# Cycle detection

Cycle detection is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Detect a cycle in an iterated function using Brent's algorithm.

Detecting cycles in iterated function sequences is a sub-problem in many computer algorithms, such as factoring prime numbers. Some such algorithms are highly space efficient, such as Floyd's cycle-finding algorithm, also called the "tortoise and the hare algorithm". A more time efficient algorithm than "tortoise and hare" is Brent's Cycle algorithm. This task will implement Brent's algorithm.

See https://en.wikipedia.org/wiki/Cycle_detection for a discussion of the theory and discussions of other algorithms that are used to solve the problem.

When testing the cycle detecting function, you need two things:

1) An iterated function

2) A starting value

The iterated function used in this example is: f(x) = (x*x + 1) modulo 255.

The starting value used is 3.

With these as inputs, a sample program output would be:

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

The output prints the first several items in the number series produced by the iterated function, then identifies how long the cycle is (6) followed by the zero-based index of the start of the first cycle (2). From this you can see that the cycle is:

101,2,5,26,167,95

## C

Translation of: Modula-2
`#include <stdio.h>#include <stdlib.h> typedef int(*I2I)(int);typedef struct {    int a, b;} Pair; Pair brent(I2I f, int x0) {    int power = 1, lam = 1, tortoise = x0, hare, mu, i;    Pair result;     hare = (*f)(x0);    while (tortoise != hare) {        if (power == lam) {            tortoise = hare;            power = power * 2;            lam = 0;        }        hare = (*f)(hare);        lam++;    }     hare = x0;    i = 0;    while (i < lam) {        hare = (*f)(hare);        i++;    }     tortoise = x0;    mu = 0;    while (tortoise != hare) {        tortoise = (*f)(tortoise);        hare = (*f)(hare);        mu++;    }     result.a = lam;    result.b = mu;    return result;} int lambda(int x) {    return (x*x + 1) % 255;} int main() {    int x0 = 3, x = 3, i;    Pair result;     printf("[3");    for (i = 1; i <= 40; ++i) {        x = lambda(x);        printf(", %d", x);    }    printf("]\n");     result = brent(lambda, x0);    printf("Cycle length = %d\nStart index = %d\nCycle = [", result.a, result.b);     x0 = 3;    x = x0;    for (i = 1; i <= result.b; ++i) {        x = lambda(x);    }    for (i = 1; i <= result.a; ++i) {        if (i > 1) {            printf(", ");        }         printf("%d", x);        x = lambda(x);    }     printf("]\n");     return 0;}`
Output:
```[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
Cycle = [101, 2, 5, 26, 167, 95]```

## C++

`struct ListNode {      int val;      ListNode *next;      ListNode(int x) : val(x), next(NULL) {} }; ListNode* Solution::detectCycle(ListNode* A) {    ListNode* slow = A;    ListNode* fast = A;    ListNode* cycleNode = 0;    while (slow && fast && fast->next)    {        slow = slow->next;        fast = fast->next->next;        if (slow == fast)        {            cycleNode = slow;            break;        }    }    if (cycleNode == 0)    {        return 0;    }    std::set<ListNode*> setPerimeter;    setPerimeter.insert(cycleNode);    for (ListNode* pNode = cycleNode->next; pNode != cycleNode; pNode = pNode->next)    setPerimeter.insert(pNode);    for (ListNode* pNode = A; true; pNode = pNode->next)    {        std::set<ListNode*>::iterator iter = setPerimeter.find(pNode);        if (iter != setPerimeter.end())         {            return pNode;        }    }}`

## C#

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

`  // First file: Cycles.cs// Author: Paul Anton Chernoch using System; namespace DetectCycles{  /// <summary>  /// Find the length and start of a cycle in a series of objects of any IEquatable type using Brent's cycle algorithm.  /// </summary>  public class Cycles<T> where T : IEquatable<T>  {    /// <summary>    /// Find the cycle length and start position of a series using Brent's cycle algorithm.    ///     ///  Given a recurrence relation X[n+1] = f(X[n]) where f() has    ///  a finite range, you will eventually repeat a value that you have seen before.    ///  Once this happens, all subsequent values will form a cycle that begins    ///  with the first repeated value. The period of that cycle may be of any length.    /// </summary>    /// <returns>A tuple where:    ///    Item1 is lambda (the length of the cycle)     ///    Item2 is mu, the zero-based index of the item that started the first cycle.</returns>    /// <param name="x0">First item in the series.</param>    /// <param name="yielder">Function delegate that generates the series by iterated execution.</param>    public static Tuple<int,int> FindCycle(T x0, Func<T,T> yielder)    {      int power, lambda;      T tortoise, hare;      power = lambda = 1;      tortoise = x0;      hare = yielder(x0);       // Find lambda, the cycle length      while (!tortoise.Equals (hare)) {        if (power == lambda) {          tortoise = hare;          power *= 2;          lambda = 0;          }        hare = yielder (hare);        lambda += 1;      }       // Find mu, the zero-based index of the start of the cycle      var mu = 0;      tortoise = hare = x0;      for (var times = 0; times < lambda; times++)         hare = yielder (hare);       while (!tortoise.Equals (hare))       {        tortoise = yielder (tortoise);        hare = yielder (hare);        mu += 1;      }       return new Tuple<int,int> (lambda, mu);    }  }} // Second file: Program.cs using System; namespace DetectCycles{	class MainClass	{		public static void Main (string[] args)		{			// A recurrence relation to use in testing			Func<int,int> sequence = (int _x) => (_x * _x + 1) % 255; 			// Display the first 41 numbers in the test series			var x = 3;			Console.Write(x);			for (var times = 0; times < 40; times++) 			{				x = sequence(x);				Console.Write(String.Format(",{0}", x));			}			Console.WriteLine(); 			// Test the FindCycle method			var cycle = Cycles<int>.FindCycle(3, sequence);			var clength = cycle.Item1;			var cstart = cycle.Item2;			Console.Write(String.Format("Cycle length = {0}\nStart index = {1}\n", clength, cstart));		}	}}  `

## D

Translation of: Java
`import std.range;import std.stdio; void main() {    brent(i => (i * i + 1) % 255, 3);} void brent(int function(int) f, int x0) {    int cycleLength;    int hare = x0;    FOUND:    for (int power = 1; ; power *= 2) {        int tortoise = hare;        for (int i = 1; i <= power; i++) {            hare = f(hare);             if (tortoise == hare) {                cycleLength = i;                break FOUND;            }        }    }     hare = x0;    for (int i = 0; i < cycleLength; i++)        hare = f(hare);     int cycleStart = 0;    for (int tortoise = x0; tortoise != hare; cycleStart++) {        tortoise = f(tortoise);        hare = f(hare);    }     printResult(x0, f, cycleLength, cycleStart);} void printResult(int x0, int function(int) f, int len, int start) {    writeln("Cycle length: ", len);    writefln("Cycle: %(%s %)", iterate(x0, f).drop(start).take(len));} auto iterate(int start, int function(int) f) {    return only(start).chain(generate!(() => start=f(start)));}`
Output:
```Cycle length: 6
Cycle: 101 2 5 26 167 95```

