Cycle detection

Cycle detection
You are encouraged to solve this task according to the task description, using any language you may know.

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

C#

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

<lang C#>

// First file: Cycles.cs

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)); } } }

</lang>

Ruby

Works with: ruby version 2.0

<lang Ruby>

1. Author: Paul Anton Chernoch
2. Purpose:
3. Find the cycle length and start position of a numerical seried using Brent's cycle algorithm.
5. Given a recurrence relation X[n+1] = f(X[n]) where f() has
6. a finite range, you will eventually repeat a value that you have seen before.
7. Once this happens, all subsequent values will form a cycle that begins
8. with the first repeated value. The period of that cycle may be of any length.
10. Parameters:
11. x0 ...... First integer value in the sequence
12. block ... Block that takes a single integer as input
13. and returns a single integer as output.
14. This yields a sequence of numbers that eventually repeats.
15. Returns:
16. Two values: lambda and mu
17. lambda .. length of cycle
18. 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
 mu = 0
 tortoise = hare = x0
 lambda.times do
   hare = yield(hare)
 end
 
 while tortoise != hare
   tortoise = yield(tortoise)
   hare = yield(hare)
   mu += 1
 end
 
 return lambda, mu

end

1. A recurrence relation to use in testing

def f(x)

 return (x * x + 1) % 255

end

1. Display the first 41 numbers in the test series

x = 3 print "#{x}" 40.times do

 x = f(x)
 print ",#{x}"

end print "\n"

1. Test the findCycle function

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

print "Cycle length = #{clength}\nStart index = #{cstart}\n"

</lang>