Member Avatar for ScienceNerd

I'm trying to solve a cryptarithmetic puzzle TOO + TOO + TOO + TOO = GOOD using a program.

My solution to this, is to use a nested loop for each unique letter (in this case T, O, G, D). The loops would systematically assign the digits from 0-9 to each letter. For example, it might first try T = 0, O = 0, G = 0, D = 0, then T = 0, O = 0, G =0, D = 1, then T = 0, O = 0, G = 0, D = 2, etc. up to T = 9, O = 9, G = 9, D = 9.

Would this be the correct way to do this? If it is, the answer I got using my program was 1642 which I know isn't correct considering there's only 3 values instead of 4 (TOO)

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Nice idea, . Brute force works as said, but you can cut the search to almost nothing by using the column-wise carries from the start. For TOO+TOO+TOO+TOO=GOOD the carries give you these constraints before you ever try T or G:

  • ones: 4O = D + 10c1
  • tens: 4O + c1 = O + 10c2 (so the ones digit repeats as O)
  • hundreds: 4T + c2 = O + 10c3, and c3 is G
    Also remember: T and G cannot be 0 (leading letters), and all letters must be different. O cannot be 0 either, or O would equal D.

Here is a small, generic solver that bakes in those carry checks first. It will find the assignment and print the mapping, then verify 4*TOO == GOOD. It runs in a handful of iterations instead of 10P4.

#include <iostream>
using namespace std;

int main() {
    for (int O = 0; O <= 9; ++O) {
        int c1 = (4*O) / 10;
        int D  = (4*O) % 10;
        if (D == O) continue;                 // distinct digits

        int tens = 4*O + c1;
        if (tens % 10 != O) continue;         // ...OO pattern
        int c2 = tens / 10;

        for (int T = 1; T <= 9; ++T) {        // T is leading
            if (T == O || T == D) continue;

            int hund = 4*T + c2;
            if (hund % 10 != O) continue;
            int G = hund / 10;                // carry to thousands
            if (G == 0 || G == T || G == O || G == D) continue;

            int TOO  = 100*T + 10*O + O;
            int GOOD = 1000*G + 100*O + 10*O + D;
            if (4*TOO == GOOD)
                cout << "T="<<T<<" O="<<O<<" G="<<G<<" D="<<D
                     << "  -->  " << 4*TOO << " == " << GOOD << "\n";
        }
    }
}

If you were getting something like 1642 earlier, double-check:

  • Build numbers as TOO = 100T + 10O + O and GOOD = 1000G + 100O + 10*O + D.
  • Enforce all-different digits and the no-leading-zero rule for T and G.
  • Add prints for c1,c2 to see why a candidate fails. ’s idea of trimming the search is spot on; the carry pruning above generalizes it.

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Member Avatar for ScienceNerd

Thanks for the reply but I need make a program using the loop to give me that answer xD

Member Avatar for VernonDozier

Thanks for the reply but I need make a program using the loop to give me that answer xD

Well if you wanted to, you could just brute-force it. There are only three letters and hence 10 * 9 * 8 possibilities:

for (int g = 0; g < 10; g++)
{
    for (int o = 0; o < 10; o++)
    {
        if (g != o)
        {
             for (int d = 0; d < 10; d++)
             {
                 if (g != d && o != d)
                 {
                      if (SentenceWorks (g, o, d))
                          // display g, o, and d
                 }
             }
        }
    }
}

SentenceWorks is a boolean function that replaces the letters with values, evaluates the terms, and sees if they are the same.

[EDIT]
Whoops. there are four letters, not three, so add a t loop. Same concept though. It's the "dumb", brute force way, bu it'll get the job done. There are probably more elegant solutions.
[/EDIT]

Member Avatar for ScienceNerd

Unfortunetly that code doesnt work for me. I don't get the correct answers with that x_x

Member Avatar for VernonDozier

Unfortunetly that code doesnt work for me. I don't get the correct answers with that x_x

It will work. Post your code.

Member Avatar for NathanOliver

this code will solve only this problem and will do it in 36 iterations. just did it for giggles to see if i could. if you want a generic code then you will have to do other things ;)

#include <iostream>
#include <stdlib.h>

using namespace std;

int main()
{
	string too = "", good = "";
	char convert[5];
	int temp;
	for (int i = 1; i < 10; i++)
	{
		for (int j = 0; j < 10; j++)
		{
			if (j == i)
				continue;
			temp = (i * 100) + (j * 10) + j;
			_itoa(temp, convert, 10);
			too = convert;
			_itoa((temp * 4), convert, 10);
			good = convert;
			if ((too[2] == good[1]) && (good[1] == good[2]) && (good[0] != good[3]) && (good[1] != good[3]) && (good.size() == 4))
			{
				cout << "SUCCESS!!!\n";
				cout << "TOO = " << temp << "\n";
				cout << "GOOD = " << (temp * 4) << "\n";
				cin.get();
				return 0;
			}
			cout << "too = " << temp << "\tgood = " << (temp * 4) << "\n";
		}
	}
	cout << "sorry this didnt work.";
	cin.get();
	return 0;
}
Member Avatar for ScienceNerd

Thanks for the suggestion!

Member Avatar for casey.cole.12576 
#include <iostream>
using namespace std;

int main()
{

    for( int t = 0; t < 10; t++ )
        for( int o = 0; o < 10; o++ )
        // put g starting at 1 so != to 0
            for( int g = 1; g < 10; g++ ) 
                for( int d = 0; d < 10; d++)
                //first if makes sure none of the letter are the same
                    if (t != o && t != d && t != g && g != o && g != d && d != o)
                    //second if checks if they are equal 
                        if ((t*100 + o*10 + o) * 4 == g*1000 + o*100 +o*10 +d){
                            cout << g << o << o << d << " = good\n";
                            cout << t << o << o << " = too";
                        } 
    return 0;
}
commented: This post is over 4 years old? :S! What is the point? -1
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