Date | May 2018 | Marks available | 1 | Reference code | 18M.3dm.hl.TZ0.3 |
Level | HL only | Paper | Paper 3 Discrete mathematics | Time zone | TZ0 |
Command term | Show that | Question number | 3 | Adapted from | N/A |
Question
Consider the complete bipartite graph \({\kappa _{3,\,3}}\).
Draw \({\kappa _{3,\,3}}\).
Show that \({\kappa _{3,\,3}}\) has a Hamiltonian cycle.
Draw \({\kappa _{3,\,2}}\) and explain why it does not have a Hamiltonian cycle.
In the context of graph theory, state the handshaking lemma.
Hence show that a graph G with degree sequence 2, 3, 3, 4, 4, 5 cannot exist.
Let T be a tree with \(v\) where \(v\) ≥ 2.
Use the handshaking lemma to prove that T has at least two vertices of degree one.
Markscheme
A1
[1 mark]
for example ADBECFA A1
Note: Accept drawing the cycle on their diagram.
Note: Accept Dirac’s theorem (although it is not on the syllabus for (a)(ii). There is no converse that could be applied for (a)(iii).
[1 mark]
A1
a Hamiltonian cycle would have to alternate between the two vertex subsets which is impossible as 2 ≠ 3 R1
Note: Award R1 for an attempt to construct a Hamiltonian cycle and an explanation of why it fails, eg, ADBEC but there is no route from C to A without re-using D or E so no cycle. There are other proofs eg, have to go in and out of A, similarly B and C giving all edges leading to a contradiction.
[2 marks]
the sum of the vertex degrees is twice the number of edges A1
[1 mark]
assume G exists
the sum 2 + 3 + 3 + 4 + 4 + 5 = 21 A1
this is odd (not even) R1
this contradicts the handshaking lemma
so G does not exist AG
[2 marks]
T has \(v - 1\) edges A1
EITHER
if \(k\) vertices have degree 1 then \(v - k\) vertices have degree ≥ 2 R1
by the handshaking lemma
\(2v - 2 \geqslant 1 \times k + 2\left( {v - k} \right)\left( { = 2v - k} \right)\) M1
this gives \(k\) ≥ 2 A1
OR
let S be the sum of vertex degrees
consider T having either no or one vertex of degree 1 R1
case 1 suppose T has no vertices of degree 1 (eg, all vertices have degrees ≥ 2)
by the handshaking lemma
\(S \geqslant 2v \ne 2\left( {v - 1} \right)\) (not possible) A1
case 2 suppose T has one vertex of degree 1 (eg, \(v - 1\) vertices have degrees ≥ 2)
by the handshaking lemma
\(S \geqslant 2\left( {v - 1} \right) + 1 \ne 2\left( {v - 1} \right)\) (not possible) A1
THEN
so T has at least two vertices of degree 1 AG
[4 marks]