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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>.