High(pdf);
// i now index in SmallPrimes
i := 0;
repeat
//search next prime that is in bounds of sieve
repeat
inc(i);
if i >= High(SmallPrimes) then
BREAK;
pr := SmallPrimes[i];
k := pr-n MOD pr;
if (k = pr) AND (n>0) then
k:= 0;
if k < SizePrDeFe then
break;
until false;
if i >= High(SmallPrimes) then
BREAK;
//no need to use higher primes
if pr*pr > n+SizePrDeFe then
BREAK;
// j is power of prime
j := CnvtoBASE(dgt,n+k,pr);
repeat
with pdf[k] do
Begin
pfpotPrim[pfMaxIdx] := pr;
pfpotMax[pfMaxIdx] := j;
pfDivCnt *= j+1;
fac := pr;
repeat
pfRemain := pfRemain DIV pr;
dec(j);
fac *= pr;
until j<= 0;
pfSumOfDivs *= (fac-1)DIV(pr-1);
inc(pfMaxIdx);
end;
k += pr;
j := IncByBaseInBase(dgt,pr);
until k >= SizePrDeFe;
until false;
//correct sum of & count of divisors
for i := 0 to High(pdf) do
Begin
with pdf[i] do
begin
j := pfRemain;
if j <> 1 then
begin
pfSumOFDivs *= (j+1);
pfDivCnt *=2;
end;
end;
end;
end;
//prime decomposition
//######################################################################
procedure Init_Check_rec(const pD:tprimeFac;var Divs,SumOfDivs:tDivisors);forward;
var
{$ALIGN 32}
PrimeDecompField:tPrimeDecompField;
{$ALIGN 32}
Divs :tDivisors;
SumOfDivs : tDivisors;
DivUsedIdx : tDivisors;
pDiv :tpDivisor;
T0: Int64;
count,rec_Cnt: NativeInt;
depth : Int32;
finished :Boolean;
procedure Check_rek_depth(SoD : Int64;i: NativeInt);
var
sum : Int64;
begin
if finished then
EXIT;
inc(rec_Cnt);
WHILE (i>0) AND (pDiv[i]>SoD) do
dec(i);
while i >= 0 do
Begin
DivUsedIdx[depth] := pDiv[i];
sum := SoD-pDiv[i];
if sum = 0 then
begin
finished := true;
EXIT;
end;
dec(i);
inc(depth);
if (i>= 0) AND (sum <= SumOfDivs[i]) then
Check_rek_depth(sum,i);
if finished then
EXIT;
// DivUsedIdx[depth] := 0;
dec(depth);
end;
end;
procedure Out_One_Sol(const pd:tprimefac;n:NativeUInt;isZK : Boolean);
var
sum : NativeInt;
Begin
if n< 7 then
exit;
with pd do
begin
writeln(OutPots(pD,n));
if isZK then
Begin
Init_Check_rec(pD,Divs,SumOfDivs);
Check_rek_depth(pfSumOfDivs shr 1-n,pFDivCnt-1);
write(pfSumOfDivs shr 1:10,' = ');
sum := n;
while depth >= 0 do
Begin
sum += DivUsedIdx[depth];
write(DivUsedIdx[depth],'+');
dec(depth);
end;
write(n,' = ',sum);
end
else
write(' no zumkeller ');
end;
end;
procedure InsertSort(pDiv:tpDivisor; Left, Right : NativeInt );
var
I, J: NativeInt;
Pivot : tItem;
begin
for i:= 1 + Left to Right do
begin
Pivot:= pDiv[i];
j:= i - 1;
while (j >= Left) and (pDiv[j] > Pivot) do
begin
pDiv[j+1]:=pDiv[j];
Dec(j);
end;
pDiv[j+1]:= pivot;
end;
end;
procedure GetDivs(const pD:tprimeFac;var Divs,SumOfDivs:tDivisors);
var
pDivs : tpDivisor;
pPot : UInt64;
i,len,j,l,p,k: Int32;
Begin
i := pD.pfDivCnt;
pDivs := @Divs[0];
pDivs[0] := 1;
len := 1;
l := 1;
with pD do
Begin
For i := 0 to pfMaxIdx-1 do
begin
//Multiply every divisor before with the new primefactors
//and append them to the list
k := pfpotMax[i];
p := pfpotPrim[i];
pPot :=1;
repeat
pPot *= p;
For j := 0 to len-1 do
Begin
pDivs[l]:= pPot*pDivs[j];
inc(l);
end;
dec(k);
until k<=0;
len := l;
end;
p := pfRemain;
If p >1 then
begin
For j := 0 to len-1 do
Begin
pDivs[l]:= p*pDivs[j];
inc(l);
end;
len := l;
end;
end;
//Sort. Insertsort much faster than QuickSort in this special case
InsertSort(pDivs,0,len-1);
pPot := 0;
For i := 0 to len-1 do
begin
pPot += pDivs[i];
SumOfDivs[i] := pPot;
end;
end;
procedure Init_Check_rec(const pD:tprimeFac;var Divs,SumOfDivs:tDivisors);
begin
GetDivs(pD,Divs,SUmOfDivs);
finished := false;
depth := 0;
pDiv := @Divs[0];
end;
procedure Check_rek(SoD : Int64;i: NativeInt);
var
sum : Int64;
begin
if finished then
EXIT;
if rec_Cnt >RECCOUNTMAX then
begin
rec_Cnt := -1;
finished := true;
exit;
end;
inc(rec_Cnt);
WHILE (i>0) AND (pDiv[i]>SoD) do
dec(i);
while i >= 0 do
Begin
sum := SoD-pDiv[i];
if sum = 0 then
begin
finished := true;
EXIT;
end;
dec(i);
if (i>= 0) AND (sum <= SumOfDivs[i]) then
Check_rek(sum,i);
if finished then
EXIT;
end;
end;
function GetZumKeller(n: NativeUint;var pD:tPrimefac): boolean;
var
SoD,sum : Int64;
Div_cnt,i,pracLmt: NativeInt;
begin
rec_Cnt := 0;
SoD:= pd.pfSumOfDivs;
//sum must be even and n not deficient
if Odd(SoD) or (SoD<2*n) THEN
EXIT(false);
//if Odd(n) then Exit(Not(odd(sum)));// to be tested
SoD := SoD shr 1-n;
If SoD < 2 then //0,1 is always true
Exit(true);
Div_cnt := pD.pfDivCnt;
if Not(odd(n)) then
if ((n mod 18) in [6,12]) then
EXIT(true);
//Now one needs to get the divisors
Init_check_rec(pD,Divs,SumOfDivs);
pracLmt:= 0;
if Not(odd(n)) then
begin
For i := 1 to Div_Cnt do
Begin
sum := SumOfDivs[i];
If (sum+1<Divs[i+1]) AND (sum<SoD) then
Begin
pracLmt := i;
BREAK;
end;
IF (sum>=SoD) then break;
end;
if pracLmt = 0 then
Exit(true);
end;
//number is practical followed by one big prime
if pracLmt = (Div_Cnt-1) shr 1 then
begin
i := SoD mod Divs[pracLmt+1];
with pD do
begin
if pfRemain > 1 then
EXIT((pfRemain<=i) OR (i<=sum))
else
EXIT((pfpotPrim[pfMaxIdx-1]<=i)OR (i<=sum));
end;
end;
Begin
IF Div_cnt <= HCN_DivCnt then
Begin
Check_rek(SoD,Div_cnt-1);
IF rec_Cnt = -1 then
exit(true);
exit(finished);
end;
end;
result := false;
end;
var
Ofs,i,n : NativeUInt;
Max: NativeUInt;
procedure Init_Sieve(n:NativeUint);
//Init Sieve i,oFs are Global
begin
i := n MOD SizePrDeFe;
Ofs := (n DIV SizePrDeFe)*SizePrDeFe;
SieveOneSieve(PrimeDecompField,Ofs);
end;
procedure GetSmall(MaxIdx:Int32);
var
ZK: Array of Uint32;
idx: UInt32;
Begin
If MaxIdx<1 then
EXIT;
writeln('The first ',MaxIdx,' zumkeller numbers');
Init_Sieve(0);
setlength(ZK,MaxIdx);
idx := Low(ZK);
repeat
if GetZumKeller(n,PrimeDecompField[i]) then
Begin
ZK[idx] := n;
inc(idx);
end;
inc(i);
inc(n);
If i > High(PrimeDecompField) then
begin
dec(i,SizePrDeFe);
inc(ofs,SizePrDeFe);
SieveOneSieve(PrimeDecompField,Ofs);
end;
until idx >= MaxIdx;
For idx := 0 to MaxIdx-1 do
begin
if idx MOD 20 = 0 then
writeln;
write(ZK[idx]:4);
end;
setlength(ZK,0);
writeln;
writeln;
end;
procedure GetOdd(MaxIdx:Int32);
var
ZK: Array of Uint32;
idx: UInt32;
Begin
If MaxIdx<1 then
EXIT;
writeln('The first odd 40 zumkeller numbers');
n := 1;
Init_Sieve(n);
setlength(ZK,MaxIdx);
idx := Low(ZK);
repeat
if GetZumKeller(n,PrimeDecompField[i]) then
Begin
ZK[idx] := n;