## Elixir

Translation of: Ruby
`defmodule Cycle_detection do  def find_cycle(x0, f) do    lambda = find_lambda(f, x0, f.(x0), 1, 1)    hare = Enum.reduce(1..lambda, x0, fn _,hare -> f.(hare) end)    mu = find_mu(f, x0, hare, 0)    {lambda, mu}  end   # Find lambda, the cycle length  defp find_lambda(_, tortoise, hare, _, lambda) when tortoise==hare, do: lambda  defp find_lambda(f, tortoise, hare, power, lambda) do    if power == lambda, do: find_lambda(f, hare, f.(hare), power*2, 1),                      else: find_lambda(f, tortoise, f.(hare), power, lambda+1)  end   # Find mu, the zero-based index of the start of the cycle  defp find_mu(_, tortoise, hare, mu) when tortoise==hare, do: mu  defp find_mu(f, tortoise, hare, mu) do    find_mu(f, f.(tortoise), f.(hare), mu+1)  endend # A recurrence relation to use in testingf = fn(x) -> rem(x * x + 1, 255) end # Display the first 41 numbers in the test seriesStream.iterate(3, &f.(&1)) |> Enum.take(41) |> Enum.join(",") |> IO.puts # Test the find_cycle function{clength, cstart} = Cycle_detection.find_cycle(3, f)IO.puts "Cycle length = #{clength}\nStart index = #{cstart}"`
Output:
```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
```

## 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 `!` at the end is mutable, and subsequent uses of that name followed by `!` change the value of the variable to the value at the top of the data stack.

`USING: formatting kernel locals make math prettyprint ; : cyclical-function ( n -- m ) dup * 1 + 255 mod ; :: brent ( x0 quot -- λ μ )    1 dup :> ( power! λ! )    x0 :> tortoise!    x0 quot call :> hare!    [ tortoise hare = not ] [        power λ = [            hare tortoise!            power 2 * power!            0 λ!         ] when        hare quot call hare!        λ 1 + λ!    ] while     0 :> μ!    x0 dup tortoise! hare!    λ [ hare quot call hare! ] times    [ tortoise hare = not ] [        tortoise quot call tortoise!        hare quot call hare!        μ 1 + μ!    ] while    λ μ ; inline 3 [ 20 [ dup , cyclical-function ] times ] { } make nip .3 [ cyclical-function ] brent"Cycle length: %d\nCycle start: %d\n" printf`
Output:
```{ 3 10 101 2 5 26 167 95 101 2 5 26 167 95 101 2 5 26 167 95 }
Cycle length: 6
Cycle start: 2
```

