Parnic
4cde56eb84
This one's part 1 destroyed me. I had a very difficult time, trying 3 separate approaches, each one of which worked for most cases but eventually fell apart on the 5th sample or my actual puzzle input. I ended up tweaking and rewriting until I landed on this which wasn't far off from what I was trying to do previously, but I had been overcomplicating things. Part 2 surprised me in that I expected a simple "ore available divided by ore needed for 1 fuel" would solve it, but of course the excess chemicals produced in any given reaction meant that it wasn't that simple. So this approach uses that estimate as a lower bound, since it always underestimates, and then bisects its way to the solution (starting at the lower bound and adding 1 each time took too long). I'm sure a smarter upper bound choice could lower the runtime of this by a bit, but runtime isn't bad enough right now for me to try any additional optimizations.
27 lines
773 B
Go
27 lines
773 B
Go
package utilities
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import (
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"math"
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)
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// Bisect takes a known-good low and known-bad high value as the bounds
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// to bisect, and a function to test each value for success or failure.
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// If the function succeeds, the value is adjusted toward the maximum,
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// and if the function fails, the value is adjusted toward the minimum.
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// The final value is returned when the difference between the success
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// and the failure is less than or equal to the acceptance threshold
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// (usually 1, for integers).
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func Bisect[T Number](low, high, threshold T, tryFunc func(val T) bool) T {
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for T(math.Abs(float64(high-low))) > threshold {
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currVal := low + ((high - low) / 2)
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success := tryFunc(currVal)
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if success {
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low = currVal
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} else {
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high = currVal
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}
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}
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return low
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}
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