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C1: The Indexing Problem: Different Proofs (3)

Reminder: This post contains 1429 words · 5 min read · by Xianbin

For the previous two different proofs, please look at the second proof and the first proof.

The Third Proof

By Yao’s minimax, we only need to construct a distribution (here we use the uniform distribution μ{0,1}n\mu\sim\{0,1\}^n) over which any deterministic algorithm cannot beat Ω(n)\Omega(n) with probability at most 1/2c1/2-c.

Let Π\Pi be the transcript of Alice with size \ell. We know that Π\Pi fully depends on the string SμS\in \mu where S=S1S2SnS = S_1S_2\ldots S_n. Notice that SiSjS_i \perp S_j for any iji\neq j.

I(Π:S)H(Π)I(\Pi: S)\leq H(\Pi) \leq \ell.

So, we only need to prove that

I(Π:S)Ω(n)I(\Pi: S) \geq \Omega(n)

Since SiSjS_i\perp S_j for any iji\neq j, by the chain rule, we have

I(Π:S)=i=1nI(Π:Si)I(\Pi:S) = \sum_{i=1}^n I(\Pi:S_i)

Therefore, we only need to prove that

I(Π:Si)=Ω(1)I(\Pi:S_i) = \Omega(1) I(Π:Si)=H(Π)H(ΠSi)=H(Si)H(SiΠ)I(\Pi:S_i) = H(\Pi)- H(\Pi \mid S_i) = H(S_i) - H(S_i \mid \Pi)

Due to the uniform distribution, H(Si)=H(1/2)=1H(S_i) = H(1/2) = 1.

Now, let us see H(SiΠ)H(S_i \mid \Pi).

H(SiΠ)=Eπ[H(SiΠ=π)]H(S_i \mid \Pi) = \mathbb{E}_\pi [H(S_i \mid \Pi=\pi)]

Let Pr[Π is wrongAlice sends message pi and Bob has input i]=rπiP_r [\Pi \text{ is wrong} \mid \text{Alice sends message pi and Bob has input i} ]= r^i_\pi

According to the condition, we have Ei,mrmi1/2c\mathbb{E}_{i,m}r^i_m \leq 1/2 -c.

Then, by Jensen’s inequality,

H(SiΠ)=Eπ[H(SiΠ=π)]H(Eπ[SiΠ=π])H(Eπ[rmi])H(S_i \mid \Pi) = \mathbb{E}_\pi [H(S_i \mid \Pi =\pi)] \leq H(\mathbb{E}_\pi[S_i\mid \Pi=\pi])\\ \leq H(\mathbb{E}_\pi[r^i_m])

The last one is due to Fano’s inequality. (It seems that reference [1] is wrong here.)

Consider all ii, we have

Ei[H(SiΠ)]=H(Ei,m[rmi])\mathbb{E}_i [H(S_i \mid \Pi)] = H(\mathbb{E}_{i,m}[r^i_m])

Then, we get that

1H(1/2c)=Ω(1)1-H(1/2-c) = \Omega(1)

Then, it is proved.

Reference

[1]. Communication Complexity 16 Sep, 2011 (@ TIFR) 10. Pointer Chasing.