I have gone through Google and Stack Overflow search, but nowhere I was able to find a clear and straightforward explanation for how to calculate time complexity.
What do I know already?
Say for code as simple as the one below:
char h = 'y'; // This will be executed 1 time
int abc = 0; // This will be executed 1 time
Say for a loop like the one below:
for (int i = 0; i < N; i++) {
Console.Write('Hello, World!!');
}
int i=0; This will be executed only once.
The time is actually calculated to i=0 and not the declaration.
i < N; This will be executed N+1 times
i++ This will be executed N times
So the number of operations required by this loop are {1+(N+1)+N} = 2N+2. (But this still may be wrong, as I am not confident about my understanding.)
OK, so these small basic calculations I think I know, but in most cases I have seen the time complexity as O(N), O(n^2), O(log n), O(n!), and many others.
How to find time complexity of an algorithm
You add up how many machine instructions it will execute as a function of the size of its input, and then simplify the expression to the largest (when N is very large) term and can include any simplifying constant factor.
For example, lets see how we simplify 2N + 2 machine instructions to describe this as just O(N).
Why do we remove the two 2s ?
We are interested in the performance of the algorithm as N becomes large.
Consider the two terms 2N and 2.
What is the relative influence of these two terms as N becomes large? Suppose N is a million.
Then the first term is 2 million and the second term is only 2.
For this reason, we drop all but the largest terms for large N.
So, now we have gone from 2N + 2 to 2N.
Traditionally, we are only interested in performance up to constant factors.
This means that we don't really care if there is some constant multiple of difference in performance when N is large. The unit of 2N is not well-defined in the first place anyway. So we can multiply or divide by a constant factor to get to the simplest expression.
So 2N becomes just N.
This is an excellent article: Time complexity of algorithm
The below answer is copied from above (in case the excellent link goes bust)
The most common metric for calculating time complexity is Big O notation. This removes all constant factors so that the running time can be estimated in relation to N as N approaches infinity. In general you can think of it like this:
statement;
Is constant. The running time of the statement will not change in relation to N.
for ( i = 0; i < N; i++ )
statement;
Is linear. The running time of the loop is directly proportional to N. When N doubles, so does the running time.
for ( i = 0; i < N; i++ ) {
for ( j = 0; j < N; j++ )
statement;
}
Is quadratic. The running time of the two loops is proportional to the square of N. When N doubles, the running time increases by N * N.
while ( low <= high ) {
mid = ( low + high ) / 2;
if ( target < list[mid] )
high = mid - 1;
else if ( target > list[mid] )
low = mid + 1;
else break;
}
Is logarithmic. The running time of the algorithm is proportional to the number of times N can be divided by 2. This is because the algorithm divides the working area in half with each iteration.
void quicksort (int list[], int left, int right)
{
int pivot = partition (list, left, right);
quicksort(list, left, pivot - 1);
quicksort(list, pivot + 1, right);
}
Is N * log (N). The running time consists of N loops (iterative or recursive) that are logarithmic, thus the algorithm is a combination of linear and logarithmic.
In general, doing something with every item in one dimension is linear, doing something with every item in two dimensions is quadratic, and dividing the working area in half is logarithmic. There are other Big O measures such as cubic, exponential, and square root, but they're not nearly as common. Big O notation is described as O ( <type> ) where <type> is the measure. The quicksort algorithm would be described as O (N * log(N )).
Note that none of this has taken into account best, average, and worst case measures. Each would have its own Big O notation. Also note that this is a VERY simplistic explanation. Big O is the most common, but it's also more complex that I've shown. There are also other notations such as big omega, little o, and big theta. You probably won't encounter them outside of an algorithm analysis course. ;)
Taken from here - Introduction to Time Complexity of an Algorithm
1. Introduction
In computer science, the time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function of the length of the string representing the input.
2. Big O notation
The time complexity of an algorithm is commonly expressed using big O notation, which excludes coefficients and lower order terms. When expressed this way, the time complexity is said to be described asymptotically, i.e., as the input size goes to infinity.
For example, if the time required by an algorithm on all inputs of size n is at most 5n3 + 3n, the asymptotic time complexity is O(n3). More on that later.
A few more examples:
1 = O(n)
n = O(n2)
log(n) = O(n)
2 n + 1 = O(n)
3. O(1) constant time:
An algorithm is said to run in constant time if it requires the same amount of time regardless of the input size.
Examples:
array: accessing any element
fixed-size stack: push and pop methods
fixed-size queue: enqueue and dequeue methods
4. O(n) linear time
An algorithm is said to run in linear time if its time execution is directly proportional to the input size, i.e. time grows linearly as input size increases.
