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.
SandMmust 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 COEFF_TYPE qbpp::int128_t
#define ENERGY_TYPE qbpp::int128_t
#include <string_view>
#define MAXDEG 2
#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 qbpp::Vector<int> K = {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(x[i] * 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);
solver.target_energy(0);
auto sol = solver.search();
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.