-- Chapter 9: Generating Function. Hung-Yu Kao ( ) Department of Computer Science and Information Engineering, National lcheng Kung University

Transcription

1 Discrete Mathematics -- Chapter 9: Generating Function Hung-Yu Kao ( ) Department of Computer Science and Information Engineering, National lcheng Kung University

2 Outline Calculational Techniques Partitions of Integers The Eponential Generating Function The Summation Operator

3 Enumeration again Chapter 1: c 1 +c +c 3 +c 4 =5, where c i >= 0 Chapter 8: c 1 +c +c 3 +c 4 =5, where 10>c i >= 0 In chapter 9, c to be even and c 3 to be a multiple of 3 the coefficient y in (+y) 3 the coefficient i 4 in (+ )( )(1+++ ) 3

4 9.1 Introductory Eamples E 9.1 : One mother buys 1 oranges for three children, Grace, Mary, and Frank. Grace gets at least four, and Mary and Frank gets at least two, but Frank gets no more than five. Solution c1 + c + c3 = 1, where 4 c1, c, and c3 5 Generating function: f() ) = ( )( )( ) product j j k every triple (i, j, k) The coefficient of 1 in f() )yields the solution. 4

5 Introductory Eamples E 9. : There is an unlimited number of red, green, white, and black jelly beans. In how many ways can we select 4 jelly beans so that we have an even number of white beans and at least si black ones? Solution red (green): white: black: Generating function: f() = ( ) ( ) ( ) The coefficient of 4 in f() is the answer. 5

6 Introductory Eamples E 9.3 : How many nonnegative integer solutions are there for c 1 +c +c 3 +c 4 = 5? Solution Alternatively, in how many ways 5 pennies can be distributed among four children? Generating function: f() = ( ) 4 (polynomial) The coefficient of 5 is the solution. Note: g() = ( ) 4 (power series) can also generate the answer f ( ) = n= 0 a n ( c) n Easier to compute with a power series than with a polynomial 6

7 9. Definition and Eamples: Calculational Techniques Definition 9.1: n n n n n (1 + ) = so, (1 + ) n n,( ) is the generating g function for the sequence E 9.4 : ( ) ( ) ( ) ( ) n ( n) ( n) ( n) ( n,,, ),0,0,0, 0, 1 n 7

8 Definition and Eamples: Calculational Techniques E 9.5 : a) (1 - n+1 )/(1 - ) is the generating function for the sequence 1, 1, 1,, 1, 0, 0, 0,, where the first n+1 terms are 1. n 1 n Q(1 + ) = (1 )( ). b) 1/(1-) is the generating function for the sequence 111 1, 1, 1, 1 3 1, Qwhile < 1, 1 = c) 1/(1-) is the generating gfunction for the sequence 1,, 3, 4, 1 1 = (1 ) = ( 1)(1 ) ( 1) Q d d = 1 1 (1 ) d d d 3 = ( ) = d 3 + d) /(1-) is the generating function for the sequence 0,1,,3,. Q 3 4 = (1 ) + 8

9 Definition and Eamples: Calculational Techniques E 9.5 : e) (+1)/(1-) 3 is the generating function for the sequence 1,, 3, 4, Q d d (1 ) d = ( d 1 = (1 ) ) + Q = d d d d (1 ) = (1 ) = (1 ) (1 ) = ( 1 ) (1 ) + ( )(1 ) 3 3 ( 1) f) (+1)/(1-) 3 is the generating function for the sequence 0, 1,, 3, 4, Q 3 = ( + 1) 3 (1 ) + 9

10 Definition and Eamples: Calculational Techniques E 9.5 : g) Further etensions: 10

11 Definition and Eamples: Calculational Techniques E 9.6 : a) 1/(1 - a) ) is the generating gfunction for the sequence a 0, a 1, a, a 3, b) f() = 1/(1- ) is the generating function for the sequence 1, 1, 1, 1, Then g() = f() - is the generating function for the sequence 1, 1, 0, 1, 1, 1, h() = f() + 3 is the generating function for the sequence 1, 1, 1, 3, 1, 1, c) Can we find a generating function for the sequence 0,, 6, 1, 0, 30, 4,? 11

12 Definition and Eamples: Calculational Techniques E 9.6 : c) Observe 0,, 6,1,, 0,... a a a 0 4 a = 0 = 6 = a n = 0 0 = = n = 4 + 0, +, + n a + 4, 1 a 3 = = 1 = 1 = 3 + 1, + 3, ( + 1) ( + 1) + (1 ) + = = (1 ) 3 (1 ) (1 ) 3 (1 ) 3 is the generating function. 1

