Solution

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Structural Analysis

The Binary String Bijection (Most Important Single Insight)

The configurations within a block of length L (non-overlapping, non-adjacent integer-length sub-blocks, including the empty config) biject naturally with binary strings of length L: map each string to the configuration whose blocks are its maximal runs of 1s. Adjacent 1s form one block; a 0 between two runs automatically provides the required gap. This gives:

\text{(Number of valid configs within a block of length } L) = 2^L

Total Sub-Castle Count: T(k, L) = (k+1)^L

Define T(k, L) = number of "towers" of at most k rows that can sit above a single block of length L (counting every block placed above it, not counting L itself). The recursion is:

T(k, L) = \sum_{b \in \{0,1\}^L} \prod_{\text{runs of length } l \text{ in } b} T(k-1, l)

because blocks in different positions are independent — the sub-blocks generated above two sibling blocks never interact (they're automatically separated by the gap inherited from the parent row).

Base case: T(0, L) = 1.

By induction (and the binary string bijection), $$ T(k-1, l) = k^l $$ , so:

T(k, L) = \sum_{b \in \{0,1\}^L} \prod_{\text{runs}} k^l = \sum_{b} k^{(\text{Number of ones in } b)} = (1 + k)^L

Therefore $$ T(k, L) = (k+1)^L $$ , a beautiful and crucial result.

Corollary: The total number of castles with bottom block w and height ≤ h (any parity) is $$ h^w $$ .

The Main Formula

Let G(w, h) = # castles with bottom block w, height ≤ h, even total block count. Since the bottom block contributes 1 block, even total ↔ odd # blocks in rows 2..h.

Define the signed count P(k, L) = sum over all towers of height ≤ k above L of $(-1)^{\text{Number of blocks in tower}}$ .

Then:

G(w, h) = \frac{h^w - P(h-1, w)}{2}

and therefore:

F(w, h) = \frac{h^w - (h-1)^w - P(h-1, w) + P(h-2, w)}{2}

Verification: $$ F(4, 2) = (16 - 1 - P(1, 4) + P(0, 4))/2 $$ . One can compute $P(1, L) = \operatorname{Re}((1+i)^{L+1})$ , giving P(1, 4) = −4, P(0, 4) = 1, so F(4, 2) = (16 − 1 + 4 + 1)/2 = 10. ✓

Computing P(k, L): The Transfer Matrix Structure

P(k, L) satisfies the same recursion as T, but with a sign for each block:

P(k, L) = \sum_{b \in \{0,1\}^L} (-1)^{\#\text{runs}(b)} \prod_{\text{runs of length } l} P(k-1, l)

Key structural fact: For fixed k, P(k, L) satisfies a linear recurrence in L whose characteristic polynomial has degree ~k. This follows from a transfer-matrix argument:

The "gap" and "in-run" states can be tracked with a state vector of dimension equal to the number of distinct exponential terms in $P(k-1, \cdot)$ (starting from P(0, L) = 1, one term, growing by one per level).
The eigenvalues of the size-(k+1) transfer matrix $$ M_k $$ determine the characteristic polynomial of the recurrence for P(k, L) in L.

This means P(k, L) as a function of L is a sum of k+1 exponentials (eigenvalues of $$ M_k $$ ), computable via matrix exponentiation.

Sub-Problem Strategy

F(10^12, 100): Large w, Fixed h

h = 100, so the transfer matrix $M_{99}$ is ~100×100.
Build $M_{99}$ iteratively: each level adds one dimension to the state.
Raise $M_{99}^{10^{12}}$ using binary (fast) matrix exponentiation in $O(100^3 \cdot \log 10^{12})$ operations mod $$ 10^9+7 $$ .
This is entirely feasible (~4×10^7 operations).

