Classical and Quantum Circuit Synthesis

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Classical and Quantum Circuit Synthesis

• An Algorithmic Approach

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Problem

• How do we synthesize efficient n-input quantum
  circuits using CNOT gates only ?
• Of the n! possible n-qubit permutations, how many
  can be implemented in this way?
• What is the longest such circuit?
• How can long chains of CNOTs be systematically
  reduced?

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Goals

• Way to synthesize quantum circuits
• Currently, reducing chains of C-Not gates is
  important
• More complex circuits (with Toffoli gates,
  etc...)
• Larger circuits (more qubits, more gates, etc...)

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Reducing C-Nots

• Want to minimize error
• Need an algorithmic approach
• Classical circuit approach ? discrete reversible
  case

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Example
wire 0 -------X--O-----X--X--O- wire 1
-O--X-----X--O--O-----X- wire 2
-X--O--O-----X-----O---- Reduces to wire 0
-O- wire 1 --- wire 2 -X-
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Classical program goals

• Synthesize optimal classical circuits
• Provides method for creating reduced chains of
  C-Not gates
• Scalable
• As many inputs/outputs as necessary
• Versatile
• Use various gate libraries (AND, OR, NOT, etc...)

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Refresher Truth Tables
AND gate
These can be represented in row format as
well X 0011 -gt 3 Y 0101 -gt 5 -------- Z
0001 -gt 1 (35)1
(Columnar format)
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How to Synthesize?
?
Inputs
Desired result
X 0011
Z 0110
Y 0101
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How to Synthesize?
Inputs
Desired result
X 0011
Gate Library AND, OR, NAND
Z 0110
Y 0101
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Tree Structure

• Solution space representable as a tree
• Inputs at head of tree
• Each level down represents addition of new gate
• Branch factor depends on number of current gates
  and allowable new gates

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Inputs X, Y
AXY
AXY
Level 1
Level 2
(Same as left)
BXY
BXY
BXA
BXA
BYA
BYA
If we have N inputs (N2 here) G gate types
(G2, we have AND and OR) Then at level L we have
a branch factor of B G (NL-1, 2) / 2
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Branch and Bound

• Depth-first search
• First do top?bottom, then left?right
• Travel down to leaves before going to siblings
• Can be implemented by a recursive function,
  passing L1 with each call

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DFS Branch Bound
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DFS Branch Bound
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DFS Branch Bound
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DFS Branch Bound
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DFS Branch Bound
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BB Synthesis
Inputs
Desired result
X 0011
AXX0011
Z 0110
Y 0101
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BB Synthesis
Inputs
Desired result
X 0011
AXX0011
BXX0011
Z 0110
Y 0101
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BB Synthesis
Inputs
Desired result
X 0011
CXX0011 CXX0011 C(XX)1100 CXY0001 ...

AXX0011
BXX0011
Z 0110
Y 0101
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BB Synthesis
Inputs
Desired result
X 0011
CXX0011 CXX0011 C(XX)1100 CXY0001 ...

AXX0011
BXX0011
Z 0110
Y 0101
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BB Synthesis
Inputs
Desired result
X 0011
CXX0011 CXX0011 C(XX)1100 CXY0001 ...

AXX0011
B(XX)0011
Z 0110
Y 0101
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Solution Found!
Inputs
Desired result
X 0011
AXY0111
B(XY)1110
CAB0110
Z 0110
Y 0101
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(No Transcript)
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Bounding Conditions

• Branch factor unnecessarily large
• Need bounding conditions
• Dont need duplicate gates (AXX,BXX)
• Symmetrical gates done only once
• Dont need XY, then YX
• Symmetrical branches of solution space can be
  bounded away
• Optimality requirement creates a bound

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Optimality Bounding

• Assume we are only allowed one more gate
• This gate cannot use all three level-two gates as
  inputs
• If added, two of the level-two gates would be
  used, one would be left dangling
• No reason to add a gate, bound now

Current circuit
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Dont-cares

• Suppose we are interested in only part of the
  output.
• Say Z0110 and Z0111 are both desirable outputs.
  We dont care about the last bit.
• Can represent output as Z 011X
• This relaxed restriction allows us to find a
  suitable match more quickly
• Quantum circuits can also have dont-cares, but
  they are different than classical ones.

