Counting Pedigrees up to Isomorphism
1. Introduction
2. Definition
3. Sex-labelled Discrete Generation Model
4. Sex Label a given unlabelled pedigree
5. Unsex-labelled Discrete Generation Model
6. Summary
Appendix
1. Introduction
When investigating problems like reconstruction of pedigrees or the probability of
data given a particular pedigree, interesting questions related to pedigrees arise.
Simulations based on a model of human population history and geography find that an
individual that is the genealogical ancestor of all living humans existed just a few
thousand years ago, as suggested in a recently published article by Prof. Hein.1 One
related fascinating question, which we are going to investigate in this project, is that
how many different pedigrees there actually are up to isomorphism if we trace back a
certain number of generations.
To do this, we will introduce some definitions.
2. Definition
Pedigree:
A pedigree is a directed graph showing the relationship between individuals according
to their parentage. Both of (2.1) and (2.2) below are examples of pedigrees.
Generation 0:
We call the most recent generation ‘generation 0’, in this project, there is only 1
individual in generation 0.
Discrete:
As we said above, the most recent generation is ‘generation 0’, then let the parents of
the individual in generation g be in generation g+1, and allow the case that more than
one generation number are assigned to one single individual. If no individual in the
pedigree belongs to more than one generation, we say the pedigree is discrete.
Pedigree (2.1) as below is an example of a discrete pedigree, while (2.2) being nondiscrete.
(2.1)
Generation 4
Generation 3
Generation 2
Generation 1
Generation 0
1
Hein, Jotun. ‘Pedigrees for all humanity’ Nature Vol 431 September 2004, Page 518-519
(2.2)
3. Sex-labelled Discrete Generation Model
To start off, we are going to concentrate on a simpler version of the original question,
in which we are only interested in the pedigrees which are sex-labelled, and are of
discrete time and generation.
In this particular case, since every individual in a pedigree is labelled by the sex, there
is a unique link between generation 0 (our starting individual) and any other
individuals in the pedigree. This is a very useful characteristic of the sex-labelled
pedigrees, in the sense that it guarantees every individual is uniquely determined,
which saves us a lot of trouble when counting. Also, the discrete time and generation
constraints make it possible for us to adopt a recursion moving back generation by
generation.
The idea of the counting is as follows:
Sampling one individual, the number of grandparents is a natural number between 2
and 4 inclusive. Similarly, the number of different ancestors g generations back in
time (the ancestors in generation g only) is a natural number between 2 to 2g inclusive.
The number of different pedigrees back to g generations can be easily calculated if we
know the number of different pedigrees of 1 generation with c number of children and
p number of parents(c is any natural number from 1 to 2g-1 inclusive, and p is any
natural number from 2 to 2g inclusive). This number, however, can be calculated
easily.
Now, let’s write the idea above out in symbolic terms. Let Ng(p) denote the number
of different pedigrees going back g generations with 1 individual in generation 0 and
p individuals in generation g, and A(c, p) denote the number of different pedigrees of
1 generation with c children and p parents. Adopting these notations, our recursions
can be shown as the following expressions:
A(c, p) 

1i , j  c
i j p
Sc, i  Sc, j
(3.1)
Ng  1( p)   Ng (c)  A(c, p)
(3.2)
c
In Equation (3.1), Sc,j, the Stirling number of second kind, is the number of different
ways to partition c elements into j sets. Here, we assume among those p parents, i of
them are female and j of them male. Since the partitions of females and males are
independent, A(c, p), the number of different pedigrees of 1 generation with c
children and p parents is just the product of Sc,i and Sc, j summing over all possible
i,j-s.
Notice that among these p parents, at least one of them is female and at most p-1 of
them are female, we can rewrite Equation (1.1):
A(c, p ) 
Min[ c , p 1]

Sc, i  Sc, p  i
(3.3)
i 1
Following this recursion algorithm, we can write a programme called ‘countp’ in
Mathematica.
Step 1. Delete any existing function with the name ‘countp’;
Step 2. Declare fuction ‘countp’ to have input ‘generation’, a integer number variable;
Step 3. Declare matrix variables ‘matrixn’, ‘matrixa’, number variable ‘g’, ‘c’, ‘p’;
‘matrixn’—the g,pth entry in matrixn is Ng(p) above.
‘matrixa’—the c,pth entry in matrixa is A(c,p) above.