Consider the following examples. Below I am linearly searching for an element, and this has a time complexity of O(n).
int find = 66;
var numbers = new int[] { 33, 435, 36, 37, 43, 45, 66, 656, 2232 };
for (int i = 0; i < numbers.Length - 1; i++)
{
if(find == numbers[i])
{
return;
}
}
More Examples:
Array: Linear Search, Traversing, Find minimum etc
ArrayList: contains method
Queue: contains method
5. O(log n) logarithmic time:
An algorithm is said to run in logarithmic time if its time execution is proportional to the logarithm of the input size.
Example: Binary Search
Recall the "twenty questions" game - the task is to guess the value of a hidden number in an interval. Each time you make a guess, you are told whether your guess is too high or too low. Twenty questions game implies a strategy that uses your guess number to halve the interval size. This is an example of the general problem-solving method known as binary search.
6. O(n2) quadratic time
An algorithm is said to run in quadratic time if its time execution is proportional to the square of the input size.
Examples:
Bubble Sort
Selection Sort
Insertion Sort
7. Some useful links
Big-O Misconceptions
Determining The Complexity Of Algorithm
Big O Cheat Sheet
Several examples of loop.
O(n) time complexity of a loop is considered as O(n) if the loop variables is incremented / decremented by a constant amount. For example following functions have O(n) time complexity.
// Here c is a positive integer constant
for (int i = 1; i <= n; i += c) {
// some O(1) expressions
}
for (int i = n; i > 0; i -= c) {
// some O(1) expressions
}
O(nc) time complexity of nested loops is equal to the number of times the innermost statement is executed. For example, the following sample loops have O(n2) time complexity
for (int i = 1; i <=n; i += c) {
for (int j = 1; j <=n; j += c) {
// some O(1) expressions
}
}
for (int i = n; i > 0; i += c) {
for (int j = i+1; j <=n; j += c) {
// some O(1) expressions
}
For example, selection sort and insertion sort have O(n2) time complexity.
O(log n) time complexity of a loop is considered as O(log n) if the loop variables is divided / multiplied by a constant amount.
for (int i = 1; i <=n; i *= c) {
// some O(1) expressions
}
for (int i = n; i > 0; i /= c) {
// some O(1) expressions
}
For example, [binary search][3] has _O(log n)_ time complexity.
O(log log n) time complexity of a loop is considered as O(log log n) if the loop variables is reduced / increased exponentially by a constant amount.
// Here c is a constant greater than 1
for (int i = 2; i <=n; i = pow(i, c)) {
// some O(1) expressions
}
//Here fun is sqrt or cuberoot or any other constant root
for (int i = n; i > 0; i = fun(i)) {
// some O(1) expressions
}
One example of time complexity analysis
int fun(int n)
{
for (int i = 1; i <= n; i++)
{
for (int j = 1; j < n; j += i)
{
// Some O(1) task
}
}
}
Analysis:
For i = 1, the inner loop is executed n times.
For i = 2, the inner loop is executed approximately n/2 times.
For i = 3, the inner loop is executed approximately n/3 times.
For i = 4, the inner loop is executed approximately n/4 times.
…………………………………………………….
For i = n, the inner loop is executed approximately n/n times.
So the total time complexity of the above algorithm is (n + n/2 + n/3 + … + n/n), which becomes n * (1/1 + 1/2 + 1/3 + … + 1/n)
The important thing about series (1/1 + 1/2 + 1/3 + … + 1/n) is around to O(log n). So the time complexity of the above code is O(n·log n).
References:
1
2
3
Time complexity with examples
1 - Basic operations (arithmetic, comparisons, accessing array’s elements, assignment): The running time is always constant O(1)
Example:
read(x) // O(1)
a = 10; // O(1)
a = 1,000,000,000,000,000,000 // O(1)
2 - If then else statement: Only taking the maximum running time from two or more possible statements.
Example:
age = read(x) // (1+1) = 2
if age < 17 then begin // 1
status = "Not allowed!"; // 1
end else begin
status = "Welcome! Please come in"; // 1
visitors = visitors + 1; // 1+1 = 2
end;
So, the complexity of the above pseudo code is T(n) = 2 + 1 + max(1, 1+2) = 6. Thus, its big oh is still constant T(n) = O(1).
3 - Looping (for, while, repeat): Running time for this statement is the number of loops multiplied by the number of operations inside that looping.
Example:
total = 0; // 1
for i = 1 to n do begin // (1+1)*n = 2n
total = total + i; // (1+1)*n = 2n
end;
writeln(total); // 1
So, its complexity is T(n) = 1+4n+1 = 4n + 2. Thus, T(n) = O(n).
