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 need to send a blob to the server with ajax, but it can end up getting somewhat large, and I'd like to decrease upload time. I've tried jszip already, but that just gave me an empty file inside the zip. I've also tried btoa(), but it turns out that the encoded value just ends up being [object Blob] instead of the actual blob data. What are my options for compressing blobs?
Here was the code I used for jszip:
var zip = new JSZip();
zip.file("recording.wav", blobFile);
var content = zip.generate();
I then appended "content" to a FormData object and sent it to the server. On the server side, I decoded the POST data (from base64). The zip file opened just fine, but recording.wav was a 0 length file.
Additionally, I've tried using the LZW implementation found here. This was the additional code I used to compress it:
var compressed;
var reader = new FileReader();
reader.onload = function(event){
compressed = LZW.compress(event.target.result);
};
reader.readAsText(blobFile);
However, decompressing it returns null.
Caveat: compressing things like audio files would be better done using an algorithm meant specifically for that type of data, perhaps something lossy. However, knowing how hard it was to find a reasonable lossless implementation as provided below, I'm very concerned that it will be hard to find a good implementation in Javascript for that type of data specifically that meets your needs.
In any case, I've had this general need for compression/decompression in Javascript as well, and I needed the same algorithm to work both client (browser) and server-side (node.js) and I needed it to work on very large files. I had checked out jszip and I also tried that LZW algorithm among at least five or six others none of which satisfied the requirements. I can't remember what the issue was with each specifically, but suffice to say it is surprisingly hard to find a good and FAST compressor/decompressor in javascript that works both server and client side and handles large files! I tried at least a dozen different implementations of various compression algorithms, and finally settled with this one - it hasn't failed me yet!
UPDATE
This is the original source:
https://code.google.com/p/jslzjb/source/browse/trunk/Iuppiter.js?r=2
By someone named Bear - thanks Bear, whoever you are, you're the best.
It is LZJB: http://en.wikipedia.org/wiki/LZJB
UPDATE 2
Corrected a problem with missing semicolon - should not give the object not a function error any longer.
This implementation stops working on data less than about 80 characters in length. So I updated the example to reflect that.
Realized the base64 encode/decode methods are in fact exposed on the object passed in for this version, so...
Currently seeing what we can do about specific blob types - what for example the best approach would be for a image versus audio etc as that would be useful for JS folks in general... will update here with what is found.
UPDATE 3
There is a much better wrapper around the original Iuppiter source from Bear than the one I posted below. It is written by cscott and on github here: https://github.com/cscott/lzjb
I'll be switching to this one, as it does streams as well.
Below is an example in Node.js of its use with a wav file. But before copying the example, let me give you the terrible news first, at least for this one wav file that I tried:
63128 Jun 19 14:09 beep-1.wav
63128 Jun 19 17:47 beep-2.wav
89997 Jun 19 17:47 beep-2.wav.compressed
So it successfully regenerated the wav (and it played). However, the compressed one appears to be larger than the original. Well shoot. In any case, might be good to try on your data, you never know, you might get lucky. Here's the code I used:
var fs = require('fs');
var lzjb = require('lzjb');
fs.readFile('beep-1.wav', function(err, wav){
// base 64 first
var encoded = wav.toString('base64');
// then utf8 - you don't want to go utf-8 directly
var data = new Buffer(encoded, 'utf8');
// now compress
var compressed = lzjb.compressFile(data, null, 9);
// the next two lines are unnecessary, but to see what kind of
// size is written to disk to compare with the original binary file
var compressedBuffer = new Buffer(compressed, 'binary');
fs.writeFile('beep-2.wav.compressed', compressedBuffer, 'binary', function(err) {});
// decompress
var uncompressed = lzjb.decompressFile(compressed);
// decode from utf8 back to base64
var encoded2 = new Buffer(uncompressed).toString('utf8');
// decode back to binary original from base64
var decoded = new Buffer(encoded2, 'base64');
// write it out, make sure it is identical
fs.writeFile('beep-2.wav', decoded, function(err) {});
});
At the end of the day, I think its going to be too difficult to achieve any level of compression on most forms of binary data that isn't clobbered by the resulting base64 encoding. The days of control characters for terminals still haunt us to this day. You could try upping to a different base, but that has its risks and issues as well.
