Generating Elementary Combinatorial Objects
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- Carmella Cannon
- 9 years ago
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1 Fall 2009
2 Combinatorial Generation Combinatorial Generation: an old subject Excerpt from: D. Knuth, History of Combinatorial Generation, in pre-fascicle 4B, The Art of Computer Programming Vol 4.
3 Combinatorial Generation Combinatorial Generation We are going to look at combinatorial generation of: Subsets k-subsets Permutations To do a sequential generation, we need to impose some order on the set of objects we are generating. Many types of ordering are possible; we will discuss two types: lexicographical ordering and minimal change ordering.
4 Combinatorial Generation Combinatorial Generation (cont d) Let S be a finite set and N = S. A rank function is a bijection rank: S {0, 1,..., N 1}. It has another bijection associated with it unrank: {0, 1,..., N 1} S. A rank function defines an ordering on S. Once an ordering is chosen, we can talk about the following types of algorithms: Successor: given an object, return its successor. Rank: given an object S S, return rank(s) Unrank: given a rank i {0, 1,..., N 1}, return unrank(n), its corresponding object.
5 Generating Subsets: Lexicographical Ordering Generating Subsets (of an n-set): Lexicographical Ordering Represent a subset of an n-set by its characteristic vector: subset X of {1,2,3} characteristic vector {1,2} [1,1,0] {3} [0,0,1] Definition The characteristic vector of a subset T X is a vector X (T ) = [x n 1, x n 2,..., x 1, x 0 ] where { 1, if n i T x i = 0, otherwise.
![X of {1,2,3} characteristic vector {1,2} [1,1,0] {3} [0,0,1] Definition The characteristic](/docs-images/49/16971694/images/page_5.jpg "vector of a subset T X is a vector X (T ) = [x n 1, x n 2,.")
6 Generating Subsets: Lexicographical Ordering Example: lexicographical ordering of subsets of a 3-set lexico rank X (T ) = [x 2, x 1, x 0 ] T 0 [0, 0, 0] 1 [0, 0, 1] {3} 2 [0, 1, 0] {2} 3 [0, 1, 1] {2, 3} 4 [1, 0, 0] {1} 5 [1, 0, 1] {1, 3} 6 [1, 1, 0] {1, 2} 7 [1, 1, 1] {1, 2, 3} Note that the order is lexicographical on X (T ) and not on T. Note that X (T ) corresponds to the binary representation of rank!
![4 [1, 0, 0] {1} 5 [1, 0, 1] {1, 3} 6 [1, 1, 0] {1, 2} 7 [1, 1, 1] {1, 2, 3} Note that the order is](/docs-images/49/16971694/images/page_6.jpg "lexicographical on X (T ) and not on T.")
7 Generating Subsets: Lexicographical Ordering Ranking More efficient implementation: SubsetLexRank (n, T ) r 0; for i 1 to n do r 2 r; if (i T ) then r r + 1; return r; Books version: if (i T ) then r r + 2 n i This is like a conversion from the binary representation to the number.
8 Generating Subsets: Lexicographical Ordering Unranking SubsetLexUnrank (n, r) T ; for i n downto 1 do if (r mod 2 = 1) then T T {i}; r r 2 ; return T ; This is like a conversion from number to its binary representation.
9 Generating Subsets: Lexicographical Ordering Successor The following algorithm is adapted for circular ranking, that is, the successor of the largest ranked object is the object of rank 0. SubsetLexSuccessor (n, T ) i 0; while (i n 1) and (n i T ) do T T \ {n i}; i i + 1; if (i n 1) then T T {n i}; return T ; This algorithm works like an increment on a binary number.
10 Generating Subsets: Lexicographical Ordering Examples: successor of a subset in lexicographical ordering 1 SubsetLexSuccessor(3, {2, 3}) = {1}. {2, 3} [0, 1, 1] {1} [1, 0, 0] 2 SubsetLexSuccessor(4, {1, 4}) = {1, 3}. {1, 4} [1, 0, 0, 1] {1, 3} [1, 0, 1, 0]
11 Generating Subsets (of an n-set) : Minimal Change Ordering Generating Subsets (of an n-set) : Minimal Change Ordering In minimal change ordering, successive sets are as similar as possible. The hamming distance between two vectors is defined as the number of bits in which the two vectors differ. Example: dist( , ) = 2. When we apply to the subsets corresponding to the binary vectors, it is equivalent to: dist(t 1, T 2 ) = T 1 T 2 = (T 1 \ T 2 ) (T 2 \ T 1 ). A Gray Code is a sequence of vectors with successive vectors having hamming distance exactly 1. Example: [00, 01, 11, 10]. We will now see a construction for one possible type of Gray Codes...
