1. Abridged Problem Statement  
Given a regular N-gon with vertices numbered 1…N in clockwise order, you know the 2D coordinates of two distinct vertices indices N1 and N2. Compute and print the coordinates of all N vertices in index order, to 6 decimal places.

2. Detailed Editorial  
We have a regular N-gon inscribed in a circle. Let its center be C and radius R. Two known vertices A and B correspond to indices a and b (1≤a,b≤N). The central angle between A and B along the circumcircle is  
  Δ = 2π·|b−a|/N.  
The straight‐line distance d = |A−B| is the chord length for that central angle. By the chord–radius relation,  
  d = 2R·sin(Δ/2) ⇒ R = d / (2·sin(Δ/2)).  

To find the center C, observe that C lies on the perpendicular bisector of segment AB, at distance h from the midpoint M of AB, where  
  h = R·cos(Δ/2) = (d/2)/tan(Δ/2) .  
Compute M = (A+B)/2 and a perpendicular direction to vector AB, e.g. rotate AB by +90°: (dx,dy)→(dy,−dx). Scale that unit-perpendicular by h/(|AB|/2) = d/(2·tan(Δ/2)) divided by (d) to get h. Adding M plus or minus that gives two possible centers; choose the one that yields consistent clockwise ordering (or simply use a formula that produces the correct orientation, as below).

Once C and R are known, compute the polar angle φ of point A relative to C:  
  φ = atan2(A.y−C.y, A.x−C.x).  
Then the i-th vertex has angle φ_i = φ + (i−a)*(+2π/N) if indices increase in the positive (counterclockwise) direction, or φ + (a−i)*(2π/N) for clockwise indexing. Here vertices are numbered clockwise, so we use φ + 2π·(a−i)/N.

Finally,  
  P_i = C + R·(cos φ_i, sin φ_i).  

All computations take O(N), trivially small for N≤150.

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

// We use complex<long double> to represent 2D points and do vector ops.
using Point = complex<long double>;

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

    int n, a, b;
    // Read N (number of vertices), and the two known indices a, b.
    cin >> n >> a >> b;

    vector<Point> points(n + 1);
    long double x, y;

    // Read coordinates of vertex a
    cin >> x >> y;
    points[a] = Point(x, y);

    // Read coordinates of vertex b
    cin >> x >> y;
    points[b] = Point(x, y);

    // Ensure a < b for convenience
    if (a > b) {
        swap(a, b);
    }

    const long double PI = acosl(-1.0L);

    // Distance between the known vertices (chord length)
    long double dist = abs(points[b] - points[a]);

    // Half the central angle between a and b: Δ/2 = π*(b-a)/n
    long double halfAngle = PI * (b - a) / n;

    // Radius from chord formula: d = 2R sin(Δ/2)
    long double radius = dist / (2.0L * sinl(halfAngle));

    // Midpoint of AB
    Point mid = (points[a] + points[b]) / 2.0L;

    // Distance from midpoint to center: h = (d/2)/tan(Δ/2)
    long double h = (dist / 2.0L) / tanl(halfAngle);

    // Vector from A to B
    long double dx = points[b].real() - points[a].real();
    long double dy = points[b].imag() - points[a].imag();

    // Compute center by moving from midpoint in the perpendicular direction (dy, -dx),
    // scaled so that its length is h. Note (dy, -dx) has length = dist.
    Point center = mid + Point(dy * (h / dist), -dx * (h / dist));

    // Compute the initial angle φ of point a relative to center
    long double phi = atan2l(points[a].imag() - center.imag(),
                              points[a].real() - center.real());

    // Generate all other points by stepping angles clockwise: +2π*(a-i)/n
    cout << fixed << setprecision(6);
    for (int i = 1; i <= n; ++i) {
        long double angle;
        if (i == a) {
            // already known
            cout << points[a].real() << " " << points[a].imag() << "\n";
        } else if (i == b) {
            // already known
            cout << points[b].real() << " " << points[b].imag() << "\n";
        } else {
            // Compute the new angle for index i
            angle = phi + 2.0L * PI * (a - i) / n;
            // Place the point on the circle
            long double px = center.real() + radius * cosl(angle);
            long double py = center.imag() + radius * sinl(angle);
            cout << px << " " << py << "\n";
        }
    }

    return 0;
}
```

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

def main():
    data = sys.stdin.read().strip().split()
    n, a, b = map(int, data[:3])
    xa, ya = map(float, data[3:5])
    xb, yb = map(float, data[5:7])

    # Ensure a < b by swapping if needed
    if a > b:
        a, b = b, a
        xa, xb = xb, xa
        ya, yb = yb, ya

    # Convert to complex for convenience
    A = complex(xa, ya)
    B = complex(xb, yb)

    # Central half-angle between A and B
    PI = math.pi
    half_angle = PI * (b - a) / n

    # Chord length
    d = abs(B - A)

    # Radius from chord relation d = 2R sin(half_angle)
    R = d / (2 * math.sin(half_angle))

    # Midpoint of AB
    M = (A + B) / 2

    # Distance from midpoint to center along perpendicular bisector
    h = (d / 2) / math.tan(half_angle)

    # Perpendicular direction: rotate AB by +90°, i.e. (dx,dy)->(dy,-dx)
    dx = B.real - A.real
    dy = B.imag - A.imag
    # Normalize the perpendicular vector length to h
    perp = complex(dy * (h / d), -dx * (h / d))

    # Choose the correct center (the one that yields clockwise numbering).
    # We take M + perp, which matches the C++ orientation.
    C = M + perp

    # Base angle φ for vertex a
    phi = math.atan2(A.imag - C.imag, A.real - C.real)

    # Precompute step for each index shift (clockwise)
    step = 2 * PI / n

    # Generate and print all vertices
    out = []
    for i in range(1, n + 1):
        if i == a:
            P = A
        elif i == b:
            P = B
        else:
            # For clockwise numbering, angle = φ + step*(a - i)
            ang = phi + step * (a - i)
            P = C + R * complex(math.cos(ang), math.sin(ang))
        out.append(f"{P.real:.6f} {P.imag:.6f}")

    sys.stdout.write("\n".join(out))

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

5. Compressed Editorial  
Compute the circumcircle of the regular N-gon using the given chord AB:  
1. Let Δ = 2π·|b−a|/N, chord length d = |A−B|.  
2. Radius R = d/(2·sin(Δ/2)).  
3. Center C lies on the perpendicular bisector of AB at distance h=(d/2)/tan(Δ/2) from the midpoint.  
4. Find φ = atan2(A.y−C.y, A.x−C.x).  
5. Generate each vertex i by φ_i=φ+2π·(a−i)/N and P_i = C + R·(cosφ_i, sinφ_i).