I am creating a simple bracket system and I need a way to check if there are a correct number of teams, OR if my program needs to compensate for bye rounds.
Right now, I am checking for "powers of two" with this function:
function validBracket(data) {
var x = data.teams.length;
return ((x != 0) && !(x & (x - 1)));
}
This works pretty well, but I am needing to know how many Bye rounds to add.
For instance, if I had 16 teams, I would not need to add anymore teams. However, if I had 12 teams, I would need the first 4 teams to get a bye round.
How can I calculate number of bye rounds to add to my bracket? And would hard-coding an array of powers of two be better?
In pseudo code, something like this is what i was thinking of:
if(validateBracket(data)) {
// Valid number of teams (power of two). Keep going.
} else {
var byeRounds = calculateByeRounds();
}
NOTE: I would rather not use an array of powers of two like below:
var powersOfTwo = [2,4,8,16,32,...];
The reasoning behind this is that I would be limiting the number of teams that could be put in the system (however, I don't think a person would have over 256 teams).
var needed = (1 << Math.ceil(Math.log2(n))) - n;
More generalized solution for extreme cases:
var needed = Math.pow(2, Math.ceil(Math.log2(n))) - n;
Related
I'm making this acres and karats calculator for my uncle to help him in his work.
I'll explain the whole idea of this thing with this example. So if you add 3.22 + 2.2 it should be = 5.42 but in this calculator 3.22 + 2.2 should = 6, because 3 acres + 2 acres = 5 acres and 22 karats + 2 karats = 1 acre, so the total would be 6 acres.
The way I'm doing it in the code is that I'm splitting a number like 3.22 to two, 3 and 22 and the other number to 2 and 2 and I add the whole numbers together and the fractions together and if the fractions are >= 24 I add one to the whole numbers and if there're fractions left from the whole calculation I leave it. For example 3.15 + 2.15 = 6.6, but I'm stuck on how I can add the numbers, there's also an error in there that I don't know how to resolve.
Anyway here's the code
function getValue(v) {
return +v.toString().match(/\.(\d*)/)[1] || 0;
}
function getTotal() {
d += Math.floor(num);
p += getValue(num);
if (p >= 24) {
p -= 24;
++d;
}
total = d + p / 100;
ptag.textContent = total;
}
I added the part of the code where I'm stuck.
Note: I'm trying to make the thing able to add multiple numbers not only two. Also I'm trying to add subtraction but I have no idea how to start working on the subtraction because I haven't even finished the addition.
If the error you are talking about is something like this:
Uncaught TypeError: Cannot read property '1' of null
It is because of your getValue function.
My suggestion is, instead of using something as complicated as
function getValue(v) {
return +v.toString().match(/\.(\d*)/)[1] || 0;
}
use
function getValue(v) {
return floor((v % 1) * 100);
}
This has the same effect as the code you wrote. Which for example, from input 3.13, returns 13.
But there are few other problems.
First, you should update your num variable every now and often, otherwise, it is always going to stay as an empty string (you only defined it on line 20, and you didn't update it after that).
Second, you should clear the d and p variable after you use. As of right now, both of these variables just keeps on increasing every time you run the getTotal function
For your question of how you can add two numbers, I suggest you to create a variable where you can store the first number that the user typed.
For example, when the user typed in 4.19 and pressed the plus button, save that 4.19 into a variable (let's say firstNum).
Then when the user pressed equal button, add the number from the current input field with the firstNum variable.
On how exactly you are going to add two different numbers, break two numbers you want to add into Acres part and Karats parts. Then add them separately, then use your getTotal.
So if the number is 3.21 and 5.18, add 3 and 5, add 21 and 18, then add both of them.
you'll get 8.39. Finally, convert 8.39 into 9.15.
Sorry if my calculation is not correct. It is my first time with this concept!
But I believe this is the way to go.
I am trying to get every single combination of elements into an array. Now I can use the method below, and remove the duplicates, but this way is far to slow for my use.
