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Pauli-Korrelations-Encoding för dat Verringern vun de Maxcut-Anförderungen

Bruuktied-Schattung: 30 Minuten op'n Eagle r3-Prozessor (HENWIES: Dat is blots 'ne Schattung. Dien Looptied kann annersweg wesen.)

Achtergrund

Dit Tutorial stellt Pauli-Korrelations-Encoding (PCE) [1] vör, een Ansatz, de dorfför dacht is, Optimerensproblemen effizient in Qubits för Quantenbereken to koderen. PCE bildt klassische Variabeln in Optimerensproblemen op Meerkörper-Pauli-Matrix-Korrelationen af, wat to een polynomielle Kompression vun de Platzanförderungen vun dat Probleem föhrt. Dörch den Insatz vun PCE warrt de Antall vun de för dat Koderen nödigen Qubits verringert, wat besünners vördeelhaftig is för kortfristige Quantengeräte mit begrenzte Qubit-Ressourcen. Dorto warrt analytisch nawesen, dat PCE inhärent Barren Plateaus afschwächt un super-polynomielle Wedderstannsfähigkeit gegen dat Phänomen beedt. Disse inbuute Egenschap ermöglicht bispellose Leestung bi Quanten-Optimerenslösers.

Överblick

De PCE-Ansatz besteiht ut dree Höövdschreden, as in Afbilldung 1 ut [1] hierünner vörstellt warrt:

  1. Koderen vun dat Optimerensprobleem in een Pauli-Korrelationsruum.
  2. Lösung vun dat Probleem mit een quanten-klassischen Optimerenslöser.
  3. Dekoderen vun de Lösung trüch in den oorsprünklichen Optimerengsruum. De PCE-Ansatz is anpassbaar an jeden Quanten-Optimerenslöser, de Pauli-Korrelationsmatrizen verarbeiden kann. pce-overview.png In Afbilldung 1 ut [1] warrt dat Max-Cut-Probleem as Bispeel för de Veranschaulikung vun den PCE-Ansatz bruukt. Dat Max-Cut-Probleem mit m=9m=9 Knoten warrt in een Pauli-Korrelationsruum kodeert, wobei dat Optimerensprobleem as Korrelationsmatrix dorstellt warrt, besünners as 2-Körper-Pauli-Matrix-Korrelationen över n=3n=3 Qubits (Q1,Q2,Q3)(Q_1, Q_2, Q_3). Knotenfarven wiest den Pauli-String, de för jeden kodeerten Knoten bruukt warrt. To'n Bispeel warrt Knoten 1, de de binäre Variable x1x_1 entsprickt, dörch den Verwachtensweert vun Z1Z2I3Z_1 \otimes Z_2 \otimes I_3 kodeert, wieldes x8x_8 dörch I1Y2Y3I_1 \otimes Y_2 \otimes Y_3 kodeert warrt. Dat entsprickt een Kompression vun de mm Variabeln vun dat Probleem in n=O(m1/2) n = O(m^{1/2}) Qubits. Allgemener ermöglicht kk-Körper-Korrelationen polynomielle Kompressionen vun de Ordnung kk. De wählte Pauli-Satz ümfaat dree Deelmengen vun gegensietig kommuteerenden Pauli-Strings, wodörch all mm Korrelationen experimentell mit blots dree Meetinstellen schatt warrn köönt.

Een Verlustfunkschoon L\mathcal{L} vun Pauli-Verwachtensweerten, de de oorsprünkliche Max-Cut-Zeelfunkschoon naahmt, warrt konstrueert. De Verlustfunkschoon warrt denn mit een quanten-klassischen Optimerenslöser as den Variational Quantum Eigensolver (VQE) optimeert.

Sobald de Optimerung afslooten is, warrt de Lösung in den oorsprünklichen Optimerengsruum trüchkodeert, wodörch de optimale Max-Cut-Lösung erreekt warrt.

