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* Added solution for Project Euler problem 109
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90 lines
3.0 KiB
Python
90 lines
3.0 KiB
Python
"""
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In the game of darts a player throws three darts at a target board which is
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split into twenty equal sized sections numbered one to twenty.
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The score of a dart is determined by the number of the region that the dart
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lands in. A dart landing outside the red/green outer ring scores zero. The black
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and cream regions inside this ring represent single scores. However, the red/green
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outer ring and middle ring score double and treble scores respectively.
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At the centre of the board are two concentric circles called the bull region, or
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bulls-eye. The outer bull is worth 25 points and the inner bull is a double,
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worth 50 points.
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There are many variations of rules but in the most popular game the players will
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begin with a score 301 or 501 and the first player to reduce their running total
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to zero is a winner. However, it is normal to play a "doubles out" system, which
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means that the player must land a double (including the double bulls-eye at the
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centre of the board) on their final dart to win; any other dart that would reduce
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their running total to one or lower means the score for that set of three darts
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is "bust".
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When a player is able to finish on their current score it is called a "checkout"
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and the highest checkout is 170: T20 T20 D25 (two treble 20s and double bull).
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There are exactly eleven distinct ways to checkout on a score of 6:
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D3
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D1 D2
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S2 D2
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D2 D1
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S4 D1
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S1 S1 D2
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S1 T1 D1
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S1 S3 D1
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D1 D1 D1
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D1 S2 D1
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S2 S2 D1
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Note that D1 D2 is considered different to D2 D1 as they finish on different
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doubles. However, the combination S1 T1 D1 is considered the same as T1 S1 D1.
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In addition we shall not include misses in considering combinations; for example,
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D3 is the same as 0 D3 and 0 0 D3.
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Incredibly there are 42336 distinct ways of checking out in total.
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How many distinct ways can a player checkout with a score less than 100?
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Solution:
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We first construct a list of the possible dart values, separated by type.
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We then iterate through the doubles, followed by the possible 2 following throws.
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If the total of these three darts is less than the given limit, we increment
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the counter.
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"""
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from itertools import combinations_with_replacement
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def solution(limit: int = 100) -> int:
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"""
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Count the number of distinct ways a player can checkout with a score
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less than limit.
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>>> solution(171)
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42336
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>>> solution(50)
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12577
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"""
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singles: list[int] = [x for x in range(1, 21)] + [25]
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doubles: list[int] = [2 * x for x in range(1, 21)] + [50]
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triples: list[int] = [3 * x for x in range(1, 21)]
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all_values: list[int] = singles + doubles + triples + [0]
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num_checkouts: int = 0
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double: int
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throw1: int
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throw2: int
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checkout_total: int
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for double in doubles:
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for throw1, throw2 in combinations_with_replacement(all_values, 2):
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checkout_total = double + throw1 + throw2
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if checkout_total < limit:
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num_checkouts += 1
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return num_checkouts
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if __name__ == "__main__":
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print(f"{solution() = }")
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