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Further showing the advantage of the pythonic way, is that the above construction remains about as clear when you refactor it for better pruning: (
(to_number(s, e, n, d),
to_number(m, o, r, e),
to_number(m, o, n, e, y))
for d, digits in choose1(digits)
for e, digits in choose1(digits)
for y, digits in choose1(digits)
if (d + e) % 10 == y
for n, digits in choose1(digits)
for r, digits in choose1(digits)
if (n + r + (d + e) // 10) % 10 == e
for o, digits in choose1(digits)
if (e + o + (n + r + (d + e) // 10) // 10) % 10 == n
for s, digits in choose1(digits, 0)
for m, digits in choose1(digits, 0)
if (s + m + (e + o + (n + r + (d + e) // 10) // 10) // 10) % 10 == o
if (s + m + (e + o + (n + r + (d + e) // 10) // 10) // 10) // 10 == m)
Ends up being several order of magnitudes faster since it partially sums each place and prunes aggressively. You can eliminate the sub-expressions, but it has noise impact on the run time (likely due to trading off temporary tuples for cheap arithmetic): (
(to_number(s, e, n, d),
to_number(m, o, r, e),
to_number(m, o, n, e, y))
for d, digits in choose1(digits)
for e, digits in choose1(digits)
for y, digits in choose1(digits)
if (d + e) % 10 == y
for n, digits in choose1(digits)
for r, digits in choose1(digits)
if (n + r + (d + e) // 10) % 10 == e
for o, digits in choose1(digits)
if (e + o + (n + r + (d + e) // 10) // 10) % 10 == n
for s, digits in choose1(digits, 0)
for m, digits in choose1(digits, 0)
if (s + m + (e + o + (n + r + (d + e) // 10) // 10) // 10) % 10 == o
if (s + m + (e + o + (n + r + (d + e) // 10) // 10) // 10) // 10 == m)
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