Recursive Backtracking in Go for Bioinformatics Applications: 1. Counting Variations

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This is the first in a series of three blog posts describing our solution to a bioinformatics problem from Rosalind.info, Problem BA1(i) (Find most frequent words with mismatches in a string). To solve this problem and generate variations of a DNA string as required, we implemented a recursive backtracking method in the Go programming language.



Table of Contents



Problem Description

The task at hand is to take a given input strand of DNA, and generate variations from it that have up to \(d\) differences (a Hamming distance of \(d\)) in the codons (base pairs).

In part 1 of this series, we walk through the construction of an analytical formula to count the number of variations of a given DNA string that can be generated, given the constraints of the problem.

In part 2 of this series, we cover several techniques to generate variations on a DNA string, and present pseudocode for the recursive backtracking method that we use here.

In part 3 of this series, we will cover our implementation of the recursive backtracking method in the Go programming language.

Useful Functions

It's always useful to review some basic mathematics useful for combinatorics applications. We'll review the factorial and binomial functions, which will show up in our final formula for the total nubmer of variations we will be generating.

Factorial Function

The factorial function for an integer \(n\) is written \(n!\) and is defined for \(n \geq 1\) as:

$$ n! = n \cdot (n-1) \cdot \dots \cdot 2 \cdot 1 $$

for example, \(5!\) would be:

$$ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 $$

Binomial Function

The binomial function has many applications in combinatorics. It is the number of ways of independently selecting \(k\) items from a set of \(n\) items, and is written:

$$ \binom{n}{k} = \dfrac{ n! }{ k! (n-k)! } $$



Counting Permutations

What we want is a formula to count the number of permutations

To derive a formula, it helps to think through the problem starting with smaller special cases, and generalize from there in terms of the problem parameters.

Deriving the Formula

The problem we're trying to solve is generating all perms with hamming distance less than or equal to d, but let's start with a simpler problem: generating all perms with hamming distance of exactly d.

Then we can just sum up over each d.

Start with a simple situation: string of dna with 3 codons. Case of hamming distance 0 too trivial, so start with case of hamming distance of 1.

There are two terms in our combinatorics formula that we need to think about:

  • Term 1: We have a certain number of codons to modify (this is fixed by the Hamming distance d that we pick). Term 1 counts up the number of ways of selecting which indices of the original DNA string to modify.

  • Term 2: Once we've picked out the indices we are going to modify, we have several variations for each index (4 total codons, so 3 variations). Term 2 is a count of the number of variations that are possible, given the choice of indices in the original DNA string to modify.

The approach here is to think about these two terms independently and separately. Each term has a formula to count the number of possibilities indexed by each. Then, because these are independent choices, the total number of combined choices is the product of these two terms.

Term 1: Picking DNA Indices

The first term in our formula for number of variations will be the term representing the number of ways of choosing which indices in the original DNA input string to edit.

Given a Hamming distance of \(d\), and the fact that we have one and only one edit (Hamming distance unit) per base pair, Term 1 counts the number of ways of picking \(d\) items from a set of \(n\) items. Order does not matter.

This problem is equivalent to having a row of \(n\) on/off switches, all in the off position, and counting the number of ways of throwing exactly \(d\) of them into the on position.

Likewise, it is equivalent to having \(d\) identical colored balls, and counting the number of ways of placing them into \(n\) slots, one ball per slot.

We can see how the problem has a kind of triangular structure. Returning to the scenario of \(n\) on/off switches:

  • If we have \(d = n\) switches to throw, or if we have \(d = 0\) switches to throw, in either case we have only 1 possible outcome.

  • If we have \(d = n-1\) switches to throw, or if we have \(d = 1\) switch to throw, either way we have \(n\) possible outcomes

  • If we have \(d = n-2\) or \(d = 2\) switches to throw, there are \(n (n-1)\) possible outcomes; etc.

In fact, this problem - choosing \(d\) things from a set of \(n\) things - is common enough that there is a special function just to describe it, and that's the binomial function (covered above).

The binomial function is defined as:

$$ \binom{n}{k} = \dfrac{ n! }{ k! (n-k)! } $$

In the scenarios posed above, the order of our choices did not matter - the balls were not numbered, the order in which we threw each switch did not affect the outcome.

If the order did matter, if the order in which the on/off switches were thrown mattered or if the balls that were placed into slots had sequential numbers on them, then we would need a different function - the expression above to count the number of outcomes would not have a \(k!\) in the denominator.

Term 1: Side Note on Ordering

If the order of the index choices does not matter, the \(k!\) term in the denominator must be included to cancel out double-counting in the situations where (for example) \(i\) is chosen first and \(j\) is chosen second, and then the situation where \(j\) is chosen first and \(i\) is chosen second.

If the \(k!\) term is present in the denominator, it says that the order in which items are selected does not matter, in which case we are generating combinations.

