Project Euler problem 191 solution

Author: darkstar

project_euler/problem_191/__init__.py

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+"""
+Prize Strings
+Problem 191
+
+A particular school offers cash rewards to children with good attendance and
+punctuality. If they are absent for three consecutive days or late on more
+than one occasion then they forfeit their prize.
+
+During an n-day period a trinary string is formed for each child consisting
+of L's (late), O's (on time), and A's (absent).
+
+Although there are eighty-one trinary strings for a 4-day period that can be
+formed, exactly forty-three strings would lead to a prize:
+
+OOOO OOOA OOOL OOAO OOAA OOAL OOLO OOLA OAOO OAOA
+OAOL OAAO OAAL OALO OALA OLOO OLOA OLAO OLAA AOOO
+AOOA AOOL AOAO AOAA AOAL AOLO AOLA AAOO AAOA AAOL
+AALO AALA ALOO ALOA ALAO ALAA LOOO LOOA LOAO LOAA
+LAOO LAOA LAAO
+
+How many "prize" strings exist over a 30-day period?
+"""
+
+
+def solution(days=30):
+    """Returns the number of possible prize strings for a particular number
+    of days, using a simple recursive function with caching to speed it up.
+
+    >>> solution()
+    1918080160
+    >>> solution(4)
+    43
+    """
+
+    # a cache to speed up calculation
+    cache = {}
+
+    # we will be using a simple recursive function
+    def calculate(days, absent, late):
+        # if we are absent twice, or late 3 consecutive days,
+        # no further prize strings are possible
+        if late == 3 or absent == 2:
+            return 0
+
+        # if we have no days left, and have not failed any other rules,
+        # we have a prize string
+        if days == 0:
+            return 1
+
+        # No easy solution, so now we need to do the recursive calculation
+
+        # First, check if the combination is already in the cache, and
+        # if yes, return the stored value from there since we already
+        # know the number of possible prize strings from this point on
+        key = (days, absent, late)
+        if key in cache:
+            return cache[key]
+
+        # if we are late (but not absent), the "absent" counter stays as
+        # it is, but the "late" counter increases by one
+        state_late = calculate(days - 1, absent, late + 1)
+
+        # if we are absent, the "absent" counter increases by 1, and the
+        # "late" counter resets to 0
+        state_absent = calculate(days - 1, absent + 1, 0)
+
+        # if we are on time, this resets the "late" counter and keeps the
+        # absent counter
+        state_ontime = calculate(days - 1, absent, 0)
+
+        # the total number of prize strings is the sum of these three
+        # possible states
+        prizestrings = state_late + state_absent + state_ontime
+
+        # now store the value we calculated in the cache
+        cache[key] = prizestrings
+        return prizestrings
+
+    return calculate(days, 0, 0)
+
+
+if __name__ == "__main__":
+    print(solution())