Problem 191: Prize Strings

Mathematical Development

Definition. Let $\Sigma = \{O, A, L\}$ and let $\mathcal{P}_n \subseteq \Sigma^n$ denote the set of prize strings of length $n$. Let $g(n)$ denote the number of strings of length $n$ over $\{O, A\}$ that contain no three consecutive $A$‘s.

Theorem 1 (Tribonacci-type recurrence for $g$). The function $g$ satisfies

with initial values $g(0) = 1$, $g(1) = 2$, $g(2) = 4$.

Proof. We partition the set of valid strings of length $n$ over $\{O, A\}$ (containing no $AAA$) by the length of the maximal suffix of $A$‘s. Every such string falls into exactly one of three disjoint cases:

• Suffix of zero $A$‘s (string ends in $O$): the prefix of length $n - 1$ is an arbitrary valid string, giving $g(n-1)$ strings.
• Suffix of exactly one $A$ (string ends in $OA$, or is $A$ itself when $n=1$): for $n \ge 2$, the character at position $n-2$ must be $O$, and the prefix of length $n - 2$ is an arbitrary valid string, giving $g(n-2)$ strings.
• Suffix of exactly two $A$‘s (string ends in $OAA$, or is $AA$ when $n = 2$): for $n \ge 3$, the character at position $n - 3$ must be $O$ (since $AAA$ is forbidden), and the prefix of length $n - 3$ is an arbitrary valid string, giving $g(n-3)$ strings.

A suffix of three or more $A$‘s is impossible by the constraint. These three cases are exhaustive and mutually exclusive, establishing $g(n) = g(n-1) + g(n-2) + g(n-3)$ for $n \ge 3$.

The base cases are verified by direct enumeration:

• $g(0) = 1$: the empty string $\varepsilon$.
• $g(1) = 2$: the strings $\{O, A\}$.
• $g(2) = 4$: the strings $\{OO, OA, AO, AA\}$. $\blacksquare$

Theorem 2 (Counting prize strings). The total number of prize strings of length $n$ is

Proof. We partition $\mathcal{P}_n$ by the number of occurrences of $L$.

Case 1 (zero $L$‘s). The string lies in $\{O, A\}^n$ with no $AAA$ substring. By definition, there are exactly $g(n)$ such strings.

Case 2 (exactly one $L$). Fix the position of $L$ at index $k \in \{0, 1, \ldots, n-1\}$ (0-indexed). The string decomposes as

The prefix is a string over $\{O, A\}$ of length $k$ containing no $AAA$ (counted by $g(k)$). The suffix is a string over $\{O, A\}$ of length $n - 1 - k$ containing no $AAA$ (counted by $g(n-1-k)$). Crucially, the $L$ at position $k$ interrupts any consecutive-$A$ run, so the prefix and suffix constraints are independent. By the multiplication principle, there are $g(k) \cdot g(n-1-k)$ valid strings with $L$ at position $k$. Summing over all $k$ gives $\sum_{k=0}^{n-1} g(k) \cdot g(n-1-k)$.

Since a prize string has at most one $L$, Cases 1 and 2 are exhaustive and disjoint. $\blacksquare$

Remark. The sum $\sum_{k=0}^{n-1} g(k) \cdot g(n-1-k)$ is a discrete convolution of $g$ with itself, evaluated at $n - 1$.

Corollary. For $n = 30$: $g(30) = 53{,}798{,}080$ and $|\mathcal{P}_{30}| = 1{,}918{,}080{,}160$.

Editorial

The constraint on absences is local, so it is natural to separate the problem into strings over {O, A} first. If g(n) counts length-n strings with no AAA, the last block of consecutive absences has length 0, 1, or 2, which gives the Tribonacci recurrence immediately. That handles every valid record with no late day at all.

The single late day is then inserted by position. If the L sits at index k, the prefix on its left and the suffix on its right are completely independent because the L breaks any run of absences. So the number of records with the L at that position is g(k) * g(n-1-k). Summing those products over all k, and then adding the zero-L case, gives the required total.

Pseudocode

Set g(0) = 1, g(1) = 2, and g(2) = 4.
For each length from 3 up to n:
    extend the no-late count with g(length-1) + g(length-2) + g(length-3).

Start the answer with g(n), which covers records containing no L.
For each possible position of the unique L:
    multiply the valid prefix count by the valid suffix count,
    and add that product to the answer.

Return the final total.

Complexity Analysis

• Time: $O(n)$ arithmetic operations — $O(n)$ for the recurrence and $O(n)$ for the convolution sum. Each operation involves integers of size $O(n \log n)$ bits, so the bit-complexity is $O(n^2 \log n)$, though for $n = 30$ all values fit in a single machine word.
• Space: $O(n)$ to store the array $g[0..n]$. This can be reduced to $O(1)$ auxiliary space by accumulating the convolution sum during the forward recurrence pass using a two-pointer technique, though the constant-factor savings are negligible for $n = 30$.

Answer

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

int main() {
    // g[n] = number of valid strings of length n over {O,A} with no "AAA"
    // Recurrence: g[n] = g[n-1] + g[n-2] + g[n-3]
    const int N = 30;
    vector<long long> g(N + 1);
    g[0] = 1; g[1] = 2; g[2] = 4;
    for (int i = 3; i <= N; i++)
        g[i] = g[i-1] + g[i-2] + g[i-3];

    // Strings with 0 L's
    long long ans = g[N];

    // Strings with exactly 1 L: place L at position k, 0-indexed
    // prefix of length k and suffix of length N-1-k must both be valid no-L strings
    for (int k = 0; k < N; k++)
        ans += g[k] * g[N - 1 - k];

    cout << ans << endl;
    return 0;
}

"""
Problem 191: Prize Strings

Count 30-character strings over {O, A, L} with at most one L
and no three consecutive A's.
"""

def solve(n=30):
    # g[i] = number of valid strings of length i with no L and no "AAA"
    g = [0] * (n + 1)
    g[0] = 1
    g[1] = 2
    g[2] = 4
    for i in range(3, n + 1):
        g[i] = g[i-1] + g[i-2] + g[i-3]

    # No L: g[n]
    ans = g[n]

    # Exactly one L at position k: g[k] * g[n-1-k]
    for k in range(n):
        ans += g[k] * g[n - 1 - k]

    return ans

answer = solve()
assert answer == 1918080160
print(answer)