Snell Tomography for Net-To-Gross Estimation Using Quantum Annealing

Home , Quantum annealing

Rahul Sarkar1 Stewart A. Levin2

1Institute for Computational and Mathematical Engineering (ICME) Stanford University

2Department of Geophysics Stanford University

SEG 2018 Annual Meeting, Anaheim Ray Tomography

Tomography ≈ localizing subsurface properties

A.Curtis (2004) Zhang et al. (2012)

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 2 / 24 Ray Tomography

Tomography ≈ localizing subsurface properties

But not always...

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 3 / 24 Total Ambiguity 1 Ray, 2 Materials I Sand – 3.0 km/s I Shale – 2.5 km/s I Experiment vertical ray I Source-receiver distance = 4.5 km I Source-receiver traveltime = 1.7 s

SAND 0.75 km SAND 1.5 km SHALE 3.0 km SHALE 2.0 km

SAND 0.375 km

SHALE 3.0 km SHALE 1.0 km SAND 1.5 km SAND 0.375 km

Many answers

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 4 / 24 No Ambiguity 1 Ray, 2 Materials I Sand – 3.0 km/s I Shale – 2.5 km/s I Experiment vertical ray I Source-receiver distance = 4.5 km I Source-receiver traveltime = 1.7 s

SAND 0.75 km SAND 1.5 km SHALE 3.0 km SHALE 2.0 km

SAND 0.375 km

SHALE 3.0 km SHALE 1.0 km SAND 1.5 km SAND 0.375 km Exactly one sand-shale ratio (1:2)!

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 5 / 24 Challenge: Can we determine the fraction of each material in the macrolayer between the source and receivers?

R1 R2

((( (((( ((((

Experimental Geometry

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 6 / 24 1D Earth Model

I Horizontal layers of unknown thicknesses but known material properties

I The receivers are placed on the top of the layers

I There is one source at the bottom of the layers

I We measure both travel time and oﬀset for every source-receiver pair

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 7 / 24 OMG! R

((( (((( ((((

2 2 2 2 2 2 2 6 0 = 4γ (γ − 1) T v1 (Z + L ) s 2 2 2 2 2 2 2 5 −4γ(γ − 1) Tv1L[(γ + 1)(L + Z ) + T v1 ] s +(γ2 − 1)2[(γ2 − 1)2Z 4 + 2(γ4 + 1)L2Z 2 + (γ2 + 1)2L4 2 2 2 2 2 4 4 4 +2(γ + 1)T v1 (L − Z ) + T v1 ] s 2 2 2 2 3 +4γ(γ − 1) Tv1L(L + 2Z ) s 2 2 2 2 2 2 2 2 2 −2(γ − 1) L [(γ + 1)(L + Z ) + T v1 ] s +(γ2 − 1)2L4

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 8 / 24 Potential of Quantum Computing Some problems that are hard to solve using today’s classical computers, have the potential to be eﬃciently solvable using a quantum computer.

Some possible use cases of quantum computing: I Large combinatoric optimization problems I Quantum chemistry & material design I Quantum cryptography I Quantum machine learning

Kandala et al., Nature 549 (2017)

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 9 / 24 Quo Vadis Quantum Computing?

Commercial general purpose quantum computers are 5-10 years out. I Today: < 100 quantum bits I Near term: ∼ 1000 quantum bits (NISQ-era) I Commercial: ∼ 1 million quantum bits

Specialized devices called quantum annealers exist today. I Today: 2000 quantum bits I Near Term: ∼ 5000–10000 quantum bits

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 10 / 24 What Are Quantum Annealers?

I Specialized quantum devices I Finds the minimum energy eigenstate for Ising Hamiltonians

t t Ht = (1 − T )Hi + T Hf Adiabatic quantum computing: If you change slowly the initial Hamiltonian to the www.dwavesys.com/quantum-computing ﬁnal Hamiltonian, the system will remain in its lowest energy state.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 11 / 24 Quadratic Unconstrained Binary Optimization (QUBO)

One possible use of quantum annealers, is to solve QUBO problems!

