Slides By: Alexander Eskin, Ilya Berdichevsky

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DSPACE

• Slides By Alexander Eskin, Ilya Berdichevsky
• (From the Course Computational Complexity by
  Ely Porat)

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Agenda

• Definition of DSPACE
• DSPACE(s(n)) ? DTIME(2O(s(n)))
• DSPACE(O(1)) DFA
• DSPACE(o(logn)) ? DSPACE(O(1))
• DSPACE(o(loglogn)) DSPACE(O(1))

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Definition of DSPACE

• Let M be a deterministic Turing Machine with 3
  tapes Input, Space (work) and Output.
• DSPACE ( s(n) )
• L ? M M(x) ?L(X) ? ?n Space(M) ?
  s(n)

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Explanation

• Input tape is read-only.
• Work tape is read-write.
• Space(M) is measured by the bounds of the
  machines position on this tape.
• Output tape is write-only.
• DSPACE ( s(n) ) is the class of all the languages
  that can be decided deterministically using s(n)
  work space.

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Theorem DSPACE(s(n)) ? DTIME(2O(s(n)))

• Proof
• Let M be a Turing Machine s.t. L(M) ?
  DSPACE(s(n)) ? Space(M) s(n)
• ? 0, 1, ? number of states for the work
  tape is ? Space 3 s(n)
• The number of possible head positions in the work
  tape is s(n).
• The number of possible head positions in the
  input tape is n.
• The number of the states of M is QM.

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• This entails
• The number of the possible configurations of M is
  s(n) ? 3s(n) ? n ? QM 2O(s(n))
• Claim M cannot do more than 2O(s(n)) steps
• Otherwise, M will go through the same
  configuration at least twice, entering an
  infinite loop.
• Therefore, M has to run in time ? 2O(s(n))

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Explanation

• If M will go through the same configuration
  twice, it will enter an infinite loop, since
• The same configuration Ci ? the same head
  position in the input tape and the same state.
• But M is deterministic, so in both cases it will
  move to the same configuration Ci1 etc.
• Since after the first time M entered Ci , it
  returned there one more time, it would do so
  after the second time and so on.

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Theorem DSPACE(O(1)) DFA

• Proof
• Claim DSPACE(O(1)) TWA (Two way automata)
• Claim TWA DFA
• Claims 1, 2 entail DSPACE(O(1)) DFA

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Explanation

• Proof of Claim 1
• Let M be a Turing machine. Space(M) O(1) ?
  there is constant number of possible contents of
  the work tape, say y. There are Q states ? M
  can be simulated by TWA with Q?y states (still
  constant of states).
• The proof of Claim 2 is beyond the scope of this
  course.

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Theorem DSPACE(o(logn)) ? DSPACE(O(1))

• Proof
• We will show
• DSPACE(log log(n)) ? DSPACE(O(1))
• Let L be a language
• ?(L) 0, 1,
• L w 00001001011 ?k ? N
  the l-th substring of w delimited by has length
  k and is the binary representation
  of l-1, 0 ? l lt 2k
• Claim L is not regular.

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• Lemma L ? DSPACE(log log(n))
• Proof Let M be a Turing Machine, doing the
  following
• Check that every block delimited by is
  k-length. This takes log(k) space.
• Check the 1st block is all 0s and the last block
  is all 1s. This takes O(1) space.
• For each two consecutive sub-strings, verify that
  they are xx1.

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Explanation

• The claim that L is not regular can be easily
  proved by the Pumping Lemma
• Step 3 above (for each two consecutive
  sub-strings, verify that they are xx1) is done
  as follows
• Start from the LSB (right-to-left)
• Check every 1 in Xm ? 0 in Xm1
• Check the 1st 0 in Xm ? 1 in Xm1

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• Back to the proof of the lemma
• Verifying that each two consecutive sub-strings
  are xx1 (step 3) takes log(k) space, since each
  time we compare 1 bit only, and need a log(k)
  counter to count k bits forward
• Steps 1-3 take together O(log(k)) space, which is
  O( log log(n) )
• We have shown L ? DSPACE(log log(n))
• But L is not regular ? DSPACE(log log(n)) ?
  DSPACE(O(1)) ? DSPACE(o(log(n)) ? DSPACE(O(1))

