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.
- Task
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
11l
F print_result(x0, f, len, start)
print(‘Cycle length = ’len)
print(‘Start index = ’start)
V i = x0
L 1..start
i = f(i)
V cycle = [0] * len
L 0.<len
cycle[L.index] = i
i = f(i)
print(‘Cycle: ’, end' ‘’)
print(cycle)
F brent(f, x0)
Int cycle_length
V hare = x0
V power = 1
L
V tortoise = hare
L(i) 1..power
hare = f(hare)
I tortoise == hare
cycle_length = i
^L.break
power *= 2
hare = x0
L 1..cycle_length
hare = f(hare)
V cycle_start = 0
V tortoise = x0
L tortoise != hare
tortoise = f(tortoise)
hare = f(hare)
cycle_start++
print_result(x0, f, cycle_length, cycle_start)
brent(i -> (i * i + 1) % 255, 3)- Output:
Cycle length = 6 Start index = 2 Cycle: [101, 2, 5, 26, 167, 95]
8086 Assembly
cpu 8086
org 100h
section .text
mov ax,3 ; Print the first 20 values in the sequence
mov bx,f
xor cx,cx
mov dx,20
call seq
mov ax,3 ; Use Brent's algorithm to find the cycle
mov bx,f
call brent
mov bp,ax ; BP = start of cycle
mov di,dx ; DI = length of cycle
mov dx,nl ; Print a newline
call prstr
mov dx,len ; Print "Length: "
call prstr
mov ax,di ; Print the length
call prnum
mov dx,ix ; Print "Index: "
call prstr
mov ax,bp ; Print the index
call prnum
mov dx,nl ; Print another newline
call prstr
mov ax,3 ; Print the cycle
mov bx,f
mov cx,bp
mov dx,di
jmp seq
;;; Brent's algorithm
;;; Input: AX = x0, BX = address of function
;;; Output: AX = mu, DX = lambda, CX, SI, BP destroyed
;;; The routine in BX must preserve BX, CX, DX, SI, BP
brent: mov bp,ax ; BP = x0
mov cx,ax ; CX = tortoise
xor dx,dx ; DX = lambda
mov si,1 ; SI = power
.lam: call bx ; Apply function (AX = hare)
inc dx ; Lambda += 1
cmp ax,cx ; Done yet?
je .mu
cmp dx,si ; Time to start a new power of two?
jne .lam
mov cx,ax ; Tortoise = hare
shl si,1 ; power *= 2
xor dx,dx ; lambda = 0
jmp .lam
.mu: mov ax,bp ; Find position of first repetition
mov cx,dx ; CX = lambda
.apply: call bx
loop .apply
mov cx,bp ; CX = tortoise
xor si,si ; SI = mu
.lmu: cmp ax,cx ; Done yet?
je .done
call bx ; hare = f(hare)
xchg ax,cx ; tortoise = f(tortoise)
call bx
xchg ax,cx
inc si
jmp .lmu
.done: mov ax,si
ret
;;; Function to use
;;; AX = (AX*AX+1) mod 255
f: mul al ; AX = AL*AL (we know anything mod 255 is <256)
inc ax ; + 1
div byte [fdsor] ; This is faster than freeing up a byte register
xchg al,ah ; Remainder into AL
xor ah,ah ; Pad to 16 bits
ret
;;; -- Utility routines --
;;; Print decimal value of AL. Destroys AX, DX.
prnum: mov dx,nbuf ; Number output buffer
xchg bx,dx
.dgt: aam ; Extract digit
add al,'0' ; Make ASCII digit
dec bx
mov [bx],al ; Store in buffer
xchg ah,al ; Continue with rest of digits
test al,al ; As long as there are digits
jnz .dgt
xchg bx,dx ; Put buffer pointer back in DX
prstr: mov ah,9 ; Print using MS-DOS call.
int 21h
ret
;;; Print DX values in sequence x, f(x), f(f(x))... starting at CX.
;;; AX = x0, BX = function. AX, CX, DX, SI destroyed.
seq: test cx,cx ; CX = 0?
jz .print
.skip: call bx ; Otherwise skip CX numbers
loop .skip
.print: mov cx,dx ; Amount of numbers to print
.loop: mov si,ax ; Keep a copy of AX (print routine trashes it)
call prnum
mov ax,si ; Restore x
call bx ; Find the next value
loop .loop
ret
section .data
fdsor: db 255 ; This 255 is used for the modulus calculation
db '...' ; Buffer for byte-sized numeric output
nbuf: db ' $'
ix: db 'Index: $'
len: db 'Length: $'
nl: db 13,10,'$'
- Output:
3 10 101 2 5 26 167 95 101 2 5 26 167 95 101 2 5 26 167 95 Length: 6 Index: 2 101 2 5 26 167 95
Ada
This implementation is split across three files. The first is the specification of a generic package:
generic
type Element_Type is private;
package Brent is
type Brent_Function is access function (X : Element_Type) return Element_Type;
procedure Brent(F : Brent_Function; X0 : Element_Type; Lambda : out Integer; Mu : out Integer);
end Brent;
The second is the body of the generic package:
package body Brent is
procedure Brent (F : Brent_Function; X0 : Element_Type; Lambda : out Integer; Mu : out Integer) is
Power : Integer := 1;
Tortoise : Element_Type := X0;
Hare : Element_Type := F(X0);
begin
Lambda := 1;
Mu := 0;
while Tortoise /= Hare loop
if Power = Lambda then
Tortoise := Hare;
Power := Power * 2;
Lambda := 0;
end if;
Hare := F(Hare);
Lambda := Lambda + 1;
end loop;
Tortoise := X0;
Hare := X0;
for I in 0..(Lambda-1) loop
Hare := F(Hare);
end loop;
while Hare /= Tortoise loop
Tortoise := F(Tortoise);
Hare := F(Hare);
Mu := Mu + 1;
end loop;
end Brent;
end Brent;
By using a generic package, this implementation can be made to work for any unary function, not just integers. As a demonstration two instances of the test function are created and two instances of the generic package. These are produced inside the main procedure:
with Brent;
with Text_Io;
use Text_Io;
procedure Main is
package Integer_Brent is new Brent(Element_Type => Integer);
use Integer_Brent;
function F (X : Integer) return Integer is
((X * X + 1) mod 255);
type Mod255 is mod 255;
package Mod255_Brent is new Brent(Element_Type => Mod255);
function F255 (X : Mod255) return Mod255 is
(X * X + 1);
lambda : Integer;
Mu : Integer;
X : Integer := 3;
begin
for I in 1..41 loop
Put(Integer'Image(X));
if I < 41 then
Put(",");
end if;
X := F(X);
end loop;
New_Line;
