I have no problem with the code I am posting below as it works fine but I was just hoping someone could explain to me why var result = 1.
The way I understand it now is that "result *= base" is the same as "result = result * base." With this being the case wouldn't the power also just be the number inputed into exponent paramenter? 1 * 10 = 10 etc... Thanks in advance!
var power = function(base, exponent) {
var result = 1; // <--- Why this???
for (var count = 0; count < exponent; count++)
result *= base;
return result;
};
console.log(power(2, 10));
Because result needs to be a number with a value before you can multiple it by another number. NaN * base === NaN and 0 * base === 0
Try removing it and see what happens:
var power = function(base, exponent) {
var result;
for (var count = 0; count < exponent; count++)
result *= base;
return result;
};
console.log(power(2, 10)); // output: NaN
The way I understand it now is that "result *= base" is the same as "result = result * base." With this being the case wouldn't the power also just be the number inputed into exponent paramenter? 1 * 10 = 10 etc...
Your result *= base is in a loop. So each time you go through the loop you are updating the result variable with a new value. So first time through the loop you have:
result = 1 * 10
And the next time you have:
result = 10 * 10
And the next:
result = 100 * 10
And so on...
If you moved the var result = 1 part inside the loop, then your function would be broken as result would get reset on each iteration.
result is initialized to base to the power zero, which happens to be 1.
Then you multiply it exponent times by base, resulting in base^exponent.
so 2^10 = 1 * 2 * 2 * 2 * 2 .. 10 times.
You have two ways of solving the problem,
Either you start from 1 and multiply the base with itself n times (like in your example), or you start from base and you multiply n-1 times:
var result = base;
for (var count = 0; count < exponent - 1; count++)
result *= base;
//this will return wrong result if you call power(2, 0)
The first case is obviously better since the base can be 0.
The shortest answer to the question is because base ^ 0 = 1
Related
So I have some numbers x = 320232 y = 2301 z = 12020305. I want to round these numbers off using JavaScript so that they become x = 320000 y = 2300 z = 12000000.
I tried Math.round and Math.floor but turns out that they only work with decimal values like
a = 3.1; Math.round(a); // Outputs 3 and not whole numbers.
So my question is can we round of whole numbers using JavaScript and If yes then how?
Edit: I want it to the round of to the starting 3 digit places as seen in the variables above. Like If there was another variable called c = 423841 It should round off to become c = 424000.
You could work with the logarithm of ten and adjust the digits.
const
format = n => v => {
if (!v) return 0;
const l = Math.floor(Math.log10(Math.abs(v))) - n + 1;
return Math.round(v / 10 ** l) * 10 ** l;
};
console.log([0, -9876, 320232, 2301, 12020305, 123456789].map(format(3)));
The solution is to first calculate how many numbers need to be rounded away, and then use that in a round.
Math.round(1234/100)*100 would round to 1200 so we can use this to round. We then only need to determan what to replace 100 with in this example.
That is that would be a 1 followed by LENGTH - 3 zeros. That number can be calculated as it is 10 to the power of LENGTH - 3, in JS: 10 ** (length - 3).
var x = 320232;
var y = 2301;
var z = 12020305;
function my_round(number){
var org_number = number;
// calculate integer number
var count = 0;
if (number >= 1) ++count;
while (number / 10 >= 1) {
number /= 10;
++count;
}
// length - 3
count = Math.round(count) - 3;
if (count < 0){
count = 0;
}
// 10 to the power of (length - 3)
var helper = 10 ** count;
return Math.round(org_number/helper)*helper;
}
alert(my_round(x));
alert(my_round(y));
alert(my_round(z));
It is not the prettiest code, though I tried to make it explainable code.
This should work:
function roundToNthPlace(input, n) {
let powerOfTen = 10 ** n
return Math.round(input/powerOfTen) * powerOfTen;
}
console.log([320232, 2301,12020305, 423841].map(input => roundToNthPlace(input, 3)));
Output: [320000, 2000, 12020000, 424000]
Given any number between 0 and 1, such as 0.84729347293923, is there a simple way to make it into 84729347293923 without string or regex manipulation? I can think of using a loop, which probably is no worse than using a string because it is O(n) with n being the number of digits. But is there a better way?
function getRandom() {
let r = Math.random();
while (Math.floor(r) !== r) r *= 10;
return r;
}
for (let i = 0; i < 10; i++)
console.log(getRandom());
Integers mod 1 = 0, non integers mod 1 != 0.
while ((r*=10) % 1);
Ok, just want to refactor my code (i realized that was bad so this is what i discovered to correctly get the value as you requested).
