Related
Thanks to Nina I have a code to compare two sentences word by word and return the number of word matches like this:
function includeWords(wanted, seen) {
var wantedMap = wanted.split(/\s+/).reduce((m, s) => m.set(s, (m.get(s) || 0) + 1), new Map),
wantedArray = Array.from(wantedMap.keys()),
count = 0;
seen.split(/\s+/)
.forEach(s => {
var key = wantedArray.find(t => s === t || s.length > 3 && t.length > 3 && (s.startsWith(t) || t.startsWith(s)));
if (!wantedMap.get(key)) return;
console.log(s, key)
++count;
wantedMap.set(key, wantedMap.get(key) - 1);
});
return count;
}
let matches = includeWords('i was sent to earth to protect you introduced', 'they\'re were protecting him i knew that i was aware introducing');
console.log('Matched words: ' + matches);
The code works fine, but there is still one issue:
What if we want to return a match for introduced and introducing too?
If you want the program to consider the words 'introduce' and 'introducing' as a match, it would amount to a "fuzzy" match (non binary logic). One simple way of doing this would require more code, the algorithm of which would possibly resemble
Take 2 words that you wish to match, tokenize into ordered list
of letters
Compare positionally the respective letters, i.e
match a[0]==b[0]? a[1]==b[1] where a[0] represents the first letter
of the first word and b[0] represents the first tokenized
letter/character potential match candidate
KEep a rolling numeric count of such positional matches. In this case it is 8 (introduc).
divide by word length of a = 8/9 call this f
divide by word length of b = 8/11 call this g
Provide a threshold value beyond which the program will consider it a match. eg. if you say anything above 70% in BOTH f and g can be
considered a match - viola, you have your answer!
Please note that there is some normalization also needed to prevent low length words from becoming false positives. you can add a constraint that the aforementioned calculation applies to words with at least 5 letters(or something to that effect!
Hope this helps!!
Regards,
SR
You could calculate similarites for a word pair and get a relation how many characters are similar bei respecting the length of the given word and the wanted pattern.
function getSimilarity(a, b) {
var i = 0;
while (i < a.length) {
if (a[i] !== b[i]) break;
i++;
}
return i / Math.max(a.length, b.length);
}
console.log(getSimilarity('abcdefghij', 'abc')); // 0.3
console.log(getSimilarity('abcdefghij', 'abcdef')); // 0.6
console.log(getSimilarity('abcdefghij', 'abcdefghij')); // 1
console.log(getSimilarity('abcdef', 'abcdefghij')); // 0.6
console.log(getSimilarity('abcdefghij', 'abcdef')); // 0.6
console.log(getSimilarity('abcdefghij', 'xyz')); // 0
console.log(getSimilarity('introduced', 'introducing')); // 0.7272727272727273
Here's a quick fix solution.
It's not intended as a complete solution.
Since the English language has more than a few quirks that would almost require an AI to understand the language.
First add a function that can compare 2 words and returns a boolean.
It'll also make it easier to test for specific words, and adapt to what's really needed.
For example, here's a function that does the simple checks that were already used.
Plus an '...ed' versus '...ing' check.
function compareWords (word1, word2) {
if (word1 === word2)
return true;
if (word1.length > 3 && word2.length > 3) {
if (word1.startsWith(word2) || word2.startsWith(word1))
return true;
if (word1.length > 4 && word2.length > 4) {
if (/(ing|ed)$/.test(word1) && word1.replace(/(ing|ed)$/, 'inged') === word2.replace(/(ing|ed)$/, 'inged'))
return true;
}
}
return false;
}
//
// tests
//
let words = [
["same", "same"],
["different", "unsame"],
["priced", "pricing"],
["price", "priced"],
["producing", "produced"],
["produced", "producing"]
];
words.forEach( (arr, idx) => {
let word1= arr[0];
let word2= arr[1];
let isSame = compareWords(word1, word2);
console.log(`[${word1}] ≈ [${word2}] : ${isSame}`);
});
Then use it in the code you already have.
...
seen.split(/\s+/)
.forEach(s => {
var key = wantedArray.find(t => compareWords(t, s));
...
Regarding string similarity, here's f.e. an older SO post that has some methods to compare strings : Compare Strings Javascript Return %of Likely
I have implemented this, it seems to work fine. any suggestions would be appreciated..
let speechResult = "i was sent to earth to introducing protect yourself introduced seen";
let expectSt = ['they were protecting him knew introducing that you i seen was aware seen introducing'];
// Create arrays of words from above sentences
let speechResultWords = speechResult.split(/\s+/);
let expectStWords = expectSt[0].split(/\s+/);
function includeWords(){
// Declare a variable to hold the count number of matches
let arr = [];
for(let a = 0; a < speechResultWords.length; a++){
for(let b = 0; b < expectStWords.length; b++){
if(similarity(speechResultWords[a], expectStWords[b]) > 69){
arr.push(speechResultWords[a]);
console.log(speechResultWords[a] + ' includes in ' + expectStWords[b]);
}
} // End of first for loop
} // End of second for loop
let uniq = [...new Set(arr)];
return uniq.length;
};
let result = includeWords();
console.log(result)
// The algorithmn
function similarity(s1, s2) {
var longer = s1;
var shorter = s2;
if (s1.length < s2.length) {
longer = s2;
shorter = s1;
}
var longerLength = longer.length;
if (longerLength == 0) {
return 1.0;
}
return (longerLength - editDistance(longer, shorter)) / parseFloat(longerLength)*100;
}
function editDistance(s1, s2) {
s1 = s1.toLowerCase();
s2 = s2.toLowerCase();
var costs = new Array();
for (var i = 0; i <= s1.length; i++) {
var lastValue = i;
for (var j = 0; j <= s2.length; j++) {
if (i == 0)
costs[j] = j;
else {
if (j > 0) {
var newValue = costs[j - 1];
if (s1.charAt(i - 1) != s2.charAt(j - 1))
newValue = Math.min(Math.min(newValue, lastValue),
costs[j]) + 1;
costs[j - 1] = lastValue;
lastValue = newValue;
}
}
}
if (i > 0)
costs[s2.length] = lastValue;
}
return costs[s2.length];
}
How can I generate some unique random numbers between 1 and 100 using JavaScript?
For example: To generate 8 unique random numbers and store them to an array, you can simply do this:
var arr = [];
while(arr.length < 8){
var r = Math.floor(Math.random() * 100) + 1;
if(arr.indexOf(r) === -1) arr.push(r);
}
console.log(arr);
Populate an array with the numbers 1 through 100.
Shuffle it.
Take the first 8 elements of the resulting array.
Modern JS Solution using Set (and average case O(n))
const nums = new Set();
while(nums.size !== 8) {
nums.add(Math.floor(Math.random() * 100) + 1);
}
console.log([...nums]);
Another approach is to generate an 100 items array with ascending numbers and sort it randomly. This leads actually to a really short and (in my opinion) simple snippet.
const numbers = Array(100).fill().map((_, index) => index + 1);
numbers.sort(() => Math.random() - 0.5);
console.log(numbers.slice(0, 8));
Generate permutation of 100 numbers and then choose serially.
Use Knuth Shuffle(aka the Fisher-Yates shuffle) Algorithm.
JavaScript:
function fisherYates ( myArray,stop_count ) {
var i = myArray.length;
if ( i == 0 ) return false;
int c = 0;
while ( --i ) {
var j = Math.floor( Math.random() * ( i + 1 ) );
var tempi = myArray[i];
var tempj = myArray[j];
myArray[i] = tempj;
myArray[j] = tempi;
// Edited thanks to Frerich Raabe
c++;
if(c == stop_count)return;
}
}
CODE COPIED FROM LINK.
EDIT:
Improved code:
function fisherYates(myArray,nb_picks)
{
for (i = myArray.length-1; i > 1 ; i--)
{
var r = Math.floor(Math.random()*i);
var t = myArray[i];
myArray[i] = myArray[r];
myArray[r] = t;
}
return myArray.slice(0,nb_picks);
}
Potential problem:
Suppose we have array of 100 numbers {e.g. [1,2,3...100]} and we stop swapping after 8 swaps;
then most of the times array will look like {1,2,3,76,5,6,7,8,...numbers here will be shuffled ...10}.
Because every number will be swapped with probability 1/100 so
prob. of swapping first 8 numbers is 8/100 whereas prob. of swapping other 92 is 92/100.
But if we run algorithm for full array then we are sure (almost)every entry is swapped.
Otherwise we face a question : which 8 numbers to choose?
The above techniques are good if you want to avoid a library, but depending if you would be alright with a library, I would suggest checking out Chance for generating random stuff in JavaScript.
Specifically to solve your question, using Chance it's as easy as:
// One line!
var uniques = chance.unique(chance.natural, 8, {min: 1, max: 100});
// Print it out to the document for this snippet so we can see it in action
document.write(JSON.stringify(uniques));
<script src="http://chancejs.com/chance.min.js"></script>
Disclaimer, as the author of Chance, I am a bit biased ;)
To avoid any long and unreliable shuffles, I'd do the following...
Generate an array that contains the number between 1 and 100, in order.