See this for example:
What is the most efficient binary to text encoding?
And this:
Why don't people use base128?
One thing though, definitely before you accept the answer, please please try it out on your blob, I've mainly used it for compressing utf-8, and I'd like to be sure it works on your specific data.
In any case, here it is!
/**
$Id: Iuppiter.js 3026 2010-06-23 10:03:13Z Bear $
Copyright (c) 2010 Nuwa Information Co., Ltd, and individual contributors.
All rights reserved.
Redistribution and use in source and binary forms, with or without
modification, are permitted provided that the following conditions are met:
1. Redistributions of source code must retain the above copyright notice,
this list of conditions and the following disclaimer.
2. Redistributions in binary form must reproduce the above copyright
notice, this list of conditions and the following disclaimer in the
documentation and/or other materials provided with the distribution.
3. Neither the name of Nuwa Information nor the names of its contributors
may be used to endorse or promote products derived from this software
without specific prior written permission.
THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE
DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT OWNER OR CONTRIBUTORS BE LIABLE
FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR
SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER
CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY,
OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE
OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
$Author: Bear $
$Date: 2010-06-23 18:03:13 +0800 (星期三, 23 六月 2010) $
$Revision: 3026 $
*/
var fastcompressor = {};
(function (k) {
k.toByteArray = function (c) {
var h = [],
b, a;
for (b = 0; b < c.length; b++) a = c.charCodeAt(b), 127 >= a ? h.push(a) : (2047 >= a ? h.push(a >> 6 | 192) : (65535 >= a ? h.push(a >> 12 | 224) : (h.push(a >> 18 | 240), h.push(a >> 12 & 63 | 128)), h.push(a >> 6 & 63 | 128)), h.push(a & 63 | 128));
return h
};
k.Base64 = {
CA: "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/",
CAS: "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789-_",
IA: Array(256),
IAS: Array(256),
init: function () {
var c;
for (c = 0; 256 > c; c++) k.Base64.IA[c] = -1, k.Base64.IAS[c] = -1;
c = 0;
for (iS = k.Base64.CA.length; c < iS; c++) k.Base64.IA[k.Base64.CA.charCodeAt(c)] = c, k.Base64.IAS[k.Base64.CAS.charCodeAt(c)] = c;
k.Base64.IA["="] = k.Base64.IAS["="] = 0
},
encode: function (c, h) {
var b, a, d, e, m, g, f, l, j;
b = h ? k.Base64.CAS : k.Base64.CA;
d = c.constructor == Array ? c : k.toByteArray(c);
e = d.length;
m = 3 * (e / 3);
g = (e - 1) / 3 + 1 << 2;
a = Array(g);
for (l = f = 0; f < m;) j = (d[f++] & 255) << 16 | (d[f++] & 255) << 8 | d[f++] & 255, a[l++] = b.charAt(j >> 18 & 63), a[l++] = b.charAt(j >> 12 & 63), a[l++] = b.charAt(j >> 6 & 63), a[l++] = b.charAt(j & 63);
f = e - m;
0 < f && (j = (d[m] &