## FreeBASIC

### Brent's algorithm

Translation of: Python
`' version 11-01-2017' compile with: fbc -s console ' define the function f(x)=(x*x +1) mod 255Function f(x As Integer) As Integer    Return (x * x +1) Mod 255End Function Sub brent(x0 As Integer, ByRef lam As Integer, ByRef mu As Integer)     Dim As Integer i, power = 1    lam = 1     ' main phase: search successive powers of two    Dim As Integer tortoise = f(x0)   ' f(x0) is the element/node next to x0.    Dim As Integer hare = f(f(x0))     While tortoise <> hare        If power = lam Then            tortoise = hare            power *= 2            lam = 0        End If        hare = f(hare)        lam += 1    Wend     ' Find the position of the first repetition of length ?    mu = 0    tortoise = x0    hare = x0    For i = 0 To lam -1        ' range(lam) produces a list with the values 0, 1, ... , lam-1        hare = f(hare)    Next    ' The distance between the hare and tortoise is now ?.     ' Next, the hare and tortoise move at same speed until they agree    While tortoise <> hare        tortoise = f(tortoise)        hare = f(hare)        mu += 1    Wend End Sub ' ------=< MAIN >=------ Dim As Integer  i, j, lam, mu, x0 = 3 brent(x0, lam, mu)Print " Brent's algorithm"Print " Cycle starts at position: "; mu; " (starting position = 0)"Print " The length of the Cycle = "; lamPrint j = f(x0)Print " Cycle: ";For i = 1 To lam + mu -1    If i >= mu Then        Print j;        If i <> (lam + mu -1) Then Print ", "; Else Print ""    End If    j = f(j)NextPrint ' empty keyboard bufferWhile Inkey <> "" : WendPrint : Print "hit any key to end program"SleepEnd`
Output:
``` Brent's algorithm
Cycle starts at position:  2 (starting position = 0)
The length of the Cycle =  6

Cycle:  101,  2,  5,  26,  167,  95```

### Tortoise and hare. Floyd's algorithm

Translation of: Wikipedia
`' version 11-01-2017' compile with: fbc -s console ' define the function f(x)=(x*x +1) mod 255Function f(x As Integer) As Integer    Return (x * x +1) Mod 255End Function Sub floyd(x0 As Integer, ByRef lam As Integer, ByRef mu As Integer)     ' Main phase of algorithm: finding a repetition x_i = x_2i.    ' The hare moves twice as quickly as the tortoise and    ' the distance between them increases by 1 at each step.    ' Eventually they will both be inside the cycle and then,    ' at some point, the distance between them will be    ' divisible by the period ?.    Dim As Integer tortoise = f(x0)   ' f(x0) is the element/node next to x0.    Dim As Integer hare = f(f(x0))     While tortoise <> hare        tortoise = f(tortoise)        hare = f(f(hare))    Wend     ' At this point the tortoise position, ?, which is also equal    ' to the distance between hare and tortoise, is divisible by    ' the period ?. So hare moving in circle one step at a time,    ' and tortoise (reset to x0) moving towards the circle, will    ' intersect at the beginning of the circle. Because the    ' distance between them is constant at 2?, a multiple of ?,    ' they will agree as soon as the tortoise reaches index µ.     ' Find the position µ of first repetition.    mu = 0    tortoise = x0    While tortoise <> hare        tortoise = f(tortoise)        hare = f(hare)         ' Hare and tortoise move at same speed        mu += 1    Wend     ' Find the length of the shortest cycle starting from x_µ    ' The hare moves one step at a time while tortoise is still.    ' lam is incremented until ? is found.    lam = 1    hare = f(tortoise)    While tortoise <> hare        hare = f(hare)        lam += 1    Wend End Sub  ' ------=< MAIN >=------ Dim As Integer  i, j, lam, mu, x0 = 3 floyd(x0, lam, mu)Print " Tortoise and hare. Floyd's algorithm"Print " Cycle starts at position: "; mu; " (starting position = 0)"Print " The length of the Cycle = "; lamPrint j = f(x0)Print " Cycle: ";For i = 1 To lam + mu -1    If i >= mu Then        Print j;        If i <> (lam + mu -1) Then Print ", "; Else Print ""    End If    j = f(j) NextPrint ' empty keyboard bufferWhile Inkey <> "" : WendPrint : Print "hit any key to end program"SleepEnd`

## Go

Translation of: Python
Translation of: Wikipedia

Run it on the go playground, or on go play space.

` package main import "fmt" func brent(f func(i int) int, x0 int) (int, int) {	var λ, µ, power, tortoise, hare int 	// Main phase: search successive powers of two.	power = 1	λ = 1	tortoise = x0	hare = f(x0) // f(x0) is the element/node next to x0. 	for tortoise != hare {		if power == λ { // Time to start a new power of two.			tortoise = hare			power *= 2			λ = 0		}		hare = f(hare)		λ++	} 	// Find the position of the first repetition of length λ.	µ = 0	tortoise, hare = x0, x0	for i := 0; i < λ; i++ {		// produces a list with the values 0,1,...,λ-1.		hare = f(hare)		// The distance between hare and tortoise is now λ.	} 	// The tortoise and the hare move at the same speed until they agree.	for tortoise != hare {		tortoise = f(tortoise)		hare = f(hare)		µ++	} 	return λ, µ} func f(i int) int {	return (i*i + 1) % 255} func main() {	x0 := 3	λ, µ := brent(f, x0)	fmt.Println("Cycle length:", λ)	fmt.Println("Cycle start index:", µ)	a := []int{}	for i := 0; i <= λ+1; i++ {		a = append(a, x0)		x0 = f(x0)	}	fmt.Println("Cycle:", a[µ:µ+λ])}`
Output:
```Cycle length: 6
Cycle start index: 2
Cycle: [101 2 5 26 167 95]```

Most of solutions given in other languages are not total. For function which does not have any cycles under iteration (i.e. f(x) = 1 + x) they will never terminate.

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.

`import Data.List (findIndex) findCycle :: Eq a => [a] -> Maybe ([a], Int, Int)findCycle lst =  do l <- findCycleLength lst     mu <- findIndex (uncurry (==)) \$ zip lst (drop l lst)     let c = take l \$ drop mu lst     return (c, l, mu) findCycleLength :: Eq a => [a] -> Maybe IntfindCycleLength [] = NothingfindCycleLength (x:xs) =  let loop _ _ _ [] = Nothing      loop pow lam x (y:ys)        | x == y     = Just lam        | pow == lam = loop (2*pow) 1 y ys        | otherwise  = loop pow (1+lam) x ys  in loop 1 1 x xs`

Examples

```λ> findCycle (cycle [1,2,3])
Just ([1,2,3],3,0)

λ> findCycle ([1..100] ++ cycle [1..12])
Just ([1,2,3,4,5,6,7,8,9,10,11,12],12,100)

λ> findCycle [1..1000]
Nothing
```
```λ> findCycle (iterate (\x -> (x^2 + 1) `mod` 255) 3)
Just ([101,2,5,26,167,95],6,2)

λ> findCycle (take 100 \$ iterate (\x -> x+1) 3)
Nothing

λ> findCycle (take 100000 \$ iterate (\x -> x+1) 3)
Nothing
```

## J

Brute force implementation:

`cdetect=:1 :0  r=. [email protected](,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:

`   255&|@(1 0 1&p.) cdetect 32 6`

(Note that 1 0 1 are the coefficients of the polynomial `1 + (0 * x) + (1 * x * x)` or, if you prefer `(1 * x ^ 0) + (0 * x ^ 1) + (1 * x ^ 2)` - 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.)