inc(idx);
end;
inc(i,2);
inc(n,2);
If i > High(PrimeDecompField) then
begin
dec(i,SizePrDeFe);
inc(ofs,SizePrDeFe);
SieveOneSieve(PrimeDecompField,Ofs);
end;
until idx >= MaxIdx;
For idx := 0 to MaxIdx-1 do
begin
if idx MOD (80 DIV 8) = 0 then
writeln;
write(ZK[idx]:8);
end;
setlength(ZK,0);
writeln;
writeln;
end;
procedure GetOddNot5(MaxIdx:Int32);
var
ZK: Array of Uint32;
idx: UInt32;
Begin
If MaxIdx<1 then
EXIT;
writeln('The first odd 40 zumkeller numbers not ending in 5');
n := 1;
Init_Sieve(n);
setlength(ZK,MaxIdx);
idx := Low(ZK);
repeat
if GetZumKeller(n,PrimeDecompField[i]) then
Begin
ZK[idx] := n;
inc(idx);
end;
inc(i,2);
inc(n,2);
If n mod 5 = 0 then
begin
inc(i,2);
inc(n,2);
end;
If i > High(PrimeDecompField) then
begin
dec(i,SizePrDeFe);
inc(ofs,SizePrDeFe);
SieveOneSieve(PrimeDecompField,Ofs);
end;
until idx >= MaxIdx;
For idx := 0 to MaxIdx-1 do
begin
if idx MOD (80 DIV 8) = 0 then
writeln;
write(ZK[idx]:8);
end;
setlength(ZK,0);
writeln;
writeln;
end;
BEGIN
InitSmallPrimes;
T0 := GetTickCount64;
GetSmall(220);
GetOdd(40);
GetOddNot5(40);
writeln;
n := 1;//8996229720;//1;
Init_Sieve(n);
writeln('Start ',n,' at ',i);
T0 := GetTickCount64;
MAX := (n DIV DELTAMAX+1)*DELTAMAX;
count := 0;
repeat
writeln('Count of zumkeller numbers up to ',MAX:12);
repeat
if GetZumKeller(n,PrimeDecompField[i]) then
inc(count);
inc(i);
inc(n);
If i > High(PrimeDecompField) then
begin
dec(i,SizePrDeFe);
inc(ofs,SizePrDeFe);
SieveOneSieve(PrimeDecompField,Ofs);
end;
until n > MAX;
writeln(n-1:10,' tested found ',count:10,' ratio ',count/n:10:7);
MAX += DELTAMAX;
until MAX>10*DELTAMAX;
writeln('runtime ',(GetTickCount64-T0)/1000:8:3,' s');
writeln;
writeln('Count of recursion 59,641,327 for 8,996,229,720');
n := 8996229720;
Init_Sieve(n);
T0 := GetTickCount64;
Out_One_Sol(PrimeDecompField[i],n,true);
writeln;
writeln('runtime ',(GetTickCount64-T0)/1000:8:3,' s');
END.
|
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #EchoLisp | EchoLisp |
;; to save space and time, we do'nt stringify Ω = 5^4^3^2 ,
;; but directly extract tail and head and number of decimal digits
(lib 'bigint) ;; arbitrary size integers
(define e10000 (expt 10 10000)) ;; 10^10000
(define (last-n big (n 20))
(string-append "..." (number->string (modulo big (expt 10 n)))))
(define (first-n big (n 20))
(while (> big e10000)
(set! big (/ big e10000))) ;; cut 10000 digits at a time
(string-append (take (number->string big) n) "..."))
;; faster than directly using (number-length big)
(define (digits big (digits 0))
(while (> big e10000)
(set! big (/ big e10000))
(set! digits (1+ digits)))
(+ (* digits 10000) (number-length big)))
(define Ω (expt 5 (expt 4 (expt 3 2))))
(last-n Ω )
→ "...92256259918212890625"
(first-n Ω )
→ "62060698786608744707..."
(digits Ω )
→ 183231
|
http://rosettacode.org/wiki/Zhang-Suen_thinning_algorithm | Zhang-Suen thinning algorithm | This is an algorithm used to thin a black and white i.e. one bit per pixel images.
For example, with an input image of:
################# #############
################## ################
################### ##################
######## ####### ###################
###### ####### ####### ######
###### ####### #######
################# #######
################ #######
################# #######
###### ####### #######
###### ####### #######
###### ####### ####### ######
######## ####### ###################
######## ####### ###### ################## ######
######## ####### ###### ################ ######
######## ####### ###### ############# ######
It produces the thinned output:
# ########## #######
## # #### #
# # ##
# # #
# # #
# # #
############ #
# # #
# # #
# # #
# # #
# ##
# ############
### ###
Algorithm
Assume black pixels are one and white pixels zero, and that the input image is a rectangular N by M array of ones and zeroes.
The algorithm operates on all black pixels P1 that can have eight neighbours.
The neighbours are, in order, arranged as:
P9 P2 P3
P8 P1 P4
P7 P6 P5
Obviously the boundary pixels of the image cannot have the full eight neighbours.
Define
A
(
P
1
)
{\displaystyle A(P1)}
= the number of transitions from white to black, (0 -> 1) in the sequence P2,P3,P4,P5,P6,P7,P8,P9,P2. (Note the extra P2 at the end - it is circular).
Define
B
(
P
1
)
{\displaystyle B(P1)}
= The number of black pixel neighbours of P1. ( = sum(P2 .. P9) )
Step 1
All pixels are tested and pixels satisfying all the following conditions (simultaneously) are just noted at this stage.
(0) The pixel is black and has eight neighbours
(1)
2
<=
B
(
P
1
)
<=
6
{\displaystyle 2<=B(P1)<=6}
(2) A(P1) = 1
(3) At least one of P2 and P4 and P6 is white
(4) At least one of P4 and P6 and P8 is white
After iterating over the image and collecting all the pixels satisfying all step 1 conditions, all these condition satisfying pixels are set to white.
Step 2
All pixels are again tested and pixels satisfying all the following conditions are just noted at this stage.
(0) The pixel is black and has eight neighbours
(1)
2
<=
B
(
P
1
)
<=
6
{\displaystyle 2<=B(P1)<=6}
(2) A(P1) = 1
(3) At least one of P2 and P4 and P8 is white
(4) At least one of P2 and P6 and P8 is white
After iterating over the image and collecting all the pixels satisfying all step 2 conditions, all these condition satisfying pixels are again set to white.
Iteration
If any pixels were set in this round of either step 1 or step 2 then all steps are repeated until no image pixels are so changed.
Task
Write a routine to perform Zhang-Suen thinning on an image matrix of ones and zeroes.
Use the routine to thin the following image and show the output here on this page as either a matrix of ones and zeroes, an image, or an ASCII-art image of space/non-space characters.
00000000000000000000000000000000
01111111110000000111111110000000
01110001111000001111001111000000
01110000111000001110000111000000
01110001111000001110000000000000
01111111110000001110000000000000
01110111100000001110000111000000
01110011110011101111001111011100
01110001111011100111111110011100
00000000000000000000000000000000
Reference
Zhang-Suen Thinning Algorithm, Java Implementation by Nayef Reza.
