#Making a diagonal using matrix

1 messages · Page 1 of 1 (latest)

vital seal
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How can I start to make a matrix that is made up of diagonals of numbers

celest scaffoldBOT
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<@&987246399047479336> please have a look, thanks.

celest scaffoldBOT
#

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vital seal
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an example of what the program should be able to do

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it takes in a user inputted number

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and then makes a diagonal nxn and starts at 0 diagonal with the numbers adjacent to both sides of it

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also a diagonal but increasing by one

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here is one more example

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if input was 2

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output should look like that

granite bison
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u can use (i-j) to get the diagonal index and put the max of (i, j) in that diagonal

granite bison
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int n = get.nextInt();
            
            int[][] mat = new int[n][n];
            
            for(int gap=-1*n; gap<=n; gap++)
            {
                int i=0, j=0;
                
                if(gap < 0) {
                    i = -1*gap;
                    j = 0;
                }
                else {
                    i = 0;
                    j = gap;
                }
                
                int val = Math.max(i, j);
                while(i<n && j<n)
                {
                    mat[i][j] = val;                    
                    i++;
                    j++;
                }
            }
            
            for(int i=0; i<n; i++) {
                for(int j=0; j<n; j++) {
                    put.write(mat[i][j]+" ");
                }
                put.write("\n");
            }
celest scaffoldBOT
# granite bison ``` int n = get.nextInt(); int[][] mat = new int[n][n];...

Detected code, here are some useful tools:

Formatted code 
int n = get.nextInt();
int [] [] mat = new int [n] [n] ;
for (int gap =  - 1 * n; gap <= n; gap++) {
  int i = 0, j = 0;
  if (gap < 0) {
    i =  - 1 * gap;
    j = 0;
  }
  else {
    i = 0;
    j = gap;
  }
  int val = Math.max(i, j);
  while (i < n && j < n) {
    mat[i] [j]  = val;
    i++;
    j++;
  }
}
for (int i = 0; i < n; i++) {
  for (int j = 0; j < n; j++) {
    put.write(mat[i] [j]  + " ");
  }
  put.write("\n");
}
granite bison
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ig, this is it, any problem, ping me

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@vital seal

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well try more, if not then only, see code : D
you should be able to do it. JUST DRY RUN

eager dust
vital seal
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I’m having a bit of trouble understanding it

granite bison
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there are many ways to print this

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this above program is printing them diagonally, so it looks good but there are many approaches, some simple ones too

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for this example, what we see?

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in row 0, 0 is present in col 0, right?

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in row 1, 0 is present in col 1?

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can i say for any row=i, 0 is present on col=i ?

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@vital seal

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dont look above code, i am explaining u a different approach to see, when looks on test case

granite bison
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so we know the location of 0

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loop from 0 th position towards right, incrementing by one

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loop from 0 th position towards left incrementing by one

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for every row, this will give the same solution, u need

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u can also think column-wise too

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this is an OBSERVATION problem only

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for the above code, if u want to understand that, i will explain, but try to program it this way

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for practise

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try to print this one

granite bison
# granite bison

@vital seal try to solve this for practise on these questions of observation

granite bison
vital seal
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Ohh

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I do see the pattern now

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I prob could’ve done this

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Ty for your help

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I’ll attempt ur example

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If I can

granite bison