Yes, but what you really wanted was ([: (ic-:ic2) >)"0
Henry Rich
On 1/22/2013 2:52 AM, Linda Alvord wrote:
Aha!
([:(ic-:ic2)"0>) 1j2 0j2
1 1
Linda
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of km
Sent: Tuesday, January 22, 2013 2:05 AM
To: [email protected]
Subject: Re: [Jprogramming] [Jprogrammingou Hermitian from triangular
Linda, rank is saying "Gotcha again!" --Kip
(ic-:ic2)&> b. 0
0 0 0
([:(ic-:ic2)>) b. 0
_ _ _
Sent from my iPad
On Jan 22, 2013, at 12:28 AM, "Linda Alvord" <[email protected]> wrote:
What is the problem here?
(ic-:ic2)&> 1j2 0j2
1 1
([:(ic-:ic2)>) 1j2 0j2
0
u@v y ↔ u v y Moreover, the monadic uses of u@v and u&v are equivalent.
Linda
-----Original Message-----
From: [email protected] [mailto:programming-
[email protected]] On Behalf Of David Ward Lambert
Sent: Monday, January 21, 2013 10:45 AM
To: programming
Subject: Re: [Jprogramming] [Jprogrammingou Hermitian from triangular
Interrupting may be a bad idea.
NB. -------- Use 13 : correctly.
NB. f=: 13 :',j./&i:/+.y' NB. no.
NB. f=: 13 : '([: , j./&i:/@+.) y' NB. yes.
ic2 =: [: , j./&i:/@+.
NB. Using a proverb is as if its definition
NB. were inserted with parenthesis.
NB. Copy ic2 definition verbatim,
NB. Surround with parentheses.
NB. Put in the arguments, then fix quotes.
13 : '([: , j./&i:/@+.) y'
[: , j./&i:/@+.
NB. The definition was tacit,
NB. 13 : returns this tacit verb unchanged
NB. -------- Equivalence of ic and ic2
NB."However, I do not understand how ic and ic2 agree when they don't!"
ic =: [: , ([: i: 9&o.) j./ [: i: 11&o.
ic2 =: [: , j./&i:/@+.
NB. ic and ic2 are identical at rank 0
(ic -: ic2)&>1j2 0j2
1 1
([: < ic)"0 ] 1j2 0j2
+----------------------------------------------------------------+----
+----------------------------------------------------------------+----
+----------------------------------------------------------------+--
---------+
|_1j_2 _1j_1 _1 _1j1 _1j2 0j_2 0j_1 0 0j1 0j2 1j_2 1j_1 1 1j1 1j2|0j_2
0j_1 0 0j1 0j2|
+----------------------------------------------------------------+----
+----------------------------------------------------------------+----
+----------------------------------------------------------------+--
---------+
([: < ic2)"0 ] 1j2 0j2
+----------------------------------------------------------------+----
+----------------------------------------------------------------+----
+----------------------------------------------------------------+--
---------+
|_1j_2 _1j_1 _1 _1j1 _1j2 0j_2 0j_1 0 0j1 0j2 1j_2 1j_1 1 1j1 1j2|0j_2
0j_1 0 0j1 0j2|
+----------------------------------------------------------------+----
+----------------------------------------------------------------+----
+----------------------------------------------------------------+--
---------+
Date: Mon, 21 Jan 2013 03:58:24 -0500
From: "Linda Alvord" <[email protected]>
To: <[email protected]>
Subject: Re: [Jprogramming] [Jprogrammingou Hermitian from triangular
Message-ID: <001d01cdf7b5$78a4cd20$69ee6760$@net>
Content-Type: text/plain; charset="koi8-r"
Since this has a mistake it is hard to reconstruct what I did wrong.
This
represents what I was doing:
ic2=: [: , j./&i:/@+.
ic2
[: , j./&i:/@+.
([: , j./&i:/@+.)2
_2 _1 0 1 2
f=: 13 :',j./&i:/+.y'
................
ic 1j2 0j2
_1j_2 _1j_1 _1 _1j1 _1j2 _1j_2 _1j_1 _1 _1j1 _1j2 0j_2 0j_1 0 0j1 0j2
0j_2
0j_1 0 0j1 0j2 1j_2 1j_1 1 1j1 1j2 1j_2 1j_1 1 1j1 1j2 0j_2 0j_1 0
0j1
0j2
0j_2 0j_1 0 0j1 0j2 0j_2 0j_1 0 0j1 0j2 0j_2 0j_1 0 0j1 0j2 0j_2 0j_1
0
0j1
0j2 0j_2 0j_1 0 0j1 0j2
ic2 1j2 0j2
_1j_2 _1j_1 _1 _1j1 _1j2 0j_2 0j_1 0 0j1 0j2 1j_2 1j_1 1 1j1 1j2 0j_2
0j_1 0
0j1 0j2 0 0 0 0 0 0 0 0 0 0
(ic-:ic2)&>1j2 0j2
1 1
However, I do not understand how ic and ic2 agree when they don't!
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