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• Theorem DSPACE(o(loglogn)) DSPACE(O(1))
• Proof
• Claim DSPACE(o(loglogn)) ? DSPACE(O(1))
• Lemma DSPACE(o(loglogn)) ? DSPACE(O(1))
• DSPACE(o(loglogn)) DSPACE(O(1)) (entailed by 1.
  and 2. above)

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Explanation

• Claim
• DSPACE(o(loglogn)) ? DSPACE(O(1))
• Proof
• o(loglogn)) ? O(1) ?
• DSPACE(o(loglogn)) ? DSPACE(O(1))

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• Claim
• Given input x x n such that M accepts x,
  then M can be on every cell on the input tape at
  most k times
• k 2S(n) ? S(n) ? QM O(2S(n))
• Proof
• k is number of possible different configuration
  of the machine.
• If M were on the cell more than k times then it
  would be in the same configuration twice and thus
  never terminate.

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Explanation

• k is the number of possible different
  configurations of the machine
• 2S(n) number of possible contents of the
  work-tape.
• S(n) number of possible locations for the r/w
  head on the work-tape.
• QM - set of states of M.
• If M were on the cell more than k times then it
  would be in the same configuration twice.
• The second time in the same configuration the
  machine will perform exactly as the first time ?
  endless loop.

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• Lemma DSPACE(o(loglogn)) ? DSPACE(O(1))
• Proof
• Definition
• semi-configuration is a configuration with
  position on the input tape replaced by the symbol
  at the current input tape position configuration
  of the machine
• For every location i on the input tape we
  consider all possible semi-configuration of M
  when passing location I
• The sequence of such a configurations is
  Ci Ci1,Ci2 Cir then it length is bounded
  by O(2S(n)) ? r

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• Proof of Lemma (contd.)
• The number of possible different sequences of
  semi-configurations of M, associated with any
  position on input tape is bounded by
• S(n) o(log log n)
• ? n0?N such that ?n ? n0,
• Well show that ?n ? n0 S(n) S(n0), thus
  L?DSPACE(S(n0)) ? DSPACE(S(1)) proving the
  theorem.

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• Proof of Lemma (contd.)
• Assume to the contrary that there exists n such
  that S(n) gt S(n0)
• Let
  and let x1?0,1n1 be such that
  SpaceM(x1) gt S(n0)
• x1 is the shortest input on which M uses more
  space then S(n0).
• The number of sequences of semi-configurations at
  any position in the input tape is
• Labeling n1 positions on the input tape by at
  most sequences means there must be at least
  three positions with the same sequence of
  semi-configurations.

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• Without loss of generality say x1 ?a?a?a?
• Each of the position with symbol a has the same
  sequence of semi-configurations attached.
• Claim
• The machine produces the same final
  semi-configuration with either ?a or ?a
  eliminated from the input.

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Explanation

• Proof of the Claim
• Consider cutting a?, leaving x1 ?a?a?
• On x1 M proceeds on the input exactly as with x1
  until it first reaches the a.
• M will make the same decision to go right or left
  on x1 as it did on x1 since all the information
  stored in the machine at the current head
  position is identical.
• If the machine goes left then its computation
  will proceed identically because it still did not
  see difference on both inputs.

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Explanation (contd.)

• Proof of the claim (contd.)
• If the machine goes right
• Say this is the ith time at the first a.
• Since the semi-configuration is the same in both
  cases then on input x1 the machine M also went
  right on the ith time seeing the second a.
• The machine proceeded and either terminated or
  came back for the (i1)th time to the second a.
• In either case on input x1 the machine M is
  going to do the same thing but now on the first a.

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Explanation (contd.)

• Proof of the claim (contd.)
• As the machine proceeds through the sequence of
  semi-configurations we see that the final
  semi-configuration on x1 will be same as for x1
• arguing each time that on x1 we will have the
  same sequence of semi-configurations.
• The case in which ?a is eliminated is identical.

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• Proof of the Lemma (contd.)
• Let x2 ?a?a? and x3 ?a?a?
• If the worst case of space usage processing x1
  was in ?a then WM(x1) WM(x2) WM(x3)
• If the worst case of space usage was in ?a then
  WM(x1) WM(x2)?WM(x3)
• If the worst case of space usage was in a? then
  WM(x1) WM(x3)?WM(x2)
• Chose x1?x2,x3 to maximize WM(x1) ? WM(x1)
  WM(x1) and x1 lt x1
• Contradiction to assumption that x1 is minimal
  length string using more then S(n0) space ? no
  such x1 exists.