Related
I have the following function
function randomNum(max, used){
newNum = Math.floor(Math.random() * max + 1);
if($.inArray(newNum, used) === -1){
console.log(newNum + " is not in array");
return newNum;
}else{
return randomNum(max,used);
}
}
Basically I am creating a random number between 1 - 10 and checking to see if that number has already been created, by adding it to an array and checking the new created number against it. I call it by adding it to a variable..
UPDATED:
for(var i=0;i < 10;i++){
randNum = randomNum(10, usedNums);
usedNums.push(randNum);
//do something with ranNum
}
This works, but in Chrome I get the following error:
Uncaught RangeError: Maximum call stack size exceeded
Which I guess it's because I am calling the function inside itself too many times. Which means my code is no good.
Can someone help me with the logic? what's a best way to make sure my numbers are not repeating?
If I understand right then you're just looking for a permutation (i.e. the numbers randomised with no repeats) of the numbers 1-10?
Maybe try generating a randomised list of those numbers, once, at the start, and then just working your way through those?
This will calculate a random permutation of the numbers in nums:
var nums = [1,2,3,4,5,6,7,8,9,10],
ranNums = [],
i = nums.length,
j = 0;
while (i--) {
j = Math.floor(Math.random() * (i+1));
ranNums.push(nums[j]);
nums.splice(j,1);
}
So, for example, if you were looking for random numbers between 1 - 20 that were also even, then you could use:
nums = [2,4,6,8,10,12,14,16,18,20];
Then just read through ranNums in order to recall the random numbers.
This runs no risk of it taking increasingly longer to find unused numbers, as you were finding in your approach.
EDIT: After reading this and running a test on jsperf, it seems like a much better way of doing this is a Fisher–Yates Shuffle:
function shuffle(array) {
var i = array.length,
j = 0,
temp;
while (i--) {
j = Math.floor(Math.random() * (i+1));
// swap randomly chosen element with current element
temp = array[i];
array[i] = array[j];
array[j] = temp;
}
return array;
}
var ranNums = shuffle([1,2,3,4,5,6,7,8,9,10]);
Basically, it's more efficient by avoiding the use of 'expensive' array operations.
BONUS EDIT: Another possibility is using generators (assuming you have support):
function* shuffle(array) {
var i = array.length;
while (i--) {
yield array.splice(Math.floor(Math.random() * (i+1)), 1)[0];
}
}
Then to use:
var ranNums = shuffle([1,2,3,4,5,6,7,8,9,10]);
ranNums.next().value; // first random number from array
ranNums.next().value; // second random number from array
ranNums.next().value; // etc.
where ranNums.next().value will eventually evaluate to undefined once you've run through all the elements in the shuffled array.
Overall this won't be as efficient as the Fisher–Yates Shuffle because you're still splice-ing an array. But the difference is that you're now doing that work only when you need it rather than doing it all upfront, so depending upon your use case, this might be better.
//random number without repetition in JavaScript, Just in one line;
//it can be used as _id;
//it not need to store or check;
const myRnId = () => parseInt(Date.now() * Math.random());
console.log(myRnId()); // any random number included timeStamp;
function Myrand(max,min){
arr=[];
for (i = 0; i < max; i++) {
x = Math.floor( Math.random() * max) + min;
if(arr.includes(x) == true){
i=i-1;
}else{
if(x>max==false){
arr.push(x);
}
}
}
return arr;
}
console.log(Myrand(5,1));
Try this:
var numbers = []; // new empty array
var min, max, r, n, p;
min = 1;
max = 50;
r = 5; // how many numbers you want to extract
for (let i = 0; i < r; i++) {
do {
n = Math.floor(Math.random() * (max - min + 1)) + min;
p = numbers.includes(n);
if(!p){
numbers.push(n);
}
}
while(p);
}
console.log(numbers.join(" - "));
let arr = [];
do {
let num = Math.floor(Math.random() * 10 + 1);
arr.push(num);
arr = arr.filter((item, index) => {
return arr.indexOf(item) === index
});
} while (arr.length < 10);
console.log(arr);
HTML
<p id="array_number" style="font-size: 25px; text-align: center;"></p>
JS
var min = 1;
var max = 90;
var stop = 6; //Number of numbers to extract
var numbers = [];
for (let i = 0; i < stop; i++) {
var n = Math.floor(Math.random() * max) + min;
var check = numbers.includes(n);
if(check === false) {
numbers.push(n);
} else {
while(check === true){
n = Math.floor(Math.random() * max) + min;
check = numbers.includes(n);
if(check === false){
numbers.push(n);
}
}
}
}
sort();
//Sort the array in ascending order
function sort() {
numbers.sort(function(a, b){return a-b});
document.getElementById("array_number").innerHTML = numbers.join(" - ");
}
DEMO
The issue is that as you approach saturation you begin to take longer and longer to generate a unique number "randomly". For instance, in the example you provided above the max is 10. Once the used number array contains 8 numbers it can potentially take a long time for the 9th and 10th to be found. This is probably where the maximum call stack error is being generated.
