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We can also define the number of triangles that vertex <math>i</math> is involved in as <math>\delta (i)</math> and, as each triangle is counted three times, we can express the number of triangles in G as <math>\delta (G) = \frac{1}{3} \sum_{i\in V} \ \delta (i)</math>.  

We can also define the number of triangles that vertex <math>i</math> is involved in as <math>\delta (i)</math> and, as each triangle is counted three times, we can express the number of triangles in G as <math>\delta (G) = \frac{1}{3} \sum_{i\in V} \ \delta (i)</math>.  

我们也可以将'''<font color="#FF8000">顶点</font>'''<math>i</math>所涉及的三角形的数量定义为<math>\delta(i)</math>，并且由于每个三角形都被计数了三次，'''<font color="#FF8000">图</font>''<math>G</math>中三角形的个数为<math>\delta (G) = \frac{1}{3} \sum_{i\in V} \ \delta (i)</math>。

我们也可以将'''<font color="#FF8000">顶点</font>'''<math>i</math>所涉及的三角形的数量定义为<math>\delta(i)</math>，并且由于每个三角形都被计数了三次，'''<font color="#FF8000">图</font>'''<math>G</math>中三角形的个数为<math>\delta (G) = \frac{1}{3} \sum_{i\in V} \ \delta (i)</math>。

Assuming that triadic closure holds, only two strong edges are required for a triple to form. Thus, the number of theoretical triples that should be present under the triadic closure hypothesis for a vertex <math>i</math> is <math>\tau (i) = \binom{d_i}{2}</math>, assuming <math>d_i \ge 2</math>. We can express <math>\tau (G) = \frac{1}{3} \sum_{i\in V} \ \tau (i)</math>.

Assuming that triadic closure holds, only two strong edges are required for a triple to form. Thus, the number of theoretical triples that should be present under the triadic closure hypothesis for a vertex <math>i</math> is <math>\tau (i) = \binom{d_i}{2}</math>, assuming <math>d_i \ge 2</math>. We can express <math>\tau (G) = \frac{1}{3} \sum_{i\in V} \ \tau (i)</math>.

Assuming that triadic closure holds, only two strong edges are required for a triple to form. Thus, the number of theoretical triples that should be present under the triadic closure hypothesis for a vertex <math>i</math> is <math>\tau (i) = \binom{d_i}{2}</math>, assuming <math>d_i \ge 2</math>. We can express <math>\tau (G) = \frac{1}{3} \sum_{i\in V} \ \tau (i)</math>.

Assuming that triadic closure holds, only two strong edges are required for a triple to form. Thus, the number of theoretical triples that should be present under the triadic closure hypothesis for a vertex <math>i</math> is <math>\tau (i) = \binom{d_i}{2}</math>, assuming <math>d_i \ge 2</math>. We can express <math>\tau (G) = \frac{1}{3} \sum_{i\in V} \ \tau (i)</math>.

假设'''<font color="#FF8000">三元闭包</font>'''性质成立，则一个'''<font color="#FF8000">三元组</font>'''仅需要两条'''<font color="#FF8000">强联系</font>'''便可形成三角形。 因此，在'''<font color="#FF8000">三元闭包</font>'''性质成立的前提下理论上'''<font color="#FF8000">顶点</font>'''<math>i</math>所涉及的三角形的数量为<math>\tau(i) = \binom{d_i}{2}</math>， 假设<math>d_i \ge 2</math>。 我们可以表示<math>\tau(G) = \frac{1}{3} \sum_{i\in V} \ \tau(i)</math>。

假设'''<font color="#FF8000">三元闭包</font>'''性质成立，则一个三元组仅需要两条'''<font color="#FF8000">强联系</font>'''便可形成三角形。 因此，在'''<font color="#FF8000">三元闭包</font>'''性质成立的前提下，理论上'''<font color="#FF8000">顶点</font>'''<math>i</math>所涉及的三角形的数量为<math>\tau(i) = \binom{d_i}{2}</math>， 假设<math>d_i \ge 2</math>。 我们可以表示<math>\tau(G) = \frac{1}{3} \sum_{i\in V} \ \tau(i)</math>。

Now, for a vertex <math>i</math> with <math>d_i \ge 2</math>, the [[clustering coefficient]] <math>c(i)</math> of vertex <math>i</math> is the fraction of triples for vertex <math>i</math> that are closed, and can be measured as <math>\frac{\delta (i)}{\tau (i)}</math>. Thus, the [[clustering coefficient]] <math>C(G)</math> of graph <math>G</math> is given by <math>C(G) = \frac {1}{N_2} \sum_{i \in V, d_i \ge 2} c(i)</math>, where <math>N_2</math> is the number of nodes with degree at least 2.

Now, for a vertex <math>i</math> with <math>d_i \ge 2</math>, the [[clustering coefficient]] <math>c(i)</math> of vertex <math>i</math> is the fraction of triples for vertex <math>i</math> that are closed, and can be measured as <math>\frac{\delta (i)}{\tau (i)}</math>. Thus, the [[clustering coefficient]] <math>C(G)</math> of graph <math>G</math> is given by <math>C(G) = \frac {1}{N_2} \sum_{i \in V, d_i \ge 2} c(i)</math>, where <math>N_2</math> is the number of nodes with degree at least 2.