Transcript Discrete Mathematics Final Review Zeph Grunschlag Review Agenda List of sections covered Review major concepts with formulae Review.

Discrete Mathematics
Final Review
Zeph Grunschlag
Review
1
Agenda
List of sections covered
Review major concepts with formulae
Review
2
Sections Covered
1.1 Logic
1.2 Tautologies/Equivalence
1.3 Predicates/Quantifiers
1.4 Sets
1.5 Set Theory Operations
1.6 Functions
1.7 Sequences/Sums
1.8 Big-O, Big-, Big-
2.1 Algorithms
2.2 Complexity
2.3 Basic Number Theory
2.4 Number Theory Algorithms
2.5 RSA
(2.6 Matrices)
Review
3.1 Proofs
3.2 Induction
3.3 Recursive Definitions
3.4 Recursion
(3.5 Program Correctness)
4.1 Counting Basics
4.2 Pigeonhole Principle
4.3 Permutations and Combinations
4.4 Discrete Probability
4.5 Probability Theory
4.6 Generalizations
5.1 Recurrence Relations
5.2 Solving Linear Recurrences
5.5 Inclusion-Exclusion
5.6 More Inclusion-Exclusion
3
Sections Covered
6.1 Relation Basics
6.2 n-ary Relation
6.3 Representing Relations
7.1 Graph Basics
7.2 Graph Terminology
7.3 Graph Representations
7.4 Connectedness
7.5 Euler and Hamilton
7.7 Planar Graphs
7.8 Coloring
Review
Sections Skipped
2.6
3.5
4.7
5.3, 5.4
6.4, 6.5, 6.6
7.6
4
Logic
A proposition is a statement that is true
or false. Atomic propositions p,q,r…
are combined to form compound
propositions using the following
logical connectives :
Review
5
Logical Connectives
Operator
Negation
Conjunction
Disjunction
Exclusive or
Conditional
Biconditional
Review
Symbol Usage






Java
not
!
and
&&
or
||
xor
(p||q)&&(!p||!q)
if,then
p?q:true
iff
(p&&q)||(!p&&!q)
6
Truth Tables
Logical operators/connectives are defined by
truth tables:
p
p
Negation truth table (unary):
F
T
T
F
Binary truth tables:
p
q
p q
T
T
F
F
T
F
T
F
T
F
F
F
pq p q p q p  q
T
T
T
F
F
T
T
F
T
F
T
T
T
F
F
T
Bit Strings
Can define logical operators on bit
strings. This is done bit by bit. No
carry-over is taken.
EG:
01 1011 0010 1001

Review
10 0010 1111 1001
11 1001 1101 0000
8
Tautologies
Logical Equivalence
A compound proposition is a tautology if
it is always true, regardless of the its
atomic propositions (e.g. p  ¬p ). If it’s
always false, it’s a contradiction (e.g.
p¬p ). If neither, it’s a contingency
(e.g. p ¬p ).
Two compound propositions p, q are
logically equivalent if p  q is a
tautology. Notation: p q .
Review
9
Contrapositive vs. Converse
Given an implication p q
the converse is q p
while the contrapositive is ¬q ¬p
Review
10
Logical Proofs
There are two basic techniques for proving
tautologies and logical equivalences:
1) Build a truth table. Verify that…
last column is all TRUE for tautology
relevant columns equal for equivalence
2) Using tables on next pages, derive…
TRUE starting from supposed tautology
1st proposition from 2nd
Review
11
Tables of Logical Equivalences
 Identity laws
Like adding 0
 Domination laws
Like multiplying by 0
 Idempotent laws
Delete redundancies
 Double negation
“I don’t like you, not”
 Commutativity
Like “x+y = y+x”
 Associativity
Like “(x+y)+z = y+(x+z)”
 Distributivity
Like “(x+y)z = xz+yz”
Review
 De Morgan
12
Tables of Logical Equivalences
 Excluded middle
 Negating creates opposite
 Definition of implication in
terms of Not and Or
Review
13
Quantifiers
Existential Quantifier
“” reads “there exists”
Universal Quantifier
“” reads “for all”
Order matters:
y x R (x,y ) and x y R (x,y ) may
not be logically equivalent.
Review
14
DeMorgan Identities
“Not all true iff one is false.”


Conjunctional version:
(p1p2…pn)  (p1p2…pn)
Universal quantifier version:
 x P(x )  x P(x )
“Not one is true iff all are false.”


