# Lecture 20

## Palindrome Lower Bound

Theorem $$D(EQ)\geq n$$

Theorem Any single-tape TM that decides $$PAL=\{ww^R:w\in\{0,1\}^*\}$$ must have running time $$\Omega(n^2)$$.

Proof Let $$M$$ be a single-tape TM that decides $$PAL$$ with running time $$t(n)$$.

Goal: Define a protocol $$\pi$$ that computes $$EQ$$ with communication cost that depends on $$t(n)$$.

Idea 1: Alice+Bob can compute $$EQ(x,y)$$ by simulating $$M$$ on $$x|y^R$$. $$M$$ accepts iff $$x0^ny^R\in PAL\Leftrightarrow x=y$$

Define $$\pi$$ follows:

1. Alice simulates $$M$$ until the tape head crosses over to position $$n+1$$ on the input tape.
2. Alice sends the current state $$q$$ to Bob. $$O(1)$$ bits of communication.
3. Bob continues the simulation until the tape head crosses back to position $$n$$.
4. Bob send state $$q$$ to Alice. $$O(1)$$ bits of communication.
5. Repeat until $$M$$ halts. $$O(1)$$ bits sent by current simulator to tell the other if $$M$$ accepted or rejected.

Let $$m(x,y)$$ be the number of times $$M$$ crosses the boundary at position $$n$$ on the tape when the input is $$xy^R$$.

So total cost of $$\pi$$ is $$O(m(x,y))$$. With previouse theorems, $$n\leq D(EQ)\leq cost(\pi)\leq O(m(x,y))\leq O(t(2n))$$ so $$t(n)\geq \Omega(n)$$.

Idea 2: Simulate $$M$$ on $$x0^{2n}y^R$$ but choose the best boundary according to the input (to minimize number of handoffs) instead of just fixing the midpoint as that boundary

1. Simulate $$M$$ on $$x0^{2n}x^R$$. Let $$i(x)$$ be the index $$n, n+1,..,3n$$ corresponding to the boundary crossed the least number of time by $$M$$ or $$x0^{2n}x^R$$. Let $$m(x)$$ be number of times $$M$$ crossed the boundary $$i(x)$$.
2. Alice sends $$i(x)$$ and $$m(x)$$ to Bob. $$O(\log n+\log(t(4n)))$$
3. Bob computes $$i(y)$$ and $$m(y)$$ by simulateing $$M$$ on $$y0^{2n}y^R$$
4. Bob sends $$1$$ bit to say wheter $$i(x)=i(y)$$ and $$m(x)=m(y)$$
5. If $$i(x)\neq i(y)$$ or $$m(x)\neq m(y)$$, then $$x\neq y$$, so we are done. Otherwise, run the simulation as before but with handoffs at boundary $$i(x)$$. $$O(m(x))$$ bits of communication.

So the total cost of $$\pi$$ is $$O(m(x)+\log n)$$.

Claim: $$m(x)\leq\frac{t(4n)}{2n}$$

Proof Each of the $$t(4n)$$ transitions of $$M$$ crosses at most $$1$$ of the boundaries $$n$$, $$n+1$$, ..., $$3n$$, so at least one of those boundaries must be crossed $$\frac{t(4n)}{2n}$$ times.

So $$n\leq D(EQ)\leq cost(\pi)\leq O(\frac{4n}{n}+\log n)$$

$\Omega(n)\leq \frac{t(4n)}{n}+\log n$ $\Omega(n^2-n\log n)\leq t(4n)$ $\Omega(n^2)\leq t(4n)$ $\Omega(n^2)\leq t(n)$