# Solving Cryptarithmetic Puzzles | Backtracking-8

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

SEND + MORE -------- MONEY --------

The goal here is to assign each letter a digit from 0 to 9 so that the arithmetic works out correctly. The rules are that all occurrences of a letter must be assigned the same digit, and no digit can be assigned to more than one letter.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend **live classes **with experts, please refer **DSA Live Classes for Working Professionals **and **Competitive Programming Live for Students**.

- First, create a list of all the characters that need assigning to pass to Solve
- If all characters are assigned, return true if puzzle is solved, false otherwise
- Otherwise, consider the first unassigned character
- for (every possible choice among the digits not in use)
make that choice and then recursively try to assign the rest of the characters

if recursion successful, return true

if !successful, unmake assignment and try another digit - If all digits have been tried and nothing worked, return false to trigger backtracking

`/* ExhaustiveSolve` `* ---------------` `* This is the "not-very-smart" version of cryptarithmetic solver. It takes` `* the puzzle itself (with the 3 strings for the two addends and sum) and a` `* string of letters as yet unassigned. If no more letters to assign` `* then we've hit a base-case, if the current letter-to-digit mapping solves` `* the puzzle, we're done, otherwise we return false to trigger backtracking` `* If we have letters to assign, we take the first letter from that list, and` `* try assigning it the digits from 0 to 9 and then recursively working` `* through solving puzzle from here. If we manage to make a good assignment` `* that works, we've succeeded, else we need to unassign that choice and try` `* another digit. This version is easy to write, since it uses a simple` `* approach (quite similar to permutations if you think about it) but it is` `* not so smart because it doesn't take into account the structure of the` `* puzzle constraints (for example, once the two digits for the addends have` `* been assigned, there is no reason to try anything other than the correct` `* digit for the sum) yet it tries a lot of useless combos regardless` `*/` `bool` `ExhaustiveSolve(puzzleT puzzle, string lettersToAssign)` `{` ` ` `if` `(lettersToAssign.empty()) ` `// no more choices to make` ` ` `return` `PuzzleSolved(puzzle); ` `// checks arithmetic to see if works` ` ` `for` `(` `int` `digit = 0; digit <= 9; digit++) ` `// try all digits` ` ` `{` ` ` `if` `(AssignLetterToDigit(lettersToAssign[0], digit))` ` ` `{` ` ` `if` `(ExhaustiveSolve(puzzle, lettersToAssign.substr(1)))` ` ` `return` `true` `;` ` ` `UnassignLetterFromDigit(lettersToAssign[0], digit);` ` ` `}` ` ` `}` ` ` `return` `false` `; ` `// nothing worked, need to backtrack` `}` |

The algorithm above actually has a lot in common with the permutations algorithm, it pretty much just creates all arrangements of the mapping from characters to digits and tries each until one works or all have been successfully tried. For a large puzzle, this could take a while.

A smarter algorithm could take into account the structure of the puzzle and avoid going down dead-end paths. For example, if we assign the characters starting from the one’s place and moving to the left, at each stage, we can verify the correctness of what we have so far before we continue onwards. This definitely complicates the code but leads to a tremendous improvement in efficiency, making it much more feasible to solve large puzzles.

Below pseudocode, in this case, has more special cases, but the same general design

- Start by examining the rightmost digit of the topmost row, with a carry of 0
- If we are beyond the leftmost digit of the puzzle, return true if no carry, false otherwise
- If we are currently trying to assign a char in one of the addends

If char already assigned, just recur on the row beneath this one, adding value into the sum

If not assigned, then- for (every possible choice among the digits not in use)

make that choice and then on row beneath this one, if successful, return true

if !successful, unmake assignment and try another digit - return false if no assignment worked to trigger backtracking

- for (every possible choice among the digits not in use)
- Else if trying to assign a char in the sum
- If char assigned & matches correct,

recur on next column to the left with carry, if success return true, - If char assigned & doesn’t match, return false
- If char unassigned & correct digit already used, return false
- If char unassigned & correct digit unused,

assign it and recur on next column to left with carry, if success return true - return false to trigger backtracking

**Source:**

http://see.stanford.edu/materials/icspacs106b/H19-RecBacktrackExamples.pdf