In the example below, I don't want to make a counter as a param. Rather, I just want to return '+ 1' each time so that what gets returned is the number of steps it takes. My issue lies with the base case. If I do return + 1, I get the correct number of steps plus one additional step so I tried just return but that delivers NaN. Is it even possible?
var numberOfSteps = function(num) {
if (num == 0) {
return;
} else {
if (num % 2 == 0) {
return 1 + numberOfSteps(num/2);
} else {
return 1 + numberOfSteps(num - 1);
}
}
};
edit : The goal is to track how many steps it takes to reduce a number to 0. If it's even, divide by 2 or else subtract by 1. Ultimately, I want to return the number of steps it takes for any given number to get reduced to 0 following those rules
I hope the point has gotten through in the long comment thread and other answers that return + 1 is equivalent to return (+1), that is, return the integer positive one. And since there are no steps to take once you've reached zero, +1 is the wrong answer. Similarly, a plain return is functionally equivalent to return undefined. But undefined is not a number, and you're going to run into problems if you later try to add 1 to it. So the solution from the comments or other answers to return the correct number of steps, which in this case 0, will fix your code.
I would like to point out another way to solve this, though:
const numberOfSteps = (n) =>
n <= 0
? 0
: 1 + numberOfSteps (n % 2 == 0 ? n / 2 : n - 1)
console .log (numberOfSteps (12))
There are superficial differences here from the other solutions, such as using an arrow function, using a conditional statement (ternary) rather than if-statements, and using <= 0 instead of < 0 to avoid possible infinite loops on negative numbers.
But the fundamental difference is that this code only has one recursive branch. I think this is a better match to the problem.
We can think of this as a function which answers "How many steps does it take to reach 0 from our input number if each step cuts even numbers in half and subtracts one from odd ones?" Well that logically leads to a base case (we're already at 0) so have to return 0, and a recursive one (we're at some positive integer) so have to add 1 to the total steps required from our next entry.
By doing this single recursive call and adding one to the result, we make it clearer what the recursion is doing.
If this is unclear, then this alternative might show what I mean:
const takeStep = (n) =>
n % 2 == 0 ? n / 2 : n - 1
const numberOfSteps = (n) =>
n <= 0
? 0
: 1 + numberOfSteps (takeStep (n))
Think you just need to return 0 when it's...zero.
var numberOfSteps = function(num) {
if (num == 0) {
return 0;
} else {
if (num % 2 == 0) {
return 1 + numberOfSteps(num/2);
} else {
return 1 + numberOfSteps(num - 1);
}
}
}
return + 1 maybe doesn't do what you think it does: it returns the number 1. + here means positive not negative, there is no addition or subtraction going on. It will also give you one too many steps.
return; by itself returns undefined, which when converted to a Number, translates to NaN, because, well, it's not a number.
Related
I use common functions like map() and clamp() to transform values.
I wonder if there is a similar mathematical function which lets me translate a rising value to an on and off state (0 and 1).
Could a square wave help?
For more context: I want the value to be 1 whilst something is animating and 0 after it finished. The input value is clamped, the functionality of the function I'm looking for is like a switch. It would be a substitute for a complex if statement. Executed in a loop I'm trying to save resources by not using performance heavy if statements.
Looking forward to your input!
This is what I came up with in the end. Not sure if this is helpful to anyone else. It converts any value between 0 and 1 to 1.
But if the input value is equal to 0 or the input value is equal to 1, the function will return 0 as well.
Resulting in the loopback / switch behavior.
It returns 1 only whilst a value is progressing.
Hope this makes sense and is correct. Happy to hear your thoughts.
Also happy to hear some further thoughts on if such a function is more performant than an if-statement.
In the end I had to solve this to prove to myself that it's possible to do with math. (of course it is):)
const loopback = (i, m = 100) => {
//returns 1 while value is progressing, returns 0 if value is 0 or 1
//based on triangle wave function: https://discourse.processing.org/t/sawtooth-wave-algorithm/27433
i = clamp(i) //normalize value to 0-1
return Math.ceil(Math.abs((i*(m)) % (m)) / m)
}
const clamp = (number, min = 0, max = 1) => {
return Math.max(min, Math.min(number, max));
}
//in my usecase, this is an alternative to following condition:
/*if(i >= 1 || i <= 0) {
return 0;
} else {
return 1;
}
I am trying to count TRAILING zeros from a recursive manner. Basically I split the final recursive result and then created a var counter that will count all the zeros.
function countingZeros(n) {
if (n < 0) {
// Termination condition to prevent infinite recursion
return;
}
// Base case
if (n === 0) {
return 1;
}
// Recursive case
let final = n * countingZeros(n -1);
let counter = 0;
String(final).split('').forEach(function(item){
item === 0 ? counter++ : counter;
});
return counter;
}
countingZeros(12) // => suppose to output 2 since there are 2 trailing zeros from 479001600 but got 0
countingZeros(6) // => suppose to get 1 since 720 is the final result.
