Chapter 6. Advanced Counting Techniques

Transcription

1 Chapter 6 Advaced Coutig Techiques

2 6.: Recurrece Relatios Defiitio: A recurrece relatio for the sequece {a } is a equatio expressig a i terms of oe or more of the previous terms of the sequece: a,a2,a3,,a -, with >= 0, ( 0 beig a oegative iteger). A sequece is called a solutio of a recurrece relatio if its terms satisfy the recurrece relatio.

3 Recurrece Example Cosider the recurrece relatio a = 2a a 2 ( 2). Which of the followig are solutios? a = 3 Yes -> 2 [3(-)] 3(-2) = 3 => a a = 2 No -> a 0 =, a = 2, a 2 = 4; a 2 = 2a a 0 = 2.2 = 3 a 2 a = 5 Yes -> a = = 5 = a

4 Iitial coditios The iitial coditios (base coditios): Specify the terms that precede the first term where the recurrece relatio takes effect. i.e., specify a 0

5 Example Applicatios Growth of bak accout Iitial Amout P 0 =$0,000 After Years= P Compoud Iterest I = % Sol: P =P - +(I/00)P - =(.)P - Usig Iteratio we get, P =(.) P 0 i.e P 30 =(.) 30 0,000=$228,922.97

6 Example Applicatio Rabbits ad Fiboacci Numbers Growth of rabbit populatio i which each rabbit yields ew oe every period startig 2 periods after its birth. P = P + P 2 (Fiboacci relatio)

7 Classic Tower of Haoi Example Problem: Get all disks from peg to peg 2. Oly move disk at a time. Never set a larger disk o a smaller oe.

8 Haoi Recurrece Relatio Let H = # moves for a stack of disks. Optimal strategy: Move top disks to spare peg. (H moves) Move bottom disk. ( move) Move top to bottom disk. (H moves) Note: H = 2H +

9 Why is H = 2H - + Oly move disk at a time. Never set a larger disk o a smaller oe.

10 Solvig Tower of Haoi RR H = 2 H + = 2 (2 H 2 + ) + = 2 2 H = 2 2 (2 H 3 + ) = 2 3 H = 2 H = (sice H =) = 2

11 6.2: Solvig Recurreces Defiitio: A liear homogeeous recurrece relatio of degree k with costat coefficiet is a recurrece relatio of the form a = c a + + c k a k, where the c i are all real, ad c k 0. The solutio is uiquely determied if k iitial coditios a 0 a k are provided

12 6.2: Solvig Recurreces.. Liear?: Right had side is sum of multiples of previous terms. Homogeous?: No terms that are NOT multiples of the a j s. Degree?: k-degree sice previous k terms are used.

13 Solvig with cost. Coefficiets Basic idea: Look for solutios of the form a = r, where r is a costat. This requires the characteristic equatio: r = c r + + c k r k, i.e., r k c r k c k = 0 (Dividig both sides by r -k ad subtractig right had side from left). The solutios (characteristic roots) ca yield a explicit formula for the sequece.

14 Solvig Theorem: Let c ad c 2 be real umbers. Suppose that r 2 c r c 2 = 0 has two distict roots r ad r 2. The the sequece {a } is a solutio of the recurrece relatio a = c a + c 2 a 2 if ad oly if a = α r + α 2 r 2 for 0, where α, α 2 are costats.

15 Theorem : Proof 2 thigs to prove Case : Roots are r & r 2, i.e., {a = α r + α 2 r 2 } a is a solutio. r & r 2 are roots of r 2 c r c 2 = 0 r 2 = c r + c 2 ; r 22 = c r 2 + c 2. c a + c 2 a 2 = c (α r - + α 2 r 2 - ) + c 2 (α r -2 + α 2 r 2-2 ) α r -2 (c r + c 2 ) + α 2 r 2-2 (c r 2 + c 2 ) α r -2 r 2 + α 2 r 2-2 r 2 2 a

