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Measuring Complexity

Measuring Complexity. Lecture 36 Section 7.1 Fri, Nov 16, 2007. Time Complexity Classes. Let t : N  R + be a function. The time complexity class TIME( t ( n )) is the set of all languages that are decidable by an O ( t ( n )) Turing machine. Examples. SHIFT is in TIME( n ).

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Measuring Complexity

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  1. Measuring Complexity Lecture 36 Section 7.1 Fri, Nov 16, 2007

  2. Time Complexity Classes • Let t : N  R+ be a function. • The time complexity class TIME(t(n)) is the set of all languages that are decidable by an O(t(n)) Turing machine.

  3. Examples • SHIFT is in TIME(n). • COPY is in TIME(n2). • INCR and DECR are in TIME(n). • If ADD uses INCR and DECR, then what is the time complexity class of ADD? • If MULT uses ADD, then what is the time complexity class of MULT?

  4. Integer Factorization • Let the input be an n-bit integer k. • Suppose a Turing machine uses the following strategy to factor the integer. • Divide k by each integer from 2 to k – 1. • Assume division requires constant time.

  5. Integer Factorization • What will be the run time of this machine?

  6. Single Tape vs. Multitape • Theorem: If a problem P is in TIME(t(n)) for k-tape Turing machine and t(n)  n, then it is in TIME(t2(n)) for a single-tape Turing machine.

  7. Single Tape vs. Multitape • Proof: • The single-tape equivalent S of the multitape machine M records the contents of the k tapes of M on its single tape. • That requires O(n) steps. • In the t(n) transitions of M, it can scan at most t(n) cells of each of its tapes. Call that the “active” portion.

  8. Single Tape vs. Multitape • For each transition of S: • S scans the tape once to determine the current symbol on each tape. • The time required by the scan is at most kt(n) = O(t(n)). • Then S simulates on each portion of its tape the action of M on each of its tapes. • That action may require shifting right all remaining cells.

  9. Single Tape vs. Multitape • Each shift requires up to time t(n). • The time required is at most t(n) O(t(n)) = O(t2(n)). • The total is O(n) + O(t2(n)) = O(t2(n)).

  10. Deterministic vs. Nondeterministic • Theorem: If a problem P is in TIME(t(n)) for a nondeterministic Turing machine, then it is in TIME(2O(t(n))) for a deterministic Turing machine.

  11. Deterministic vs. Nondeterministic • Proof: • Let N be a nondeterministic Turing machine that decides P. • Consider a tree whose branches represent the different possible execution paths of N. • Let b be the largest number of branches at any node of the tree.

  12. Deterministic vs. Nondeterministic • No path through the tree can have length greater than t(n). • So the total number of leaves is no greater than bt(n). • A deterministic Turing machine D will do a breadth-first search of the tree, searching for the accept state.

  13. Deterministic vs. Nondeterministic • For every node searched, D begins at the root node and follows the path to that node. • That requires time at most t(n). • The number of nodes is less than twice the number of leaves. • So there are at most 2bt(n) nodes.

  14. Deterministic vs. Nondeterministic • The total search time required is at most O(t(n)(2bt(n))) = O(bO(t(n))). • So D is in TIME(bO(t(n))). • Finally, D is a 3-tape machine. • It can be simulated by a single-tape machine that is in TIME((bO(t(n)))2) = TIME(2O(t(n))).

  15. Example • The factoring problem can be solved nondeterministically in O(n) time, where n equals the length of the number. (How?) • Therefore, the factoring problem can be solved deterministically in O(2O(n)) time.

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