Develop Reference

javascript external called function does not return value [duplicate]

Im solving a codewars problem and im pretty sure i've got it working:
function digital_root(n) {
// ...
n = n.toString();
if (n.length === 1) {
return parseInt(n);
} else {
let count = 0;
for (let i = 0; i < n.length; i++) {
//console.log(parseInt(n[i]))
count += parseInt(n[i]);
}
//console.log(count);
digital_root(count);
}
}
console.log(digital_root(942));
Essentially it's supposed to find a "digital root":
A digital root is the recursive sum of all the digits in a number.
Given n, take the sum of the digits of n. If that value has two
digits, continue reducing in this way until a single-digit number is
produced. This is only applicable to the natural numbers.
So im actually getting the correct answer at the end but for whatever reason on the if statement (which im watching the debugger run and it does enter that statement it will say the return value is the correct value.
But then it jumps out of the if statement and tries to return from the main digital_root function?
Why is this? shouldn't it break out of this when it hits the if statement? Im confused why it attempt to jump out of the if statement and then try to return nothing from digital_root so the return value ends up being undefined?

You're not returning anything inside else. It should be:
return digital_root(count);
^^^^^^^
Why?
digital_root is supposed to return something. If we call it with a one digit number, then the if section is executed, and since we return from that if, everything works fine. But if we provide a number composed of more than one digit then the else section get executed. Now, in the else section we calculate the digital_root of the count but we don't use that value (the value that should be returned). The line above could be split into two lines of code that makes it easy to understand:
var result = digital_root(count); // get the digital root of count (may or may not call digital_root while calculating it, it's not owr concern)
return result; // return the result of that so it can be used from the caller of digital_root

Code review
My remarks is code comments below
// javascript generally uses camelCase for function names
// so this should be digitalRoot, not digital_root
function digital_root(n) {
// variable reassignment is generally frowned upon
// it's somewhat silly to convert a number to a string if you're just going to parse it again
n = n.toString();
if (n.length === 1) {
// you should always specify a radix when using parseInt
return parseInt(n);
} else {
let count = 0;
for (let i = 0; i < n.length; i++) {
//console.log(parseInt(n[i]))
count += parseInt(n[i]);
}
// why are you looping above but then using recursion here?
// missing return keyword below
digital_root(count);
}
}
console.log(digital_root(942));
Simple recursive solution
With some of those things in mind, let's simplify our approach to digitalRoot...
const digitalRoot = n =>
n < 10 ? n : digitalRoot(n % 10 + digitalRoot((n - n % 10) / 10))
console.log(digitalRoot(123)) // => 6
console.log(digitalRoot(1234)) // 10 => 1
console.log(digitalRoot(12345)) // 15 => 6
console.log(digitalRoot(123456)) // 21 => 3
console.log(digitalRoot(99999999999)) // 99 => 18 => 9
Using reduce
A digital root is the recursive sum of all the digits in a number. Given n, take the sum of the digits of n. If that value has two digits, continue reducing in this way until a single-digit number is produced. This is only applicable to the natural numbers.
If you are meant to use an actual reducing function, I'll show you how to do that here. First, we'll make a toDigits function which takes an integer, and returns an Array of its digits. Then, we'll implement digitalRoot by reducing those those digits using an add reducer initialized with the empty sum, 0
// toDigits :: Int -> [Int]
const toDigits = n =>
n === 0 ? [] : [...toDigits((n - n % 10) / 10), n % 10]
// add :: (Number, Number) -> Number
const add = (x,y) => x + y
// digitalRoot :: Int -> Int
const digitalRoot = n =>
n < 10 ? n : digitalRoot(toDigits(n).reduce(add, 0))
console.log(digitalRoot(123)) // => 6
console.log(digitalRoot(1234)) // 10 => 1
console.log(digitalRoot(12345)) // 15 => 6
console.log(digitalRoot(123456)) // 21 => 3
console.log(digitalRoot(99999999999)) // 99 => 18 => 9

its a recursive function the code should be somewhat like this
function digital_root(n) {
// ...
n=n.toString();
if(n.length === 1){
return parseInt(n);
}
else
{
let count = 0;
for(let i = 0; i<n.length;i++)
{
//console.log(parseInt(n[i]))
count+=parseInt(n[i]);
}
//console.log(count);
return digital_root(count);
}
}
you should return the same function instead of just calling it to get the correct call stack

