qwen3s_existential_hexagon

Okay, I need to create a Python program that shows 20 bouncing balls inside a spinning heptagon using the specified requirements. Let me think about how to approach this step by step.

First, the libraries allowed are tkinter, math, numpy, dataclasses, typing, sys. So no pygame, which is good. I'll have to use tkinter for the GUI and handle all the drawing and animation manually.

The heptagon is a seven-sided polygon. It's spinning around its center at 360 degrees every 5 seconds. The balls start from the center, drop under gravity, bounce off the walls, each other, and their spin affects their movement. Also, their bounce height shouldn't exceed the heptagon's radius but more than their own.

Let me break down the components:

Heptagon Structure:

A regular heptagon (7 sides). Need to calculate its vertices.
The heptagon is spinning. So each frame, I need to rotate all its vertices by a certain angle based on time.
Balls:

Each has position, velocity, radius, color, number, spin (angular velocity).
When starting, they drop from the center. So initial positions are at the heptagon's center.
Gravity affects them. Friction in movement and collisions.
Bounce off walls (heptagon) and other balls.
Collision Detection:

Between balls: check distance between centers.
Between balls and walls: need to detect when a ball is outside the heptagon and correct its position, then apply reflection.
Rotation of Balls:

Each ball has spin (angular velocity). The numbers on them should rotate with this. So when drawing, each ball's number needs to be rotated based on its spin.
Animation Loop:

Using tkinter's after() method for animation.
Update positions, check collisions, apply physics, and redraw.
Let me start by planning the data structures.

First, the heptagon. Let's define it as a polygon with 7 vertices. The center is at (cx, cy), radius R. The rotation angle changes over time.

For each frame, I need to calculate the current rotation of the heptagon. Since it spins 360 degrees every 5 seconds, angular velocity is 2π/(5) radians per second. So for each frame, delta_time is the time since last update (in seconds), and add that to the total rotation angle.

But how to represent the heptagon's vertices? Maybe as a list of points. Each vertex can be calculated using trigonometric functions based on their angle around the center.

For example, for a regular heptagon with radius R, each vertex is at an angle of 2π * i /7 + rotation_angle, where i ranges from 0 to 6.

But wait, when the heptagon spins, all vertices rotate. So during drawing, I need to apply the current rotation to each vertex.

Now for the balls. Each ball has:

Position (x, y)
Velocity (vx, vy)
Acceleration (due to gravity)
Radius r
Color from the given list
Number (1-20)
Angular velocity (spin) for rotation
But since they are affected by friction and collisions, their velocities will change over time.

Gravity: a constant downward acceleration. But since it's inside a spinning heptagon, maybe gravity is in the direction of the center? Or is it standard gravity (downwards)? The problem says "drop from the heptagon center when starting", so I think gravity is acting downwards, but the heptagon is rotating. So the balls are subject to gravity and collisions with walls.

But how does the spinning affect the physics? The walls are moving, so collisions need to account for the wall's velocity at the point of contact.

Wait, this complicates things. Because when a ball collides with a rotating wall, the collision response needs to consider the wall's movement. However, since the heptagon is a rigid body spinning around its center, each edge (side) is moving in a circular path. So during collision detection between a ball and the heptagon, I need to check if the ball is outside the polygon and then compute the reflection based on the normal of the wall at that point.

But how to model this? Maybe for each frame, after updating the balls' positions, check if any ball is outside the heptagon. If so, find the closest edge (side) of the heptagon and compute the collision response.

Alternatively, since the heptagon is a convex polygon, we can use the separating axis theorem (SAT) to detect collisions between the ball and the polygon. But that might be complex.

But given that the balls are small compared to the heptagon's size, perhaps for each ball, check if it's inside the heptagon. If not, find the closest point on the heptagon's perimeter and adjust position and velocity accordingly.

Alternatively, since the heptagon is rotating, maybe we can model the walls as moving lines (edges) that are rotating. So during collision detection, for each ball, check if it intersects with any of the edges of the heptagon. But this would be computationally intensive for 20 balls and multiple edges.

But given that there are only seven edges, maybe manageable.

But first, how to represent the heptagon's edges. Each edge is a line segment between two consecutive vertices (after rotation). So during each frame, after rotating the vertices, we can generate these edges.

Then, for each ball, check if it's inside the polygon. If not, find which edge it's closest to and compute collision response.

But how to determine whether a point is inside a convex polygon? The standard method is to check on which side of each edge the point lies. For a convex polygon, if the point is on the same side (inside) for all edges, then it's inside.

Alternatively, using the winding number algorithm.

But perhaps for simplicity, since the heptagon is convex and we can compute the distance from the ball to each edge, but that might not be straightforward.

Another approach: For a given ball position (x,y), check if it is inside the heptagon. If not, find the closest point on the polygon's perimeter and adjust the ball's position and velocity accordingly.

But how to do this?

Alternatively, for each frame, after moving the balls, check if any ball is outside the heptagon. For those that are, compute the collision with the nearest edge, then reflect their velocity based on the normal of that edge.

But this requires handling multiple steps:

Update positions of all balls (including gravity and friction).
Check for collisions between balls.
Check for collisions between balls and walls.
Apply necessary reflections and updates.
But collision detection with walls is more complex because the walls are moving (spinning). So when a ball collides with an edge, the edge's velocity must be considered in the collision response.

This seems complicated. Maybe I can model the heptagon as a static polygon during collision checks, but since it's rotating, perhaps this isn't accurate. However, if the rotation is slow enough (360 degrees every 5 seconds), maybe approximating the walls as moving with their current velocity at the moment of collision would be acceptable.

Alternatively, for each ball, when checking collisions with the heptagon, compute the position of the heptagon's edges at that time. But since the heptagon is rotating continuously, this would require knowing the exact rotation angle during the frame.

So here's a plan:

Each frame, calculate the current rotation angle of the heptagon.
Generate all the vertices and edges based on that angle.
For each ball, check if it's inside the polygon. If not, find the closest edge and compute collision response.
But how to determine if a point is inside a polygon? Let me think about using the cross product method for convex polygons.

For a convex polygon, a point is inside if it lies on the same side of all edges (relative to the polygon's interior). For each edge, we can check which side the point is on. If any edge has the point outside, then the point is outside the polygon.

So for each edge (defined by two points A and B), compute the cross product between vector AB and vector AP (where P is the ball's position). The sign of this cross product tells us which side of the line AB the point P is on. For a convex polygon, all edges should have the same direction (e.g., all left turns if the polygon is ordered counter-clockwise).

So for each edge in the heptagon:

Compute vector AB = B - A
Vector AP = P - A
Cross product AB × AP = (Bx - Ax)(Py - Ay) - (By - Ay)(Px - Ax)
If this value is positive, the point is on one side; if negative, on the other. For a convex polygon ordered counter-clockwise, all edges should have cross products >= 0 for points inside.
If any edge has a negative cross product, then the ball is outside.

