68 lines
2.6 KiB
Python
68 lines
2.6 KiB
Python
input_file = "input.txt"
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adapters = []
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with open(input_file) as adapter_ratings:
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line = adapter_ratings.readline()
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while line and line != "\n":
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adapters.append(int(line))
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line = adapter_ratings.readline()
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adapters.sort()
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computer_adapter = max(adapters) + 3
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# List of couples separated by three, which must be present to have a working chain
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fixed_points = []
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one_diffs = 0
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three_diffs = 1 # As the computer adapter is always the highest one we got plus three
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# The charging outlet is 0, compute the difference now
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if adapters[0] == 1:
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one_diffs += 1
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elif adapters[0] == 3:
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fixed_points.append((adapters[0], 0))
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three_diffs += 1
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for index in range(len(adapters)-1):
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current_difference = adapters[index+1] - adapters[index]
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if current_difference == 1:
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one_diffs += 1
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elif current_difference == 3:
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if (adapters[index], index) not in fixed_points:
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fixed_points.append((adapters[index], index))
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fixed_points.append((adapters[index+1], index+1))
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three_diffs += 1
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print(f"The product of one and three differences is : {one_diffs}*{three_diffs} = {one_diffs*three_diffs}")
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# The last adapter is always needed as it is the only link possible to the computer adapter
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if (adapters[-1], len(adapters)-1) not in fixed_points:
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fixed_points.append((adapters[-1], len(adapters)-1))
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# Compute the distance separating each fixed point. If they are not successive, it means that there are possible
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# permutations between those two fixed points. Store the count of numbers we can choose from
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separation_distance = []
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for index in range(len(fixed_points) - 1):
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if fixed_points[index + 1][1] - fixed_points[index][1] > 1:
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separation_distance.append(fixed_points[index + 1][1] - fixed_points[index][1] - 1)
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# Distance between 0 and the first fixed point
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separation_distance.insert(0, fixed_points[0][1])
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total_combinations = 1
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# This would probably not work in a general case other than this puzzle, as I only take into account small possible
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# separations, which is all I have got.
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for separation in separation_distance:
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# If we have three numbers, it means that we have at least a separation of 4 jolts between the two fixed points,
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# which is not possible to cross if we do not choose any number between those two. So, do not count this choice.
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if separation == 3:
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# Number of subsets of a set is 2^n where n is the number of elements in the set.
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total_combinations *= 2**3 - 1
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else:
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total_combinations *= 2**separation
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print(f"The number of combinations is : {total_combinations}")
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