This part I understand:
btf1=: 3 : ',(i.-n)+/n*i.n=. %:#y'
btf2=: 13 : ',(i.-n)+/n*i.n=. %:#y'
(btf1 -: btf2)"0 *:i.10
1 1 1 1 1 1 1 1 1 1
5!:4 <'btf1'
-- 3
-- : -+- ,:',(i.-n)+/n*i.n=. %:#y'
5!:4 <'btf2'
-- [:
│ -- [:
│ +- ,
│ │ -- [:
+----+ ------+- i.
│ │ │ L- -
│ │ │
--+ L----+- / --- +
│ │
│ │ -- ]
│ L-----+- *
│ L- i.
│ -- [:
L----+- %:
L- #
Roger, I don't understand what you are doing in this next section:
n=sqrt(m);
q=m-n;
for(i=n-1;i>=0;--i) for(j=i;j<=q;j+=n) *r++=j;
1 What is m?
2 What is for?
3 Is this C (I don't know C and it isn't J)
Linda
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Roger Hui
Sent: Monday, January 28, 2013 7:10 PM
To: Programming forum
Subject: Re: [Jprogramming] reverse table row indices using indices
n=sqrt(m);
q=m-n;
for(i=n-1;i>=0;--i) for(j=i;j<=q;j+=n) *r++=j;
On Mon, Jan 28, 2013 at 4:06 PM, Roger Hui <[email protected]>wrote:
> From that:
>
> btf1=: 3 : ',(i.-n)+/n*i.n=. %:#y'
> (btf -: btf1)"0 *:i.10
> 1 1 1 1 1 1 1 1 1 1
>
> A simpler C implementation derives:
>
> n=sqrt(m);
> for(i=n-1;i>=0;--i) for(j=0;j<m;j+=n) *r++=i+j;
>
>
>
>
> On Mon, Jan 28, 2013 at 3:47 PM, Brian Schott
<[email protected]>wrote:
>
>> I think I got it.
>>
>> btf =: monad define NB. Brians transpose flip
>> n=. %:#y
>> r=. i. 0
>> for_j. i. n do.
>> for_k. i. n do.
>> r =. r, (_2*j)+(n-1)+j+n*k
>> NB. smoutput i1
>> end.
>> end.
>> r
>> )
>>
>> btf i. 9
>> 2 5 8 1 4 7 0 3 6
>> ---------------------------------------------------------------------
>> - For information about J forums see
>> http://www.jsoftware.com/forums.htm
>>
>
>
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