16 Theorem : Proof 2 thigs to prove Case 2: a is a solutio a = α r + α 2 r 2 for some α & α 2. a is a solutio a 0 = C 0 = α + α 2. C = α r + α 2 r 2 α = (C C 0 r 2 )/(r - r 2 ) α 2 = C 0 α = (C 0 r -C )/(r - r 2 ) Works oly if r r 2 a = α r + α 2 r 2 works for 2 iitial coditios Sice the iitial coditios uiquely determie the sequece, a = α r + α 2 r 2

17 Example Solve the recurrece a = a + 2a 2 give the iitial coditios a 0 = 2, a = 7. A = r r = r - + 2r-2 r2 = r +2 Solutio: Use theorem c =, c 2 = 2 Characteristic equatio: r 2 r 2 = 0 Solutios: r = [ ( )±(( ) 2 4 ( 2)) /2 ] / 2 = (±9 /2 )/2 = (±3)/2, so r = 2 or r =. So a = α 2 + α 2 ( ).

18 Example Cotiued To fid α ad α 2, solve the equatios for the iitial coditios a 0 ad a : a 0 = 2 = α α 2 ( ) 0 a = 7 = α 2 + α 2 ( ) Simplifyig, we have the pair of equatios: 2 = α + α 2 7 = 2α α 2 which we ca solve easily by substitutio: α 2 = 2 α ; 7 = 2α (2 α ) = 3α 2; 9 = 3α ; α = 3; α 2 = -. Fial aswer: a = 3 2 ( )

19 The Case of Degeerate Roots Theorem2: Let c ad c2 be real umbers with c2 0. Suppose that r 2 c r c 2 = 0 has oly oe root r 0. A sequece {a } is a solutio of the recurrece relatio a =c a - + c 2 a -2 if ad oly if a = α r 0 + α 2 r 0, for all 0, for some costats α, α 2.

20 k-liear Homogeeous Recurrece Relatios with Costat Coefficiets Theorem3: Let c,c2,.ck be real umbers. Suppose the C.E. If this has k distict roots r i, the the solutios to the recurrece are of the form: k a = α r i= i i r i= if ad oly if k i i= for all 0, where the α i are costats. k k c r i = a 0 = k c i a i

21 Theorem 3: Example Let a = 6a - a -2 +6a -3 ; a 0 =2,a =5, & a 2 =5. C.E.: r 3 6r 2 + r 6; Roots =,2, & 3. Solutio: a = α. + α α 3.3 a 0 = 2 = α + α 2 + α 3 a = 5 = α + α α 3.3 a 2 = 5 = α + α α 3.9 α = ; α 2 = ; α 3 = 2. a =

22 Degeerate t-roots Theorem 4: Suppose there are t roots r,,r t with multiplicities m,,m t. m i >= for i = t. The: for all 0, where all the α are costats. = = = t i i m j j j i r a i 0 α,

23 Theorem 4: Example E.g., Roots of C.E. are 2, 2, 2, 5, 5, & 9. Solutio: (α,0 + α, + α,2 2).2 + (α 2,0 + α 2, ).5 + α 3,0 9

24 Liear NoHomogeeous Recurrece Relatios with Costat Coefficiets Liear NoHomogeeous RRs with costat coefficiets may (ulike LiHoReCoCos) cotai some terms F() that deped oly o (ad ot o ay a i s). Geeral form: a = c a + + c k a k + F() The associated homogeeous recurrece relatio (associated LiHoReCoCo).

25 Solutios of LiNoReCoCos Theorem 5: If a (p) is a particular solutio to the LiNoReCoCo, the k a = cia i + F( ) i= The all its solutios are of the form: a = a (p) + a (h), where a (h) is a solutio to the associated homogeeous RR a k = cia i= i

26 Theorem 5: Proof {a (p) } is a particular solutio: a (p) = c a - (p) + c 2 a -2 (p) +..+c k a -k (p) + F() () Let b be a 2 d solutio: b = c b - + c 2 b c k b -k + F() (2) (2) (): b a (p) = c (b - -a - (p) ) + c 2 (b -2 -a -2 (p) ) + + c k (b -k -a -k (p) ) Hece, {b a (p) } is a solutio of the associated homogeeous RR, say {a (h) }. b = a (p) + a (h).

