Draw a random partition of the input vector xs from the uniform distribution on partitions. This proceeds by randomizing the placement of each breakpoint, in other words by walking a random path in a perfect binary tree. O(n) for a vector length n.

This function preserves the order of the input list.

Partition a vector v according to groupings provided by the bits bs. If the first set bit in bs is at a position larger than the last index of v, this returns [v]. More generally, bits set at positions after the last index of v do not contribute to the grouping. bs == 0 always results in [v].

See partitionFromBits for other examples.

Examples

>>> import qualified Data.Vector as V
>>> partitionFromBits 5 (V.fromList [0..2::Int])
[[0],[1],[2]]
>>> partitionFromBits 13 (V.fromList [0..2::Int])
[[0],[1],[2]]
>>> partitionFromBits 4 (V.fromList [0..2::Int])
[[0,1],[2]]
>>> partitionFromBits 8 (V.fromList [0..2::Int])
[[0,1,2]]

Select a partition of [0..n-1] into disjoint cycle graphs, uniformly over the n! possibilities. The sampler relies on the fact that such partitions are isomorphic with the permutations of [0..n-1] via the map sending a permutation sigma to the edge set {(i, sigma(i))}_0^{n-1}. In other words, the cycle partition graphs are isomorphic with the rotation matrices.

Therefore, this function simply calls uniformPermutation and tuples the result with its indices. The returned value is a vector of edges. O(n), since uniformPermutation implements the Fisher-Yates shuffle.

uniformPermutation uses in-place mutation, so this function must be run in a PrimMonad context.

Examples

>>> import System.Random.Stateful
>>> import qualified RandomCycle.Vector as RV
>>> import Data.Vector (Vector)
>>> runSTGen_ (mkStdGen 1305) $ RV.uniformCyclePartition 4 :: Vector (Int, Int)
[(0,1),(1,3),(2,2),(3,0)]

Uniform selection of a cycle partition graph of [0..n-1] elements, conditional on an edge-wise predicate. See uniformCyclePartition for details on the sampler.

O(n/p), where p is the probability that a uniformly sampled cycle partition graph (over all n! possible) satisfies the conditions. This can be highly non-linear since p in general is a function of n.

Since this is a rejection sampling method, the user is asked to provide a counter for the maximum number of sampling attempts in order to guarantee termination in cases where the edge predicate has probability of success close to zero.

Note this will return pure Nothing if given a number of vertices that is non-positive, in the third argument, unlike uniformCyclePartition which will throw an error.

Examples

>>> import System.Random.Stateful
>>> import qualified RandomCycle.Vector as RV
>>> import Data.Vector (Vector)
>>> -- No self-loops
>>> rule = uncurry (/=)
>>> n = 5
>>> maxit = n * 1000
>>> runSTGen_ (mkStdGen 3) $ RV.uniformCyclePartitionThin maxit rule n
Just [(0,2),(1,3),(2,0),(3,4),(4,1)]