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					              Wild Circuits
      Investigating the Limits of
        MIN/MAX/AVG Circuits
                  Brendan Juba

  Faculty Advisor: Manuel Blum
Graduate Mentor: Ryan Williams
Definitions: MIN/MAX/AVG Circuits
                                            unsatisfied
                                                               satisfied
 We are given a circuit, C, with
  feedback, operating on real numbers                         1 0
  from the closed interval [0,1].
 C contains                                         0           MIN

   MIN, MAX, or AVG gates with two inputs
   “Inputs” to the circuit that are hard-wired to             AVG

    either 0 or 1.                                                     0
 |C| denotes the number of gates of C                        MAX

   Here, |C| = 3
 When the output of a gate is the                   0
  appropriate function of its inputs, we
  say that the gate is satisfied                  satisfied
Definitions: MIN/MAX/AVG Circuits
 Settings of the gate outputs from the interval
  [0,1] are value vectors                                         1 0
    A value vector for C, v  [0,1]|C|
    The ith entry, vi, is the output of the ith gate.
 We may also consider an update function,
                                                                    MIN


  F: [0,1]|C|  [0,1]|C|
    We are interested in two varieties: single-gate               AVG

     update functions and circuit-wide update
     functions:
                                                                  MAX
        A single gate update function replaces the output
         of a single designated gate with the correct output
         value.
           • We will call iterating over the single gate update
             functions “gate-by-gate update”
        The circuit-wide update function simultaneously
         replaces the output of every gate with a value that
         is correct with respect to the old values
 Definition: Stable Circuit Problem
 A vector v is stable iff every gate is satisfied. (F(v) = v)
    We wish to find these stable vectors.
 Gate-by-gate update from 0 clearly obtains a stable vector
  in the limit. This is the minimum stable solution
                 1 0                                   1 0
    stable                            unstable

                    MIN                                  MIN
             0                                     0

                  AVG                                   AVG
       1/2                                   1/2

                 MAX                                   MAX




                          1/2                                  0
Stable Circuit Decision Problem
 We designate some ith gate of C in the minimum stable
  solution, s, and ask, “is si ≥ 1/2?”
 We can reduce the function problem to this decision
  problem. We can find 2|C| bits of any si, which may be
  shown to be sufficient.
    Suppose there is a gate xk: sxk ≥ 1/2 iff the kth bit of si is a 1
        If sxk ≥ 1/2, we assume its binary decimal representation is
         .100…0 with 0s in positions 2 through 2k-1
        Otherwise, it is .011…1 with 1s in positions 2 through 2k-1
    Depending on whether sxk ≥ 1/2, we add a new gate, xk+1:
     AVG(xk,1/2-1/22k), if sxk = .100…0
     AVG(xk,1/2+1/22k), if sxk = .011…1
        In the former case, sxk+1=(.1)(.100…0 + .011…11)
        In the latter, sxk+1=(.1)(.011…1 + .100…01)
    In the solution for the modified circuit, xk+1 clearly has the
     desired properties. The largest construction is O(|C|2) gates.
Unique Solution Circuits
                                                     x       0
 Replace any wire from x to y in the
  circuit with the construction on the
  right using m AVG gates
                                                         AVG
 This circuit has a unique solution
  (Shapley, 1953)
  Suppose the original circuit-wide update              AVG

   function is F, stable solutions are u and v
  If ||u-v||∞= c, then it is easy to see
   c = ||u-v|| = (1-1/2m)||F(u)-F(v)|| ≤ (1-1/2m)c
  Clearly, c = 0.                                       AVG

 These solutions turn out to be
  arbitrarily close to the minimum                       y

  stable solutions (for appropriate m).
Stable Circuit is in NPco-NP
(Condon, 1992)
 A nondeterministic machine M can, in
  polynomial time, on input circuit C, for gate i
   Build a suitably close unique solution circuit C’
   Guess the solution to C’
   Verify the guessed vector is a solution
   Accept or reject, respectively, precisely when the value
    of gate i is above 1/2 (since C’ was close to C, either i is
    above 1/2 in neither, or it is above 1/2 in both)
Stable Circuit is P-hard
 If we use no AVG gates, the wires of the circuit
  will only carry 0 or 1
 It is immediate that we may use MIN as AND,
  MAX as OR
 For any circuit with fixed inputs, we can
  construct a “complement” circuit
  Switch 0 inputs with 1 inputs
  Switch MIN gates with MAX gates
 We can now negate by crossing a wire between
  the original and complement circuits
  (In this AVG-free case, deciding the output is in P, too)
Observation and Motivation
 If we apply gate-by-gate or circuit-wide update
  on arbitrary starting value vectors, we can obtain
  “interesting” circuits
  One such “interesting” circuit is a binary counter
  We do not necessarily obtain the stable configurations
   of our circuits this way -- this is not Stable Circuit
 Can we obtain such circuits under gate-by-gate
  update from 0?
  If so, the minimum stable solution is the configuration of
   the device after an unbounded amount of time!
“Leapfrog” circuits
 We assign each wire a “threshold” wire and
  interpret its value relative to that threshold
   Above threshold: T
   Below threshold: F
 It is already clear that we still have AND and OR
 There is also a construction for NOT (next slide)
   If there are W wires which we wish to interpret relative to
    the same threshold, this gadget takes Θ(W) gates
 NB: The circuits are still monotone!
   As we update, a value may seem to rise or fall, as we
    follow it across different wires through the circuit
   The value on any particular wire only rises as the gates
    of the circuit are updated
 NOT Gadget
 th         x0      x1         x2

