Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
C++ program for Solving Cryptarithmetic Puzzles
In the crypt-arithmetic problem, some letters are used to assign digits to it. Like ten different letters are holding digit values from 0 to 9 to perform arithmetic operations correctly. There are two words are given and another word is given as answer of addition for those two words. As an example we can say that two words ‘BASE’ and ‘BALL’, and the result is ‘GAMES’. Now if we try to add BASE and BALL by their symbolic digits, we will get the answer GAMES.
NOTE − There must be ten letters maximum, otherwise it cannot be solved.
Input
This algorithm will take three words.
Output
It will show which letter holds which number from 0 – 9.
For this case it is like this.
| Letter | A | B | E | G | L | M | S |
| Values | 4 | 2 | 1 | 0 | 5 | 9 | 6 |
Algorithm
For this problem, we will define a node, which contains a letter and its corresponding values.
isValid(nodeList, count, word1, word2, word3)
Input - A list of nodes, the number of elements in the node list and three words.
Output - True if the sum of the value for word1 and word2 is same as word3 value.
Begin m := 1 for each letter i from right to left of word1, do ch := word1[i] for all elements j in the nodeList, do if nodeList[j].letter = ch, then break done val1 := val1 + (m * nodeList[j].value) m := m * 10 done m := 1 for each letter i from right to left of word2, do ch := word2[i] for all elements j in the nodeList, do if nodeList[j].letter = ch, then break done val2 := val2 + (m * nodeList[j].value) m := m * 10 done m := 1 for each letter i from right to left of word3, do ch := word3[i] for all elements j in the nodeList, do if nodeList[j].letter = ch, then break done val3 := val3 + (m * nodeList[j].value) m := m * 10 done if val3 = (val1 + val2), then return true return false End
permutation(nodeList, count, n, word1, word2, word3)
Input - List of nodes, number of items in the list, number of assigned letters and three words.
Output - True when all letters are assigned with values correctly to solve the sum.
Begin if n letters are assigned, then for all digits i from 0 to 9, do if digit i is not used, then nodeList[n].value := i if isValid(nodeList, count, word1, word2, word3) = true for all items j in the nodeList, do show the letter and corresponding values. done return true done return fasle for all digits i from 0 to 9, do if digit i is not used, then nodeList[n].value := i mark as i is used if permutation(nodeList, count, n+1, word1, word2, word3), return true otherwise mark i as not used done return false End
Example
#include <iostream>
#include<vector<
using namespace std;
vector<int< use(10); //set 1, when one character is assigned previously
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;
for (i = s1.length() - 1; i >= 0; i--){ //find number for first string
char ch = s1[i];
for (j = 0; j < count; j++)
if (nodeList[j].letter == ch) //when ch is present, break the loop
break;
val1 += m * nodeList[j].value;
m *= 10;
}
m = 1;
for (i = s2.length() - 1; i >= 0; i--){ //find number for second string
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;
for (i = s3.length() - 1; i >= 0; i--){ //find number for third string
char ch = s3[i];
for (j = 0; j < count; j++)
if (nodeList[j].letter == ch)
break;
val3 += m * nodeList[j].value;
m *= 10;
}
if (val3 == (val1 + val2)) //check whether the sum is same as 3rd string or not
return 1;
return 0;
}
bool permutation(int count, node* nodeList, int n, string s1, string s2, string s3) {
if (n == count - 1){ //when values are assigned for all characters
for (int i = 0; i < 10; i++){
if (use[i] == 0){ // for those numbers, which are not used
nodeList[n].value = i; //assign value i
if (isValid(nodeList, count, s1, s2, s3) == 1){ //check validation
cout << "Solution found: ";
for (int j = 0; j < count; j++) //print code, which are assigned
cout << " " << nodeList[j].letter << " = "
<< nodeList[j].value;
return true;
}
}
}
return false;
}
for (int i = 0; i < 10; i++){
if (use[i] == 0){ // for those numbers, which are not used
nodeList[n].value = i; //assign value i and mark as not available for future use
use[i] = 1;
if (permutation(count, nodeList, n + 1, s1, s2, s3)) //go for next characters
return true;
use[i] = 0; //when backtracks, make available again
}
}
return false;
}
bool solvePuzzle(string s1, string s2,string s3) {
int uniqueChar = 0; //Number of unique characters
int len1 = s1.length();
int len2 = s2.length();
int len3 = s3.length();
vector<int> freq(26); //There are 26 different characters
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++)
if (freq[i] > 0) //whose frequency is > 0, they are present
uniqueChar++;
if (uniqueChar > 10) { //as there are 10 digits in decimal system
cout << "Invalid strings";
return 0;
}
node nodeList[uniqueChar];
for (int i = 0, j = 0; i < 26; i++) { //assign all characters found in three strings
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";
}Output
Solution found: A = 4 B = 2 E = 1 G = 0 L = 5 M = 9 S = 6
Updated on: 30-Jul-2019
842 Views
Advertisements