255) << 10 | (2 == f ? (d[e - 1] & 255) << 2 : 0), a[g - 4] = b.charAt(j >> 12), a[g - 3] = b.charAt(j >> 6 & 63), a[g - 2] = 2 == f ? b.charAt(j & 63) : "=", a[g - 1] = "=");
return a.join("")
},
decode: function (c, h) {
var b, a, d, e, m, g, f, l, j, p, q, n;
b = h ? k.Base64.IAS : k.Base64.IA;
c.constructor == Array ? (d = c, m = !0) : (d = k.toByteArray(c), m = !1);
e = d.length;
g = 0;
for (f = e - 1; g < f && 0 > b[d[g]];) g++;
for (; 0 < f && 0 > b[d[f]];) f--;
l = "=" == d[f] ? "=" == d[f - 1] ? 2 : 1 : 0;
a = f - g + 1;
j = 76 < e ? ("\r" == d[76] ? a / 78 : 0) << 1 : 0;
e = (6 * (a - j) >> 3) - l;
a = Array(e);
q = p = 0;
for (eLen = 3 * (e / 3); p < eLen;) n = b[d[g++]] << 18 | b[d[g++]] <<
12 | b[d[g++]] << 6 | b[d[g++]], a[p++] = n >> 16 & 255, a[p++] = n >> 8 & 255, a[p++] = n & 255, 0 < j && 19 == ++q && (g += 2, q = 0);
if (p < e) {
for (j = n = 0; g <= f - l; j++) n |= b[d[g++]] << 18 - 6 * j;
for (b = 16; p < e; b -= 8) a[p++] = n >> b & 255
}
if (m) return a;
for (n = 0; n < a.length; n++) a[n] = String.fromCharCode(a[n]);
return a.join("")
}
};
k.Base64.init();
NBBY = 8;
MATCH_BITS = 6;
MATCH_MIN = 3;
MATCH_MAX = (1 << MATCH_BITS) + (MATCH_MIN - 1);
OFFSET_MASK = (1 << 16 - MATCH_BITS) - 1;
LEMPEL_SIZE = 256;
k.compress = function (c) {
var h = [],
b, a = 0,
d = 0,
e, m, g = 1 << NBBY - 1,
f, l, j = Array(LEMPEL_SIZE);
for (b = 0; b < LEMPEL_SIZE; b++) j[b] =
3435973836;
c = c.constructor == Array ? c : k.toByteArray(c);
for (b = c.length; a < b;) {
if ((g <<= 1) == 1 << NBBY) {
if (d >= b - 1 - 2 * NBBY) {
f = b;
for (d = a = 0; f; f--) h[d++] = c[a++];
break
}
g = 1;
m = d;
h[d++] = 0
}
if (a > b - MATCH_MAX) h[d++] = c[a++];
else if (e = (c[a] + 13 ^ c[a + 1] - 13 ^ c[a + 2]) & LEMPEL_SIZE - 1, l = a - j[e] & OFFSET_MASK, j[e] = a, e = a - l, 0 <= e && e != a && c[a] == c[e] && c[a + 1] == c[e + 1] && c[a + 2] == c[e + 2]) {
h[m] |= g;
for (f = MATCH_MIN; f < MATCH_MAX && c[a + f] == c[e + f]; f++);
h[d++] = f - MATCH_MIN << NBBY - MATCH_BITS | l >> NBBY;
h[d++] = l;
a += f
} else h[d++] = c[a++]
}
return h
};
k.decompress = function (c,
h) {
var b, a = [],
d, e = 0,
m = 0,
g, f, l = 1 << NBBY - 1,
j;
b = c.constructor == Array ? c : k.toByteArray(c);
for (d = b.length; e < d;) {
if ((l <<= 1) == 1 << NBBY) l = 1, f = b[e++];
if (f & l)
if (j = (b[e] >> NBBY - MATCH_BITS) + MATCH_MIN, g = (b[e] << NBBY | b[e + 1]) & OFFSET_MASK, e += 2, 0 <= (g = m - g))
for (; 0 <= --j;) a[m++] = a[g++];
else break;
else a[m++] = b[e++]
}
if (!("undefined" == typeof h ? 0 : h)) {
for (b = 0; b < m; b++) a[b] = String.fromCharCode(a[b]);
a = a.join("")
}
return a
}
})(fastcompressor);
And if memory serves... here's how you use it:
var compressed = fastcompressor.compress("0123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789"); // data less than this length poses issues.
var decompressed = fastcompressor.decompress(compressed);
Rgds....Hoonto/Matt
Also, what I've posted is minified but beautified, and very slightly adapted for ease-of-use. Check the link in the update above for the original stuff.
