Quantum Processes in Graph Computing

Quantum Processes in Graph Computing
Dr. Marko A. Rodriguez
DataStax and Apache TinkerPop
Rodriguez, M.A., Watkins, J.H., "Quantum Walks with Gremlin," GraphDay '16, pp. 1-16, Austin, Texas, January 2016.
http://arxiv.org/abs/1511.06278

Part 1
Wave Dynamics in Graphs

However, vertices don't "move."
The graph topology is the only space.
There is no "space" outside the graph.
"Then how do you model an oscillating vertex?"
By computing the amount of "energy"
at the vertex over time.
"But traversers are indivisible objects!
Energy must be able to be
divided indeﬁnitely and take negative values."
Lets look at Gremlin Sacks.

Gremlin Sacks
(a data structure local to the traverser)
(each traverser can carry an object along its way)
The traverser is not divisible, but that
double ﬂoating point number in his sack is!

1.0
0.0
-1.0
1.0
Don't worry about the algorithm, just watch the visualization for now.

-0.5 0.5
0.0
-1.0
1.0

-0.5 0.250.25
0.0
-1.0
1.0

-0.125 0.125-0.3750.375
0.0
-1.0
1.0

0.375-0.3125 -0.3125
0.0
-1.0
1.0

0.15625 -0.156250.34375-0.34375
0.0
-1.0
1.0

1 2 30 4 5 6
1.0
g.withSack(1.0,sum).V(3).
repeat(
union(
sack(mult).by(-0.5).out('left'),
sack(mult).by(0.5).out('right')
)
).times(3)
An initial energy of 1.0 perturbs the graph.

1 2 30 4 5 6
repeat(
union(
)
).times(3)
-0.5 0.51.0
out('left') out('right')
The energy diffuses left and right along the graph.

1 2 30 4 5 6
repeat(
union(
)
).times(3)
-0.5 0.5
1
tim
e
The total energy in the system is always 1.

Energy waves reverberate and thus, prior "left energy" can go right.
1 2 30 4 5 6
repeat(
union(
)
).times(3)
-0.5 0.250.25 -0.5 0.5
out('left') out('right') out('left') out('right')

1 2 30 4 5 6
repeat(
union(
)
).times(3)
-0.5 0.250.25 -0.5 0.5
out('left') out('right') out('left') out('right')
(-0.5 * 0.5) + (0.5 * -0.5)
-0.5
-0.25 + -0.25
Energy constructively and destructively interferes.

1 2 30 4 5 6
repeat(
union(
)
).times(3)
-0.5 0.250.25
2
tim
es

1 2 30 4 5 6
repeat(
union(
)
).times(3)
-0.125 0.125-0.3750.375 -0.5 0.250.25
out('left') out('left') out('left')out('right') out('right') out('right')

1 2 30 4 5 6
repeat(
union(
)
).times(3)
3
tim
es
-0.125 0.125-0.3750.375

A wave is a diffusion of energy within a space.
The space is the graph.
The energy is contained within the traversers (via their sacks).
When a traverser splits, its sack energy is divided amongst its children
traversers' sacks.
When traversers merge, their sack energy is summed to a
single traverser sack.
If two energy sacks with the same parity are merged,
constructive interference.
If a negative and a positive energy sack are merged,
destructive interference.

Part 2
Quantum Mechanics
via the
Copenhagen Interpretation

1
2
3
0 4 5
gremlin> g.V(0)
==>v[0]

1
2
3
0 4 5
gremlin> g.V(0).out()
==>v[1]

1
2
3
0 4 5
gremlin> g.V(0).out().out()
==>??
?
?
"Two outgoing adjacent vertices,
what should I do?"

1
2
3
0 4 5
==>v[2]
==>v[3]

1
2
3
0 4 5
==>v[2]
==>v[3]
2 particles are created from 1.
In graph computing: "There are two legal paths, so take both."
In physical computing: "Violates conservation of energy law (matter is created)."

1
2
3
0 4 5
gremlin> g.V(0).out().out().out()
==>v[4]
==>v[4]
In graph computing: "Two length 3 paths lead to vertex 4."
In physical computing: "The twin particles coexist at vertex 4."

