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Chapter 5 Relations and Functions

Discrete and Combinatorial Mathematics R. P. Grimaldi , 5 th edition, 2004. Chapter 5 Relations and Functions. Relations. For sets A and B, its Cartesian product ( cross product ) is the set of all ordered pairs (a,b) where a  A and b  B. In symbols A  B = {(a, b) | a  A and b  B}

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Chapter 5 Relations and Functions

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  1. Discrete and Combinatorial MathematicsR. P. Grimaldi,5th edition, 2004 Chapter 5 Relations and Functions

  2. Relations For sets A and B, its Cartesian product (cross product) is the set of all ordered pairs (a,b) where aA and bB. • In symbols AB = {(a, b) | aA and bB} • Example: A = {1, 2, 3} and B = {a, b} AB = {(1,a),(1,b),(2,a),(2,b),(3,a),(3,b)}, A = . For sets A and B, any subset of AB is called a binary relation from A to B. • R = {(1,a), (1,b), (2,b), (3,a)} is a relation from A to B. • (1,a)R also denotes 1Ra. For finite sets A and B with |A| = m and |B| = n, there are 2mn relations from A to B, including the empty relation and the relation AB itself.

  3. Product on Set Operations Theorem 5.1 For any sets A, B, C   : (a) A  (B  C) = (A  B)  (A  C) (b) A  (B  C) = (A  B)  (A  C) (c) (A  B)  C = (A  C)  (B  C) (d) (A  B)  C = (A  C)  (B  C) Proof of (a). (a,b)A  (B  C)  aA and bB  C  aA and bB and bC  (aA and bB) and (aA and bC)  (a,b)A  B and (a,b)A  C  (a,b)(A  B)  (A  C) #

  4. Properties of Relations Let R be a relation on a set A, i.e., R is a subset of the Cartesian product AA. • R is reflexive(反身) if (x,x)R, for all xA • A = {1,2,3} reflexive: R = {(1,1),(2,2),(3,3)} not reflexive: R = {(1,1),(1,2),(2,2),(2,3)} • R is symmetric(對稱) if (x,y)R  (y,x) R, for all x, yA • A = {1,2,3} symmetric: R = {(1,2),(2,1),(3,3)} not symmetric: R = {(1,1),(1,2),(2,2)}

  5. Properties of Relations (cont.) • R is transitive(遞移) if (x,y)R and (y,z)R  (x,z) R • A = {1,2,3} transitive: R = {(1,2),(2,3),(1,3),(3,1),(2,1),(1,1),(2,2),(3,2),(3,3)} not transitive: R = {(1,2),(2,3),(3,3)} • R is antisymmetric(反對稱) if for all x,yX such that xy, if (x,y) R  (y,x) R • A = {1,2,3} antisymmetric: R = {(1,2),(2,3),(1,3),(1,1)} not antisymmetric: R = {(1,2),(2,1),(3,3)}

  6. Relation Matrices Let X, Y be sets and R a relation from X to Y • Write the matrix M = (mij) of the relation as follows: • Rows of M = elements of X • Columns of M = elements of Y • Element mi,j = 0 if the element of X in row i and the element of Y in column j are not related • Element mi,j = 1 if the element of X in row i and the element of Y in column j are related

  7. Example Let X = {1, 2, 3}, Y = {a, b, c, d} Let R = {(1,a), (1,d), (2,a), (2,b), (2,c)} The matrix M of the relation R is M =

  8. Matrices • Let E = (eij)mn, F = (fij)mn be two mn (0,1)-matrices. We say E  F if eij  fij for all i,j. • For nZ+, (0,1)-matrix In = (ij)nn where ij = 1 if i = j, otherwise ij = 0. • Let A = (aij)mn be a (0,1)-matrix. The transpose of A, denoted by Atr, is the matrix (a*ji)nm where a*ji = aij for all i,j.

  9. Properties of Relation Matrices Theorem 7.2 Given a set A with |A| = n and a relation R on A, Let M denote the relation matrix for R. Then (a) R is reflexiveiff In  M; (All terms mii in the main diagonal of M are 1.) (b) R is symmetric iff M = Mtr. (aij = aji for all i and j.) (c) R is transitive iff M2  M. (cij in M2 is nonzero then entry mij in M is also nonzero.) (d) R is antisymmetric iff MMtr  In.

