Cycle detection: Difference between revisions

 

;Task:

=={{header|11l}}==

{{trans|D}}

Int cycle_length

V hare = x0

print_result(x0, f, cycle_length, cycle_start)

 

{{out}}

<pre>

Cycle: [101, 2, 5, 26, 167, 95]

</pre>

 

=={{header|C}}==

{{trans|Modula-2}}

#include <stdlib.h>

 

 

return 0;

{{out}}

<pre>[3, 10, 101, 2, 5, 26, 167, 95, 101, 2, 5, 26, 167, 95, 101, 2, 5, 26, 167, 95, 101, 2, 5, 26, 167, 95, 101, 2, 5, 26, 167, 95, 101, 2, 5, 26, 167, 95, 101, 2, 5]

Cycle = [101, 2, 5, 26, 167, 95]</pre>

 

This solution uses generics, so may find cycles of any type of data, not just integers.

 

 

// First file: Cycles.cs

}

 

 

=={{header|D}}==

{{trans|Java}}

import std.stdio;

 

auto iterate(int start, int function(int) f) {

return only(start).chain(generate!(() => start=f(start)));

{{out}}

<pre>Cycle length: 6

=={{header|Elixir}}==

{{trans|Ruby}}

def find_cycle(x0, f) do

lambda = find_lambda(f, x0, f.(x0), 1, 1)

# Test the find_cycle function

{clength, cstart} = Cycle_detection.find_cycle(3, f)

 

{{out}}

=={{header|Factor}}==

This is a strict translation of the Python code from the Wikipedia article. Perhaps a more idiomatic version could be added in the future, although there is value in showing how Factor's lexical variables differ from most languages. A variable binding with <code>!</code> at the end is mutable, and subsequent uses of that name followed by <code>!</code> change the value of the variable to the value at the top of the data stack.

 

: cyclical-function ( n -- m ) dup * 1 + 255 mod ;

3 [ 20 [ dup , cyclical-function ] times ] { } make nip .

3 [ cyclical-function ] brent

{{out}}

<pre>

Cycle length: 6

Cycle start: 2

</pre>

 

===Brent's algorithm===

{{trans|Python}}

' compile with: fbc -s console

 

Print : Print "hit any key to end program"

Sleep

{{out}}

<pre> Brent's algorithm

===Tortoise and hare. Floyd's algorithm===

{{trans|Wikipedia}}

' compile with: fbc -s console

 

Print : Print "hit any key to end program"

Sleep

 

=={{header|Go}}==

 

Run it on the [https://play.golang.org/p/unOtxuwZfg go playground], or on [https://goplay.space/#S1pQZSuJci go play space].

package main

 

}

fmt.Println("Cycle:", a[µ:µ+λ])

 

{{out}}

Cycle start index: 2

Cycle: [101 2 5 26 167 95]</pre>

 

=={{header|Haskell}}==

Haskellers, being able to handle conceptually infinite structures, traditionally consider totality as an important issue. The following solution is total for total inputs (modulo totality of iterated function) and allows nontermination only if input is explicitly infinite.

 

 

findCycle :: Eq a => [a] -> Maybe ([a], Int, Int)

| pow == lam = loop (2*pow) 1 y ys

| otherwise = loop pow (1+lam) x ys

 

'''Examples'''

Nothing

</pre>

 

=={{header|J}}==

Brute force implementation:

 

r=. ~.@(,u@{:)^:_ y

n=. u{:r

(,(#r)-])r i. n

 

In other words: iterate until we stop getting new values, find the repeated value, and see how it fits into the sequence.

Example use:

 

 

(Note that 1 0 1 are the coefficients of the polynomial <code>1 + (0 * x) + (1 * x * x)</code> or, if you prefer <code>(1 * x ^ 0) + (0 * x ^ 1) + (1 * x ^ 2)</code> - it's easier and probably more efficient to just hand the coefficients to p. than it is to write out some other expression which produces the same result.)