Generate a random number between 1 and 100
Look up the number at this index in the array and store in your results
Remove the elemnt from the array, making it one shorter
Repeat from step 2, but use 99 as the upper limit of the random number
Repeat from step 2, but use 98 as the upper limit of the random number
Repeat from step 2, but use 97 as the upper limit of the random number
Repeat from step 2, but use 96 as the upper limit of the random number
Repeat from step 2, but use 95 as the upper limit of the random number
Repeat from step 2, but use 94 as the upper limit of the random number
Repeat from step 2, but use 93 as the upper limit of the random number
Voila - no repeated numbers.
I may post some actual code later, if anybody is interested.
Edit: It's probably the competitive streak in me but, having seen the post by #Alsciende, I couldn't resist posting the code that I promised.
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">
<html>
<head>
<title>8 unique random number between 1 and 100</title>
<script type="text/javascript" language="Javascript">
function pick(n, min, max){
var values = [], i = max;
while(i >= min) values.push(i--);
var results = [];
var maxIndex = max;
for(i=1; i <= n; i++){
maxIndex--;
var index = Math.floor(maxIndex * Math.random());
results.push(values[index]);
values[index] = values[maxIndex];
}
return results;
}
function go(){
var running = true;
do{
if(!confirm(pick(8, 1, 100).sort(function(a,b){return a - b;}))){
running = false;
}
}while(running)
}
</script>
</head>
<body>
<h1>8 unique random number between 1 and 100</h1>
<p><button onclick="go()">Click me</button> to start generating numbers.</p>
<p>When the numbers appear, click OK to generate another set, or Cancel to stop.</p>
</body>
I would do this:
function randomInt(min, max) {
return Math.round(min + Math.random()*(max-min));
}
var index = {}, numbers = [];
for (var i=0; i<8; ++i) {
var number;
do {
number = randomInt(1, 100);
} while (index.hasOwnProperty("_"+number));
index["_"+number] = true;
numbers.push(number);
}
delete index;
This is a very generic function I have written to generate random unique/non-unique integers for an array. Assume the last parameter to be true in this scenario for this answer.
/* Creates an array of random integers between the range specified
len = length of the array you want to generate
min = min value you require
max = max value you require
unique = whether you want unique or not (assume 'true' for this answer)
*/
function _arrayRandom(len, min, max, unique) {
var len = (len) ? len : 10,
min = (min !== undefined) ? min : 1,
max = (max !== undefined) ? max : 100,
unique = (unique) ? unique : false,
toReturn = [], tempObj = {}, i = 0;
if(unique === true) {
for(; i < len; i++) {
var randomInt = Math.floor(Math.random() * ((max - min) + min));
if(tempObj['key_'+ randomInt] === undefined) {
tempObj['key_'+ randomInt] = randomInt;
toReturn.push(randomInt);
} else {
i--;
}
}
} else {
for(; i < len; i++) {
toReturn.push(Math.floor(Math.random() * ((max - min) + min)));
}
}
return toReturn;
}
Here the 'tempObj' is a very useful obj since every random number generated will directly check in this tempObj if that key already exists, if not, then we reduce the i by one since we need 1 extra run since the current random number already exists.
In your case, run the following
_arrayRandom(8, 1, 100, true);
That's all.
Shuffling the numbers from 1 to 100 is the right basic strategy, but if you need only 8 shuffled numbers, there's no need to shuffle all 100 numbers.
I don't know Javascript very well, but I believe it's easy to create an array of 100 nulls quickly. Then, for 8 rounds, you swap the n'th element of the array (n starting at 0) with a randomly selected element from n+1 through 99. Of course, any elements not populated yet mean that the element would really have been the original index plus 1, so that's trivial to factor in. When you're done with the 8 rounds, the first 8 elements of your array will have your 8 shuffled numbers.
var arr = []
while(arr.length < 8){
var randomnumber=Math.ceil(Math.random()*100)
if(arr.indexOf(randomnumber) === -1){arr.push(randomnumber)}
}
document.write(arr);
shorter than other answers I've seen
Implementing this as a generator makes it pretty nice to work with. Note, this implementation differs from ones that require the entire input array to be shuffled first.
This sample function works lazily, giving you 1 random item per iteration up to N items you ask for. This is nice because if you just want 3 items from a list of 1000, you don't have to touch all 1000 items first.
// sample :: Integer -> [a] -> [a]
const sample = n => function* (xs) {
let ys = xs.slice(0);
let len = xs.length;
while (n > 0 && len > 0) {
let i = (Math.random() * len) >> 0;
yield ys.splice(i,1)[0];
n--; len--;
}
}
// example inputs
let items = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
let numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
// get 3 random items
for (let i of sample(3) (items))
console.log(i); // f g c
// partial application
const lotto = sample(3);
for (let i of lotto(numbers))
console.log(i); // 3 8 7
// shuffle an array
const shuffle = xs => Array.from(sample (Infinity) (xs))
console.log(shuffle(items)) // [b c g f d e a]
I chose to implement sample in a way that does not mutate the input array, but you could easily argue that a mutating implementation is favourable.
For example, the shuffle function might wish to mutate the original input array. Or you might wish to sample from the same input at various times, updating the input each time.
// sample :: Integer -> [a] -> [a]
const sample = n => function* (xs) {
let len = xs.length;
while (n > 0 && len > 0) {
let i = (Math.random() * len) >> 0;
yield xs.splice(i,1)[0];
n--; len--;
}
}
// deal :: [Card] -> [Card]
const deal = xs => Array.from(sample (2) (xs));
// setup a deck of cards (13 in this case)
// cards :: [Card]
let cards = 'A234567890JQK'.split('');
// deal 6 players 2 cards each
// players :: [[Card]]
let players = Array.from(Array(6), $=> deal(cards))
console.log(players);
// [K, J], [6, 0], [2, 8], [Q, 7], [5, 4], [9, A]
// `cards` has been mutated. only 1 card remains in the deck
console.log(cards);
// [3]
sample is no longer a pure function because of the array input mutation, but in certain circumstances (demonstrated above) it might make more sense.
Another reason I chose a generator instead of a function that just returns an array is because you may want to continue sampling until some specific condition.
Perhaps I want the first prime number from a list of 1,000,000 random numbers.
"How many should I sample?" – you don't have to specify
"Do I have to find all the primes first and then select a random prime?" – Nope.
Because we're working with a generator, this task is trivial
const randomPrimeNumber = listOfNumbers => {
for (let x of sample(Infinity) (listOfNumbers)) {
if (isPrime(x))
return x;
}
return NaN;
}
This will continuously sample 1 random number at a time, x, check if it's prime, then return x if it is. If the list of numbers is exhausted before a prime is found, NaN is returned.
Note:
This answer was originally shared on another question that was closed as a duplicate of this one. Because it's very different from the other solutions provided here, I've decided to share it here as well
var numbers = [];
for (let i = 0; i < 8; i++) {
let a = true,
n;
while(a) {
n = Math.floor(Math.random() * 100) + 1;
a = numbers.includes(n);
}
numbers.push(n);
}
console.log(numbers);
Same permutation algorithm as The Machine Charmer, but with a prototyped implementation. Better suited to large number of picks. Uses js 1.7 destructuring assignment if available.
// swaps elements at index i and j in array this
// swapping is easy on js 1.7 (feature detection)
Array.prototype.swap = (function () {
var i=0, j=1;
try { [i,j]=[j,i]; }
catch (e) {}
if(i) {
return function(i,j) {
[this[i],this[j]] = [this[j],this[i]];
return this;
}
} else {
return function(i,j) {
var temp = this[i];
this[i] = this[j];
this[j] = temp;
return this;
}
}
})();
// shuffles array this
Array.prototype.shuffle = function() {
for(var i=this.length; i>1; i--) {
this.swap(i-1, Math.floor(i*Math.random()));
}
return this;
}
// returns n unique random numbers between min and max
function pick(n, min, max) {
var a = [], i = max;
while(i >= min) a.push(i--);
return a.shuffle().slice(0,n);
}
pick(8,1,100);
Edit:
An other proposition, better suited to small number of picks, based on belugabob's answer. To guarantee uniqueness, we remove the picked numbers from the array.
// removes n random elements from array this
// and returns them
Array.prototype.pick = function(n) {
if(!n || !this.length) return [];
var i = Math.floor(this.length*Math.random());
return this.splice(i,1).concat(this.pick(n-1));
}
// returns n unique random numbers between min and max
function pick(n, min, max) {
var a = [], i = max;
while(i >= min) a.push(i--);
return a.pick(n);
}
pick(8,1,100);
for arrays with holes like this [,2,,4,,6,7,,]
because my problem was to fill these holes. So I modified it as per my need :)
the following modified solution worked for me :)
var arr = [,2,,4,,6,7,,]; //example
while(arr.length < 9){
var randomnumber=Math.floor(Math.random()*9+1);
var found=false;
for(var i=0;i<arr.length;i++){
if(arr[i]==randomnumber){found=true;break;}
}
if(!found)
for(k=0;k<9;k++)
{if(!arr[k]) //if it's empty !!MODIFICATION
{arr[k]=randomnumber; break;}}
}
alert(arr); //outputs on the screen
The best earlier answer is the answer by sje397. You will get as good random numbers as you can get, as quick as possible.
