Step by Step solution about How to resolve the algorithm Partition function P step by step in the Julia programming language

The provided code computes the partition function for a given integer n, which represents the number of distinct ways in which a positive integer n can be represented as a sum of positive integers. Here's a breakdown of the code:

1. Function partDiffDiff(n): This function computes the difference between two consecutive partition numbers. It returns (n+1)÷2 if n is odd and n+1 if n is even.

2. Function partDiff(n) with @memoize: This function computes the partition number p(n), which represents the number of ways to represent n as a sum of positive integers. The @memoize decorator is used to cache the results of the function, so that subsequent calls with the same input will return the cached value.

3. Function partitionsP(n) with @memoize: This function computes the partition function p(n) using the pentagonal number theorem. It considers consecutive partitions p(i) and adds or subtracts them to the partial sum, depending on the value of i. The result is then returned as a BigInt type. The @memoize decorator is used here as well to cache the results.

4. Main Code: In the main part of the code, the value of n is set to 6666. The @time decorator is used to measure the execution time of the partitionsP(n) function. The result is then printed to the console.

Here are some of the key implementation details:

• The function partDiff uses the recurrence relation p(n) = p(n-1) + p(n-2) + p(n-5) + ... to compute the partition number p(n).

• The function partitionsP uses the pentagonal number theorem, which states that if the pentagonal numbers P(k) are defined as P(k) = k * (3 * k - 1) / 2, then p(n) = (-1)^k * P(k) for odd k such that P(k) <= n and p(n) = 0 otherwise.

This code provides a method for efficiently computing the partition function p(n) for large values of n. The memoization technique used in the partDiff and partitionsP functions significantly improves the performance of the computation.

using Memoize

function partDiffDiff(n::Int)::Int
    isodd(n) ? (n+1)÷2 : n+1
end

@memoize function partDiff(n::Int)::Int
    n<2 ? 1 : partDiff(n-1)+partDiffDiff(n-1)
end

@memoize function partitionsP(n::Int)
    T=BigInt
    if n<2
        one(T)
    else
        psum = zero(T)
        for i ∈ 1:n
            pd = partDiff(i)
            if pd>n
                break
            end
            if ((i-1)%4)<2
                psum += partitionsP(n-pd)
            else
                psum -= partitionsP(n-pd)
            end
        end
        psum
    end
end

n=6666
@time println("p($n) = ", partitionsP(n))