Integer_Brent.Brent(F'Access, 3, Lambda, Mu);
Put_Line("Cycle Length: " & Integer'Image(Lambda));
Put_Line("Start Index : " & Integer'Image(Mu));
Mod255_Brent.Brent(F255'Access, 3, Lambda, Mu);
Put_Line("Cycle Length: " & Integer'Image(Lambda));
Put_Line("Start Index : " & Integer'Image(Mu));
Put("Cycle : ");
X := 3;
for I in 0..(Mu + Lambda) loop
if Mu <= I and I < (Lambda + Mu) then
Put(Integer'Image(X));
end if;
X := F(X);
end loop;
end Main;
- 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 Length: 6 Start Index : 2 Cycle : 101 2 5 26 167 95
APL
brent←{
f←⍺⍺
lam←⊃{
l p t h←⍵
p=l: 1 (p×2) h (f h) ⋄ (l+1) p t (f h)
}⍣{=/2↓⍺} ⊢ 1 1 ⍵ (f ⍵)
mu←⊃{
(⊃⍵+1),f¨1↓⍵
}⍣{=/1↓⍺} ⊢ 0 ⍵ (f⍣lam⊢⍵)
mu lam
}
task←{
seq←{f←⍺⍺ ⋄ (⊃⍺)↓{⍵,f⊃⌽⍵}⍣(1-⍨+/⍺)⊢⍵}
⎕←0 20 ⍺⍺ seq ⍵ ⍝ First 20 elements
⎕←(↑'Index' 'Length'),⍺⍺ brent ⍵ ⍝ Index and length of cycle
⎕←(⍺⍺ brent ⍺⍺ seq⊢)⍵ ⍝ Cycle
}
(255 | 1 + ⊢×⊢) task 3
- Output:
3 10 101 2 5 26 167 95 101 2 5 26 167 95 101 2 5 26 167 95 Index 2 Length 6 101 2 5 26 167 95
BCPL
get "libhdr"
// Brent's algorithm. 'fn' is a function pointer,
// 'lam' and 'mu' are output pointers.
let brent(fn, x0, lam, mu) be
$( let power, tort, hare = 1, x0, fn(x0)
!lam := 1
until tort = hare do
$( if power = !lam then
$( tort := hare
power := power * 2
!lam := 0
$)
hare := fn(hare)
!lam := !lam + 1
$)
tort := x0
hare := x0
for i = 1 to !lam do
hare := fn(hare)
!mu := 0
until tort = hare do
$( tort := fn(tort)
hare := fn(hare)
!mu := !mu + 1
$)
$)
// Print fn^m(x0) to fn^n(x0)
let printRange(fn, x0, m, n) be
$( for i = 0 to n-1 do
$( if m<=i then writef("%N ", x0)
x0 := fn(x0)
$)
wrch('*N')
$)
let start() be
$( let lam, mu = 0, 0
// the function to find a cycle in
let f(x) = (x*x + 1) rem 255
// print the first 20 values
printRange(f, 3, 0, 20)
// find the cycle
brent(f, 3, @lam, @mu)
writef("Length: %N*NIndex: %N*N", lam, mu)
// print the cycle
printRange(f, 3, mu, mu+lam)
$)- Output:
3 10 101 2 5 26 167 95 101 2 5 26 167 95 101 2 5 26 167 95 Length: 6 Index: 2 101 2 5 26 167 95
BQN
_Brent← {F _𝕣 x0:
p←l←1
(I ← {p=l?
l↩1 ⋄ p×↩2 ⋄ 𝕩I𝕩 ;
l+↩1 ⋄ 𝕨I𝕩
}⍟≠⟜F) x0
m←0
{m+↩1 ⋄ 𝕨𝕊⍟≠○F𝕩}⟜(F⍟l) x0
l‿m‿(F⍟(m+↕l)x0)
}
- Example use:
(255|1+ט)_Brent 3 ⟨ 6 2 ⟨ 101 2 5 26 167 95 ⟩ ⟩
C
#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_sharp
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));
}
}
}
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;
}
}
}
CLU
% Find a cycle in f starting at x0 using Brent's algorithm
brent = proc [T: type] (f: proctype (T) returns (T), x0: T)
returns (int,int)
where T has equal: proctype (T,T) returns (bool)
pow: int := 1
lam: int := 1
tort: T := x0
hare: T := f(x0)
while tort ~= hare do
if pow = lam then
tort := hare
pow := pow * 2
lam := 0
end
hare := f(hare)
lam := lam + 1
end
tort := x0
hare := x0
for i: int in int$from_to(1,lam) do
hare := f(hare)
end
mu: int := 0
while tort ~= hare do
tort := f(tort)
hare := f(hare)
mu := mu + 1
end
return(lam, mu)
end brent
% Iterate over a function starting at x0 for N steps,
% starting at a given index
iterfunc = iter [T: type] (f: proctype (T) returns (T), x0: T, n, start: int)
yields (T)
for i: int in int$from_to(1, start) do
x0 := f(x0)
end
for i: int in int$from_to(1, n) do
yield(x0)
x0 := f(x0)
end
end iterfunc
% Iterated function
step_fn = proc (x: int) returns (int)
return((x*x + 1) // 255)
end step_fn
start_up = proc ()
po: stream := stream$primary_output()
% Print the first 20 items of the sequence
for i: int in iterfunc[int](step_fn, 3, 20, 0) do
stream$puts(po, int$unparse(i) || " ")
end
stream$putl(po, "")
% Find a cycle
lam, mu: int := brent[int](step_fn, 3)
% Print the length and index
stream$putl(po, "Cycle length: " || int$unparse(lam))
stream$putl(po, "Start index: " || int$unparse(mu))
% Print the cycle
stream$puts(po, "Cycle: ")
for i: int in iterfunc[int](step_fn, 3, lam, mu) do
stream$puts(po, int$unparse(i) || " ")
end
end start_up- 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 Cycle: 101 2 5 26 167 95
Cowgol
include "cowgol.coh";
typedef N is uint8;
interface BrentFn(x: N): (r: N);
sub Brent(f: BrentFn, x0: N): (lam: N, mu: N) is
var power: N := 1;
var tortoise := x0;
var hare := f(x0);
while tortoise != hare loop
if power == lam then
tortoise := hare;
power := power << 1;
lam := 0;
end if;
hare := f(hare);
lam := lam + 1;
end loop;
tortoise := x0;
hare := x0;
var i: N := 0;
while i < lam loop
i := i + 1;
hare := f(hare);
end loop;
mu := 0;
while tortoise != hare loop
tortoise := f(tortoise);
hare := f(hare);
mu := mu + 1;
end loop;
end sub;
sub PrintRange(f: BrentFn, x0: N, start: N, end_: N) is
var i: N := 0;
while i < end_ loop
if i >= start then
print_i32(x0 as uint32);
print_char(' ');
end if;
i := i + 1;
x0 := f(x0);
end loop;
print_nl();
end sub;
sub x2_plus1_mod255 implements BrentFn is
r := ((x as uint16 * x as uint16 + 1) % 255) as uint8;
end sub;
PrintRange(x2_plus1_mod255, 3, 0, 20);
var length: N;
var start: N;
(length, start) := Brent(x2_plus1_mod255, 3);
print("Cycle length: "); print_i32(length as uint32); print_nl();
print("Cycle start: "); print_i32(start as uint32); print_nl();
PrintRange(x2_plus1_mod255, 3, start, length+start);- 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: 2 101 2 5 26 167 95
D
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
EasyLang
func f x .