Model QUBO Problem minimize x T Q x subject to x ∈ {0, 1}n

I Many other problems can be reduced to the QUBO form, for example problems with linear equality and inequality constraints.

I In our problem x assigns materials to layers to match traveltimes and oﬀsets.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 12 / 24 Problem Formulation Horizontal constant thickness layers : We will assume that we have a ﬁnite number N of horizontal sublayers with same thickness δ. So total thickness is Nδ.

Finite material set : There are only a ﬁnite number of materials M, with ﬁxed known velocities given by v1,..., vM .

Finite number of rays : We have a ﬁnite number of rays with corresponding travel time measurements are given by T1,..., TK , and the oﬀsets are given by L1,..., LK .

R1 R2

(( (((( ((( S Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 13 / 24 Layer Assignment

Introduce binary variables : (1 if sublayer i is material j xij = 0 otherwise, for all i = 1,..., N and j = 1,..., M.

Introduce constraints : To ensure that each sublayer gets assigned to one and only one material we will additionally need the family of constraints

M X xij = 1 , for all i = 1,..., N. j=1

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 14 / 24 Setup Objective Function

Least Squares Objective Function for kth Ray

2 2  N M   N M  1 X X X X Jk = Lk − αkj xij  + Tk − βkj xij  v 2 max i=1 j=1 i=1 j=1

th Assuming that the k ray has ray-parameter pk , we have deﬁned: th I αkj : Horizontal distance traveled by k ray with ray-parameter pk in a layer with velocity vj . th I βkj : Time spent by k ray with ray-parameter pk in a layer with velocity vj .

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 15 / 24 Alternating Minimization

Continuous Optimization Problem:

I Hold xij ’s ﬁxed, and minimize over pk ’s

I Trivial ray tracing

Discrete Optimization Problem:

I Hold pk ’s ﬁxed, and minimize over xij ’s

I Combinatorial challenge

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 16 / 24 Classical Computer Trials

I Take total vertical thickness of 1 km. I Number of materials M = 3. I Materials : I Sandstone — 3.0 km / s, Shale — 2.5 km / s, Salt — 4.6 km / s I Number of sublayers was varied from N = 2,..., 32. I Number of rays was varied from K = 2, 4, 8. I For each combination of N and K, 50 independent problem instances were created. For each instance, uniformly spaced pk ’s were used to generate the “true” data. I Each instance was solved 50 times and statistics were gathered.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 17 / 24 Performance of the Algorithm

Reduction in objective function Number of iterations 150 Nrays = 2 Nrays = 2 6 N = 4 140 Nrays = 4 rays N = 8 Nrays = 8 rays 130 5

120 4 110

100 3

Number of iterations 2 80

Percentage decrease in objective 70 5 10 15 20 25 30 5 10 15 20 25 30 Number of layers Number of layers

Pretty good reduction in objective The alternating algorithm converges quite function. fast in a small number of iterations.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 18 / 24 Impact of Number of Rays on Sand-Shale Ratio I 2 materials, 6 layers I Diﬀerent sand-shale ratios I Vary number of rays: K = 1, 2, 3, 4, 5

True Sand Shale Observed Sand Shale Ratio Ratio K = 1 K = 2 K = 3 K = 4 K = 5 5.0 5.0 5.0 5.0 5.0 5.0 2.0 2.0 2.0 2.0 2.0 2.0 1.0 1.0 1.0 1.0 1.0 1.0 0.5 0.5 0.5 0.5 0.5 0.5 0.2 0.2 0.2 0.2 0.2 0.2

With just 2 materials, we were able to recover the true sand-shale ratios exactly on average. With more materials, increasing the number of rays helps, but our computational experiments suggest that the eﬀect is not monotonic.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 19 / 24 How Big Today?

