Problem 237: Tours on a 4 x n Playing Board

Mathematical Foundation

Theorem (Transfer Matrix Representation). The count $T(n)$ of Hamiltonian paths from the top-left to the bottom-left corner of a $4 \times n$ grid can be expressed as

where $\mathbf{M}$ is a $k \times k$ transfer matrix encoding valid column-to-column transitions and $\mathbf{e}_{\mathrm{start}}, \mathbf{e}_{\mathrm{end}}$ are boundary state vectors.

Proof. Decompose the grid into columns. At the boundary between columns $i$ and $i+1$, the partial Hamiltonian path induces a connectivity pattern among the 4 row positions, specifying which pairs of boundary crossings are connected through the visited portion. This connectivity pattern forms a finite state. The transfer matrix $\mathbf{M}$ has entry $M_{ij}$ equal to the number of valid fillings of a single column that transition from boundary state $i$ to boundary state $j$ while maintaining the Hamiltonian property. The product $\mathbf{M}^{n-1}$ accounts for $n-1$ column transitions in an $n$-column grid. $\square$

Lemma (Linear Recurrence from Characteristic Polynomial). The sequence $T(n)$ satisfies a linear recurrence of order $k$ (the dimension of $\mathbf{M}$), with characteristic polynomial equal to $\det(xI - \mathbf{M})$.

Proof. By the Cayley-Hamilton theorem, $\mathbf{M}$ satisfies its own characteristic polynomial: $p(\mathbf{M}) = 0$ where $p(x) = \det(xI - \mathbf{M})$. Since $T(n)$ is a linear function of $\mathbf{M}^{n-1}$, it inherits the same recurrence relation defined by the coefficients of $p(x)$. $\square$

For the $4 \times n$ board, careful enumeration of boundary states yields a transfer matrix of dimension $k \approx 6$. The resulting linear recurrence (verified by the initial values including $T(10) = 2329$) allows computation via matrix exponentiation modulo $10^8$.

Theorem (Matrix Exponentiation Modular Arithmetic). For any integer $m$ and modulus $M$, the matrix power $\mathbf{M}^m \bmod M$ can be computed in $O(k^3 \log m)$ operations using repeated squaring, where each entry is reduced modulo $M$ at every step.

Proof. Write $m$ in binary: $m = \sum_{j} b_j 2^j$. Then $\mathbf{M}^m = \prod_{j: b_j=1} \mathbf{M}^{2^j}$. Each squaring requires $O(k^3)$ multiplications, and there are $\lfloor \log_2 m \rfloor + 1$ squarings. Modular reduction at each step preserves correctness since $(AB) \bmod M = ((A \bmod M)(B \bmod M)) \bmod M$ for matrix entries. $\square$

Editorial

T(n) = number of Hamiltonian paths on a 4 x n grid from top-left (0,0) to bottom-left (3,0), visiting every cell exactly once. Given T(10) = 2329, find T(10^12) mod 10^8. The sequence satisfies the linear recurrence: T(n) = 2T(n-1) + 2T(n-2) - 2*T(n-3) + T(n-4) with initial values T(1)=1, T(2)=1, T(3)=4, T(4)=8. This was found by: 1. Computing small values via brute-force backtracking. 2. Finding the minimal recurrence via Gaussian elimination. We use matrix exponentiation to compute T(10^12) mod 10^8. We construct the k x k transfer matrix M for 4 x n Hamiltonian paths. We then (top-left to bottom-left), determined by enumerating boundary states. Finally, compute M^(10^12 - 1) mod 10^8 via repeated squaring.

Pseudocode

Construct the k x k transfer matrix M for 4 x n Hamiltonian paths
(top-left to bottom-left), determined by enumerating boundary states
Compute M^(10^12 - 1) mod 10^8 via repeated squaring
Extract T(10^12) from appropriate entry

Complexity Analysis

• Time: $O(k^3 \log n)$ where $k \approx 6$ is the transfer matrix dimension and $n = 10^{12}$. This gives roughly $O(6^3 \cdot 40) = O(8640)$ arithmetic operations.
• Space: $O(k^2)$ for storing the matrix.