return (x * x + 1) mod 255
.
proc brent x0 . lam mu tab[] .
power = 1
lam = 1
tort = x0
hare = f x0
tab[] = [ tort hare ]
while tort <> hare
if power = lam
tort = hare
power = power * 2
lam = 0
.
hare = f hare
tab[] &= hare
lam += 1
.
tort = x0
hare = x0
for i = 1 to lam
hare = f hare
.
# one based
mu = 1
while tort <> hare
tort = f tort
hare = f hare
mu += 1
.
.
x0 = 3
brent x0 le start tab[]
print tab[]
print "cycle length: " & le & " start index: " & start
- Output:
[ 3 10 101 2 5 26 167 95 101 2 5 26 167 95 ] cycle length: 6 start index: 3
Elixir
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)
end
end
# A recurrence relation to use in testing
f = fn(x) -> rem(x * x + 1, 255) end
# Display the first 41 numbers in the test series
Stream.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
FOCAL
01.10 S X0=3;T %3
01.20 S X=X0;F I=1,20;T X;D 2
01.30 D 3;T !" START",M,!,"LENGTH",L,!
01.40 S X=X0;F I=1,M;D 2
01.50 F I=1,L;T X;D 2
01.60 T !
01.70 Q
02.01 C -- X = X*X+1 MOD 255
02.10 S X=X*X+1
02.20 S X=X-255*FITR(X/255)
03.01 C -- BRENT'S ALGORITHM ON ROUTINE 2
03.10 S X=X0;S P=1;S L=1;S T=X0;D 2
03.20 I (T-X)3.3,3.6,3.3
03.30 I (P-L)3.5,3.4,3.5
03.40 S T=X;S P=P*2;S L=0
03.50 D 2;S L=L+1;G 3.2
03.60 S T=X0;S X=X0;F I=1,L;D 2
03.70 S H=X;S M=0
03.80 I (T-H)3.9,3.99,3.9
03.90 S X=T;D 2;S T=X;S X=H;D 2;S H=X;S M=M+1;G 3.8
03.99 R- Output:
= 3= 10= 101= 2= 5= 26= 167= 95= 101= 2= 5= 26= 167= 95= 101= 2= 5= 26= 167= 95 START= 2 LENGTH= 6 = 101= 2= 5= 26= 167= 95
Forth
Works with gforth.
: cycle-length { x0 'f -- lambda } \ Brent's algorithm stage 1
1 1 x0 dup 'f execute
begin 2dup <> while
2over = if
2swap nip 2* 0
2swap nip dup
then
'f execute rot 1+ -rot
repeat 2drop nip ;
: iterations ( x f n -- x )
>r swap r> 0 ?do over execute loop nip ;
: cycle-start { x0 'f lambda -- mu } \ Brent's algorithm stage 2
0 x0 dup 'f lambda iterations
begin 2dup <> while
swap 'f execute swap 'f execute rot 1+ -rot
repeat 2drop ;
: find-cycle ( x0 'f -- mu lambda ) \ Brent's algorithm
2dup cycle-length dup >r cycle-start r> ;
\ --- usage ---
: .cycle { start len x0 'f -- }
x0
start 1- 0 do 'f execute loop
len 0 do 'f execute dup . loop
drop ;
: f(x) dup * 1+ 255 mod ;
3 ' f(x) find-cycle
." The cycle starts at offset " over . ." and has a length of " dup . cr
." The cycle is " 3 ' f(x) .cycle cr
bye
- Output:
The cycle starts at offset 2 and has a length of 6 The cycle is 101 2 5 26 167 95
FreeBASIC
Brent's algorithm
' version 11-01-2017
' compile with: fbc -s console
' define the function f(x)=(x*x +1) mod 255
Function f(x As Integer) As Integer
Return (x * x +1) Mod 255
End 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 = "; lam
Print
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)
Next
Print
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
- 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
' version 11-01-2017
' compile with: fbc -s console
' define the function f(x)=(x*x +1) mod 255
Function f(x As Integer) As Integer
Return (x * x +1) Mod 255
End 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 = "; lam
Print
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)
Next
Print
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
Go
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]
Groovy
import java.util.function.IntUnaryOperator
class CycleDetection {
static void main(String[] args) {
brent({ i -> (i * i + 1) % 255 }, 3)
}
static void brent(IntUnaryOperator f, int x0) {
int cycleLength = -1
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) {
printf("Cycle length: %d%nCycle: ", len)
int n = x0
for (int i = 0; i < start + len; i++) {
n = f.applyAsInt(n)
if (i >= start) {
printf("%s ", n)
}
}
println()
}
}
- Output:
Cycle length: 6 Cycle: 2 5 26 167 95 101
Haskell
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 Int
findCycleLength [] = Nothing
findCycleLength (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=. ~.@(,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 3
2 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
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
jq
Works with gojq, the Go implementation of jq
def floyd(f; x0):
{tort: (x0|f)}
| .hare = (.tort|f)
| until( .tort == .hare;
.tort |= f
| .hare |= (f|f) )
| .mu = 0
| .tort = x0
| until( .tort == .hare;
.tort |= f
| .hare |= f
| .mu += 1)
| .lambda = 1
| .hare = (.tort|f)
| until (.tort == .hare;
.hare |= f
| .lambda += 1 )
| {lambda, mu} ;
def task(f; x0):
def skip($n; stream):
foreach stream as $s (0; .+1; select(. > $n) | $s);
floyd(f; x0)
| .,
"Cycle:",
skip(.mu; limit((.lambda + .mu); 3 | recurse(f)));The specific function and task
def f: (.*. + 1) % 255;
task(f;3)- Output:
{
"lambda": 6,
"mu": 2
}
Cycle:
3
10
101
2
5
26
167
95
Julia
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, μ
end
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) |>
collect
println("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, cycleTab
end
local f = function (x) return (x * x + 1) % 255 end
local x0 = 3
local 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
Mathematica /Wolfram Language
s = {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};
{transient, repeat} = FindTransientRepeat[s, 2];
Print["Starting index: ", Length[transient] + 1]
Print["Cycles: ", Floor[(Length[s] - Length[transient])/Length[repeat]]]
- Output:
Starting index: 3 Cycles: 6
Modula-2
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 255
END 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;
ReadChar
END CycleDetection.
