# Introduction

The VCF format encodes genotypes by the index of the enumeration of genotypes given ploidy number and alleles. This allows for direct access to a value associated with a genotype within an array when one works with genotype likelihoods.

# Motivation

Plants species exhibit a diverse number of ploidy, for example, the strawberry is an octoploid and the pear is a triploid.

Copy number variations in somatic variant calling also leads to variable ploidy to consider when genotyping a locus.

While there are explicit functions that could be googled for handling haploid and diploid cases. It seems difficult to find the closed forms for the general case. This wiki fills in that need.

# The number of genotypes given a ploidy and alleles

\begin{align} F(P,A) = \binom{P+A-1}{A-1} \\ \end{align}

where P is the ploidy number and A is the number of alleles.

# Getting the index of a genotype in an enumerated list given a ploidy and alleles

\begin{align} G(a_1,.. , a_P) = \sum_{k=1}^P \binom{k+a_k-1}{a_k-1} \end{align}

where P is the number of ploidy, $a_1$, $a_2$ .. $a_P$ are the alleles in numeric encoding (0 to A-1) and are ordered (e.g. AB and ABCCCC are ordered but ACB is not ordered).

This is well defined because:

\begin{align} \binom{n}{r} = \begin{cases} \frac{n!}{(n-r)!r!} &, r \le n, r\ge0, n\ge0 \\ 0 & \text{otherwise} \end{cases} \end{align}

Because $a_k$ may be 0, we will see cases of $\binom{k-1}{-1}$ when $a_k=0$. This is alright because of the definition of $\binom{n}{r}$ which defines this case as 0. But to make it more sensible, we can define the function equivalently as:

\begin{align} G(a_1,.. , a_P) = \sum_{k=1}^P \binom{k+a_k-1}{k} \end{align}

So when $a_k=0$, the binomial coefficient reads as $\binom{k-1}{k}$ which equals 0 since there are 0 ways to choose k items from k-1 items.

# Getting the genotypes from a genotype index and a given ploidy

  The genotype index is computed by a summation of a series which is
monotonically decreasing.  This allows you to compute the inverse function
from index to the ordered genotypes by using a "water rapids algorithm" with
decreasing height of each mini water fall.

std::vector<int32_t> bcf_ip2g(int32_t genotype_index, uint32_t no_ploidy)
{
std::vector<int32_t> genotype(no_ploidy, 0);
int32_t pth = no_ploidy;
int32_t max_allele_index = genotype_index;
int32_t leftover_genotype_index = genotype_index;
while (pth>0)
{
for (int32_t allele_index=0; allele_index <= max_allele_index; ++allele_index)
{
int32_t i = choose(pth+allele_index-1, pth);
if (i>=leftover_genotype_index || allele_index==max_allele_index)
{
if (i>leftover_genotype_index) --allele_index;
leftover_genotype_index -= choose(pth+allele_index-1, pth);
--pth;
max_allele_index = allele_index;
genotype[pth] = allele_index;
break;
}
}
}
return genotype;
}

 todo:: describe in a human understandable fashion.


# Simple cases

Ploidy Alleles No. of Genotypes Index
1 A $F(1, A) = \binom{1+A-1}{A-1} = A$ $G(a_1) = F(1, a_1) = a_1$
2 A $F(2,A) = \binom{2+A-1}{A-1} = \binom{A+1}{2}$ $G(a_1,a_2) = F(1, a_1) + F(2, a_2) = a_1 + \binom{a_2+1}{2}$

# Derivation for counting the number of genotypes

There must always be P observed alleles and there can only be at most A alleles. This can be modeled by P+A-1 points where you choose A-1 points to be dividers that separate the alleles. Thus the number of ways you can observe this is $\binom{P+A-1}{A-1}$ which is equivalent to $\binom{P+A-1}{P}$.

# Derivation for getting the index of a genotype in an enumerated list

## Observation of nested patterns

An important observation here is that for the enumeration of A alleles for a given P ploidy, the enumeration of A-1 alleles for P ploidy is a subsequence.

Index A=4,P=3 A=3,P=3 A=2,P=3 A=1,P=3
0

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19

AAA

AAB
ABB
BBB
AAC
ABC
BBC
ACC
BCC
CCC
ABD
BBD
ACD
BCD
CCD
BDD
CDD
DDD

AAA

AAB
ABB
BBB
AAC
ABC
BBC
ACC
BCC
CCC

AAA

AAB
ABB
BBB

AAA

Another important observation here is that for a genotype $(a_1, a_2, .. a_P)$, The $F(P, a_p)$th genotype to $(a_1, a_2, .. a_P)$ all end with $a_p$, this does not help distinguish the order, so we need to only examine the genotype $(a_1,..., a_{P-1})$. The sub genotype $(a_1,..., a_{P-1})$ is also ordered else $(a_1,..., a_{P})$ is not ordered.

The nested genotype sequence is in red. The blue sequence shows the sequence of genotypes enumerated without involving $a_P$.

Index A=4,P=3 A=4,P=2 A=4,P=1
0

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19

AAA

AAB
ABB
BBB
AAC
ABC
BBC
ACC
BCC
CCC
ABD
BBD
ACD
BCD
CCD
BDD
CDD
DDD

AAA

AAB
ABB
BBB
AAC
ABC
BBC
ACC
BCC
CCC
ABD
BBD
ACD
BCD
CCD
BDD
CDD
DDD

AAA

AAB
ABB
BBB
AAC
ABC
BBC
ACC
BCC
CCC
ABD
BBD
ACD
BCD
CCD
BDD
CDD
DDD

The above 2 observations are the key to breaking down the enumeration recursively.

## Derivation

$a_1, ... a_P$ is ordered and indexed 0 to A-1.

\begin{align} G(a_1,.. , a_P) &= \| \{\text{genotypes of } a_P \text{ alleles for ploidy P}\} \| + G(a_1,.. , a_{P-1})\\ &= F(P, a_P) + G(a_1,..,a_{P-1}) \\ &= F(P, a_P) + F(P-1, a_{P-1}) + G(a_1,..,a_{P-2}) \\ &= F(P, a_P) + F(P-1, a_{P-1}) + ... + F(1,a_1) \\ &= \sum_{k=1}^P F(k, a_k) \\ &= \sum_{k=1}^P \binom{k+a_k-1}{a_k-1} \end{align}

This algorithm is demonstrated in the following table to obtain the index of the genotype C/C/D.

Index Iteration 0 Iteration 1 Iteration 2
0

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

AAA

AAB
ABB
BBB
AAC
ABC
BBC
ACC
BCC
CCC
ABD
BBD
ACD
BCD
CCD

AAA

AAB
ABB
BBB
AAC
ABC
BBC
ACC
BCC
CCC

AA
AB
BB
AC
BC
CC

AA
AB
BB

A
B
C

Function call G(CCD) F(3,3) G(CC) F(2,2) G(C) = F(1, 2)
value returned 10 3 2 index=10+3+2=15 (QED!)

# Algorithm for enumerating the genotypes given ploidy and alleles

The below code is for enumerating genotypes and can be used to test the above equations.

   uint32_t no = 0 // some global variable
void print_genotypes(uint32_t A, uint32_t P, std::string genotype)
{
if (genotype.size()==P)
{
std::cerr << no << ") " << genotype << "\n";
++no;
}
else
{
for (uint32_t a=0; a<A; ++a)
{
std::string s(1,(char)(a+65));
s.append(genotype);
print_genotypes(a+1, P, s);
}
}
}


# Acknowledgement

To Petr Danecek for double checking this.