Holographic Approximation: Difference between revisions
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\frac{Z_0(T,e)}{Z_1(T,e)}\\ | \frac{Z_0(T,e)}{Z_1(T,e)}\\ | ||
&= | &= | ||
\ | \left( | ||
\sum_{\ell=0}^k\sum_{S\in{[k]\choose \ell}}f_\ell \prod_{i\in S}Z_{1}(T_i,e_i)\prod_{i\in [k]\setminus S}Z_0(T_i,e_i) | \sum_{\ell=0}^k\sum_{S\in{[k]\choose \ell}}f_\ell \prod_{i\in S}Z_{1}(T_i,e_i)\prod_{i\in [k]\setminus S}Z_0(T_i,e_i) | ||
\right) | |||
\Bigg / | |||
\left( | |||
\sum_{\ell=0}^k\sum_{S\in{[k]\choose \ell}}f_{\ell+1} \prod_{i\in S}Z_1(T_i,e_i)\prod_{i\in [k]\setminus S}Z_0(T_i,e_i) | \sum_{\ell=0}^k\sum_{S\in{[k]\choose \ell}}f_{\ell+1} \prod_{i\in S}Z_1(T_i,e_i)\prod_{i\in [k]\setminus S}Z_0(T_i,e_i) | ||
\right) | |||
\end{align} | \end{align} | ||
</math> | </math> |
Revision as of 20:42, 23 April 2012
Holant Problem
Recursion on tree
[math]\displaystyle{ Z_b(G,e)=\#\{\sigma_G\mid \sigma_G(e)=b\}=\sum_{\sigma\in[q]^G\atop\sigma(e)=b}wt(\sigma) }[/math]
[math]\displaystyle{ \begin{align} Z_0(T,e) &= \sum_{\ell=0}^k\sum_{S\in{[k]\choose \ell}}f_\ell \prod_{i\in S}Z_{1}(T_i,e_i)\prod_{i\in [k]\setminus S}Z_0(T_i,e_i)\\ Z_1(T,e) &= \sum_{\ell=0}^k\sum_{S\in{[k]\choose \ell}}f_{\ell+1} \prod_{i\in S}Z_1(T_i,e_i)\prod_{i\in [k]\setminus S}Z_0(T_i,e_i) \end{align} }[/math]
[math]\displaystyle{ \begin{align} R_T &= \frac{Z_0(T,e)}{Z_1(T,e)}\\ &= \left( \sum_{\ell=0}^k\sum_{S\in{[k]\choose \ell}}f_\ell \prod_{i\in S}Z_{1}(T_i,e_i)\prod_{i\in [k]\setminus S}Z_0(T_i,e_i) \right) \Bigg / \left( \sum_{\ell=0}^k\sum_{S\in{[k]\choose \ell}}f_{\ell+1} \prod_{i\in S}Z_1(T_i,e_i)\prod_{i\in [k]\setminus S}Z_0(T_i,e_i) \right) \end{align} }[/math]