## Java

Works with: Java version 8
`import java.util.function.*;import static java.util.stream.IntStream.*; public class CycleDetection {     public static void main(String[] args) {        brent(i -> (i * i + 1) % 255, 3);    }     static void brent(IntUnaryOperator f, int x0) {        int cycleLength;        int hare = x0;        FOUND:        for (int power = 1; ; power *= 2) {            int tortoise = hare;            for (int i = 1; i <= power; i++) {                hare = f.applyAsInt(hare);                 if (tortoise == hare) {                    cycleLength = i;                    break FOUND;                }            }        }         hare = x0;        for (int i = 0; i < cycleLength; i++)            hare = f.applyAsInt(hare);         int cycleStart = 0;        for (int tortoise = x0; tortoise != hare; cycleStart++) {            tortoise = f.applyAsInt(tortoise);            hare = f.applyAsInt(hare);        }         printResult(x0, f, cycleLength, cycleStart);    }     static void printResult(int x0, IntUnaryOperator f, int len, int start) {        System.out.printf("Cycle length: %d%nCycle: ", len);        iterate(x0, f).skip(start).limit(len)                .forEach(n -> System.out.printf("%s ", n));    }}`
```Cycle length: 6
Cycle: 101 2 5 26 167 95```

## Julia

Works with: Julia version 0.6

Following the Wikipedia article:

`using IterTools function floyd(f, x0)    local tort = f(x0)    local hare = f(tort)    while tort != hare        tort = f(tort)        hare = f(f(hare))    end     local μ = 0    tort = x0    while tort != hare        tort = f(tort)        hare = f(hare)        μ += 1    end     λ = 1    hare = f(tort)    while tort != hare        hare = f(hare)        λ += 1    end     return λ, μend f(x) = (x * x + 1) % 255 λ, μ = floyd(f, 3)cycle = iterate(f, 3) |>    x -> Iterators.drop(x, μ) |>    x -> Iterators.take(x, λ) |>    collectprintln("Cycle length: ", λ, "\nCycle start index: ", μ, "\nCycle: ", join(cycle, ", "))`
Output:
```Cycle length: 6
Cycle start index: 2
Cycle: 101, 2, 5, 26, 167, 95```

## Kotlin

`// version 1.1.2 typealias IntToInt = (Int) -> Int fun brent(f: IntToInt, x0: Int): Pair<Int, Int> {    // main phase: search successive powers of two    var power = 1    var lam = 1    var tortoise = x0    var hare = f(x0)  // f(x0) is the element/node next to x0.    while (tortoise != hare) {        if (power == lam) {  // time to start a new power of two?            tortoise = hare            power *= 2            lam = 0        }        hare = f(hare)        lam++    }     // Find the position of the first repetition of length 'lam'    var mu = 0    tortoise = x0    hare = x0    for (i in 0 until lam) hare = f(hare)     // The distance between the hare and tortoise is now 'lam'.    // Next, the hare and tortoise move at same speed until they agree    while (tortoise != hare) {        tortoise = f(tortoise)        hare = f(hare)        mu++    }    return Pair(lam, mu)} fun main(args: Array<String>) {    val f = { x: Int -> (x * x + 1) % 255 }    // generate first 41 terms of the sequence starting from 3    val x0 = 3    var x = x0    val seq = List(41) { if (it > 0) x = f(x) ; x }    println(seq)    val (lam, mu) = brent(f, x0)    val cycle = seq.slice(mu until mu + lam)    println("Cycle length = \$lam")    println("Start index  = \$mu")    println("Cycle        = \$cycle")}`
Output:
```[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
Cycle        = [101, 2, 5, 26, 167, 95]
```

