# C++ program for Solving Cryptarithmetic Puzzles

Newspapers and magazines often have crypt-arithmetic puzzles of the form:
Examples:

```Input : s1 = SEND, s2 = "MORE", s3 = "MONEY"
Output : One of the possible solution is:
D=1 E=5 M=0 N=3 O=8 R=2 S=7 Y=6
Explanation:
The above values satisfy below equation :
SEND
+ MORE
--------
MONEY
--------
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

It is strongly recommended to refer Backtracking | Set 8 (Solving Cryptarithmetic Puzzles) for approach of this problem.

The idea is to assign each letter a digit from 0 to 9 so that the arithmetic works out correctly. A permutation is a recursive function which calls a check function for every possible permutation of integers.
Check function checks whether the sum of first two numbers corresponding to first two string is equal to the third number corresponding to third string. If the solution is found then print the solution.

 `// CPP program for solving cryptographic puzzles ` `#include ` `using` `namespace` `std; ` ` `  `// vector stores 1 corresponding to index ` `// number which is already assigned ` `// to any char, otherwise stores 0 ` `vector<``int``> use(10); ` ` `  `// structure to store char and its corresponding integer ` `struct` `node ` `{ ` `    ``char` `c; ` `    ``int` `v; ` `}; ` ` `  `// function check for correct solution ` `int` `check(node* nodeArr, ``const` `int` `count, string s1, ` `                               ``string s2, string s3) ` `{ ` `    ``int` `val1 = 0, val2 = 0, val3 = 0, m = 1, j, i; ` ` `  `    ``// calculate number corresponding to first string ` `    ``for` `(i = s1.length() - 1; i >= 0; i--) ` `    ``{ ` `        ``char` `ch = s1[i]; ` `        ``for` `(j = 0; j < count; j++) ` `            ``if` `(nodeArr[j].c == ch) ` `                ``break``; ` ` `  `        ``val1 += m * nodeArr[j].v; ` `        ``m *= 10; ` `    ``} ` `    ``m = 1; ` ` `  `    ``// calculate number corresponding to second string ` `    ``for` `(i = s2.length() - 1; i >= 0; i--) ` `    ``{ ` `        ``char` `ch = s2[i]; ` `        ``for` `(j = 0; j < count; j++) ` `            ``if` `(nodeArr[j].c == ch) ` `                ``break``; ` ` `  `        ``val2 += m * nodeArr[j].v; ` `        ``m *= 10; ` `    ``} ` `    ``m = 1; ` ` `  `    ``// calculate number corresponding to third string ` `    ``for` `(i = s3.length() - 1; i >= 0; i--) ` `    ``{ ` `        ``char` `ch = s3[i]; ` `        ``for` `(j = 0; j < count; j++) ` `            ``if` `(nodeArr[j].c == ch) ` `                ``break``; ` ` `  `        ``val3 += m * nodeArr[j].v; ` `        ``m *= 10; ` `    ``} ` ` `  `    ``// sum of first two number equal to third return true ` `    ``if` `(val3 == (val1 + val2)) ` `        ``return` `1; ` ` `  `    ``// else return false ` `    ``return` `0; ` `} ` ` `  `// Recursive function to check solution for all permutations ` `bool` `permutation(``const` `int` `count, node* nodeArr, ``int` `n, ` `                 ``string s1, string s2, string s3) ` `{ ` `    ``// Base case ` `    ``if` `(n == count - 1) ` `    ``{ ` ` `  `        ``// check for all numbers not used yet ` `        ``for` `(``int` `i = 0; i < 10; i++) ` `        ``{ ` ` `  `            ``// if not used ` `            ``if` `(use[i] == 0) ` `            ``{ ` ` `  `                ``// assign char at index n integer i ` `                ``nodeArr[n].v = i; ` ` `  `                ``// if solution found ` `                ``if` `(check(nodeArr, count, s1, s2, s3) == 1) ` `                ``{ ` `                    ``cout << ``"\nSolution found: "``; ` `                    ``for` `(``int` `j = 0; j < count; j++) ` `                        ``cout << ``" "` `<< nodeArr[j].c << ``" = "` `                             ``<< nodeArr[j].v; ` `                    ``return` `true``; ` `                ``} ` `            ``} ` `        ``} ` `        ``return` `false``; ` `    ``} ` ` `  `    ``for` `(``int` `i = 0; i < 10; i++) ` `    ``{ ` ` `  `        ``// if ith integer not used yet ` `        ``if` `(use[i] == 0) ` `        ``{ ` ` `  `            ``// assign char at index n integer i ` `            ``nodeArr[n].v = i; ` ` `  `            ``// mark it as not available for other char ` `            ``use[i] = 1; ` ` `  `            ``// call recursive function ` `            ``if` `(permutation(count, nodeArr, n + 1, s1, s2, s3)) ` `                ``return` `true``; ` ` `  `            ``// backtrack for all other possible solutions ` `            ``use[i] = 0; ` `        ``} ` `    ``} ` `    ``return` `false``; ` `} ` ` `  `bool` `solveCryptographic(string s1, string s2, ` `                                   ``string s3) ` `{ ` `    ``// count to store number of unique char ` `    ``int` `count = 0; ` ` `  `    ``// Length of all three strings ` `    ``int` `l1 = s1.length(); ` `    ``int` `l2 = s2.length(); ` `    ``int` `l3 = s3.length(); ` ` `  `    ``// vector to store frequency of each char ` `    ``vector<``int``> freq(26); ` ` `  `    ``for` `(``int` `i = 0; i < l1; i++) ` `        ``++freq[s1[i] - ``'A'``]; ` ` `  `    ``for` `(``int` `i = 0; i < l2; i++) ` `        ``++freq[s2[i] - ``'A'``]; ` ` `  `    ``for` `(``int` `i = 0; i < l3; i++) ` `        ``++freq[s3[i] - ``'A'``]; ` ` `  `    ``// count number of unique char ` `    ``for` `(``int` `i = 0; i < 26; i++) ` `        ``if` `(freq[i] > 0) ` `            ``count++; ` ` `  `    ``// solution not possible for count greater than 10 ` `    ``if` `(count > 10) ` `    ``{ ` `        ``cout << ``"Invalid strings"``; ` `        ``return` `0; ` `    ``} ` ` `  `    ``// array of nodes ` `    ``node nodeArr[count]; ` ` `  `    ``// store all unique char in nodeArr ` `    ``for` `(``int` `i = 0, j = 0; i < 26; i++) ` `    ``{ ` `        ``if` `(freq[i] > 0) ` `        ``{ ` `            ``nodeArr[j].c = ``char``(i + ``'A'``); ` `            ``j++; ` `        ``} ` `    ``} ` `    ``return` `permutation(count, nodeArr, 0, s1, s2, s3); ` `} ` ` `  `// Driver function ` `int` `main() ` `{ ` `    ``string s1 = ``"SEND"``; ` `    ``string s2 = ``"MORE"``; ` `    ``string s3 = ``"MONEY"``; ` ` `  `    ``if` `(solveCryptographic(s1, s2, s3) == ``false``) ` `        ``cout << ``"No solution"``; ` `    ``return` `0; ` `} `

Output:

```Solution found:  D=1 E=5 M=0 N=3 O=8 R=2 S=7 Y=6

```

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