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

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