I am a beginner in code world. I have troubles understanding recursion in JavaScript especially when it needs two or more looping. Like I want to print rectangle using recursion. I don't know completely how to make a base case, condition when it still executed. For examples, these codes below I use to print rectangle or holey rectangle.
function box(num) {
for (let i = 0; i < num; i++) {
let str = ''
for (let j = 0; j < num; j++) {
str += '*'
}
console.log(str)
}
}
box(5)
function holeBox (num) {
for(let i = 0; i < num; i++){
let str = ''
for(let j = 0; j < num; j++){
if(i == 0 || i == num -1 || j == 0 || j == num - 1) {
str += '*'
} else {
str += ' '
}
}
console.log(str)
}
}
holeBox (5)
Please help me to understand recursion, an explanation would be great. My goals are not only to solve those codes but also to understand how recursion works. I've searched there's no good source to learn recursion, or I just too dumb to understand. Thanks in advance
To understand how recursion works, just think of how you can split up what you want to accomplish into smaller tasks, and how the function can complete one of those tasks, and then call itself to do the next- and so on until it is finished. I personally don't think printing boxes is the best way to learn recursion, so imagine you wanted to search an array for a specific value; ignore JavaScript's indexOf()/find() functions or similar for now.
To do this using loops, its easy, just iterate over the array, and check every value:
//Returns the index of the first occurrence of a value in an array, or -1 if nothing is found
function search(needle, haystack) {
for (let i = 0; i < haystack.length; i++) {
if (haystack[i] == needle) return i;
}
return -1;
}
Doing this using recursion is easy as well:
function recursiveSearch(needle, haystack, i) {
if (i > (haystack.length - 1)) return -1; //check if we are at the end of the array
if (haystack[i] == needle) return i; //check if we've found what we're looking for
//if we haven't found the value yet and we're not at the end of the array, call this function to look at the next element
return recursiveSearch(needle, haystack, i + 1);
}
These functions do the same thing, just differently. In the recursive function, the two if statements are the base cases. The function:
Tests if the current element is out of bounds of the array (meaning we've already searched every element), and if so, returns -1
Tests if the current element is what we're looking for, and if so, returns the index
If neither of the statements above apply, we call this function recursively to check the next element
Repeat this until one of the base cases kicks in.
Note that recursive functions are usually called from other helper functions so that you don't have to pass the initial parameters to call the function. For example, the recursiveSearch() function above would be private, and it would be called by another function like this:
function search(needle, haystack) {
return recursiveSearch(needle, haystack, 0);
}
so that we don't have to include the third parameter when we call it, thus decreasing confusion.
Yes, even your box code can be turn into recursion but I don't think it will help you understand the concept of recursion.
If you really have to:
function getBox(arr, size) {
let length = arr.length;
if (length == size)
return arr; // recursion stop rule - if the size reached
for (let i = 0; i < length; i++)
arr[i].push("*"); // fill new size for all row
arr.push(new Array(length + 1).fill("*")); // add new row
return getBox(arr, size); // recursive call with arr bigger in 1 row
}
However, I believe #Gumbo answer explain the concept better then this...
Related
I hope everyone is having an amazing day so far.
I am asking for help on an exercise that I am stuck on. I have gone through and researched all over but still am not getting it right.
The task is:
Create a for loop in which an iterator starting from 0 up to the iteratorMax value (not included) (the second parameter to your function) is incremented by 1. Add the iterator to num(the first parameter to your function) for each incrementation. you must return the number at the end of your function.
I was given this code to start with:
function IncrementNumber(num, iteratorMax){
//Add For loop here
//return value here
}
I have tried coding it a number of different ways but still not correct.
for example:
function IncrementNumber(num, iteratorMax){
for (let i = 0; i < iteratorMax; i++) {
return(i);
}
}
I have tried to declare "i" but that seems to be incorrect also.
I really appreciate any help or hints to what I am missing or doing wrong in my code.
Thank you!
For the IncrementNumber(value, N) method, when value is 0, this algorithm returns the result of the equation:
Total = N x (N-1) / 2
function IncrementNumber(num, iteratorMax){
for(let i = 0 ; i < iteratorMax ; ++i){
num += i;
console.log(`i: ${i}\tnumber: ${num}`);
}
return num;
}
console.log(`Result: ${IncrementNumber(0, 10)}`);
If you call return within the loop, the loop will only be executed once.
You need to move the return statement outside of the loop.
Whenever calling return, this will cause the current function to exit.
function IncrementNumber(num, iteratorMax){
for (let i = 0; i < iteratorMax; i++) {
num += i;
}
return num;
}
I'm trying to write a Javascript function that counts the vowels in a string by calling another function inside that function, but when I test it in the console it returns 0.
