Dinesman's multiple-dwelling problem: Difference between revisions
Dinesman's multiple-dwelling problem (view source)
Revision as of 13:55, 28 February 2023
, 1 year agoAdded AppleScript.
Thundergnat (talk | contribs) m (syntax highlighting fixup automation) |
(Added AppleScript.) |
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0 Smith
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=={{header|AppleScript}}==
<syntaxhighlight lang="applescript">on Dinesman()
set output to {}
(* American floor numbering used in these comments to match AppleScript's 1-based indices. *)
-- Baker's not on the fifth floor.
repeat with Baker from 1 to 4
-- Cooper's not on the first floor. Nor on the fifth as Miller's somewhere above him.
-- Fletcher's also not on these floors, so both are in the middle three. They're also
-- at least two floors apart, so one must be on the second and the other on the fourth.
repeat with Cooper from 2 to 4 by 2
if (Cooper ≠ Baker) then
set Fletcher to 6 - Cooper
-- Miller's somewhere above Cooper.
if (Fletcher ≠ Baker) then repeat with Miller from (Cooper + 1) to 5
-- Try to fit Smith in somewhere not adjacent to Fletcher.
if ((Miller ≠ Fletcher) and (Miller ≠ Baker)) then repeat with Smith from 1 to 5
if ((Smith is not in {Baker, Cooper, Fletcher, Miller}) and ¬
((Fletcher - Smith > 1) or (Smith - Fletcher > 1))) then
tell {missing value, missing value, missing value, missing value, missing value}
set {item Baker, item Cooper, item Fletcher, item Miller, item Smith} to ¬
{"Baker", "Cooper", "Fletcher", "Miller", "Smith"}
set end of output to {bottomToTop:it}
end tell
end if
end repeat
end repeat
end if
end repeat
end repeat
return {numberOfSolutions:(count output), solutions:output}
end Dinesman
return Dinesman()</syntaxhighlight>
{{output}}
<syntaxhighlight lang="applescript">{numberOfSolutions:1, solutions:{{bottomToTop:{"Smith", "Cooper", "Baker", "Fletcher", "Miller"}}}}</syntaxhighlight>
=={{header|AutoHotkey}}==
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