Proof sketch:
0. Working with G is equivalent to working with
a matrix of indices into the ravelled values.
For example:
G=: 5 5 [EMAIL PROTECTED] 2e9
U=: , G
H=: i.5 5
G -: H { U
1
H
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
1. Since G is symmetric, the matrix H simplifies to
H=: (<.|:) H
H
0 1 2 3 4
1 6 7 8 9
2 7 12 13 14
3 8 13 18 19
4 9 14 19 24
2. R dp G dp |:R is equivalent to rot G where
rot=: (1&|.)@:(1&|."1) . By (2) G -: rot G .
Therefore G -: rot^:n G for all n.
3. An index matrix for rot G obtains by (<.rot) H .
In general:
G -: H { U
(rot G) -: ((<.rot) H) { U
(rot^:n G) -: ((<.rot)^:n H) { U
4. cd G computes the list of counter-diagonals of a
square matrix, where
cd=: (|. +/ ])@[EMAIL PROTECTED] (</.&, /: {./.&,[EMAIL PROTECTED]) ]
For example,
i. 5 5
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
cd i.5 5
+--+-----+--------+----------+------------+---------+------+---+-+
|20|15 21|10 16 22|5 11 17 23|0 6 12 18 24|1 7 13 19|2 8 14|3 9|4|
+--+-----+--------+----------+------------+---------+------+---+-+
5. Assert (n{.D) -: |.(-n){.D=. cd (<.rot)^:n H
That is, after n applications of (<.rot),
the length n prefix of the counter diagonals of
the resultant matrix equals the reverse of the
length n suffix. Therefore, after #H applications
of (<.rot), the list of counter diagonals matches
its reversal:
D=: cd (<.rot)^:(#H) H
(-: |.) D
1
6. After #H applications of (<.rot), each counter
diagonal has a unitary nub:
1 = [EMAIL PROTECTED]&> D
1 1 1 1 1 1 1 1 1
7. A matrix is mirror-image symmetric if its
counter diagonals satisfy (a) reversal invariance
(b) each opened atom has a unitary nub
(-: |.) D=: cd H
1 = [EMAIL PROTECTED]&> D
----- Original Message -----
From: John Wilson <[EMAIL PROTECTED]>
Date: Tuesday, January 2, 2007 11:45 am
Subject: [Jgeneral] Array-oriented proof?
> Let G be a square matrix, and
> R =: |: (1&|."1) (=&i.) #G NB. rotation
> M =: |: ( |."1) (=&i.) #G NB. mirror-image
> dp =: +/ . *
>
> Prove the following statement:
>
> If
> (1) G -: |:G NB. G is symmetric
> (2) G -: R dp G dp |:R NB. G is rotationally symmetric
> then
> (3) G -: M dp G dp |:M NB. G is mirror-image symmetric
>
> I can prove this using subscripts (i.e., involving
> expressions like (<i;j){G ), but is there a proof
> involving just the arrays G, R and M?
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
