Tree sort – sorting by binary search tree. Time complexity – O(n²). In such a tree, each node has numbers on the left less than the node, on the right greater than the node, when coming from the root and printing values from left to right, we get a sorted list of numbers. Amazing, right?
Let’s consider the binary search tree diagram:
Derrick Coetzee (public domain)
Try manually reading the numbers starting from the second to last left node in the lower left corner, for each node on the left – a node on the right.
It will look like this:
- The second to last node on the bottom left is 3.
- It has a left branch – 1.
- We take this number (1)
- Next we take the vertex itself 3 (1, 3)
- On the right is branch 6, but it contains branches. Therefore, we read it in the same way.
- Left branch of node 6 is number 4 (1, 3, 4)
- The node itself is 6 (1, 3, 4, 6)
- Right 7 (1, 3, 4, 6, 7)
- We go up to the root node – 8 (1,3, 4,6, 7, 8)
- Print everything on the right by analogy
- We get the final list – 1, 3, 4, 6, 7, 8, 10, 13, 14
To implement the algorithm in code, two functions are required:
- Building a Binary Search Tree
- Print out the binary search tree in the correct order
A binary search tree is assembled in the same way as it is read, each node is assigned a number on the left or right, depending on whether it is smaller or larger.
Example in Lua:
function Node:new(value, lhs, rhs)
output = {}
setmetatable(output, self)
self.__index = self
output.value = value
output.lhs = lhs
output.rhs = rhs
output.counter = 1
return output
end
function Node:Increment()
self.counter = self.counter + 1
end
function Node:Insert(value)
if self.lhs ~= nil and self.lhs.value > value then
self.lhs:Insert(value)
return
end
if self.rhs ~= nil and self.rhs.value < value then
self.rhs:Insert(value)
return
end
if self.value == value then
self:Increment()
return
elseif self.value > value then
if self.lhs == nil then
self.lhs = Node:new(value, nil, nil)
else
self.lhs:Insert(value)
end
return
else
if self.rhs == nil then
self.rhs = Node:new(value, nil, nil)
else
self.rhs:Insert(value)
end
return
end
end
function Node:InOrder(output)
if self.lhs ~= nil then
output = self.lhs:InOrder(output)
end
output = self:printSelf(output)
if self.rhs ~= nil then
output = self.rhs:InOrder(output)
end
return output
end
function Node:printSelf(output)
for i=0,self.counter-1 do
output = output .. tostring(self.value) .. " "
end
return output
end
function PrintArray(numbers)
output = ""
for i=0,#numbers do
output = output .. tostring(numbers[i]) .. " "
end
print(output)
end
function Treesort(numbers)
rootNode = Node:new(numbers[0], nil, nil)
for i=1,#numbers do
rootNode:Insert(numbers[i])
end
print(rootNode:InOrder(""))
end
numbersCount = 10
maxNumber = 9
numbers = {}
for i=0,numbersCount-1 do
numbers[i] = math.random(0, maxNumber)
end
PrintArray(numbers)
Treesort(numbers)
Важный нюанс что для чисел которые равны вершине придумано множество интересных механизмов подцепления к ноде, я же просто добавил счетчик к классу вершины, при распечатке числа возвращаются по счетчику.
Ссылки
https://gitlab.com/demensdeum/algorithms/-/tree/master/sortAlgorithms/treesort
Источники
Convert Sorted Array to Binary Search Tree (LeetCode 108. Algorithm Explained) – YouTube
Sorting algorithms/Tree sort on a linked list – Rosetta Code
How to handle duplicates in Binary Search Tree? – GeeksforGeeks
Tree Sort | GeeksforGeeks – YouTube
Bucket Sort
Bucket Sort – sorting by buckets. The algorithm is similar to counting sort, with the difference that the numbers are collected in “buckets”-ranges, then the buckets are sorted using any other, sufficiently productive, sorting algorithm, and the final chord is the unfolding of the “buckets” one by one, resulting in a sorted list.
The algorithm’s time complexity is O(nk). The algorithm works in linear time for data that obeys a uniform distribution law. To put it simply, the elements must be in a certain range, without “spikes”, for example, numbers from 0.0 to 1.0. If among such numbers there are 4 or 999, then such a series is no longer considered “even” according to the yard laws.