jsFiddle Demo showing iteration count being maxed
By iterating inside of your recursion, you can see that a large amount of execution occurs when the array is completely saturated, but the function is called. In this scenario, the function should exit.
jsFiddle Demo with early break
if( used.length >= max ) return undefined;
And one last way to accomplish both the iteration checks and the infinite recursion would be like this jsFiddle Demo:
function randomNum(max, used, calls){
if( calls == void 0 ) calls = 0;
if( calls++ > 10000 ) return undefined;
if( used.length >= max ) return undefined;
var newNum = Math.floor(Math.random() * max + 1);
if($.inArray(newNum, used) === -1){
return newNum;
}else{
return randomNum(max,used,calls);
}
}
<!DOCTYPE html>
<html>
<body>
<h2>JavaScript Math.random()</h2>
<p>Math.random() returns a random number between 0 (included) and 1 (excluded):</p>
<p id="demo"></p>
<script>
var storeArray = []
function callRamdom(){
var randomNumber = Math.floor(Math.random() * 5);
return randomNumber;
}
function randomStore(){
var localValue = callRamdom()
var status = false;
for(i=0;i<5; i++){
var aa = storeArray[i];
if(aa!=localValue){
console.log(storeArray[i]+"hhhhh"+ localValue);
if(i==4){
status=true;
}
}
else
break;
}
if(status==true){
storeArray.push(localValue);
}
if(storeArray.length!=5){
randomStore();
}
return storeArray;
}
document.getElementById("demo").innerHTML = randomStore();
</script>
</body>
</html>
while(randArr.length < SIZEOFARRAY){
val = Math.floor((Math.random() * RANGEOFVALUES));
if(randArr.indexOf(val) < 0){
randArr.push(val);
}
}
You can change SIZEOFARRAY to the size of the array you wish to use
and also change RANGEOFVALUES to the range of values you wish to randomize
const GenerateRandomNumbers = (max) => {
let orderNumbers = new Set();
for (let i = 1; ;i++){
let random = Math.floor(Math.random() * max + 1) ;
orderNumbers.add(random);
if (orderNumbers.size == max){
break;
}
}
return orderNumbers;}
function randomNumbers(max) {
function range(upTo) {
var result = [];
for(var i = 0; i < upTo; i++) result.push(i);
return result;
}
function shuffle(o){
for(var j, x, i = o.length; i; j = Math.floor(Math.random() * i), x = o[--i], o[i] = o[j], o[j] = x);
return o;
}
var myArr = shuffle(range(max));
return function() {
return myArr.shift();
};
}
Built a little test, try this on jsfiddle:
var randoms = randomNumbers(10),
rand = randoms(),
result = [];
while(rand != null) {
result.push(rand);
rand = randoms();
}
console.log(result);
Shuffle function courtesy of dzone.com.
This is how I achieve it using underscore.js
To get n integers from min to max values. Where n is the size argument.
var randomNonRepeatingIntFromInterval = function(min, max, size) {
var values = [];
while (values.length < size) {
values.push(Math.floor(Math.random() * ( max - min + 1) + min));
values = _.uniq(values);
}
return values;
}
Sorry this is a new answer to an old question, but this can be done more efficiently with a map. What you're after is random selection rather than non-repeating random. Non-repeating random is nonsensical.
