I am trying to understand, how recursion works. I got the basic idea, but the details remain unclear. Here is a simple example in javascript:
function sumTo(n){
if (n > 1){
return n + sumTo(n-1)
} else {
return n
}
}
sumTo(3);
It is supposed to count all numbers in 3 and the result is 6 (1 + 2 + 3 = 6), but I don't get, how it works.
OK, we start from if condition. 3 > 1, so we return n and call the function again, but what will be inside if brackets?
Does it look like this:
3 + sumTo(2) // 3 - 1 = 2
Or we don't do anything with n, and wait for the next function:
3 + sumTo(n - 1) // n will come later
I was told that the last function will return 1 to the upper one, but I don't get what it will do with this 1.
If there is some step-by-step explanation for ultimate dummies, please share.
I know there is a lot of similar questions, but found no help.
UPD: It looks like I finally found out how it works, spending two days and asking everyone I could. I'll try to explain for ultimate dummies like me. That's gonna be long, but I hope somebody with the same troubles will find this post and it will be useful.
At first I'd like to show another example of recursion, which is a little easier. I found it at http://www.integralist.co.uk/posts/js-recursion.html and changed values from (1, 10) to (2, 3).
function sum(x, y) {
if (y > 0) {
return sum(x + 1, y - 1);
} else {
return x;
}
}
sum(2, 3);
When we start the function, we check the if condition y > 0. Y is 3, so the condition is true. So we return sum(x + 1, y - 1), i. e. sum(2 + 1, 3 - 1), i. e sum(3, 2).
Now we need to count sum(3, 2). Again, we go to the beginning and start from condition y > 0. Y is 2, so the condition is true. So we return sum(x + 1, y - 1), i. e. sum(3 + 1, 2 - 1), i. e sum(4, 1).
Now we need to count sum(4, 1). One more time, we check the condition y > 0. Y is 1, the condition is true. We return sum(x + 1, y - 1), i. e. sum(4 + 1, 1 - 1), i. e sum(5, 0).
Now we need to count sum(5, 0). We check the condition y > 0 and it is false. According to the if-else in the function, we return x, which is 5. So, sum(2, 3) returns 5.
Now let's do the same for sumTo();
function sumTo(n){
if (n > 1){
return n + sumTo(n-1)
} else {
return n
}
}
sumTo(3);
Starting from sumTo(3). Checking the n > 1 condition: 3 > 1 is true, so we return n + sumTo(n - 1), i. e. 3 + sumTo(3 - 1), i. e. 3 + sumTo(2).
To go on, we need to count sumTo(2).
To do this, we again start the function from checking the n > 1 condition: 2 > 1 is true, so we return n + sumTo(n - 1), i. e. 2 + sumTo(2 - 1), i. e. 2 + sumTo(1).
To go on, we need to count sumTo(1).
To do this, we again start the function and check the n > 1 condition. 1 > 1 is false, so, according to if-else, we return n, i. e. 1. Thus, sumTo(1) results in 1.
Now we pass the result of sumTo(1) to the upper function, sumTo(2), where we earlier stated that we needed sumTo(1) to go on.
SumTo(2) returns n + sumTo(n-1), i. e. 2 + sumTo(2 - 1), i. e. 2 + sumTo(1), i. e. 2 + 1. Thus, sumTo(2) results in 3.
Now we pass the result of sumTo(2) to the upper function, sumTo(3), where we earlier stated that we needed sumTo(2) to go on.
SumTo(3) returns n + sumTo(n-1), i. e. 3 + sumTo(3 - 1), i. e. 3 + sumTo(2), i. e. 3 + 3. Thus, sumTo(3) finally results in 6. So, sumTo(3) returns 6.
My mistake was that everywhere I tried to insert 3 instead of n, while n was decreasing to 1.
Thanks to everyone, who responded to this question.
You can understand like
sumTo(3);
returns => 3 + sumTo(2) // n is greater than 1
returns => 2 + sumTo(1)
returns => 1 // 1 as n is not greater than 1
That's right, sumTo(n) waits until sumTo(n-1) is completed.