Example of implementation in Julia:
buckets = Vector{Vector{Int}}()
for i in 0:bucketsCount - 1
bucket = Vector{Int}()
push!(buckets, bucket)
end
maxNumber = maximum(numbers)
for i in 0:length(numbers) - 1
bucketIndex = 1 + Int(floor(bucketsCount * numbers[1 + i] / (maxNumber + 1)))
push!(buckets[bucketIndex], numbers[1 + i])
end
for i in 0:length(buckets) - 1
bucketIndex = 1 + i
buckets[bucketIndex] = sort(buckets[bucketIndex])
end
flat = [(buckets...)...]
print(flat, "\n")
end
numbersCount = 10
maxNumber = 10
numbers = rand(1:maxNumber, numbersCount)
print(numbers,"\n")
bucketsCount = 10
bucketSort(numbers, bucketsCount)
На производительность алгоритма также влияет число ведер, для большего количества чисел лучше взять большее число ведер (Algorithms in a nutshell by George T. Heineman)
Ссылки
https://gitlab.com/demensdeum/algorithms/-/tree/master/sortAlgorithms/bucketSort
Источники
https://www.youtube.com/watch?v=VuXbEb5ywrU
https://www.youtube.com/watch?v=ELrhrrCjDOA
https://medium.com/karuna-sehgal/an-introduction-to-bucket-sort-62aa5325d124
https://www.geeksforgeeks.org/bucket-sort-2/
https://ru.wikipedia.org/wiki/%D0%91%D0%BB%D0%BE%D1%87%D0%BD%D0%B0%D1%8F_%D1%81%D0%BE%D1%80%D1%82%D0%B8%D1%80%D0%BE%D0%B2%D0%BA%D0%B0
https://www.youtube.com/watch?v=LPrF9yEKTks
https://en.wikipedia.org/wiki/Bucket_sort
https://julialang.org/
https://www.oreilly.com/library/view/algorithms-in-a/9780596516246/ch04s08.html
Radixsort
Radix Sort is a radix sort. The algorithm is similar to counting sort in that there is no comparison of elements, instead, elements are grouped *symbolically* into *buckets*, the bucket is selected by the index of the current number-symbol. Time complexity is O(nd).
It works something like this:
- As input we receive the numbers 6, 12, 44, 9
- Let’s create 10 list buckets (0-9) into which we will add/sort numbers bitwise.
Next:
- Start a loop with counter i until the maximum number of characters in the number
- According to the index i from right to left we get one symbol for each number, if there is no symbol, we consider it zero
- Convert the symbol to a number
- We select a bucket by index – number, put the number there in full
- After we finish iterating over the numbers, we transform all the buckets back into a list of numbers
- We get numbers sorted by rank
- Repeat until all the digits are gone
Radix Sort example in Scala:
import scala.util.Random.nextInt
object RadixSort {
def main(args: Array[String]) = {
var maxNumber = 200
var numbersCount = 30
var maxLength = maxNumber.toString.length() - 1
var referenceNumbers = LazyList.continually(nextInt(maxNumber + 1)).take(numbersCount).toList
var numbers = referenceNumbers
var buckets = List.fill(10)(ListBuffer[Int]())
for( i <- 0 to maxLength) { numbers.foreach( number => {
var numberString = number.toString
if (numberString.length() > i) {
var index = numberString.length() - i - 1
var character = numberString.charAt(index).toString
var characterInteger = character.toInt
buckets.apply(characterInteger) += number
}
else {
buckets.apply(0) += number
}
}
)
numbers = buckets.flatten
buckets.foreach(x => x.clear())
}
println(referenceNumbers)
println(numbers)
println(s"Validation result: ${numbers == referenceNumbers.sorted}")
}
}
The algorithm also has a version for parallel execution, for example on a GPU; there is also a bit sorting variant, which is probably very interesting and truly breathtaking!
Links
https://gitlab .com/demensdeum/algorithms/-/blob/master/sortAlgorithms/radixSort/radixSort.scala
Sources
https://ru.wikipedia.org/wiki/%D0%9F%D0%BE%D1%80%D0%B0%D0%B7%D1%80%D1%8F%D 0%B4%D0%BD%D0%B0%D1%8F_%D1%81%D0%BE%D1%80%D1%82%D0%B8%D1%80%D0%BE%D0%B2%D0% BA%D0%B0
https://www.geeksforgeeks.org/radix-sort/
https://www.youtube.com/watch?v=toAlAJKojos
https://github.com/gyatskov/radix-sort
Heapsort
Heapsort is a pyramid sort. The time complexity of the algorithm is O(n log n), fast, right? I would call this sorting – sorting of falling stones. It seems to me that the easiest way to explain it is visually.
