AMS 301 HOMEWORK # 1
Prof. Tucker
Section 1.1: for 3,9,15,23--see answer in back of text; 6a) 5 vertices, b) 9 vertices;
18. c,e,k},{b,h,i};
20.
.
Correction: In 20b) figure, circle in first column should be one vertex higher
Section 1.2: 5adh, 7, see answer in back: 6 c) no, 5-degree and degree-3 vertices form a triangle only in right graph, f)
no, only right graph has triangles, h) no, only left graph has triangles.
Section 1.3: 1, See answer in back for 1c: 2b) vertex = person, edge = joining siblings, yes, c) no model, no; 6. n is odd
and so parity of deg. 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;
12a) (A,Z)-(C,X)-(E,Y)-(G,X)-(F,W)-(H,U)-(I,V)-(M,M)
AMS 301 HOMEWORK # 2
Prof. Tucker
Section 1.4: All odd problems-- see answers in back of text
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); 16a) if edge E is incident to two vertices with degrees d, d' (both even degree), then E
is incident to (d-1) + (d'-1) = d+d' - 2 other edges, and the associated vertex for E in line graph has degree d+d' - 2
(even degree), b) If every vertex has odd degree in G, then d+d' - 2 (as defined in part a) ) will be even;
AMS 301 HOMEWORK # 3 Prof. Tucker
Section 2.2: 3, 7, 9 see answers in back of text 4. each graph has many Hamilton paths- e) no circuit: rule 1 at a & f;
now b,h,i,e each need 1 more edge but c, g, d need 6 edges collectively connecting to b,h,i,e;
h) at m,by symmetry use k-m-i, delete m-g; at j, by rule 2, must use exactly one of k-j and j-i; by sym. use j-k (and jd) delete j-i; at k, delete e-k and k-l; forces d-e-f and f-l-a; delete d-c and f-a; subcircuit a-b-c-i-m-k-j-d-e-f-l-g-a;
l) by rule 1 use a-m-b, c-n-d, g-k-j; at g by symmetry use g-f, delete g-h, forcing i-h-c; at c, delete c-b, forcing b-f; at
f, delete e-f, forcing j-e-a, and subciruit a-e-j-k-g-f-m-a;
m) use h-i-j (rule 1), two choices at h (vertical edge or h-g)- vertical edge, by symmetry choose h-b and delete h-g, hm (rule 3), by rule 1 at g and m use a-g-l-m-o, delete l-n, at n use o-n-q, delete o-p, at p use q-p-j, delete q-k and j-k
(now k has degree 1), similar reasoning when h-g is used at h;
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; 16. vertices = banquets, edges = overlapping rooms, colors = days of the week.
Section 2.4: 7 see answer in back of text.
AMS 301 HOMEWORK # 4
Prof. Tucker
Section 3.1: 1a, 13, 25, 29 see back of book. 2. 21 vertices; 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; 10. l ≤ mh  (by Corollary
part b) n = (ml - 1)/(m - 1) ≤ (mh+1 - 1)/(m - 1); 14. Since adding any edge (x,y) forms a circuit, there is already
a path between x and y; so the graph is connected and circuit-free;
Section 3.2: 1, 5, 19 see back of book. 4. connected, lots of spanning trees. 16b) (0,0-(0,5)-(7,5)-(7,0)-(2,5).