論文提要:
圖論上，談到兩點相似(similar)，是這樣定義的:若a，b是圖G上的兩點，且存在一箇定義於V (G)的某排列ϕ，滿足:(1)ϕ∈Aut (0) 及(2) ϕ(a) =b. 則稱G 之a. b兩點相似。由定義吾人固然知必有一ϕ∈Aut(O). 滿足ϕ(a) =b. 如果G之a. b兩點相似的話。然而，有人不禁會問:若G上任二點a. b相似，則是否存在一ϕ∈Aut (G) .同時滿足ϕ (a)=b 且ϕ (b) =a 呢7?(本文稱G為對稱相似(symmetrically similar). 若答案為真時。)
作者在研究GRAPH RECONSTRUCTION CONJECTURE時，亦產生相同的疑問。本文乃作者試就此一問題，將樹型圖(Tree)加以研究，發現：凡是具有有限點的樹形圖(finite tree) 皆具備此特性。
首先，吾人知: 任意樹形圖乃一不具環路的聯結圖( acyclic cconnected graph ).且任二不同點，a. b僅可決定出唯一的(a-b)路逕(path)。
本文先針對此路徑觀察出二項特性(1)當(a-b)為偶路徑，c為其中點，且ϕ (a) = b.ϕ∈Aut (G) 時，ϕ作用在c之後不變。(2)當(a-b)為奇路徑，y. z為其二中點，且ϕ (a) = b，ϕ∈Aut(G)時，ϕ作用在y. z 之後對調(即ϕ (y)=z 且φ(z)=y) 。其證明包含在定理2. 1. 7 及定理2. 1. 10 立中。
其次，以定理2. 1. 7 及定理2. 1. 10 為基礎，本文將證明出確實存在一ϕ∈Aut (G) .同時滿足ϕ (a) = b 且ϕ (b) = a. 此處G為一樹形圖。
由於找不到反例，本文將給一箇Conjecture 作為總結，即任一有限simple圖皆具對稱相似性。

[Introduction]
[Introduction]
It's well known in graph theory that if u and v are two vertices of a graph G and for some automophism ϕ of G. ϕ (u)=v then u and v are said to be similar. A graph G is said to be vertex-symmetric (edge-symmetric) if every pair of its vertices (edges) are similar.(see [5]. pp. 171-175)
Several properties concerning these similarities have been dealt with in some papers. For example,
(1) If u and v are similar, then G-u≅G-v. (see [6])
(2) Not every regular edge-symmetric graph is vertex-symmetric. (see [4])
Especially, (1) is frequently occurred in the articles about the famous problem--graph reconstruction conjecture. e.g. [2]. [7] and [8]. And probably, the research of these similarities of graphs is a key to solve this reconstruction conjecture.
In this paper, we will investigate a special kind of similarity of trees. That is, we'll assert that if u and v are two similar vertices of a finite tree T. then there is an automorphism ¢ of T satisfying: ¢(u) = v and ¢(v) = u. ( In such case. we say that T is symmetrically similar (Definition 2.1.2) or T preserves symmetric similarity . )
In Part I. we only give the fundamental definitions and concepts about graph theory as the preliminaries for our main subject. And in general, we follow the terminology and notations of [1]. [3] and [5].
In Part II, we'll present some lemmas. e.g. Lemma 2.1.4. Lemma 2.1.5. Lemma 2.1.6. Lemma 2.1.8. Lemma 2.1.9 and Lemma 2.2.1 and we will prove:
Theorem 2.1.7 Let T be a tree. aT ̃b and (a-b): a=X_0, X_1,…, X_(d-1), X_d=c, X_(d+1),…, X_(ad-1), X_ad, X_(ad+1)=b be an odd path of T where d<0. If ϕ(a)=b, ϕ∈Aut(T), then ϕ∈(X_1)= X_(d-1),, i=0, 1,…, d.
Theorem 2.1.10 Let T be a tree. AT ̃b and (a-b): a=X_0, X_1,…, X_(d-1), X_d=y, X_(d+1)=z,…, X_(ad-1), X_ad, X_(ad+1)=b be an odd path of T where d>0. If ϕ(a)=b, ϕ∈Aut(T), then ϕ∈(X_1)= X_(d-1), i=0, 1,…, d.
Theorem 2.2.2 Let T be a tree. AT ̃b and (a-b): a=X_0, X_1,…, X_(d-1), X_d=c, X_(d+1),…, X_(ad-1), X_ad=b is an even path of T where d>1, then ∃ψ∈Aut(T) s.t. (1) ψ(a)=b, ψ(b)=a and (2) ψ(x)=x, ∀x∈{w∈V(T)│Tc<w>≠Tc<a>}, Tc<w>≠Tc<b>.
Theorem 2.2.3 Let T be a tree. AT ̃b and (a-b): a=X_0, X_1,…, X_(d-1), X_d=y, X_(d+1)=z,…, X_ad, X_(ad+1)=b be an odd path of T where d>0. then ∃ψ∈Aut(T) s.t. (1) ψ(a)=b, ψ(b)=a
Since the length of any (a-b) path of the tree T is either even or odd, imploying Theorem 2.2.2-(1) 2.2.3. we immediately have:
Theorem 2.2.4 Every Tree is symmetrically similar
At last, we give a conjecture as a conclusion for this paper, i.e. Conjecture: Every simple graph is symmetrically similar.

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