Here is my first function that works fine and recognizes if a string is a vowel:
function isVowel(ch){
var pattern = /[aeiouAEIOU]/
return pattern.test(ch);
};
For the second function none of my ideas have worked. Here are a few examples of what I have tried so far:
This one returns me a 0:
function countVowels(str){
var count = 0;
for(var i; i <= str.length; ++i){
if(isVowel(i)){
++count;
}
}
return count;
};
I also tried the above, but removing the .length after str in the for() area.
Another example, but this one gives me an error:
function countVowels(str){
var count = 0
var pattern = /[aeiouAEIOU]/
for(var i = 1; i <= str.length(pattern); ++i){
if(isVowel(i)){
++count;
}
}
return count;
};
I've tried various other functions as well, but for the sake of keeping this post relatively short I won't continue to post them. I'm quite new to Javascript and I'm not sure what I'm doing wrong. Any help would be greatly appreciated!
Try using .match() with the g attribute on your String.
g: global
i: case insensitive
Regexp documentation
function countVowels(ch){
return ch.match(/[aeiouy]/gi).length;
}
var str = "My string";
alert(countVowels(str)); // 2
Although Robiseb answer is the way to go, I want to let you know why you code is not working (I'm referring your first attempt). Basically you made two mistakes in the loop:
As CBroe stated, you are passing i to your isVowel function. i is a integer representing the index of the loop, not the actual character inside the string. To get the character you can do str.substr(i, 1), what means "give me one character from the position i inside the string".
You are not giving a initial value to the i variable. When you create a variable, it is undefined, so you can not increment it.
alert(countVowels("hello"));
function countVowels(str) {
var count = 0;
for (var i = 0; i <= str.length; ++i) {
if (isVowel(str.substr(i, 1))) {
count++;
}
}
return count;
};
function isVowel(ch) {
var pattern = /[aeiouAEIOU]/
return pattern.test(ch);
};
UPDATE: You will see that other answers use other methods to select the character inside the string from the index. You actually have a bunch of different options. Just for reference:
str.slice(i,i+1);
str.substring(i,i+1);
str.substr(i,1));
str.charAt(i);
str[i];
i is the index, not the character. It should be:
if (isVowel(str[i])) {
count++;
}
Also, str.length(pattern) is wrong. length is a property, not a function, so it should just be str.length.
You forgot to assign the value 0 to i variable
And parameter for isVowel is the character, not the index of string
Here information about the JS language: https://stackoverflow.com/tags/javascript/info
function isVowel(ch){
var pattern = /[aeiouAEIOU]/
return pattern.test(ch);
}
function countVowels(str){
var count = 0;
// you forgot to assign the value to i variable
for(var i = 0; i < str.length; i++){
// isVowel(str[i]), not isVowel(i)
if(isVowel(str[i])){
count++;
}
}
return count;
}
console.log(countVowels('forgot'))
Obviously you should do it this way:
function isVowel(c){
var lc = c.toLowerCase();
if(lc === 'y'){
return (Math.floor(Math.random() * 2) == 0);
}
return ['a','e','i','o','u'].indexOf(lc) > -1;
}
function countVowels(s){
var i = 0;
s.split('').each(function(c){
if(isVowel(c)){
i++;
}
});
return i;
}
console.log(countVowels("the quick brown fox jumps over the lazy dog"));
Which, although less efficient and less useful than other answers, at least has the entertaining property of returning a different count 50% of the time, because sometimes Y.
I have got a homework where i am supposed to write a pseudo random number generator in JavaScript. This is the code bit i wrote
var k = 0;
var slump = function(n, k) {
if (k < 10) {
console.log("stop");
}
else {
k++;
console.log((5*n + 1) % 8);
return slump((5*n + 1) % 8, k);
}
};
slump(0);
k is supposed to hold the amount of times the function has been called. But instead of just running the function ten times, it just keeps running. Is there any way to get around this?
You have two subtly different options here, depending on how idiomatic and clever you'd like to get.
The classic implementation, with a slight tweak as JS doesn't support default parameters, would be to use something like:
var finalDepth = 0;
function slump(n, k) {
k = k || 0; // Set to 0 if falsy (null, undef, or 0)
if (logic) {
finalDepth = k; // Record the depth on the last call
} else {
return slump((5*n + 1) % 8, k + 1);
}
}
This will very simply record the deepest the stack has been, and hang onto the value until the next call.
