Problem 524: First Sort II

Mathematical Foundation

Theorem. For all $n \ge 1$,

Proof. From Problem 523, we established the recurrence $E(n) - E(n-1) = (n+1)/6$ with $E(1) = 0$. Summing the telescoping series:

$\square$

Lemma (Modular Inverse of 12). Let $p = 10^9 + 7$. Then $12^{-1} \equiv 83333334 \pmod{p}$.

Proof. We verify directly: $12 \times 83333334 = 999999\,408 + 408 = 1\,000\,000\,008 = p + 1 \equiv 1 \pmod{p}$. Since $p$ is prime and $\gcd(12, p) = 1$ (as $p$ is odd and not divisible by 3), the inverse exists and is unique. $\square$

Theorem (Modular Evaluation). With $p = 10^9 + 7$ and $N = 10^{18}$:

Proof. Since $p$ is prime and $12 \not\equiv 0 \pmod{p}$, the fraction $(n-1)(n+4)/12$ is well-defined modulo $p$ as the product $(n-1)(n+4) \cdot 12^{-1} \bmod p$. By Fermat’s little theorem, $12^{-1} \equiv 12^{p-2} \pmod{p}$. $\square$

Lemma (Reduction of $10^{18} \bmod p$). Since $p = 10^9 + 7$, we have $10^9 \equiv -7 \pmod{p}$, hence $10^{18} = (10^9)^2 \equiv 49 \pmod{p}$.

Proof. $10^{18} - 49 = (10^9 - 7)(10^9 + 7) = (10^9 - 7) \cdot p \equiv 0 \pmod{p}$. $\square$

Corollary. The final computation:

• $a = 10^{18} - 1 \equiv 48 \pmod{p}$,
• $b = 10^{18} + 4 \equiv 53 \pmod{p}$,
• $c = 48 \times 53 = 2544$,
• $E(10^{18}) \equiv 2544 \times 83333334 \pmod{p}$.

Computing: $2544 \times 83333334 = 212\,000\,017\,296$. Dividing by $p$: $212\,000\,017\,296 = 211 \times 1\,000\,000\,007 + r$, where $211 \times 1\,000\,000\,007 = 211\,000\,001\,477$, so $r = 212\,000\,017\,296 - 211\,000\,001\,477 = 1\,000\,015\,819$.

Editorial

E(n) = (n-1)(n+4)/12, the expected number of insertion steps to sort a random permutation using the “first sort” algorithm. Compute E(10^18) mod (10^9 + 7).

Pseudocode

    MOD = 10^9 + 7
    n = 10^18
    a = (n - 1) mod MOD // = 48
    b = (n + 4) mod MOD // = 53
    inv12 = power(12, MOD - 2, MOD) // = 83333334
    Return (a * b mod MOD) * inv12 mod MOD

Complexity Analysis

• Time: $O(\log p)$ for the single modular exponentiation to compute $12^{-1}$. All other operations are $O(1)$.
• Space: $O(1)$.

Answer

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

/*
 * Problem 524: First Sort II
 *
 * E(n) = (n-1)(n+4) / 12, the expected number of insertion steps
 * to sort a random permutation via the "first sort" algorithm.
 *
 * Compute E(10^18) mod (10^9 + 7).
 *
 * Key insight: 10^18 mod p = 49 (since 10^9 ≡ -7 mod p, so 10^18 ≡ 49).
 * Then (49 - 1)(49 + 4) / 12 = 48 * 53 / 12 = 2544 / 12 = 212 mod p.
 * Wait, 2544 / 12 = 212 exactly, so E ≡ 212 (mod p)? Let me recheck.
 *
 * Actually, modular division: 2544 * inv(12) mod p.
 * inv(12) = pow(12, p-2, p) = 83333334.
 * 2544 * 83333334 mod (10^9 + 7).
 */

const long long MOD = 1e9 + 7;

long long power(long long base, long long exp, long long mod) {
    long long result = 1;
    base %= mod;
    while (exp > 0) {
        if (exp & 1) result = result * base % mod;
        base = base * base % mod;
        exp >>= 1;
    }
    return result;
}

int main() {
    long long n = 1000000000000000000LL; // 10^18
    long long p = MOD;

    long long a = (n - 1) % p;
    long long b = (n + 4) % p;
    long long inv12 = power(12, p - 2, p);
    long long ans = a % p * (b % p) % p * inv12 % p;

    cout << ans << endl;

    // Verification for small cases
    // E(5) = (4)(9)/12 = 3
    long long e5 = 4LL * 9 % p * inv12 % p;
    assert(e5 == 3);

    // E(4) = (3)(8)/12 = 2
    long long e4 = 3LL * 8 % p * inv12 % p;
    assert(e4 == 2);

    return 0;
}

"""
Problem 524: First Sort II

E(n) = (n-1)(n+4)/12, the expected number of insertion steps
to sort a random permutation using the "first sort" algorithm.

Compute E(10^18) mod (10^9 + 7).
"""

from fractions import Fraction
from itertools import permutations

MOD = 10**9 + 7

# --- Method 1: Closed-form modular computation ---
def solve_closed_form(n: int, mod: int):
    """Compute E(n) = (n-1)(n+4)/12 mod p."""
    a = (n - 1) % mod
    b = (n + 4) % mod
    inv12 = pow(12, mod - 2, mod)
    return a * b % mod * inv12 % mod

# --- Method 2: Exact rational computation (verification) ---
def E_exact(n: int) -> Fraction:
    """Compute E(n) exactly as a fraction."""
    return Fraction(n - 1, 1) * Fraction(n + 4, 1) / 12

# --- Method 3: Monte Carlo simulation (verification) ---
def E_simulate(n: int, trials: int = 100000) -> float:
    """Estimate E(n) by simulating the sorting algorithm."""
    import random
    total_steps = 0
    for _ in range(trials):
        perm = list(range(1, n + 1))
        random.shuffle(perm)
        steps = 0
        while True:
            # Find first descent
            found = -1
            for i in range(1, len(perm)):
                if perm[i] < perm[i - 1]:
                    found = i
                    break
            if found == -1:
                break
            # Remove and reinsert
            val = perm.pop(found)
            # Binary search for correct position in sorted prefix
            lo, hi = 0, found
            while lo < hi:
                mid = (lo + hi) // 2
                if perm[mid] < val:
                    lo = mid + 1
                else:
                    hi = mid
            perm.insert(lo, val)
            steps += 1
        total_steps += steps
    return total_steps / trials

# --- Verify small cases ---
for n in range(1, 10):
    exact = E_exact(n)
    formula = Fraction(n - 1, 1) * Fraction(n + 4, 1) / 12
    assert exact == formula, f"E({n}) mismatch"

assert E_exact(1) == 0
assert E_exact(2) == Fraction(1, 2)
assert E_exact(4) == 2
assert E_exact(5) == 3

# Verify modular computation
for n in [1, 2, 5, 10, 100, 10**6]:
    exact_mod = int(E_exact(n)) % MOD if E_exact(n).denominator == 1 else \
                (E_exact(n).numerator * pow(E_exact(n).denominator, MOD - 2, MOD)) % MOD
    computed = solve_closed_form(n, MOD)
    assert exact_mod == computed, f"Mod mismatch at n={n}"

# --- Compute the answer ---
ans = solve_closed_form(10**18, MOD)
print(ans)