Where _a is the collection, and r is not part of the collection, we lambda the random value r:
function aRandom(f){
var r = Math.random();
aRandom._a[r] ? aRandom(f) : f(r,aRandom._a[r] = 1);
}
aRandom._a = {};
//usage:
aRandom(function(r){ console.log(r) });
Redefine aRandom._a when the browser gets sluggish. To avoid eventual sluggishness, one should really use an UUID generation algo with sufficient entropy so that chances of repeat are effectively zero, rather than brute forcing differentiability. I chose the function name aRandom because latin prefix A- means "away from." Since the more it's used, the further away from random the output. The function produces one million unique values in 2100 ms on a Macbook.
Advantage of the above solution is no need to limit the set. As well, multiple callers can use it at the same time and assume their values are different from all other callers. This is handy for such things as noise jittering distributions with insured no overlaps.
However, it can be modified to return integers as well, so as to restrict ram use to the length supplied:
function aRandom(f,c){
var r = Math.floor(Math.random()*c);
aRandom._a[r] ? aRandom(f,c) : f(r,aRandom._a[r] = 1);
}
aRandom._a = {};
//usage:
var len = 10;
var resultset = [];
for(var i =0; i< len; i++){
aRandom(function(r){ resultset.push(r); }, len);
}
console.log(resultset);
randojs.com makes this a simple one-liner:
randoSequence(1, 10)
This will return an array of numbers from 1 through 10 in random order.
You just need to add the following to the head of your html document, and you can do pretty much whatever you want with randomness easily. Random values from arrays, random jquery elements, random properties from objects, and even preventing repetitions as I've shown here.
<script src="https://randojs.com/1.0.0.js"></script>
Just one solution for reference
const fiveNums = () => {
const ranNum = () => Math.floor(Math.random() * (10 + 1));
let current;
let arr = [];
while(arr.length < 5) {
if(arr.indexOf(current = ranNum()) === -1) {
arr.push(current);
}
}
return arr;
};
fiveNums();
In case no permutation is wanted and/or length shall be variable, here is a solution for non repeating randomized lists/arrays without if-statements:
Shuffle function:
Input:
Array or object(list) of arbitrary length
optional: last key to be filtered (Array: index number, List: String of key)
Output:
random Key
to get your random item use myArrayOrList[key]
// no repeat if old_key is provided
function myShuffle(arr_or_list, old_key = false) {
var keys = Array.from(Object.keys(arr_or_list)); //extracts keys
if (old_key != false) {
keys.splice(keys.indexOf(old_key), 1); // removes old_key from keys
};
var randomKey = keys[Math.floor(Math.random() * keys.length)]; // get random key
return randomKey;
}
//test:
const a = [10, 20, 30, 40, 50, 60];
const b = {
"a": 10,
"bla-bla bla": 20,
"d": 30,
"c": 40
};
var oldKeys_a = [];
var oldKeys_b = [];
oldKeys_a[0] = myShuffle(a);
oldKeys_b[0] = myShuffle(b);
var i;
for (i = 1; i < 10; i++) {
oldKeys_a[i] = myShuffle(a, oldKeys_a[i - 1]);
oldKeys_b[i] = myShuffle(b, oldKeys_b[i - 1]);
}
alert('oldKeys_a: ' + oldKeys_a + '; oldKeys_b: ' + oldKeys_b)
//random...
//>>> oldKeys_a: 1,3,0,0,5,0,4,5,2,3; oldKeys_b: d,a,d,bla-bla bla,a,c,d,bla-bla bla,a,d <<<
Non-repeating range random number generation with the recursive patterns.
const getRandom = (max, memo) => {
if (max != memo.length) {
const pos = Math.floor(Math.random() * max);
if (memo.includes(pos)) {
return getRandom(max, memo);
} else {
return pos;
}
}
}
const random = [];
const range = 6;
for (let index = 0; index < range; index++) {
random.push(getRandom(range, random))
}
console.log('random', random) // random (6) [5, 3, 0, 2, 1, 4]
function getRandomNumberNoRepeat(length){
let numberPick = [0,1,2,3,4,5,6,7,8,9]
return numberPick.sort(() => Math.random() -0.5).slice(0, length)}
console.log(getRandomNumberNoRepeat(3));
let display = document.getElementById("container");
let myArray = [];
let randomiser = (min, max, vals) => {
while (myArray.length < vals) {
let randNum = Math.floor(Math.random() * (max - min + 1) + min);
if (!myArray.includes(randNum)) {
myArray.push(randNum);
}
}
return (display.textContent = myArray.join(" - "));
};
randomiser(1, 35, 7);
Here is one where you can specify how many number you need.