If you want to be slightly more JS-like, you can use closure to keep track of the calls:
function createGenerator() {
var counter = 0;
return {
slump: function (n) {
++counter; // Closure captures counter, counter persists between slump calls but is unique for each createGenerator
if (logic) {
// stop
} else {
return slump((5*n + 1) % 8, k + 1);
}
},
getCounter: function () { return counter; }
}
}
You may be able to use some of the features from ES6 iterators (or generators) to make this more clever.
The function parameter k is uninitialized, therefore not a number. this means in particular that the termination test k < 10 fails as well as the k++ statement doesn't change k' s value. so slump gets always called with the same value for parameter k and the recursion never stops.
Whenever you write a recursive function, you need to make sure that:
There's a base case (in your case, when the console.log statement runs)
The function proceeds towards the base case, and
The function works, assuming the success of the recursive call.
You're running into a problem with the second part; you increment k, but that doesn't bring you any closer to the part where k < 10. In short, you probably want to switch that test around and make sure you're initially calling the function with the right number of arguments. (Aadit M Shah pointed out that you're calling it with one, and it expects two, meaning that it ends up undefined when you call it.)
Either way, iteration would definitely work better here:
var n = 0;
for(var i = 0; i < 10; i++) {
n = (5 * n + 1) % 8;
console.log(n);
}
Below is just a section of my code but I know it's problematic because I can't get it to return any value except 'undefined'. I have been over this for hours and cannot figure it out.
I want to be able to input a number and have its factors pushed to an array. I have tested it by alerting the first item in the array and I get nothing. I'm sure this is a pretty easy but I just can't figure it out. Here is the code:
var numberInQuestion = prompt("Of what number are you wanting to find the largest prime factor?");
//determine factors and push to array for later use
var factorsArray = [];
function factors(numberInQuestion){
for(var i = 2; i < numberInQuestion-1; i++){
if(numberInQuestion % i === 0){
return factorsArray.push[i];
} else {
continue;
}
}
};
factors(numberInQuestion);
alert(factorsArray[0]);
Thanks for any help!
you can only return one value
you must use (), not [] for calling push
factorsArray should be local to factors (put the definition inside the function)
the else { continue; } is useless
Here is the fully corrected code:
var numberInQuestion = prompt("Of what number are you wanting to find the factors of?");
//determine factors
function factors(numberInQuestion){
var factorsArray = []; // make it local
for (var i = 2; i < numberInQuestion-1; i++){
if(numberInQuestion % i === 0){
factorsArray.push(i); // use (), and don't return here
} // no need for else { continue; } because it's a loop anyway
}
return factorsArray; // return at the end
};
var result = factors(numberInQuestion); // assign the result to a variable
alert(result);
Here's a JSFiddle.
You have an error in your pushing syntax. Correct syntax for pushing is -
factorsArray.push(i);
Also returning immediately from the function after finding the first divisor will not give you the full list. You probably want to return after you've found out all the divisors.
Taking all of the above into consideration, you should rewrite your function as follow -
function factors(numberInQuestion){
for(var i = 2; i < numberInQuestion - 1; i++){
if(numberInQuestion % i === 0) {
factorsArray.push(i);
}
}
}
and you will be OK.
You've coded this so that when you find the first factor your function returns immediately. Just get rid of the return keyword in that statement. (What "return" means in JavaScript and other similar languages is to immediately exit the function and resume from where the function was called.)
Oh, also, you call functions (like .push()) with parentheses, not square brackets.
The function should not return when pushing to the array. Return the array after executing the loop. The else clause is also unnecessary.
var numberInQuestion = prompt("Of what number are you wanting to find the largest prime factor?");
function factors(numberInQuestion){
var factorsArray = [];
for(var i = 2; i < numberInQuestion-1; i++){
if(numberInQuestion % i === 0 && isPrime(i)){
factorsArray.push(i);
}
}
return factorsArray;
};
var factors = factors(numberInQuestion);
alert(factors[factors.length-1]);
//From: http://stackoverflow.com/questions/11966520/how-to-find-prime-numbers
function isPrime (n)
{
if (n < 2) return false;
var q = Math.sqrt (n);
for (var i = 2; i <= q; i++)
{
if (n % i == 0)
{
return false;
}
}
return true;
}
Given the purpose of the example two items must be considered
The code does not determine if the number is actually prime. The code will return the smallest factor possible since the loop starts at two and increments, then returns the first element in the array. The largest factor would actually be the last element in the array. I have corrected the example to find the greatest prime factor. You can test it via this fiddle: http://jsfiddle.net/whKGB/1/
I have trouble dealing with my for loops now, I'm trying to compare two datum, basically it will compare 2 items, then it will write the matches and the mismatches on the webpage.
I managed to write the matches on the webpage, it was working good. But there's a bug in my mismatch compare.