Anförderungen

Bevör du mit dit Tutorial anfängst, sorg dorfför, dat du dit Folgende installeert hest:

  • Qiskit SDK v1.0 oder höger, mit Visualiseerung-Ünnerstütten
  • Qiskit Runtime v0.22 oder höger (pip install qiskit-ibm-runtime)

Inrichten

# Added by doQumentation — required packages for this notebook
!pip install -q networkx numpy qiskit qiskit-ibm-runtime rustworkx scipy
from itertools import combinations

import numpy as np
import rustworkx as rx
from scipy.optimize import minimize

from qiskit.circuit.library import efficient_su2
from qiskit.transpiler.preset_passmanagers import generate_preset_pass_manager
from qiskit.quantum_info import SparsePauliOp
from qiskit_ibm_runtime import EstimatorV2 as Estimator
from qiskit_ibm_runtime import QiskitRuntimeService
from qiskit_ibm_runtime import Session
from rustworkx.visualization import mpl_draw

service = QiskitRuntimeService()
backend = service.least_busy(
operational=True, simulator=False, min_num_qubits=127
)
def calc_cut_size(graph, partition0, partition1):
"""Calculate the cut size of the given partitions of the graph."""

cut_size = 0
for edge0, edge1 in graph.edge_list():
if edge0 in partition0 and edge1 in partition1:
cut_size += 1
elif edge0 in partition1 and edge1 in partition0:
cut_size += 1
return cut_size

Schritt 1: Klassische Ingaven op een Quantenprobleem afbilden

Max-Cut-Probleem

Dat Max-Cut-Probleem is een kombinatorisch Optimerensprobleem, dat op een Graaf G=(V,E)G = (V, E) defineert is, wobei VV de Menge vun de Knoten un EE de Menge vun de Kanten is. Dat Zeel is dat, de Knoten in twee Mengen SS un VSV \setminus S to partitioneren, so dat de Antall vun de Kanten twüschen de twee Mengen maximeert warrt. För een utföhrliche Beschrieven vun dat Max-Cut-Probleem kiek na dat Tutorial "Quantum approximate optimization algorithm". Dorto warrt dat Max-Cut-Probleem as Bispeel in dat Tutorial "Advanced Techniques for QAOA" bruukt. In disse Tutorials warrt de QAOA-Algorithmus för de Lösung vun dat Max-Cut-Probleem insett.

Graaf -> Hamiltonian

Dit Tutorial bruukt een Tofall-Graaf mit 1000 Knoten.

De Probleemgrötte is mööglicherwiese swaar to visualiseren, dorüm is hierünner een Graaf mit 100 Knoten dorstellt. (De direkte Dorstellung vun een Graaf mit 1.000 Knoten würr to dicht maken, üm wat to sehn!) De Graaf, mit den wi arbeiden doot, is teihnmaal grötter.

mpl_draw(rx.undirected_gnp_random_graph(100, 0.1, seed=42))

Output of the previous code cell

num_nodes = 1000  # Number of nodes in graph
graph = rx.undirected_gnp_random_graph(num_nodes, 0.1, seed=42)
import networkx as nx

nx_graph = nx.Graph()
nx_graph.add_nodes_from(range(num_nodes))
for edge in graph.edge_list():
nx_graph.add_edge(edge[0], edge[1])
curr_cut_size, partition = nx.approximation.one_exchange(nx_graph, seed=1)
print(f"Initial cut size: {curr_cut_size}")
Initial cut size: 28075

Wi kodeert den Graaf mit 1000 Knoten in 2-Körper-Pauli-Matrix-Korrelationen över 100 Qubits. De Graaf warrt as Korrelationsmatrix dorstellt, wobei jede Knoten dörch een Pauli-String kodeert warrt. Dat Vörteken vun den Verwachtensweert vun den Pauli-String gifft de Partition vun den Knoten an. To'n Bispeel warrt Knoten 0 dörch een Pauli-String kodeert, 0=I19...I2X1X0\prod_0 = I_{19} \otimes ... I_2 \otimes X_1 \otimes X_0. Dat Vörteken vun den Verwachtensweert vun dissen Pauli-String gifft de Partition vun Knoten 0 an. Wi defineert een Pauli-Korrelations-Encoding (PCE) relativ to \prod as

xisgn(i)x_i \coloneqq \textit{sgn}(\langle\prod_i \rangle)

wobei xix_i de Partition vun Knoten ii is un iψiψ\langle \prod_i \rangle \coloneqq \langle \psi |\prod_i| \psi \rangle de Verwachtensweert vun den Pauli-String is, de Knoten ii över een Quantentostand ψ|\psi \rangle kodeert. Nu kodeert wi den Graaf mit PCE in een Hamiltonian. Wi deelt de Knoten in dree Mengen op: S1S_1, S2S_2 un S3S_3. Denn kodeert wi de Knoten in jede Menge mit de Pauli-Strings XX, YY bzw. ZZ.

