From ed146f7fe1d10ab57a87e0be8d510c54f0e227d3 Mon Sep 17 00:00:00 2001 From: Andrew Ekstedt Date: Wed, 6 Dec 2023 06:19:22 +0000 Subject: [PATCH] day 6 fancy solution + cleanup --- day06/sol.py | 47 ++++++++++++++++++++++++++++++++++++++++++----- 1 file changed, 42 insertions(+), 5 deletions(-) diff --git a/day06/sol.py b/day06/sol.py index 6b2ae04..fd5d776 100644 --- a/day06/sol.py +++ b/day06/sol.py @@ -1,11 +1,12 @@ +from math import sqrt sample = { - "time": [7, 15, 30], + "time": [7, 15, 30], "distance": [9, 40, 200], } input = { -"time": [ 48, 87, 69, 81,], -"distance": [ 255, 1288, 1117, 1623,], + "time": [ 48, 87, 69, 81,], + "distance": [ 255, 1288, 1117, 1623,], } @@ -21,18 +22,54 @@ def solve(data): d = (t-i)*v if d > best: ways += 1 + part1 *= ways return part1 + +def fancy_solve(data): + time = data["time"] + dist = data["distance"] + + for t, best in zip(time, dist): + # the distance traveled if the boat is released at time i + # is equal to (t-i)*i + # we want to know the range of values of i for which this + # exceeds the best time + # (t-i)*i > best + # a little rearranging gets us the quadratic equation + # i^2 - ti + best <= 0 + # which we can determine the crossing points for + # using the quadratic formula (the good one, not the + # one you learned in school) + h = t/2 + s = sqrt(h*h - best) + #print(h-s, h+s) + + # in general these will not be at integer values, + # so we probe the floor and ceiling of each crossing + # point to determine exactly where the condition is met + a = int(h-s) + b = int(h+s) + if (t-a)*a <= best: + a += 1 + if (t-b)*b > best: + b += 1 + + ways = b - a + return ways + + print(solve(sample)) print(solve(input)) def part2(data): - return {"time": [int("".join(str(x) for x in data["time"]))], + return {"time": [int("".join(str(x) for x in data["time"]))], "distance": [int("".join(str(x) for x in data["distance"]))]} print(solve(part2(sample))) -print(solve(part2(input))) +print(fancy_solve(part2(sample))) +print(fancy_solve(part2(input)))