Solving Cryptarithmetic Puzzles

Previous
Quiz
Next

What is Crypt-arithmetic Puzzle?

A Crypt-arithmetic puzzle, also known as a cryptogram, is a type of mathematical puzzle in which we assign digits to alphabetical letters or symbols. The end goal is to find the unique digit assignment to each letter so that the given mathematical operation holds true. In this puzzle, the equation performing an addition operation is the most commonly used. However, it also involves other arithmetic operations, such as subtraction, multiplication etc.

The rules for a Crypt-arithmetic puzzle are as follows −

  • We can use digits from 0 to 9 only to represent a unique alphabetical letter in the puzzle.

  • The same digit cannot be assigned to different letters in the whole equation.

  • The resulting equation formed by replacing letters with digits should be mathematically correct.

Input Output Scenario

Suppose the given equation is −

Input:
             B A S E
             B A L L
           ----------
           G A M E S

In the above equation, the words namely "BASE" and "BALL" are added to produce "GAMES". The algorithm will associate each letter of the given words with a unique number from 0 to 9. For the above input, the ouput should be −

Cryptarithmetic

Solving Crypt-arithmetic Puzzles using Backtracking Approach

The naive approach to solving the cryptarithmetic problem is by taking one letter from each operand starting from the left-hand side and assigning digits from 0 to 9 one by one. After assigning the digits, check the validity of the arithmetic expression. However, this method is inefficient for larger operands.

To solve a crypt-arithmetic problem using the backtracking approach, follow the below steps −

  • First, identify all the unique characters from the given arithmetic expression.

  • Next, try assigning digits to the letters. If duplicacy is found backtrack and unassign. This way all possible combinations of digits for each letter will be generated.

  • Now, replace the letters with digits and check if the expression is true.

Example

In the following example, we will practically demonstrate how to solve the cryptarithmetic problem.

C C++ Java Python 
#include <stdio.h>
#include <string.h>
//set 1, when one character is assigned previously
int use[10] = {0};
// structure
struct node {
   char letter;
   int value;
};
int isValid(struct node* nodeList, const int count, char* s1, char* s2, char* s3) {
   int val1 = 0, val2 = 0, val3 = 0, m = 1, j, i;
   //find number for first string
   for (i = strlen(s1) - 1; i >= 0; i--) {    
      char ch = s1[i];
      for (j = 0; j < count; j++)
         //when ch is present, break the loop
         if (nodeList[j].letter == ch)       
            break;
      val1 += m * nodeList[j].value;
      m *= 10;
   }
   m = 1;
   //find number for second string
   for (i = strlen(s2) - 1; i >= 0; i--) {   
      char ch = s2[i];
      for (j = 0; j < count; j++)
         if (nodeList[j].letter == ch)
            break;
      val2 += m * nodeList[j].value;
      m *= 10;
   }
   m = 1;
   //find number for third string
   for (i = strlen(s3) - 1; i >= 0; i--) {    
      char ch = s3[i];
      for (j = 0; j < count; j++)
         if (nodeList[j].letter == ch)
            break;
      val3 += m * nodeList[j].value;
      m *= 10;
   }
   //check whether the sum is same as 3rd string or not
   if (val3 == (val1 + val2))    
      return 1;
   return 0;
}
int permutation(int count, struct node* nodeList, int n, char* s1, char* s2, char* s3) {
   //when values are assigned for all characters
   if (n == count - 1) {     
      for (int i = 0; i < 10; i++) {
         // for those numbers, which are not used 
         if (use[i] == 0) {   
            //assign value i 
            nodeList[n].value = i; 
            //check validation
            if (isValid(nodeList, count, s1, s2, s3) == 1) { 
               printf("Solution found: ");
               //print code, which are assigned
               for (int j = 0; j < count; j++)    
                  printf(" %c = %d", nodeList[j].letter, nodeList[j].value);
               return 1;
            }
         }
      }
      return 0;
   }
   for (int i = 0; i < 10; i++) {
      // for those numbers, which are not used
      if (use[i] == 0) {    
         //assign value i and mark as not available for future use 
         nodeList[n].value = i;    
         use[i] = 1;
         //go for next characters
         if (permutation(count, nodeList, n + 1, s1, s2, s3) == 1)    
            return 1;
         //when backtracks, make available again 
         use[i] = 0; 
      }
   }
   return 0;
}
int solvePuzzle(char* s1, char* s2, char* s3) {
   //Number of unique characters
   int uniqueChar = 0; 
   int len1 = strlen(s1);
   int len2 = strlen(s2);
   int len3 = strlen(s3);
   //There are 26 different characters
   int freq[26] = {0}; 
   for (int i = 0; i < len1; i++)
      ++freq[s1[i] - 'A'];
   for (int i = 0; i < len2; i++)
      ++freq[s2[i] - 'A'];
   for (int i = 0; i < len3; i++)
      ++freq[s3[i] - 'A'];
   for (int i = 0; i < 26; i++)
      //whose frequency is > 0, they are present
      if (freq[i] > 0)     
         uniqueChar++;
   //as there are 10 digits in decimal system
   if (uniqueChar > 10) { 
      printf("Invalid strings");
      return 0;
   }
   struct node nodeList[uniqueChar];
   //assign all characters found in three strings  
   for (int i = 0, j = 0; i < 26; i++) {     
      if (freq[i] > 0) {
         nodeList[j].letter = (char)(i + 'A');
         j++;
      }
   }
   return permutation(uniqueChar, nodeList, 0, s1, s2, s3);
}
int main() {
   char s1[] = "BASE";
   char s2[] = "BALL";
   char s3[] = "GAMES";
   if (solvePuzzle(s1, s2, s3) == 0)
      printf("No solution");
   return 0;
}