  10. Property (c) M = M2= • M232 0 •  m31m12+m32m22+m33m32+m34m42  0 • There exists k such thatm3kmk2  0 m3k  0andmk2  0m32 0. #

  11. Order Relations Let X be a set and R a relation on X. • R is a partial order on X if R is reflexive, antisymmetric and transitive. (e.g. “”) LetR be a partial order on X. • R is atotal order (linear order) on X if every pair of elements x,yX satisfies xRy or yRx. (e.g. topological sorting)

  12. Topological Sorting a d  c a b c d e b e

  13. Inverse of a Relation Given a relation R from X to Y, its inverse R-1 is the relation from Y to X defined by R-1 = { (y,x) | (x,y)  R } Example: If R = {(1,a), (1,d), (2,a), (2,b), (2,c)} then R-1= {(a,1), (d,1), (a,2), (b,2), (c,2)}

  14. Equivalence Relations Let X be a set and R a relation on X. R is an equivalence relation on X ifR is reflexive, symmetric and transitive. (e.g. “=”)

  15. Equivalence Classes Let R be an equivalence relation on a set A. For each x A, the equivalence class of x, denoted by [x] ={ yA | yRx }. Theorem 7.6 If R is an equivalence relation on a set A, and x, y  A, then (a) x  [x]; (b) xRy iff [x] = [y]; and (c) [x] = [y] or [x]∩[y] = .

  16. Proof of Theorem 7.6 • By the reflexive property of R. • () Let w[x]. xRy and wRx  wRy. (R is transitive.)  w[y]  [x]  [y]. R is symmetric  yRx  …  [y]  [x].  [x] = [y]. () Let [x] = [y]. x[x]  x[y]  xRy.

  17. Proof of Theorem 7.6 (cont.) (c) Assume [x]  [y] and [x]∩[y]  . Let v  A with v  [x] and v  [y].  vRx and vRy  xRv and vRy (symmetric)  xRy (transitive)  [x] = [y]. (by (b))  ∴ [x] = [y] or [x]∩[y] = . #

  18. Partitions A partitionS on a set X is a family {A1, A2,…, An} of subsets of X, such that • A1∪A2∪A3∪…∪An = X • Aj∩Ak= for every j,k with jk, 1<j,k<n. Example: if X = {integers}, E = {even integers) and O = {odd integers}, then S = {E, O} is a partition of X.

  19. Partitions and Equivalence Relations Theorem 7.7 If A is a set, then (a) any equivalence relation R on A induces a partition of A, and (b) any partition of A gives rise to an equivalence relation R on A. Proof. Part (a) follows from (a) and (c) of Theorem 7.6. Part (b): Let S be a partition of A. Define xRy if x, y are in the same set T for TS. Let xA.  x  T for some T  S. ∴xRx and R is reflexive. Suppose xRy.  x,y  T for some T  S. ∴yRx and R is symmetric. Suppose xRy and yRz.  x,y,z  T for some T  S. ∴xRz and R is transitive. #

  20. Functions For nonempty sets X, Y, a function, or mapping, f from X to Y, denoted by f : X  Y, is a relation from X to Y such that if two pairs (x,y) and (x,y’)  f, then y = y’. • Example: Dom(f) = X = {a, b, c, d}, Rng(f) = {1, 3, 5} f(a) = f(b) = 3, f(c) = 5, f(d) = 1.

  21. How many functions? Let X,Y be nonempty sets with |X| = m and |Y| = n. Define the functions with domain X. x1 y1 x2 y2 xm yn For each xi we select one of the n elements of Y.  There are nm functions from X to Y.

  22. One-to-one functions A function f : X  Y is one-to-one, injective  for each y  Y there exists at most one x  X with f(x) = y. Alternative definition: f : X  Y is one-to-one  for each pair of distinct elements x1, x2  X there exist two distinct elements y1, y2  Y such that f(x1) = y1 and f(x2) = y2. Examples: • The function f(x) = 2x from the set of real numbers to itself is one-to-one • The function f : R  R defined by f(x) = x2 is not one-to-one, since for every real number x, f(x) = f(-x).

  23. How many one-to-one functions? Let X,Y be nonempty sets with |X| = m and |Y| = n, where m  n. x1 y1 x2 y2 xm yn There are n choices for x1, n-1 choices for x2, n-2 choices for x3, and so on.  There are P(n,m) one-to-one functions from X to Y.

  24. Onto functions A function f : A  B is onto ,or surjective  for each b  B there exists at least one a  A with f(a) = b, i.e. Rng(f) = B. Example: • The function f: R  R defined by f(x) = x3 is an onto function. • The function g: R  R defined by g(x) = x2 is not an onto function.