Generate random & unique 4 digit codes without brute force

I'm building an app and in one of my functions I need to generate random & unique 4 digit codes. Obviously there is a finite range from 0000 to 9999 but each day the entire list will be wiped and each day I will not need more than the available amount of codes which means it's possible to have unique codes for each day. Realistically I will probably only need a few hundred codes a day.
The way I've coded it for now is the simple brute force way which would be to generate a random 4 digit number, check if the number exists in an array and if it does, generate another number while if it doesn't, return the generated number.
Since it's 4 digits, the runtime isn't anything too crazy and I'm mostly generating a few hundred codes a day so there won't be some scenario where I've generated 9999 codes and I keep randomly generating numbers to find the last remaining one.
It would also be fine to have letters in there as well instead of just numbers if it would make the problem easier.
Other than my brute force method, what would be a more efficient way of doing this?
Thank you!

Since you have a constrained number of values that will easily fit in memory, the simplest way I know of is to create a list of the possible values and select one randomly, then remove it from the list so it can't be selected again. This will never have a collision with a previously used number:
function initValues(numValues) {
const values = new Array(numValues);
// fill the array with each value
for (let i = 0; i < values.length; i++) {
values[i] = i;
}
return values;
}
function getValue(array) {
if (!array.length) {
throw new Error("array is empty, no more random values");
}
const i = Math.floor(Math.random() * array.length);
const returnVal = array[i];
array.splice(i, 1);
return returnVal;
}
// sample code to use it
const rands = initValues(10000);
console.log(getValue(rands));
console.log(getValue(rands));
console.log(getValue(rands));
console.log(getValue(rands));
This works by doing the following:
Generate an array of all possible values.
When you need a value, select one from the array with a random index.
After selecting the value, remove it from the array.
Return the selected value.
Items are never repeated because they are removed from the array when used.
There are no collisions with used values because you're always just selecting a random value from the remaining unused values.
This relies on the fact that an array of integers is pretty well optimized in Javascript so doing a .splice() on a 10,000 element array is still pretty fast (as it can probably just be memmove instructions).
FYI, this could be made more memory efficient by using a typed array since your numbers can be represented in 16-bit values (instead of the default 64 bits for doubles). But, you'd have to implement your own version of .splice() and keep track of the length yourself since typed arrays don't have these capabilities built in.
For even larger problems like this where memory usage becomes a problem, I've used a BitArray to keep track of previous usage of values.
Here's a class implementation of the same functionality:
class Randoms {
constructor(numValues) {
this.values = new Array(numValues);
for (let i = 0; i < this.values.length; i++) {
this.values[i] = i;
}
}
getRandomValue() {
if (!this.values.length) {
throw new Error("no more random values");
}
const i = Math.floor(Math.random() * this.values.length);
const returnVal = this.values[i];
this.values.splice(i, 1);
return returnVal;
}
}
const rands = new Randoms(10000);
console.log(rands.getRandomValue());
console.log(rands.getRandomValue());
console.log(rands.getRandomValue());
console.log(rands.getRandomValue());

Knuth's multiplicative method looks to work pretty well: it'll map numbers 0 to 9999 to a random-looking other number 0 to 9999, with no overlap:
const hash = i => i*2654435761 % (10000);
const s = new Set();
for (let i = 0; i < 10000; i++) {
const n = hash(i);
if (s.has(n)) { console.log(i, n); break; }
s.add(n);
}
To implement it, simply keep track of an index that gets incremented each time a new one is generated:
const hash = i => i*2654435761 % (10000);
let i = 1;
console.log(
hash(i++),
hash(i++),
hash(i++),
hash(i++),
hash(i++),
);
These results aren't actually random, but they probably do the job well enough for most purposes.