My solution is very similar to his solution. However, sometimes you want the random numbers in random order, and that is why I decided to post an answer. In addition, I provide a general function.
function selectKOutOfN(k, n) {
if (k>n) throw "k>n";
var selection = [];
var sorted = [];
for (var i = 0; i < k; i++) {
var rand = Math.floor(Math.random()*(n - i));
for (var j = 0; j < i; j++) {
if (sorted[j]<=rand)
rand++;
else
break;
}
selection.push(rand);
sorted.splice(j, 0, rand);
}
return selection;
}
alert(selectKOutOfN(8, 100));
Here is my ES6 version I cobbled together. I'm sure it can be a little more consolidated.
function randomArray(i, min, max) {
min = Math.ceil(min);
max = Math.floor(max);
let arr = Array.from({length: i}, () => Math.floor(Math.random()* (max - min)) + min);
return arr.sort();
}
let uniqueItems = [...new Set(randomArray(8, 0, 100))]
console.log(uniqueItems);
How about using object properties as a hash table? This way your best scenario is to only randomize 8 times. It would only be effective if you want a small part of the range of numbers. It's also much less memory intensive than Fisher-Yates because you don't have to allocate space for an array.
var ht={}, i=rands=8;
while ( i>0 || keys(ht).length<rands) ht[Math.ceil(Math.random()*100)]=i--;
alert(keys(ht));
I then found out that Object.keys(obj) is an ECMAScript 5 feature so the above is pretty much useless on the internets right now. Fear not, because I made it ECMAScript 3 compatible by adding a keys function like this.
if (typeof keys == "undefined")
{
var keys = function(obj)
{
props=[];
for (k in ht) if (ht.hasOwnProperty(k)) props.push(k);
return props;
}
}
var bombout=0;
var checkArr=[];
var arr=[];
while(arr.length < 8 && bombout<100){
bombout++;
var randomNumber=Math.ceil(Math.random()*100);
if(typeof checkArr[randomNumber] == "undefined"){
checkArr[randomNumber]=1;
arr.push(randomNumber);
}
}
// untested - hence bombout
if you need more unique you must generate a array(1..100).
var arr=[];
function generateRandoms(){
for(var i=1;i<=100;i++) arr.push(i);
}
function extractUniqueRandom()
{
if (arr.length==0) generateRandoms();
var randIndex=Math.floor(arr.length*Math.random());
var result=arr[randIndex];
arr.splice(randIndex,1);
return result;
}
function extractUniqueRandomArray(n)
{
var resultArr=[];
for(var i=0;i<n;i++) resultArr.push(extractUniqueRandom());
return resultArr;
}
above code is faster:
extractUniqueRandomArray(50)=>
[2, 79, 38, 59, 63, 42, 52, 22, 78, 50, 39, 77, 1, 88, 40, 23, 48, 84, 91, 49, 4, 54, 93, 36, 100, 82, 62, 41, 89, 12, 24, 31, 86, 92, 64, 75, 70, 61, 67, 98, 76, 80, 56, 90, 83, 44, 43, 47, 7, 53]
Adding another better version of same code (accepted answer) with JavaScript 1.6 indexOf function. Do not need to loop thru whole array every time you are checking the duplicate.
var arr = []
while(arr.length < 8){
var randomnumber=Math.ceil(Math.random()*100)
var found=false;
if(arr.indexOf(randomnumber) > -1){found=true;}
if(!found)arr[arr.length]=randomnumber;
}
Older version of Javascript can still use the version at top
PS: Tried suggesting an update to the wiki but it was rejected. I still think it may be useful for others.
This is my personal solution :
<script>
var i, k;
var numbers = new Array();
k = Math.floor((Math.random()*8));
numbers[0]=k;
for (var j=1;j<8;j++){
k = Math.floor((Math.random()*8));
i=0;
while (i < numbers.length){
if (numbers[i] == k){
k = Math.floor((Math.random()*8));
i=0;
}else {i++;}
}
numbers[j]=k;
}
for (var j=0;j<8;j++){
alert (numbers[j]);
}
</script>
It randomly generates 8 unique array values (between 0 and 7), then displays them using an alert box.
function getUniqueRandomNos() {
var indexedArrayOfRandomNo = [];
for (var i = 0; i < 100; i++) {
var randNo = Math.random();
indexedArrayOfRandomNo.push([i, randNo]);
}
indexedArrayOfRandomNo.sort(function (arr1, arr2) {
return arr1[1] - arr2[1]
});
var uniqueRandNoArray = [];
for (i = 0; i < 8; i++) {
uniqueRandNoArray.push(indexedArrayOfRandomNo[i][0]);
}
return uniqueRandNoArray;
}
I think this method is different from methods given in most of the answers, so I thought I might add an answer here (though the question was asked 4 years ago).
We generate 100 random numbers, and tag each of them with numbers from 1 to 100. Then we sort these tagged random numbers, and the tags get shuffled randomly. Alternatively, as needed in this question, one could do away with just finding top 8 of the tagged random numbers. Finding top 8 items is cheaper than sorting the whole array.
One must note here, that the sorting algorithm influences this algorithm. If the sorting algorithm used is stable, there is slight bias in favor of smaller numbers. Ideally, we would want the sorting algorithm to be unstable and not even biased towards stability (or instability) to produce an answer with perfectly uniform probability distribution.
This can handle generating upto 20 digit UNIQUE random number
JS
var generatedNumbers = [];
function generateRandomNumber(precision) { // input --> number precision in integer
if (precision <= 20) {
var randomNum = Math.round(Math.random().toFixed(precision) * Math.pow(10, precision));
if (generatedNumbers.indexOf(randomNum) > -1) {
if (generatedNumbers.length == Math.pow(10, precision))
return "Generated all values with this precision";
return generateRandomNumber(precision);
} else {
generatedNumbers.push(randomNum);
return randomNum;
}
} else
return "Number Precision shoould not exceed 20";
}
generateRandomNumber(1);
jsFiddle
This solution uses the hash which is much more performant O(1) than checking if the resides in the array. It has extra safe checks too. Hope it helps.
function uniqueArray(minRange, maxRange, arrayLength) {
var arrayLength = (arrayLength) ? arrayLength : 10
var minRange = (minRange !== undefined) ? minRange : 1
var maxRange = (maxRange !== undefined) ? maxRange : 100
var numberOfItemsInArray = 0
var hash = {}
var array = []
if ( arrayLength > (maxRange - minRange) ) throw new Error('Cannot generate unique array: Array length too high')
while(numberOfItemsInArray < arrayLength){
// var randomNumber = Math.floor(Math.random() * (maxRange - minRange + 1) + minRange)
// following line used for performance benefits
var randomNumber = (Math.random() * (maxRange - minRange + 1) + minRange) << 0
if (!hash[randomNumber]) {
hash[randomNumber] = true
array.push(randomNumber)
numberOfItemsInArray++
}
}
return array
}
document.write(uniqueArray(1, 100, 8))
You can also do it with a one liner like this:
[...((add, set) => add(set, add))((set, add) => set.size < 8 ? add(set.add(Math.floor(Math.random()*100) + 1), add) : set, new Set())]
getRandom (min, max) {
return Math.floor(Math.random() * (max - min)) + min
}
getNRandom (min, max, n) {
const numbers = []
if (min > max) {
return new Error('Max is gt min')
}
if (min === max) {
return [min]
}
if ((max - min) >= n) {
while (numbers.length < n) {
let rand = this.getRandom(min, max + 1)
if (numbers.indexOf(rand) === -1) {
numbers.push(rand)
}
}
}
if ((max - min) < n) {
for (let i = min; i <= max; i++) {
numbers.push(i)
}
}
return numbers
}
Using a Set is your fastest option. Here is a generic function for getting a unique random that uses a callback generator. Now it's fast and reusable.
// Get a unique 'anything'
let unique = new Set()
function getUnique(generator) {
let number = generator()
while (!unique.add(number)) {
number = generator()
}
return number;
}
// The generator. Return anything, not just numbers.
const between_1_100 = () => 1 + Math.floor(Math.random() * 100)
// Test it
for (var i = 0; i < 8; i++) {
const aNumber = getUnique(between_1_100)
}
// Dump the 'stored numbers'
console.log(Array.from(unique))
This is a implementation of Fisher Yates/Durstenfeld Shuffle, but without actual creation of a array thus reducing space complexity or memory needed, when the pick size is small compared to the number of elements available.
To pick 8 numbers from 100, it is not necessary to create a array of 100 elements.
Assuming a array is created,
From the end of array(100), get random number(rnd) from 1 to 100
Swap 100 and the random number rnd
Repeat step 1 with array(99)
If a array is not created, A hashMap may be used to remember the actual swapped positions. When the second random number generated is equal to the one of the previously generated numbers, the map provides the current value in that position rather than the actual value.
const getRandom_ = (start, end) => {
return Math.floor(Math.random() * (end - start + 1)) + start;
};
const getRealValue_ = (map, rnd) => {
if (map.has(rnd)) {
return getRealValue_(map, map.get(rnd));
} else {
return rnd;
}
};
const getRandomNumbers = (n, start, end) => {
const out = new Map();
while (n--) {
const rnd = getRandom_(start, end--);
out.set(getRealValue_(out, rnd), end + 1);
}
return [...out.keys()];
};
console.info(getRandomNumbers(8, 1, 100));
console.info(getRandomNumbers(8, 1, Math.pow(10, 12)));
console.info(getRandomNumbers(800000, 1, Math.pow(10, 15)));
Here is an example of random 5 numbers taken from a range of 0 to 100 (both 0 and 100 included) with no duplication.
let finals = [];
const count = 5; // Considering 5 numbers
const max = 100;
for(let i = 0; i < max; i++){
const rand = Math.round(Math.random() * max);
!finals.includes(rand) && finals.push(rand)
}
finals = finals.slice(0, count)
Requirement: Algorithm to generate all possible combinations of a set , without duplicates , or recursively calling function to return results.
