1 / 41

Chapter 5

Chapter 5. Counting. Contents. Combinatorics ( 組合學 ) The study of arrangement of objects Enumeration: The counting of objects with certain properties i.e., Find the size of the set {x | x satisfies property P} A branch of combinatorics Permutation: Ordered arrangements of objects

adora
Download Presentation

Chapter 5

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 5 Counting

  2. Contents • Combinatorics (組合學) • The study of arrangement of objects • Enumeration: • The counting of objects with certain properties • i.e., Find the size of the set {x | x satisfies property P} • A branch of combinatorics • Permutation: • Ordered arrangements of objects • Combination: • Unordered arrangement of objects • Problems that can be solved by counting techniques: • Allowable passwords on a computer system • Gambling games • Probability problems

  3. 5.1 The basics of counting • Problem: [# allowable passwds] • Properties on a passwd: 1. 6,7 or 8 chars in length 2. alphabet: S ={0,..,9} U {A,B,..,Z} 3. contains at least one digit. ==> How many passwds are there ? • Sol: • #passwd = #passwd of length 6 + ..7 +.. 8 • #passwd of length k = |Sk \ Sk w/o digit | • = |Sk| - |Sk w/o digit| =|S|k - (|S|-10)k • Hence #passwd = (368 - 268) + (367-267)+(366-266).

  4. Basic counting principles: • The sum rule: • Task t1 can be done in n1 ways • task t2 can be done in n2 ways and • t1 and t2 cannot be done at the same time • ==>There are n1+n2 ways to do either task. Ex1: • Box1 : n1 balls ; Box 2: n2 balls, • ==> there are n1 + n2 possible outcomes to take a ball from either box1 or box 2. Ex 2: • 37 teachers; 83 students to be chosen as a representative ==> there are 37+83 possible choices.

  5. Generalized sum rule • Task t1 can be done in n1 ways ... Task tk can be done in nk ways ==> There are (n1+...+nk) ways to do a task among t1,..,tn. Ex2: List A: 23 projects B: 15 projects C: 19 projects ==> There are 23 + 19 + 15=57 projects that can be chosen • Formalization: [in terms of set notation] • A1,..,Ak : k disjoint finite sets • |A1 U A2 ...U Ak | = Si=1,k |Ai|

  6. The product rule • T = (T1,..,Tk) T1 can be done in n1 ways, ... Tk can be done in nk ways ==> there are n1 x n2 x...x nk = Pi=1,k ni ways to do the task. Ex 4: Labeling a chair with a upper-case letter and a number in [1,100]. ==> there are at most 26 x 100 chairs with different labels. Ex 5: 32 computers, each containing 24 ports ==> there are 32 x 24 ways to choose a port. (why production rule ?) Task = (choose a computer, chose a port )

  7. Example product rule Ex 6. # of bit strings of length 7 = ? Ex 8: # functions from {1...,m} to {1,..,n} =? Ex 8': #of partial functions from {1,..m} to {1,...,n} = ? Ex 9: # of 1-1 functions from {1,..,m} to {1,...,n} (m  n) = ? Ex11: Value of k = ? after execution of the code. k := 0; for i1 = 1 to n1 for i2 = 1 to n2 ... for im = 1 to nm {k = k+1 ; print 0}; print k // ==> k = Ps=1,..m ns = # of 0 printed out.

  8. Ex 12. use product rule to prove |P(S)| = 2|S| sol: S = {e1,e2,...,en}, at the ith step choose or not choose ei ==> 2x2...x2 = 2n possible choices, each determining a unique subset. Formalization: - |A1xA2x...xAn| = |A1|x...x|An|. - ie. = |{(x1,...,xn) | xiAi } |

  9. More complicated counting problems • require both rules • Ex13: identifier(name of variable) requirement: • a string of one or two alphanumeric chars • lower and upper cases treated as identical • beginning with a letter • different from 5 reserved words: {if, Do, for ,..} • ==> # different identifiers available = ? • sol: • #id = #id of size 1 + #id of size 2 - 5 --- sum rule • #id1 = 26 • #id2 = 26 x (10+26) --- product rule • #id = 26 + 26x36 - 5 = 957.