So, sumTo(3) waits for sumTo(2),
sumTo(2) waits for sumTo(1),
then sumTo(1) returns 1,
sumTo(2) returns 2 + 1
and sumTo(3) returns 3 + 2 + 1
Showing work:
sumTo(4) = (4 + 3) + (2 + 1) = 10 // 4 + sumTo(3). function called four times
sumTo(3) = (3 + 2) + 1 = 6 // 3 + sumTo(2). called three times
sumTo(2) = (2 + 1) = 3 // 2 + sumTo(1). called twice
sumTo(1) = (1) = 1 // called once
It will probably be easier for you to wrap your head around if you think of it backwards, from the ground up instead of from the top down. Like this:
sumTo(1) = 1 + sumTo(0) = 1
sumTo(2) = 2 + sumTo(1) = 3
sumTo(3) = 3 + sumTo(2) = 6
sumTo(4) = 4 + sumTo(3) = 10
Notice how now you can keep adding to the list, and it will be easy to calculate the previous one because you're just adding the two sums.
The chain of events is as follows:
sumTo(3) adds 3 and calls sumTo(2) which also calls sumTo(1) and returns 3, giving you a grand total of 6.
Does that make sense? I'll gladly elaborate or clarify if anyone has questions.
An important question to understand is why to use recursion and when to use recursion. A good discussion on the subject can be found here: When to use recursion?
A classic example of recursion is the fibonacci sequence. Another good example would be traversing a directory of files on your computer, for example if you wanted to search every file inside of a folder that contains other folders. You can also use recursion to factor exponents.
Consider an easier example of recursion:
function multiplyBy10(i) {
if ( !i ) return 0;
return 10+multiplyBy10(i-1);
}
This function will multiply the number by 10 using recursion. It's good to get a grasp on when recursion is practical because there are times it makes things easier for you. However, it's best to keep things simple and not confuse yourself wherever possible. :)
Related
Can anyone explain this to me? I'm having trouble wrapping my head around this concept of having function within a function.
function factorialize(num) {
if (num === 0) {
return 1;
}
return num * factorialize(num-1);
}
factorialize(10);
This is not a loop right? Why does the function call itself continuously? How does it know when to stop? Wouldn't it just continue to factorialize negative numbers to infinity?
Appreciate the help and guidance as always.
This is not a loop right?
Nope, it is not. It is a recursive function, a function that calls itself until a certain condition is met. In this case, is when num is equals to 0.
Why does the function call itself continuously?
Because the function is called in the return value of this function. It will continue calling the function itself, until num equals 0. In which case the function will exit and return 1.
How does it know when to stop?
The condition if (num === 0). The variable num gets subtracted if num isn't equal to 0 as stated in the code return num * factorialize(num-1);. Notice that the function returns fresh function call but with the parameter num - 1. When num becomes 0. The function returns 1.
Wouldn't it just continue to factorialize negative numbers to
infinity?
Because we have this if (num === 0) condition. So the num parameter decreases each time the code return num * factorialize(num-1); gets called, and eventually the above condition gets fulfilled and the function exits.
We can break it down step by step:
factorialize(10) gets called.
num is not 10, so return num * factorialize(num - 1). In this case num is 10 and num - 1 is 9, so we are actually returning the following: 10 * factorialize(10 - 1).
factorialize(9) gets called, then repeat step (1) and (2) until you get factorialize(0).
When you reach factorialize(0), it will return 1. So the whole function effectively returns 10 * 9 * 8 * ... 1 * 1.
Makes sense?
Recursion is a technique for iterating over an operation by having a function call itself repeatedly until it arrives at a result. You can read about it here
Lets visualize:
factorialize(10)
10*factorialize(9)
9*factorialize(8)
8*factorialize(7)
7 *factorialize(6)
6 *factorialize(5)
5 *factorialize(4)
4 *factorialize(3)
3 *factorialize(2)
2 *factorialize(1)
1 *factorialize(0)
1 (if num==0 return 1)
1
2
6
24
120
720
5040
40320
362880
3628800
Hope this helps!
