func makeUmemo()

in internal/stats/udist.go [165:297]


func makeUmemo(twoU, n1 int, t []int) []map[ukey]float64 {
	// Another candidate for a fast implementation is van de Wiel,
	// "The split-up algorithm: a fast symbolic method for
	// computing p-values of distribution-free statistics". This
	// is what's used by R's coin package. It's a comparatively
	// recent publication, so it's presumably faster (or perhaps
	// just more general) than previous techniques, but I can't
	// get my hands on the paper.
	//
	// TODO: ~40% of this function's time is spent in mapassign on
	// the assignment lines in the two loops and another ~20% in
	// map access and iteration. Improving map behavior or
	// replacing the maps altogether with some other constant-time
	// structure could double performance.
	//
	// TODO: The worst case for this function is when there are
	// few ties. Yet the best case overall is when there are *no*
	// ties. Can we get the best of both worlds? Use the fast
	// algorithm for the most part when there are few ties and mix
	// in the general algorithm just where we need it? That's
	// certainly possible for sub-problems where t[:k] has no
	// ties, but that doesn't help if t[0] has a tie but nothing
	// else does. Is it possible to rearrange the ranks without
	// messing up our computation of the U statistic for
	// sub-problems?

	K := len(t)

	// Compute a coefficients. The a slice is indexed by k (a[0]
	// is unused).
	a := make([]int, K+1)
	a[1] = t[0]
	for k := 2; k <= K; k++ {
		a[k] = a[k-1] + t[k-2] + t[k-1]
	}

	// Create the memo table for the counts function, A. The A
	// slice is indexed by k (A[0] is unused).
	//
	// In "The Mann Whitney Distribution Using Linked Lists", they
	// use linked lists (*gasp*) for this, but within each K it's
	// really just a memoization table, so it's faster to use a
	// map. The outer structure is a slice indexed by k because we
	// need to find all memo entries with certain values of k.
	//
	// TODO: The n1 and twoU values in the ukeys follow strict
	// patterns. For each K value, the n1 values are every integer
	// between two bounds. For each (K, n1) value, the twoU values
	// are every integer multiple of a certain base between two
	// bounds. It might be worth turning these into directly
	// indexible slices.
	A := make([]map[ukey]float64, K+1)
	A[K] = map[ukey]float64{ukey{n1: n1, twoU: twoU}: 0}

	// Compute memo table (k, n1, twoU) triples from high K values
	// to low K values. This drives the recurrence relation
	// downward to figure out all of the needed argument triples.
	//
	// TODO: Is it possible to generate this table bottom-up? If
	// so, this could be a pure dynamic programming algorithm and
	// we could discard the K dimension. We could at least store
	// the inputs in a more compact representation that replaces
	// the twoU dimension with an interval and a step size (as
	// suggested by Cheung, Klotz, not that they make it at all
	// clear *why* they're suggesting this).
	tsum := sumint(t) // always ∑ t[0:k]
	for k := K - 1; k >= 2; k-- {
		tsum -= t[k]
		A[k] = make(map[ukey]float64)

		// Construct A[k] from A[k+1].
		for A_kplus1 := range A[k+1] {
			rkLow := maxint(0, A_kplus1.n1-tsum)
			rkHigh := minint(A_kplus1.n1, t[k])
			for rk := rkLow; rk <= rkHigh; rk++ {
				twoU_k := A_kplus1.twoU - rk*(a[k+1]-2*A_kplus1.n1+rk)
				n1_k := A_kplus1.n1 - rk
				if twoUmin(n1_k, t[:k], a) <= twoU_k && twoU_k <= twoUmax(n1_k, t[:k], a) {
					key := ukey{n1: n1_k, twoU: twoU_k}
					A[k][key] = 0
				}
			}
		}
	}

	// Fill counts in memo table from low K values to high K
	// values. This unwinds the recurrence relation.

	// Start with K==2 base case.
	//
	// TODO: Later computations depend on these, but these don't
	// depend on anything (including each other), so if K==2, we
	// can skip the memo table altogether.
	if K < 2 {
		panic("K < 2")
	}
	N_2 := t[0] + t[1]
	for A_2i := range A[2] {
		Asum := 0.0
		r2Low := maxint(0, A_2i.n1-t[0])
		r2High := (A_2i.twoU - A_2i.n1*(t[0]-A_2i.n1)) / N_2
		for r2 := r2Low; r2 <= r2High; r2++ {
			Asum += mathChoose(t[0], A_2i.n1-r2) *
				mathChoose(t[1], r2)
		}
		A[2][A_2i] = Asum
	}

	// Derive counts for the rest of the memo table.
	tsum = t[0] // always ∑ t[0:k-1]
	for k := 3; k <= K; k++ {
		tsum += t[k-2]

		// Compute A[k] counts from A[k-1] counts.
		for A_ki := range A[k] {
			Asum := 0.0
			rkLow := maxint(0, A_ki.n1-tsum)
			rkHigh := minint(A_ki.n1, t[k-1])
			for rk := rkLow; rk <= rkHigh; rk++ {
				twoU_kminus1 := A_ki.twoU - rk*(a[k]-2*A_ki.n1+rk)
				n1_kminus1 := A_ki.n1 - rk
				x, ok := A[k-1][ukey{n1: n1_kminus1, twoU: twoU_kminus1}]
				if !ok && twoUmax(n1_kminus1, t[:k-1], a) < twoU_kminus1 {
					x = mathChoose(tsum, n1_kminus1)
				}
				Asum += x * mathChoose(t[k-1], rk)
			}
			A[k][A_ki] = Asum
		}
	}

	return A
}