Disclaimer:
This is copy-paste from my answer to another question here. The code was in turn ported from yet another question here.
Utilities:
function isPrime(n) {
if (n <= 1) return false;
if (n <= 3) return true;
if (n % 2 == 0 || n % 3 == 0) return false;
for (let i = 5; i * i <= n; i = i + 6) {
if (n % i == 0 || n % (i + 2) == 0) return false;
}
return true;
}
function findNextPrime(n) {
if (n <= 1) return 2;
let prime = n;
while (true) {
prime++;
if (isPrime(prime)) return prime;
}
}
function getIndexGeneratorParams(spaceSize) {
const N = spaceSize;
const Q = findNextPrime(Math.floor(2 * N / (1 + Math.sqrt(5))))
const firstIndex = Math.floor(Math.random() * spaceSize);
return [firstIndex, N, Q]
}
function getNextIndex(prevIndex, N, Q) {
return (prevIndex + Q) % N
}
Usage
// Each day you bootstrap to get a tuple of these parameters and persist them throughout the day.
const [firstIndex, N, Q] = getIndexGeneratorParams(10000)
// need to keep track of previous index generated.
// it’s a seed to generate next one.
let prevIndex = firstIndex
// calling this function gives you the unique code
function getHashCode() {
prevIndex = getNextIndex(prevIndex, N, Q)
return prevIndex.toString().padStart(4, "0")
}
console.log(getHashCode());
Explanation
For simplicity let’s say you want generate non-repeat numbers from 0 to 35 in random order. We get pseudo-randomness by polling a "full cycle iterator"†. The idea is simple:
have the indexes 0..35 layout in a circle, denote upperbound as N=36
decide a step size, denoted as Q (Q=23 in this case) given by this formula‡
Q = findNextPrime(Math.floor(2 * N / (1 + Math.sqrt(5))))
randomly decide a starting point, e.g. number 5
start generating seemingly random nextIndex from prevIndex, by
nextIndex = (prevIndex + Q) % N
So if we put 5 in we get (5 + 23) % 36 == 28. Put 28 in we get (28 + 23) % 36 == 15.
This process will go through every number in circle (jump back and forth among points on the circle), it will pick each number only once, without repeating. When we get back to our starting point 5, we know we've reach the end.
†: I'm not sure about this term, just quoting from this answer
‡: This formula only gives a nice step size that will make things look more "random", the only requirement for Q is it must be coprime to N

This problem is so small I think a simple solution is best. Build an ordered array of the 10k possible values & permute it at the start of each day. Give the k'th value to the k'th request that day.
It avoids the possible problem with your solution of having multiple collisions.

Need help writing code to convert decimal to binary without the use of the toString

I'm trying to create my own decimal to binary converter with the method of decrementing the inputted variable (decimal value), by dividing it by 2 and storing the remainder (like 2nd grade math remainder), which is always either 0 or 1. Each of the remainder values i thin should be stored in an array and I think maybe put in backwards so that the most significant digit is first in the array (this is because when decrementing the remainer values are filled in left to right). Soooo yea i dont really know how to store the remainder values in an array using a function
Thanks in advance and if something is confusing then feel free to ask because im not even sure if this is the best method of doing this its just what i came up with
function decimalToBinary(num) {
var bin = 0;
while (num > 0) {
bin = num % 2 + bin;
num >>= 1; // basically /= 2 without remainder if any
}
alert("That decimal in binary is " + bin);
}

Your code is almost correct. The main problem is that bin starts out as 0; when you add a digit, they are added numerically, so your code ends up just counting the binary 1s: in this manner, 10 is initial 0, and +1+0+1+0, resulting in 2. You want to handle it as a string: ""+1+0+1+0 results in 1010. So, the only needed change is:
var bin = "";
If you want to solve it using arrays, with minimal changes to your code, it would be:
function decimalToBinary(num) {
var bin = [];
while (num > 0) {
bin.unshift(num % 2);
num >>= 1; // basically /= 2 without remainder if any
}
alert("That decimal in binary is " + bin.join(''));
}
Here, I use .unshift to add an element to the head of the array (and renumbering the remaining elements); .join() to collect them all into a string.
Or this:
function decimalToBinary(num) {
var bin = [];
while (num > 0) {
bin[bin.length] = num % 2;
num >>= 1; // basically /= 2 without remainder if any
}
alert("That decimal in binary is " + bin.reverse().join(''));
}
This is not as good, but illustrates some more things you can do with arrays: taking their length, setting an arbitrary element, and flipping them around.

