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Solve the no connection puzzle: Difference between revisions
→{{header|M2000 Interpreter}}
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=={{header|M2000 Interpreter}}==
Addition a reducing algorithm for double connections, if A has D then from D we remove A unless D has only A so we remove D from A, unless A has only D.
<lang M2000 Interpreter>
Module no_connection_puzzle {
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Append Connections, "E":="ABDFGH","F":="HEB", "G":="CDE","H":="DEF"
\\ eliminate double connections
While con {
m$=eval$(con, con^)
c$=eval$(con)
for i=1 to len(C$) {
d$=mid$(c$,i,1)
r$=Filter$(Connections$(d$), m$)
if r$="" then {
c1$=filter$(c$, d$)
if c1$<>"" then Return connections, m$:=c1$: c$=c1$
} else {
Return connections, d$:=r$
}
}
}
Inventory Holes
For i=0 to 7 : Append Holes, Chr$(65+i):=i : Next i
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4 6
</pre >
=={{header|Mathematica}}==
This one simply takes all permutations of the pegs and filters out invalid solutions.
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