The input is a list of numbers, for example:
5, 0, 7, 2, 3, 9, 4
From left to right, a data structure is made – a binary tree, or as I call it – a pyramid. Pyramid elements can have a maximum of two child elements, but only one top element.
Let’s make a binary tree:
⠀⠀5
⠀0⠀7
2 3 9 4
If you look at the pyramid for a long time, you can see that it is simply numbers from an array, following one after another, the number of elements on each floor is multiplied by two.
Next comes the most interesting part, we sort the pyramid from the bottom up, using the falling pebbles method (heapify). Sorting could start from the last floor (2 3 9 4), but there is no point because there is no floor below to fall to.
Therefore, we begin to drop elements from the penultimate floor (0 7)
⠀⠀5
⠀0⠀7
2 3 9 4
The first element to fall is chosen from the right, in our case it is 7, then we look at what is under it, and under it is 9 and 4, nine is greater than four, and nine is also greater than seven! We drop 7 on 9, and lift 9 to the place of 7.
⠀⠀5
⠀0⠀9
2 3 7 4
Next, we understand that the seven has nowhere to fall lower, we move on to the number 0, which is on the penultimate floor on the left:
⠀⠀5
⠀0⠀9
2 3 7 4
We look at what’s underneath it – 2 and 3, two is less than three, three is greater than zero, so we swap zero and three:
⠀⠀5
⠀3⠀9
2 0 7 4
When you reach the end of the floor, go to the floor above and drop everything there if you can.
The result will be a data structure – a heap, namely a max heap, since the largest element is at the top:
⠀⠀9
⠀3⠀7
2 0 5 4
If you return to the array representation, you will get a list:
[9, 3, 7, 2, 0, 5, 4]
From this we can conclude that by swapping the first and last elements, we will get the first number in the final sorted position, namely 9 should be at the end of the sorted list, swap them:
[4, 3, 7, 2, 0, 5, 9]
Let’s look at a binary tree:
⠀⠀4
⠀3⠀7
2 0 5 9
We have a situation where the bottom of the tree is sorted, we just need to drop 4 to the correct position, we repeat the algorithm, but we do not take into account the already sorted numbers, namely 9:
⠀⠀4
⠀3⠀7
2 0 5 9
⠀⠀7
⠀3⠀4
2 0 5 9
⠀⠀7
⠀3⠀5
2 0 4 9
It turned out that, having dropped 4, we raised the next largest number after 9 – 7. We swap the last unsorted number (4) and the largest number (7)
⠀⠀4
⠀3⠀5
2 0 7 9
It turns out that now we have two numbers in the correct final position:
4, 3, 5, 2, 0, 7, 9
Then we repeat the sorting algorithm, ignoring those already sorted, and as a result we get a heap of the following type:
⠀⠀0
⠀2⠀3
4 5 7 9
Or as a list:
0, 2, 3, 4, 5, 7, 9
Implementation
The algorithm is usually divided into three functions:
- Creating a heap
- Heapify Algorithm
- Replace the last unsorted element with the first
The heap is created by going through the penultimate row of the binary tree using the heapify function, from right to left until the end of the array. Then the first number swap is made in the loop, after which the first element falls/stays in place, as a result of which the largest element gets to the first place, the loop is repeated with the participants decreasing by one, since after each pass at the end of the list there are sorted numbers.