JS Zip will work fine just correct your syntax..
function create_zip() {
var zip = new JSZip();
zip.add("recording.wav", blobfile);//here you have to give blobFile in the form of raw bits >> convert it in json notation.. or stream ..
zip.add("hello2.txt", "Hello Second World\n");//this is optional..
content = zip.generate();
location.href="data:application/zip;base64," + content;
}
you can add multiple files too..
Just zip.file will become zip.add
and then zip.generate() will do the rest.. as you have done,
or refer old Post its last part of JavaScript, and NativeBridge will be helpful if you can utilize, in this post user records using Objective C that you can ignore, but sends this object using JavaScript and Socket that you can/may utilize..
I hope this will do ... :)
What is the best way of implementing a bit array in JavaScript?
Here's one I whipped up:
UPDATE - something about this class had been bothering me all day - it wasn't size based - creating a BitArray with N slots/bits was a two step operation - instantiate, resize. Updated the class to be size based with an optional second paramter for populating the size based instance with either array values or a base 10 numeric value.
(Fiddle with it here)
/* BitArray DataType */
// Constructor
function BitArray(size, bits) {
// Private field - array for our bits
this.m_bits = new Array();
//.ctor - initialize as a copy of an array of true/false or from a numeric value
if (bits && bits.length) {
for (var i = 0; i < bits.length; i++)
this.m_bits.push(bits[i] ? BitArray._ON : BitArray._OFF);
} else if (!isNaN(bits)) {
this.m_bits = BitArray.shred(bits).m_bits;
}
if (size && this.m_bits.length != size) {
if (this.m_bits.length < size) {
for (var i = this.m_bits.length; i < size; i++) {
this.m_bits.push(BitArray._OFF);
}
} else {
for(var i = size; i > this.m_bits.length; i--){
this.m_bits.pop();
}
}
}
}
/* BitArray PUBLIC INSTANCE METHODS */
// read-only property - number of bits
BitArray.prototype.getLength = function () { return this.m_bits.length; };
// accessor - get bit at index
BitArray.prototype.getAt = function (index) {
if (index < this.m_bits.length) {
return this.m_bits[index];
}
return null;
};
// accessor - set bit at index
BitArray.prototype.setAt = function (index, value) {
if (index < this.m_bits.length) {
this.m_bits[index] = value ? BitArray._ON : BitArray._OFF;
}
};
// resize the bit array (append new false/0 indexes)
BitArray.prototype.resize = function (newSize) {
var tmp = new Array();
for (var i = 0; i < newSize; i++) {
if (i < this.m_bits.length) {
tmp.push(this.m_bits[i]);
} else {
tmp.push(BitArray._OFF);
}
}
this.m_bits = tmp;
};
// Get the complimentary bit array (i.e., 01 compliments 10)
BitArray.prototype.getCompliment = function () {
var result = new BitArray(this.m_bits.length);
for (var i = 0; i < this.m_bits.length; i++) {
result.setAt(i, this.m_bits[i] ? BitArray._OFF : BitArray._ON);
}
return result;
};
// Get the string representation ("101010")
BitArray.prototype.toString = function () {
var s = new String();
for (var i = 0; i < this.m_bits.length; i++) {
s = s.concat(this.m_bits[i] === BitArray._ON ? "1" : "0");
}
return s;
};
// Get the numeric value
BitArray.prototype.toNumber = function () {
var pow = 0;
var n = 0;
for (var i = this.m_bits.length - 1; i >= 0; i--) {
if (this.m_bits[i] === BitArray._ON) {
n += Math.pow(2, pow);
}
pow++;
}
return n;
};
/* STATIC METHODS */
// Get the union of two bit arrays
BitArray.getUnion = function (bitArray1, bitArray2) {
var len = BitArray._getLen(bitArray1, bitArray2, true);
var result = new BitArray(len);
for (var i = 0; i < len; i++) {
result.setAt(i, BitArray._union(bitArray1.getAt(i), bitArray2.getAt(i)));
}
return result;
};
// Get the intersection of two bit arrays
BitArray.getIntersection = function (bitArray1, bitArray2) {
var len = BitArray._getLen(bitArray1, bitArray2, true);
var result = new BitArray(len);
for (var i = 0; i < len; i++) {
result.setAt(i, BitArray._intersect(bitArray1.getAt(i), bitArray2.getAt(i)));
}
return result;