I decided to write this code and make a chrome extension for it to use when playing the lotto. haha. Just a learning experience for me and thanks to all who contributed. I read all posts and I'm better of today than I was yesterday in understanding shuffle and random non repeating numbers.
You don't really want a lost of random numbers. Truly random numbers must be able to repeat.
Truly random number are like throwing dice. Any number can come up next.
Shuffled numbers are like drawing playing cards. Each number can come up only once.
What you are really asking for is to shuffle a list of numbers and then use the first so many numbers from the shuffled list.
Think of making a list of numbers in order, and then using the random number generator to randomly select a number from a copy of that list. Each time, put the selected number at the end of your new list and remove it from the copy of the old list, shortening that list. When you are done, the new list will contain the shuffled numbers and the copy of the old list will be empty.
Alternately, you can take the number selected and use it immediately, shortening the copy of the list by removing the used number. Because you have removed the number from the list, it can't come up again.
I have code that I am trying to refactor. Im new to javascript so Im tring to make more readable code using functions in libraries like underscore.
The function below can detect when string
contains 3 or more ordered characters such as (234, efg, LmN)
and
when string contains 3 or more repeated (lll, 444, MMm, ###)
const input = "Dfdf123125";
const myStr = input.toLowerCase();
const n = 3;
let isRepeating = false;
let isSequential = false;
for (let i = 0; i < myStr.length; i++) {
if (i + (n - 1) <= myStr.length) {
let isRepeatingTemp = false;
let isSequentialTemp = false;
for (let j = i; j < i + n; j++) {
(myStr.charCodeAt(i) === myStr.charCodeAt(j)) ? isRepeatingTemp = true: isRepeatingTemp = false;
(myStr.charCodeAt(i) === myStr.charCodeAt(j) - (n - 1)) ? isSequentialTemp = true : isSequentialTemp = false;
}
if (isRepeatingTemp) isRepeating = true;
if (isSequentialTemp) isSequential = true;
}
}
Im trying to to see if I can optimize this and make it more readable with underscore and/or even make time/space complexity better. I know this can also be done with regx but im trying to get it done without it.
Instead of the inner for loop, I chunked the string to n using Array.prototype.slice() to see ahead n characters. I used Array.prototype.indexOf() to find if it's sequential based off the abc and num constants(ref). To see if it's repeating, I used Array.prototype.every() that loops through the chunk and check if they're similar and return a boolean based on the expression.
The result gives the output of each instance found, and if it was sequential or repeating.
const input = "Dfdf123125";
function RepSeq(str, n) {
var rep = false;
var seq = false;
var result = [];
const num = '0123456789';
const abc = 'abcdefghijklmnopqrstuvqxyz';
if (str.length < n) return false;
for (var i = 0; i < str.length; i++) {
if (i + n > str.length) break;
var chunk = str.slice(i, i + n);
var seqABC = abc.indexOf(chunk) > -1;
var seq123 = num.indexOf(chunk) > -1;
if (seq123 || seqABC) {
seq = true;
result.push(chunk);
}
if ([...chunk].every(v => v.toLowerCase() === chunk[0].toLowerCase())) {
rep = true;
result.push(chunk);
}
}
return {
repetition: rep,
sequential: seq,
out: result
};
}
console.log(RepSeq(input, 3));
// Output:
// {
// out: ["123"],
// repetition: false,
// sequential: true
// }
With this method, we're peeking at the string one block(i+n) at a time. Ex(n=3):
1. [Dfd]f123125
2. D[fdf]123125
3. Df[df1]23125
4. Dfd[f12]3125
5. Dfdf[123]125 - Sequential!
6. Dfdf1[231]25
7. Dfdf12[312]5
8. Dfdf123[125]
What is the most efficient way to write a javascript loop to calculate the number of occurrences of 7's (as an example number) that will be encountered in counting from 1 to 100?
