Remainder Problem

The following problem can be solved using QUBO++. Find the minimum non-negative integer $x$ such that

the remainder when $x$ is divided by 3 is 2,
the remainder when $x$ is divided by 5 is 3, and
the remainder when $x$ is divided by 7 is 5.

Since 3, 5, and 7 are pairwise coprime, it is enough to search $x$ within one period:

\[0\leq x \leq 3\times 5\times 7 -1\]

Introduce non-negative integers $d_3$, $d_5$, and $d_7$ (quotients) and rewrite the remainder conditions as linear equalities:

\[\begin{aligned} x - 3d_3 &= 2 \\ x - 5d_5 &=3 \\ x - 7d_7 &= 5 \end{aligned}\]

We want to minimize $x$ subject to these constraints. From the range $x$ above, we can bound the quotient variables as

\[\begin{aligned} 0&\leq d_3 \leq 5\times 7-1 \\ 0&\leq d_5 \leq 3\times 7-1 \\ 0&\leq d_7 \leq 3\times 5-1 \end{aligned}\]

QUBO++ praogram

The following program finds a solution $x$ for this remainder problem:

#include <qbpp/qbpp.hpp>
#include <qbpp/easy_solver.hpp>

int main() {
  auto x = 0 <= qbpp::var_int("x") <= 3 * 5 * 7 - 1;
  auto d3 = 0 <= qbpp::var_int("d3") <= 5 * 7 - 1;
  auto d5 = 0 <= qbpp::var_int("d5") <= 3 * 7 - 1;
  auto d7 = 0 <= qbpp::var_int("d7") <= 3 * 5 - 1;
  auto c3 = x - 3 * d3 == 2;
  auto c5 = x - 5 * d5 == 3;
  auto c7 = x - 7 * d7 == 5;
  auto f = x + 1000 * (c3 + c5 + c7);
  f.simplify_as_binary();

  auto solver = qbpp::easy_solver::EasySolver(f);
  auto sol = solver.search({{"time_limit", 1.0}});

  std::cout << "x = " << sol(x) << std::endl;
  std::cout << sol(x) << " - 3 * " << sol(d3) << " = " << sol(*c3) << std::endl;
  std::cout << sol(x) << " - 5 * " << sol(d5) << " = " << sol(*c5) << std::endl;
  std::cout << sol(x) << " - 7 * " << sol(d7) << " = " << sol(*c7) << std::endl;
}

The three constraints are represented as c3, c5, and c7. Each of them is converted into a QUBO penalty term that becomes 0 when the corresponding equality holds.

We then minimize x with a large penalty weight (1000) so that satisfying the constraints is prioritized over reducing x.

Finally, the Easy Solver searches for a low-energy solution of f within the time limit (1.0 second), and the obtained values are printed as follows:

x = 68
- 3 * 22 = 2
- 5 * 13 = 3
- 7 * 9 = 5

Therefore,

\[\begin{aligned} x &\equiv 68 & (\bmod 105) \end{aligned}\]

and the minimum solution is $x=68$.

剰余問題

以下の問題はQUBO++を用いて解くことができます。次の条件を満たす最小の非負整数 $x$ を求めます:

$x$ を3で割った余りが2
$x$ を5で割った余りが3
$x$ を7で割った余りが5

3、5、7は互いに素であるため、1周期内で $x$ を探索すれば十分です:

\[0\leq x \leq 3\times 5\times 7 -1\]

非負整数 $d_3$、$d_5$、$d_7$（商）を導入し、剰余条件を線形等式として書き直します:

\[\begin{aligned} x - 3d_3 &= 2 \\ x - 5d_5 &=3 \\ x - 7d_7 &= 5 \end{aligned}\]

これらの制約の下で $x$ を最小化したいです。上記の $x$ の範囲から、商の変数は以下のように制限できます:

\[\begin{aligned} 0&\leq d_3 \leq 5\times 7-1 \\ 0&\leq d_5 \leq 3\times 7-1 \\ 0&\leq d_7 \leq 3\times 5-1 \end{aligned}\]

QUBO++ プログラム

以下のプログラムは、この剰余問題の解 $x$ を求めます:

#include <qbpp/qbpp.hpp>
#include <qbpp/easy_solver.hpp>

int main() {
  auto x = 0 <= qbpp::var_int("x") <= 3 * 5 * 7 - 1;
  auto d3 = 0 <= qbpp::var_int("d3") <= 5 * 7 - 1;
  auto d5 = 0 <= qbpp::var_int("d5") <= 3 * 7 - 1;
  auto d7 = 0 <= qbpp::var_int("d7") <= 3 * 5 - 1;
  auto c3 = x - 3 * d3 == 2;
  auto c5 = x - 5 * d5 == 3;
  auto c7 = x - 7 * d7 == 5;
  auto f = x + 1000 * (c3 + c5 + c7);
  f.simplify_as_binary();

  auto solver = qbpp::easy_solver::EasySolver(f);
  auto sol = solver.search({{"time_limit", 1.0}});

  std::cout << "x = " << sol(x) << std::endl;
  std::cout << sol(x) << " - 3 * " << sol(d3) << " = " << sol(*c3) << std::endl;
  std::cout << sol(x) << " - 5 * " << sol(d5) << " = " << sol(*c5) << std::endl;
  std::cout << sol(x) << " - 7 * " << sol(d7) << " = " << sol(*c7) << std::endl;
}

3つの制約は c3、c5、c7 として表現されています。それぞれは、対応する等式が成り立つときに0になるQUBOペナルティ項に変換されます。

次に、制約の充足を $x$ の削減よりも優先するために、大きなペナルティ重み（1000）を用いて x を最小化します。

最後に、Easy Solverが制限時間（1.0秒）内で f の低エネルギー解を探索し、得られた値は以下のように出力されます:

x = 68
- 3 * 22 = 2
- 5 * 13 = 3
- 7 * 9 = 5

したがって、

\[\begin{aligned} x &\equiv 68 & (\bmod 105) \end{aligned}\]

最小の解は $x=68$ です。