1
2
3
0 4 5
gremlin> g.V(0).out().out().out().out()
==>v[5]
==>v[5]
In graph computing: "Two length 3 paths lead to vertex 5."
In physical computing: "The twin particles coexist at vertex 5."

1
2
3
0 4 5
What does a physical particle do when it is faced with a choice?
?
?

1
2
3
0 4 5
The particle takes both options!
The indivisible "particle" divides!

1
2
3
0 4 5
Two particles can't be "seen," but we know both options are taken.
When we try to see both "particles", they instantly become one particle.
"Why do you say the particle took both paths?
If you ﬁnd it only on a single path, then wasn't only one path taken?"
Wave interference!
Explanation coming soon...
"Wave Function Collapse"

The smallest "quanta" allowed is the particle.
Only one particle is ever directly observed.
However, prior to observation, the particle splits into pieces (a wave).
When we observe the system, the pieces becomes one again (a particle).
Other Interpretations of Quantum Mechanics
Multi-World Interpretation
Pilot Wave Interpretation
photon
electron
bacteria!
an isolated system

Part 3
The Quantum Wave Function

Classical Particle
(A single unique "thing" exists)
The Particle Diffuses over Space as a Wave
(The particle is in a quantum superposition)
The Wave Function is Perturbed
(e.g., light tries to "see" the particle)
The Wave Function Collapses to a Classical Particle
(The wave turns into a particle)

Classical Particle
(A single unique "thing" exists)
The Particle Diffuses over Space as a Wave
(The particle is in a quantum superposition)
The Wave Function is Perturbed
(e.g., light tries to "see" the particle)
The Wave Function Collapses to a Classical Particle
(The wave turns into a particle)
ObservedObservedInferred

48 49 5047 51 52 53
A 101 vertex line graph with the center being vertex 50.
Space

Classical Initial Step
The particle has a deﬁnite location and a spin/potential to the left.
48 49 5047 51 52 53
[1, 0]
[1,0]50
left-ness
right-ness
For each option (degree of freedom),
we need a vector component.
In the article, we call it "traversal superposition."

0.00.20.40.60.81.0
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53
The probability of seeing the particle at vertex 50 is 100%.
[1, 0]
12
+ 02
= 1
Wave Function to Probability Distribution

Coin Step
48 49 5047 51 52 53
H =
1
√
2
1 1
1 −1
The particle's spin is put into superposition.
[1, 0] →
1
√
2
,
1
√
2
Ithastheoptiontogo
leftorrightsoitgoesboth!

Coin Step
48 49 5047 51 52 53
H =
1
√
2
1 1
1 −1
[1, 0] →
1
√
2
,
1
√
2
"right particles" going right ﬂip their sign.
"left particles" going left or right keep their sign.

Coin Step
48 49 5047 51 52 53
H =
1
√
2
1 1
1 −1
[1, 0] →
1
√
2
,
1
√
2
1
√
2
1 1
1 −1
· [1, 0] =
1√
2
1√
2
1√
2
− 1√
2
· [1, 0] =
1
√
2
,
1
√
2

Coin Step
48 49 5047 51 52 53
H =
1
√
2
1 1
1 −1
1 ·
1
√
2
+ 0 ·
1
√
2
, 1 ·
1
√
2
+ 0 · −
1
√
2
=
1
√
2
,
1
√
2
[1, 0] →
1
√
2
,
1
√
2
1
√
2
1 1
1 −1
· [1, 0] =
1√
2
1√
2
1√
2
− 1√
2
· [1, 0] =
1
√
2
,
1
√
2

Coin Step
48 49 5047 51 52 53
1
√
2
,
1
√
2

The "particle" spinning left goes left and
the "particle" spinning right goes right.
Shift Step
48 49 5047 51 52 53
1
√
2
, 0 0,
1
√
2
1
√
2
,
1
√
2

48 49 5047 51 52 53
1
√
2
, 0 0,
1
√
2
Shift Step

P(v) = |c0|2
+ |c1|2
48 49 5047 51 52 53
1
√
2
, 0 0,
1
√
2
1
√
2
2
+ 02
= 0.5 02
+
1
√
2
2
= 0.5
The probability of locating the classical particle each vertex
is a function of the total wave amplitude at that vertex.
50%-chance 50%-chance

0.00.10.20.30.40.5
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53
1
√
2
, 0 0,
1
√
2
1
√
2
2
+ 02
= 0.5 02
+
1
√
2
2
= 0.5
The probability of seeing the particle at vertex 49 or 51 is 50%.