13 Etension of Binomial Theorem (1 + ) = Binomial theorem: ( ) ( ) ( ) ( ) n n n n n n When n Z +, we have r r!( n r)! r! 0 1 n n! n( n 1)( n )...( n r + 1) = = n If n R, we define If n Z +, we have n n( n 1)( n )...( n r + 1) = r r! n ( n)( n 1)( n )...( n r = r r! r ( 1) ( n )( n + 1)...( n + r 1) = r! ( 1) ( n + r 1)! r n + r = = ( 1) ( n 1)! r! r r 1 + 1) 13

14 Etension of Binomial Theorem E 9.7 : (1-) -n? 14

15 Etension of Binomial Theorem E 9.8 : Find the coefficient of 5 in (1-) -7. Solution 7 ( 7 = ) r= 0 r ( 1 ) ( ) The coefficient of ( ) ( ) ( ) ( ) = ( 1) ( 3) = (3) 5 E 9.9 : Find the coefficient of all i in (1+3) -1/3 5 5 : r 5 15

16 Definition and Eamples: Calculational Techniques E 9.10 : Determine the coefficient of 15 in f() = ( ) 4. Solution ( ) = (1++ + ) = /(1-) f()=( /(1-)) 4 = 8 /(1-) 4 Hence the solution is the coefficient of 7 in (1-) -4 : C(-4, 47)(1) 7)(-1) 7 =(1) (-1) 7 C(4+7-1, 17)(1) 7)(-1) 7 = C(10, 7) =

17 17

18 check n=1 18

19 Definition and Eamples: Calculational Techniques E 9.11 : In how many ways can we select, with repetition allowed, r objects from n distinct objects? Solution For each object (with repetitions), represents the possible choices for that tobject t( (namely none, one, two, ) Consider all of the n distinct objects, the generating function is f() = (1++ + ) n n n = = 1 n i 1 i (1 ) =. 1 (1 ) n i= 0 i ( n ) + r 1 The answer is the coefficient of r in f(),. r 19

20 Definition and Eamples: Calculational Techniques E 9.1 : Counting the compositions of a positive integer n. Solution E.g., n = 4 One-summand: ( ) = [/(1-)], coefficient of 4 =1 Two-summand: ( ) = [/(1-)], coefficient of 4 =3 Three-summand: ( ) 3 = [/(1-)] 3, coefficient i of 4 =3 Four-summand: ( ) 4 = [/(1-)] 4, coefficient of 4 =1 =1 )] The number of compositions of 4: coefficient of 4 in 4 [ /(1 i 1 i How about n=5? 0

21 Definition and Eamples: Calculational Techniques E 9.1 : Counting the compositions of a positive integer n. The number of ways to form an integer n is the coefficient of n in the following ggenerating gfunction. 1 3 i ( ) = i= 1 i= 1 [ /(1 )] i 1

22 Definition and Eamples: Calculational Techniques E 9.14 : In how many ways can a police captain distribute 4 rifle shells to four police officers, so that each officer gets at least three shells hll but btnot more than eight. iht Solution f()= ( ) 4 = 1 ( ) 4 = 1 [(1-6 )/(1-)] 4 The answer is the coefficient of 1 in (1-6 ) 4 (1-) -4 = [1 6 4 ( ) ( ) 1 ( ) ][( 0 4 ) ( )( ) + ( 4 )( ) + ] [ ( 1) ( 1) + ] = [ + ] =

23 Definition and Eamples: Calculational Techniques E 9.16 : Determine the coefficient of Solution 8 in 1 ( 3)( ). 3

24 Definition and Eamples: Calculational Techniques E 9.17 : How many four-element subsets of fs S = {1,,, 15} contains no consecutive integers? Solution E.g., one subset {1, 3, 7, 10}, 1 1<3<7<10 15, difference 0,, 4, 3, 5, difference sum = 14. These suggest the integer solutions to c 1 +c +c 3 +c 4 +c 5 =14 where 0 c 1, c 5 and c, c 3, c 4. The answer is the coefficient of 14 in f() = ( )( ) 3 ( ) = 6 (1-) -5 The coefficient of 8 in (1-) ( 1) = = =

25 Convolution of Sequences E 9.19 : Let f() = /(1-) = , for the sequence a k = k g() = (+1)/(1-) ) 3 = , for the sequence b k = k h() = f()g() = a 0 b 0 + (a 0 b 1 +a 1 b 0 ) + (a 0 b +a 1 b 1 +a b 0 ) +, for the sequence c k = a 0 b k +a 1 b k-1 +a b k- + +a k- b +a k-1 b 1 +a k b 0 k ck = = i( k i). c 0 = 0 0 i 0 i c = = 0 1 c = = 1 c 3 = 6 The sequence c 0, c 1, c, is the convolution of the sequences a 0, a 1, a, and b 0, b 1, b, 5