Answer

#include <bits/stdc++.h>
using namespace std;

/*
 * Problem 237: Tours on a 4 x n Playing Board
 *
 * T(n) = number of Hamiltonian paths on 4 x n grid from (0,0) to (3,0).
 * Find T(10^12) mod 10^8.
 *
 * Recurrence (found by brute-force + Gaussian elimination):
 *   T(n) = 2*T(n-1) + 2*T(n-2) - 2*T(n-3) + T(n-4)
 *   T(1)=1, T(2)=1, T(3)=4, T(4)=8
 *
 * Verified: T(10) = 2329.
 *
 * Use 4x4 companion matrix exponentiation modulo 10^8.
 */

typedef long long ll;
typedef vector<vector<ll>> Mat;

const ll MOD = 1e8;

Mat mat_mul(const Mat& A, const Mat& B) {
    int n = A.size();
    Mat C(n, vector<ll>(n, 0));
    for (int i = 0; i < n; i++)
        for (int k = 0; k < n; k++) {
            if (A[i][k] == 0) continue;
            for (int j = 0; j < n; j++)
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;
        }
    return C;
}

Mat mat_pow(Mat M, ll p) {
    int n = M.size();
    Mat R(n, vector<ll>(n, 0));
    for (int i = 0; i < n; i++) R[i][i] = 1;
    while (p > 0) {
        if (p & 1) R = mat_mul(R, M);
        M = mat_mul(M, M);
        p >>= 1;
    }
    return R;
}

int main(){
    // Companion matrix for T(n) = 2T(n-1) + 2T(n-2) - 2T(n-3) + T(n-4)
    Mat M = {
        {2, 2, MOD - 2, 1},
        {1, 0, 0, 0},
        {0, 1, 0, 0},
        {0, 0, 1, 0}
    };

    ll N = 1000000000000LL; // 10^12

    // State at n=4: [T(4), T(3), T(2), T(1)] = [8, 4, 1, 1]
    Mat Mk = mat_pow(M, N - 4);

    vector<ll> state = {8, 4, 1, 1};
    ll ans = 0;
    for (int j = 0; j < 4; j++)
        ans = (ans + Mk[0][j] * state[j]) % MOD;

    cout << ans << endl;

    return 0;
}

"""
Problem 237: Tours on a 4 x n Playing Board

T(n) = number of Hamiltonian paths on a 4 x n grid from top-left (0,0) to
bottom-left (3,0), visiting every cell exactly once.

Given T(10) = 2329, find T(10^12) mod 10^8.

The sequence satisfies the linear recurrence:
    T(n) = 2*T(n-1) + 2*T(n-2) - 2*T(n-3) + T(n-4)

with initial values T(1)=1, T(2)=1, T(3)=4, T(4)=8.

This was found by:
1. Computing small values via brute-force backtracking.
2. Finding the minimal recurrence via Gaussian elimination.

We use matrix exponentiation to compute T(10^12) mod 10^8.
"""

MOD = 10**8

def mat_mul(A, B, mod):
    """Multiply two square matrices modulo mod."""
    n = len(A)
    C = [[0]*n for _ in range(n)]
    for i in range(n):
        for k in range(n):
            if A[i][k] == 0:
                continue
            for j in range(n):
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod
    return C

def mat_pow(M, p, mod):
    """Compute M^p modulo mod using binary exponentiation."""
    n = len(M)
    R = [[1 if i == j else 0 for j in range(n)] for i in range(n)]
    while p > 0:
        if p & 1:
            R = mat_mul(R, M, mod)
        M = mat_mul(M, M, mod)
        p >>= 1
    return R

# Recurrence: T(n) = 2*T(n-1) + 2*T(n-2) - 2*T(n-3) + T(n-4)
# Initial: T(1)=1, T(2)=1, T(3)=4, T(4)=8

# Verify small values
T = [0, 1, 1, 4, 8]
for i in range(5, 20):
    T.append(2*T[-1] + 2*T[-2] - 2*T[-3] + T[-4])
assert T[10] == 2329, f"T(10) = {T[10]}, expected 2329"

# Companion matrix:
# [T(n)  ]   [2  2 -2  1] [T(n-1)]
# [T(n-1)] = [1  0  0  0] [T(n-2)]
# [T(n-2)]   [0  1  0  0] [T(n-3)]
# [T(n-3)]   [0  0  1  0] [T(n-4)]

M = [
    [2, 2, MOD - 2, 1],
    [1, 0, 0, 0],
    [0, 1, 0, 0],
    [0, 0, 1, 0]
]

N = 10**12

# State at n=4: [T(4), T(3), T(2), T(1)] = [8, 4, 1, 1]
Mk = mat_pow(M, N - 4, MOD)

state = [8, 4, 1, 1]
ans = 0
for j in range(4):
    ans = (ans + Mk[0][j] * state[j]) % MOD

print(ans)