Review
Disjunctional version:
(p1p2…pn)  (p1p2…pn)
Existential quantifier version:
 x P(x )  x P(x )
15
Logical English
Logical Puzzles
Precise English statements can be expressed in
terms of logical constructs. In cases of English
puzzles, this can be useful for solving.
EG: Can there be a man that shaves exactly all
the people that don’t shave themselves.
Let S(x,y) = “x shaves y”. Asking if following
statement satisfiable:
 x  y S(y,y )  S(x,y)
Not satisfiable (trying plugging in y = x )1
Review
16
Standard Numerical Sets
Natural numbers: N = { 0, 1, 2, 3, 4, … }
Integers: Z = { … -3, -2, -1, 0, 1, 2, 3, … }
Positive integers: Z+ = {1, 2, 3, 4, 5, … }
Real numbers: R
Rational numbers: Q (decimal numbers with
repeating decimal expansion)
Review
17
Set Notation
element : 3  S
not an element : 3  S
subset : S  T
strict subset : S T
empty (or null) set : {} or 
cardinality : |{1, -13, 4, -13, 1}| = 3
Review
18
Set Theoretic Operations
power set : P (S ) or 2s

2nd version reminds us that |2s| = 2|s|
Cartesian product :


A B = { (a, b) | a  A and b  B }
|A1A2…An| = |A1||A2| … |An|
complement :
A={x|xA}
union :
A B = { x | x  A  x  B }
intersection : A B = { x | x  A  x  B }
difference :
A-B={x|xA  xB}
symmetric difference:AB={ x | xA  xB }
Review
19
Venn Diagrams
Complement
A
Intersection
U
U
A
A
Union
Review
B
Disjoint Union
U
AB
A
AB
B
U

A
A B
B
20
Venn Diagrams
Difference
Symmetric Difference
AB
U
A-B
B
A
U
B
A
Review
21
Sets as Bit-Strings
If elements ordered, each set can be viewed as
bit-string s. Operators act on bit-strings:
complement A  1’s complement -s
union A B  disjunction s  t
intersection A B  conjunction s  t
difference A - B  s  (-t )
symmetric difference AB  s  t
EG: If U = {1,2,3,4}, A = {2}=0100,
B = {1,2,4}=1101 then:
AB = 0100  1101 = 1001 = {1,4}
Review
22
Functions
One-to-One


NO:


NO:
YES:
NO:
 reverse:


NO:
 Reverse
YES:
Onto

Bijection (1-to-1 and onto)
YES:
 reverse:
Review
23
Function Notation
Composition : f g (a) = g ( f (a) )
n
Exponentiation :
f n (x ) = f f f f  … f (x )
Floor and Ceiling : x  , x 
f : A  B --function from A to B


domain
codomain
Range : f (A )
Review
24
Sequences and Strings
A sequence is a function N  S. A finite
sequence in S is a function from first n
numbers to S. Notation {ai } or just ai usually
used instead of f (i ). Finite sequences
expressed as n-tuples, e.g. (1,2,3,4,5).
A string is just a finite sequence where S is a set
of characters. Strings denoted by putting
characters together, e.g. 12345. Empty string
denoted by . Can concatenate strings u and
v to obtain u · v (or usually just uv ). EG: ·v
= v ·  = v. Reverse of w is denoted by w R .
Review
25
Summations
Sum of first n
numbers: n
n(n  1)
i

2
i 1
Sum of first n k ’th
powers: n
k
k 1
i


(
n
)

Sum of a sequence
n from 0 to n :
a
i 0
i
 a0  a1  a2  ...  an
Sum of all the
number in S :
x
xS
i 1
Double sum:
2
3
  ij
i 0
Review
j 1
Geometric sum:
ar n1 - a
ar  a  ar  ar  ...  ar 

r -1
i 0
n
i
2
n
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Cardinality and Countability
Countable:
1.
2.
Finite, or
0 -same cardinality as N: there is a bijection
with the natural numbers.
Lemmas: S is countable if…


…there is an onto function from another
countable set
…there is a 1-to-1 function to another countable
set
EG: Z is countable because it is enumerated by
the sequence
i  i  1
ai  -(-1) 