"Character Recognition Systems: A Guide for Students and Practitioners" By Mohamed Cheriet, Nawwaf Kharma, Cheng-Lin Liu, Ching Suen
| #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | nB[mat_] := Delete[mat // Flatten, 5] // Total;
nA[mat_] := Module[{l},
l = Flatten[mat][[{2, 3, 6, 9, 8, 7, 4, 1, 2}]];
Total[Map[If[#[[1]] == 0 && #[[2]] == 1, 1, 0] &,
Partition[l, 2, 1]]]
];
iW1[mat_] :=
Module[{l = Flatten[mat]},
If[Apply[Times, l[[{2, 6, 8}]]] + Apply[Times, l[[{4, 6, 8}]]] ==
0, 0, 1]];
iW2[mat_] :=
Module[{l = Flatten[mat]},
If[Apply[Times, l[[{2, 6, 4}]]] + Apply[Times, l[[{4, 2, 8}]]] ==
0, 0, 1]];
check[i_, j_, dat_, t_] := Module[{mat, d = Dimensions[dat], r, c},
r = d[[1]];
c = d[[2]];
If[i > 1 && i < r && j > 1 && j < c,
mat = dat[[i - 1 ;; i + 1, j - 1 ;; j + 1]];
If[dat[[i, j]] == 1 && nA[mat] == 1 && 2 <= nB[mat] <= 6 &&
If[t == 1, iW1[mat], iW2[mat]] == 0, 0, dat[[i, j]]],
dat[[i, j]]
]];
iter[dat_] :=
Module[{i =
Flatten[Outer[List, Range[Dimensions[dat][[1]]],
Range[Dimensions[dat][[2]]]], 1], tmp},
tmp = Partition[check[#[[1]], #[[2]], dat, 1] & /@ i,
Dimensions[dat][[2]]];
Partition[check[#[[1]], #[[2]], tmp, 2] & /@ i,
Dimensions[tmp][[2]]]];
FixedPoint[iter, dat] |
http://rosettacode.org/wiki/Zhang-Suen_thinning_algorithm | Zhang-Suen thinning algorithm | This is an algorithm used to thin a black and white i.e. one bit per pixel images.
For example, with an input image of:
################# #############
################## ################
################### ##################
######## ####### ###################
###### ####### ####### ######
###### ####### #######
################# #######
################ #######
################# #######
###### ####### #######
###### ####### #######
###### ####### ####### ######
######## ####### ###################
######## ####### ###### ################## ######
######## ####### ###### ################ ######
######## ####### ###### ############# ######
It produces the thinned output:
# ########## #######
## # #### #
# # ##
# # #
# # #
# # #
############ #
# # #
# # #
# # #
# # #
# ##
# ############
### ###
Algorithm
Assume black pixels are one and white pixels zero, and that the input image is a rectangular N by M array of ones and zeroes.
The algorithm operates on all black pixels P1 that can have eight neighbours.
The neighbours are, in order, arranged as:
P9 P2 P3
P8 P1 P4
P7 P6 P5
Obviously the boundary pixels of the image cannot have the full eight neighbours.
Define
A
(
P
1
)
{\displaystyle A(P1)}
= the number of transitions from white to black, (0 -> 1) in the sequence P2,P3,P4,P5,P6,P7,P8,P9,P2. (Note the extra P2 at the end - it is circular).
Define
B
(
P
1
)
{\displaystyle B(P1)}
= The number of black pixel neighbours of P1. ( = sum(P2 .. P9) )
Step 1
All pixels are tested and pixels satisfying all the following conditions (simultaneously) are just noted at this stage.
(0) The pixel is black and has eight neighbours
(1)
2
<=
B
(
P
1
)
<=
6
{\displaystyle 2<=B(P1)<=6}
(2) A(P1) = 1
(3) At least one of P2 and P4 and P6 is white
(4) At least one of P4 and P6 and P8 is white
After iterating over the image and collecting all the pixels satisfying all step 1 conditions, all these condition satisfying pixels are set to white.
Step 2
All pixels are again tested and pixels satisfying all the following conditions are just noted at this stage.
(0) The pixel is black and has eight neighbours
(1)
2
<=
B
(
P
1
)
<=
6
{\displaystyle 2<=B(P1)<=6}
(2) A(P1) = 1
(3) At least one of P2 and P4 and P8 is white
(4) At least one of P2 and P6 and P8 is white
After iterating over the image and collecting all the pixels satisfying all step 2 conditions, all these condition satisfying pixels are again set to white.
Iteration
If any pixels were set in this round of either step 1 or step 2 then all steps are repeated until no image pixels are so changed.
Task
Write a routine to perform Zhang-Suen thinning on an image matrix of ones and zeroes.
Use the routine to thin the following image and show the output here on this page as either a matrix of ones and zeroes, an image, or an ASCII-art image of space/non-space characters.
00000000000000000000000000000000
01111111110000000111111110000000
01110001111000001111001111000000
01110000111000001110000111000000
01110001111000001110000000000000
01111111110000001110000000000000
01110111100000001110000111000000
01110011110011101111001111011100
01110001111011100111111110011100
00000000000000000000000000000000
Reference
Zhang-Suen Thinning Algorithm, Java Implementation by Nayef Reza.
"Character Recognition Systems: A Guide for Students and Practitioners" By Mohamed Cheriet, Nawwaf Kharma, Cheng-Lin Liu, Ching Suen
| #Nim | Nim | import math, sequtils, strutils
type
Bit = 0..1
BitMatrix = seq[seq[Bit]] # Two-dimensional array of 0/1.
Neighbors = array[2..9, Bit] # Neighbor values.
const Symbols = [Bit(0): '.', Bit(1): '#']
func toBitMatrix(s: openArray[string]): BitMatrix =
## Convert an array of 01 strings into a BitMatrix.
for row in s:
assert row.allCharsInSet({'0', '1'})
result.add row.mapIt(Bit(ord(it) - ord('0')))
proc `$`(m: BitMatrix): string =
## Return the string representation of a BitMatrix.
for row in m:
echo row.mapIt(Symbols[it]).join()
# Templates to allow using double indexing.
template `[]`(m: BitMatrix; i, j: Natural): Bit = m[i][j]
template `[]=`(m: var BitMatrix; i, j: Natural; val: Bit) = m[i][j] = val
func neighbors(m: BitMatrix; i, j: int): Neighbors =
## Return the array of neighbors.
[m[i-1, j], m[i-1, j+1], m[i, j+1], m[i+1, j+1],
m[i+1, j], m[i+1, j-1], m[i, j-1], m[i-1, j-1]]
func transitions(p: Neighbors): int =
## Return the numbers of transitions from P2 to P9.
for (i, j) in [(2, 3), (3, 4), (4, 5), (5, 6),
(6, 7), (7, 8), (8, 9), (9, 2)]:
result += ord(p[i] == 0 and p[j] == 1)
func thinned(m: BitMatrix): BitMatrix =
## Return a thinned version of "m".
const Pair1 = [2, 8]
const Pair2 = [4, 6]
let rowMax = m.high
let colMax = m[0].high
result = m
while true:
var changed = false
for step in 1..2:
let (p1, p2) = if step == 1: (Pair1, Pair2) else: (Pair2, Pair1)
var m = result
for i in 1..<rowMax:
for j in 1..<colMax:
# Check criteria.
if m[i, j] == 0: # criterion 0.
continue
let p = m.neighbors(i, j)
if sum(p) notin 2..6: # criterion 1.
continue
if transitions(p) != 1: # criterion 2.
continue
if p[p1[0]] + p[p2[0]] + p[p2[1]] == 3 or # criterion 3.
p[p1[1]] + p[p2[0]] + p[p2[1]] == 3: # criterion 4.
continue
# All criteria satisfied. Store a 0 in "result".
result[i, j] = 0
changed = true
if not changed: break
when isMainModule:
const Input = ["00000000000000000000000000000000",
"01111111110000000111111110000000",
"01110001111000001111001111000000",
"01110000111000001110000111000000",
"01110001111000001110000000000000",
"01111111110000001110000000000000",
"01110111100000001110000111000000",
"01110011110011101111001111011100",
"01110001111011100111111110011100",
"00000000000000000000000000000000"]
let input = Input.toBitMatrix()
let output = input.thinned()
echo "Input image:"
echo input
echo()
echo "Output image:"
echo output |
http://rosettacode.org/wiki/Zeckendorf_number_representation | Zeckendorf number representation | Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.