## 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.

`function brent (f, x0)    local tortoise, hare, mu = x0, f(x0), 0    local cycleTab, power, lam = {tortoise, hare}, 1, 1    while tortoise ~= hare do        if power == lam then            tortoise = hare            power = power * 2            lam = 0        end        hare = f(hare)        table.insert(cycleTab, hare)        lam = lam + 1    end    tortoise, hare = x0, x0    for i = 1, lam do hare = f(hare) end    while tortoise ~= hare do        tortoise = f(tortoise)        hare = f(hare)        mu = mu + 1    end    return lam, mu, cycleTabend local f = function (x) return (x * x + 1) % 255 endlocal x0 = 3local cycleLength, startIndex, sequence = brent(f, x0)print("Sequence:", table.concat(sequence, " "))print("Cycle length:", cycleLength)print("Start Index:", startIndex)`
Output:
```Sequence:       3 10 101 2 5 26 167 95 101 2 5 26 167 95
Cycle length:   6
Start Index:    2```

## Modula-2

Translation of: Kotlin
`MODULE CycleDetection;FROM FormatString IMPORT FormatString;FROM Terminal IMPORT WriteString,WriteLn,ReadChar; TYPE IntToInt = PROCEDURE(INTEGER) : INTEGER;TYPE Pair =    RECORD        a,b : INTEGER;    END; PROCEDURE Brent(f : IntToInt; x0 : INTEGER) : Pair;VAR power,lam,tortoise,hare,mu,i : INTEGER;BEGIN    (* main phase: search successive powers of two *)    power := 1;    lam := 1;    tortoise := x0;    hare := f(x0);  (* f(x0) is the element/node next to x0. *)    WHILE tortoise # hare DO        (* time to start a new power of two? *)        IF power = lam THEN            tortoise := hare;            power := power * 2;            lam := 0        END;        hare := f(hare);        INC(lam)    END;     (* Find the position of the first repetition of length 'lam' *)    mu := 0;    tortoise := x0;    hare := x0;    i := 0;    WHILE i < lam DO        hare := f(hare);        INC(i)    END;     (* The distance between the hare and tortoise is now 'lam'.       Next, the hare and tortoise move at same speed until they agree *)    WHILE tortoise # hare DO        tortoise := f(tortoise);        hare := f(hare);        INC(mu)    END;     RETURN Pair{lam,mu}END Brent; PROCEDURE Lambda(x : INTEGER) : INTEGER;BEGIN    RETURN (x * x + 1) MOD 255END Lambda; VAR    buf : ARRAY[0..63] OF CHAR;    x0,x,i : INTEGER;    result : Pair;BEGIN    x0 := 3;    x := x0;     WriteString("[3");    FOR i:=1 TO 40 DO        x := Lambda(x);        FormatString(", %i", buf, x);        WriteString(buf)    END;    WriteString("]");    WriteLn;     result := Brent(Lambda, x0);    FormatString("Cycle length = %i\nStart index  = %i\nCycle        = [", buf, result.a, result.b);    WriteString(buf);     x0 := 3;    x := x0;    FOR i:=1 TO result.b DO        x := Lambda(x)    END;    FOR i:=1 TO result.a DO        IF i > 1 THEN            WriteString(", ")        END;         FormatString("%i", buf, x);        WriteString(buf);        x := Lambda(x)    END;     WriteString("]");    WriteLn;     ReadCharEND CycleDetection.`

## ooRexx

`/* REXX */x=3list=xDo i=1 By 1  x=f(x)  p=wordpos(x,list)  If p>0 Then Do    list=list x    Say 'list ='list '...'    Say 'Start index = ' (wordpos(x,list)-1) '(zero based)'    Say 'Cycle length =' (words(list)-p)    Say 'Cycle        =' subword(list,p,(words(list)-p))    Leave    End  list=list x  EndExitf: Return (arg(1)**2+1)//255 `
Output:
```list =3 10 101 2 5 26 167 95 101 ...
Start index =  2 (zero based)
Cycle length = 6
Cycle        = 101 2 5 26 167 95
```

## Perl

Translation of: Perl 6
`use utf8; sub cyclical_function { (\$_ * \$_ + 1) % 255 } sub brent {    my(\$f, \$x0) = @_;    my \$power = 1;    my \$λ = 1;    my \$tortoise = \$x0;    my \$hare = &\$f(\$x0);    while (\$tortoise != \$hare) {        if (\$power == \$λ) {            \$tortoise = \$hare;            \$power *= 2;            \$λ = 0;        }        \$hare = &\$f(\$hare);        \$λ += 1;    }     my \$μ = 0;    \$tortoise = \$hare = \$x0;    \$hare = &\$f(\$hare) for 0..\$λ-1;     while (\$tortoise != \$hare) {        \$tortoise = &\$f(\$tortoise);        \$hare = &\$f(\$hare);        \$μ += 1;    }    return \$λ, \$μ;} my ( \$l, \$s ) = brent( \&cyclical_function, 3 ); sub show_range {    my(\$start,\$stop) = @_;    my \$result;    my \$x = 3;    for my \$n (0..\$stop) {        \$result .= "\$x " if \$n >= \$start;        \$x = cyclical_function(\$x);    }    \$result;} print show_range(0,19) . "\n";print "Cycle length \$l\n";print "Cycle start index \$s\n";print show_range(\$s,\$s+\$l-1) . "\n";`
Output:
```3 10 101 2 5 26 167 95 101 2 5 26 167 95 101 2 5 26 167 95
Cycle length 6
Cycle start index 2
101 2 5 26 167 95```

## Perl 6

Works with: Rakudo version 2016-01

Pretty much a line for line translation of the Python code on the Wikipedia page.