  10. The inclusion-Exclusin principle: Ex15:#bit strings of length 8 beginning with 1 or ending with 0 = ? sol: #bitString-b1 = 27; #bitString-e0 = 27; #bitString-b1&-e0 = 26 => #bitString-b1|e0 = 27 + 27 - 26. • Set formalization: |A U B | = |A| + |B| - |A Ç B| Analogy: Box 1: n balls of diff colors Box 2: m balls of diff colors k: #colors appearing in both Box 1 and Box 2. ==> #possible colors for a ball chosen = ?

  11. 0 0 1 1 0 0 0 1 1 0 0 1 1 Irregular cases: tree diagram. • #bit strings of length 4 w/o two consecutive 1s.

  12. 5.2 The pigeonhole principle • Theorem: If k+1 objects are put into k boxes, then there is at least one box containing more than one objects. pf: Let xi be #objects box i contains. If xi £ 1 for all i => k+1 = Si=1..k xi£ k, a contradiction!. Ex1: At least two persons have the same birthday in 367 people. Analogy: people =>object, date => box. Ex2: 27 words => two beginning with the same letter. Ex3: exam. scale: 0~100 => min # of students to guarantee occurrence of receiving same scores on students = ?

  13. Generalized pigeonhole principle • Theorem 2: N objects placed on k boxes => some box contains at least [N/k] (ceiling) objects. Pf: let Xi = #objects box i contains if xi < [N/k] for all i => S xi k([N/k]-1) < K (N/k +1 -1) = N, impossible!. Ex4: 100 people => there are at least [100/12] = 9 born in the same month. Ex5: min # students in a course to guarantee at least 6 receive the same grade, if there are 5 possible grades. sol: find min N s.t. [N/5] = 6.

  14. Elegant applications Ex7: Baseball team: 1. play 1 game at least a day. 2. play no more than 45 games a month (30 days) => $ an interval [n, n+m] during which the team play exactly 14 games. sol: Let ak = # games played in [1,k] => 1  a1 < a2< ... < a30 45 and 15  a1+14 < a2+14 < ... <a30+14  59. All 60 terms: a1,..,a30,a1+14,..,a30+14 are among [1, 59] => $ term aj and ai+14 s.t., aj = ai + 14. This means 14 games were played in [i+1,j]

  15. More elegant application Ex8; a1,..., an+1: n+1 positive integers £ 2n. => $ i,j s.t., ai | aj. sol: Write each ak as 2 pkx qk (Pk 0; qk: odd) ak odd => ak = 20 x ak ak even = 1101000 => pk = 3; qk = 13. => there are n+1 items: q1,..,qk+1, each odd and £ 2n. => $ i < j s.t. qi = qj = q. => ai = 2pi q; aj = 2pj q => aj = ai 2 pj-pi => ai | aj.

  16. more applcations Def: a1,..,an: a sequence of numbers. aj1,aj2,...,ajm is a subsequence iff 1£ j1<j2<..< jm£ n. Theorem 3: Every seq of n2 + 1distinct numbers contains a subseq of n+1 numbers that is either strictly increasing or decreasing. Example: 8,11,9,1,4,6,12,10,5,7 : 32+ 1 numbers. => $ incr. or decr. subseq of length 4: 1 4 6 12, and more... Pf: a1,a2,... an2+1 :any seq of distinct n2+ 1 numbers. Ik: the length of the longest incr. subseq beginning from ak. Dk: the length of the longest decr. subseq beginning from ak. Note: for 1 £ i,j £ n2 + 1, we have i=j iff (Ii,Di) = (Ij,Dj). Now suppose no incr. nor decr. subseq of length n+1. => each 1£ Di,Ii£ n for i = 1,…, n2+1. => only n2 distinct (D,I)’s => $ 1£s < t£ n2+1 s.t. (Ds,Is) = (Dt, It) => contradiction!.

  17. Ramsey theory Ex 10: A group of 6 people. Each pair are either friends or enemies. => $ three mutual friends or mutual enemies. Pf: Let A : anyone of the group. => among other 5: at least 3 are all A's friend --- By PigeonHole Prinp.--- (1) (say,B,C,D) or A's enemies --- (2) case 1: if $ a friend pair (say, BC) => A,B,C are mutual friends. o/w: B,C,D are mutual enemies. case 2: Similar to case 1.

