1. Abridged Problem Statement  
Given two sets of n points in the plane representing the same constellation, up to a rigid translation and rotation, find the smallest non-negative rotation angle (in radians, ≤ π) that, after translating both sets to their centroids, aligns one set to the other.

2. Detailed Editorial  
We need to recover the relative rotation between two point‐clouds that are identical up to translation and rotation. The high-level steps are:

 1. Translation Invariance via Centroids  
    – Compute the centroid (average of x’s, average of y’s) of each point set.  
    – Subtract its centroid from each point to recenter each cloud at the origin.  

 2. Encoding Each Star by Polar Coordinates  
    – For each centered point, compute its polar angle in [0,2π) and its distance from the origin.  
    – Sort the points by (angle, distance).  
      This produces two sorted sequences of 2D vectors, `A` and `B`, that correspond “in order” around the circle.

 3. Trying All Alignments  
    – Imagine that vector `A[0]` must match some `B[i]` after rotation.  
    – For each i=0…n−1:  
       • Check that the lengths |A[0]| and |B[i]| agree (within EPS) or else skip.  
       • Let Δ = angle(A[0]) − angle(B[i]), normalized to [0,2π).  
       • Rotate all vectors of B by +Δ and cyclically shift so that B[i] aligns with A[0].  
       • Check elementwise that A[j] ≈ rotated_B[(i+j)%n] for all j.  
       • Also check the “flip” rotation 2π−Δ (sometimes the smallest positive rotation is the complement).  
       • Keep the minimum valid rotation.

 4. Edge Cases  
    – n=1: Any single point can only differ by translation; the rotation is defined as 0.  
    – Guarantee: No two stars coincide, so angles and distances are well-defined.

Complexity: O(n log n) for sorting + O(n^2) checks in the worst case, which is acceptable up to n=1000 in optimized C++.

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

// Epsilon for floating-point comparisons
const double EPS = 1e-9;
const double PI  = acos(-1.0);

// Normalize an angle to the range [0, 2*PI)
double normalize_angle(double a) {
    // fmod might return negative
    a = fmod(a, 2*PI);
    if (a < 0) a += 2*PI;
    return a;
}

// Rotate a complex number (vector) by angle 'ang'
complex<double> rotate_vec(const complex<double>& v, double ang) {
    // exp(i*ang) = cos(angle) + i*sin(angle)
    return v * polar(1.0, ang);
}

// Attempt to match A against B after rotating B by 'rot'
// with a cyclic shift of 'offset'. Returns true if all match.
bool match_with_rotation(
    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) {
        // rotate B[ (offset + j) % n ]
        complex<double> bv = rotate_vec(B[(offset+j) % n], rot);
        // Compare real and imag parts to A[j]
        if (fabs(bv.real() - A[j].real()) > EPS ||
            fabs(bv.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 constellation
    for(int i = 0; i < n; i++){
        double x, y;
        cin >> x >> y;
        P1[i] = complex<double>(x, y);
    }
    // Read second constellation
    for(int i = 0; i < n; i++){
        double x, y;
        cin >> x >> y;
        P2[i] = complex<double>(x, y);
    }

    // If there's only one star, rotation is 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);

    // Build (angle, distance, vector) tuples for each point set
    auto build_data = [&](const vector<complex<double>>& P, complex<double> cent){
        vector< tuple<double,double,complex<double>> > data;
        for(auto& pt : P) {
            complex<double> v = pt - cent; 
            double dist = abs(v);
            double ang  = (dist < EPS) ? 0.0 : normalize_angle(arg(v));
            data.emplace_back(ang, dist, v);
        }
        // Sort by (angle asc, distance asc)
        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_data(P1, c1);
    auto D2 = build_data(P2, c2);

    // Extract only the sorted vectors
    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;  // we will minimize over [0, 2π)

    // Try aligning A[0] with each B[i]
    for(int i = 0; i < n; i++){
        double lenA = abs(A[0]);
        double lenB = abs(B[i]);
        // lengths must match
        if (fabs(lenA - lenB) > EPS) continue;
        // compute candidate rotation Δ = angle(A[0]) - angle(B[i])
        double angA = normalize_angle(arg(A[0]));
        double angB = normalize_angle(arg(B[i]));
        double delta = normalize_angle(angA - angB);