I am expecting to get 2 in return as the counter must return but instead I got 0. Any idea what am I missing on my function? How should I fix it?
I think you're working too hard. First of all, in response to a comment, you don't actually need to calculate the factorial, since all you really need is to count factors of 5 and of 2. And since there are many more factors of 2, your real answer is just counting factors of 5. But each factor of 5 must be a factor of one of {1, 2, 3, ... n}, so we just have to add up the highest powers of five that evenly divide into each of {1, 2, 3, ... n}.
We can do that with some simple recursion:
const fiveFactors = (n, acc = 0) => (n % 5 == 0)
? fiveFactors(n / 5, acc + 1)
: acc
const factZeros = (n, acc = 0) => (n > 0)
? factZeros(n - 1, acc + fiveFactors(n))
: acc
factZeros(1000) //=> 249
Note that both functions are eligible for tail-call optimization.
Also, although this does involve a double recursion, it's not really ill-performant. Four out of five times, the internal recursion stops on the first call, and of the remainder, four out of five stop on the second call, and so on.
You are trying to count the number of zeroes using string functions(i will assume, you forgot to include the factorial method. Correct flow could have been- you first pass the input to a factorial method and pass output from factorial method to countingZeros method). Anyways as stated already in other answer, you don't really need to calculate the factorial product to count the trailing zeroes.
Here a sample to count the number of trailing zeroes in n!
temp = 5;
zeroes = 0;
//counting the sum of multiples of 5,5^2,5^3....present in n!
while(n>=temp){
fives = n/temp;
zeroes = zeroes + fives;
temp = temp*5;
}
printf("%d",zeroes);
Note that each multiple of 5 in the factorial product will contribute 1 to the number of trailing zeros. On top of this, each multiple of 25 will contribute an additional 1 to the number of trailing zeros. Then, each multiple of 125 will contribute another 1 to the number of trailing zeros, and so on.
Here's a great link to understand the concept behind this:
https://brilliant.org/wiki/trailing-number-of-zeros/
I'm reading a JavaScript book, I'm new at this, so I got to the part of recursion and I get how recursion works but is not that part is hard to me is the math part.
This is the code:
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);
}
lets say I pass to the function 50 as a value right
isEven(50);
That give me true... how come 50 == 0 is true or 75 == 1 is false... I really don't get it.
When you pass 50 to this function, it goes to the last block (else) and is executed recursively. When you write return isEven(n - 2);, you execute isEven(48), and then the next time isEven(46) and so on right upto the point where you reach isEven(0).
This calls the first if block and you get true as your output. The true is returned by isEven(0), followed by isEven(2) (becuase when you executed isEven(2) you ended up going to return isEven(0)) and this bubbles up the stack to finally return the output of isEven(50) as true.
This code is kind of a long way of checking whether a number is even. It relies on the fact that if a number is even and you continually subtract 2 from it, you'll eventually reach 0. Otherwise, if your number is odd, subtracting 2 repeatedly will eventually reach 1.
So if the number you pass is greater than 0, it will send the next number (n - 2) back to the function again, until it either reaches 0 or 1. Then we stop.
If the number is negative, we just flip the sign and do the same process.
This is an interesting way of applying recursion to an issue that could be solved as simply as:
return n % 2 == 0
Basically, it works by having two base cases: n == 0 and n == 1. Both of these are "known", so they just return true or false. If the number is negative, do a recursive call with the sign reversed (isEven(-n)). Otherwise, you just subtract 2 (isEven(n - 2)). This works because any odd number minus 2 is still an odd number, and even number minus 2 is still even. So, you simply keep subtracting 2 from n until n fits one of the base cases (n == 0 or n == 1).
if (n == 0) // base case
return true;
else if (n == 1) // base case
return false;
else if (n < 0) // reverse sign, call recursively
return isEven(-n);
else // default, subtracts 2, calls recursively
return isEven(n - 2);
Use this tool to visualize your code and you'll see where you're getting hung up.
Click the Forward > button after the page loads to step through your code.
You can simplify the function by using % operator
let isEven = n => !(n % 2);
When I try to do 8067 % 80.67 I get 80.66999999999983, instead of 0 beacuse of known floating point javascript behaviour.
So I went and made a function for this, to avoid floating point javascript errors.
function math(a, b) {
var left = Math.abs(a),
times = 1,
abs = a >= 0 ? 1 : -1;
while (Math.abs(a) >= b * times) {
left -= b;
times++;
}
return (a - (b * (times - 1))) * abs;
}
http://jsfiddle.net/s5w3C/
So my question is: is this usefull, ie a good tool to use instead of %? is there cases where this will also give falsy results like the modulus % oprator.
I am looking for a tools to calculate % consistently.
