Detailed mathematical approach

Problem Summary

We want to tile a \(3\times 3\times n\) tower with \(2\times 1\times 1\) blocks, allowing all axis-aligned orientations, and count the number of valid towers modulo

$M=100000007.$

Let \(f(n)\) be that count. The challenge is that the requested index is not merely large: it is

$N=10^{10000}.$

So even a fast layer-by-layer dynamic program is still hopeless unless we compress the sequence very aggressively.

Mathematical Approach

1) A parity observation.

The tower volume is \(9n\). Each block covers 2 unit cubes. Therefore \(f(n)=0\) whenever \(9n\) is odd, i.e. whenever \(n\) is odd. This is not enough to solve the problem, but it is a good first sanity check on the sequence.

2) Slice the tower by layers.

Process the tower one \(3\times 3\) layer at a time. A single layer has 9 cells, so we encode a layer state by a 9-bit mask

$s\in[0,2^9).$

A bit of \(s\) is 1 if that cell is already occupied by a domino that started in the previous layer and sticks into the current one. Since there are 9 cells, the total number of masks is

$2^9=512.$

3) What one transition means.

Fix an incoming mask \(s\). We must finish filling the current \(3\times 3\) layer completely. While doing that, a domino can be placed in exactly three ways:

1) entirely inside the current layer, horizontally along columns,

2) entirely inside the current layer, vertically along rows,

3) across the layer boundary, meaning one cube is in the current layer and the other cube will occupy the same cell in the next layer.

The third type creates a 1 in the outgoing mask \(t\).

4) DFS generates the transfer counts.

The code uses a depth-first search on the current layer. At each step it picks the first free cell and branches over the legal placements listed above. When the layer becomes fully covered, we have produced exactly one transition

$s\to t.$

Let \(T_{s,t}\) be the number of ways this can happen.

Because the layer is only \(3\times 3\), this DFS is finite and can be done once for all 512 incoming masks.

5) Transfer-matrix dynamic programming.

Let \(v_k[s]\) be the number of ways to build the first \(k\) layers and finish with outgoing mask \(s\). Then

$v_{k+1}[t]=\sum_{s=0}^{511} v_k[s]\,T_{s,t}\pmod M.$

The initial condition is

$v_0[0]=1,\qquad v_0[s]=0\ \text{for }s\ne 0,$

because before building anything there is no pending occupancy from a previous layer.

The quantity we actually want is the fully closed tower count

$f(k)=v_k[0],$

since at the top of a valid tower no domino may stick out into an imaginary next layer.

6) Why a linear recurrence must exist.

The vector \(v_k\) lives in a 512-dimensional vector space over \(\mathbb Z_M\), and it evolves by multiplication with the same fixed \(512\times 512\) matrix \(T\). Therefore the scalar sequence \(f(k)=v_k[0]\) is automatically a linear recurrence sequence. In principle its recurrence length is at most 512, although the minimal recurrence can be smaller.

7) Recover the minimal recurrence with Berlekamp-Massey.

The program generates the first 1301 terms

$f(0),f(1),\dots,f(1300)$

using the 512-state DP. Then Berlekamp-Massey reconstructs the shortest recurrence

$f(n)=\sum_{j=1}^{L} c_j f(n-j)\pmod M.$

This is the crucial compression step: after that, we never need the 512-state DP again.

8) Small checkpoints.

The C++ code validates the recovered recurrence against known terms:

$f(2)=229,\qquad f(4)=117805,\qquad f(10)=96149360,$

and also much later values

$f(10^3)=24806056,\qquad f(10^6)=30808124.$

These checkpoints are important: they verify both the transfer-DP part and the recurrence-evaluation part.

9) Why giant matrix exponentiation is replaced by Kitamasa-style polynomial arithmetic.

Once we know the recurrence, computing \(f(N)\) is equivalent to reducing the monomial \(x^N\) modulo the characteristic polynomial

$P(x)=x^L-\sum_{j=1}^{L} c_j x^{L-j}.$

If

$x^N \equiv a_0+a_1x+\cdots+a_{L-1}x^{L-1}\pmod{P(x)},$

then

$f(N)=a_0f(0)+a_1f(1)+\cdots+a_{L-1}f(L-1)\pmod M.$

The code computes these coefficients by repeated squaring and polynomial combination, which is the standard Kitamasa viewpoint.

10) Handling \(N=10^{10000}\).

The exponent does not fit into any machine integer type. So the code stores it as the decimal string

$1000\ldots 0$

with 10000 zeros, repeatedly divides that string by 2, and records the remainders. This produces the bits of \(N\) in least-significant-bit-first order, which is exactly what binary exponentiation needs.

Algorithm

1) Precompute all layer transitions \(T_{s,t}\) by DFS over a \(3\times 3\) slice.

2) Run the 512-state DP long enough to obtain a prefix of \(f(n)\).

3) Apply Berlekamp-Massey to extract the minimal linear recurrence.

4) Validate the recurrence on additional known terms.

5) Convert \(N=10^{10000}\) from decimal to binary bits.

6) Evaluate \(f(N)\) modulo \(M\) using Kitamasa-style polynomial squaring.

Complexity Analysis

The transition precomputation is finite and bounded by the 512 masks. The prefix DP is also small. If the minimal recurrence length is \(L\), the final evaluation costs roughly

$O(L^2\log N)$

modular operations and

$O(L)$

memory. That is what makes the astronomical index \(10^{10000}\) manageable.

Checks And Final Result

The code checks

$f(2)=229,\quad f(4)=117805,\quad f(10)=96149360,\quad f(10^3)=24806056,\quad f(10^6)=30808124.$

For the required exponent

$N=10^{10000},$

the final answer is

$f(N)=96972774.$

Further Reading

1. Problem page: https://projecteuler.net/problem=324
2. Berlekamp-Massey algorithm: https://en.wikipedia.org/wiki/Berlekamp-Massey_algorithm
3. Linear recurrences / Kitamasa: https://cp-algorithms.com/algebra/linear-recurrence.html