num_qubits = 100

list_size = num_nodes // 3
node_x = [i for i in range(list_size)]
node_y = [i for i in range(list_size, 2 * list_size)]
node_z = [i for i in range(2 * list_size, num_nodes)]

print("List 1:", node_x)
print("List 2:", node_y)
print("List 3:", node_z)
List 1: [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, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 240, 241, 242, 243, 244, 245, 246, 247, 248, 249, 250, 251, 252, 253, 254, 255, 256, 257, 258, 259, 260, 261, 262, 263, 264, 265, 266, 267, 268, 269, 270, 271, 272, 273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285, 286, 287, 288, 289, 290, 291, 292, 293, 294, 295, 296, 297, 298, 299, 300, 301, 302, 303, 304, 305, 306, 307, 308, 309, 310, 311, 312, 313, 314, 315, 316, 317, 318, 319, 320, 321, 322, 323, 324, 325, 326, 327, 328, 329, 330, 331, 332]
List 2: [333, 334, 335, 336, 337, 338, 339, 340, 341, 342, 343, 344, 345, 346, 347, 348, 349, 350, 351, 352, 353, 354, 355, 356, 357, 358, 359, 360, 361, 362, 363, 364, 365, 366, 367, 368, 369, 370, 371, 372, 373, 374, 375, 376, 377, 378, 379, 380, 381, 382, 383, 384, 385, 386, 387, 388, 389, 390, 391, 392, 393, 394, 395, 396, 397, 398, 399, 400, 401, 402, 403, 404, 405, 406, 407, 408, 409, 410, 411, 412, 413, 414, 415, 416, 417, 418, 419, 420, 421, 422, 423, 424, 425, 426, 427, 428, 429, 430, 431, 432, 433, 434, 435, 436, 437, 438, 439, 440, 441, 442, 443, 444, 445, 446, 447, 448, 449, 450, 451, 452, 453, 454, 455, 456, 457, 458, 459, 460, 461, 462, 463, 464, 465, 466, 467, 468, 469, 470, 471, 472, 473, 474, 475, 476, 477, 478, 479, 480, 481, 482, 483, 484, 485, 486, 487, 488, 489, 490, 491, 492, 493, 494, 495, 496, 497, 498, 499, 500, 501, 502, 503, 504, 505, 506, 507, 508, 509, 510, 511, 512, 513, 514, 515, 516, 517, 518, 519, 520, 521, 522, 523, 524, 525, 526, 527, 528, 529, 530, 531, 532, 533, 534, 535, 536, 537, 538, 539, 540, 541, 542, 543, 544, 545, 546, 547, 548, 549, 550, 551, 552, 553, 554, 555, 556, 557, 558, 559, 560, 561, 562, 563, 564, 565, 566, 567, 568, 569, 570, 571, 572, 573, 574, 575, 576, 577, 578, 579, 580, 581, 582, 583, 584, 585, 586, 587, 588, 589, 590, 591, 592, 593, 594, 595, 596, 597, 598, 599, 600, 601, 602, 603, 604, 605, 606, 607, 608, 609, 610, 611, 612, 613, 614, 615, 616, 617, 618, 619, 620, 621, 622, 623, 624, 625, 626, 627, 628, 629, 630, 631, 632, 633, 634, 635, 636, 637, 638, 639, 640, 641, 642, 643, 644, 645, 646, 647, 648, 649, 650, 651, 652, 653, 654, 655, 656, 657, 658, 659, 660, 661, 662, 663, 664, 665]
List 3: [666, 667, 668, 669, 670, 671, 672, 673, 674, 675, 676, 677, 678, 679, 680, 681, 682, 683, 684, 685, 686, 687, 688, 689, 690, 691, 692, 693, 694, 695, 696, 697, 698, 699, 700, 701, 702, 703, 704, 705, 706, 707, 708, 709, 710, 711, 712, 713, 714, 715, 716, 717, 718, 719, 720, 721, 722, 723, 724, 725, 726, 727, 728, 729, 730, 731, 732, 733, 734, 735, 736, 737, 738, 739, 740, 741, 742, 743, 744, 745, 746, 747, 748, 749, 750, 751, 752, 753, 754, 755, 756, 757, 758, 759, 760, 761, 762, 763, 764, 765, 766, 767, 768, 769, 770, 771, 772, 773, 774, 775, 776, 777, 778, 779, 780, 781, 782, 783, 784, 785, 786, 787, 788, 789, 790, 791, 792, 793, 794, 795, 796, 797, 798, 799, 800, 801, 802, 803, 804, 805, 806, 807, 808, 809, 810, 811, 812, 813, 814, 815, 816, 817, 818, 819, 820, 821, 822, 823, 824, 825, 826, 827, 828, 829, 830, 831, 832, 833, 834, 835, 836, 837, 838, 839, 840, 841, 842, 843, 844, 845, 846, 847, 848, 849, 850, 851, 852, 853, 854, 855, 856, 857, 858, 859, 860, 861, 862, 863, 864, 865, 866, 867, 868, 869, 870, 871, 872, 873, 874, 875, 876, 877, 878, 879, 880, 881, 882, 883, 884, 885, 886, 887, 888, 889, 890, 891, 892, 893, 894, 895, 896, 897, 898, 899, 900, 901, 902, 903, 904, 905, 906, 907, 908, 909, 910, 911, 912, 913, 914, 915, 916, 917, 918, 919, 920, 921, 922, 923, 924, 925, 926, 927, 928, 929, 930, 931, 932, 933, 934, 935, 936, 937, 938, 939, 940, 941, 942, 943, 944, 945, 946, 947, 948, 949, 950, 951, 952, 953, 954, 955, 956, 957, 958, 959, 960, 961, 962, 963, 964, 965, 966, 967, 968, 969, 970, 971, 972, 973, 974, 975, 976, 977, 978, 979, 980, 981, 982, 983, 984, 985, 986, 987, 988, 989, 990, 991, 992, 993, 994, 995, 996, 997, 998, 999]
def build_pauli_correlation_encoding(pauli, node_list, n, k=2):
pauli_correlation_encoding = []
for idx, c in enumerate(combinations(range(n), k)):
if idx >= len(node_list):
break
paulis = ["I"] * n
paulis[c[0]], paulis[c[1]] = pauli, pauli
pauli_correlation_encoding.append(("".join(paulis)[::-1], 1))