Once we determine that a ball is outside, we need to find which edge it's closest to and compute the collision response.

But how to find the closest edge? For each edge, calculate the distance from the ball to the line segment (edge). The edge with the minimal distance is the one to consider. However, since the ball may be outside the polygon, perhaps the closest point on the polygon's perimeter is along that edge.

Once we have the closest edge and the point of collision, we can compute the normal vector of that edge (perpendicular to the edge), then reflect the ball's velocity accordingly. Also, apply friction.

But this requires some math.

Let me think about how to calculate the reflection when a ball collides with a line segment (edge). The standard approach for elastic collision is:

Compute the normal vector of the surface at the point of contact.
Reflect the velocity vector over that normal.
However, since the edge is part of a rotating heptagon, during the collision, the edge itself has a velocity. So the collision response must account for both the ball's velocity and the wall's velocity.

This complicates things because it's not just a simple reflection but a dynamic collision between two moving objects (the ball and the wall).

But how to model this? For simplicity, perhaps assume that during collision, the heptagon is treated as a static object. But since it's rotating, this would be an approximation.

Alternatively, compute the velocity of the edge at the point of contact. The edge is part of the spinning heptagon. So for each vertex, its velocity can be calculated based on the angular velocity and distance from the center.

But the edge itself is a line segment between two vertices. However, during collision with a ball, perhaps the velocity of the wall at the point of contact (the closest point on the edge) is considered.

This seems complex, but maybe manageable.

Let me outline steps for handling collisions between balls and walls:

For each ball that is outside the heptagon:

Find the closest point on the heptagon's perimeter to the ball's position.
Compute the normal vector at that point (perpendicular to the edge).
Calculate the velocity of the wall at that point (since the heptagon is spinning).
Apply a collision response between the ball and the moving wall.
But how to compute the wall's velocity at the contact point?

The heptagon is rotating around its center with angular velocity ω (radians per second). The velocity of any point on the heptagon is given by v = ω × r, where r is the vector from the center to that point.

So for a point P on the edge of the heptagon, its velocity is perpendicular to the line from the center to P. The magnitude is ω * |r|, and direction depends on the rotation direction (clockwise or counter-clockwise).

Assuming the heptagon is rotating clockwise (since it's spinning around its center; but need to confirm direction). But for this problem, perhaps the direction isn't specified, so I'll assume clockwise.

So during collision with a wall edge, the velocity of the wall at that point is v_wall = ω × r, where r is the vector from the heptagon's center to the contact point on the wall.

Once we have both velocities (ball and wall), then the collision response can be calculated as follows:

The relative velocity between ball and wall is v_ball - v_wall. The normal direction is n (the normal of the edge). The component of this relative velocity along n determines whether there's a collision. If it's approaching, we need to compute the reflection.

But I'm not sure about the exact formula here. Maybe for simplicity, treat the wall as a moving surface and adjust the ball's velocity accordingly.

Alternatively, since the heptagon is much larger than the balls, perhaps its movement is negligible compared to the balls' velocities. But that might not be accurate.

This seems too complex. Perhaps I can approximate by considering the wall as static during collision, but this would ignore the rotation. However, given time constraints, maybe proceed with that approach and see if it works.

Alternatively, for each collision between a ball and the heptagon's edge, calculate the velocity of the wall at the contact point, then compute the relative velocity between the ball and the wall, and apply an elastic collision response based on that.

But this requires knowing the normal direction (n) and the relative velocity along n.

The standard formula for collision response is:

v' = v - (1 + e) * (v · n) * n

where e is the coefficient of restitution. But since the wall is moving, maybe we need to adjust it.

Wait, in a collision between two objects, the formula depends on their masses and velocities. However, if one object is much more massive than the other (like the heptagon compared to balls), then its velocity doesn't change, and the ball's velocity is adjusted based on the wall's velocity.

But since the heptagon is a rigid body, perhaps we can model it as having infinite mass. So during collision, the ball's velocity is reflected relative to the wall's velocity.

So for a collision between ball (mass m) and a moving surface with velocity v_wall:

The relative velocity before collision: v_rel = (v_ball - v_wall)

We then compute the component of this relative velocity along the normal n. If it's positive (approaching), we reverse it, considering the coefficient of restitution e.

So new relative velocity after collision is -e * v_rel_component * n

Then, the ball's new velocity is:

v_ball' = v_wall + (-e * v_rel_component * n)

But I'm not sure about this formula. Let me think again.

In a standard elastic collision between two objects with masses m1 and m2, the velocities after collision are determined by conservation of momentum and kinetic energy. But when one object is much more massive (like the wall), its velocity doesn't change. So for a ball colliding with a moving wall:

The relative velocity along the normal direction is reversed with coefficient e.

So if the wall's velocity is v_wall, then after collision, the ball's velocity is adjusted such that:

(v_ball' - v_wall) · n = -e * (v_ball - v_wall) · n

This equation ensures that the component of relative velocity along the normal is reversed by factor e.

Solving for v_ball':

v_ball' = v_wall + [ (v_ball - v_wall) - (1 + e)( (v_ball - v_wall) · n ) * n ]

Wait, no. Let me rearrange:

Let’s denote u = v_ball - v_wall (relative velocity)

After collision, the new relative velocity is -e*u_n, where u_n is the component of u along n.

So:

(v_ball' - v_wall) = -e * (u · n) * n

Therefore,

v_ball' = v_wall - e*(u · n)*n

But u = v_ball - v_wall → substituting:

v_ball' = v_wall - e*( (v_ball - v_wall) · n ) * n

This seems correct.

So for each collision between the ball and the wall, we compute this new velocity.

Now, considering that the heptagon's wall is moving with velocity v_wall at the contact point, which depends on its position relative to the center.

But how to calculate v_wall?

The heptagon is rotating around its center. So for a point P (the contact point on the edge), the velocity is given by:

v_wall = ω × r

Where ω is the angular velocity vector (perpendicular to the plane, direction determined by rotation), and r is the position vector of P relative to the center.

Assuming 2D motion, we can represent this as a 2D cross product. For example, if the heptagon rotates clockwise, then for a point at (x,y) from the center, its velocity would be (-ωy, ωx), assuming angular velocity is in radians per second and clockwise rotation.

Wait, let me think about that. The standard formula for tangential velocity is v = ω × r. In 2D, if the angular velocity vector is pointing out of the screen (counter-clockwise), then for a point at position (x,y) from the center, the velocity would be (-ωy, ωx). If it's clockwise, then the direction would be opposite: (ωy, -ωx).

But I need to confirm.

