Still trying to understand L: - it looks so useful,
but meanwhile, how do you get an even more irregular tiling such as:
+-----------+-----+
| 1 2 3 4| 4 5|
| 8 9 10 11|11 12|
|15 16 17 18|18 19|
+-----------+-----+
|29 30 31 32|32 33|
+-----------+-----+
|15 16 17 18|18 19|
|22 23 24 25|25 26|
+-----------+-----+
NB. ie a 3 x 2 tiling ?
This almost does the job:
[t1 =. <@,:"1/ ,"0"0 1"1/ >(0 3,.4 1,. 2 2);1 4,:4 2
+---+---+---+
|0 1|0 4|0 0|
|3 4|3 2|3 0|
+---+---+---+
|4 1|4 4|4 0|
|1 4|1 2|1 0|
+---+---+---+
|2 1|2 4|2 0|
|2 4|2 2|2 0|
+---+---+---+
NB. or use Raul's latest posting as for t
NB. ?
T ];.0~L:0 t1
+-----------+-----++
| 1 2 3 4| 4 5||
| 8 9 10 11|11 12||
|15 16 17 18|18 19||
+-----------+-----++
|29 30 31 32|32 33||
+-----------+-----++
|15 16 17 18|18 19||
|22 23 24 25|25 26||
+-----------+-----++
So the current approach is fine for equal numbers of across and down
tiles, less good
for unequal ones. I suppose we can prune t1 to remove columns and/or
rows having
all zero lengths, or remove all empty tiles. In this case, the fact
that there are 3
columns and two columns in the specification might be used to filter the
result.
Cheers,
Mike
On 24/09/2021 17:54, R.E. Boss wrote:
This is the way to go for even very irregular tiling.
Let the starting points and their lengths be given per axis, like in
(0 4,:3 1);1 4,:4 2
+---+---+
|0 4|1 4|
|3 1|4 2|
+---+---+
and
[ T=.i.7 7
0 1 2 3 4 5 6
7 8 9 10 11 12 13
14 15 16 17 18 19 20
21 22 23 24 25 26 27
28 29 30 31 32 33 34
35 36 37 38 39 40 41
42 43 44 45 46 47 48
First make de tiling by indices
t=.<@,:"1/ ,"0"0 1"1/ >(0 4,:3 1);1 4,:4 2
+---+---+
|0 1|0 4|
|3 4|3 2|
+---+---+
|4 1|4 4|
|1 4|1 2|
+---+---+
then
T ];.0~L:0 t
+-----------+-----+
| 1 2 3 4| 4 5|
| 8 9 10 11|11 12|
|15 16 17 18|18 19|
+-----------+-----+
|29 30 31 32|32 33|
+-----------+-----+
All credits to Miller.
R.E. Boss
-----Original Message-----
From: Programming <[email protected]> On Behalf Of R.E.
Boss
Sent: vrijdag 24 september 2021 16:02
To: [email protected]
Subject: Re: [Jprogramming] irregular tillings
This is probably the closest one could get, building the irregular tiling first
in indices and then applying ;.0 Thanks.
R.E. Boss
-----Original Message-----
From: Programming
<[email protected]<mailto:[email protected]>>
On Behalf Of Raul Miller
Sent: vrijdag 24 september 2021 11:36
To: Programming forum
<[email protected]<mailto:[email protected]>>
Subject: Re: [Jprogramming] irregular tillings
Not easily (you could produce a larger result and discard the unwanted content).
However, you could do this:
t=: 0 2 5 e.~i.7
T=: i.7 7
T];.0~L:0<@,."1/~(}:,.2-~/\])I.t,1
+-----+--------+-----+
|0 1 |2 3 4 | 5 6|
|7 8 |9 10 11 |12 13|
+-----+--------+-----+
|14 15|16 17 18|19 20|
|21 22|23 24 25|26 27|
|28 29|30 31 32|33 34|
+-----+--------+-----+
|35 36|37 38 39|40 41|
|42 43|44 45 46|47 48|
+-----+--------+-----+
I hope this helps,
--
Raul
On Fri, Sep 24, 2021 at 4:32 AM R.E. Boss
<[email protected]<mailto:[email protected]>> wrote:
see https://code.jsoftware.com/wiki/Vocabulary/semidot3#dyadic
Let
[t=.;2 3 2<@{."(0)1
1 0 1 0 0 1 0
T=:i.7 7
then what I want is produced with
>,.&,.&.>/('';~t)<;.1 L:_ 0 ('';t)<;.1 T
+-----+--------+-----+
|0 1 |2 3 4 | 5 6|
|7 8 |9 10 11 |12 13|
+-----+--------+-----+
|14 15|16 17 18|19 20|
|21 22|23 24 25|26 27|
|28 29|30 31 32|33 34|
+-----+--------+-----+
|35 36|37 38 39|40 41|
|42 43|44 45 46|47 48|
+-----+--------+-----+
Is this (also) possible with <;.3 ?
R.E. Boss
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