#include <iostream>
#include <vector>
using namespace std;
//set 1, when one character is assigned previously
vector<int> use(10);        
struct node {
   char letter;
   int value;
};
int isValid(node* nodeList, const int count, string s1, string s2, string s3) {
   int val1 = 0, val2 = 0, val3 = 0, m = 1, j, i;
   //find number for first string
   for (i = s1.length() - 1; i >= 0; i--) {    
      char ch = s1[i];
      for (j = 0; j < count; j++)
         //when ch is present, break the loop
         if (nodeList[j].letter == ch)       
            break;
      val1 += m * nodeList[j].value;
      m *= 10;
   }
   m = 1;
   //find number for second string
   for (i = s2.length() - 1; i >= 0; i--) {   
      char ch = s2[i];
      for (j = 0; j < count; j++)
         if (nodeList[j].letter == ch)
            break;
      val2 += m * nodeList[j].value;
      m *= 10;
   }
   m = 1;
   //find number for third string
   for (i = s3.length() - 1; i >= 0; i--) {    
      char ch = s3[i];
      for (j = 0; j < count; j++)
         if (nodeList[j].letter == ch)
            break;
      val3 += m * nodeList[j].value;
      m *= 10;
   }
   //check whether the sum is same as 3rd string or not
   if (val3 == (val1 + val2))    
      return 1;
   return 0;
}
bool permutation(int count, node* nodeList, int n, string s1, string s2, string s3) {
   //when values are assigned for all characters        
   if (n == count - 1) {     
      for (int i = 0; i < 10; i++) {
         // for those numbers, which are not used 
         if (use[i] == 0) {   
             //assign value i 
            nodeList[n].value = i; 
            //check validation
            if (isValid(nodeList, count, s1, s2, s3) == 1) { 
               cout << "Solution found: ";
               //print code, which are assigned
               for (int j = 0; j < count; j++)    
                  cout << " " << nodeList[j].letter << " = " << nodeList[j].value;
               return true;
            }
         }
      }
      return false;
   }
   for (int i = 0; i < 10; i++) {
      // for those numbers, which are not used 
      if (use[i] == 0) {    
         //assign value i and mark as not available for future use 
         nodeList[n].value = i;    
         use[i] = 1;
         //go for next characters
         if (permutation(count, nodeList, n + 1, s1, s2, s3))    
            return true;
         //when backtracks, make available again    
         use[i] = 0; 
      }
   }
   return false;
}
bool solvePuzzle(string s1, string s2,string s3) {
   //Number of unique characters    
   int uniqueChar = 0; 
   int len1 = s1.length();
   int len2 = s2.length();
   int len3 = s3.length();
   //There are 26 different characters
   vector<int> freq(26); 
   for (int i = 0; i < len1; i++)
      ++freq[s1[i] - 'A'];
   for (int i = 0; i < len2; i++)
      ++freq[s2[i] - 'A'];
   for (int i = 0; i < len3; i++)
      ++freq[s3[i] - 'A'];

