Solutions - MA3604 Discrete Mathematics - 2020
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Solutions - MA3604 Discrete Mathematics - 2020
1. (a) i. The adjacency matrix corresponding to the labelled graph is
For an adjacency matrix M the entry (Mn)ij is the number of paths of length n starting at vertex (i) and ending at vertex (j) in the corresponding labelled graph. Taking n = 4 in our case we nd
Therefore there are eight paths of length four starting at vertex-2 and ending at vertex-3.
ii. The same graph labelled under is
and the new adjacency matrix is
Writing (i) = (P(i)), then P is the permutation
P(1) = 2, P(2) = 3, P(3) = 1.
Let U be the permutation matrix with entries Uij = i,P(j):
We write M for the adjacency matrix of the same graph, now with vertices labelled according to . It was shown during the Course that M = UTM U. Then
This agrees with the adjacency matrix found earlier.
(b) The i-th diagonal entry of M2 is MijMji (sum over j implied). For each j, this is the number of edges from vertex i to vertex j, times the number of edges from vertex j
back to vertex i.
Each of these terms must be zero for (M2)ii to be zero. We deduce that if there is an edge from vertex i to vertex j, then there is no edge from j back to i.
2. (a) The transition matrix is
Stationary states (x,y, 1 x y) are left eigenvectors of W having eigenvalue 1. Therefore
The only solution to this is x = 3/7, y = 2/7. Therefore the sole stationary state of the given graph is (3/7, 2/7, 2/7).
(b) Start by nding the spectrum of W. Its characteristic polynomial is (1/6+x)(1/2 x)(1 x), so the eigenvalues of W are 1/6, 1/2 and 1, all having algebraic multi- plicity one. [2] Because all eigenvalues are distinct the matrix is diagonalisable. The Jordan form J will be diagonal, with the eigenvalues as diagonal entries. Nevertheless we still need to calculate the eigenvectors in order to have the change-of-basis matrix that takes
Jn to Wn and so to know the large-n behaviour of the latter.
Let us nd the eigenvectors associated with eigenvalue -1/6.
A basic solution to this is u = ( -2, 1, 1). There are no more independent eigenvectors
associated to eigenvalue 1/6, nor is u in Im(W + I/6).
Nex we fnd the eigenvectors associated with eigenvalue 1/2.
A basic solution to this is v = (0, 1, -1). There are no more independent eigenvec-
tors associated to eigenvalue 1/2 and v Im(W -I/2), so we are done with this eigenvalue.
As for eigenvalue 1, we already found eigenvector w = (3/7, 2/7, 2/7), which is not
in Im (W -I).
The matrix that eects a change of basis from the original one to the ordered vector set (u;v;w) is
The Jordan form of W is
|
As discussed during the Course, a Jordan block carrying an eigenvalue of modulus less than one converges to a zero matrix upon raising it to increasingly higher powers.
(c) It follows from the above that
Calling this limit W∞, any initial state (x,y, 1 x y) will converge under time
evolution to the state (x,y, 1 x y)W∞ .
In our case
(x,y, 1 x y)W∞ = (3/7, 2/7, 2/7),
as was to be expected, (3/7, 2/7, 2/7) being the only stationary state of the given
graph.
3. (a) Start with the transition matrix of the graph:
Let be the probability that a random walk on the graph starts at vertex-1 and
returns to it in exactly n steps. Inspection of the graph gives
For n 3 we proceed as explained in the Course. The probabilities of the rst step to lead to vertices 2 or 3 are W12 or W13 respectively. Transitions between vertices 2 and 3 correspond to the submatrix of W which excludes row and column one. The last step, from vertices 2 or 3 back to vertex 1, has probabilities W21 or W31 respectively. Therefore, for n 3 we have
Note that this result is actually valid also for n = 2, so generally:
(b) If the sum of all for n 1 equals one then any random walk on the graph,
which starts from vertex-1, will return to it after some number of steps.
Therefore the answer is in the armative.
(c) The expected (or average) time of rst return for a random walk on the graph is
That is, the average number of steps required for a random walk which starts at vertex-1 to return to it for the rst time is 5/2.
4. (a) The degree formula for vertices gives 5V5 + 6V6 = 2|E|,
in which Vn is the number of vertices of degree n. The degree formula for faces gives
3|F| = 2|E|.
Substitution of these into the Euler formula, with |V| = V5 + V6, results in
which simplies into V5 = 12.
(b) By the degree formula for faces, the average degree face is f = 2|E|/|F|. For a trivalent graph 2|E| = 3|V|, and because the graph is planar and connected we have also the Euler formula |V| |E| + |F| = 2. Combining these formulas we nd that
f = 6 12/|F|.
The larger a graph of this kind is, the larger |F| becomes, and f tends to 6.
(c) i. By the Kuratowski Theorem, a graph admits planar representations if and only if it contains no subgraphs isomorphic to a subdivision of K5 or K3,3 (it is not required that the theorem be stated.)
If the rightmost (or the leftmost) vertex and all the edges attached to it are removed, the resulting subgraph is isomorphic to K3,3 (this must be shown in sucient detail.) Therefore the graph admits no planar representations.
ii. If the middle horizontal edge is removed, the resulting subgraph is isomorphic to a subdivision of K5. Therefore this graph admits no planar representations either.
2022-01-14