`sub cyclical-function (\x) { (x * x + 1) % 255 }; my ( \$l, \$s ) = brent( &cyclical-function, 3 ); put join ', ', (3, -> \x { cyclical-function(x) } ... *)[^20], '...';say "Cycle length \$l.";say "Cycle start index \$s.";say (3, -> \x { cyclical-function(x) } ... *)[\$s .. \$s + \$l - 1]; sub brent (&f, \$x0) {    my \$power = my \$λ = 1;    my \$tortoise = \$x0;    my \$hare = f(\$x0);    while (\$tortoise != \$hare) {        if \$power == \$λ {            \$tortoise = \$hare;            \$power *= 2;            \$λ = 0;        }        \$hare = f(\$hare);        \$λ += 1;    }     my \$μ = 0;    \$tortoise = \$hare = \$x0;    \$hare = f(\$hare) for ^\$λ;     while (\$tortoise != \$hare) {        \$tortoise = f(\$tortoise);        \$hare = f(\$hare);        \$μ += 1;    }    return \$λ, \$μ;}`
Output:
```3, 10, 101, 2, 5, 26, 167, 95, 101, 2, 5, 26, 167, 95, 101, 2, 5, 26, 167, 95, ...
Cycle length 6.
Cycle start index 2.
(101 2 5 26 167 95)```

## Phix

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

`function f(integer x)    return mod(x*x+1,255)end function function brent(integer x0)integer pow2 = 1, len = 1, pos = 1integer tortoise = x0,        hare = f(x0)sequence s = {tortoise,hare} -- (kept for output only)     -- main phase: search successive powers of two    while tortoise!=hare do        if pow2 = len then            tortoise = hare            pow2 *= 2            if pow2>16 then                return {s,0,0}            end if            len = 0        end if        hare = f(hare)        s &= hare        len += 1    end while     -- Find the position of the first repetition of length len    tortoise = x0    hare = x0    for i=1 to len do        hare = f(hare)    end for    -- The distance between the hare and tortoise is now len.     -- Next, the hare and tortoise move at same speed until they agree    while tortoise<>hare do        tortoise = f(tortoise)        hare = f(hare)        pos += 1    end while     return {s,len,pos}end function sequence sinteger len, pos, x0 = 3{s,len,pos} = brent(x0)printf(1," Brent's algorithm\n")?sprintf(1," Cycle starts at position: %d (1-based)\n",{pos})printf(1," The length of the Cycle = %d\n",{len})?s[pos..pos+len-1]`
Output:
``` Brent's algorithm
{3,10,101,2,5,26,167,95,101,2,5,26,167,95}
Cycle starts at position: 3 (1-based)
The length of the Cycle = 6
{101,2,5,26,167,95}
```

## Python

Function from the Wikipedia article:

`import itertools def brent(f, x0):    # main phase: search successive powers of two    power = lam = 1    tortoise = x0    hare = f(x0)  # f(x0) is the element/node next to x0.    while tortoise != hare:        if power == lam:  # time to start a new power of two?            tortoise = hare            power *= 2            lam = 0        hare = f(hare)        lam += 1     # Find the position of the first repetition of length lam    mu = 0    tortoise = hare = x0    for i in range(lam):    # range(lam) produces a list with the values 0, 1, ... , lam-1        hare = f(hare)    # The distance between the hare and tortoise is now lam.     # Next, the hare and tortoise move at same speed until they agree    while tortoise != hare:        tortoise = f(tortoise)        hare = f(hare)        mu += 1     return lam, mu def iterate(f, x0):    while True:        yield x0        x0 = f(x0) if __name__ == '__main__':    f = lambda x: (x * x + 1) % 255    x0 = 3    lam, mu = brent(f, x0)    print("Cycle length: %d" % lam)    print("Cycle start index: %d" % mu)    print("Cycle: %s" % list(itertools.islice(iterate(f, x0), mu, mu+lam)))`
Output:
```Cycle length: 6
Cycle start index: 2
Cycle: [101, 2, 5, 26, 167, 95]
```

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

`import itertools def brent_length(f, x0):    # main phase: search successive powers of two    hare = x0    power = 1    while True:        tortoise = hare        for i in range(1, power+1):            hare = f(hare)            if tortoise == hare:                return i        power *= 2 def brent(f, x0):    lam = brent_length(f, x0)     # Find the position of the first repetition of length lam    mu = 0    hare = x0    for i in range(lam):    # range(lam) produces a list with the values 0, 1, ... , lam-1        hare = f(hare)    # The distance between the hare and tortoise is now lam.     # Next, the hare and tortoise move at same speed until they agree    tortoise = x0    while tortoise != hare:        tortoise = f(tortoise)        hare = f(hare)        mu += 1     return lam, mu def iterate(f, x0):    while True:        yield x0        x0 = f(x0) if __name__ == '__main__':    f = lambda x: (x * x + 1) % 255    x0 = 3    lam, mu = brent(f, x0)    print("Cycle length: %d" % lam)    print("Cycle start index: %d" % mu)    print("Cycle: %s" % list(itertools.islice(iterate(f, x0), mu, mu+lam)))`
Output:
```Cycle length: 6
Cycle start index: 2
Cycle: [101, 2, 5, 26, 167, 95]
```

## Racket

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

` #lang racket/base ;; returns (values lambda mu)(define (brent f x0)  ;; main phase: search successive powers of two  (define λ    (let main-phase ((power 1)                     (λ 1)                     (tortoise x0)                     (hare (f x0))) ;; f(x0) is the element/node next to x0.      (cond [(= hare tortoise) λ]            ;; time to start a new power of two?            [(= power λ) (main-phase (* power 2) 1 hare (f hare))]            [else (main-phase power (add1 λ) tortoise (f hare))])))   (values   λ   ;; Find the position of the first repetition of length λ   (let race ((µ 0)              (tortoise x0)              ;; The distance between the hare and tortoise is now λ.              (hare (for/fold ((hare x0)) ((_ (in-range λ))) (f hare))))     ;; Next, the hare and tortoise move at same speed until they agree     (if (= tortoise hare) µ (race (add1 µ) (f tortoise) (f hare)))))) (module+ test  (require rackunit racket/generator)  (define (f x) (modulo (+ (* x x) 1) 255))  (define (make-generator f x0)    (generator () (let loop ((x x0)) (yield x) (loop (f x)))))   (define g (make-generator f 3))   (define l (for/list ((_ 20)) (g)))  (check-equal? l '(3 10 101 2 5 26 167 95 101 2 5 26 167 95 101 2 5 26 167 95))    (displayln l)  (let-values (([µ λ] (brent f 3)))    (printf "Cycle length = ~a~%Start Index = ~a~%" µ λ))) `
Output:
```(3 10 101 2 5 26 167 95 101 2 5 26 167 95 101 2 5 26 167 95)
Cycle length = 6
Start Index = 2
```

## REXX

### Brent's algorithm

`/*REXX program detects a cycle in an iterated function  [F]  using Brent's algorithm.   */init= 3;   @= init                                  do  until length(@) > 79;    @= @ f(word(@, words(@) ) )                                  end   /*until*/say '  original list='    @  ...                          /*display original number list*/call Brent init                                           /*invoke Brent algorithm for F*/parse var result cycle idx                                /*get 2 values returned from F*/say 'numbers in list='    words(@)                        /*display number of numbers.  */say '  cycle length ='    cycle                           /*display the cycle    to term*/say '  start index  ='     idx     "  ◄─── zero index"    /*   "     "  index     "   " */say 'cycle sequence ='   subword(@, idx+1, cycle)         /*   "     "  sequence  "   " */exit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/Brent: procedure; parse arg x0 1 tort;   pow=1;     #=1   /*TORT  is set to value of X0.*/       hare= f(x0)                                        /*get 1st value for the func. */                   do  while tort\==hare                   if pow==#  then do;  tort= hare        /*set value of TORT to HARE.  */                                        pow = pow + pow   /*double the value of  POW.   */                                        #   = 0           /*reset  #  to zero  (lambda).*/                                   end                   hare= f(hare)                   #= #+1                                 /*bump the lambda count value.*/                   end   /*while*/       hare= x0                   do #;          hare= f(hare)           /*generate number of F values.*/                   end   /*j*/       tort= x0                                           /*find position of the 1st rep*/                   do mu=0  while tort \== hare           /*MU  is a  zero─based  index.*/                                  tort= f(tort)                                  hare= f(hare)                   end   /*mu*/       return # mu/*──────────────────────────────────────────────────────────────────────────────────────*/f:     return ( arg(1) **2  +  1)   //   255     /*this defines/executes the function F.*/`

output   using the defaults:

