Add solution for Project Euler problem 191

project_euler/problem_191/sol1.py

@@ -0,0 +1,105 @@
+"""
+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?
+
+References:
+    - The original Project Euler project page:
+      https://projecteuler.net/problem=191
+"""
+
+
+cache = {}
+
+
+def _calculate(days: int, absent: int, late: int) -> int:
+    """
+    A small helper function for the recursion, mainly to have
+    a clean interface for the solution() function below.
+
+    It should get called with the number of days (corresponding
+    to the desired length of the 'prize strings'), and the
+    initial values for the number of consecutive absent days and
+    number of total late days.
+
+    >>> _calculate(days=4, absent=0, late=0)
+    43
+    >>> _calculate(days=30, absent=2, late=0)
+    0
+    >>> _calculate(days=30, absent=1, late=0)
+    98950096
+    """
+
+    # 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]
+
+    # now we calculate the three possible ways that can unfold from
+    # this point on, depending on our attendance today
+
+    # 1) 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)
+
+    # 2) 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)
+
+    # 3) if we are on time, this resets the "late" counter and keeps the
+    # absent counter
+    state_ontime = _calculate(days - 1, absent, 0)
+
+    prizestrings = state_late + state_absent + state_ontime
+
+    cache[key] = prizestrings
+    return prizestrings
+
+
+def solution(days: int = 30) -> int:
+    """
+    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
+    """
+
+    return _calculate(days, absent=0, late=0)
+
+
+if __name__ == "__main__":
+    print(solution())