To keep track of the current position,Nwrites the name of the path ofM(x)currently being traversed on track 1 of the working tape

(1)

Theorem

L⊆NL⊆P⊆NP⊆PSPACE.

Proof: The inclusionsL⊆NLandP⊆NPare immediately clear, since every DTM, by definition, is a special NTM.

To prove thatNP⊆PSPACE, letM be any NTM running in timep(n)for somep∈IPol.

To define a DTMNthat decidesL(M)in polynomial space, we take advantage of an important property of the complexity resource space:

Unlike the time resource,space is reusable.

Using the same space on its working tape again and again, DTMN performs adepth-first searchthrough the computation tree ofM(x).

(2)

Construct DTMN, running on inputx of lengthn, as follows:

In addition to its input tape,N has one working tape subdivided into three tracks.

Recall that every polynomial is space-constructible. N first marks the spacep(n)on its working tape, i.e., exactlyp(n)cells.

Then,N systematically traverses the computation tree ofM(x) (assumed to be a full binary tree) by adepth-first search.

To keep track of the current position,Nwrites the name of the path ofM(x)currently being traversed on track 1 of the working tape.

A path name can be encoded in binary by a lengthp(n)string s∈ {0,1}^∗: thei^thbit ofs represents thei^thbranching ofM(x).

The leftmost bits ofsmay be marked, so as to indicate that this initial part of the path has already been processed.

(3)

Initially,Nwrites the path name 0^p(n)on track 1 and it writes the start configuration ofM(x)on track 2.

Then,N starts the search by simulating the computation ofM(x) along the path whose name, says, is currently written on track 1.

N writes two successive configurations occurring along this path alternately on the tracks 2 and 3.

LetC0`_M C1`_M · · · `_M C_p(n)be the sequence of configurations corresponding tos.

(4)

Foriwith 1≤i<p(n), consider the three successive configurationsCi−1,C_i, andC_i+1, where

(a) Ci−1`_M C_i `_MC_i+1,

(b) Ci is the first configuration alongsas yet unvisited, (c) the firsti−1 bits ofs are currently marked, and (d) track 2, say, currently containsCi−1.

Then,N writesC_i on track 3, deleting the former content of track 3 and marking thei^th bit ofson track 1.

N thus proceeds, alternately switching the roles of track 2 and track 3, until the current path is completely processed and all bits ofsare marked.

If an accepting configuration is reached, thenN halts and acceptsx.

(5)

Ifshas been processed without reaching an accepting configuration, then

Nadds a one in binary to the content of track 1 and

keeps traversing the as yet unvisited configurations on that new path.

If all the paths ofM(x)have been processed without success, thenN rejectsx.

It follows thatL(N) =L(M). SinceNworks deterministically in polynomial space,

NP⊆PSPACE.

(6)

To prove the remaining inclusion,NL⊆P, letMb be any NTM operating in logarithmic space, and letx be any input of lengthn.

Note that the above systematic search through the entire computation tree doesnotwork here:

In fact, any log-space bounded NTM can spend as much asq(n)steps along each path, whereq∈IPol, which results in a total of 2^q(n)

potential computation paths. Why?

This is becauseDSPACE(s)⊆DTIME(2^ILin(s))for, say,s = loggives computation paths of length2^c·logⁿ=2^logⁿ^c =n^c =q(n).

Thus, a polynomial-time bounded DTM has no chance of checking every path ofM(xb )up to its full length ofq(n)steps.

(7)

Fortunately, however, an argument analogous to the proof that DSPACE(s)⊆DTIME(2^ILin(s)) fors≥log

shows that there exists a constantc, depending onMb only, such that the number of distinct configurations ofM(xb )is bounded above by

2^c·logⁿ=2^logⁿ^c =n^c. (1)

That is, many of the configurations in the full depthq(n)computation tree ofM(x)b must occur repeatedly.

Hence, one can construct a DTMNb that, on inputx, searches through only a polynomial-size part of the computation tree ofM(xb ), truncating any path as soon as some configuration is encountered twice.

(8)

To this end,Nb uses three working tapes.

