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Problem solution · Java

CCC 1999 P5 - Letter Arithmetic

CCC 1999 P5 - Letter Arithmetic: 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
184 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 1999 P5 - Letter Arithmetic, 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

  • 184 lines of Java from the credited upstream file ccc99s5.java.
  • The implementation visibly relies on sequence storage.
  • 11 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 1999 P5 - Letter Arithmetic · JavaJava
Use this to learn the idea, then write your own version.
// CCC 1999// Problem 5: Letter Arithmethic // I can see NO way any student could solve this problem// (having never seen it before), plus 4 other problems in 3 hours!// BUT here's a solution for what it's worth. // This converts SEND + MORE = MONEY// into the numbers that make it work. // exhaustive search will be used. all permutations of 0,1,2,3,4,5,6,7,8,9// will be tried. // file input and output.// input:  number of triple, then the triples themselves import java.awt.*;import hsa.*; public class P5LetterArithmetic{    static Console cc;    static public int [] a = new int [10]; // one permutation to be used    static public String [] s = new String [3]; // the 3 strings    static public String nodups;    static public int cn = 9; // 0.. 9 digits are used    static public int pn;     // number of unique letters    static public byte [] value = new byte [255]; // value of 'A', etc...    static public boolean stop;    static public TextInputFile fi = new TextInputFile ("letter.in");    static public TextOutputFile fo = new TextOutputFile ("letter.out");     public static void main (String [] args)    {	cc = new Console ();	int count; 	count = fi.readInt ();	for (int i = 1 ; i <= count ; i++)	{	    s [0] = fi.readString ();	    s [1] = fi.readString ();	    s [2] = fi.readString ();	    stop = false;	    nodups = findUnique (s);	    pn = nodups.length () - 1; // sets the length of the permutations	    doit (s, nodups);	}    }      // Given a string, it returns all the unique letters in the string    public static String findUnique (String [] s)    {	String u = new String ();	int k;	u = "";	for (int j = 0 ; j < 3 ; j++)	    for (int i = 0 ; i < s [j].length () ; i++)	    {		k = 0;		while (k < u.length () && u.charAt (k) != s [j].charAt (i))		    k++;		if (k >= u.length ())		    u = u + s [j].charAt (i);	    }	return u;    }      public static void doit (String [] s, String u)    {	// need to generate all combinations of u.length() numbers	// from the set 0 to 9 AND THEN all the permutations of each set!	// UGH!!!	choose (0, pn);     }      // this will generate all combinations of r numbers from the set 0,1,..9    // call with 0, r-1    public static void choose (int b, int c)    {	if (c == -1)	    permute (0);	else if (!stop)	    for (int i = b ; i < cn - c + 1 ; i++)	    {		a [c] = i;		choose (i + 1, c - 1);	    }    }      // generates all the permutations of the array a.    // array a has values 0, 1, 2, ..., pn (inclusive)    public static void permute (int i)    {	int t;	if (i > pn)	    process ();	else if (!stop)	{	    permute (i + 1);	    for (int j = i + 1 ; j <= pn ; j++)	    {		t = a [j];		a [j] = a [i];		a [i] = t;		permute (i + 1);		t = a [j];		a [j] = a [i];		a [i] = t;	    }	}    }      public static void process ()    {	// move the combination of numbers (a) into the value array	for (int i = 0 ; i < nodups.length () ; i++)	    value [nodups.charAt (i)] = (byte) a [i]; 	// if the permutation is okay, print it!	// and stop the permutation recursion process	if (okay ())	{	    for (byte j = 0 ; j < s [0].length () ; j++)	    {		cc.print (value [s [0].charAt (j)]);		fo.print (value [s [0].charAt (j)]);	    }	    cc.println ();	    fo.println ();	    for (byte j = 0 ; j < s [1].length () ; j++)	    {		cc.print (value [s [1].charAt (j)]);		fo.print (value [s [1].charAt (j)]);	    }	    cc.println ("");	    fo.println ();	    for (byte j = 0 ; j < s [2].length () ; j++)	    {		cc.print (value [s [2].charAt (j)]);		fo.print (value [s [2].charAt (j)]);	    }	    cc.println ("");	    fo.println ();	    cc.println ("");	    fo.println ();	    stop = true;	}    }      // Checks if the values for the permutation, using    // combination of s[0] + s[1] = s[2] actually works    public static boolean okay ()    {	int carry = 0;	int t;	int j = s [0].length () - 1;	int k = s [1].length () - 1;	int i = s [2].length () - 1;	boolean fine = true;	while (i >= 0 && fine)	{	    t = carry;	    if (j >= 0)		t = t + value [s [0].charAt (j--)];	    if (k >= 0)		t = t + value [s [1].charAt (k--)];	    carry = t / 10;	    t = t % 10;	    fine = t == value [s [2].charAt (i--)];	}	return fine && carry == 0;    }}

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