Moyo_Ruby_adagrams #114
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Moyo_Ruby_adagrams #114
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added 8 commits
March 22, 2023 21:55
| import random | ||
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| # Function to draw 10 random letters from the letter pool | ||
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| def draw_letters(): |
| def draw_letters(): | ||
| pass | ||
| LETTER_POOL = { |
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Since this dictionary is never changed, it would be good to have it declared outside of the function so that it doesn't have to be recreated each time the function is called.
| # Create a list of letters with the frequency as specified in the pool | ||
| letter_list = [] | ||
| for letter, count in LETTER_POOL.items(): | ||
| letter_list.extend([letter] * count) |
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| return hand | ||
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| # Function to check if the given word can be formed using the available letters | ||
| def uses_available_letters(word, letter_bank): |
| return False | ||
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| # If all letters are in the letter bank copy, the word is valid | ||
| return True | ||
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| def score_word(word): |
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| def score_word(word): | ||
| pass | ||
| word = word.upper() | ||
| letter_scores = { |
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Similar to the letter pool above, this dictionary can be declared outside of the function.
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| # If the scores are equal, apply tie-breaking rules: | ||
| score == highest_word[1] and ( | ||
| # If the current word has 10 letters and the highest_word doesn't, choose the current word | ||
| (len(word) == 10 and len(highest_word[0]) != 10) or | ||
| # If neither word has 10 letters, choose the word with fewer letters | ||
| (len(highest_word[0]) != 10 and len(word) < len(highest_word[0])) | ||
| ) | ||
| ): | ||
| # Update the highest_word with the current word and its score | ||
| highest_word = (word, score) | ||
| # If the scores are equal and the word lengths are also equal, use the order in the supplied list | ||
| elif score == highest_word[1] and len(word) == len(highest_word[0]): | ||
| # Check the index of the current word and highest_word in the original list | ||
| # If the index of the current word is lower, update highest_word with the current word and its score | ||
| highest_word = (word_list.index(word) < word_list.index(highest_word[0])) and (word, score) or highest_word |
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These statements are a bit hard to read. For the sake of keeping the code readable, it might be worth it to have if and elif statements here to lay out the logic rather than trying to get everything together with ands and ors.
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OMG this was so hard