Detailed mathematical approach

Problem Summary

The problem defines a sequence \(D(n)\) associated with counting amoeba configurations in a three-dimensional grid. The computational goal used by the implementations is to evaluate \(D(10000)\) modulo \(10^9\), while several smaller values are checked exactly before the large modular run.

A direct enumeration of all relevant 3D shapes would grow far too quickly. The implemented method therefore replaces explicit shape generation with a layered dynamic program whose states encode admissible boundary profiles and whose scalar recurrences extract the final sequence value.

Mathematical Approach

For exposition, denote the two triangular state families by \(A_{n,k}(m)\) and \(B_{n,k}(m)\), and denote the scalar sequences by \(P(m)\) and \(Q(m)\). The desired quantity is

$D(n)=Q(n-1).$

The implementations never materialize the underlying amoebas one by one. Instead, they count how many partial configurations can end in each admissible profile and then close the whole system with two one-dimensional recurrences.

Step 1: A triangular threshold controls when a layer can be active

The first structural quantity is the triangular threshold

$T(n)=\frac{(n+1)(n+2)}{2}.$

This is the smallest size parameter at which the level indexed by \(n\) can contribute. Consequently, for every \(n\ge 1\),

$m<T(n)\implies A_{n,k}(m)=B_{n,k}(m)=0.$

So when the outer loop is processing a fixed \(m\), only levels with \(T(n)\le m\) are relevant. Because \(T(n)\sim n^2/2\), the largest active level is only \(O(\sqrt m)\).

Step 2: Boundary conditions collapse the level \(n=0\) into a scalar base stream

Negative sizes are impossible, so every quantity evaluated at \(m<0\) is defined to be \(0\). At level \(n=0\), the two state families do not store a full row of profile data. Instead they reduce to the same scalar stream:

$A_{0,0}(m)=B_{0,0}(m)=P(m),$

while

$A_{0,k}(m)=B_{0,k}(m)=0\qquad (k\ne 0).$

This identification is essential because the higher-level recurrences refer to \(n-1\), so the whole system must know what happens when the index reaches zero. The initial scalar condition is

$Q(0)=1,$

and all other missing values are supplied automatically by the zero rules for negative shifts.

Step 3: Every active profile obeys a coupled six-term recurrence

For \(1\le k\le n\), define the folded index and the edge multiplier by

$r(n,k)=\begin{cases} k,&k<n,\\ k-1,&k=n, \end{cases} \qquad \lambda(n,k)=\begin{cases} 1,&k<n,\\ 2,&k=n. \end{cases}$

Then for every active layer \(m\ge T(n)\), the two profile families satisfy

$\begin{aligned} A_{n,k}(m)=&\,A_{n,k}(m-n-2)+B_{n+1,1}(m-n-3)+A_{n+1,k+1}(m-n-3)\\ &+A_{n-1,r(n,k)}(m-n-1)+B_{n,1}(m-n-2)+A_{n,r(n,k)+1}(m-n-2), \end{aligned}$

$\begin{aligned} B_{n,k}(m)=&\,B_{n,k}(m-n-2)+B_{n+1,k+1}(m-n-3)+\lambda(n,k)\,A_{n+1,1}(m-n-3)\\ &+B_{n-1,r(n,k)}(m-n-1)+B_{n,r(n,k)+1}(m-n-2)+\lambda(n,k)\,A_{n,1}(m-n-2). \end{aligned}$

These are exactly the transitions evaluated by the implementations. The special case \(k=n\) is the only point where the coefficient \(2\) appears, so the edge of the triangular index set has a slightly different combinatorial weight from the interior.

Step 4: Two scalar recurrences close the dynamic system

Once all \(A\) and \(B\) states at size \(m\) have been filled, the implementations update the scalar layer by

$P(m)=Q(m-1)+4P(m-2)+2A_{1,1}(m-3)+B_{1,1}(m-3),$

and, for \(m\ge 1\),

$Q(m)=3Q(m-1)+3P(m-2).$

The second recurrence produces the sequence whose shifted values are the final answers. Therefore

$D(n)=Q(n-1).$

From the dynamic-programming point of view, \(P(m)\) is the bridge between the boundary level \(n=0\) and the genuinely two-parameter profile states with \(n\ge 1\).