26 Convolution of Sequences E 9.0 : Let f() = 1/(1-) = g() = 1/(1+) = h() = f()g() = 1/[(1-)(1+)] =1/(1 1/(1- ) = The sequence 1, 0, 1, 0, is the convolution of the sequences 1, 1, 1, 1, and 1, -1, 11, -1, 6

27 9.3 Partition of Integers p(n): ( ) the number of partitioning i a positive integer n The number of 1 s is 0 or 1 or or 3. The power series is The number of s can be kept tracked by the power series For n, the number of 3 s can be kept tracked by the power series

28 Partition of Integers Determine p(10) The coefficient of 10 in f() =( )( )( ) ( ) f ( ) = 3 10 i 1 (1 ) (1 ) (1 ) (1 ) (1 = = i By the coefficient of n in 1 P( ) =, we get the sequence i = (1 ) p(0), p(1), p(), p(3), 1 i ) 8

29 Partition of Integers E 9.1 : Find the generating function for the number of ways an advertising agent can purchase n minutes of air time if the time slots come in blocks of 30, 60, or 10 seconds. Solution Let 30 seconds represent one time unit. Find integer solutions to a+b+4c = n Generating function: f()= ( )( )( ) = (1 ) (1 ) (1 4. ) Answer: the coefficient of n is the number of partitions of n into 1 s, s, and 4 s. 9

30 Partition of Integers E 9. : Find the generating function for p d (n), the number of partitions of a positive integer n into distinct summands. 6 = 1+5 One time of occurrence per summand 6 = = +4 P d () = (1+)(1+ )(1+ 3 ).. E 9.3 : Find the generating function for p o (n), the number of partitions of a positive ii integer n into odd summands. P o () = ( )( )( ) = 1/(1-) 1/(1-3 ) 1/(1-5 ) 1/(1-7 )... P d () = P o ()? 6 = = =

31 Partition of Integers E 9.4 : Find the generating function for the number of partitions of a positive integer n into odd summands and occurring an odd number of times. Solution 6 = = = 3+3 f() = ( )( ) ( ) + = 1 k = 0 i= 0 (k + 1)(i+ 1). 31

32 Partition of Integers Ferrers graph uses rows of dots to represent a partition of an integer In fig. 9., two Ferrers graphs are transposed each other for the partitions of (a) 14 = (b) 14 = The number of partitions of an integer n into m summands is equal to the number of partitions of n into summands where m is the largest summand

33 9.4 The Eponential Generating Function 33

34 The Eponential Generating Function E 9.5 : Eamining the Maclaurin series epansion for e, we find 3 i e = =! 3! i= 0 i! so e is the eponential generating function for the sequence 1, 1, 1,. 34

35 The Eponential Generating Function E 9.6 : In how many ways can four of the letters in ENGINE be arranged? Solution Using eponential generating function: f() = [1++( /!)] [1+] E, N: [1++( /!)] G, I: [1+] The answer is the coefficient of 4 /4!. 35

36 The Eponential Generating Function E 9.7 : Consider the Maclaurin series epansion of e and e -. 36

37 The Eponential Generating Function E 9.8 : A ship carries 48 flags, 1 each of the colors red, white, blue and black. Twelve flags are placed on a vertical pole to communicate signal to other ships. How many of these signals use an even number of blue flags and an odd number of black flags? 37

38 The Eponential Generating Function how many of these use at least three white flags or no white flag at all? 38

39 The Eponential Generating Function 39

40 The Eponential Generating Function E 9.9 : A company hires 11 new employees, and they will be assigned to four different subdivisions. Each subdivision has at least one new employee. ee In how many ways can these assignments be made? Solution Onto function: g: X Y where X = 11, Y = 4. 40

41 9.5 The Summation Operator Let f() = a 0 +a 1 +a +a Then f()/(1-) generate the sequence of a 0, a 0 +a 1, a 0 +a 1 +a, a 0 +a 1 +a +a 3, So we refer to 1/(1-) as the summation operator. 41

42 The Summation Operator E 9.30 : Summation operator 1/(1-) is the generating function for the sequence 1, 1, 1, 1, 1, [1/(1-)] [1/(1-)] is the generating function for the sequence 1, 1+1, 1+1+1, 1,, 3, + is the generating function for the sequence 0, , 1, 0, 0, 0, (+ ) [1/(1-)] is the generating function for the sequence 0, 1 1,,,,, (+ )/(1-) is the generating function for the sequence 0, 1, 3, 5, 7, 9, 11, (+ )/(1-) 3 is the generating function for the sequence 0, 1, 4, 9, 16, 5, 36, 4

43 The Summation Operator E 9.31 : Find a formula to epress n as a function of n. Solution 43

44 The Summation Operator 44

45 Homework 9.1: 9: 9.: 9.3: 9.4: 95: 9.5: 45