Dinesman's multiple-dwelling problem: Difference between revisions
Dinesman's multiple-dwelling problem (view source)
Revision as of 19:13, 25 June 2011
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<pre>[2, 1, 3, 4, 0]</pre>
=={{header|J}}==
This problem asks us to pick from one of several possibilities. We can represent these possibilities as a permutation of the residents' initials, arranged in order from lowest floor to top floor:
<lang j>possible=: ((i.!5) A. i.5) { 'BCFMS'</lang>
Additionally, we are given a variety of constraints which eliminate some possibilities:
<lang j>possible=: (#~ 'B' ~: {:"1) possible NB. Baker not on top floor
possible=: (#~ 'C' ~: {."1) possible NB. Cooper not on bottom floor
possible=: (#~ 'F' ~: {:"1) possible NB. Fletcher not on top floor
possible=: (#~ 'F' ~: {."1) possible NB. Fletcher not on bottom floor
possible=: (#~ </@i."1&'CM') possible NB. Miller on higher floor than Cooper
possible=: (#~ 0 = +/@E."1~&'SF') possible NB. Smith not immediately below Fletcher
possible=: (#~ 0 = +/@E."1~&'FS') possible NB. Fletcher not immediately below Smith
possible=: (#~ 0 = +/@E."1~&'CF') possible NB. Cooper not immediately below Fletcher
possible=: (#~ 0 = +/@E."1~&'FC') possible NB. Fletcher not immediately below Cooper</lang>
The answer is thus:
<lang j> possible
SCBFM</lang>
(bottom floor) Smith, Cooper, Baker, Fletcher, Miller (top floor)
=={{header|PureBasic}}==
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