modifications in T denoted by ^^^^
T=: 3 : 0
NB. count number of repetitions
t0=. ;<@i."0 #/.~ +/\1,2~:/\y
NB. identify where number is 1 and next number is 2 and identify last number
NB.
^^^^^^^^^^^^^^^^^^^^^^^
NB. identify where number is 2 or more and
NB. (next number is 0) or (number is last), add last identifiers
t10=.(1 (_1})(2=1|.t0)*.t0=1.)+.(0=1|.!.0 t0)*.t0>1
NB. ^^^^^^^^
NB. box string accordingly, add spaces after each token in a box,
NB. except for the last token
t1=. (' '}:@,@,.~])&.> y <;.2~ t10
NB. make (#t1) $' <span>';'</span> '
t2=.(#t1) $' <span>';'</span> '
NB. stitch the last 2 together and ravel
NB. if t is odd delete last box, and raze
;}:^:(2|#t1),t1,.t2
)
t3=: 'abccdeeeeefefefaaaffg'
T t3
a b c c d e e <span>e e e</span> f e f e f a a <span>a</span> f f g
R.E. Boss
> -----Oorspronkelijk bericht-----
> Van: [EMAIL PROTECTED] [mailto:programming-
> [EMAIL PROTECTED] Namens Sherlock, Ric
> Verzonden: zaterdag 12 juli 2008 1:22
> Aan: Programming forum
> Onderwerp: RE: [Jprogramming] Programming brain-teaser
>
> ---R.E. Boss wrote:
> > NB. if t is odd, delete last box and raze
> > ;}:^:(2|#t1),t1,.t2
> > a b c c d e e <span>e e e</span> f e f e f a a <span>a</span> f f
> > <span>f</span>
> >
> >
> > T=: 3 : 0
> > t0=. ;<@i."0 #/.~ +/\1,2~:/\y
> > t1=. (' '}:@,@,.~])&.> y <;.2~((2=1|.t0)*.t0=1.)+.(0=1|.!.0 t0)*.t0>1
> > t2=.(#t1) $' <span>';'</span> '
> > ;}:^:(2|#t1),t1,.t2
> > )
>
> Both your solution and mine need fixing to handle cases with odd numbers
> of groups. Raul's is fine.
> t3=. 'abccdeeeeefefefaaaffg'
>
> bookend space group1 t3
> a b c c d e e <span>e e e</span> f e f e f a a <span>a</span> f f g <span>
> T t3
> a b c c d e e <span>e e e</span> f e f e f a a <span>a</span>
> teas <"0 t3
> a b c c d e e <span>e e e</span> f e f e f a a <span>a</span> f f g
> ----------------------------------------------------------------------
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