© Copyright 1986-2002 by the University of Washington. Written by Joseph Felsenstein. Permission is granted to copy this document provided that no fee is charged for it and that this copyright notice is not removed.
MIX is a general parsimony program which carries out the Wagner and Camin-Sokal parsimony methods in mixture, where each character can have its method specified separately. The program defaults to carrying out Wagner parsimony.
The Camin-Sokal parsimony method explains the data by assuming that changes 0 --> 1 are allowed but not changes 1 --> 0. Wagner parsimony allows both kinds of changes. (This under the assumption that 0 is the ancestral state, though the program allows reassignment of the ancestral state, in which case we must reverse the state numbers 0 and 1 throughout this discussion). The criterion is to find the tree which requires the minimum number of changes. The Camin-Sokal method is due to Camin and Sokal (1965) and the Wagner method to Eck and Dayhoff (1966) and to Kluge and Farris (1969).
Here are the assumptions of these two methods:
That these are the assumptions of parsimony methods has been documented in a series of papers of mine: (1973a, 1978b, 1979, 1981b, 1983b, 1988b). For an opposing view arguing that the parsimony methods make no substantive assumptions such as these, see the papers by Farris (1983) and Sober (1983a, 1983b), but also read the exchange between Felsenstein and Sober (1986).
The input for MIX is the standard input for discrete characters programs, described above in the documentation file for the discrete-characters programs. States "?", "P", and "B" are allowed.
The options are selected using a menu:
Mixed parsimony algorithm, version 3.6a3 Settings for this run: U Search for best tree? Yes X Use Mixed method? No P Parsimony method? Wagner J Randomize input order of species? No. Use input order O Outgroup root? No, use as outgroup species 1 T Use Threshold parsimony? No, use ordinary parsimony A Use ancestral states in input file? No W Sites weighted? No M Analyze multiple data sets? No 0 Terminal type (IBM PC, ANSI, none)? (none) 1 Print out the data at start of run No 2 Print indications of progress of run Yes 3 Print out tree Yes 4 Print out steps in each character No 5 Print states at all nodes of tree No 6 Write out trees onto tree file? Yes Are these settings correct? (type Y or the letter for one to change) |
The options U, X, J, O, T, A, and M are the usual User Tree, miXed methods, Jumble, Outgroup, Ancestral States, and Multiple Data Sets options, described either in the main documentation file or in the Discrete Characters Programs documentation file. The user-defined trees supplied if you use the U option must be given as rooted trees with two-way splits (bifurcations). The O option is acted upon only if the final tree is unrooted and is not a user-defined tree. One of the important uses of the the O option is to root the tree so that if there are any characters in which the ancestral states have not been specified, the program will print out a table showing which ancestral states require the fewest steps. Note that when any of the characters has Camin-Sokal parsimony assumed for it, the tree is rooted and the O option will have no effect.
The option P toggles between the Camin-Sokal parsimony criterion and the default Wagner parsimony criterion. Option X invokes mixed-method parsimony. If the A option is invoked, the ancestor is not to be counted as one of the species.
The F (Factors) option is not available in this program, as it would have no effect on the result even if that information were provided in the input file.
Output is standard: a list of equally parsimonious trees, which will be printed as rooted or unrooted depending on which is appropriate, and, if the user chooses, a table of the number of changes of state required in each character. If the Wagner option is in force for a character, it may not be possible to unambiguously locate the places on the tree where the changes occur, as there may be multiple possibilities. If the user selects menu option 5, a table is printed out after each tree, showing for each branch whether there are known to be changes in the branch, and what the states are inferred to have been at the top end of the branch. If the inferred state is a "?" there will be multiple equally-parsimonious assignments of states; the user must work these out for themselves by hand.
If the Camin-Sokal parsimony method is invoked and the Ancestors option is also used, then the program will infer, for any character whose ancestral state is unknown ("?") whether the ancestral state 0 or 1 will give the fewest state changes. If these are tied, then it may not be possible for the program to infer the state in the internal nodes, and these will all be printed as ".". If this has happened and you want to know more about the states at the internal nodes, you will find helpful to use MOVE to display the tree and examine its interior states, as the algorithm in MOVE shows all that can be known in this case about the interior states, including where there is and is not amibiguity. The algorithm in MIX gives up more easily on displaying these states.
If the A option is not used, then the program will assume 0 as the ancestral state for those characters following the Camin-Sokal method, and will assume that the ancestral state is unknown for those characters following Wagner parsimony. If any characters have unknown ancestral states, and if the resulting tree is rooted (even by outgroup), a table will also be printed out showing the best guesses of which are the ancestral states in each character. You will find it useful to understand the difference between the Camin-Sokal parsimony criterion with unknown ancestral state and the Wagner parsimony criterion.
If the U (User Tree) option is used and more than one tree is supplied, the program also performs a statistical test of each of these trees against the best tree. This test, which is a version of the test proposed by Alan Templeton (1983) and evaluated in a test case by me (1985a). It is closely parallel to a test using log likelihood differences invented by Kishino and Hasegawa (1989), and uses the mean and variance of step differences between trees, taken across characters. If the mean is more than 1.96 standard deviations different then the trees are declared significantly different. The program prints out a table of the steps for each tree, the differences of each from the highest one, the variance of that quantity as determined by the step differences at individual sites, and a conclusion as to whether that tree is or is not significantly worse than the best one. It is important to understand that the test assumes that all the binary characters are evolving independently, which is unlikely to be true for many suites of morphological characters.
