Detailed mathematical approach

Problem Summary

Problem 762 asks for a large-index counting value associated with amoebas in a two-dimensional grid, reported modulo \(10^9\). The C++, Python, and Java implementations show that the combinatorial count can be encoded by a single sequence \((C_n)\), so the computational task is to evaluate that sequence at \(n=100000\).

Instead of simulating every possible grid evolution, the solution uses the fact that the sequence is completely determined by nine initial values and an order-8 linear recurrence. That turns a difficult counting problem into a short modular recurrence computation.

Mathematical Approach

Let \(C_n\) denote the required counting sequence modulo \(10^9\). The implementations expose the complete recurrence data directly.

Step 1: Initial Values

The first nine terms are

$C_0=1,\ C_1=1,\ C_2=2,\ C_3=4,\ C_4=9,\ C_5=20,\ C_6=46,\ C_7=105,\ C_8=243.$

These seed values anchor the whole sequence. Once they are known, every later term is forced by the same linear rule.

Step 2: Order-8 Linear Recurrence

For every \(n\ge 9\), the counting sequence satisfies

$C_n \equiv 2C_{n-1}+2C_{n-2}-C_{n-3}-3C_{n-4}-5C_{n-5}+4C_{n-6}-2C_{n-7}+4C_{n-8}\pmod{10^9}.$

This means the entire history is compressed into the most recent eight terms. After the recurrence has been derived, no direct enumeration of amoeba states is needed.

Step 3: State-Vector Interpretation

Package the recent values into an 8-dimensional state vector

$v_n=\begin{pmatrix}C_{n-7}\\C_{n-6}\\C_{n-5}\\C_{n-4}\\C_{n-3}\\C_{n-2}\\C_{n-1}\\C_n\end{pmatrix}\qquad (n\ge 8).$

Then one step of the sequence is a linear transition

$v_{n+1}\equiv \begin{pmatrix} 0&1&0&0&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1\\ 4&-2&4&-5&-3&-1&2&2 \end{pmatrix} v_n \pmod{10^9}.$

So the problem becomes a finite-dimensional linear dynamical system. The associated characteristic polynomial is

$r^8-2r^7-2r^6+r^5+3r^4+5r^3-4r^2+2r-4.$

The implementation does not need to solve this polynomial explicitly; iterating the recurrence is enough.

Step 4: Modular Arithmetic and Sign Handling

Several coefficients are negative, so each new term is reduced immediately modulo \(10^9\):

$C_n \leftarrow C_n \bmod 10^9.$

If the reduced value is negative, adding one copy of the modulus returns it to the standard residue range \(0\le C_n<10^9\).

Because the recurrence uses only addition, subtraction, and multiplication by fixed small integers, this normalization preserves the exact sequence in \(\mathbb{Z}/10^9\mathbb{Z}\).

Worked Example

The first nontrivial term is \(C_9\). Substituting the seed values gives

$\begin{aligned} C_9&=2C_8+2C_7-C_6-3C_5-5C_4+4C_3-2C_2+4C_1\\ &=2\cdot 243+2\cdot 105-46-3\cdot 20-5\cdot 9+4\cdot 4-2\cdot 2+4\cdot 1\\ &=561. \end{aligned}$

Applying the same rule once more,

$\begin{aligned} C_{10}&=2C_9+2C_8-C_7-3C_6-5C_5+4C_4-2C_3+4C_2\\ &=2\cdot 561+2\cdot 243-105-3\cdot 46-5\cdot 20+4\cdot 9-2\cdot 4+4\cdot 2\\ &=1301. \end{aligned}$

This confirms that the recurrence is being advanced correctly before moving on to the much larger target index.

How the Code Works

The C++, Python, and Java implementations all follow the same pattern. They store the nine seed values first. If the requested index is smaller than \(9\), the answer is returned directly from that initial table.

Otherwise the implementation keeps a rolling window containing the most recent nine terms. Each loop iteration forms the next term from the fixed coefficient pattern, reduces the result modulo \(10^9\), shifts the window one position to the left, and appends the new term on the right.

When the loop reaches \(n=100000\), the rightmost entry of the window is the desired answer. The runtime is small because the heavy combinatorial reasoning has already been condensed into the recurrence itself.

Complexity Analysis

Let \(N\) be the target index. Every iteration performs a constant amount of arithmetic, so the running time is \(O(N)\). The rolling window has fixed length, so the memory usage is \(O(1)\).

For this problem size, a direct linear pass to \(N=100000\) is easily fast enough, so no additional acceleration technique is required.

Footnotes and References

1. Project Euler problem page: https://projecteuler.net/problem=762
2. Linear recurrence with constant coefficients: Wikipedia - Linear recurrence with constant coefficients
3. Companion matrix: Wikipedia - Companion matrix
4. Modular arithmetic: Wikipedia - Modular arithmetic
5. Recurrence relation: Wikipedia - Recurrence relation