      AVG



             MAX



             MIN        MIN        MIN




AVG                 MAX        MAX


            ~x0    x1         x2         th   x0   x1   x2
      th
Caveats
 Assumptions:
  1. All values above [below] threshold are equal
  2. The values th, T, and F are all different
  3. We may specify the update order for the gates of the
     circuit
 Take each in turn:
  1. Everything starts from zero, the property is preserved
     by all three gates
  2. We can push th above zero by means of an AVG gate
      With feedback, we must pass the other wires to be
       interpreted relative to that threshold through similar
       constructions so as to maintain relative values
  3. Update order doesn’t change the solution we approach
Two-bit Counter Circuit
                        1      AVG
                                     1
 x0                     x1     th
              NOT
              NOT


            MIN          MIN



                  MAX
       1                 1
      AVG          AVG




                                     0
                                         x0   x1   th
Two-bit Counter Circuit
                        1      AVG
                                     17/32

 x0                     x1     th
              NOT
              NOT


            MIN          MIN



                  MAX
       1                 1
      AVG          AVG



                                     1/2
                                             x0   x1   th
Two-bit Counter Circuit
                        1      AVG
                                     781/
                                     1024
 x0                     x1     th
              NOT
              NOT


            MIN          MIN



                  MAX
       1                 1
      AVG          AVG



                                     195/
                                     256    x0   x1   th
Two-bit Counter Circuit
                        1            7217/
                               AVG   8192
 x0                     x1     th
              NOT
              NOT


            MIN          MIN



                  MAX
       1                 1
      AVG          AVG


                                     28867/
                                     32768
                                              x0   x1   th
Serving Suggestions
                                       carry-in
 The counter generalizes to                                    xi

  n bits easily                                    NOT
   The n-bit counter takes Θ(n2)                  NOT
    gates, due to the size of the
    NOT gadgets                           MIN      MIN        MIN
                                                                     carry-
 We may also build a                              MAX
                                                                     out
  gadget such that, if its
                                                   xi
  input is ever above
  threshold, a wire in the                        MAX     1
                                        input
  gadget stays above                                AVG

  threshold forever:                (the internal wire must also pass
                                    through the NOT gadgets)
NP-hardness

                                             x1               x2 x3 th, etc.
 Let any boolean formula be given…
 Ex: (x1~x2x3)  (~x1~x2x3)                         NOT

 Since we have AND, OR, and NOT                         NOT
 gates, we can easily translate any
 formula into a circuit which has an         MAX               MAX
 output above its threshold iff the
 formula is satisfied by the                      MAX         MAX
 assignment from the input wires, as
 we have on the right.                                  MIN


                                          (x1~x2x3)(~x1~x2x3)

 By attaching xi to the ith bit of the counter, we try all possible
 assignments, allowing us to encode answers to SAT on a wire. The
 number of gates in this circuit is quadratic in the length of the formula.
Entering PSPACE
 We can still do better: using the counter, we will
  decide whether quantified boolean formulas are            x1
  valid
 Assume WLOG that the quantifiers alternate: odd
  variables are universal, even existential
 Observe that the counter “walks” along the              x0    x0
  leaves of a tree of assignments, left to right.     
 Suppose that at the bottom we evaluate the
  quantifier-free part of the formula on the specified 00 01 10    11
  assignment.
 Now suppose at every  level of the tree, we have one bit of memory
  for the left branch
     Set it to T when the branch is T, reset it to F when leaving that subtree.
 Pass T up the tree when
     We see T at either branch at an  level
     We see T at the right branch of a  level with the left branch bit already
      set to T.
 T is passed up from the top of the tree iff we have a TQBF.
Quantifier Circuit: xi xi-1 A
xi Carry-out: xi     v i0       A   • IH: the wire labeled A will carry T iff the
                                      shorter boolean formula with alternating
          NOT                         quantifiers, A, is satisfied by the current
                                      assignment to xi-1,…,xn from the counter
                   MIN
                                    • vi0 is a register that holds the evaluation of
          NOT                         xi-1A when xi is F, while xi+1,…,xn remain
                                      fixed
                                        • When there is a carry out of xi, xi+1 has
                                          altered, (new branch) so we reset vi0 to F
          MIN
                    MIN             • If vi0 holds T, and when xi is T, for some xi-1 A
                                      carries T, then the wire labeled xixi-1A
                     MAX
                                      carries T. Otherwise, the wire remains F.
           MIN                      • Observe that the wire xixi-1A will carry T iff
xi                       v i0
                                      xixi-1A is satisfied by the current
        xixi-1A                     assignment to xi+1,…,xn, so the IH is satisfied
End of the Line: Thwarted by
PSPACE
 In the limit, the separation between T and F
  shrinks as the internal wires approach 1.
   It is not immediately clear how to recover the values of
    any wire from the minimum stable solution
 Recall: finding values in the limit (the minimum
  stable solution) is known to be in NPco-NP
 Answers to PSPACE-hard problems (QSAT)
  may be encoded on the wires as we update
   Since the “space” in Leapfrog circuits is bounded by the
    number of gates, it is doubtful that such circuits can
    solve anything harder
 In the limit, it is impossible to distinguish the
  values in Leapfrog circuits unless NP =
 Stoppable NOT Gadget
 th          x0        check         x1             x2          This gadget behaves
      AVG
                                                                identically to the regular
                                                                NOT, unless check is set
                 MAX                                            high, in which case, all
                                                                outputs are set high.
                 MIN       MIN           MIN         MIN
                                                                Gadgets such as this
                          MAX                                   suggest that the problem
AVG                                                             with our Leapfrog counter
           MAX                           MAX        MAX
                                                                was in the AVG gates we
      th                         check         x1          x2   used to “power” it from 0.
                 ~x0               GAME OVER
                                 Thank you for playing
                                           CAPCOM

				
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