If the U (User Tree) option is used and more than one tree is supplied, the program also performs a statistical test of each of these trees against the best tree. This test, which is a version of the test proposed by Alan Templeton (1983) and evaluated in a test case by me (1985a). It is closely parallel to a test using log likelihood differences invented by Kishino and Hasegawa (1989), and uses the mean and variance of step differences between trees, taken across characters. If the mean is more than 1.96 standard deviations different then the trees are declared significantly different. The program prints out a table of the steps for each tree, the differences of each from the highest one, the variance of that quantity as determined by the step differences at individual characters, and a conclusion as to whether that tree is or is not significantly worse than the best one. It is important to understand that the test assumes that all the binary characters are evolving independently, which is unlikely to be true for many suites of morphological characters.
If there are more than two trees, the test done is an extension of the KHT test, due to Shimodaira and Hasegawa (1999). They pointed out that a correction for the number of trees was necessary, and they introduced a resampling method to make this correction. In the version used here the variances and covariances of the sums of steps across characters are computed for all pairs of trees. To test whether the difference between each tree and the best one is larger than could have been expected if they all had the same expected number of steps, numbers of steps for all trees are sampled with these covariances and equal means (Shimodaira and Hasegawa's "least favorable hypothesis"), and a P value is computed from the fraction of times the difference between the tree's value and the lowest number of steps exceeds that actually observed. Note that this sampling needs random numbers, and so the program will prompt the user for a random number seed if one has not already been supplied. With the two-tree KHT test no random numbers are used.
In either the KHT or the SH test the program prints out a table of the number of steps for each tree, the differences of each from the lowest one, the variance of that quantity as determined by the differences of the numbers of steps at individual characters, and a conclusion as to whether that tree is or is not significantly worse than the best one.
At the beginning of the program is a constant, maxtrees, the maximum number of trees which the program will store for output.
The program is descended from earlier programs SOKAL and WAGNER which have long since been removed from the PHYLIP package, since MIX has all their capabilites and more.
5 6
Alpha 110110
Beta 110000
Gamma 100110
Delta 001001
Epsilon 001110
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Mixed parsimony algorithm, version 3.6a3
5 species, 6 characters
Wagner parsimony method
Name Characters
---- ----------
Alpha 11011 0
Beta 11000 0
Gamma 10011 0
Delta 00100 1
Epsilon 00111 0
4 trees in all found
+--Epsilon
+-----4
! +--Gamma
+--2
! ! +--Delta
--1 +-----3
! +--Beta
!
+-----------Alpha
remember: this is an unrooted tree!
requires a total of 9.000
steps in each character:
0 1 2 3 4 5 6 7 8 9
*-----------------------------------------
0! 2 2 2 1 1 1
From To Any Steps? State at upper node
( . means same as in the node below it on tree)
1 1?011 0
1 2 no .?... .
2 4 maybe .0... .
4 Epsilon yes 0.1.. .
4 Gamma no ..... .
2 3 yes .?.00 .
3 Delta yes 001.. 1
3 Beta maybe .1... .
1 Alpha maybe .1... .
+--------Gamma
!
+--2 +--Epsilon
! ! +--4
! +--3 +--Delta
--1 !
! +-----Beta
!
+-----------Alpha
remember: this is an unrooted tree!
requires a total of 9.000
steps in each character:
0 1 2 3 4 5 6 7 8 9
*-----------------------------------------
0! 1 2 1 2 2 1
From To Any Steps? State at upper node
( . means same as in the node below it on tree)
1 1?011 0
1 2 no .?... .
2 Gamma maybe .0... .
2 3 maybe .?.?? .
3 4 yes 001?? .
4 Epsilon maybe ...11 .
4 Delta yes ...00 1
3 Beta maybe .1.00 .
1 Alpha maybe .1... .
+--------Epsilon
+--4
! ! +-----Gamma
! +--2
--1 ! +--Delta
! +--3
! +--Beta
!
+-----------Alpha
remember: this is an unrooted tree!
requires a total of 9.000
steps in each character:
0 1 2 3 4 5 6 7 8 9
*-----------------------------------------
0! 2 2 2 1 1 1
From To Any Steps? State at upper node
( . means same as in the node below it on tree)
1 1?011 0
1 4 maybe .0... .
4 Epsilon yes 0.1.. .
4 2 no ..... .
2 Gamma no ..... .
2 3 yes ...00 .
3 Delta yes 0.1.. 1
3 Beta yes .1... .
1 Alpha maybe .1... .
+--------Gamma
+--2
! ! +-----Epsilon
! +--4
--1 ! +--Delta
! +--3
! +--Beta
!
+-----------Alpha
remember: this is an unrooted tree!
requires a total of 9.000
steps in each character:
0 1 2 3 4 5 6 7 8 9
*-----------------------------------------
0! 2 2 2 1 1 1
From To Any Steps? State at upper node
( . means same as in the node below it on tree)
1 1?011 0
1 2 maybe .0... .
2 Gamma no ..... .
2 4 maybe ?.?.. .
4 Epsilon maybe 0.1.. .
4 3 yes ?.?00 .
3 Delta yes 0.1.. 1
3 Beta yes 110.. .
1 Alpha maybe .1... .
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