Project Euler/502/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

1. Verify the binary string bijection and $T(k,L)=(k+1)^{L}$ empirically for small values.
2. Derive and validate the main formula $F(w,h)=(h^{w}-(h-1)^{w}-P(h-1,w)+P(h-2,w))/2$.
3. 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.
4. Solve F(10^12, 100) using matrix exponentiation in L.
5. Compute P(k, 100) for many k, apply Berlekamp-Massey, determine recurrence order, solve F(100, 10^12).
6. Tackle F(10000, 10000) — likely the hardest; use both directions and the structure discovered in steps 4–5 to find the right approach.