I'm trying to resolve a small programming challenge, which is calculating the nth number of the Golomb's sequence (see this for more help). I've written a simple solution, but it may have any problem, because the number at 2500000 position is 10813 but my program gives me 10814.
var golomb = (function(){
var cache = [null, 1];
const o = 0.5 * (1 + Math.sqrt(5)); // Golden ratio
return function(n){
return cache[n] || (function(){
return Math.round(Math.pow(o, 2-o) * Math.pow(n, o-1));
})();
};
})();
var num = golomb(process.argv[2]);
console.log(num);
Maybe, the golden ratio needs more lenght than JavaScript gives. Someone can help? Thanks.
For what it's worth, here is a function based on the recurrence relation, with a cache, that gives the correct answer pretty quickly
var golomb = (function() {
var cache = [null, 1];
return function(n) {
var i;
for (i=cache.length;i<n;i++) cache[i]=golomb(i);
return cache[n] || (cache[n]=1+golomb(n-golomb(golomb(n-1))));
}
})();
check it up on jsFiddle
Sgmonda, the formula you got from Wolfram Alpha isn't an exact solution. I've actually complained to them, since I like the Golomb sequence. The recurrence relation is exact but slow, even if you cache it. Heh, that's why the programming challenge is a challenge.
From the wikipedia article:
Colin Mallows has given an explicit recurrence relation:
a(1) = 1;
a(n + 1) = 1 + a(n + 1 − a(a(n)))
You need to implement your solution in this iterative method that uses integers.
A quick attempt at trying to implement that gives:
function golomb(n) {
if(n == 1) return 1;
else return 1 + golomb(n − golomb(golomb(n-1)));
}
Related
Problem statement: I'm trying to get string > binary without using the inbuilt method in javascript.
This is a piece of program where a string input (like "ABC") is accepted, then it is translated to an array of equivalent code value ([65,66,67]).
Function binary() will change a number to binary. But I'm unable to join them together to loop through all the contents. Please help. (I'm a noob, please forgive my bad code and bad explanation)
var temp3 = [65,66,67];
var temp2 = [];
var r;
for(i=0;i<temp3.length;i++) {
var r = temp3[i];
temp2.push(binary(r));
}
function binary(r) {
if (r === 0) return;
temp2.unshift(r % 2);
binary(Math.floor(r / 2));
return temp2;
}
console.log(temp2);
I think this is a cleaner version of this function. It should work for any non-negative integers, and would be easy enough to extend to the negatives. If we have a single binary digit (0 or 1) and hence are less than 2, we just return the number converted to a string. Otherwise we call recursively on the floor of half the number (as yours does) and append the final digit.
const binary = (n) =>
n < 2
? String (n)
: binary (Math.floor (n / 2)) + (n % 2)
console.log (binary(22)) //=> '10110'
console.log ([65, 66, 67] .map (binary)) //=> ['1000001', '1000010', '1000011']
In your function you have this code
var r = temp3[i];
I don't see any temp3 variable anywhere in your code above so I'd imagine that could be causing some issues.
I am not sure my issue is related to programming or related to concept of LLL algorithm and what has been mentioned on Wikipedia.
I decided to implement LLL algorithm as it has been written on Wikipedia (step-by-step / line-by-line) to actually learn the algorithm and make sure it is truly working but I am getting unexpected or invalid results.
So, I used JavaScript (programming language) and node.js (JavaScript engine) to implement it and this is the git repository to get the complete code.
Long story short, value of K gets out of range, for example when we have only 3 vectors (array size is 3, thus maximum value of index would be 2), but k becomes 3 and it is nonsense.
My code is step-by-step (line-by-line) implementation of the algorithm mentioned on Wikipedia and what I did was only implementing it. So I don't what is the issue.