Heapsort example in Ruby:
module Colors
BLUE = "\033[94m"
RED = "\033[31m"
STOP = "\033[0m"
end
def heapsort(rawNumbers)
numbers = rawNumbers.dup
def swap(numbers, from, to)
temp = numbers[from]
numbers[from] = numbers[to]
numbers[to] = temp
end
def heapify(numbers)
count = numbers.length()
lastParentNode = (count - 2) / 2
for start in lastParentNode.downto(0)
siftDown(numbers, start, count - 1)
start -= 1
end
if DEMO
puts "--- heapify ends ---"
end
end
def siftDown(numbers, start, rightBound)
cursor = start
printBinaryHeap(numbers, cursor, rightBound)
def calculateLhsChildIndex(cursor)
return cursor * 2 + 1
end
def calculateRhsChildIndex(cursor)
return cursor * 2 + 2
end
while calculateLhsChildIndex(cursor) <= rightBound
lhsChildIndex = calculateLhsChildIndex(cursor)
rhsChildIndex = calculateRhsChildIndex(cursor)
lhsNumber = numbers[lhsChildIndex]
biggerChildIndex = lhsChildIndex
if rhsChildIndex <= rightBound
rhsNumber = numbers[rhsChildIndex]
if lhsNumber < rhsNumber
biggerChildIndex = rhsChildIndex
end
end
if numbers[cursor] < numbers[biggerChildIndex]
swap(numbers, cursor, biggerChildIndex)
cursor = biggerChildIndex
else
break
end
printBinaryHeap(numbers, cursor, rightBound)
end
printBinaryHeap(numbers, cursor, rightBound)
end
def printBinaryHeap(numbers, nodeIndex = -1, rightBound = -1)
if DEMO == false
return
end
perLineWidth = (numbers.length() * 4).to_i
linesCount = Math.log2(numbers.length()).ceil()
xPrinterCount = 1
cursor = 0
spacing = 3
for y in (0..linesCount)
line = perLineWidth.times.map { " " }
spacing = spacing == 3 ? 4 : 3
printIndex = (perLineWidth / 2) - (spacing * xPrinterCount) / 2
for x in (0..xPrinterCount - 1)
if cursor >= numbers.length
break
end
if nodeIndex != -1 && cursor == nodeIndex
line[printIndex] = "%s%s%s" % [Colors::RED, numbers[cursor].to_s, Colors::STOP]
elsif rightBound != -1 && cursor > rightBound
line[printIndex] = "%s%s%s" % [Colors::BLUE, numbers[cursor].to_s, Colors::STOP]
else
line[printIndex] = numbers[cursor].to_s
end
cursor += 1
printIndex += spacing
end
print line.join()
xPrinterCount *= 2
print "\n"
end
end
heapify(numbers)
rightBound = numbers.length() - 1
while rightBound > 0
swap(numbers, 0, rightBound)
rightBound -= 1
siftDown(numbers, 0, rightBound)
end
return numbers
end
numbersCount = 14
maximalNumber = 10
numbers = numbersCount.times.map { Random.rand(maximalNumber) }
print numbers
print "\n---\n"
start = Time.now
sortedNumbers = heapsort(numbers)
finish = Time.now
heapSortTime = start - finish
start = Time.now
referenceSortedNumbers = numbers.sort()
finish = Time.now
referenceSortTime = start - finish
print "Reference sort: "
print referenceSortedNumbers
print "\n"
print "Reference sort time: %f\n" % referenceSortTime
print "Heap sort: "
print sortedNumbers
print "\n"
if DEMO == false
print "Heap sort time: %f\n" % heapSortTime
else
print "Disable DEMO for performance measure\n"
end
if sortedNumbers != referenceSortedNumbers
puts "Validation failed"
exit 1
else
puts "Validation success"
exit 0
end
Without visualization, this algorithm is not easy to understand, so the first thing I recommend is to write a function that will print the current form of the binary tree.
Links
https://gitlab.com/demensdeum/algorithms/-/blob/master/sortAlgorithms/heapsort/heapsort.rb
Sources
http://rosettacode.org/wiki/Sorting_algorithms/Heapsort
https://www.youtube.com/watch?v=LbB357_RwlY
https://habr.com/ru/company/ otus/blog/460087/
https://ru.wikipedia.org/wiki/Пыramidal_sorting
https://neerc.ifmo.ru/wiki /index.php?title=Heap_sort
https://wiki5.ru/wiki/Heapsort
https://wiki.c2.com/?HeapSort p>
https://ru.wikipedia.org/wiki/Tree (data structure)
https://ru.wikipedia.org/wiki/Heap (data structure)
https://www.youtube.com/watch?v=2DmK_H7IdTo
https://www.youtube.com/watch?v=kU4KBD4NFtw
https://www.youtube.com/watch?v=DU1uG5310x0
https://www.youtube.com/watch?v =BzQGPA_v-vc
https://www.geeksforgeeks.org/ array-representation-of-binary-heap/
https://habr.com/ru/post/112222/ a>
https://www.cs.usfca. edu/~galles/visualization/BST.html
https://www.youtube.com/watch?v=EQzqHWtsKq4
https://ru.wikibrief.org/wiki/Heapsort
https://www.youtube.com/watch?v=GUUpmrTnNbw
Quicksort
Quicksort is a divide-and-conquer sorting algorithm. Recursively, we sort the array of numbers in parts, placing the numbers in smaller and larger order from the selected pivot element, and insert the pivot element itself into the gap between them. After several recursive iterations, we get a sorted list. Time complexity is O(n2).
