Recursive even function issue with understanding (Javascript) - javascript

The problem is very simple, I have a function from 'Javascript Allonge' book, and having a hard time in understanding it.
The function is called even, and it's as follows:
var even = function(num) {
return (num === 0) || !(even(num -1));
}
it checks whether the number is even or not, but I do not understand how. It calls itself recursively, and technically, always reaches zero, no? How does that work?

This is based on an inductive definition of numbers being odd or even - a number, n is 'even' when the number before it, n - 1 is odd. This thinking naturally makes sense - 4 is even if 3 is odd.
And so the function even is defined as:
1. even(0) is true - because 0 is even
2. even(n) is the negation of even(n - 1)
Another way to think of it is imagine even(4) being called step by step. By hand, replace the even(4) with result of evaluation it with your function:
even(4)
= !(even(3))
= !(!even(2))
= !(!(!even(1))
= !(!(!(!even(0)))
= !(!(!(!true))
= true
// ...even(4) == true

Well, divide and conquer.
First of all you have two expressions to calculate. The first one is just stopping the recursion at the point when the number is 0, this is easy.
The second expression
!(even(num -1))
is a bit more complicated. It always starts with a call even(num -1) and then negates it.
For the first element !(even(num -1) === true), so now you see that for every second element starting from 1 (1, 3, 5 etc.) it will return false, for others the inverted value.

But there is a negation on the recursive call to even, so that, for every activation record on the stack, the return value gets inverted:
0 --> true
1 --> true --> false
2 --> true --> false --> true

This function indeed calls itself recursively, but with an added inversion each time. return (num === 0) returns true when num equals zero. If num is greater than zero, it calculates the result of even(num-1) and inverses it.
So, with num = 1, the function returns the inverse of even(0), which is true, making the end result false. With num = 2, the function returns the inverse of even(1), which we've just shown to be false, making the end result true. The same principle applies for all integers.

Related

Recursions Javascript - Freecodecamp

beginner here. Was working through a code question on fcc and came across some javascript code I can't seem to understand.
function sum(arr, n) {
// Only change code below this line
if (n <= 0) {
return 0;
} else {
return sum(arr, n - 1) + arr[n - 1]
}
// Only change code above this line
}
console.log(sum([2, 3, 4, 5], 3))
// Console then spits out "9"
I understand how the second have of the return statement "arr[n - 1]" produces
4", but I'm unsure of how the first half "sum(arr, n - 1)" comes up with the number "5" that when added to 4, gives me the number "9" in the console.
I've narrowed it down to "sum(arr, 2)", but I can't seem to understand what the code is doing?
Thank you in advance!
sum(arr,2) calls the same function "sum" with the same arr parameter and a limit that's 2 instead of 3.
While the first call (inside console.log) is supposed to add a number of parameters (3), the second call (inside the first) should add all parameters but the last one (2), which is added explicitely.
That's the point of recursion: a function that calls itself until a specific condition is met. The condition in this case is if there is no element left to be added (n<=0). Up to this all numbers appear as "arr[n-1]" in the chain of calls and are added to 0 (which is the return value in the case of the ending condition). So the actual sum is calculated as all recursive calls return one after the other.

What's the difference in returning 0 or 1 on a javascript sorting function? [duplicate]