2


Review
27
Uncountability of R
Cantor’s Diabolical Diagonal
If R were countable could list all numbers in
(0,1) by a sequence:
r1 , r2 , r3 , r4 , r5 , r6 , r7, …
Cantor’s diabolical diagonalization creates a
number “revil” between 0 and 1 which is not
on the list, contradicting countability
assumption. Let ri,j be the j ’th decimal digit
in the fractional part of i’ th number ri .
Define digits of revil by the following rule:
The j ’th digit of revil is 5 if rj,j  5. Otherwise
the j’ ’th digit is 4.
revil is an anti-diagonal of the seqencs {rn }
Review
28
Cantor's Diagonalization
Example
 Decimal expansions of ri 
r1
r2
r3
r4
r5
r6
r7
0.
0.
0.
0.
0.
0.
0.
1
1
2
7
0
5
7
2
5
5
8
1
5
6
3
1
4
9
1
5
7
4
1
2
0
0
5
9
5
1
0
6
1
5
5
6
1
9
2
0
5
4
7
1
0
3
1
5
4
0.
5
4
5
5
5
4
5 29
:
r
evil
Review
Big-O, Big-, Big-
DEF: Let f and g be functions with domain R0 or N
and codomain R.
f (x ) = O ( g (x ) ) if there are constants C and k
such  x > k, |f (x )|  C  |g (x )|
f (x ) = (g (x )) iff g (x ) = O (f (x ))
f (x ) = (g (x )) iff f (x )=O(g(x))  f (x)=(g(x))
Evaluate lim f ( x)  a Limit exists…
x  g ( x )


…if and only if f (x ) = O ( g (x ) )
…and positive, then f (x ) = (g (x ))
Polynomials are big- of largest term
Sums are big-O of biggest term
Non-zero constants are irrelevant
30
Asymptotically Incomparable
Functions
4
4
x 10
Some functions are incomparable.
EG: |x 2 sin(x)| vs. 5x 1.5 :
3.5
y = x2
3
2.5
y = 5x 1.5
2
y = |x 2 sin(x)|
1.5
1
0.5
0
0
Review
20
40
60
80
100
120
140
160
180
200
31
Division, mod function
Divisor (or factor): If a = b·c then b | a
How many pos. no.’s ≤ N divisible by d ?
Answer:
N/d
Identities:



a|b  a|c  a|(b + c )
a|b  a|bc
a|b  b|c  a|c
Division Algorithm: Given dividend a and
divisor d there is a unique quotient q with
remainder r  [0,d-1] satisfying: a = dq + r

Review
mod function : With a,d,q,r as above, define
a mod d = r
32
Primality, gcd, lcd
Prime : A number divisible only by itself and 1
Composite : A number that isn’t prime.
Primality Test : x is composite iff it’s divisible by some
prime ≤ x
Fundamental Theorem of Arithmetic : Every number
can be decomposed uniquely into a product of prime
numbers.
Greatest common divisor (gcd): Biggest number which
divides both x and y.

THEOREM: There are integers s,t such that gcd(x,y) = sx+ty
Least common multiple (lcm): Smallest number divisibl
by both x and y

lcm(x,y) = xy / gcd(x,y)
Relatively Prime : two numbers with no common
factors. Equivalently: gcd(x,y ) = 1
Review
33
Euclidean and Extended
Euclidean Algorithm
INPUT : integers m, n
(m > 0, n ≥ 0 )
OUTPUT : gcd(m,n)
integer euclid(m,n)
x = m, y = n
while(y > 0)
r = x mod y
x=y
y=r
return x
Review
INPUT : integers m, n
(m > 0, n ≥ 0 )
OUTPUT : s, t such that
sm+tn = gcd(m,n)
(int.,int.) ext_euclid(m,n)
r = m mod n
//remainder
q = m/n
//quotient
(s’,t’ ) = ext_euclid(n,r)
return ( t’ , s’-t’q )
34
Modular Congruence
Often confused with mod function.
a  a’ (mod b) is a relation on Z, not a function!
DEF: a  a’ (mod b) ) iff b | (a – a’ ).