The code below would find every possible combination for 2 digits below 4. Now in the code I actually want to use this for, the least possible code would be 6 for loops (within each other) with the amount being 18 (rememeber this is the minimum).
The code below would execute amount^[amount of for loops], or amount^2 which in this case is 16. That means that in the code I want to use this for, it executes 18^6 times, or 34 million times. And this is the minimum, which would get much higher.
After trying to run my code (with 6 foor loops in which amount = 18), it crashed my browser... My question is: Is there any faster and more efficient (not elegant. I don't care how elegant it is) in which my browser won't crash?
Note: This question is not a duplicate question. All the other questions simply ask for a way to do this, however I already have a way. I am just trying to make it more efficient and faster so that it actually works correctly.
let combinations = [];
let amount = 4;
for (let a = 0; a < amount; a++) {
for (let b = 0; b < amount; b++) {
combinations.push(`${a}${b}`);
}
}
console.log(combinations);
Below is a snippet providing a possible example for how my code would work.
let possibilities = [];
let amount = 6; //Amount is set by me, so don't worry about it being incorrect
for (let a = 0; a < amount; a++) {
for (let b = 0; b < amount; b++) {
possibilities.push(a + b);
}
}
possibilities = [...new Set(possibilities)]; //Removes duplicates
possibilities.sort((a, b) => b - a); //Sorts in descending order
possibilities = possibilities.slice(0, 3); //Gets top 3 values
console.log(possibilities);
Ok, as discussed in the comments, if you need top 3 values for a particular amount, you could just do something simple like below:
let amount = 6;
let highest = amount - 1,second_highest = amount - 2,third_highest = amount - 3;
let possibilities = [
highest + highest,
highest + second_highest,
highest + third_highest
];
console.log(possibilities);
I don't know the any better solution for this, but yes there are some conditions you need to check first.
If(amount <= 0) return 'Invalid amount, Please enter a valid amount"
So if somebody enters a negative or zero value your loop will goes into infinite loop, and make the situation more worst.
if(amount === 1) return '1 possible combination'
As amount less than 1 is 0 only and combinations for 0 is 1 only, you need not to parse whole loop for 6 digits or n digits for 0 so it will get solve in complexity of 1 instead of N(no. of digits).
And for amount greater then 1 you can create manual loops, like here you created 2 loops for 2 digits, you create 6 loops for 6 digits, better create dynamic logic for this to create number of loops automatically.
You need to consider 1111, 1112 this type of combinations as well right?
Or if only 1234, 2134, 2314 this kind of scenarios are required? This can be done in very less complexity.
For duplication you can store combinations as a key value pair. and then Object.Keys will be your combinations.
Pretty much the title; I'm looking for a way to take a number (say 78, for example) and divide it into the minimum number of equal parts that don't exceed another number (say 25, for example). The number of parts doesn't matter, it should just be as small as possible - this means that the resulting numbers should all be as large as possible.
For some context, the reason I want to do this is to work with even text splitting. The messaging service I'm working with only allows 25 lines per message, so I split my text into multiple messages, but I don't want to have messages with only one or two lines. To this end, I'd like to split the text every n lines, where n is as close to 25 as possible and creates as even of a split across the messages as possible. The number of resulting messages doesn't matter at all, though, so it's not a concern.
After some blind bumbling, I managed to solve the issue.
out = [list]
if(list.split("\n").length > 25) {
n = list.split("\n").length
x = Math.floor(n / 25)
y = n / x
while(y > 25) {
x += 1
y = n / x
}
out = []
while(list) {
l25 = list.split(/\r?\n/, Math.ceil(y)).join("\n")
list = list.slice(l25.length)
out.push(l25)
}
}
// Send each value in out as a separate message
Where list is the string to split and out is the resulting array of strings.
I'm in the process of coding an application that does the following:
Generates a random number with 4 digits.
Changes it once per calendar day.
Won't change that full day. Only once in a day.