12 Generating Subsets (of an n-set) : Minimal Change Ordering Construction for Binary Reflected Gray Codes n=1 n= n=
13 Generating Subsets (of an n-set) : Minimal Change Ordering In general, build G n as follows: G n 0 G 0 n G n-1 n-1 G n-1 n-1 n G 0 1 More formally, we define G n inductively as follows: Theorem (2.1) G 1 = [0, 1] G n = [0G n 1 0,, 0G n 1 2 n 1 1, 1Gn 1 2 n 1 1, 1Gn 1 0 ] For any n 1, G n is a gray code. Exercise: prove this theorem by induction on n.
14 Generating Subsets (of an n-set) : Minimal Change Ordering Successor Examples: G 3 = [000, 001, 011, 010, 110, 111, 101, 100] G 4 = [0000, 0001, 0011, 0010, 0110, 0111, 0101, 0100, Rules for calculating successor: 1100, 1101, 1111, 1110, 1010, 1011, 1001, 1000]. If vector has even weight (even number of 1 s): flip last bit. If vector has odd weight (odd number of 1 s): from right to left, flip bit after the first 1.
![successor: 1100, 1101, 1111, 1110, 1010, 1011, 1001, 1000].](/docs-images/49/16971694/images/page_14.jpg "If vector has even weight (even number of 1 s): flip last bit.")
15 Generating Subsets (of an n-set) : Minimal Change Ordering GrayCodeSuccessor (n, T ) if ( T is even) then U T {n}; (flip last bit) else j n; while (j T ) and (j > 0) do j j 1; if (j = 1) then U ; (I changed for circular order) else U T {j 1}; return U;
16 Generating Subsets (of an n-set) : Minimal Change Ordering Ranking and Unranking r b 3 b 2 b 1 b 0 bin.rep. r a 2 a 1 a 0 G 3 r Set b 3 = 0 in the example above. We need to relate (b n b n 1... b 0 ) and (a n 1 a n 2,... a 0 ). Lemma (Lemma 1.) Let P (n): For 0 r 2 n 1, a j b j + b j+1 (mod 2), for all 0 j n 1. Then, P (n) holds for all n 1.
17 Generating Subsets (of an n-set) : Minimal Change Ordering Lemma (Lemma 1.) Let P (n): For 0 r 2 n 1, a j b j + b j+1 (mod 2), for all 0 j n 1. Then, P (n) holds for all n 1. Proof: We will prove P (n) by induction on n. Basis: P (1) holds, since a 0 = b 0 and b 1 = 0. Induction step: Assume P (n 1) holds. We will prove P (n) holds. Case 1. r 2 n 1 1 (first half of G n ). Note that b n 1 = 0 = a n 1 and b n = 0, which implies By induction, a n 1 = 0 = b n 1 + b n. (1) a j b j + b j+1 (mod 2), for all0 j n 2. (2) Equations (1) and (2) imply P (n).
18 Generating Subsets (of an n-set) : Minimal Change Ordering Proof of Lemma 1 (cont d) Case 2. 2 n r 2 n 1 (second half of G n ). Note that b n 1 = 1 = a n 1 and b n = 0, which implies a n 1 1 b n 1 + b n (mod 2). (3) Now, G n r = 1G n 1 2 n 1 r = 1a n 2a n 3... a 1 a 0. The binary representation of 2 n 1 r is 0(1 b n 2 )(1 b n 3 )... (1 b 1 )(1 b 0 ). By induction hypothesis we know that, for all 0 j n 2, a j (1 b j ) + (1 b j+1 mod 24) Equations (3) and (5) imply P (n). b j + b j+1 (mod 25)
19 Generating Subsets (of an n-set) : Minimal Change Ordering Lemma (Lemma 2.) Let n 1, 0 r 2 n 1. Then, n 1 b j a i (mod 2), for all 0 j n 1. i=j Proof: n 1 a i i=j n 1 b i + b i+1 (mod 2) [By Lemma 1] i=j b j + 2b j b n 1 + b n (mod 2) b j + b n (mod 2) b j (mod 2) [Since b n = 0].