        // Test both Δ and its complement 2π-Δ
        for(int t = 0; t < 2; ++t) {
            double rot = (t == 0 ? delta : normalize_angle(2*PI - delta));
            // As we sorted, B[i] was at index i; after rotation we shift by -i
            if (match_with_rotation(A, B, rot, i)) {
                double cand = min(rot, 2*PI - rot);
                if (cand < best) best = cand;
            }
        }
    }

    // Output result in [0, π]
    if (best > PI) best = 2*PI - best;
    cout << fixed << setprecision(11) << best << "\n";
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
import math

# Tolerance for floating-point comparisons
EPS = 1e-9

def normalize_angle(a):
    """Normalize angle a to [0, 2π)."""
    a %= 2*math.pi
    if a < 0:
        a += 2*math.pi
    return a

def rotate_vector(v, ang):
    """Rotate 2D vector v = (x,y) by angle ang."""
    c, s = math.cos(ang), math.sin(ang)
    return (v[0]*c - v[1]*s, v[0]*s + v[1]*c)

def get_sorted_vectors(points):
    """
    Center points at their centroid, compute (angle, dist, vector),
    sort by (angle, dist), and return the list of vectors only.
    """
    n = len(points)
    # Find centroid
    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
        dist = math.hypot(dx, dy)
        # If at centroid (should not happen if n>1), define angle=0
        if dist < EPS:
            ang = 0.0
        else:
            ang = normalize_angle(math.atan2(dy, dx))
        data.append((ang, dist, (dx, dy)))

    # Sort by angle, then by distance
    data.sort(key=lambda t: (t[0], t[1]))
    # Return just the shifted vectors in sorted order
    return [vec for _,_,vec in data]

def match_after_rotation(A, B, rot, offset):
    """
    Check if rotating B by rot and cyclically shifting by offset
    makes it coincide with A elementwise.
    """
    n = len(A)
    for i in range(n):
        bx, by = B[(offset+i) % n]
        rx, ry = rotate_vector((bx, by), rot)
        ax, ay = A[i]
        if abs(rx - ax) > EPS or abs(ry - ay) > EPS:
            return False
    return True

def solve():
    n = int(input())
    pts1 = [tuple(map(float, input().split())) for _ in range(n)]
    pts2 = [tuple(map(float, input().split())) for _ in range(n)]

    # Single point ⇒ rotation = 0
    if n == 1:
        print("0.00000000000")
        return

    A = get_sorted_vectors(pts1)
    B = get_sorted_vectors(pts2)

    best = 2*math.pi

    # Try matching A[0] to each B[i]
    for i in range(n):
        ax, ay = A[0]
        bx, by = B[i]
        lenA = math.hypot(ax, ay)
        lenB = math.hypot(bx, by)
        # lengths must agree
        if abs(lenA - lenB) > EPS:
            continue

        # compute the rotation that aligns B[i]→A[0]
        angA = math.atan2(ay, ax)
        angB = math.atan2(by, bx)
        delta = normalize_angle(angA - angB)

        # Check both delta and its complement
        for rot in (delta, normalize_angle(2*math.pi - delta)):
            if match_after_rotation(A, B, rot, i):
                cand = min(rot, 2*math.pi - rot)
                best = min(best, cand)

    # Ensure output ≤ π
    if best > math.pi:
        best = 2*math.pi - best

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

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

5. Compressed Editorial  
- Translate both clouds to their centroids for translation invariance.  
- Convert each centered point to (angle in [0,2π), distance).  
- Sort by angle then distance to get cyclically ordered vectors A, B.  
- For each possible pairing A[0]↔B[i] with equal length, compute candidate rotation Δ = arg(A[0])−arg(B[i]) (mod 2π).  
- Check if rotating B by Δ (or 2π−Δ) and shifting by i aligns all vectors to A.  
- Keep the minimum non-negative rotation ≤ π.