Switching a 1D array to at 2D array, any advice?

carlosk711 · 10-22-2012, 07:18 AM

(I originally posted in the wrong forum, sorry)
I want it to be a 4x4 array the reads out 1-4 at random with repeating

EX)
4312
3124
1243
2431

My current program reads out a 1-10 at random without repeating
Can someone help me modify it?

Code:

#include <iostream>
#include <cstdlib>
#include <time.h>
using namespace std;
int main ()
{
    int A[10];
    srand (time(NULL) );
    int count = 0;
    while (count <10){
    int k = (rand() % 10) + 1;
    bool Placed = false;
    for (int i = 0; (i < count); i++){
    if (A[i] == k){
    Placed = true;
}
}
    if (!Placed){
    A[count] = k;
    count ++;
}
}


    for (int i=0; i<10; i++)
{
    cout << A[i] << " ";
}
    cout << endl;
    return 0;
}

millgates · 10-22-2012, 08:01 AM

So what's the problem? Why don't you just create another loop to run the algorithm as many times as you need? BTW, the algorithm you use to generate permutation is not very efficient. It would probably be better to generate a sorted array first and than shuffle its elements randomly.

Edit: Oh, and don't double post. If you create a thread in the wrong forum, the way to go is to kindly ask a moderator to move it (You can do it using the "Report" button below the post)

carlosk711 · 10-22-2012, 08:43 AM

I need to switch it to a 4x4 array regardless, how would you suggest I do it? Write a completely new program?

millgates · 10-22-2012, 09:13 AM

Is this a homework? Do you understand how the original program works? The change required to make it do what you want is quite simple.
I am not sure what exactly is the problem for you.
You can declare a 4x4 array like this:

Code:

int A[4][4];

Alternatively, you may create a one dimensional array of size 16

Code:

int A[16];

where Aij = A[4 * i + j]

You can then create a loop:

Code:

for (unsigned j = 0; j < numberOfRows; j++) {
    // do whatever you do in the first program, just with A[j][whatever] instead of A[whatever]
    // and the loops will go from 0 to 3 instead of 0 - 9
}

carlosk711 · 10-22-2012, 10:13 AM

It's not homework, I just want to practice 2-D arrays, and by getting examples it will help me

carlosk711 · 10-22-2012, 10:20 AM

So how should my final program look? I tried to change mine but I'm not following

EDIT: And yes, I created the original, so I understand it

millgates · 10-22-2012, 10:44 AM

Quote:

Originally Posted by carlosk711

I tried to change mine but I'm not following

What exactly do you not understand? What your program does is to create a random permutation of numbers in the 1D array A. Now what you want to do is create a 2D 4x4 array (so that's an array of four 1D arrays). So you you iterate over a new index, say j from 0 to 3 and for each such j you do with A[j][i] exactly what you did with A[i] in your first program. So all you have to do is put this part of your program:

Code:

    int count = 0;
    while (count <10){
    int k = (rand() % 10) + 1;
    bool Placed = false;
    for (int i = 0; (i < count); i++){
    if (A[i] == k){
    Placed = true;
}
}
    if (!Placed){
    A[count] = k;
    count ++;
}
}


    for (int i=0; i<10; i++)
{
    cout << A[i] << " ";
}
    cout << endl;

inside a loop over j
and add [j] to every occurence of A in the program. If you get stuck, just show us what you've got so far.

danielbmartin · 10-22-2012, 10:47 AM

Quote:

Originally Posted by carlosk711

My current program reads out a 1-10 at random without repeating
Can someone help me modify it?

If you have the desired string of 16 numbers,
you may reshape it into a 4x4 with sed ...

Code:

echo "4312312412432431"  \
|sed -e :a -e 's/\(.*[0-9]\)\([0-9]\{4\}\)/\1\n\2/;ta'

Daniel B. Martin

millgates · 10-22-2012, 11:51 AM

Quote:

Originally Posted by danielbmartin

If you have the desired string of 16 numbers,
you may reshape it into a 4x4 with sed ...

Code:

echo "4312312412432431"  \
|sed -e :a -e 's/\(.*[0-9]\)\([0-9]\{4\}\)/\1\n\2/;ta'

Daniel B. Martin

I believe the original poster's intention is to practice using 2D arrays in C++, not formatting a string. But I may be wrong...
Anyway, that's a very nice sed there

BTW, how about just

Code:

sed 's/[0-9]\{4\}/&\n/g;s/\n$//'

carlosk711: Ehm, now when I think about it, do you want to just print four random permutations of {1..4} or do you want a matrix with elements unique in both rows and columns?

Elv13 · 10-22-2012, 12:45 PM

From the other thread:

As this is C++, using static array is not necessary, use Vectors or Lists so you have a the find() function.

As for the 2D static array, a neat trick is to keep your existing code intact (but use 14 as size instead of 10). Then make a new array of size 4 and type int*. Place the newArr[0] = A, newArr[1] = newArr[1] = A+4; and so on. With that, you can have both a continuous and a 2D view of the same dataset. Again, it is a little trick, not something that would work in a pure algorithm point of view. It work because of the way C does static array pointer.

danielbmartin · 10-22-2012, 01:10 PM

[QUOTE=millgates;4812310]... how about just

Code:

sed 's/[0-9]\{4\}/&\n/g;s/\n$//'

Excellent! Your sed is better because it has no loop.

Daniel B. Martin

carlosk711 · 10-22-2012, 01:12 PM

That's interesting, I'll have to try it out! But I doubt my professor would accept it on an exam. Right now I'm just practicing using 2-D arrays.

AND to respond earlier a question on a practice exam says to create a 2-D array that will take in numbers 1-4 at random and read out each row, without repetition.

for example:

2134
1243
4312
3124

EDIT: Yes it's for C++ using basic programming

carlosk711 · 10-22-2012, 03:09 PM

NEW PROBLEM:
I've written it so the rows work
Now I need each column to read 1-4 without repeating as well

Heres the code where only the rows work:

Code:

#include <iostream>
#include <cstdlib>
#include <time.h>
using namespace std;
int main ()
{
    int A[4][4];
    srand (time(NULL) );
    int j = 0;
    while (j < 4)
    {
    int count = 0;
    while (count <4)
        {
        int k = (rand() % 4) + 1;
        bool Placed = false;
        for (int i = 0; (i < count); i++)
            {
            if (A[j][i] == k)
                {
                Placed = true;
                }
            }
        if (!Placed)
            {
            A[j][count] = k;
            count ++;
            }
        }
    for (int i=0; i<4; i++)
    {
        cout << A[j][i] << " ";
    }
    cout << endl;
    j++;
}
    return 0;
}

millgates · 10-22-2012, 03:30 PM

Ok. So what I would do is to initialize the array with a trivial set of numbers:

Code:

This is a state where no number is repeated in any row or column. It is obvious that if I swap any two rows, the resulting state will also be free of repetitions. Analogically, swapping any two columns will result in a non-repetitive permutation. So, I would randomly shuffle rows and then columns (not necessarily in that particular order). The result would then be a random matrix of numbers without repetition.