Each time the function "factorialize" runs your argument "num" get decremented by
1. So num will run from 10 to 1 and when 1 is returned it will reach its terminal case 0. It is only terminal because of your "if" clause ( your base case ) when num === 0, you return 1, and your recursive "factorizlize" is never reached, so it is never executed.
I am not really good with analogies. However, If it still doesn't make sense let me know.
Every iterative problem can be solved by using recursion. try to look at this code
var start = 1;
var end = 10;
var increment = 1;
document.write("loop output: ")
for(var num = start; num <= end; num = num + increment){
document.write(num);
document.write(" ");
}
document.write("\n function output:\n")
function iteration(num) {
//stop condition statement
if (num > end) {
return 1;
}
// inside code block
document.write(num);
document.write(" ");
//incremention
iteration(num+increment);
}
iteration(start);
It uses internal functional frame stack for recursion purpose. for more about recursion look here . little practice is need to understand this concept.Hope make sense. Happy Coding !
Much apologies for the vague title, but I need to elaborate. Here is the code in question, which I read on http://ariya.ofilabs.com/2013/07/prime-numbers-factorial-and-fibonacci-series-with-javascript-array.html:
function isPrime(i) {
return (i > 1) && Array.apply(0, Array(1 + ~~Math.sqrt(i))).
every(function (x, y) {
console.log(x + ' ' + i % y);
return (y < 2) || (i % y !== 0)
});
}
isPrime(23);
isPrime(19);
isPrime(188);
Just for fun, I added those logs so we can see some output:
undefined NaN
undefined 0
undefined 1
undefined 2
undefined 3
undefined NaN
undefined 0
undefined 1
undefined 1
undefined 3
undefined NaN
undefined 0
undefined 0
This is the first time I have every seen apply and every, so bear with me, but my understanding is that apply basically calls the Array function, where the first argument is the substitution for its this and the second is the output...Never would think that would be useful, but this function seems to work, so...
Here, they seem to be creating an array of length equal to the square root of the number in question. I suppose that makes sense because the square root would be the largest possible factor of the number in question.
OK, so from here, if we were to log that array for, say, the first number, it would look like this:
> var i = 23;
undefined
> Array.apply(0, Array(1 + ~~Math.sqrt(i)));
[ undefined, undefined, undefined, undefined, undefined ]
Great, so it is an array of five undefined. Ok, fine, so from here, the every method is supposed to check whether every element in that array passes the callback function test (or whatever).
The Microsoft documentation specifies three possible arguments for the every method:
value
index
array
Therefore, in this example x is the value, i.e. undefined, and y is the index.
Our output agrees with that conclusion. However, I'm still fuzzy about nested return statements (if the lowest one returns, does its parent also return?), the || operator here (if the first test passes, does the every loop stop?), and just generally how this works.
EDIT
the log should be with an x, not a y. my mistake:
console.log(y + ' ' + i % y); -> console.log(x + ' ' + i % y);
EXPLANATION
So, how did I come across this code, you ask? Well, of course, the simplest way to check for a prime in Java would be like this:
public static boolean isPrime(double num) {
for (double i = 2.0; i < sqrt(num); i++) {
if (num % i == 0.0) {
return true;
}
}
return false;
}
or Python
def isPrime(num):
x = 2
isPrime = True
while x < math.sqrt(num):
if num % x == 0:
isPrime = False
break
x = x + 1
return isPrime
or js
function isPrime(n) {
for (var i = 2.0; i < Math.sqrt(n); i++) {
if (n % i === 0.0) {
return false;
}
}
return true;
}
But say I wanted to check for the largest prime factor of a number like 600851475143 These looping methods would take too long, right? I think this "hack", as we are describing it, may be even less efficient, because it is using arrays instead of integers or floats, but even still, I was just looking for a more efficient way to solve that problem.
The code in that post is basically crap. Teaching people to write code while simultaneously using hacks is garbage. Yes, hacks have their place (optimization), but educators should demonstrate solutions that don't depend on them.
Hack 1
// the 0 isn't even relevant here. it should be null
Array.apply(0, Array(1 + ...))