4 - Nested loop (looping inside looping): Since there is at least one looping inside the main looping, running time of this statement used O(n^2) or O(n^3).
Example:
for i = 1 to n do begin // (1+1)*n = 2n
for j = 1 to n do begin // (1+1)n*n = 2n^2
x = x + 1; // (1+1)n*n = 2n^2
print(x); // (n*n) = n^2
end;
end;
Common running time
There are some common running times when analyzing an algorithm:
O(1) – Constant time
Constant time means the running time is constant, it’s not affected by the input size.
O(n) – Linear time
When an algorithm accepts n input size, it would perform n operations as well.
O(log n) – Logarithmic time
Algorithm that has running time O(log n) is slight faster than O(n). Commonly, algorithm divides the problem into sub problems with the same size. Example: binary search algorithm, binary conversion algorithm.
O(n log n) – Linearithmic time
This running time is often found in "divide & conquer algorithms" which divide the problem into sub problems recursively and then merge them in n time. Example: Merge Sort algorithm.
O(n2) – Quadratic time
Look Bubble Sort algorithm!
O(n3) – Cubic time
It has the same principle with O(n2).
O(2n) – Exponential time
It is very slow as input get larger, if n = 1,000,000, T(n) would be 21,000,000. Brute Force algorithm has this running time.
O(n!) – Factorial time
The slowest!!! Example: Travelling salesman problem (TSP)
It is taken from this article. It is very well explained and you should give it a read.
When you're analyzing code, you have to analyse it line by line, counting every operation/recognizing time complexity. In the end, you have to sum it to get whole picture.
For example, you can have one simple loop with linear complexity, but later in that same program you can have a triple loop that has cubic complexity, so your program will have cubic complexity. Function order of growth comes into play right here.
Let's look at what are possibilities for time complexity of an algorithm, you can see order of growth I mentioned above:
Constant time has an order of growth 1, for example: a = b + c.
Logarithmic time has an order of growth log N. It usually occurs when you're dividing something in half (binary search, trees, and even loops), or multiplying something in same way.
Linear. The order of growth is N, for example
int p = 0;
for (int i = 1; i < N; i++)
p = p + 2;
Linearithmic. The order of growth is n·log N. It usually occurs in divide-and-conquer algorithms.
Cubic. The order of growth is N3. A classic example is a triple loop where you check all triplets:
int x = 0;
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
for (int k = 0; k < N; k++)
x = x + 2
Exponential. The order of growth is 2N. It usually occurs when you do exhaustive search, for example, check subsets of some set.
Loosely speaking, time complexity is a way of summarising how the number of operations or run-time of an algorithm grows as the input size increases.
Like most things in life, a cocktail party can help us understand.
O(N)
When you arrive at the party, you have to shake everyone's hand (do an operation on every item). As the number of attendees N increases, the time/work it will take you to shake everyone's hand increases as O(N).
Why O(N) and not cN?
There's variation in the amount of time it takes to shake hands with people. You could average this out and capture it in a constant c. But the fundamental operation here --- shaking hands with everyone --- would always be proportional to O(N), no matter what c was. When debating whether we should go to a cocktail party, we're often more interested in the fact that we'll have to meet everyone than in the minute details of what those meetings look like.
O(N^2)
The host of the cocktail party wants you to play a silly game where everyone meets everyone else. Therefore, you must meet N-1 other people and, because the next person has already met you, they must meet N-2 people, and so on. The sum of this series is x^2/2+x/2. As the number of attendees grows, the x^2 term gets big fast, so we just drop everything else.
O(N^3)
You have to meet everyone else and, during each meeting, you must talk about everyone else in the room.
O(1)
The host wants to announce something. They ding a wineglass and speak loudly. Everyone hears them. It turns out it doesn't matter how many attendees there are, this operation always takes the same amount of time.
O(log N)
The host has laid everyone out at the table in alphabetical order. Where is Dan? You reason that he must be somewhere between Adam and Mandy (certainly not between Mandy and Zach!). Given that, is he between George and Mandy? No. He must be between Adam and Fred, and between Cindy and Fred. And so on... we can efficiently locate Dan by looking at half the set and then half of that set. Ultimately, we look at O(log_2 N) individuals.
O(N log N)
You could find where to sit down at the table using the algorithm above. If a large number of people came to the table, one at a time, and all did this, that would take O(N log N) time. This turns out to be how long it takes to sort any collection of items when they must be compared.
Best/Worst Case
You arrive at the party and need to find Inigo - how long will it take? It depends on when you arrive. If everyone is milling around you've hit the worst-case: it will take O(N) time. However, if everyone is sitting down at the table, it will take only O(log N) time. Or maybe you can leverage the host's wineglass-shouting power and it will take only O(1) time.
