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5.2 A Time Hierarchy

Lower Bounds on Time ComplexityTractability and Intractability

Intuitively, it seems obvious that some problems require more time to solve than others. Thefollowing result confirms this intuitive assessment while implying the existence of a timehierarchy for the class of language recognition problems.

Definitions A function T(n) is said to be time-constructible if there exists a T(n) time-

bounded, deterministic Turing machine that for each n has an input of length n on which it makesexactly T(n) moves. The function is said to befully time-constructible if there exists adeterministic Turing machine that makes exactly T(n) moves on each input of length n. Afunction S(n) is said to bespace-constructible if there exists an S(n) space-bounded,deterministic Turing machine that for each n has an input of length n on which it requires exactlyS(n) space. The function is said to befully space-constructible if there exists a deterministicTuring machine that requires exactly S(n) space on each input of length n.
http://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese3.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese3.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese3.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese1.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese1.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese1.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese1.html#tailtheory-bk-fivese1.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese1.html#tailtheory-bk-fivese1.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese1.html#tailtheory-bk-fivese1.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese2.html#tailtheory-bk-fivese2.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese2.html#tailtheory-bk-fivese2.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese2.html#tailtheory-bk-fivese2.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-five.html#theory-bk-fivese2.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-five.html#theory-bk-fivese2.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-five.html#theory-bk-fivese2.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk.html#Q2-60002-8http://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk.html#Q2-60002-8http://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese2.html#Q1-60002-9http://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese2.html#Q1-60002-9http://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese2.html#Q1-60002-10http://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese2.html#Q1-60002-10http://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese2.html#Q1-60002-10http://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese2.html#Q1-60002-9http://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk.html#Q2-60002-8http://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-five.html#theory-bk-fivese2.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese2.html#tailtheory-bk-fivese2.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese1.html#tailtheory-bk-fivese1.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese1.htmlhttp://www.cse.ohio-state.edu/~gurari/theory-bk/theory-bk-fivese3.html


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Example 5.2.1 The deterministic Turing machine M in Figure5.2.1

Figure 5.2.1A T(n) = 2n time-bounded, deterministic Turing machine.

makes exactly t(x) = |x| + (number of 1's in x) moves on a given input x. t(x) = 2|x| when xcontains no 0's, and t(x) < 2|x| when x contains 0's.

The existence of M implies that T(n) = 2n is a time-constructible function, because

a. M is 2n time-bounded, andb. For each n there exists the input 1n of length n on which M makes exactly 2n moves.

The existence of the deterministic Turing machine M does not imply that 2n is fully time-constructible, because M does not make exactly 2n moves on each input of length n. However,M can be modified to show that 2n is a fully time-constructible function.

Convention In this section Mx denotes a Turing machine that is represented by the string x ofthe following form. If x = 1jx0 for some j 0 and for some standard binary representation x0 of adeterministic Turing machine M, then Mx denotes M. Otherwise, Mx denotes a deterministicTuring machine that accepts no input. The string x is said to be apadded binary representationof Mx.

Theorem 5.2.1 Consider any function T1(n) and any fully time-constructible function T2(n),that for each c > 0 have an nc such that T2(n) c(T1(n))

2 for all n nc. Then there is a languagewhich is inDTIME(T2(n)) but not inDTIME(T1(n)).

Proof Let T1(n) and T2(n) be as in the statement of the theorem. Let U be a universal Turingmachine similar to the universal Turing transducer in the proof of Lemma5.1.1. The maindifference is that here U assumes an input (M, x) in which M is represented by a padded binaryrepresentation instead of a standard binary representation. U starts each computation by goingover the "padding" 1j until it reaches the first 0 in the input. Then U continues with itscomputation in the usual manner while ignoring the padding. U uses a third auxiliary work tapefor keeping track of the distance of its input head from the end of the padding. The result isshown by diagonalization over the language L = { v | v is in {0, 1}*, and U does not accept (Mv,v) in T2(|v|) time }.
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L is obtained from the diagonal of the table Tuniversal (see Figure5.2.2).