// ** important
// {b} set of vectors are denoted by this.matrix_before
// {b*} set of vectors are denoted by this.matrix_after
calculate_LLL() {
this.matrix_after = new gs(this.matrix_before, false).matrix; // initialize after vectors: perform Gram-Schmidt, but do not normalize
var flag = false; // invariant
var k = 1;
while (k <= this.dimensions && !flag) {
for (var j = k - 1; j >= 0; j--) {
if (Math.abs(this.mu(k, j)) > 0.5) {
var to_subtract = tools.multiply(Math.round(this.mu(k, j)), this.matrix_before[j], this.dimensions);
this.matrix_before[k] = tools.subtract(this.matrix_before[k], to_subtract, this.dimensions);
this.matrix_after = new gs(this.matrix_before, false).matrix; // update after vectors: perform Gram-Schmidt, but do not normalize
}
}
if (tools.dot_product(this.matrix_after[k], this.matrix_after[k], this.dimensions) >= (this.delta - Math.pow(this.mu(k, k - 1), 2)) * tools.dot_product(this.matrix_after[k - 1], this.matrix_after[k - 1], this.dimensions)) {
if (k + 1 >= this.dimensions) { // invariant: there is some issue, something is wrong
flag = true; // invariant is broken
console.log("something bad happened ! (1)");
}
k++;
// console.log("if; k, j");
// console.log(k + ", " + j);
} else {
var temp_matrix = this.matrix_before[k];
this.matrix_before[k] = this.matrix_before[k - 1];
this.matrix_before[k - 1] = temp_matrix;
this.matrix_after = new gs(this.matrix_before, false).matrix; // update after vectors: perform Gram-Schmidt, but do not normalize
if (k === Math.max(k - 1, 1) || k >= this.dimensions || Math.max(k - 1, 1) >= this.dimensions) { // invariant: there is some issue, something is wrong
flag = true; // invariant is broken
console.log("something bad happened ! (2)");
}
k = Math.max(k - 1, 1);
// console.log("else; k, j");
// console.log(k + ", " + j);
}
console.log(this.matrix_before);
console.log("\n");
} // I added this flag variable to prevent getting exceptions and terminate the loop gracefully
console.log("final: ");
console.log(this.matrix_before);
}
// calculated mu as been mentioned on Wikipedia
// mu(i, j) = <b_i, b*_j> / <b*_j, b*_j>
mu(i, j) {
var top = tools.dot_product(this.matrix_before[i], this.matrix_after[j], this.dimensions);
var bottom = tools.dot_product(this.matrix_after[j], this.matrix_after[j], this.dimensions);
return top / bottom;
}
Here is the screenshot of the algorithm that is on Wikipedia:
Update #1: I added more comments to the code to clarify the question hoping that someone would help.
Just in case you are wondering about the already available implementation of the code, you can type: LatticeReduce[{{0,1},{2,0}}] wolfram alpha to see how this code suppose to behave.
Update #2: I cleaned up the code more and added a validate function to make Gram Schmidt code is working correctly, but still code fails and value of k exceeds number of dimensions (or number of vectors) which doesn't make sense.
The algorithm description in Wikipedia uses rather odd notation -- the vectors are numbered 0..n (rather than, say, 0..n-1 or 1..n), so the total number of vectors is n+1.
The code you've posted here treats this.dimensions as if it corresponds to n in the Wikipedia description. Nothing wrong with that so far.
However, the constructor in the full source file on GitHub sets this.dimensions = matrix[0].length. Two things about this look wrong. The first is that surely matrix[0].length is more like m (the dimension of the space) than n (the number of vectors, minus 1 for unclear reasons). The second is that if it's meant to be n then you need to subtract 1 because the number of vectors is n+1, not n.
So if you want to use this.dimensions to mean n, I think you need to initialize it as matrix.length-1. With the square matrix in your test case, using matrix[0].length-1 would work, but I think the code will then break when you feed in a non-square matrix. The name dimensions is kinda misleading, too; maybe just n to match the Wikipedia description?
Or you could call it something like nVectors, let it equal matrix.length, and change the rest of the code appropriately, which just means an adjustment in the termination condition for the main loop.
I am developing a JavaScript application and I needed a recursive algorithm for the longest common subsequence, so I went here and tried this one out.
It goes like this:
function lcs(a, b) {
var aSub = a.substr(0, a.length - 1);
var bSub = b.substr(0, b.length - 1);
if (a.length === 0 || b.length === 0) {
return '';
} else if (a.charAt(a.length - 1) === b.charAt(b.length - 1)) {
return lcs(aSub, bSub) + a.charAt(a.length - 1);
} else {
var x = lcs(a, bSub);
var y = lcs(aSub, b);
return (x.length > y.length) ? x : y;
}
}
It worked fine with the few test cases i tried until now, but I found that it loops on the following test case:
a: This entity works ok
b: This didn't work ok but should after
It also loops with:
a: This entity works ok
b: This didn't work as well
which at some point should get in the middle branch.
I have noticed that it is a translation of a Java version (here) of the same algorithm. It goes like this:
public static String lcs(String a, String b){
int aLen = a.length();
int bLen = b.length();
if(aLen == 0 || bLen == 0){
return "";
}else if(a.charAt(aLen-1) == b.charAt(bLen-1)){
return lcs(a.substring(0,aLen-1),b.substring(0,bLen-1))
+ a.charAt(aLen-1);
}else{
String x = lcs(a, b.substring(0,bLen-1));
String y = lcs(a.substring(0,aLen-1), b);
return (x.length() > y.length()) ? x : y;
}
}
I supposed that the JavaScript translation was wrong assuming that String.substr() and String.substring() were the same (which they aren't).
To be sure that it wasn't the case, I tried the Java one on the same test case here.
Guess what? Also the java version does not end.
I am struggling to debug it, as it is recursive.