![i=j Proof: n 1 a i i=j n 1 b i + b i+1 (mod 2) [By Lemma 1] i=j b j + 2b](/docs-images/49/16971694/images/page_19.jpg "j+1 +.")
20 Generating Subsets (of an n-set) : Minimal Change Ordering Let n 1, 0 r 2 n 1. We haved proved the following properties hold, for all 0 j n 1, n 1 b j a i (mod 2). i=j a j b j + b j+1 (mod 2), The first property is used for ranking: GrayCodeRank (n, T ) r 0; b 0; for i n 1 downto 0 do if ((n i) T ) then ( if a i = 1) b 1 b; ( b i = b i+1 ) r 2r + b; return r;
21 Generating Subsets (of an n-set) : Minimal Change Ordering The second property is used for unranking: GrayCodeUnrank (n, r) T ; b r mod 2; r r 2 ; for i 0 to n 1 do b r mod 2 if (b b ) then T T {n i}; b b; r r 2 ; return T ;
22 Generating k-subsets: Lexicographical Ordering Generating k-subsets (of an n-set): Lexicographical Ordering Example: k = 3, n = 5. rank T T 0 {1, 2, 3} [1, 2, 3] 1 {1, 2, 4} [1, 2, 4] 2 {1, 2, 5} [1, 2, 5] 3 {1, 3, 4} [1, 3, 4] 4 {1, 3, 5} [1, 3, 5] 5 {1, 4, 5} [1, 4, 5] 6 {2, 3, 4} [2, 3, 4] 7 {2, 3, 5} [2, 3, 5] 8 {2, 4, 5} [2, 4, 5] 9 {3, 4, 5} [3, 4, 5]
23 Generating k-subsets: Lexicographical Ordering Successor Example/idea: n = 10, Successor({..., 5, 8, 9, 10})={..., 6, 7, 8, 9} ksubsetlexsuccessor ( T, k, n) U T ; i k; while (i 0) and (t i = n k + i) do i i 1; if (i = 0) then U = [1, 2,..., k]; else for j i to k do u j (t i + 1) + j i; return U;
24 Generating k-subsets: Lexicographical Ordering Ranking How many subsets preceed T = [t 1, t 2,..., t k ]? all sets [X,...] with 1 X t 1 1 all sets [t 1, X,...] with t X t 2 1. all sets [t 1,..., t k 1, X,...] with t k X t k 1 Thus, rank(t ) = k i=1 t i 1 j=t i 1 +1 ( ) n j. k i ( t 1 1 ( n j j=1 k 1) ) ( t 2 1 ( n j j=t 1 +1 k 2) ) ( t k 1 j=t k 1 +1 ( n j k (k 1)) )
25 Generating k-subsets: Lexicographical Ordering ksubsetlexrank ( T, k, n) r 0; t 0 0; for i 1 to k do for j t i to t i 1 do r r + ( n j k i) ; return r;
26 Generating k-subsets: Lexicographical Ordering Unranking t 1 = x x 1 j=1 t 2 = x x 1 etc. ( n j j=t 1 +1 ( n j k 1 k 1) r < x ) j=1 r t1 1 ( n j k 2 ksubsetlexunrank (r, k, n) x 1; for i 1 to k do while (r ( n x k i) ) do r r ( n x k i) ; x x + 1; t i x; x x + 1; return T ; j=1 ) ( n j k 1) < x j=t 1 +1 ( n j ) k 1
27 Generating k-subsets (of an n-set): Minimal Change Ordering Generating k-subsets : Minimal Change Ordering The minimum Hamming distance possible between k-subsets is 2. Revolving Door Ordering It is based on Pascal s Identity: ( n k = n 1 + n 1 k k 1). We define the sequence of k-subsets A n,k based on A n 1,k and the reverse of A n 1,k 1, as follows: [ ] A n,k = A n 1,k 0,..., A n 1,k {n},..., An 1,k 1 0 {n}, A n,0 = [ ] for 1 k n 1 A n,n = [{1, 2,..., n}] (, n 1 k ) 1 An 1,k 1 ( k 1) 1
28 Generating k-subsets (of an n-set): Minimal Change Ordering Example: Bulding A 5,3 using A 4,3 and A 4,2 A 4,3 = [{1, 2, 3}, {1, 3, 4}, {2, 3, 4}, {1, 2, 4}] A 4,2 = [{1, 2}, {2, 3}, {1, 3}, {3, 4}, {2, 4}, {1, 4}] A 5,3 = [{1, 2, 3}, {1, 3, 4}, {2, 3, 4}, {1, 2, 4}, {1, 4, 5}, {2, 4, 5}, {3, 4, 5}, {1, 3, 5}, {2, 3, 5}, {1, 2, 5}]
29 Generating k-subsets (of an n-set): Minimal Change Ordering To see that the revolving door ordering is a minimal change ordering, prove: 1 A n,k = {1, 2,..., k 1, n}. ( n k) 1 2 A n,k 0 = {1, 2,..., k}. 3 For any n, k, 1 k n, A n,k is a minimal ordering of S n k.
