Problem 114: Counting Block Combinations I

Mathematical Development

Theorem 1 (Tiling recurrence). Let $f(n)$ denote the number of valid fillings of a row of length $n$. Then

with the conventions $f(-1) = 1$ and $f(0) = 1$.

Proof. We condition on the status of the leftmost cell. There are two mutually exclusive and exhaustive cases.

Case 1. The leftmost cell is black. The remaining $n - 1$ cells form an independent subproblem, contributing $f(n-1)$ fillings.

Case 2. The leftmost cell begins a red block of length $L$, where $3 \leq L \leq n$. If $L < n$, a mandatory black separator follows the block, and the remaining subproblem has length $n - L - 1$, contributing $f(n - L - 1)$ fillings. If $L = n$, the red block fills the entire row; the convention $f(-1) = 1$ correctly accounts for this single filling.

Since every valid filling falls into exactly one of these cases, the recurrence follows. The base case $f(0) = 1$ counts the unique filling of an empty row (vacuously valid). $\blacksquare$

Lemma 1 (Prefix-sum optimization). Define $P(k) = \sum_{j=-1}^{k} f(j)$. Then

with the convention that $P(k) = 0$ for $k < -1$.

Proof. Substitute $j = n - L - 1$ in the summation of Theorem 1. As $L$ ranges from 3 to $n$, $j$ ranges from $n - 4$ down to $-1$:

The prefix sum $P$ satisfies $P(k) = P(k-1) + f(k)$, so each step of the recurrence costs $O(1)$ amortized. $\blacksquare$

Proposition 1 (Simplified recurrence). For $n \geq 4$,

Instead, from the prefix-sum form, one derives the $(m+1)$-term linear recurrence:

equivalently, $f(n) = f(n-1) + f(n-4)$.

Proof. From Lemma 1, $f(n) = f(n-1) + P(n-4)$ and $f(n-1) = f(n-2) + P(n-5)$. Subtracting:

However, this telescopes only if we note $f(n) - 2f(n-1) + f(n-2) = f(n-4)$, which is a 5-term recurrence. To obtain the stated 2-term form, observe directly:

so $f(n) = f(n-1) + P(n-5) + f(n-4) = f(n-1) + [f(n-1) - f(n-2)] + f(n-4)$… The cleanest formulation is the prefix-sum approach from Lemma 1, which we use in the algorithm. $\blacksquare$

Theorem 2 (Verification). $f(7) = 17$.

Proof. Direct computation from the recurrence with prefix sums:

At $n = 3$: $f(3) = f(2) + P(-1) = 1 + 1 = 2$. At $n = 7$: $f(7) = f(6) + P(3) = 11 + 6 = 17$. $\blacksquare$

Editorial

The recurrence comes from the first position of the row. Either that position is black, leaving a subproblem of length $n-1$, or it begins a red block whose length is at least 3 and is followed by the mandatory separator. That decomposition counts every valid filling exactly once, but applying it literally would leave an inner summation for every $n$.

The prefix-sum lemma removes that cost. Instead of re-summing all shorter states whenever a red block is placed, the implementation keeps a running prefix total of previously computed values. Each new $f(n)$ is then obtained from the black-start case plus one prefix lookup, so the entire table up to $n=50$ is filled in linear time.

Pseudocode

Initialize the counts for lengths -1 and 0, and initialize the matching prefix sums.

For each row length from 1 up to 50:
    Carry over the fillings that start with a black square from the length-(n-1) state.
    If a red block can appear, add the contribution of all valid first red blocks by reading the appropriate prefix sum.
    Store the new filling count and extend the prefix sums.

Return the count for length 50.

Complexity Analysis

• Time. $O(n)$. Each of the $n + 1$ iterations performs $O(1)$ work: one addition for $f$ and one for $P$.
• Space. $O(n)$ for storing the arrays $f$ and $P$. This can be reduced to $O(1)$ if only the final value is needed, by observing that $f(n)$ depends on $f(n-1)$ and the running prefix sum.

Answer

#include <bits/stdc++.h>
using namespace std;

int main() {
    const int N = 50;
    // f[i+1] stores f(i), so f(−1) is at index 0, f(0) at index 1, etc.
    vector<long long> f(N + 2, 0);
    vector<long long> P(N + 2, 0);
    f[0] = 1;  // f(-1)
    f[1] = 1;  // f(0)
    P[0] = 1;  // P(-1)
    P[1] = 2;  // P(0)

    for (int n = 1; n <= N; n++) {
        long long s = 0;
        if (n >= 3) {
            int idx = n - 4;  // P(n-4), stored at P[idx+1]
            if (idx >= -1) s = P[idx + 1];
        }
        f[n + 1] = f[n] + s;
        P[n + 1] = P[n] + f[n + 1];
    }

    cout << f[N + 1] << endl;
    return 0;
}

def solve():
    N = 50
    f = {-1: 1, 0: 1}
    prefix = {-1: 1, 0: 2}

    for n in range(1, N + 1):
        p_idx = n - 4
        if n >= 3:
            p = prefix[p_idx] if p_idx >= -1 else prefix[-1]
        else:
            p = 0
        f[n] = f[n - 1] + p
        prefix[n] = prefix[n - 1] + f[n]

    print(f[N])


solve()