Scheme:
- We start by getting the list of elements from the outside, the sorting boundaries. In the first step, the sorting boundaries will be from the beginning to the end.
- We check that the boundaries of the beginning and end do not intersect, if this happens, then it’s time to finish
- We select some element from the list, call it the reference
- Move it to the right at the end to the last index so it doesn’t get in the way
- Create a counter of *smaller numbers* that is still equal to zero
- We go through the list in a loop from left to right, up to the last index where the reference element is located, not inclusive
- Each element is compared with the reference
- If it is less than the reference, then we swap them by the index of the counter of smaller numbers. We increment the counter of smaller numbers.
- When the cycle reaches the reference element, we stop and swap the reference element with the element with the counter of smaller numbers.
- We run the algorithm separately for the left smaller part of the list, and separately for the right larger part of the list.
- Eventually all recursive iterations will start to stop due to the check in point 2
- Get a sorted list
Quicksort was invented by scientist Charles Anthony Richard Hoare at Moscow State University, who, having learned Russian, studied computer translation and probability theory at the Kolmogorov School. In 1960, due to the political crisis, he left the Soviet Union.
Example implementation in Rust:
use rand::Rng;
fn swap(numbers: &mut [i64], from: usize, to: usize) {
let temp = numbers[from];
numbers[from] = numbers[to];
numbers[to] = temp;
}
fn quicksort(numbers: &mut [i64], left: usize, right: usize) {
if left >= right {
return
}
let length = right - left;
if length <= 1 {
return
}
let pivot_index = left + (length / 2);
let pivot = numbers[pivot_index];
let last_index = right - 1;
swap(numbers, pivot_index, last_index);
let mut less_insert_index = left;
for i in left..last_index {
if numbers[i] < pivot {
swap(numbers, i, less_insert_index);
less_insert_index += 1;
}
}
swap(numbers, last_index, less_insert_index);
quicksort(numbers, left, less_insert_index);
quicksort(numbers, less_insert_index + 1, right);
}
fn main() {
let mut numbers = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0];
let mut reference_numbers = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0];
let mut rng = rand::thread_rng();
for i in 0..numbers.len() {
numbers[i] = rng.gen_range(-10..10);
reference_numbers[i] = numbers[i];
}
reference_numbers.sort();
println!("Numbers {:?}", numbers);
let length = numbers.len();
quicksort(&mut numbers, 0, length);
println!("Numbers {:?}", numbers);
println!("Reference numbers {:?}", reference_numbers);
if numbers != reference_numbers {
println!("Validation failed");
std::process::exit(1);
}
else {
println!("Validation success!");
std::process::exit(0);
}
}
If nothing is clear, I suggest watching a video by Rob Edwards from the University of San Diego https://www.youtube.com/watch?v=ZHVk2blR45Q it shows the essence and implementation of the algorithm in the simplest way, step by step.
Links
https://gitlab.com/demensdeum /algorithms/-/tree/master/sortAlgorithms/quickSort
Sources
https://www.youtube.com/watch?v =4s-aG6yGGLU
https://www.youtube.com/watch?v=ywWBy6J5gz8
https://www.youtube.com/watch?v=Hoixgm4-P4M
https://ru.wikipedia.org/wiki/Быстрая_сортировка
https://www.youtube.com/watch?v=Hoixgm4-P4M
https://www.youtube.com/watch?v=XE4VP_8Y0BU
https://www.youtube.com/watch?v=MZaf_9IZCrc
https://www.youtube.com/watch?v=ZHVk2blR45Q
http://rosettacode.org/wiki/Sorting_algorithms/Quicksort
https://www.youtube.com/watch?v=4s-aG6yGGLU
https://www.youtube.com/watch?v=dQw4w9WgXcQ
https://www.youtube.com/watch?v=maibrCbZWKw
https://www.geeksforgeeks.org/quick-sort/
https://www.youtube.com/watch?v=uXBnyYuwPe8
Binary Insertion Sort
Binary Insertion Sort is a variant of insertion sort in which the insertion position is determined using binary search. The time complexity of the algorithm is O(n2)
The algorithm works like this:
- Starts a loop from zero to the end of the list
- In the loop, a number is selected for sorting, the number is saved in a separate variable
- Binary search finds the index to insert this number compared to the numbers to the left
- Once the index is found, the numbers on the left are shifted one position to the right, starting with the insertion index. The process will erase the number you want to sort.