Recall that the first six distinct Fibonacci numbers are: 1, 2, 3, 5, 8, 13.
The decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.
10100 is not the only way to make 11 from the Fibonacci numbers however; 0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that no two consecutive Fibonacci numbers can be used which leads to the former unique solution.
Task
Generate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.
The intention in this task to find the Zeckendorf form of an arbitrary integer. The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task.
Also see
OEIS A014417 for the the sequence of required results.
Brown's Criterion - Numberphile
Related task
Fibonacci sequence
| #Factor | Factor | USING: formatting kernel locals make math sequences ;
:: fib<= ( n -- seq )
1 2 [ [ dup n <= ] [ 2dup + [ , ] 2dip ] while drop , ]
{ } make ;
:: zeck ( n -- str )
0 :> s! n fib<= <reversed>
[ dup s + n <= [ s + s! 49 ] [ drop 48 ] if ] "" map-as ;
21 <iota> [ dup zeck "%2d: %6s\n" printf ] each |
http://rosettacode.org/wiki/Zeckendorf_number_representation | Zeckendorf number representation | Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.
Recall that the first six distinct Fibonacci numbers are: 1, 2, 3, 5, 8, 13.
The decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.
10100 is not the only way to make 11 from the Fibonacci numbers however; 0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that no two consecutive Fibonacci numbers can be used which leads to the former unique solution.
Task
Generate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.
The intention in this task to find the Zeckendorf form of an arbitrary integer. The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task.
Also see
OEIS A014417 for the the sequence of required results.
Brown's Criterion - Numberphile
Related task
Fibonacci sequence
| #Forth | Forth | : fib<= ( n -- n )
>r 0 1 BEGIN dup r@ <= WHILE tuck + REPEAT drop rdrop ;
: z. ( n -- )
dup fib<= dup . -
BEGIN ?dup WHILE
dup fib<= dup [char] + emit space . -
REPEAT ;
: tab 9 emit ;
: zeckendorf ( -- )
21 0 DO
cr i 2 .r tab i z.
LOOP ; |
http://rosettacode.org/wiki/100_doors | 100 doors | There are 100 doors in a row that are all initially closed.
You make 100 passes by the doors.
The first time through, visit every door and toggle the door (if the door is closed, open it; if it is open, close it).
The second time, only visit every 2nd door (door #2, #4, #6, ...), and toggle it.
The third time, visit every 3rd door (door #3, #6, #9, ...), etc, until you only visit the 100th door.
Task
Answer the question: what state are the doors in after the last pass? Which are open, which are closed?
Alternate:
As noted in this page's discussion page, the only doors that remain open are those whose numbers are perfect squares.
Opening only those doors is an optimization that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.
| #C.2B.2B | C++ | #include <iostream>
int main()
{
bool is_open[100] = { false };
// do the 100 passes
for (int pass = 0; pass < 100; ++pass)
for (int door = pass; door < 100; door += pass+1)
is_open[door] = !is_open[door];
// output the result
for (int door = 0; door < 100; ++door)
std::cout << "door #" << door+1 << (is_open[door]? " is open." : " is closed.") << std::endl;
return 0;
} |
http://rosettacode.org/wiki/Arrays | Arrays | This task is about arrays.
For hashes or associative arrays, please see Creating an Associative Array.
For a definition and in-depth discussion of what an array is, see Array.
Task
Show basic array syntax in your language.
Basically, create an array, assign a value to it, and retrieve an element (if available, show both fixed-length arrays and
dynamic arrays, pushing a value into it).
Please discuss at Village Pump: Arrays.
Please merge code in from these obsolete tasks:
Creating an Array
Assigning Values to an Array
Retrieving an Element of an Array
Related tasks
Collections
Creating an Associative Array
Two-dimensional array (runtime)
| #ERRE | ERRE | DIM A%[100] ! integer array
DIM S$[50] ! string array
DIM R[50] ! real array
DIM R#[70] ! long real array
|
http://rosettacode.org/wiki/Arithmetic/Complex | Arithmetic/Complex | A complex number is a number which can be written as:
a
+
b
×
i
{\displaystyle a+b\times i}
(sometimes shown as:
b
+
a
×
i
{\displaystyle b+a\times i}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are real numbers, and
i
{\displaystyle i}
is √ -1
Typically, complex numbers are represented as a pair of real numbers called the "imaginary part" and "real part", where the imaginary part is the number to be multiplied by
i
{\displaystyle i}
.
Task
Show addition, multiplication, negation, and inversion of complex numbers in separate functions. (Subtraction and division operations can be made with pairs of these operations.)
Print the results for each operation tested.
Optional: Show complex conjugation.
By definition, the complex conjugate of
a
+
b
i
{\displaystyle a+bi}
is
a
−
b
i
{\displaystyle a-bi}
Some languages have complex number libraries available. If your language does, show the operations. If your language does not, also show the definition of this type.
| #Pop11 | Pop11 | lvars a = 1.0 +: 1.0, b = 2.0 +: 5.0 ;
a+b =>
a*b =>
1/a =>
a-b =>
a-a =>
a/b =>
a/a =>
;;; The same, but using exact values
1 +: 1 -> a;
2 +: 5 -> b;
a+b =>
a*b =>
1/a =>
a-b =>
a-a =>
a/b =>
a/a => |
http://rosettacode.org/wiki/Arithmetic/Rational | Arithmetic/Rational | Task
Create a reasonably complete implementation of rational arithmetic in the particular language using the idioms of the language.
Example
Define a new type called frac with binary operator "//" of two integers that returns a structure made up of the numerator and the denominator (as per a rational number).
Further define the appropriate rational unary operators abs and '-', with the binary operators for addition '+', subtraction '-', multiplication '×', division '/', integer division '÷', modulo division, the comparison operators (e.g. '<', '≤', '>', & '≥') and equality operators (e.g. '=' & '≠').
Define standard coercion operators for casting int to frac etc.
If space allows, define standard increment and decrement operators (e.g. '+:=' & '-:=' etc.).
Finally test the operators:
Use the new type frac to find all perfect numbers less than 219 by summing the reciprocal of the factors.
Related task
Perfect Numbers
| #Ruby | Ruby |
for candidate in 2 .. 2**19
sum = Rational(1, candidate)
for factor in 2 .. Integer.sqrt(candidate)
if candidate % factor == 0
sum += Rational(1, factor) + Rational(1, candidate / factor)
end
end
if sum.denominator == 1
puts "Sum of recipr. factors of %d = %d exactly %s" %
[candidate, sum.to_i, sum == 1 ? "perfect!" : ""]
end
end |
http://rosettacode.org/wiki/Arithmetic-geometric_mean | Arithmetic-geometric mean |
This page uses content from Wikipedia. The original article was at Arithmetic-geometric mean. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Write a function to compute the arithmetic-geometric mean of two numbers.
The arithmetic-geometric mean of two numbers can be (usefully) denoted as
a
g
m
(
a
,
g
)
{\displaystyle \mathrm {agm} (a,g)}
, and is equal to the limit of the sequence:
a
0
=
a
;
g
0
=
g
{\displaystyle a_{0}=a;\qquad g_{0}=g}
a
n
+
1
=
1
2
(
a
n
+
g
n
)
;
g
n
+
1
=
a
n
g
n
.
{\displaystyle a_{n+1}={\tfrac {1}{2}}(a_{n}+g_{n});\quad g_{n+1}={\sqrt {a_{n}g_{n}}}.}
Since the limit of
a
n
−
g
n
{\displaystyle a_{n}-g_{n}}
tends (rapidly) to zero with iterations, this is an efficient method.
