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CSC 250

Theory of Computation

Smith Computer Science



Lecture Notes 30: More About Computational Complexity


Outline

This class we'll discuss:

Time Complexity Classes





Some Typical Classes



Check out examples Here







Activity 1 [2 minutes] How would you do this faster?:
(Wait; then Click)

 Also...What complexity is this?








Activity 2 [2 minutes] Is PALINDROME in \( \mathcal{O} (n^2) \)?:
(Wait; then Click)

 Also...What complexity is this?




Example:

SEL = {<M, A> | M sorts integer array A using SELECTION-SORT }



Activity 3 [2 minutes] Is \(SEL\) in \( \mathcal{O} (n^2) \)?:
(Wait; then Click)

 First: Build an algorithm that solves this in \( \mathcal{O} (n^2) \)... (this would mean YES!)
THEN, you may ask if \(SEL\) might belong to a lower complexity class ... (don't try...yet)












This is an example \( < G, s, t > \in PATH\)



Activity 4 [2 minutes] How would you Prove this?:
(Wait; then Click)

 You must build a Decider that solves it in \( \mathcal{O} (n^k) \) (Polynomial time)

Use Breadth-First-Search!
  1. On input \(< G, s, t>\)
  2. Mark all nodes rechable from the current node (starting at s and moving to the next in a running queue)
  3. Continue to expand until we reach \(t\) or \(\nexists\) another node
  4.   if we reach \(t\) : ACCEPT
  5. REJECT


What is the complexity!!!???

Complexity analysis:
Worst case: traversing every edge.
How many edges? ...

Another approach:
At most \(\mid V \mid\) phases are executed.
Each phase takes polynomial time to explore marked nodes and their outgoing edges.
(Imagine \(\mid V \mid\) nodes in a straight line VS \(\mid V \mid\) nodes all connected to each other)




















Activity 5 [2 minutes] How would you Prove this?:
(Wait; then Click)

 AutoTM_t = { |M| is a TM that accepts in ≤ t(|M|) steps }

Claim 1: AutoTM_t is decidable

Proof of Claim 1:
On input , simulate M on for t(|M|) simulated steps ADWID.



Claim 2: AutoTM_t can’t be decided by any basic TM in ≤ t(n) steps
Proof of claim 2: (everyone’s favorite contradiction argument)
Assume AutoTM_t is decided in time ≤ t(n) by some TM \(M_A\)
This implies that the complement is also decided by some other TM \( M_\overline{A} \) (just swap the accepting and rejecting states of \(M_A\) )



Thus, for every Turing machine M:

\(< M>\) ∈ \(\overline{AutoTM_t}\) iff \( M_\overline{A} \) accepts in time ≤ t(|M|)
And by taking the complement of the definition of AutoTM_t:

\(< M>\) ∈ \(AutoTM_t\) iff \( M_{\overline{A}} \) does not accept \(< M>\) in time ≤ t(|M|).

What happens if we plug in \( M_\overline{A} \) for M in both statements: \( M_\overline{A} \) ∈ \(\overline{AutoTM_t}\) iff \( M_\overline{A} \) accepts in time ≤ t(|\( M_\overline{A} \)|)
\( M_\overline{A} \) ∈ AutoTM_tc iff \( M_\overline{A} \) does not accept in time ≤ t(|M|).

– Contradiction!


















Homework


[Due for everyone]
TODO

[Optional]
TODO