private Integer ProcessRegion(Integer w, Integer h, Integer a1, Integer b1, Integer a2, Integer b2, Integer x0, Integer y0)
{
// The hyperbola is defined by H(x, y): x*y = n.
// Line L1 has -slope m1 = a1/b1.
// Line L2 has -slope m2 = a2/b2.
// Both lines pass through P0 = (x0, y0).
// The region is a parallelogram with the left side bounded L1,
// the bottom bounded by L2, with width w (along L2) and height h
// (along L1). The lower-left corner is P0 (the intersection of
// L2 and L1) and represents (u, v) = (0, 0).
// Both w and h are counted in terms of lattice points, not length.
// For the purposes of counting, the lattice points on lines L1 and L2
// have already been counted.
// Note that a1*b2 - b1*a2 = 1 because
// m2 and m1 are Farey neighbors, e.g. 1 & 2 or 3/2 & 2 or 8/5 & 5/3
// The equations that define (u, v) in terms of (x, y) are:
// u = b1*(y-y0)+a1*(x-x0)
// v = b2*(y-y0)+a2*(x-x0)
// And therefore the equations that define (x, y) in terms of (u, v) are:
// x = x0-b1*v+b2*u
// y = y0+a1*v-a2*u
// Since all parameters are integers and a1*b2 - b1*a2 = 1,
// every lattice point in (x, y) is a lattice point in (u, v)
// and vice-versa.
// Geometrically, the UV coordinate system is the composition
// of a translation and two shear mappings. The UV-based hyperbola
// is essentially a "mini" hyperbola that resembles the full
// hyperbola in that:
// - The equation is still a hyperbola (although it is now a quadratic in two variables)
// - The endpoints of the curve are roughly tangent to the axes
// We process the region by "lopping off" the maximal isosceles
// right triangle in the lower-left corner and then process
// the two remaining "slivers" in the upper-left and lower-right,
// which creates two smaller "micro" hyperbolas, which we then
// process recursively.
// When we are in the region of the original hyperbola where
// the curvature is roughly constant, the deformed hyperbola
// will in fact resemble a circular arc.
// A line with -slope = 1 in UV-space has -slope = (a1+a2)/(b1+b2)
// in XY-space. We call this m3 and the line defining the third side
// of the triangle as L3 containing point P3 tangent to the hyperbola.
// This is all slightly complicated by the fact that diagonal that
// defines the region that we "lop off" may be broken and shifted
// up or down near the tangent point. As a result we actually have
// P3 and P4 and L3 and L4.
// We can measure work in units of X because it is the short
// axis and it ranges from cbrt(n) to sqrt(n). If we did one
// unit of work for each X coordinate we would have an O(sqrt(n))
// algorithm. But because there is only one lattice point on a
// line with slope m per the denominator of m in X and because
// the denominator of m roughly doubles for each subdivision,
// there will be less than one unit of work for each unit of X.
// As a result, each iteration reduces the work by about
// a factor of two resulting in 1 + 2 + 4 + ... + sqrt(r) steps
// or O(sqrt(r)). Since the sum of the sizes of the top-level
// regions is O(sqrt(n)), this gives a O(n^(1/4)) algorithm for
// nearly constant curvature.
// However, since the hyperbola is increasingly non-circular for small
// values of x, the subdivision is not nearly as beneficial (and
// also not symmetric) so it is only worthwhile to use region
// subdivision on regions where cubrt(n) < n < sqrt(n).
// The sqrt(n) bound comes from symmetry and the Dirichlet
// hyperbola method (which we also use). The cubrt(n)
// bound comes from the fact that the second deriviative H''(x)
// exceeds one at (2n)^(1/3) ~= 1.26*cbrt(n). Since we process
// regions with adjacent integral slopes at the top level, by the
// time we get to cbrt(n), the size of the region is at most
// one, so we might as well process those values using the
// naive approach of summing y = n/x.