```  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 ...
numbers in list= 27
cycle length = 6
start index  = 2   ◄─── zero index
cycle sequence = 101 2 5 26 167 95
```

### sequential search algorithm

`/*REXX program detects a cycle in an iterated function  [F]  using a sequential search. */x=3;  @=x                                                  /*initial couple of variables*/                        do  until cycle\==0;   x= f(x)     /*calculate another number.  */                        cycle= wordpos(x, @)               /*This could be a repeat.    */                        @= @ x                             /*append number to   @  list.*/                        end   /*until*/#= words(@) - cycle                                        /*compute the cycle length.  */say '  original list='  @ ...                              /*display the sequence.      */say 'numbers in list='  words(@)                           /*display number of numbers. */say '  cycle length ='  #                                  /*display the cycle   to term*/say '  start index  ='  cycle - 1     "  ◄─── zero based"  /*   "     "  index    "   " */say 'cycle sequence ='  subword(@, cycle, #)               /*   "     "  sequence "   " */exit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/f:   return ( arg(1) **2  +  1)   //   255       /*this defines/executes the function F.*/`
output:
```  original list= 3 10 101 2 5 26 167 95 101 ...
numbers in list= 9
cycle length = 6
start index  = 2   ◄─── zero based
cycle sequence = 101 2 5 26 167 95
```

### hash table algorithm

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

`/*REXX program detects a cycle in an iterated function  [F]  using a hash table.        */x=3;  @=x;  !.=.;  !.x=1                         do n=1+words(@);  x= f(x);   @= @ x  /*add the number to list. */                         if !.x\==.  then leave               /*A repeat?   Then leave. */                         !.x= n                               /*N:  numbers in  @  list.*/                         end   /*n*/say '  original list=' @ ...                                  /*maybe display the list. */say 'numbers in list=' n                                      /*display number of nums. */say '  cycle length =' n - !.x                                /*   "      "    cycle.   */say '  start index  =' !.x - 1      '  ◄───  zero based'      /*   "      "    index.   */say 'cycle sequence =' subword(@, !.x, n - !.x)               /*   "      "   sequence. */exit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/f:   return ( arg(1) **2  +  1)   //   255       /*this defines/executes the function F.*/`
output   is identical to the 2nd REXX version.

### robust hash table algorithm

This REXX version implements a   robust   hash table algorithm, which means that when the divisor
(in the   F   function)   is fairly large, the cycle length can be very large, making the creation of the
iterated function sequence.   This becomes problematic when the mechanism to append a number to
the sequence becomes time consuming because the sequence contains many hundreds of thousands
of numbers.

This REXX version allows the divisor for the   F   function to be specified, which can be chosen to stress
test the hash table algorithm.   A divisor which is   two raised to the 49th power   was chosen;   it
generates a cyclic sequence that contains over 1.5 million numbers.

`/*REXX program detects a  cycle  in an  iterated function  [F]  using a robust hashing. */parse arg power .                                      /*obtain optional args from C.L. */if power=='' | power=","  then power=8                 /*Not specified?  Use the default*/numeric digits 500                                     /*be able to handle big numbers. */divisor= 2**power - 1                                  /*compute the divisor, power of 2*/numeric digits max(9, length(divisor) * 2 + 1)         /*allow for the  square  plus one*/say '       power ='  power                            /*display the power   to the term*/say '     divisor ='  "{2**"power'}-1 = '    divisor   /*   "     "  divisor. "  "    " */sayx=3;    @=x;   @@=;      m=100;    !.=.;     !.x=1     /*M:  maximum numbers to display.*/                do n=1+words(@);    x= f(x);  @@[email protected]@ x               if n//2000==0  then do;   @[email protected] @@;   @@=;  end   /*Is a 2000th N?  Rejoin.*/               if !.x\==.     then leave               /*is this number a repeat?  Leave*/               !.x= n               end   /*n*/                             /*N:   is the size of   @   list.*/@= space(@ @@)                                         /*append residual numbers to  @  */if n<m  then say '  original list=' @ ...              /*maybe display the list to term.*/             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.*/if n<m  then say 'cycle sequence =' subword(@, !.x, n- !.x) /*maybe display the sequence*/exit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/f:   return ( arg(1) **2  +  1)   //   255       /*this defines/executes the function F.*/`
output   when the input (power of two) used is:     49
```       power = 49
divisor = {2**49}-1 =  562949953421311

