* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Machine Models - Columbia University
Survey
Document related concepts
Mathematics of radio engineering wikipedia , lookup
Big O notation wikipedia , lookup
Positional notation wikipedia , lookup
Wiles's proof of Fermat's Last Theorem wikipedia , lookup
Law of large numbers wikipedia , lookup
Large numbers wikipedia , lookup
Georg Cantor's first set theory article wikipedia , lookup
List of prime numbers wikipedia , lookup
Elementary mathematics wikipedia , lookup
Principia Mathematica wikipedia , lookup
Four color theorem wikipedia , lookup
Collatz conjecture wikipedia , lookup
Transcript
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 pq 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: (p1p2…pn) (p1p2…pn) Universal quantifier version: x P(x ) x P(x ) “Not one is true iff all are false.” Review Disjunctional version: (p1p2…pn) (p1p2…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 } |A1A2…An| = |A1||A2| … |An| complement : A={x|xA} union : A B = { x | x A x B } intersection : A B = { x | x A x B } difference : A-B={x|xA xB} symmetric difference:AB={ x | xA xB } Review 19 Venn Diagrams Complement A Intersection U U A A Union Review B Disjoint Union U AB A AB B U A A B B 20 Venn Diagrams Difference Symmetric Difference AB 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 AB s t EG: If U = {1,2,3,4}, A = {2}=0100, B = {1,2,4}=1101 then: AB = 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 xS i 1 Double sum: 2 3 ij i 0 Review j 1 Geometric sum: ar n1 - a ar a ar ar ... ar r -1 i 0 n i 2 n 26 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 R0 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 | - | AC | - | 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 ) sE Random variable : a real function X : S R p( s ) X ( s ) Expectation of X : E( X ) sS 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 (EF ) / p (F ) Independent events : p(EF )=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 eE 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) eE deg (e) eE 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