- The previously saved number is inserted at the insertion index
- At the end of the loop, the entire list will be sorted
During binary search, a situation is possible when the number will not be found, but the index is not returned. Due to the peculiarity of binary search, the number closest to the desired one will be found, then to return the index it will be necessary to compare it with the desired one, if the desired one is less, then the desired one should be on the left by index, and if it is greater or equal, then on the right.
Go code:
import (
"fmt"
"math/rand"
"time"
)
const numbersCount = 20
const maximalNumber = 100
func binarySearch(numbers []int, item int, low int, high int) int {
for high > low {
center := (low + high) / 2
if numbers[center] < item { low = center + 1 } else if numbers[center] > item {
high = center - 1
} else {
return center
}
}
if numbers[low] < item {
return low + 1
} else {
return low
}
}
func main() {
rand.Seed(time.Now().Unix())
var numbers [numbersCount]int
for i := 0; i < numbersCount; i++ {
numbers[i] = rand.Intn(maximalNumber)
}
fmt.Println(numbers)
for i := 1; i < len(numbers); i++ { searchAreaLastIndex := i - 1 insertNumber := numbers[i] insertIndex := binarySearch(numbers[:], insertNumber, 0, searchAreaLastIndex) for x := searchAreaLastIndex; x >= insertIndex; x-- {
numbers[x+1] = numbers[x]
}
numbers[insertIndex] = insertNumber
}
fmt.Println(numbers)
}
Links
Sources
https://www.geeksforgeeks.org/binary-insertion- sort/
https://www.youtube.com/watch?v=-OVB5pOZJug
Shell Sort
Shell Sort is a variant of insertion sorting with preliminary combing of the array of numbers.
We need to remember how insertion sort works:
1. The loop starts from zero to the end of the loop, thus the array is divided into two parts
2. For the left part, the second cycle is started, comparing elements from right to left, the smaller element on the right is omitted until a smaller element on the left is found
3. At the end of both cycles, we get a sorted list
Once upon a time, computer scientist Donald Shell was puzzled by how to improve the insertion sort algorithm. He came up with the idea of also going through the array with two cycles, but at a certain distance, gradually reducing the “comb” until it turns into a regular insertion sort algorithm. Everything is really that simple, no pitfalls, we add another cycle to the two cycles on top, in which we gradually reduce the size of the “comb”. The only thing you will need to do is check the distance when comparing, so that it does not go beyond the array.
A really interesting topic is the choice of the sequence of changing the comparison length at each iteration of the first cycle. It is interesting because the performance of the algorithm depends on it.
A table of known variants and time complexity can be found here: https://en.wikipedia.org/wiki/Shellsort#Gap_sequences
Different people were involved in calculating the ideal distance, apparently they were so interested in the topic. Couldn’t they just run Ruby and call the fastest sort() algorithm?
In general, these strange people wrote dissertations on the topic of calculating the distance/gap of the “comb” for the Shell algorithm. I simply used the results of their work and checked 5 types of sequences, Hibbard, Knuth-Pratt, Chiura, Sedgewick.
