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package org.hammerlab.math
import scala.collection.immutable.SortedSet
import scala.math.{ exp, log, max, min }
/**
* Some utilities for generating exponential sequences of integers that can be used as e.g. histogram-bucket boundaries.
*/
object Steps {
/**
* Divide [0, maxDepth] into N geometrically-evenly-spaced steps (of size ≈maxDepth^(1/N)).
*
* Until the k-th step is bigger than k, the whole number k is used in its stead.
*/
def geometricEvenSteps(maxDepth: Int, N: Int = 100): SortedSet[Int] = {
val logMaxDepth = log(maxDepth)
SortedSet(
0 ::
(
for {
i ← 1 until N
} yield
min(
maxDepth,
max(
i,
exp(
(i - 1) * logMaxDepth / (N - 2)
)
.toInt
)
)
)
.toList: _*
)
}
/**
* Produce a set of "round numbers" between 0 and a provided N, inclusive.
*
* Coverage is relatively dense but the total number of sampled/returned integers is still O(log(N)) in the input N;
* specifically, 35 integers are returned in each factor-of-10 window (detailed below).
*
* The absolute difference between consecutive integers is non-decreasing over the entire range and, (after the [0,10]
* interval), no two consecutive integers returned are more than 10% different from one another.
*
*
* 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, base case: include all of [0, 10].
* 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, step by one from 10% to 20% of the next power of 10 (100 here).
* 20, 22, 24, 26, 28,
* 30, 32, 34, 36, 38,
* 40, 42, 44, 46, 48, step by two from 20% to 50% of the next power of 10.
* 50, 55,
* 60, 65,
* 70, 75,
* 80, 85,
* 90, 95, step by five from 50% to 100% of the next power of 10.
*
* …then repeat the [10, 95] portion, multiplied by powers of 10:
*
* 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, this is 10x the "steps by one" section above.
* 200, 220, 240, 260, 280,
* 300, 320, 340, 360, 380, likewise, 10x the "steps by two" from above.
*
* …etc.
*/
def roundNumbers(maxDepth: Int): SortedSet[Int] =
SortedSet(
(
(0 until 10) ++
RoundNumbers(
(10 until 20) ++ (20 until 50 by 2) ++ (50 until 100 by 5),
maxDepth,
10
)
.toSeq
): _*
)
}