I have written a custom Decimal to Binary method:
function toBinary (input) {
let options = [1];
let max = 0;
let i = 1;
while(i) {
max = Math.pow(2, i);
if (max > input) break;
options.push(max);
i++;
}
let j = options.length;
let result = new Array(j);
result.fill("0");
while(j >= 0) {
if (options[j] <= input) {
result[j] = "1"
input = input - options[j];
}
j--;
}
return [...result].reverse().join("");
}
//Test the toBin method with built-in toString(2)
toBinary(100) === (100).toString(2) // true
toBinary(1) === (1).toString(2) // true
toBinary(128) === (128).toString(2) // true

Fibonacci Sequence - Find the number of digits - JavaScript

So, I have successfully written the Fibonacci sequence to create an array with the sequence of numbers, but I need to know the length (how many digits) the 500th number has.
I've tried the below code, but its finding the length of the scientific notation (22 digits), not the proper 105 it should be returning.
Any ideas how to convert a scientific notation number into an actual integer?
var fiblength = function fiblength(nth) {
var temparr = [0,1];
for(var i = 2; i<=nth; i++){
var prev = temparr[temparr.length-2],
cur = temparr[temparr.length-1],
next = prev + cur;
temparr.push(next);
}
var final = temparr[temparr.length-1].toString().length;
console.log(temparr[temparr.length-1]);
return final;
};
a = fiblength(500);
console.log(a);

Why not use the simple procedure of dividing the number by 10 until the number is less than 1.
Something as simple as this should work (a recursive def obv works as well)
function getDigits(n) {
var digits = 0;
while(n >= 1) {
n/=10;
digits += 1;
}
return digits;
}
getDigits(200);//3
getDigits(3.2 * 10e20);//=>22

Here's a solution in constant time:
function fiblength(n) {
return Math.floor((n>1)?n*.2089+.65051:1);
}
Let's explain how I arrived to it.
All previous solutions will probably not work for N>300 unless you have a BigNumber library in place. Also they're pretty inneficient.
There is a formula to get any Fibonacci number, which uses PHI (golden ratio number), it's very simple:
F(n) = ABS((PHI^n)/sqrt(5))
Where PHI=1.61803399 (golden ratio, found all over the fibonacci sequence)
If you want to know how many digits a number has, you calculate the log base 10 and add 1 to that. Let's call that function D(n) = log10(n) + 1
So what you want fiblength to be is in just the following function
fiblength(n) = D(F(n)) // number of digits of a fibonacci number...
Let's work it out, so you see what the one liner code will be like once you use math.
Substitute F(n)
fiblength(n) = D(ABS((PHI^n)/sqrt(5)))
Now apply D(n) on that:
fiblength(n) = log10(ABS((PHI^n)/sqrt(5))) + 1
So, since log(a/b) = log(a) - log(b)
fiblength(n) = log10(ABS((PHI^n))) - log10(sqrt(5))) + 1
and since log(a^n) = n * log(a)
fiblength(n) = n*log10(PHI) - log10(sqrt(5))) + 1
Then we evaluate those logarithms since they're all on constants
and add the special cases of n=0 and n=1 to return 1
function fiblength(n) {
return Math.floor((n>1)?n*.2089+.65051:1);
}
Enjoy :)
fiblength(500) => 105 //no iterations necessary.

Most of the javascript implementations, internally use 64 bit numbers. So, if the number we are trying to represent is very big, it uses scientific notation to represent those numbers. So, there is no pure "javascript numbers" based solution for this. You may have to look for other BigNum libraries.
As far as your code is concerned, you want only the 500th number, so you don't have to store the entire array of numbers in memory, just previous and current numbers are enough.
function fiblength(nth) {
var previous = 0, current = 1, temp;
for(var i = 2; i<=nth; i++){
temp = current;
current = previous + current;
previous = temp;
}
return current;
};

My Final Solution
function fiblength(nth) {
var a = 0, b = 1, c;
for(var i=2;i<=nth;i++){
c=b;
b=a+b;
a=c;
}
return Math.floor(Math.log(b)/Math.log(10))+1;
}
console.log(fiblength(500));
Thanks for the help!!!

