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CSCI 2670 Introduction to Theory of Computing

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CSCI 2670Introduction to Theory of Computing

November 15, 2005

- Today
- Comparing complexity of different computing models

- This week
- Finish 7.1 and do 7.2

November 15, 2005

- Office hours changed tomorrow
- 11:30 to 1:00 instead of 11:00 – 12:30

November 15, 2005

- We can examine an algorithm to determine how long it will take to halt on an input of length n
- The amount of time to complete is called the algorithms complexity class
Definition: Let t:N→N be a function. The time complexity class, TIME(t(n)), is defined as follows.

TIME(t(n))={L|L is a language decided by an O(t(n)) time TM}

November 15, 2005

- Finding minimum element in a set
- Amount of time depends on the structure of the input
- If set is a sorted array?
- O(1)

- If set is an unsorted array?
- O(n)

- If set is a balanced sorted tree?
- O(log n)

November 15, 2005

- The complexity of our algorithms depends on assumptions about our data & other model assumptions
- The complexity of an algorithm can vary depending on the machine we use
- Recall we assume reasonable encoding of all numbers
- Any integer of value v should take O(log v) space

November 15, 2005

- Example, let L={w | w is a palindrome}
- How long will it take us to decide L on a standard TM?
- Go back and forth crossing off matching symbols at beginning and end
- O(n2)

- How long will it take us to decide L on a 2-tape TM?
- Copy string
- Compare symbols reading forward on tape 1 and backward on tape 2
- O(n)

November 15, 2005

Theorem: Let t(n) be a function, where t(n) n. Then every t(n) time multitape TM has an equivalent O(t2(n)) time single-tape TM

Proof idea: Consider structure of equivalent single-tape TM. Analyzing behavior shows each step on multi-tape machine takes O(t(n)) on single tape machine

November 15, 2005

0 1 ~ ~ ~ ~ ~ ~

a a a ~ ~ ~ ~ ~

M

a b ~ ~ ~ ~ ~ ~

# 0 1 # a a a # a b # ~ ~

S

Equivalent machines

November 15, 2005

Simulating k-tape behavior

- Single tape start string is
#w#~#...#~#

- Each move proceeds as follows:
- Start at leftmost slot
- Scan right to (k+1)st # to find symbol at each virtual tape head
- Make second pass making updates indicated by k-tape transition function
- When a virtual head moves onto a #, shift string to right

November 15, 2005

- Analyzing simulation of k-tape machine
- Each step on single-tape machine has two phases
- Scan tape
- Perform operations

- How long does first phase take?
- Length of string on tape
- Each portion has O(t(n)) length (this occurs if tape heads only move right)

November 15, 2005

- How long does second phase take?
- Perform k steps
- Each step may require a right shift

- Each step takes O(t(n)) time
- Total of k steps is O(t(n)) because k is a constant

- Perform k steps
- What’s the total time?
- O(t(n)) steps each take O(t(n)) time
- Total time is O(t2(n))

November 15, 2005

Definition: Let P be a non-deterministic Turing machine. The running time of P is the function f:N→N, where f(n) is the maximum number of steps that P uses on any branch of its computation in any input of length n.

November 15, 2005

f(n)

…

…

accept

accept/reject

reject

Deterministic

Non-deterministic

November 15, 2005

Theorem: Let t(n) be a function where t(n)n. Then every t(n) time non-deterministic single-tape Turing machine has an equivalent 2O(t(n)) time deterministic single-tape Turing machine.

November 15, 2005

Proof: Given a non-deterministic TM, P, running in t(n) time, construct a 3-tape deterministic TM that simulates P.

The height of the tree is at most t(n). Assume the maximum number of branches in the tree is b. Therefore, the number of leaves in the tree is O(bt(n)).

November 15, 2005

Claim: The total number of nodes is less than twice the number of leaves – i.e. O(bt(n)).

(Σ0≤k≤n bk) bn = (1 – bn+1)/(1-b) bn

< – bn+1/(1-b) bn

= - b / (1-b)

≤ 2

November 15, 2005

Deterministic TM does a breadth-first search of the non-deterministic TM’s tree.

Total time to search tree is O(t(n)) to travel from the root to a leaf × O(bt(n)), the number of leaves.

O(t(n)bt(n)) = O(2log_2 t(n) 2(log_2 b)t(n)) = O(2O(t(n)))

November 15, 2005

Are we done?

No! We constructed a 3-tape TM with running time O(2O(t(n)))

Single-tape TM will take O((2O(t(n)))2)=O(22O(t(n)))=O(2O(t(n)))

Are we done?

Yes!

November 15, 2005