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Dinesman's multiple-dwelling problem: Difference between revisions
Dinesman's multiple-dwelling problem (view source)
Revision as of 02:16, 26 June 2011
, 13 years ago→Icon and Unicon
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}</lang>
<pre>[2, 1, 3, 4, 0]</pre>
=={{header|Icon}} and {{header|Unicon}}==
<lang Icon>invocable all
global nameL, nameT, rules
procedure main() # Dinesman
nameT := table()
nameL := ["Baker", "Cooper", "Fletcher", "Miller", "Smith"]
rules := [ [ distinct ],
[ "~=", "Baker", top() ],
[ "~=", "Cooper", bottom() ],
[ "~=", "Fletcher", top() ],
[ "~=", "Fletcher", bottom() ],
[ ">", "Miller", "Cooper" ],
[ notadjacent, "Smith", "Fletcher" ],
[ notadjacent, "Fletcher", "Cooper" ],
[ showsolution ],
[ stop ] ]
if not solve(1) then
write("No solution found.")
end
procedure dontstop() # use if you want to search for all solutions
end
procedure showsolution() # show the soluton
write("The solution is:")
every write(" ",n := !nameL, " lives in ", nameT[n])
return
end
procedure eval(n) # evaluate a rule
r := copy(rules[n-top()])
every r[i := 2 to *r] := rv(r[i])
if get(r)!r then suspend
end
procedure rv(x) # return referenced value if it exists
return \nameT[x] | x
end
procedure solve(n) # recursive solver
if n > top() then { # apply rules
if n <= top() + *rules then
( eval(n) & solve(n+1) ) | fail
}
else # setup locations
(( nameT[nameL[n]] := bottom() to top() ) & solve(n + 1)) | fail
return
end
procedure distinct(a,b) # ensure each name is distinct
if nameT[n := !nameL] = nameT[n ~== key(nameT)] then fail
suspend
end
procedure notadjacent(n1,n2) # ensure n1,2 are not adjacent
if not adjacent(n1,n2) then suspend
end
procedure adjacent(n1,n2) # ensure n1,2 are adjacent
if abs(n1 - n2) = 1 then suspend
end
procedure bottom() # return bottom
return if *nameL > 0 then 1 else 0
end
procedure top() # return top
return *nameL
end</lang>
Output:<pre>The solution is:
Baker lives in 3
Cooper lives in 2
Fletcher lives in 4
Miller lives in 5
Smith lives in 1</pre>
=={{header|J}}==
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