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.
I'm optimizing the compiler of a language to JavaScript, and found a very interesting, if not frustrating, case:
function add(n,m) {
return n === 0 ? m : add(n - 1, m) + 1;
};
var s = 0;
for (var i = 0; i < 100000; ++i) {
s += add(4000, 4000);
}
console.log(s);
It takes 2.3s to complete on my machine[1]. But if I make a very small change:
function add(n,m) {
return (() => n === 0 ? m : add(n - 1, m) + 1)();
};
var s = 0;
for (var i = 0; i < 100000; ++i) {
s += add(4000, 4000);
}
console.log(s);
It completes in 1.1s. Notice the only difference is the addition of an immediately invoked lambda, (() => ...)(), around the return of add. Why does this added call make my program two times faster?
[1] MacBook Pro 13" 2020, 2.3 GHz Quad-Core Intel Core i7, Node.js v15.3.0
Interesting! From looking at the code, it seems fairly obvious that the IIFE-wrapped version should be slower, not faster: in every loop iteration, it creates a new function object and calls it (which the optimizing compiler will eventually avoid, but that doesn't kick in right away), so generally just does more work, which should be taking more time.
The explanation in this case is inlining.
A bit of background: inlining one function into another (instead of calling it) is one of the standard tricks that optimizing compilers perform in order to achieve better performance. It's a double-edged sword though: on the plus side, it avoids calling overhead, and can often enable further optimizations, such as constant propagation, or elimination of duplicate computation (see below for an example). On the negative side, it causes compilation to take longer (because the compiler does more work), and it causes more code to be generated and stored in memory (because inlining a function effectively duplicates it), and in a dynamic language like JavaScript where optimized code typically relies on guarded assumptions, it increases the risk of one of these assumptions turning out to be wrong and a large amount of optimized code having to be thrown away as a result.
Generally speaking, making perfect inlining decisions (not too much, not too little) requires predicting the future: knowing in advance how often and with which parameters the code will be executed. That is, of course, impossible, so optimizing compilers use various rules/"heuristics" to make guesses about what might be a reasonably good decision.
One rule that V8 currently has is: don't inline recursive calls.
That's why in the simpler version of your code, add will not get inlined into itself. The IIFE version essentially has two functions calling each other, which is called "mutual recursion" -- and as it turns out, this simple trick is enough to fool V8's optimizing compiler and make it sidestep its "don't inline recursive calls" rule. Instead, it happily inlines the unnamed lambda into add, and add into the unnamed lambda, and so on, until its inlining budget runs out after ~30 rounds. (Side note: "how much gets inlined" is one of the somewhat-complex heuristics and in particular takes function size into account, so whatever specific behavior we see here is indeed specific to this situation.)
In this particular scenario, where the involved functions are very small, inlining helps quite a bit because it avoids call overhead. So in this case, inlining gives better performance, even though it is a (disguised) case of recursive inlining, which in general often is bad for performance. And it does come at a cost: in the simple version, the optimizing compiler spends only 3 milliseconds compiling add, producing 562 bytes of optimized code for it. In the IIFE version, the compiler spends 30 milliseconds and produces 4318 bytes of optimized code for add. That's one reason why it's not as simple as concluding "V8 should always inline more": time and battery consumption for compiling matters, and memory consumption matters too, and what might be acceptable cost (and improve performance significantly) in a simple 10-line demo may well have unacceptable cost (and potentially even cost overall performance) in a 100,000-line app.
Now, having understood what's going on, we can get back to the "IIFEs have overhead" intuition, and craft an even faster version:
function add(n,m) {
return add_inner(n, m);
};
function add_inner(n, m) {
return n === 0 ? m : add(n - 1, m) + 1;
}
On my machine, I'm seeing:
simple version: 1650 ms
IIFE version: 720 ms
add_inner version: 460 ms
Of course, if you implement add(n, m) simply as return n + m, then it terminates in 2 ms -- algorithmic optimization beats anything an optimizing compiler could possibly accomplish :-)
Appendix: Example for benefits of optimization. Consider these two functions:
function Process(x) {
return (x ** 2) + InternalDetail(x, 0, 2);
}
function InternalDetail(x, offset, power) {
return (x + offset) ** power;
}
(Obviously, this is silly code; but let's assume it's a simplified version of something that makes sense in practice.)
When executed naively, the following steps happen:
evaluate temp1 = (x ** 2)
call InternalDetail with parameters x, 0, 2
evaluate temp2 = (x + 0)
evaluate temp3 = temp2 ** 2
return temp3 to the caller
evaluate temp4 = temp1 + temp3
return temp4.
If an optimizing compiler performs inlining, then as a first step it will get:
function Process_after_inlining(x) {
return (x ** 2) + ( (x + 0) ** 2 );
}
which allows two simplifications: x + 0 can be folded to just x, and then the x ** 2 computation occurs twice, so the second occurrence can be replaced by reusing the result from the first:
function Process_with_optimizations(x) {
let temp1 = x ** 2;
return temp1 + temp1;
}
So comparing with the naive execution, we're down to 3 steps from 7:
evaluate temp1 = (x ** 2)
evaluate temp2 = temp1 + temp1
return temp2
I'm not predicting that real-world performance will go from 7 time units to 3 time units; this is just meant to give an intuitive idea of why inlining can help reduce computational load by some amount.
