http://bit.ly/2NGRiIX ]]>

This is not done by MO or wikipedia. The lean theorem prover might be taking a step this direction.

I would prefer for such a project to reach through from mathematical idea to known engineering applications, or at least to PDE.

]]>I would start with mozilla MDN and take serious looks at codepen.io.

]]>Claim: Let be a bipartite graph on two vertex sets , each of size , with edges. Then the number of homomorphisms of into (with the convention that the vertices belong to and the vertices belong to is at most .

Proof: Let be a random variable taking values in , distributed uniformly on the edges of . It has entropy . It can be sampled by sampling from the marginal distribution of and then sampling from the conditional marginal of . Next, consider the random variable with sampled as before, and sampled as before, conditioned on , but independently. Since the ‘s are chosen independently we have

where I’ve used the trivial bound .

Next, swap and repeat; consider the random variable , with the above distribution. It can be sampled by taking from its marginal (I’m referring to the joint marginal of the ‘s), and then choosing from the conditional marginal. So do this, but choose from the conditional marginal independently.

We now get, similarly to before,

where I’ve used what seems like an incredibly wasteful bound .

To summarize: the random variable is supported on vertex sets which support a homomorphism of into , but not necessarily uniformly. Denoting the set of all homomorphisms by this gives

as required.

I’m currently teaching entropy techniques in my probabilistic-method course at Weizmann, and stumbled over this thread.

I want to point out that in the proof for the number of $C_4$’s there’s no need to do any calculations or considerations regarding the degree distribution. Here’s a proof for any complete bipartite graph that exemplifies this.

Claim: Let $G$ be a bipartite graph on two vertex sets $V_x,V_y$, each of size $n$, with $\alpha n^2$ edges. Then the number of homomorphisms of $K_{s,t}$ into $G$ (with the convention that the $s$ vertices belong to $V_x$ and the $t$ vertices belong to $V_y$ is at most $n^{s+t}\alpha^{st}$.

Proof: Let (X,Y) be a random variable taking values in $V_x \times V_y$, distributed uniformly on the edges of $G$. It has entropy $\log(\alpha n^2)$. It can be sampled by sampling $X=x_1$ from the marginal distribution of $X$ and then sampling $Y=y_1$ from the conditional marginal of $Y|X=x_1$. Next, consider the random variable $(X_1,Y_1,….Y_t)$ with $X_1=x_1$ sampled as before, and $Y_1=y_1,….Y_n=y_n$ sampled as before, conditioned on $X_1=x_1$, but independently. Since the $Y$’s are chosen independently we have

$H(X_1,Y_1,….Y_n) = H(X_1) + \sum H(Y_i |X_1) = tH(X,Y) – (t-1)H(X) \ge \log (\alpha^t n^{t+1})$, where I’ve used the trivial bound $H(X) \le \log(n)$.

Next, swap and repeat; consider the random variable $((Y_1,….Y_t),X_1)$, with the above distribution. It can be sampled by taking $(Y_1,….Y_t)$ from its marginal (I’m referring to the joint marginal of the $Y$’s), and then choosing $X_1$ from the conditional marginal. So do this, but choose $X_1,….X_s$ from the conditional marginal independently.

We now get, similarly to before,

$H(Y_1,…Y_t,X_1,….X_n)= H(Y_1,…Y_t) + \sum H(X_j| Y_1,…Y_t)=

sH(Y_1,….,Y_t,X_1) – (s-1)H(Y_1,….Y_t) \ge \log (\alpha^{st} n^{s+t}$, where I’ve used what seems like an incredibly wasteful bound $H(Y_1,…Y_t) \le \log(n^t)$.

To summarize: the random variable $(X_1,…X_s,Y_1,…Y_t)$ is supported on vertex sets which support a homomorphism of $K_s,t$ into $G$, but not necessarily uniformly. Denoting the set of all homomorphisms by $\Sigma$ this gives

$ \log (\alpha^{st} n^{s+t} \le H(X_1,….X_s,Y_1,…,Y_t) \le \log (|\Sigma)$ as required. ]]>

Are you a fellow of Trinity College? What do you say about their decision to pull out of USS and UCU’s call to boycott them?

]]>A combinatorial approach to density Hales-Jewett | Gowers

]]>http://bit.ly/2KAMaDk ]]>

“The one that uses the Möbius strip and the torus is called the Klein bottle.”

No, this’s not the Klein bottle (genus 2), but Dyck’s surface (genus 3).

In general, glue an n-holed torus with m copies of Möbius strip, you get a nonorientable two-dimensional manifold of genus 2n+m

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