zhengruifeng commented on a change in pull request #27758: [SPARK-31007][ML]
KMeans optimization based on triangle-inequality
URL: https://github.com/apache/spark/pull/27758#discussion_r386809482
##########
File path:
mllib/src/main/scala/org/apache/spark/mllib/clustering/DistanceMeasure.scala
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@@ -234,6 +342,39 @@ private[spark] object EuclideanDistanceMeasure {
}
private[spark] class CosineDistanceMeasure extends DistanceMeasure {
+
+ /**
+ * @return Radii of centers. If distance between point x and center c is
less than
+ * the radius of center c, then center c is the closest center to
point x.
+ * For Cosine distance, it is similar to Euclidean distance.
However, here
+ * radian/angle is used instead of Cosine distance: for center c,
finding
+ * its closest center, computing the radian/angle between them,
halving the
+ * radian/angle, and converting it back to Cosine distance at the
end.
+ */
+ override def computeRadii(centers: Array[VectorWithNorm]): Array[Double] = {
+ val k = centers.length
+ if (k == 1) {
+ Array(Double.NaN)
+ } else {
+ val distances = Array.fill(k)(Double.PositiveInfinity)
+ var i = 0
+ while (i < k) {
+ var j = i + 1
+ while (j < k) {
+ val d = distance(centers(i), centers(j))
+ if (d < distances(i)) distances(i) = d
+ if (d < distances(j)) distances(j) = d
+ j += 1
+ }
+ i += 1
+ }
+
+ // d = 1 - cos(x)
+ // r = 1 - cos(x/2) = 1 - sqrt((cos(x) + 1) / 2) = 1 - sqrt(1 - d/2)
+ distances.map(d => 1 - math.sqrt(1 - d / 2))
Review comment:
Yes, Cosine distance doesn't obey the triangle inequality, but the following
lemma should be available to apply:
given a point x, and let b and c be centers. If angle(x, b)<angle(b,c)/2,
then angle(x,b)<angle(x,c),
cos_distance(x,b)=1-cos(x,b)<cos_distance(x,c)=1-cos(x,c)
That is because: [PRINCIPLES FROM
GEOMETRY](http://www.angelfire.com/nt/navtrig/B1.html)
> Each side of a spherical triangle is less than the sum of the other two.
angle(x,b) + angle(x,c) > angle(b,c)
angle(x,b) < angle(b,c)/2
=> angle(x,c) > angle(b,c)/2 > angle(x,b)
=> cos_distance(x,c) > cos_distance(x,b)
angle(x,b) < angle(b,c)/2
<=> cos(x,b) > sqrt{ (cos(b,c) + 1)/2 }
<=> cos_distance(x,b) < 1 - sqrt{ (cos(b,c) + 1)/2 } = 1 - sqrt{ 1 -
cos_distance(b,c) / 2 }
=> Give two centers b and c, if point x has cos_distance(x,b) < 1 - sqrt{ 1
- cos_distance(b,c) / 2 }, then point x belongs to center b.
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