Tape 1 is again subdivided into three tracks and is used the same way the working tape of DTMN is used in the above proof of

NP⊆PSPACE.

That is, track 1 again keeps track of the current position in the search, by storing the current path name and marking how much of it has already been processed.

Tracks 2 and 3 again store two successive configurations, alternately producing the next one along the current path whose name is written on track 1.

(9)

In addition, DTMNb has two more tapes.

Tape 2 keeps a list of every new configuration ofM(xb )as yet found:

Whenever a new configuration, call itC, is produced on either track 2 or track 3 of tape 1, it is compared with each configuration currently contained in tape 2 to check whether or notCindeed is new.

If so,C is added to the list on tape 2;

otherwise, the current path ofM(x)b can safely be truncated.

Tape 3 contains a binary counter of lengthc·logn, wherecis the constant from Equation (1).

This counter is incremented by one each time a new configuration is added to tape 2.

(10)

If an accepting configuration ofM(xb )is found in the course of this process, thenNb halts and acceptsx.

If the search throughM(xb )is completed without success or if the counter ontape 3is full, i.e., the maximum number ofn^c distinct configurations is found and none of them is accepting, thenNb rejectsx.

It follows thatL(N) =b L(M).b

(11)

To estimate the time needed, note thatM(xb )has at mostn^c configurations of lengthO(logn)each.

Thus, the number of steps needed to compare the current configuration on tape 1 with the entire content oftape 2is in O(n^c·logn)and thus in

O(n^c+1).

Hence, comparing each of the at mostn^c possible configurations on tape 1 with the content oftape 2requires altogether at most

O(n^2c+1) steps.

(12)

Similarly, the process of producing new configurations on tape 1 and copying them ontotape 2requires at most

O(n^c+1)

steps.

Incrementing the counter on tape 3 at mostn^c times can be done in parallel.

Summing up,N(xb )needs time no more than polynomial inn.

Hence,NL⊆P. q

(13)

Corollary (to the proof)

1 If t is space-constructible, then

NTIME(t)⊆DSPACE(t).

2 If s≥logis constructible in time2^ILin(s), then NSPACE(s)⊆DTIME(2^ILin(s)).

This corollary immediately implies that:

NTIME(t)⊆DTIME(2^ILin(t⁾) and NSPACE(s)⊆DSPACE(2^ILin(s)), which upperbounds, for both time and space, the costs of trading nondeterminism for determinism.

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What about the lower bounds?

Is an exponential increase in the resources really necessary in order to deterministically simulate nondeterminism?

For the time resource, the answer to this question is not known.

For the space resource, the answer is no!

Savitch’s Theorem shows that for the resource space, already a quadratic increase is enough to deterministically simulate nondeterminism.

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Theorem (Savitch’s Theorem)

If s≥log is space-constructible, then

NSPACE(s)⊆DSPACE(s²).

(16)

Proof:

LetAbe any set inNSPACE(s), and letMbe some NTM acceptingAand working in spaces(n)on inputs of lengthn.

Goal: Construct a DTMN decidingAin spaceO(s²).

The constant implicit in theOnotation can safely be neglected by the Linear Compression Theorem.

Sinces≥log,MacceptsAin time t(n)≤2^c·s(n)

for some constantc that depends on the program ofM only and can be easily determined.

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ACCEPT_M is the unique accepting configuration ofM on any input, and

START_M(x)is the unique start configuration ofMon inputx. Letk =c·s(n), where we assume that 2^k−1<t(n)≤2^k, which impliesdlogt(n)e=k.

Suppose that the configurations ofM(x)are suitably encoded by strings over a fixed alphabet, and that all such strings have the same lengthd·s(n), for some constantd.

Enumerate all strings of lengthd·s(n)as C₁,C₂, . . . ,Cm

in the lexicographical ordering. Note that not all stringsC_i may encode syntactically correct configurations ofM(x).

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x ∈A ⇐⇒ START_M(x)`²_M^k ACCEPT_M (2)

⇐⇒ (∃i) h

START_M(x)`²_M^k−1 C_i andC_i `²_M^k−1 ACCEPT_Mi .

