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