P587: Added explanation and slightly simplified code in all language solutions.

Author: nayuki

java/p587.java

@@ -14,6 +14,73 @@ public static void main(String[] args) {
 	}
 
 
+	/* 
+	 * Start by defining the coordinate system in a convenient way. The position and scale of the diagram don't
+	 * matter because we only care about the ratio of areas, not the absolute areas. So, let the bottom left
+	 * of the diagram be the origin (x = 0, y = 0), and let each circle to have a radius of 1.
+	 * 
+	 * The leftmost circle is centered at (1, 1), and its equation is (x - 1)^2 + (y - 1)^2 = 1.
+	 * The diagonal line has slope = s = 1 / n (for any positive n), and the line's equation is y = s * x.
+	 * From basic geometry, the area of the blue L-section is 1 - pi / 4.
+	 * 
+	 * Let's find the x-coordinate where the diagonal line intersects the first circle.
+	 * Take the equation of the circle and substitute y = s * x for the line:
+	 * 
+	 *   (x - 1)^2 + (s*x - 1)^2 = 1.
+	 *   (x^2 - 2x + 1) + (s^2 x^2 - 2s*x + 1) = 1.
+	 *   (1 + s^2)x^2 + (-2 - 2s)x + 1 = 0.
+	 * 
+	 * We can apply the quadratic formula with a = 1 + s^2, b = -2 - 2s, c = 1. There are two solutions for x,
+	 * and we only want the smaller value. Thus, let X = (-b - sqrt(b^2 - 4ac)) / (2a). Or equivalently
+	 * with more numerical stability (using the Citardauq formula), X = (2c) / (-b + sqrt(b^2 - 4ac)).
+	 * 
+	 * The orange concave triangle can be divided into two parts by a vertical line:
+	 * 
+	 * - The left part is a proper triangle, whose area is easily seen as x * y / 2 = X^2 * s / 2.
+	 * 
+	 * - The right part is the region between the circle and the baseline. Let's re-express
+	 *   the circle's equation in terms of y, and only keep the lower semicircle:
+	 *   
+	 *     (x - 1)^2 + (y - 1)^2 = 1.
+	 *     (y - 1)^2 = 1 - (x - 1)^2.
+	 *     y - 1 = -sqrt(1 - (x - 1)^2).
+	 *     y = 1 - sqrt(1 - (x - 1)^2).
+	 *     y = 1 - sqrt(1 - (x^2 - 2x + 1)).
+	 *     y = 1 - sqrt(2x - x^2).
+	 *   
+	 *   Now, the indefinite integral of f(x) = 1 - sqrt(2x - x^2) with respect to x
+	 *   is F(x) = (x - 1) - [sqrt(2x - x^2) * (x - 1) + asin(x - 1)] / 2.
+	 *   Finding this integral is not obvious, but verifying it is a fairly straightforward
+	 *   mechanical procedure involving differentiation and simplification.
+	 *   
+	 *   The area of the right part is the integral of f(x) for x from X to 1, because the start is
+	 *   the x-coordinate where line meets the circle, and the end is where the circle meets the baseline.
+	 *   Hence the area is equal to F(1) - F(X).
+	 * 
+	 * All in all, for any given n, the area of the orange concave triangle is X^2 * s / 2 + F(1) - F(X).
+	 * The rest of the algorithm is a brute-force search with n = 1, 2, 3, ... until the ratio condition is met.
+	 * 
+	 * Additional notes:
+	 * - Intuitively, as n increases and the slope gets smaller, the area of the orange concave triangle should strictly
+	 *   decrease. This statement is in fact true, but proving it involves a big pile of differentiation and algebra.
+	 *   0. We need to show that X (which is the x-coordinate of the line-circle intersection) increases with n.
+	 *      We'd differentiate X with respect to n, and get an expression that is always positive for any positive n.
+	 *   1. Because X increases with n, the area of the right part, with its always-positive integrand, must decrease.
+	 *   2. As for the left part, we'd differentiate X^2 * s / 2 with respect to n, and get a huge messy formula.
+	 *      It turns out this formula is negative for all n > 1. Hence the area of this triangle also decreases with n.
+	 *   After we prove that increasing n leads to decreasing orange area, we could use
+	 *   binary search to find the minimum value of n needed to meet the ratio requirement.
+	 * - The use of floating-point arithmetic, for basic arithmetic operations (+ - * /) and irrational functions (sqrt,
+	 *   asin) alike, is inherently difficult or impossible to prove the correctness of. Furthermore, the algorithms
+	 *   for irrational functions are hard to understand and beyond the scope of this problem, and the error bounds for
+	 *   all operations are difficult to reason about.
+	 *   It should be possible to solve this particular problem using only integer arithmetic in a provably correct way.
+	 *   The basic idea would be to round the result of each operation both down and up to an integer fraction,
+	 *   keep track of pessimistic intervals that are guaranteed to contain the true value, accept a comparison only
+	 *   if the intervals don't overlap, and recompute everything at a higher precision if a comparison is inconclusive.
+	 *   Note: Because it doesn't seem easy to compute pi and asin(), it might be better to
+	 *   approximate integrals directly using the Darboux definition of lower and upper sums.
+	 */
 	public String run() {
 		double lSectionArea = 1 - Math.PI / 4;
 		for (int i = 1; ; i++) {
@@ -22,8 +89,7 @@ public String run() {
 			double b = -2 * (slope + 1);
 			double c = 1;
 			double x = (2 * c) / (-b + Math.sqrt(b * b - 4 * a * c));
-			double concaveTriangleArea = x * (1 - Math.sqrt((-x + 2) * x)) / 2;
-			concaveTriangleArea += integral(1) - integral(x);
+			double concaveTriangleArea = (x * x * slope / 2) + (integral(1) - integral(x));
 			if (concaveTriangleArea / lSectionArea < 0.001)
 				return Integer.toString(i);
 			if (i == Integer.MAX_VALUE)
@@ -35,7 +101,7 @@ public String run() {
 	// The indefinite integral of (1 - sqrt(2x - x^2)) dx.
 	private static double integral(double x) {
 		double t = x - 1;
-		return t - (Math.sqrt(1 - t * t) * t + Math.asin(t)) / 2;
+		return t - (Math.sqrt(x * (2 - x)) * t + Math.asin(t)) / 2;
 	}
 