hamiltonian = []
for pauli, weight in pauli_correlation_encoding:
hamiltonian.append(SparsePauliOp.from_list([(pauli, weight)]))

return hamiltonian

pauli_correlation_encoding_x = build_pauli_correlation_encoding(
"X", node_x, num_qubits
)
pauli_correlation_encoding_y = build_pauli_correlation_encoding(
"Y", node_y, num_qubits
)
pauli_correlation_encoding_z = build_pauli_correlation_encoding(
"Z", node_z, num_qubits
)

Schritt 2: Probleem för de Utfören op Quantenhardware optimeren

Quantenschaltkreis

Hier warrt de Tostand ψ|\psi \rangle mit θ\mathbf{\theta} parametriseert, un wi optimeert disse Parameter θ\mathbf{\theta} mit een variationellen Ansatz. Dit Tutorial bruukt den efficient_su2-Ansatz för unsen variationellen Algorithmus wegen sien Utdrucksfähigkeit un eenfache Implementerung. Wi bruukt ok de relaxeerte Verlustfunkschoon, de later in dit Tutorial inföhrt warrt. As Ergebnis köönt wi grootskalige Problemen mit weniger Qubits un geringere Schaltkreisdeepte angahn.

# Build the quantum circuit
qc = efficient_su2(num_qubits, ["ry", "rz"], reps=2)
# Optimize the circuit

pm = generate_preset_pass_manager(optimization_level=3, backend=backend)
qc = pm.run(qc)

Verlustfunkschoon

För de Verlustfunkschoon L\mathcal{L} bruukt wi een Relaxation vun de Max-Cut-Zeelfunkschoon as in [1] beschreven, de as V(x)(i,j)EWi,j(1xixj)\mathcal{V}(\mathbf{x}) \coloneqq \sum_{(i, j) \in E} W_{i, j}(1-x_i x_j) defineert is. Hier betekent Wi,jW_{i, j} dat Gewicht vun de Kant (i,j)(i, j) un xix_i de Partition vun Knoten ii. De Verlustfunkschoon L\mathcal{L} is geven dörch:

L(i,j)EWi,jtanh(αi)tanh(αj)+L(reg)\mathcal{L}\coloneqq \sum_{(i, j) \in E} W_{i, j} \text{tanh} (\alpha \langle\prod_i \rangle) \text{tanh} (\alpha \langle\prod_j \rangle) + \mathcal{L}^{(\text{reg})}

wobei de Max-Cut-Zeelfunkschoon dörch glatte hyperbolische Tangenten vun de Verwachtensweerten vun de Pauli-Strings ersett warrt, de de Knoten kodeert. De Regulariserungsterm L(reg)\mathcal{L}^{(\text{reg})} un de Skalerungsfaktor α\alpha, proportional to de Antall vun de Qubits, warrt inföhrt, üm de Leestung vun den Löser to verbettern.

De Regulariserungsterm is defineert as:

L(reg)\mathcal{L}^{(\text{reg})} is defineert as L(reg)βν[1miVtanh(αi)2]2\mathcal{L}^{(\text{reg})} \coloneqq \beta \nu \lbrack \frac{1}{m} \sum_{i \in V} \text{tanh} (\alpha \langle\prod_i \rangle)^2 \rbrack ^2

wobei β=1/2\beta=1/2, ν=E/2+(m1)/4\nu = |E|/2 + (m -1) /4 un mm de Antall vun de Knoten in den Graaf is.

def loss_func_estimator(x, ansatz, hamiltonian, estimator, graph):
"""
Calculates the specified loss function for the given ansatz, Hamiltonian, and graph.

The expectation values of each Pauli string in the Hamiltonian are first obtained
by running the ansatz on the quantum backend. These expectation values are then
passed through the nonlinear function tanh(alpha * prod_i). The loss function is
subsequently computed from these transformed values.
"""
job = estimator.run(
[
(ansatz, hamiltonian[0], x),
(ansatz, hamiltonian[1], x),
(ansatz, hamiltonian[2], x),
]
)
result = job.result()

# calculate the loss function
node_exp_map = {}
idx = 0
for r in result:
for ev in r.data.evs:
node_exp_map[idx] = ev
idx += 1

loss = 0
alpha = num_qubits
for edge0, edge1 in graph.edge_list():
loss += np.tanh(alpha * node_exp_map[edge0]) * np.tanh(
alpha * node_exp_map[edge1]
)

regulation_term = 0
for i in range(len(graph.nodes())):
regulation_term += np.tanh(alpha * node_exp_map[i]) ** 2
regulation_term = regulation_term / len(graph.nodes())
regulation_term = regulation_term**2
beta = 1 / 2
v = len(graph.edges()) / 2 + (len(graph.nodes()) - 1) / 4
regulation_term = beta * v * regulation_term

loss = loss + regulation_term

global experiment_result
print(f"Iter {len(experiment_result)}: {loss}")
experiment_result.append({"loss": loss, "exp_map": node_exp_map})
return loss

Schritt 3: Utfören mit Qiskit Primitives

In dit Tutorial sett wi max_iter=50 för de Optimerenssleep för Demonstrationstwecke. Wenn wi de Antall vun de Iterationen erhögen, köönt wi betere Ergebnissen verwachten.