Assuming the host is unavailable, we can say that the Inigo-finding algorithm has a lower-bound of O(log N) and an upper-bound of O(N), depending on the state of the party when you arrive.
Space & Communication
The same ideas can be applied to understanding how algorithms use space or communication.
Knuth has written a nice paper about the former entitled "The Complexity of Songs".
Theorem 2: There exist arbitrarily long songs of complexity O(1).
PROOF: (due to Casey and the Sunshine Band). Consider the songs Sk defined by (15), but with
V_k = 'That's the way,' U 'I like it, ' U
U = 'uh huh,' 'uh huh'
for all k.
For the mathematically-minded people: The master theorem is another useful thing to know when studying complexity.
O(n) is big O notation used for writing time complexity of an algorithm. When you add up the number of executions in an algorithm, you'll get an expression in result like 2N+2. In this expression, N is the dominating term (the term having largest effect on expression if its value increases or decreases). Now O(N) is the time complexity while N is dominating term.
Example
For i = 1 to n;
j = 0;
while(j <= n);
j = j + 1;
Here the total number of executions for the inner loop are n+1 and the total number of executions for the outer loop are n(n+1)/2, so the total number of executions for the whole algorithm are n + 1 + n(n+1/2) = (n2 + 3n)/2.
Here n^2 is the dominating term so the time complexity for this algorithm is O(n2).
Other answers concentrate on the big-O-notation and practical examples. I want to answer the question by emphasizing the theoretical view. The explanation below is necessarily lacking in details; an excellent source to learn computational complexity theory is Introduction to the Theory of Computation by Michael Sipser.
Turing Machines
The most widespread model to investigate any question about computation is a Turing machine. A Turing machine has a one dimensional tape consisting of symbols which is used as a memory device. It has a tapehead which is used to write and read from the tape. It has a transition table determining the machine's behaviour, which is a fixed hardware component that is decided when the machine is created. A Turing machine works at discrete time steps doing the following:
It reads the symbol under the tapehead.
Depending on the symbol and its internal state, which can only take finitely many values, it reads three values s, σ, and X from its transition table, where s is an internal state, σ is a symbol, and X is either Right or Left.
It changes its internal state to s.
It changes the symbol it has read to σ.
It moves the tapehead one step according to the direction in X.
Turing machines are powerful models of computation. They can do everything that your digital computer can do. They were introduced before the advent of digital modern computers by the father of theoretical computer science and mathematician: Alan Turing.
Time Complexity
It is hard to define the time complexity of a single problem like "Does white have a winning strategy in chess?" because there is a machine which runs for a single step giving the correct answer: Either the machine which says directly 'No' or directly 'Yes'. To make it work we instead define the time complexity of a family of problems L each of which has a size, usually the length of the problem description. Then we take a Turing machine M which correctly solves every problem in that family. When M is given a problem of this family of size n, it solves it in finitely many steps. Let us call f(n) the longest possible time it takes M to solve problems of size n. Then we say that the time complexity of L is O(f(n)), which means that there is a Turing machine which will solve an instance of it of size n in at most C.f(n) time where C is a constant independent of n.
Isn't it dependent on the machines? Can digital computers do it faster?
Yes! Some problems can be solved faster by other models of computation, for example two tape Turing machines solve some problems faster than those with a single tape. This is why theoreticians prefer to use robust complexity classes such as NL, P, NP, PSPACE, EXPTIME, etc. For example, P is the class of decision problems whose time complexity is O(p(n)) where p is a polynomial. The class P do not change even if you add ten thousand tapes to your Turing machine, or use other types of theoretical models such as random access machines.
A Difference in Theory and Practice
It is usually assumed that the time complexity of integer addition is O(1). This assumption makes sense in practice because computers use a fixed number of bits to store numbers for many applications. There is no reason to assume such a thing in theory, so time complexity of addition is O(k) where k is the number of bits needed to express the integer.
Finding The Time Complexity of a Class of Problems
The straightforward way to show the time complexity of a problem is O(f(n)) is to construct a Turing machine which solves it in O(f(n)) time. Creating Turing machines for complex problems is not trivial; one needs some familiarity with them. A transition table for a Turing machine is rarely given, and it is described in high level. It becomes easier to see how long it will take a machine to halt as one gets themselves familiar with them.
Showing that a problem is not O(f(n)) time complexity is another story... Even though there are some results like the time hierarchy theorem, there are many open problems here. For example whether problems in NP are in P, i.e. solvable in polynomial time, is one of the seven millennium prize problems in mathematics, whose solver will be awarded 1 million dollars.