// Finally, at some point the region becomes small enough and we
// can just count points under the hyperbola using whichever axis
// is shorter. This is quite a bit harder than computing y = n/x
// because the transformations we are using result in a general
// quadratic in two variables. Nevertheless, with some
// preliminary calculations, each value can be calculated with
// a few additions, a square root and a division.
// Sum the lattice points.
var sum = (Integer)0;
// Process regions on the stack.
while (true)
{
// Process regions iteratively.
while (true)
{
// Nothing left process.
if (w <= 0 || h <= 0)
{
break;
}
// Check whether the point at (w, 1) is inside the hyperbola.
if ((b2 * w - b1 + x0) * (a1 - a2 * w + y0) <= n)
{
// Remove the first row.
sum += w;
x0 -= b1;
y0 += a1;
--h;
if (h == 0)
{
break;
}
}
// Check whether the point at (1, h) is inside the hyperbola.
if ((b2 - b1 * h + x0) * (a1 * h - a2 + y0) <= n)
{
// Remove the first column.
sum += h;
x0 += b2;
y0 -= a2;
--w;
if (w == 0)
{
break;
}
}
// Invariants for the remainder of the processing of the region:
// H(u,v) at v=h, 0 <= u < 1
// H(u,v) at u=w, 0 <= v < 1
// -du/dv at v=h >= 0
// -dv/du at u=w >= 0
// In other words: the hyperbola is less than one unit away
// from the axis at P1 and P2 and the distance from the axis
// to the hyperbola increases monotonically as you approach
// (u, v) = (0, 0).
Debug.Assert((b2 - b1 * h + x0) * (a1 * h - a2 + y0) > n);
Debug.Assert((b2 * w - b1 + x0) * (a1 - a2 * w + y0) > n);
Debug.Assert(b2 * a1 - a2 * b1 == 1);
// Find the pair of points (u3, v3) and (u4, v4) below H(u,v) where:
// -dv/du at u=u3 >= 1
// -dv/du at u=u4 <= 1
// u4 = u3 + 1
// Specifically, solve:
// (a1*(v+c2)-a2*(u+c1))*(b2*(u+c1)-b1*(v+c2)) = n at dv/du = -1
// Then u3 = floor(u) and u4 = u3 + 1.
// Note that there are two solutions, one negative and one positive.
// We take the positive solution.
// We use the identity (a >= 0, b >= 0; a, b, elements of Z):
// floor(b*sqrt(a/c)) = floor(sqrt(floor(b^2*a/c)))
// to enable using integer arithmetic.
// Formula:
// u = (a1*b2+a2*b1+2*a1*b1)*sqrt(n/(a3*b3))-c1
var c1 = a1 * x0 + b1 * y0;
var c2 = a2 * x0 + b2 * y0;
var a3 = a1 + a2;
var b3 = b1 + b2;
var coef = a1 * b2 + b1 * a2;
var denom = 2 * a1 * b1;
var sqrtcoef = coef + denom;
var u3 = IntegerMath.FloorSquareRoot(sqrtcoef * sqrtcoef * n / (a3 * b3)) - c1;
var u4 = u3 + 1;
// Finally compute v3 and v4 from u3 and u4 by solving
// the hyperbola for v.
// Note that there are two solutions, both positive.
// We take the smaller solution (nearest the u axis).
// Formulas:
// v = ((a1*b2+a2*b1)*(u+c1)-sqrt((u+c1)^2-4*a1*b1*n))/(2*a1*b1)-c2
// u = ((a1*b2+a2*b1)*(v+c2)-sqrt((v+c2)^2-4*a2*b2*n))/(2*a2*b2)-c1
var uc1 = u3 + c1;
var a = uc1 * uc1 - 2 * denom * n;
var b = uc1 * coef;
var v3 = u3 != 0 ? (b - IntegerMath.CeilingSquareRoot(a)) / denom - c2 : h;
var v4 = (b + coef - IntegerMath.CeilingSquareRoot(a + 2 * uc1 + 1)) / denom - c2;
Debug.Assert(u3 < w);
// Compute the V intercept of L3 and L4. Since the lines are diagonal the intercept
// is the same on both U and V axes and v13 = u03 and v14 = u04.