The majority , if not all of the Answers provided at Permutations in JavaScript? recursively call a function from within a loop or other function to return results.
Example of recursive function call within loop
function p(a, b, res) {
var b = b || [], res = res || [], len = a.length;
if (!len)
res.push(b)
else
for (var i = 0; i < len
// recursive call to `p` here
; p(a.slice(0, i).concat(a.slice(i + 1, len)), b.concat(a[i]), res)
, i++
);
return res
}
p(["a", "b", "c"]);
The current Question attempts to create the given permutation in a linear process , relying on the previous permutation.
For example , given an array
var arr = ["a", "b", "c"];
to determine the total number of possible permutations
for (var len = 1, i = k = arr.length; len < i ; k *= len++);
k should return 6 , or total number of possible permutations of arr ["a", "b", "c"]
With the total number of individual permutations determined for a set , the resulting array which would contain all six permutations could be created and filled using Array.prototype.slice() , Array.prototype.concat() and Array.prototype.reverse()
var res = new Array(new Array(k));
res[0] = arr;
res[1] = res[0].slice(0,1).concat(res[0].slice(-2).reverse());
res[2] = res[1].slice(-1).concat(res[1].slice(0,2));
res[3] = res[2].slice(0,1).concat(res[2].slice(-2).reverse());
res[4] = res[3].slice(-2).concat(res[3].slice(0,1));
res[5] = res[4].slice(0,1).concat(res[4].slice(-2).reverse());
Attempted to reproduce results based on the pattern displayed at the graph for An Ordered Lexicographic Permutation Algorithm based on one published in Practical Algorithms in C++ at Calculating Permutations and Job Interview Questions .
There appears to be a pattern that could be extended if the input set was , for example
["a", "b", "c", "d", "e"]
where 120 permutations would be expected.
An example of an attempt at filling array relying only on previous permutation
// returns duplicate entries at `j`
var arr = ["a", "b", "c", "d", "e"], j = [];
var i = k = arr.length;
arr.forEach(function(a, b, array) {
if (b > 1) {
k *= b;
if (b === i -1) {
for (var q = 0;j.length < k;q++) {
if (q === 0) {
j[q] = array;
} else {
j[q] = !(q % i)
? array.slice(q % i).reverse().concat(array.slice(0, q % i))
: array.slice(q % i).concat(array.slice(0, q % i));
}
}
}
}
})
however have not yet been able to make the necessary adjustments at parameters for .slice() , .concat() , .reverse() at above js to step from one permutation to the next ; while only using the previous array entry within res to determine current permutation , without using recursive.
Noticed even , odd balance of calls and tried to use modulus % operator and input array .length to either call .reverse() or not at ["a", "b", "c", "d", "e"] array , though did not produce results without duplicate entries.
The expected result is that the above pattern could be reduced to two lines called in succession for the duration of the process until all permutations completed, res filled ; one each for call to .reverse() , call without .reverse() ; e.g., after res[0] filled
// odd , how to adjust `.slice()` , `.concat()` parameters
// for array of unknown `n` `.length` ?
res[i] = res[i - 1].slice(0,1).concat(res[i - 1].slice(-2).reverse());
// even
res[i] = res[1 - 1].slice(-1).concat(res[i - 1].slice(0,2));
Question: What adjustments to above pattern are necessary , in particular parameters , or index , passed .slice() , .concat() to produce all possible permutations of a given set without using a recursive call to the currently processing function ?
var arr = ["a", "b", "c"];
for (var len = 1, i = k = arr.length; len < i; k *= len++);
var res = new Array(new Array(k));
res[0] = arr;
res[1] = res[0].slice(0, 1).concat(res[0].slice(-2).reverse());
res[2] = res[1].slice(-1).concat(res[1].slice(0, 2));
res[3] = res[2].slice(0, 1).concat(res[2].slice(-2).reverse());
res[4] = res[3].slice(-2).concat(res[3].slice(0, 1));
res[5] = res[4].slice(0, 1).concat(res[4].slice(-2).reverse());
console.log(res);
Edit, Update
Have found a process to utilize pattern described above to return permutations in lexicographic order for an input up to .length 4 , using a single for loop. Expected results are not returned for array with .length of 5.
The pattern is based on the second chart at "Calculating Permutations and Job Interview Questions"[0].
Would prefer not to use .splice() or .sort() to return results, though used here while attempting to adhere to last "rotate" requirement at each column. The variable r should reference the index of the first element of the next permutation, which it does.
The use of .splice() , .sort() could be included if their usage followed the pattern at the chart ; though at js below, they actually do not.
Not entirely certain that the issue with js below is only the statement following if (i % (total / len) === reset) , though that portion required the most investment of time; yet still does not return expected results.
Specifically, now referring to the chart, at rotating , for example 2 to index 0, 1 to index 2. Attempted to achieve this by using r , which is a negative index, to traverses from right to left to retrieve next item that should be positioned at index 0 of adjacent "column".
At next column, 2 would be placed at index 2 , 3 would be placed at index 0. This is portion, as far as have been able to grasp or debug, so far, is the area where error is occurring.
Again, returns expected results for [1,2,3,4], though not for [1,2,3,4,5]
var arr = [1, 2, 3, 4];
for (var l = 1, j = total = arr.length; l < j ; total *= l++);
for (var i = 1
, reset = 0
, idx = 0
, r = 0
, len = arr.length
, res = [arr]
; i < total; i++) {
// previous permutation
var prev = res[i - 1];
// if we are at permutation `6` here, or, completion of all
// permutations beginning with `1`;
// setting next "column", place `2` at `index` 0;
// following all permutations beginning with `2`, place `3` at
// `index` `0`; with same process for `3` to `4`
if (i % (total / len) === reset) {
r = --r % -(len);
var next = prev.slice(r);
if (r === -1) {
// first implementation used for setting item at index `-1`
// to `index` 0
// would prefer to use single process for all "rotations",
// instead of splitting into `if` , `else`, though not there, yet
res[i] = [next[0]].concat(prev.slice(0, 1), prev.slice(1, len - 1)
.reverse());
} else {
// workaround for "rotation" at from `index` `r` to `index` `0`
// the chart does not actually use the previous permutation here,
// but rather, the first permutation of that particular "column";
// here, using `r` `,i`, `len`, would be
// `res[i - (i - 1) % (total / len)]`
var curr = prev.slice();
// this may be useful, to retrieve `r`,
// `prev` without item at `r` `index`
curr.splice(prev.indexOf(next[0]), 1);
// this is not optiomal
curr.sort(function(a, b) {
return arr.indexOf(a) > arr.indexOf(b)
});
// place `next[0]` at `index` `0`
// place remainder of sorted array at `index` `1` - n
curr.splice(0, 0, next[0])
res[i] = curr
}
idx = reset;
} else {
if (i % 2) {
// odd
res[i] = prev.slice(0, len - 2).concat(prev.slice(-2)
.reverse())
} else {
// even
--idx
res[i] = prev.slice(0, len - (len - 1))
.concat(prev.slice(idx), prev.slice(1, len + (idx)))
}
}
}
// try with `arr` : `[1,2,3,4,5]` to return `res` that is not correct;
// how can above `js` be adjusted to return correct results for `[1,2,3,4,5]` ?
console.log(res, res.length)
Resources:
Generating Permutation with Javascript
(Countdown) QuickPerm Head Lexicography:
(Formally Example_03 ~ Palindromes)
Generating all Permutations [non-recursive]
(Attempt to port to from C++ to javascript jsfiddle http://jsfiddle.net/tvvvjf3p/)
Calculating Permutation without Recursion - Part 2
permutations of a string using iteration
iterative-permutation
Permutations by swapping
Evaluation of permutation algorithms
Permutation algorithm without recursion? Java
Non-recursive algorithm for full permutation with repetitive elements?
String permutations in Java (non-recursive)
Generating permutations lazily
How to generate all permutations of a list in Python
Can all permutations of a set or string be generated in O(n log n) time?
Finding the nth lexicographic permutation of ‘0123456789’
Combinations and Permutations
Here is a simple solution to compute the nth permutation of a string:
function string_nth_permutation(str, n) {
var len = str.length, i, f, res;
for (f = i = 1; i <= len; i++)
f *= i;
if (n >= 0 && n < f) {
for (res = ""; len > 0; len--) {
f /= len;
i = Math.floor(n / f);
n %= f;
res += str.charAt(i);
str = str.substring(0, i) + str.substring(i + 1);
}
}
return res;
}
The algorithm follows these simple steps:
first compute f = len!, there are factorial(len) total permutations of a set of len different elements.
as the first element, divide the permutation number by (len-1)! and chose the element at the resulting offset. There are (len-1)! different permutations that have any given element as their first element.
remove the chosen element from the set and use the remainder of the division as the permutation number to keep going.
perform these steps with the rest of the set, whose length is reduced by one.