27 Example Fid all solutios to a = 3a +2. Which solutio has a = 3? To solve this LiNoReCoCo, solve its associated LiHoReCoCo equatio: a = 3a, ad its solutios are a (h) = α3, where α is a costat. By Theorem 5, the solutios to the origial problem are all of the form a = a (p) + α3. So, all we eed to do is fid oe a (p) that works.

28 Trial Solutios Sice F()=2, i.e it is liear so a reasoable trial solutio is a liear fuctio i, say p = c + d. The the equatio a = 3a +2 becomes, c+d = 3(c( )+d) + 2, (for all ) Simplifyig, we get ( 2c+2) + (3c 2d) = 0 (collect terms) So c = ad d = 3/2. So a (p) = 3/2 is a solutio.

29 Fidig a Desired Solutio From Theorem 5, we kow that all geeral solutios to our example are of the form: a = 3/2 + α3. Solve this for α for the give case, a = 3: 3 = 3/2 + α3 α = /6 The aswer is a = 3/2 + (/6)3

30 Theorem 6 Suppose {a } satisfies LiNoRR: k = ca i i + i= t t i Ad F()= bt i s ; b t ad s are real os. i= 0 Whe s is ot a root of C.E. of the associated LiHoRR, there is a particular solutio of the form t t i pt i s i= 0 a F( )

31 Theorem 6 cotiue.. Whe s is a root of this C.E. ad its multiplicity is m, there is a particular solutio of the form t m i= 0 p t i t i This factor m esures that the proposed particular solutio will ot already be a solutio of the associated LiHoRR. s

32 Theorem 6: Example a = 6a - 9a -2 + F(); F() = 3 C.E.: r 2 6r + 9 => (r-3)2 = 0; r = 3. Solutio: By Theorem 6, s = 3, multiplicity m = 2. Solutio is of the form p 0 2 3, where p 0 is polyomial costat.

33 6.3: Divide & Coquer R.R.s May types of problems are solvable by reducig a problem of size ito some umber a of idepedet subproblems, each of size /b, where a ad b>. The time complexity to solve such problems is give by a recurrece relatio: f() = a f(/b) + g() where g() is the extra operatios required. This is called a divide-ad-coquer recurrece relatio

34 Divide+Coquer Examples Biary search: Break list ito subproblem (smaller list) (so a=) of size /2 (so b=2). So f() =.f(/2)+2 (g()=2 costat) Merge sort: Break list of legth ito 2 sublists (a=2), each of size /2 (so b=2), the merge them, i g() = operatios So M() = 2M(/2) +

35 Fast Multiplicatio Example This algorithm splits each of two 2-bit itegers ito two bits blocks. Thus, from multiplicatios of two 2-bit itegers, it is reduced to oly three multiplicatios of bit itegers, plus shifts ad additios To fid the product ab of two 2-digitbase 2 umbers, a=(a 2- a 2-2 a 0 ) 2 ad b=(b 2- b 2-2 b 0 ) 2, first, we break them i half: a=2 A +A 0, b=2 B +B 0,

36 Derivatio of Fast Multiplicatio ab = = = = = (2 2 2 (2 ( A A B A B ( A B + ( A B + + ( A 2 2 A )(2 2 ) A B 0 0 ) A B A 0 A B + 0 ( A B 0 0 A B )( B B A B (2 (2 + + B B + ) A B + ) A B 0 0 ) ) A B 0 + ( A B A B 0 ) A B A B A B + + ) + ) (Multiply out 0 polyomials) Zero ( A B 0 0 A B (Factor last polyomial) Three multiplicatios, each with -digit umbers 0 0 ))

37 Recurrece Rel. for Fast Mult. Notice that the time complexity f() of the fast multiplicatio algorithm obeys the recurrece: f(2)=3f()+c() i.e., f()=3f(/2)+c() So a=3, b=2. Time to do the eeded adds & subtracts of -digit ad 2-digit umbers