var r3 = u3 + v3;
var r4 = u4 + v4;
Debug.Assert(IntegerMath.Abs(r3 - r4) <= 1);
// Count points horizontally or vertically if one axis collapses (or is below our cutoff)
// or if the triangle exceeds the bounds of the rectangle.
if (u3 <= smallRegionCutoff || v4 <= smallRegionCutoff || r3 > h || r4 > w)
{
if (h > w)
{
sum += CountPoints(w, c1, c2, coef, denom);
}
else
{
sum += CountPoints(h, c2, c1, coef, 2 * a2 * b2);
}
break;
}
// Add the triangle defined L1, L2, and smaller of L3 and L4.
var size = IntegerMath.Min(r3, r4);
sum += size * (size - 1) / 2;
// Adjust for the difference (if any) between L3 and L4.
if (r3 != r4)
{
sum += r3 > r4 ? u3 : v4;
}
// Push left region onto the stack.
stack.Push(new Region(u3, h - r3, a1, b1, a3, b3, x0 - b1 * r3, y0 + a1 * r3));
// Process right region iteratively (no change to a2 and b2).
w -= r4;
h = v4;
a1 = a3;
b1 = b3;
x0 = x0 + b2 * r4;
y0 = y0 - a2 * r4;
}
// Any more regions to process?
if (stack.Count == 0)
{
break;
}
// Pop a region off the stack for processing.
var region = stack.Pop();
w = region.w;
h = region.h;
a1 = region.a1;
b1 = region.b1;
a2 = region.a2;
b2 = region.b2;
x0 = region.x0;
y0 = region.y0;
}
// Return the sum of lattice points in this region.
return(sum);
}
private Integer ProcessRegion(Integer w, Integer h, Integer a1, Integer b1, Integer a2, Integer b2, Integer x0, Integer y0)
{
// The hyperbola is defined by H(x, y): x*y = n.
// Line L0 has slope m0 = -a2/b2.
// Line L1 has slope m1 = -a1/b1.
// Both lines pass through P01 = (x0, y0).
// The region is a parallelogram with the left side bounded L1,
// the bottom bounded by L0, with width w (along L0) and height h
// (along L1). The lower-left corner is P01 (the intersection of
// L0 and L1) and represents (u, v) = (0, 0).
// Both w and h are counted in terms of lattice points, not length.
// For the purposes of counting, the lattice points on lines L0 and L1
// have already been counted.
// Note that b2*a1 - a2*b1 = 1 because
// m0 and m1 are Farey neighbors, e.g. 1 & 2 or 3/2 & 2 or 8/5 & 5/3
// The equations that define (u, v) in terms of (x, y) are:
// u = b1*(y-y0)+a1*(x-x0)
// v = b2*(y-y0)+a2*(x-x0)
// And therefore the equations that define (x, y) in terms of (u, v) are:
// x = x0-b1*v+b2*u
// y = y0+a1*v-a2*u
// Since all parameters are integers and b2*a1 - a2*b1 = 1,
// every lattice point in (x, y) is a lattice point in (u, v)
// and vice-versa.
// Geometrically, the UV coordinate system is the composition
// of a translation and two shear mappings. The UV-based hyperbola
// is essentially a "mini" hyperbola that resembles the full
// hyperbola in that:
// - The equation is still a hyperbola (although it is now a quadratic in two variables)
// - The endpoints of the curve are roughly tangent to the axes
// We process the region by "lopping off" the maximal isosceles
// right triangle in the lower-left corner and then processing
// the two remaining "slivers" in the upper-left and lower-right,
// which creates two smaller "micro" hyperbolas, which we then
// process recursively.
// When we are in the region of the original hyperbola where
// the curvature is roughly constant, the deformed hyperbola
// will in fact resemble a circular arc.