This algorithm is very simple and has interesting properties:
It computes the n-th permutation directly.
If the set is ordered, the permutations are generated in lexicographical order.
It works even if set elements cannot be compared to one another, such as objects, arrays, functions...
Permutation number 0 is the set in the order given.
Permutation number factorial(a.length)-1 is the last one: the set a in reverse order.
Permutations outside this range are returned as undefined.
It can easily be converted to handle a set stored as an array:
function array_nth_permutation(a, n) {
var b = a.slice(); // copy of the set
var len = a.length; // length of the set
var res; // return value, undefined
var i, f;
// compute f = factorial(len)
for (f = i = 1; i <= len; i++)
f *= i;
// if the permutation number is within range
if (n >= 0 && n < f) {
// start with the empty set, loop for len elements
for (res = []; len > 0; len--) {
// determine the next element:
// there are f/len subsets for each possible element,
f /= len;
// a simple division gives the leading element index
i = Math.floor(n / f);
// alternately: i = (n - n % f) / f;
res.push(b.splice(i, 1)[0]);
// reduce n for the remaining subset:
// compute the remainder of the above division
n %= f;
// extract the i-th element from b and push it at the end of res
}
}
// return the permutated set or undefined if n is out of range
return res;
}
clarification:
f is first computed as factorial(len).
For each step, f is divided by len, giving exacty the previous factorial.
n divided by this new value of f gives the slot number among the len slots that have the same initial element. Javascript does not have integral division, we could use (n / f) ... 0) to convert the result of the division to its integral part but it introduces a limitation to sets of 12 elements. Math.floor(n / f) allows for sets of up to 18 elements. We could also use (n - n % f) / f, probably more efficient too.
n must be reduced to the permutation number within this slot, that is the remainder of the division n / f.
We could use i differently in the second loop, storing the division remainder, avoiding Math.floor() and the extra % operator. Here is an alternative for this loop that may be even less readable:
// start with the empty set, loop for len elements
for (res = []; len > 0; len--) {
i = n % (f /= len);
res.push(b.splice((n - i) / f, 1)[0]);
n = i;
}
I think this post should help you. The algorithm should be easily translatable to JavaScript (I think it is more than 70% already JavaScript-compatible).
slice and reverse are bad calls to use if you are after efficiency. The algorithm described in the post is following the most efficient implementation of the next_permutation function, that is even integrated in some programming languages (like C++ e.g.)
EDIT
As I iterated over the algorithm once again I think you can just remove the types of the variables and you should be good to go in JavaScript.
EDIT
JavaScript version:
function nextPermutation(array) {
// Find non-increasing suffix
var i = array.length - 1;
while (i > 0 && array[i - 1] >= array[i])
i--;
if (i <= 0)
return false;
// Find successor to pivot
var j = array.length - 1;
while (array[j] <= array[i - 1])
j--;
var temp = array[i - 1];
array[i - 1] = array[j];
array[j] = temp;
// Reverse suffix
j = array.length - 1;
while (i < j) {
temp = array[i];
array[i] = array[j];
array[j] = temp;
i++;
j--;
}
return true;
}
One method to create permutations is by adding each element in all of the spaces between elements in all of the results so far. This can be done without recursion using loops and a queue.
JavaScript code:
function ps(a){
var res = [[]];
for (var i=0; i<a.length; i++){
while(res[res.length-1].length == i){
var l = res.pop();
for (var j=0; j<=l.length; j++){
var copy = l.slice();
copy.splice(j,0,a[i]);
res.unshift(copy);
}
}
}
return res;
}
console.log(JSON.stringify(ps(['a','b','c','d'])));
Here could be another solution, inspired from the Steinhaus-Johnson-Trotter algorithm:
function p(input) {
var i, j, k, temp, base, current, outputs = [[input[0]]];
for (i = 1; i < input.length; i++) {
current = [];
for (j = 0; j < outputs.length; j++) {
base = outputs[j];
for (k = 0; k <= base.length; k++) {
temp = base.slice();
temp.splice(k, 0, input[i]);
current.push(temp);
}
}
outputs = current;
}
return outputs;
}
// call
var outputs = p(["a", "b", "c", "d"]);
for (var i = 0; i < outputs.length; i++) {
document.write(JSON.stringify(outputs[i]) + "<br />");
}
Here's a snippet for an approach that I came up with on my own, but naturally was also able to find it described elsewhere:
generatePermutations = function(arr) {
if (arr.length < 2) {
return arr.slice();
}
var factorial = [1];
for (var i = 1; i <= arr.length; i++) {
factorial.push(factorial[factorial.length - 1] * i);
}
var allPerms = [];
for (var permNumber = 0; permNumber < factorial[factorial.length - 1]; permNumber++) {
var unused = arr.slice();
var nextPerm = [];
while (unused.length) {
var nextIndex = Math.floor((permNumber % factorial[unused.length]) / factorial[unused.length - 1]);
nextPerm.push(unused[nextIndex]);
unused.splice(nextIndex, 1);
}
allPerms.push(nextPerm);
}
return allPerms;
};
Enter comma-separated string (e.g. a,b,c):
<br/>
<input id="arrInput" type="text" />
<br/>
<button onclick="perms.innerHTML = generatePermutations(arrInput.value.split(',')).join('<br/>')">
Generate permutations
</button>
<br/>
<div id="perms"></div>
Explanation
Since there are factorial(arr.length) permutations for a given array arr, each number between 0 and factorial(arr.length)-1 encodes a particular permutation. To unencode a permutation number, repeatedly remove elements from arr until there are no elements left. The exact index of which element to remove is given by the formula (permNumber % factorial(arr.length)) / factorial(arr.length-1). Other formulas could be used to determine the index to remove, as long as it preserves the one-to-one mapping between number and permutation.
Example
The following is how all permutations would be generated for the array (a,b,c,d):
# Perm 1st El 2nd El 3rd El 4th El
0 abcd (a,b,c,d)[0] (b,c,d)[0] (c,d)[0] (d)[0]
1 abdc (a,b,c,d)[0] (b,c,d)[0] (c,d)[1] (c)[0]
2 acbd (a,b,c,d)[0] (b,c,d)[1] (b,d)[0] (d)[0]
3 acdb (a,b,c,d)[0] (b,c,d)[1] (b,d)[1] (b)[0]
4 adbc (a,b,c,d)[0] (b,c,d)[2] (b,c)[0] (c)[0]
5 adcb (a,b,c,d)[0] (b,c,d)[2] (b,c)[1] (b)[0]
6 bacd (a,b,c,d)[1] (a,c,d)[0] (c,d)[0] (d)[0]
7 badc (a,b,c,d)[1] (a,c,d)[0] (c,d)[1] (c)[0]
8 bcad (a,b,c,d)[1] (a,c,d)[1] (a,d)[0] (d)[0]
9 bcda (a,b,c,d)[1] (a,c,d)[1] (a,d)[1] (a)[0]
10 bdac (a,b,c,d)[1] (a,c,d)[2] (a,c)[0] (c)[0]
11 bdca (a,b,c,d)[1] (a,c,d)[2] (a,c)[1] (a)[0]
12 cabd (a,b,c,d)[2] (a,b,d)[0] (b,d)[0] (d)[0]
13 cadb (a,b,c,d)[2] (a,b,d)[0] (b,d)[1] (b)[0]
14 cbad (a,b,c,d)[2] (a,b,d)[1] (a,d)[0] (d)[0]
15 cbda (a,b,c,d)[2] (a,b,d)[1] (a,d)[1] (a)[0]
16 cdab (a,b,c,d)[2] (a,b,d)[2] (a,b)[0] (b)[0]
17 cdba (a,b,c,d)[2] (a,b,d)[2] (a,b)[1] (a)[0]
18 dabc (a,b,c,d)[3] (a,b,c)[0] (b,c)[0] (c)[0]
19 dacb (a,b,c,d)[3] (a,b,c)[0] (b,c)[1] (b)[0]
20 dbac (a,b,c,d)[3] (a,b,c)[1] (a,c)[0] (c)[0]
21 dbca (a,b,c,d)[3] (a,b,c)[1] (a,c)[1] (a)[0]
22 dcab (a,b,c,d)[3] (a,b,c)[2] (a,b)[0] (b)[0]
23 dcba (a,b,c,d)[3] (a,b,c)[2] (a,b)[1] (a)[0]
Note that each permutation # is of the form:
(firstElIndex * 3!) + (secondElIndex * 2!) + (thirdElIndex * 1!) + (fourthElIndex * 0!)
which is basically the reverse process of the formula given in the explanation.
I dare to add another answer, aiming at answering you question regarding slice, concat, reverse.
The answer is it is possible (almost), but it would not be quite effective. What you are doing in your algorithm is the following:
Find the first inversion in the permutation array, right-to-left (inversion in this case defined as i and j where i < j and perm[i] > perm[j], indices given left-to-right)
place the bigger number of the inversion
concatenate the processed numbers in reversed order, which will be the same as sorted order, as no inversions were observed.
concatenate the second number of the inversion (still sorted in accordsnce with the previos number, as no inversions were observed)
This is mainly, what my first answer does, but in a bit more optimal manner.
