Math Problem: Find Three Integers

The following math problem can be solved using PyQBPP.

Problem

Find integers $x$, $y$, $z$ that satisfy:

\[\begin{aligned} \frac{1}{x}+\frac{1}{y}+\frac{1}{z} = 1\\ 1 < x < y < z \end{aligned}\]

PyQBPP program

Since PyQBPP can handle polynomial expressions, we first rewrite the constraints. Multiplying both sides of the first constraint by $xyz$ yields:

\[xy+yz+zx - xyz = 0\]

The strict inequalities $x<y<z$ can be encoded as

\[\begin{aligned} 1 &\leq y-x \\ 1 &\leq z-y \end{aligned}\]

The following PyQBPP program formulates these constraints as a HUBO expression and solves it using the Exhaustive Solver:

import pyqbpp as qbpp

x = qbpp.between(qbpp.var_int("x"), 1, 10)
y = qbpp.between(qbpp.var_int("y"), 1, 10)
z = qbpp.between(qbpp.var_int("z"), 1, 10)

c1 = x * y + y * z + z * x - x * y * z == 0
c2 = qbpp.between(y - x, 1, 9)
c3 = qbpp.between(z - y, 1, 9)

f = c1 + c2 + c3
f.simplify_as_binary()
solver = qbpp.ExhaustiveSolver(f)
result = solver.search({"best_energy_sols": 0})

seen = set()
for sol in result.sols():
    key = (sol(x), sol(y), sol(z))
    if key not in seen:
        seen.add(key)
        xv, yv, zv = key
        print(f"(x,y,z) = ({xv}, {yv}, {zv})")

The three constraints are encoded as c1, c2, and c3, and combined into a single objective f. The Exhaustive Solver searches for optimal solutions of f and prints the resulting $(x,y,z)$ tuples.

Because f introduces auxiliary variables during binary simplification, the same $(x,y,z)$ assignment may appear multiple times in the returned solution set. Therefore, we use a set to remove duplicates before printing.

This program produces the following output:

(x,y,z) = (2, 3, 6)

This indicates that the problem has exactly one solution in the searched range, namely $(x,y,z)=(2,3,6)$.

数学問題：3つの整数を求める

以下の数学問題をPyQBPPを用いて解くことができます。

問題

以下を満たす整数 $x$、$y$、$z$ を求めてください：

\[\begin{aligned} \frac{1}{x}+\frac{1}{y}+\frac{1}{z} = 1\\ 1 < x < y < z \end{aligned}\]

PyQBPPプログラム

PyQBPPは多項式を扱えるため、まず制約を書き換えます。 最初の制約の両辺に $xyz$ を掛けると：

\[xy+yz+zx - xyz = 0\]

狭義の不等式 $x<y<z$ は以下のようにエンコードできます：

\[\begin{aligned} 1 &\leq y-x \\ 1 &\leq z-y \end{aligned}\]

以下のPyQBPPプログラムは、これらの制約をHUBO式として定式化し、Exhaustive Solverを用いて解きます：

import pyqbpp as qbpp

x = qbpp.between(qbpp.var_int("x"), 1, 10)
y = qbpp.between(qbpp.var_int("y"), 1, 10)
z = qbpp.between(qbpp.var_int("z"), 1, 10)

c1 = x * y + y * z + z * x - x * y * z == 0
c2 = qbpp.between(y - x, 1, 9)
c3 = qbpp.between(z - y, 1, 9)

f = c1 + c2 + c3
f.simplify_as_binary()
solver = qbpp.ExhaustiveSolver(f)
result = solver.search({"best_energy_sols": 0})

seen = set()
for sol in result.sols():
    key = (sol(x), sol(y), sol(z))
    if key not in seen:
        seen.add(key)
        xv, yv, zv = key
        print(f"(x,y,z) = ({xv}, {yv}, {zv})")

3つの制約は c1、c2、c3 としてエンコードされ、単一の目的関数 f にまとめられます。 Exhaustive Solverが f の最適解を探索し、得られた $(x,y,z)$ のタプルを出力します。

f はバイナリ簡約化の際に補助変数を導入するため、同じ $(x,y,z)$ の割り当てが返される解集合に複数回現れる場合があります。 そのため、出力前に set を使って重複を除去しています。

このプログラムの出力は以下の通りです：

(x,y,z) = (2, 3, 6)

これは、探索範囲内でこの問題がちょうど1つの解 $(x,y,z)=(2,3,6)$ を持つことを示しています。