48 49 5047 51 52 53
H =
1
√
2
1 1
1 −1
Each "particles'" spin is put into superposition.
Coin Step
1
2
,
1
2
1
2
, −
1
2

48 49 5047 51 52 53
Shift Step
1
2
,
1
2
The "particles" spinning left go left and
the "particles" spinning right go right.
0, −
1
2
1
2
, 0
1
2
,
1
2
1
2
, −
1
2

48 49 5047 51 52 53
Shift Step
1
2
,
1
2
0, −
1
2
1
2
, 0
Constructive Interference

0.00.10.20.30.40.5
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53
1
2
,
1
2
0, −
1
2
1
2
, 0
1
2
2
+ 02
=
1
4
1
2
2
+
1
2
2
=
1
2
02
+ −
1
2
2
=
1
4

48 49 5047 51 52 53
Coin Step
H =
1
√
2
1 1
1 −1
The "particles'" spin is put into superposition.
1
2
√
2
,
1
2
√
2
−
1
2
√
2
,
1
2
√
2
1
√
2
, 0

48 49 5047 51 52 53
Shift Step
The "particles" spinning left go left and
the "particles" spinning right go right.
1
2
√
2
, 0
1
√
2
,
1
2
√
2
−
1
2
√
2
, 0 0,
1
2
√
2
Destructive InterferenceConstructive Interference

0.00.10.20.30.40.50.6
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53
1
2
√
2
, 0
1
√
2
,
1
2
√
2
−
1
2
√
2
, 0 0,
1
2
√
2
1
2
√
2
2
+ 02
=
1
8
02
+
1
2
√
2
2
=
1
8
−
1
2
√
2
2
+ 02
=
1
8
1
√
2
2
+
1
2
√
2
2
=
5
8

0.00.10.20.30.40.50.6
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53

0.00.10.20.30.40.5
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53

0.00.10.20.30.4
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53

0.000.050.100.150.200.250.30
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53

0.000.050.100.150.200.25
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53

0.00.10.20.3
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53

0.000.050.100.15
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
13 14 1512 16 17 18

0.000.050.100.15
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
Wave Function is Perturbed by Light
13 14 1512 16 17 18

0.000.050.100.15
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
Wave Function Collapse
13 14 1512 16 17 18
sample

Classical Particle
13 14 1512 16 17 18
[1, 0]
0.00.20.40.60.81.0
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97

0.000.050.100.15
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
13 14 1512 16 17 18
Could
have
been
thisvertex.
Again,itsa
probabilitydistribution.

0.000.050.100.15
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
13 14 1512 16 17 18
Could not be at this vertex because there
is no wave energy at that vertex.

g.withSack([1,0],sackSum).V(50).
repeat(
sack(hadamard).
union(
sack(project).by(constant([1,0])).out('left'),
sack(project).by(constant([0,1])).out('right')
)
).times(50).
group().by(id).by(sack().map(norm)).
unfold().sample(1).by(values)
48 49 5047 51 52 53

48 49 5047 51 52 53
[1, 0]
repeat(
sack(hadamard).
union(
)
).times(50).
Classical Initial Step
sackSum = { a,b -> [a[0] + b[0],a[1] + b[1]] }

hadamard = { a,b ->
[(1/Math.sqrt(2)) * (a[0] + a[1]), (1/Math.sqrt(2)) * (a[0] - a[1])]
}
repeat(
sack(hadamard).
union(
)
).times(50).
Coin Step
H =
1
√
2
1 1
1 −1
48 49 5047 51 52 53
1
√
2
,
1
√
2

project = { a,b -> [a[0] * b[0], a[1] * b[1]] }
repeat(
sack(hadamard).
union(
)
).times(50).
Shift Step
48 49 5047 51 52 53
1
√
2
, 0 0,
1
√
2
1
√
2
,
1
√
2

repeat(
sack(hadamard).
union(
)
).times(50).
Diffuse the wave function for 50 steps
48 49 5047 51 52 53