NOTE: As the question says that "given any number between 0 and 1", this solution only works for values between 0 and 1:
window.onload = ()=>{
function getLen(num){
let currentNumb = num;
let integratedArray = [];
let realLen = 0;
/*While the number is not an integer, we will multiply the copy of the original
*value by ten, and when the loop detects that the number is already an integer
*the while simply breaks, in this process we are storing each transformations
*of the number in an array called integratedArray*/
while(!(Number.isInteger(currentNumb))){
currentNumb *= 10;
integratedArray.push(currentNumb);
}
/*We iterate over the array and compare each value of the array with an operation
*in which the resultant value should be exactly the same as the actual item of the
*array, in the case that both are equal we assign the var realLen to i, and
*in case that the values were not the same, we simply breaks the loop, if the
*values are not the same, this indicates that we found the "trash numbers", so
*we simply skip them.*/
for(let i = 0; i < integratedArray.length; i++){
if(Math.floor(integratedArray[i]) === Math.floor(num * Math.pow(10, i + 1))){
realLen = i;
}else{
break;
}
}
return realLen;
}
//Get the float value of a number between 0 and 1 as an integer.
function getShiftedNumber(num){
//First we need the length to get the float part of the number as an integer
const len = getLen(num);
/*Once we have the length of the number we simply multiply the number by
*(10) ^ numberLength, this eliminates the comma (,), or point (.), and
*automatically transforms the number to an integer in this case a large integer*/
return num * (Math.pow(10, len));
}
console.log(getShiftedNumber(0.84729347293923));
}
So the explanation is the next:
Because we want to convert this number without using any string, regex or any another thing, first we need to get the length of the number, this is a bit hard to do without using string conversions... so i did the function getLen for this purpose.
In the function getLen, we have 3 variables:
currentNumb: This var is a copy of the original value (the original number), this value help us to found the length of the number and we can do some transforms to this value whitout changing the original reference of the number.
We need to multiply this value any times is needed to transform the number to an integer and then multiplyng this value by ten to ten.
with the help of a while (this method makes the number a false integer).
NOTE: I saw "False integer" because when i was making the tests i realized that in the number is being adding more digits than normal... (Very very strange), so this stupid but important thing makes neccesary the filter of these "trash numbers", so later we proccess them.
integratedArray: This array stores the values of the result of the first while operations, so the last number stored in this array is an integer, but this number is one of the "fake integers", so with this array we need to iterate later to compare what of those stored values are different to the original value multiplied by (10 * i + 1), so here is the hint:
In this case the first 12 values of this array are exactly the same with the operation of Math.floor(num * Math.pow(10, i + 1))), but in the 13th value of the array these values are not the same so... yes!, there are those "trash numbers" that we were searching for.
realLen: This is the variable where we will store the real length of the number converting the float part of this number in an integer.
Some binary search approach:
Its useless if avarage length < 8;
It contains floating point issues.
But hey it is O(log n) with tons of wasted side computations - i guess if one counts them its event worse than just plain multiplication.
I prefer #chiliNUT answer. One line stamp.
function floatToIntBinarySearch(number){
const max_safe_int_length = 16;
const powers = [
1,
10,
100,
1000,
10000,
100000,
1000000,
10000000,
100000000,
1000000000,
10000000000,
100000000000,
1000000000000,
10000000000000,
100000000000000,
1000000000000000,
10000000000000000
]
let currentLength = 16
let step = 16
let _number = number * powers[currentLength]
while(_number % 1 != 0 || (_number % 10 | 0) == 0){
step /= 2
if( (_number % 10 | 0) == 0 && !(_number % 1 != 0)){
currentLength = currentLength - step;
} else {
currentLength = step + currentLength;
}
if(currentLength < 1 || currentLength > max_safe_int_length * 2) throw Error("length is weird: " + currentLength)
_number = number * powers[currentLength]
console.log(currentLength, _number)
if(Number.isNaN(_number)) throw Error("isNaN: " + ((number + "").length - 2) + " maybe greater than 16?")
}
return number * powers[currentLength]
}
let randomPower = 10 ** (Math.random() * 10 | 0)
let test = (Math.random() * randomPower | 0) / randomPower
console.log(test)
console.log(floatToIntBinarySearch(test))
I want the user to enter a number and print back the amount of digits of that number.
I know that I can use length, but my homework asking for while loop.
This is what I have so far:
var num;
var count = 0;
num = prompt('Enter number: ');
function counter(x, y) {
while (x > 0) {
y++;
x /= 10;
}
return y;
}
var result = counter(num, count);
console.log(result);
When I give the number 3456 (example), I get back the number 328. I want it to print back the number 4.
