Tidy up structure of briefcase
I had a spare fifteen minutes and decided that I should tidy up my monorepo. The work of tidying up is not finished; this is a small step in the right direction. TL;DR - Created a tools directory - Created a scratch directory (see README.md for more information) - Added README.md to third_party - Renamed delete_dotfile_symlinks -> symlinkManager - Packaged symlinkManager as an executable symlink-mgr using buildGo
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scratch/data_structures_and_algorithms/optimal-stopping.py
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scratch/data_structures_and_algorithms/optimal-stopping.py
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from random import choice
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from math import floor
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# Applying Chapter 1 from "Algorithms to Live By", which describes optimal
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# stopping problems. Technically this simulation is invalid because the
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# `candidates` function takes a lower bound and an upper bound, which allows us
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# to know the cardinal number of an individual candidates. The "look then leap"
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# algorithm is ideal for no-information games - i.e. games when upper and lower
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# bounds aren't known. The `look_then_leap/1` function is ignorant of this
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# information, so it behaves as if in a no-information game. Strangely enough,
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# this algorithm will pick the best candidate 37% of the time.
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#
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# Chapter 1 describes two algorithms:
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# 1. Look-then-leap: ordinal numbers - i.e. no-information games. Look-then-leap
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# finds the best candidate 37% of the time.
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# 2. Threshold: cardinal numbers - i.e. where upper and lower bounds are
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# known. The Threshold algorithm finds the best candidate ~55% of the time.
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#
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# All of this and more can be studied as "optimal stopping theory". This applies
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# to finding a spouse, parking a car, picking an apartment in a city, and more.
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# candidates :: Int -> Int -> Int -> [Int]
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def candidates(lb, ub, ct):
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xs = list(range(lb, ub + 1))
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return [choice(xs) for _ in range(ct)]
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# look_then_leap :: [Integer] -> Integer
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def look_then_leap(candidates):
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best = candidates[0]
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seen_ct = 1
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ignore_ct = floor(len(candidates) * 0.37)
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for x in candidates[1:]:
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if ignore_ct > 0:
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ignore_ct -= 1
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best = max(best, x)
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else:
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if x > best:
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print('Choosing the {} candidate.'.format(seen_ct))
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return x
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seen_ct += 1
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print('You may have waited too long.')
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return candidates[-1]
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candidates = candidates(1, 100, 100)
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print(candidates)
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print(look_then_leap(candidates))
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