Figure

5.2.2

Hypothetical table Tuniversal indicating membership in the language { (Mw, u) | Uaccepts (Mw, u) in T2(|u|) time }.

In the table Tuniversal the entry at row Mw and column u is equal to 1 if U accepts (Mw, u) in T2(|u|)time, and it is equal to 0 if U does not. The proof relies on the observation that each O(T1(n))time-bounded, deterministic Turing machine Mx0 that accepts L has also a padded representation

x for which U can simulate the whole computation of Mx on x in T2(|x|) time. Consequently, Mxaccepts x if and only if U does not accept (Mx, x) or, equivalently, if and only if Mx does notaccept x.

Specifically, for the purpose of showing that L is not inDTIME(T1(n)), assume to the contrarythat L is in the class. Under this assumption, there is a dT1(n) time-bounded, deterministic Turingmachine M that accepts L, for some constant d. Let x0 be a standard binary representation of M,and c be the corresponding constant cM implied by Lemma5.1.1for the representation x0 of M.Let x = 1jx0 for some j that satisfies j + c(dT1(j + |x0|))

2 T2(j + |x0|), that is, x = 1jx0 for a large

enough j to allow U sufficient time T2(|x|) for simulating the whole computation of Mx on inputx. Such a value j exists because for big enough j the following inequalities hold.

j + c (j + |x 0|) + c

T1(j + |x0|) + c

(1 + cd )

T2(j + |x0|)
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Consider the string x = 1jx0. By definition, |x| = j + |x0| and so j + c(dT1(|x|))2 T2(|x|). Moreover,

x is a padded binary representation of M. For the string x one of the following two cases musthold. However, neither of them can hold, so implying the desired contradiction to the assumptionthat L is inDTIME(T1(n)).

Case 1 x is in L. The assumption together with L = L(Mx) imply that Mx accepts x in dT1(|x|)time. In such a case, by Lemma5.1.1U accepts (Mx, x) in j + c(dT1(|x|))

2 T2(|x|) time.On the other hand, x in L together with the definition of L imply that U does not accept xin T2(|x|) time. The contradiction implies that this case cannot hold.

Case 2

x is not in L. The assumption together with L = L(Mx) imply that Mx does not accept x. Insuch a case, U does not accept (Mx, x) either. On the other hand, x not in L together withthe definition of L imply that U accepts (Mx, x). The contradiction implies that this casecannot hold either.

To show that L is inDTIME(T2(n)) consider the deterministic four auxiliary-work-tape Turing

machine M that on input x proceeds according to the following algorithm.Step 1M stores (Mx, x) in its first auxiliary work tape. That is, M stores the string x, followed bythe separator 01, followed by the representation 011 of the left endmarker , followed byx, followed by the representation 0111 of the right endmarker $. In addition, M enclosesthe sequence of strings above between the "left endmarker" and the "right endmarker"

, respectively.

Step 2M computes the value of T2(|x|) and stores it in the second auxiliary work tape.

Step 3

M follows the moves of U on the content of its first auxiliary work tape, that is, on (Mx,x). M uses its third and fourth auxiliary work tapes for recording the content of the twoauxiliary work tapes of U. During the simulation M interprets as the left endmarker ,and as the right endmarker $. M halts in an accepting configuration if it determines thatU does not reach an accepting state in T2(|x|) moves. Otherwise, M halts in anonaccepting configuration.

By construction, the Turing machine M is of O(T2(|x|)) time complexity. The fully time-constructibility of T2(n) is required for Step 2.

Example 5.2.2 Let T1(n) = nkand T2(n) = 2

n. T1(n) and T2(n) satisfy the conditions ofTheorem5.2.1. Therefore the classDTIME(2n) properly contains the classDTIME(nk).

