Classical and Quantum Circuit Synthesis An Algorithmic Approach
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Classical and Quantum Circuit Synthesis
An Algorithmic Approach
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?
Goals
Way to synthesize quantum circuitsCurrently, reducing chains of C-Not gates
is importantMore complex circuits (with Toffoli gates,
etc...)Larger circuits (more qubits, more gates,
etc...)
Reducing C-Nots
Want to minimize error
Need an algorithmic approach
Classical circuit approach discrete reversible case
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-
Example
Classical program goals
Synthesize optimal classical circuitsProvides method for creating reduced
chains of C-Not gates
ScalableAs many inputs/outputs as necessary
VersatileUse various gate libraries (AND, OR, NOT,
etc...)
Refresher – Truth Tables
X Y Z=X^Y
0 0 0
0 1 0
1 0 0
1 1 1
AND gate
(Columnar format)
These can be represented in row format as well:
X = 0011 -> 3Y = 0101 -> 5--------Z = 0001 -> 1
(3^5)==1
How to Synthesize?
Inputs
X = 0011
Y = 0101
Desired result
Z = 0110
?
How to Synthesize?
Inputs
X = 0011
Y = 0101
Desired result
Z = 0110Gate Library: AND, OR, NAND
Tree Structure
Solution space representable as a tree Inputs at head of tree Each level down represents addition of
new gateBranch factor depends on number of
current gates and allowable new gates
Inputs X, Y
A=X*Y A=X+Y
B=X*Y B=X+Y B=X*A B=X+A B=Y*A B=Y+A
Level 1
Level 2(Same as left)
If we have
N inputs (N=2 here)
G gate types (G=2, we have AND and OR)
Then at level L we have a branch factor of
B = G * (N+L-1, 2) / 2
Branch and Bound
Depth-first searchFirst do topbottom, then leftrightTravel down to leaves before going to
siblingsCan be implemented by a recursive
function, passing L+1 with each call
DFS Branch & Bound
DFS Branch & Bound
DFS Branch & Bound
DFS Branch & Bound
DFS Branch & Bound
B&B Synthesis
Inputs
X = 0011
Y = 0101
Desired result
Z = 0110A=X*X=0011
B&B Synthesis
Inputs
X = 0011
Y = 0101
Desired result
Z = 0110A=X*X=0011 B=X*X=0011
B&B Synthesis
Inputs
X = 0011
Y = 0101
Desired result
Z = 0110A=X*X=0011 B=X*X=0011
C=X*X=0011
C=X+X=0011
C=(X*X)’=1100
C=X*Y=0001
...
B&B Synthesis
Inputs
X = 0011
Y = 0101
Desired result
Z = 0110A=X*X=0011 B=X+X=0011
C=X*X=0011
C=X+X=0011
C=(X*X)’=1100
C=X*Y=0001
...
B&B Synthesis
Inputs
X = 0011
Y = 0101
Desired result
Z = 0110A=X*X=0011 B=(X*X)’=0011
C=X*X=0011
C=X+X=0011
C=(X*X)’=1100
C=X*Y=0001
...
Solution Found!
Inputs
X = 0011
Y = 0101
Desired result
Z = 0110A=X+Y=0111 B=(X*Y)’=1110 C=A*B=0110
Bounding Conditions
Branch factor unnecessarily large
Need bounding conditionsDon’t need duplicate gates (A=X*X,B=X*X)Symmetrical gates done only once
Don’t need X*Y, then Y*XSymmetrical branches of solution space
can be bounded awayOptimality requirement creates a bound
Optimality Bounding
Current circuitAssume 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
Don’t-caresSuppose we are interested in only part of the output. Say Z=0110 and Z=0111 are both desirable
outputs. We don’t 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 don’t-cares, but they are different than classical ones.
Multiple Outputs
We may desire more than one outputEx: 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.
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
Compact Representation (Revisited)
For a three qubit function, our inputs (A, B, C) will have these values:
A B C V
0 0 0 0
0 0 1 1
0 1 0 2
0 1 1 3
1 0 0 4
1 0 1 5
1 1 0 6
1 1 1 7
A = 00001111 = 15
B = 00110011 = 51
C = 01010101 = 85
Permutation Representation
We want to synthesize the outputs A’, B’, and C’ from the inputs A, B, and C
A B C V
0 0 0 0
0 0 1 1
0 1 0 2
0 1 1 3
1 0 0 4
1 0 1 5
1 1 0 6
1 1 1 7
A’ B’ C’ V
1 0 0 4
0 1 0 2
1 1 0 6
1 1 1 7
0 1 1 3
0 0 0 0
1 0 1 5
0 0 1 1
To be reversible, all of the values of V must be distinct. This allows us to treat the outputs as permutations of the inputs.
Permutation Representation
A B C V
0 0 0 0
0 0 1 1
0 1 0 2
0 1 1 3
1 0 0 4
1 0 1 5
1 1 0 6
1 1 1 7
15 51 85
A’ B’ C’ V’
1 0 0 4
0 1 0 2
1 1 0 6
1 1 1 7
0 1 1 3
0 0 0 0
1 0 1 5
0 0 1 1
178 120 27
So the problem:
(15,51,85)->(178,120,27)
Can be expressed as:
01234567 -> 42673051
Simplified Circuit Representation
A
B
C
A’
B’
C’
(A, B) (B, C)
Simplified Circuit Representation
A
B
C
A’
B’
C’
A1
B1
C1
A2
C2
B2
(A, B) (B, C)
A B C V
0 0 0 0
0 0 1 1
0 1 0 2
0 1 1 3
1 0 0 4
1 0 1 5
1 1 0 6
1 1 1 7
A1 B1 C1 V1
0 0 0 0
0 0 1 1
0 1 0 2
0 1 1 3
1 1 0 6
1 1 1 7
1 0 0 4
1 0 1 5
(A, B) (B, C)
A2 B2 C2 V2
0 0 0 0
0 0 1 1
0 1 1 3
0 1 0 2
1 1 1 7
1 1 0 6
1 0 0 4
1 0 1 5
A’ B’ C’ V’
0 0 0 0
0 0 1 1
0 1 1 3
0 1 0 2
1 1 1 7
1 1 0 6
1 0 0 4
1 0 1 5
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.
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---
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--
Why?
Recall that C-Nots act like XOR gates. That is, |x>|y> |x>|x XOR y>What does this code do?A = A xor B
B = A xor BA = A xor B
This is a variable swap. The chains on the previous page represent wire swaps.
How many C-Nots?
Experimental results:Of the 40320 (=8!) permutations, exactly
168 are achievable. 2 of them require 6 C-Not gates24 require 560 require 451 require 324 require 2 (6 * 6 - duplicates)6 require 1 (3 control choices * 2 targets)1 requires 0
Why only 168?
Not all of the 8! permutations are achievableThe 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
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.
Future Plans... Make Program:
More scalable Currently using 32-bit integers to store wire
values. 2^5 = 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.
Potential improvements
ScalabilityUse STL bit_vector, or bit_set... fully
expandable
SpeedSince all the gates and the functions are
reversible, the whole circuit is too.Functions can be synthesized in reverse
Reversible Synthesis
Recall that C-Not acts like XORC-Not is its own inverse (going left to right
is same as right to left)
Synthesis can be done rightleftWe already know how to achieve the input
values with several gates, because we’ve done this leftright before
This allows us to have a LUT when we get near enough to the inputs
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