If you’re a dedicated Unreal Tournament 99 fan like me, you’ll want to run the game on the latest operating systems and hardware. I have successfully run Unreal Tournament 99 on an M1 Macbook Pro running macOS Ventura 13.0.1.
To run the game under macOS for the M1 processor you need:
Create an Unreal Tournament folder in ~/Library/Application Support/
Copy the Windows versions of the Maps, Sounds, Textures, Music folder to ~/Library/Application Support/Unreal Tournament
Delete the files LadderFonts.utx, UWindowFonts.utx from the folder ~/Library/Application Support/Unreal Tournament/Textures
Run UnrealTournament.app from /Applications, enjoy the frags!
The penultimate step is needed to display the correct fonts, the original ones are displayed too small.
After starting, configure the screen resolution, keyboard, font size in the GUI, and other necessary settings.
Windows 11
Also, for dessert, the launch of Unreal Tournament 99 on Windows 11, the game works immediately after installation, without additional shamanism, but there are problems with displaying the GUI, the performance of an outdated D3D renderer. Therefore, it is better to use the patched version.
The launch process is very similar to that for macOS:
I recently bought a very inexpensive Getorix GK-45X USB keyboard with RGB backlight. After connecting it to a Macbook Pro on an M1 processor, it became clear that the RGB backlight was not working. Even by pressing the magic combination Fn + Scroll Lock, it was not possible to turn on the backlight, only the backlight level of the MacBook screen changed.
There are several solutions to this problem, namely OpenRGB (does not work), HID LED Test (does not work). Only the kvmswitch utility worked: https://github.com/stoutput/OSX-KVM
You need to download it from the github and allow it to run from the terminal in the Security panel of the System Settings.
As I understood from the description, after launching the utility sends pressing Fn + Scroll Lock, thus turning on/off the backlight on the keyboard.
Tree sort – binary search tree sort. Time complexity – O(n²). In such a tree, each node has numbers less than the node on the left, more than the node on the right, when coming from the root and printing the values from left to right, we get a sorted list of numbers. Surprising huh?
Try to manually read the numbers starting from the penultimate left node of the lower left corner, for each node on the left – a node – on the right.
It will turn out like this:
The penultimate node at the bottom left is 3.
She has a left branch – 1.
Take this number (1)
Next, take the vertex 3 (1, 3) itself
To the right is branch 6, but it contains branches. Therefore, we read it in the same way.
Left branch of node 6 number 4 (1, 3, 4)
Node 6 itself (1, 3, 4, 6)
Right 7 (1, 3, 4, 6, 7)
Go up to the root node – 8 (1,3, 4 ,6, 7, 8)
Print everything on the right by analogy
Get the final list – 1, 3, 4, 6, 7, 8, 10, 13, 14
To implement the algorithm in code, you need two functions:
Building a binary search tree
Printing the binary search tree in the correct order
They assemble a binary search tree in the same way as they read it, a number is attached to each node on the left or right, depending on whether it is less or more.
Lua example:
Node = {value = nil, lhs = nil, rhs = nil}
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)
An important nuance is that for numbers that are equal to the vertex, a lot of interesting mechanisms for hooking to the node have been invented, but I just added a counter to the vertex class, when printing, the numbers are returned by the counter.
Bucket Sort – bucket sorting. The algorithm is similar to sorting by counting, with the difference that the numbers are collected into “buckets”-ranges, then the buckets are sorted using any other, sufficiently productive, sorting algorithm, and the final chord is the expansion of the “buckets” one by one, resulting in a sorted list.
The time complexity of the algorithm is O(nk). The algorithm runs in linear time for data that obeys a uniform distribution. To put it simply, the elements must be in a certain range, without “splashes”, for example, numbers from 0.0 to 1.0. If among such numbers there are 4 or 999, then such a series, according to the laws of the yard, is no longer considered “even”.
Implementation example in Julia:
function bucketSort(numbers, bucketsCount)
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)
The performance of the algorithm is also affected by the number of buckets, for more numbers it is better to take a larger number of buckets (Algorithms in a nutshell by George T. Heineman)
Radix Sort – radix sort. The algorithm is similar to counting sort in that there is no comparison of elements, instead elements are *character-by-character* grouped into *buckets* (buckets), the bucket is selected by the index of the current number-character. Time complexity – O(nd).
Works like this:
The input will be the numbers 6, 12, 44, 9
Let’s create 10 buckets of lists (0-9) into which we will add/sort numbers bit by bit.
Further:
Run a loop with counter i up to the maximum number of characters in the number
At index i from right to left we get one character for each number, if there is no character, then we consider it to be zero
The character is converted to a number
Select a bucket by index – number, put the whole number there
After finishing iterating over numbers, convert all buckets back to a list of numbers
Get numbers sorted by digit
Repeat until all digits run out
Radix Sort example in Scala:
import scala.collection.mutable.ListBuffer
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 the GPU; there is also a variant of bit sort, which is probably very interesting and truly breathtaking!
Heapsort – heap sort. Time complexity – O(n log n), fast eh? 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, with 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 these are just numbers from the array, going one after another, the number of elements in each floor is multiplied by two.
Then the fun begins, we sort the pyramid from bottom to top, using the falling pebbles method (heapify). Sorting could be started from the last floor (2 3 9 4 ), but it makes no sense because there is no floor below where one could fall.