   for (int i = 0; i < 26; i++)
      //whose frequency is > 0, they are present
      if (freq[i] > 0)     
         uniqueChar++;
   //as there are 10 digits in decimal system
   if (uniqueChar > 10) { 
      cout << "Invalid strings";
      return 0;
   }
   node nodeList[uniqueChar];
   //assign all characters found in three strings    
   for (int i = 0, j = 0; i < 26; i++) {     
      if (freq[i] > 0) {
         nodeList[j].letter = char(i + 'A');
         j++;
      }
   }
   return permutation(uniqueChar, nodeList, 0, s1, s2, s3);
}
int main() {
   string s1 = "BASE";
   string s2 = "BALL";
   string s3 = "GAMES";
   if (solvePuzzle(s1, s2, s3) == false)
      cout << "No solution";
}

public class Main {
   // Set 1 when one character is assigned previously    
   int[] use = new int[10]; 
   class Node {
      char letter;
      int value;
   }
   public int isValid(Node[] nodeList, int count, String s1, String s2, String s3) {
      int val1 = 0, val2 = 0, val3 = 0;
      int m = 1;
      int j, i;
      //find number for first string
      for (i = s1.length() - 1; i >= 0; i--) {
         char ch = s1.charAt(i);
         for (j = 0; j < count; j++) {
            // when ch is present, break the loop 
            if (nodeList[j].letter == ch) {
               break;
            }
         }
         val1 += m * nodeList[j].value;
         m *= 10;
      }
      m = 1;
       //find number for second string
      for (i = s2.length() - 1; i >= 0; i--) { 
         char ch = s2.charAt(i);
         for (j = 0; j < count; j++) {
            if (nodeList[j].letter == ch) {
               break;
            }
         }
         val2 += m * nodeList[j].value;
         m *= 10;
      }
      m = 1;
      //find number for third string
      for (i = s3.length() - 1; i >= 0; i--) { 
         char ch = s3.charAt(i);
         for (j = 0; j < count; j++) {
            if (nodeList[j].letter == ch) {
               break;
            }
         }
         val3 += m * nodeList[j].value;
         m *= 10;
      }
      //check whether the sum is same as 3rd string or not
      if (val3 == (val1 + val2)) { 
         return 1;
      }
      return 0;
   }
   public int permutation(int count, Node[] nodeList, int n, String s1, String s2, String s3) {
      //when values are assign 
      if (n == count - 1) { 
         // for those numbers, which are not used 
         for (int i = 0; i < 10; i++) { 
            if (use[i] == 0) {
               //assign value i    
               nodeList[n].value = i; 
               if (isValid(nodeList, count, s1, s2, s3) == 1) {
                  System.out.print("Solution found:");
                  //print code, which are assigned
                  for (int j = 0; j < count; j++) { 
                     System.out.print(" " + nodeList[j].letter + " = " + nodeList[j].value);
                  }
                  return 1;
               }
            }
         }
         return 0;
      }
      // for those numbers, which are not used
      for (int i = 0; i < 10; i++) { 
         if (use[i] == 0) {
            //assign value i and mark as not available for future use
            nodeList[n].value = i;
            use[i] = 1;
            if (permutation(count, nodeList, n + 1, s1, s2, s3) == 1) {
               //go for next characters
               return 1; 
            }
            //when backtracks, make available again
            use[i] = 0; 
         }
      }
      return 0;
   }
   public int solvePuzzle(String s1, String s2, String s3) {
      //Number of unique characters
      int uniqueChar = 0; 
      int len1 = s1.length();
      int len2 = s2.length();
      int len3 = s3.length();
      // There are 26 different characters
      int[] freq = new int[26]; 
      for (int i = 0; i < len1; i++) {
         freq[s1.charAt(i) - 'A']++;
      }
      for (int i = 0; i < len2; i++) {
         freq[s2.charAt(i) - 'A']++;
      }
      for (int i = 0; i < len3; i++) {
         freq[s3.charAt(i) - 'A']++;
      }
      //whose frequency is > 0, they are present
      for (int i = 0; i < 26; i++) { 
         if (freq[i] > 0) {
             uniqueChar++;
         }
      }
      //as there are 10 digits in decimal system
      if (uniqueChar > 10) { 
         System.out.println("Invalid strings");
         return 0;
      }
      Node[] nodeList = new Node[uniqueChar];
      int j = 0;
      for (int i = 0; i < 26; i++) {
         //assign all characters found in three strings 
         if (freq[i] > 0) { 
             nodeList[j] = new Node();
             nodeList[j].letter = (char) (i + 'A');
             j++;
         }
      }
      return permutation(uniqueChar, nodeList, 0, s1, s2, s3);
   }
   public static void main(String[] args) {
      Main main = new Main();
      String s1 = "BASE";
      String s2 = "BALL";
      String s3 = "GAMES";
      if (main.solvePuzzle(s1, s2, s3) == 0) {
         System.out.println("No solution");
      }
   }
}