When sorting an array of numbers in JavaScript, I accidentally used < instead of the usual - -- but it still works. I wonder why?
Example:
var a = [1,3,2,4]
a.sort(function(n1, n2){
return n1<n2
})
// result is correct: [4,3,2,1]
And an example array for which this does not work (thanks for Nicolas's example):
[1,2,1,2,1,2,1,2,1,2,1,2]
This sort works on your input array due to its small size and the current implementation of sort in Chrome V8 (and, likely, other browsers).
The return value of the comparator function is defined in the documentation:
If compareFunction(a, b) is less than 0, sort a to an index lower than
b, i.e. a comes first.
If compareFunction(a, b) returns 0, leave a and
b unchanged with respect to each other, but sorted with respect to all
different elements.
If compareFunction(a, b) is greater than 0, sort b to an index lower than a, i.e. b comes first.
However, your function returns binary true or false, which evaluate to 1 or 0 respectively when compared to a number. This effectively lumps comparisons where n1 < n2 in with n1 === n2, claiming both to be even. If n1 is 9 and n2 is 3, 9 < 3 === false or 0. In other words, your sort leaves 9 and 3 "unchanged with respect to each other" rather than insisting "sort 9 to an index lower than 3".
If your array is shorter than 11 elements, Chrome V8's sort routine switches immediately to an insertion sort and performs no quicksort steps:
// Insertion sort is faster for short arrays.
if (to - from <= 10) {
InsertionSort(a, from, to);
return;
}
V8's insertion sort implementation only cares if the comparator function reports b as greater than a, taking the same else branch for both 0 and < 0 comparator returns:
var order = comparefn(tmp, element);
if (order > 0) {
a[j + 1] = tmp;
} else {
break;
}
Quicksort's implementation, however, relies on all three comparisons both in choosing a pivot and in partitioning:
var order = comparefn(element, pivot);
if (order < 0) {
// ...
} else if (order > 0) {
// ...
}
// move on to the next iteration of the partition loop
This guarantees an accurate sort on arrays such as [1,3,2,4], and dooms arrays with more than 10 elements to at least one almost certainly inaccurate step of quicksort.
Update 7/19/19: Since the version of V8 (6) discussed in this answer, implementation of V8's array sort moved to Torque/Timsort in 7.0 as discussed in this blog post and insertion sort is called on arrays of length 22 or less.
The article linked above describes the historical situation of V8 sorting as it existed at the time of the question:
Array.prototype.sort and TypedArray.prototype.sort relied on the same Quicksort implementation written in JavaScript. The sorting algorithm itself is rather straightforward: The basis is a Quicksort with an Insertion Sort fall-back for shorter arrays (length < 10). The Insertion Sort fall-back was also used when Quicksort recursion reached a sub-array length of 10. Insertion Sort is more efficient for smaller arrays. This is because Quicksort gets called recursively twice after partitioning. Each such recursive call had the overhead of creating (and discarding) a stack frame.
Regardless of any changes in the implementation details, if the sort comparator adheres to standard, the code will sort predictably, but if the comparator doesn't fulfill the contract, all bets are off.
After my initial comment, I wondered a little bit about how easy it is to find arrays for which this sorting method fails.
I ran an exhaustive search on arrays of length up to 8 (on an alphabet of size the size of the array), and found nothing. Since my (admittedly shitty) algorithm started to be too slow, I changed it to an alphabet of size 2 and found that binary arrays of length up to 10 are all sorted properly. However, for binary arrays of length 11, many are improperly sorted, for instance [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0].
// Check if 'array' is properly sorted using the "<" comparator function
function sortWorks(array) {
let sorted_array = array.sort(function(n1, n2) {
return n1 < n2;
});
for (let i=0; i<sorted_array.length-1; i++) {
if (sorted_array[i] < sorted_array[i+1]) return false;
}
return true;
}
// Get a list of all arrays of length 'count' on an alphabet of size 'max_n'.
// I know it's awful, don't yell at me :'(
var arraysGenerator;
arraysGenerator = function (max_n, count) {
if (count === 0) return [[]];
let arrays = arraysGenerator(max_n, count-1);
let result = [];
for (let array of arrays) {
for (let i = 0; i<max_n; i++) {
result.push([i].concat(array));
}
}
return result;
}
// Check if there are binary arrays of size k which are improperly sorted,
// and if so, log them
function testArraysOfSize(k) {
let arrays = arraysGenerator(2, k);
for (let array of arrays) {
if (!sortWorks(array)) {
console.log(array);
}
}
}
I'm getting some weird false-positives for some reason though, not sure where my mistake is.
EDIT:
After checking for a little while, here's a partial explanation on why OP's "wrong" sorting method works for lengths <=10 and for lengths >=11: it looks like (at least some) javascript implementations use InsertionSort if the array length is short (length <= 10) and QuickSort otherwise. It looks like QuickSort actively uses the "-1" outputs of the compare function while InsertionSort does not and relies only on the "1" outputs.
Source: here, all thanks to the original author.
If we analyze what's being done, it seems that this is mostly luck as in this case, 3 and 2 are considered to be "the same" and should be interchangeable. I suppose in such cases, the JS engines keep the original order for any values that have been deemed equal:
let a = [1, 3, 2, 4];
a.sort((n1, n2) => {
const result = n1 < n2;
if (result < 0) {
console.log(`${n1} comes before ${n2}`);
} else if (result > 0) {
console.log(`${n2} comes before ${n1}`);
} else {
console.log(`${n1} comes same position as ${n2}`);
}
return result;
})
console.log(a);
As pointed out in the comments, this isn't guaranteed to work ([1,2,1,2,1,2,1,2,1,2,1,2] being a counter-example).
Depending on exact sort() implementation, it is likely that it never checks for -1. It is easier and faster, and it makes no difference (as sorting is not guaranteed to be stable anyway, IIRC).
If the check sort() makes internally is compareFunction(a, b) > 0, then effectively true is interpreted as a > b, and false is interpreted as a <= b. And then your result is exactly what one would expect.
Of course the key point is that for > comparison true gets covered to 1 and false to 0.
Note: this is all speculation and guesswork, I haven't confirmed it experimentally or in browser source code - but it's reasonably likely to be correct.
Sort function expects a comparator which returns a number (negative, zero, or positive).
Assuming you're running on top of V8 engine (Node.js, Chrome etc.), you can find in array implementation that the returned value is compared to 0 (yourReturnValue > 0). In such case, the return value being casted to a number, so:
Truthy values are converted to 1
Falsy values are converted to 0
So based the documentation and the above, your function will return a sorted array in descending order in your specific case, but might brake in other cases since there's no regard to -1 value.