30 Generating k-subsets (of an n-set): Minimal Change Ordering Ranking The ranking algorithm is based on the following fact (prove it as an exercise): rank(t ) = k i=1 ( 1) k i (( ti i ) ) { k i=1 1 = ( 1)k i( t ii ) [, k even k i=1 ( 1)k i( t ii ) ] 1, k odd Hint: Prove the first equality by induction and the second, directly. KsubsetRevDoorRank( T, k) r (k mod 2); s 1; for i k downto 1 do r r + s ( t ii ) s s; return r;
31 Generating k-subsets (of an n-set): Minimal Change Ordering Ranking Algorithm Example 1 rank(136) = r =? Look at ( ( 6 3) = +20, then 3 ( 2) = 3, then 1 1) = +1 then ( ( 3 2) = ) = 1 1 r = ( 6 3) = +20
32 Generating k-subsets (of an n-set): Minimal Change Ordering Ranking Algorithm Example 2 rank(245) = r =? Look at ( ( 5 3) = +10, then 4 ( 2) = 6, then 2 1) = +2 then ( 4 2) = ( 2 1) = ( 5 3) =
33 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking IDEA/ Example: n = 7, k = 4, r = 8 4 T, 5, 6, 7 T 5 T, 6, 7 T 6 T, 7 T 7 T ( 4 4 = = = 15 7 ) 4 = 21
34 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking IDEA/ Example: n = 7, k = 4, r = 8 4 T, 5, 6, 7 T 5 T, 6, 7 T 6 T, 7 T 7 T ( 4 4 = = = 15 7 ) 4 = 21 We can determine the largest element in the set: r = 8 implies {,,, 6}.
35 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking IDEA/ Example: n = 7, k = 4, r = 8 4 T, 5, 6, 7 T 5 T, 6, 7 T 6 T, 7 T 7 T ( 4 4 = = = 15 7 ) 4 = 21 We can determine the largest element in the set: r = 8 implies {,,, 6}. Now, solve it recursively for n = 5, k = 3, r = ( 6 4) r 1 = 6.
36 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking IDEA/ Example: n = 7, k = 4, r = 8 4 T, 5, 6, 7 T 5 T, 6, 7 T 6 T, 7 T 7 T ( 4 4 = = = 15 7 ) 4 = 21 We can determine the largest element in the set: r = 8 implies {,,, 6}. Now, solve it recursively for n = 5, k = 3, r = ( 6 4) r 1 = 6. KsubsetRevDoorUnrank(r, k, n) x n; for i k downto 1 do While ( x i) > r do x x 1; t i x + 1 r ( ) x+1 i r 1; return T ;
37 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking Algorithm: Example 1 n = 6, k = 3, r = 12, T = [?,?,?]
