Day 14: Restroom Redoubt
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Haskell, alternative approach
The x and y coordinates of robots are independent. 101 and 103 are prime. So, the pattern of x coordinates will repeat every 101 ticks, and the pattern of y coordinates every 103 ticks.
For the first 101 ticks, take the histogram of x-coordinates and test it to see if it’s roughly randomly scattered by performing a chi-squared test using a uniform distrobution as the basis. [That code’s not given below, but it’s a trivial transliteration of the formula on wikipedia, for instance.] In my case I found a massive peak at t=99.
Same for the first 103 ticks and y coordinates. Mine showed up at t=58.
You’re then just looking for solutions of t = 101m + 99, t = 103n + 58 [in this case]. I’ve a library function, maybeCombineDiophantine, which computes the intersection of these things if any exist; again, this is basic wikipedia stuff.
day14b ls = let rs = parse ls size = (101, 103) positions = map (\t -> process size t rs) [0..] -- analyse x coordinates. These should have period 101 xs = zip [0..(fst size)] $ map (\rs -> map (\(p,_) -> fst p) rs & C.count & chi_squared (fst size)) positions xMax = xs & sortOn snd & last & fst -- analyse y coordinates. These should have period 103 ys = zip [0..(snd size)] $ map (\rs -> map (\(p,_) -> snd p) rs & C.count & chi_squared (snd size)) positions yMax = ys & sortOn snd & last & fst -- Find intersections of: t = 101 m + xMax, t = 103 n + yMax ans = do (s,t) <- maybeCombineDiophantine (fromIntegral (fst size), fromIntegral xMax) (fromIntegral (snd size), fromIntegral yMax) pure $ minNonNegative s t in trace ("xs distributions: " ++ show (sortOn snd xs)) $ trace ("ys distributions: " ++ show (sortOn snd ys)) $ trace ("xMax = " ++ show xMax ++ ", yMax = " ++ show yMax) $ trace ("answer could be " ++ show ans) $ ans
Very cool, taking a statistical approach to discern random noise from picture.
Thanks. It was the third thing I tried - began by looking for mostly-symmetrical, then asked myself “what does a christmas tree look like?” and wiring together some rudimentary heuristics. When those both failed (and I’d stopped for a coffee) the alternative struck me. It seems like a new avenue into the same diophantine fonisher that’s pretty popular in these puzzles - quite an interesting one.
This day’s puzzle is clearly begging for some inventive viaualisations.
Very nice!
I should add - it’s perfectly possible to draw pictures which won’t be spotted by this test, but in this case as it happens the distributions are exceedingly nonuniform at the critical point.