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量子エンジニア(アニーリング式)第3回の勉強会メモです。第1回、第2回のメモは以下。

方程式制約

x_0 + x_1 + x_2 = 3
x_0 + x_1 + x_2 - 3 = 0
H = 0
H += (q[0] + q[1] + q[2] - 3)**2

報酬とペナルティ

2個の量子ビットが同時に1になったら報酬

H = 0
H += -(q[0]*q[1])

2個の量子ビットが同時に1になったらペナルティ

H = 0
H += (q[0]*q[1])

Reference

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