pce = []
pce.append(
[op.apply_layout(qc.layout) for op in pauli_correlation_encoding_x]
)
pce.append(
[op.apply_layout(qc.layout) for op in pauli_correlation_encoding_y]
)
pce.append(
[op.apply_layout(qc.layout) for op in pauli_correlation_encoding_z]
)
# Run the optimization using Session

with Session(backend=backend) as session:
estimator = Estimator(mode=session)

experiment_result = []

def loss_func(x):
return loss_func_estimator(
x, qc, [pce[0], pce[1], pce[2]], estimator, graph
)

np.random.seed(42)
initial_params = np.random.rand(qc.num_parameters)
result = minimize(
loss_func, initial_params, method="COBYLA", options={"maxiter": 50}
)
print(result)
Iter 0: 16659.649201600296
Iter 1: 12104.242957555361
Iter 2: 6541.137221994661
Iter 3: 6650.6188244671985
Iter 4: 7033.193518185085
Iter 5: 6743.687931793412
Iter 6: 6223.574718684094
Iter 7: 6457.3302709535965
Iter 8: 6581.316449107595
Iter 9: 6365.761052029896
Iter 10: 6415.872673527322
Iter 11: 6421.996561600348
Iter 12: 6636.372822791712
Iter 13: 6965.174320702346
Iter 14: 6774.236562696287
Iter 15: 6393.837617108355
Iter 16: 6234.311401676519
Iter 17: 6518.192237615901
Iter 18: 6559.933925068997
Iter 19: 6646.157979243488
Iter 20: 6573.726111605048
Iter 21: 6190.642092901959
Iter 22: 6653.06500163594
Iter 23: 6545.713700369988
Iter 24: 6399.996441760465
Iter 25: 6115.959687941808
Iter 26: 6665.915093554849
Iter 27: 6832.882201259893
Iter 28: 6541.392749578919
Iter 29: 6813.3456910443165
Iter 30: 6460.800944368402
Iter 31: 6359.635437029245
Iter 32: 6040.891641882451
Iter 33: 6573.930674936448
Iter 34: 6668.031753293785
Iter 35: 6450.002712889748
Iter 36: 6519.8298811058075
Iter 37: 6467.134502398199
Iter 38: 6655.284651397334
Iter 39: 6371.168353987336
Iter 40: 6480.337259347923
Iter 41: 6339.256786764425
Iter 42: 6588.635046825541
Iter 43: 6617.677964971322
Iter 44: 6469.0441600679205
Iter 45: 6567.874244906106
Iter 46: 6217.899975264532
Iter 47: 6783.481394627947
Iter 48: 6813.371853626112
Iter 49: 6506.5871531488765
message: Maximum number of function evaluations has been exceeded.
success: False
status: 2
fun: 6040.891641882451
x: [ 1.375e+00 1.951e+00 ... 1.923e-01 4.087e-02]
nfev: 50
maxcv: 0.0

Schritt 4: Nabearbeiten un Trüchgaav vun dat Ergebnis in dat wünschte klassische Formaat

De Partitionen vun de Knoten warrt dörch Utweerden vun dat Vörteken vun de Verwachtensweerten vun de Pauli-Strings bestimmt, de de Knoten kodeert.

# Calculate the partitions based on the final expectation values
# If the expectation value is positive, the node belongs to partition 0 (par0)
# Otherwise, the node belongs to partition 1 (par1)

par0, par1 = set(), set()