  18. 5.3 permutations and combinations • [Permutation] S={x1,...,xn } : a set with n elements. 1. Any ordered listing (arrangement) of all elements of S: ie. a1,...,an s.t. {a1,..,an} = S is called a permutation of S. 2. Any ordered listings of r distinct elements of S: a1,...,ar is called an r-permutation of S. Ex: S= {1,2,3} ==> 3,1,2 is a perm of S; 1,3 is a 2-perm of S.

  19. # of r-permutations Theorem 1: #r-perm in a set with n elements (r  n ) is P(n,r) = n (n-1)...(n-r+1) = n!/(n-r)! pf: a1,..ar: any r-perm. ==> n ways to choose a1; n-1 ways to choose a2, ... n-r+1 ways to choose ar ==> P(n,r) possible choices. Ex3: 8 runners compete for 3 medals. #ways to award medals = ? Sol: gold medal --- 8 ways Silver ---> 7 ways; bronze --> 6 ways ==> P(8,3) = 8x7x6. Ex4: Visiting 8 cities beginning from a specific city. P(7,7) = 7 ! = 5040 ways.

  20. Combinations • S = {X1,...,Xn }: a set . An r-combination of S is an unordered selection of r elements from S. (i.e., any subset of S of length r) Ex: S= {1,2,3,4} ==> {1,3,4} = {4,3,1}, {1,2,3} are 3-comb. of S. Theorem 2: |S|= n, r £ n => # r-combinations of S = C(n,r) = ? Sol: Since P(n,r) =C(n,r) (r!), ----- to find a perm, first choose r elements from S and then permute them. C(n,r) = n! / r!(n-r)! Ex12: #ways to choose 5 among 10 persons to attend a game. Sol: C(10,5) Corollary 2: C(n,r) = C(n,n-r). -- a subset with r elements present = a subset with n-r elements not present. Ex14: #bit strings of length n containing exactly r 1's = C(n, r). = #bit stings of length n containing (n-r) 0's = C(n, n-r)

  21. 4.5 Binominal Coefficients • C(n,r) is called a binominal coefficient since it is the coefficient in the expansion of binominal expression (x+y)n. • Theorem 1: (x+y)n = Sk=0,n C(n,k)xkyn-k. Pf: (x+y)n = (x1+y1)(x2+y2)...(xn+yn). Each xkyn-k corresponds to one k-combination in {x1,..,xn}. There are C(n,k) k-comb, so the coeff of xkyn-k is C(n,k). Ex4: What is the coeff of x12y13 in (2x-3y)25 ?

  22. Corollaries of the binominal theorem Corollary 1: Sk=0,n C(n,k) = 2 n pf: 2 n = (1+1)n = Sk=0,nC(n,k)1k1n-k =Sk=0,nC(n,k). Corollary 2: Sk=0,n (-1)k C(n,k)= 0 pf: 0 = (-1 +1)n = Sk=0,n C(n,k)(-1)k 1 n-k = 0 Corollary 2' : (n,0) + (n,2) + … (n, 2n/2 ) = (n,1) + (n,3) + …(n, 2 (n-1)/2 +1). Corollary 3: Sk=0,n C(n,k)2k= 3n. Pf: Show it yourself.

  23. Pascal's identity Theorem 2: [Pascal's identity]: C(n+1,k) = C(n,k) + C(n, k-1). Def1: A combinatorial proof of an identity s = t is a proof that uses counting arguments to show that both sides of the identity correspond to different ways of counting the same set of objects. Pf: 1. by def. (omitted) 2. Combinatorial proof: Let S: any set of n+1 elements; a: any element of S. Then C(n+1, k) = # k-comb of S = #k-comb of S containing a + #k-comb of S w/o a = C(n, k-1) + C(n, k). Pascal's Triangle: (0,0) 1 (1,0) (1,1) 1 1 (2,0) (2,1) (2,2) 1 2 1 (3,0)(3,1)(3,2)(3,3) 1 3 3 1 ---- ------------------------ -- ----- -- --- -- - --

  24. Binomial coefficients(cont'd) Corollary 1: Sk=0,n C(n,k) = 2n. Pf: combinatorial proof: S: any set with n elements. => #subsets of S = 2n = Sk=0,n (#subsets of S with k elements) = Sk=0,n C(n,k) Theorem 3: Vandermonde's identity: [generalization of Thm 2] m,n ³ 0 ; r £ m + n. ==> C(m+n, r) = St=0,r C(m,t) C(n,r-t) Pf: Let |S| = m, |T| = n and S and T are disjoint. ==> C(m+n,r) = #r-comb. of SUT. = St=0,r (#t-combinations of S x #(r-t)-comb of T) = St=0,r C(m,t) C(n,r-t) Note: for Theorem 2: C(1+n,k) = C(1,0)C(n,k) +C(1,1)C(n,k-1) +… = C(n,k)+C(n,k-1) where (m,n,r) = (1,n, k).

