Is the converse true? That is, if G is mirror-image
symmetric, then is it necessarily the case that G is
symmetric and rotationally symmetric? Answer: No.
Again we can work with a matrix of indices:
G=: 5 5 [EMAIL PROTECTED] 2e9
U=: , G
H=: i.5 5
G -: H { U
1
M dp G dp |:M is equivalent to flip G where
flip=: |.@:(|."1) . By definition mirror-image
symmetric means G -: flip G . Therefore
G -: flip^:n G for all n.
An index matrix for flip G obtains by (<.flip) H .
In general:
G -: H { U
(flip G) -: ((<.flip) H) { U
(flip^:n G) -: ((<.flip)^:n H) { U
(flip^:_ G) -: ((<.flip)^:_ H) { U
So:
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
] K=: (<.flip)^:_ H
0 1 2 3 4
5 6 7 8 9
10 11 12 11 10
9 8 7 6 5
4 3 2 1 0
This last result is mirror-image symmetric but not
symmetric or rotationally symmetric:
(-: |:) K
0
(-: rot) K
0
----- Original Message -----
From: Roger Hui <[EMAIL PROTECTED]>
Date: Thursday, January 4, 2007 11:42 am
Subject: Re: [Jgeneral] Array-oriented proof?
> 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?
>
>
> -------------------------------------------------------------------
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>
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