38 Theorem Let f() = af(/b) + c wheever is divisible by b, where a, b is a iteger greater tha, ad c is a positive real umber. The { log f () ( b a is O ) O(log) if a > if a = Furthermore, whe =b k, where k is a positive iteger, a log b f() = C + C 2. where C = f() + c/(a-) ad C 2 = -c/(a-)

39 Theorem: Proof = b k ; f() = akf() + a = : f() = f() + ck; Sice = b k, k = log b f() = f() + clog b. is a power of b or ot: f() is O(log), whe a =. a > : (from Theorem, Sectio 3.2) f() = a k f() + c(a k )/(a ) = a k [f() + c/(a )] c/(a ) = C log b a + C 2, sice a k = alog b = log b a f() is O( log b a ) k j= 0 j a c = a k f () + c k j= 0 a j

40 MASTER Theorem Let f() = af(/b) + c d wheever =b k, where a, b>, ad c ad d are real umber, with c positive ad d oegative. The f() is O ( d ) O ( d log ) logb a o( ) if a<b d if a=b d if a>b d

41 Example We kow that umber of comparisos used by the merge sort to sort a list of elemets is less the M(), where M()=2M(/2) + From Master Theorem we fid that M() is O(log)

42 6.4 Geeratig Fuctios Geeratig fuctios: Used to represet sequeces efficietly by codig the terms of a sequece as coefficiets of powers of a variable x i a formal power series Defiitio: Let S = {a0, a, a2, a3,...} be a (ifiite) sequece of real umbers. The the geeratig fuctio G(x), of S is the series G(x)=a 0 +a x+ +a k x k + = k =0 a k k x

43 Example of GF Let S = {,,,,... }. The G(x) =+ x + x x k +... = =0 k k x if S=,,,,, By Theorem of Sectio 3.2, (x 6 -)/(x-) = + x + x x 5 G(x) =(x 6 -)/(x-) is the geeratig fuctio of sequece,,,,,

44 Theorems of GF Theorem : Let f(x) = a k ad g(x)= b k k x k x. The, ( a k + b k ) x f(x) + g(x)= ad f(x)g(x)= k =0 k =0 k k = 0 j= 0 a j b k j x k k k =0

45 Example A expressio for the geeratig fuctio of the sequece S= {,,,,,,...} is /(- x) /(- x) = + x + x 2 + x 3 + Let f(x) = /(-x)2; f(x) = What are the coefficiets of a k? From theorem, we have ( + x + x 2 +..) x ( + x + x 2 k +..) x =0 k x k =0 =0 k k a k x k /(- x) 2 k = x = k = 0 j= 0 k= k x 0 ( k + ) x k

46 Permutatios & Combiatios Permutatio: A ordered arragemet of objects from a set of distict objects. P(,r): distict objects; r umber of objects i the permutatio P(,r) = (-)(-2) (-r+) P(,r) =!/(-r)! Example: Selectig first 3 prize wiers from 00 differet people 00 x 99 x 98 = 970,200. Combiatio: Uordered selectio of r elemets from a set. C(,r) =!/[r!(-r)!]; is o-egative, 0 r Proof: r-permutatio is ordered r-combiatios.

47 Permutatios & Combiatios Proof: r-permutatio is ordered r-combiatios.: P(,r) = C(,r) x P(r,r) C(,r) = P(,r) / P(r,r) C(,r) = [!/(-r)!]/[r!/(r-r)!] =!/[r!(-r)!] Corollary : C(,r) = C(,-r) if & r are o-egative ad r. Proof: C(,-r) =!/[(-r)!(- (-r))!] =!/[(-r)!r!] Note: C(,) = C(,0) =. Example: Choosig 5 players from 0 member team. C(0,5) = 0!/(5!5!) = 252.