Therefore, we start dropping elements from the penultimate floor (0 7)
⠀⠀5
⠀0⠀7
2 3 9 4
The first element to fall is selected on the right, in our case it is 7, then we look at what is under it, and below it are 9 and 4, nine is more than four, so also nine is more than seven! We drop 7 on 9, and raise 9 to place 7.
⠀⠀5
⠀0⠀9
2 3 7 4
Further, we understand that the seven has nowhere to fall below, go to the number 0, which is located on the penultimate floor on the left:
⠀⠀5
⠀0⠀9
2 3 7 4
We look at what is under it – 2 and 3, two is less than three, three is greater than zero, so we change zero and three in places:
⠀⠀5
⠀3⠀9
2 0 7 4
When you get to the end of the floor, go to the floor above and drop everything there if you can.
The result is a data structure – a heap (heap), namely max heap, because at the top is the largest element:
⠀⠀9
⠀3⠀7
2 0 5 4
If you return it to an array representation, you get a list:
[9, 3, 7, 2, 0, 5, 4]
From this we can conclude that by swapping the first and last element, we will get the first number in the final sorted position, namely 9 should be at the end of the sorted list, swap:
[4, 3, 7, 2, 0, 5, 9]
Let’s look at the binary tree:
⠀⠀4
⠀3⠀7
2 0 5 9
The result is a situation in which the lower part of the tree is sorted, you just need to drop 4 to the correct position, repeat the algorithm, but 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 we, having dropped 4, raised the next largest number after 9 – 7. Swap the last unsorted number (4) and the largest number (7)
⠀⠀4
⠀3⠀5
2 0 7 9
It turned out that now we have two numbers in the correct final position:
4, 3, 5, 2, 0, 7, 9
Next, we repeat the sorting algorithm, ignoring those already sorted, as a result we get heap:
⠀⠀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:
Heap creation
Sifting algorithm (heapify)
Replacing the last unsorted element and the first one
A heap is created by traversing the penultimate row of the binary tree using the heapify function, from right to left to the end of the array. Then the first replacement of numbers is made in the cycle, after which the first element falls / remains in place, as a result of which the largest element falls into first place, the cycle repeats with a decrease in participants by one, because after each pass, the sorted numbers remain at the end of the list.
Heapsort example in Ruby:
DEMO = true
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 view of the binary tree.
Recently, it turned out that users of modern Nvidia GPUs under Arch Linux do not need to use the bumblebee package at all, for example, for me it did not detect an external monitor when connected. I recommend removing the bumblebee package and all related packages, and installing prime using the instructions on the Arch Wiki.
Next, to launch all games on Steam and 3D applications, add prime-run, for Steam this is done like this prime-run %command% in additional launch options.
To check the correctness, you can use glxgears, prime-run glxgears. https://bbs.archlinux.org/viewtopic.php? pid=2048195#p2048195
Quicksort is a divide-and-conquer sorting algorithm. Recursively, in parts, parse the array of numbers, setting the numbers in the smaller and larger order from the selected pivot element, insert the pivot element itself into the hot-point between them. After a few recursive iterations, you’ll end up with a sorted list. Time complexity O(n2).
Scheme:
We start with the fact that we get a list of elements outside, sorting boundaries. In the first step, the sort boundaries will be from start to finish.
Check that the boundaries of the beginning and end do not intersect, if this happens, then it’s time to finish
Select some element from the list, call it pivot
Shift to the right to the end to the last index, so as not to interfere
Create a counter of *smaller numbers* while equal to zero
Loop through the list from left to right, up to and including the last index where the anchor element is located
Each element is compared to the pivot
If it is less than the reference, then we swap it in places according to the index of the counter of smaller numbers. Increment the counter of smaller numbers.
When the cycle reaches the anchor element, we stop, swap the anchor element with the element by the counter of smaller numbers.
We run the algorithm separately for the left smaller part of the list, and separately for the right most part of the list.
As a result, all recursive iterations will start to stop due to the check in paragraph 2
Get sorted list
Quicksort was invented by the scientist Charles Antony Richard Hoare at Moscow State University, having learned Russian, he studied computer translation, as well as probability theory at the Kolmogorov school. In 1960, due to a political crisis, he left the Soviet Union.
An example implementation in Rust:
extern crate rand;
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 you don't understand anything, I suggest watching a video by Rob Edwards from the San Diego State University https://www.youtube.com/watch?v=ZHVk2blR45Q it most simply, step by step, shows the essence and implementation of the algorithm.
We use cookies on our website. By clicking “Accept”, you consent to the use of ALL the cookies. Мы используем куки на сайте. Нажимая "ПРИНЯТЬ" вы соглашаетесь с этим.
This website uses cookies to improve your experience while you navigate through the website. Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. We also use third-party cookies that help us analyze and understand how you use this website. These cookies will be stored in your browser only with your consent. You also have the option to opt-out of these cookies. But opting out of some of these cookies may affect your browsing experience.
Necessary cookies are absolutely essential for the website to function properly. This category only includes cookies that ensures basic functionalities and security features of the website. These cookies do not store any personal information.
Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. It is mandatory to procure user consent prior to running these cookies on your website.
