This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: AMS 301.1 Fi rst Test, Fall 2009, SOLUT IONS 1. Graphs are isomorphic.See Figure 1.19 in text. 2. Test A: LB = 18 (subtracting first from rows). Don't use (1,3), LB=22; use (1.3) (deleting row 1 & col 3 and setting (entry (3,1) = infinity), LB = 19. Test B: LB = 14 (subtracting first from rows). Don't use (1,3), LB = 19; use (1,3) (deleting row 1 & col 3 and setting entru (3,1) = infinity), LB= 16 3. Test A: connected; test B: not connected 4. Test A and 5. Test B: vertices equal actuaries/statisticians, edges = personaly conflicts, colors = hotels; does 6 colors suffice? 5. Test A and 4. Test B: a) not possible, violates e <=3v-6; b) possible, v = (Test A, 12), (Test B, 10)-- remember that all vertices musst have deg 3; c) many possibilities, e.g., t riangle with trailing path (like a kite), 6. As explained in class, by symmetry and need to use at least one vertical edge, you can assume you start with edge (c,h), at c by symmetry choose b-c & delete c-d, forcing at d, e-d- i. Two cases at b or h or e or i. We consider cases at h Case 1. Use h-f, delete h-j => at j use e-j-g, at e delete e-a, => at a use b-a-f, subcricuit a-b-c- h-f-a. Case 2. Use h-j, delete h-f => at f, use a-f-i, at i delete i-g => at g use b-g-j, subcircuit b-c-h-j- g-b. 7. a) Many possibilities. b) If a cutset K did not intersect a tree T, then G-K would still be connected since G-K would contain a spanning tree. c) If G1 and G2 are the two components of G-K, and if circuit C starts in G1, then any time C uses an edge of K, it moves over to G2. I t must eventually return to G1, using a second edge of K. Thus crossing to G2 and returning to G1 cna occur many times, but each time the circuit uses 2 edges of K....
View Full Document
- Spring '08