# QAOA for MaxCut

In the Pennylane instance, a Max cut problem of 4 qubits was shown. Here, I want to try to study 5 qubits, but there is a problem in the output results. The results are still four qubit results, and I am trying to change【xtick_labels = list(map(lambda x: format(x, “04b”), xticks))】to【xtick_labels = list(map(lambda x: format(x, “05b”), xticks))】But it’s just adding a |0> before each bit, which is problematic. How should I modify it？

Best wishes!

``````# Put code here
import pennylane as qml
from pennylane import numpy as np

np.random.seed(42)

n_wires = 5
graph = [(0, 1), (0, 3), (0, 2), (0, 4)]

# unitary operator U_B with parameter beta
def U_B(beta):
for wire in range(n_wires):
qml.RX(2 * beta, wires=wire)
def U_C(gamma):
for edge in graph:
wire1 = edge
wire2 = edge
qml.CNOT(wires=[wire1, wire2])
qml.RZ(2 * gamma, wires=wire2)
qml.CNOT(wires=[wire1, wire2])

def bitstring_to_int(bit_string_sample):
bit_string = "".join(str(bs) for bs in bit_string_sample)
return int(bit_string, base=2)

dev = qml.device("lightning.qubit", wires=n_wires, shots=1)

@qml.qnode(dev)
def circuit(gammas, betas, edge=None, n_layers=1):
# apply Hadamards to get the n qubit |+> state
for wire in range(n_wires):
# p instances of unitary operators
for i in range(n_layers):
U_C(gammas[i])
U_B(betas[i])
if edge is None:
# measurement phase
return qml.sample()
# during the optimization phase we are evaluating a term
# in the objective using expval
H = qml.PauliZ(edge) @ qml.PauliZ(edge)
return qml.expval(H)
def qaoa_maxcut(n_layers=1):
print("\np={:d}".format(n_layers))

# initialize the parameters near zero
init_params = 0.01 * np.random.rand(2, n_layers, requires_grad=True)

# minimize the negative of the objective function
def objective(params):
gammas = params
betas = params
neg_obj = 0
for edge in graph:
# objective for the MaxCut problem
neg_obj -= 0.5 * (1 - circuit(gammas, betas, edge=edge, n_layers=n_layers))
return neg_obj

# optimize parameters in objective
params = init_params
steps = 30
for i in range(steps):
params = opt.step(objective, params)
if (i + 1) % 5 == 0:
print("Objective after step {:5d}: {: .7f}".format(i + 1, -objective(params)))

# sample measured bitstrings 100 times
bit_strings = []
n_samples = 100
for i in range(0, n_samples):
bit_strings.append(bitstring_to_int(circuit(params, params, edge=None, n_layers=n_layers)))

# print optimal parameters and most frequently sampled bitstring
counts = np.bincount(np.array(bit_strings))
most_freq_bit_string = np.argmax(counts)
print("Optimized (gamma, beta) vectors:\n{}".format(params[:, :n_layers]))
print("Most frequently sampled bit string is: {:04b}".format(most_freq_bit_string))

return -objective(params), bit_strings
bitstrings1 = qaoa_maxcut(n_layers=1)
bitstrings2 = qaoa_maxcut(n_layers=2)
import matplotlib.pyplot as plt

xticks = range(0, 16)
xtick_labels = list(map(lambda x: format(x, "05b"), xticks))
bins = np.arange(0, 17) - 0.5

fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(8, 4))
plt.subplot(1, 2, 1)
plt.title("n_layers=1")
plt.xlabel("bitstrings")
plt.ylabel("freq.")
plt.xticks(xticks, xtick_labels, rotation="vertical")
plt.hist(bitstrings1, bins=bins)

plt.subplot(1, 2, 2)
plt.title("n_layers=2")
plt.xlabel("bitstrings")
plt.ylabel("freq.")
plt.xticks(xticks, xtick_labels, rotation="vertical")
plt.hist(bitstrings2, bins=bins)
plt.tight_layout()
plt.show()

# Put full error message here``````

Hello @ming !

I checked your script and it seems that the results are compatible with a 5 qubits problem. However, on this line, the variable `most_freq_bit_string` is being formatted as a length-4 binary string by writing `:04b`, as Isaac pointed out in a previous post you made:

I think you are getting the impression the result is 4 qubits from here. I would suggest as debug exercise to check out in the decimal form and see if makes sense Another problem I’ve found in you code is this part:

The number of ticks is not compatible with the range of your results, since you have 2^5 possible binary string outcomes. Try fixing the range of the bar plot to match the dimension of your problem I am sure this will help to better visualize your results. 