class Main:
    #Set 1 when one character is assigned previously
    use = [0] * 10  
    class Node:
        def __init__(self):
            self.letter = ''
            self.value = 0

    def isValid(self, nodeList, count, s1, s2, s3):
        val1 = 0
        val2 = 0
        val3 = 0
        m = 1
        j = 0
        i = 0
        #find number for first string
        for i in range(len(s1) - 1, -1, -1):  
            ch = s1[i]
            for j in range(count):
                #when ch is present, break the loop
                if nodeList[j].letter == ch:  
                    break
            val1 += m * nodeList[j].value
            m *= 10

        m = 1
        #find number for the second string
        for i in range(len(s2) - 1, -1, -1):  
            ch = s2[i]
            for j in range(count):
                if nodeList[j].letter == ch:
                    break
            val2 += m * nodeList[j].value
            m *= 10

        m = 1
        #find number for the third string
        for i in range(len(s3) - 1, -1, -1):  
            ch = s3[i]
            for j in range(count):
                if nodeList[j].letter == ch:
                    break
            val3 += m * nodeList[j].value
            m *= 10
        #check whether the sum is the same as the 3rd string or not
        if val3 == (val1 + val2):  
            return 1
        return 0

    def permutation(self, count, nodeList, n, s1, s2, s3):
        #when values are assigned
        if n == count - 1:  
            for i in range(10):
                #for those numbers, which are not used
                if self.use[i] == 0:  
                    #assign value i
                    nodeList[n].value = i  
                    if self.isValid(nodeList, count, s1, s2, s3) == 1:
                        print("Solution found:", end='')
                         #print code, which is assigned
                        for j in range(count): 
                            print(f" {nodeList[j].letter} = {nodeList[j].value}", end='')
                        return 1
            return 0

        for i in range(10):
            #for those numbers, which are not used
            if self.use[i] == 0:  
                #assign value i and mark as not available for future use
                nodeList[n].value = i  
                self.use[i] = 1
                if self.permutation(count, nodeList, n + 1, s1, s2, s3) == 1:  
                    #go for the next characters
                    return 1
                 #when backtracking, make available again    
                self.use[i] = 0 
        return 0

    def solvePuzzle(self, s1, s2, s3):
        #Number of unique characters
        uniqueChar = 0  
        len1 = len(s1)
        len2 = len(s2)
        len3 = len(s3)
        #There are 26 different characters
        freq = [0] * 26  

        for i in range(len1):
            freq[ord(s1[i]) - ord('A')] += 1

        for i in range(len2):
            freq[ord(s2[i]) - ord('A')] += 1

        for i in range(len3):
            freq[ord(s3[i]) - ord('A')] += 1

        for i in range(26):
            #whose frequency is > 0, they are present
            if freq[i] > 0:  
                uniqueChar += 1
        #as there are 10 digits in the decimal system
        if uniqueChar > 10:  
            print("Invalid strings")
            return 0

        nodeList = [self.Node() for _ in range(uniqueChar)]
        j = 0

        for i in range(26):
            #assign all characters found in three strings
            if freq[i] > 0:  
                nodeList[j].letter = chr(i + ord('A'))
                j += 1

        return self.permutation(uniqueChar, nodeList, 0, s1, s2, s3)

if __name__ == "__main__":
    main = Main()
    s1 = "BASE"
    s2 = "BALL"
    s3 = "GAMES"
    if main.solvePuzzle(s1, s2, s3) == 0:
        print("No solution")

Output

Solution found:  A = 4 B = 2 E = 1 G = 0 L = 5 M = 9 S = 6
Print Page
Previous
Next
Advertisements