I didn't really inspect the algorithm for correctness, but if you care about efficiency, this is a bad idea. Basically, the larger the input, the slower your code will execute.
I think any fix will only work to a certain level of accuracy and for certain sized numbers. Perhaps something like the following will be sufficient:
function nearlyMod(a, b) {
var precision = ('' + b).split('.').length;
var estimate = (a % b).toFixed(precision);
return estimate == b ? 0 : +estimate;
}
console.log(nearlyMod(8067, 80.66)); // 1
console.log(nearlyMod(8067, 80.67)); // 0
console.log(nearlyMod(8067, 80.68)); // 79.68
It tests if the result is an even divisor within the precision of the original number. If so, it returns 0, otherwise it returns a number to the same precision (which may or may not be what you want).
The result is always a number (the value returned from toFixed is a string, hence +estimate).
A better name might be "roundedMod" or similar.
What is the best way to prevent divide by 0 in javascript that is accepting user inputs.
If there is no particular way to achieve this what would be the best way to handle such a situation so as to not prevent other scripts from executing?
Any insights are much appreciated.
There is no way to do that with the normal / and /= operators.
The best way to do what you want is with guards:
function notZero(n) {
n = +n; // Coerce to number.
if (!n) { // Matches +0, -0, NaN
throw new Error('Invalid dividend ' + n);
}
return n;
}
and then do division like
numerator / notZero(denominator)
Alternatively you can always guard the output
function dividend(numerator, denominator) {
var quotient = numerator / denominator;
if (quotient !== quotient) { throw new Error(numerator + " / " + denominator); }
return quotient;
}
but that loses the readability and expressiveness of /=.
Off the top of my head you could:
Check the user input to see if the denominator is zero (or evaluates to zero, depending on what your script actually does).
Check if the result of the action isFinite() and if not then handle appropriately.
what would be the best way to handle such a situation so as to not prevent other scripts from executing
Division by zero doesn't seem to prevent other scripts from execution in JavaScript:
var a = 20;
var b = 0;
var result = a/b;
console.log(result); // returns Infinity
If you want something different to happen in case of division by zero, you could use
function divideIfNotZero(numerator, denominator) {
if (denominator === 0 || isNaN(denominator)) {
return null;
}
else {
return numerator / denominator;
}
}
Hope this is useful
(denominator != 0 ? numerator/denominator : Infinity)
or whatever value you want to put at the end.
Greetings.
To prevent (unwanted) execution
Always verify critical user input and/or results
Use logic and/or callbacks you can prevent to execute
On HTML forms etc. you can use i.e. return false; as value to stop submission.
Why not just check if the denominator is zero?
if(x != 0) z = y / x;
You can also check if the result is Infinity:
3 / 0 == Infinity
Results in true;
(Only tested in chrome.)
A bit different than stopping execution, but the ternary operator is a pretty slick way to customize variable assignment.
var one = 1,
zero = 0,
customValue = 1;
var quotient = zero===0 ? customValue : one / zero;
This way, by setting the customVariable to the integer of your choice, you can expect a predictable result when division by zero occurs.
The best way is contextual. But here's the easiest:
function myFunction( input ){
input = 0 ? 0.0001 : input; // same as if( input == 0 ){ input = 0.0001; }
return 1 / input;
}
Basically if the input is zero, turn it into a very small number before using as a denominator. Works great for integers, since after your division you can round them back down.
A couple caveats prevent this from being universal:
It could cause false positives if your input accepts really small numbers
It won't trigger any error-handling code, if you need to do something special if zero is entered
So it's best for general-purpose, non-critical cases. For example, if you need to return the result of a complex calculation and don't care if the answer is accurate to N digits (determined by 0.0001 vs. 0.00000001, etc.); you just don't want it to break on a divide-by-zero.
As another answer suggested, you could also create a reusable global function.
function divisor( n ){ return ( n = 0 ? 0.0001 : n ); }
function myFunction( input ){ return 1 / divisor( input ); }
Possible improvements:
function divisor( n, orError ){
if( typeof n == 'undefined' || isNaN( n ) || !n ){
if( orError ){ throw new Error( 'Divide by zero.' ); }
return 0.000000000000001;
}else{ return 0 + n; }
}
This would take any value (null, number, string, object) and if invalid or zero, return the failsafe zero-like value. It would also coerce the output to a number just in case it was a string and you were doing something odd. All this would ensure that your divisor function always worked. Finally, for cases where you wanted to handle such errors yourself, you could set the second parameter to true and use a try/catch.
Set a cap on what the value for the numerator can be and set the numerator to that value when the denominator equals 0.
This is a faster approach yet is confusing
let divisor;
let dividend;
let result =(dividend/divisor) || 0
if the result for instance if you are calculating percentage is infinite you can give it 0 as value;
const progress = goal == 0 ? 0 : total/goal