Lower Bounds on Time Complexity

In addition to implying the existence of a time hierarchy for the language recognition problems,Theorem5.2.1can be used to show lower bounds on the time complexity of some problems.Specifically, consider any two functions T1(n) and T2(n) that satisfy the conditions ofTheorem5.2.1. Assume that each membership problem Ki for a language inDTIME(T2(n)) canbe reduced by a T3(n) time-bounded, deterministic Turing transducer Mi to some fixed problemK (see Figure5.2.3).
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Figure

5.2.3A set of Turing transducers M1, M2, . . . for reducing the problems K1, K2, . . . inDTIME(T2(n)) to a given language recognition problem K. Each Mi on instance x ofKi provides an instance y of K, where K has the answer yes for y if and only if Ki hasthe answer yes for x.

In addition, assume that each such Mi on input x of length n provides an output y of length f(n) atmost. Then the membership problems for the languages inDTIME(T2(n)) are decidable in T3(n)+ T(f(n)) time if K can be solved in T(n) time. In such a case, a lower bound for the timecomplexity T(n) of K is implied, since by Theorem5.2.1the classDTIME(T2(n)) contains a

problem that requires more than cT1(n) time for each constant c, that is, the inequality T3(n) +T(f(n)) > cT1(n) must hold for infinitely many n's. The lower bound is obtained by substituting mfor f(n) to obtain the inequality T(m) > cT1(f

-1(m)) - T3(f-1(m)) or, equivalently, the inequality

T(n) > cT1(f-1(n)) - T3(f

-1(n)).

Example 5.2.3 Consider the time bounds T1(n) = 2an, T2(n) = 2

bn for b > 2a, and T3(n) = f(n) =n log n. For such a choice, T3(n) + T(f(n)) > cT1(n) implies that n log n + T(n log n) > c2

an. Bysubstituting m for n log n it follows that T(m) > c2an - m = c2am/log n - m c2am/log m - m 2dm/log mor, equivalently, that T(n) > 2dn/log n for some constant d.

The approach above for deriving lower bounds is of special interest in the identification ofintractable problems, that is, problems that require impractical amounts of resources to solve.Such an identification can save considerable effort that might otherwise be wasted in trying tosolve intractable problems.

Tractability and Intractability

In general, a problem is considered to be tractable if it is of polynomial time complexity. This isbecause its time requirements grow slowly with input length. Conversely, problems of
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exponential time complexity are considered to be intractable, because their time requirementsgrow rapidly with input length and so can be practically solved only for small inputs. Forinstance, an increase by a factor of 2 in n, increases the value of a polynomial p(n) of degree k byat most a factor of 2k. On the other hand, such an increase at least squares the value of 2p(n).

The application of the approach above in the identification of intractable problems employspolynomially time-bounded reductions.

A problem K1 is said to bepolynomially time reducible to a problem K2 if there existpolynomially time-bounded, deterministic Turing transducers Tfand Tg that for each instance I1of K1 satisfy the following conditions (see Figure5.2.4).

Figure

5.2.4

Reduction by polynomially time-bounded, deterministic Turing transducers TfandTg.

a. Tfon input I1 gives an instance I2 of K2.b. K1 has a solution S1 at I1 if and only if K2 has a solution S2 at I2, where S1 is the output of

Tg on input S2.

In the case that K1 and K2 are decision problems, with no loss of generality it can be assumedthat Tg computes the identity function g(S) = S, that is, that Tg on input S2 outputs S1 = S2.

A given complexity class Cof problems can be used to show the intractability of a problem K byshowing that the following two conditions hold.

a. Ccontains some intractable problems.

b. Each problem in Cis polynomially time reducible to K, that is, K is at least as hard tosolve as any problem in C.

Once a problem K is determined to be intractable, it then might be used to show the intractabilityof some other problems by showing that K is polynomially time reducible to . In such a case,the easier K is the easier the reductions are, and the larger the class of such applicable problems

is.
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The observation above sparks our interest in the "easiest" intractable problems K, and in thecomplexity classes Cwhose intractable problems are all "easiest" intractable problems.

In what follows, a problem K is said to be a C-hard problem with respect to polynomial timereductions, or just a C-hard problem when the polynomial time reductions are understood, if

every problem in the class Cis polynomially time reducible to the problem K. The problem K issaid to be C-complete if it is a C-hard problem in C.

Our interest here is in the cases that C=NPand C=PSPACE.