What's the difference between Math.random() >= 0.5 and Math.random() - 0.5

I want to generate an array of length n, and the elements of the array are the random integer between 2 and 32. I use the function follow but I find that 17 will always be the first element of the returned array. What's more, when I change to sort function to sort(() => Math.random() - 0.5), it works well.
So I am confused that what's the difference betweenMath.random() >= 0.5 and Math.random() - 0.5? And how the difference affects the sort() function?
const fn = (n) => {
let arr = [];
for (let i = 2; i < 33; i++) {
arr.push(i);
}
return arr.sort(() => Math.random() >= 0.5).slice(0, n)
}
You're not using sort for it's intended purpose, and the results are therefore unpredictable, weird and can vary between browser implementations. If you wish to shuffle an array, here is a far better function.
The function passed into Array.sort() should accept two arguments x and y and returns a negative value if x < y, zero if x = y, or a positive value if x > y.
In your first try, you use sort(() => Math.random() >= 0.5), which returns a boolean; this is going to be cast to a 0 or a 1. This then means that your function is telling the sorter that whatever first argument you pass in is always going to be equal to or greater than whatever second argument you pass in. It just happens that 17 is passed in as the second argument every time your function is called; you tell the browser that it is therefore less than or equal to every other element in the array, and thus it will get put at the beginning of the array.
Your second attempt, with sort(() => Math.random() - 0.5), returns with equal probability that the first number is greater than the second, or vice versa, which makes the shuffle work much better. However, because of the unreliability of the whole thing there's zero assurance that the shuffle will work in all browsers or be particularly random. Please use the "real" shuffle algorithm linked above.
Source: http://www.ecma-international.org/ecma-262/6.0/#sec-array.prototype.sort
For js sort, the param is the compare function, that need to be return 3 values: negative, zero, positive for less than, equal and greater than.
If you use >=, it only returns boolean.