Hack 2
// This is just Math.floor(x), but trying to be clever
~~x
Hack 3
// this is an outright sin; totally unreadable code
// I bet most people don't know the binding precedence of % over +
y + ' ' + i % y
// this is evaluated as
y + ' ' + (i % y)
// example
2 + ' ' + (5 % 2) //=> "2 1"
I'm still fuzzy about nested return statements (if the lowest one returns, does its parent also return?),
No. A return only return the function the statement exists in
the || operator here (if the first test passes, does the every loop stop?)
No. Array.prototype.every will return false as soon as the callback returns a false. If a false is never returned from the callback, .every will return `true.
function isEven(x) { return x % 2 === 0; }
[2,4,5,6].every(isEven); //=> false, stops at the 5
[2,4,6].every(isEven); //=> true
Here's an example of .every short circuiting
[1,2,3,4,5,6].every(x=> {console.log(x, x<4); return x<4;});
// 1 true
// 2 true
// 3 true
// 4 false
//=> false
See how it stops once the callback returns false? Elements 5 and 6 aren't even evaluated.
... and just generally how this works.
&& kind of works like Array.prototype.every and || kind of works like Array.prototype.some.
&& will return false as soon as the first false is encountered; in other words, it expects every arguments to be true.
|| will return true as soon as the first true is encountered; in other words, it expects only some argument to be true.
Relevant: short circuit evaluation
Right now I'm going through Codecademy's recursion track and I'm confused how to interpret this correct code.
// Create an empty array called "stack"
var stack = []
// Here is our recursive function
function power(base, exponent) {
// Base case
if ( exponent === 0 ) {
return 1;
}
// Recursive case
else {
stack[exponent-1] = base * power(base, exponent - 1); //confused by start of this line
return stack[exponent-1];
}
}
power(3,3)
console.log(stack) // [3,9,27]
If exponent-1 becomes 2 then 1 then 0, why does 3 become the element at the 0th position in the array rather than at the 2nd position (and so on) ?
I'd really appreciate any help.
On the first pass, exponent is 3, so you will store a value at stack[2]. But that value is not calculated until the recursive call has completed with power(3,2)...power(3, 1).
So the assignment to stack[3-1] is preceded by the one to stack [3-2], which in turn is preceded by the one to stack[3-2]
Notice that there are four different exponents in the execution of this call, one in each scope of the power invocation.
call power(3, 3)
exponent' = 3
is not 0
calls power(3, 2)
exponent'' = 2
is not 0
calls power(3, 1)
exponent''' = 1
is not 0
calls power(3, 0)
exponent'''' = 0
is 0
returns 1
multiplies the return value by 3
assigns it to stack[exponent'''-1]: stack[0] = 3
returns 3
multiplies the return value by 3
assigns it to stack[exponent''-1]: stack[1] = 9
returns 9
multiplies the return value by 3
asigns it to stack[exponent'-1]: stack[2] = 27
returns 27
logs the value of stack
Indeed the stack is built "backwards", after returning from the recursive calls, not before entering them. If you want to have a better representation of the call stack, you can try to add
callstack.push(exponent);
in the first line of your function body. After the execution of your script, the callstack would look like [3, 2, 1, 0] as you might have expected.
It helps if you write it out in a table:
power(3, 3) stack[2] = 3 * power(3, 2)
power(3, 2) stack[1] = 3 * power(3, 1)
power(3, 1) stack[0] = 3 * power(3, 0)
power(3, 0) return 1
Then substitute the values:
stack[0] = 3 * 1
stack[1] = 3 * 3
stack[2] = 3 * 9
This question already has answers here:
How does this recursion work?
(11 answers)
Closed 7 years ago.
This example is from Eloquant javascript and I can't get how it works. I got how basically the recursion works though from other basic examples. Can anyone explain to me know how this works?