I tried:
function my_doubt()
{
var place = document.getElementById("my_div")
place.innerHTML=Math.floor((Math.random()*100)+1);
}
I'm getting a random number with Math.random(). However, I'm rather clueless about how to generate a different number for each day. What are some common approaches for tackling this problem?
Note: It doesn't have to be really random. A pseudo - random number is also OK.
You need to seed the random number generator with a number derived from the current date, for example "20130927" for today.
You haven't been clear about your requirements, so I don't know how random you need (do you have requirements for how uniform of a distribution you need?).
This will generate a random looking 4 digit number which may be good enough for your requirements, but if you perform an analysis you'll find the number isn't actually very random:
function rand_from_seed(x, iterations){
iterations = iterations || 100;
for(var i = 0; i < iterations; i++)
x = (x ^ (x << 1) ^ (x >> 1)) % 10000;
return x;
}
var random = rand_from_seed(~~((new Date)/86400000)); // Seed with the epoch day.
Now that your question is a bit more reasonable, clear and nicer in tone. I can give you a way to get the same result on the client-side. However as others mentioned, to maintain consistency, you probably want to maintain the number on the server to ensure consistency.
var oneDayInMs = 1000*60*60*24;
var currentTimeInMs = new Date().getTime(); // UTC time
var timeInDays = Math.floor(currentTimeInMs / oneDayInMs);
var numberForToday = timeInDays % 9999;
console.log(numberForToday);
// zero-filling of numbers less than four digits might be optional for you
// zero-filled value will be a string to maintain its leading 0s
var fourDigitNumber = numberForToday.toString();
while(fourDigitNumber.length < 4)
{
fourDigitNumber = 0+fourDigitNumber;
}
console.log(fourDigitNumber);
// remember that this number rotates every and is unique for 10000 days
1)create a random number in javascript
2)store in cookie that will expire after one day
3)get value from cookie, if it does not exist goto 1
I’m having problems generating normally distributed random numbers (mu=0 sigma=1)
using JavaScript.
I’ve tried Box-Muller's method and ziggurat, but the mean of the generated series of numbers comes out as 0.0015 or -0.0018 — very far from zero!! Over 500,000 randomly generated numbers this is a big issue. It should be close to zero, something like 0.000000000001.
I cannot figure out whether it’s a method problem, or whether JavaScript’s built-in Math.random() generates not exactly uniformly distributed numbers.
Has someone found similar problems?
Here you can find the ziggurat function:
http://www.filosophy.org/post/35/normaldistributed_random_values_in_javascript_using_the_ziggurat_algorithm/
And below is the code for the Box-Muller:
function rnd_bmt() {
var x = 0, y = 0, rds, c;
// Get two random numbers from -1 to 1.
// If the radius is zero or greater than 1, throw them out and pick two
// new ones. Rejection sampling throws away about 20% of the pairs.
do {
x = Math.random()*2-1;
y = Math.random()*2-1;
rds = x*x + y*y;
}
while (rds === 0 || rds > 1)
// This magic is the Box-Muller Transform
c = Math.sqrt(-2*Math.log(rds)/rds);
// It always creates a pair of numbers. I'll return them in an array.
// This function is quite efficient so don't be afraid to throw one away
// if you don't need both.
return [x*c, y*c];
}
If you generate n independent normal random variables, the standard deviation of the mean will be sigma / sqrt(n).
In your case n = 500000 and sigma = 1 so the standard error of the mean is approximately 1 / 707 = 0.0014. The 95% confidence interval, given 0 mean, would be around twice this or (-0.0028, 0.0028). Your sample means are well within this range.
Your expectation of obtaining 0.000000000001 (1e-12) is not mathematically grounded. To get within that range of accuracy, you would need to generate about 10^24 samples. At 10,000 samples per second that would still take 3 quadrillon years to do...this is precisely why it's good to avoid computing things by simulation if possible.
On the other hand, your algorithm does seem to be implemented correctly :)