This line:
x /= 10;
Should be changed to:
x = Math.floor(x / 10);
The logic assumes integer division: 1234 is supposed to become 123, 12, 1 and 0. JavaScript does not have built in integer division so you need to use Math.floor to emulate it. Complete example with some fixes:
function countDigits(num) {
var count = 0;
while (num > 0) {
num = Math.floor(num / 10);
count++;
}
return count;
}
var num;
do {
num = Number(prompt("Enter number:"));
} while (Number.isNaN(num));
num = Math.abs(num); // just in case you want to handle -ve numbers
var result = countDigits(num);
console.log(result);
The problem is that the division operation will eventually end up converting x to a float and you'll have something like:
x / 10 === 0.1;
x / 10 === 0.01;
x / 10 === 0.001;
....
if you always parse (round) the result of the division to an integer, you'll get the expected result.
var num;
var count = 0;
num = prompt('Enter number: ');
function counter(x, y) {
while (x > 0) {
y++;
x = parseInt(x / 10);
}
return y;
}
var result = counter(num, count);
console.log(result);
You could check againt a number by taking the power of a decimal count.
function counter(value) {
var decimals = 0;
do {
decimals++;
} while (value >= 10 ** decimals)
return decimals;
}
console.log(counter(0));
console.log(counter(1));
console.log(counter(7));
console.log(counter(42));
console.log(counter(999));
console.log(counter(1000));
console.log(counter(1001));
First of all you should convert the input into a number, preferably using the Number function (using unary + has the same effect).
Secondly a division like 5 / 10 will return 0.5 which is bigger than 0. You should instead check if the number is bigger than or equal to 1.
function counter(num) {
num = Math.abs(num) / 10;
var count = 1;
while (num >= 1) {
count++;
num /= 10;
}
return count;
}
console.log(counter(+prompt('Enter number: ')));
You could also use a do while loop and avoid having an extra division outside the loop.
As others have pointed out, y doesn't need to be a parameter, it can be a local variable. But that's not your problem; let's add some extra logging to your loop:
function counter(x) {
let y=0;
while (x > 0) {
console.log("x=" + x + ", y=" + y);
y++;
x /= 10;
}
return y;
}
counter(3456);
The output looks like this:
x=3456, y=0
x=345.6, y=1
x=34.56, y=2
x=3.4560000000000004, y=3
x=0.3456, y=4
x=0.03456, y=5
...
You wanted the loop to stop at 0.3456, but that's still more than 0. (This mistake actually gives you a chance to learn something extra: can you explain why the loop ever finishes at all?)
Hopefully this will give you enough of a hint to complete the homework assignment - remember that debugging is an extremely important part of programming.
Please don't use cycles to measure length of an integer...
Use math instead! Logarithm will do much better job for you.
function numberLength(number) {
return Math.floor(Math.log10(Math.abs(number))) + 1
}
console.log(numberLength(YOUR_NUMBER));
This code returns NaN when the input is 0. I think it depends on your philosophy what length the 0 should have, so I am leaving that case unhandled.
I have a number which I need to divide into 5 parts. However, I want each part to be a random number. But when all the parts are added together, they equal the original number. I am unsure of how to do this with JavaScript. Furthermore, I don't want the min of the divided parts to be 0 or 1, I want to set the min myself.
For example, the number is 450. I want the divided parts to be no less than 60. So to start, the array would be [60,60,60,60,60]. But I want to randomize so that they all add up to 450. What would be the best way to go about doing this?
Thank you!
This is what I've tried so far:
let i = 0;
let number = 450;
let numArray = [];
while(i <= 5){
while(number > 0) {
let randomNum = Math.round(Math.random() * number) + 1;
numArray.push(randomNum);
number -= randomNum;
}
i += 1;
}
let your number be N, and let pn be the nth part. To get 5 parts:
p1 = random number between 0 and N
p2 = random number between 0 and N - p1
p3 = random number between 0 and N - p2 - p1
p4 = random number between 0 and N - p3 - p2 - p1
p5 = N - p4 - p3 - p2 - p1
Edit 2017
To make it seem more random, shuffle the numbers after you generate them
Edit 2020
I guess some code wouldn't hurt. Using ES7 generators:
function* splitNParts(num, parts) {
let sumParts = 0;
for (let i = 0; i < parts - 1; i++) {
const pn = Math.ceil(Math.random() * (num - sumParts))
yield pn
sumParts += pn
}
yield num - sumParts;
}
Fiddle Link
Sum the five minimums (eg min = 60) up:
var minSum = 5 * min
Then get the difference between your original number (orNumber = 450) and minSum.
var delta = orNumber - minSum
Now you get 4 different random numbers in the range from 0 to exclusive 1.
Sort these numbers ascending.
Foreach of these randoms do the following:
Subtract it from the last one (or zero for the first)
Multiply this number with the delta and you get one of the parts.
The last part is the delta minus all other parts.
Afterwards you just have to add your min to all of the parts.