Demonstrate the function by calculating:
a
g
m
(
1
,
1
/
2
)
{\displaystyle \mathrm {agm} (1,1/{\sqrt {2}})}
Also see
mathworld.wolfram.com/Arithmetic-Geometric Mean
| #UNIX_Shell | UNIX Shell | function agm {
float a=$1 g=$2 eps=${3:-1e-11} tmp
while (( abs(a-g) > eps )); do
print "debug: a=$a\tg=$g"
tmp=$(( (a+g)/2.0 ))
g=$(( sqrt(a*g) ))
a=$tmp
done
echo $a
}
agm $((1/sqrt(2))) 1 |
http://rosettacode.org/wiki/Arithmetic-geometric_mean | Arithmetic-geometric mean |
This page uses content from Wikipedia. The original article was at Arithmetic-geometric mean. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Write a function to compute the arithmetic-geometric mean of two numbers.
The arithmetic-geometric mean of two numbers can be (usefully) denoted as
a
g
m
(
a
,
g
)
{\displaystyle \mathrm {agm} (a,g)}
, and is equal to the limit of the sequence:
a
0
=
a
;
g
0
=
g
{\displaystyle a_{0}=a;\qquad g_{0}=g}
a
n
+
1
=
1
2
(
a
n
+
g
n
)
;
g
n
+
1
=
a
n
g
n
.
{\displaystyle a_{n+1}={\tfrac {1}{2}}(a_{n}+g_{n});\quad g_{n+1}={\sqrt {a_{n}g_{n}}}.}
Since the limit of
a
n
−
g
n
{\displaystyle a_{n}-g_{n}}
tends (rapidly) to zero with iterations, this is an efficient method.
Demonstrate the function by calculating:
a
g
m
(
1
,
1
/
2
)
{\displaystyle \mathrm {agm} (1,1/{\sqrt {2}})}
Also see
mathworld.wolfram.com/Arithmetic-Geometric Mean
| #VBA | VBA | Private Function agm(a As Double, g As Double, Optional tolerance As Double = 0.000000000000001) As Double
Do While Abs(a - g) > tolerance
tmp = a
a = (a + g) / 2
g = Sqr(tmp * g)
Debug.Print a
Loop
agm = a
End Function
Public Sub main()
Debug.Print agm(1, 1 / Sqr(2))
End Sub |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #J | J | 0 ^ 0
1 |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Java | Java | System.out.println(Math.pow(0, 0)); |
http://rosettacode.org/wiki/Zig-zag_matrix | Zig-zag matrix | Task
Produce a zig-zag array.
A zig-zag array is a square arrangement of the first N2 natural numbers, where the
numbers increase sequentially as you zig-zag along the array's anti-diagonals.
For a graphical representation, see JPG zigzag (JPG uses such arrays to encode images).
For example, given 5, produce this array:
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24
Related tasks
Spiral matrix
Identity matrix
Ulam spiral (for primes)
See also
Wiktionary entry: anti-diagonals
| #APL | APL | zz ← {⍵⍴⎕IO-⍨⍋⊃,/{(2|⍴⍵):⌽⍵⋄⍵}¨(⊂w)/¨⍨w{↓⍵∘.=⍨∪⍵}+/[1]⍵⊤w←⎕IO-⍨⍳×/⍵} ⍝ General zigzag (any rectangle)
zzSq ← {zz,⍨⍵} ⍝ Square zigzag
zzSq 5
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24 |
http://rosettacode.org/wiki/Yellowstone_sequence | Yellowstone sequence | The Yellowstone sequence, also called the Yellowstone permutation, is defined as:
For n <= 3,
a(n) = n
For n >= 4,
a(n) = the smallest number not already in sequence such that a(n) is relatively prime to a(n-1) and
is not relatively prime to a(n-2).
The sequence is a permutation of the natural numbers, and gets its name from what its authors felt was a spiking, geyser like appearance of a plot of the sequence.
Example
a(4) is 4 because 4 is the smallest number following 1, 2, 3 in the sequence that is relatively prime to the entry before it (3), and is not relatively prime to the number two entries before it (2).
Task
Find and show as output the first 30 Yellowstone numbers.
Extra
Demonstrate how to plot, with x = n and y coordinate a(n), the first 100 Yellowstone numbers.
Related tasks
Greatest common divisor.
Plot coordinate pairs.
See also
The OEIS entry: A098550 The Yellowstone permutation.
Applegate et al, 2015: The Yellowstone Permutation [1].
| #Delphi | Delphi |
program Yellowstone_sequence;
{$APPTYPE CONSOLE}
uses
System.SysUtils,
Boost.Generics.Collection,
Boost.Process;
function gdc(x, y: Integer): Integer;
begin
while y <> 0 do
begin
var tmp := x;
x := y;
y := tmp mod y;
end;
Result := x;
end;
function Yellowstone(n: Integer): TArray<Integer>;
var
m: TDictionary<Integer, Boolean>;
a: TArray<Integer>;
begin
m.Init;
SetLength(a, n + 1);
for var i := 1 to 3 do
begin
a[i] := i;
m[i] := True;
end;
var min := 4;
for var c := 4 to n do
begin
var i := min;
repeat
if not m[i, false] and (gdc(a[c - 1], i) = 1) and (gdc(a[c - 2], i) > 1) then
begin
a[c] := i;
m[i] := true;
if i = min then
inc(min);
Break;
end;
inc(i);
until false;
end;
Result := copy(a, 1, length(a));
end;
begin
var x: TArray<Integer>;
SetLength(x, 100);
for var i in Range(100) do
x[i] := i + 1;
var y := yellowstone(High(x));
writeln('The first 30 Yellowstone numbers are:');
for var i := 0 to 29 do
Write(y[i], ' ');
Writeln;
//Plotting
var plot := TPipe.Create('gnuplot -p', True);
plot.WritelnA('unset key; plot ''-''');
for var i := 0 to High(x) do
plot.WriteA('%d %d'#10, [x[i], y[i]]);
plot.WritelnA('e');
writeln('Press enter to close');
Readln;
plot.Kill;
plot.Free;
end. |
http://rosettacode.org/wiki/Yahoo!_search_interface | Yahoo! search interface | Create a class for searching Yahoo! results.
It must implement a Next Page method, and read URL, Title and Content from results.
| #AutoHotkey | AutoHotkey | test:
yahooSearch("test", 1)
yahooSearch("test", 2)
return
yahooSearch(query, page)
{
global
start := ((page - 1) * 10) + 1
filedelete, search.txt
urldownloadtofile, % "http://search.yahoo.com/search?p=" . query
. "&b=" . start, search.txt
fileread, content, search.txt
reg = <a class="yschttl spt" href=".+?" >(.+?)</a></h3></div><div class="abstr">(.+?)</div><span class=url>(.+?)</span>
index := found := 1
while (found := regexmatch(content, reg, self, found + 1))
{
msgbox % title%A_Index% := fix(self1)
content%A_Index% := fix(self2)
url%A_Index% := fix(self3)
}
}
fix(url)
{
if pos := instr(url, "</a></h3></div>")
StringLeft, url, url, pos - 1
url := regexreplace(url, "<.*?>")
return url
} |
http://rosettacode.org/wiki/Arithmetic_evaluation | Arithmetic evaluation | Create a program which parses and evaluates arithmetic expressions.
Requirements
An abstract-syntax tree (AST) for the expression must be created from parsing the input.
The AST must be used in evaluation, also, so the input may not be directly evaluated (e.g. by calling eval or a similar language feature.)
The expression will be a string or list of symbols like "(1+3)*7".
The four symbols + - * / must be supported as binary operators with conventional precedence rules.
Precedence-control parentheses must also be supported.
Note
For those who don't remember, mathematical precedence is as follows:
Parentheses
Multiplication/Division (left to right)
Addition/Subtraction (left to right)
C.f
24 game Player.
Parsing/RPN calculator algorithm.