};
// Get the difference between to bit arrays
BitArray.getDifference = function (bitArray1, bitArray2) {
var len = BitArray._getLen(bitArray1, bitArray2, true);
var result = new BitArray(len);
for (var i = 0; i < len; i++) {
result.setAt(i, BitArray._difference(bitArray1.getAt(i), bitArray2.getAt(i)));
}
return result;
};
// Convert a number into a bit array
BitArray.shred = function (number) {
var bits = new Array();
var q = number;
do {
bits.push(q % 2);
q = Math.floor(q / 2);
} while (q > 0);
return new BitArray(bits.length, bits.reverse());
};
/* BitArray PRIVATE STATIC CONSTANTS */
BitArray._ON = 1;
BitArray._OFF = 0;
/* BitArray PRIVATE STATIC METHODS */
// Calculate the intersection of two bits
BitArray._intersect = function (bit1, bit2) {
return bit1 === BitArray._ON && bit2 === BitArray._ON ? BitArray._ON : BitArray._OFF;
};
// Calculate the union of two bits
BitArray._union = function (bit1, bit2) {
return bit1 === BitArray._ON || bit2 === BitArray._ON ? BitArray._ON : BitArray._OFF;
};
// Calculate the difference of two bits
BitArray._difference = function (bit1, bit2) {
return bit1 === BitArray._ON && bit2 !== BitArray._ON ? BitArray._ON : BitArray._OFF;
};
// Get the longest or shortest (smallest) length of the two bit arrays
BitArray._getLen = function (bitArray1, bitArray2, smallest) {
var l1 = bitArray1.getLength();
var l2 = bitArray2.getLength();
return l1 > l2 ? smallest ? l2 : l1 : smallest ? l2 : l1;
};
CREDIT TO #Daniel Baulig for asking for the refactor from quick and dirty to prototype based.
I don't know about bit arrays, but you can make byte arrays easy with new features.
Look up typed arrays. I've used these in both Chrome and Firefox. The important one is Uint8Array.
To make an array of 512 uninitialized bytes:
var arr = new UintArray(512);
And accessing it (the sixth byte):
var byte = arr[5];
For node.js, use Buffer (server-side).
EDIT:
To access individual bits, use bit masks.
To get the bit in the one's position, do num & 0x1
The Stanford Javascript Crypto Library (SJCL) provides a Bit Array implementation and can convert different inputs (Hex Strings, Byte Arrays, etc.) to Bit Arrays.
Their code is public on GitHub: bitwiseshiftleft/sjcl. So if you lookup bitArray.js, you can find their bit array implementation.
A conversion from bytes to bits can be found here.
Something like this is as close as I can think of. Saves bit arrays as 32 bit numbers, and has a standard array backing it to handle larger sets.
class bitArray {
constructor(length) {
this.backingArray = Array.from({length: Math.ceil(length/32)}, ()=>0)
this.length = length
}
get(n) {
return (this.backingArray[n/32|0] & 1 << n % 32) > 0
}
on(n) {
this.backingArray[n/32|0] |= 1 << n % 32
}
off(n) {
this.backingArray[n/32|0] &= ~(1 << n % 32)
}
toggle(n) {
this.backingArray[n/32|0] ^= 1 << n % 32
}
forEach(callback) {
this.backingArray.forEach((number, container)=>{
const max = container == this.backingArray.length-1 ? this.length%32 : 32
for(let x=0; x<max; x++) {
callback((number & 1<<x)>0, 32*container+x)
}
})
}
}
let bits = new bitArray(10)
bits.get(2) //false
bits.on(2)
bits.get(2) //true
bits.forEach(console.log)
/* outputs:
false
false
true
false
false
false
false
false
false
false
*/
bits.toggle(2)
bits.forEach(console.log)
/* outputs:
false
false
false
false
false
false
false
false
false
false
*/
bits.toggle(0)
bits.toggle(1)
bits.toggle(2)
bits.off(2)
bits.off(3)
bits.forEach(console.log)
/* outputs:
true
true
false
false
false
false
false
false
false
false
*/
2022
As can be seen from past answers and comments, the question of "implementing a bit array" can be understood in two different (non-exclusive) ways:
an array that takes 1-bit in memory for each entry
an array on which bitwise operations can be applied
As #beatgammit points out, ecmascript specifies typed arrays, but bit arrays are not part of it. I have just published #bitarray/typedarray, an implementation of typed arrays for bits, that emulates native typed arrays and takes 1 bit in memory for each entry.