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5.2 A Time Hierarchy

Lower Bounds on Time ComplexityTractability and Intractability

Intuitively, it seems obvious that some problems require more time to solve than others. Thefollowing result confirms this intuitive assessment while implying the existence of a timehierarchy for the class of language recognition problems.

Definitions A function T(n) is said to be time-constructible if there exists a T(n) time-bounded, deterministic Turing machine that for each n has an input of length n on which it makesexactly T(n) moves. The function is said to befully time-constructible if there exists adeterministic Turing machine that makes exactly T(n) moves on each input of length n. Afunction S(n) is said to bespace-constructible if there exists an S(n) space-bounded,deterministic Turing machine that for each n has an input of length n on which it requires exactlyS(n) space. The function is said to befully space-constructible if there exists a deterministicTuring machine that requires exactly S(n) space on each input of length n.

Example 5.2.1 The deterministic Turing machine M in Figure5.2.1
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Figure 5.2.1A T(n) = 2n time-bounded, deterministic Turing machine.

makes exactly t(x) = |x| + (number of 1's in x) moves on a given input x. t(x) = 2|x| when xcontains no 0's, and t(x) < 2|x| when x contains 0's.

The existence of M implies that T(n) = 2n is a time-constructible function, because

a. M is 2n time-bounded, andb. For each n there exists the input 1n of length n on which M makes exactly 2n moves.

The existence of the deterministic Turing machine M does not imply that 2n is fully time-constructible, because M does not make exactly 2n moves on each input of length n. However,M can be modified to show that 2n is a fully time-constructible function.

Convention In this section Mx denotes a Turing machine that is represented by the string x ofthe following form. If x = 1jx0 for some j 0 and for some standard binary representation x0 of adeterministic Turing machine M, then Mx denotes M. Otherwise, Mx denotes a deterministicTuring machine that accepts no input. The string x is said to be a padded binary representationof Mx.

Theorem 5.2.1 Consider any function T1(n) and any fully time-constructible function T2(n),that for each c > 0 have an nc such that T2(n) c(T1(n))

2 for all n nc. Then there is a languagewhich is inDTIME(T2(n)) but not inDTIME(T1(n)).

Proof Let T1(n) and T2(n) be as in the statement of the theorem. Let U be a universal Turingmachine similar to the universal Turing transducer in the proof of Lemma5.1.1. The maindifference is that here U assumes an input (M, x) in which M is represented by a padded binaryrepresentation instead of a standard binary representation. U starts each computation by goingover the "padding" 1j until it reaches the first 0 in the input. Then U continues with itscomputation in the usual manner while ignoring the padding. U uses a third auxiliary work tapefor keeping track of the distance of its input head from the end of the padding. The result isshown by diagonalization over the language L = { v | v is in {0, 1}*, and U does not accept (Mv,v) in T2(|v|) time }.

L is obtained from the diagonal of the table Tuniversal (see Figure5.2.2).
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Figure

5.2.2Hypothetical table Tuniversal indicating membership in the language { (Mw, u) | Uaccepts (Mw, u) in T2(|u|) time }.

In the table Tuniversal the entry at row Mw and column u is equal to 1 if U accepts (Mw, u) in T2(|u|)time, and it is equal to 0 if U does not. The proof relies on the observation that each O(T1(n))time-bounded, deterministic Turing machine Mx0 that accepts L has also a padded representationx for which U can simulate the whole computation of Mx on x in T2(|x|) time. Consequently, Mxaccepts x if and only if U does not accept (Mx, x) or, equivalently, if and only if Mx does notaccept x.