Parsing/RPN to infix conversion.
| #Scala | Scala |
package org.rosetta.arithmetic_evaluator.scala
object ArithmeticParser extends scala.util.parsing.combinator.RegexParsers {
def readExpression(input: String) : Option[()=>Int] = {
parseAll(expr, input) match {
case Success(result, _) =>
Some(result)
case other =>
println(other)
None
}
}
private def expr : Parser[()=>Int] = {
(term<~"+")~expr ^^ { case l~r => () => l() + r() } |
(term<~"-")~expr ^^ { case l~r => () => l() - r() } |
term
}
private def term : Parser[()=>Int] = {
(factor<~"*")~term ^^ { case l~r => () => l() * r() } |
(factor<~"/")~term ^^ { case l~r => () => l() / r() } |
factor
}
private def factor : Parser[()=>Int] = {
"("~>expr<~")" |
"\\d+".r ^^ { x => () => x.toInt } |
failure("Expected a value")
}
}
object Main {
def main(args: Array[String]) {
println("""Please input the expressions. Type "q" to quit.""")
var input: String = ""
do {
input = readLine("> ")
if (input != "q") {
ArithmeticParser.readExpression(input).foreach(f => println(f()))
}
} while (input != "q")
}
}
|
http://rosettacode.org/wiki/Zeckendorf_arithmetic | Zeckendorf arithmetic | This task is a total immersion zeckendorf task; using decimal numbers will attract serious disapprobation.
The task is to implement addition, subtraction, multiplication, and division using Zeckendorf number representation. Optionally provide decrement, increment and comparitive operation functions.
Addition
Like binary 1 + 1 = 10, note carry 1 left. There the similarity ends. 10 + 10 = 101, note carry 1 left and 1 right. 100 + 100 = 1001, note carry 1 left and 2 right, this is the general case.
Occurrences of 11 must be changed to 100. Occurrences of 111 may be changed from the right by replacing 11 with 100, or from the left converting 111 to 100 + 100;
Subtraction
10 - 1 = 1. The general rule is borrow 1 right carry 1 left. eg:
abcde
10100 -
1000
_____
100 borrow 1 from a leaves 100
+ 100 add the carry
_____
1001
A larger example:
abcdef
100100 -
1000
______
1*0100 borrow 1 from b
+ 100 add the carry
______
1*1001
Sadly we borrowed 1 from b which didn't have it to lend. So now b borrows from a:
1001
+ 1000 add the carry
____
10100
Multiplication
Here you teach your computer its zeckendorf tables. eg. 101 * 1001:
a = 1 * 101 = 101
b = 10 * 101 = a + a = 10000
c = 100 * 101 = b + a = 10101
d = 1000 * 101 = c + b = 101010
1001 = d + a therefore 101 * 1001 =
101010
+ 101
______
1000100
Division
Lets try 1000101 divided by 101, so we can use the same table used for multiplication.
1000101 -
101010 subtract d (1000 * 101)
_______
1000 -
101 b and c are too large to subtract, so subtract a
____
1 so 1000101 divided by 101 is d + a (1001) remainder 1
Efficient algorithms for Zeckendorf arithmetic is interesting. The sections on addition and subtraction are particularly relevant for this task.
| #PicoLisp | PicoLisp | (seed (in "/dev/urandom" (rd 8)))
(de unpad (Lst)
(while (=0 (car Lst))
(pop 'Lst) )
Lst )
(de numz (N)
(let Fibs (1 1)
(while (>= N (+ (car Fibs) (cadr Fibs)))
(push 'Fibs (+ (car Fibs) (cadr Fibs))) )
(make
(for I (uniq Fibs)
(if (> I N)
(link 0)
(link 1)
(dec 'N I) ) ) ) ) )
(de znum (Lst)
(let Fibs (1 1)
(do (dec (length Lst))
(push 'Fibs (+ (car Fibs) (cadr Fibs))) )
(sum
'((X Y) (unless (=0 X) Y))
Lst
(uniq Fibs) ) ) )
(de incz (Lst)
(addz Lst (1)) )
(de decz (Lst)
(subz Lst (1)) )
(de addz (Lst1 Lst2)
(let Max (max (length Lst1) (length Lst2))
(reorg
(mapcar + (need Max Lst1 0) (need Max Lst2 0)) ) ) )
(de subz (Lst1 Lst2)
(use (@A @B)
(let
(Max (max (length Lst1) (length Lst2))
Lst (mapcar - (need Max Lst1 0) (need Max Lst2 0)) )
(loop
(while (match '(@A 1 0 0 @B) Lst)
(setq Lst (append @A (0 1 1) @B)) )
(while (match '(@A 1 -1 0 @B) Lst)
(setq Lst (append @A (0 0 1) @B)) )
(while (match '(@A 1 -1 1 @B) Lst)
(setq Lst (append @A (0 0 2) @B)) )
(while (match '(@A 1 0 -1 @B) Lst)
(setq Lst (append @A (0 1 0) @B)) )
(while (match '(@A 2 0 0 @B) Lst)
(setq Lst (append @A (1 1 1) @B)) )
(while (match '(@A 2 -1 0 @B) Lst)
(setq Lst (append @A (1 0 1) @B)) )
(while (match '(@A 2 -1 1 @B) Lst)
(setq Lst (append @A (1 0 2) @B)) )
(while (match '(@A 2 0 -1 @B) Lst)
(setq Lst (append @A (1 1 0) @B)) )
(while (match '(@A 1 -1) Lst)
(setq Lst (append @A (0 1))) )
(while (match '(@A 2 -1) Lst)
(setq Lst (append @A (1 1))) )
(NIL (match '(@A -1 @B) Lst)) )
(reorg (unpad Lst)) ) ) )
(de mulz (Lst1 Lst2)
(let (Sums (list Lst1) Mulz (0))
(mapc
'((X)
(when (= 1 (car X))
(setq Mulz (addz (cdr X) Mulz)) )
Mulz )
(mapcar
'((X)
(cons
X
(push 'Sums (addz (car Sums) (cadr Sums))) ) )
(reverse Lst2) ) ) ) )
(de divz (Lst1 Lst2)
(let Q 0
(while (lez Lst2 Lst1)
(setq Lst1 (subz Lst1 Lst2))
(setq Q (incz Q)) )
(list Q (or Lst1 (0))) ) )
(de reorg (Lst)
(use (@A @B)
(let Lst (reverse Lst)
(loop
(while (match '(@A 1 1 @B) Lst)
(if @B
(inc (nth @B 1))
(setq @B (1)) )
(setq Lst (append @A (0 0) @B) ) )
(while (match '(@A 2 @B) Lst)
(inc
(if (cdr @A)
(tail 2 @A)
@A ) )
(if @B
(inc (nth @B 1))
(setq @B (1)) )
(setq Lst (append @A (0) @B)) )
(NIL
(or
(match '(@A 1 1 @B) Lst)
(match '(@A 2 @B) Lst) ) ) )
(reverse Lst) ) ) )
(de lez (Lst1 Lst2)
(let Max (max (length Lst1) (length Lst2))
(<= (need Max Lst1 0) (need Max Lst2 0)) ) )
(let (X 0 Y 0)
(do 1024
(setq X (rand 1 1024))
(setq Y (rand 1 1024))
(test (numz (+ X Y)) (addz (numz X) (numz Y)))
(test (numz (* X Y)) (mulz (numz X) (numz Y)))
(test (numz (+ X 1)) (incz (numz X))) )
(do 1024
(setq X (rand 129 1024))
(setq Y (rand 1 128))
(test (numz (- X Y)) (subz (numz X) (numz Y)))
(test (numz (/ X Y)) (car (divz (numz X) (numz Y))))
(test (numz (% X Y)) (cadr (divz (numz X) (numz Y))))
(test (numz (- X 1)) (decz (numz X))) ) )
(bye) |
http://rosettacode.org/wiki/Zeckendorf_arithmetic | Zeckendorf arithmetic | This task is a total immersion zeckendorf task; using decimal numbers will attract serious disapprobation.
The task is to implement addition, subtraction, multiplication, and division using Zeckendorf number representation. Optionally provide decrement, increment and comparitive operation functions.
Addition
Like binary 1 + 1 = 10, note carry 1 left. There the similarity ends. 10 + 10 = 101, note carry 1 left and 1 right. 100 + 100 = 1001, note carry 1 left and 2 right, this is the general case.