Example:
function numberOccurences(targetNumber, minNumber, maxNumber) {
var count = 0;
for (i = minNumber; i < maxNumber; i++) {
count = count + (i.toString().split(targetNumber).length - 1);
}
return count;
}
var result = numberOccurences(7,1,100);
This will do it without looking at the actual numbers. Sorry, no loop, but you did ask for effeciency. If you really want to use a loop, make the recursion an iteration.
function digitOccurences(digit, min, max, base) {
if (typeof base != "number") base = 10;
return digitOccurencesPlus(digit, max, base, 1, 0) - digitOccurencesPlus(digit, min, base, 1, 0);
function digitOccurencesPlus(digit, N, base, pow, rest) {
if (N == 0) return 0;
var lastDigit = N%base,
prevDigits = (N-lastDigit)/base;
var occsInLastDigit = pow*(prevDigits+(lastDigit>digit));
var occsOfLastInRest = rest * (lastDigit==digit);
// console.log(prevDigits+" "+lastDigit, rest, occsInLastDigit, occsOfLastInRest);
return occsInLastDigit + occsOfLastInRest + digitOccurencesPlus(digit, prevDigits, base, pow*base, pow*lastDigit+rest);
}
}
This is an interesting problem, and already has similar answers for other languages. Maybe you could try to make this one in javascript: Count the number of Ks between 0 and N
That solution is for occurences from 0 to n, but you could easily use it to calculate from a to b this way:
occurences(a,b)= occurences(0,b)-occurences(0,a)
This is much faster (x6) than my original function...JSPERF
function numberOccurences2(targetNumber, minNumber, maxNumber) {
var strMe = "";
for (i = minNumber; i < maxNumber; i++) {
strMe = strMe.concat(i);
}
var re = new RegExp(targetNumber,"g");
var num1 = strMe.length;
var num2 = strMe.replace(re, "").length;
num2 = num1- num2;
return (num2);
}
There has to be a faster way still...
In how many ways can a group of 9 people work in 3 disjoint subgroups of 2, 3 and 4 persons? How can I generates all the possibilities via backtracking with javascript.
Example:
Gs = group([aldo,beat,carla,david,evi,flip,gary,hugo,ida],[2,2,5]);
console.log(Gs); // [[aldo,beat],[carla,david],[evi,flip,gary,hugo,ida]], ...
Note that I do not want permutations of the group members; i.e. [[aldo,beat],...] is the same solution as [[beat,aldo],...]. However, there's a difference between [[aldo,beat],[carla,david],...] and [[carla,david],[aldo,beat],...].
*No libraries please.
If you only need the number of ways a group of 9 people can be divided into 3 subgroups of 2, 3 and 4 people each then that's easy to compute mathematically using C (the function to calculate the number of combinations).
First you have 9 people out of which you need to select 2 people. Hence you do C(9, 2).
Next you have 7 people out of which you need to select 3 people. Hence you do C(7, 3).
Finally you have 4 people out of which you need to select 4 people. Hence you do C(4, 4). However C(n, n) is always 1.
Hence the number of ways to divide a group of 9 people into 3 subgroups of 2, 3 and 4 people is C(9, 2) * C(7, 3) * C(4, 4). This can be simplified to C(9, 2) * C(7, 3), which is 36 * 35 which is 1260.
We can write a function to compute this for us:
function ways(n) {
var l = arguments.length, w = 1;
for (var i = 1; i < l; i++) {
var m = arguments[i];
w *= combinations(n, m);
n -= m;
}
return w;
}
To make this function work we need to define the function combinations:
function combinations(n, k) {
return factorial(n) / factorial(n - k) / factorial(k);
}
Finally we need to define the function for factorial:
function factorial(n) {
var f = n;
while (--n) f *= n;
return f;
}
Then we compute the number of ways as follows:
alert(ways(9, 2, 3)); // 1260
You can see the demo here: http://jsfiddle.net/bHSuh/
Note that we didn't need to specify the last subgroup of 4 people because that is implied.
