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# 数据科学代写|复杂网络代写Complex Network代考|PCS810

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## 数据科学代写|复杂网络代写Complex Network代考|Graph Partitioning Using the Cavity Method

The statistical mechanics formulation of the q-partitioning problem is done via the following ferromagnetic Potts Hamiltonian:
$$\mathcal{H}F({\sigma})=-\sum{i \neq j} J_{i j} \delta\left(\sigma_i, \sigma_j\right),$$
where $J_{i j}$ is the ${0,1}$ adjacency matrix of the graph and $\sigma_i$ denotes the Potts spin variable with $\sigma_i \in{1,2, \ldots, q}$. Once one finds the ground state under the constraint $\sum_i \delta\left(\sigma_i, \tau\right)=N / q$ for all $\tau \in{1,2, \ldots, q}$, one can write the total number of cut edges $C$ in the system using the ground state energy $E_g$ of the above Hamiltonian (6.1):
$$C_q=M+E_g=M\left(\frac{q-1}{q}-Q_q\right) .$$
Note the difference to (5.5). Also note that the modularity of the q-partition $Q_q$ can be expressed via Hamiltonian (6.1) as
$$Q_q=-\frac{\mathcal{H}_F}{M}-\frac{1}{q} .$$
This expression is only valid for magnetization zero, i.e., an exact q-partition.

## 数据科学代写|复杂网络代写Complex Network代考|Cavity Method at Zero Temperature

The ground state energy of (6.1) can be calculated by applying the cavity method at zero temperature following the approach presented by Mezard and Parisi [8] in the formulation for a Potts model as presented by Braunstein et al. $[9,10]$ for coloring random graphs. The energy of a system of $N$ spins is written as dependent on a “cavity spin” $\sigma_1$ via the “cavity field” $\boldsymbol{h}1$ : $$E^N\left(\sigma_1\right)=A-\sum{\tau=1}^q h_1^\tau \delta\left(\tau, \sigma_1\right)$$
Note that $h_1^\tau$ takes only integer values, if $J_{i j}$ is composed of only ${0,1}$. The components of the cavity field $\boldsymbol{h}i$ denote the change in energy of the system with a change in spin $i$. In general, these are different from the “effective fields” $\sum_j J{i j} \sigma_j$ acting on spin $\sigma_i$, which are used to calculate the magnetization. Adding a new spin $\sigma_0$ connected to $\sigma_1$, the energy of the now $N+1$ spin system is a function of both $\sigma_1$ and $\sigma_0$ :
$$E^{N+1}\left(\sigma_1, \sigma_0\right)=A-\sum_{\tau=1}^q h_1^\tau \delta\left(\tau, \sigma_1\right)-J_{10} \delta\left(\sigma_1, \sigma_0\right) .$$
One can now write this expression in such a way that it only depends on the newly added cavity spin $\sigma_0$ :
$$E^{N+1}\left(\sigma_0\right)=\min {\sigma_1} E^{N+1}\left(\sigma_1, \sigma_0\right) \equiv A-w\left(\boldsymbol{h}_1\right)-\sum{\tau=1}^q \hat{u}^\tau\left(J_{10}, \boldsymbol{h}_1\right) \delta\left(\tau, \sigma_0\right) .$$
The functions $w$ and $\hat{u}$ take the following form:
\begin{aligned} w(\boldsymbol{h}) & =\max \left(h^1, \ldots, h^q\right), \ \hat{u}^\tau(J, \boldsymbol{h}) & =\max \left(h^1, \ldots, h^\tau+J, \ldots, h^q\right)-w(\boldsymbol{h}) . \end{aligned}
From (6.8) one sees that $\hat{u}^\tau(\boldsymbol{h})$ is one, whenever the $\tau$ th component of $\boldsymbol{h}$ is maximal with respect to all other components in $\boldsymbol{h}$ and zero otherwise. Due to possible degeneracy in the components of $\boldsymbol{h}$, the vector $\hat{u}(\boldsymbol{h})$ may have more than one non-zero entry and is never completely zero.

## 数据科学代写|复杂网络代写Complex Network代考|Graph Partitioning Using the Cavity Method

q划分问题的统计力学公式是通过以下铁磁波茨哈密顿量来完成的:
$$\mathcal{H}F({\sigma})=-\sum{i \neq j} J_{i j} \delta\left(\sigma_i, \sigma_j\right),$$

$$C_q=M+E_g=M\left(\frac{q-1}{q}-Q_q\right) .$$

$$Q_q=-\frac{\mathcal{H}_F}{M}-\frac{1}{q} .$$

## 数据科学代写|复杂网络代写Complex Network代考|Cavity Method at Zero Temperature

(6.1)的基态能量可以按照Mezard和Parisi[8]在Braunstein等人$[9,10]$为随机图上色提出的Potts模型公式中提出的方法，在零温度下应用空腔法计算。一个$N$自旋系统的能量被写成依赖于“腔自旋”$\sigma_1$通过“腔场”$\boldsymbol{h}1$: $$E^N\left(\sigma_1\right)=A-\sum{\tau=1}^q h_1^\tau \delta\left(\tau, \sigma_1\right)$$

$$E^{N+1}\left(\sigma_1, \sigma_0\right)=A-\sum_{\tau=1}^q h_1^\tau \delta\left(\tau, \sigma_1\right)-J_{10} \delta\left(\sigma_1, \sigma_0\right) .$$

$$E^{N+1}\left(\sigma_0\right)=\min {\sigma_1} E^{N+1}\left(\sigma_1, \sigma_0\right) \equiv A-w\left(\boldsymbol{h}1\right)-\sum{\tau=1}^q \hat{u}^\tau\left(J{10}, \boldsymbol{h}_1\right) \delta\left(\tau, \sigma_0\right) .$$
$w$和$\hat{u}$函数的形式如下:
\begin{aligned} w(\boldsymbol{h}) & =\max \left(h^1, \ldots, h^q\right), \ \hat{u}^\tau(J, \boldsymbol{h}) & =\max \left(h^1, \ldots, h^\tau+J, \ldots, h^q\right)-w(\boldsymbol{h}) . \end{aligned}

## MATLAB代写

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