expected_number_boolean_combinations_n_eq_5.py

"""
CALCULATION FOR NUMBER OF EXPECTED EXPRESSIONS NOTATION README ->

CASE: 5 SYSTEMS

I'm using the following notation
    # N_X_Ys is the number of combinations present using Y Systems, selected from X Total Systems. 
    # FOR EXAMPLE N_5_2s is the number of combinations using 2 systems, selected from 5 total systems
    
    # N_X_Y_PQs is the number of combinations present using Y Systems by adding system combinations of P and Q elements. 
    # FOR EXAMPLE N_5_4_31s is the number of combinations using 4 systems formed by joining a combination of 3 systems and a combination of 1 system. 

Author: Michael v. Heinz
"""

def n_expected_expressions():    
    TOTAL = 5 # This is the total number of systems we're using.
    def n_5_s(TOTAL):
        X  = 1 # This is the # of combinations present using X SYSTEMS

        N_5_1s = math.factorial(TOTAL)/(math.factorial(TOTAL-X)*math.factorial(X))
        print(N_5_1s)

        return N_5_1s

    N_5_1s = n_5_s(TOTAL)

    # B Lets look at possible combinations using 2 systems. 
    def n_5_2s(TOTAL):
        X = 2
        N_5_2s = math.factorial(TOTAL)/(math.factorial(TOTAL-X)*math.factorial(X)) * 2 # multiply by 2 to account for binary operations
        print(N_5_2s)

        return N_5_2s

    N_5_2s = n_5_2s(TOTAL)

    # C Lets look at possible combinations using 3 systems
    def n_5_3s(N_5_2s, N_5_1s):
        X=3
        N_5_3s = N_5_2s * N_5_1s * (3/5) # we take cartesian product of combinations of 2s and combinations of 1s
        # we multiply by 3/5 because we have only 3 options left since 2 are already used in the combination of 2 systems
        N_5_3s = N_5_3s * 2 # we multiply by 2 to account for binary operations
        print(N_5_3s)

        return N_5_3s

    N_5_3s = n_5_3s(N_5_2s, N_5_1s)

    def n_5_4s(N_5_3s, N_5_2s, N_5_1s):
        # Lets Look at possible combinations of 4 systems drawing from set of 5 systems
        # We can get to 4 systems with 3 systems + 1 system 
        # We can also get to 4 systems with 2 systems + 2 systems
        X=4
        #(1) Let's start with 3 systems + 1 system
        # Lets start by taking the cartesian product of N_5_3s and N_5_1s and multiplying by (2/5)
        N_5_4_31s = N_5_3s * N_5_1s * (2/5) # we multiply by (2/5) in this case b/c we have two choices left from the 5
        N_5_4_31s = N_5_4_31s * 2 # we multiply by 2 because we have 2 options for operator between the 3 systems and 1 system

        print(N_5_4_31s)

        #(2) Now lets look at 2 systems + 2 sytems to make 4. THis will be a bit more challenging. We have to draw from our
        # N_5_2s
        N_5_4_22s = N_5_2s * N_5_2s * (3/10) * (1/2) * 2 # We first find the cartesian product of combos of 2 with combos of 2
        # Then we multiply by (1/2) because we don't care about order
        # Then we multiply by (3/10), which is derived from C(3,2)/C(5,2)
        # Then we multiply by 2 to account for choice of binary operators
        N_5_4s = N_5_4_22s + N_5_4_31s
        print(N_5_4_22s)

        print(N_5_4s)  

        return N_5_4s

    N_5_4s = n_5_4s(N_5_3s, N_5_2s, N_5_1s)

    def n_5_5_total(N_5_4s, N_5_3s, N_5_2s, N_5_1s):
        # Lets Look at possible combinations of 5 systems drawing from set of 5 systems
        # We can get to 5 systems with 4 systems + 1 system 
        # We can also get to 5 systems with 3 systems + 2 systems

        #(1) Let's start with 4 systems + 1 system
        # Lets start by taking the cartesian product of N_5_4s and N_5_1s 
            # RECALL THAT N_5_4s includes combinations of 2s and 2s, as well as combinations of 3s and 1s. We need to 
            #take both into account. 
        N_5_5_41s = N_5_4s * N_5_1s * (1/5) * 2 # we multiply by 1/5 because we only have 1 element left in the N_5_1s 
        # that has not already been used in the N_5_4s. We multiply by the 2 to account for binary operation. 
        print(N_5_5_41s)
        # (2) Now lets move to 2 systems + 3 systems
        N_5_5_32s = N_5_3s * N_5_2s * 1/10 * 2 # we multiply by 1/10 because there is only one possible combination in the 2s 
        print(N_5_5_32s)
        # that will fit with the 3s (ie, C(2,2)/C(5,2)= (1/10)); we multiply by 2 to account for binary operators
        N_5_5s = N_5_5_32s+N_5_5_41s
        print(N_5_5s)

        return N_5_1s + N_5_2s + N_5_3s + N_5_4s + N_5_5s

    print('Total:', n_5_5_total(N_5_4s, N_5_3s, N_5_2s, N_5_1s))