Dinesman's multiple-dwelling problem: Difference between revisions

=={{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)