// A line with -slope = 1 in UV-space has -slope = (a2+a1)/(b2+b1)
// in XY-space. We call this m2 and the line defining the third side
// of the triangle as L2 contain point P2 tangent to the hyperbola.
// This is all slightly complicated by the fact that diagonal that
// defines the region that we "lop off" may be broken and shifted
// up or down near the tangent point. As a result we actually have
// P2a and P2b and L2a and L2b.
// We can measure work in units of X because it is the short
// axis and it ranges from cbrt(n) to sqrt(n). If we did one
// unit of work for each X coordinate we would have an O(sqrt(n))
// algorithm. But because there is only one lattice point on a
// line with slope m per the denominator of m in X and because
// the denominator of m roughly doubles for each subdivision,
// there will be less than one unit of work for each unit of X.
// As a result, each iteration reduces the work by about
// a factor of two resulting in 1 + 2 + 4 + ... + sqrt(r) steps
// or O(sqrt(r)). Since the sum of the sizes of the top-level
// regions is O(sqrt(n)), this gives a O(n^(1/4)) algorithm for
// nearly constant curvature.
// However, since the hyperbola is increasingly non-circular for small
// values of x, the subdivision is not nearly as beneficial (and
// also not symmetric) so it is only worthwhile to use region
// subdivision on regions where cubrt(n) < n < sqrt(n).
// The sqrt(n) bound comes from symmetry and the Dirichlet
// hyperbola method, which we also use. The cubrt(n)
// bound comes from the fact that the second deriviative H''(x)
// exceeds one at (2n)^(1/3) ~= 1.26*cbrt(n). Since we process
// regions with adjacent integral slopes at the top level, by the
// time we get to cbrt(n), the size of the region is at most
// one, so we might as well process those values using the
// naive approach of summing y = n/x.
// Finally, at some point the region becomes small enough and we
// can just count points under the hyperbola using whichever axis
// is shorter. This is quite a bit harder than computing y = n/x
// because the transformations we are using result in a general
// quadratic in two variables. Nevertheless, with some
// preliminary calculations, each value can be calculated with
// a few additions, a square root and a division.
// Sum the lattice points.
var sum = (Integer)0;
// Process regions on the stack.
while (true)
{
// Process regions iteratively.
while (true)
{
// Nothing left process.
if (w <= 0 || h <= 0)
{
break;
}
// Check whether the point at (w, 1) is inside the hyperbola.
if ((b2 * w - b1 + x0) * (a1 - a2 * w + y0) <= n)
{
// Remove the first row.
sum += w;
x0 -= b1;
y0 += a1;
--h;
if (h == 0)
{
break;
}
}
// Check whether the point at (1, h) is inside the hyperbola.
if ((b2 - b1 * h + x0) * (a1 * h - a2 + y0) <= n)
{
// Remove the first column.
sum += h;
x0 += b2;
y0 -= a2;
--w;
if (w == 0)
{
break;
}
}
// Invariants for the remainder of the processing of the region:
// H(u,v) at v=h, 0 <= u < 1
// H(u,v) at u=w, 0 <= v < 1
// -du/dv at v=h >= 0
// -dv/du at u=w >= 0
// In other words: the hyperbola is less than one unit away
// from the axis at P0 and P1 and the distance from the axis
// to the hyperbola increases monotonically as you approach
// (u, v) = (0, 0).
Debug.Assert((b2 - b1 * h + x0) * (a1 * h - a2 + y0) > n);
Debug.Assert((b2 * w - b1 + x0) * (a1 - a2 * w + y0) > n);
Debug.Assert(b2 * a1 - a2 * b1 == 1);
// Find the pair of points (u2a, v2a) and (u2b, v2b) below H(u,v) where:
// -dv/du at u=u2a >= 1
// -dv/du at u=u2b <= 1
// u2b = u2a + 1
// Specifically, solve:
// (x0 - b1*v + b2*u)*(y0 + a1*v - a2*u) = n at dv/du = -1
// and solve for the line tan = u + v tangent passing through that point.