The problem is because the resulting number was converted into a string before any meaningful calculations could be made. Here's how it could have been solved in the original code:
var fiblength = function fiblength(nth) {
var temparr = [0,1];
for(var i = 2; i<=nth; i++){
var prev = temparr[temparr.length-2],
cur = temparr[temparr.length-1],
next = prev + cur;
temparr.push(next);
}
var x = temparr[temparr.length-1];
console.log(x);
var length = 1;
while (x > 1) {
length = length + 1;
x = x/10;
}
return length;
};
console.log ( fiblength(500) );

Generate unique number within range (0 - X), keeping a history to prevent duplicates

I ran into the challenge where I need a function that returns a random number within a given range from 0 - X. Not only that, but I require the number returned to be unique; not duplicating numbers that have already been returned on previous calls to the function.
Optionally, when this is done (e.g. the range has been 'exhausted'), just return a random number within the range.
How would one go about doing this?

This should do it:
function makeRandomRange(x) {
var used = new Array(x),
exhausted = false;
return function getRandom() {
var random = Math.floor(Math.random() * x);
if (exhausted) {
return random;
} else {
for (var i=0; i<x; i++) {
random = (random + 1) % x;
if (random in used)
continue;
used[random] = true;
return random;
}
// no free place found
exhausted = true;
used = null; // free memory
return random;
}
};
}
Usage:
var generate = makeRandomRange(20);
var x1 = generate(),
x2 = generate(),
...
Although it works, it has no good performance when the x-th random is generated - it searches the whole list for a free place. This algorithm, a step-by-step Fisher–Yates shuffle, from the question Unique (non-repeating) random numbers in O(1)?, will perform better:
function makeRandomRange(x) {
var range = new Array(x),
pointer = x;
return function getRandom() {
pointer = (pointer-1+x) % x;
var random = Math.floor(Math.random() * pointer);
var num = (random in range) ? range[random] : random;
range[random] = (pointer in range) ? range[pointer] : pointer;
return range[pointer] = num;
};
}
(Demo at jsfiddle.net)
Extended version which does only generate one "group" of unique numbers:
function makeRandomRange(x) {
var range = new Array(x),
pointer = x;
return function getRandom() {
if (range) {
pointer--;
var random = Math.floor(Math.random() * pointer);
var num = (random in range) ? range[random] : random;
range[random] = (pointer in range) ? range[pointer] : pointer;
range[pointer] = num;
if (pointer <= 0) { // first x numbers had been unique
range = null; // free memory;
}
return num;
} else {
return Math.floor(Math.random() * x);
}
};
}
(Demo)

You got some great programming answer. Here's one with a more theoretical flavor to complete your panorama :-)
Your problem is called "sampling" or "subset sampling" and there are several ways you could do this. Let N be the range you are sampling frame (i.e., N=X+1) and M be the size of your sample (the number of elements you want to pick).
if N is much larger than M, you'll want to use an algorithm such as the one suggested by Bentley and Floyd in his column "Programming Pearls: a sample of brilliance" (temporarily available without ACM's lock screen here), I really recommend this as they explicitly give code and discuss in terms of hash tables, etc.; there a few neat tricks in there
if N is within the same range as M, then you might want to use the Fisher-Yates shuffle but stop after only M steps (instead of N)
if you don't really know then the algorithm on page 647 of Devroye's book on random generation is pretty fast.

I wrote this function. It keeps its own array with a history of generated numbers, preventing initial duplicates, continuing to output a random number if all numbers in the range have been outputted once:
// Generates a unique number from a range
// keeps track of generated numbers in a history array
// if all numbers in the range have been returned once, keep outputting random numbers within the range
var UniqueRandom = { NumHistory: new Array(), generate: function(maxNum) {
var current = Math.round(Math.random()*(maxNum-1));
if (maxNum > 1 && this.NumHistory.length > 0) {
if (this.NumHistory.length != maxNum) {
while($.inArray(current, this.NumHistory) != -1) { current = Math.round(Math.random()*(maxNum-1)); }
this.NumHistory.push(current);
return current;
} else {
//unique numbers done, continue outputting random numbers, or we could reset the history array (NumHistory = [];)
return current;
}
} else {
//first time only
this.NumHistory.push(current);
return current;
}
}
};
Here's a working Fiddle
I hope this is of use to someone!
Edit: as pointed out by Pointy below, it might get slow with a large range (here is a
fiddle, going over a range from 0-1000, which seems to run fine). However; I didn't require a very large range, so perhaps this function is indeed not suited if you look to generate and keep track of an enormous range.

You may try generating the number using the current date and time value which would make it unique. To make it within the range, you may have to use some mathematical function.