Example
Consider the permutation 9,10, 11, 8, 7, 6, 5, 4 ,3,2,1
The first inversion right-to-left is 10, 11.
And really the next permutation is:
9,11,1,2,3,4,5,6,7,8,9,10=9concat(11)concat(rev(8,7,6,5,4,3,2,1))concat(10)
Source code
Here I include the source code as I envision it:
var nextPermutation = function(arr) {
for (var i = arr.length - 2; i >= 0; i--) {
if (arr[i] < arr[i + 1]) {
return arr.slice(0, i).concat([arr[i + 1]]).concat(arr.slice(i + 2).reverse()).concat([arr[i]]);
}
}
// return again the first permutation if calling next permutation on last.
return arr.reverse();
}
console.log(nextPermutation([9, 10, 11, 8, 7, 6, 5, 4, 3, 2, 1]));
console.log(nextPermutation([6, 5, 4, 3, 2, 1]));
console.log(nextPermutation([1, 2, 3, 4, 5, 6]));
The code is avaiable for jsfiddle here.
A fairly simple C++ code without recursion.
#include <vector>
#include <algorithm>
#include <iterator>
#include <iostream>
#include <string>
// Integer data
void print_all_permutations(std::vector<int> &data) {
std::stable_sort(std::begin(data), std::end(data));
do {
std::copy(data.begin(), data.end(), std::ostream_iterator<int>(std::cout, " ")), std::cout << '\n';
} while (std::next_permutation(std::begin(data), std::end(data)));
}
// Character data (string)
void print_all_permutations(std::string &data) {
std::stable_sort(std::begin(data), std::end(data));
do {
std::copy(data.begin(), data.end(), std::ostream_iterator<char>(std::cout, " ")), std::cout << '\n';
} while (std::next_permutation(std::begin(data), std::end(data)));
}
int main()
{
std::vector<int> v({1,2,3,4});
print_all_permutations(v);
std::string s("abcd");
print_all_permutations(s);
return 0;
}
We can find next permutation of a sequence in linear time.
Here is an answer from #le_m. It might be of help.
The following very efficient algorithm uses Heap's method to generate all permutations of N elements with runtime complexity in O(N!):
function permute(permutation) {
var length = permutation.length,
result = [permutation.slice()],
c = new Array(length).fill(0),
i = 1, k, p;
while (i < length) {
if (c[i] < i) {
k = i % 2 && c[i];
p = permutation[i];
permutation[i] = permutation[k];
permutation[k] = p;
++c[i];
i = 1;
result.push(permutation.slice());
} else {
c[i] = 0;
++i;
}
}
return result;
}
console.log(JSON.stringify(permute([1, 2, 3, 4])));
You can use a stack to go through permutations.
This approach is ideal when dealing with trees or other problems while not leaning on recursion.
You will need to make adjustments to not have any duplicate values.
type permutation = [string, string[]]
function p(str: string): string[]{
const res: string[] = []
const stack: permutation[] = [["", str.split('')]]
while(stack.length){
const [head, tail] = stack.pop()
if(!tail.length){
res.push(head)
continue
}
for(let i = 0; i < tail.length; i++){
let newTail = tail.slice()
newTail.splice(i, 1)
stack.push([head + tail[i], newTail])
}
}
return res
}
I am going through the Odin Project and part of that is doing questions 1-3 in the Euler project. I am stumped on question 2:
"By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms."
I am so frustrated! What am I doing wrong? Here's what I have so far. Thanks!
function f() {
var fib = [];
fib.push(1,2,3);
var i = fib.length;
var total = 0;
while(fib[i] < 4000000) {
var x = fib[i-2] + fib [i-1];
if(x % 2 == 0) {
total += x;
}
} return total;
}
console.log(f());
The fibonacci sequences starts 1, 1, 2, not 1, 2, 3.
Also, your solution looks like it will work, but you are storing every number in the sequence instead of just the last two, so this will gobble memory comparatively.
As #DLeh notes, the fibonacci sequence starts with 1,1,2 - not 1,2,3. However, that doesn't change the result of summing the even valued items. The problem you're having is that at this point:
while(fib[i] < 4000000) {
fib[i] is undefined, so the function immediately exits with the total staying at 0. Also within the while loop, you're not pushing the next item in the sequence into your array. The below code fixes both of these problems:
function f() {
var fib = [];
fib.push(1,1);
var i = fib.length;
var total = 0;
while(fib[i-1] < 4000000) {
var x = fib[i-2] + fib [i-1];
fib.push(x);
i = fib.length;
if(x % 2 == 0) {
total += x;
}
} return total;
}
console.log(f()); //4613732
#DLeh also pointed out that you're storing more numbers than needed, this solution works without using the array:
function f() {
var f1 = 1;
var f2 = 1;
var total = 0;
while (f2 < 4000000) {
var t = f1 + f2;
if (t % 2 == 0)
total += t;
f1 = f2;
f2 = t;
}
return total;
}
console.log(f()); //4613732
Just for grins, note that you can do this problem without any use of %, and just + operations. Every third value in the sequence is even. That is, 2 is followed by 3 (odd), and then 3 + 2 is 5 (odd), but that sum of two odd numbers gets us back to even (8) and the cycle repeats.
Thus:
function evenFibTotal(limit) {
var a = 1, b = 1, c = 2, total = 0;
while (c < limit) {
total += c;
a = b + c;
b = a + c;
c = a + b;
}
return total;
}
On each iteration, the second trailing value is set to the next value in the sequence (b + c), and that plus the current one is the first trailing value, and finally the next even Fibonacci number is the sum of those two.
(There's also the closed solution but it's no fun :)
How can I generate some unique random numbers between 1 and 100 using JavaScript?
For example: To generate 8 unique random numbers and store them to an array, you can simply do this:
var arr = [];
while(arr.length < 8){
var r = Math.floor(Math.random() * 100) + 1;
if(arr.indexOf(r) === -1) arr.push(r);
}
console.log(arr);
Populate an array with the numbers 1 through 100.
Shuffle it.
Take the first 8 elements of the resulting array.
Modern JS Solution using Set (and average case O(n))
const nums = new Set();
while(nums.size !== 8) {
nums.add(Math.floor(Math.random() * 100) + 1);
}
console.log([...nums]);
Another approach is to generate an 100 items array with ascending numbers and sort it randomly. This leads actually to a really short and (in my opinion) simple snippet.
const numbers = Array(100).fill().map((_, index) => index + 1);
numbers.sort(() => Math.random() - 0.5);
console.log(numbers.slice(0, 8));
Generate permutation of 100 numbers and then choose serially.
Use Knuth Shuffle(aka the Fisher-Yates shuffle) Algorithm.
JavaScript:
function fisherYates ( myArray,stop_count ) {
var i = myArray.length;
if ( i == 0 ) return false;
int c = 0;
while ( --i ) {
var j = Math.floor( Math.random() * ( i + 1 ) );
var tempi = myArray[i];
var tempj = myArray[j];
myArray[i] = tempj;
myArray[j] = tempi;
// Edited thanks to Frerich Raabe
c++;
if(c == stop_count)return;
}
}
CODE COPIED FROM LINK.
EDIT:
Improved code:
function fisherYates(myArray,nb_picks)
{
for (i = myArray.length-1; i > 1 ; i--)
{
var r = Math.floor(Math.random()*i);
var t = myArray[i];
myArray[i] = myArray[r];
myArray[r] = t;
}
return myArray.slice(0,nb_picks);
}
Potential problem:
Suppose we have array of 100 numbers {e.g. [1,2,3...100]} and we stop swapping after 8 swaps;
then most of the times array will look like {1,2,3,76,5,6,7,8,...numbers here will be shuffled ...10}.
Because every number will be swapped with probability 1/100 so
prob. of swapping first 8 numbers is 8/100 whereas prob. of swapping other 92 is 92/100.
But if we run algorithm for full array then we are sure (almost)every entry is swapped.
Otherwise we face a question : which 8 numbers to choose?
The above techniques are good if you want to avoid a library, but depending if you would be alright with a library, I would suggest checking out Chance for generating random stuff in JavaScript.
Specifically to solve your question, using Chance it's as easy as:
// One line!
var uniques = chance.unique(chance.natural, 8, {min: 1, max: 100});
// Print it out to the document for this snippet so we can see it in action
document.write(JSON.stringify(uniques));
<script src="http://chancejs.com/chance.min.js"></script>
Disclaimer, as the author of Chance, I am a bit biased ;)
To avoid any long and unreliable shuffles, I'd do the following...
Generate an array that contains the number between 1 and 100, in order.
Generate a random number between 1 and 100
Look up the number at this index in the array and store in your results
Remove the elemnt from the array, making it one shorter
Repeat from step 2, but use 99 as the upper limit of the random number
Repeat from step 2, but use 98 as the upper limit of the random number
Repeat from step 2, but use 97 as the upper limit of the random number
Repeat from step 2, but use 96 as the upper limit of the random number
Repeat from step 2, but use 95 as the upper limit of the random number
Repeat from step 2, but use 94 as the upper limit of the random number
Repeat from step 2, but use 93 as the upper limit of the random number
Voila - no repeated numbers.
I may post some actual code later, if anybody is interested.