Related
I found some very helpful solutions to writing a Javascript function that recursively checks whether a string is a palindrome here. I would like to know what the time and space complexity of the following solution would be. Can you explain how each line contributes to big O?
function isPalindrome(string) {
if (string.length < 2) return true;
if (string[0] === string[string.length - 1]) {
return isPalindrome(string.slice(1, string.length - 1))
}
return false;
}
At first it would seem that your function is O(n), because you call it recursively n/2 times. However, in each call you also use string.slice, which has a complexity of O(n). Because of that, your function is actually O(n^2)
You recursively call the function n/2 times, where n is the length of the string, since you remove first and last entry of the string at each iteration.
Therefore, the complexity would be O(n/2) = O(n), and you have at most 3 operations at each function call. That would multiply all this by a constant, which does not change the complexity.
EDIT: as noted in comments and another answer, one of these operations is string.slice. You need to check for the complexity of this, because it will multiply as well and can change overall complexity. If slice is O(1) constant, then overall you'll have O(n). If slice is O(n), then overall that would be O(n^2).
For space complexity, you create a lot of arrays. I'll let you the details, but I'm tempted to say it's O(n^2) at first sight.
Sketch of calculations : first array is size n, then n-2, n-4, sum that all up with some math summation formulas.
hint : n + (n-1) + (n-2) + ... is n * (n + 1) / 2 which is O(n^2), that should give enough "feel" that this is O(n^2) too.
I'm using the standard Fisher-Yates algorithm to randomly shuffle a deck of cards in an array. However, I'm unsure if this will actually produce a true distribution of all possible permutations of a real-world shuffled deck of cards.
V8's Math.random only has 128-bits of internal state. Since there are 52 cards in a deck, 52 factorial would require 226-bits of internal state to generate all possible permutations.
However, I'm unsure if this applies when using Fisher-Yates since you aren't actually generating each possible but just getting one position randomly out of 52.
function shuffle(array) {
var m = array.length, t, i;
while (m) {
i = Math.floor(Math.random() * m--);
t = array[m];
array[m] = array[i];
array[i] = t;
}
return array;
}
In general, if a pseudorandom number generator admits fewer than 52 factorial different seeds, then there are some permutations that particular PRNG can't choose when it shuffles a 52-item list, and Fisher-Yates can't change that. (The set of permutations a particular PRNG can choose can be different from the set of permutations another PRNG can choose, even if both PRNGs are initialized with the same seed.) See also this question.
Note that although the Math.random algorithm used by V8 admits any of about 2^128 seeds at the time of this writing, no particular random number algorithm is mandated by the ECMAScript specification of Math.random, which states only that that method uses an "implementation-dependent algorithm or strategy" to generate random numbers (see ECMAScript sec. 20.2.2.27).
A PRNG's period can be extended with the Bays-Durham shuffle, which effectively increases that PRNG's state length (see Severin Pappadeux's answer). However, if you merely initialize the Bays-Durham table entries with outputs of the PRNG (rather than use the seed to initialize those entries), it will still be the case that that particular PRNG (which includes the way in which it initializes those entries and selects those table entries based on the random numbers it generates) can't choose more permutations than the number of possible seeds to initialize its original state, because there would be only one way to initialize the Bays-Durham entries for a given seed — unless, of course, the PRNG actually shuffles an exorbitant amount of lists, so many that it generates more random numbers without cycling than it otherwise would without the Bays-Durham shuffle.
For example, if the PRNG is 128 bits long, there are only 2^128 possible seeds, so there are only 2^128 ways to initialize the Bays-Durham shuffle, one for each seed, unless a seed longer than 128 bits extends to the Bays-Durham table entries and not just the PRNG's original state. (This is not to imply that the set of permutations that PRNG can choose is always the same no matter how it selects table entries in the Bays-Durham shuffle.)
EDIT (Aug. 7): Clarifications.
EDIT (Jan. 7, 2020): Edited.
You are right. With 128 bits of starting state, you can only generate at most 2128 different permutations. It doesn't matter how often you use this state (call Math.random()), the PRNG is deterministic after all.
Where the number of calls to Math.random() actually matter is when
each call would draw some more entropy (e.g. from hardware random) into the system, instead of relying on the internal state that is initialised only once
the entropy of a single call result is so low that you don't use the entire internal state over the run of the algorithm
Well, you definitely need RNG with 226bits period for all permutation to be covered, #PeterO answer is correct in this regard. But you could extend period using Bays-Durham shuffle, paying by effectively extending state of RNG. There is an estimate of the period of the B-D shuffled RNG and it is
P = sqrt(Pi * N! / (2*O))
where Pi=3.1415..., N is B-D table size, O is period of the original generator. If you take log2 of the whole expression, and use Stirling formula for factorial, and assume P=2226 and O=2128, you could get estimate for N, size of the table in B-D algorithm. From back-of-the envelope calculation N=64 would be enough to get all your permutations.
