Boolean Algebra
CMU Spring’14 18-760 VLSI CAD
[100 pts] Homework 1
Out Wed, Jan 22; Due Mon, Feb 3 (by noon in HH1112)
1.
Properties of Boolean Difference [15 pts]
(i) Use Boolean algebra and the basic properties of Shannon cofactors from the notes to show that this identity is true. Again, f and g are functions of x1,x2,...xn, and x refers to some arbitrary variable in x1,x2,...xn.
∂ ( f + g)
∂g
∂f # ∂f ∂g &
= f • ⊕ g• ⊕% • (
∂x
∂x
∂x $ ∂x ∂x '
Hints: (a) Notice that there are no “x” variables in the functions on the left hand side of this equation (since we cofactored them out), yet there are “x” variables on the right side, since there is both an f function and a g function there. The only way this can be true is if the equation works when x=0 and it also works when x=1. So, the first thing to do is to set x to a constant on the right hand side, and then simplify. So--you have to show this equation works for both x=0 and x=1.
(b) If you first blast all of the EXORs down into their sum of products (SOP) form, this is not necessarily the easiest way to do the derivation. Use what you know about cofactors, i.e., the
“cofactor of an EXOR is the EXOR of the cofactors” for all simplifications, and so on.
(c) It’s helpful to recall that AND distributes over EXOR, i.e. a • (b ⊕ c) = ab ⊕ ac
(d) It’s also helpful to recall that a ⊕ a = 0 and a ⊕ a = 1 for any Boolean expression “a”.
(e) It’s probably easiest to simplify the right hand side until it looks like the left, and not the other way around.
(ii) Use ordinary Bolean algebra to show that this identity is true: if function f does not depend on variable x, then:
∂( f + g)
∂g
= f•
∂x
∂x
(Note: think carefully about what it means that f does not depend on x. This means something very specific in terms of the Shannon decomposition.)
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2.
Shannon Expansion(s) [10 pts]
So far, we know that the Shannon expansion identity says: F = x Fx + x’ Fx’ Are there any other ways of
[100 pts] Homework 1
Out Wed, Jan 22; Due Mon, Feb 3 (by noon in HH1112)
1.
Properties of Boolean Difference [15 pts]
(i) Use Boolean algebra and the basic properties of Shannon cofactors from the notes to show that this identity is true. Again, f and g are functions of x1,x2,...xn, and x refers to some arbitrary variable in x1,x2,...xn.
∂ ( f + g)
∂g
∂f # ∂f ∂g &
= f • ⊕ g• ⊕% • (
∂x
∂x
∂x $ ∂x ∂x '
Hints: (a) Notice that there are no “x” variables in the functions on the left hand side of this equation (since we cofactored them out), yet there are “x” variables on the right side, since there is both an f function and a g function there. The only way this can be true is if the equation works when x=0 and it also works when x=1. So, the first thing to do is to set x to a constant on the right hand side, and then simplify. So--you have to show this equation works for both x=0 and x=1.
(b) If you first blast all of the EXORs down into their sum of products (SOP) form, this is not necessarily the easiest way to do the derivation. Use what you know about cofactors, i.e., the
“cofactor of an EXOR is the EXOR of the cofactors” for all simplifications, and so on.
(c) It’s helpful to recall that AND distributes over EXOR, i.e. a • (b ⊕ c) = ab ⊕ ac
(d) It’s also helpful to recall that a ⊕ a = 0 and a ⊕ a = 1 for any Boolean expression “a”.
(e) It’s probably easiest to simplify the right hand side until it looks like the left, and not the other way around.
(ii) Use ordinary Bolean algebra to show that this identity is true: if function f does not depend on variable x, then:
∂( f + g)
∂g
= f•
∂x
∂x
(Note: think carefully about what it means that f does not depend on x. This means something very specific in terms of the Shannon decomposition.)
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2.
Shannon Expansion(s) [10 pts]
So far, we know that the Shannon expansion identity says: F = x Fx + x’ Fx’ Are there any other ways of