It is explained on the link below.
http://eloquentjavascript.net/03_functions.html
function findSolution(target){
function find(start, history){
if (start == target)
return history;
else if (start > target) {
return null;
}
else
return find(start + 5, "(" + history + " + 5)") ||
find(start * 3, "(" + history + " * 3)")
}
return find(1, "1");
}
findSolution(24));
//answer
//"(((1 * 3) + 5) * 3)"
First you need to understand how the returns work for the inner function.
return func(A) || func(B)
If func(A) is not null it's returned, otherwise func(B)
Now our target is 24 and the starting point would be find(1, 1) (the return value of the outer function which triggers the recursive function)
find(1, 1) may have two possible outputs:
A: find(6, (1+5))
or
B find(3, (1*3))
B will be on hold unless we get a null return from A (or reach the target), so we continue with A, the return will be:
Aa: find(11, ((1+5)+5))
or
Ab: find(15, ((1+5)*3))
Again Ab on hold and we continue with Aa which will return:
Aa1: find(16, (((1+5)+5)+5))
Aa2: find(33, (((1+5)+5)*3))
Aa2 is bigger than target (24) so it will never continue. So we continue with Aa1 which will return 21 and then 25 which passes the target and so returns null. So we go back to Ab which returns 45 > target. So A returns null anyway and now we try B:
Ba: find(8, ((1*3)+5))
Bb: find(9, ((1*3)*3))
Continue with Ba which then returns:
Ba1: find(13, (((1*3)+5)+5))
Ba2: find(24, (((1*3)+5)*3))
We continue with Ba1 which will return 18 and then 23 and then 28 which is bigger than target and so null. So go back to Ba2 which is 24 == target and so finishes the calculation and returns its history which is:
(((1*3)+5)*3)
jsfiddle DEMO
You can see the log of all the end points (when null reached or finally when a match found).
EDIT:
This is how the function will execute:
//instead of find I use f
findSolution(24) >
f(1, 1) >
f(6, (1+5)) >
f(11, ((1+5)+5)) >
f(16, (((1+5)+5)+5)) >
f(21, ((((1+5)+5)+5)+5)) >
f(26, (((((1+5)+5)+5)+5)+5)) > NULL
f(63, (((((1+5)+5)+5)+5)*3)) > NULL
f(48, ((((1+5)+5)+5)*3)) > NULL
f(33, (((1+5)+5)*3)) > NULL
f(18, ((1+5)*3)) >
f(23, (((1+5)*3)+5)) >
f(28, ((((1+5)*3)+5)+5)) > NULL
f(69, ((((1+5)*3)+5)*3)) > NULL
f(54, (((1+5)*3)+5)) > NULL
f(3, (1*3)) >
f(8, ((1*3)+5)) >
f(13, (((1*3)+5)+5)) >
f(18, ((((1*3)+5)+5)+5)) >
f(23, (((((1*3)+5)+5)+5)+5)) >
f(28, ((((((1*3)+5)+5)+5)+5)+5)) > NULL
f(69, ((((((1*3)+5)+5)+5)+5)*3)) > NULL
f(54, (((((1*3)+5)+5)+5)*3)) > NULL
f(39, ((((1*3)+5)+5)*3)) > NULL
f(24, (((1*3)+5)*3)) > MATCH
1. function findSolution(target){
2. function find(start, history){
3. if (start == target)
4. return history;
5. else if (start > target) {
6. return null;
7. }
8. else
9. return find(start + 5, "(" + history + " + 5)") ||
10. find(start * 3, "(" + history + " * 3)")
11. }
12. return find(1, "1");
13. }
14.
15. findSolution(24);
When line 15 calls the function findSolution, 24 is passed in for target. Inside the function findSolution, another function find is defined. This function is then called using find(1, "1") on line 12, which starts the recursion. On the first pass of this recursion, a check is made to see if the recursion is completed. This happens when start is equal to target (line 3), which is the value given previously for findSolution (on line 15). In this case, it won't on the first pass, so it proceeds to the next part on line 5. On this line it checks if the start value has gone beyond the target, and if so, returns null. If NOT, then it proceeds to line 9, which recursively calls find with the current start value (1 on the first pass), plus 5 (so 6), and if fails (null is returned [see line 6]), the second call on line 10 occurs because of the || operator. This tries start*3, and on the first pass, this is 1*3, or 3. This keeps going until it has gone as far as it can adding to or multiplying the start value.
In a nutshell, the recursion keeps going via find while start does not reach the target. It ends when either the start value is an exact match to target (line 3), or if the target cannot be reached because of the test on line 5, which doesn't allow start + 5 or start * 3 to go beyond it.
