ch024.xhtml

Playing Poker (Part 5)

In the last few puzzles we identified four-of-a-kind and full house. Much of the logic for this puzzle will be similar to those, but obviously tweaked somewhat for the next cases.

All you have left in our poker regex family is to identify three-of-a-kind, a pair, and two pairs. As before, you may assume that tests for various hands will run in descending order of strength. So, for example, if your test for a pair will incidentally detect a hand that has four-of-a-kind, that is not a problem since it indeed ipso facto has a pair.

Create these three functions in this puzzle:

is_three_of_kind(hand)
is_two_pairs(hand)
is_pair()

Before you turn the page…

Remember that three is more than two, but less than four.

Identifying two- or three-of-a-kind is a lot like identifying four-of-a-kind, just with fewer repetitions. We could do it without sorting the hand, but doing so, as with our full house solution, is a bit easier.

>>> def is_three_of_kind(hand):
...     try:
...         hand = prettify(hand)
...     except:
...         pass  # Already pretty
...     hand = cardsort(hand)
...     hand = re.sub(r'[^AKQJT98765432]', '', hand)
...     pat = r'(.)\1{2}'  # No begin/end markers
...     match = re.search(pat, hand)
...     return match.group(1) if match else False
...
...
>>> is_three_of_kind('AS 6H QH 6S 2D')
False
>>> is_three_of_kind('AS 6H QH 6S 6D')
'6'

Identifying a pair is basically identical. We simply need to settle for one copy of a card number rather than two copies.

def is_pair(hand):
    try:
        hand = prettify(hand)
    except:
        pass  # Already pretty
    hand = cardsort(hand)
    hand = re.sub(r'[^AKQJT98765432]', '', hand)
    pat = r'(.)\1'  # No begin/end markers
    match = re.search(pat, hand)
    return match.group(1) if match else False

Matching two pairs is actually a little trickier. Remember that for a full house we matched either two of one number followed by three of the other, or matched the reverse, three then two. However, the “gap” of an unmatched number can occur in more different ways in this case. Thinking about it, two pairs might look like any of the following (even assuming sorting):

X X _ Y Y
_ X X Y Y
X X Y Y _

The unmatched number cannot occur in sorted positions 2 or 4 since that leaves only three cards to the other side of the unmatched number (and we have stipulated sorted order of the hand).

As elsewhere, we return the helpful “truthy” value that might be used later in comparing hands of the same type (namely, the two numbers of the pairs, in sorted order).

>>> def is_two_pairs(hand):
...     try:
...         hand = prettify(hand)
...     except:
...         pass  # Already pretty
...     hand = cardsort(hand)
...     hand = re.sub(r'[^[AKQJT98765432]', '', hand)
...     # Three ways to match with unmatched number
...     pat = (r"(.)\1.(.)\2|"
...            r".(.)\3(.)\4|"
...            r"(.)\5(.)\6.")
...     match = re.search(pat, hand)
...     if not match:
...         return False
...     else:
...         return ''.join(n for n in match.groups() if n)
...
>>> is_two_pairs('AH 6S 3H AD 6C')
'A6'
>>> is_two_pairs('AH 6S 3H AD 3C')
'A3'
>>> is_two_pairs('AH 6S 3H KD 3C')
False

The remainder of your poker game program is left for a further exercise. The rest of what you’d need to do won’t have much to do with regular expressions, simply usual program flow and data structures.