Question

a cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits such that the given arithmetic equation consisting of letters holds true. given a cryptarithm as an array of strings crypt, count the number of its valid solutions. the solution is valid if each letter represents a different digit, and the leading digit of any multi-digit number is not zero. crypt has the following structure: [word1, word2, word3], which stands for the word1 word2

Answer 1

Using the knowledge in computational language in C++ it is possible to write a code that  cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits

Writting the code:

#include <bits/stdc++.h>

using namespace std;

// chracter to digit mapping, and the inverse

// (if you want better performance: use array instead of unordered_map)

unordered_map<char, int> c2i;

unordered_map<int, char> i2c;

int ans = 0;

// limit: length of result

int limit = 0;

// digit: index of digit in a word, widx: index of a word in word list, sum: summation of all word[digit]  

bool helper(vector<string>& words, string& result, int digit, int widx, int sum) {

   if (digit == limit) {

       ans += (sum == 0);

       return sum == 0;

   }

   // if summation at digit position complete, validate it with result[digit].

   if (widx == words.size()) {

       if (c2i.count(result[digit]) == 0 && i2c.count(sum%10) == 0) {

           if (sum%10 == 0 && digit+1 == limit) // Avoid leading zero in result

               return false;

           c2i[result[digit]] = sum % 10;

           i2c[sum%10] = result[digit];

           bool tmp = helper(words, result, digit+1, 0, sum/10);

           c2i.erase(result[digit]);

           i2c.erase(sum%10);

           ans += tmp;

           return tmp;

       } else if (c2i.count(result[digit]) && c2i[result[digit]] == sum % 10){

           if (digit + 1 == limit && 0 == c2i[result[digit]]) {

               return false;

           }

           return helper(words, result, digit+1, 0, sum/10);

       } else {

           return false;

       }

   }

   // if word[widx] length less than digit, ignore and go to next word

   if (digit >= words[widx].length()) {

       return helper(words, result, digit, widx+1, sum);

   }

   // if word[widx][digit] already mapped to a value

   if (c2i.count(words[widx][digit])) {

       if (digit+1 == words[widx].length() && words[widx].length() > 1 && c2i[words[widx][digit]] == 0)

           return false;

       return helper(words, result, digit, widx+1, sum+c2i[words[widx][digit]]);

   }

   // if word[widx][digit] not mapped to a value yet

   for (int i = 0; i < 10; i++) {

       if (digit+1 == words[widx].length() && i == 0 && words[widx].length() > 1) continue;

       if (i2c.count(i)) continue;

       c2i[words[widx][digit]] = i;

       i2c[i] = words[widx][digit];

       bool tmp = helper(words, result, digit, widx+1, sum+i);

       c2i.erase(words[widx][digit]);

       i2c.erase(i);

   }

   return false;

}

void isSolvable(vector<string>& words, string result) {

   limit = result.length();

   for (auto &w: words)

       if (w.length() > limit)

           return;

   for (auto&w:words)

       reverse(w.begin(), w.end());

   reverse(result.begin(), result.end());

   int aa = helper(words, result, 0, 0, 0);

}

int main()

{

   ans = 0;

   vector<string> words={"GREEN" , "BLUE"} ;

   string result = "BLACK";

   isSolvable(words, result);

   cout << ans << "\n";

   return 0;

}

See more about C++ code at brainly.com/question/19705654

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