This idea can be realized in spaceO(s²).

(19)

On inputx of lengthn, DTMNworks as follows:

1 N constructss(n)and computesk =c·s(n) =dlogt(n)e.

2 N generates the following pattern on its work tape:

START_M(x)

| {z }

d·s(n)cells

#

kblocks

z }| {

2· · ·2

| {z }

d·s(n)cells

# 2· · ·2

| {z }

d·s(n)cells

# · · · # 2· · ·2

| {z }

d·s(n)cells

#ACCEPT_M

| {z }

d·s(n)cells

where#is a special symbol separating thesek+2 blocks.

3 N(x)simulates the computation ofM(x)deterministically within thesek+2 blocks, reusing the same space on its tape over and over again.

4 N acceptsx if and only if it reaches theACCEPT_M configuration.

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We prove thatN’s simulation ofM(x)succeeds for eachx ∈A:

Ifx ∈A, thenN’s simulation of

START_M(x)`²_M^k ACCEPT_M

succeeds in the space marked above.

Ifx 6∈A, thenNrejectsx. The proof is by induction onk.

k =0: Ncan check whether or not

START_M(x)`²_M⁰ ACCEPT_M

by simulatingM(x)for one step.

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(k −1) 7→k:

N systematically cycles through all lengthd·s(n)strings C₁,C₂, . . . ,C_m,

which potentially encode configurations ofM(x), searching for one C_i that satisfies (2).

IfC_i is the string currently being checked,N first checks whether or notC_i is a syntactically correct configuration ofM(x).

If not,Nproceeds by examining the next string,C_i+1.

(22)

Otherwise (i.e., ifC_i is a syntactically correct configuration ofM(x)),N writesC_i onto the(k +1)^stblock of its working tape:

START_M(x)

| {z }

d·s(n)cells

#

k−1 blocks

z }| {

2· · ·2

| {z }

d·s(n)cells

# · · · # 2· · ·2

| {z }

d·s(n)cells

# C_i

| {z }

d·s(n)cells

#ACCEPT_M

| {z }

d·s(n)cells

and checks whether or not

START_M(x)`²_M^k−1 C_i,

which is possible by the induction hypothesis.

If this test fails,N erasesC_i from the(k+1)^st block and proceeds by examining the next string,C_i+1.

(23)

Otherwise (i.e., ifSTART_M(x)`²_M^k−1 C_i),N erasesC_i from the (k +1)^stblock of its working tape after having it copied onto the second block:

START_M(x)

| {z }

d·s(n)cells

# C_i

| {z }

d·s(n)cells

#

k−1 blocks

z }| {

2· · ·2

| {z }

d·s(n)cells

# · · · # 2· · ·2

| {z }

d·s(n)cells

#ACCEPT_M

| {z }

d·s(n)cells

and checks whether or not

C_i `²_M^k−1 ACCEPT_M,

which again is possible by the induction hypothesis.

If this test fails,N erasesC_i from the second block and proceeds by examining the next string,C_i+1.

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Otherwise (i.e., ifC_i has passed both tests,START_M(x)`²_M C_i andC_i `²_M^k⁻¹ ACCEPT_M, as required by (2)),Nacceptsx and halts.

If none of the potential configurationsC_i ofM(x)passes both tests, thenN rejectsx and halts.

Clearly,L(N) =A.

Sincek =c·s(n)and there arek+2 blocks of sized·s(n)each, N(x)works in spaceO((s(n))²).

The Linear Compression Theorem then impliesA∈DSPACE(s²),

which proves Savitch’s theorem. q

(25)

Corollary (from Savitch’s Theorem)

1 PSPACE=NPSPACE.

2 NL⊂PSPACE.

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Theorem (Immerman and Szelepcs ´enyi) If s≥log is space-constructible, then

NSPACE(s) =coNSPACE(s).

without proof

Corollary (from Immerman and Szelepcs ´enyi’s Theorem)

1 NLINSPACE=coNLINSPACE.

2 NL=coNL.