Wild Circuits

1
Wild Circuits

• Investigating the Limits of MIN/MAX/AVG
  CircuitsBrendan Juba
• Faculty Advisor Manuel BlumGraduate Mentor
  Ryan Williams

2
Definitions MIN/MAX/AVG Circuits
unsatisfied
satisfied

• We are given a circuit, C, with feedback,
  operating on real numbers from the closed
  interval 0,1.
• C contains
• MIN, MAX, or AVG gates with two inputs
• Inputs to the circuit that are hard-wired to
  either 0 or 1.
• C denotes the number of gates of C
• Here, C 3
• When the output of a gate is the appropriate
  function of its inputs, we say that the gate is
  satisfied

1
0
0
MIN
AVG
0
MAX
0
satisfied
3
Definitions MIN/MAX/AVG Circuits

• Settings of the gate outputs from the interval
  0,1 are value vectors
• A value vector for C, v ? 0,1C
• The ith entry, vi, is the output of the ith gate.
• This is an implicit ordering of the gates of C
• We may also consider an update function, F
  0,1C ? 0,1C
• 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

1
0
MIN
AVG
MAX
4
Definition Stable Circuit Problem

• A vector v is stable iff every gate is satisfied.
  (F(v) v)
• Gate-by-gate update from the vector 0 obtains a
  stable vector in the limit. This is the minimum
  stable solution
• We wish to find the minimum stable solution

1
0
1
0
unstable
stable
MIN
MIN
0
0
1/2
1/2
AVG
AVG
MAX
MAX
1/2
0
5
Definition STABLE CIRCUIT (Decision Problem)

• We are given a circuit C, and some designated ith
  gate. In the minimum stable solution of C, s,
  is si 1/2?
• If we can efficiently solve this decision
  problem, we can efficiently solve the function
  problem we can find 2C bits of any si, which
  may be shown to be sufficient.

• Inductively suppose we know the first k-1 bits of
  si to be v
• Modify C
• (1-1/2k-v) requires k gates

(1-1/2k-v)
AVG
ith gate

• In the minimum stable solution, this new AVG
  gates output is above 1/2 iff the kth bit of si
  is a 1, so the decision problem tells us the kth
  bit of si
• Ex Suppose v .011010, si .0110101 (k 7)
  thenAVG(si,1-1/2k-v) (.0110101 .1001011)/2
  .10000000
• If si .0110100 then AVG(si,1-1/2k-v)
  (.0110100 .1001011)/2 .01111111

6
Previously, on STABLE CIRCUIT

• STABLE CIRCUIT is in NP?co-NP (Condon, 1992)
• We can modify our circuits to have a unique
  solution that is identical to the minimum stable
  solution up to the 2Cth bit
• This unique solution can be guessed and checked
• STABLE CIRCUIT is P-hard
• MONOTONE CIRCUIT is a special case

7
Observations and Motivations

• Our original motivation was to show STABLE
  CIRCUIT was hard for some class beyond P
• If we apply gate-by-gate update to arbitrary
  starting value vectors, we can obtain
  interesting circuits
• We do not necessarily obtain stable
  configurations of our circuits -- this is not
  Stable Circuit
• If we apply gate-by-gate update to the value
  vector 0, can we still obtain interesting
  circuits?
• If so, the minimum stable solution is the
  configuration of the device after an unbounded
  amount of time!

8
Can we obtain interesting circuits starting
from 0?

• YES

9
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
  T(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

10
NOT Gadget
th
x1
x0
x2
AVG
AVG
AVG
MAX
MAX
MAX
AVG
AVG
AVG
MIN
MIN
MIN
MIN
MIN
MIN
MIN
MIN
MIN
MAX
MAX
MAX
MAX
MAX
MAX
x0
th
x0
x1
x2
th
x1
x2
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Caveats

• Assumptions
• All values above below threshold are equal
• th has a value distinct from all other inputs
• We may specify the update order for the gates of
  the circuit
• Take each in turn
• Everything starts from zero and the property is
  preserved by our AND, OR, and NOT gates
• We can push th above zero by means of an AVG gate
• With feedback, we must also pass the other wires
  through AVG gates to preserve relative values
• Update order doesnt change the solution we
  approach

12
Two-bit Counter Circuit
1
1
AVG
AVG
AVG
AVG
x0
x1
th
NOT
NOT
MIN
MIN
MIN
MIN
MIN
MIN
MAX
MAX
MAX
1
1
AVG
AVG
AVG
AVG
AVG
AVG
0
x0
x1
th
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Two-bit Counter Circuit
1
17/32
AVG
x0
x1
th
NOT
NOT
MIN
MIN
MIN
MIN
MIN
MIN
MAX
MAX
MAX
1
1
AVG
AVG
1/2
x0
x1
th
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Two-bit Counter Circuit
1
781/ 1024
AVG
x0
x1
th
NOT
NOT
MIN
MIN
MIN
MIN
MIN
MIN
MAX
MAX
MAX
1
1
AVG
AVG
195/256
x0
x1
th
15
Two-bit Counter Circuit
1
7217/8192
AVG
x0
x1
th
NOT
NOT
MIN
MIN
MIN
MIN
MIN
MIN
MAX
MAX
MAX
1
1
AVG
AVG
28867/32768
x0
x1
th
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Serving Suggestions

