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· 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) over which any deterministic algorithm cannot beat Ω(n) with probability at most 1/2−c.
Let Π be the transcript of Alice with size ℓ. We know that Π fully depends on the string S∈μ where S=S1S2…Sn. Notice that Si⊥Sj for any i=j.
I(Π:S)≤H(Π)≤ℓ.
So, we only need to prove that
I(Π:S)≥Ω(n)
Since Si⊥Sj for any i=j, by the chain rule, we have
I(Π:S)=i=1∑nI(Π:Si)
Therefore, we only need to prove that
I(Π:Si)=Ω(1)
I(Π:Si)=H(Π)−H(Π∣Si)=H(Si)−H(Si∣Π)
Due to the uniform distribution, H(Si)=H(1/2)=1.
Now, let us see H(Si∣Π).
H(Si∣Π)=Eπ[H(Si∣Π=π)]
Let Pr[Π is wrong∣Alice sends message pi and Bob has input i]=rπi
According to the condition, we have Ei,mrmi≤1/2−c.
Then, by Jensen’s inequality,
H(Si∣Π)=Eπ[H(Si∣Π=π)]≤H(Eπ[Si∣Π=π])≤H(Eπ[rmi])
The last one is due to Fano’s inequality. (It seems that reference [1] is wrong here.)
Consider all i, we have
Ei[H(Si∣Π)]=H(Ei,m[rmi])
Then, we get that
1−H(1/2−c)=Ω(1)
Then, it is proved.
Reference
[1]. Communication Complexity 16 Sep, 2011 (@ TIFR) 10. Pointer Chasing.