0.000.050.100.15
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
norm = { Math.pow(it.get()[0],2) +
Math.pow(it.get()[1],2) }
repeat(
sack(hadamard).
union(
)
).times(50).
13 14 1512 16 17 18
P(v) = |c0|2
+ |c1|2

repeat(
sack(hadamard).
union(
)
).times(50).
Wave Function Collapse (Classical Particle Manifested)
13 14 1512 16 17 18
0.000.050.100.15
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
sample

repeat(
sack(hadamard).
union(
)
).times(50).
A Quantum Walk on a Line with Gremlin

H =
1
√
2
1 1
1 −1
Y =
1
√
2
1 i
i 1
Balanced Unitary OperatorUnbalanced Unitary Operator
0.000.050.100.15
vertices
probability
0 6 12 20 28 36 44 52 60 68 76 84 92 100
0.000.020.040.060.080.10
vertices
probability
0 6 12 20 28 36 44 52 60 68 76 84 92 100
Any energy split operator can be used as long as:
1. The operation conserves the total energy in the system.
2. It is possible to reverse the operation.
In general, a unitary operator is used to diffuse energy.
complexnumbers
i2
=−1
U · U∗
= I
nofriction
noinformation
loss

The probability of seeing the particle at a particular vertex after
50 iterations for both a quantum walk and a classical walk.
Quantum walks can explore more graph in a shorter amount of time!
0.000.020.040.060.080.10
vertices
probability
0 6 12 20 28 36 44 52 60 68 76 84 92 100
√
50 = 7.07
50 = 50
√
n
n
Y =
1
√
2
1 i
i 1

Part 4
The Double Slit Experiment

Thomas Young (1773-1829)
Question
"Is light a wave or a particle?"
continuous light source double-slit screen photoelectric ﬁlm

Hypothesis
"If light is a wave, then the ﬁlm will show a
constructive and destructive interference pattern."

Result
"Light must be a wave because it has interference and
refraction patterns like other natural waves such as
water waves."

G. I. Taylor (1886-1975)
single photon light source double-slit screen photoelectric ﬁlm
Question
"What happens when only a single photon
of light is emitted?"
The most indivisible amount of light.

G. I. Taylor (1886-1975)
Hypothesis
"The photons must act like bullets
and hit directly behind the slits."
50%
chance
50% chance

G. I. Taylor (1886-1975)
Experiment
"Shoot photon #1!"

G. I. Taylor (1886-1975)
Experiment
"Shoot photon #2!"

G. I. Taylor (1886-1975)
Experiment
"Shoot photon #3!"

G. I. Taylor (1886-1975)
Experiment
"Shoot photon #4!"

Fast forward to photon number 20.

G. I. Taylor (1886-1975)
Experiment
"Shoot photon #20!"

G. I. Taylor (1886-1975)
Result
"A single quantum of light (a photon) turns into a
wave after emission and then is localized back to a
particle at the ﬁlm based on the wave function."

single photon light source
double-slit screen
photoelectric ﬁlm

Quantum Processes in Graph Computing

"Particle" takes both options.

Probability distribution generated from the wave function.

Wave function collapses back to a classical particle.

0.0e+001.0e-052.0e-05
screen vertices
probability
0 2 4 6 8 10 12 14 16 18 20
Com
puted
using
G
rem
lin

g.withSack([0,0,1,0],sackSum).V(10).
repeat(
sack(grover).
union(
choose(out('left'),
out('left').sack(project).by(constant([1,0,0,0])),
identity().sack(project).by(constant([1,0,0,0])).
sack(flip).by(constant("lr"))),
choose(out('right'),
out('right').sack(project).by(constant([0,1,0,0])),
choose(out('up'),
out('up').sack(project).by(constant([0,0,1,0])),
sack(flip).by(constant("ud"))),
choose(out('down'),
out('down').sack(project).by(constant([0,0,0,1])),
sack(flip).by(constant("ud"))))).
times(22).

repeat(
sack(grover).
union(
choose(out('left'),
choose(out('up'),
choose(out('down'),
times(22).
Four degrees of freedom means you have a four entry energy vector.

repeat(
sack(grover).
union(
choose(out('left'),
choose(out('up'),
choose(out('down'),
times(22).
Bottom-middle vertex. The single photon light source.