Equivalently: a  a’ (mod b) ) iff a mod b = a’ mod b
Identities. Allow arithmetic in “(mod b) – world” to make sense


a mod b  a (mod b)
Suppose a  a’ (mod b) and c  c’ (mod b). Then:
 a+c  (a’+c’ )(mod b)
 ac  a’c’ (mod b)
 a k  a’ k (mod b)
Modular Inverse:



If a and b are relatively prime, can invert a in “(mod b) – world”. x is
an inverse of a if ax  1 (mod b).
If x also in range [1,b -1], use notation x = a -1 (mod b)
Compute using extended Euclidean algorithm
1. Find s,t such that sa + tb = 1
2. Therefore sa = 1 – tb so mod-b we have
sa  1 (mod b)
3. a -1 (mod b) = s mod b is the inverse of a modulo b
Review
35
Number Systems
A base-b number is a string
u = ak ak-1 ak-2 … a2 a1 a0
with the ai taken from
{0,1,2,3,…,b-2,b-1}.
u represents the number
(u )b = akbk+ak-1bk-1+…+a1b + a0
When b > 10, use capital letters:
A=10, B=11, C=12, etc.
Standard bases:
Binary (base-2)
Octal (base-8)
Decimal (base-10)
Hexadecimal (base-16)
Review
Base-b representation and/or
conversion algorithm :
INPUT :
positive integer n
positive integer b // the base
string represent(n,b)
q = n, i = 0
while( q > 0 )
ui = q mod b
q = q/b
i = i +1
return ui ui-1 ui-2 … u2 u1 u0
36
Arithmetical Algorithms
Addition and Multiplication
Addition in any base b :
Binary multiplication:
string add(xk xk-1…x1x0,
bitstring multiply(xk xk-1…x1x0,
yk yk-1…y1y0)
yk yk-1…y1y0 , int b)
x = xk xk-1…x1x0
carry = 0, xk+1 = yk+1 = 0
p=0
for(i = 0 to k+1)
for(i = 0 to k+1)
digitSum=carry+xi
+yi
if(yi == 1)
p = add(p , x << i )
zi =digitSum mod b
carry =digitSum /b  return p
return zk+1zk zk-1…z1z0
Review
37
1’s and 2’s Complement
1’s Complement
Fix k bits.
Represent numbers |x | < 2k-1
Most significant bit tells the
sign


0 –positive
1 –negative
Positive numbers the same as
standard binary expansion
Negate numbers by flipping
all bits
Adding numbers:
1.
2.
Review
Add as usual.
If there was carryover, add 1
2’s Complement
Fix k bits.
Represent no.s in [-2(k-1),2(k-1))
Most significant bit tells the
sign


0 –positive
1 –negative
Positive numbers the same as
standard binary expansion
Negate numbers by flipping all
bits then adding 1
Adding numbers:
1.
2.
Add as usual.
Drop any carryover!
38
Fast Modular Exponentiation
Fermat’s Little Theorem
To exponentiate quickly mod-N :
Use repeated squaring
technique coded on the right
Simplify as much as possible
before any squaring

(-7)2 mod 23 much easier
than 392 mod 23
May find cyclical pattern
Use Fermat’s Little Theorem1
when possible:


Review
a n  a n mod p-1(mod p)
EG: 923 mod 23
= 9(23 mod 22) mod 23
= 91 mod 23 = 9
INPUT : integers m,e,N s.t e,N >0
fastExp(m,e,N)
unun-1 un-2 … u2 u1 u0 =
representInBinary(e)
sqPow0= m mod N
for( i = 0 to n-1)
sqPowi+1 = sqPowi 2 mod N
pow = 1
for(i = 0 to n)
if (ui == 1 )
pow = pow·sqPowi mod N
return pow
39
Chinese Remainder Theorem
Solve1 x  [0, m1m2m3)
x  a1 (mod m1)
x  a2 (mod m2)
x  a3 (mod m3)
1. y1=a2·a3 , y2=a1·a3,
y3=a1·a2
2. z1=y1·(y1-1·mod m1),
z2=y2·(y2-1·mod m2),
z3=y3·(y3-1·mod m3)
3. z = a1z1+a2z2+a3z3
4. x = z mod m1m2m3
Review
Theorem says solution
exists and is unique in
range [0, m1m2m3)
assuming m1 , m2 , m3
relatively prime.
Useful for carrying
arithmetic out relative a
big number N = m1m2m3
1.
2.
Carry out relative each mi
Patch solutions together by
Chinese remaindering
Proves that RSA
decryption works
40
Encryption and Decryption
Including RSA and Caesar
1. Messages are converted into blocks of numbers.
2. Each number block m is scrambled using some
modular function f (m).
Examples of f (m):

