{1,2,3,...}, and d

iii. The event “at least one juror has a birthday in December” is the complement of theevent “no jurors have a birthday in December”, which is the subject of part iv. Wefind that the required probability is( ) 11 121 − .12iv. The event “no jurors have a birthday in December” consists of the 11 12 outcomesthat don’t contain a 12 in any component. This event has probability11 12 ( ) 11 1212 12 = .1216. (a) Ordering the vertices as a, b, c, d, the adjacency matrix is⎡⎤0 1 1 0A G = ⎢ 1 0 1 0⎥⎣ 1 1 0 1 ⎦ .0 0 1 0(b) Ordering the vertices as above, **and** the edges as e 1 , e 2 , e 3 , e 4 , the incidence matrix is⎡⎤1 1 0 0M G = ⎢ 1 0 1 0⎥⎣ 0 1 1 1 ⎦ .0 0 0 1(c) We need the (1, 3) element of A 3 G. We have⎡⎤ ⎡0 1 1 0A 2 G = ⎢ 1 0 1 0⎥⎣ 1 1 0 1 ⎦ ·⎢⎣0 0 1 0A 3 G =⎡⎢⎣2 1 1 11 2 1 11 1 3 01 1 0 1⎤⎥⎦ ·⎡⎢⎣0 1 1 01 0 1 01 1 0 10 0 1 00 1 1 01 0 1 01 1 0 10 0 1 0⎤ ⎡⎥⎦ = ⎢⎣⎤ ⎡⎥⎦ = ⎢⎣2 1 1 11 2 1 11 1 3 01 1 0 1? ? 4 ?? ? ? ?4 ? ? ?? ? ? ?where ‘?’ denotes some value not of immediate interest. We see that there are 4 pathsof length 3 from vertex a to vertex c in G.17. Let the graph have n vertices. The ith row r i (which is equal to the ith column c i ) of theadjacency matrix is of the form r i = (r i1 , r i2 , . . . , r in ) where{ 1 if vj is a neighbour of vr ij =i ,0 otherwise.The total number of ones in r i is equal to the degree of v i . The (i, i)th element of A 2 is theinner product of r i with c i , i.e., the inner product of r i with itself, which has valuen∑r i · r i = rij 2 = |{j : v j is a neighbor of v i }| = deg v i .j=1⎤⎥⎦ ,⎤⎥⎦ ,6

18. (a) A graph has an Euler circuit if **and** only if every vertex has even degree. The degree ofa vertex in K m,n is either m or n. Thus K m,n has an Euler circuit if **and** only if both m**and** n are even.(b) Let {v 1 , v 2 , . . . , v m } **and** {w 1 , w 2 , . . . , w n } be the two vertex classes in K m,n . By symmetry,without loss of generality a Hamiltonian circuit in K m,n would be of the formv 1 -w 1 -v 2 -w 2 -· · ·, ending in · · ·-v m -w m -v 1 , **and** so exists only if m = n **and** n > 1. Amongthe complete bipartite graphs, only K 2,2 , K 3,3 , K 4,4 , . . . have Hamiltonian circuits.(c) Note that K m,n contains K 3,3 as a subgraph whenever both m **and** n are greater than orequal to 3. By Kuratowski’s Theorem, no such graph can be planar. If m = 1, K m,n isa tree, which can be drawn as shown in Fig. 1(a), **and** is clearly planar. If m = 2, K m,ncan be drawn as shown in Fig. 1(b), **and** is also clearly planar. Since K m,n is isomorphicto K n,m , we find that K m,n is planar if **and** only if at least one of m **and** n is less than 3.(a)(b)Figure 1: K m,n is planar when (a) m = 1 **and** (b) m = 219. We have the following table, which shows the operation of Dijstra’s algorithm. Initially eachvertex is located at infinite distance from a, which is denoted (-,-), except vertex a itself,which is located at distance zero from a. At each step we choose the nearest vertex v ∗ , **and**update the distance to the neighbours of v ∗ according to the update equationL k (w) = min{L k−1 (w), L k−1 (v ∗ ) + wt(v ∗ , w)}.Once a vertex has been chosen, an optimum path to that vertex is known, **and** it doesn’t needto be considered again. This is indicated with an asterisk in the table.The operation ofStep v ∗ a b c d e f g h i j k l m z0 (0,-) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-) Dijkstra’s algorithmis shown in1 a * (1,a) (3,a) (-,-) (2,a) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-)2 b * * (3,a) (4,b) (2,a) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-) the table. We see3 e * * (3,a) (3,e) * (-,-) (6,e) (6,e) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-) that it costs 9 units4 c * * * (3,e) * (4,c) (6,e) (5,c) (-,-) (-,-) (-,-) (-,-) (-,-) (-,-)5 d * * * * * (4,c) (6,e) (5,c) (4,d) (-,-) (-,-) (-,-) (-,-) (-,-) to get from a to z,6 f * * * * * * (6,e) (5,c) (4,d) (-,-) (7,f) (-,-) (-,-) (-,-) **and** one should get7 i * * * * * * (5,i) (5,c) * (-,-) (7,f) (7,i) (-,-) (-,-) to z via m, to m8 g * * * * * * * (5,c) * (6,g) (7,f) (6,g) (-,-) (-,-) via j, to j via g, to9 h * * * * * * * * * (6,g) (6,h) (6,g) (-,-) (-,-)g via i, to i via d,10 j * * * * * * * * * * (6,h) (6,g) (7,j) (10,j)11 k * * * * * * * * * * * (6,g) (7,j) (10,j) to d via e, to e via12 l * * * * * * * * * * * * (7,j) (10,j) a, so that the best13 m * * * * * * * * * * * * * (9,m) path is:14 z * * * * * * * * * * * * * *a-e-d-i-g-j-m-z.7

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