Specifically, for the purpose of showing that L is not inDTIME(T1(n)), assume to the contrarythat L is in the class. Under this assumption, there is a dT1(n) time-bounded, deterministic Turingmachine M that accepts L, for some constant d. Let x0 be a standard binary representation of M,and c be the corresponding constant cM implied by Lemma5.1.1for the representation x0 of M.Let x = 1jx0 for some j that satisfies j + c(dT1(j + |x0|))

2 T2(j + |x0|), that is, x = 1jx0 for a large

enough j to allow U sufficient time T2(|x|) for simulating the whole computation of Mx on inputx. Such a value j exists because for big enough j the following inequalities hold.

j + c (j + |x 0|) + c

T1(j + |x0|) + c

(1 + cd )

T2(j + |x0|)

Consider the string x = 1jx0. By definition, |x| = j + |x0| and so j + c(dT1(|x|))2 T2(|x|). Moreover,

x is a padded binary representation of M. For the string x one of the following two cases musthold. However, neither of them can hold, so implying the desired contradiction to the assumptionthat L is inDTIME(T1(n)).
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Case 1x is in L. The assumption together with L = L(Mx) imply that Mx accepts x in dT1(|x|)time. In such a case, by Lemma5.1.1U accepts (Mx, x) in j + c(dT1(|x|))

2 T2(|x|) time.On the other hand, x in L together with the definition of L imply that U does not accept xin T2(|x|) time. The contradiction implies that this case cannot hold.

Case 2 x is not in L. The assumption together with L = L(Mx) imply that Mx does not accept x. Insuch a case, U does not accept (Mx, x) either. On the other hand, x not in L together withthe definition of L imply that U accepts (Mx, x). The contradiction implies that this casecannot hold either.

To show that L is inDTIME(T2(n)) consider the deterministic four auxiliary-work-tape Turingmachine M that on input x proceeds according to the following algorithm.

Step 1

M stores (Mx, x) in its first auxiliary work tape. That is, M stores the string x, followed bythe separator 01, followed by the representation 011 of the left endmarker , followed byx, followed by the representation 0111 of the right endmarker $. In addition, M encloses

the sequence of strings above between the "left endmarker" and the "right endmarker", respectively.

Step 2

M computes the value of T2(|x|) and stores it in the second auxiliary work tape.

Step 3M follows the moves of U on the content of its first auxiliary work tape, that is, on (Mx,x). M uses its third and fourth auxiliary work tapes for recording the content of the twoauxiliary work tapes of U. During the simulation M interprets as the left endmarker ,and as the right endmarker $. M halts in an accepting configuration if it determines thatU does not reach an accepting state in T2(|x|) moves. Otherwise, M halts in anonaccepting configuration.

By construction, the Turing machine M is of O(T2(|x|)) time complexity. The fully time-

constructibility of T2(n) is required for Step 2.

Example 5.2.2 Let T1(n) = nkand T2(n) = 2

n. T1(n) and T2(n) satisfy the conditions ofTheorem5.2.1. Therefore the classDTIME(2n) properly contains the classDTIME(nk).

Lower Bounds on Time Complexity

In addition to implying the existence of a time hierarchy for the language recognition problems,Theorem5.2.1can be used to show lower bounds on the time complexity of some problems.Specifically, consider any two functions T1(n) and T2(n) that satisfy the conditions ofTheorem5.2.1. Assume that each membership problem Ki for a language inDTIME(T2(n)) canbe reduced by a T3(n) time-bounded, deterministic Turing transducer Mi to some fixed problemK (see Figure5.2.3).
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Figure

5.2.3

A set of Turing transducers M1, M2, . . . for reducing the problems K1, K2, . . . in

DTIME(T2(n)) to a given language recognition problem K. Each Mi on instance x ofKi provides an instance y of K, where K has the answer yes for y if and only if Ki hasthe answer yes for x.

In addition, assume that each such Mi on input x of length n provides an output y of length f(n) atmost. Then the membership problems for the languages inDTIME(T2(n)) are decidable in T3(n)+ T(f(n)) time if K can be solved in T(n) time. In such a case, a lower bound for the timecomplexity T(n) of K is implied, since by Theorem5.2.1the classDTIME(T2(n)) contains aproblem that requires more than cT1(n) time for each constant c, that is, the inequality T3(n) +T(f(n)) > cT1(n) must hold for infinitely many n's. The lower bound is obtained by substituting m

for f(n) to obtain the inequality T(m) > cT1(f

-1

(m)) - T3(f

-1

(m)) or, equivalently, the inequalityT(n) > cT1(f-1(n)) - T3(f

-1(n)).