 }

mathematica/p587.mathematica

@@ -7,19 +7,87 @@
  *)
 
 
+(* 
+ * Start by defining the coordinate system in a convenient way. The position and scale of the diagram don't
+ * matter because we only care about the ratio of areas, not the absolute areas. So, let the bottom left
+ * of the diagram be the origin (x = 0, y = 0), and let each circle to have a radius of 1.
+ * 
+ * The leftmost circle is centered at (1, 1), and its equation is (x - 1)^2 + (y - 1)^2 = 1.
+ * The diagonal line has slope = s = 1 / n (for any positive n), and the line's equation is y = s * x.
+ * From basic geometry, the area of the blue L-section is 1 - pi / 4.
+ * 
+ * Let's find the x-coordinate where the diagonal line intersects the first circle.
+ * Take the equation of the circle and substitute y = s * x for the line:
+ * 
+ *   (x - 1)^2 + (s*x - 1)^2 = 1.
+ *   (x^2 - 2x + 1) + (s^2 x^2 - 2s*x + 1) = 1.
+ *   (1 + s^2)x^2 + (-2 - 2s)x + 1 = 0.
+ * 
+ * We can apply the quadratic formula with a = 1 + s^2, b = -2 - 2s, c = 1. There are two solutions for x,
+ * and we only want the smaller value. Thus, let X = (-b - sqrt(b^2 - 4ac)) / (2a). Or equivalently
+ * with more numerical stability (using the Citardauq formula), X = (2c) / (-b + sqrt(b^2 - 4ac)).
+ * 
+ * The orange concave triangle can be divided into two parts by a vertical line:
+ * 
+ * - The left part is a proper triangle, whose area is easily seen as x * y / 2 = X^2 * s / 2.
+ * 
+ * - The right part is the region between the circle and the baseline. Let's re-express
+ *   the circle's equation in terms of y, and only keep the lower semicircle:
+ *   
+ *     (x - 1)^2 + (y - 1)^2 = 1.
+ *     (y - 1)^2 = 1 - (x - 1)^2.
+ *     y - 1 = -sqrt(1 - (x - 1)^2).
+ *     y = 1 - sqrt(1 - (x - 1)^2).
+ *     y = 1 - sqrt(1 - (x^2 - 2x + 1)).
+ *     y = 1 - sqrt(2x - x^2).
+ *   
+ *   Now, the indefinite integral of f(x) = 1 - sqrt(2x - x^2) with respect to x
+ *   is F(x) = (x - 1) - [sqrt(2x - x^2) * (x - 1) + asin(x - 1)] / 2.
+ *   Finding this integral is not obvious, but verifying it is a fairly straightforward
+ *   mechanical procedure involving differentiation and simplification.
+ *   
+ *   The area of the right part is the integral of f(x) for x from X to 1, because the start is
+ *   the x-coordinate where line meets the circle, and the end is where the circle meets the baseline.
+ *   Hence the area is equal to F(1) - F(X).
+ * 
+ * All in all, for any given n, the area of the orange concave triangle is X^2 * s / 2 + F(1) - F(X).
+ * The rest of the algorithm is a brute-force search with n = 1, 2, 3, ... until the ratio condition is met.
+ * 
+ * Additional notes:
+ * - Intuitively, as n increases and the slope gets smaller, the area of the orange concave triangle should strictly
+ *   decrease. This statement is in fact true, but proving it involves a big pile of differentiation and algebra.
+ *   0. We need to show that X (which is the x-coordinate of the line-circle intersection) increases with n.
+ *      We'd differentiate X with respect to n, and get an expression that is always positive for any positive n.
+ *   1. Because X increases with n, the area of the right part, with its always-positive integrand, must decrease.
+ *   2. As for the left part, we'd differentiate X^2 * s / 2 with respect to n, and get a huge messy formula.
+ *      It turns out this formula is negative for all n > 1. Hence the area of this triangle also decreases with n.
+ *   After we prove that increasing n leads to decreasing orange area, we could use
+ *   binary search to find the minimum value of n needed to meet the ratio requirement.
+ * - The use of floating-point arithmetic, for basic arithmetic operations (+ - * /) and irrational functions (sqrt,
+ *   asin) alike, is inherently difficult or impossible to prove the correctness of. Furthermore, the algorithms
+ *   for irrational functions are hard to understand and beyond the scope of this problem, and the error bounds for
+ *   all operations are difficult to reason about.
+ *   It should be possible to solve this particular problem using only integer arithmetic in a provably correct way.
+ *   The basic idea would be to round the result of each operation both down and up to an integer fraction,
+ *   keep track of pessimistic intervals that are guaranteed to contain the true value, accept a comparison only
+ *   if the intervals don't overlap, and recompute everything at a higher precision if a comparison is inconclusive.
+ *   Note: Because it doesn't seem easy to compute pi and asin(), it might be better to
+ *   approximate integrals directly using the Darboux definition of lower and upper sums.
+ *)
+
 LSectionArea = 1 - Pi / 4;
 