However I believe that you want to generate each possible way. This is the sort of thing that the amb operator is perfect for. So the first thing we'll do is write the amb operator in JavaScript:
function amb(options, callback) {
var length = options.length;
for (var i = 0; i < length; i++) {
try {
callback(options[i]); // try the next option
return; // no problem, quit
} catch (e) {
continue; // problem, next
}
}
throw new Error("amb tree exhausted"); // throw a tantrum
}
Next we'll write a function which picks a given set of items from a list of indices:
function pick(list, items) {
var length = list.length, selected = [], rest = [];
for (var i = 0; i < length; i++) {
if (items.indexOf(i) < 0) rest.push(list[i]);
else selected.push(list[i]);
}
return [selected, rest];
}
We also need a function which will generate a list of indices:
function getIndices(length) {
var indices = [];
for (var i = 0; i < length; i++)
indices.push(i);
return indices;
}
Finally we'll implement the group function recursively:
function group(options, divisions) {
var subgroup = [], groups = [], n = 0;
var indices = getIndices(options.length);
var division = divisions.shift(), remaining = divisions.length;
try {
amb(indices, select);
} catch (e) {
return groups;
}
function select(index) {
subgroup.push(index);
if (++n < division) {
try { amb(indices.slice(index + 1), select); }
catch (e) { /* we want to continue processing */ }
} else {
var subgroups = pick(options, subgroup);
if (remaining) {
var children = group(subgroups.pop(), divisions.slice());
var length = children.length;
for (var i = 0; i < length; i++)
groups.push(subgroups.concat(children[i]));
} else groups.push(subgroups);
}
n--;
subgroup.pop();
throw new Error;
}
}
Now you can use it as follows:
var groups = group([
"aldo", "beat", "carla",
"david", "evi", "flip",
"gary", "hugo", "ida"
], [2, 3]);
Notice again that you didn't need to specify the last subgroup of 4 people since it's implied.
Now let's see whether the output is as we expected it to be:
console.log(groups.length === ways(9, 2, 3)); // true
There you go. There are exactly 1260 ways that a group of 9 people can be divided into 3 subgroups of 2, 3 and 4 people each.
Now I know that my group function looks a little daunting but it's actually really simple. Try to read it and understand what's going on.
Imagine that you're the boss of 9 people. How would you divide them into 3 subgroups of 2, 3 and 4 people? That's exactly the way my group function works.
If you still can't understand the logic after a while then I'll update my answer and explain the group function in detail. Best of luck.
BTW I just realized that for this problem you don't really need amb. You may simply use forEach instead. The resulting code would be faster because of the absence of try-catch blocks:
function group(options, divisions) {
var subgroup = [], groups = [], n = 0;
var indices = getIndices(options.length);
var division = divisions.shift(), remaining = divisions.length;
indices.forEach(select);
return groups;
function select(index) {
subgroup.push(index);
if (++n < division) indices.slice(index + 1).forEach(select);
else {
var subgroups = pick(options, subgroup);
if (remaining) {
var children = group(subgroups.pop(), divisions.slice());
var length = children.length;
for (var i = 0; i < length; i++)
groups.push(subgroups.concat(children[i]));
} else groups.push(subgroups);
}
subgroup.pop();
n--;
}
}
Since we don't use amb anymore the execution time of the program has decreased tenfold. See the result for yourself: http://jsperf.com/amb-vs-foreach
Also I've finally created a demo fiddle of the above program: http://jsfiddle.net/Ug6Pb/
i am sure there are faster formulas, but i was never that great at math, and this seems to work if i understand the problem correctly:
function combo(r, ops){
function unq(r){return r.filter(function(a,b,c){return !this[a] && (this[a]=1);},{}); }
var combos={}, pairs=[];
r.forEach(function(a,b,c){
combos[a]=r.filter(function not(a){return a!=this && !combos[a]}, a);
});
Object.keys(combos).forEach(function(k){
combos[k].forEach(function(a){
pairs.push([k, a]+'');
});
});
return unq(unq(
pairs.map(function(a){
return unq(a.split(",")).sort();
})).map(function(a){
return a.length==ops && a;
}).filter(Boolean))
.sort();
}//end combo
var r="aldo,beat,carla,david,evi,flip,gary,hugo,ida".split(",");
// find groups of different lengths:
combo(r, 2) // 2 folks == 36 combos
combo( combo(r, 2), 3) // 3 folks == 84 combos
combo( combo( combo(r, 2), 3), 4) // 4 folks == 126 combos
i didn't bother to recursive-ize the function since you only need to go 4-in and a lispy invocation works, but if i had to go further, i'd want to write one additional outer wrapper to sandwich the calls...
the core implementation of the backtracking algorithm is simple (see function doBacktrack below). Usually the complexity is in the details of the specific backtracking problem
the following is my implementation of a backtracking algorithm for your problem. it is based on the backtracking algorithm description in Steven Skiena's Algorithm Design Manual (or what I remember of it).