  25. Corollaies of Theorem 3: Corollary 4: C(2n,n) = Sk=0,n C(n,k)2. Pf: C(2n,n) = C(n+n, n) = Sk=0,n C(n, k) C(n,n-k) = Sk=0,n C(n,k) C(n,k) Corollary 5: C(n+1,r+1) = Sk=r,n C(k,r). pf: By combinatorial argument. C(n+1, r+1) = #of bit strings of length (n+1) containing r+1 1's. = Sk'=r+1,n+1 #bit strings of length (n+1) with the last 1 appearing at position k'. = Sk'=r+1,n+1 C(k'-1, r) = C(r,r) + C(r+1,r) + … + C(n,r) = Sk=r,n C(k, r)

  26. 5.5 generalized permutations and combinations • Permutations with repetition: Def: S = {x1,..,Xn}: a set. An r-permutation of S with repetition is any seq a1,..,ar s.t. {a1,..,ar} Í S. Theorem 1: #r-perms in a set with n elements = nr. Ex1: #strings of length n in the English alphabet = 26n.

  27. Multisets • A multiset (bag) is a set allowing multiple occurrences of its elements. Ex: S={1,1,2,3,4,4,4,5} is a multiset in which 1 occurs twice,4 occurs 3 times. Note: S ¹ {1,2,3,4}. Properties of multisets: A1. [Membership:] For each multiset and each object x: #(S,x) =def number of occurrences of x in S. ==> 1. #(S,x) is called the multiplicity of x in S. 2. x is a memebr of S iff #(S,x) > 0. 3. Every multiset is uniquely identified by its membership function. A2. Empty multiset {}: for all object x, #({},x) = 0.

  28. Properties of multisets A3: Cardinality: • |S| = # of occurrences of elements of S • = Sx  U #(S, x). • For simplicity, a finite multiset S can be represented as {X1a1, X2a2,...,Xnan}, where X1,..,Xn are all objects of U s.t. #(S,Xj) =aj > 0. A4: [inclusion and equality] • AÍ B iff for all x, #(A,x) £ #(B,x). • A = B iff AÍ B and BÍ A iff #(A,x) = #(B,x) for all x. • Ex: {1,2, 3} Í {1,1, 2,3 }. Facts: A = B => |A| = |B|; A Í B => |A| £ |B|.

  29. Multiset operations A4: operations: • A U B = S with #(S,x) = max(#(A,x), #(B,x)) • A Ç B = S with #(S,x) = min(#(A,x), #(B,x)) • A \ B = S with #(S,x) = #(A,x) - #(AÇB,x) • Note: complement is not defined in multisets. • Ex: A = {1,2,2,2,3,5,5}; B = {3,4,4,1,1} • ==> A U B = ? A Ç B = ? A \ B = ? A5: Bag space: • D: an (ordinary) set domain • Dn =def the set of all multisets whose elements are in D s.t. no elements occur more than n times. • (i.e., S in Dn iff #(S,x) £ n for all objects x.) • Note: If bag space is given => #(~S, x) = n - #(S,x). • If n = 1 => all multisets become ordinary sets.

  30. Combinations with repetition Def: [r-combination with repetition] • An r-combination with repetition of a set S is a multiset of size r with elements got from S. • i.e., any multiset of size r from the bag space Sr. Theorem 2:#r-combinations with repetition from a set with n objects = #ways to choose r objects from n kinds of objects H(n,r) = C(n+r-1,r) --- = C(n+r-1, n-1) Pf: 1. ex: #ways to choose 4 elements from {1,2,3}. ==> 1 2 2 3 <==> 0 | 0 0 | 0 000|0| <==> 111 2 there are H(3,4) = C(6,4) ways to put two bars on 6 possible positions. Hence the solu is C(6,2) = 15. In general the problem is equ. to #ways to put n-1 bars on r+n-1 positions = C(n+r-1, n-1) = C(n+r-1, r)

