1) Abridged problem statement
- You are given N points Mi = (mxi, myi), i = 1..N, which are the midpoints of the polygon edges (possibly self-intersecting). Edge i connects vertex Vi to Vi+1 (with VN+1 = V1).
- Decide if there exists a polygon V1..VN whose edge midpoints are exactly the given Mi. If yes, print YES and one valid sequence of vertices (any one); otherwise print NO.
- Coordinates have at most 3 decimal places; output coordinates must have at least 3 decimal places.

2) Detailed editorial
- Let Vi = (Xi, Yi) denote the unknown vertex i. Given Mi = (mxi, myi), the midpoint relation for each edge i is:
  (Xi + Xi+1)/2 = mxi
  (Yi + Yi+1)/2 = myi
  equivalently Xi + Xi+1 = 2 mxi and Yi + Yi+1 = 2 myi, with indices modulo N.
- The X and Y coordinates are independent. Solve the same 1D system twice (for x and y).

Solving the 1D system:
- Define the recurrence Xi+1 = 2 mxi − Xi for i = 1..N−1. This expresses all Xi in terms of X1.
- The last equation must also hold: XN + X1 = 2 mxN. Substituting the recurrence into this last equation yields two cases:
  - N odd:
    Summing equations with alternating signs:
      (X1 + X2) − (X2 + X3) + (X3 + X4) − ... + (XN + X1) = 2(mx1 − mx2 + ... + mxN)
    The left side telescopes to 2X1, hence X1 = mx1 − mx2 + mx3 − ... + mxN.
    After X1 is known, reconstruct X2..XN by Xi+1 = 2 mxi − Xi. This yields a unique solution.
  - N even:
    The same alternating sum makes the left side 0, so a necessary and sufficient consistency condition is:
      mx1 − mx2 + mx3 − ... − mxN = 0
    (equivalently sum_{i=1}^N (-1)^{i+1} mxi = 0).
    If it fails, no solution exists. If it holds, there are infinitely many solutions; pick any X1 (e.g., X1 = 0), then reconstruct Xi by Xi+1 = 2 mxi − Xi.

- Apply the same logic independently to y-coordinates.
- Overall:
  - For odd N: always exactly one solution in both X and Y; reconstruct directly.
  - For even N: the alternating sums for both X and Y must be zero. If either fails, answer NO. If both pass, choose any starting X1,Y1 (e.g., 0,0) and reconstruct.

- Numeric considerations:
  - Inputs have ≤ 3 decimal digits; doubles are fine. Validate equalities with a small epsilon (e.g., 1e-6). Print with fixed 3 decimals (or more) to satisfy “at least 3 decimals”.

Complexity:
- O(N) time and O(N) memory.

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-6;

int n;
vector<double> mx, my;

void read() {
    cin >> n;
    mx.resize(n);
    my.resize(n);

    for(int i = 0; i < n; i++) {
        cin >> mx[i] >> my[i];
    }
}

vector<double> solve_system(vector<double> mids) {
    int n = mids.size();

    double x1;
    if(n % 2 == 1) {
        x1 = 0;
        for(int i = 0; i < n; i++) {
            if(i % 2 == 0) {
                x1 += mids[i];
            } else {
                x1 -= mids[i];
            }
        }
    } else {
        x1 = 0;
    }

    vector<double> x(n);
    x[0] = x1;
    for(int i = 0; i < n - 1; i++) {
        x[i + 1] = 2 * mids[i] - x[i];
    }

    if(abs(x[n - 1] + x[0] - 2 * mids[n - 1]) > eps) {
        return {};
    }

    return x;
}

void solve() {
    // The two main observations are:
    //   (1) The hint in the problem statement is that we don't care about
    //       edge intersections.
    //   (2) The midpoint of (x1, y1) and (x2, y2) is ((x1+x2)/2,
    //       (y1+y2)/2), and so the X and Y dimensions are independent.
    //
    // Therefore, we can solve the following system:
    //   x1 + x2 = 2 * mx1
    //   x2 + x3 = 2 * mx2
    //     ...
    //   xn + x1 = 2 * mxn
    //
    // (and the analogous one for y).
    //
    // To solve this system, we can notice that we can split into two cases by
    // parity:
    //   (1) N is odd. Then we have either 0 or 1 solutions. If there is a
    //       single solution, we can get x1 and then recover everything. To
    //       get x1, we will get:
    //          2 * x1 = 2 * (mx1 - mx2 + mx3 - mx4 + ... + mxn).
    //          x1 = (mx1 - mx2 + mx3 - mx4 + ... + mxn)
    //   (2) N is even. Then we have either 0 or infinitely many solutions. If
    //   that's the case, we will try x1 = 0.
    //
    // After we have decided on x1, recovering is simple by x{i+1} = 2 * mxi -
    // xi. Now we have the x1...xn candidate. We want to re-validate that this
    // is a solution to this system.

    vector<double> x = solve_system(mx);
    vector<double> y = solve_system(my);

    if(x.empty() || y.empty()) {
        cout << "NO" << endl;
        return;
    }

    cout << "YES" << endl;
    for(int i = 0; i < n; i++) {
        cout << setprecision(3) << fixed << x[i] << " " << y[i] << endl;
    }
}

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 with detailed comments
import sys
from math import isclose

EPS = 1e-6  # floating-point tolerance

def solve_system(mids):
    """
    Solve the 1D system:
      x_i + x_{i+1} = 2 * mids[i], for i=0..n-1 (indices modulo n)
    Returns a list x[0..n-1] if solvable, else [].
    """
    n = len(mids)
    # Choose/derive x1 depending on parity
    if n % 2 == 1:
        # Unique solution for odd n:
        # x1 = m1 - m2 + m3 - ... + mN (using 0-based indices)
        x1 = 0.0
        for i, m in enumerate(mids):
            if i % 2 == 0:
                x1 += m
            else:
                x1 -= m
    else:
        # Even n: either no or infinitely many solutions; pick arbitrary x1
        x1 = 0.0

    # Reconstruct full sequence via recurrence
    x = [0.0] * n
    x[0] = x1
    for i in range(n - 1):
        x[i + 1] = 2.0 * mids[i] - x[i]

    # Validate the last equation x[n-1] + x[0] == 2*mids[n-1]
    if abs((x[-1] + x[0]) - 2.0 * mids[-1]) > EPS:
        return []
    return x

def main():
    data = sys.stdin.read().strip().split()
    it = iter(data)
    n = int(next(it))
    mx = [0.0] * n
    my = [0.0] * n
    for i in range(n):
        mx[i] = float(next(it))
        my[i] = float(next(it))

    X = solve_system(mx)
    Y = solve_system(my)

    if not X or not Y:
        print("NO")
        return

    print("YES")
    for i in range(n):
        # Print with at least 3 decimals; here exactly 3
        print(f"{X[i]:.3f} {Y[i]:.3f}")

if __name__ == "__main__":
    main()

5) Compressed editorial
- Model: For each i, Vi = (Xi, Yi), Mi = (mxi, myi). Midpoint constraints give Xi + Xi+1 = 2 mxi and Yi + Yi+1 = 2 myi (indices modulo N).
- The X and Y systems are independent. Solve 1D: define recurrence Xi+1 = 2 mxi − Xi.
- Odd N: unique solution. From alternating sum: X1 = m1 − m2 + ... + mN. Reconstruct all Xi; same for Yi.
- Even N: consistency condition: m1 − m2 + ... − mN = 0 (for x) and similarly for y. If both hold, infinite solutions; pick any X1,Y1 (e.g., 0) and reconstruct; else NO.
- Validate final equation with epsilon; print any valid solution. Complexity O(N).