Lecture 5
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The “WHY” slide
Logistics
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HW1 due today
HW2 available today, due Wed 1/21
Office Hours this week:
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µ Me: Friday 10:00-11:00 CSE 668
µ TA (Josh): Friday 3:30 CSE 002/3
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Last lecture
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1
NAND/NOR more common/efficient
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It is always good to remember logical/theoretical concepts
visually. This is one way to remember the NAND/NOR
conversion and De Morgan’s laws easily.
Logic Simplification
Converting to use NAND and NOR
Minimizing functions using Boolean cubes
CSE370, Lecture 5
NAND and NOR are more efficient gates than AND or OR (and
therefore more common). Your computer is built almost exclusively
on NAND and NOR gates. It is good to knowhow to convert any
logic circuits to a NAND/NOR circuit.
Pushing bubbles
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Logic gates and truth tables
Implementing logic functions
Canonical forms
Today’s lecture
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Converting to use NAND and NOR
If you are building a computer or a cool gadget, you want to
optimize on size and efficiency. Having extra unnecessary
operations/gates is not great. We teach nice techniques to allow
logic simplifications.
CSE370, Lecture 5
2
CMOS NAND and NOR Gates
CMOS logic gates are more common and efficient
in the inverted forms
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NAND, NOR, NOT
Even though Canonical forms discussed so far used
AND/OR, NAND/NOR preferred for real hardware
implementation
X
Y
Z
X
Y
Z
X
0
0
1
1
Y
0
1
0
1
NOR
Z
1
1
1
0
NAND
CSE370, Lecture 5
3
NAND and NOR (truth table)
CSE370, Lecture 5
NAND and NOR (logic gates)
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(X + Y)' = X' • Y'
NOR is equivalent to AND
with inputs complemented
X
0
0
1
1
Y
0
1
0
1
X'
1
1
0
0
Y'
1
0
1
0
de Morgan's
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(X + Y)' X' • Y'
1
1
0
0
0
0
0
0
■
■
■
■
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(X • Y)' = X' + Y'
NAND is equivalent to OR
with inputs complemented
CSE370, Lecture 5
X
0
0
1
1
Y
0
1
0
1
X'
1
1
0
0
4
Y'
1
0
1
0
(X • Y)' X' + Y'
1
1
1
1
1
1
0
0
Standard form:
Inverted:
AND with complemented inputs ≡ NOR
OR with complemented inputs ≡ NAND
OR ≡ NAND with complemented inputs
AND ≡ NOR with complemented inputs
NOR
OR
5
A'B' = (A + B)'
A + B = (A'B')'
CSE370, Lecture 5
NOR
OR
A' + B' = (AB)'
(AB) = (A' + B')'
pushing
the
bubble
NAND
NAND
AND
AND
6
Converting to use NAND/NOR
Converting to use NAND/NOR (con’t)
Introduce inversions ("bubbles")
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Z = AB+CD
µ Conserve inversions
µ Do not alter logic function
Z = AB + CD
Example
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Example: AND/OR network to NOR/NOR
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Introduce bubbles in pairs
AND/OR to NAND/NAND
= (A'+B')'+(C'+D')’
= (A'+B')'+(C'+D')'
= [(A'+B')'+(C'+D')’]’’
= [(A'+B')(C'+D')]'
= {[(A'+B')'+(C'+D')']'}'
= [(AB)'(CD)']'
A
A
B
NAND
B
B
Z
C
C
D
D
NAND
Z
C
Z
Z
NOR
C’
D
CSE370, Lecture 5
B’
NOR
NAND
7
Converting to use NAND/NOR (con’t)
D’
CSE370, Lecture 5
8
Converting to use NAND/NOR (con’t)
Example: AND/OR network to NOR/NOR
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A’
NOR
A
Example: OR/AND to NAND/NAND
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Z = AB+CD
= (A'+B')'+(C'+D')’
= [(A'+B')'+(C'+D')’]’’
A
= {[(A'+B')'+(C'+D')']'}'
B
A
A
B
B
NAND
Z
C
D
D
NOR
Converting to use NAND/NOR (con’t)
B
C
D
D
Example: OR/AND to NOR/NOR
A
NOR
NAND
Z
B
Z
B
C
C
D
D
NOR
Z
NOR
NAND
CSE370, Lecture 5
10
A
NAND
C
D’
CSE370, Lecture 5
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A
Z
D
Z
NAND
Converting to use NAND/NOR(con’t)
Example: OR/AND to NAND/NAND
B
C’
Z
9
A
C
