Documentation

The basic idea is to concat all the Yes's and skip over the No's. The one wrinkle is if everything is No, then in that case return an arbitrary one of the No's. This can be done in linear time in the length of the list. Call that length n. Check for all No. This takes time proportional to n. If it is true return one of them. If it is not true. Concat all the Yes, and skip all the No. We find the first No (if it exist), and all the Yes by partitioning the list This is similar in spirit to Constrained.GenT.catGEs, but doesn't require a a Monad to escape on the first No.

Given count, return a list if pairs, that add to count splitsOf 6 --> [(1,5),(2,4),(3,3)]. Note we don't return reflections like (5,1) and (4,2), as they have the same information as (1,5) and (2,4).

Given a Path, find a representative solution, ans, for that path, such that 1) (length ans) == count, 2) (sum ans) == total 3) (all p ans) is True What is a path? Suppose i==5, then we recursively explore every way to split 5 into split pairs that add to 5. I.e. (1,4) (2,3), then we split each of those. Here is a picture of the graph of all paths for i==5. A path goes from the root '5' to one of the leaves. Note all leaves are count == '1 (where the solution is '[total]'). To solve for 5, we could solve either of the sub problems rooted at 5: [1,4] or [2,3]. In pickAll we will try to solve both, but in pick1, we only attempt 1 of those sub problems. 5 | [1,4] | | | [1,3] | | | | | [1,2] | | | | | [1,1] | | | [2,2] | | | | | [1,1] | | | [1,1] | [2,3] | | | [1,2] | | | [1,1] [1,1] In pickAll will explore a path for every split of count so if it returns (No _), we can be somewhat confidant that no solution exists. Note that count of 1 and 2, are base cases. When count is greater than 1, we need to sample from [smallest..total], so smallest better be less that or equal to total

If the sample is small enough, then enumerate all of it, otherwise take a fair sample.

Generates a fair sample of numbers between smallest and largest. makes sure there are numbers of all sizes. Controls both the size of the sample and the precision (how many powers of 10 are covered) Here is how we generate one sample when we call (fair (-3455) (10234) 12 3 True) raw = [(-9999,-1000),(-999,-100),(-99,-10),(-9,-1),(0,9),(10,99),(100,999),(1000,9999),(10000,99999)] ranges = [(-3455,-1000),(-999,-100),(-99,-10),(-9,-1),(0,9),(10,99),(100,999),(1000,9999),(10000,10234)] count = 4 largePrecision = [(10000,10234),(1000,9999),(100,999)] smallPrecision = [(-3455,-1000),(-999,-100),(-99,-10)] answer generated = [10128,10104,10027,10048,4911,7821,5585,2157,448,630,802,889] isLarge==True means be biased towards the large end of the range, isLArge==False means be biased towards the small end of the range,

like (logBase10 n), except negative answers mean negative numbers, rather than fractions less than 1.