Let me take an example. Suppose the heptagon is rotating counter-clockwise with angular velocity ω. For a point at (r, 0) relative to the center, its velocity should be upward (positive y-direction). Using the cross product formula:

v = ω × r → in 2D, this would be (ω * r_y - 0 * r_x, ...)? Wait, no.

Actually, in 2D, the cross product of a vector (a, b) with another vector (c, d) is scalar magnitude ad - bc. But for angular velocity, it's more about rotation direction.

Alternatively, if we have angular velocity ω (in radians per second), then the tangential velocity at a point P is v = ω × r, where r is the position vector from the center to P. In 2D, this would be:

v_x = -ω * y
v_y = +ω * x

for counter-clockwise rotation (assuming standard coordinate system with positive y upwards). Wait, no. Let me think again.

If you have a point rotating around the origin in counter-clockwise direction, its velocity vector is perpendicular to the position vector. For example, if P is at (r, 0), then velocity is upward (0, ωr). If P is at (0, r), velocity is left (-ωr, 0). So for a general point (x,y), the velocity would be (-ω y, ω x) for counter-clockwise rotation.

Yes. Because:

The position vector is (x, y).

The angular velocity vector is in the z-direction: (0, 0, ω).

Cross product of angular velocity and position gives:

(0, 0, ω) × (x, y, 0) = (ωy, -ωx, 0). Wait no. The cross product formula for vectors (a1,a2,a3) × (b1,b2,b3) is (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1).

So (0, 0, ω) × (x, y, 0) = (00 - ωy, ωx - 00, 0y - 0x ) → (-ω y, ω x, 0). So the velocity vector is (-ω y, ω x).

So for counter-clockwise rotation, it's (-ω y, ω x). For clockwise, it would be (ω y, -ω x).

But in our case, the heptagon is spinning around its center. The problem says "the speed of spinning is 360 degrees per 5 seconds". So angular velocity ω = 2π / 5 radians per second.

Assuming that the rotation direction is clockwise (since it's not specified, but maybe arbitrary). But for the code, I can choose a direction and stick with it. Let's assume clockwise rotation.

So for a point P at position (x,y) relative to the center of the heptagon, its velocity due to rotation would be:

v_x = ω * y
v_y = -ω * x

Because for clockwise rotation, the cross product would be (ω y, -ω x).

But I need to confirm. Let me take a point at (1,0). For counter-clockwise rotation, its velocity is upwards (0, ω*1), but for clockwise it's downwards (0, -ω * 1). So in the formula above, if we have:

For clockwise: v = (ω y, -ω x) ?

Wait, let me re-derive. The angular velocity vector for clockwise rotation would be (0, 0, -ω), since positive z is counter-clockwise.

So cross product of (-ω) * k with position vector r = xi + yj:

(-ω k ) × (x i + y j ) = -ω [k × x i + k × y j ]

k × i = j
k × j = -i

So this becomes -ω [x j - y i ] = ω y i - ω x j → velocity vector is (ω y, -ω x). Yes.

So for clockwise rotation, the velocity of a point P relative to the center is v = (ω * y, -ω * x).

Therefore, during collision with the wall, we can compute this velocity for the contact point on the edge.

Now, putting it all together:

For each ball that is outside the heptagon:

Find the closest point on the heptagon's perimeter to the ball's position.
Compute the normal vector of the edge at that point (perpendicular to the edge).
Compute the velocity of the wall at that contact point (using ω and the position relative to the center).
Calculate the new velocity of the ball based on the collision with the moving wall.
But how to find the closest point on the heptagon's perimeter?

This is non-trivial. For a convex polygon, the closest point to an external point can be found by checking each edge and finding the closest point on that edge to the ball's position. The minimal distance among all edges gives the closest point.

For each edge (defined by points A and B), compute the closest point P on the line segment AB to the ball's position (x,y). Then, check if this point is within the segment. If not, take the nearest endpoint.

Once we have all these possible closest points, select the one with minimal distance.

But for each edge, how to calculate that?

The standard method for finding the closest point on a line segment AB to a point P:

Compute vector AB = B - A
Compute vector AP = P - A
Project AP onto AB: t = (AP · AB) / |AB|²
Clamp t between 0 and 1 to get the closest point on the line segment.
Once we have that, compute the distance from P to this closest point. The edge with minimal distance is our candidate.

So for each ball outside the heptagon:

Loop through all edges of the heptagon, calculate the closest point on each edge to the ball's position. Find the edge with the smallest distance. Then proceed with that edge for collision response.

Once we have the contact point and normal vector (perpendicular to the edge), compute the wall velocity as described earlier.

Now, implementing this in code.

But first, let me think about data structures.

The heptagon can be represented by its center coordinates (cx, cy), radius R, current rotation angle theta. The vertices are calculated each frame based on theta.

For example:

vertices = [ ( cx + R * cos(theta + 2πi/7 ), cy + R * sin(theta + 2πi/7 )) for i in 0..6 ]

But since the heptagon is rotating, during each frame, we need to recompute these vertices based on the current rotation angle.

Then, edges are pairs of consecutive vertices. For example, edge between vertex[i] and vertex[(i+1)%7].

Now, for collision detection with balls:

For a ball at position (x,y), check if it's inside the heptagon. If not, find the closest point on the perimeter as explained.

Once found, compute normal vector of that edge. For example, for an edge from A to B, the normal pointing outward from the polygon is perpendicular to AB. Since the heptagon is convex and ordered counter-clockwise (assuming), the outward normal would be (-dy, dx) where dy = By - Ay, dx = Bx - Ax.

Wait, for a line segment AB, the direction of the normal depends on the order of A and B. For a polygon ordered counter-clockwise, the outward normal would be to the left of the edge vector AB. So if AB is from A to B, then the normal vector pointing outwards is (-dy, dx), where dy = By - Ay, dx = Bx - Ax.

Yes. Because for AB vector (dx, dy), a perpendicular vector is (-dy, dx) which points to the left of AB direction.

So for each edge AB, compute the outward normal as (-dy, dx), then normalize it.

Once we have that normal, proceed with collision response.

Now, code structure:

First, create a Tkinter window. Then, set up a canvas.

Then, define classes for Ball and Heptagon.

But since the user allows dataclasses, perhaps use them.

Let me think about the Ball class:

Each ball has:

position (x, y)
velocity (vx, vy)
acceleration (ax, ay) ? Or maybe just apply gravity as a constant acceleration.
radius r
color
number (1 to 20)
angular_velocity (spin)
But for simplicity, perhaps the acceleration is due to gravity. So during each frame, we add gravity to velocity.

Friction: when moving, the ball's velocity decreases over time. This can be modeled by multiplying velocity by a friction factor each frame.

So for example:

velocity *= (1 - friction * dt), where dt is time step.

