哈工大深圳研究生院组合数学部分作业题答案习题课6,8

Homework 613．Let A={A1,A2,A3,A4,A5,A6} whereA1={1,2} A2={2,3} A3={3,4} A4={4,5} A5={5,6} A6={6,1}Determine the number of different SDR’s that A has. Generalize to n sets.Solution: When we choose 1 in A 1, if we choose 3 in A 2, we can choose 4 only in A 3, we can choose 5 only in A 4, and we can choose 6 only in A 5, however, we can choose 1 only in A 6，it will contract with 1 in A 1Hence, we can choose only 2 in A .2 if we choose 1 in A 1,3 in A 3,4 in A 4,5 in A 5,6 in A When we choose 2 in A 6.1, we can only choose 3 in A 3, 4 in A 4, 5 in A 5, 1 in A 1That is SDR . Hence, there are only two SDRs in A . 1={1,2,3,4,5,6}and SDR 2Similarity, we can generalize to n sets. There are only two SDRs in n-sets the same.={2,3,4,5,6,1}That is SDR 1={1,2,…,n} and SDR 2={2,3,4,…,n,1}23. Use the deferred acceptance algorithm to obtain both the women-optimal and men-optimal stable complete marriage for the preferential ranking matrix.Conclude that for the given preferential ranking matrix there is onlyone stable complete marriage.a b c dAA BB CC DD�1,32,3 1,44,13,24,33,32,2 2,21,4 4,12,23,44,13,11,4�Solution：(1) women-optimalThe results of the algorithm are as follows:1)A choose a, B choose a, C choose b, D choose d; a rejects B2)B choose d; d rejects D.3)D choose b; b rejects C.4)C choose a; a rejects A.5)A choose c.In 5), there are no rejections, and(2) men-optimalThe results of the algorithm are as follows:1)a choose D, b choose B, c choose D, d choose C; D rejects a.2)a choose C, C rejects d.3)d choose B, B rejects b.4)b choose D, D reject c.5)c choose A.In 5), there are no rejections, anda C,b D,c A,d B.Conclude that, for given preferential ranking matrix, there is only one stable complete marriage.3. Consider an m-by-n chessboard where at least one of m and n is even. The board has an equal number of white and black squares. Show that if m and n are at least 2 and if exactly one white and exactly one black square are forbidden, the resulting board has a perfect cover with dominoes.Solution:Let n be even, there are even numbers of columns. From top row, if the top row has no squares forbidden, then that row can be covered by dominoes. So we can remove this row. Repeat this till the new top row has a forbidden square. Do the same thing from the bottom row. So we can assume that forbidden squares are on the top and bottom rows.Let’s consider the first two columns from the left. If there is nofirst two columns, they can be covered by forbidden square in thedominoes and we can remove them. Repeat this till there is a forbidden square in the first two columns. Do the same from the right side so we ca n also assume the forbidden squares lie in the first two columns and the last two columns. After the above have been done, we can assume that chessflip board is of one of the following three situations (with rotation or ofover the chess board if needed). In each of the case, there are even numbers of columns.(A) (B) (C)So the two forbidden squares have one black and the other white, there are even numbers of columns. We draw the figures, in the cases (A) and (B), there has to be odd number of rows while in the case (C), there are even number of rows. Then we divide the board into pieces as shown in the following such each piece is rectangular with at least one side being even, thus can be covered by dominoes respectively.So the entire board has a perfect cover by dominoes.(A) (B) (C)4. Determine the max-matching and the min-cover of the right graph by applying the matching algorithm. We choose the red edges and obtain a matching M1.Find a minimum edge cover for the right graph.Answer:Now we get the matching )},(),,{(44121y x y x M = and U 131,x x = {}. (i) The vertices 31,x x are labeled (*).x 1y 1x x y 2 y 3 y 4y 5(ii) Scan the vertices in U 154321,,,,y y y y y in turn, and label with (1x ), since all vertices incident to 3x .already have a label, no vertex of Y get label (3x ).