Example 5.2.3 Consider the time bounds T1(n) = 2an, T2(n) = 2

bn for b > 2a, and T3(n) = f(n) =n log n. For such a choice, T3(n) + T(f(n)) > cT1(n) implies that n log n + T(n log n) > c2

an. Bysubstituting m for n log n it follows that T(m) > c2an - m = c2am/log n - m c2am/log m - m 2dm/log mor, equivalently, that T(n) > 2dn/log n for some constant d.

The approach above for deriving lower bounds is of special interest in the identification ofintractable problems, that is, problems that require impractical amounts of resources to solve.Such an identification can save considerable effort that might otherwise be wasted in trying tosolve intractable problems.

Tractability and Intractability

In general, a problem is considered to be tractable if it is of polynomial time complexity. This isbecause its time requirements grow slowly with input length. Conversely, problems ofexponential time complexity are considered to be intractable, because their time requirementsgrow rapidly with input length and so can be practically solved only for small inputs. For
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instance, an increase by a factor of 2 in n, increases the value of a polynomial p(n) of degree k byat most a factor of 2k. On the other hand, such an increase at least squares the value of 2p(n).

The application of the approach above in the identification of intractable problems employspolynomially time-bounded reductions.

A problem K1 is said to bepolynomially time reducible to a problem K2 if there existpolynomially time-bounded, deterministic Turing transducers Tfand Tg that for each instance I1of K1 satisfy the following conditions (see Figure5.2.4).

Figure

5.2.4Reduction by polynomially time-bounded, deterministic Turing transducers TfandTg.

a. Tfon input I1 gives an instance I2 of K2.b. K1 has a solution S1 at I1 if and only if K2 has a solution S2 at I2, where S1 is the output of

Tg on input S2.

In the case that K1 and K2 are decision problems, with no loss of generality it can be assumedthat Tg computes the identity function g(S) = S, that is, that Tg on input S2 outputs S1 = S2.

A given complexity class Cof problems can be used to show the intractability of a problem K byshowing that the following two conditions hold.

a. Ccontains some intractable problems.b. Each problem in Cis polynomially time reducible to K, that is, K is at least as hard to

solve as any problem in C.

Once a problem K is determined to be intractable, it then might be used to show the intractabilityof some other problems by showing that K is polynomially time reducible to . In such a case,the easier K is the easier the reductions are, and the larger the class of such applicable problems

is.

The observation above sparks our interest in the "easiest" intractable problems K, and in thecomplexity classes Cwhose intractable problems are all "easiest" intractable problems.
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In what follows, a problem K is said to be a C-hard problem with respect to polynomial timereductions, or just a C-hard problem when the polynomial time reductions are understood, ifevery problem in the class Cis polynomially time reducible to the problem K. The problem K issaid to be C-complete if it is a C-hard problem in C.

Our interest here is in the cases that C=NPand C=PSPACE.

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ranbir [email protected]

ranbir [email protected]

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1. Theory of computation - Wikipedia, the free encyclopedia

en.wikipedia.org/wiki/Theory_of_computation

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In theoretical computer science and mathematics, the theory of computation is the branch that

deals with how efficiently problems can be solved on a model of...