38 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking Algorithm: Example 1 n = 6, k = 3, r = 12, T = [?,?,?] ( 6 3 = 20,... [12]..., 5 3 = 10,..., 4 3 = 4,..., 3 ) 3 = 1
39 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking Algorithm: Example 1 n = 6, k = 3, r = 12, T = [?,?,?] ( 6 3 = 20,... [12]..., 5 3 = 10,..., 4 3 = 4,..., 3 ) 3 = 1 n = 5, k = 2, r = = 7, T = [?,?, 6]
40 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking Algorithm: Example 1 n = 6, k = 3, r = 12, T = [?,?,?] ( 6 3 = 20,... [12]..., 5 3 = 10,..., 4 3 = 4,..., 3 ) 3 = 1 n = 5, k = 2, r = = 7, T = [?,?, 6] ( 5 2 = 10,... [7]..., 4 2 = 6,..., 3 2 = 3,..., 2 ) 2 = 1
41 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking Algorithm: Example 1 n = 6, k = 3, r = 12, T = [?,?,?] ( 6 3 = 20,... [12]..., 5 3 = 10,..., 4 3 = 4,..., 3 ) 3 = 1 n = 5, k = 2, r = = 7, T = [?,?, 6] ( 5 2 = 10,... [7]..., 4 2 = 6,..., 3 2 = 3,..., 2 ) 2 = 1 n = 4, k = 1, r = = 2, T = [?, 5, 6]
42 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking Algorithm: Example 1 n = 6, k = 3, r = 12, T = [?,?,?] ( 6 3 = 20,... [12]..., 5 3 = 10,..., 4 3 = 4,..., 3 ) 3 = 1 n = 5, k = 2, r = = 7, T = [?,?, 6] ( 5 2 = 10,... [7]..., 4 2 = 6,..., 3 2 = 3,..., 2 ) 2 = 1 n = 4, k = 1, r = = 2, T = [?, 5, 6] ( 4 1 = 4,..., 3 1 = 3,... [2],..., 2 1 = 2,..., 1 ) 1 = 1
43 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking Algorithm: Example 1 n = 6, k = 3, r = 12, T = [?,?,?] ( 6 3 = 20,... [12]..., 5 3 = 10,..., 4 3 = 4,..., 3 ) 3 = 1 n = 5, k = 2, r = = 7, T = [?,?, 6] ( 5 2 = 10,... [7]..., 4 2 = 6,..., 3 2 = 3,..., 2 ) 2 = 1 n = 4, k = 1, r = = 2, T = [?, 5, 6] ( 4 1 = 4,..., 3 1 = 3,... [2],..., 2 1 = 2,..., 1 ) 1 = 1 T = [3, 5, 6]
44 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking Algorithm: Example 2 n = 6, k = 3, r = 7, T = [?,?,?]
45 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking Algorithm: Example 2 n = 6, k = 3, r = 7, T = [?,?,?] ( 6 3 = 20,..., 5 3 = 10,...[7]..., 4 3 = 4,..., 3 ) 3 = 1
46 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking Algorithm: Example 2 n = 6, k = 3, r = 7, T = [?,?,?] ( 6 3 = 20,..., 5 3 = 10,...[7]..., 4 3 = 4,..., 3 ) 3 = 1 n = 4, k = 2, r = = 2, T = [?,?, 5]
47 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking Algorithm: Example 2 n = 6, k = 3, r = 7, T = [?,?,?] ( 6 3 = 20,..., 5 3 = 10,...[7]..., 4 3 = 4,..., 3 ) 3 = 1 n = 4, k = 2, r = = 2, T = [?,?, 5] ( 4 2 = 6,..., 3 2 = 3,... [2]..., 2 ) 2 = 1
48 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking Algorithm: Example 2 n = 6, k = 3, r = 7, T = [?,?,?] ( 6 3 = 20,..., 5 3 = 10,...[7]..., 4 3 = 4,..., 3 ) 3 = 1 n = 4, k = 2, r = = 2, T = [?,?, 5] ( 4 2 = 6,..., 3 2 = 3,... [2]..., 2 ) 2 = 1 n = 2, k = 1, r = = 0, T = [?, 3, 5]
49 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking Algorithm: Example 2 n = 6, k = 3, r = 7, T = [?,?,?] ( 6 3 = 20,..., 5 3 = 10,...[7]..., 4 3 = 4,..., 3 ) 3 = 1 n = 4, k = 2, r = = 2, T = [?,?, 5] ( 4 2 = 6,..., 3 2 = 3,... [2]..., 2 ) 2 = 1 n = 2, k = 1, r = = 0, T = [?, 3, 5] ( 2 1 = 2,..., 1 ) 1 = 1,... [0]
50 Generating k-subsets (of an n-set): Minimal Change Ordering Unranking Algorithm: Example 2 n = 6, k = 3, r = 7, T = [?,?,?] ( 6 3 = 20,..., 5 3 = 10,...[7]..., 4 3 = 4,..., 3 ) 3 = 1 n = 4, k = 2, r = = 2, T = [?,?, 5] ( 4 2 = 6,..., 3 2 = 3,... [2]..., 2 ) 2 = 1 n = 2, k = 1, r = = 0, T = [?, 3, 5] ( 2 1 = 2,..., 1 ) 1 = 1,... [0] T = [1, 3, 5]
51 Generating k-subsets (of an n-set): Minimal Change Ordering Successor Let T = [1, 2, 3,..., j 1, t j,...], where j = min{i : t i i}. Consider fours cases for computing successor: Case A: k j (mod 2) Case A1: if tj+1 = t j + 1 then move j to the right, and remove t j + 1. Example: Successor({1, 2, 3, 7, 8, 12}) = {{1, 2, 3, 4, 7, 12}. Case A2: if tj+1 t j + 1 then move j to the left, and add t j + 1. Example: Successor({1, 2, 3, 7, 10, 12}) = {1, 2, 7, 8, 10, 12}. Case B: k j (mod 2) Case B1: if j > 1 then increment t j 1 and (if exists) t j 2. Example: Successor({1, 2, 3, 7, 10}) = {1, 3, 4, 7, 10}. Case B2: if j = 1 then decrement t1 Example: Successor{7, 9, 10, 12}) = {6, 9, 10, 12}.