I've not added pruning to the algorithm (because it's already taken me longer than I thought it would do :) ) but if you want to improve its performance just add a reasonable implementation for the function done() to prevent continuing with the processing of candidates that can be inferred to not be viable solutions
function backtrack() {
var people =
['aldo','beat','carla','david','evi','flip','gary','hugo','ida'];
var initial_state =
[[], [], []];
var groups =
[2, 3, 4];
var data =
{groups: groups, people: people, people_idx_for_name: {}};
people.forEach(function(e, i) {
data['people_idx_for_name'][e] = i;
});
var solutions = [];
doBacktrack(initial_state, solutions, data);
return solutions;
}
function doBacktrack(candidate, solutions, data) {
// console.log('processing: ' + candidate);
if (isSolution(candidate, data)) {
processSolution(candidate, solutions);
}
if (done(candidate, solutions, data)) {
return;
}
var new_candidates = calculateNewCandidates(candidate, data);
for (var i=0; i<new_candidates.length; i++) {
doBacktrack(new_candidates[i], solutions, data);
}
}
function calculateNewCandidates(candidate, data) {
var groups = data['groups'];
var i = 0;
while (i<groups.length && candidate[i].length == groups[i]) { i++; }
if (i < groups.length) {
//determine list of not yet selected people
var not_yet_selected = determineNotYetSelectedPeople(candidate, data, i);
var results = [];
for (var j=0; j<not_yet_selected.length; j++) {
var candidate_copy = candidate.slice(0);
for (var k=0; k<candidate_copy.length; k++) {
candidate_copy[k] = candidate_copy[k].slice(0);
}
candidate_copy[i].push(not_yet_selected[j])
results.push(candidate_copy);
}
return results;
} else {
return [];
}
}
function determineNotYetSelectedPeople(candidate, data, group) {
var people = data['people'];
var people_idx_for_name = data['people_idx_for_name'];
var selected_people = {};
var results = [];
var max = -Number.MAX_VALUE;
candidate.forEach(function(candidate_group, i) {
candidate_group.forEach(function(already_selected_person_name) {
var already_selected_person_idx = people_idx_for_name[already_selected_person_name];
if (max < already_selected_person_idx && i==group) { max = already_selected_person_idx; }
selected_people[already_selected_person_name] = true;
});
});
for (var i=0; i<people.length; i++) {
if (!selected_people[people[i]] && i > max) { results.push(people[i]); }
}
return results;
}
function isSolution(candidate, data) {
var groups = data['groups'];
for (var i=0; i<groups.length; i++) {
if (candidate[i].length != groups[i]) {return false;}
}
return true;
}
function processSolution(candidate, solutions) {
var solution = [];
candidate.forEach(function(e) {
var l = [];
solution.push(l);
e.forEach(function(f) {
l.push(f);
});
});
solutions.push(solution);
}
//use this to improve performance with prunning if possible
function done() {
return false;
}
var solutions = backtrack();
console.log(solutions);
console.log(solutions.length);
For a project I'm working on, I needed a Javascript function that would return a random number, in a given range, without repeating itself until the whole range is 'depleted'. As there was no such thing around, I have managed to create it myself.
The function will also require an id to be passed. This way, if you require multiple random numbers, each with their own history, the id keeps track of them all.
The function works, however I need some advice;
is this the 'proper' way to achieve what I want to achieve?
how fast will inArray() perform with very big ranges (maxNum) values? I have a feeling that large numbers will slow the function down, as it is randomizing numbers until it generates a number that is still 'valid' (i.e. not in the history array). But I can't figure out another way to do this..