To count combinations, use the "n choose k" function. See the Combination article on Wolfram MathWorld.

If the \(k!\) term is not present in the denominator, it says that the order in which items are selected does matter, in which case we are generating permutations.

To count permutations, use the "n pick k" function. See the Permutation article on Wolfram MathWorld.

Term 2: Modifying DNA Codons

Once we've selected the \(d\) indices in the original DNA string that we are going to modify, we have to count the number of ways those base pairs can be modified.

We have \(d\) base pairs to modify, and \(c = 4\) total codons (ATGC). Each base pair that we are modifying has \(c-1\) possible codons that it we can swap it out with, and each choice is independent, so the number of possibile outcomes (Term 2) is:

$$ (c-1)^{d} $$



Final Counting Formula

To write the final formula for counting the number of variations \(V\) of a given DNA string of length \(n\) that are a Hamming distance of less than or equal to \(d\), with \(c\) possible codons (A, T, G, C), we will need to sum over Hamming distances from 0 to \(d\):

$$ V = \sum_{k = 0}^{d} \binom{n}{k} (c-1)^{k} $$

Implementing in Go

Now, let's look at how we would implement this counting formula in Go. This will be useful, since programs run much faster when they are able to allocate all the sapce they need in memory ahead of time. Counting the number of variations of our DNA input string will allow us to do just that.

Binomial and Factorial Functions in Go

We'll start with binomial and factorial functions in Go: continuing with our theme of recursion, we implement a recursive factorial function.

// Compute the factorial of an integer.
func Factorial(n int) int {
    if n < 2 {
        // base case
        return 1
    } else {
        // recursive case
        return n * Factorial(n-1)
    }
}

The factorial function will behave correctly for the case of \(n=1\) and \(n=0\), and will return 1 if \(n\) is negative (which is reasonable behavior for our purposes.)

The binomial function utilizes the factorial function:

// Returns value of the binomial coefficient Binom(n, k).
func Binomial(n, k int) int {

    result := 1

    // Since C(n, k) = C(n, n-k)
    if k > (n - k) {
        k = n - k
    }

    // Calculate value of:
    //
    // ( n * (n-1) * ... * (n-k+1) )
    // -----------------------------
    //   ( k * (k-1) * ... * 1 )
    // 
    for i := 0; i < k; i++ {
        result *= n - i
        result /= i + 1
    }

    return result
}

(Note that we might want to add some additional error checks to the Binomial() function.)

Variations Counting Function in Go

Now we can put everything together into a function to count the number of "Hamming neighbors" - variations on a given DNA string that are a Hamming distance of up to \(d\) away from the original DNA string.

To count the number of Hamming neighbors, we implement the formula above. We leave out the error checks on the parameter values here, for brevity.

The inputs are:

  • \(n\) - length of DNA input string
  • \(d\) - maximum Hamming distance
  • \(c\) - number of codons (4, ATGC)
// Given an input string of DNA of length n,
// a maximum Hamming distance of d,
// and a number of codons c, determine
// the number of Hamming neighbors of
// distance less than or equal to d
// using a combinatorics formula.
func CountHammingNeighbors(n, d, c int) (int, error) {

    // We require the following:
    // n > 0
    // d >= 0
    // c > 0

    // Use combinatorics to calculate number of variations
    nv := 0
    for dd := 0; dd <= d; dd++ {

        // Binomial(n,d) => number of ways we can
        //                  pick codons to edit
        next_term := Binomial(n, dd)

        // (c-1)^d => number of ways that the codons
        //            we picked to edit can be edited
        for j := 0; j < dd; j++ {
            next_term *= (c - 1)
        }
        nv += next_term
    }
    return nv, nil
}

We can run this with a few values of k and d to verify it returns the expected values:

For kmer AAA k = 3:
d = 0, count = 1
d = 1, count = 10
d = 2, count = 37
d = 3, count = 64

for a kmer of length 3, we can compute the first 3 values (1, 10, 37) by hand. The last value, when \(d = k\), is a special case where every base pair in the DNA strand can be changed to any codon. Since there are 4 possible codons, this leads to \(4^k = 2^{2k}\) possibilities.

For \(d = k = 3\), we have \(2^6 = 64\) possible DNA strings.

Now, moving on to \(k=5\):

For kmer AAAAA k = 5:
d = 0, count = 1
d = 1, count = 16
d = 2, count = 106
d = 3, count = 376
d = 4, count = 781
d = 5, count = 1024

We can calculate 1 and 16 by hand, verifying those two numbers. As before, the case of \(k = d = 5\) gives a total of \(4^5 = 2^{10} = 1024\) possible DNA strings.



Next: Recursive Backtracking in Go for Bioinformatics Applications: 2. Generating Variations

Tags:    go    golang    rosalind    bioinformatics    recursion    backtracking    strings    combinatorics