Nim
Translation of Wikipedia Python program.
import strutils, sugar
func brent(f: int -> int; x0: int): (int, int) =
# Main phase: search successive powers of two.
var
power, λ = 1
tortoise = x0
hare = f(x0)
while tortoise != hare:
if power == λ:
# Time to start a new power of two.
tortoise = hare
power *= 2
λ = 0
hare = f(hare)
inc λ
# Find the position of the first repetition of length λ.
tortoise = x0
hare = x0
for i in 0..<λ:
hare = f(hare)
# The distance between the hare and tortoise is now λ.
# Next, the hare and tortoise move at same speed until they agree.
var μ = 0
while tortoise != hare:
tortoise = f(tortoise)
hare = f(hare)
inc μ
result = (λ, μ)
when isMainModule:
func f(x: int): int = (x * x + 1) mod 255
let x0 = 3
let (λ, μ) = brent(f, x0)
echo "Cycle length: ", λ
echo "Cycle start index: ", μ
var cycle: seq[int]
var x = x0
for i in 0..<λ+μ:
if i >= μ: cycle.add x
x = f(x)
echo "Cycle: ", cycle.join(" ")
- Output:
Cycle length: 6 Cycle start index: 2 Cycle: 101 2 5 26 167 95
ooRexx
/* REXX */
x=3
list=x
Do 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
End
Exit
f: 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
use utf8;
sub cyclical_function { ($_[0] * $_[0] + 1) % 255 }
sub brent {
my($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
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 = 1 integer 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 s integer len, pos, x0 = 3 {s,len,pos} = brent(x0) printf(1," Brent's algorithm\n") ?s printf(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}
PL/M
100H:
/* CP/M CALLS */
BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS;
EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT;
PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT;
/* THIS IS A HACK TO CALL A FUNCTION GIVEN ITS ADDRESS */
APPLY: PROCEDURE (FN, ARG) ADDRESS;
DECLARE (FN, ARG) ADDRESS;
INNER: PROCEDURE (X) ADDRESS; /* THIS SETS UP THE ARGUMENTS IN MEMORY */
DECLARE X ADDRESS;
GO TO FN; /* THEN WE JUMP TO THE ADDRESS GIVEN */
END INNER;
RETURN INNER(ARG);
END APPLY;
/* THIS IS ANOTHER HACK - THE SINGLE 0 (WHICH IS AN 8080 NOP) WILL BE
STORED AS A CONSTANT RIGHT BEFORE THE FUNCTION.
PL/M-80 DOES NOT ALLOW THE PROGRAMMER TO DIRECTLY GET THE ADDRESS
OF A FUNCTION, BUT THE ADDRESS OF THIS CONSTANT WILL BE RIGHT IN
FRONT OF IT AND CAN BE USED INSTEAD. */
DECLARE F$ADDR DATA (0);
/* F(X) FROM THE TASK */
F: PROCEDURE (X) ADDRESS;
DECLARE X ADDRESS;
RETURN (X*X + 1) MOD 255;
END F;
/* PRINT A NUMBER */
PRINT$NUMBER: PROCEDURE (N);
DECLARE S (7) BYTE INITIAL ('..... $');
DECLARE (N, P) ADDRESS, C BASED P BYTE;
P = .S(5);
DIGIT:
P = P - 1;
C = N MOD 10 + '0';
N = N / 10;
IF N > 0 THEN GO TO DIGIT;
CALL PRINT(P);
END PRINT$NUMBER;
/* PRINT F^M(X0) TO F^N(X0) */
PRINT$RANGE: PROCEDURE (FN, X0, M, N);
DECLARE (FN, X0, M, N, I) ADDRESS;
DO I=0 TO N-1;
IF I>=M THEN CALL PRINT$NUMBER(X0);
X0 = APPLY(FN,X0);
END;
CALL PRINT(.(13,10,'$'));
END PRINT$RANGE;
/* BRENT'S ALGORITHM */
BRENT: PROCEDURE (FN, X0, MU$A, LAM$A);
DECLARE (FN, X0, MU$A, LAM$A) ADDRESS;
DECLARE (MU BASED MU$A, LAM BASED LAM$A) ADDRESS;
DECLARE (TORT, HARE, POW, I) ADDRESS;
POW, LAM = 1;
TORT = X0;
HARE = APPLY(FN,X0);
DO WHILE TORT <> HARE;
IF POW = LAM THEN DO;
TORT = HARE;
POW = SHL(POW, 1);
LAM = 0;
END;
HARE = APPLY(FN,HARE);
LAM = LAM + 1;
END;
TORT, HARE = X0;
DO I=1 TO LAM;
HARE = APPLY(FN,HARE);
END;
MU = 0;
DO WHILE TORT <> HARE;
TORT = APPLY(FN,TORT);
HARE = APPLY(FN,HARE);
MU = MU + 1;
END;
END BRENT;
DECLARE (MU, LAM) ADDRESS;
/* PRINT THE FIRST 20 VALUES IN THE SEQUENCE */
CALL PRINT$RANGE(.F$ADDR, 3, 0, 20);
/* FIND THE CYCLE */
CALL BRENT(.F$ADDR, 3, .MU, .LAM);
CALL PRINT(.'LENGTH: $');
CALL PRINT$NUMBER(LAM);
CALL PRINT(.'START: $');
CALL PRINT$NUMBER(MU);
CALL PRINT(.(13,10,'$'));
/* PRINT THE CYCLE */
CALL PRINT$RANGE(.F$ADDR, 3, MU, MU+LAM);
CALL EXIT;
EOF- Output:
3 10 101 2 5 26 167 95 101 2 5 26 167 95 101 2 5 26 167 95 LENGTH: 6 START: 2 101 2 5 26 167 95
Python
Procedural
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]
Functional
In functional terms, the problem lends itself at first sight (see the Haskell version), to a reasonably compact recursive definition of a cycleLength function:
'''Cycle detection by recursion.'''
from itertools import (chain, cycle, islice)
from operator import (eq)
# cycleFound :: Eq a => [a] -> Maybe ([a], Int, Int)
def cycleFound(xs):
'''Just the first cycle found, with its length
and start index, or Nothing if no cycle seen.
