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: \(x_{I(S),0} = x_{I(M),0}= 0\)

PyQBPP program for SEND+MORE=MONEY

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

import pyqbpp as qbpp

LETTERS = "SENDMORY"
L = len(LETTERS)

def I(c):
    return LETTERS.index(c)

x = qbpp.var("x", shape=(L, 10))

onehot = qbpp.sum(qbpp.vector_sum(x) == 1)

different = 0
for i in range(L - 1):
    for j in range(i + 1, L):
        different += qbpp.sum(x[i] * x[j])

send = 0
more = 0
money = 0
for k in range(10):
    send += k * (1000 * x[I('S')][k] + 100 * x[I('E')][k] + 10 * x[I('N')][k] + x[I('D')][k])
    more += k * (1000 * x[I('M')][k] + 100 * x[I('O')][k] + 10 * x[I('R')][k] + x[I('E')][k])
    money += k * (10000 * x[I('M')][k] + 1000 * x[I('O')][k] + 100 * x[I('N')][k] + 10 * x[I('E')][k] + x[I('Y')][k])

equal = (send + more - money == 0)

P = 10000
f = P * (onehot + different) + equal
f.simplify_as_binary()

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

solver = qbpp.EasySolver(g)
sol = solver.search(target_energy=0)

full_sol = qbpp.Sol(f).set(sol, ml)

print(f"onehot = {full_sol(onehot)}")
print(f"different = {full_sol(different)}")
print(f"equal = {full_sol(equal)}")

val = [next((k for k in range(10) if full_sol(x[i][k]) == 1), -1) for i in range(L)]

def digit_str(d):
    return "*" if d < 0 else str(d)

print("SEND + MORE = MONEY")
print(f"{digit_str(val[I('S')])}{digit_str(val[I('E')])}{digit_str(val[I('N')])}{digit_str(val[I('D')])} + "
      f"{digit_str(val[I('M')])}{digit_str(val[I('O')])}{digit_str(val[I('R')])}{digit_str(val[I('E')])} = "
      f"{digit_str(val[I('M')])}{digit_str(val[I('O')])}{digit_str(val[I('N')])}{digit_str(val[I('E')])}{digit_str(val[I('Y')])}")

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 dict 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 assignments ml via qbpp.Sol(f).set(sol, ml) to produce full_sol for the original objective f.

Finally, the one-hot rows of full_sol(x) are decoded into digits by scanning each row for the index k with value 1 (or -1 if none is found), 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.