<p> AMS 301 HOMEWORK # 1 Prof. Tucker Section 1.1: for 1,9,23--see answer in back of text; 6a) 5 vertices, b) 9 vertices; 14. Many possibilities; 16a) (a,b), (a,g) and (c,d), (d,e) and (c,d), (e,f) and (d,e),(e,f), b) {b,g}, {b,f}, {c,f}, {c,e}, {d,f}, c) 3 (edges incident to any vertex) </p><p>20a) 20c) . Block surveillance vertices = • Corner surveillance vertices = O . 22: 3 indep. sets. Section 1.2: 5bck, 7, see answer in back: 4. Not isomorphic; 6 c) no, 5-degree and degree-3 vertices form a triangle only in right graph, d) no, complement of left graph is 2 4-circuits, g) no, only right graph has triangles. Section 1.3: 1a) 8, b) 13, c) 8, 12, 16, 24, 48 (Answer in back is WRONG); 6. n-1 is odd and so parity of deg. in G is same in complement. Ans: n-1, 10. Make graph with vertex = team, edge = game. Within a given conference, the subgraph consists of 13 vertices, each of degree 11— contradicting Corollary; 12b) (A,Z)-(B,X)-(C,Y)-(D,W)-(C,U)-(B,V)-(A,T)-(B,R)-(C,S)-(M,M)</p><p>AMS 301 HOMEWORK # 2 Prof. Tucker Section 1.4: All odd problems-- see answers in back of text. 4n = (sum of degrees) = 2e and so e = 2n; Euler's formula gives 10 = r = e - v + 2 or 10 = 2n - n + 2 => n = 8. 16. K11 has 11x10/2 =55 edges, but graph and its compl must each have ≤ 3x11 - 6 = 27 edges. 20. Replace 3 by k in the proof of the Corollary. </p><p>Section 2.1: 3,7 see answer in back of text. 2.a) n odd, b) yes, K2, c) when r and s are even; 4. 2 times (to link two pairs of the 6 odd-degree vertices; 10. No, build graph with vertex for each chessboard square and an edge joining two squares that are linked by a knight's move, the 8 vertices corresponding to squares on edge of board one away from corner have degree 3;</p><p>AMS 301 HOMEWORK # 3 Prof. Tucker Section 2.2: 3, 7, 9 see answers in back of 4. each graph has many Hamilton paths- b) rule 1 at d, f forces subcircuit a-d-g-f-a; j) by sym., at j, use c-j-d; to visit a, b, require subpath c-a-b-d or c-b-a-d => subcircuit; k) a-e-f, c-k-j forced; by sym at f pick f-n, delete f-g, at g use o-g-h; 2 cases at n, 1) use n-o and del. o-i, o-l; use m-l-j-i-k; 3 edges at j; 2) use n-m, subcase 2i) use m-l, del. m-d, at d use a-d-c, del. a-b, c-b, b has 1 edge, subcase 2ii) use m-d, del. m-l, at l use j-l-o; at o, del. o-i, at i use j-i-h, j deg 3;</p><p>40) 3 cases at f: 1) use b-f-i, del. f-c, f-j, at c, j, use a-c-g, g-j-k, del. g-h, at h use a-h-k, subcir. a-c-g-j-k-h-a; 2) use b-f-c, del. f-i, f-j, at i,j use e-i-k, k-j-g, del. k-d, k-h, at d,h use e-d-a, g-h-a, subcir. a-h-g-j-k-i-e-d-a; 3) use b-f-j, del. f-c, f-I, at c, i use a-c-g, e-i-k; two subcases at b, use b-a or b-e (details left to reader). 16. No help. Section 2.3: 1, 15 see back of book. 14a) vertices = exp., edges = time overlap, colors = observatories, b) 4 colors (with 3 colors, A,D, G must be the same color. Section 2.4: 7 see answer in back of text.</p><p>AMS 301 HOMEWORK # 4 Prof. Tucker Section 3.1: 1a, 11, 13, 21, 25, 29 see back of book. 4. no circuits no K3,3 or K5 configuration (or just draw tree as in Figure 3.1 in planar fashion); 6. r =1, and so 1 = r = e - v + 2  e = v - 1; Section 3.2: 1, 5, 23 see back of book. 4. connected, lots of spanning trees. 16b) (0,0-(0,5)-(7,5)-(7,0)-(2,5). </p>