Thread: Bingo Card Logic (no repetition)

Hi everybody, good afternoon.

• I made a code to the problem below (in blue), but it seems that my logic is in some hole. Already I revised, and do not understand why I still have repeated numbers. Can anyone give me strength? Thank you!!

• The logic that I thought code is as follows: I fill in a field of the matrix, with "i" and "j" and compare it to previous ones through another FOR with "k" and "z". If this field that just fill is equal to a previous field, I go back a position (j = j-1) and start the FOR again. To do so we made use FOR`s 4 (one inside the other, to enter the matrix and make the comparison), four variables (i, j, k, z) and an "aux" that serves as a precondition for my stop.

• Important: I made the code with a 4x4 matrix to test it. Then I will replicate for a 99x99 matrix.

• The problem: Make a program to automatically generate numbers between 0 and 99 for a bingo card. Knowing that each card should contain 5 lines and 5 numbers, manages this data so as not to have repeated numbers in the chart.

• The Code:

Code:

Code:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>


int main () {


	int M[4][4], i,j,k,z, aux = 0;


	srand(time(NULL));


	for(i=0;i<4;i++){
		for(j=0;j<4;j++){
		


			aux = 0;
			M[i][j] = i;
	
			for(k = i; k >= 0; k--){


				if(aux == 1){
					break;
				}


				for(z = j; z >= 0; z--){
					if((M[k][z] == M[i][j]) && ((k != i) && (z != j))){
						aux = 1;
						j = j - 1;
						break;
					}
				}
			}
		}
	}
			


	printf("\n\n\n\n");


	for(i=0;i<4;i++){
		for(j=0;j<4;j++){
			printf("%d\t", M[i][j]);
		}
		printf("\n");
	}


	return 0;
}

I will follow your advice, Salem.

In fact, I did what you said before. But I will try to do it again...

Thanks!

Originally Posted by Salem

Generally, the approach to take is to try and solve the problem just using a pencil and paper.

Solve such problems as
- how to pick unique numbers in the range 0 to 99
- how to randomly position 'n' of 0 to 99 on the grid

If you've no idea how you would solve it on paper, using a series of repeatable steps, then just randomly writing code in hopes of stumbling on a solution won't work.

Hint: consider how to implement a shuffle.

Salem,


I have just made my logic again (with my head fresh ) and worked ...


I really would like to know how you think when you are building algorithms.


I am starting to do this now (I am learning in C) and something that are trivial in human thinking are not so trivial in computer thinking.


For example, I spent some time to build a code that could make the Product (C) of two matriz (A * B = C), once the rows and columns dont go in parallels.

Originally Posted by Salem

Generally, the approach to take is to try and solve the problem just using a pencil and paper.

Solve such problems as
- how to pick unique numbers in the range 0 to 99
- how to randomly position 'n' of 0 to 99 on the grid

If you've no idea how you would solve it on paper, using a series of repeatable steps, then just randomly writing code in hopes of stumbling on a solution won't work.

Hint: consider how to implement a shuffle.

Code:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>


int main () {




	int M[5][5], i,j,k,z,aux = 1;


	srand(time(NULL));


	for(i=0;i<5;i++){
		for(j=0;j<5;j++){
			
			aux = 1;
			M[i][j] = rand() % 100;
			for(k=0;k<5;k++){
				for(z=0;z<5;z++){


					if((i == k && j == z) || (((k*5)+z) > ((i*5)+j))){
						aux = 1;
					}
					else if(M[i][j] == M[k][z]){
						aux = 0;
						break;
					}
				}
			
				if(aux == 0){
					j = j -1;
					break;
				}
			}	
		}
	}






	printf("Matriz Completa:\n");
	for(i=0;i<5;i++){
		for(j=0;j<5;j++){
			printf("%d\t", M[i][j]);
		}
		printf("\n");
	}


	return 0;
}