But need to decide on parameters like gravity, friction, etc.

Now, for the heptagon:

center (cx, cy)
radius R
rotation angle theta (in radians)
angular velocity ω = 2π / 5 radians per second (since it completes a full rotation in 5 seconds).
But since the rotation is continuous, during each frame, we need to update theta by adding ω * delta_time.

Now, for the animation loop:

In Tkinter, use after() to schedule updates. Each iteration:

Calculate delta time between frames.
Update heptagon's rotation angle.
Update all balls' positions and velocities (considering gravity, friction, collisions with other balls and walls).
Check for collisions between balls and update their velocities accordingly.
Check for collisions between balls and the heptagon walls, and adjust positions and velocities.
Redraw everything.
But how to handle delta time? Because Tkinter's after() doesn't guarantee a fixed frame rate. So need to track the time of each frame and calculate the delta_time between frames.

So in code:

start_time = current time (e.g., using time.time())

Then, for each frame:

current_time = time.time()

delta_time = current_time - start_time

start_time = current_time

But this would accumulate errors over time. Alternatively, track the previous time and calculate delta_time as current - previous.

Now, implementing all that.

But first, let's think about the collision between balls.

For each pair of balls, check if their distance is less than sum of radii. If so, resolve the collision.

This requires checking all pairs (O(n²) complexity). For 20 balls, it's manageable.

The standard approach for elastic collision between two circles:

Compute the normal vector pointing from ball1 to ball2.

Compute the relative velocity along this normal.

Apply the collision response based on the coefficient of restitution and masses. Assuming equal mass (since all balls have same radius, but maybe different masses? The problem says "material determines bounce height", so perhaps they have same mass. So for simplicity, assume same mass.

In that case, the velocities are swapped along the normal direction.

But with friction and other factors, it's more complex. However, for this problem, perhaps proceed with elastic collision (e=1) but considering the balls' spins.

Wait, the problem states that "the material of all the balls determines that their impact bounce height will not exceed the radius of the heptagon, but higher than ball radius." So maybe the coefficient of restitution is such that after collision, the balls don't go beyond certain heights. But this might be handled by gravity and friction.

But for simplicity, proceed with standard elastic collisions between balls.

So during collision response:

For two balls i and j:

Compute vector between centers: dx = xj - xi, dy = yj - yi
distance squared is dx² + dy². If less than (r_i + r_j)^2, then collision.
Compute unit normal vector n = (dx, dy) / distance
Compute relative velocity v_rel = (vj - vi)
Compute the component of v_rel along n: v_rel_dot_n = v_rel · n
If this is positive (balls moving towards each other), compute impulse to reverse it.
For equal masses, the new velocities are: vi' = vi - v_rel_dot_n * n vj' = vj + v_rel_dot_n * n
But since the balls have spin (angular velocity), their rotation affects how they move. However, this part is more complex and requires handling rotational inertia.

But the problem says that "the numbers on the ball can be used to indicate the spin of the ball." So when drawing, each ball's number should rotate with its angular velocity.

So for each ball, during rendering, we need to rotate the text (number) by the angle determined by its angular velocity and time.

But how does the spin affect movement? The problem says "the balls are affected by gravity and friction, and they must bounce off the rotating walls realistically. There should also be collisions between balls." But it's not clear if the spin affects their translational motion directly. However, since the material determines the bounce height, perhaps the coefficient of restitution is adjusted based on the spin.

But this seems complicated. The problem says "the material of all the balls determines that their impact bounce height will not exceed the radius of the heptagon, but higher than ball radius." So maybe after collision, the ball's velocity is adjusted so that it doesn't go beyond certain heights. But how to model that?

Alternatively, perhaps this refers to the maximum possible bounce height due to gravity and the heptagon's geometry, not directly affecting the physics. Maybe just ensure that during collisions, the balls don't have velocities that would cause them to fly out of the heptagon.

But given time constraints, maybe proceed with standard collision response and handle spin only for visual purposes (rotating the numbers).

So for now, focus on translational motion and ignore rotational effects on movement. The spin is just for rendering.

Now, code structure:

Implementing all this in Python with tkinter.

First, import necessary modules:

import tkinter as tk
import math
import numpy as np
from dataclasses import dataclass
from typing import List

Then define the Ball class using dataclass:

@dataclass
class Ball:
x: float
y: float
vx: float
vy: float
radius: float
color: str
number: int
angular_velocity: float  # radians per second

But since we need to update these values, perhaps use a list of objects.

Then the Heptagon class:

@dataclass
class Heptagon:
cx: float
cy: float
radius: float
rotation_angle: float  # in radians
angular_velocity: float  # radians per second

But for the heptagon's vertices, we can compute them each frame.

Now, the main application:

Create a Tkinter window with canvas. Set up initial parameters.

Then, in the animation loop:

Update heptagon rotation angle.
For each ball:
Apply gravity (add acceleration to velocity)
Apply friction (multiply velocity by (1 - friction * delta_time))
Update position: x += vx * delta_time, y += vy * delta_time
Check for collisions between balls and resolve them
Check for collisions between balls and heptagon walls, resolve them
Redraw all elements
But need to handle the order of operations. For example, after moving balls, check for collisions with walls, then with other balls.

Wait, but collision detection with walls may require adjusting positions (since a ball may have moved outside the heptagon). So during each frame:

Update velocities and positions based on physics (gravity, friction).
Check for collisions between balls and resolve them.
Check for collisions between balls and heptagon walls, adjust positions and velocities as needed.
But this order is important. Because if a ball collides with another ball and then also with the wall, we need to handle both.

Alternatively, check all possible collisions in each frame.

But perhaps first handle ball-ball collisions, then ball-wall collisions.

Now, implementing collision between balls:

For all pairs of balls (i < j):

dx = xj - xi
dy = yj - yi
distance_squared = dxdx + dydy
if distance_squared < (r_i + r_j)^2:
resolve collision

But need to calculate the exact positions and velocities.

Once resolved, adjust their velocities accordingly.

Now, for ball-wall collisions:

For each ball, check if it's outside the heptagon. If yes, find closest point on perimeter, compute normal vector, wall velocity, then apply collision response.

Now, implementing the code step by step.

First, define constants and initial setup.

But how to determine the size of the heptagon? It should be large enough to contain all balls. Since balls start from center and bounce around, perhaps set the radius R such that when balls are at maximum distance (say, 2*R), but need to ensure that during collisions, they don't go beyond.

But for initial setup, maybe choose a heptagon with radius larger than the maximum possible ball movement. For example, let's say the heptagon has radius 300 pixels, and balls have radius 10 pixels. So the heptagon is large enough.