(iii) Scan the vertices 5431,,,y y y y labeled in (ii), and label 2x with (1y ), label 4x with (4y ).(*)x 1y 1(1x )x(*)x 3x y 2 y 3(1x ) y 4(1x )y 5(1x )(*)x 1 y1x (*)x 3x y 2 y 3 y 4y 5(iv) We scan the vertices 2x and 4x labeled in (iii), and label 2y with (2x ).(v) Scan the vertices 2y labeled in (iv), and find that no new labels are possible.We have achieved breakthrough. We find the 1M -augmenting path11221x y x y r = using the labels as a guide. Then )},(),,(),,{(4411222y x y x y x M =(*)x 1y 1(1x )(1y )x 2(*)x 3(4y )x y 2(2x ) y 3(1x ) y 4(1x )y 5(1x )(*)x 1 y 1(1x )(1y )x 2(*)x 3(4y )x y 2 y 3(1x ) y 4(1x )y 5(1x )and }{32x U =.(vi) The vertices 3x are labeled (*).(vii) Scan the vertices in U 25y in turn, and label with (3x )x 1y 1x 2(*)x 3x 4y 2 y 3 y 4y 5x 1y 1x 2x 3x 4y 2 y 3 y 4y 5(Viii) Scan the vertex 5y labeled in (vii), and find that no new labels are possible.We have achieved breakthrough. We find the 2M -augmenting path352x y r = using the labels as a guide. Then)},(),,(),,(),,{(441122533y x y x y x y x M =is a matching of four edges.Now we can get thatx 1 y 1x 2x 3x 4y 2 y 3 y 4y 5x 1 y 1x 2(*)x 3x 4y 2 y 3 y 4y 5(3x )the max-matching )},(),,(),,(),,{(44112253y x y x y x y x M = and the min-cover= {54321,,,,y y y y y }.2) Find a minimum edge cover for the right graph.Follow the steps above; we can get a max-matching)},(),,(),,(),,{(44112253y x y x y x y x M =.And we can find that there are also some vertices uncovered by the max-matching. Obviously, the vertex 3y Now we can construct a subgraph composed of edges incident to the vertex is uncovered.3y , we find the max-matching of the subgraph, and add it to the max-matching of M, then we can get a minimum edge cover.Fig The subgraphThe max-matching of the subgraph is {(x 1,y 3)}. Now we can get a minimum edgecover={(x 1,y 3)}∪M ={(x 1,y 3)(x 3,y 5)(x 2,y 2)(x 1,y 1)(x 4,y 4)}x 1y 1x 2x 3x 4y 2 y 3y 4y 5x1y1x2 x3 x4y2 y3 y4y5Fig The minimum edge cover第八次作业16. Apply the algorithm for the GCD in Section 10.1 to 15 and 46, and then use the results to determine the multiplicative inverse of 15 in Z46Answer:.The result for Computing the GCD of 15 and 46 are displayed in the following:Now ,we write as a linear combination of 15 and 46：1=46-3*15From the above, 1 is the GCD of 15 and 46.Thus, we can get the multiplicative inverse of 15 in Z4615:-121. Determine the complementary design of the BIBD with parameters bb=vv=77,kk=rr=33,λλ=11in Section 10.2= – 3 = 43.•b: the number of blocks;•v: the number of varieties;•k: the number of varieties in each block;• r : the number of blocks containing each variety• λλ: the number of blocks containing each pair of varieties. Answer:Apply ()11−−=k v r λ to this case, now we have b =v =7,k =r =3,λ=1.Hence, we can get ()()311717111≠=−−×=−−=k v r λ. Thus, we can design such a complementary.We can get the complementary design of the BIBD withparametersb’ = b = 7, v’ = v = 7, k’ = v-k = 4, r’ = b-r = 4 ,213272'=+×−=+−=λλr b .28. Show that BB={00,11,33,99}is a difference set in Z13 Answer:, and use thisdifference set as a starter block to construct an SBIBD. Identify the parameters of the block design.We compute the subtraction table and obtain:We can find that non-zero integers in Z13}{9,3,1,0=B occur exactly once as adifference, hence is a difference set in Z1332. Use Theorem 10.3.2 to construct a Steiner triple system of index 1 having 21 varieties..