Arie Girshson
306236399
Exercise 1: Perfect Matching
Notations:
𝑜(𝐺) – Number of odd components of graph G.
Theorem:
G has a perfect matching if and only if: 𝑜(𝐺 − 𝑆) ≤ |𝑆| 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑆 ⊂ 𝑉(𝐺).
Proof:
Suppose 𝐺 has a perfect matching 𝑀. Let 𝑆 ⊂ 𝑉(𝐺), and 𝐺1 , 𝐺2 , … , 𝐺𝑛 be the odd components of 𝐺 − 𝑆.
Because 𝐺𝑖 is odd, there must be some vertex 𝑢𝑖 of 𝐺𝑖 matched inside 𝑀 with some vertex 𝑣𝑖 of 𝑆
(Because G has perfect matching).
𝑮𝟏
𝒖𝟏
𝑮𝟐
𝒖𝟐
..…
𝑮𝒏
..…
𝒖𝒏
𝑬𝒗𝒆𝒏 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕𝒔 𝒐𝒇 𝑮 − 𝑺
𝑶𝒅𝒅 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕𝒔 𝒐𝒇 𝑮 − 𝑺
𝑮
𝒗𝟏 𝒗𝟐
Since {𝑣1 , 𝑣2 , … , 𝑣𝑛 } ⊆ 𝑆:
..…
𝒗𝒏
..…
𝑺
𝑜(𝐺 − 𝑆) = 𝑛 = |{𝑣1 , 𝑣2 , … , 𝑣𝑛 }| ≤ |𝑆|.
Conversely, suppose 𝐺 satisfies the schematic above, but has no perfect matching. In this case, 𝐺 is a
spanning sub-graph of a maximal graph 𝐺 ∗ having no perfect matching.
Since 𝐺 − 𝑆 is a spanning sub-graph of 𝐺 ∗ − 𝑆 (as denoted above) we get : 𝑜(𝐺 ∗ − 𝑆) ≤ 𝑜(𝐺 − 𝑆)
and 𝑜(𝐺 ∗ − 𝑆) ≤ |𝑆| 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑆 ⊂ 𝑉(𝐺 ∗ ).
Reminder: Setting 𝑆 = 0 → 𝑜(𝐺 ∗ ) = 0 → 𝑣(𝐺 ∗ ) is even.
Arie Girshson
306236399
Denote by 𝑈 the set of vertices of degree 𝑣 − 1 in 𝐺 ∗ . If 𝑈 = 𝑉, 𝐺 ∗ has a perfect matching, therefore,
assume 𝑈 ≠ 𝑉 and show that 𝐺 ∗ − 𝑈 is a disjoint union of complete graphs.
Suppose, to the contrary, that some component of 𝐺 ∗ − 𝑈 is not complete, meaning in this component
there are vertices 𝑥, 𝑦, 𝑧 that 𝑥𝑦 ∈ 𝐸(𝐺 ∗ ), 𝑦𝑧 ∈ 𝐸(𝐺 ∗ ), 𝑥𝑧 ∉ 𝐸(𝐺 ∗ ). Moreover, since 𝑦 ∉ 𝑈, there is a
vertex 𝑤 in 𝐺 ∗ − 𝑈 such that 𝑦𝑤 ∉ 𝐸(𝐺 ∗ ).
𝒚
𝒘
𝒙
𝒛
Since 𝐺 ∗ is a maximal graph (but not perfect matching), 𝐺 ∗ + 𝑒 has a perfect matching for all 𝑒 ∉ 𝐸(𝐺 ∗ ).
Let 𝑀1 and 𝑀2 be perfect matching each, in 𝐺 ∗ + 𝑥𝑧 and 𝐺 ∗ + 𝑦𝑤, respectively.
Denote by 𝐻 the sub-graph of 𝐺 ∗ ∪ {𝑥𝑧, 𝑦𝑤} induced by 𝑀1 ∆𝑀2 .
Since each vertex of 𝐻 has degree two, 𝐻 is a disjoint union of even cycles, since edges of 𝑀1 alternate
with edges of 𝑀2 around them.
Distinguish two cases:
Case 1: 𝑥𝑧 and 𝑦𝑤 are in different components of 𝐻. Then if 𝑦𝑤 is in cycle 𝐶 of 𝐻, the edges of 𝑀1 in 𝐶
with the edges of 𝑀2 not in 𝐶 constitute a perfect matching in 𝐺 ∗ , contradicting the definition of 𝐺 ∗ .
Case 1 Illustration:
𝒚
𝒙
𝒘
𝒛
𝑀1 − 𝑅𝑒𝑑 𝐹𝑢𝑙𝑙 𝐿𝑖𝑛𝑒
𝑀2 − 𝐺𝑟𝑒𝑒𝑛 𝐷𝑎𝑠ℎ𝑒𝑑 𝐿𝑖𝑛𝑒
Arie Girshson
306236399
Case 2: 𝑥𝑧 and 𝑦𝑤 are in the same component 𝐶 of 𝐻. By symmetry of 𝑥 and 𝑧, assume that the vertices
𝑥, 𝑦, 𝑤 and 𝑧 occur in that order on 𝐶. Then the edges of 𝑀1 in the section 𝑦𝑤 … 𝑧 of 𝐶, together with
the edge 𝑦𝑧 and the edges of 𝑀2 not in the section 𝑦𝑤 … 𝑧 of 𝐶 constitute a perfect matching in 𝐺 ∗ , and
again contradicting the definition of 𝐺 ∗ .
Case 2 Illustration:
𝒘
𝒚
𝑀1 − 𝑅𝑒𝑑 𝐹𝑢𝑙𝑙 𝐿𝑖𝑛𝑒
𝑀2 − 𝐺𝑟𝑒𝑒𝑛 𝐷𝑎𝑠ℎ𝑒𝑑 𝐿𝑖𝑛𝑒
𝒛
𝒙
According to the contradictions of cases 1 and 2, it follows that 𝐺 ∗ − 𝑈 is a disjoint union of complete
graphs, and according to the following schematic (already shown above):
𝒚
𝒙
𝒘
𝒛
𝑜(𝐺 ∗ − 𝑈) ≤ 𝑈. Therefore, at most |𝑈| of the components of 𝐺 ∗ − 𝑈 are odd, then 𝐺 ∗ has a perfect
matching – one vertex in each odd component of 𝐺 ∗ − 𝑈 is matched with a vertex of 𝑈, while the
remaining vertices in 𝑈, and in components of 𝐺 ∗ − 𝑈, are matched as follows:
..
…𝑮∗ − 𝑼
𝑶𝒅𝒅 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕𝒔 𝒐𝒇
..
… 𝒐𝒇 𝑮∗ − 𝑼
𝑬𝒗𝒆𝒏 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕𝒔
𝑮
…..
…
𝑼
Arie Girshson
306236399
Since 𝐺 ∗ was assumed to have no perfect matching, the desired contradiction is obtained.
Therefore, 𝐺 have perfect matching.