Now, code:

Initialize Tkinter window:

root = tk.Tk()
canvas = tk.Canvas(root, width=800, height=600)
canvas.pack()

Then, define colors as given:

colors = [
"#f8b862", "#f6ad49", "#f39800", "#f08300", "#ec6d51",
"#ee7948", "#ed6d3d", "#ec6800", "#ec6800", "#ee7800",
"#eb6238", "#ea5506", "#ea5506", "#eb6101", "#e49e61",
"#e45e32", "#e17b34", "#dd7a56", "#db8449", "#d66a35"
]

Then create 20 balls:

balls = []
for i in range(20):
ball = Ball(
x=canvas.winfo_width()/2,
y=canvas.winfo_height()/2,
vx=0.0,
vy=0.0,
radius=10,
color=colors[i],
number=i+1,
angular_velocity=0.0
)
balls.append(ball)

But need to adjust initial positions and velocities.

Wait, the problem says: "All balls drop from the heptagon center when starting." So initially, their position is at the center of the heptagon (cx, cy), but they start with zero velocity? Or maybe some initial velocity?

No, the description says "drop", so perhaps they start with zero velocity. But then gravity will make them fall.

But the heptagon's center is at (canvas width/2, canvas height/2). So for example:

heptagon = Heptagon(
cx=400,
cy=300,
radius=300,
rotation_angle=0.0,
angular_velocity= 2 * math.pi / 5  # radians per second
)

But need to make sure that the heptagon is large enough.

Now, for gravity: let's assume a constant acceleration downward (e.g., 9.8 m/s², but scaled for pixels). However, since we are working in pixels and time steps may vary, it's better to use arbitrary units.

Let me choose gravity as 0.5 pixels per frame squared. Or perhaps define a gravity value that makes the balls fall realistically.

But during each frame, velocity is updated with acceleration (gravity) multiplied by delta_time.

Friction: for example, multiply velocity by (1 - friction * delta_time). Let's say friction is 0.01 per second.

Now, in code:

gravity = 0.5  # arbitrary units
friction_coefficient = 0.01  # per second

But during each frame, the delta_time is calculated.

Now, for the animation loop:

def animate():
global heptagon, balls, last_time

current_time = time.time()
delta_time = current_time - last_time
last_time = current_time

# Update heptagon rotation
heptagon.rotation_angle += heptagon.angular_velocity * delta_time

# Update balls' velocities and positions
for ball in balls:
    # Apply gravity
    ball.vy += gravity * delta_time

    # Apply friction
    ball.vx *= (1 - friction_coefficient * delta_time)
    ball.vy *= (1 - friction_coefficient * delta_time)

    # Update position
    ball.x += ball.vx * delta_time
    ball.y += ball.vy * delta_time

# Check for collisions between balls
for i in range(len(balls)):
    for j in range(i+1, len(balls)):
        b1 = balls[i]
        b2 = balls[j]
        dx = b2.x - b1.x
        dy = b2.y - b1.y
        distance_squared = dx*dx + dy*dy
        min_distance = b1.radius + b2.radius
        if distance_squared < min_distance * min_distance:
            # Collision, resolve it
            distance = math.sqrt(distance_squared)
            # Normalize the vector between centers
            nx = dx / distance
            ny = dy / distance

            # Compute relative velocity
            v_rel_x = (b2.vx - b1.vx)
            v_rel_y = (b2.vy - b1.vy)

            # Dot product of relative velocity and normal
            v_rel_dot_n = v_rel_x * nx + v_rel_y * ny

            # If approaching, resolve collision
            if v_rel_dot_n > 0:
                # Elastic collision for equal masses
                impulse = v_rel_dot_n
                b1.vx -= impulse * nx
                b1.vy -= impulse * ny
                b2.vx += impulse * nx
                b2.vy += impulse * ny

# Check for collisions between balls and heptagon walls
for ball in balls:
    if not is_inside_heptagon(ball, heptagon):
        # Find closest point on perimeter
        closest_point = find_closest_point_on_heptagon(ball, heptagon)
        # Compute normal vector of the edge at that point
        edge_normal = compute_edge_normal(closest_point, heptagon)
        # Compute wall velocity at contact point
        wall_velocity = compute_wall_velocity(closest_point, heptagon)
        # Apply collision response
        resolve_collision_with_wall(ball, closest_point, edge_normal, wall_velocity)

# Redraw everything
canvas.delete("all")
draw_heptagon(heptagon)
for ball in balls:
    draw_ball(ball)
root.after(16, animate)  # ~60 FPS
But need to implement the helper functions: is_inside_heptagon, find_closest_point_on_heptagon, compute_edge_normal, compute_wall_velocity, resolve_collision_with_wall.

Implementing is_inside_heptagon:

def is_inside_heptagon(ball, heptagon):
# Check if ball's position (ball.x, ball.y) is inside the heptagon
# Assuming heptagon is convex and ordered counter-clockwise

# Get vertices of heptagon
vertices = get_heptagon_vertices(heptagon)
for i in range(len(vertices)):
    A = vertices[i]
    B = vertices[(i+1)%7]
    # Compute cross product (B - A) × (P - A)
    dx = B[0] - A[0]
    dy = B[1] - A[1]
    px = ball.x - A[0]
    py = ball.y - A[1]
    cross = dx * py - dy * px
    if cross < 0:
        return False
return True
But need to implement get_heptagon_vertices:

def get_heptagon_vertices(heptagon):
cx, cy = heptagon.cx, heptagon.cy
R = heptagon.radius
theta = heptagon.rotation_angle
vertices = []
for i in range(7):
angle = theta + 2 * math.pi * i / 7
x = cx + R * math.cos(angle)
y = cy + R * math.sin(angle)
vertices.append( (x, y) )
return vertices

Now, find_closest_point_on_heptagon:

def find_closest_point_on_heptagon(ball, heptagon):
# Find the closest point on the heptagon's perimeter to ball's position
vertices = get_heptagon_vertices(heptagon)
min_distance_squared = float('inf')
closest_point = None

for i in range(len(vertices)):
    A = vertices[i]
    B = vertices[(i+1)%7]

    # Find closest point on segment AB to ball's position
    closest = closest_point_on_segment(ball.x, ball.y, A[0], A[1], B[0], B[1])
    distance_squared = (ball.x - closest[0])**2 + (ball.y - closest[1])**2

    if distance_squared < min_distance_squared:
        min_distance_squared = distance_squared
        closest_point = closest

return closest_point
def closest_point_on_segment(px, py, ax, ay, bx, by):
# Compute the closest point on segment AB to point P (px, py)
# Using vector projection
abx = bx - ax
aby = by - ay
apx = px - ax
apy = py - ay

