# lab10.py
# Skeleton by D. Gries, L. Lee, S. Marschner, and W. White
# April 6, 2014
"""Loop exercises."""

import card2
import sys

def isprime(p):
    """Returns: True if p a prime, False otherwise.

    p is prime if it is >= 2 and divisible by only 1 and itself

    Precondition: p is a positive int"""
    # Battle plan:
    # * if p<=2, return the truth value of p==2
    # * otherwise, use a while-loop to see if some integer in 2..p-1 divides p:
    #   if one is found, return False
    #   if none are found, return True
    # Invariant: p is not divisible by integers in 2..k-1.
    # So, k marks the beginning of the "unchecked" integers.


    if p <= 2:
        return p==2
    else:
        # Return False if some integer in 2..p-1 divides p

        k = None # TODO: FIX. For what k is 2..k-1 the empty range?
                 # It's true that nothing in the empty range divides p.

        # inv: p is not divisible by integers in 2..k-1
        while k < p:
            if p % k == 0:
                pass # TODO: FIX THIS
            k=k # TODO: FIX.  If we know this k doesn't divide p, then we have
                # to change the beginning of the "unchecked" integers.

        # Post: no integer in 2..p-1 divides p
        return 14 #TODO:  FIX


# How many draws, on average,  until you get a two of a kind?
def exp_pair(n, verbose=False):
    """Returns: number of draws, averaged over n trials, that it takes to
    get two of a kind from a full deck.

    If verbose, prints out the ranks that were drawn and the number of draws.
    Pre: n a positive int, verbose a boolean. """
    # Use a for-loop to run multiple trials; use a while-loop for each trial.
    # The invariant for the while-loop should be:
    #   * <found> is the list of "unmatched" ranks found so far
    #   * <newrank> is the rank of the *next* card drawn, which has not
    #      been checked against <found> yet.
    # So, when we get a case where <newrank> is in <found>, then we hit
    # a two-of-a-kind, and the number of

    # Make sure you take <verbose> into account correctly.

    results = [] # list of number of draws for trials run so far

    for t in range(n):
        deck = card2.full_deck()
        found = None # TODO; FIX THIS # list of ranks found so far

        newrank = deck.pop().rank # Don't change this line
        if verbose:
            print newrank

        while True: # TODO: FIX THIS: replace it with the boolean opposite of
                    # when we know that we can stop looking for a pair
                    # Note that we needn't worry about running out of cards,
                    # since a full deck has pairs in it, for sure.
            pass

            # update <found> and <newrank> according to the invariant


        # If the invariant has been maintained, then the number of draws to get
        # a pair is len(found) + 1 (all the unmatched draws plus the last one).

        if verbose:
            print 'number of draws: ', len(found) + 1
        results.append(len(found) + 1)



    return sum(results)/float(n)






if __name__ == '__main__':

    # usage: python lab10.py   which sets n to a default value, verbosity off
    # or
    # python lab10.py n     which sets n as given, verbosity off
    # or
    # python lab10.py n <anything at all> which sets n as given, verbosity on

    args_given = (len(sys.argv) > 2)

    n = (int(sys.argv[1]) if args_given else 10000)
    verbose = (False if not args_given else bool(sys.argv[2]))

    print ('In ' + str(n) + ' trials,' +
         ' the avg number of draws to get a pair was ' + str(exp_pair(n, verbose)))