The script:
var UniqueRandom = {
NumHistory: [],
generate: function (maxNum, id) {
if (!this.NumHistory[id]) this.NumHistory[id] = [];
if (maxNum >= 1) {
var current = Math.round(Math.random() * (maxNum - 1)), x = 0;
if (maxNum > 1 && this.NumHistory[id].length > 0) {
if (this.NumHistory[id].length !== maxNum) {
while ($.inArray(current, this.NumHistory[id]) !== -1) {
current = Math.round(Math.random() * (maxNum - 1));
x = x + 1;
}
this.NumHistory[id].push(current);
} else {
//reset
this.NumHistory[id] = [current];
}
} else {
//first time only
this.NumHistory[id].push(current);
}
return current;
} else {
return maxNum;
}
},
clear: function (id) {
this.NumHistory[id] = [];
}
};
usage would be: (100 being the range (0-100) and the_id being.. well, the id)
UniqueRandom.NumHistory[100, 'the_id']
I have set up a Fiddle with a demo.
It's not best practice. Imo it would be better to instantiate an object per series of numbers that needs to be generated.
I'd suggest generating an array of all possible values and shuffling it. Then you can just pop of it.
I took Jack's code and adapted it to work with the popping array method.
function fisherYates ( myArray ) {
var i = myArray.length;
if ( i == 0 ) return false;
while ( --i ) {
var j = Math.floor( Math.random() * ( i + 1 ) );
var tempi = myArray[i];
var tempj = myArray[j];
myArray[i] = tempj;
myArray[j] = tempi;
}
}
function RandomGenerator(maxNum) {
this.max = maxNum;
this.initRandomArray();
}
RandomGenerator.prototype.generate = function() {
// if no more numbers available generate new array
if( this.left === 0 ) this.initRandomArray();
this.last = this.arr.pop();
this.left = this.arr.length;
this.history.push( this.last );
return this.last;
}
RandomGenerator.prototype.initRandomArray = function() {
this.arr = [];
this.history = [];
this.last = null;
this.left = this.max;
for( var i = 0; i < this.max; i++ ) {
this.arr.push( i );
}
fisherYates( this.arr );
}
var mygen = new RandomGenerator(100);
console.log( mygen.generate() );
I got the fisherYates algorithm from here.
The approach of generating a new random number if it is already found in a history object will result in unnecessary looping.
Fiddle here
I tend to think that it is indeed not most efficient. I dont immediately get the //first time only.
Further, you can make code more readable by skipping the else return .. and writing the condition to be the opposite, e.g.:
if (maxNum >= 1) {
//code
} else {
return maxNum;
}
becomes
if (maxNum < 1) { // or maybe even if maxNum == 0
return maxNum;
}
//code
Also your x variable seems to be redundant.
I would probably implement it like this, using actual instances of random generators. This keeps the history of each generator separated.
function RandomGenerator(maxNum)
{
this.max = maxNum;
this.history = {};
this.histn = 0;
}
// generate random number in range [0..maxNum)
RandomGenerator.prototype.generate = function()
{
var value;
if (this.histn == this.max ) {
return false;
}
do {
value = Math.floor(Math.random() * this.max );
} while (this.history[value]);
this.history['' + value] = 1;
++this.histn;
return value;
}
var mygen = new RandomGenerator(100);
console.log(mygen.generate());
In my implementation I'm choosing a plain object for the history instead of an array; testing whether a value has been generated before is done by testing a property instead of $.inArray().
I agree with Alex that in most use cases, you'd want to generate an array of all values, shuffle them, and then pop them as you need them instead.
Here is an example:
var getShuffledUniqueRandoms = function(count, suffix) {
var values = [];
for (var i = 1; i < count+1; i++) {
values.push(i + suffix);
}
// Shuffle function originally from:
//+ Jonas Raoni Soares Silva
//# http://jsfromhell.com/array/shuffle [v1.0]
return (function(o){ //v1.0
for(var j, x, i = o.length; i; j = parseInt(Math.random() * i), x = o[--i], o[i] = o[j], o[j] = x);
return o;
})(values);
}
var values = getShuffledUniqueRandoms(10, "index");
$('button').click(function() {
if (values.length == 0) {
$('body').append('<p>Out of values.</p>');
} else {
$('body').append('<p>' + values.pop() + '</p>');
}
});
FIDDLE
With this algorithm, it has a bigger upfront cost, but at least it has a known time it'll take to complete (roughly O(n)).
With the algorithm where you are constantly checking to see if a random value is in an array, it'll get slower and slower with each new iteration.
Now if your data set is always relatively small, your algorithm could work a little better, but anything larger than 10 or so and it starts losing it's edge.