# Compute t: projection of AP onto AB
t = (apx * abx + apy * aby) / (abx**2 + aby**2)
t = max(0, min(1, t))  # clamp between 0 and 1

closest_x = ax + t * abx
closest_y = ay + t * aby
return (closest_x, closest_y)
Next, compute_edge_normal:

def compute_edge_normal(closest_point, heptagon):
# Find the edge that contains the closest point, then compute its outward normal
vertices = get_heptagon_vertices(heptagon)
for i in range(len(vertices)):
A = vertices[i]
B = vertices[(i+1)%7]

    # Check if the closest_point lies on this segment AB
    # For simplicity, assume that it's part of this edge (since we found it via checking all edges)
    dx = B[0] - A[0]
    dy = B[1] - A[1]
    normal_x = -dy
    normal_y = dx
    # Normalize the normal vector
    length = math.hypot(normal_x, normal_y)
    if length == 0:
        continue
    normal_x /= length
    normal_y /= length
    return (normal_x, normal_y)

# If no edge found (should not happen)
return (0, 1)  # default up
Wait, but how to determine which edge the closest_point belongs to? Because in find_closest_point_on_heptagon, we check all edges and select the one with minimal distance. So for that particular closest_point, it's on the corresponding edge.

But in the code above, when we compute the normal vector, we need to know which edge it is. However, in the current implementation of find_closest_point_on_heptagon, we don't track which edge the closest point belongs to. So this function would return the normal for the last edge checked, which may not be correct.

This is a problem. Need to modify find_closest_point_on_heptagon to also return the edge (A and B) that contains the closest point.

So modifying the function:

def find_closest_point_on_heptagon(ball, heptagon):
vertices = get_heptagon_vertices(heptagon)
min_distance_squared = float('inf')
closest_point = None
closest_edge_A = None
closest_edge_B = None

for i in range(len(vertices)):
    A = vertices[i]
    B = vertices[(i+1)%7]

    # Find closest point on segment AB to ball's position
    closest = closest_point_on_segment(ball.x, ball.y, A[0], A[1], B[0], B[1])
    distance_squared = (ball.x - closest[0])**2 + (ball.y - closest[1])**2

    if distance_squared < min_distance_squared:
        min_distance_squared = distance_squared
        closest_point = closest
        closest_edge_A = A
        closest_edge_B = B

return (closest_point, closest_edge_A, closest_edge_B)
Then, compute_edge_normal can take the edge A and B:

def compute_edge_normal(A, B):
dx = B[0] - A[0]
dy = B[1] - A[1]
normal_x = -dy
normal_y = dx
length = math.hypot(normal_x, normal_y)
if length == 0:
return (0.0, 1.0)  # default up
normal_x /= length
normal_y /= length
return (normal_x, normal_y)

Now, the resolve_collision_with_wall function:

def resolve_collision_with_wall(ball, closest_point, edge_normal, wall_velocity):
# Compute collision response between ball and moving wall

# Ball's position is at closest_point + some offset? Or is it outside?
# Assume that the ball's center is outside the heptagon. So we need to move it inside.

# First, adjust the ball's position to be on the perimeter
# The distance between ball's center and the closest point is the radius of the ball plus the distance from the edge.
# Wait, no: the ball is outside the heptagon. The closest_point is on the perimeter. So the distance between ball's center and closest_point should be equal to the ball's radius.

# But due to movement, it might have passed through. So we need to move it back along the normal direction by (distance - ball.radius) ?

# Compute vector from closest_point to ball's position
dx = ball.x - closest_point[0]
dy = ball.y - closest_point[1]
distance = math.hypot(dx, dy)

if distance == 0:
    return

# Normalize the direction
dir_x = dx / distance
dir_y = dy / distance

# Move the ball's position to be exactly at the perimeter (distance from center is radius)
# So new_position = closest_point + ball.radius * dir
# But need to check if that's correct.

# Alternatively, since the ball's center is outside the heptagon, we can move it back along the normal direction by (distance - ball.radius) ?

# For example:
# The distance between ball and perimeter is (distance - 0), but the ball has radius r. So to make it touch the perimeter, move it back by (distance - r).

# But I'm not sure. Maybe need to calculate how much to push it back.

# Let's assume that the closest_point is on the heptagon's perimeter, and the ball's center is outside. The distance between them is d = distance from ball to perimeter.

# To make the ball touch the perimeter, we move its center towards the normal direction by (d - r) ?

# For example: if the ball's center is at distance d from the perimeter, then moving it back along the normal direction by (d - r) would place it so that the distance between center and perimeter is r.

# So new_position = closest_point + (dir_x * (d - r)) ?

# But need to calculate this properly.

# Let me think: The ball's radius is r. To make sure that the ball is just touching the heptagon, the distance from the ball center to the perimeter should be exactly r.

# So if current distance from ball to perimeter is d, then we need to move it back by (d - r) along the normal direction.

# But how to compute this?

# The vector from the closest_point to the ball's position is (dx, dy). The unit vector in that direction is (dir_x, dir_y).

# So new_position = ball.x - (distance - ball.radius) * dir_x
# new_position_y = ball.y - (distance - ball.radius) * dir_y

# But need to check if distance > ball.radius. If not, no need.

if distance <= ball.radius:
    return  # already inside or touching

# Compute the amount to move back
correction = distance - ball.radius
ball.x -= correction * dir_x
ball.y -= correction * dir_y

# Now, apply collision response for velocity

# Compute wall_velocity at closest_point (which is on the heptagon)
# The wall_velocity is computed based on the heptagon's rotation and position of closest_point.

# For that, compute the vector from the heptagon's center to the closest_point
cx = heptagon.cx
cy = heptagon.cy
px = closest_point[0]
py = closest_point[1]

dx = px - cx
dy = py - cy

# The wall_velocity is (ω * dy, -ω * dx) for clockwise rotation.
# Assuming the heptagon's angular velocity is stored in heptagon.angular_velocity.

omega = heptagon.angular_velocity
vx_wall = omega * dy  # since it's clockwise: v_x = ω * y, v_y = -ω * x ?
vy_wall = -omega * dx

# Now compute the relative velocity between ball and wall
# v_rel = (ball.vx - vx_wall, ball.vy - vy_wall)
v_rel_x = ball.vx - vx_wall
v_rel_y = ball.vy - vy_wall

# Compute the component of this relative velocity along the normal direction
n_x, n_y = edge_normal  # which is the outward normal of the edge
v_rel_dot_n = v_rel_x * n_x + v_rel_y * n_y

if v_rel_dot_n > 0:
    # The ball is moving towards the wall. Apply collision response.
    e = 0.8  # coefficient of restitution, adjust as needed
    impulse = - (1 + e) * v_rel_dot_n
    ball.vx -= impulse * n_x
    ball.vy -= impulse * n_y
But wait, in the earlier formula for collision with moving wall:

v_ball' = v_wall + [ (v_ball - v_wall) - (1 + e)( (v_ball - v_wall) · n ) * n ]

Which simplifies to:

v_ball' = v_wall - e*( (v_ball - v_wall) · n ) * n

So in code:

impulse = -e * v_rel_dot_n
ball.vx = vx_wall + impulse * n_x
ball.vy = vy_wall + impulse * n_y

But I need to recheck the formula.

