Click here to Skip to main content
Click here to Skip to main content

Tagged as

Chinese Remainder Problem

, 17 Oct 2008 CPOL
Rate this:
Please Sign up or sign in to vote.
Solve the Chinese remainder problem cleverly
ChineseRemainderProblem.png

Introduction

Around A.D. 100, the Chinese mathematician Sun-Tsu solved the problem of finding those integers x that leave remainders 2, 3, and 2 when divided by 3, 5, and 7 respectively. One such solution is x = 23; all solutions are of the form 23 + 105k for arbitrary integers k.

The Chinese remainder problem says that integers a,b,c are pairwise coprime, N leaves remainders r1, r2, r3 when divided by a, b, c respectively, finding N. The problem can be described by the following equation:

N = a * x + r<sub>1</sub> = b * y + r<sub>2</sub> = c * z + r<sub>3</sub>

How can we solve this problem and how to compute it by a computer?

Background

Traditionally this problem is solved by Chinese remainder theorem, using the following approach:
Find numbers n1, n2, n3 such that:
n1 mod a = 1 and n1 mod bc = 0
n2 mod b = 1 and n2 mod ac = 0
n3 mod c = 1 and n3 mod ab = 0
Then N1 = n1*r1 + n2*r2 + n3*r3, N0 = N1 mod abc, N = N0 + abc*n where n=0,1,2...

For example, we have N = 3x+2 = 5y+3 = 7z+2
n1 = 35u = 3v+1 = 70 when u=2, v=23
n2 = 21u = 5v+1 = 21 when u=1, v=4
n3 = 15u = 7v+1 = 15 when u=1, v=2
N1 = 70*2 + 21*3 + 15*2 = 233
N0 = 233 mod (3*5*7) = 23

A New Solution

Now I present a method that can compute the smallest solution N0 directly, it is easy to carry out both by human and by computer.

Firstly we solve a * x + r1 = b * y + r2, it can be rewritten as ax = by + r.
Notice that if x0,y0 is a solution of ax = by + 1, then x = bn + rx0,y = an + ry0 is a solution of ax = by + r for any integer n.
Because a(bn+rx0) = b(an+ry0)+r
=> abn + arx0 = ban + bry0 + r
=> ax0 = by0 + 1

So the problem come down to solve ax = by + 1.
If a > b, we can solve (a-b)x = by' + 1, y' = y - x first, say the solution is x,y', then y = y'+x.
Similarly this can be done if a < b.
Repeat the procedure, a and b become smaller and smaller.
When a = 1, we can let y = 1 and x = b + 1, else if b = 1 we can let x = 1 and y = a - 1.

Since x = bn + rx0,y = an + ry0 we can let x1 = rx0 mod b and y1 = ry0 mod a.
x1,y1 will be the smallest numbers fulfil a * x + r1 = b * y + r2.
Let r4 = ax1 + r1 = by1 + r2, we can see that solutions are of the form abu + r4.
Then we solve abu + r4 = cz + r3 and get u1,z1.
Finally N0 = abu1 + r4 = cz1 + r3.

For example, we solve N = 3x+2 = 5y+3 = 7z+2 again.
Firstly solve 3x=5y+1
=> 3(x-y)=2y+1
=> x-y=2(y-(x-y))+1
=> x0-y0=3,2y0-x0=1
=> x0=7,y0=4
=> r4 =3*7+2=23
Then we solve 15u+23=7z+2 => 7z=15u+21 (here r=21, we are ready to know ru0 mod 7 = 0),
7z=15u+1 => 7(z-u)=8u+1 => 7(z-2u)=u+1 => u0=6,z0=13 => u1 = 21*6 mod 7 = 0 => N0=15*0+23=23.

Using the Code

Here is a C# program that implements the above algorithm:

/// <summary>
/// Solve Chinese Remainder Problem
/// </summary>
public class RemainderProblem
{
    /// <summary>
    /// Get Greatest Common Divisor
    /// </summary>
    public static int GCD(int a, int b)
    {
        if (a > b)
        {
            int t = b;
            b = a;
            a = t;
        }
        int r = b % a;
        while (r != 0)
        {
            b = a;
            a = r;
            r = b % a;
        }
        return a;
    }
    
    /// <summary>
    /// Get Least Common Multiple
    /// </summary>
    public static int LCM(int a, int b)
    {
        int gcd = GCD(a, b);
        return a * b / gcd;
    }
    
    /// <summary>
    /// Solve ax=by+1
    /// </summary>
    public static void Solve(int a, int b, out int x, out int y)
    {
        if (b == 1)
        {
            x = 1;
            y = a - 1;
        }
        else if (a == 1)
        {
            y = 1;
            x = b + 1;
        }
        else if (b > a)
        {
            int subx;
            Solve(a, b - a, out subx, out y);
            x = y + subx;
        }
        else if (a > b)
        {
            int suby;
            Solve(a - b, b, out x, out suby);
            y = x + suby;
        }
        else
        {
            throw new Exception(String.Format
		("The equation {0}x={1}y+1 has no integer solution.", a, b));
        }
    }
    