  25. How many onto functions? Let X,Y be nonempty sets with |X| = 4 and |Y| = 3. w 1 x 2 y 3 z • There are 34 functions from X to Y. • There are 24 functions from X to {1,2} ({1,3},{2,3}). • The set of these 24 functions contains the functions • from X to {1} and the functions from X to {2}. •  There are C(3,3)34 – C(3,2)24 + C(3,1)14 functions.

  26. How many onto functions? Let X,Y be nonempty sets with |X| = m and |Y| = n. • There are C(n,n)nm – C(n,n-1)(n-1)m + C(n,n-2)(n-2)m – • … + (-1)n-1 C(n,1) 1m = • onto functions from X to Y. Stirling number of the second kind, denoted by S(m,n): The number of ways to distribute the m distinct objects into n identical containers, with no container left empty, is

  27. Bijective functions A function f : X Y is bijective  f is one-to-one and onto. • Examples: • A linear function f(x) = ax + b is a bijective function from the set of real numbers to itself • The function f(x) = x3 is bijective from the set of real numbers to itself. • Question: Let X,Y be nonempty sets with |X| = n and |Y| = n. How many bijective functions are there from X to Y?

  28. Inverse Given a function y = f(x), the inverse f -1 is the set {(y, x) | y = f(x)}. • The inverse f -1 of f is not necessarily a function. • Example: if f(x) = x2, then f -1 (4) = ± 41/2= ± 2, not a unique value and therefore f -1 is not a function. • However, if f is a bijective function, it can be shown that f -1 is a function.

  29. Composite Functions If f : A  B and g : B  C, we define the composite function, denoted by g ◦ f : A  C, by g ◦ f (a) = g(f(a)) for each a  A. • Example: f(x) = x2 -1, g(x) = 3x + 5. Then f ◦ g(x) = f(g(x)) = f(3x + 5) = (3x + 5)2 - 1 • Composite functions are associative: f ◦ (g ◦h) = (f ◦ g) ◦ h, • But, in general, it is not commutative: f ◦ g  g ◦ f.

  30. Binary Operations For any nonempty sets A, B, any function f : A x A  B is called a binary operation on A. If B  A then the binary operation is said to closed (on A). • Examples • The function f : Z x Z  Z, f(a,b) = a – b, is a closed binary operation on Z. • The function g : Z+x Z+ Z, g(a,b) = a – b, is a binary operation on Z+, but it is not closed.

  31. Unary Operations A function g : A  A is called aunary operation on A. • Examples: The function h : R+  R+ defined by h(a) = 1/a is a unary operation on R+.

  32. The Pigeonhole Principle If m pigeons occupy n pigeonholes and m > n, then at least one pigeonhole has two or more pigeons roosting in it.

  33. Generalized Form If X and Y are finite sets with |X| = n, |Y| = m and k = n/m, then there are at least k values a1, a2,…, ak  X such that f(a1) = f(a2) = … f(ak). • Example: n = 5, m = 3 k = n/m = 5/3 = 2.

  34. Example 1 Any subset of size 6 from the set S = {1, 2, 3, …, 9} must contains two elements whose sum is 10. Proof. Pigeons: the elements of a 6-element subset of S. Pigeonholes: the subsets {1,9},{2,8},{3,7},{4,6},{5}. By the pigeonhole principal, there are at least two elements whose sum is 10. #

  35. Example 2 Prove that if 101 integers are selected from the set S = {1, 2, 3, …, 200}, then there are two integers such that one divides the other. Proof.The fundamental theorem of arithmetic: Every positive integer (except 1) can be represented in exactly one way apart from rearrangement as a product of one or more primes.  xS [x = 2ky], with k  0, and gcd(2,y) = 1.  yT={1,3,5,…,199}. Pigeons: 101 selected integers. Pigeonholes: 100 elements of T.   two selected integers 2my and 2ny. #

  36. Example 3 Let m  Z+ with m odd. Prove that there exists a positive integer n such that m divides 2n – 1. Proof. Pick m+1 positive integers: 21–1, 22–1, 23–1, …, 2m–1, 2m+1–1.   s, t  Z+ [(2s–1) mod m = (2t–1) mod m], 1s<tm+1. (By the pigeonhole principal and the division algorithm)  Let 2s–1 = q1m+r and 2t–1 = q2m+r.  2t–2s = (q2 – q1)m = 2s (2t-s–1)  m is odd.  m| (2t-s–1)  n = t-s. #

  37. Brainstorm 假設有一座小鎮，(1) 鎮上每個人頭髮的數目都不同﹔(2) 沒有人的頭髮數目剛好是500﹔(3) 鎮的人口比鎮裏任何一人頭髮的數目為多。 問題：鎮上最多有多少人？

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