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Multiple Outputs

• We may desire more than one output
• Ex Half adder. Has two bits as input, two bits
  as output (sum, carry)
• Optimal multi-output circuit may be different
  than circuit created by merging the individual
  optimal ones.
• Causes an approximately linear increase in
  synthesis time.

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Reversible Synthesis

• To maintain reversibility, must have same number
  of qubits at every stage
• Fan out must be 1, so there is no expansion in
  the number of gates at any stage
• Both the input/output functions and all the gates
  must be reversible

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Compact Representation (Revisited)

• For a three qubit function, our inputs (A, B, C)
  will have these values

A 00001111 15 B 00110011 51 C 01010101
85
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Permutation Representation

• We want to synthesize the outputs A, B, and C
  from the inputs A, B, and C

To be reversible, all of the values of V must be
distinct. This allows us to treat the outputs as
permutations of the inputs.
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Permutation Representation
So the problem (15,51,85)-gt(178,120,27) Can be
expressed as 01234567 -gt 42673051
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Simplified Circuit Representation
A
A
B
B
C
C
(A, B)
(B, C)
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Simplified Circuit Representation
A
A
A1
A2
B
B
B1
B2
C
C
C1
C2
(A, B)
(B, C)
35
(A, B)
(B, C)
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Results

• Question given a 3-Qubit system, what is the
  longest possible irreducible chain of C-Nots?
• Note that some longest chain must exist because
  there are finitely many (8!) permutations.

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Bounding

• At each stage, we store the value on each wire.
• If all wire values are the same, bound
• If two gates can be permuted w/o affecting the
  circuit, only try one of them
• -O-O- -O-O-
• -X--- ---X-
• ---X- -X---

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Answer

• Longest chains
• wire 0 --O--X--O-----------
• wire 1 -----------X--O--X--
• wire 2 --X--O--X--O--X--O--
• AND
• wire 0 -----------X--O--X--
• wire 1 --O--X--O-----------
• wire 2 --X--O--X--O--X--O--

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Why?

• Recall that C-Nots act like XOR gates. That is,
  xgtygt ? xgtx XOR ygt
• What does this code do?
• A A xor B
• B A xor B
• A A xor B
• This is a variable swap. The chains on the
  previous page represent wire swaps.

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How many C-Nots?

• Experimental results
• Of the 40320 (8!) permutations, exactly 168 are
  achievable.
• 2 of them require 6 C-Not gates
• 24 require 5
• 60 require 4
• 51 require 3
• 24 require 2 (6 6 - duplicates)
• 6 require 1 (3 control choices 2 targets)
• 1 requires 0

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Why only 168?

• Not all of the 8! permutations are achievable
• The permutation must be even. Recall every C-Not
  caused two pairs of values to be swapped
• Exactly half of the 8! permutations are even
• Various other factors reduce this more

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Parity of a Permutation
The number of line crossings determines the
parity of a permutation. There are four in this
case, and so the permutation is even.
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Future Plans... Make Program

• More scalable
• Currently using 32-bit integers to store wire
  values. 25 32 only 5 wires at a time now.
• More versatile
• Currently only using C-Not gates. Allow n-input
  Toffoli gates, other gates
• Quicker
• Determine methods for bounding better, exploit
  characteristics of quantum circuits.

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Potential improvements

• Scalability
• Use STL bit_vector, or bit_set... fully
  expandable
• Speed
• Since all the gates and the functions are
  reversible, the whole circuit is too.
• Functions can be synthesized in reverse

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Reversible Synthesis

• Recall that C-Not acts like XOR
• C-Not is its own inverse (going left to right is
  same as right to left)
• Synthesis can be done right?left
• We already know how to achieve the input values
  with several gates, because weve done this
  left?right before
• This allows us to have a LUT when we get near
  enough to the inputs