    /// <summary>
    /// Solve ax = by + c
    /// </summary>
    public static void Solve(int a, int b, int c, out int x1, out int y1)
    {
        /* if 
               a * x0 = b * y0 + 1
           then
               x = b * t + c * x0
               y = a * t + c * y0
           satisfies
               a * x  = b * y  + c
           so
               x1 = (c * x0) mod b
               y1 = (c * y0) mod a
         */
        int d = GCD(a, b);
        if (d > 1)
        {
            if (c % d != 0)
            {
                throw new Exception(String.Format
		("The equation {0}x={1}y+{2} has no integer solution.", a, b, c));
            }
            a = a / d;
            b = b / d;
            c = c / d;
        }
        int x0, y0;
        Solve(a, b, out x0, out y0);
        x1 = (c * x0) % b;
        y1 = (c * y0) % a;
    }
    
    /// <summary>
    /// Solve a * x + r1 = b * y + r2
    /// </summary>
    public static void Solve(int a, int b, int r1, int r2, out int x1, out int y1)
    {
        if (r2 > r1)
        {
            Solve(a, b, r2 - r1, out x1, out y1);
        }
        else if (r1 > r2)
        {
            Solve(b, a, r1 - r2, out y1, out x1);
        }
        else
        {
            x1 = b;
            y1 = a;
        }
    }
    
    /// <summary>
    /// Solve a * x + r1 = b * y + r2 = c * z + r3 = n
    /// n0 = n mod t
    /// </summary>
    public static void Solve(int a, int b, int c, int r1, 
			int r2, int r3, out int n0, out int t)
    {
        int x1, y1;
        Solve(a, b, r1, r2, out x1, out y1);
        // let r4 = a * x1 + r1 = b * y1 + r2
        // to satisfy n = a * x + r1 = b * y + r2
        // we can see n = a * b * u + r4
        int r4 = a * x1 + r1;
        int x2, y2;
        Solve(a * b, c, r4, r3, out x2, out y2);
        n0 = c * y2 + r3;
        t = LCM(LCM(a, b), c);
        //n0 = n % t;
    }
}

In the form, we implement the button call as:

private void buttonSolve_Click(object sender, EventArgs e)
{
    try
    {
        int a = int.Parse(textBoxA.Text);
        int b = int.Parse(textBoxB.Text);
        int c = int.Parse(textBoxC.Text);
        int r1 = int.Parse(textBox1.Text);
        int r2 = int.Parse(textBox2.Text);
        int r3 = int.Parse(textBox3.Text);
        int n0, t;
        RemainderProblem.Solve(a, b, c, r1, r2, r3, out n0, out t);
        textBoxN.Text = n0.ToString();
        textBoxT.Text = t.ToString();
        listBoxN.Items.Clear();
        for (int i = 0; i < 20; i++)
        {
            listBoxN.Items.Add(n0 + t * i);
        }
        listBoxN.Items.Add("...");
    }
    catch (Exception error)
    {
        MessageBox.Show(error.Message, "No solution");
    }
}

Points of Interest

The algorithm can be easily extended to solve the generalized Chinese remainder problem:

N = a<sub>1</sub> * x<sub>1</sub> + r<sub>1</sub> = a<sub>2</sub> * x<sub>2</sub> + r<sub>2</sub> = ... = a<sub>n</sub> * x<sub>n</sub> + r<sub>n </sub>

History

  • 2008-09-24 Initial submission
  • 2008-10-18 Updated so that the condition a,b,c are pairwise coprime is no longer required
  • 2008-10-19 Fixed the error that if a,b,c are not coprime then the period t should be the least common multiple of a,b,c

License

This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Share

About the Author

Liu Junfeng
Software Developer (Senior) Beyondsoft SH
China China
No Biography provided

Comments and Discussions

 
Question//n0 = n % t; PinmemberAndrei Zakharevich20-Sep-13 12:41 
QuestionBuggy? PinmemberLono17524-Jan-12 10:38 
AnswerRe: Buggy? PinmemberLiu Junfeng22-Mar-12 20:46 
GeneralRe: Buggy? PinmemberLono17522-Mar-12 20:59 
GeneralMy vote of 5 Pinmemberjfriedman15-Feb-11 16:05 
Excellent Title. Excellent Layout. Excellent Explination.
Question韩信点兵? PinmemberHenry Liang24-Sep-08 8:00 
AnswerRe: 韩信点兵? PinmemberLiu Junfeng24-Sep-08 18:09 

General General    News News    Suggestion Suggestion    Question Question    Bug Bug    Answer Answer    Joke Joke    Rant Rant    Admin Admin   

Use Ctrl+Left/Right to switch messages, Ctrl+Up/Down to switch threads, Ctrl+Shift+Left/Right to switch pages.

| Advertise | Privacy | Terms of Use | Mobile
Web04 | 2.8.141223.1 | Last Updated 17 Oct 2008
Article Copyright 2008 by Liu Junfeng
Everything else Copyright © CodeProject, 1999-2014
Layout: fixed | fluid