# Proof of Grinberg’s theorem

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**Question:**

Proof of Grinberg’s theorem

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**Technifi Expert’s Answer:**

Let G be a Hamiltonian graph and F a cycle basis of G. 𝑓 𝑖 𝑓 𝑖 ∈ F is referred to a cycle (an interior face in Grinberg theorem) of order i 𝑖 ≤ V . 𝑓 𝑖 denotes the number of 𝑓 𝑖 and F the number of F. Therefore, we have 𝑓 = 𝑓3 + 𝑓4 +⋯+ 𝑓 V ,V𝑖 ∈ V 𝑓 𝑖 . Since the union of the whole cycles (both vertices and edges) in a cycle basis is the given graph itself, then the number of all vertices of the given graph equals to the cardinality of this union, that is V = V3 ∪V4 ∪⋯∪V V . By inclusion-exclusion principle, we have that

V = V𝑖 V 𝑖=3 − |V𝑖 ∩V𝑗 V 3≤i<𝑗≤ V |

+ |V𝑖 ∩V𝑗 V 3≤i<𝑗<𝑘≤ V ∩V𝑘|− ⋯+ −1 V −1|V3 ∩ V4 ∩V5⋯∩V V |………………………. 3.2

Note that in G every Hamiltonian cycle can be represented by the symmetric difference of a set of cycles in a cycle basis of G and the union of every pair of disjoint cycles is null. Let V𝑎 ∩V𝑏 denote the item of that the value of the intersection of every pair of joint cycles is 2. Then, we rewrite the equality (3.2),

V = V𝑖 V 𝑖=3 − |V𝑖 ∩ V𝑗 V 3≤𝑖<𝑗≤ V |……………………….3.3

Since V𝑖 ∩ V𝑗 = 2 and the number of pair of joint cycles is F −1, then we have |V𝑖 ∩V𝑗 V 3≤𝑖<𝑗≤ V | = 2( F − 1). Using 𝑓3 + 𝑓4 +⋯+ 𝑓 V to replace F , we obtain |V𝑖 ∩V𝑗 V 3≤𝑖<𝑗≤ V | = 2 𝑓3 + 𝑓4 + ⋯+ 𝑓 V − 1 = 2 𝑓 𝑖 V 𝑖=3 − 1 ……………………….3.4

Furthermore, V𝑖 V 𝑖=3 is the sum of all subsets of vertices, that is

V𝑖 V 𝑖=3 = V3 + V4 +⋯+ V V ………………………. 3.5

where V3 = 3 𝑓3 , V4 = 4 𝑓4 , ⋯, V V = V ∙ 𝑓 V ; so the equality (3.5) can be rewritten as follows,

V𝑖 V 𝑖=3 = 3 𝑓3 +4 𝑓4 +⋯+ V ∙ 𝑓 V = 𝑖 𝑓 𝑖 V 𝑖=3 ………………………. 3.6

Using the equality (3.4) and (3.6) to substitute the correspondent items in the equality (3.3), we derive

4 / 5

𝑖 𝑓 𝑖 V 𝑖=3 −2 𝑓 𝑖 V 𝑖=3 −1 = V ………………………. 3.7

Rearranging the expression gives:

𝑖𝑓 𝑖 −2𝑓 𝑖 = V − 2 V 𝑖=3 ………………………. 3.8

Particularly, in the case of a planar graph, we can replace 𝑓 𝑖 with interior faces 𝑓′𝑖 and obtain,

𝑖𝑓′𝑖 − 2𝑓′𝑖 = V − 2 V 𝑖=3 ………………………. 3.9

Hence, the equality (3.9) is same to the equality (1.1) that completes the proof.

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