• The counter generalizes to n bits easily
• The n-bit counter takes T(n2) gates, due to the
  size of the NOT gadgets
• We now have our counter
• We next investigate the power of Leapfrog
  circuits, using the counter
• First, we will need to make precise what we mean
  by Leapfrog circuits

carry-in
xi
NOT
NOT
MIN
MIN
MIN
carry-out
MAX
xi
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Definition LEAPFROG

• Let LEAPFROG be the following problem Given a
  circuit C and designated gates i and th, consider
  the sequence of vectors v1, v2, obtained during
  gate-by-gate update of C from 0 in the order of
  the gate indices of CIs there an index t such
  that vti gt vtth?
• LEAPFROG captures our notion of what Leapfrog
  circuits compute

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LEAPFROG vs. STABLE CIRCUIT

• NB Not the same problem!!
• But, STABLE CIRCUIT obviously reduces to LEAPFROG
  (include a gate that outputs constant
  1/2-1/22C)
• Is LEAPFROG hard?
• YES -- we will see in a moment
• Does LEAPFROG reduce to STABLE CIRCUIT?
• If yes, then STABLE CIRCUIT is also hard.

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LEAPFROG is hard! (NP-hard)
Let any boolean formula be given Ex
(x1?x2?x3) ? (x1?x2?x3) Since we have AND,
OR, and NOT gates, formulas easily translate into
circuits.
x1
x2
x3
th, etc.
If we attach xi to the ith bit of the counter, we
try all possible assignments, allowing us to
reduce SAT to LEAPFROG. The number of gates in
these SAT circuits is quadratic in the length of
the formula.
(x1?x2?x3)?(x1?x2?x3)
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LEAPFROG is really hard! (PSPACE-hard)

• We can still do better using the counter, we
  will decide whether quantified boolean formulas
  are valid (Reducing TQBF to LEAPFROG)
• Assume WLOG that the quantifiers alternate odd
  variables are universal, even ones are
  existential
• Leaves in this tree correspond to assignments
• The counter walks along the leaves, left to right
  
• At the bottom we evaluate the quantifier-free
  part of the formula on the specified assignment.

x1
?
x0
x0
?
00
01
10
11

• Each ? level of the tree has 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
• 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.

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Quantifier Circuit ?xi (?xi-1 A)
A
xi
vi0
Carry-out xi

• IH the wire A will be T iff the shorter formula
  with alternating quantifiers, A, is satisfied by
  the assignment to xn,,xi-1 from the counter
• vi0 is our bit of memory storing the value of
  (Axi F) (the left branch) under the fixed
  assignment to xn,,xi1
• When there is a carry out of xi, xi1 has
  altered, so we reset vi0 to F
• If vi0 (Axi F) T and (Axi T) T (on
  the right branch), then the wire labeled ?xi?xi-1
  A is set to T. Otherwise, the wire remains F.
• Notice we try both settings of xi-1 for each
  branch. The wire ?xi ?xi-1 A is T iff ?xi ?xi-1 A
  is satisfied by the assignment to xn,,xi1, so
  the Inductive Hypothesis is satisfied

NOT
MIN
MIN
MIN
NOT
MIN
MIN
MIN
MIN
MIN
MIN
MAX
MAX
MAX
MIN
MIN
MIN
xi
vi0
?xi ?xi-1 A
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End of the Line Thwarted by PSPACE

• Recall finding values in the limit (the minimum
  stable solution) is known to be in NP?co-NP
• Answers to PSPACE-hard problems (TQBF) may be
  encoded on the wires as we update
• Since circuits of AND/OR/NOT gates can be
  evaluated in PSPACE, we would need to drastically
  alter our model to solve anything harder
• Hence, unless NP PSPACE, LEAPFROG does not
  reduce to STABLE CIRCUIT
• Thus, in general, Leapfrog circuits
  (specifically, our counter) cannot be stopped

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Open problems

• How hard is STABLE CIRCUIT?
• We had also succeeded in placing the function
  version in PLS, but still no hardness results
• Is Stable Circuit PLS-complete?
• Is STABLE CIRCUIT in P?
• How hard is LEAPFROG, actually?
• Trivially RE, but this says rather little
• Is LEAPFROG decidable?