Comparison of logical variables vs qubits

120 Qubits : Nrays = 2 Qubits : Nrays = 4

Qubits : Nrays = 8 100 Logical Variables

20 Number of qubits / logical variables

2 4 6 8 10 12 14 16 Number of layers

Due to lack of full connectivity of the D-Wave annealer, mapping the problem to the hardware leads to an increase in the number of eﬀective qubits.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 20 / 24 Stochasticity of Annealing

Histogram of solutions 250

200

150 Counts 100

50 minimum objective value

0 1.00 0.75 0.50 0.25 0.00 0.25 0.50 0.75 1.00 Normalized objective function value

The quantum annealing process has inbuilt stochasticity. Solutions are recovered close to the lowest energy “ground state” of the objective function or Hamiltonian.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 21 / 24 Our Thoughts

I Quantum annealer did provide good solutions.

I For number of binary variables beyond 100, solving the problem through enumeration is nearly impossible for a classical computer.

I This method can also be useful for a kind of “uncertainty quantiﬁcation”, as you can potentially generate an ensemble of solutions that satisfy a given error tolerance.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 22 / 24 Acknowledgments

I We thank Peter L. McMahon for many discussions on quantum computing.

I We thank QC Ware Corp. and D-Wave Systems Inc. for giving us permissions to publish the quantum computing results.

I This work was partially supported with funding from the aﬃliates of the Stanford Exploration Project.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 23 / 24 Q&A

Thank You

Questions?

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 24 / 24 What Problems Can Quantum Computers Solve Eﬃciently?

BQP = Bounded-error quantum polynomial time

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 24 / 24 QUBO to Ising

Model QUBO Problem minimize x T Q x subject to x ∈ {0, 1}n

Conversion to Ising Hamiltonian

Change of variables: si = 2xi − 1 X X minimize Jij si sj + hi si i>j i subject to s ∈ {−1, 1}n

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 24 / 24 An Optimization Problem With a Lot of Binary Variables Leads to a non-convex “mixed integer program” : Sum of Squares Optimization Problem

K X minimize Jk k=1

subject to xij ∈ {0, 1} , ∀ i ∈ {1,..., N}, ∀ j ∈ {1,..., M}, M X xij = 1 , ∀ i ∈ {1,..., N}. j=1

I Continuous variables : The ray-parameters p1,..., pK I Binary variables : Layer assignments x11,..., xNM I These are typically hard problems

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 24 / 24 Problem Structure

I The continuous problem is separable over each ray! So each Jk can be minimized independently.

2 2  N M   N M  1 X X X X Jk = Lk − αkj xij  + Tk − βkj xij  v 2 max i=1 j=1 i=1 j=1 K X Objective Function : Jk k=1

I The discrete problem is a Quadratic Binary Optimization Problem. Number of binary variables is NM.

Question: Can we reduce the number of binary variables ?

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 24 / 24 Towards a Better Combinatorial Optimization Problem: Symmetry of Composition

Symmetry of Composition: Given an assignment of layers, the objective function is unchanged under arbitrary permutation of layers (this holds because of the horizontal layer assumption). This is a key observation.

Key idea: Just count number of times a material repeats. Let yj denote number of times material j occurs.

I Bound constraint : 0 ≤ yj ≤ N, for all j = 1,..., M. PM I Sum constraint : j=1 yj = N.

Good but not enough! Only M integer variables, but each variable can take N + 1 integer values.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 24 / 24 Towards a Better Combinatorial Optimization Problem: Binary Encoding Trick

Binary Encoding Trick:

Since yj ’s are positive uniformly spaces integers, work with the binary representation of each variable.

r X l−1 yj = bjl 2 , ∀ j = 1,..., M , l=1

where r = blog2 Nc + 1, and bjl ∈ {0, 1}.

So now we have Mr ≈ M log2 N binary variables. This is a log N compression.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 24 / 24 Equivalent Combinatorial Optimization Problem

Equivalent Optimization Problem

 2 K  M r  X 1 X X l−1 minimize  Lk − αkj bjl 2   + v 2  k=1 max j=1 l=1  2 K  M r  X  X X l−1  Tk − βkj bjl 2   k=1 j=1 l=1

subject to bjl ∈ {0, 1} , ∀ j ∈ {1,..., M} , ∀ l ∈ {1,..., r} , M r X X l−1 bjl 2 = N. j=1 l=1

Still a non-convex mixed integer optimization problem.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 24 / 24 Alternating Minimization Algorithm