History-Branches-Models of computation-References
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dels_of_computationhttp://en.wikipedia.org/wiki/Theory_of_computation#Models_of_computationhttp://en.wikipedia.org/wiki/Theory_of_computation#Models_of_computationhttp://en.wikipedia.org/wiki/Theory_of_computation#Referenceshttp://en.wikipedia.org/wiki/Theory_of_computation#Referenceshttp://en.wikipedia.org/wiki/Theory_of_computation#Referenceshttp://en.wikipedia.org/wiki/Theory_of_computation#Referenceshttp://en.wikipedia.org/wiki/Theory_of_computation#Models_of_computationhttp://en.wikipedia.org/wiki/Theory_of_computation#Brancheshttp://en.wikipedia.org/wiki/Theory_of_computation#Historyhttps://www.google.co.in/search?biw=1280&bih=628&q=related:en.wikipedia.org/wiki/Theory_of_computation+theory+of+computation&tbo=1&sa=X&ei=K2v-UeXVL5DKrAfm14HAAw&sqi=2&ved=0CCoQHzAAhttp://webcache.googleusercontent.com/search?q=cache:vzbse6Zw27gJ:en.wikipedia.org/wiki/Theory_of_computation+&cd=1&hl=kn&ct=clnk&gl=inhttp://en.wikipedia.org/wiki/Theory_of_computationhttps://www.google.co.in/support/websearch/bin/answer.py?answer=179386&hl=knhttps://www.google.co.in/search?q=theory+of+computati&biw=1280&bih=628&source=lnt&tbs=li:1&sa=X&ei=K2v-UeXVL5DKrAfm14HAAw&sqi=2&ved=0CB4QpwUoAQhttps://www.google.co.in/search?q=theory+of+computation&biw=1280&bih=628&source=lnt&tbs=qdr:y&sa=X&ei=K2v-UeXVL5DKrAfm14HAAw&sqi=2&ved=0CBoQpwUoBQhttps://www.google.co.in/search?q=theory+of+computation&biw=1280&bih=628&source=lnt&tbs=qdr:m&sa=X&ei=K2v-UeXVL5DKrAfm14HAAw&sqi=2&ved=0CBkQpwUoBAhttps://www.google.co.in/search?q=theory+of+computation&biw=1280&bih=628&source=lnt&tbs=qdr:w&sa=X&ei=K2v-UeXVL5DKrAfm14HAAw&sqi=2&ved=0CBgQpwUoAwhttps://www.google.co.in/search?q=theory+of+computation&biw=1280&bih=628&source=lnt&tbs=qdr:d&sa=X&ei=K2v-UeXVL5DKrAfm14HAAw&sqi=2&ved=0CBcQpwUoAghttps://www.google.co.in/search?q=theory+of+computation&biw=1280&bih=628&source=lnt&tbs=qdr:h&sa=X&ei=K2v-UeXVL5DKrAfm14HAAw&sqi=2&ved=0CBYQpwUoAQ


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2. An Introduction to the Theory of Computationwww.cse.ohio-state.edu/~gurari/ theory-bk/theory-

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An Introduction to the Theory of Computation. Eitan Gurari, Ohio State University Computer

Science Press, 1989, ISBN 0-7167-8182-4. Copyright Eitan M.

3. NPTEL :: Computer Science and Engineering - Theory of Computation

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NPTEL >> Computer Science and Engineering >> Theory of Computation (Video ) Under

Review >> Lecture-01-Theory of Computation...

4. [PDF]

Theory of Computation - Computer Science and Engineering

www.cse.iitd.ernet.in/~sak/courses/toc/toc-recursive-function- theory

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18 2009 - Theory of computation is of course a very broad and deep area, and it is ... on

Theory of Computation in India follow the classic text by Hopcroft ...

5. Mod-01 Lec-01 Lecture-01-Theory of Computation - YouTube:5 61:05

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26 2012 - nptelhrd
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Theory of Computation by Prof. Somenath Biswas, Department of Computer Science &

Engineering,IIT Kanpur...

6. Computer Sc - Theory of Computation - YouTube5

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Theory of Computation by Prof.Kamala Krithivasan,Department of Computer Science and

Engineering,I...

7. Theory of Computationwww.aduni.org/courses/ theory

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ADUni.org is the website of the alumni of ArsDigita University (ADU). ADU was a one-year,

intensive post-baccalaureate program in Computer Science based ...

8. [PDF]

Introduction to theory of computation

csustan.csustan.edu/~tom/Lecture-Notes/ Computation/computation

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gg T theory of computation traditionally deals with processing an input

string of symbols into an output string of symbols. Note that in the.

9. Theory of Computationtheoryofcomputationg

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8 2012 - Theory of Computation blog about DFA, NFA, PDA, Context Free Grammar,

Turing Machine, Pumping Lemma, GATE Papers, Material, ...
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19/19

10. homepage | Theory of Computation

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Research areas include algorithms, complexity theory, computation and biology, cryptography

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