Can't get it how the function works

So basically i just started learning JS and had a little exercise which was basically making a function to check if the number is even without using modular arithmetic. When i was done with it i just wanted to compare mine with the answer and i couldn't really get how does it work.
function isEven(n) {
if (n == 0)
return true;
else if (n == 1)
return false;
else if (n < 0)
return isEven(-n);
else
return isEven(n - 2);
}
I'm not sure how the part (n-2) works does it somehow puts the number in a loop and basically does n-=2 until the number gets 1 or 0?
Let's take a look at what is going on behind the scenes when this function is run:
isEven(8)
// isEven(8) Is 8 even?
// isEven(6) Is 6 even?
// isEven(4) Is 4 even?
// isEven(2) Is 2 even?
// isEven(0) Is 0 even? --> Yes, the function immediately returns true
// so I know the one at the top, 8, is even
And so on. For any even number, it eventually gets to 0. For any odd number it eventually gets to 1.
If the number is negative, the function makes it positive and runs itself again against the positive value, if it's > 1, it will run itself again until the number is 1 or 0, decreasing the number by 2 on each iteration. It's called recursion.
When n <0 it puts a minus before the n in the function and runs it again. If Else then the function is called again but n is decreased by 2. This runs till you get true or false.

Is there a better way of writing v = (v == 0 ? 1 : 0); [closed]