Because it reproduces the behaviour of native typed arrays, it does not include any bitwise operations though. So, I have also published #bitarray/es6, which extends the previous with bitwise operations.
I wouldn't debate what is the best way of implementing bit array, as per the asked question, because "best" could be argued at length, but those are certainly some way of implementing bit arrays, with the benefit that they behave like native typed arrays.
import BitArray from "#bitarray/es6"
const bits1 = BitArray.from("11001010");
const bits2 = BitArray.from("10111010");
for (let bit of bits1.or(bits2)) console.log(bit) // 1 1 1 1 1 0 1 0
You can easily do that by using bitwise operators. It's quite simple.
Let's try with the number 75.
Its representation in binary is 100 1011. So, how do we obtain each bit from the number?
You can use an AND "&" operator to select one bit and set the rest of them to 0. Then with a Shift operator, you remove the rest of 0 that doesn't matter at the moment.
Example:
Let's do an AND operation with 4 (000 0010)
0100 1011 & 0000 0010 => 0000 0010
Now we need to filter the selected bit, in this case, was the second-bit reading right to left.
0000 0010 >> 1 => 1
The zeros on the left are no representative. So the output will be the bit we selected, in this case, the second one.
var word=75;
var res=[];
for(var x=7; x>=0; x--){
res.push((word&Math.pow(2,x))>>x);
}
console.log(res);
The output:
Expected:
In case you need more than a simple number, you can apply the same function for a byte. Let's say you have a file with multiple bytes. So, you can decompose that file in a ByteArray, then each byte in the array in a BitArray.
Good luck!
#Commi's implementation is what I ended up using .
I believe there is a bug in this implementation. Bits on every 31st boundary give the wrong result. (ie when index is (32 * index - 1), so 31, 63, 95 etc.
I fixed it in the get() method by replacing > 0 with != 0.
get(n) {
return (this.backingArray[n/32|0] & 1 << n % 32) != 0
}
The reason for the bug is that the ints are 32-bit signed. Shifting 1 left by 31 gets you a negative number. Since the check is for >0, this will be false when it should be true.
I wrote a program to prove the bug before, and the fix after. Will post it running out of space.
for (var i=0; i < 100; i++) {
var ar = new bitArray(1000);
ar.on(i);
for(var j=0;j<1000;j++) {
// we should have TRUE only at one position and that is "i".
// if something is true when it should be false or false when it should be true, then report it.
if(ar.get(j)) {
if (j != i) console.log('we got a bug at ' + i);
}
if (!ar.get(j)) {
if (j == i) console.log('we got a bug at ' + i);
}
}
}
2022
We can implement a BitArray class which behaves similar to TypedArrays by extending DataView. However in order to avoid the cost of trapping the direct accesses to the numerical properties (the indices) by using a Proxy, I believe it's best to stay in DataView domain. DataView is preferable to TypedArrays these days anyway as it's performance is highly improved in recent V8 versions (v7+).