Anyone has any idea on what is going wrong with it?
As others have pointed out in the comments, the program itself is correct. The issue you are experiencing is due that, in this implementation, the code has an exponential time complexity, and therefore takes A LONG time to run with your example input. If you let it run for a LONG time, it will return the correct result.
As others have also pointed out in the comments, LCS between two Strings is solvable with a lower time complexity using dynamic programming, which will solve it much quicker. Refer to the internet for more help (wikipedia ) or, better, try to solve it yourself thinking about the fact that there are, for each String of length n, exactly N^2 substrings. You can trivially solve it in N^2*M^2 (n m are the lengths of the two strings) by just checking if any substring of a is present in b. Ask yourself if you can do better for exercise? If yes how, if no, why.
UPDATED:
Using javascript or jQuery, how can I convert a number into it's different variations:
eg:
1000000
to...
1,000,000 or 1000K
OR
1000
to...
1,000 or 1K
OR
1934 and 1234
to...
1,934 or -2K (under 2000 but over 1500)
or
1,234 or 1k+ (over 1000 but under 1500)
Can this is done in a function?
Hope this make sense.
C
You can add methods to Number.prototype, so for example:
Number.prototype.addCommas = function () {
var intPart = Math.round(this).toString();
var decimalPart = (this - Math.round(this)).toString();
// Remove the "0." if it exists
if (decimalPart.length > 2) {
decimalPart = decimalPart.substring(2);
} else {
// Otherwise remove it altogether
decimalPart = '';
}
// Work through the digits three at a time
var i = intPart.length - 3;
while (i > 0) {
intPart = intPart.substring(0, i) + ',' + intPart.substring(i);
i = i - 3;
}
return intPart + decimalPart;
};
Now you can call this as var num = 1000; num.addCommas() and it will return "1,000". That's just an example, but you'll find that all the functions create will involve converting the numbers to strings early in the process then processing and returning the strings. (The separating integer and decimal part will probably be particularly useful so you might want to refactor that out into its own method.) Hopefully this is enough to get you started.
Edit: Here's how to do the K thing... this one's a bit simpler:
Number.prototype.k = function () {
// We don't want any thousands processing if the number is less than 1000.
if (this < 1000) {
// edit 2 May 2013: make sure it's a string for consistency
return this.toString();
}
// Round to 100s first so that we can get the decimal point in there
// then divide by 10 for thousands
var thousands = Math.round(this / 100) / 10;
// Now convert it to a string and add the k
return thousands.toString() + 'K';
};
Call this in the same way: var num = 2000; num.k()
Theoretically, yes.
As TimWolla points out, it requires a lot of logic.
Ruby on Rails have a helper for presenting time with words. Have a look at the documentation. The implementation for that code is found on GitHub, and could give you some hint as how to go about implementing this.
I agree with the comment to reduce the complexity by choosing one format.
Hope you find some help in my answer.
[5, 4, 4, 6].indexOfArray([4, 6]) // 2
['foo', 'bar', 'baz'].indexOfArray(['foo', 'baz']) // -1
I came up with this:
Array.prototype.indexOfArray = function(array) {
var m = array.length;
var found;
var index;
var prevIndex = 0;
while ((index = this.indexOf(array[0], prevIndex)) != -1) {
found = true;
for (var i = 1; i < m; i++) {
if (this[index + i] != array[i]) {
found = false;
}
}
if (found) {
return index;
}
prevIndex = index + 1
}
return index;
};
Later I have find wikipedia calls it Naïve string search:
In the normal case, we only have to look at one or two characters for each wrong position to see that it is a wrong position, so in the average case, this takes O(n + m) steps, where n is the length of the haystack and m is the length of the needle; but in the worst case, searching for a string like "aaaab" in a string like "aaaaaaaaab", it takes O(nm) steps.
Can someone write a faster indexOfArray method in JavaScript?
The algorithm you want is the KMP algorithm (http://en.wikipedia.org/wiki/Knuth%E2%80%93Morris%E2%80%93Pratt_algorithm) used to find the starting index of a substring within a string -- you can do exactly the same thing for an array.
I couldn't find a javascript implementation, but here are implementations in other languages http://en.wikibooks.org/wiki/Algorithm_implementation/String_searching/Knuth-Morris-Pratt_pattern_matcher -- it shouldn't be hard to convert one to js.
FWIW: I found this article a good read Efficient substring searching It discusses several variants of Boyer-Moore although it's not in JavaScript. The Boyer-Moore-Horspool variant (by Timo Raita’s -- see first link for link) was going to be my "suggestion" for a potential practical speed gain (does not reduce big-O though -- big-O is upper limit only!). Pay attention to the Conclusion at the bottom of the article and the benchmarks above.
I'm mainly trying to put up opposition for the Knuth-Morris-Pratt implementation ;-)