How to call opt.step function with two vars sets

PennyLane Help
1

From the tutorial, we know: var = opt.step(lambda v: cost(v, X, Y), var)
My question is: if I have two vars, say. var_1, var_2, and they have different sizes. How do I call the step function?

2

Hi @cubicgate!

The optimizers provided with PennyLane all work best if the trainable parameters are a single NumPy array.

If this is not the case, there are tricks you can do (such as reshaping/flattening/concatenation) to make sure that the final parameters passed to your cost function are a flattened array:

import pennylane as qml
from pennylane import numpy as np, expval, var

dev = qml.device("default.qubit", wires=3)

@qml.qnode(dev)
def circuit(var1, var2):
    qml.Hadamard(wires=0)
    qml.Hadamard(wires=1)

    # layer 1
    qml.RX(var1[0], wires=0)
    qml.RZ(var1[1], wires=1)
    qml.CNOT(wires=(0, 1))

    # layer 2
    qml.RX(var2[0], wires=0)
    qml.RY(var2[1], wires=1)
    qml.RZ(var2[2], wires=2)

    qml.CNOT(wires=(0, 1))
    return expval(qml.PauliY(0)), var(qml.PauliZ(1))

var1 = np.array([0.54, -0.12])
var2 = np.array([-0.6543, 0.123, 1.95])

opt = qml.GradientDescentOptimizer(0.1)

def cost(params):
    """Trains the output of the circuit such
    that the parameters for each layer
    result in the expectation <Y> on wire
    0 is a magnitude of 2 different from
    the variance var(PauliZ) on wire 1
    """
    var1 = params[:2]
    var2 = params[2:]
    res = circuit(var1, var2)
    return np.abs(res[0] - res[1] + 2)

params = np.concatenate([var1, var2])

for i in range(100):
    params = opt.step(cost, params)
    print("Cost:", cost(params))

print("Final circuit value:", circuit(params))
print("Final parameters:", params)
3

Alternatively, you may wish to use the more advanced PyTorch or TF interfaces to the PennyLane QNode. These allow the QNode to be compatible with PyTorch/TensorFlow tensors and optimizers, which are much more flexible than the optimizers provided by PennyLane.

For example, below is the exact same program as my previous post, this time written using PyTorch instead of NumPy:

import torch
from torch.autograd import Variable

import numpy as np

import pennylane as qml
from pennylane import expval, var

dev = qml.device("default.qubit", wires=3)

@qml.qnode(dev, interface="torch")
def circuit(var1, var2):
    qml.Hadamard(wires=0)
    qml.Hadamard(wires=1)

    # layer 1
    qml.RX(var1[0], wires=0)
    qml.RZ(var1[1], wires=1)
    qml.CNOT(wires=(0, 1))

    # layer 2
    qml.RX(var2[0], wires=0)
    qml.RY(var2[1], wires=1)
    qml.RZ(var2[2], wires=2)

    qml.CNOT(wires=(0, 1))
    return expval(qml.PauliY(0)), var(qml.PauliZ(1))

var1 = Variable(torch.tensor([0.54, -0.12]), requires_grad=True)
var2 = Variable(torch.tensor([-0.6543, 0.123, 1.95]), requires_grad=True)

def cost(var1, var2):
    """Trains the output of the circuit such
    that the parameters for each layer
    result in the expectation <Y> on wire
    0 is a magnitude of 2 different from
    the variance var(PauliZ) on wire 1
    """
    res = circuit(var1, var2)
    return torch.abs(res[0] - res[1] + 2)

opt = torch.optim.Adam([var1, var2], lr=0.1)

for i in range(100):
    opt.zero_grad()
    loss = cost(var1, var2)
    loss.backward()
    opt.step()
    print("Cost:", loss)

var1_final, var2_final = opt.param_groups[0]['params']

print("Final circuit value:", circuit(var1_final, var2_final))
print("Final parameters:", var1_final, var2_final)

Note that the PyTorch optimizer supports two trainable tensors of different size, so there is no need to concatenate them.

4

Thanks Josh for your code and explanation!

6

hi, did you solved the problem

7

Hi @JEEVARATHINAM! This is now fixed in the latest versions of PennyLane :slightly_smiling_face:

If you have a cost function of the form

def cost(x, y, z):
    ...

then you can do

x, y, z = opt.step(cost, x, y, z)
8

Thanks.

I was trying to save this model.

https://pennylane.readthedocs.io/en/user-docs-refactor/tutorials/pennylane_quantum_neural_net.html

May I know how to do it

9

Hi @JEEVARATHINAM! Since this question is unrelated to this thread, could you please open a new topic on the Forum homepage :slightly_smiling_face: Thank you!

This will make your question (and any answers) more visible to our moderators and other readers.

1 Like