I have found an elegant and efficient solution to the problem
posed in the appended msg: given y which is computed by
y=: (c{a.) (256|m b.)&.(a.&i.) a.
or
y=: a. (256|m b.)&.(a.&i.) c{a.
where atom c e. i.256 and atom m e. 16+i.16, then ff y are
2 integers such that
y -: (c1{a.) (256|m1 b.)&.(a.&i.) a. [ 'c1 m1'=. ff y
ff=: 3 : 0 " 1
assert. ((,256)-:$y) *. 2=3!:0 y
'i j'=. a.i.0 255{y
if. i=0 do. cm=. j,17
elseif. j=i do. cm=. i,19
elseif. j=255 do. cm=. i,23
elseif. j=255-i do. cm=. i,22
elseif. j=0 do. cm=. i,18
elseif. i=255 do. cm=. j,27
elseif. 1 do. assert. 0 [ 'y is not generated by m b.' end.
assert. y -: (c{a.) (256|m b.)&.(a.&i.) a. [ 'c m'=. cm
cm
)
For example:
'c m'=: 0 16+?256 16
c
110
m
19
y=: c (256|m b.)&.(a.&i.) a.
y=: (c{a.) (256|m b.)&.(a.&i.) a.
ff y
110 19
y -: (110{a.) (256|19 b.)&.(a.&i.) a.
1
'c m'=: 0 16+?256 16
c
54
m
25
y=: a. (256|m b.)&.(a.&i.) c{a.
ff y
201 22
y -: (201{a.) (256|22 b.)&.(a.&i.) a.
1
ff a.{~ ?~256
|assertion failure: ff
| 0['y is not generated by m b.'
Exhaustive test on all possibilities:
$ ff (16+i.16) 4 : 'y (256|x b.)&.(a.&i.) a.'"0/a.
16 256 2
$ ff (16+i.16) 4 : 'a. (256|x b.)&.(a.&i.) y '"0/a.
16 256 2
----- Original Message -----
From: "Roger Hui" <[EMAIL PROTECTED]>
To: Programming forum <[email protected]>
Date: Tuesday, September 25, 2007 12:06:35 HKT
Subject: Re: [Jprogramming] Decoding
Suppose
c=: ?256
d=: 16+?16
e=: a.{~ c (d b.) i.256
NB. e is used as: map=: e {~ a. i. ]
Can you recover c and d given e? c and d are not
necessarily unique so it suffices to compute c1 and d1
such that e -: a.{~ c1 (d1 b.) i.256 .
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