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How to Generate a random array in C ++ , with 100 different numbers

Dec 15, 2014 at 5:47pm
a73417 (5)
hi guys, i'm learning c++ and I have problems how to create a random board[10][10], in C++ with 100 differents numbers.


my actual code is this, but I don't know why it don't work :
int x,y,x2,y2;
int valid;

for (x=0;x<N;x++)
for(y=0; y<N; y++){
do{
tabul[x][y]=rand()%100;
for (x2=0;x2<N;x2++)
for(y2=0; y2<N; y2++){
if (x==x2 && y==y2 && tabul[x][y]==tabul[x2][y2] ) valid=0;
}

}while(!valid);




I aprecciate your time and your help!
Dec 15, 2014 at 6:01pm
anup30 (968)
first code allowing duplicates.
Dec 15, 2014 at 6:12pm
a73417 (5)
but the first code make a random and then checks if it has the same number and if it has, make an random, no?
Dec 15, 2014 at 6:22pm
anup30 (968)
no, i meant, program it without checking. fill the array with randoms (same number can appear multiple times)
Dec 15, 2014 at 6:24pm
a73417 (5)
exactly but what need I do to check if it duplicates and make an random till has no number twice?
Dec 15, 2014 at 6:35pm
anup30 (968)
if you program as arr1[100] instead of arr2[10][10] it'll be easier

if you must need 2d array you can copy arr1 to arr2 - its easier than directly create by checking arr2
Dec 15, 2014 at 7:36pm
Lowest0ne (1536)
Fill the array in sequentially and then shuffle it.
Dec 15, 2014 at 8:14pm
minomic (226)
> Fill the array in sequentially and then shuffle it.

According to me, this is the best strategy. You are guaranteed that every number is unique and you add randomization through shuffling.
Implementation should be fairly straight-forward:
- a for-loop to fill the array
- a random number of exchanges which involve two randomly selected indexes
That's all.

If you need a matrix structure (i.e. a bi-dimensional array), you can do as we said with a mono-dimensional array, then you can fill a matrix with the values you obtain from the array.
Last edited on Dec 15, 2014 at 8:14pm
Dec 15, 2014 at 8:24pm
anup30 (968)
> Fill the array in sequentially and then shuffle it.

> +1 for this example

> but learning how to check is important. as many times they wont be inherited from a serial.

> the above way technically isn't random. if you see first 99 numbers you can surely tell what 100th number is !
Dec 16, 2014 at 5:09pm
a73417 (5)
i think i found the solution:

int x,y,x2,y2,i;

int v[100];
for (i=0; i<100; i++)
v[i]=i;


for( x2=0; x2<N; x2++)

for ( y2=0; y2<N; y2++){

do{ i=rand()%100;

} while(v[i]==-1);

tabul[y2][x2]=v[i];
v[i]=-1;
}
Dec 16, 2014 at 5:28pm
minomic (226)
Ok, I haven't tried your code but is seems ok. I just want you to notice one thing: you say

do {
pick a random number
} while (this number is not good)

Well... since rand is random (actually pseudo-random but anyway...) you are not guaranteed that this code will be able to terminate. Imagine this situation: you have your array which is

n -1 -1 -1 ....................................... -1

where n is the last element you have to put in the matrix and is located at position 0. Then you are not guaranteed that sooner or later 0 will be picked by the rand function. Actually this happens (probably after a while) but it is not 100% reliable.

Instead the solution we were discussing before is this one:

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#include <iostream>
#include <cstdlib>
#include <ctime>

void swap(int array[], int pos1, int pos2) {
	int tmp = array[pos1];
	array[pos1] = array[pos2];
	array[pos2] = tmp;
}

int main() {

	srand(time(NULL));

	int numbers[100];
	int matrix[10][10];

	// fill in the numbers
	for (int i = 0; i < 100; ++i) {
		numbers[i] = i;
	}

	// a random number of swaps
	int repeat = rand() % 100 + 100;

	// perform the swaps...
	for (int i = 0; i < repeat; ++i) {
		// ... with random positions
		swap(numbers, rand() % 100, rand() % 100);
	}

	// fill in the matrix
	for (int i = 0; i < 10; ++i) {
		for (int j = 0; j < 10; ++j) {
			matrix[i][j] = numbers[i * 10 + j];
		}
	}

	std::cout << "This is the matrix:" << std::endl << std::endl;

	// print the matrix
	for (int i = 0; i < 10; ++i) {
		for (int j = 0; j < 10; ++j) {
			std::cout << matrix[i][j] << "  ";
		}
		std::cout << std::endl;
	}

	return 0;
}
Edit & run on cpp.sh


And here there is no guessing: it starts, works and completes. Always! :)
Last edited on Dec 16, 2014 at 5:45pm
Dec 17, 2014 at 11:15am
a73417 (5)
thanks a lot guys ! :)
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