Earlier, we had:

(v_ball' - v_wall) · n = -e (v_ball - v_wall) · n

So solving for v_ball':

v_ball' = v_wall + [ -e (v_ball - v_wall) · n ] * n

Which is:

v_ball' = v_wall - e * (v_rel · n ) * n

Where v_rel = v_ball - v_wall.

Thus, in code:

n_x, n_y = edge_normal
v_rel_dot_n = (ball.vx - vx_wall)*n_x + (ball.vy - vy_wall)*n_y

if v_rel_dot_n > 0:
e = 0.8
impulse = -e * v_rel_dot_n
ball.vx = vx_wall + impulse * n_x
ball.vy = vy_wall + impulse * n_y

But need to make sure that the normal vector is pointing outward from the heptagon.

Assuming that compute_edge_normal returns the outward normal, this should be correct.

Now, implementing all these functions.

But in Python, how to pass the heptagon to resolve_collision_with_wall? Because in the current code structure, the function would need access to the heptagon's parameters. So perhaps modify the function to take heptagon as an argument.

So:

def resolve_collision_with_wall(ball, closest_point, edge_normal, heptagon):
# ... previous code ...

# Compute wall_velocity
cx = heptagon.cx
cy = heptagon.cy
px = closest_point[0]
py = closest_point[1]

dx = px - cx
dy = py - cy

omega = heptagon.angular_velocity
vx_wall = omega * dy  # for clockwise rotation
vy_wall = -omega * dx

# ... rest of the code ...
But in the previous code, the function 'resolve_collision_with_wall' is called with ball, closest_point, edge_normal, and heptagon.

Now, putting it all together.

Another thing to consider: when moving the ball back into the heptagon, we need to ensure that it doesn't penetrate further. But this adjustment is done by moving along the normal direction.

Now, for drawing:

def draw_heptagon(heptagon):
vertices = get_heptagon_vertices(heptagon)
# Draw lines between consecutive vertices
for i in range(len(vertices)):
x1, y1 = vertices[i]
x2, y2 = vertices[(i+1)%7]
canvas.create_line(x1, y1, x2, y2, fill="black")

def draw_ball(ball):
# Draw the ball as a circle
x = ball.x
y = ball.y
r = ball.radius
canvas.create_oval(x - r, y - r, x + r, y + r, fill=ball.color, outline="")
# Draw the number, rotated with angular_velocity
# For simplicity, assume that the angular velocity is in radians per second, and we rotate the text accordingly.
# But how to compute the rotation angle for this frame?
# The rotation angle is angular_velocity * delta_time (since last update)
# However, since the animation loop uses time.time(), need to track the current time.

But wait, in the animate function, we have delta_time. So during drawing, the ball's spin is updated based on that.

But how to track the rotation angle of each ball for rendering?

Each ball has an angular_velocity (radians per second). The rotation angle can be tracked as a separate variable.

So perhaps add a 'rotation_angle' attribute to the Ball class:

@dataclass
class Ball:
x: float
y: float
vx: float
vy: float
radius: float
color: str
number: int
angular_velocity: float  # radians per second
rotation_angle: float  # current rotation angle for drawing

Then, during each frame, update the rotation_angle:

ball.rotation_angle += ball.angular_velocity * delta_time

But in the animate function, after updating velocities and positions, we also need to update the rotation angles.

So in the animate function:

for ball in balls:
# ... previous code ...
ball.rotation_angle += ball.angular_velocity * delta_time

Then, when drawing the number, rotate it by ball.rotation_angle.

But how to do that with tkinter's create_text? Tkinter's text doesn't support rotation directly. So need to use a canvas item and transform it.

Alternatively, draw the text as a polygon or use other methods. But this is complex.

An alternative approach: for each ball, when drawing, calculate the position of the number (centered on the ball) and rotate the text accordingly.

But since tkinter doesn't support rotation, perhaps approximate by using a polygon or using PIL to generate an image with rotated text, then display it as an image. But that may be complicated.

Alternatively, for simplicity, just draw the number without rotation. However, the problem states that "the numbers on the ball can be used to indicate the spin of the ball." So this needs to be addressed.

But given time constraints and complexity, perhaps proceed with drawing the number normally, but note that it's not rotated. Alternatively, for a quick solution, use the text as is.

Alternatively, find a way to rotate the text using tkinter's create_text with rotation. After some research, I recall that in Tkinter, you can't directly rotate text, but you can use the Canvas' create_polygon method to draw rotated text by converting each character to a polygon and rotating it. However, this is complex.

Alternatively, for simplicity, skip the rotation of numbers and just display them normally, noting that it's not perfect but meets the problem's requirements as much as possible.

But since the user specified that "the numbers on the ball can be used to indicate the spin of the ball," perhaps we need to find a way to rotate the text.

An alternative idea: use the canvas' create_text with an angle. However, Tkinter does not support this directly. But there's a workaround using the 'angle' option in some versions? Or maybe using the 'transform' parameter?

After checking, it seems that Tkinter's Canvas.create_text does not support rotation. So to achieve this, one must use other methods.

An alternative is to use the PIL library to generate an image of the text with rotation and then display it as a photoimage in Tkinter. However, the user specified allowed libraries include tkinter, math, numpy, dataclasses, typing, sys. PIL (Pillow) is not listed. So cannot use that.

Thus, for this problem, perhaps proceed without rotating the numbers, but note that it's a limitation.

So, for draw_ball function:

def draw_ball(ball):
x = ball.x
y = ball.y
r = ball.radius
canvas.create_oval(x - r, y - r, x + r, y + r, fill=ball.color, outline="")
# Draw the number in the center
canvas.create_text(x, y, text=str(ball.number), fill="black")

Now, putting it all together.

But need to handle the initial rotation_angle for each ball. In the Ball class, initialize rotation_angle as 0.0.

Now, testing this code would require handling various edge cases, but given time constraints, proceed with writing the full code.

Potential issues:

The heptagon's vertices are computed correctly.
Collision detection between balls and walls is accurate.
Handling of ball positions when they are outside the heptagon.
Correct application of velocity changes during collisions.
Proper handling of delta_time for physics calculations.
Another thing to consider: the initial position of the balls. They start at the center of the heptagon, but if the heptagon is spinning, their initial position may be inside or not? Well, since they are at the center, which is inside the heptagon.

Now, putting all code together into a single Python file.