'''
return bind(cycleLength(xs))(
lambda n: bind(
findIndex(uncurry(eq))(zip(xs, xs[n:]))
)(lambda iStart: Just(
(xs[iStart:iStart + n], n, iStart)
))
)
# cycleLength :: Eq a => [a] -> Maybe Int
def cycleLength(xs):
'''Just the length of the first cycle found,
or Nothing if no cycle seen.
'''
def go(pwr, lng, x, ys):
if ys:
y, *yt = ys
return Just(lng) if x == y else (
go(2 * pwr, 1, y, yt) if (
lng == pwr
) else go(pwr, 1 + lng, x, yt)
)
else:
return Nothing()
return go(1, 1, xs[0], xs[1:]) if xs else Nothing()
# TEST ----------------------------------------------------
# main :: IO ()
def main():
'''Reports of any cycle detection.'''
print(
fTable(
'First cycle detected, if any:\n'
)(fst)(maybe('No cycle found')(
showCycle
))(
compose(cycleFound)(snd)
)([
(
'cycle([1, 2, 3])',
take(1000)(cycle([1, 2, 3]))
), (
'[0..100] + cycle([1..8])',
take(1000)(
chain(
enumFromTo(0)(100),
cycle(enumFromTo(1)(8))
)
)
), (
'[1..500]',
enumFromTo(1)(500)
), (
'f(x) = (x*x + 1) modulo 255',
take(1000)(iterate(
lambda x: (1 + (x * x)) % 255
)(3))
)
])
)
# DISPLAY -------------------------------------------------
# showList :: [a] -> String
def showList(xs):
''''Compact stringification of a list,
(no spaces after commas).
'''
return ''.join(repr(xs).split())
# showCycle :: ([a], Int, Int) -> String
def showCycle(cli):
'''Stringification of cycleFound tuple.'''
c, lng, iStart = cli
return showList(c) + ' (from:' + str(iStart) + (
', length:' + str(lng) + ')'
)
# GENERIC -------------------------------------------------
# Just :: a -> Maybe a
def Just(x):
'''Constructor for an inhabited Maybe (option type) value.'''
return {'type': 'Maybe', 'Nothing': False, 'Just': x}
# Nothing :: Maybe a
def Nothing():
'''Constructor for an empty Maybe (option type) value.'''
return {'type': 'Maybe', 'Nothing': True}
# bind (>>=) :: Maybe a -> (a -> Maybe b) -> Maybe b
def bind(m):
'''bindMay provides the mechanism for composing a
sequence of (a -> Maybe b) functions.
If m is Nothing, it is passed straight through.
If m is Just(x), the result is an application
of the (a -> Maybe b) function (mf) to x.'''
return lambda mf: (
m if m.get('Nothing') else mf(m.get('Just'))
)
# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
def compose(g):
'''Right to left function composition.'''
return lambda f: lambda x: g(f(x))
# enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
'''Integer enumeration from m to n.'''
return lambda n: list(range(m, 1 + n))
# findIndex :: (a -> Bool) -> [a] -> Maybe Int
def findIndex(p):
'''Just the first index at which an
element in xs matches p,
or Nothing if no elements match.'''
def go(xs):
try:
return Just(next(
i for i, x in enumerate(xs) if p(x)
))
except StopIteration:
return Nothing()
return lambda xs: go(xs)
# fst :: (a, b) -> a
def fst(tpl):
'''First member of a pair.'''
return tpl[0]
# fTable :: String -> (a -> String) ->
# (b -> String) ->
# (a -> b) -> [a] -> String
def fTable(s):
'''Heading -> x display function -> fx display function ->
f -> value list -> tabular string.'''
def go(xShow, fxShow, f, xs):
w = max(map(compose(len)(xShow), xs))
return s + '\n' + '\n'.join([
xShow(x).rjust(w, ' ') + ' -> ' + fxShow(f(x)) for x in xs
])
return lambda xShow: lambda fxShow: (
lambda f: lambda xs: go(
xShow, fxShow, f, xs
)
)
# iterate :: (a -> a) -> a -> Gen [a]
def iterate(f):
'''An infinite list of repeated
applications of f to x.'''
def go(x):
v = x
while True:
yield v
v = f(v)
return lambda x: go(x)
# maybe :: b -> (a -> b) -> Maybe a -> b
def maybe(v):
'''Either the default value v, if m is Nothing,
or the application of f to x,
where m is Just(x).'''
return lambda f: lambda m: v if m.get('Nothing') else (
f(m.get('Just'))
)
# snd :: (a, b) -> b
def snd(tpl):
'''Second member of a pair.'''
return tpl[1]
# take :: Int -> [a] -> [a]
# take :: Int -> String -> String
def take(n):
'''The prefix of xs of length n,
or xs itself if n > length xs.'''
return lambda xs: (
xs[0:n]
if isinstance(xs, list)
else list(islice(xs, n))
)
# uncurry :: (a -> b -> c) -> ((a, b) -> c)
def uncurry(f):
'''A function over a tuple
derived from a default or
curried function.'''
return lambda xy: f(xy[0], xy[1])
# concat :: [[a]] -> [a]
# concat :: [String] -> String
def concat(xxs):
'''The concatenation of all the elements in a list.'''
xs = list(chain.from_iterable(xxs))
unit = '' if isinstance(xs, str) else []
return unit if not xs else (
''.join(xs) if isinstance(xs[0], str) else xs
)
# MAIN ---
if __name__ == '__main__':
main()
- Output:
First cycle detected, if any:
cycle([1, 2, 3]) -> [1,2,3] (from:0, length:3)
[0..100] + cycle([1..8]) -> [1,2,3,4,5,6,7,8] (from:101, length:8)
[1..500] -> No cycle found
f(x) = (x*x + 1) modulo 255 -> [101,2,5,26,167,95] (from:2, length:6)
But recursion scales poorly in Python, and the version above, while good for lists of a few hundred elements, will need reworking for longer lists and better use of space.
