A8.2 Circuits, expressions and truth tables

Write sum_of_products(outputs). Its input outputs is a list of eight integers, each 0 or 1: the Q column of a truth table for inputs A, B and C, with the rows in binary order, so outputs[0] is the row A=0 B=0 C=0 and outputs[7] is A=1 B=1 C=1. It returns a string: Q = followed by the minterms joined with OR . Each minterm is in brackets, lists A, B and C in that order joined with AND , and writes an input that is 0 in that row as NOT and a space before the letter. If no row is 1, it returns Q = 0. For example, [0, 0, 1, 0, 0, 0, 0, 1] gives Q = (NOT A AND B AND NOT C) OR (A AND B AND C). Print the result for these three tables, one per line, in this order: [0, 1, 1, 1, 1, 1, 1, 1], then [0, 0, 1, 0, 0, 0, 1, 1], then [0, 0, 0, 0, 0, 0, 0, 0].

# the two lines every program starts with: the commands, then the robot
from bugbot import *
connect()

def sum_of_products(outputs):
    return "Q = "

print(sum_of_products([0, 1, 1, 1, 1, 1, 1, 1]))
print(sum_of_products([0, 0, 1, 0, 0, 0, 1, 1]))
print(sum_of_products([0, 0, 0, 0, 0, 0, 0, 0]))

The hint students can ask for: Row number i, written as three binary digits, is the values of A, B and C on that row. For each row whose output is 1, turn each digit into a letter or NOT and a letter, join them with AND inside brackets, then join the terms with OR. Think about what to print when no row is 1.

A solution

# the two lines every program starts with: the commands, then the robot
from bugbot import *
connect()

def sum_of_products(outputs):
    names = ["A", "B", "C"]
    terms = []
    for i in range(8):
        if outputs[i] == 1:
            bits = format(i, "03b")
            literals = []
            for k in range(3):
                if bits[k] == "1":
                    literals.append(names[k])
                else:
                    literals.append("NOT " + names[k])
            terms.append("(" + " AND ".join(literals) + ")")
    if len(terms) == 0:
        return "Q = 0"
    return "Q = " + " OR ".join(terms)

stop_rule = [0, 1, 1, 1, 1, 1, 1, 1]
led_rule = [0, 0, 1, 0, 0, 0, 1, 1]
never = [0, 0, 0, 0, 0, 0, 0, 0]
print(sum_of_products(stop_rule))
print(sum_of_products(led_rule))
print(sum_of_products(never))

Any program that meets the task's checks is marked correct in the simulator; this is one way, not the only way.