 (* The indefinite integral of (1 - sqrt(2x - x^2)) dx. *)
 TheIntegral[x_] := Block[{t = x - 1},
-  t - (Sqrt[1 - t^2] * t + ArcSin[t]) / 2]
+  t - (Sqrt[x * (2 - x)] * t + ArcSin[t]) / 2]
 
 ConcaveTriangleArea[i_] := Block[{
     slope = 1 / i,
     a = slope^2 + 1,
     b = -2 * (slope + 1),
     c = 1,
     x = (2 * c) / (-b + Sqrt[b^2 - 4 * a * c])},
-  x * (1 - Sqrt[(-x + 2) * x]) / 2 + TheIntegral[1] - TheIntegral[x]]
+  (x^2 * slope / 2) + (TheIntegral[1] - TheIntegral[x])]
 
 i = 1;
 While[ConcaveTriangleArea[i] / LSectionArea >= 1 / 1000,

python/p587.py

@@ -9,11 +9,76 @@
 import itertools, math
 
 
+# Start by defining the coordinate system in a convenient way. The position and scale of the diagram don't
+# matter because we only care about the ratio of areas, not the absolute areas. So, let the bottom left
+# of the diagram be the origin (x = 0, y = 0), and let each circle to have a radius of 1.
+# 
+# The leftmost circle is centered at (1, 1), and its equation is (x - 1)^2 + (y - 1)^2 = 1.
+# The diagonal line has slope = s = 1 / n (for any positive n), and the line's equation is y = s * x.
+# From basic geometry, the area of the blue L-section is 1 - pi / 4.
+# 
+# Let's find the x-coordinate where the diagonal line intersects the first circle.
+# Take the equation of the circle and substitute y = s * x for the line:
+# 
+#   (x - 1)^2 + (s*x - 1)^2 = 1.
+#   (x^2 - 2x + 1) + (s^2 x^2 - 2s*x + 1) = 1.
+#   (1 + s^2)x^2 + (-2 - 2s)x + 1 = 0.
+# 
+# We can apply the quadratic formula with a = 1 + s^2, b = -2 - 2s, c = 1. There are two solutions for x,
+# and we only want the smaller value. Thus, let X = (-b - sqrt(b^2 - 4ac)) / (2a). Or equivalently
+# with more numerical stability (using the Citardauq formula), X = (2c) / (-b + sqrt(b^2 - 4ac)).
+# 
+# The orange concave triangle can be divided into two parts by a vertical line:
+# 
+# - The left part is a proper triangle, whose area is easily seen as x * y / 2 = X^2 * s / 2.
+# 
+# - The right part is the region between the circle and the baseline. Let's re-express
+#   the circle's equation in terms of y, and only keep the lower semicircle:
+#   
+#     (x - 1)^2 + (y - 1)^2 = 1.
+#     (y - 1)^2 = 1 - (x - 1)^2.
+#     y - 1 = -sqrt(1 - (x - 1)^2).
+#     y = 1 - sqrt(1 - (x - 1)^2).
+#     y = 1 - sqrt(1 - (x^2 - 2x + 1)).
+#     y = 1 - sqrt(2x - x^2).
+#   
+#   Now, the indefinite integral of f(x) = 1 - sqrt(2x - x^2) with respect to x
+#   is F(x) = (x - 1) - [sqrt(2x - x^2) * (x - 1) + asin(x - 1)] / 2.
+#   Finding this integral is not obvious, but verifying it is a fairly straightforward
+#   mechanical procedure involving differentiation and simplification.
+#   
+#   The area of the right part is the integral of f(x) for x from X to 1, because the start is
+#   the x-coordinate where line meets the circle, and the end is where the circle meets the baseline.
+#   Hence the area is equal to F(1) - F(X).
+# 
