Question

(a) Draw the graphs Ks,2 and K5,3 using the standard arrangement 2. For example, K5,2 should have a row of 5 vertices above a row of 2 vertices, and the edges connect each vertex in the top row to each vertex in the bottom row. (b) Draw K52 as a plane graph, i.e., with no edges crossing. (c) Complete the following table, recalling E is the number of edges in a graph and V is the number of vertices. (Strictly speaking, this is incorrect use of notation as E and V are sets, and we should use E and |VI to indicate the number of edges and vertices.) K5,2 K5,3 (d) Carefully prove Ks,3 is nonplanar using Kuratowski's Theorem. (e) Carefully prove K5,3 is nonplanar using Theorem 4.4.5 3. Number the functions 1 → 5 from slowest to fastest, for long term growth. Functionn 銘10th! (m) 3n Number

2 (a) Draw the graphs K5,2 and K5,3 using the standard arrangement.

For example, K5,2 should have a row of 5 vertices above a row of 2 vertices, and the edges connect each vertex in the top row to each vertex in the bottom row.

(b) Draw K5,2 as a plane graph, i.e., with no edges crossing.

(c) Complete the following table, recalling E is the number of edges in a graph and V is the number of vertices. (Strictly speaking, this is an incorrect use of notation as E and V are sets, and we should use |E| and |V | to indicate the number of edges and vertices.)

(d) Carefully prove K5,3 is nonplanar using Kuratowski’s Theorem.

(e) Carefully prove K5,3 is nonplanar using Theorem 4.4.5.

3. Number the functions 1 → 5 from slowest to fastest, for long term growth.

Answer 1

2. a) Pic Uploaded

Dcte 3 2 (o 2 JS s,3 -5,2

b)The graph K5 is non-planar. Proof. let us assume that K5 is planar. Let the vertices of K5 be denoted by 1, 2, 3, 4 and 5. Then, in any planar drawing of K5, the cycle [123451], of K5, must appear as a cycle. Let us draw it in the form of a pentagon. Then the edge {1, 4} will either lie completely inside the pentagon or completely outside it. Without loss of generality, let us assume that the edge {1, 4} lies completely inside the pentagon. Therefore, both the edges {2, 5} and {3, 5} will lie completely outside the pentagon (or else they will intersect the edge {1, 4}). Now, the edges {1, 3} and {2, 4}. If {1, 3} is drawn inside the pentagon than the edge {2, 4} cannot be drawn without intersecting either the edge {1, 3} or the edge {3, 5}. A similar argument is valid if the edge {2, 4} is drawn inside the pentagon.

c) E= v(v-1)/2

	V	E	3V-6	2V-4
K5,2	7	21	15	10
K5,3	8	28	18	12

d)we have that a planar graph must satisfy e ≤ 3v − 6. Note that for K5,3, e = 28 and v = 8. Since 28 is not less than equal to 18, it must be that K5 is not planar.