52 Generating k-subsets (of an n-set): Minimal Change Ordering For each case, prove Rank(successor(T )) Rank(T ) = 1. Proof of case A1: Successor({1, 2, 3, 7, 8, 12}) = {{1, 2, 3, 4, 7, 12}. rank(successor(t )) rank(t ) = ( ) ( ) j = ( 1) k j + ( 1) k j 1 tj j j + 1 ( ) ( ) ( 1) k j tj ( 1) k j 1 tj + 1 j j + 1 ( ( )) j tj + 1 tj tj = + = = 1. j j + 1 j + 1 j Prove other cases A2, B1, B2, similarly.
53 Generating k-subsets (of an n-set): Minimal Change Ordering Successor Algorithm: Case A2 { }
54 Generating k-subsets (of an n-set): Minimal Change Ordering Successor Algorithm: Case A2 { } { } (1237 is the last of A 7,4 )
55 Generating k-subsets (of an n-set): Minimal Change Ordering Successor Algorithm: Case A2 { } { } (1237 is the last of A 7,4 ) {??? } (now the block ending in 8 starts)
56 Generating k-subsets (of an n-set): Minimal Change Ordering Successor Algorithm: Case A2 { } { } (1237 is the last of A 7,4 ) {??? } (now the block ending in 8 starts) { } (before 8, put last of A 7,3 )
57 Generating k-subsets (of an n-set): Minimal Change Ordering Successor Algorithm: Case A1 { }
58 Generating k-subsets (of an n-set): Minimal Change Ordering Successor Algorithm: Case A1 { } { } 1237 is the last of A 7,4 ; can t + + 7
59 Generating k-subsets (of an n-set): Minimal Change Ordering Successor Algorithm: Case A1 { } { } 1237 is the last of A 7,4 ; can t {???? 7 12 } is 1st ending on 8 of A 8,5
60 Generating k-subsets (of an n-set): Minimal Change Ordering Successor Algorithm: Case A1 { } { } 1237 is the last of A 7,4 ; can t {???? 7 12 } is 1st ending on 8 of A 8,5 { } pred in A 8,5 is last of A 7,5, so 12347
61 Generating k-subsets (of an n-set): Minimal Change Ordering Successor Algorithm: Case B1 { }
62 Generating k-subsets (of an n-set): Minimal Change Ordering Successor Algorithm: Case B1 { } { } is last of A 7,5
63 Generating k-subsets (of an n-set): Minimal Change Ordering Successor Algorithm: Case B1 { } { } is last of A 7,5 {??? } pred is second to last of A 7,5
64 Generating k-subsets (of an n-set): Minimal Change Ordering Successor Algorithm: Case B1 { } { } is last of A 7,5 {??? } pred is second to last of A 7,5 { } i.e. successor(1234) = 1245
65 Generating k-subsets (of an n-set): Minimal Change Ordering Successor Algorithm: Case B2 { }
66 Generating k-subsets (of an n-set): Minimal Change Ordering Successor Algorithm: Case B2 { } { } leftmost possible change is in 1st
67 Generating k-subsets (of an n-set): Minimal Change Ordering Successor Algorithm: Case B2 { } { } leftmost possible change is in 1st { } given direction, do 6
