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* Added solution for Project Euler problem 144
* updating DIRECTORY.md
* Better variable names
* updating DIRECTORY.md
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102 lines
3.8 KiB
Python
102 lines
3.8 KiB
Python
"""
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In laser physics, a "white cell" is a mirror system that acts as a delay line for the
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laser beam. The beam enters the cell, bounces around on the mirrors, and eventually
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works its way back out.
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The specific white cell we will be considering is an ellipse with the equation
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4x^2 + y^2 = 100
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The section corresponding to −0.01 ≤ x ≤ +0.01 at the top is missing, allowing the
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light to enter and exit through the hole.
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The light beam in this problem starts at the point (0.0,10.1) just outside the white
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cell, and the beam first impacts the mirror at (1.4,-9.6).
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Each time the laser beam hits the surface of the ellipse, it follows the usual law of
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reflection "angle of incidence equals angle of reflection." That is, both the incident
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and reflected beams make the same angle with the normal line at the point of incidence.
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In the figure on the left, the red line shows the first two points of contact between
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the laser beam and the wall of the white cell; the blue line shows the line tangent to
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the ellipse at the point of incidence of the first bounce.
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The slope m of the tangent line at any point (x,y) of the given ellipse is: m = −4x/y
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The normal line is perpendicular to this tangent line at the point of incidence.
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The animation on the right shows the first 10 reflections of the beam.
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How many times does the beam hit the internal surface of the white cell before exiting?
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"""
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from math import isclose, sqrt
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def next_point(
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point_x: float, point_y: float, incoming_gradient: float
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) -> tuple[float, float, float]:
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"""
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Given that a laser beam hits the interior of the white cell at point
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(point_x, point_y) with gradient incoming_gradient, return a tuple (x,y,m1)
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where the next point of contact with the interior is (x,y) with gradient m1.
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>>> next_point(5.0, 0.0, 0.0)
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(-5.0, 0.0, 0.0)
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>>> next_point(5.0, 0.0, -2.0)
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(0.0, -10.0, 2.0)
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"""
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# normal_gradient = gradient of line through which the beam is reflected
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# outgoing_gradient = gradient of reflected line
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normal_gradient = point_y / 4 / point_x
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s2 = 2 * normal_gradient / (1 + normal_gradient * normal_gradient)
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c2 = (1 - normal_gradient * normal_gradient) / (
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1 + normal_gradient * normal_gradient
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)
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outgoing_gradient = (s2 - c2 * incoming_gradient) / (c2 + s2 * incoming_gradient)
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# to find the next point, solve the simultaeneous equations:
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# y^2 + 4x^2 = 100
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# y - b = m * (x - a)
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# ==> A x^2 + B x + C = 0
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quadratic_term = outgoing_gradient ** 2 + 4
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linear_term = 2 * outgoing_gradient * (point_y - outgoing_gradient * point_x)
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constant_term = (point_y - outgoing_gradient * point_x) ** 2 - 100
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x_minus = (
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-linear_term - sqrt(linear_term ** 2 - 4 * quadratic_term * constant_term)
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) / (2 * quadratic_term)
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x_plus = (
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-linear_term + sqrt(linear_term ** 2 - 4 * quadratic_term * constant_term)
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) / (2 * quadratic_term)
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# two solutions, one of which is our input point
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next_x = x_minus if isclose(x_plus, point_x) else x_plus
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next_y = point_y + outgoing_gradient * (next_x - point_x)
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return next_x, next_y, outgoing_gradient
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def solution(first_x_coord: float = 1.4, first_y_coord: float = -9.6) -> int:
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"""
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Return the number of times that the beam hits the interior wall of the
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cell before exiting.
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>>> solution(0.00001,-10)
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1
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>>> solution(5, 0)
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287
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"""
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num_reflections: int = 0
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point_x: float = first_x_coord
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point_y: float = first_y_coord
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gradient: float = (10.1 - point_y) / (0.0 - point_x)
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while not (-0.01 <= point_x <= 0.01 and point_y > 0):
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point_x, point_y, gradient = next_point(point_x, point_y, gradient)
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num_reflections += 1
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return num_reflections
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if __name__ == "__main__":
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print(f"{solution() = }")
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