+# All in all, for any given n, the area of the orange concave triangle is X^2 * s / 2 + F(1) - F(X).
+# The rest of the algorithm is a brute-force search with n = 1, 2, 3, ... until the ratio condition is met.
+# 
+# Additional notes:
+# - Intuitively, as n increases and the slope gets smaller, the area of the orange concave triangle should strictly
+#   decrease. This statement is in fact true, but proving it involves a big pile of differentiation and algebra.
+#   0. We need to show that X (which is the x-coordinate of the line-circle intersection) increases with n.
+#      We'd differentiate X with respect to n, and get an expression that is always positive for any positive n.
+#   1. Because X increases with n, the area of the right part, with its always-positive integrand, must decrease.
+#   2. As for the left part, we'd differentiate X^2 * s / 2 with respect to n, and get a huge messy formula.
+#      It turns out this formula is negative for all n > 1. Hence the area of this triangle also decreases with n.
+#   After we prove that increasing n leads to decreasing orange area, we could use
+#   binary search to find the minimum value of n needed to meet the ratio requirement.
+# - The use of floating-point arithmetic, for basic arithmetic operations (+ - * /) and irrational functions (sqrt,
+#   asin) alike, is inherently difficult or impossible to prove the correctness of. Furthermore, the algorithms
+#   for irrational functions are hard to understand and beyond the scope of this problem, and the error bounds for
+#   all operations are difficult to reason about.
+#   It should be possible to solve this particular problem using only integer arithmetic in a provably correct way.
+#   The basic idea would be to round the result of each operation both down and up to an integer fraction,
+#   keep track of pessimistic intervals that are guaranteed to contain the true value, accept a comparison only
+#   if the intervals don't overlap, and recompute everything at a higher precision if a comparison is inconclusive.
+#   Note: Because it doesn't seem easy to compute pi and asin(), it might be better to
+#   approximate integrals directly using the Darboux definition of lower and upper sums.
 def compute():
 	# The indefinite integral of (1 - sqrt(2x - x^2)) dx.
 	def integral(x):
 		t = x - 1.0
-		return t - (math.sqrt(1.0 - t**2) * t + math.asin(t)) / 2.0
+		return t - (math.sqrt(x * (2.0 - x)) * t + math.asin(t)) / 2.0
 
 	lsectionarea = 1.0 - math.pi / 4.0
 	for i in itertools.count(1):
@@ -22,8 +87,7 @@ def integral(x):
 		b = -2.0 * (slope + 1.0)
 		c = 1.0
 		x = (2.0 * c) / (-b + math.sqrt(b * b - 4 * a * c))
-		concavetrianglearea = x * (1.0 - math.sqrt((-x + 2.0) * x)) / 2.0
-		concavetrianglearea += integral(1.0) - integral(x)
+		concavetrianglearea = (x**2 * slope / 2) + (integral(1.0) - integral(x))
 		if concavetrianglearea / lsectionarea < 0.001:
 			return str(i)