Edit: It's probably the competitive streak in me but, having seen the post by #Alsciende, I couldn't resist posting the code that I promised.
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">
<html>
<head>
<title>8 unique random number between 1 and 100</title>
<script type="text/javascript" language="Javascript">
function pick(n, min, max){
var values = [], i = max;
while(i >= min) values.push(i--);
var results = [];
var maxIndex = max;
for(i=1; i <= n; i++){
maxIndex--;
var index = Math.floor(maxIndex * Math.random());
results.push(values[index]);
values[index] = values[maxIndex];
}
return results;
}
function go(){
var running = true;
do{
if(!confirm(pick(8, 1, 100).sort(function(a,b){return a - b;}))){
running = false;
}
}while(running)
}
</script>
</head>
<body>
<h1>8 unique random number between 1 and 100</h1>
<p><button onclick="go()">Click me</button> to start generating numbers.</p>
<p>When the numbers appear, click OK to generate another set, or Cancel to stop.</p>
</body>
I would do this:
function randomInt(min, max) {
return Math.round(min + Math.random()*(max-min));
}
var index = {}, numbers = [];
for (var i=0; i<8; ++i) {
var number;
do {
number = randomInt(1, 100);
} while (index.hasOwnProperty("_"+number));
index["_"+number] = true;
numbers.push(number);
}
delete index;
This is a very generic function I have written to generate random unique/non-unique integers for an array. Assume the last parameter to be true in this scenario for this answer.
/* Creates an array of random integers between the range specified
len = length of the array you want to generate
min = min value you require
max = max value you require
unique = whether you want unique or not (assume 'true' for this answer)
*/
function _arrayRandom(len, min, max, unique) {
var len = (len) ? len : 10,
min = (min !== undefined) ? min : 1,
max = (max !== undefined) ? max : 100,
unique = (unique) ? unique : false,
toReturn = [], tempObj = {}, i = 0;
if(unique === true) {
for(; i < len; i++) {
var randomInt = Math.floor(Math.random() * ((max - min) + min));
if(tempObj['key_'+ randomInt] === undefined) {
tempObj['key_'+ randomInt] = randomInt;
toReturn.push(randomInt);
} else {
i--;
}
}
} else {
for(; i < len; i++) {
toReturn.push(Math.floor(Math.random() * ((max - min) + min)));
}
}
return toReturn;
}
Here the 'tempObj' is a very useful obj since every random number generated will directly check in this tempObj if that key already exists, if not, then we reduce the i by one since we need 1 extra run since the current random number already exists.
In your case, run the following
_arrayRandom(8, 1, 100, true);
That's all.
Shuffling the numbers from 1 to 100 is the right basic strategy, but if you need only 8 shuffled numbers, there's no need to shuffle all 100 numbers.
I don't know Javascript very well, but I believe it's easy to create an array of 100 nulls quickly. Then, for 8 rounds, you swap the n'th element of the array (n starting at 0) with a randomly selected element from n+1 through 99. Of course, any elements not populated yet mean that the element would really have been the original index plus 1, so that's trivial to factor in. When you're done with the 8 rounds, the first 8 elements of your array will have your 8 shuffled numbers.
var arr = []
while(arr.length < 8){
var randomnumber=Math.ceil(Math.random()*100)
if(arr.indexOf(randomnumber) === -1){arr.push(randomnumber)}
}
document.write(arr);
shorter than other answers I've seen
Implementing this as a generator makes it pretty nice to work with. Note, this implementation differs from ones that require the entire input array to be shuffled first.
This sample function works lazily, giving you 1 random item per iteration up to N items you ask for. This is nice because if you just want 3 items from a list of 1000, you don't have to touch all 1000 items first.
// sample :: Integer -> [a] -> [a]
const sample = n => function* (xs) {
let ys = xs.slice(0);
let len = xs.length;
while (n > 0 && len > 0) {
let i = (Math.random() * len) >> 0;
yield ys.splice(i,1)[0];
n--; len--;
}
}
// example inputs
let items = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
let numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
// get 3 random items
for (let i of sample(3) (items))
console.log(i); // f g c
// partial application
const lotto = sample(3);
for (let i of lotto(numbers))
console.log(i); // 3 8 7
// shuffle an array
const shuffle = xs => Array.from(sample (Infinity) (xs))
console.log(shuffle(items)) // [b c g f d e a]
I chose to implement sample in a way that does not mutate the input array, but you could easily argue that a mutating implementation is favourable.
For example, the shuffle function might wish to mutate the original input array. Or you might wish to sample from the same input at various times, updating the input each time.
// sample :: Integer -> [a] -> [a]
const sample = n => function* (xs) {
let len = xs.length;
while (n > 0 && len > 0) {
let i = (Math.random() * len) >> 0;
yield xs.splice(i,1)[0];
n--; len--;
}
}
// deal :: [Card] -> [Card]
const deal = xs => Array.from(sample (2) (xs));
// setup a deck of cards (13 in this case)
// cards :: [Card]
let cards = 'A234567890JQK'.split('');
// deal 6 players 2 cards each
// players :: [[Card]]
let players = Array.from(Array(6), $=> deal(cards))
console.log(players);
// [K, J], [6, 0], [2, 8], [Q, 7], [5, 4], [9, A]
// `cards` has been mutated. only 1 card remains in the deck
console.log(cards);
// [3]
sample is no longer a pure function because of the array input mutation, but in certain circumstances (demonstrated above) it might make more sense.
Another reason I chose a generator instead of a function that just returns an array is because you may want to continue sampling until some specific condition.
Perhaps I want the first prime number from a list of 1,000,000 random numbers.
"How many should I sample?" – you don't have to specify
"Do I have to find all the primes first and then select a random prime?" – Nope.
Because we're working with a generator, this task is trivial
const randomPrimeNumber = listOfNumbers => {
for (let x of sample(Infinity) (listOfNumbers)) {
if (isPrime(x))
return x;
}
return NaN;
}
This will continuously sample 1 random number at a time, x, check if it's prime, then return x if it is. If the list of numbers is exhausted before a prime is found, NaN is returned.
Note:
This answer was originally shared on another question that was closed as a duplicate of this one. Because it's very different from the other solutions provided here, I've decided to share it here as well
var numbers = [];
for (let i = 0; i < 8; i++) {
let a = true,
n;
while(a) {
n = Math.floor(Math.random() * 100) + 1;
a = numbers.includes(n);
}
numbers.push(n);
}
console.log(numbers);
Same permutation algorithm as The Machine Charmer, but with a prototyped implementation. Better suited to large number of picks. Uses js 1.7 destructuring assignment if available.
// swaps elements at index i and j in array this
// swapping is easy on js 1.7 (feature detection)
Array.prototype.swap = (function () {
var i=0, j=1;
try { [i,j]=[j,i]; }
catch (e) {}
if(i) {
return function(i,j) {
[this[i],this[j]] = [this[j],this[i]];
return this;
}
} else {
return function(i,j) {
var temp = this[i];
this[i] = this[j];
this[j] = temp;
return this;
}
}
})();
// shuffles array this
Array.prototype.shuffle = function() {
for(var i=this.length; i>1; i--) {
this.swap(i-1, Math.floor(i*Math.random()));
}
return this;
}
// returns n unique random numbers between min and max
function pick(n, min, max) {
var a = [], i = max;
while(i >= min) a.push(i--);
return a.shuffle().slice(0,n);
}
pick(8,1,100);
Edit:
An other proposition, better suited to small number of picks, based on belugabob's answer. To guarantee uniqueness, we remove the picked numbers from the array.
// removes n random elements from array this
// and returns them
Array.prototype.pick = function(n) {
if(!n || !this.length) return [];
var i = Math.floor(this.length*Math.random());
return this.splice(i,1).concat(this.pick(n-1));
}
// returns n unique random numbers between min and max
function pick(n, min, max) {
var a = [], i = max;
while(i >= min) a.push(i--);
return a.pick(n);
}
pick(8,1,100);
for arrays with holes like this [,2,,4,,6,7,,]
because my problem was to fill these holes. So I modified it as per my need :)
the following modified solution worked for me :)
var arr = [,2,,4,,6,7,,]; //example
while(arr.length < 9){
var randomnumber=Math.floor(Math.random()*9+1);
var found=false;
for(var i=0;i<arr.length;i++){
if(arr[i]==randomnumber){found=true;break;}
}
if(!found)
for(k=0;k<9;k++)
{if(!arr[k]) //if it's empty !!MODIFICATION
{arr[k]=randomnumber; break;}}
}
alert(arr); //outputs on the screen
The best earlier answer is the answer by sje397. You will get as good random numbers as you can get, as quick as possible.
