Calculate the number of trees in which Fitch parsimony will reconstruct m steps, where a leaves are labelled with one state, and b leaves are labelled with a second state.

Carter1(m, a, b)

Log2Carter1(m, a, b)

LogCarter1(m, a, b)

Arguments

m

Number of steps.

a, b

Number of leaves labelled 0 and 1.

Details

Implementation of theorem 1 from Carter et al. (1990)

References

Carter M, Hendy M, Penny D, Székely LA, Wormald NC (1990). “On the distribution of lengths of evolutionary trees.” SIAM Journal on Discrete Mathematics, 3(1), 38--47. doi:10.1137/0403005 .

Steel MA (1993). “Distributions on bicoloured binary trees arising from the principle of parsimony.” Discrete Applied Mathematics, 41(3), 245--261. doi:10.1016/0166-218X(90)90058-K .

Steel M, Charleston M (1995). “Five surprising properties of parsimoniously colored trees.” Bulletin of Mathematical Biology, 57(2), 367--375. doi:10.1016/0092-8240(94)00051-D .

(Steel M, Goldstein L, Waterman MS (1996). “A central limit theorem for the parsimony length of trees.” Advances in Applied Probability, 28(4), 1051--1071. doi:10.2307/1428164 . )

Other profile parsimony functions: PrepareDataProfile(), StepInformation(), WithOneExtraStep(), profiles

Examples

# The character 0 0 0 1 1 1
Carter1(1, 3, 3) # Exactly one step
#> [1] 9
Carter1(2, 3, 3) # Two steps (one extra step)
#> [1] 54

# Number of trees that the character can map onto with exactly _m_ steps
# if non-parsimonious reconstructions are permitted:
cumsum(sapply(1:3, Carter1, 3, 3))
#> [1]   9  63 105

# Three steps allow the character to map onto any of the 105 six-leaf trees.