•Hints:• 1. construct two Steiner triple systems B1 and B2• 2. construct B based on B with 3 and 7varieties, respectively.1and B2. Answer:1）Construct two Steiner triple systems B1 and B2 with 3 and 7 varieties, respectively.Let X={a0,a1,a2}and Y={b0,b1,b2,b3,b4,b5,b6}be two sets of varieties.Let B1={a0,a1,a2}and B2=�{b0,b1,b3},{b1,b2,b4}, {b2,b3,b5},{b3,b4,b6},{b4,b5,b0},{b5,b6,b1},{b6,b0,b2}�be the Steiner triple systems of X and Y, respectively.2）Construct B based on B1 and B2.We define a set B of triples of the elements of X. Let {c ir, c js, c kt} be a set of 3 elements of X. then {c ir, c js, c kt} is a triple of B iff one of the following holds:i) r = s = tIf r=s=t=1, we can get {0,3,9}, {3,6,12}, {6,9,15}, {9,12,18}, {12,15,0},{15,18,3}, {18,0,6};If r=s=t=2, we can get {1,4,10}, {4,7,13}, {7,10,16}, {10,13,19},{13,16,1}, {16,19,4}, {20,1,7};If r=s=t=3, we can get {2,5,11}, {5,8,14}, {8,11,17}, {11,14,20},{14,17,2}, {17,20,5}, {20,2,8};ii) i = j = k, We can get {0, 1, 2}, {3, 4, 5}, {6, 7, 8},{9, 10, 11}, {12, 13,14}, {15, 16, 17}, {18, 19, 20};(iii) i, j and k are all different and {a i, a j, a k} is a triple of B1, and r, s and t are all different and {b r, b s, b t} is a triple of B2. Put another way, c ir, c js, and c kt are in 3 different rows and 3 different columns of the array, and the rows in which they lie correspond to a triple of B1 and the columns inwhich they lie correspond to a triple of B2 If {a .i, a j, a k If {a }={0, 1, 3}，we can get {0, 4, 11}, {0, 10, 5}, {3, 1, 11}, {3, 10, 2}, {9, 1, 5}, {9, 4, 2};i, a j, a k If {a }={1, 2, 4}, we can get {3, 7, 14}, {3, 13, 8}, {6, 4, 14}, {6, 13, 5}, {12, 4, 8}, {12, 7, 5};i, a j, a k If{a }={2, 3, 5}，we can get {6, 10, 17}, {6, 16, 11}, {9, 7, 17}, {9, 16,8},{15,7,11},{15,10, 8};i, a j, a k If {a }={3,4,6}, we can get {9,13,20},{9,19,14},{12,10, 20},{12,19,11},{18,10,14},{18,13,11};i, a j, a k If {a }={4, 5, 0}，we can get {12,16,2}, {12, 1,17}, {15,13,2},{15,1,14}, {0,13,17}, {0,16,14};i, a j, a k If {a }={5,6, 1}，we can get {15, 19, 5},{15,4,20},{18, 16, 5},{18, 4, 17},{3,16,20},{3,19,17};i, a j, a k}={6,0, 2}，we can get {18, 1, 8}, {18, 7, 2}, {0, 19, 8}, {0, 7,20}, {6, 19, 2}, {6, 1, 20};52. Construct a completion of the 3-by-6 Latin rectangleAnswer:According to theorem 10.4.11(390页), this 3-by-6 Latin rectangle has a completion. Let X={x 0, x 1, x 2, x 3, x 4, x 5} representing the 6 elements, and Y= {y 0, y 1, y 2, y 3, y 4, y 5} representing the columns. An edge (x i, yj ) in denotes element i doesn’t appear in column j. Then we construct a regular of degree 3 Bipartite graph of X and Y ，G=(X, Δ, Y) as follows.y 2x 0x 2x 3x 4y 1y 3x 1y 5y 4y0x 5(1)Computing the perfect matching using the matching algorithm (312页) and the results are as follows. The red lines represent the maximum matching. （找出一个完美匹配即可，不必写出寻找的步骤，建议通过观察得出完美匹配最好，匹配算法太麻烦了）0 1 2 3 4 5 4 3 1 5 2 0 54312y 2x 0x 2x 3x 4y 1y 3x 1y 5y 4y0x 5So a new row is{x 2, x 0, x 4, x 1, x 5, x 3Similarly, we can draw the graph for the above matrix as follows.} could be added to the original matrix as follows.�01243134552054324012153� y 2x 0x 2x 3x 4y 1y 3x 1y 5y 4y0x 5(2)Computing the perfect matching and the results are as follows. The red lines represent the maximum matching.y 2x 0x 2x 3x 4y 1y 3x 1y 5y 4y0x 5Hence a new row is{x 1, x 5, x 0, x 2, x 3, x 4Similarly, we can draw the graph for the above matrix as follows.} and the new Latin rectangle is⎣⎢⎢⎢⎡012431345520543204150012153234⎦⎥⎥⎥⎤ y 2x 0x 2x 3x 4y 1y 3x 1y 5y 4y0x 5From the graph, we can easily know that all the lines construct a perfect matching. So we the final answer as follows⎣⎢⎢⎢⎢⎡012431543335520012204150325153234401⎦⎥⎥⎥⎥⎤58. Construct a completion of the semi-Latin squareAnswer ：思路：构造二分图，找完美匹配（匹配算法）。