original list= 3 10 101 2 5 26 167 95 101 ...
numbers in list= 9
cycle length = 6
start index  = 2   ◄─── zero based
cycle sequence = 101 2 5 26 167 95
```

### variant of the hash table algorithm

There is more information in the "hash table"
and f has no "side effect".

`/*REXX pgm detects a cycle in an iterated function [F]               */x=3; list=x; p.=0; p.x=1Do q=2 By 1  x=f(x)                           /* the next value                 */  list=list x                      /* append it to the list          */  If p.x>0 Then Do                 /* x was in the list              */    cLen=q-p.x                     /* cycle length                   */    Leave    End  p.x=q                            /* note position of x in the list */  EndSay 'original list='  list ...Say 'cycle length ='  cLen                   /*display the cycle len */Say 'start index  ='  p.x-1 "  (zero based)" /*   "     "  index.    */Say 'the sequence ='  subword(list,p.x,cLen) /*   "     "  sequence. */Exit/*-------------------------------------------------------------------*/f: Return (arg(1)**2+1)// 255;                /*define the function F*/`

## Ruby

Works with: ruby version 2.0
`# Author: Paul Anton Chernoch# Purpose:#   Find the cycle length and start position of a numerical seried using Brent's cycle algorithm.## Given a recurrence relation X[n+1] = f(X[n]) where f() has# a finite range, you will eventually repeat a value that you have seen before.# Once this happens, all subsequent values will form a cycle that begins# with the first repeated value. The period of that cycle may be of any length.## Parameters:#   x0 ...... First integer value in the sequence#   block ... Block that takes a single integer as input #             and returns a single integer as output.#             This yields a sequence of numbers that eventually repeats.# Returns:#   Two values: lambda and mu#   lambda .. length of cycle#   mu ...... zero-based index of start of cycle#def findCycle(x0)  power = lambda = 1  tortoise = x0  hare = yield(x0)   # Find lambda, the cycle length  while tortoise != hare    if power == lambda      tortoise = hare      power *= 2      lambda = 0    end    hare = yield(hare)    lambda += 1  end   # Find mu, the zero-based index of the start of the cycle  hare = x0  lambda.times { hare = yield(hare) }   tortoise, mu = x0, 0  while tortoise != hare    tortoise = yield(tortoise)    hare = yield(hare)    mu += 1  end   return lambda, muend # A recurrence relation to use in testingdef f(x) (x * x + 1) % 255 end # Display the first 41 numbers in the test seriesputs (1..40).reduce(){|acc,_| acc << f(acc.last)}.join(",") # Test the findCycle functionclength, cstart = findCycle(3) { |x| f(x) }puts "Cycle length = #{clength}\nStart index = #{cstart}"`
Output:
```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
```

## Sidef

Translation of: Perl 6
`func brent (f, x0) {    var power = 1    var λ = 1    var tortoise = x0    var hare = f(x0)     while (tortoise != hare) {        if (power == λ) {            tortoise = hare            power *= 2            λ = 0        }        hare = f(hare)        λ += 1    }     var μ = 0    tortoise = x0    hare = x0    { hare = f(hare) } * λ     while (tortoise != hare) {        tortoise = f(tortoise)        hare = f(hare)        μ += 1    }     return (λ, μ)} func cyclical_function(x) { (x*x + 1) % 255 } var (l, s) = brent(cyclical_function, 3) var seq = gather {    var x = 3    { take(x); x = cyclical_function(x) } * 20} say seq.join(', ')+', ...' say "Cycle length #{l}.";say "Cycle start index #{s}."say [seq[s .. (s + l - 1)]]`
Output:
```3, 10, 101, 2, 5, 26, 167, 95, 101, 2, 5, 26, 167, 95, 101, 2, 5, 26, 167, 95, ...
Cycle length 6.
Cycle start index 2.
[101, 2, 5, 26, 167, 95]
```

## zkl

Algorithm from the Wikipedia

`fcn cycleDetection(f,x0){ // f(int), x0 is the integer starting value of the sequence  # main phase: search successive powers of two  power:=lam:=1;  tortoise,hare:=x0,f(x0);  # f(x0) is the element/node next to x0.  while(tortoise!=hare){     if(power==lam){  # time to start a new power of two?	tortoise,lam=hare,0;	power*=2;     }     hare=f(hare);     lam+=1;  }  # Find the position of the first repetition of length λ  mu:=0; tortoise=hare=x0;  do(lam){ hare=f(hare) } # The distance between the hare and tortoise is now λ   # Next, the hare and tortoise move at same speed till they agree  while(tortoise!=hare){     tortoise,hare=f(tortoise),f(hare);     mu+=1;  }   return(lam,mu);}`
`cycleDetection(fcn(x){ (x*x + 1)%255 }, 3).println(" == cycle length, start index");`
Output:
```L(6,2) == cycle length, start index
```