content_image/394410.jpg content_image/394411.jpg content_image/394412.jpg content_image/394414.jpg content_image/394413.jpg content_image/394415.jpg
So the time needed for dynamic balance is 

In the simulations, both for the binary and the ternary cases, if the final value σ ′ i results in the denominator of β (see Eq. (3)) being zero, we propose it happens with probability zero, while the other values happen with probability one (see Eq. (17) for example). In this man- ner, no matter what M (0) (or M (0) and ρ (0)) is, the 

<img src='content_image/394410.jpg'>

network will reach the dynamically balanced state. 

When each node has three spin values, the variations of the densities of each triad relation with time are as follows, 

## 2. The generalized model 

<img src='content_image/394411.jpg'>

When the network gets dynamically balanced, it satisfies 

where i = 1 , 2 , 3 , ..., 10, along with the normalization 

Although Eqs. (A11) are very hard to solve, from ap- pendix A 1 and the symmetric spin transition mechanism, 

<img src='content_image/394412.jpg'>

<img src='content_image/394414.jpg'>

we expect n + ≃ n − and M ( ∞ ) ≃ 0 in the dynamically balanced state. Defining n + ≃ n − = x and n 0 = y in the dynamically balanced state, the densities of each type of the triangles are 

where l, m and n are the numbers of positive, neutral and negative nodes in a$_{i}$ respectively. C$_{l}$mn is the corre- sponding combinatorial number. For example C$_{201}$ = 3. Because near M = 0, there are fluctuations. The spin transition probabilities are ( M ̸ = 0) 

<img src='content_image/394413.jpg'>

<img src='content_image/394415.jpg'>