If we start by refactoring the recursion into the form of a higher order (but still recursive) until p f x function, we can then reimplement the internals of until itself to avoid recursion, without losing the benefits of compositional structure:
Recursive until:
# until :: (a -> Bool) -> (a -> a) -> a -> a
def until(p):
'''The result of repeatedly applying f until p holds.
The initial seed value is x.'''
def go(f, x):
return x if p(x) else go(f, f(x))
return lambda f: lambda x: go(f, x)
cycleLength refactored in terms of until:
# cycleLength :: Eq a => [a] -> Maybe Int
def cycleLength(xs):
'''Just the length of the first cycle found,
or Nothing if no cycle seen.'''
# f :: (Int, Int, Int, [Int]) -> (Int, Int, Int, [Int])
def f(tpl):
pwr, lng, x, ys = tpl
y, *yt = ys
return (2 * pwr, 1, y, yt) if (
lng == pwr
) else (pwr, 1 + lng, x, yt)
# p :: (Int, Int, Int, [Int]) -> Bool
def p(tpl):
_, _, x, ys = tpl
return (not ys) or x == ys[0]
if xs:
_, lng, x, ys = until(p)(f)(
(1, 1, xs[0], xs[1:])
)
return (
Just(lng) if (x == ys[0]) else Nothing()
) if ys else Nothing()
else:
return Nothing()
Iterative reimplementation of until:
# until_ :: (a -> Bool) -> (a -> a) -> a -> a
def until_(p):
'''The result of repeatedly applying f until p holds.
The initial seed value is x.'''
def go(f, x):
v = x
while not p(v):
v = f(v)
return v
return lambda f: lambda x: go(f, x)
and now it all works again, with the structure conserved but recursion removed.
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:
'''Cycle detection without recursion.'''
from itertools import (chain, cycle, islice)
from operator import (eq)
# cycleFound :: Eq a => [a] -> Maybe ([a], Int, Int)
def cycleFound(xs):
'''Just the first cycle found, with its length
and start index, or Nothing if no cycle seen.
'''
return bind(cycleLength(xs))(
lambda n: bind(
findIndex(uncurry(eq))(zip(xs, xs[n:]))
)(lambda iStart: Just(
(xs[iStart:iStart + n], n, iStart)
))
)
# cycleLength :: Eq a => [a] -> Maybe Int
def cycleLength(xs):
'''Just the length of the first cycle found,
or Nothing if no cycle seen.'''
# f :: (Int, Int, Int, [Int]) -> (Int, Int, Int, [Int])
def f(tpl):
pwr, lng, x, ys = tpl
y, *yt = ys
return (2 * pwr, 1, y, yt) if (
lng == pwr
) else (pwr, 1 + lng, x, yt)
# p :: (Int, Int, Int, [Int]) -> Bool
def p(tpl):
_, _, x, ys = tpl
return (not ys) or x == ys[0]
if xs:
_, lng, x, ys = until(p)(f)(
(1, 1, xs[0], xs[1:])
)
return (
Just(lng) if (x == ys[0]) else Nothing()
) if ys else Nothing()
else:
return Nothing()
# TEST ----------------------------------------------------
# main :: IO ()
def main():
'''Reports of any cycle detection.'''
print(
fTable(
'First cycle detected, if any:\n'
)(fst)(maybe('No cycle found')(
showCycle
))(
compose(cycleFound)(snd)
)([
(
'cycle([1, 2, 3])',
take(10000)(cycle([1, 2, 3]))
), (
'[0..10000] + cycle([1..8])',
take(20000)(
chain(
enumFromTo(0)(10000),
cycle(enumFromTo(1)(8))
)
)
), (
'[1..10000]',
enumFromTo(1)(10000)
), (
'f(x) = (x*x + 1) modulo 255',
take(10000)(iterate(
lambda x: (1 + (x * x)) % 255
)(3))
)
])
)
# DISPLAY -------------------------------------------------
# showList :: [a] -> String
def showList(xs):
''''Compact stringification of a list,
(no spaces after commas).
'''
return ''.join(repr(xs).split())
# showCycle :: ([a], Int, Int) -> String
def showCycle(cli):
'''Stringification of cycleFound tuple.'''
c, lng, iStart = cli
return showList(c) + ' (from:' + str(iStart) + (
', length:' + str(lng) + ')'
)
# GENERIC -------------------------------------------------
# Just :: a -> Maybe a
def Just(x):
'''Constructor for an inhabited Maybe (option type) value.'''
return {'type': 'Maybe', 'Nothing': False, 'Just': x}
# Nothing :: Maybe a
def Nothing():
'''Constructor for an empty Maybe (option type) value.'''
return {'type': 'Maybe', 'Nothing': True}
# bind (>>=) :: Maybe a -> (a -> Maybe b) -> Maybe b
def bind(m):
'''bindMay provides the mechanism for composing a
sequence of (a -> Maybe b) functions.
If m is Nothing, it is passed straight through.
If m is Just(x), the result is an application
of the (a -> Maybe b) function (mf) to x.'''
return lambda mf: (
m if m.get('Nothing') else mf(m.get('Just'))
)
# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
def compose(g):
'''Right to left function composition.'''
return lambda f: lambda x: g(f(x))
# concat :: [[a]] -> [a]
# concat :: [String] -> String
def concat(xxs):
'''The concatenation of all the elements in a list.'''
xs = list(chain.from_iterable(xxs))
unit = '' if isinstance(xs, str) else []
return unit if not xs else (
''.join(xs) if isinstance(xs[0], str) else xs
)
# enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
'''Integer enumeration from m to n.'''
return lambda n: list(range(m, 1 + n))
# findIndex :: (a -> Bool) -> [a] -> Maybe Int
def findIndex(p):
'''Just the first index at which an
element in xs matches p,
or Nothing if no elements match.'''
def go(xs):
try:
return Just(next(
i for i, x in enumerate(xs) if p(x)
))
except StopIteration:
return Nothing()
return lambda xs: go(xs)
# fst :: (a, b) -> a
def fst(tpl):
'''First member of a pair.'''
return tpl[0]
# fTable :: String -> (a -> String) ->
# (b -> String) ->
# (a -> b) -> [a] -> String
def fTable(s):
'''Heading -> x display function -> fx display function ->
f -> value list -> tabular string.'''
def go(xShow, fxShow, f, xs):
w = max(map(compose(len)(xShow), xs))
return s + '\n' + '\n'.join([
xShow(x).rjust(w, ' ') + ' -> ' + fxShow(f(x)) for x in xs
])
return lambda xShow: lambda fxShow: (
lambda f: lambda xs: go(
xShow, fxShow, f, xs
)
)
# iterate :: (a -> a) -> a -> Gen [a]
def iterate(f):
'''An infinite list of repeated
applications of f to x.'''
