## 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, <= pi) 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, 2*pi) 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 delta = angle(A[0]) - angle(B[i]), normalized to [0, 2*pi).
  - Rotate all vectors of B by +delta 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 2*pi - delta (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

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

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

int n;
vector<pair<double, double>> stars1, stars2;

void read() {
    cin >> n;
    stars1.resize(n);
    stars2.resize(n);
    cin >> stars1;
    cin >> stars2;
}

double normalize_angle(double angle) {
    if(angle < 0) {
        angle += 2 * PI;
    }

    angle = fmod(angle, 2 * PI);
    return angle;
}

pair<double, double> rotate_vector(pair<double, double> v, double angle) {
    double cos_a = cos(angle);
    double sin_a = sin(angle);
    return {v.first * cos_a - v.second * sin_a,
            v.first * sin_a + v.second * cos_a};
}

vector<pair<double, double>> get_all_vectors(
    const vector<pair<double, double>>& stars
) {
    int m = stars.size();
    double cx = 0, cy = 0;
    for(auto& s: stars) {
        cx += s.first;
        cy += s.second;
    }

    cx /= m;
    cy /= m;

    vector<tuple<double, double, pair<double, double>>> star_data;
    for(int i = 0; i < m; i++) {
        double dx = stars[i].first - cx;
        double dy = stars[i].second - cy;

        double angle;
        if(fabs(dx) < EPS && fabs(dy) < EPS) {
            angle = numeric_limits<double>::infinity();
        } else {
            angle = normalize_angle(atan2(dy, dx));
        }

        double distance = sqrt(dx * dx + dy * dy);
        star_data.push_back({angle, distance, {dx, dy}});
    }

    sort(star_data.begin(), star_data.end(), [](const auto& a, const auto& b) {
        if(get<0>(a) != get<0>(b)) {
            return get<0>(a) < get<0>(b);
        }
        return get<1>(a) < get<1>(b);
    });

    vector<pair<double, double>> vectors;
    for(auto& item: star_data) {
        vectors.push_back(get<2>(item));
    }

    return vectors;
}

bool vectors_match(
    const vector<pair<double, double>>& vectors1,
    const vector<pair<double, double>>& vectors2, double rotation_angle,
    int offset
) {
    int m = vectors1.size();
    vector<pair<double, double>> rotated(m);
    for(int i = 0; i < m; i++) {
        rotated[i] = rotate_vector(vectors2[i], rotation_angle);
    }

    for(int i = 0; i < m; i++) {
        auto v1 = vectors1[i];
        auto v2 = rotated[(i + offset) % m];
        if(fabs(v1.first - v2.first) > EPS ||
           fabs(v1.second - v2.second) > EPS) {
            return false;
        }
    }

    return true;
}

void solve() {
    // Each photo is a set of stars whose centroid is translation invariant, so
    // we represent each star by its vector from the centroid. Within a photo we
    // sort these vectors by (polar angle, length), giving a canonical cyclic
    // ordering that is the same up to rotation.
    //
    // A rotation maps vectors1 onto vectors2 with some cyclic offset. We anchor
    // on vectors1[0] and try matching it against each vectors2[i] of equal
    // length, which fixes a candidate rotation angle = angle(v1) - angle(v2).
    // For each candidate we verify the whole set matches under that rotation
    // and cyclic offset (checking both the angle and its 2*pi complement), then
    // keep the smallest valid angle. The answer lies in [0, pi], so for each
    // accepted rotation we also consider 2*pi - angle.

    if(n == 1) {
        cout << fixed << setprecision(11) << 0.0 << '\n';
        return;
    }

    auto vectors1 = get_all_vectors(stars1);
    auto vectors2 = get_all_vectors(stars2);

    double min_angle = numeric_limits<double>::infinity();

    for(int i = 0; i < n; i++) {
        auto v1 = vectors1[0];
        auto v2 = vectors2[i];

        double v1_len = sqrt(v1.first * v1.first + v1.second * v1.second);
        double v2_len = sqrt(v2.first * v2.first + v2.second * v2.second);

        if(v1_len < EPS || v2_len < EPS) {
            continue;
        }

        if(fabs(v1_len - v2_len) > EPS) {
            continue;
        }

        double v1_angle = atan2(v1.second, v1.first);
        double v2_angle = atan2(v2.second, v2.first);
        double rotation_angle = normalize_angle(v1_angle - v2_angle);

        if(vectors_match(vectors1, vectors2, rotation_angle, i) ||
           vectors_match(vectors1, vectors2, 2 * PI - rotation_angle, i)) {
            min_angle = min(min_angle, rotation_angle);
            min_angle = min(min_angle, 2 * PI - rotation_angle);
        }
    }

    cout << fixed << setprecision(11) << min_angle << '\n';
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

---

## 4. Python Solution

```python
import math

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

def normalize_angle(a):
    """Normalize angle a to [0, 2*pi)."""
    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 <= pi
    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, 2*pi), 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 delta = arg(A[0]) - arg(B[i]) (mod 2*pi).
- Check if rotating B by delta (or 2*pi - delta) and shifting by i aligns all vectors to A.
- Keep the minimum non-negative rotation <= pi.
