13 lines
679 B
Markdown
13 lines
679 B
Markdown
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My approach to this problem is pretty crude. Perhaps I will later come back and try to find a better algorithm.
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Currently we iterate (brute-force) through all possible numbers and count the chains that end with one.
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The `next()` function is implemented like this:
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~~~~~~~~~~~~~~~~~~~
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0\>:55+%:*\ :#v_$$
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^/+55g05+p05<
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~~~~~~~~~~~~~~~~~~~
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We also remember in an 8x71 cache all previously found numbers so we can abort some sequences before we reach `1` or `89`.
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This is the main optimization from pure brute-force in this program.
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We can prove that an 568-element cache is enough because no number in the sequence (except the first) can be greater than `9^2 * 7` (` = 567`)
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