48 Biomial Coefficiets & Theorem C(,r): deoted by r C(,r): Biomial coefficiet these umbers occur as co-efficiets i the expasio of powers of biomial expressios such as (a+b) Biomial Theorem: x, y are variables; oegative iteger. (x + y) = j=0 C(,j).x -j.y j = C(,0)x + C(,)x - y + C(,2)x -2 y 2 +.C(,-)xy - + C(,)y Example: (x + y) 2 = (x+y)(x+y) = xx + xy + xy + yy = x 2 + 2xy + y 2

49 Biomial Coefficiets & Theorem Proof: Cout the umber of terms of form x -j.y j Choose xs from the sums (so that the other terms are ys) Coefficiet of x -j y j is C(,-j) = C(,j) Example : Expasio of (x + y)4 = j=0 4 C(4,j)x 4-j y j = C(4,0)x4 + C(4,)x 3 y + C(4,2)x 2 y 2 + C(4,3)xy 3 + C(4,4)y4 = x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + y 4 Example 2: Coeffieciet of x 2 y 3 i (x+y) 25 = C(25,3) = 25!/(3! 2!) = 5,200,300

50 Biomial Theorem: Corollary Corollary : is o-egative k=0 C(,k) = 2 Proof: 2 = ( + ) = k=0 C(,k) k -k = k=0 C(,k) Basically, k=0 C(,k) is the total umber of subsets of a set with elemets 2

51 Biomial Theorem: Corollary 2 Corollary 2: is positive k=0 (-) k C(,k) = 0 Proof: 0 = 0 = ((-) + ) = k=0 C(,k)(-) k -k = k=0 C(,k)(-) k C(,0) + C(,2) + C(,4) +.. = C(,) + C(,3) + C(,5) +

52 Biomial Theorem: Corollary 3 Corollary 3: is positive k=0 2 k C(,k) = 3 Proof: 3 = (+2) = k=0 C(,k)(2) k -k = k=0 C(,k)2 k

53 Permutatios With Repetitios Chapter 4.5 Theorem : Number of r-permutatios of a set of objects with repetitios allowed is r. Theorem 2: There are C(+r-, r) r-combiatios from a set with elemets whe repetitio of elemets is allowed.

54 Exteded Biomial Coefficiets Defiitio: Let u be a real umber ad k a o-egative iteger. The the exteded biomial coefficiet (EBC) is: ( u ) { = u( u )...( u k + )/ k! k if k>0, if k=0. e.g:ebc(-2,3) = (-2)(-3)(-4) / 3! = -4. EBC(0.5,3)=(0.5)(-0.5)(-.5) / 3! = /6

55 Example Whe the top parameter (u) is egative iteger, the EBC ca be expressed i terms of ordiary biomial coefficiets. Usig earlier defiitio: EBC = (-)(--).(--r+)/r! = (-) r (+) (+r-)/r! = (-) r (+r-)!/(r!*(-)!). Therefore, r + r r = ( ) = ( ) C ( + r, r) r, Z+ r r

56 The Exteded Biomial Theorem Let x be a real umber with x < ad let u be a real umber. The (+x) u = k =0 ( u ) x k k

57 Example Fid GF for (+x) - ad (-x) - By EB Theorem, it follows that (+x) - = k = 0 We kow that ( ) k x k ( ) r = ( ) C( + r, r) r (+x) - = k = 0 ( ) k C ( + k, k ) x k Replacig x with x we get (-x) - = k = 0 C ( + k, k ) x k

58 Coutig Problems ad Geeratig Fuctios Geeratig Fuctios ca be used to solve a wide variety of coutig problems. I particular they ca be used to cout the umber of combiatios of various types.

59 Example Use GF to fid the umber of ways to select r objects of differet kids {a r } if we must select at least oe object of each kid? Chapter 4, theorem. Each of the kids of objects cotributes the factor (x+x 2 +x 3 +) to the GF G(x), Hece G(x)= (x+x 2 +x 3 +) =x (+ x+x 2 +x 3 +) =x /(-x) Usig EB Theorem, we have G(x) = x. (-x) - = x r = 0 ( ) r ( ) = x r x = r 0 C( + r, r) x r G(x)= + ( +, ) r C r r x r=0

60 Cotiue Substitutig t = +r, we get G(x)= Replacig t by r as the idex of summatio, we get G(x)= t = r= C ( t, t ) x C ( r, r ) x t r Hece, there are C(r-,r-) ways to select r objects of differet kids if we must select at least oe object of each kid.

