<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

446. Rotation Estimation
Time limit per test: 0.75 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard

Mr. Nod is an astrologist and has defined a new constellation. He took two photos of the constellation to foretell a future of his friend. The constellation consists of n stars. The shape of the constellation in these photos are the same, but the angle of them are different because these photos were taken on a different day. He foretells a future by the difference of the angle of them. Your job is to write a program to calculate the difference of the angle of two constellation.
Input
The input is given in the following format:
n
x1,1 y1,1
...
x1,n y1,n
x2,1 y2,1
...
x2,n y2,n
The first line of the input contains a positive integer n (n ≤ 1000). The next n lines contain two real numbers x1,i and y1,i (|x1,i|, |y1,i| ≤ 100), where (x1,i, y1,i) denotes the coordinates of the i-th star of the constellation in the first photo. The next n lines contain two real numbers x2,i and y2,i (|x2,i|, |y2,i| ≤ 100), where (x2,i, y2,i) denotes the coordinates of the i-th star of the constellation in the second photo. Note that the ordering of the stars does not matter for the sameness. It is guaranteed that distance between every pair of stars in each photo is larger than 10-5.
Output
You should print a non-negative real number which is the difference of the angle of the constellation in the first photo and in the second photo. The difference should be in radian, and should not be negative. If there are two or more solutions, you should print the smallest one, i.e. your solution should be an angle between 0 and pi radians, inclusive. The difference may be printed with any number of digits after decimal point, provided the absolute error does not exceed 10-7.
Example(s)
sample input
sample output
3
0.0 0.0
1.0 1.0
0.0 1.0
3.0 3.0
2.0 2.0
3.0 2.0
3.14159265359

<|response|>
1. Abridged Problem Statement  
You are given two sets of n points in the plane that represent the same constellation but differ by an unknown translation and rotation. Compute the smallest non-negative rotation angle (in radians, up to π) that, after translating both sets to their centroids, aligns one set to the other within an absolute error of 1e-7.

2. Key Observations  
- A rigid motion in the plane can be decomposed into a translation followed by a rotation.  
- Translating both point sets to their centroids makes them centered at the origin, removing translation.  
- A rotation around the origin can be represented by multiplying each centered point by a complex exponential or by applying a 2×2 rotation matrix.  
- To match points without knowing correspondence, sort each centered set by polar angle (in [0,2π)) and then by distance from the origin. This imposes a cyclic order around the circle.  
- Once sorted, matching reduces to aligning the first vector of set A with some vector of set B of the same length. The difference of their angles is the candidate rotation.  
- You must also consider the complementary rotation (2π minus that angle) because matching might occur in the “flipped” direction around the circle.  
- Check each candidate by rotating and cyclically shifting B and comparing all points. Keep the minimum rotation in [0,π].

3. Full Solution Approach  
1. Read n and the two point sets P1 and P2.  
2. If n = 1, the rotation is defined to be 0.  
3. Compute the centroids c1 and c2 of P1 and P2.  
4. Subtract centroids: for each point p, form the vector v = p − c.  
5. For each centered vector v, compute its distance r = |v| and its polar angle θ = atan2(v.y, v.x) normalized to [0,2π).  
6. Create arrays of tuples (θ, r, v) for both sets, then sort each array by increasing θ, breaking ties by r. Extract the third element v into arrays A and B of complex numbers (or pairs).  
7. Initialize best_angle = 2π.  
8. For each index i from 0 to n−1:  
   a. If |A[0]| and |B[i]| differ by more than EPS, skip.  
   b. Let α = arg(A[0]) and β = arg(B[i]), both in [0,2π). Compute Δ = normalize(α − β).  
   c. For each candidate rot in {Δ, normalize(2π − Δ)}:  
      i. Rotate every B[j] by rot.  
      ii. Cyclically shift the rotated B so that index i moves to index 0.  
      iii. Compare each resulting vector to A[j] within EPS in both x and y (or real and imaginary).  
      iv. If all match, let cand = min(rot, 2π−rot) and update best_angle = min(best_angle, cand).  
9. If best_angle > π, set best_angle = 2π − best_angle.  
10. Print best_angle with sufficient precision (e.g. 11 decimal places).  

Time complexity is O(n log n) for sorting plus O(n²) in the worst case for alignment checks, acceptable for n≤1000 in optimized code.

4. C++ Implementation with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

const double EPS = 1e-9;
const double PI  = acos(-1.0);

// Normalize angle a into [0, 2π)
double normalize(double a) {
    a = fmod(a, 2*PI);
    if (a < 0) a += 2*PI;
    return a;
}

// Rotate a complex vector v by angle ang
complex<double> rotateVec(const complex<double>& v, double ang) {
    return v * polar(1.0, ang);
}

// Check if rotating B by rot and cyclically shifting by offset
// makes it equal to A elementwise (within EPS)
bool matchAll(const vector<complex<double>>& A,
              const vector<complex<double>>& B,
              double rot, int offset) {
    int n = A.size();
    for (int j = 0; j < n; ++j) {
        complex<double> bv = B[(offset + j) % n];
        complex<double> rv = rotateVec(bv, rot);
        if (fabs(rv.real() - A[j].real()) > EPS ||
            fabs(rv.imag() - A[j].imag()) > EPS)
            return false;
    }
    return true;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;
    vector<complex<double>> P1(n), P2(n);