import time
import random
from functools import reduce
import math
DEMO_MODE = False
if input("Demo Mode Y/N? ").upper() == "Y":
DEMO_MODE = True
class Colors:
BLUE = '\033[94m'
RED = '\033[31m'
END = '\033[0m'
def swap(list, lhs, rhs):
list[lhs], list[rhs] = list[rhs], list[lhs]
return list
def colorPrintoutStep(numbers: List[int], lhs: int, rhs: int):
for index, number in enumerate(numbers):
if index == lhs:
print(f"{Colors.BLUE}", end = "")
elif index == rhs:
print(f"{Colors.RED}", end = "")
print(f"{number},", end = "")
if index == lhs or index == rhs:
print(f"{Colors.END}", end = "")
if index == lhs or index == rhs:
print(f"{Colors.END}", end = "")
print("\n")
input(">")
def ShellSortLoop(numbers: List[int], distanceSequence: List[int]):
distanceSequenceIterator = reversed(distanceSequence)
while distance:= next(distanceSequenceIterator, None):
for sortArea in range(0, len(numbers)):
for rhs in reversed(range(distance, sortArea + 1)):
lhs = rhs - distance
if DEMO_MODE:
print(f"Distance: {distance}")
colorPrintoutStep(numbers, lhs, rhs)
if numbers[lhs] > numbers[rhs]:
swap(numbers, lhs, rhs)
else:
break
def ShellSort(numbers: List[int]):
global ShellSequence
ShellSortLoop(numbers, ShellSequence)
def HibbardSort(numbers: List[int]):
global HibbardSequence
ShellSortLoop(numbers, HibbardSequence)
def ShellPlusKnuttPrattSort(numbers: List[int]):
global KnuttPrattSequence
ShellSortLoop(numbers, KnuttPrattSequence)
def ShellPlusCiuraSort(numbers: List[int]):
global CiuraSequence
ShellSortLoop(numbers, CiuraSequence)
def ShellPlusSedgewickSort(numbers: List[int]):
global SedgewickSequence
ShellSortLoop(numbers, SedgewickSequence)
def insertionSort(numbers: List[int]):
global insertionSortDistanceSequence
ShellSortLoop(numbers, insertionSortDistanceSequence)
def defaultSort(numbers: List[int]):
numbers.sort()
def measureExecution(inputNumbers: List[int], algorithmName: str, algorithm):
if DEMO_MODE:
print(f"{algorithmName} started")
numbers = inputNumbers.copy()
startTime = time.perf_counter()
algorithm(numbers)
endTime = time.perf_counter()
print(f"{algorithmName} performance: {endTime - startTime}")
def sortedNumbersAsString(inputNumbers: List[int], algorithm) -> str:
numbers = inputNumbers.copy()
algorithm(numbers)
return str(numbers)
if DEMO_MODE:
maximalNumber = 10
numbersCount = 10
else:
maximalNumber = 10
numbersCount = random.randint(10000, 20000)
randomNumbers = [random.randrange(1, maximalNumber) for i in range(numbersCount)]
ShellSequenceGenerator = lambda n: reduce(lambda x, _: x + [int(x[-1]/2)], range(int(math.log(numbersCount, 2))), [int(numbersCount / 2)])
ShellSequence = ShellSequenceGenerator(randomNumbers)
ShellSequence.reverse()
ShellSequence.pop()
HibbardSequence = [
0, 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 4095,
8191, 16383, 32767, 65535, 131071, 262143, 524287, 1048575,
2097151, 4194303, 8388607, 16777215, 33554431, 67108863, 134217727,
268435455, 536870911, 1073741823, 2147483647, 4294967295, 8589934591
]
KnuttPrattSequence = [
1, 4, 13, 40, 121, 364, 1093, 3280, 9841, 29524, 88573, 265720,
797161, 2391484, 7174453, 21523360, 64570081, 193710244, 581130733,
1743392200, 5230176601, 15690529804, 47071589413
]
CiuraSequence = [
1, 4, 10, 23, 57, 132, 301, 701, 1750, 4376,
10941, 27353, 68383, 170958, 427396, 1068491,
2671228, 6678071, 16695178, 41737946, 104344866,
260862166, 652155416, 1630388541
]
SedgewickSequence = [
1, 5, 19, 41, 109, 209, 505, 929, 2161, 3905,
8929, 16001, 36289, 64769, 146305, 260609, 587521,
1045505, 2354689, 4188161, 9427969, 16764929, 37730305,
67084289, 150958081, 268386305, 603906049, 1073643521,
2415771649, 4294770689, 9663381505, 17179475969
]
insertionSortDistanceSequence = [1]
algorithms = {
"Default Python Sort": defaultSort,
"Shell Sort": ShellSort,
"Shell + Hibbard" : HibbardSort,
"Shell + Prat, Knutt": ShellPlusKnuttPrattSort,
"Shell + Ciura Sort": ShellPlusCiuraSort,
"Shell + Sedgewick Sort": ShellPlusSedgewickSort,
"Insertion Sort": insertionSort
}
for name, algorithm in algorithms.items():
measureExecution(randomNumbers, name, algorithm)
reference = sortedNumbersAsString(randomNumbers, defaultSort)
for name, algorithm in algorithms.items():
if sortedNumbersAsString(randomNumbers, algorithm) != reference:
print("Sorting validation failed")
exit(1)
print("Sorting validation success")
exit(0)
In my implementation, for a random set of numbers, the fastest gaps are Sedgewick and Hibbard.
mypy
I would also like to mention the static typing analyzer for Python 3 – mypy. It helps to cope with the problems inherent in languages with dynamic typing, namely, it eliminates the possibility of slipping something where it shouldn’t.