  31. More precise proof of Theorem 2 (skipped): Pf: Let S = {1,2,..., n}, T = {1,2,..n, .., n+r-1} TT : the set of all (n-1)-combinations of T (i.e., subsets of T with n-1 elements ). SS: the set of all multi subsets of Sr with r elements. note: |SS| = #r-combinations with repetitions from S. Define f: TT -> SS s.t. if ordered listing of X = x1,..,xn-1, then #(f(X), k) = xk - xk-1 -1 (where x0 = 0 and x n = n+r.) Fact: f is 1-1 and onto. Hence |SS| = |TT| = C(n+r-1, r). 1. onto: for any element Y = {1a1,2a2,...,nan} of SS let x1 = 1+a1, x2 = x1 +1+ a2,.., xn-1 = xn-2 + 1+ an-1 ==> X = {x1,..,xn-1} is a (n-1)-comb and f(X) = Y. 2. 1-1: If X = {x1,..,xn-1} ¹ Y = {y1,..,yn-1} => let (xj, yj) be the first of all pairs at which X and Y disagree => #(f(X), j) =xj - x j-1 -1¹ yj - y j-1 -1 = #f(f(Y), j) ==> f(X) ¹ f(Y)

  32. Applications Ex6:#solutions in x1 + x2 + x3 = 11, where x1,x2,x3 are non-negative integer. Sol: Each solu. corresponds to a way of selecting 11 items from {x1,x2,x3}. (i.e., {x1a1 x2a2 x3a3 } with a1 + a2 + a3 = 11 ). ==> #solu = H(3,11) = C(11+3-1, 2) =13 12 /2 = 78. Ex6': The same as Ex6 but with constraints: x1³ 1, x2³ 2, x3³ 3. solu: Let y1 = x1 -1, y2 = x2 -2, y3 = x3 -3. y1 + y2 + y3 = x1 + x2 + x3 - 6 = 11 -6 = 5 --- (1) with constraints: y1,y2,y3 ³ 0 ---(2) #solu of (x1,x2,x3) = #solu(y1,y2,y3) in (1,2) = H(3,5) = C(5+3-1, 2) = 21.

  33. Examples Ex7: What is the value of k after execution ? k := 0; for i1 := 1 to n for i2 := 1 to i1 .... for im := 1 to im-1 k := k+1. sol: k is increased by 1 for each iteration (i1,i2,..,im) with 1£ im£ im-1 ...i2£ i1 £ n . k = the size of the set {(i1,..,im) | 1£ im£ im-1 ...i2£ i1 £ n } = H(n,m) = C(n+m-1, m). [ = #m-comb with rep. from {1,..,n} ] Ex7’ : What is the # of increasing (and strictly increasing) integer functions f: {1,…,m}  {1,..,n} ? Sol: H(n,m) = C(m+n-1, m) and C(n,m), resprctively. Ex: if m = 4 and select {2, 32, 5}  (I4,I3,I2,I1) = (2, 3,3 5)

  34. Permutations of sets with indistinguished objects. • Permutations of multisets: • S = {x1a1,...,xnan } with S ai = m. • ==> #permutations of S = m!/(a1! a2!... an!) • ==> #r-permutations of S = ? (0 < r < m+1) Ex 8: #ways to reorder the letters of the word: "success" sol: S = {s3,c2,e,u} => #perm = 7! / 3!2! pf: a1 objects of type x1 can be placed among the m positions in c(m,a1) ways. a2 objects of type x2 ... m-a1 pos .. in c(m-a1,a2) ways, ... an objects of type xn can be placed in C(an,an) ways. ==> #perm = C(m,a1)xC(m-a1,a2 )x …xC(an+an-1, an-1)xC(an,an) = m! / Pai!

  35. Examples Theorem 4: #ways to distribute n distinguished objects into k boxes so that box i contains ni objects = n! /n1!...nk!. • Same as #way to permute S={Box1n1,Box2n2,…,Boxknk} in n positions pf: C(n,n1) ways to put n1 objects into box 1 C(n-n1,n2) ways to put n2 objects into box 2,... C(nk, nk) ways to put nk objects into box k ==> totally n! /n1! ...nk! ways Ex9: 52 cards to 4 players, each with 5 cards. • Same as #ways to permute {E5,S5,W5,N5, Rest32} ==> C(52,5) x C(52-5,5) xC(52-10,5) x C(52-15,5). c.f: #ways to distribute n indistinguishable objects into k distinguished boxes = H(k, n).