A 4x4 unitary operator called the Grover coin. R =
1
2




−1 1 1 1
1 −1 1 1
1 1 −1 1
1 1 1 −1




repeat(
sack(grover).
union(
choose(out('left'),
choose(out('up'),
choose(out('down'),
times(22).

repeat(
sack(grover).
union(
choose(out('left'),
choose(out('up'),
choose(out('down'),
times(22).
Diffuse the wave in all four directions on the lattice.

repeat(
sack(grover).
union(
choose(out('left'),
choose(out('up'),
choose(out('down'),
times(22).
If there is a "wall," reﬂect the energy back. No decoherence/friction.
The
article
presents
the
details
around
reﬂection.

repeat(
sack(grover).
union(
choose(out('left'),
choose(out('up'),
choose(out('down'),
times(22).
As before, generate the probability distribution and sample it (i.e. collapse the wave form).

repeat(
sack(grover).
union(
choose(out('left'),
choose(out('up'),
choose(out('down'),
times(22).
A Quantum Walk on a Bounded Lattice with Gremlin

x-axis vertices
y-axisvertices
5
10
15
20
5 10 15 20
Step 22

0.0e+001.0e-052.0e-05
screen vertices
probability
0 2 4 6 8 10 12 14 16 18 20
x-axis vertices
y-axisvertices
5
10
15
20
5 10 15 20
Step 22

In classical computing, a bit is either 0 or 1.
0
[1, 0]
1
In quantum computing, a qubit can be 0, 1, or a superposition of both.
0
[0, 1]
1
1
√
2
, −
1
√
2

NOT
In classical computing, a gate takes a {0,1}* and outputs {0,1}*.
0 1
In quantum computing, a quantum gate takes a superposition of {0,1}*
and outputs a superposition of {0,1}*.
NOT
X =
0 1
1 0
1
√
2
, −
1
√
2
−
1
√
2
,
1
√
2
f(0)→
1
f(0,1)→
1,0
Two
function
evaluations
forthe
costofone!

U
Every quantum gate is a unitary operator.
U · U∗
= I
Not all gates have to be the same operator.
YH
H =
1
√
2
1 1
1 −1
Y =
1
√
2
1 i
i 1
H Y RH
Gates are connected by "wires."
R =
1
2




−1 1 1 1
1 −1 1 1
1 1 −1 1
1 1 1 −1




R
2-qubits!
Its all just vertices and
edges!

A quantum algorithm (circuit) can evaluate
all possible inputs at once.breadth-first search
depth-first search
The algorithm's purpose is to guide the wave function to yield the
correct answer to the problem with a high-probability (near 100%) at
the time of sampling (i.e. wave function collapse).
5-qubit register, where each qubit is both a 0 and 1 (thus, all 32 possible states at once).
1 0 1 0 0
We can never observe a superposition of outputs.
Thus, our quantum algorithms must ensure wave collapse to the correct classical state.
* Read about the pilot-wave interpretation of quantum mechanics.
Breadth-first memory requirements not needed!
* Read about the multi-world interpretation of quantum mechanics.
Depth-first speed enabled!

breadth-ﬁrst search
depth-ﬁrst search
1 0 1 0 0
0.03
0
1
0
1
All 32 states/values exist simultaneously.
Only one state/value can exist at observation.
1/32
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
25
=
32
Total wave energy is split equally amongst all states/values.
If the wave function collapses at this point in time, only one number (0-31) will manifest itself in reality with each being equally likely.
A quantum algorithm leverages simultaneous states, but can ultimately only yield one output state (a collapsed wave).
Thus, if you don't want a random answer, the algorithm must focus the wave energy on the correct solution.
Waves are controlled via constructive and destructive interference as the various parallel computations can interact with one another.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
1 0 1 0 0

+
Quantum Circuit
Graph
Quantum Wave Function
Traversal
Quantum Graph Computing
Structure Process
U
U
U

Fin.
Rodriguez, M.A., Watkins, J.H., "Quantum Walks with Gremlin," GraphDay '16, pp. 1-16, Austin, Texas, January 2016.
http://arxiv.org/abs/1511.06278

Quantum Processes in Graph Computing

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