I understand that "a" solution is:
function Factorial(number)
{
if(number == 0 || number == 1){
return 1;
}
return number * Factorial(number -1);
}
I want to understand what exactly is going on. I understand what is going on all the way to the last part when number == 1.
If we were to take a simple example of say 3!
3 is not equal to 0 or 1 so we return 3 * Factorial(2)
2 is not equal to 0 or 1 so we return 2 * Factorial(1)
1 is equal to 1 so we return 1
How do we know when to stop? Is it the fact that we return 1 that tells the function to stop?
If that is the case, why does the function not stop when we first return 3 * Factorial(2)? Is it because it's returning a function so that it must continue until it no longer returns a function?
Thank you
I think you have to understand the logic of factorial.
So factorial is just products, indicated by an exclamation mark ie, if you write
0! = 1
1! = 1
2! = 2*1
3! = 3*2*1
4! = 4*3*2*1
5! = 5*4*3*2*1
Hope you find the pattern, so you can write the above factorials as:
0! = 1
1! = 1
2! = 2*1!
3! = 3*2!
4! = 4*3!
5! = 5*4!
So in your function you are using the similar logic.
Now in your function
if(number == 0 || number == 1)
{
return 1;
}
The above logic is to cover the first two cases i.e, 0! and 1! And
return number * Factorial(number -1);
is for the rest of the numbers.
So you are using the recursive technique of solving the factorial problem.
To understand recursion, lets take a number say 5 i.e., we want find the value of 5!.
Then first your function will check
if(number == 0 || number == 1)
which is not satisfied, then it moves to the next line ie,
return number * Factorial(number -1);
which gives
5*Factorial(5-1) which is equal to 5*Factorial(4)
Now on subsequent calls to your Factorial function it will return the value like below:
5*(4*Factorial(4-1)) which is equal to 5*(4*Factorial(3))
5*(4*(3*Factorial(3-1)) which is equal to 5*(4*(3*Factorial(2)))
5*(4*(3*(2*Factorial(2-1)))) which is equal to 5*(4*(3*(2*Factorial(1))))
Now when it returns factorial(1) then your condition
if(number == 0 || number == 1)
is satisfied and hence you get the result as:
5*4*3*2*1 = 120
On a side note:
Beware that factrial is used only for positive integers.
Recursion relies on what is called a base case:
if(number == 0 || number == 1){
return 1;
}
That if statement is called your base case. The base case defines when the recursion should stop. Note how you are returning 1 not returning the result of a call to the function again such as Factorial(number -1)
If the conditions for your base case are not met (i.e. if number is NOT 1 or 0) then you proceed to call the function again with number * Factorial(number - 1)
If that is the case, why does the function not stop when we first return 3 * Factorial(2)?
Your simple example of 3! can be elaborated like this :
return 3 * Factorial(2)
will then be replaced by
return 3 * (2 * Factorial(1))
which then will be replaced by
return 3 * (2 * 1) // = 6 Hence 6 is returned at last and recursion ends.
How do we know when to stop?
When all your Factorial(value) is replaced by a returned value we stop.
Is it the fact that we return 1 that tells the function to stop?
In a way, yes. Because it is the last returned value.
It is called recursion.
This function is called like this
var result = Factorial(3);
in Factorial function
First time
return 3*Factorial(2);
Now here return statement doesnt get executed insted Factorial is called again..
so second time
return 2*Factorial(1);
again in Factorial(1)
Third time
return 1;
So go to second
return 2*1;
Next to first
return 3*(2*1);
Finally
var result = 3*2*1 = 6.
The function is recursing (calling itself) - and taking one from "number" each time.
Eventually (as long as its a positive integer you call it with, otherwise you'll probably get an infinite loop) you'll always hit the condition (number == 1) so instead of recursing further, it'll return 1 rather than call the function again.
Then, once you have hit the bottom (1), it'll start to run the function and work back up the other way along the function call stack, using the previous result each time:
1 = 1
(2*1) = 2
(3*2) = 6
(4*6) = 24
etc
So the final return statement from the function will return the required result