Solve the no connection puzzle: Difference between revisions
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m →annotated solutions: used a more idiomatic way of creating a number of blanks that doesn't require counting. |
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Tested 12094 positions and did 20782 swaps.</pre> |
Tested 12094 positions and did 20782 swaps.</pre> |
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=={{header|Elixir}}== |
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{{trans|Ruby}} |
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This solution uses HLPsolver from [[Solve_a_Hidato_puzzle#Elixir | here]] |
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<lang elixir># It solved if connected A and B, connected G and H (according to the video). |
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# require HLPsolver |
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adjacent = for i <- -2..2, j <- -2..2, not(i in -1..1 and j in -1..1), do: {i,j} |
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layout = ~S""" |
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A - B |
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/|\ /|\ |
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/ | X | \ |
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/ |/ \| \ |
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C - D - E - F |
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\ |\ /| / |
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\ | X | / |
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\|/ \|/ |
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G - H |
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""" |
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board = """ |
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. 0 0 . |
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0 1 0 0 |
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. 0 0 . |
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""" |
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HLPsolver.solve(board, adjacent, false) |
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|> Enum.sort |> Enum.map(fn {_,cell} -> cell.value end) |
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|> Enum.zip(~w[A B C D E F G H]) |
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|> Enum.reduce(layout, fn {n,c},acc -> String.replace(acc, c, to_string(n)) end) |
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|> IO.puts</lang> |
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{{out}} |
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<pre> |
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4 - 6 |
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/|\ /|\ |
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/ | X | \ |
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/ |/ \| \ |
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7 - 1 - 8 - 2 |
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\ |\ /| / |
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\ | X | / |
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\|/ \|/ |
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3 - 5 |
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</pre> |
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=={{header|Go}}== |
=={{header|Go}}== |
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