UPDATE
Ok, here is an example implementation of RNG extended with B-D shuffle. First, I implemented in Javascript Xorshift128+, using BigInt, which is apparently default RNG in V8 engine as well. Compared with C++ one, they produced identical output for first couple of dozen calls. 128bits seed as two 64bits words. Windows 10 x64, NodeJS 12.7.
const WIDTH = 2n ** 64n;
const MASK = WIDTH - 1n; // to keep things as 64bit values
class XorShift128Plus { // as described in https://v8.dev/blog/math-random
_state0 = 0n;
_state1 = 0n;
constructor(seed0, seed1) { // 128bit seed as 2 64bit values
this._state0 = BigInt(seed0) & MASK;
this._state1 = BigInt(seed1) & MASK;
if (this._state0 <= 0n)
throw new Error('seed 0 non-positive');
if (this._state1 <= 0n)
throw new Error('seed 1 non-positive');
}
next() {
let s1 = this._state0;
let s0 = this._state1;
this._state0 = s0;
s1 = ((s1 << 23n) ^ s1 ) & MASK;
s1 ^= (s1 >> 17n);
s1 ^= s0;
s1 ^= (s0 >> 26n);
this._state1 = s1;
return (this._state0 + this._state1) & MASK; // modulo WIDTH
}
}
Ok, then on top of XorShift128+ I've implemented B-D shuffle, with table of size 4. For your purpose you'll need table more than 84 entries, and power of two table is much easier to deal with, so let's say 128 entries table (7bit index) shall be good enough. Anyway, even with 4 entries table and 2bit index we need to know which bits to pick to form index. In original paper B-D discussed picking them from the back of rv as well as from front of rv etc. Here is where B-D shuffle needs another seed value - telling algorithm to pick say, bits from position 2 and 6.
class B_D_XSP {
_xsprng;
_seedBD = 0n;
_pos0 = 0n;
_pos1 = 0n;
_t; // B-D table, 4 entries
_Z = 0n;
constructor(seed0, seed1, seed2) { // note third seed for the B-D shuffle
this._xsprng = new XorShift128Plus(seed0, seed1);
this._seedBD = BigInt(seed2) & MASK;
if (this._seedBD <= 0n)
throw new Error('B-D seed non-positive');
this._pos0 = findPosition(this._seedBD); // first non-zero bit position
this._pos1 = findPosition(this._seedBD & (~(1n << this._pos0))); // second non-zero bit position
// filling up table and B-D shuffler
this._t = new Array(this._xsprng.next(), this._xsprng.next(), this._xsprng.next(), this._xsprng.next());
this._Z = this._xsprng.next();
}
index(rv) { // bit at first position plus 2*bit at second position
let idx = ((rv >> this._pos0) & 1n) + (((rv >> this._pos1) & 1n) << 1n);
return idx;
}
next() {
let retval = this._Z;
let j = this.index(this._Z);
this._Z = this._t[j];
this._t[j] = this._xsprng.next();
return retval;
}
}
Use example is as follow.
let rng = new B_D_XSP(1, 2, 4+64); // bits at second and sixth position to make index
console.log(rng._pos0.toString(10));
console.log(rng._pos1.toString(10));
console.log(rng.next());
console.log(rng.next());
console.log(rng.next());
Obviously, third seed value of say 8+128 would produce different permutation from what is shown in the example, you could play with it.
Last step would be to make 226bit random value by calling several (3 of 4) times B-D shuffled rng and combine 64bit values (and potential carry over) to make 226 random bits and then convert them to the deck shuffle.
I have a problem which requires a string to be transformed into another one by appending copies of its' initial value to itself. The problem allows to remove single characters at some places.
Explanation
let x = "abba"; // First string
let y = "aba" // Second initial string
y("aba") => remove last "a" => y("ab") => y+initialY = "ab"+"aba" =>
y("ababa") => remove char at index 2 => y("abba") => y == x => sucess
My algorithm successfully solves the problem:
let x = "abbbbcccaaac"
let y = "abc"
let xArr = x.split('')
let yArr = y.split('')
let count = 0;
for (let i = 0; i < xArr.length; i++) {
if(yArr[i] == undefined) {
yArr = yArr.concat(y.split(''));
count++;
}
if(xArr[i] != yArr[i]) {
yArr.splice(i, 1);
i--;
}
}
console.log("Output is:", yArr.join(''))
console.log("Appends in order to transform:", count)
The algorithm works as intended, however, I am uncertain regarding its time and space complexity and most importantly - efficiency.