Step 4. Define matrixn to have ‘generation’ number of rows and 2^‘geneation’
number of columns with 1,2th entry being 1 and all the other entries being 0;
Step 5. Define matrixa to have 2^(‘generation’ – 1) number of rows and 2^‘geneation’
number of columns with all entries being 0;
Step 6. Assign ‘g’ value 2;
Step 7. If g<= ‘generation’, let c = 1;
Otherwise, go to Step 12;
Step 8. If c<= 2^(g-1), let p = 2;
Otherwise, go to Step 11;
Step 9. If p<=2c, let matrixa[[c,p]] = Summation of Sc,j * Sc,p-j over integer values
for j from 1 to the minimum of c and p-1,
let matrixn[[g,p]] =matrixn[[g,p]] + matrixn[[g-1,c]]*matrixa[[c,p]],
p=p+1;
repeat Step 9;
Otherwise, go to Step 10;
Step 10. Let c=c+1, then go to Step 8;
Step 11. Let g=g+1, then go to Step 7;
Step 12. Sum the last row of matrixn and return the value.
(Matrix[[m,n]] denote the m,n th entry in matrix called ‘Matrix’)
Using the programme we wrote in Mathematica, we have got the following result:
Generation 2
1
1
1
2
3
0
2
Number of Parents
4
5
6
0
0
0
1
0
0
7
0
0
8
0
0
Total
1
4
4
3
28
84
98
52
12
1
279
Number of different sex labelled pedigrees (#)
2
3
4
5
6
Generation 1
1
4
279
2.8766*10^7 2.81597*10^20 7.40974*10^52
#
8
9
10
Generation 7
#
2.76739*10^131 2.88199*10^317 3.52352*10^749 3.89498*10^1737
From the numbers about, we can see that the number of different sex labelled
pedigrees grows very fast with generations. If we trace for 3 generations, then there
are 279 sex-labelled pedigrees up to isomorphism, and if we trace back one generation
more, the number of pedigrees up to isomorphism rises up to 2.8766*10^7! The
recursion programme in Mathematica we used above can calculate up to 10
generations. Beyond that, calculations get very slow.
Tracing back each number of generations, I plotted the log of number of different
pedigrees which have p distinct parents in the earliest generation against p, getting a
log distribution of the number of different pedigrees having a certain number of
parents in the last generation for each generation.
2
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971
978
977
976
975
982
981
980
979
986
985
984
983
990
989
988
987
1003
1002
1001
1000
995
994
993
992
991
1007
1006
1005
1004
999
998
997
996
1011
1010
1009
1008
1015
1014
1013
1012
1019
1018
1017
1016
1023
1022
1021
1020
1024
10 generation
According the bar chart above, we can tell that all of them are of an ‘n’ shape, and the
position of maximum value shifts to the left as the number of generation increases.
The following table gives the position (the number of different parents in the earliest
generation) where there are most different pedigrees for tracing back different
generations.
Generation
Position for most
pedigrees
Most number of
Parents
1
2
2
3
3
5
4
8
5
13
2
4
8
16
32
Generation
Position for most
pedigrees
Most number of
Parents
6
22
7
39
8
68
9
120
10
214
64
128
256
512
1024
4. Sex Label a given unlabelled pedigree
Now since we have an idea how many sex labelled pedigrees up to isomorphism
tracing back to any generation from no more than 10. A natural question to ask would
be how many ways there are to sex label a given pedigree if it is not sex labelled.
Consider the number of ways to sex-label a pedigree with c children and p parents
with 1 generation back only, we notice that there are only two ways to sex-label such
a pedigree if it is not possible to partition all the parents and children into two or more
none empty set such that any two children in different sets share no parent and any
two parents in two different sets have no common child. We call the children and
parents group with the above characteristics ‘closed’.
For any given pedigree with only 2 generations (tracing back only 1 generation) of
finite number of parents and children, it is possible to check the number of closed
groups. Since sex-labelling all these different closed groups are independent, and
there are precisely 2 ways to sex-label each of the closed group, the number of ways
to sex-label a pedigree with 2 generation is 2 to the power of number of different
proper closed groups. To say a closed group is proper, we mean it is able to sex-label
this closed group. Naturally, we call a pedigree that is not able to be sex-labelled
improper. An example of an improper closed group is 3 parents in which any two of
them share a child.
In order to distinguish the improper cases, we can adopt the following algorithm. For
any pedigree with two generations, c children and p parents, we label the parents from
1 to p. Start from parent 1 and assign it value 0, and then assign all the parents who
share at least one children with parent 1 the value 1. Suppose parent m is the one with
smallest ordinal number sharing kid(s) with parent 1(1<m<=p), we move on the
consider parent m. Take all parents who share at least one child with parent m, and in
which, we assign parents who have no sex value assigned yet the opposite sex value
of parent m (0 in this case). For those who share child(ren) with parent m but have got
a sex value already, we check whether their assigned sex is indeed the opposite sex
from parent m, if it fails for any of them, this group is improper.