68 Generating k-subsets (of an n-set): Minimal Change Ordering KsubsetRevDoorSuccessor( T, k, n) t k+1 n + 1; j 1; While (j k) and (t j = j) do j j + 1; if (k j (mod 2)) then if (j = 1) then t 1 t 1 1; (Case B2) else (Case B1) t j 1 j; t j 2 j 1; else if (t j+1 t j + 1) then (Case A2) t j 1 t j ; t j t j + 1 else (Case A1) t j+1 t j ; t j j; return T ;
69 Generating Permutations: Lexicographical Ordering Generating Permutations: Lexicographical Ordering A permutation is a bijection Π : {1, 2,..., n} {1, 2,..., n}. We represent it by a list: Π = [Π[1], Π[2],..., Π[n]]. Lexicographical Ordering: n = 3 rank permutation 0 [1, 2, 3] 1 [1, 3, 2] 2 [2, 1, 3] 3 [2, 3, 1] 4 [3, 1, 2] 5 [3, 2, 1]
70 Generating Permutations: Lexicographical Ordering Successor Example: Π = [3, 5, 4, 7, 6, 2, 1] Let i = index right before a decreasing suffix = 3. Let j= index of the successor of π[i] in {Π[i + 1],..., Π[n]} π[i] = 4, successor of π[i] = 4 in {7, 6, 2, 1} is 6, π[5] = 6, so j = 5. Swap Π[i] and Π[j], and reverse {Π[i + 1],..., Π[n]}. Successor(Π) = [3, 5, 6, 1, 2, 4, 7]
71 Generating Permutations: Lexicographical Ordering Note that: i = max{l : Π[l] < Π[l + 1]} j = max{l : Π[l] > Π[i]}. For the algorithm, we add: Π[0] = 0. PermLexSuccessor(n, Π) Π[0] 0; i n 1; while (Π[i] > Π[i + 1]) do i i 1; if (i = 0) then return Π = [1, 2,..., n] j n; while (Π[j] < Π[i]) do j j 1; t Π[j]; Π[j] Π[i]; Π[i] t; (swap Π[i] and Π[j]) // In-place reversal of Π[i + 1],..., Π[n]: for h i + 1 to n i 2 do t Π[h]; Π[h] Π[n + i + 1 h]; Π[n + i + 1 h] t; return Π;
72 Generating Permutations: Lexicographical Ordering Ranking How many permutations come before Π = [3, 5, 1, 2, 4]? the ones of the form Π = [1,...] (there are (n 1)! = 24 of them) the ones of the form Π = [2,...] (there are (n 1)! = 24 of them) plus the rank of [5, 1, 2, 4] as a permutation of {1, 2, 4, 5}, which is the standard rank of [4, 1, 2, 3]. So, Rank([3, 5, 1, 2, 4]) = 2 4!+ Rank([4, 1, 2, 3]) = 2 4! + 3 3!+ Rank([1, 2, 3]) = 2 4! + 3 3! + 0 2!+ Rank([1, 2]) = 2 4! + 3 3! + 0 2! + 0 1!+Rank([1]) = 2 4! + 3 3! + 0 2! + 0 1! + 0 = 66
73 Generating Permutations: Lexicographical Ordering General Formula: Rank([1], 1) = 0, Rank(Π, n) = (Π[1] 1) (n 1)!+ Rank(Π, n 1), where { Π Π[i + 1] 1, if Π[i + 1] > Π[1] [i] = Π[i + 1], if Π[i + 1] < Π[1]
74 Generating Permutations: Lexicographical Ordering PermLexRank(n, Π) r 0; Π Π; for j 1 to n 1 do (Note: correction from book: n n 1) r r + (Π [j] 1) (n j)! for i j + 1 to n do if (Π [i] > Π [j]) then Π [i] = Π [i] 1; return r;
75 Generating Permutations: Lexicographical Ordering Unranking Unranking uses the factorial representation of r. Let 0 r n! 1. Then, (d n 1, d n 2,..., d 1 ) is the factorial representation of r if r = n 1 i=1 d i i!, where 0 d i i. (Exercise: prove that such r has a unique factorial representation.)
