Math Puzzle: SEND MORE MONEY

SEND + MORE = MONEY is a famous alphametic puzzle: assign a decimal digit to each letter so that \(\text{SEND}+\text{MORE}=\text{MONEY}\)

The constraints are:

  • The digits assigned to letters are all distinct.
  • S and M must not be 0.

QUBO++ formulation

We assign a unique index to each letter as follows:

index 0 1 2 3 4 5 6 7
letter S E N D M O R Y

Let $I(\alpha)$ denote the index of letter $\alpha$ ($\in \lbrace S,E,N,D,M,O,R,Y\rbrace$). We use an $8\times 10$ binary matrix $X=(x_{i,j})$ $(0\leq i\leq 7, 0\leq j\leq 9)$ to represent the digit assigned to each letter: $x_{I(\alpha),j}=1$ if and only if letter $\alpha$ is assigned digit $j$.

One-hot constraints (each letter takes exactly one digit)

Each row of $X$ must be one-hot:

\[\begin{aligned} \text{onehot} &=\sum_{i=0}^{7}\Bigl(\sum_{j=0}^{9}x_{i,j}=1\Bigr) \\ &=\sum_{i=0}^{7}\Bigl(1-\sum_{j=0}^{9}x_{i,j}\Bigr)^2 \end{aligned}\]

The value of $\text{onehot}$ is minimized to 0 if and only if every row is one-hot.

All-different constraints (no two letters share the same digit)

Digits must be distinct across letters, i.e., no two rows choose the same column: \(\begin{aligned} \text{different} &=\sum_{0\leq i<j\leq 7}\sum_{k=0}^9x_{i,k}x_{j,k} \end{aligned}\)

Encoding the words as linear expressions

The values of $\text{SEND}$, $\text{MORE}$, and $\text{MONEY}$ are represented by:

\[\begin{aligned} \text{SEND} &= 1000\sum_{k=0}^9 kx_{I(S),k}+ 100\sum_{k=0}^9 kx_{I(E),k}+ 10\sum_{k=0}^9 kx_{I(N),k}+\sum_{k=0}^9 kx_{I(D),k}\\ &= \sum_{k=0}^9k(1000x_{I(S),k}+100x_{I(E),k}+10x_{I(N),k}+x_{I(D),k})\\ \text{MORE} &= 1000\sum_{k=0}^9 kx_{I(M),k}+ 100\sum_{k=0}^9 kx_{I(O),k}+ 10\sum_{k=0}^9 kx_{I(R),k}+\sum_{k=0}^9 kx_{I(E),k}\\ &= \sum_{k=0}^9k(1000x_{I(M),k}+100x_{I(O),k}+10x_{I(R),k}+x_{I(E),k})\\ \text{MONEY} &= 10000\sum_{k=0}^9 kx_{I(M),k}+1000\sum_{k=0}^9 kx_{I(O),k}+ 100\sum_{k=0}^9 kx_{I(N),k}+ 10\sum_{k=0}^9 kx_{I(E),k}+\sum_{k=0}^9 kx_{I(Y),k}\\ &= \sum_{k=0}^9k(10000x_{I(M),k}+ 1000x_{I(O),k}+100x_{I(N),k}+10x_{I(E),k}+x_{I(Y),k}) \end{aligned}\]

Equality constraint

We enforce the equation by penalizing the residual:

\[\begin{aligned} \text{equal} &= \Bigl(\text{SEND}+\text{MORE} = \text{MONEY}\Bigr) \\ &= \Bigl(\text{SEND}+\text{MORE} - \text{MONEY}\Bigr)^2 \end{aligned}\]

Combined objective

All constraints are combined into a single objective:

\[\begin{aligned} f & = P\cdot (\text{onehot}+\text{different})+\text{equal} \end{aligned}\]

where P is a sufficiently large constant to prioritize feasibility (onehot and different). In principle, if all terms are nonnegative and each becomes 0 exactly when its constraint holds, then any solution with $f=0$ satisfies all constraints. In practice, choosing a larger P often helps heuristic solvers.

In this case, there is no need to prioritize them and we can set $P=1$, because $\text{equal}\geq 0$ always holds and $f$ takes a minimum value of 0 only if $\text{onehot}=\text{different}=\text{equal}=0$ holds. However, a large constant $P$ helps solvers to find the optimal solution.