=={{header|Java}}==

{{works with|Java|8}}

import static java.util.stream.IntStream.*;

 

.forEach(n -> System.out.printf("%s ", n));

}

 

<pre>Cycle length: 6

Cycle: 101 2 5 26 167 95</pre>

 

=={{header|Julia}}==

Following the Wikipedia article:

 

 

function floyd(f, x0)

x -> Iterators.take(x, λ) |>

collect

 

{{out}}

 

=={{header|Kotlin}}==

 

typealias IntToInt = (Int) -> Int

println("Start index = $mu")

println("Cycle = $cycle")

 

{{out}}

=={{header|Lua}}==

Fairly direct translation of the Wikipedia code, except that the sequence is stored in a table and passed back as a third return value.

local tortoise, hare, mu = x0, f(x0), 0

local cycleTab, power, lam = {tortoise, hare}, 1, 1

print("Sequence:", table.concat(sequence, " "))

print("Cycle length:", cycleLength)

{{out}}

<pre>Sequence: 3 10 101 2 5 26 167 95 101 2 5 26 167 95

Cycle length: 6

Start Index: 2</pre>

 

=={{header|Modula-2}}==

{{trans|Kotlin}}

FROM FormatString IMPORT FormatString;

FROM Terminal IMPORT WriteString,WriteLn,ReadChar;

 

ReadChar

 

=={{header|ooRexx}}==

x=3

list=x

End

Exit

{{out}}

<pre>

 

=={{header|Perl}}==

 

sub cyclical_function { ($_[0] * $_[0] + 1) % 255 }

print "Cycle length $l\n";

print "Cycle start index $s\n";

{{out}}

<pre>3 10 101 2 5 26 167 95 101 2 5 26 167 95 101 2 5 26 167 95

Cycle start index 2

101 2 5 26 167 95</pre>

 

=={{header|Phix}}==

Translation of the Wikipedia code, but using the more descriptive len and pos, instead of lambda and mu, and adding a limit.

{{out}}

<pre>

{101,2,5,26,167,95}

</pre>

 

=={{header|Python}}==

===Procedural===

Function from the Wikipedia article:

 

def brent(f, x0):

print("Cycle length: %d" % lam)

print("Cycle start index: %d" % mu)

{{out}}

<pre>

 

A modified version of the above where the first stage is restructured for clarity:

 

def brent_length(f, x0):

print("Cycle length: %d" % lam)

print("Cycle start index: %d" % mu)

{{out}}

<pre>Cycle length: 6

{{Trans|Haskell}}

{{Works with|Python|3.7}}

 

from itertools import (chain, cycle, islice)

# MAIN ---

if __name__ == '__main__':

{{Out}}

<pre>First cycle detected, if any:

 

Recursive ''until'':

def until(p):

'''The result of repeatedly applying f until p holds.

def go(f, x):

return x if p(x) else go(f, f(x))

 

''cycleLength'' refactored in terms of ''until'':

def cycleLength(xs):

'''Just the length of the first cycle found,

) if ys else Nothing()

else:

 

Iterative reimplementation of ''until'':

def until_(p):

'''The result of repeatedly applying f until p holds.

v = f(v)

return v

 

 

The Python no longer falls out of the tree at the sight of an ouroboros, and we can happily search for cycles in lists of several thousand items:

{{Works with|Python|3.7}}

 

from itertools import (chain, cycle, islice)

# MAIN ---

if __name__ == '__main__':

{{Out}}

<pre>First cycle detected, if any:

[1..10000] -> No cycle found

f(x) = (x*x + 1) modulo 255 -> [101,2,5,26,167,95] (from:2, length:6)</pre>

 

=={{header|Racket}}==

 

I feel a bit bad about overloading λ, but it&rsquo;s in the spirit of the algorithm.

