[REBOL] Re: Monday Puzzle
From: jelinem1:nationwide at: 24-Jul-2001 8:22
Here's a "brute force" solution. All else I can think of are either a
variation of this, or require more code to set up the 'y block.
>> f: does [a: [9] either 9 = a/1 [a/1: 0][a/1: 1 + a/1]]
>> y: []
== []
>> append y :f
== [func [][a: [9] either 9 = a/1 [a/1: 0] [a/1: 1 + a/1]]]
>> append/only y y
== [func [][a: [9] either 9 = a/1 [a/1: 0] [a/1: 1 + a/1]] [...]]
>> ; ----------------------------------------------------
>> loop 20 [prin [y/1 " "] y: second y] print ""
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9
>> loop 20 [prin [length? y " "] y: second y] print ""
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
A nice diversion from work.
- Michael Jelinek
From: "Larry Palmiter" <[larry--ecotope--com]>@rebol.com on 07/23/2001 10:29
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Subject: [REBOL] Monday Puzzle
Hi all,
Here's a puzzle. I made a block named y of length 2 which shows the
following behavior:
>> type? y
== block!
>> length? y
== 2
>> y/1
== 0
>> loop 20 [prin [y/1 " "] y: second y] print ""
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9
Here's a hint!
>> loop 20 [prin [length? y " "] y: second y] print ""
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
How was the block constructed and what does the MOLD of the block look
like?
Enjoy
-Larry