Algorithm 1 Alternating minimization algorithm procedure Alternating QUBO

// Random assignment xij ← 0 , for all i = 1,..., N and j = 1,..., M for i = 1 to N do j ← Randomly choose from the set {1,..., M} xij ← 1 PN Compute yj = xij , for all j = 1,..., M i=1 Compute bjl as binary representation of yj , for all j = 1,..., M and l = 1,..., r

// Alternating minimization while Not converged do pk ← arg min Jk , for all k = 1,..., K, and with all bjl ﬁxed. pk bjl ← solution of QUBO with all pk ﬁxed. Pr l−1 Compute yj = bjl 2 , for all j = 1,..., M l=1

return pk , yj for all k = 1,..., K and j = 1,..., M

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 24 / 24 Discrete Search Over Ray-Parameters

Algorithm 2 Global ray parameter search 1: procedure Discretized global search (k, P) 2: ∆p ← 1 P vmax 3: S ← {n∆p : n = 0,..., P − 1} 2 2 M ! M ! 1 P δpvj yj P δyj 4: pk ← arg min 2 Lk − + Tk − vmax p 2 2 p 2 2 p∈S j=1 1−p vj j=1 vj 1−p vj 5: return pk

P controls level of discretization.

This might seem scary, but is not precisely because ray-parameter search is independent over each ray.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 24 / 24 Continuous Search Over Ray-Parameters

Algorithm 3 Newton ray parameter search 1: procedure Continuous local search (k, γ ∈ [0, 1]) p 2 2 2: pmin ← Lk /(vmax Z + Lk ), pmax ← 1/vmax pmin+pmax 3: pk ← 2 4: while Not converged do 2 2 2 0 ((1−γ)(T (pk )−Tk ) +γ(L(pk )−Lk ) /vmax) 5: pk ← pk − 2 2 2 00 ((1−γ)(T (pk )−Tk ) +γ(L(pk )−Lk ) /vmax) 6: pk ← max(pmin, min(pmax, pk ))

7: return pk

γ ∈ [0, 1] controls how travel time and oﬀset terms are weighed in the search process.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 24 / 24 Classical Results: Wall Clock Time to Convergence

Time to solution

Nrays = 2

1.75 Nrays = 4

Nrays = 8 1.50

1.25

1.00

0.75

Time in seconds 0.50

0.25

5 10 15 20 25 30 Number of layers

Time to convergence increases with number of layers.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 24 / 24 Classical Results: Mean Layer Assignment Error

Layer assignment solution (average error) 0.6 Nrays = 2 1

| Nrays = 4 | ) 0.5 Nrays = 8 d e v l o

s 0.4 x

l 0.3 a u t c a

x 0.2 ( t a 1 m

N 0.1 | | 1 N

0.0 5 10 15 20 25 30 Number of layers

With the right scaling, errors are constant over number of layers.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 24 / 24 Classical Results: Mean Ray-Parameter Error

Ray parameter solution (average error)

N = 2 0.010 rays

1 Nrays = 4 |

| Nrays = 8 d

e 0.008 v l o s p 0.006 l a u t

c 0.004 a p | | s a

e 0.002 1 m N

0.000 5 10 15 20 25 30 Number of layers

With the right scaling, errors are constant over number of layers.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 24 / 24 Preliminary Quantum Solve Trials

I Same experimental setup as in the classical case.

I We varied N = 2,..., 16, and K = 2, 4, 8.

I For each combination of N and K, 10 independent problem instances were created. For each instance, uniformly spaced pk ’s were used to generate the “true” data.

I We solved each instance of the QUBO problem on the D-WAVE 2000Q quantum annealer 1000 times and gathered statistics on the total time to solution.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 24 / 24 Quantum Annealing Results: Time to Solution Time to solution

800 Nrays = 2 Nrays = 4

700 Nrays = 8

600

500

400

300 Time in seconds

200

100

2 4 6 8 10 12 14 16 Number of layers

The average time per anneal cycle is about same as the time taken to solve the optimization problem on a classical computer.

Rahul Sarkar Snell Tomography SEG 2018 Annual Meeting, Anaheim 24 / 24