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I want to toggle a variable between 0 and 1. If it's 0 I want to set it to 1, else if it's 1 I want to set it to 0.
This is such a fundamental operation that I write so often I'd like to investigate the shortest, clearest possible way of doing it. Here's my best so far:
v = (v == 0 ? 1 : 0);
Can you improve on this?
Edit: the question is asking how to write the above statement in the fewest characters while retaining clarity - how is this 'not a real question'? This wasn't intended to be a code-golf exercise, though some interesting answers have come out of people approaching it as golf - it's nice to see golf being used in a constructive and thought-provoking manner.
You can simply use:
v = 1 - v;
This of course assumes that the variable is initialised properly, i.e. that it only has the value 0 or 1.
Another method that is shorter but uses a less common operator:
v ^= 1;
Edit:
To be clear; I never approached this question as code golf, just to find a short way of doing the task without using any obscuring tricks like side effects of operators.
Since 0 is a false value and 1 is a true value.
v = (v ? 0 : 1);
If you are happy to use true and false instead of numbers
v = !v;
or if they must be numbers:
v = +!v; /* Boolean invert v then cast back to a Number */
v = (v + 1) % 2 and if you need to cycle through more values just change 2 for (n + 1). Say you need to cycle 0,1,2 just do v = (v + 1) % 3.
You could write a function for it and use it like:
v = inv(v)
If you don't care about any possibility other than 1:
v = v ? 0 : 1;
In the above case, v will end up being 1 if v is 0, false, undefined or null. Take care using this kind of approach - v will be 0 even if v is "hello world".
Lines like v = 1 - v, or v ^= 1 or v= +!v will all get the job done, but they constitute what I would refer to as hacks. These are not beautiful lines of code, but cheap tricks to have the intended effect. 1 - v does not communicate "toggle the value between 0 and 1". This makes your code less expressive and introduces a place (albeit a small one) where another developer will have to parse your code.
Having instead a function like v = toggle(v) communicates the intent at the quickest glance.
(Honesty and mathematical integrity - given the number of votes on this "answer" - have led me to edit this answer. I held off as long as possible because it was intended as a short quip and not as anything "deep" so putting in any explanation seemed counter to the purpose. However, the comments are making it clear that I should be clear to avoid misunderstanding.)
My original answer:
The wording of this part of the specification:
If it's 0, I want to set it to 1, else set it to 0.
implies that the most accurate solution is:
v = dirac_delta(0,v)
First, the confession: I did get my delta functions confused. The Kronecker delta would have been slightly more appropriate, but not by much as I wanted something that was domain-independent (the Kronecker delta is mainly used just for integers). But I really shouldn't have used delta functions at all, I should have said:
v = characteristic_function({0},v)
Let me clarify. Recall that a function is a triple, (X,Y,f), where X and Y are sets (called the domain and codomain respectively) and f is a rule that assigns an element of Y to each element of X. We often write the triple (X,Y,f) as f: X &rightarrow; Y. Given a subset of X, say A, there is a characteristic function which is a function χA: X &rightarrow; {0,1} (it can also be thought of as a function to a larger codomain such as &Nopf; or &Ropf;). This function is defined by the rule:
χA(x) = 1 if x &in; A and χA(x) = 0 if x ∉ A.
If you like truth tables, it's the truth table for the question "Is the element x of X an element of the subset A?".
So from this definition, it's clear that the characteristic function is what is needed here, with X some big set containing 0 and A = {0}. That's what I should have written.
And so to delta functions. For this, we need to know about integration. Either you already know it, or you don't. If you don't, nothing I can say here will tell you about the intricacies of the theory, but I can give a one sentence summary. A measure on a set X is in essence "that which is needed to make averages work". That is to say that if we have a set X and a measure μ on that set then there is a class of functions X &rightarrow; &Ropf;, called measurable functions for which the expression ∫X f dμ makes sense and is, in some vague sense, the "average" of f over X.
Given a measure on a set, one can define a "measure" for subsets of that set. This is done by assigning to a subset the integral of its characteristic function (assuming that this is a measurable function). This can be infinite, or undefined (the two are subtly different).
There are lots of measures around, but there are two that are important here. One is the standard measure on the real line, &Ropf;. For this measure, then ∫&Ropf; f dμ is pretty much what you get taught in school (is calculus still taught in schools?): sum up little rectangles and take smaller and smaller widths. In this measure, the measure of an interval is its width. The measure of a point is 0.
Another important measure, which works on any set, is called the point measure. It is defined so that the integral of a function is the sum of its values:
∫X f dμ = ∑x &in;X f(x)
This measure assigns to each singleton set the measure 1. This means that a subset has finite measure if and only if it is itself finite. And very few functions have finite integral. If a function has a finite integral, it must be non-zero only on a countable number of points. So the vast majority of functions that you probably know do not have finite integral under this measure.
And now to delta functions. Let's take a very broad definition. We have a measurable space (X,μ) (so that's a set with a measure on it) and an element a &in; X. We "define" the delta function (depending on a) to be the "function" δa: X &rightarrow; &Ropf; with the property that δa(x) = 0 if x ≠ a and ∫X δa dμ = 1.
The most important fact about this to get a-hold of is this: The delta function need not be a function. It is not properly defined: I have not said what δa(a) is.
What you do at this point depends on who you are. The world here divides in to two categories. If you are a mathematician, you say the following:
Okay, so the delta function might not be defined. Let's look at its hypothetical properties and see if we can find a proper home for it where it is defined. We can do that, and we end up with distributions. These are not (necessarily) functions, but are things that behave a little like functions, and often we can work with them as if they were functions; but there are certain things that they don't have (such as "values") so we need to be careful.
If you are not a mathematician, you say the following:
Okay, so the delta function might not be properly defined. Who says so? A bunch of mathematicians? Ignore them! What do they know?
Having now offended my audience, I shall continue.
The dirac delta is usually taken to be the delta function of a point (often 0) in the real line with its standard measure. So those who are complaining in the comments about me not knowing my deltas are doing so because they are using this definition. To them, I apologise: although I can wriggle out of that by using the Mathematician's defence (as popularised by Humpty Dumpty: simply redefine everything so that it is correct), it is bad form to use a standard term to mean something different.
But there is a delta function which does do what I want it to do and it is that which I need here. If I take a point measure on a set X then there is a genuine function δa : X &rightarrow; &Ropf; which satisfies the criteria for a delta function. This is because we are looking for a function X &rightarrow; &Ropf; which is zero except at a and such that the sum of all of its values is 1. Such a function is simple: the only missing piece of information is its value at a, and to get the sum to be 1 we just assign it the value 1. This is none other than the characteristic function on {a}. Then:
∫X δa dμ = ∑x &in; X δa(x) = δa(a) = 1.
So in this case, for a singleton set, the characteristic function and the delta function agree.
In conclusion, there are three families of "functions" here:
The characteristic functions of singleton sets,
The delta functions,
The Kronecker delta functions.
The second of these is the most general as any of the others is an example of it when using the point measure. But the first and third have the advantage that they are always genuine functions. The third is actually a special case of the first, for a particular family of domains (integers, or some subset thereof).
So, finally, when I originally wrote the answer I wasn't thinking properly (I wouldn't go so far as to say that I was confused, as I hope I've just demonstrated I do know what I'm talking about when I actually think first, I just didn't think very much). The usual meaning of the dirac delta is not what is wanted here, but one of the points of my answer was that the input domain was not defined so the Kronecker delta would also not have been right. Thus the best mathematical answer (which I was aiming for) would have been the characteristic function.
I hope that that is all clear; and I also hope that I never have to write a mathematical piece again using HTML entities instead of TeX macros!
in general whenever you need to toggle between two values , you can just subtract the current value from the sum of the two toggle values :
0,1 -> v = 1 - v
1,2 -> v = 3 - v
4,5 -> v = 9 - v
You could do
v = Math.abs(--v);
The decrement sets the value to 0 or -1, and then the Math.abs converts -1 to +1.
If it must be the integer 1 or 0, then the way you're doing it is fine, though parentheses aren't needed. If these a are to be used as booleans, then you can just do:
v = !v;
v = v == 0 ? 1 : 0;
Is enough !
List of solutions
There are three solutions I would like to propose. All of them convert any value to 0 (if 1, true etc.) or 1 (if 0, false, null etc.):
v = 1*!v
v = +!v
v = ~~!v
and one additional, already mentioned, but clever and fast (although works only for 0s and 1s):
v = 1-v
Solution 1
You can use the following solution:
v = 1*!v
This will first convert the integer to the opposite boolean (0 to True and any other value to False), then will treat it as integer when multiplying by 1. As a result 0 will be converted to 1 and any other value to 0.
As a proof see this jsfiddle and provide any values you wish to test: jsfiddle.net/rH3g5/
