Re: Help

```The task is not strictly defined: the two matrices Tri provided are
different in size.  Therefore, I treated it as a general problem.  Here is
a 1-line solution for any size matrix:

print, (bindgen(k*m+m-1, x) mod m) eq 0

Here,
m is the length of the cycle within a row,
k is an integer multiplier to obtain the necessary row length,
x - arbitrary number of rows.

Example with m=4, k=3, x=5:
IDL> print, (bindgen(3*4+3, 5) mod 4) eq 0
1   0   0   0   1   0   0   0   1   0   0   0   1   0   0
0   1   0   0   0   1   0   0   0   1   0   0   0   1   0
0   0   1   0   0   0   1   0   0   0   1   0   0   0   1
0   0   0   1   0   0   0   1   0   0   0   1   0   0   0
1   0   0   0   1   0   0   0   1   0   0   0   1   0   0

Resulting matrix can be trimmed in any fashion you desire.  Or, maybe I
could just type matrices in, if they were needed precisely and only in the
way Tri sent them?...
Cheers,
Pavel

Tri VU KHAC wrote:

> Hi folks,
>
> I'm looking for an efficient way to produce the matrix of type (I woulk
> like to be able to avoid the FOR statement)
>
> 0 1 0 1 0 1 0 1
> 1 0 1 0 1 0 1 0
> 0 1 0 1 0 1 0 1
> 1 0 1 0 1 0 1 0
>
> OR
>
> 1 0 0 1 0 0 1
> 0 1 0 0 1 0 0
> 0 0 1 0 0 1 0
> 1 0 0 1 0 0 1
>
> ETC.
>
> Best regards,
> Tri.

```

• References:
• Help
• From: Tri VU KHAC