My solution is very similar to his solution. However, sometimes you want the random numbers in random order, and that is why I decided to post an answer. In addition, I provide a general function.
function selectKOutOfN(k, n) {
if (k>n) throw "k>n";
var selection = [];
var sorted = [];
for (var i = 0; i < k; i++) {
var rand = Math.floor(Math.random()*(n - i));
for (var j = 0; j < i; j++) {
if (sorted[j]<=rand)
rand++;
else
break;
}
selection.push(rand);
sorted.splice(j, 0, rand);
}
return selection;
}
alert(selectKOutOfN(8, 100));
Here is my ES6 version I cobbled together. I'm sure it can be a little more consolidated.
function randomArray(i, min, max) {
min = Math.ceil(min);
max = Math.floor(max);
let arr = Array.from({length: i}, () => Math.floor(Math.random()* (max - min)) + min);
return arr.sort();
}
let uniqueItems = [...new Set(randomArray(8, 0, 100))]
console.log(uniqueItems);
How about using object properties as a hash table? This way your best scenario is to only randomize 8 times. It would only be effective if you want a small part of the range of numbers. It's also much less memory intensive than Fisher-Yates because you don't have to allocate space for an array.
var ht={}, i=rands=8;
while ( i>0 || keys(ht).length<rands) ht[Math.ceil(Math.random()*100)]=i--;
alert(keys(ht));
I then found out that Object.keys(obj) is an ECMAScript 5 feature so the above is pretty much useless on the internets right now. Fear not, because I made it ECMAScript 3 compatible by adding a keys function like this.
if (typeof keys == "undefined")
{
var keys = function(obj)
{
props=[];
for (k in ht) if (ht.hasOwnProperty(k)) props.push(k);
return props;
}
}
var bombout=0;
var checkArr=[];
var arr=[];
while(arr.length < 8 && bombout<100){
bombout++;
var randomNumber=Math.ceil(Math.random()*100);
if(typeof checkArr[randomNumber] == "undefined"){
checkArr[randomNumber]=1;
arr.push(randomNumber);
}
}
// untested - hence bombout
if you need more unique you must generate a array(1..100).
var arr=[];
function generateRandoms(){
for(var i=1;i<=100;i++) arr.push(i);
}
function extractUniqueRandom()
{
if (arr.length==0) generateRandoms();
var randIndex=Math.floor(arr.length*Math.random());
var result=arr[randIndex];
arr.splice(randIndex,1);
return result;
}
function extractUniqueRandomArray(n)
{
var resultArr=[];
for(var i=0;i<n;i++) resultArr.push(extractUniqueRandom());
return resultArr;
}
above code is faster:
extractUniqueRandomArray(50)=>
[2, 79, 38, 59, 63, 42, 52, 22, 78, 50, 39, 77, 1, 88, 40, 23, 48, 84, 91, 49, 4, 54, 93, 36, 100, 82, 62, 41, 89, 12, 24, 31, 86, 92, 64, 75, 70, 61, 67, 98, 76, 80, 56, 90, 83, 44, 43, 47, 7, 53]
Adding another better version of same code (accepted answer) with JavaScript 1.6 indexOf function. Do not need to loop thru whole array every time you are checking the duplicate.
var arr = []
while(arr.length < 8){
var randomnumber=Math.ceil(Math.random()*100)
var found=false;
if(arr.indexOf(randomnumber) > -1){found=true;}
if(!found)arr[arr.length]=randomnumber;
}
Older version of Javascript can still use the version at top
PS: Tried suggesting an update to the wiki but it was rejected. I still think it may be useful for others.
This is my personal solution :
<script>
var i, k;
var numbers = new Array();
k = Math.floor((Math.random()*8));
numbers[0]=k;
for (var j=1;j<8;j++){
k = Math.floor((Math.random()*8));
i=0;
while (i < numbers.length){
if (numbers[i] == k){
k = Math.floor((Math.random()*8));
i=0;
}else {i++;}
}
numbers[j]=k;
}
for (var j=0;j<8;j++){
alert (numbers[j]);
}
</script>
It randomly generates 8 unique array values (between 0 and 7), then displays them using an alert box.
function getUniqueRandomNos() {
var indexedArrayOfRandomNo = [];
for (var i = 0; i < 100; i++) {
var randNo = Math.random();
indexedArrayOfRandomNo.push([i, randNo]);
}
indexedArrayOfRandomNo.sort(function (arr1, arr2) {
return arr1[1] - arr2[1]
});
var uniqueRandNoArray = [];
for (i = 0; i < 8; i++) {
uniqueRandNoArray.push(indexedArrayOfRandomNo[i][0]);
}
return uniqueRandNoArray;
}
I think this method is different from methods given in most of the answers, so I thought I might add an answer here (though the question was asked 4 years ago).
We generate 100 random numbers, and tag each of them with numbers from 1 to 100. Then we sort these tagged random numbers, and the tags get shuffled randomly. Alternatively, as needed in this question, one could do away with just finding top 8 of the tagged random numbers. Finding top 8 items is cheaper than sorting the whole array.
One must note here, that the sorting algorithm influences this algorithm. If the sorting algorithm used is stable, there is slight bias in favor of smaller numbers. Ideally, we would want the sorting algorithm to be unstable and not even biased towards stability (or instability) to produce an answer with perfectly uniform probability distribution.
This can handle generating upto 20 digit UNIQUE random number
JS
var generatedNumbers = [];
function generateRandomNumber(precision) { // input --> number precision in integer
if (precision <= 20) {
var randomNum = Math.round(Math.random().toFixed(precision) * Math.pow(10, precision));
if (generatedNumbers.indexOf(randomNum) > -1) {
if (generatedNumbers.length == Math.pow(10, precision))
return "Generated all values with this precision";
return generateRandomNumber(precision);
} else {
generatedNumbers.push(randomNum);
return randomNum;
}
} else
return "Number Precision shoould not exceed 20";
}
generateRandomNumber(1);
jsFiddle
This solution uses the hash which is much more performant O(1) than checking if the resides in the array. It has extra safe checks too. Hope it helps.
function uniqueArray(minRange, maxRange, arrayLength) {
var arrayLength = (arrayLength) ? arrayLength : 10
var minRange = (minRange !== undefined) ? minRange : 1
var maxRange = (maxRange !== undefined) ? maxRange : 100
var numberOfItemsInArray = 0
var hash = {}
var array = []
if ( arrayLength > (maxRange - minRange) ) throw new Error('Cannot generate unique array: Array length too high')
while(numberOfItemsInArray < arrayLength){
// var randomNumber = Math.floor(Math.random() * (maxRange - minRange + 1) + minRange)
// following line used for performance benefits
var randomNumber = (Math.random() * (maxRange - minRange + 1) + minRange) << 0
if (!hash[randomNumber]) {
hash[randomNumber] = true
array.push(randomNumber)
numberOfItemsInArray++
}
}
return array
}
document.write(uniqueArray(1, 100, 8))
You can also do it with a one liner like this:
[...((add, set) => add(set, add))((set, add) => set.size < 8 ? add(set.add(Math.floor(Math.random()*100) + 1), add) : set, new Set())]
getRandom (min, max) {
return Math.floor(Math.random() * (max - min)) + min
}
getNRandom (min, max, n) {
const numbers = []
if (min > max) {
return new Error('Max is gt min')
}
if (min === max) {
return [min]
}
if ((max - min) >= n) {
while (numbers.length < n) {
let rand = this.getRandom(min, max + 1)
if (numbers.indexOf(rand) === -1) {
numbers.push(rand)
}
}
}
if ((max - min) < n) {
for (let i = min; i <= max; i++) {
numbers.push(i)
}
}
return numbers
}
Using a Set is your fastest option. Here is a generic function for getting a unique random that uses a callback generator. Now it's fast and reusable.
// Get a unique 'anything'
let unique = new Set()
function getUnique(generator) {
let number = generator()
while (!unique.add(number)) {
number = generator()
}
return number;
}
// The generator. Return anything, not just numbers.
const between_1_100 = () => 1 + Math.floor(Math.random() * 100)
// Test it
for (var i = 0; i < 8; i++) {
const aNumber = getUnique(between_1_100)
}
// Dump the 'stored numbers'
console.log(Array.from(unique))
This is a implementation of Fisher Yates/Durstenfeld Shuffle, but without actual creation of a array thus reducing space complexity or memory needed, when the pick size is small compared to the number of elements available.
To pick 8 numbers from 100, it is not necessary to create a array of 100 elements.
Assuming a array is created,
From the end of array(100), get random number(rnd) from 1 to 100
Swap 100 and the random number rnd
Repeat step 1 with array(99)
If a array is not created, A hashMap may be used to remember the actual swapped positions. When the second random number generated is equal to the one of the previously generated numbers, the map provides the current value in that position rather than the actual value.
const getRandom_ = (start, end) => {
return Math.floor(Math.random() * (end - start + 1)) + start;
};
const getRealValue_ = (map, rnd) => {
if (map.has(rnd)) {
return getRealValue_(map, map.get(rnd));
} else {
return rnd;
}
};
const getRandomNumbers = (n, start, end) => {
const out = new Map();
while (n--) {
const rnd = getRandom_(start, end--);
out.set(getRealValue_(out, rnd), end + 1);
}
return [...out.keys()];
};
console.info(getRandomNumbers(8, 1, 100));
console.info(getRandomNumbers(8, 1, Math.pow(10, 12)));
console.info(getRandomNumbers(800000, 1, Math.pow(10, 15)));
Here is an example of random 5 numbers taken from a range of 0 to 100 (both 0 and 100 included) with no duplication.
let finals = [];
const count = 5; // Considering 5 numbers
const max = 100;
for(let i = 0; i < max; i++){
const rand = Math.round(Math.random() * max);
!finals.includes(rand) && finals.push(rand)
}
finals = finals.slice(0, count)