for i in experiment_result[-1]["exp_map"]:
if experiment_result[-1]["exp_map"][i] >= 0:
par0.add(i)
else:
par1.add(i)
print(par0, par1)
{0, 1, 4, 8, 9, 10, 12, 13, 14, 15, 16, 18, 25, 27, 31, 32, 34, 36, 38, 39, 40, 41, 44, 46, 47, 48, 49, 50, 51, 52, 57, 60, 61, 62, 63, 64, 65, 66, 68, 71, 79, 81, 82, 86, 88, 91, 92, 93, 94, 95, 96, 99, 100, 105, 106, 107, 112, 114, 115, 121, 123, 129, 133, 134, 145, 147, 161, 165, 166, 168, 171, 173, 184, 185, 187, 188, 192, 193, 194, 196, 197, 198, 202, 205, 206, 207, 208, 209, 210, 211, 215, 217, 218, 219, 220, 221, 225, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 238, 241, 242, 243, 244, 246, 247, 248, 249, 251, 252, 253, 255, 256, 257, 258, 259, 261, 262, 264, 265, 266, 268, 269, 270, 272, 273, 275, 276, 277, 278, 279, 281, 283, 284, 285, 286, 288, 292, 293, 294, 299, 300, 303, 305, 306, 307, 308, 310, 312, 313, 314, 316, 317, 319, 321, 326, 327, 328, 333, 336, 338, 340, 341, 342, 344, 345, 346, 349, 351, 352, 353, 356, 357, 360, 361, 362, 363, 364, 366, 368, 370, 374, 378, 379, 380, 381, 382, 383, 384, 386, 387, 388, 389, 390, 391, 393, 394, 395, 396, 397, 398, 404, 405, 406, 409, 411, 413, 415, 416, 418, 421, 425, 426, 427, 428, 429, 433, 434, 435, 437, 444, 450, 456, 457, 458, 459, 462, 463, 465, 467, 469, 470, 472, 476, 479, 484, 487, 489, 492, 493, 497, 498, 499, 502, 506, 508, 513, 516, 517, 518, 519, 521, 523, 526, 527, 528, 531, 532, 533, 535, 536, 537, 539, 540, 541, 542, 543, 544, 545, 547, 549, 550, 552, 557, 562, 563, 564, 565, 567, 568, 569, 570, 571, 572, 573, 576, 578, 579, 580, 583, 585, 587, 588, 589, 591, 595, 596, 597, 600, 602, 603, 604, 605, 606, 607, 608, 609, 610, 612, 618, 619, 623, 624, 625, 626, 627, 628, 630, 632, 636, 637, 640, 644, 646, 649, 652, 656, 657, 658, 659, 661, 662, 663, 664, 667, 669, 670, 671, 672, 674, 675, 676, 677, 678, 679, 680, 681, 682, 683, 684, 685, 686, 687, 688, 689, 690, 692, 693, 694, 695, 696, 698, 700, 701, 703, 706, 707, 708, 709, 712, 713, 714, 716, 717, 718, 719, 721, 722, 723, 724, 725, 726, 728, 730, 731, 733, 734, 735, 737, 739, 740, 741, 743, 744, 746, 748, 750, 751, 752, 753, 754, 758, 760, 761, 762, 763, 764, 765, 766, 774, 778, 780, 782, 787, 795, 800, 802, 803, 808, 809, 812, 818, 822, 825, 827, 834, 836, 840, 843, 845, 847, 850, 853, 854, 857, 858, 863, 864, 865, 866, 867, 868, 869, 870, 872, 873, 874, 875, 876, 878, 880, 881, 882, 883, 884, 885, 887, 888, 889, 890, 891, 893, 894, 895, 896, 898, 901, 902, 903, 904, 905, 907, 908, 910, 911, 912, 913, 914, 