CS 260-01 Chapter 1 Notes

Chapter 1 Discrete Structures Notes

Additional hints on solving logic problems. Truth tables may be used on their own or to show that statements are logically equivalent. They are also used to show tautologies, contradictions and contingencies.

If you are asked to produce a truth table, you need only show what the truth value of the statement is given for each of the combinations of values for the variables. (See Table 2 in section 1.2)
To show that two statements are logically equivalent, you need to show the columns giving their truth values agree. (See Table 4 in section 1.2. Column 5 agrees with column 8) One way to do this is to do a truth table for the left hand side and then do one for the right hand side. If they agree, the statements are logically equivalent. (Logically equivalent statements are those with <=>.) The double allow indicated that the statements are logically equivalent. In addition to doing a truth table, you need to state that since the statement on the left hand side produces the same truth values as the statement of the right hand side, the two statements are logically equivalent.

You can also show that statements are logically equivalent using the rules below to show that the right and left sides reduce to the same thing. (See example 5 in section 2.)

To show that an implication is a

tautology, you need to show that the the truth value for every combination of input values is TRUE. (p^q)->p is a tautology since the truth value for the expression is true for all combinations of input values. In addition to the truth table, you need to state since the value of the statement is true for all combinations of values for the input variables, the statement is a tautology.

p	q	p^q	(p^q)->p
T	T	T	T
T	F	F	T
F	T	F	T
F	F	F	T

contradiction, you need to show that the truth value for every combination of input values is FALSE.
contingency, you need to show that the truth value for the combination of input values has at least on TRUE and one FALSE

Variables are letters such as p,q and r. Values are TRUE and FALSE.

The Relationship of Logic, Sets and Boolean Algebra. As you can see from the table below, there is a correspondence among the symbols of logic, sets and Boolean Algebra. All the laws that apply to one, generally apply to the other two. For Logic and Boolean Algebra, there is a one to one relationship. Sets are a bit more complicated since they can take more than two values; however, they generally have similar laws. Click here for truth tables.

Logic	Sets	Boolean	Comments
^	Intersection	*	AND
V	Union	+	OR
-	Ä	`	NOT
T	U	1	TRUE
F	Ø	0

	Basic Laws
Name	Logical Equivalencies	Boolean Identities	Set Identities
Identity laws	p ^ T <-> p p V F <-> p	x*1=1 x+0=x	A U Ø = A A ^ U = A
Domination	p V T <-> T p ^ F <-> F	x+1=1 x*0=0	A U U = U A ^ Ø = Ø
Idempotent	p V P <-> p p ^ p <-> p	x+x = x x*x = x	A U A = A A ^ A = A
Double negation	~p(~p) <-> p	~(~x) = x	~(~A)=A
Commutative	p V q <-> q V p p ^ q <-> q ^ p	x+y = y+x xy = yx	A U B = B U A A ^ B = B ^ A
Associative	(pVq)Vr<->pV(qVr) (p^q)^r<->p^(q^r)	(x+y)+z = x+(y+z) (xy)z = x(Yz)	(AUB)UC=AU(BUC) (A^B)^C=A^(B^C)
Distributive	pV(q^r)<->(pVq)^(pVr) p^(qVr)<->(p^q)V(p^r)	x+yz = (x+y)(x+z) x(y+z) = xy+xz	A U (B^C) = (A^B) U (A^C) A ^ (BUC) = (A U B) ^ (A U C)
DeMorgan	~(p^q)<->~pV~q ~(pVq)<->~p^~q	~(xy) = ~x+~y ~(x+y) = ~(xy)	~(A U B) = ~A ^ ~B ~(A ^ B) = ~A U ~ B

What is Boolean Algebra?

Boolean Algebra is an algebra that consists of a set of two values, {0,1} along with the set of operators {`,V,^}. The operators have the following outcomes.

+	0	1	*	1	`
0	0	1	0	0	0	1
1	1	0	1	1	1	0

Boolean Algebra has many uses.Some of these are found in
- Designing switches
- Doing math on a microprocessor
- Representing sets.
- Finite state machines
- Encrypting algorithms
Boolean Algebra can form functions which can be easily evaluated. See table above.
- given: f(x,y)=x+y+x*y, f(0,1)=0+1+0*1=0+1+0=1

Symmetric Difference - (note there should be a circle around the + sign) A + B is the set that contains the elements that are in either A or B but not both.
- A + B = (AUB) - (A^B) elements that are in either A or B or both, minus the elements that are in both A and B.
- A + B = (A-B) U (B-A) elements that are in A but not B, or elements that are in B but not A
- A + B = B + A commutative
- A + (B+C) = (A + B) +C associative
- A+ A = Ø inverse
- A + Ø = A identity
- A under the operator + forms a group. You can also show that union and intersection form a group.
Difference Operator (-) A - B contains those elements that are in A but not B.
- A - B is generally not equal to B-A. (What if A=B?)
- (A - B) - C is generally not equal to A - (B - C)
- A - B = A - the compliment of B
Example of the use of sets, logic and functions using SAS.

Functions:

Summmations and Sequences

The sum of i where i = 1 to n (the sum of the first n numbers)

write out	1 2 3 4 5 6 7
reverse	7 6 5 4 3 2 1
add	8 8 8 8 8 8 8 (7 8's)
this suggests the formula	(7*8)/2 = the SUM
or, more generally	[n(n+1)]/2

Computer Versions of Sequences and Summations
Problem	Program	Results
The sequence from 1 to 100 of {b_n} where b_n = (-1)ⁿ (Ex. 2)	DO I = 1 TO 100 PRINT(B(I)**N) END	{-1,1,-1,1...}
Summation of 1/j where j= 1 to 100 (Ex 8)	SUM=0 DO J=1 TO 100 SUM=SUM + 1 / (A(J) END	SUM=1+1/2,+1/3+1/4+1/5.....+1/100
Summation of i from 1 to 4 of the summation of j from 1 to 3 of ij (ex. 13) [Double Summation]*	SUM=0 DO I=1 TO 4 DO J=1 TO 3 SUM + SUM + I * J END END	SUM=11 + 12 + 13 + 21 + 22 + 23... = 60

Complexity of Algorithms:

The big-O notation - used in estimating the relative complexity of various algorithms.

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