# Lecture 15

## Definitions

Definition The (worst-case) running time of a single-tape TM $$M$$ is the function $$t:\mathbb{N}\rightarrow\mathbb{N}\cup\infty$$ where $$t(n)$$ is the maximum number of transitions followed by $$M$$ before halting over all inputs of length $$n$$.

Definition The class of languages $$TIME(t(n))$$ for some function $$t:\mathbb{N}\rightarrow\mathbb{N}$$ is the set of languages $$L$$ such that there exist a single-tape TM $$M$$ with running time $$O(t(n))$$ that decides $$L$$.

## Some Examples

### Regular Language

Theorem If $$L$$ is regular, then $$L\in TIME(n)$$

Proof From any DFA that recognizes $$L$$, we can build a corresponding single-tape TM that halts after $$n+1$$ transitions

### $$L=\{0^k1^k:k\geq 0\}$$

Theorem If $$L=\{0^k1^k:k\geq 0\}$$ is in $$TIME(n^2)$$

Proof TM needs $$\frac{n}{2}$$ transitions to skip $$0$$s and check the corresponding $$1$$.

Proof Sketch Erase every second $$0$$ and $$1$$, check parities. Each pass takes $$O(n)$$ transitions. There are $$O(\log n)$$ passes.

Theorem Any single-tape TM that decides $$L$$ has running time $$\Omega (n\log n)$$

Proof Bonus

Theorem there is a 2-tape TM that decides $$L$$ in time $$O(n)$$.

Proof Use the second tape as a stack.

Corollary $$TIME(t(n))\neq TIME^{multi}(t(n))$$

Theorem For every language $$L$$ that can be decided by $$k$$-tape TM with running time $$t(n)$$ , there is a $$2-tape$$ $$TM$$ with running time $$O(t(n)\log(t(n)))$$.

### Palindrome

Theorem The language $$PAL=\{xx^R:x\in\{0,1\}^*\}$$ can be decided by $$2$$-tape TM with running time $$O(n)$$ but every $$1$$-tape TM that decides $$PAL$$ has running time $$\Omega(n^2)$$.

## Linear Speedup Theorem

Theorem For any constant $$c>0$$, if there is a multi-tape TM $$M$$ that decides $$L$$ in time $$t(n)$$, there is also a multi-tape TM $$M'$$ that decides $$L$$ in time $$ct(n)+2n$$.

Question Is $$TIME^{multi}(t(n))=TIME^{multi}(n)$$ for every $$t:\mathbb{N}\rightarrow\mathbb{N}$$? No

### Weak Linear Speedup Theorem

If there is a single-tape TM $$M$$ that decides $$L$$ in time $$t(n)$$, there is a $$2-tape$$ TM $$M'$$ that decides $$L$$ in time $$\frac{1}{2}t(n)+2n$$.

Proof We can choose $$\Gamma'=\Gamma^3$$

$(x_1,x_2,x_3),(x_3,x_4,x_5),...$

Compressing the input takes time $$n$$ and going back to then beginning take time $$\frac{n}{2}$$.

Compressing all the transistions of $$M$$ within a single window into $$1$$ transition means $$M'$$ makes at most $$\frac{1}{2}t(n)$$ extra transitions.