61 Usig GF to Solve Recurrece Relatios We ca fid the solutio to a recurrece relatio ad its iitial coditios by fidig a explicit formula for the associated geeratig fuctio.

62 Example Solve the R.R., a k =3a k- for k=,2,3 ad iitial coditio a 0 =2. a = 3a a = 2. k k 0 k k j 0 k 0 k 0 j 0 k= k= j= 0 Gx () = a+ ax = a+ 3a x = a+ 3x ax = a+ 3 xgx () a Gx a x x 0 k k k k () = = 0 3 = (2 3) 3x k= 0 k= 0 Uses idetity/(-ax) = Cosequetly, a k =2. 3 k =0 k a k x k

63 6.5 The priciple of Iclusio-Exclusio If A, B ad C are fiite sets the A B = A + B A B

64 Cotiue If A, B ad C are fiite sets the A B C = A + B + C - A B - B C - A C - A B C A B A B A C B C C

65 Geeral Iclusio-Exclusio Theorem : Let A,,A be fiite sets. The A A 2 A =... ) (... 2 k j i k j i j i j i i i A A A A A A A A A I I I I I I + + +

66 Geeral Theorem: Proof A A 2 A : Show a elemet i the uio is couted oly oce i the right had side. Let a be a member of exactly r sets (A, ). a is couted C(r,) times by st summatio, C(r,2) by 2 d summatio of the itersectios, ad C(r,m) times by m th summatio ivolvig m sets A i. a is couted exactly C(r,) C(r,2) + C(r,3) - + (- ) r+ C(r,r). By Corollary 2 of Sectio 4.4, C(r,0) - C(r,) + C(r,2) - C(r,3) - + (-) r C(r,r) = 0. C(r,0) = C(r,) C(r,2) + C(r,3) - + (-) r+ C(r,r). a is couted oly oce.

67 6.6 Applicatios of Iclusio-Exclusio Alterative Form: Let A i be the subset cotaiig the elemets that have Property P i. Writig these quatities i terms of sets, we have A i I Ai 2 I... I Aik = N( Pi Pi 2... P ik If the umber of elemets with oe of the properties P,,P is deoted by N(P P 2 P ) the, N(P P 2 P )=N- A A2... A )

68 Cotiue From the iclusioexclusio priciple, we see that N(P P 2 P )=N )... ( ) (... ) ( ) ( 2 k j i k j i j i j i i i P PP N PP P N PP N P N + + +

69 The Number of Oto Fuctios Let m ad be positive itegers with (m ). How may ways to assig m differet jobs to differet employees if every employee is assiged at least oe job? Assume m = 5, = 4, followig Theorem (ext slide): 4 5 C(4,) C(4,2).2 5 C(4,3) 5 = 240

70 The Number of Oto Fuctios Theorem: Let m ad be positive itegers with (m ). The, there are m - C (,) ( -) m + C (,2) ( -2) m + +(-) i C (,i ) (-i ) m + + (-) - C (,-) m oto fuctios from a set with m elemets to a set with elemets

71 Deragemets A deragemet is a permutatio of objects that leaves o object i its origial positio. Theorem: The umber of deragemets of a set with elemets is D =!! + 2! 3! + 4! + L+! ( )

72 Theorem Proof Property P i : A permutatio that fixes elemet i. Number of deragemets: umber of permutatios havig oe of the properties Pi for i =,2,,. D = N(P,P 2,,P ) (from iclusio-exclusio) = N - N ( P ) + N( PP ) -... i i i< j i j N: umber of permutatios of elemets. So, N = P(). Similarly: N(Pi) = P(-); N(Pi,Pj) = P(-2). Isertig these values i the expressio for D, we get the formula.