Finally, since $\text{S}$ and $\text{M}$ must not be 0, we fix the binary variables as follows: \(x_{I(S),0} = x_{I(M),0}= 0\)

QUBO++ program for SEND+MORE=MONEY

The following QUBO++ program implements the QUBO formulation above and finds a solution using EasySolver:

#define INTEGER_TYPE_C128E128

#include <string_view>

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

constexpr std::string_view LETTERS = "SENDMORY";
constexpr size_t L = LETTERS.size();

constexpr size_t I(char c) {
  for (size_t i = 0; i < LETTERS.size(); ++i) {
    if (LETTERS[i] == c) return i;
  }
  return L;
}

const auto K = qbpp::int_array({0, 1, 2, 3, 4, 5, 6, 7, 8, 9});

int main() {
  auto x = qbpp::var("x", L, 10);


  auto onehot = qbpp::sum(qbpp::vector_sum(x) == 1);

  auto different = qbpp::toExpr(0);
  for (size_t i = 0; i < L - 1; ++i) {
    for (size_t j = i + 1; j < L; ++j) {
      different += qbpp::sum(qbpp::row(x, i) * qbpp::row(x, j));
    }
  }

  auto send = qbpp::sum((x[I('S')] * 1000 + x[I('E')] * 100 + x[I('N')] * 10 + x[I('D')]) * K);
  auto more = qbpp::sum((x[I('M')] * 1000 + x[I('O')] * 100 + x[I('R')] * 10 + x[I('E')]) * K);
  auto money = qbpp::sum((x[I('M')] * 10000 + x[I('O')] * 1000 + x[I('N')] * 100 + x[I('E')] * 10 + x[I('Y')]) * K);


  auto equal = send + more - money == 0;

  qbpp::coeff_t P = 10000;
  auto f = P * (onehot + different) + equal;

  f.simplify_as_binary();

  qbpp::MapList ml = {{x[I('S')][0], 0}, {x[I('M')][0], 0}};
  auto g = qbpp::replace(f, ml);

  g.simplify_as_binary();
  auto solver = qbpp::easy_solver::EasySolver(g);
  auto sol = solver.search({{"target_energy", 0}});

  auto full_sol = qbpp::Sol(f).set(sol).set(ml);

  std::cout << "onehot = " << full_sol(onehot) << std::endl;
  std::cout << "different = " << full_sol(different) << std::endl;
  std::cout << "equal = " << full_sol(equal) << std::endl;

  auto val = qbpp::onehot_to_int(full_sol(x));

  auto str = [](int d) -> std::string {
    return (d < 0) ? "*" : std::to_string(d);
  };

  std::cout << "SEND + MORE = MONEY" << std::endl;
  std::cout << str(val[I('S')]) << str(val[I('E')]) << str(val[I('N')])
            << str(val[I('D')]) << " + " << str(val[I('M')]) << str(val[I('O')])
            << str(val[I('R')]) << str(val[I('E')]) << " = " << str(val[I('M')])
            << str(val[I('O')]) << str(val[I('N')]) << str(val[I('E')])
            << str(val[I('Y')]) << std::endl;
}

In this program, LETTERS assigns an integer index to each letter in "SENDMORY", which is used to implement $I(\alpha)$. We define an L$\times$10 matrix x of binary variables (here $L=8$). The expressions onehot, different, and equal are computed according to the formulation and combined into a single objective f with a penalty weight P.

We use a qbpp::MapList object ml to fix x[I('S')][0] and x[I('M')][0] to 0, and create a reduced expression g by applying this replacement. The solver is run on g, and the resulting assignment sol is merged with the fixed mapping ml to produce full_sol for the original objective f.

Finally, qbpp::onehot_to_int(full_sol(x)) converts the one-hot rows into digits, and the program prints the obtained solution. This program produces the following output:

onehot = 0
different = 0
equal = 0
SEND + MORE = MONEY
9567 + 1085 = 10652

This confirms that all constraints are satisfied and the correct solution is obtained.