    // Read first photo
    for (int i = 0; i < n; ++i) {
        double x, y;
        cin >> x >> y;
        P1[i] = {x, y};
    }
    // Read second photo
    for (int i = 0; i < n; ++i) {
        double x, y;
        cin >> x >> y;
        P2[i] = {x, y};
    }

    // Special case: single point ⇒ rotation = 0
    if (n == 1) {
        cout << fixed << setprecision(11) << 0.0 << "\n";
        return 0;
    }

    // Compute centroids
    complex<double> c1{0,0}, c2{0,0};
    for (int i = 0; i < n; ++i) {
        c1 += P1[i];
        c2 += P2[i];
    }
    c1 /= double(n);
    c2 /= double(n);

    // Helper to build sorted (angle, radius, vector) list
    auto build = [&](const vector<complex<double>>& P,
                     const complex<double>& c) {
        vector< tuple<double,double,complex<double>> > data;
        data.reserve(n);
        for (auto& p : P) {
            complex<double> v = p - c;
            double r = abs(v);
            double ang = (r < EPS ? 0.0 : normalize(arg(v)));
            data.emplace_back(ang, r, v);
        }
        sort(data.begin(), data.end(),
             [](auto &a, auto &b){
                 if (get<0>(a) != get<0>(b))
                     return get<0>(a) < get<0>(b);
                 return get<1>(a) < get<1>(b);
             });
        return data;
    };

    auto D1 = build(P1, c1);
    auto D2 = build(P2, c2);

    // Extract only the vectors in sorted order
    vector<complex<double>> A(n), B(n);
    for (int i = 0; i < n; ++i) {
        A[i] = get<2>(D1[i]);
        B[i] = get<2>(D2[i]);
    }

    double best = 2*PI;

    // Try matching A[0] to each B[i]
    for (int i = 0; i < n; ++i) {
        double rA = abs(A[0]);
        double rB = abs(B[i]);
        if (fabs(rA - rB) > EPS) continue;

        double angA = normalize(arg(A[0]));
        double angB = normalize(arg(B[i]));
        double delta = normalize(angA - angB);

        // Test delta and its complement
        for (int t = 0; t < 2; ++t) {
            double rot = (t == 0 ? delta : normalize(2*PI - delta));
            if (matchAll(A, B, rot, i)) {
                double cand = min(rot, 2*PI - rot);
                if (cand < best) best = cand;
            }
        }
    }

    // Ensure result is ≤ π
    if (best > PI) best = 2*PI - best;

    cout << fixed << setprecision(11) << best << "\n";
    return 0;
}
```

5. Python Implementation with Detailed Comments  
```python
import math
import sys
input = sys.stdin.readline

EPS = 1e-9
PI  = math.pi

def normalize(a):
    "Normalize angle a into [0, 2π)."
    a %= 2*PI
    if a < 0:
        a += 2*PI
    return a

def rotate(vx, vy, ang):
    "Rotate vector (vx,vy) by angle ang."
    c = math.cos(ang)
    s = math.sin(ang)
    return (vx*c - vy*s, vx*s + vy*c)

def build(points):
    "Center by centroid, compute (angle, radius, vector), sort, return list of vectors."
    n = len(points)
    cx = sum(x for x,y in points) / n
    cy = sum(y for x,y in points) / n
    data = []
    for x,y in points:
        dx, dy = x - cx, y - cy
        r = math.hypot(dx, dy)
        ang = 0.0 if r < EPS else normalize(math.atan2(dy, dx))
        data.append((ang, r, (dx, dy)))
    data.sort(key=lambda t: (t[0], t[1]))
    return [v for _,_,v in data]

def match_all(A, B, rot, offset):
    "Check if rotating B by rot and shifting by offset matches A."
    n = len(A)
    for j in range(n):
        bx, by = B[(offset + j) % n]
        rx, ry = rotate(bx, by, rot)
        ax, ay = A[j]
        if abs(rx - ax) > EPS or abs(ry - ay) > EPS:
            return False
    return True

def main():
    n = int(input())
    P1 = [tuple(map(float, input().split())) for _ in range(n)]
    P2 = [tuple(map(float, input().split())) for _ in range(n)]

    if n == 1:
        print("0.00000000000")
        return

    A = build(P1)
    B = build(P2)

    best = 2*PI

    # Align A[0] with each B[i]
    for i in range(n):
        ax, ay = A[0]
        bx, by = B[i]
        rA = math.hypot(ax, ay)
        rB = math.hypot(bx, by)
        if abs(rA - rB) > EPS:
            continue
        angA = normalize(math.atan2(ay, ax))
        angB = normalize(math.atan2(by, bx))
        delta = normalize(angA - angB)

        # Test both delta and its complement
        for rot in (delta, normalize(2*PI - delta)):
            if match_all(A, B, rot, i):
                cand = min(rot, 2*PI - rot)
                if cand < best:
                    best = cand

    # Ensure answer is in [0, π]
    if best > PI:
        best = 2*PI - best

    print(f"{best:.11f}")

if __name__ == "__main__":
    main()
```