// Then u2a = floor(u) and u2b = u2a + 1.
// Finally compute v2a and v2b from u2a and u2b using the tangent line
// which may result in a value too small by at most one.
// Note that there are two solutions, one negative and one positive.
// We take the positive solution.
// We use the identities (a >= 0, b >= 0, c > 0; a, b, c elements of Z):
// floor(b*sqrt(a)/c) = floor(floor(sqrt(b^2*a))/c)
// floor(b*sqrt(a*c)/c) = floor(sqrt(b^2*a/c))
// to enable using integer arithmetic.
// Formulas:
// a3b3 = b3*a3, mxy1 = b1*y0+a1*x0, mxy2 = b3*y0+a3*x0
// u = floor((2*b1*a3+1)*sqrt(a3b3*n)/a3b3-mxy1)
// v = floor(-u+2*sqrt(a3b3*n)-mxy2)
var a3 = a1 + a2;
var b3 = b1 + b2;
var a3b3 = a3 * b3;
var mxy1 = a1 * x0 + b1 * y0;
var mxy2 = a3 * x0 + b3 * y0;
var sqrtcoef = 2 * b1 * a3 + 1;
var tan = IntegerMath.FloorSquareRoot(2 * 2 * a3b3 * n) - mxy2;
var u2a = IntegerMath.FloorSquareRoot(sqrtcoef * sqrtcoef * n / a3b3) - mxy1;
var v2a = u2a != 0 ? tan - u2a : h;
var u2b = u2a < w ? u2a + 1 : w;
var v2b = tan - u2b;
// Check for under-estimate of v2a and/or v2b.
if (u2a != 0)
{
var v2aplus = v2a + 1;
if ((b2 * u2a - b1 * v2aplus + x0) * (a1 * v2aplus - a2 * u2a + y0) <= n)
{
++v2a;
}
}
var v2bplus = v2b + 1;
if ((b2 * u2b - b1 * v2bplus + x0) * (a1 * v2bplus - a2 * u2b + y0) <= n)
{
++v2b;
}
// Compute the V intercept of L2a and L2b. Since the lines are diagonal the intercept
// is the same on both U and V axes and v12a = u02a and v12b = u02b.
var v12a = u2a + v2a;
var v12b = u2b + v2b;
Debug.Assert(IntegerMath.Abs(v12a - v12b) >= 0 && IntegerMath.Abs(v12a - v12b) <= 1);
// Count points horizontally or vertically if one axis collapses (or is below our cutoff)
// or if the triangle exceeds the bounds of the rectangle.
if (u2a <= smallRegionCutoff || v2b <= smallRegionCutoff || v12a > w || v12b > h)
{
if (h > w)
{
sum += CountPoints(true, w, a2, b2, a1, b1, x0, y0);
}
else
{
sum += CountPoints(false, h, a1, b1, a2, b2, x0, y0);
}
break;
}
// Add the triangle defined L0, L1, and smaller of L2a and L2b.
var v12 = IntegerMath.Min(v12a, v12b);
sum += v12 * (v12 - 1) / 2;
// Adjust for the difference (if any) between L2a and L2b.
if (v12a != v12b)
{
sum += v12a > v12b ? u2a : v2b;
}
// Push left region onto the stack.
stack.Push(new Region(u2a, h - v12a, a1, b1, a3, b3, x0 - b1 * v12a, y0 + a1 * v12a));
// Process right region iteratively (no change to a2 and b2).
w -= v12b;
h = v2b;
a1 = a3;
b1 = b3;
x0 = x0 + b2 * v12b;
y0 = y0 - a2 * v12b;
}
// Any more regions to process?
if (stack.Count == 0)
{
break;
}
// Pop a region off the stack for processing.
var region = stack.Pop();
w = region.w;
h = region.h;
a1 = region.a1;
b1 = region.b1;
a2 = region.a2;
b2 = region.b2;
x0 = region.x0;
y0 = region.y0;
}
// Return the sum of lattice points in this region.
return(sum);
}