Is this algorithm in O(n) time complexity where n is the length of x?
If this is not O(n), can the problem be solved in O(n) time?
Does .concat(), .splice() or .split() somehow change the time complexity since they are nested in a for loop? What if they weren't, do they still change the time complexity of an algorithm and by how much?
Given the rules of this problem, is this an efficient way to solve it?
What is the space complexity of this algorithm?
Normally a question like this is quite difficult to give a definite answer to, because different implementations of Javascript have different time complexities for basic array operations (such as creating a new array of size n). Javascript arrays will typically be implemented either as dynamic arrays or hashtables, and these data structures have different performance characteristics.
So, there is no definitive time complexity for splice to remove one element from an array. What we can say is that removing one element takes linear time for a dynamic array, and as #Ry- points out in the comments, also linear time for a hashtable, because of the need to renumber the later indices. We can also say that it's highly likely one of these two data structures is used, and no sensible implementation will take more than linear time to do splice.
Either way, the worst case for your algorithm is when x = 'aa...aa' and y = 'abb...bb', i.e. x is n copies of 'a', and y is 'a' followed by (m - 1) copies of 'b'.
For a dynamic array or a hashtable, then the time complexity for just the splice operations is O(nm²). This is because the outer loop iterates O(nm) times (note the i-- inside the loop, which happens every time the letter 'b' needs to be removed), and the splice operation requires shifting or renumbering O(m) elements in yArr after index i.
But suppose some more exotic data structure is used which supports removing an element in sub-linear time (e.g. a skip list). In that case, the above only gives O(nm) times the complexity of the "remove" operation. But we haven't counted concat yet; this creates a new data structure and copies every item into it, which will still take linear time. concat is called O(n) times and takes an average of O(n + m) time per call, so the complexity of just the concat operations is O(n² + nm).
So the time complexity is very likely O(n² + nm²), and certainly at least O(n² + nm); not linear.
The space complexity is O(n), since the length of yArr is never more than twice as long as xArr.
I am trying to figure out the runtime complexity of the function below. It finds the prime factors of given number.
function test(n){
n = parseInt(n);
var factors = [];
for(var i = 2; i <= n; i++){
while((n % i) === 0){
factors.push(i);
n /= i;
}
}
return factors;
}
while loop's execution depends on the value of i and I have no idea how can I calculate it or should I. This is not O(N2) right?
It is clear that your outer for loop is O(n). The tricky part here is your inner while loop. This loop is computing the number of prime factors for a given value of n. So if we had an explicit function, or at least upper bound, to computing prime factors we could just multiply by n to get the overall bound.
I found the following two helpful questions on our sister site, Mathematics Stack Exchange:
https://math.stackexchange.com/questions/938204/upper-bound-number-of-distinct-prime-factors
https://math.stackexchange.com/questions/1972003/upper-bound-for-count-of-unique-prime-divisors
They both reveal that an upper bound is:
log(n) / log( log(n) )
Because your loops are directly nested, we can just multiply the two complexities to arrive at:
O(n*log(n) / log(log(n)) )
I am looking for a solution to calculate the transmitted bytes per second of a repeatedly invoked function (below). Due to its inaccuracy, I do not want to simply divide the transmitted bytes by the elapsed overall time: it resulted in the inability to display rapid speed changes after running for a few minutes.
The preset (invoked approximately every 50ms):
function uploadProgress(loaded, total){
var bps = ?;
$('#elem').html(bps+' bytes per second');
};
How to obtain the average bytes per second for (only) the last n seconds and is it a good idea?
What other practices for calculating a non-flickering but precise bps value are available?
Your first idea is not bad, it's called a moving average, and providing you call your update function in regular intervals you only need to keep a queue (a FIFO buffer) of a constant length:
var WINDOW_SIZE = 10;
var queue = [];
function updateQueue(newValue) {
// fifo with a fixed length
queue.push(newValue);
if (queue.length > WINDOW_SIZE)
queue.shift();
}
function getAverageValue() {
// if the queue has less than 10 items, decide if you want to calculate
// the average anyway, or return an invalid value to indicate "insufficient data"
if (queue.length < WINDOW_SIZE) {
// you probably don't want to throw if the queue is empty,
// but at least consider returning an 'invalid' value in order to
// display something like "calculating..."
return null;
}
// calculate the average value
var sum = 0;
for (var i = 0; i < queue.length; i++) {
sum += queue[i];
}
return sum / queue.length;
}
// calculate the speed and call `updateQueue` every second or so
var updateTimer = setInterval(..., 1000);
An even simpler way to avoid sudden changes in calculated speed would be to use a low-pass filter. A simple discrete approximation of the PT1 filter would be:
Where u[k] is the input (or actual value) at sample k, y[k] is the output (or filtered value) at sample k, and T is the time constant (larger T means that y will follow u more slowly).