f (m) = (m+h) mod N with constants1 h,N
Decrypt  g (n) = (n - h) mod N

f (m) = (cm ) mod N with constants c,N
Decrypt  g (n) = (c-1n ) mod N

f (m)=(cm+h) mod N with constants c,h,N
Decrypt  g (n) = c-1 (n - h) mod N

f (m) = m e mod N with constants2 e,N
Decrypt  g (n) = m d mod N [d = e -1 (mod (p-1)(q -1))]
Review
41
Basic Proof Techniques
Prove statement P Q
Direct Proof
Useful Tricks
Try to use algebraic
Assume P true, show Q true
form of definition and
Indirect Proof
reduce algebraically
Assume Q false, show P false
Write rational numbers
Reductio Ad Absurdum1
(in Q) p/q with p,q
Assume P Q true,
relatively prime
show P Q true
Disrefutation of x P (x):
Just find a counterexample
Review
42
Mathematical Induction
Well Ordering Property: Non-empty subsets S of
N have a smallest element.
Simple Induction:
If the following hold:
1)
[basis] P (0) is true
2)
[induction] n P(n)P(n+1) is true
Then n P(n) is true
Strong Induction:
If the following hold:
[basis] P (0) (sometimes need more base cases)
2)
[strong induction] n [P (0)P (1) … P (n)]  P(n+1)
Then: n P(n) is true
1)
Review
43
Recursive Definitions
Factorial
Fibonacci sequence
Binomial Coefficients
“n-choose-k ”
Addition of nonnegative integers
1, if n  0
n!  
n(n - 1)! , if n  1
n, if k  0 or 1
f (n)  
 f (n - 2)  f (n - 1), if k  2
0, if k < 0 or k  n

C (n, k )  1, if k  n  0
C (n - 1, k - 1)  C (n - 1, k ), otherwise

m, if n  0

m  n  m  1, if n  1
(m  (n - 1))  1, if n  1

0, if n  0
 n -1
a



i
i 1
 ai  an , if n  0
 i 1
1, if n  0
n

ai   n -1 

i 1
  ai   an , if n  0
 i 1 
n
Summation Notation
Product Notation
Review
44
More Recursion:
Sets, Relations, Graphs
Can define sets and relations recursively as well.
Set of quantities payable with dimes and quarters:


BASE: 0  S
RECURSE: u + 10, u + 25  pal if u  pal
Set of palindromes :


BASE:
 , 0, 1  pal
RECURSE: 0u 0, 1u 1  pal if u  pal
Connectedness relation R in a simple graph G=(E,V ):


Review
BASE:
V  R (every vertex is connected to itself)
RECURSE: If (a,b)  R and {b,c}  E then (a,c)  R (if a is
connected to b, and there is an edge from b to c, then a is
connected to c)
45
Basic Counting Rules
Sum Rule : If A and B are
disjoint, then
|A  B| = |A|+|B|
Product Rule


Set theoretic:
|A  B| = |A||B|
Combinatorial: If have n
stages, where i’ th stage
allows ai different situation,
total number of combinations
is a1·a2·a3···an
 EG: No. of ways to order 5
cards. No. of choices for 1st
card 52, 2nd choice 51 cards
remaining, etc. for total of
52·51·50·49·48 orderings
 EG: Method 1 for counting
anagrams
Review
Division Rule: IfS is partitioned
into cells of size d then each cell
contains |S |/d elements

Combinatorial: If over-counting
by a factor of d for each case of
interest, divide by d
 EG: No. of 5 cards hands.
52·51·50·49·48 would be the
answer if order matters. Since
order irrelevant, and there are 5!
different orderings of 5 elements,
answer is 52·51·50·49·48 / 5!
 EG: Method 2 for counting
anagrams
46
Inclusion-Exclusion
2 sets:
|A  B | = |A|+|B |- |A B |
3 sets:
| A B C |  | A|  | B |  | C | - | A B | - | AC | - | B C |  | A B C |
n - sets:
| A1  A2    An |

| A1 |  | A2 |    | An |
- | A1  A2 | - | A1  A3 | - 
 | A1  A2  A3 |  | A1  A2  A4 |  

 ( -1)n -1 | A1  A2    An |
Review
47
Pigeonhole Principle
Simple pigeonhole:
If N+1 objects are placed into N boxes, there
is at least one box containing 2 objects.
Generalized pigeonhole:
If N objects are placed into k boxes, there is
at least one box containing N/k  objects.
Review
48
r-permutations
r-combinations
r-permutations P (n,r )



Number of r-tuples in a set of size n
Order matters
P (n,r ) =n r = n (n-1)(n-2)(n-3)···(n-r+1)
r-combinations C (n,r )