#lang racket/base

 

(let-values (([µ λ] (brent f 3)))

(printf "Cycle length = ~a~%Start Index = ~a~%" µ λ)))

 

{{out}}

Start Index = 2

</pre>

 

=={{header|REXX}}==

===Brent's algorithm===

end /*until*/

call Brent init /*invoke Brent algorithm for F*/

parse var result cycle idx /*get 2 values returned from F*/

say ' cycle length =' cycle /*display the cycle to term*/

say ' start index =' idx " ◄─── zero index" /* " " index " " */

exit /*stick a fork in it, we're all done. */

/*──────────────────────────────────────────────────────────────────────────────────────*/

end

hare= f(hare)

end /*while*/

hare= x0

return # mu

/*──────────────────────────────────────────────────────────────────────────────────────*/

<pre>

original list= 3 10 101 2 5 26 167 95 101 2 5 26 167 95 101 2 5 26 167 95 101 2 5 26 167 95 101 ...

 

===sequential search algorithm===

say ' start index =' cycle - 1 " ◄─── zero based" /* " " index " " */

exit /*stick a fork in it, we're all done. */

/*──────────────────────────────────────────────────────────────────────────────────────*/

{{out|output|:}}

<pre>

===hash table algorithm===

This REXX version is a lot faster &nbsp; (than the sequential search algorithm) &nbsp; if the &nbsp; ''cycle length'' &nbsp; and/or &nbsp; ''start index'' &nbsp; is large.

exit /*stick a fork in it, we're all done. */

/*──────────────────────────────────────────────────────────────────────────────────────*/

{{out|output|text=&nbsp; is identical to the 2^nd REXX version.}} <br><br>

 

<br>test the hash table algorithm. &nbsp; A divisor which is &nbsp; <big> ''two raised to the 49^th power'' </big> &nbsp; was chosen; &nbsp; it

<br>generates a cyclic sequence that contains over 1.5 million numbers.

parse arg power . /*obtain optional args from C.L. */

if power=='' | power="," then power=8 /*Not specified? Use the default*/

say ' divisor =' "{2**"power'}-1 = ' divisor /* " " divisor. " " " */

say

 

if !.x\==. then leave /*is this number a repeat? Leave*/

!.x= n

say 'numbers in list=' n /*display number of numbers. */

say ' cycle length =' n - !.x /*display the cycle to the term. */

say ' start index =' !.x - 1 " ◄─── zero based" /*show the index.*/

exit /*stick a fork in it, we're all done. */

/*──────────────────────────────────────────────────────────────────────────────────────*/

{{out|output|text= &nbsp; when the input (power of two) used is: &nbsp; &nbsp; <tt> 49 </tt>}}

<pre>

There is more information in the "hash table"<br>

and f has no "side effect".

x=3; list=x; p.=0; p.x=1

Do q=2 By 1

Exit

/*-------------------------------------------------------------------*/

 

=={{header|Ruby}}==

{{works with|ruby|2.0}}

 

# Purpose:

# Find the cycle length and start position of a numerical seried using Brent's cycle algorithm.

# Test the findCycle function

clength, cstart = findCycle(3) { |x| f(x) }

 

{{out}}

Cycle length = 6

Start index = 2

</pre>

 

=={{header|Sidef}}==

var power = 1

var λ = 1

say "Cycle length #{l}.";

say "Cycle start index #{s}."

{{out}}

Cycle length 6.

Cycle start index 2.

=={{header|Visual Basic .NET}}==

{{trans|C#}}

 

Function FindCycle(Of T As IEquatable(Of T))(x0 As T, yielder As Func(Of T, T)) As Tuple(Of Integer, Integer)

End Sub

 

{{out}}

<pre>3,10,101,2,5,26,167,95,101,2,5,26,167,95,101,2,5,26,167,95,101,2,5,26,167,95,101,2,5,26,167,95,101,2,5,26,167,95,101,2,5

Cycle length = 6

Start index = 2</pre>

 

=={{header|zkl}}==

Algorithm from the Wikipedia

# main phase: search successive powers of two

power:=lam:=1;

}

return(lam,mu);

{{out}}

<pre>