NOR
CSE370, Lecture 5
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B’
Z
NAND
NOR
C
A’
11
CSE370, Lecture 5
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Example of bubble pushing: before pushing
Example of bubble pushing: NAND/NAND
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CSE370, Lecture 5
13
Example of bubble pushing: NOR/NOR
CSE370, Lecture 5
Goal: Minimize two-level logic expression
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Algebraic simplification
■ not an systematic procedure
■ hard to know when we reached the minimum
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Just program it!! Computer-aided design tools
■ require very long computation times (NP hard)
■ heuristic methods employed – "educated guesses”
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Visualization methods are useful
■ our brain is good at figuring simple things out
■ many real-world problems are solvable by hand
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CSE370, Lecture 5
15
Key tool: The Uniting Theorem
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The uniting theorem → A(B’+B) = A
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The approach:
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CSE370, Lecture 5
16
Boolean cubes
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Visualization tool for the uniting theorem
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Find some variables don’t change (the A’s above) and others
do (the B’s above)
Eliminate the changing variables (the B’s)
n input variables = n-dimensional "cube"
01
0
B
0
1
0
1
F
1
1
0
0
2-cube
Y
X
00
A
0
0
1
1
11
1
1-cube
10
X
1011
A has the same value in both “on-set” rows
⇒ keep A
111
Y Z
101
0011
1111
0111
4-cube
0010
3-cube
B has a different value in the two rows
⇒ eliminate B
011
000
X
100
Y Z
0000
W
X
1000
1100
0100
F = A'B'+A'B = A'(B+B') = A'
CSE370, Lecture 5
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CSE370, Lecture 5
18
Mapping truth tables onto Boolean cubes
Example using Boolean cube
◆ Binary full-adder carry-out logic
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ON set = solid nodes
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OFF set = empty nodes
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Look for on-set adjacent
to each other
A
0
0
1
1
B
0
1
0
1
F
1
0
1
0
F
Sub-cube (a line) comprises
two nodes
11
01
B
10
00
A
0
0
0
0
1
1
1
1
A varies within the sub-cube;
B does not
A
On-set is covered by the OR of three 1-subcubes
B
0
0
1
1
0
0
1
1
Cin
0
1
0
1
0
1
0
1
Cout
0
0
0
1
0
1
1
1
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Another example using Boolean cube
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Cin
0
1
0
1
0
1
0
1
D
0
0
0
1
01
1
1
1
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A
A
A
A
A
AB(Cin'+Cin)
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0-cube
1-cube
2-cube
3-cube
(a
(a
(a
(a
single node) yields a term in 3 literals
line of two nodes) yields a term in 2 literals
plane of four nodes) yields a term in 1 literal
cube of eight nodes) yields a constant term "1"
(A'+A)BCin
111
011
B
000
Cin
101
A(B+B')(Cin+Cin‘)
111
110
010
000
F(A,B,C) = ∑m(4,5,6,7)
On-set forms a square (a 2-D cube)
B C
A
101
A
100
A is asserted (true) and unchanging
B and C vary
This sub-cube represents the literal A
D = BCin+A
CSE370, Lecture 5
101
In a 3-cube (three variables):
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On-set is covered by the OR of one 2-D subcubes and
one 3-D subcubes
B
0
0
1
1
0
0
1
1
000
Cin
CSE370, Lecture 5
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A
0
0
0
0
1
1
1
1
A(B+B')Cin
B
M-dimensional cubes in n-dimensional space
Changed one bit from the previous function
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111
Cout = BCin+AB+ACin
This sub-cube represents B'
CSE370, Lecture 5
(A'+A)BCin
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CSE370, Lecture 5
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