Let’s have a look at an example. For pedigree (4.1), consider the generation with 1,2,3
as parents and 4,5,6 as children. Assign parent 1 with value 0, and then note that both
4,5 are children of parent 1.For child 4, 2 is the other parent and for child 5, 3 is the
other parent, so we assign value 1 to both parent 2 and parent 3. Then we move on the
parent 2, note that both 4 and 6 are children of parent 2. Since we have done with
child 4, we only need to consider child 6.Parent 3 is the other parent of child 6. The
value assigned to parent 3 was 1, but the value of parent 2 is 1 as well, contradiction!
We cannot sex-label pedigree (4.1).
(4.1)
1
2
4
5
3
6
For similar pedigrees as in Section 3, where there is only one individual in generation
0, we can use the method above to figure out the number of ways to sex-label them.
But we cannot simply adopt it if there are two or more individuals in generation 0.
This is because the individuals in generation 0 are identical and will cause further
counting repetitions if we don’t adjust the method as above. Also, note that this only
works for pedigrees of discrete generation, since for non-discrete pedigrees, there may
only be one way to sex-label the parents of some individuals.
5. Unsex-labelled Discrete Generation Model
For the case that all individuals are not sex-labelled, but generation discrete, is it
possible to make a similar recursion to calculate the number of different pedigrees up
to isomorphism? The idea of the original attempt is that each time we trace back a
generation, we check on every individual and keep a track on the isomorphic groups
all these individuals form, using a list. For instance, if we go back 2 generations and
the pedigree looks like the following:
(5.1)
Generation 2
Generation 1
Generation 0
We can write it as:
generation 0: 1,0,0,0,…
generation 1: 0,1,0,0,…
generation 2: 1,1,0,0,…
In generation 0, there is only one individual, so there is one isomorphic group with
only 1 element, and no other groups at all, denote it 1,0,0,0,…
In generation 1, there are 2 individuals in total, and they are isomorphic, so that we
write it as 0,1,0,0,… to denote that there are 1 group of 2 isomorphic elements and no
other groups in this generation.
In generation 2, there are 3 individuals in total, two and only two of which are
isomorphic. So, we denote it as 1,1,0,0,… meaning that there are 1 group of 1 element,
and 1 group of 2 elements in this generation.
Now we can use a similar formula as before to do the recursion:
Ng  1( p)   Ng (c)  A(c, p)
(5.2)
c
Now, c and p are not the number of children and parents, but the lists of isomorphic
groups for the child generation and the parent generation, as our example above. Ng(c)
is the number of different unsex-labelled pedigrees there are up to isomorphism up to
generation g. A(c,p) is the number of different cases that individuals as listed in c can
have parents as listed in p.
Everything seems to work out so far, but then, we discovered a vital problem in this
model. The method to form the list, our way to keep tracks of isomorphic groups is
not informative enough. It may not be sufficient to identify the isomorphic grouping
relation in generation n if we only look at how the pedigree rises from generation n-1
to generation n.
For example, if using the method as above for the six individuals in Generation 3 in
pedigree (5.3), we will get a list of 0,1,0,1,…, 3 and 4 in a group and the rest in the
other.
(5.3)
1
2
3
4
5
6
Generation 3
Generation 2
Generation 1
Generation 0
It seems that individual 1,2,5,6 are all in one isomorphic group. However, it is too
early a stage to say so. If the pedigree up to generation 4 is as pedigree (5.4), then
individual
(5.4)
Generation 4
Generation 3
1
Generation 2
Generation 1
Generation 0
(5.5)
1
2
3
4
5
6. Summary
In this project, we have investigated several questions related to pedigree counting.
For discrete pedigrees with only 1 individual in generation 0, 2e succeeded in finding
a recursion to calculate the number of unsex-labelled pedigrees up to isomorphism
and in finding an algorithm for calculating the ways of sex label a given pedigree. The
number of different sex-labelled pedigrees rises significantly along the number of
generations. It exceeds 1700 digits if we want to go back to 10 generations.
For the unsex-labelled pedigrees, counting is rather hard. Although the number of
unsex-labelled pedigrees should be smaller than the one of sex-labelled pedigrees, it is
a lot harder to identify if two pedigrees are isomorphic and to find a computable
algorithm. We could not adopt the similar recursion as in sex-labelled case and write a
computer programme up.
Appendix
N[countp[4],6]
2.87966 107
N[countp[5],6]
2.81597 1020
N[countp[6],6]
7.40974 1052
N[countp[7],6]
2.76739 10131
N[countp[8],6]
2.88199 10317
N[countp[9],6]
3.52352 10749
N[countp[10],6]
3.89498 101737