Just like TypedArrays, BitArray will have a predetermined length at construction time. I just include a few methods in the below snippet. The popcnt property very efficiently returns the total number of 1s in BitArray. Unlike normal arrays popcnt is a highly sought after functionality for BitArrays. So much so that Web Assembly and even modern CPU's have a dedicated pop count instruction. Apart from these you can easily add methods like .forEach(), .map() etc. if need be.
class BitArray extends DataView{
constructor(n,ab){
if (n > 1.5e10) throw new Error("BitArray size can not exceed 1.5e10");
super(ab instanceof ArrayBuffer ? ab
: new ArrayBuffer(Number((BigInt(n + 31) & ~31n) >> 3n))); // Sets ArrayBuffer.byteLength to multiples of 4 bytes (32 bits)
}
get length(){
return this.buffer.byteLength << 3;
}
get popcount(){
var m1 = 0x55555555,
m2 = 0x33333333,
m4 = 0x0f0f0f0f,
h01 = 0x01010101,
pc = 0,
x;
for (var i = 0, len = this.buffer.byteLength >> 2; i < len; i++){
x = this.getUint32(i << 2);
x -= (x >> 1) & m1; //put count of each 2 bits into those 2 bits
x = (x & m2) + ((x >> 2) & m2); //put count of each 4 bits into those 4 bits
x = (x + (x >> 4)) & m4; //put count of each 8 bits into those 8 bits
pc += (x * h01) >> 56;
}
return pc;
}
// n >> 3 is Math.floor(n/8)
// n & 7 is n % 8
and(bar){
var len = Math.min(this.buffer.byteLength,bar.buffer.byteLength),
res = new BitArray(len << 3);
for (var i = 0; i < len; i += 4) res.setUint32(i,this.getUint32(i) & bar.getUint32(i));
return res;
}
at(n){
return this.getUint8(n >> 3) & (1 << (n & 7)) ? 1 : 0;
}
or(bar){
var len = Math.min(this.buffer.byteLength,bar.buffer.byteLength),
res = new BitArray(len << 3);
for (var i = 0; i < len; i += 4) res.setUint32(i,this.getUint32(i) | bar.getUint32(i));
return res;
}
not(){
var len = this.buffer.byteLength,
res = new BitArray(len << 3);
for (var i = 0; i < len; i += 4) res.setUint32(i,~(this.getUint32(i) >> 0));
return res;
}
reset(n){
this.setUint8(n >> 3, this.getUint8(n >> 3) & ~(1 << (n & 7)));
}
set(n){
this.setUint8(n >> 3, this.getUint8(n >> 3) | (1 << (n & 7)));
}
slice(a = 0, b = this.length){
return new BitArray(b-a,this.buffer.slice(a >> 3, b >> 3));
}
toggle(n){
this.setUint8(n >> 3, this.getUint8(n >> 3) ^ (1 << (n & 7)));
}
toString(){
return new Uint8Array(this.buffer).reduce((p,c) => p + ((BigInt(c)* 0x0202020202n & 0x010884422010n) % 1023n).toString(2).padStart(8,"0"),"");
}
xor(bar){
var len = Math.min(this.buffer.byteLength,bar.buffer.byteLength),
res = new BitArray(len << 3);
for (var i = 0; i < len; i += 4) res.setUint32(i,this.getUint32(i) ^ bar.getUint32(i));
return res;
}
}
Just do like
var u = new BitArray(12);
I hope it helps.
Probably [definitely] not the most efficient way to do this, but a string of zeros and ones can be parsed as a number as a base 2 number and converted into a hexadecimal number and finally a buffer.
const bufferFromBinaryString = (binaryRepresentation = '01010101') =>
Buffer.from(
parseInt(binaryRepresentation, 2).toString(16), 'hex');
Again, not efficient; but I like this approach because of the relative simplicity.
Thanks for a wonderfully simple class that does just what I need.
I did find a couple of edge-case bugs while testing:
get(n) {
return (this.backingArray[n/32|0] & 1 << n % 32) != 0
// test of > 0 fails for bit 31
}
forEach(callback) {
this.backingArray.forEach((number, container)=>{
const max = container == this.backingArray.length-1 && this.length%32
? this.length%32 : 32;
// tricky edge-case: at length-1 when length%32 == 0,
// need full 32 bits not 0 bits
for(let x=0; x<max; x++) {
callback((number & 1<<x)!=0, 32*container+x) // see fix in get()
}
})
My final implementation fixed the above bugs and changed the backArray to be a Uint8Array instead of Array, which avoids signed int bugs.