But also need to handle the initial setup and start the animation loop.

Also, need to import time for delta_time calculations.

So, in code:

import tkinter as tk
import math
import time
from dataclasses import dataclass
from typing import List

@dataclass
class Ball:
x: float
y: float
vx: float
vy: float
radius: float
color: str
number: int
angular_velocity: float  # radians per second
rotation_angle: float  # current rotation angle for drawing

@dataclass
class Heptagon:
cx: float
cy: float
radius: float
rotation_angle: float  # in radians
angular_velocity: float  # radians per second

def get_heptagon_vertices(heptagon):
cx, cy = heptagon.cx, heptagon.cy
R = heptagon.radius
theta = heptagon.rotation_angle
vertices = []
for i in range(7):
angle = theta + 2 * math.pi * i / 7
x = cx + R * math.cos(angle)
y = cy + R * math.sin(angle)
vertices.append( (x, y) )
return vertices

def is_inside_heptagon(ball, heptagon):
vertices = get_heptagon_vertices(heptagon)
for i in range(len(vertices)):
A = vertices[i]
B = vertices[(i+1)%7]
dx = B[0] - A[0]
dy = B[1] - A[1]
px = ball.x - A[0]
py = ball.y - A[1]
cross = dx * py - dy * px
if cross < 0:
return False
return True

def closest_point_on_segment(px, py, ax, ay, bx, by):
abx = bx - ax
aby = by - ay
apx = px - ax
apy = py - ay

t = (apx * abx + apy * aby) / (abx**2 + aby**2)
t = max(0, min(1, t))
closest_x = ax + t * abx
closest_y = ay + t * aby
return (closest_x, closest_y)
def find_closest_point_on_heptagon(ball, heptagon):
vertices = get_heptagon_vertices(heptagon)
min_distance_squared = float('inf')
closest_point = None
closest_edge_A = None
closest_edge_B = None

for i in range(len(vertices)):
    A = vertices[i]
    B = vertices[(i+1)%7]

    closest = closest_point_on_segment(ball.x, ball.y, A[0], A[1], B[0], B[1])
    distance_squared = (ball.x - closest[0])**2 + (ball.y - closest[1])**2

    if distance_squared < min_distance_squared:
        min_distance_squared = distance_squared
        closest_point = closest
        closest_edge_A = A
        closest_edge_B = B

return (closest_point, closest_edge_A, closest_edge_B)
def compute_edge_normal(A, B):
dx = B[0] - A[0]
dy = B[1] - A[1]
normal_x = -dy
normal_y = dx
length = math.hypot(normal_x, normal_y)
if length == 0:
return (0.0, 1.0)
normal_x /= length
normal_y /= length
return (normal_x, normal_y)

def resolve_collision_with_wall(ball, closest_point, edge_normal, heptagon):
# Move ball back into the heptagon
dx = ball.x - closest_point[0]
dy = ball.y - closest_point[1]
distance = math.hypot(dx, dy)
if distance <= ball.radius:
return

dir_x = dx / distance
dir_y = dy / distance
correction = distance - ball.radius
ball.x -= correction * dir_x
ball.y -= correction * dir_y

# Compute wall velocity at closest_point
cx = heptagon.cx
cy = heptagon.cy
px = closest_point[0]
py = closest_point[1]

dx_wall = px - cx
dy_wall = py - cy

omega = heptagon.angular_velocity
vx_wall = omega * dy_wall  # clockwise rotation
vy_wall = -omega * dx_wall

# Compute relative velocity
v_rel_x = ball.vx - vx_wall
v_rel_y = ball.vy - vy_wall

n_x, n_y = edge_normal
v_rel_dot_n = v_rel_x * n_x + v_rel_y * n_y

if v_rel_dot_n > 0:
    e = 0.8  # coefficient of restitution
    impulse = -e * v_rel_dot_n
    ball.vx = vx_wall + impulse * n_x
    ball.vy = vy_wall + impulse * n_y
def draw_heptagon(heptagon, canvas):
vertices = get_heptagon_vertices(heptagon)
for i in range(len(vertices)):
x1, y1 = vertices[i]
x2, y2 = vertices[(i+1)%7]
canvas.create_line(x1, y1, x2, y2, fill="black")

def draw_ball(ball, canvas):
x = ball.x
y = ball.y
r = ball.radius
canvas.create_oval(x - r, y - r, x + r, y + r, fill=ball.color, outline="")
# Draw number (without rotation)
canvas.create_text(x, y, text=str(ball.number), fill="black")

def main():
root = tk.Tk()
root.title("Bouncing Balls in Spinning Heptagon")
canvas_width = 800
canvas_height = 600
canvas = tk.Canvas(root, width=canvas_width, height=canvas_height)
canvas.pack()

# Initialize heptagon
heptagon = Heptagon(
    cx=canvas_width / 2,
    cy=canvas_height / 2,
    radius=300,
    rotation_angle=0.0,
    angular_velocity= (2 * math.pi) / 5  # 360 degrees per 5 seconds
)

# Initialize balls
colors = [
    "#f8b862", "#f6ad49", "#f39800", "#f08300", "#ec6d51",
    "#ee7948", "#ed6d3d", "#ec6800", "#ec6800", "#ee7800",
    "#eb6238", "#ea5506", "#ea5506", "#eb6101", "#e49e61",
    "#e45e32", "#e17b34", "#dd7a56", "#db8449", "#d66a35"
]
balls = []
for i in range(20):
    ball = Ball(
        x=heptagon.cx,
        y=heptagon.cy,
        vx=0.0,
        vy=0.0,
        radius=10,
        color=colors[i],
        number=i+1,
        angular_velocity=0.0,  # initial spin
        rotation_angle=0.0
    )
    balls.append(ball)

gravity = 0.5  # arbitrary units
friction_coefficient = 0.01  # per second

last_time = time.time()

def animate():
    nonlocal last_time
    current_time = time.time()
    delta_time = current_time - last_time
    last_time = current_time

    # Update heptagon rotation
    heptagon.rotation_angle += heptagon.angular_velocity * delta_time

    # Update balls' velocities and positions
    for ball in balls:
        # Apply gravity
        ball.vy += gravity * delta_time

        # Apply friction
        ball.vx *= (1 - friction_coefficient * delta_time)
        ball.vy *= (1 - friction_coefficient * delta_time)

        # Update position
        ball.x += ball.vx * delta_time
        ball.y += ball.vy * delta_time

        # Update rotation angle for drawing
        ball.rotation_angle += ball.angular_velocity * delta_time

    # Check collisions between balls
    for i in range(len(balls)):
        for j in range(i+1, len(balls)):
            b1 = balls[i]
            b2 = balls[j]
            dx = b2.x - b1.x
            dy = b