def go(x):
v = x
while True:
yield v
v = f(v)
return lambda x: go(x)
# maybe :: b -> (a -> b) -> Maybe a -> b
def maybe(v):
'''Either the default value v, if m is Nothing,
or the application of f to x,
where m is Just(x).'''
return lambda f: lambda m: v if m.get('Nothing') else (
f(m.get('Just'))
)
# snd :: (a, b) -> b
def snd(tpl):
'''Second member of a pair.'''
return tpl[1]
# take :: Int -> [a] -> [a]
# take :: Int -> String -> String
def take(n):
'''The prefix of xs of length n,
or xs itself if n > length xs.'''
return lambda xs: (
xs[0:n]
if isinstance(xs, list)
else list(islice(xs, n))
)
# uncurry :: (a -> b -> c) -> ((a, b) -> c)
def uncurry(f):
'''A function over a tuple
derived from a default or
curried function.'''
return lambda xy: f(xy[0], xy[1])
# until :: (a -> Bool) -> (a -> a) -> a -> a
def until(p):
'''The result of repeatedly applying f until p holds.
The initial seed value is x.'''
def go(f, x):
v = x
while not p(v):
v = f(v)
return v
return lambda f: lambda x: go(f, x)
# MAIN ---
if __name__ == '__main__':
main()
- Output:
First cycle detected, if any:
cycle([1, 2, 3]) -> [1,2,3] (from:0, length:3)
[0..10000] + cycle([1..8]) -> [1,2,3,4,5,6,7,8] (from:10001, length:8)
[1..10000] -> No cycle found
f(x) = (x*x + 1) modulo 255 -> [101,2,5,26,167,95] (from:2, length:6)
Quackery
[ stack ] is fun ( --> s )
[ stack ] is pow ( --> s )
[ stack ] is len ( --> s )
[ fun put
1 pow put
1 len put
dup fun share do
[ 2dup != while
len share
pow share = if
[ nip dup
pow share
pow tally
0 len replace ]
fun share do
1 len tally
again ]
2drop
fun release
pow release
len take ] is cyclelen ( n x --> n )
[ 0 temp put
dip [ fun put dup ]
times [ fun share do ]
[ 2dup != while
fun share do
dip [ fun share do ]
1 temp tally
again ]
2drop
fun release
temp take ] is cyclepos ( n x n --> n )
[ 2dup cyclelen
dup dip cyclepos ] is brent ( n x --> n n )
[ 2dup
20 times
[ over echo sp
tuck do swap ]
cr cr
2drop
brent
say "cycle length is "
echo cr
say "cycle starts at "
echo ] is task ( n x --> )
3 ' [ 2 ** 1+ 255 mod ] task- Output:
3 10 101 2 5 26 167 95 101 2 5 26 167 95 101 2 5 26 167 95 cycle length is 6 cycle starts at 2
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
Raku
(formerly Perl 6)
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)
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 when 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*/
say ' original list=' $ ... /*display the sequence. */
say 'numbers in list=' words($) /*display number of numbers. */
say ' cycle length =' words($) - cycle /*display the cycle to term*/
say ' start index =' cycle - 1 " ◄─── zero based" /* " " index " " */
say 'cycle sequence =' subword($, cycle, words($)-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= 1; x= 3; $= x /*assign initial value. */
do #=1+words($); x= f(x); $= $ x /*add the number to list.*/
if !.x\==. then leave /*A repeat? Then leave.*/
!.x= # /*N: numbers in $ list.*/
end /*#*/
say ' original list=' $ ... /*maybe display the list.*/
say 'numbers in list=' # /*display number of nums.*/
say ' cycle length =' # - !.x /* " " cycle. */
say ' start index =' !.x - 1 " ◄─── zero based" /* " " index. */
say 'cycle sequence =' subword($, !.x, # - !.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. " " " */
say
x=3; $=x; $$=; m=100; !.=.; !.x=1 /*M: maximum numbers to display.*/
do n=1+words($); x= f(x); $$=$$ x
if n//2000==0 then do; $=$ $$; $$=; end /*Is a 2000th N? Rejoin.*/
if !.x\==. then leave /*is this number a repeat? Leave*/
!.x= n
end /*n*/ /*N: is the size of $ list.*/
$= space($ $$) /*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=1
Do 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 */
End
Say '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*/
RPL
Translation of the Brent algorithm given in Wikipedia.
≪ 1 1 0 → f x0 power lam mu ≪ x0 DUP f EVAL @ Main phase: search successive powers of two WHILE DUP2 ≠ REPEAT IF power lam == THEN @ time to start a new power of two? SWAP DROP DUP 2 'power' STO* 0 'lam' STO END f EVAL 1 'lam' STO+ END DROP2 x0 DUP @ Find the position of the first repetition of length λ 0 lam 1 - START f EVAL NEXT @ distance between the hare and tortoise is now λ WHILE DUP2 ≠ REPEAT @ the hare and tortoise move at same speed until they agree f EVAL SWAP f EVAL SWAP 1 'mu' STO+ END DROP2 lam mu ≫ ≫ 'CYCLEB' STO
≪ SQ 1 + 255 MOD ≫ 0 CYCLEB
- Output:
2: 6 1: 2
Ruby
# 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, mu
end
# A recurrence relation to use in testing
def f(x) (x * x + 1) % 255 end
# Display the first 41 numbers in the test series
puts (1..40).reduce([3]){|acc,_| acc << f(acc.last)}.join(",")
# Test the findCycle function
clength, 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
Scala
Procedural
- Output:
Best seen in running your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).
object CycleDetection extends App {
def brent(f: Int => Int, x0: Int): (Int, Int) = {
// main phase: search successive powers of two
// f(x0) is the element/node next to x0.
var (power, λ, μ, tortoise, hare) = (1, 1, 0, x0, f(x0))
while (tortoise != hare) {
if (power == λ) { // time to start a new power of two?
tortoise = hare
power *= 2
λ = 0
}
hare = f(hare)
λ += 1
}
// Find the position of the first repetition of length 'λ'
tortoise = x0
hare = x0
for (i <- 0 until λ) hare = f(hare)
// The distance between the hare and tortoise is now 'λ'.