L18
Number of r-tuples in a set of size n
Value of (n,r )-element in Pascal triangle
Order irrelevant (so can pre-sort)
C (n,r ) = n r / r ! = n ! / ( r ! · (n - r)! )
C (n,r ) =C (n,n-r ) (symmetry of Pascal)
49
Counting Formulas
Perms and Combos
Anagrams : with n letters with repetition
numbers a1, a2 , a3 , …. , ak number of
anagrams is n ! / (a1! a2 ! a3 ! … ak !)
Functions : Domain size m, codomain size n




General :
Bijections (m =n ):
One-to-one (m ≤n ):
Onto (m ≥n):
nm
n!
P (n,m )
nm - C (n,1)·(n -1)m + C (n,2)·(n -2)m + …
+(-1)iC (n,i )·(n-i )m +…+ (-1)n-1C (n,n-1)·1m

L18
Derangements (Domain=Codomain, m=n, f (i ) = i ):
 1 1 1 1
n 1 
Dn  n!1 -  -     - 1 
n! 
 1! 2! 3! 4!
50
Counting Formulas
Stars and Bars
Number of solutions in N of x1+x2+…+xk+1=n


C (n+k,k)
Also the number of different arrangement of n ’s
and k |’s in a string
Further constraints



Review
If variable xi ≥ k, change RHS to n-k
If variable xi < k use Inclusion-Exclusion on
constraint xi ≥ k
If have inequality x1+x2+…+xk+1≤ n, add dummy
variable and count solutions to x1+x2+…+xk+1+y =n
51
Cards
52 cards per deck (not including Jokers)
4 suits :
Hearts
Diamonds Spades
Clubs
13 ranks :
L18
52
Glossary of Poker Hands
Straight flush
Five cards in sequence in the
same suit. A royal straight
flush (A-K-Q-J-10 in same
suit) is the highest non-joker
poker hand possible.
Four of a kind
Four cards of the same rank.
Full house
Three of a kind and a pair.
When matching full houses,
the one with the higher three
of a kind wins.
Flush
Five cards of the same suit.
L18
Straight
Any five cards in sequence but
not all of the same suit.
Three of a kind
Three of the same rank with
two unmatched cards.
Two pairs
Two cards of one rank with two
cards of a different rank with
one dissimilar card. When
matching pairs occurs between
players, the one with the
higher fifth card wins.
One pair
Any two cards of the same
rank.
53
Probability
If all outcomes equally likely the probability of
event E in sample space S is the ratio
p (E ) = |E | / |S |
If each outcome s has probability p (s )
probability of event E is
p( E )   p( s )
sE
Random variable : a real function X : S  R
p( s )  X ( s )
Expectation of X : E( X )  
sS

Sum rule: E(X1+X2+…+Xn ) = E(X1)+ E(X2 )+…+E(Xn )
Bernoulli experiments : Suppose A has
probability p of occurring. Probability that A
occurs exactly k times in n trials is
k · (1-p)n-k ·C (n,k )
p
L18
54
Conditional Probability
Independence
If E and F are events and p (F ) > 0 then
the conditional probability of E given F
is defined by: p (E |F ) = p (EF ) / p (F )
Independent events : p(EF )=p(E )·p(F )
Review
55
Solving Recurrence Relations
First Order Equations
an expressed in terms of only the
previous term: an =f (an -1)
 Method 1) Solve a1 =f (a0), a2
=f (a1), a3 =f (a2), etc. directly.
Careful not to simplify carelessly
to so can see proper
generalization for an
 Method 2) Telescope, reverse
of #1. Write an =f (an-1), plug
in an -1 =f (an-2) getting an in
terms of an-2. Repeat, careful
with careless simplification.
EG: Solve an = 2an-1, a0= 3:
an=2an-1=22an-2=23an-3
=… =2ian-i=…=2na0
Plug in a0= 3 for final answer:
an = 3·2n
Linear Equations with Constant
Coefficients
Homogeneous (all terms involve
an-i):
1. Consider characteristic
2.
equation1
Roots give general solution2
an = Ar1n +Br2n
If repeated roots need to
multiply each repeat by a
bigger power of n
3. Solve for constants A,B
using initial conditions
Non-homogeneous :
1. Find a particular solution3 yp
2. Add y to homogenous
general solution.
3. Solve for constants using
initial conditions.
Relations
n -ary relation on sets (A1,A2, … ,An) is a
subset of A1×A2× … ×An.
binary relations : n = 2