915, 916, 917, 918, 920, 921, 923, 925, 926, 928, 929, 930, 932, 934, 935, 936, 938, 939, 941, 943, 945, 946, 947, 948, 949, 953, 955, 956, 957, 958, 959, 961, 966, 975, 978, 980, 983, 988, 990, 996, 999} {2, 3, 5, 6, 7, 11, 17, 19, 20, 21, 22, 23, 24, 26, 28, 29, 30, 33, 35, 37, 42, 43, 45, 53, 54, 55, 56, 58, 59, 67, 69, 70, 72, 73, 74, 75, 76, 77, 78, 80, 83, 84, 85, 87, 89, 90, 97, 98, 101, 102, 103, 104, 108, 109, 110, 111, 113, 116, 117, 118, 119, 120, 122, 124, 125, 126, 127, 128, 130, 131, 132, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 146, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 160, 162, 163, 164, 167, 169, 170, 172, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 186, 189, 190, 191, 195, 199, 200, 201, 203, 204, 212, 213, 214, 216, 222, 223, 224, 237, 239, 240, 245, 250, 254, 260, 263, 267, 271, 274, 280, 282, 287, 289, 290, 291, 295, 296, 297, 298, 301, 302, 304, 309, 311, 315, 318, 320, 322, 323, 324, 325, 329, 330, 331, 332, 334, 335, 337, 339, 343, 347, 348, 350, 354, 355, 358, 359, 365, 367, 369, 371, 372, 373, 375, 376, 377, 385, 392, 399, 400, 401, 402, 403, 407, 408, 410, 412, 414, 417, 419, 420, 422, 423, 424, 430, 431, 432, 436, 438, 439, 440, 441, 442, 443, 445, 446, 447, 448, 449, 451, 452, 453, 454, 455, 460, 461, 464, 466, 468, 471, 473, 474, 475, 477, 478, 480, 481, 482, 483, 485, 486, 488, 490, 491, 494, 495, 496, 500, 501, 503, 504, 505, 507, 509, 510, 511, 512, 514, 515, 520, 522, 524, 525, 529, 530, 534, 538, 546, 548, 551, 553, 554, 555, 556, 558, 559, 560, 561, 566, 574, 575, 577, 581, 582, 584, 586, 590, 592, 593, 594, 598, 599, 601, 611, 613, 614, 615, 616, 617, 620, 621, 622, 629, 631, 633, 634, 635, 638, 639, 641, 642, 643, 645, 647, 648, 650, 651, 653, 654, 655, 660, 665, 666, 668, 673, 691, 697, 699, 702, 704, 705, 710, 711, 715, 720, 727, 729, 732, 736, 738, 742, 745, 747, 749, 755, 756, 757, 759, 767, 768, 769, 770, 771, 772, 773, 775, 776, 777, 779, 781, 783, 784, 785, 786, 788, 789, 790, 791, 792, 793, 794, 796, 797, 798, 799, 801, 804, 805, 806, 807, 810, 811, 813, 814, 815, 816, 817, 819, 820, 821, 823, 824, 826, 828, 829, 830, 831, 832, 833, 835, 837, 838, 839, 841, 842, 844, 846, 848, 849, 851, 852, 855, 856, 859, 860, 861, 862, 871, 877, 879, 886, 892, 897, 899, 900, 906, 909, 919, 922, 924, 927, 931, 933, 937, 940, 942, 944, 950, 951, 952, 954, 960, 962, 963, 964, 965, 967, 968, 969, 970, 971, 972, 973, 974, 976, 977, 979, 981, 982, 984, 985, 986, 987, 989, 991, 992, 993, 994, 995, 997, 998}