It wrote all the data on the webpage X times, here's my JS code:
function testItems(i1, i2) {
var newArray = [];
var newArray2 = [];
var count = 0;
var count2 = 0;
for(var i = 0; i < i1.length; i++) {
for(var j = 0; j < i2.length; j++) {
if(i1[i] == i2[j]) {
newArray.push(i1[i]);
count++;
} if (i1[i] !== i2[j]) {
newArray2.push(i1[i]);
count2++;
}
}
}
count-=2;
count2-=2
writeHTML(count,count2, newArray, newArray2);
}
The result was horrible for the mismatches:
alt text http://www.picamatic.com/show/2009/03/01/07/44/2523028_672x48.jpg
I was expecting it to show the mistakes, not all the strings.
The issue you're seeing is because of the nested for loop. You are essentially doing a cross-compare: for every item in i1, you are comparing it to every item in i2 (remember that j starts again at 0 every time i advances... the two loops don't run in parallel).
Since I understand from the comments below that you want to be able to compare one array to the other, even if the items in each are in a different order, I've edited my original suggestion. Note that the snippet below does not normalize differences in case between the two arrays... don't know if that's a concern. Also note that it only compares i1 against i2... not both i1 to i2 and i2 to i1, which would make the task a little more challenging.
function testItems(i1, i2) {
var newArray = [];
var newArray2 = [];
for (var i = 0; i < i1.length; i++) {
var found = false;
for (var j = 0; j < i2.length; j++) {
if (i1[i] == i2[j]) found = true;
}
if (found) {
newArray.push(i1[i])
} else {
newArray2.push(i1[i])
}
}
}
As an alternative, you could consider using a hash table to index i1/i2, but since the example of strings in your comment include spaces and I don't know if you're using any javascript helper libraries, it's probably best to stick with the nested for loops. The snippet also makes no attempt to weed out duplicates.
Another optimization you might consider is that your newArray and newArray2 arrays contain their own length property, so you don't need to pass the count to your HTML writer. When the writer receives the arrays, it can ask each one for the .length property to know how large each one is.
Not directly related to the question but you should see this:
Google techtalks about javascript
Maybe it will enlighten you :)
Couple of things about your question. First you should use '!=' instead of '!==' to check inequality. Second I am not sure why you are doing decreasing counts by 2, suggests to me that there may be duplicates in the array?! In any case your logic was wrong which was corrected by Jarrett later, but that was not a totally correct/complete answer either. Read ahead.
Your task sounds like "Given two set of arrays i1 & i2 to find i1 {intersection} i2 and i1{dash} {UNION} i2{dash}) (Group theory notation). i.e. You want to list common elements in newArray and uncommon elements in newArray2.
You need to do this.
1) Remove duplicates in both the arrays. (For improving the program efficiency later on) (This is not a MUST to get the desired result - you can skip it)
i1 = removeDuplicate(i1);
i2 = removeDuplicate(i2);
(Implementation for removeDuplicate not given).
2) Pass through i1 and find i1{dash} and i1 {intersection} i2.
var newArray = [];
var newArray2 = [];
for (var i = 0; i < i1.length; i++)
{
var found = false;
for (var j = 0; j < i2.length; j++)
{
if (i1[i] == i2[j])
{
found = true;
newArray.push(i1[i]); //add to i1 {intersection} i2.
count++;
break; //once found don't check the remaining items
}
}
if (!found)
{
newArray2.push(i1[i]); //add i1{dash} to i1{dash} {UNION} i2{dash}
count2++;[
}
}
3) Pass through i2 and append i2{dash} to i1{dash}
for(var x=0; x<i2.length; x++)
{
var found = false;
//check in intersection array as it'd be faster than checking through i1
for(var y=0; y<newArray.length; y++) {
if( i2[x] == newArray[y])
{
found = true;
break;
}
}
if(!found)
{
newArray2.push(i2[x]); //append(Union) a2{dash} to a1{dash}
count2++;
}
}
writeHTML(count,count2, newArray, newArray2);
I have a feeling that this has to do with your second comparison using "!==" instead of "!="
"!==" is the inverse of "===", not "==". !== is a more strict comparison which does not do any type casting.
For instance (5 != '5') is false, where as (5 !== '5') is true. This means it's possible that you could be pushing to both arrays in the nested loop, since if(i1[i] == i2[j]) and if(i1[i] !== i2[j]) could both be true at the same time.
The fundamental problem here is that a pair of nested loops is NOT the right approach.
You need to walk a pointer through each dataset. ONE loop that advances both as needed.
Note that figuring out which to advance in case of a mismatch is a much bigger problem than simply walking them through. Finding the first mismatch isn't a problem, getting back on track after finding it is quite difficult.