User - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-24T20:13:23Zhttp://mathoverflow.net/feeds/user/31016http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/123029/reference-request-gluing-pasting-perfect-graphs-on-a-cliqueReference request: gluing/pasting perfect graphs on a clique?astar2013-02-26T20:21:08Z2013-02-26T23:59:41Z
<p>What's the correct reference please for the proof that pasting/gluing two perfect graphs on a clique results in another perfect graph?</p>
http://mathoverflow.net/questions/120602/convergence-rate-of-iterated-nonlinear-equationsConvergence rate of iterated nonlinear equations?astar2013-02-02T16:46:13Z2013-02-04T23:51:22Z
<p>For $i=1, \dots, n$ ($n$ could be large) we have variables $x_i$ and $y_i$ relating to probability bounds s.t. $x_i, y_i \geq 0, x_i+y_i \leq 1 \; \forall i$. Each $i$ has a constant $\theta_i$, and some of the pairs $(i,j)$ have a nonzero constant $w_{ij}$ (weight of interaction, positive or negative) in which case say $j \in N(i)$.</p>
<p>Let $W_i=\sum_{j \in N(i): w_{ij}>0} w_{ij}$ (sum of positive edges), $V_i=-\sum_{j \in N(i): w_{ij}<0} w_{ij}$ (minus sum of negative edges), $\alpha_{ij} =\exp (|W_{ij}|) -1$.</p>
<p>We initialize all $x_i, y_i =0$ then iteratively update as shown below via intermediate variables ${L_i,U_i}$. All iterations are monotonically nondecreasing, $x_i, y_i$ are bounded above hence converge, empirically typically rapidly. </p>
<blockquote>
<p>Can we prove something on the rate of convergence? (any pointers appreciated)</p>
</blockquote>
<p>The problem may be easier if we restrict all $w_{ij}>0$, which would still be very helpful, in which case $V_i=0$ and just rewriting the pseudocode below, each iteration sets e.g. new $x_i \leftarrow \left(1 + \frac{\exp(-\theta_i)}{\prod_{j \in N(i)} \left(1 + \frac{\alpha_ij x_j}{1+\alpha_{ij}(1-y_i)(1-x_j)} \right) } \right)^{-1}$.</p>
<pre><code>REPEAT until convergence {
FOR each i {
Li, Ui=1 // Initialize this pass
FOR each j \in N(i) {
IF w_{ij}>0 {
Li *= 1+ \frac{\alpha_{ij} x_j} {1+\alpha_{ij} (1-y_i) (1-x_j)}
Ui *= 1+ \frac{\alpha_{ij} y_j} {1+\alpha_{ij} (1-x_i) (1-y_j)} }
ELSE {
Li *= 1+ \frac{\alpha_{ij} y_j} {1+\alpha_{ij} (1-y_i) (1-y_j)}
Ui *= 1+ \frac{\alpha_{ij} x_j} {1+\alpha_{ij} (1-x_i) (1-x_j)} }
}
x_i = 1 / (1+exp(-\theta_i + V_i)/Li)
y_i = 1 / (1+exp(\theta_i + W_i)/Ui)
}
</code></pre>