As experienced programmers say, “static typing is not needed when you have a team of professionals”, someday we will all become professionals, we will write code in complete unity and understanding with machines, but for now you can use such utilities and languages with static typing.
Links
https://gitlab.com/demensdeum /algorithms/-/tree/master/sortAlgorithms/shellSort
http://mypy-lang.org/
Sources
https://dl.acm.org/doi/10.1145/368370.368387
https://en.wikipedia.org/wiki/Shellsort
http://rosettacode.org/wiki/Sorting_algorithms/Shell_sort
https://ru.wikipedia.org/wiki/Сортировка_Шелла
https://neerc.ifmo.ru/wiki/index.php?title=Сортировка_Шелла
https://twitter.com/gvanrossum/status/700741601966985216
Double Selection Sort
Double Selection Sort is a type of selection sort that should be twice as fast. The vanilla algorithm goes through a double loop over a list of numbers, finds the minimum number and swaps it with the current digit pointed to by the loop at the level above. Double selection sort looks for the minimum and maximum number, then replaces the two digits pointed to by the loop at the level above – two numbers on the left and right. This whole orgy ends when the cursors of the numbers to be replaced meet in the middle of the list, as a result, sorted numbers are obtained to the left and right of the visual center.
The time complexity of the algorithm is similar to Selection Sort – O(n2), but there is supposedly a 30% speedup.
Borderline state
Already at this stage, you can imagine the moment of collision, for example, when the number of the left cursor (minimum number) will point to the maximum number in the list, then the minimum number is rearranged, the maximum number rearrangement immediately breaks. Therefore, all implementations of the algorithm contain a check for such cases, replacing the indices with correct ones. In my implementation, one check was enough:
maximalNumberIndex = minimalNumberIndex;
}
Реализация на Cito
Cito – язык либ, язык транслятор. На нем можно писать для C, C++, C#, Java, JavaScript, Python, Swift, TypeScript, OpenCL C, при этом совершенно ничего не зная про эти языки. Исходный код на языке Cito транслируется в исходный код на поддерживаемых языках, далее можно использовать как библиотеку, либо напрямую, исправив сгенеренный код руками. Эдакий Write once – translate to anything.
Double Selection Sort на cito:
{
public static int[] sort(int[]# numbers, int length)
{
int[]# sortedNumbers = new int[length];
for (int i = 0; i < length; i++) {
sortedNumbers[i] = numbers[i];
}
for (int leftCursor = 0; leftCursor < length / 2; leftCursor++) {
int minimalNumberIndex = leftCursor;
int minimalNumber = sortedNumbers[leftCursor];
int rightCursor = length - (leftCursor + 1);
int maximalNumberIndex = rightCursor;
int maximalNumber = sortedNumbers[maximalNumberIndex];
for (int cursor = leftCursor; cursor <= rightCursor; cursor++) { int cursorNumber = sortedNumbers[cursor]; if (minimalNumber > cursorNumber) {
minimalNumber = cursorNumber;
minimalNumberIndex = cursor;
}
if (maximalNumber < cursorNumber) {
maximalNumber = cursorNumber;
maximalNumberIndex = cursor;
}
}
if (leftCursor == maximalNumberIndex) {
maximalNumberIndex = minimalNumberIndex;
}
int fromNumber = sortedNumbers[leftCursor];
int toNumber = sortedNumbers[minimalNumberIndex];
sortedNumbers[minimalNumberIndex] = fromNumber;
sortedNumbers[leftCursor] = toNumber;
fromNumber = sortedNumbers[rightCursor];
toNumber = sortedNumbers[maximalNumberIndex];
sortedNumbers[maximalNumberIndex] = fromNumber;
sortedNumbers[rightCursor] = toNumber;
}
return sortedNumbers;
}
}
Links
https://gitlab.com/demensdeum /algorithms/-/tree/master/sortAlgorithms/doubleSelectionSort
https://github.com/pfusik/cito
Sources
https://www.researchgate.net/publication/330084245_Improved_Double_Selection_Sort_using_Algorithm
http://algolab.valemak.com/selection-double
https://www.geeksforgeeks.org/sorting-algorithm-slightly-improves-selection-sort/