数学パズル: SEND MORE MONEY

SEND + MORE = MONEY は有名な覆面算パズルです。各文字に10進数の数字を割り当てて、次の等式を成り立たせます: \(\text{SEND}+\text{MORE}=\text{MONEY}\)

制約は以下の通りです:

  • 各文字に割り当てられる数字はすべて異なる。
  • S と M は0であってはならない。

QUBO++による定式化

各文字に以下のように一意のインデックスを割り当てます:

index 0 1 2 3 4 5 6 7
letter S E N D M O R Y

$I(\alpha)$ を文字 $\alpha$ ($\in \lbrace S,E,N,D,M,O,R,Y\rbrace$) のインデックスとします。 各文字に割り当てられる数字を表すために、$8\times 10$ のバイナリ行列 $X=(x_{i,j})$ $(0\leq i\leq 7, 0\leq j\leq 9)$ を使用します。文字 $\alpha$ に数字 $j$ が割り当てられるとき、かつそのときに限り $x_{I(\alpha),j}=1$ とします。

ワンホット制約（各文字はちょうど1つの数字を取る）

$X$ の各行はワンホットでなければなりません:

\[\begin{aligned} \text{onehot} &=\sum_{i=0}^{7}\Bigl(\sum_{j=0}^{9}x_{i,j}=1\Bigr) \\ &=\sum_{i=0}^{7}\Bigl(1-\sum_{j=0}^{9}x_{i,j}\Bigr)^2 \end{aligned}\]

$\text{onehot}$ の値は、すべての行がワンホットであるとき、かつそのときに限り、最小値0になります。

全異なり制約（2つの文字が同じ数字を共有しない）

数字は文字間で異なっていなければなりません。すなわち、2つの行が同じ列を選んではいけません: \(\begin{aligned} \text{different} &=\sum_{0\leq i<j\leq 7}\sum_{k=0}^9x_{i,k}x_{j,k} \end{aligned}\)

単語の線形式としての符号化

$\text{SEND}$、$\text{MORE}$、$\text{MONEY}$ の値は以下のように表されます:

\[\begin{aligned} \text{SEND} &= 1000\sum_{k=0}^9 kx_{I(S),k}+ 100\sum_{k=0}^9 kx_{I(E),k}+ 10\sum_{k=0}^9 kx_{I(N),k}+\sum_{k=0}^9 kx_{I(D),k}\\ &= \sum_{k=0}^9k(1000x_{I(S),k}+100x_{I(E),k}+10x_{I(N),k}+x_{I(D),k})\\ \text{MORE} &= 1000\sum_{k=0}^9 kx_{I(M),k}+ 100\sum_{k=0}^9 kx_{I(O),k}+ 10\sum_{k=0}^9 kx_{I(R),k}+\sum_{k=0}^9 kx_{I(E),k}\\ &= \sum_{k=0}^9k(1000x_{I(M),k}+100x_{I(O),k}+10x_{I(R),k}+x_{I(E),k})\\ \text{MONEY} &= 10000\sum_{k=0}^9 kx_{I(M),k}+1000\sum_{k=0}^9 kx_{I(O),k}+ 100\sum_{k=0}^9 kx_{I(N),k}+ 10\sum_{k=0}^9 kx_{I(E),k}+\sum_{k=0}^9 kx_{I(Y),k}\\ &= \sum_{k=0}^9k(10000x_{I(M),k}+ 1000x_{I(O),k}+100x_{I(N),k}+10x_{I(E),k}+x_{I(Y),k}) \end{aligned}\]

等式制約

残差にペナルティを課すことで等式を強制します:

\[\begin{aligned} \text{equal} &= \Bigl(\text{SEND}+\text{MORE} = \text{MONEY}\Bigr) \\ &= \Bigl(\text{SEND}+\text{MORE} - \text{MONEY}\Bigr)^2 \end{aligned}\]

結合目的関数

すべての制約を単一の目的関数にまとめます:

\[\begin{aligned} f & = P\cdot (\text{onehot}+\text{different})+\text{equal} \end{aligned}\]

ここで P は実行可能性（onehot と different）を優先するための十分に大きな定数です。 原理的には、すべての項が非負であり、各項がその制約が成り立つときにちょうど0になるならば、$f=0$ となる任意の解はすべての制約を満たします。 実際には、より大きな P を選ぶことがヒューリスティックソルバーに有効なことが多いです。

この場合、優先順位をつける必要はなく $P=1$ と設定できます。 なぜなら $\text{equal}\geq 0$ が常に成り立ち、$f$ は $\text{onehot}=\text{different}=\text{equal}=0$ のときにのみ最小値0を取るからです。 ただし、大きな定数 $P$ はソルバーが最適解を見つけるのに役立ちます。

