Help Please

[scardoso]

Dont know how to replace each letter as s number can someone help me ... this is the question i was given ...

In cryptarithmetic puzzles, mathematical equations are written using letters. Each letter can be a digit from 0 to 9, but no two letters can be the same. Here is a sample problem:

SEND + MORE = MONEY

A solution to the puzzle is S = 9, R = 8, O = 0, M = 1, Y = 2, E = 5, N = 6, D = 7.

Write a program that finds a solution to the cryptarithmetic puzzle of the following:

TOO + TOO + TOO + TOO = GOOD

The simplest technique 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 to 9 to each letter. For example, it might first try T=0,O=0,G=0,D=0, thenT=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. In the loop body, test that each variable is unique and that the equation is satisfied. Output the values for the letters that satisfy the equation.

[condor]

Why don't you try to write yourself that "simplest technique" about you say? I think it's better for you than to do someone this task.

EDIT:
I solved this puzzle: I wrote the program and I know the unique(!) solution: (T, O, G, D) = (4, 9, 1, 6)!

[condor]

Let's come up with a helping hand: using numbers in base 10 we have the equation:

4*(100*T + 10*O + O) = 1000*G + 100*O + 10*O + D;

After reducing like terms we have:

1000*G - 400*T + 66*O + D = 0.

The program will have to test this equation is satisfied and that variables are different each other. That's all!

[AeonFlux1212]

you could also just write

if(!(T == O || T == G || T == D || O == G || O == D || G == D))...

or declare a vector of exclusions that you push_back() with the value of each for loop and have another loop in each loop that checks if the value is in the exclusion vector, but I prefer the first technique... :-)

[edit] In fact, you need to initialise an exclusion vector like
std::vector<unsigned int> exclusions = {0, 9, 8, 1, 2, 5, 6, 7}
[/edit]

[scardoso]

Thanks condor ... I thought there was a specific function i needed to replace each letter with a number but what you show there makes sense as in putting it in its slots in numerical value and adding it together ... Appreciate the help.

[condor]

@scardoso

You're welcome! If you have some troubles with the code dare to ask us.

[scardoso]

So I think I got it up until repeating numbers ... I can't figure out how to make it so it doesnt finish the loop if one of the numbers are repeated ... Any ideas?

#include <iostream>
#include <stdio.h>

using namespace std;

int main()

{

int T=0,O=0,G=0,D=0, answer=0, good=1;

while (answer != good && T < 9)
{
O = 0;

T++;

answer = 4 * (T * 100 + O * 10 + O);
good = G * 1000 + O * 100 + O * 10 + D;

printf("second loop %i .... %i\n", answer, good);

while (answer != good && O < 9)
{
G = 0;

O++;

answer = 4 * (T * 100 + O * 10 + O);
good = G * 1000 + O * 100 + O * 10 + D;

printf("third loop %i .... %i\n", answer, good);

while (answer != good && G < 9)
{
D = 0;

G++;

answer = 4 * (T * 100 + O * 10 + O);
good = G * 1000 + O * 100 + O * 10 + D;

printf("fourth loop %i .... %i\n", answer, good);

while (answer != good && D < 9)
{

D++;

answer = 4 * (T * 100 + O * 10 + O);
good = G * 1000 + O * 100 + O * 10 + D;

printf("fifth loop %i .... %i\n", answer, good);
}}}}

printf("T = %i O = %i G = %i D = %i", T,O,G,D);

}

[AeonFlux1212]

I found that there are 24 solutions to this puzzle, and none of them are Condor's...

@Condor Where do you get this 1000*, 100* stuff, how's it supposed to solve the puzzle?

Me, I applied the nested loop previously discussed, and just verified that the values were different from each other AND that the equation was satisfied...
it comes up to 24 solutions

[edit] @Condor ok... I get it... the letters become actual numbers... right... damn! I'll try again...

[AeonFlux1212]

I end up with (T, O, G, D) = (1, 6, 0, 4) and (T, O, G, D) = (4, 9, 1, 6) indeed...
nice!

But I guess T and G can't be equal to zero... so there really is one solution

[scardoso]

How did you show to by pass if the numbers are the same ones and still satisfy the equation @AeonFlux1212 and @Condor

[condor]

Here's my solution:

#include <iostream>

using namespace std;

int main()
{	
	for(int t = 1; t < 10; ++t)//t can't be 0 (it is the MSD = Most Significant Digit)!
	for(int o = 0; o < 10; ++o)
	for(int g = 1; g < 10; ++g)//also g can't be 0 ! (MSD)
	for(int d = 0; d < 10; ++d)
	if((1000 * g - 400 * t + 66 * o +  d) == 0)//if equation is satisfied
	if(g != t && g != o && g != d && t != o && t != d && o != d)//different each other
	{																
		cout << "\nSolution" << ": T = " << t << ", O = " << o << 
		", G = " << g << ", D = " << d << "\n\n";
		cout << "Check: ";
		cout << " TOO + TOO + TOO + TOO = GOOD" << '\n';
		cout << "\t" << t << o << o << " + " << t << o << o <<
		" + " << t << o << o << " + " <<t << o << o << " = " <<
		g << o << o << d << '\n';
	}
	return 0;
}

Output:

Solution: T = 4, O = 9, G = 1, D = 6

Check:  TOO + TOO + TOO + TOO = GOOD
	499 + 499 + 499 + 499 = 1996

[AeonFlux1212]

If you want another example...

#include <iostream>

using namespace std;

int main()
{
    unsigned int T, O, G, D;

    unsigned int numberOfPermutations = 0;

    for(T = 0; T < 10; T++)
    {
        for(O = 0; O < 10; O++)
        {
            for(G = 0; G < 10; G++)
            {
                for(D = 0; D < 10; D++)
                {
                    if(!(T == 0 || G == 0) && !(T == O || T == G || T == D || O == G || O == D || G == D) && 1000*G - 400*T + 66*O + D == 0)
                    {
                        cout << "T = " << T << ", O = " << O << ", G = " << G << ", D = " << D << endl;

                        numberOfPermutations++;
                    }
                }
            }
        }
    }

    cout << endl
         << "numberOfPermutations = " << numberOfPermutations << endl << endl;

    return 0;
}

It's in line 11 and 12 in condor's version and in line 19 in mine.

[scardoso]

Thanks a lot guys awesome help ... definitely showed me a lot of tricks to narrow down my programming as well ... Thanks for the help @Condor and @@AeonFlux1212

[condor]

@scardoso

You're welcome! If you consider all OK please mark this thread as Solved.