How to resolve the algorithm Sorting algorithms/Merge sort step by step in the Erlang programming language

Published on 12 May 2024 09:40 PM
#Erlang

How to resolve the algorithm Sorting algorithms/Merge sort step by step in the Erlang programming language

Table of Contents

Problem Statement

The   merge sort   is a recursive sort of order   nlog(n). It is notable for having a worst case and average complexity of   O(nlog(n)),   and a best case complexity of   O(n)   (for pre-sorted input). The basic idea is to split the collection into smaller groups by halving it until the groups only have one element or no elements   (which are both entirely sorted groups). Then merge the groups back together so that their elements are in order. This is how the algorithm gets its   divide and conquer   description.

Write a function to sort a collection of integers using the merge sort.

The merge sort algorithm comes in two parts: The functions in pseudocode look like this:

Note:   better performance can be expected if, rather than recursing until   length(m) ≤ 1,   an insertion sort is used for   length(m)   smaller than some threshold larger than   1.   However, this complicates the example code, so it is not shown here.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Sorting algorithms/Merge sort step by step in the Erlang programming language

Source code in the erlang programming language

mergeSort(L) when length(L) == 1 -> L;
mergeSort(L) when length(L) > 1 ->
    {L1, L2} = lists:split(length(L) div 2, L),
    lists:merge(mergeSort(L1), mergeSort(L2)).


pMergeSort(L) when length(L) == 1 -> L;
pMergeSort(L) when length(L) > 1 ->
    {L1, L2} = lists:split(length(L) div 2, L),
    spawn(mergesort, pMergeSort2, [L1, self()]),
    spawn(mergesort, pMergeSort2, [L2, self()]),
    mergeResults([]).

pMergeSort2(L, Parent) when length(L) == 1 -> Parent ! L;
pMergeSort2(L, Parent) when length(L) > 1 ->
    {L1, L2} = lists:split(length(L) div 2, L),
    spawn(mergesort, pMergeSort2, [L1, self()]),
    spawn(mergesort, pMergeSort2, [L2, self()]),
    Parent ! mergeResults([]).


merge_sort(List) -> m(List, erlang:system_info(schedulers)).

m([L],_) -> [L]; 
m(L, N) when N > 1  -> 
    {L1,L2} = lists:split(length(L) div 2, L),
    {Parent, Ref} = {self(), make_ref()},
    spawn(fun()-> Parent ! {l1, Ref, m(L1, N-2)} end), 
    spawn(fun()-> Parent ! {l2, Ref, m(L2, N-2)} end), 
    {L1R, L2R} = receive_results(Ref, undefined, undefined),
    lists:merge(L1R, L2R);
m(L, _) -> {L1,L2} = lists:split(length(L) div 2, L), lists:merge(m(L1, 0), m(L2, 0)).

receive_results(Ref, L1, L2) ->
    receive
        {l1, Ref, L1R} when L2 == undefined -> receive_results(Ref, L1R, L2);
        {l2, Ref, L2R} when L1 == undefined -> receive_results(Ref, L1, L2R);
        {l1, Ref, L1R} -> {L1R, L2};
        {l2, Ref, L2R} -> {L1, L2R}
    after 5000 -> receive_results(Ref, L1, L2)
    end.

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