git-subtree-dir: users/wpcarro git-subtree-mainline:464bbcb15cgit-subtree-split:24f5a642afChange-Id: I6105b3762b79126b3488359c95978cadb3efa789
		
			
				
	
	
		
			49 lines
		
	
	
	
		
			1.8 KiB
		
	
	
	
		
			Python
		
	
	
	
	
	
			
		
		
	
	
			49 lines
		
	
	
	
		
			1.8 KiB
		
	
	
	
		
			Python
		
	
	
	
	
	
| from random import choice
 | |
| from math import floor
 | |
| 
 | |
| # Applying Chapter 1 from "Algorithms to Live By", which describes optimal
 | |
| # stopping problems. Technically this simulation is invalid because the
 | |
| # `candidates` function takes a lower bound and an upper bound, which allows us
 | |
| # to know the cardinal number of an individual candidates. The "look then leap"
 | |
| # algorithm is ideal for no-information games - i.e. games when upper and lower
 | |
| # bounds aren't known. The `look_then_leap/1` function is ignorant of this
 | |
| # information, so it behaves as if in a no-information game. Strangely enough,
 | |
| # this algorithm will pick the best candidate 37% of the time.
 | |
| #
 | |
| # Chapter 1 describes two algorithms:
 | |
| # 1. Look-then-leap: ordinal numbers - i.e. no-information games. Look-then-leap
 | |
| #    finds the best candidate 37% of the time.
 | |
| # 2. Threshold: cardinal numbers - i.e. where upper and lower bounds are
 | |
| #    known. The Threshold algorithm finds the best candidate ~55% of the time.
 | |
| #
 | |
| # All of this and more can be studied as "optimal stopping theory". This applies
 | |
| # to finding a spouse, parking a car, picking an apartment in a city, and more.
 | |
| 
 | |
| 
 | |
| # candidates :: Int -> Int -> Int -> [Int]
 | |
| def candidates(lb, ub, ct):
 | |
|     xs = list(range(lb, ub + 1))
 | |
|     return [choice(xs) for _ in range(ct)]
 | |
| 
 | |
| 
 | |
| # look_then_leap :: [Integer] -> Integer
 | |
| def look_then_leap(candidates):
 | |
|     best = candidates[0]
 | |
|     seen_ct = 1
 | |
|     ignore_ct = floor(len(candidates) * 0.37)
 | |
|     for x in candidates[1:]:
 | |
|         if ignore_ct > 0:
 | |
|             ignore_ct -= 1
 | |
|             best = max(best, x)
 | |
|         else:
 | |
|             if x > best:
 | |
|                 print('Choosing the {} candidate.'.format(seen_ct))
 | |
|                 return x
 | |
|         seen_ct += 1
 | |
|     print('You may have waited too long.')
 | |
|     return candidates[-1]
 | |
| 
 | |
| 
 | |
| candidates = candidates(1, 100, 100)
 | |
| print(candidates)
 | |
| print(look_then_leap(candidates))
 |