The results are as follows:
-123 will convert to integer 0,
-10 will convert to integer 0,
-1 will convert to integer 0,
0 will convert to integer 1,
1 will convert to integer 0,
2 will convert to integer 0,
60 will convert to integer 0,
Solution 2
As mblase75 noted, jAndy had some other solution that works as mine:
v = +!v
It also first makes boolean from the original value, but uses + instead of 1* to convert it into integer. The result is exactly the same, but the notation is shorter.
Solution 3
The another approach is to use ~~ operator:
v = ~~!v
It is pretty uncommon and always converts to integer from boolean.
To sum up another answer, a comment and my own opinion, I suggest combining two things:
Use a function for the toggle
Inside this function use a more readable implementation
Here is the function which you could place in a library or maybe wrap it in a Plugin for another Javascript Framework.
function inv(i) {
if (i == 0) {
return 1
} else {
return 0;
}
}
And the usage is simply:
v = inv(v);
The advantages are:
No code Duplication
If you or anybody read this again in the future, you will understand your code in a minimum of time.
This is missing:
v = [1, 0][v];
It works as round robin as well:
v = [2, 0, 1][v]; // 0 2 1 0 ...
v = [1, 2, 0][v]; // 0 1 2 0 ...
v = [1, 2, 3, 4, 5, 0][v]; // 0 1 2 3 4 5 ...
v = [5, 0, 1, 2, 3, 4][v]; // 0 5 4 3 2 1 0 ...
Or
v = {0: 1, 1: 0}[v];
The charme of the last solution, it works with all other values as well.
v = {777: 'seven', 'seven': 777}[v];
For a very special case, like to get a (changing) value and undefined, this pattern may be helpful:
v = { undefined: someValue }[v]; // undefined someValue undefined someValue undefined ...
I don't know why you want to build your own booleans? I like the funky syntaxes, but why not write understandable code?
This is not the shortest/fastest, but the most clearest (and readable for everyone) is using the well-known if/else state:
if (v === 0)
{
v = 1;
}
else
{
v = 0;
}
If you want to be really clear, you should use booleans instead of numbers for this. They are fast enough for most cases. With booleans, you could just use this syntax, which will win in shortness:
v = !v;
Another form of your original solution:
v = Number(v == 0);
EDIT: Thanks TehShrike and Guffa for pointing out the error in my original solution.
I would make it more explicit.
What does v mean?
For example when v is some state. Create an object Status. In DDD an value object.
Implement the logic in this value object. Then you can write your code in a more functional way which is more readable. Switching status can be done by creating a new Status based on the current status. Your if statement / logic is then encapsulated in your object, which you can unittest. An valueObject is always immutable, so it has no identity. So for changing it's value you have to create a new one.
Example:
public class Status
{
private readonly int _actualValue;
public Status(int value)
{
_actualValue = value;
}
public Status(Status status)
{
_actualValue = status._actualValue == 0 ? 1 : 0;
}
//some equals method to compare two Status objects
}
var status = new Status(0);
Status = new Status(status);
Since this is JavaScript, we can use the unary + to convert to int:
v = +!v;
This will logical NOT the value of v (giving true if v == 0 or false if v == 1). Then we convert the returned boolean value into its corresponding integer representation.
Another way to do it:
v = ~(v|-v) >>> 31;
One more:
v=++v%2
(in C it would be simple ++v%=2)
ps. Yeah, I know it's double assignment, but this is just raw rewrite of C's method (which doesn't work as is, cause JS pre-increment operator dosen't return lvalue.
If you're guaranteed your input is either a 1 or a 0, then you could use:
v = 2+~v;
Just for kicks: v = Math.pow(v-1,v) also toggles between 1 and 0.
define an array{1,0}, set v to v[v], therefore v with a value of 0 becomes 1, and vica versa.
Another creative way of doing it, with v being equal to any value, will always return 0 or 1
v = !!v^1;
If possible values for v are only 0 and 1, then for any integer x, the expression:
v = Math.pow((Math.pow(x, v) - x), v);
will toggle the value.
I know this is an ugly solution and the OP was not looking for this...but I was thinking about just another solution when I was in the loo :P
Untested, but if you're after a boolean I think var v = !v will work.
Reference: http://www.jackfranklin.co.uk/blog/2011/05/a-better-way-to-reverse-variables
v=!v;
will work for v=0 and v=1; and toggle the state;
If there are just two values, as in this case(0, 1), i believe it's wasteful to use int. Rather go for boolean and work in bits. I know I'm assuming but in case of toggle between two states boolean seems to be ideal choice.
v = Number(!v)
It will type cast the Inverted Boolean value to Number, which is the desired output.
Well, As we know that in javascript only that Boolean comparison will also give you expected result.
i.e. v = v == 0 is enough for that.
Below is the code for that:
var v = 0;
alert("if v is 0 output: " + (v == 0));
setTimeout(function() {
v = 1;
alert("if v is 1 Output: " + (v == 0));
}, 1000);
JSFiddle: https://jsfiddle.net/vikash2402/83zf2zz0/
Hoping this will help you :)

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