Dinesman's multiple-dwelling problem: Difference between revisions

=={{header|XPL0}}==

<lang XPL0>include c:\cxpl\codes;

int B, C, F, M, S;

for B:= 1 to 4 do \Baker does not live on top (5th) floor

for C:= 2 to 5 do \Cooper does not live on bottom floor

if C#B then \Cooper & Baker live on different floors

for F:= 2 to 4 do \Fletcher doesn't live on top or bottom

if F#B & F#C & F#C-1 & F#C+1 then \ and she's not adjacent to Cooper

for M:= 1 to 5 do

if M#F & M#B & M>C then \Miller lives above Cooper

for S:= 1 to 5 do \Smith is not adjacent to Fletcher

if S#M & S#F & S#C & S#B & S#F-1 & S#F+1 then \show

[Text(0, "Baker "); IntOut(0, B); CrLf(0); \all

Text(0, "Cooper "); IntOut(0, C); CrLf(0); \possible

Text(0, "Fletcher "); IntOut(0, F); CrLf(0); \solutions

Text(0, "Miller "); IntOut(0, M); CrLf(0);

Text(0, "Smith "); IntOut(0, S); CrLf(0);

]</lang>

Output:

<pre>

Baker 3

Cooper 2

Fletcher 4

Miller 5

Smith 1

</pre>