Archived Notes on Grading: HW14

When you give the sum-of-products (or product-of-sums) expansion of a boolean function, you should not simplify it [since algebraic manipulation ruins the (very elegant) syntactic structure of the formula, turning a correct answer into an incorrect answer].

[Please recall that the sum-of-products expansion is defined syntactically, not semantically; i.e., it's not enough for your answer to just have the same truth table as the function F, it must also have a [very specific :)] syntactic structure. Example: for part c, !x!y is not the sum-of-products expansion, despite the fact that it is semantically equivalent to the correct answer !x!y!z + !x!yz.]

[Why is z necessary in part c? Since F is of degree 3--i.e., since it has three inputs, x, y, and z--Rosen's definition mandates that every minterm in its sum-of-products expansion must have exactly 3 literals--one x literal, one y literal, and one z literal. Every maxterm in its product-of-sums expansion must likewise have exactly 3 literals. These syntactic restrictions ensure that every minterm in the expansion is true for one (and only one) setting of the three inputs and, likewise, that every maxterm is false for one (and only one) setting.]

One common error was building the product-of-sums expression from the maxterms that correspond to the 1's in the truth table, instead of using the maxterms that correspond to the 0's. In particular, for part c, (x + y + z) (x + y + !z) is incorrect [consider the case x = y = z = 1]; (!x + !y + !z) (!x + !y + z) is also incorrect [consider the case x = z = 1 and y = 0].

[Reminder: in general, there is a one-to-one correspondence between the 1's in (the output column of) a boolean function's truth table and the minterms in its sum-or-products expansion. Likewise, there is a one-to-one correspondence between the 0's in the function's truth table and the maxterms in its product-of-sums expansion. Example: for part d, the truth table is...

    x y z  F
    - - -  -
    0 0 0  1
    0 0 1  1
    0 1 0  1
    0 1 1  1
    1 0 0  1
    1 0 1  1
    1 1 0  1
    1 1 1  0

In this case, there are seven 1's in the output column (the column labeled F), so there must be exactly seven minterms in the sum-of-products expansion. Likewise, there's one 0 in the output column, so there must be exactly one maxterm in the product-of-sums expansion.]

Also, please note that, although Rosen only asked for the sum-of-products expansion, on Handout #37 we asked for both the sum-of-products expansion and the product-of-sums expansion. In other words, you needed to construct a total of 4 expansions...

• the sum-of-products expansion of part c's F
• the product-of-sums expansion of part c's F
• the sum-of-products expansion of part d's F
• the product-of-sums expansion of part d's F

...in order to get full credit for 37C.

[If you see "sum-of-products?" on your paper, that means that you were missing one (or more) of the sum-of-products expansions. Likewise, if you see "product-of-sums?" on your paper, that means that you were missing one (or more) of the product-of-sums expansions.]