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## Playing Poker (Part 3)

In this puzzle let’s continue with matching poker hands. We handled straights and flushes in the last puzzle (and straight flushes by obvious combination). There are some other types of hands to consider now.

The next several types of hand have containing relationships among them. That is, just like a straight flush is both a straight and a flush, four-of-a-kind is trivially also three-of-a-kind and a pair. A full house is both three-of-a-kind and a pair. However, for our purposes, we will simply assume the various tests are performed in appropriate descending order of strength. The first successful test will be the classified type of the hand.

For the next few puzzles, therefore, write these functions:

• `is_four_of_kind(hand)`
• `is_full_house(hand)`
• `is_three_of_kind(hand)`
• `is_two_pairs(hand)`
• `is_pair()`

This and the next few puzzles cover the various functions. See if you can solve all of them (possibly using shared functionality) before looking at the discussion.

Before you turn the page…

You better cheat, cheat, if you can’t win.

If we have a four-of-a-kind, then the kind must occur in either the first or second card. In fact, if we retain our assumption that the cards are completely ordered, then the four can only occur as the initial four or the final four. But the following implementation does not rely on that ordering:

``````>> def is_four_of_kind(hand):
...     hand = re.sub(r'[^AKQJT98765432]', '', hand)
...     pat = r'^.?(.)(.*\1){3}'
...     match = re.search(pat, hand)
...     # Return the card number as truthy value
...     return match.group(1) if match else False
...
>>> is_four_of_kind('6H 6D 6S 6C 3S') # sorted
'6'
>>> is_four_of_kind('6♦ 3♠ 6♥ 6♠ 6♣') # not sorted
'6'
>>> is_four_of_kind('6H 6D 6S 4C 3S') # not four-of-kind
False``````

The first step is to remove everything that isn’t a card number. Then we either match nothing or the first character of the simplified hand. In the zero-width case, the following group will get the number of the first card. In the one-width case, the group will capture the second card.

The group simply grabs one character, then we must find 3 more copies of that group, but allow any prefix before each repetition. If we promised that the hand was always ordered, the extra stuff before the backreference would not be needed, but it does no harm in being zero width.