F(100, 10^12): Fixed w, Large h

Directly computing the k-direction recursion is non-trivial because the map $(P(k-1,1),\ldots,P(k-1,100)) \to (P(k,1),\ldots,P(k,100))$ is polynomial (degree up to 50), not linear.
However: for fixed L ≤ 100, P(k, L) as a function of k satisfies a linear recurrence in k (shown empirically: $$ P(k,2) = (-1)^k(k+1) $$ satisfies $(\lambda+1)^2 = 0$ ; $$ P(k,3) = (-1)^k(k+1)^2 $$ satisfies $(\lambda+1)^3 = 0$ ).
Strategy: Compute $P(0, 100), P(1, 100), \ldots, P(N, 100)$ for small N using the recursive formula, then apply Berlekamp-Massey to find the minimal linear recurrence in k, then use matrix exponentiation to jump to $k = 10^{12} - 1$ .
The recurrence order may be around 2w in the worst case — for w = 100 this is a 200×200 matrix, perfectly manageable.

F(10000, 10000): Both Large

Neither direction alone is tractable in the obvious way. Likely requires combining both directions.
One approach: use the L-direction matrix (size ~10000, which is borderline) to get a closed-form expression for P(k, 10000) as a function of k, then use the k-direction recurrence.
Alternatively, the Cayley-Hamilton theorem and minimal polynomial of the transfer matrix for each level can be used to prove that the linear recurrence in both directions has bounded order.
The Berlekamp-Massey algorithm is essential here: compute enough values of F(10000, h) for small h and detect the minimal linear recurrence, then exponentiate.

Algorithmic and Mathematical Tools

Core Combinatorics

The binary string bijection and its signed version (tracking parity via $(-1)^{\#\text{runs}}$ )
Independence of sub-configurations across sibling blocks (essential for all decompositions)
Inclusion-exclusion for "exactly height h" from "at most height h"
Signed generating functions: the standard $$ A^+ = (T + P)/2 $$ , $$ A^- = (T - P)/2 $$ decomposition

Linear Algebra and Polynomial Methods

Transfer matrix method (from combinatorics and statistical mechanics)
Fast matrix exponentiation: $O(d^3 \log N)$ for a d×d matrix to the N-th power mod p
Berlekamp-Massey algorithm: recover a minimal-length linear recurrence from a sequence of computed values — the key tool for discovering structure without full mathematical proof
Cayley-Hamilton theorem: the transfer matrix satisfies its own characteristic polynomial, so powers reduce to a linear combination of at most d previous terms
Perron-Frobenius theorem: for large k, P(k, w) is dominated by the largest-magnitude eigenvalue of the transfer matrix; this gives the asymptotic and helps verify correctness

Number Theory and Modular Arithmetic

All arithmetic mod $$ p = 10^9+7 $$ (a prime, so inverses and division by 2 are clean)
Modular inverses for the division by 2 in the main formula
Chinese Remainder Theorem as a fallback if intermediate calculations require multiple moduli

Generating Functions

For a single block of length L: the ordinary GF for configs weighted by $\prod_{\text{blocks}} x_l$ (tracking what lengths are generated) is a transfer-matrix computation over binary strings
Product formula: since siblings are independent, the GF for a full row factors over blocks
Functional equations (kernel method in analytic combinatorics) for extracting closed forms from GF recursions

Verification Strategies

F(4,2) = 10, F(13,10) = 3729050610636, F(10,13) = 37959702514 — use these at each intermediate stage to catch bugs early
Cross-check $F(w, h) + F_{\text{odd}}(w, h) = h^w - (h-1)^w$ (all castles of exactly height h)
Sanity-check $$ T(k, L) = (k+1)^L $$ numerically for small values before any deeper computation

Suggested Exploration Sequence

Verify the binary string bijection and $$ T(k,L) = (k+1)^L $$ empirically for small values.
Derive and validate the main formula $$ F(w,h) = (h^w - (h-1)^w - P(h-1,w) + P(h-2,w))/2 $$ .
Build the L-direction transfer matrix for P(k, L): implement the recurrence, confirm the matrix size grows as k+2, and verify P(k, L) matches direct brute-force for small values.
Solve F(10^12, 100) using matrix exponentiation in L.
Compute P(k, 100) for many k, apply Berlekamp-Massey, determine recurrence order, solve F(100, 10^12).
Tackle F(10000, 10000) — likely the hardest; use both directions and the structure discovered in steps 4–5 to find the right approach.

Project Euler/502/Solution

From charlesreid1

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