CPMagical CP Library

Problem solution · Java

CCC 2001 S5 - Post's Correspondence

CCC 2001 S5 - Post's Correspondence: a Java solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Depth-first search
Source
CCCSolutions
Length
118 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For CCC 2001 S5 - Post's Correspondence, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 118 lines of Java from the credited upstream file ccc01s5.java.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks detected.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from CCCSolutions by CCCSolutions contributors · Milliken Mills High School and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCCC 2001 S5 - Post's Correspondence · JavaJava
Use this to learn the idea, then write your own version.
// CCC 2001// Problem S5: Post's Correspondence Problem // given a value of m and two sequences of strings,// is there a combination of strings from the first group that equals the// same combination from the second group.// That is: is there a sequence of ai1+ai2+ai3...+aik = bi1+bi2+bi3+...+bik// where 0 < k < m // a more or less direct recursive approach does the trick. // File I/O// Input: m and n and the strings for the two groups// Ouput: k, and i1, i2, ..., ik // CAUTION: This takes a LONG time to solve post4.in, but it does work :-) import java.awt.*;import hsa.*; public class S5Post{    static Console c;    static String [] a, b;    static int m, n, k;    static int [] iarray;     public static void main (String [] args)    {	c = new Console (); 	a = new String [40];	b = new String [40];	iarray = new int [40]; 	TextInputFile fi = new TextInputFile ("post3.in");	TextOutputFile fo = new TextOutputFile ("post3.out"); 	// read info	m = fi.readInt ();	n = fi.readInt ();	for (int i = 0 ; i < n ; i++)	    a [i] = fi.readString ();	for (int i = 0 ; i < n ; i++)	    b [i] = fi.readString (); 	// the array with the answer, iarray has good values from 0 to k	// inclusive and due to Java which counts 0,1,2.. all numbers	// are printed +1 so they count 1,2,3...	if (Post ("", "", 0))	{	    c.println (k + 1);	    for (int f = 0 ; f <= k ; f++)		c.println (iarray [f] + 1);	}	else	    c.println ("No solution.");    }         // In pseudo English:    // ta and tb are the two strings you've created so far using a & b arrays    // p counts the number of strings added    //     // if you've added more than m strings     //      you're toast (can't do it)         // else if the two strings are equal (and not just empty)    //      you win!     // else if the strings aren't equal so far    //      you're toast    // else    //      add all the posssible strings from and a and b, stopping if any     //      actually work!    //      the iarray keeps the list of i's used to create the strings       static boolean Post (String ta, String tb, int p)    {	boolean okay;	if (p > m)	    return false;	else if (ta.equals (tb) && p > 0)	    return true;	else if (!equalSoFar (ta, tb))	    return false;	else	{	    int i = 0;	    boolean done = false;	    while (i < n && !done)	    {		k = p;		iarray [k] = i;		done = Post (ta + a [i], tb + b [i], p + 1);		i++;	    }	    return done;	}    }     // returns true if the shorter of two string a and b equals the     // first part of the the longer string.     // eg "aaab" and "aaabab" is true    //    "aaab" and "aab" is false    static boolean equalSoFar (String a, String b)    {	if (a.equals (b))	    return true;	else if (a.length () < b.length ())	    return b.startsWith (a);	else if (a.length () > b.length ())	    return a.startsWith (b);	else	    return false;    }}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