Occurrences of 11 must be changed to 100. Occurrences of 111 may be changed from the right by replacing 11 with 100, or from the left converting 111 to 100 + 100;
Subtraction
10 - 1 = 1. The general rule is borrow 1 right carry 1 left. eg:
abcde
10100 -
1000
_____
100 borrow 1 from a leaves 100
+ 100 add the carry
_____
1001
A larger example:
abcdef
100100 -
1000
______
1*0100 borrow 1 from b
+ 100 add the carry
______
1*1001
Sadly we borrowed 1 from b which didn't have it to lend. So now b borrows from a:
1001
+ 1000 add the carry
____
10100
Multiplication
Here you teach your computer its zeckendorf tables. eg. 101 * 1001:
a = 1 * 101 = 101
b = 10 * 101 = a + a = 10000
c = 100 * 101 = b + a = 10101
d = 1000 * 101 = c + b = 101010
1001 = d + a therefore 101 * 1001 =
101010
+ 101
______
1000100
Division
Lets try 1000101 divided by 101, so we can use the same table used for multiplication.
1000101 -
101010 subtract d (1000 * 101)
_______
1000 -
101 b and c are too large to subtract, so subtract a
____
1 so 1000101 divided by 101 is d + a (1001) remainder 1
Efficient algorithms for Zeckendorf arithmetic is interesting. The sections on addition and subtraction are particularly relevant for this task.
| #Python | Python | import copy
class Zeckendorf:
def __init__(self, x='0'):
q = 1
i = len(x) - 1
self.dLen = int(i / 2)
self.dVal = 0
while i >= 0:
self.dVal = self.dVal + (ord(x[i]) - ord('0')) * q
q = q * 2
i = i -1
def a(self, n):
i = n
while True:
if self.dLen < i:
self.dLen = i
j = (self.dVal >> (i * 2)) & 3
if j == 0 or j == 1:
return
if j == 2:
if (self.dVal >> ((i + 1) * 2) & 1) != 1:
return
self.dVal = self.dVal + (1 << (i * 2 + 1))
return
if j == 3:
temp = 3 << (i * 2)
temp = temp ^ -1
self.dVal = self.dVal & temp
self.b((i + 1) * 2)
i = i + 1
def b(self, pos):
if pos == 0:
self.inc()
return
if (self.dVal >> pos) & 1 == 0:
self.dVal = self.dVal + (1 << pos)
self.a(int(pos / 2))
if pos > 1:
self.a(int(pos / 2) - 1)
else:
temp = 1 << pos
temp = temp ^ -1
self.dVal = self.dVal & temp
self.b(pos + 1)
self.b(pos - (2 if pos > 1 else 1))
def c(self, pos):
if (self.dVal >> pos) & 1 == 1:
temp = 1 << pos
temp = temp ^ -1
self.dVal = self.dVal & temp
return
self.c(pos + 1)
if pos > 0:
self.b(pos - 1)
else:
self.inc()
def inc(self):
self.dVal = self.dVal + 1
self.a(0)
def __add__(self, rhs):
copy = self
rhs_dVal = rhs.dVal
limit = (rhs.dLen + 1) * 2
for gn in range(0, limit):
if ((rhs_dVal >> gn) & 1) == 1:
copy.b(gn)
return copy
def __sub__(self, rhs):
copy = self
rhs_dVal = rhs.dVal
limit = (rhs.dLen + 1) * 2
for gn in range(0, limit):
if (rhs_dVal >> gn) & 1 == 1:
copy.c(gn)
while (((copy.dVal >> ((copy.dLen * 2) & 31)) & 3) == 0) or (copy.dLen == 0):
copy.dLen = copy.dLen - 1
return copy
def __mul__(self, rhs):
na = copy.deepcopy(rhs)
nb = copy.deepcopy(rhs)
nr = Zeckendorf()
dVal = self.dVal
for i in range(0, (self.dLen + 1) * 2):
if ((dVal >> i) & 1) > 0:
nr = nr + nb
nt = copy.deepcopy(nb)
nb = nb + na
na = copy.deepcopy(nt)
return nr
def __str__(self):
dig = ["00", "01", "10"]
dig1 = ["", "1", "10"]
if self.dVal == 0:
return '0'
idx = (self.dVal >> ((self.dLen * 2) & 31)) & 3
sb = dig1[idx]
i = self.dLen - 1
while i >= 0:
idx = (self.dVal >> (i * 2)) & 3
sb = sb + dig[idx]
i = i - 1
return sb
# main
print "Addition:"
g = Zeckendorf("10")
g = g + Zeckendorf("10")
print g
g = g + Zeckendorf("10")
print g
g = g + Zeckendorf("1001")
print g
g = g + Zeckendorf("1000")
print g
g = g + Zeckendorf("10101")
print g
print
print "Subtraction:"
g = Zeckendorf("1000")
g = g - Zeckendorf("101")
print g
g = Zeckendorf("10101010")
g = g - Zeckendorf("1010101")
print g
print
print "Multiplication:"
g = Zeckendorf("1001")
g = g * Zeckendorf("101")
print g
g = Zeckendorf("101010")
g = g + Zeckendorf("101")
print g |
http://rosettacode.org/wiki/Zumkeller_numbers | Zumkeller numbers | Zumkeller numbers are the set of numbers whose divisors can be partitioned into two disjoint sets that sum to the same value. Each sum must contain divisor values that are not in the other sum, and all of the divisors must be in one or the other. There are no restrictions on how the divisors are partitioned, only that the two partition sums are equal.
E.G.
6 is a Zumkeller number; The divisors {1 2 3 6} can be partitioned into two groups {1 2 3} and {6} that both sum to 6.
10 is not a Zumkeller number; The divisors {1 2 5 10} can not be partitioned into two groups in any way that will both sum to the same value.
12 is a Zumkeller number; The divisors {1 2 3 4 6 12} can be partitioned into two groups {1 3 4 6} and {2 12} that both sum to 14.
Even Zumkeller numbers are common; odd Zumkeller numbers are much less so. For values below 10^6, there is at least one Zumkeller number in every 12 consecutive integers, and the vast majority of them are even. The odd Zumkeller numbers are very similar to the list from the task Abundant odd numbers; they are nearly the same except for the further restriction that the abundance (A(n) = sigma(n) - 2n), must be even: A(n) mod 2 == 0
Task
Write a routine (function, procedure, whatever) to find Zumkeller numbers.
Use the routine to find and display here, on this page, the first 220 Zumkeller numbers.
Use the routine to find and display here, on this page, the first 40 odd Zumkeller numbers.
Optional, stretch goal: Use the routine to find and display here, on this page, the first 40 odd Zumkeller numbers that don't end with 5.