Step 5: The same recurrence explains the rolling-memory optimization

Every right-hand side only looks backward by offsets such as \(m-n-1\), \(m-n-2\), \(m-n-3\), \(m-1\), \(m-2\), and \(m-3\). No transition needs the full history of all earlier layers.

If

$n_{\max}(M)=\max\{n:T(n)\le M\},$

then all look-backs for the final target size \(M\) fit inside a circular buffer of length \(n_{\max}(M)+8\). That is why the program can keep the quadratic-time recurrence while avoiding quadratic storage in \(m\).

Worked Example: The first nontrivial values

The initial layers can be computed by hand directly from the recurrence.

At \(m=0\), no triangular state is active, so

$P(0)=0,\qquad Q(0)=1.$

At \(m=1\), still no \(n\ge 1\) state survives because \(T(1)=3\). Hence

$P(1)=Q(0)=1,\qquad Q(1)=3Q(0)=3.$

At \(m=2\), the same argument gives

$P(2)=Q(1)+4P(0)=3,\qquad Q(2)=3Q(1)+3P(0)=9.$

At \(m=3\), the first triangular level appears because \(T(1)=3\). For the only index pair \((n,k)=(1,1)\), we have \(r(1,1)=0\) and \(\lambda(1,1)=2\). Using the boundary rule \(A_{0,0}(1)=B_{0,0}(1)=P(1)=1\), every invalid or negatively shifted term vanishes, so

$A_{1,1}(3)=1,\qquad B_{1,1}(3)=1.$

The scalar update then yields

$P(3)=Q(2)+4P(1)=9+4=13,$

$Q(3)=3Q(2)+3P(1)=27+3=30.$

Therefore

$D(1)=1,\qquad D(2)=3,\qquad D(3)=9,\qquad D(4)=30.$

This small example shows exactly how the first nonzero triangular state enters when the threshold is met.

How the Code Works

The C++, Python, and Java implementations all evaluate the same recurrence. The compiled solvers allocate two triangular state tables for admissible profile pairs \((n,k)\), together with two one-dimensional arrays for the scalar layer. The outer loop advances the size parameter \(m\) from \(0\) to the target, clears the current column of the circular history buffer, fills every active profile state, and finally updates the two scalar sequences.

Before the dynamic program starts, the code computes the largest active level \(n_{\max}\) from the triangular threshold. That value determines both how many profile rows are needed and how large the circular buffer must be. Because the recurrence reaches one row above the current level, the allocation includes a small safety margin beyond \(n_{\max}\).

The C++ implementation is used in two arithmetic modes: exact integer evaluation for small verification checkpoints and modular evaluation for the large final query. The Java implementation keeps the modular variant, which is enough for the published answer. The Python implementation is a thin launcher that builds and runs the compiled solver and then extracts the final numeric output, so the three language versions share the same mathematical core.

The embedded checkpoints are

$D(10)=44499,\qquad D(20)=9204559704,\qquad D(32)=22037102049132222,$

and

$D(100)\equiv 780166455\pmod{10^9}.$

These values verify that the recurrence has been wired correctly before the computation of \(D(10000)\bmod 10^9\).

Complexity Analysis

Let \(M\) be the largest size parameter reached by the program, so the final query uses \(M=n-1\). The active range satisfies \(T(n_{\max})\le M<T(n_{\max}+1)\), hence \(n_{\max}=\Theta(\sqrt M)\).

For each \(m\), the implementation visits every admissible pair \((n,k)\) with \(1\le n\le n_{\max}\) and \(1\le k\le n\). The number of profile updates per layer is therefore

$\sum_{n=1}^{n_{\max}} n=\frac{n_{\max}(n_{\max}+1)}{2}=O(n_{\max}^2),$

which makes the total running time

$O(M\,n_{\max}^2)=O(M^2).$

For memory, the circular buffer length is \(O(n_{\max})\), and the total number of stored profile entries is

$O\!\left(n_{\max}\sum_{n=1}^{n_{\max}} n\right)=O(n_{\max}^3)=O(M^{3/2}).$

The two scalar sequences add only \(O(n_{\max})\) more cells, so they do not change the leading asymptotic term.

Footnotes and References

1. Project Euler problem page: https://projecteuler.net/problem=763
2. Dynamic programming: Wikipedia — Dynamic programming
3. Triangular number: Wikipedia — Triangular number
4. Recurrence relation: Wikipedia — Recurrence relation
5. Circular buffer: Wikipedia — Circular buffer