  36. 5.6 Generating permutations & combinations • Sometimes permutations and combinations need to be generated, instead of just being counted! • Ex1: [salesman problems] • visiting 6 cities without duplication with minimal instances. • ==> try all possible permutations of the cities and find one with minimal instances. • Ex2: S={x1,...,x6}, find a subset {y1,..,yj} of S s.t. Syk = 100. • sol: find each subset (combination) T of S and check their sum.

  37. Lexicographical ordering of permutations • S={1,2,...,n} • a = a1,a2,..,as ; • b = b1,..,bt : two permutations of S [, possibly with repetitions ]. • Define a < b iff either • 1. s < t and a is a prefix of b (i.e. b = a,b(s+1) ,.. bt) or • 2. $ 1 £ k £ min(s,t) s.t. ak < bk and aj = bj for all j < k. Ex: a = 1 3 5 7 2 2 • b = 1 3 5 2 4 6 7 8 • c = 1 3 5 ==> what are their lexico. order ? • show that the lexico order < is a total order. • 1. a < b and b < c => a < c (transitive) • 2.  a ~ a < a (irreflexive) • 3. ab,either a < b or b < a or a = b. (comparable) • (T, <): a total order ; a, b : two elements of T • [a,b] = {x in T | a £ x £ b}; successor(a) = b iff a < b and for all x > a, b £ x.

  38. Generating permutation • Problem: S = {1,..,n} : a set. • a = a1,...,an : a permutation w/o repetition of S. • S(a) = the next permutation following a = ? Ex: S = [1,6] ==> 2 3 4 1 5 6 --- not max in {5,6} --> 2341 65 2 3 4 1 6 5 --- 65 max in {6,5}; 165 not max in {1,5,6} --> 2 3 4 516 2 3 4 6 5 1 ---> 2 3 5 1 4 6 Ex2: s( 3 6 2 5 4 1) = 3 6 4 1 2 5. Rules: a = a1 a2 ... an be a permutation of S= {1,..,n}. k: the largest index s.t. ak < ak+1>,...> an is in decreasing order. If k = 0 ==> a is the max permutation. o/w. Let b = min{x | x > ak /\ x in {ak+1,..,an}} and bk+1,.., bn be the least perm. of {ak,..an} \{b} ==> s(a) = a1 ... a k-1 b, bk+1 ... bn.

  39. Finding next permutation • procedure s(a) // find next perm following a=a1,..,an. // 1. find largest j s.t. aj < a j+1 j := n-1; a0 := 0; while aj > aj+1 do j := j - 1; if j = 0 return(no successor); // 2. Find smallest ak > aj among { a j+1 > a j+2> .. > an } k := n; while aj > ak do k := k-1; //3. exchange aj, ak aj <-> ak; //4. sorting aj+1 ...an sorting(aj+1,...,an); return(a1,..,an)

  40. generating combinations • S={x1,..,xn} • Each combination(i.e., subset) T of S can be represented by a bit strings t=b1 b2... bn with the correspondence that • xi T iff bi = 1. • next combination of T = combination with string coding next to that of T. • Ex: 10100100111 ---> 10100101000 (i.e. add one). • Procedure next(t) // t = b1b2,,,bn • { i := n; • while bi =1 do • {bi := 0; i := i-1} • bi := 1 }.

  41. r-combinations of a set • S={1,...,n}; T1, T2: two r-combinations of S ( r  n) • Define T1 < T2 iff increasing sorting listing of T1 < that of T2 in lexicographical order. Eg: T1={1,2,6}, T2={3, 5, 2} be two 3-combinations => T1 < T2 since 1 2 6 < 2 3 5 • Problem: how to find the next largest r-combination of any given r-combination? ex: S = {1,...,6}, T = {1,2,5,6} => next-4-com(T) = ? 56 : the largest 2-com, 256: not the largest 3-com ==> replace 256 by 345, which is the next 3-comb > 256. ==> next(T) = 1345 rule: T= {a1,...,ar}  {n-r+1,...,n}, which is the largest r-combination => let j be the largest index s.t. aj ≠ n-r +j. => next(T)={a1,..,aj-1, aj+1, aj+2,...aj+(r-j+1)}. Ex: S={1,...,9}; T={1 3 5 8 9} => next-5-comb(T) = {1,3,6,7,8}

More Related