See Also
OEIS:A083207 - Zumkeller numbers to get an impression of different partitions OEIS:A083206 Zumkeller partitions
OEIS:A174865 - Odd Zumkeller numbers
Related Tasks
Abundant odd numbers
Abundant, deficient and perfect number classifications
Proper divisors , Factors of an integer | #Perl | Perl | use strict;
use warnings;
use feature 'say';
use ntheory <is_prime divisor_sum divisors vecsum forcomb lastfor>;
sub in_columns {
my($columns, $values) = @_;
my @v = split ' ', $values;
my $width = int(80/$columns);
printf "%${width}d"x$columns."\n", @v[$_*$columns .. -1+(1+$_)*$columns] for 0..-1+@v/$columns;
print "\n";
}
sub is_Zumkeller {
my($n) = @_;
return 0 if is_prime($n);
my @divisors = divisors($n);
return 0 unless @divisors > 2 && 0 == @divisors % 2;
my $sigma = divisor_sum($n);
return 0 unless 0 == $sigma%2 && ($sigma/2) >= $n;
if (1 == $n%2) {
return 1
} else {
my $Z = 0;
forcomb { $Z++, lastfor if vecsum(@divisors[@_]) == $sigma/2 } @divisors;
return $Z;
}
}
use constant Inf => 1e10;
say 'First 220 Zumkeller numbers:';
my $n = 0; my $z;
$z .= do { $n < 220 ? (is_Zumkeller($_) and ++$n and "$_ ") : last } for 1 .. Inf;
in_columns(20, $z);
say 'First 40 odd Zumkeller numbers:';
$n = 0; $z = '';
$z .= do { $n < 40 ? (!!($_%2) and is_Zumkeller($_) and ++$n and "$_ ") : last } for 1 .. Inf;
in_columns(10, $z);
say 'First 40 odd Zumkeller numbers not divisible by 5:';
$n = 0; $z = '';
$z .= do { $n < 40 ? (!!($_%2 and $_%5) and is_Zumkeller($_) and ++$n and "$_ ") : last } for 1 .. Inf;
in_columns(10, $z); |
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #Elixir | Elixir | defmodule Arbitrary do
def pow(_,0), do: 1
def pow(b,e) when e > 0, do: pow(b,e,1)
defp pow(b,1,acc), do: acc * b
defp pow(b,p,acc) when rem(p,2)==0, do: pow(b*b,div(p,2),acc)
defp pow(b,p,acc), do: pow(b,p-1,acc*b)
def test do
s = pow(5,pow(4,pow(3,2))) |> to_string
l = String.length(s)
prefix = String.slice(s,0,20)
suffix = String.slice(s,-20,20)
IO.puts "Length: #{l}\nPrefix:#{prefix}\nSuffix:#{suffix}"
end
end
Arbitrary.test |
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #Emacs_Lisp | Emacs Lisp | (let* ((integer-width (* 65536 16)) ; raise bignum limit from 65536 bits to avoid overflow error
(answer (number-to-string (expt 5 (expt 4 (expt 3 2)))))
(length (length answer)))
(message "%s has %d digits"
(if (> length 40)
(format "%s...%s"
(substring answer 0 20)
(substring answer (- length 20) length))
answer)
length)) |
http://rosettacode.org/wiki/Zhang-Suen_thinning_algorithm | Zhang-Suen thinning algorithm | This is an algorithm used to thin a black and white i.e. one bit per pixel images.
For example, with an input image of:
################# #############
################## ################
################### ##################
######## ####### ###################
###### ####### ####### ######
###### ####### #######
################# #######
################ #######
################# #######
###### ####### #######
###### ####### #######
###### ####### ####### ######
######## ####### ###################
######## ####### ###### ################## ######
######## ####### ###### ################ ######
######## ####### ###### ############# ######
It produces the thinned output:
# ########## #######
## # #### #
# # ##
# # #
# # #
# # #
############ #
# # #
# # #
# # #
# # #
# ##
# ############
### ###
Algorithm
Assume black pixels are one and white pixels zero, and that the input image is a rectangular N by M array of ones and zeroes.
The algorithm operates on all black pixels P1 that can have eight neighbours.
The neighbours are, in order, arranged as:
P9 P2 P3
P8 P1 P4
P7 P6 P5
Obviously the boundary pixels of the image cannot have the full eight neighbours.
Define
A
(
P
1
)
{\displaystyle A(P1)}
= the number of transitions from white to black, (0 -> 1) in the sequence P2,P3,P4,P5,P6,P7,P8,P9,P2. (Note the extra P2 at the end - it is circular).
Define
B
(
P
1
)
{\displaystyle B(P1)}
= The number of black pixel neighbours of P1. ( = sum(P2 .. P9) )
Step 1
All pixels are tested and pixels satisfying all the following conditions (simultaneously) are just noted at this stage.
(0) The pixel is black and has eight neighbours
(1)
2
<=
B
(
P
1
)
<=
6
{\displaystyle 2<=B(P1)<=6}
(2) A(P1) = 1
(3) At least one of P2 and P4 and P6 is white
(4) At least one of P4 and P6 and P8 is white
After iterating over the image and collecting all the pixels satisfying all step 1 conditions, all these condition satisfying pixels are set to white.
Step 2
All pixels are again tested and pixels satisfying all the following conditions are just noted at this stage.
(0) The pixel is black and has eight neighbours
(1)
2
<=
B
(
P
1
)
<=
6
{\displaystyle 2<=B(P1)<=6}
(2) A(P1) = 1
(3) At least one of P2 and P4 and P8 is white
(4) At least one of P2 and P6 and P8 is white
After iterating over the image and collecting all the pixels satisfying all step 2 conditions, all these condition satisfying pixels are again set to white.
Iteration
If any pixels were set in this round of either step 1 or step 2 then all steps are repeated until no image pixels are so changed.
Task
Write a routine to perform Zhang-Suen thinning on an image matrix of ones and zeroes.
Use the routine to thin the following image and show the output here on this page as either a matrix of ones and zeroes, an image, or an ASCII-art image of space/non-space characters.
00000000000000000000000000000000
01111111110000000111111110000000
01110001111000001111001111000000
01110000111000001110000111000000
01110001111000001110000000000000
01111111110000001110000000000000
01110111100000001110000111000000
01110011110011101111001111011100
01110001111011100111111110011100
00000000000000000000000000000000
Reference
Zhang-Suen Thinning Algorithm, Java Implementation by Nayef Reza.
"Character Recognition Systems: A Guide for Students and Practitioners" By Mohamed Cheriet, Nawwaf Kharma, Cheng-Lin Liu, Ching Suen
| #Perl | Perl | use List::Util qw(sum min);
$source = <<'END';
............................................................
..#################...................#############.........
..##################...............################.........
..###################............##################.........
..########.....#######..........###################.........
....######.....#######.........#######.......######.........
....######.....#######........#######.......................
....#################.........#######.......................
....################..........#######.......................
....#################.........#######.......................
....######.....#######........#######.......................
....######.....#######........#######.......................
....######.....#######.........#######.......######.........
..########.....#######..........###################.........
..########.....#######.######....##################.######..
..########.....#######.######......################.######..
..########.....#######.######.........#############.######..
............................................................
END
for $line (split "\n", $source) {
push @lines, [map { 1 & ord $_ } split '', $line]
}
$v = @lines;
$h = @{$lines[0]};
push @black, @$_ for @lines;
@p8 = ((-$h-1), (-$h+0), (-$h+1), # flatland distances to 8 neighbors.
0-1, 0+1,
$h-1, $h+0, $h+1)[1,2,4,7,6,5,3,0]; # (in cycle order)
# Candidates have 8 neighbors and are known black
@cand = grep { $black[$_] } map { my $x = $_; map $_*$h + $x, 1..$v-2 } 1..$h-2;
do {
sub seewhite {
my($w1,$w2) = @_;
my(@results);
sub cycles { my(@neighbors)=@_; my $c; $c += $neighbors[$_] < $neighbors[($_+1)%8] for 0..$#neighbors; return $c }
sub blacks { my(@neighbors)=@_; sum @neighbors }
@prior = @cand; @cand = ();
for $p (@prior) {
@n = @black[map { $_+$p } @p8];
if (cycles(@n) == 1 and 2 <= sum(blacks(@n)) and sum(blacks(@n)) <= 6 and min(@n[@$w1]) == 0 and min(@n[@$w2]) == 0) {
push @results, $p;
} else {
push @cand, $p
}
}
return @results;
}
@goners1 = seewhite [0,2,4], [2,4,6]; @black[@goners1] = 0 x @goners1;
@goners2 = seewhite [0,2,6], [0,4,6]; @black[@goners2] = 0 x @goners2;
} until @goners1 == 0 and @goners2 == 0;
while (@black) { push @thinned, join '', qw<. #>[splice(@black,0,$h)] }
print join "\n", @thinned; |
http://rosettacode.org/wiki/Zeckendorf_number_representation | Zeckendorf number representation | Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.
Recall that the first six distinct Fibonacci numbers are: 1, 2, 3, 5, 8, 13.
The decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.
10100 is not the only way to make 11 from the Fibonacci numbers however; 0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that no two consecutive Fibonacci numbers can be used which leads to the former unique solution.
Task
Generate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.
The intention in this task to find the Zeckendorf form of an arbitrary integer. The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task.
Also see
OEIS A014417 for the the sequence of required results.
Brown's Criterion - Numberphile
Related task
Fibonacci sequence
| #Fortran | Fortran | F(N) = ((1 + SQRT(5))**N - (1 - SQRT(5))**N)/(SQRT(5)*2**N) |
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