76 Generating Permutations: Lexicographical Ordering Examples: 1 Unrank(15, 4) = [3, 2, 4, 1] 15 = 2 3! + 1 2! + 1 1!, put d 0 = Unrank(8, 4) = [2, 3, 1, 4] 8 = 1 3! + 1 2! + 0 1!, Unrank(21, 4) = [4, 2, 3, 1] 21 = 3 3! + 1 2! + 1 1!,
77 Generating Permutations: Lexicographical Ordering Justification: Π[1] = d n because exactly d n 1 blocks of size (n 1)! come before Π. Then, Π[2], Π[3],..., Π[n] is computed from permutation Π, as follows: r = r d n 1 (n 1)! Π =Unrank(r, n 1), { Π Π[i] = [i 1], if Π [i 1] < Π[1] Π [i 1] + 1, if Π [i 1] > Π[1] for 2 i n PermLexUnrank(r, n) Π[n] 1; for j 1 to n 1 do r mod (j+1)! d j! ; // calculates d j r r d j!; Π[n j] d + 1; for i n j + 1 to n do if (Π[i] > d) then Π[i] Π[i] + 1; return Π;
78 Generating permutations: Minimal Change Ordering Generating permutations: Minimal Change Ordering Minimal change for permutations: two permutatiosn must differ by adjacent transposition. The Trotter-Johnson algorithm follows the following ordering: T 1 = [[1]] T 2 = [[1, 2], [2, 1]] T 3 = [[1, 2, 3], [1, 3, 2], [3, 1, 2][3, 2, 1], [2, 3, 1], [2, 1, 3]]
79 Generating permutations: Minimal Change Ordering How to build T 3 using T 2 : See picture for T 4 in page 58 of the textbook.
80 Generating permutations: Minimal Change Ordering Ranking Let Π = [Π[1],..., Π[k 1], Π[k] = n, Π[k + 1],..., Π[n]]. Thus, Π is built from Π by inserting n, where Π = [Π[1],..., Π[k 1], Π[k + 1],..., Π[n]]. Rank(Π, n) = n Rank(Π, n 1) + E, { n k, if Rank(Π E =, n 1) is even k 1, if Rank(Π, n 1) is odd Example: Rank([3, 4, 2, 1], 4) = 4 Rank([3, 2, 1], 3) + E = (2 1) = 13.
81 Generating permutations: Minimal Change Ordering PermTrotterJohnsonRank(Π, n) r 0; for j 2 to n do k 1; i 1; while (Π[i] j) do if (Π[i] < j) then k k + 1; i i + 1; if (r 0 mod 2) then r j r + j k; else r j r + k 1; return r;
82 Generating permutations: Minimal Change Ordering Unranking Based on similar recursive principle. Let r = r n, Π =Unrank(r, n 1). Let k = r n r. Insert n into Π in position: k + 1, n k, if r is odd if r is even
83 Generating permutations: Minimal Change Ordering PermTrotterJohnsonUnrank(n, r) Π[1] 1; r 2 0; for j 2 to n do r 1 r j! n! ; // rank of Π when restricted to {1, 2,..., j} k r 1 j r 2 ; if (r 2 is even) then for i j 1 downto j k do Π[i + 1] Π[i]; Π[j k] j; else for i j 1 downto k + 1 do Π[i + 1] Π[i]; Π[k + 1] j; r 2 r 1 ; return Π;
84 Generating permutations: Minimal Change Ordering Successor There are four cases to analyse: Rank(Π ) is even If possible, move left: Successor([1, 4, 2, 3])=([4, 1, 2, 3]) If n is in first position, get successor of the remaining permutation: Successor([4, 1, 2, 3])=([4, 1, 3, 2]), Rank(Π ) is odd If possible, move right: Successor([3, 4, 2, 1])=([3, 2, 4, 1]) If n is in last position, get successor of the remaining permutation: Successor([3, 2, 1, 4])=([2, 3, 1, 4]). We need to be able to determine the parity of Rank(Π ).
85 Generating permutations: Minimal Change Ordering The parity of a permutation is the parity of the number of interchanges necessary for transforming the permutation into [1, 2,..., n]. Π = [5, 1, 3, 4, 2] is an even permutation since 2 steps are sufficient to convert it into [1, 2, 3, 4, 5]. Note that: parity of Rank(Π ) = parity of Π, since in the Trotter-Johnson algorithm [1, 2,..., n] has rank 0, and each swap increases the rank by 1. It is easy to compute the parity of a permutation in Θ(n 2 ): PermParity(n, Π) = {(i, j) : Π[i] > Π[j], 1 i j n}. See the textbook for a Θ(n) algorithm.
86 Generating permutations: Minimal Change Ordering PermTrotterJohnsonSuccessor(n, Π) s 0; done false; m n; for i 1 to n do ρ[i] Π[i]; while (m > 1) and (not done) do d 1; while (ρ[d] m) do d d + 1; for i d to m 1 do ρ[i] ρ[i + 1]; par PermParity(m 1, ρ); if (par = 1) then if (d = m) then m m 1; else swap Π[s + d], Π[s + d + 1] done true; else if (d = 1) then m m 1; s s + 1 else swap Π[s + d], Π[s + d + 1] done true; if (m = 1) then return [1, 2,..., n] else return Π;
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