This function generates random numbers from 0 to 1, adds them together to figure out what they need to be multiplied by to provide the correct range. It has the benefit of all the numbers being fairly distributed.
function divvy(number, parts, min) {
var randombit = number - min * parts;
var out = [];
for (var i=0; i < parts; i++) {
out.push(Math.random());
}
var mult = randombit / out.reduce(function (a,b) {return a+b;});
return out.map(function (el) { return el * mult + min; });
}
var d = divvy(450, 6, 60)
console.log(d);
console.log("sum - " + d.reduce(function(a,b){return a+b}));
You can use a do..while loop to subtract a minimum number from original number, keep a copy of original number for subtraction at conclusion of loop to push the remainder to the array
let [n, total, m = n] = [450, 0];
const [min, arr] = [60, []];
do {
n -= min; // subtract `min` from `n`
arr.push(n > min ? min : m - total); // push `min` or remainder
total += arr[arr.length - 1]; // keep track of total
} while (n > min);
console.log(arr);
To randomize output at resulting array select a number greater than min and less than n to create a random number within a specific range
let [n, total, m = n] = [450, 0];
const [min, arr, range = min + min / 2] = [60, []];
do {
let r = Math.random() * (range - min) + min; // random number in our range
n -= r; // subtract `min` from `n`
arr.push(n > min ? r : m - total); // push `r` or remainder
total += arr[arr.length - 1]; // keep track of total
} while (n > min);
console.log(arr);
I made a longer version for beginners.
const n = 450;
const iterations = 5;
const parts = [];
// we'll use this to store what's left on each iteration
let remainder = n;
for (let i = 1; i <= iterations; i += 1) {
// if it's the last iteration, we should just use whatever
// is left after removing all the other random numbers
// from our 450
if (i === iterations) {
parts.push(remainder);
break;
}
// every time we loop, a random number is created.
// on the first iteration, the remainder is still 450
const part = Math.round(Math.random() * remainder);
parts.push(part);
// we must store how much is left after our random numbers
// are deducted from our 450. we will use the lower number
// to calculate the next random number
remainder -= part;
}
// let's print out the array and the proof it still adds up
const total = totalFromParts(parts);
console.log(parts);
console.log('Total is still ' + total);
// this function loops through each array item, and adds it to the last
// just here to test the result
function totalFromParts(parts) {
return parts.reduce((sum, value) => sum + value, 0);
}
There are much more efficient ways to code this, but in the interest of explaining the logic of solving the problem, this walks through that step by step, transforming the values and explaining the logic.
// Set start number, number of fragments
// minimum fragment size, define fragments array
var n = 450
var x = 5
var minNumber = 60
var fragment = n / x
// stuff array with equal sized fragment values
var fragments = []
for (i = 0; i < x; i++) {
fragments[i] = fragment;
}
document.write("fragments: " + fragments);
var delta = [];
// iterate through fragments array
// get a random number each time between the fragment size
// and the minimum fragment sized defined above
// for even array slots, subtract the value from the fragment
// for odd array slots, add the value to the fragment
// skip the first [0] value
for (i = 1; i< x; i++) {
delta[i] = Math.floor(Math.random() * (fragment - minNumber));
document.write("<br />delta: " + delta[i]);
if((i % 2) == 1) {
fragments[i] -= delta[i]
}
else {
fragments[i] += delta[i]
}
}
// set the initial fragment value to 0
fragments[0] = 0
// defines a function we can use to total the array values
function getSum(total, num) {
return total + num;
}
// get the total of the array values, remembering the first is 0
var partialTotal = fragments.reduce(getSum)
document.write("<br />partial sum: " + partialTotal);
// set the first array value to the difference between
// the total of all the other array values and the original
// number the array was to sum up to
fragments[0] = (n - partialTotal)
// write the values out and profit.
document.write("<br />fragments: " + fragments);
var grandTotal = fragments.reduce(getSum)
document.write("<br />Grand total: " + grandTotal);
https://plnkr.co/edit/oToZe7LGpQS4dIVgYHPi?p=preview
Just wondering if anyone can tell me why this returns 100 and not 120? It should calculate the total number of the factor.
function factorialize(num) {
for(var i = 1; i <= num; i++ ) {
var fact = i+i;
total = fact * fact;
}
return total;
}
factorialize(5);
This is not the correct way to calculate the factorial. What is happening in your code is, the last time the line total = fact * fact; is run, fact has a value of 10 (because i is 5), so 10 * 10 becomes 100 and that is what it returns.
If you are trying to calculate the factorial, use this code:
function factorialize(num) {
var total = 1; // Initialize the total. 0! = 1.
for(var i = 1; i <= num; i++ ) {
total = total * i; // Add the current index to the factors by multiplying it by the current total.
}
return total;
}
TLDR is you're overwriting all the values of fact. var is scoped to a function in JS. Eventually you reach i = 5, which eventually sets fact to (5+5) * (5+5) which is 100.