adventofcode2023/day06/sol.py

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from math import sqrt
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sample = {
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"time": [7, 15, 30],
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"distance": [9, 40, 200],
}
input = {
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"time": [ 48, 87, 69, 81,],
"distance": [ 255, 1288, 1117, 1623,],
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}
def solve(data):
time = data["time"]
dist = data["distance"]
part1 = 1
for t, best in zip(time, dist):
ways = 0
for i in range(t):
v = i
d = (t-i)*v
if d > best:
ways += 1
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part1 *= ways
return part1
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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
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print(solve(sample))
print(solve(input))
def part2(data):
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return {"time": [int("".join(str(x) for x in data["time"]))],
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"distance": [int("".join(str(x) for x in data["distance"]))]}
print(solve(part2(sample)))
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print(fancy_solve(part2(sample)))
print(fancy_solve(part2(input)))