Wi köönt de Cut-Grötte vun dat Max-Cut-Probleem mit de Partitionen vun de Knoten bereken.

cut_size = calc_cut_size(graph, par0, par1)
print(f"Cut size: {cut_size}")
Cut size: 24682

Sobald dat Trainen afslooten is, föhrt wi een Runne Single-Bit-Swap-Söök dörch, üm de Lösung as klassischen Nabearbeitensschritt to verbettern. Bi dissen Prozess tauscht wi de Partitionen vun twee Knoten un beweert de Cut-Grötte. Wenn de Cut-Grötte verbettert warrt, behölt wi den Tuusch. Wi wedderhohlt dissen Prozess för all möglichen Knotenporen, de dörch een Kant verbunnen sünd.

best_bits = []
cur_bits = []

for i in experiment_result[-1]["exp_map"]:
if experiment_result[-1]["exp_map"][i] >= 0:
cur_bits.append(1)
else:
cur_bits.append(0)
print(cur_bits)
[1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1]
# Swap the partitions and calculate the cut size
best_cut = 0
for edge0, edge1 in graph.edge_list():
swapped_bits = cur_bits.copy()
swapped_bits[edge0], swapped_bits[edge1] = (
swapped_bits[edge1],
swapped_bits[edge0],
)

cur_partition = [set(), set()]
for i, bit in enumerate(swapped_bits):
if bit > 0:
cur_partition[0].add(i)
else:
cur_partition[1].add(i)
cut_size = calc_cut_size(graph, cur_partition[0], cur_partition[1])
if best_cut < cut_size:
best_cut = cut_size
best_bits = swapped_bits

print(best_cut, best_bits)
24733 [1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1]

Referenzen

[1] Sciorilli, M., Borges, L., Patti, T. L., Garcia-Martin, D., Camilo, G., Anandkumar, A., & Aolita, L. (2024). Towards large-scale quantum optimization solvers with few qubits. arXiv preprint arXiv:2401.09421.

Tutorial-Ümfraag

Bitte maak bi disse korte Ümfraag mit, üm Feedback to dit Tutorial to geven. Dien Erkenntnissen helpt uns, unse Inholten un Brukerfarung to verbettern.

Link to de Ümfraag