// Next, the hare and tortoise move at same speed until they agree
while (tortoise != hare) {
tortoise = f(tortoise)
hare = f(hare)
μ += 1
}
(λ, μ)
}
def cycle = loop.slice(μ, μ + λ)
def f = (x: Int) => (x * x + 1) % 255
// Generator for the first terms of the sequence starting from 3
def loop: LazyList[Int] = 3 #:: loop.map(f(_))
val (λ, μ) = brent(f, 3)
println(s"Cycle length = $λ")
println(s"Start index = $μ")
println(s"Cycle = ${cycle.force}")
}
Functional
import scala.annotation.tailrec
object CycleDetection {
def brent(f: Int => Int, x0: Int): (Int, Int) = {
val lambda = findLambda(f, x0)
val mu = findMu(f, x0, lambda)
(lambda, mu)
}
def cycle(f: Int => Int, x0: Int): Seq[Int] = {
val (lambda, mu) = brent(f, x0)
(1 until mu + lambda)
.foldLeft(Seq(x0))((list, _) => f(list.head) +: list)
.reverse
.drop(mu)
}
def findLambda(f: Int => Int, x0: Int): Int = {
findLambdaRec(f, tortoise = x0, hare = f(x0), power = 1, lambda = 1)
}
def findMu(f: Int => Int, x0: Int, lambda: Int): Int = {
val hare = (0 until lambda).foldLeft(x0)((x, _) => f(x))
findMuRec(f, tortoise = x0, hare, mu = 0)
}
@tailrec
private def findLambdaRec(f: Int => Int, tortoise: Int, hare: Int, power: Int, lambda: Int): Int = {
if (tortoise == hare) {
lambda
} else {
val (newTortoise, newPower, newLambda) = if (power == lambda) {
(hare, power * 2, 0)
} else {
(tortoise, power, lambda)
}
findLambdaRec(f, newTortoise, f(hare), newPower, newLambda + 1)
}
}
@tailrec
private def findMuRec(f: Int => Int, tortoise: Int, hare: Int, mu: Int): Int = {
if (tortoise == hare) {
mu
} else {
findMuRec(f, f(tortoise), f(hare), mu + 1)
}
}
def main(args: Array[String]): Unit = {
val f = (x: Int) => (x * x + 1) % 255
val x0 = 3
val (lambda, mu) = brent(f, x0)
val list = cycle(f, x0)
println("Cycle length = " + lambda)
println("Start index = " + mu)
println("Cycle = " + list.mkString(","))
}
}
- Output:
Cycle length = 6 Start index = 2 Cycle = 101,2,5,26,167,95
Sidef
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]
Visual Basic .NET
Module Module1
Function FindCycle(Of T As IEquatable(Of T))(x0 As T, yielder As Func(Of T, T)) As Tuple(Of Integer, Integer)
Dim power = 1
Dim lambda = 1
Dim tortoise As T
Dim hare As T
tortoise = x0
hare = yielder(x0)
' Find lambda, the cycle length
While Not tortoise.Equals(hare)
If power = lambda Then
tortoise = hare
power *= 2
lambda = 0
End If
hare = yielder(hare)
lambda += 1
End While
' Find mu, the zero-based index of the start of the cycle
Dim mu = 0
tortoise = x0
hare = x0
For times = 1 To lambda
hare = yielder(hare)
Next
While Not tortoise.Equals(hare)
tortoise = yielder(tortoise)
hare = yielder(hare)
mu += 1
End While
Return Tuple.Create(lambda, mu)
End Function
Sub Main()
' A recurrence relation to use in testing
Dim sequence = Function(_x As Integer) (_x * _x + 1) Mod 255
' Display the first 41 numbers in the test series
Dim x = 3
Console.Write(x)
For times = 0 To 39
x = sequence(x)
Console.Write(",{0}", x)
Next
Console.WriteLine()
' Test the FindCycle method
Dim cycle = FindCycle(3, sequence)
Console.WriteLine("Cycle length = {0}", cycle.Item1)
Console.WriteLine("Start index = {0}", cycle.Item2)
End Sub
End Module
- 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
Wren
Working from the code in the Wikipedia article:
var brent = Fn.new { |f, x0|
var lam = 1
var power = 1
var tortoise = x0
var hare = f.call(x0)
while (tortoise != hare) {
if (power == lam) {
tortoise = hare
power = power * 2
lam = 0
}
hare = f.call(hare)
lam = lam + 1
}
tortoise = hare = x0
for (i in 0...lam) hare = f.call(hare)
var mu = 0
while (tortoise != hare) {
tortoise = f.call(tortoise)
hare = f.call(hare)
mu = mu + 1
}
return [lam, mu]
}
var f = Fn.new { |x| (x*x + 1) % 255 }
var x0 = 3
var x = x0
var seq = List.filled(21, 0) // limit to first 21 terms say
for (i in 0..20) {
seq[i] = x
x = f.call(x)
}
var res = brent.call(f, x0)
var lam = res[0]
var mu = res[1]
System.print("Sequence = %(seq)")
System.print("Cycle length = %(lam)")
System.print("Start index = %(mu)")
System.print("Cycle = %(seq[mu...mu+lam])")
- Output:
Sequence = [3, 10, 101, 2, 5, 26, 167, 95, 101, 2, 5, 26, 167, 95, 101, 2, 5, 26, 167, 95, 101] Cycle length = 6 Start index = 2 Cycle = [101, 2, 5, 26, 167, 95]
XPL0
func F(X); \Function to be iterated
int X;
return rem((X*X + 1)/255);
def X0 = 3;
int Lam, Mu, Tortoise, Hare, Power, I, X;
[\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
[if Power = Lam then \time to start a new Power of two?
[Tortoise:= Hare;
Power:= Power*2;
Lam:= 0;
];
Hare:= F(Hare);
Lam:= Lam+1;
];
\Find the position of the first repetition of length Lam
Mu:= 0;
Tortoise:= X0; Hare:= X0;
\Range(Lam) produces a list with the values 0, 1, ... , Lam-1
for I:= 0 to Lam-1 do
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 do
[Tortoise:= F(Tortoise);
Hare:= F(Hare);
Mu:= Mu+1;
];
Text(0, "Cycle length: "); IntOut(0, Lam); CrLf(0);
Text(0, "Cycle start index: "); IntOut(0, Mu); CrLf(0);
Text(0, "Cycle: [");
X:= X0;
for I:= 0 to Mu+Lam-2 do
[if I >= Mu then
[IntOut(0, X); Text(0, ", ")];
X:= F(X);
];
IntOut(0, X); Text(0, "]^m^j");
]- Output:
Cycle length: 6 Cycle start index: 2 Cycle: [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