That would be translated to something like:
var speed = null;
var TIME_CONSTANT = 5;
function updateSpeed(newValue) {
if (speed === null) {
speed = newValue;
} else {
speed += (newValue - speed) / TIME_CONSTANT;
}
}
function getFilteredValue() {
return speed;
}
Both solutions will give similar results (for your purpose at least), and the latter one seems a bit simpler (and needs less memory).
Also, I wouldn't update the value that fast. Filtering will only turn "flickering" into "swinging" at a refresh rate of 50ms. I don't think anybody expects to have an upload speed shown at a refresh rate of more than once per second (or even a couple of seconds).
A simple low-pass filter is ok for just making sure that inaccuracies don't build up. But if you think a little deeper about measuring transfer rates, you get into maintaining separate integer counters to do it right.
If you want it to be an exact count, note that there is a simplification available. First, when dealing with rates, arithmetic mean of them is the wrong thing to apply to bytes/sec (sec/byte is more correct - which leads to harmonic mean). The other problem is that they should be weighted. Because of this, simply keeping int64 running totals of bytes versus observation time actually does the right thing - as stupid as it sounds. Normally, you are weighting by 1/n for each w. Look at a neat simplification that happens when you weigh by time:
(w0*b0/t0 + w1*b1/t1 + w2*b2/t2 + ...)/(w0+w1+w2+...)
totalBytes/totalWeight
(b0+b1+b2+...)/(w0+w1+w2+...)
So just keep separate (int64!) totals of bytes and milliseconds. And only divide them as a rendering step to visualize the rate. Note that if you instead used the harmonic mean (which you should do for rates - because you are really averaging sec/byte), then that's the same as the time it takes to send a byte, weighted by how many bytes there were.
1 / (( w0*t0/b0 + w1*t1/b0 + ... )/(w0+w1+w2+...)) =
totalBytes/totalTime
So arithmetic mean weighted by time is same as harmonic mean weighted by bytes. Just keep a running total of bytes in one var, and time in another. There is a deeper reason that this simplistic count actually the right one. Think of integrals. Assuming no concurrency, this is literally just total bytes transferred divided by total observation time. Assume that the computer actually takes 1 step per millisecond, and only sends whole bytes - and that you observe the entire time interval without gaps. There are no approximations.
Notice that if you think about an integral with (msec, byte/msec) as the units for (x,y), the area under the curve is the bytes sent during the observation period (exactly). You will get the same answer no matter how the observations got cut up. (ie: reported 2x as often).
So by simply reporting (size_byte, start_ms,stop_ms), you just accumulate (stop_ms-start_ms) into time and accumulate size_byte per observation. If you want to partition these rates to graph in minute buckets, then just maintain the (byte,ms) pair per minute (of observation).
Note that these are rates experienced for individual transfers. The individual transfers may experience 1MB/s (user point of view). These are the rates that you guarantee to end users.
You can leave it here for simple cases. But doing this counting right, allows for more interesting things.
From the server point of view, load matters. Presume that there were two users experiencing 1MB/s simultaneously. For that statistic, you need to subtract out the double-counted time. If 2 users do 1MB/s simultaneously for 1s, then that's 2MB/s for 1s. You need to effectively reconstruct time overlaps, and subtract out the double-counting of time periods. Explicitly logging at the end of a transfer (size_byte,start_ms,stop_ms) allows you to measure interesting things:
The number of outstanding transfers at any given time (queue length distribution - ie: "am I going to run out of memory?")
The throughput as a function of the number of transfers (throughput for a queue length - ie: "does the website collapse when our ad shows on TV?")
Utilization - ie: "are we overpaying our cloud provider?"
In this situation, all of the accumulated counters are exact integer arithmetic. Subtracting out the double-counted time suddenly gets you into more complicated algorithms (when computed efficiently and in real-time).
Use a decaying average, then you won't have to keep the old values around.
UPDATE: Basically it's a formula like this:
average = new_value * factor + average_old * (100 - factor);
You don't have to keep any old values around, they're all in the there at smaller and smaller proportions. You have to choose a value for factor that are appropriate to the mix of new and old values you want, and how often the average gets updated.
This is how the Unix "load average" is calculated I believe.