Notations for (a,b)  R
 Infix notation : aRb
 Prefix functional notation : R(a,b) has value 1
 EG: R = “<”:


Review
(3,2)<  <(3,2) = 0  (3 < 2) = 0
Composition : aRb and bSc  a (S R )c
57
Relations
Relations on A

Representations
 Adjacency matrix
 Digraph

Properties
 Reflexive : every element self-related
 Symmetric : (a,b)  R  (b,a)  R
 Transitive :(a,b),(b,c) R (a,c)  R

Exponentiation : R n  
R 
R 
R



n times
Review
58
Projection and Join
n-Join of R with S



Jn (R,S )
Each element in R whose last n coordinates match
with first n coordinates of S, create an element of join
EG, 2-join: (0,1,1)
2-join (0,1,1,1)
(1,1,1)
Projection of R to sub-coordinates



Review
Pi1,i2,…, ik (R )
Keep only the indexed coordinates
EG: (0,1,1)
1,3 projection
(0,1)
59
Graphs
Undirected graphs



Simple graphs : No loops, or multiple edges
Multigraphs : Multiple edges allowed. No loops.
Pseudographs : Loops and multiple edges allowed.
Directed graphs


Digraphs : Loops allowed
Directed multigraph: multiple edges also allowed
Graph families




Complete graphs Kn
Cycles Cn
Wheels Wn
Cubes Qn
n
1
2
3
4
5
Degrees and Handshaking
Vertex degree :


Undirected graphs: contribution of 1 for each simple
edge, 2 for each loop
1
 Edge-Vertex Handshaking: | E | 
deg( e)

2 eE
Directed graphs:
 in-degree deg-: the number of edges that stick in
 out-degree deg+:the number of edges that stick out
 Handshaking Theorem: | E | 

deg
 (e) 
eE
deg
 (e)
eE
Face degree (undirected planar graphs):


Review
The number of edges surrounding a face
Edge-Face Handshaking: | E |  1
deg( F )

2
F R
61
Coloring and Bipartiteness
n –colorable : can use n colors so that no
two adjacent vertices have same color
Chromatic number : smallest number n
for which it is n –colorable
Bipartite : same as 2-colorable

Complete Bipartite Graphs
K2,3
K4,5
4-Color Theorem: Any planar graph is 4colorable.
Review
62
Graph Isomorphism
Isomorphism from G1=(V1,E1) to G2=(V2,E2). A
function on vertices satisfying:
f is bijective
1)
2)


number of edges bet. u,v in G1 is same as bet. f
(u), f (v ) in G2
To prove isomorphism, describe f explicitly
To prove non-isomorphism, show that a graph
invariant differs for G1 vs. G2




Review
Number of vertices, edges
Degrees, degree numbers
Non-isomorphic subgraphs
Girth, connectivity
63
Paths and Connectivity
Path : a sequence of edges in graph with incident
consecutive edges




In simple graph: Any sequence of vertices with adjacent
consecutive vertices
Simple path: no repeated edge
Circuit : first and last vertex the same
Length of path: number of edges traversed
Connectedness

Undirected graph





Review
Standard: every vertex pair has a connecting path
2-Connected: Connected and no cut vertices
Connected Components: subgraphs that are connected and as
large as possible
Weakly connected digraph: if without orientation, connected
Strongly connected digraph: given any vertices (a , b) there
is a path from a to b and vice versa
64
Euler and Hamilton Paths
Euler path



Simple path containing all edges in graph
Euler circuit : Euler path that’s a circuit
THM: Euler path or circuit iff there are ≤ 2
vertices of odd degree.
Hamilton path



Review
A path visiting every vertex in G exactly once.
Hamilton circuit : allowed to start vertex twice.
No simple all-inclusive theorem for Hamilton paths
(NP-complete problem).
65
Planar Graphs
Planar if can be drawn in
plane without any crossing
edges.

To prove planar, find a way
to redraw graph without
crossing edges:


To prove non planar use one
of:
 Kurotowski’s theorem (see
text)
 Euler formula method (see
next column)
Review
Euler Formula Method



Euler’s Formula for number
of regions:
r = |E | - |V | + 2
Girth g: smallest cycle length
in G
Face-Edge handshaking +
girth :
1
| E |  rg
2

Often last formula forces too
many edges, showing that
planarity assumption was
false
66