Footnote: to illustrate how tricky all this stuff is, consider that replacing x + 0 with just x is not always possible in JavaScript, even when the compiler knows that x is always a number: if x happens to be -0, then adding 0 to it changes it to +0, which may well be observable program behavior ;-)
I've got two numbers that I want to compare. The numbers in the following example are the result of 26^26 computed in two different systems. One of which is my javascript code.
However, when comparing the two numbers I end up with something like this:
AssertionError [ERR_ASSERTION]: 4.0329146112660565e+26 == 4.0329146112661e+26
They're obviously not equal, but theoretically they should.
What's the proper way to perform equality on big numbers in javascript (even if it's an approximation)?
If what you're trying to do is determine if two numbers are practically equivalent you'll have to come up with your margin of error. One way to do this is to compute the difference between the numbers and then determine if that difference is significant or not.
So, taking your numbers from before, we could evaluate the difference between these numbers through subtraction. Since we don't really care about the sign of this difference, I'll go ahead and get the absolute value of the difference.
Math.abs(4.0329146112660565e+26 - 4.0329146112661e+26) === 4329327034368
(Sidenote: Now is not the time to explain why, but the == operator in JavaScript has confusing and error-prone behavior, use === when you want to compare values.)
That difference is a HUGE number, but related to how big our numbers are in the first place, it's rather insignificant. Intuitively, I'm tempted to divide the difference by the smallest of our original numbers like so:
4329327034368 / 4.0329146112660565e+26 === 1.0734983136696987e-14
That looks like a pretty small number. Repeat that same operation with a bunch of values and you should be able to determine what you want your margin of error to be. Then, all you'll have to do is perform the same operations with arbitrary numbers and see if that "difference ratio" is small enough for you.
function similar(a, b) {
let diff = Math.abs(a - b);
let smallest = Math.min(Math.abs(a), Math.abs(b));
let ratio = diff / smallest;
return ratio < MARGIN_OF_ERROR;
}
Now I just came up with that way of determining the importance of the difference between two numbers. It might not be a very smart way to compute it, it might be appropriate to some situations and not to others. But the general idea is that you'll have to make a function that determines if two values are close enough with your own definition of "close".
Be aware though, JavaScript is one of the worst languages you can be doing math in. Integers become imprecise when they go beyond Number.MAX_SAFE_INT (which seems to be 9007199254740991 according to Chrome, not sure if it varies between browsers or if that's a standardized constant).
Update: If your target engine is es2020 or above, you can use the new BigInt javascript primitive, for numbers higher than Number.MAX_SAFE_INTEGER
BigInt(4.0329146112660565e+26) === BigInt(4.0329146112661e+26)
//false
See more information in MDN
var a = 4.0329146112660565e+26;
var b = 4.0329146112661e+26;
a = Math.round(a/10e+20)*10e+20
b = Math.round(b/10e+20)*10e+20
a == b;
I would suggest to use one of big numbers library:
big.js (https://www.npmjs.com/package/big.js)
Example:
var x = new Big('4.0329146112660565e+26');
var y = new Big('4.0329146112661e+26');
// Should print false
console.log('Comparision result' + x.eq(y));
big-numbers (https://www.npmjs.com/package/big-numbers)
Example:
var x = bn.of('4.0329146112660565e+26');
var y = bn.of('4.0329146112661e+26');
// Should print false
console.log('Comparision result' + x.equals(y));
This output should be true.but it give false always.
Number.isInteger(parseFloat('134965.83') / parseFloat('0.01'))
Floating point arithmetic in Javascript is broken and in general as well.
It has nothing to do with division, it will return false if you don't do division since you are checking float value.
Number.isInteger(parseFloat('134965.83') / parseFloat('0.01')) translates to Number.isInteger(13496582.999999998) -> false
Check these examples.
Number.isInteger(parseFloat('134965.83') )// outputs false without division
As per the spec
If Type(argument) is not Number, return false.
If floor(abs(argument)) ≠ abs(argument), return false.
This happens because the outcome of the division is not an integer but something like 13496582.999999998.
Some floating numbers require a very small precision that is limited by the data type used. For example, the number 1/3 can never be expressed entirely as 0.333333333333 because there is a limitation to the data type size. Therefore there will always be a tiny rounding error involved in floating operations.
Edit: In response to the comment asking for a recommendation on how to deal eith this, actually there are several possibilities. It depends on the context and on accuracy required.
In short, to overcome this use a very small constant Number.EPSILON (see also this) and use it in comparisons. Disclaimer: this is just a quick example, read extensively the implications on the links provided.
var myIsInteger = function(n) {
return (n % 1) > Number.EPSILON
};
We effectively check that the residual of the division with 1 is within the constant.
parseFloat('134965.83') / parseFloat('0.01') = 13496582.999999998
And when Number.isInteger(13496582.999999998) will always return false