1. CS1022 Computer Programming & Principles
Lecture 2
Boolean Algebra

2. Plan of lecture Simplifying Boolean expressions Karnaugh maps
Logic circuits
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3. Simplifying Boolean expressions (1)
Given a Boolean expression, can we find a “simpler” equivalent expression?
“Simpler” = “fewer symbols”
Technique applied to disjunctive normal form (DNF)
Even though it might be more complex than original
Process: original expr.  DNF  simplified expr.
Technique restricted to at most 3 Boolean variables
Extension to more variables is possible
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4. Simplifying Boolean expressions (2)
To simplify, let’s suppress “”
Just like “” in algebra can be dropped (e.g., 2y = 2  y)
For instance, the expression in disjunctive normal form
(p q r)  (p q  r)  (p  q r)
Will be represented as
pqr  pqr  pqr
We can simplify the above expression as follows:
pqr  pqr  pqr 
(prq  prq)  pqr
Comm.& assoc. Laws
pr(q q)  pqr
Distr. Law
pr  pqr
Since q q = 1
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5. Simplifying Boolean expressions (3)
Compare pr  pqr and pqr  pqr  pqr
They are equivalent – all changes done via a “law”
It is much simpler than original expression
Very important: simplification can be automated
Each step in the process is an operation
Sequence of steps terminates at some point
Notice:
First step puts together two minterms which differed by one symbol
Second step reduced two terms to one
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6. Karnaugh maps (1) Simplification done via a Karnaugh map
“Device” invented in the 1950s to help circuit design
Uses a visual display to indicate which pairs of minterms can be merged as a simpler expression
For Boolean expressions with 3 variables (p, q and r)
Karnaugh map is a table with 2 rows and 4 columns
Columns labelled with all possible disjunctions of p and q and their negations
Rows labelled with r and r pq
pq
pq
pq r r
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7. Karnaugh maps (2) pq pq pq pq r 1 r
As we move from column to column precisely one change occurs in the column label
Cells correspond to 8 different minterms stemming from 3 Boolean variables
For a given Boolean expression, we write “1” in each cell whose row/column expression appears
For instance, pqr  pqr  pqr has map pq
pq
pq
pq r 1 r
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8. Karnaugh maps (3) pq pq pq pq r 1 r
Required simplification suggested by “clusters” of 1s
Shaded in green here
There is only one such cluster in the example
Corresponds to terms we combined using the laws pq
pq
pq
pq r 1 r
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9. Karnaugh maps (3) Relabelling columns preserves adjacency
Some clusters may be “hidden”
NB relabelling must be consistent with requirement that adjacent columns differ by one symbol only
Example: Karnaugh map of pqr  pqr  pqr
Relabelling columns produces an alternative Karnaugh map and reveals a pair of adjacent minterms pq
pq
pq
pq r 1 r
pq
pq
pq pq r 1 r
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10. Karnaugh maps (4) pq pq pq pq r 1 r Hence, pqr  pqr  pqr 
pqr  pr(q q)
pqr  pr
pq
pq
pq pq r 1 r
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11. Karnaugh maps (5) Let’s simplify pqr  pqr  pqr  pqr  pqr
It contains a cluster of 4 (A) and a pair (B)
There are no hidden pairs
We must try relabelling exhaustively pq
pq
pq
pq r 1 r
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12. Karnaugh maps (6) Cluster (A) corresponds to pq pq pq pq r 1 r
pqr  pqr  pqr  pqr 
(p  p)qr  (p  p)qr
qr  qr
q(r  r) q
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13. Karnaugh maps (7)
Cluster (B) corresponds to So, simplified expression is q  pr pq
pq
pq
pq r 1 r
pqr  pqr 
pr(q  q)
pr
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14. Logic circuits (1)
One of the main application of Boolean algebra is the design of binary devices
They accept a (finite) number of inputs
They produce a (finite) number of outputs
Binary: only two possible values for each input & output
Circuits as “black boxes”: ? p1 q1 p2 q2 p3 q3
...
... pn qm
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15. Logic circuits (2)
Devices are built of logic gates which perform basic Boolean operations (, , and ) a b
a  b
OR gate a b
a  b
AND gate a a
NOT gate a b
(a  b)
NAND gate
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16. Logic circuits (3) Interconnecting gates produces a logical circuit
A logical circuit evaluates a Boolean expression
Example: What is the final output of this circuit? 4 3 2 1 p q r 5 6 7
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17. Logic circuits (4) Example: What is the final output of this circuit?pq
pq
pqr
pqr
pqr
pqr  pqr 4 3 2 1 p q r 5 6 7 4 3 2 1 p q r 5 6 7 4 3 2 1 p q r 5 6 7 4 3 2 1 p q r 5 6 7
pqr  pqr  pqr
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18. Logic circuits (5) Solution: pqr  pqr  pqr Gate Inputs Output 12
p, q
pq 3
pq, r
pqr 4
pq, r
pqr 5
pq, r
pqr 6
pqr, pqr
pqr  pqr 7
pqr  pqr, pqr
pqr  pqr  pqr
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19. Logic circuits (6)
Logic circuits can be simplified if we allow AND and OR gates to have more than 2 inputs
More dramatic simplifications w/ Karnaugh maps
2 clusters (one hidden)
We can simplify things:
pqr  pqr  pq(r  r)  pq
pqr  pqr  (q  q)pr  pr
That is, pqr  pqr  pqr  pq  pr
We can further simplify this as p(q  r) pq
pq
pq
pq r 1 r
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20. Logic circuits (7)
With p(q  r), we simplify previous circuit to obtain p q r
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21. Further reading
R. Haggarty. “Discrete Mathematics for Computing”. Pearson Education Ltd (Chapter 9)
Wikipedia’s entry on Boolean algebra
Wikibooks entry on Boolean algebra
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