最後に、$\text{S}$ と $\text{M}$ は0であってはならないため、バイナリ変数を以下のように固定します: \(x_{I(S),0} = x_{I(M),0}= 0\)

SEND+MORE=MONEYのQUBO++プログラム

以下のQUBO++プログラムは、上記のQUBO定式化を実装し、EasySolverを使って解を求めます:

#define INTEGER_TYPE_C128E128

#include <string_view>

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

constexpr std::string_view LETTERS = "SENDMORY";
constexpr size_t L = LETTERS.size();

constexpr size_t I(char c) {
  for (size_t i = 0; i < LETTERS.size(); ++i) {
    if (LETTERS[i] == c) return i;
  }
  return L;
}

const auto K = qbpp::int_array({0, 1, 2, 3, 4, 5, 6, 7, 8, 9});

int main() {
  auto x = qbpp::var("x", L, 10);


  auto onehot = qbpp::sum(qbpp::vector_sum(x) == 1);

  auto different = qbpp::toExpr(0);
  for (size_t i = 0; i < L - 1; ++i) {
    for (size_t j = i + 1; j < L; ++j) {
      different += qbpp::sum(qbpp::row(x, i) * qbpp::row(x, j));
    }
  }

  auto send = qbpp::sum((x[I('S')] * 1000 + x[I('E')] * 100 + x[I('N')] * 10 + x[I('D')]) * K);
  auto more = qbpp::sum((x[I('M')] * 1000 + x[I('O')] * 100 + x[I('R')] * 10 + x[I('E')]) * K);
  auto money = qbpp::sum((x[I('M')] * 10000 + x[I('O')] * 1000 + x[I('N')] * 100 + x[I('E')] * 10 + x[I('Y')]) * K);


  auto equal = send + more - money == 0;

  qbpp::coeff_t P = 10000;
  auto f = P * (onehot + different) + equal;

  f.simplify_as_binary();

  qbpp::MapList ml = {{x[I('S')][0], 0}, {x[I('M')][0], 0}};
  auto g = qbpp::replace(f, ml);

  g.simplify_as_binary();
  auto solver = qbpp::easy_solver::EasySolver(g);
  auto sol = solver.search({{"target_energy", 0}});

  auto full_sol = qbpp::Sol(f).set(sol).set(ml);

  std::cout << "onehot = " << full_sol(onehot) << std::endl;
  std::cout << "different = " << full_sol(different) << std::endl;
  std::cout << "equal = " << full_sol(equal) << std::endl;

  auto val = qbpp::onehot_to_int(full_sol(x));

  auto str = [](int d) -> std::string {
    return (d < 0) ? "*" : std::to_string(d);
  };

  std::cout << "SEND + MORE = MONEY" << std::endl;
  std::cout << str(val[I('S')]) << str(val[I('E')]) << str(val[I('N')])
            << str(val[I('D')]) << " + " << str(val[I('M')]) << str(val[I('O')])
            << str(val[I('R')]) << str(val[I('E')]) << " = " << str(val[I('M')])
            << str(val[I('O')]) << str(val[I('N')]) << str(val[I('E')])
            << str(val[I('Y')]) << std::endl;
}

このプログラムでは、LETTERS が "SENDMORY" の各文字に整数インデックスを割り当てており、$I(\alpha)$ の実装に使用されています。 L$\times$10 のバイナリ変数行列 x を定義します（ここで $L=8$）。 式 onehot、different、equal は定式化に従って計算され、ペナルティ重み P とともに単一の目的関数 f にまとめられます。

qbpp::MapList オブジェクト ml を使用して x[I('S')][0] と x[I('M')][0] を0に固定し、この置換を適用して縮小された式 g を作成します。 ソルバーは g に対して実行され、得られた割り当て sol は固定マッピング ml とマージされて、元の目的関数 f に対する full_sol が生成されます。

最後に、qbpp::onehot_to_int(full_sol(x)) がワンホット行を数字に変換し、プログラムは得られた解を出力します。 このプログラムは以下の出力を生成します:

onehot = 0
different = 0
equal = 0
SEND + MORE = MONEY
9567 + 1085 = 10652

これにより、すべての制約が満たされ、正しい解が得られたことが確認できます。