AMS303/HW5/HW5.md

HW5

Chapter 4

• 5. 2. It cannot contain a K_5 because there is only three sets of vertices and each of the sets is an independent set. By ingnoring the connection between, any two of r, s, t can be combined together to form a bipartite graph. Let's set V(n) to be the number of vertices inside a set n, then in order to avoid K_{3,3}, the equation below must be satisfied. MIN(MAX(V(r), V(s), V(t)), (SUM(V(r), V(s), V(t))-MAX(V(r), V(s), V(t))))<3
• 8. 1. K_5 with one extra edge between any two vertices.
     2. It contains a subgraph of K_5
• 10. 1. The only way to draw K_4 in planar is bounded by a circle of 3. And the only way to draw K_{2,3} is a circle of even number. Since the outer bound does NOT match with the number of nodes, it must not be outerplanar.
      2. A graph homeomorphic or contradictable to K_4 or K_{2,3} is non-outerplanar because we can find a set of nodes and paths between them form a K_4 or K_{2,3} by igoring the extra nodes between two nodes. Similarly, for graph contains K_4 or K_{2,3}, we can ignore the extra nodes/edges that is not belong to K_4 or K_{2,3}. Thus, we can get a graph of K_4 or K_{2,3}. As we proved that K_4 or K_{2,3} is non-outerplanar, the graph contains a subgraph of homeomorphic or contradictable to K_4 or K_{2,3} is non-outerplanar.
• 13. Graph	n	m	f
      W_8	8	14	8
      O	6	12	8
      4-13	9	15	8
      K_{2,7}	9	14	7

• 14. i.
      ii.
• 15. 1. see b
      2. for a planar graph with girth x, since each face is bounded by a circle, so the number of edges of each face must be at least x. Thus we can get these formula: n-m+f=2 and f\leq\frac{2m}{x}
         combining these together, we can get n-2=m-f -> $n-2\geq\frac{x-2m}{x}$->$\frac{1}{m}\geq\frac{x-2}{x(n-2)}$->m\leq\frac{x(n-2)}{x-2}. By replacing x with f, we can get m\leq\frac{5(n-2)}{3}, and for Petersen graph, n=10, m=15, which violates the equation.
• 16. 1. for a polyhedral, set the number of pentagon is x and hexagon is y, f=x+y, 5x+6y=2m, m\leq3f-6, then we can get 5x+6y\leq6(x+y)-12, x\geq12
      2. if there are exactly 3 degree on each vertex, m=3f-6, then x=12
• 17. 1. $m\leq3f-6,f<12,sum(bdy)=2m$->sum(bdy)\leq6f-12, and \frac{sum(bdy)}{f} is the higher bound of the lowest boundary. For f<12, \frac{sum(bdy)}{f}<5
      2. if that the case, \frac{sum(bdy)}{f}=5 so its upper bound is no longer 4. Dodecahedron
• 18. 1. m=3f-6,sum(bdy)=2m, we can get sum(bdy)=6f-12 -> 12=6f-sum(bdy). If we look at each face f and the average value 6, if it is bounded by 3 edges, then the different is 6-3=3, as well as 4 is 2, 5 is 1, 6 is 0... proved.
      2. C_5 = 12
      3. if not, then LHS is negative but RHS is positive, contradiction
• 19. m for a graph with total number of 11 vertices is 55. But because m\leq3n-6, G and G* must have 27 or less edges, thus contradiction
      
• 24. 1. obvious, sum(deg) = sum(bdy) = 2*edges
      2. divide d_2 on both side, we can get 2r+2v-2e=2 which is exactly Euler's formula times 2 on both sides.
      3. 2d_1+2d_2-d_1d_2>0 -> d_1d_2-2d_1-2d_2<0 -> d_1d_2-2d_1-2d_2+4<4 -> (d_1-2)(d_2-2)<4
      4. (3,3),(3,4),(4,3),(3,5),(5,3)