Assume that you have a function f: S → S and any initial value x₀ ∈ S
(in this visualization, we are restricted to f(x) = (A*x^2 + B*x + C) % M and x₀ % M hence the function f has domain and range ∈ [0..M-1]).

The sequence of iterated function values:
{x₀, x₁ = f(x₀), x₂ = f(x₁), ..., x_i = f(x_i-1), ...}
must eventually use the same value twice (cycle), i.e. a ≠ b such that x_a = x_b.
Afterwards, the sequence must continue by repeating the cycle of values from x_a to x_b-1.

Let mu (μ) be the smallest index and let lambda (λ) (the cycle length) be the smallest positive integer such that x_μ = x_μ+λ.

The cycle-finding problem: Find such μ and λ, given f and x₀.

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You can define custom function f(x) = (A*x² + B*x + C) % M here.

1. Random: The function will be f(x) = (x² - 1) % M and only M and the x₀ are randomly generated.
2. Custom: You can specify the coefficients A, B, C of f(x) (ranging from -999 to 999), the modulo value M (ranging from 10 to 1000) and the initial value x₀ (ranging from 0 to M-1).

You can also set custom x₀, which must be ∈ [0..M-1].

Pro-tip: Since you are not logged-in, you may be a first time visitor who are not aware of the following keyboard shortcuts to navigate this e-Lecture mode: [PageDown] to advance to the next slide, [PageUp] to go back to the previous slide, [Esc] to toggle between this e-Lecture mode and exploration mode.

Floyd's Tortoise-Hare Cycle-Finding is one algorithm that can solve this problem efficiently in both time and space complexities.

It just requires O(μ+λ) time and O(1) space to do the job.

Another pro-tip: We designed this visualization and this e-Lecture mode to look good on 1366x768 resolution or larger (typical modern laptop resolution in 2017). We recommend using Google Chrome to access VisuAlgo. Go to full screen mode (F11) to enjoy this setup. However, you can use zoom-in (Ctrl +) or zoom-out (Ctrl -) to calibrate this.

This is the visualization of Floyd's Tortoise-Hare Cycle-Finding algorithm.

The shape of rho (ρ) of the sequence of iterated function values defined by f(x) and x₀ nicely visualizes μ and λ.

VisuAlgo uses green vertices to represent the tortoise (t) and orange vertices to represent the hare (h).

We start from x₀.

While tortoise's pointer != hare's pointer, we advance tortoise/hare by one step/two steps to their next values by calling f(tortoise)/f(f(hare)).

We reset hare back to x₀ and keep tortoise at its current position.

Then, we iteratively advance both pointers to their next values by one step as this maintains the gap of tortoise and hare by kλ.

The first time both pointers are equal tells us the value of μ.

Again, we let tortoise stays in its current position and set hare next to it.

Then, we iteratively move hare to the next value by one step.

The first time both pointers are equal tells us the value of λ.

The animation of this algorithm should clear up most questions involving this algorithm.

We will expand this e-Lecture slides to contain more information about this algorithm later.

Control the animation with the player controls! Keyboard shortcuts are:

Return to 'Exploration Mode' to start exploring!

Note that if you notice any bug in this visualization or if you want to request for a new visualization feature, do not hesitate to drop an email to the project leader: Dr Steven Halim via his email address: stevenhalim at gmail dot com.