I hope it is ok that I am posting this question here even though I have also posted it on other sites. If I have failed to follow proper protocols, I apologize and please let me know right away so I may remove the post and learn my lessons.
I've been a front end developer for over a year now. I went to school to learn web development, and I consider myself a somewhat capable coder when it comes to simple JavaScript stuff.
But when it comes to writing any type of Fibonacci function I cannot do it. It is as if I have a piece missing from my brain that would understand how to deal with this simple sequence of numbers.
Here is a piece of a working code that I'm pretty sure I got from a John Resig book or somewhere online:
fibonacci = (function () {
var cache = {};
return function (n) {
var cached = cache[n];
if (cached) return cached;
if (n <= 1) return n;
console.log(n);
return (cache[n] = fibonacci(n - 2) + fibonacci(n - 1));
};
}());
When I call this function with 10 as the argument, I get this sequence: 10,8,6,4,2,3,5,7,9
Here's what I understand:
fibonnaci is assigned an immediately invoked function expression (or self executing blah blah blah), to which a cache is initiated with whatever argument was passed.
If the argument was already in the cache, we just return it and live our lives in everlasting peace.
If the argument is 1 or less, that is also the end of the function, everlasting peace ensues once more.
But if neither of these conditions exist, then the function returns this statement that makes me feel as if I am just a monkey in a human suit.
What I'd like to do is generate the first 10 fibonnaci numbers in the correct sequence, because if I can do that, then I'll feel like I at least understand it.
So when the first two conditions fail, the code creates a new cache variable and sets it equal to the result of the fibonacci function with whatever argument was passed minus 2, and then it adds the result minus 1.... now for my questions
Question 1: How does the function know what fibonacci(n-2) is if fibonacci(n) has never been computed?
Question 2: Are recursive functions linear, or what order do they follow?
Question 3: If I can't understand this, do I have any hope of becoming a programmer?
Thank you for your time.
After getting past this block, I changed the function a bit to see if I could hold on to the result in a variable and output it, just to see what happens, and I got some really unexpected results.
Here's the change:
fibonacci = (function () {
var cache = {};
return function (n) {
var cached = cache[n];
if (cached) {
console.log(cached);
return cached;
}
if (n <= 1) {
console.log(n);
return n;
}
console.log(n);
var result = (cache[n] = fibonacci(n - 2) + fibonacci(n - 1));
console.log(result);
return result;
};
}());
Here's the resulting pattern:
10,8,6,4,2,0,1,1,3,1,1,2,3,5,2,3,5,8,7,5,8,13,21,9,13,21,34,55
Any help with why this happens?
Well, let's start with what you understand (or say you understand):
fibonnaci is assigned an immediately invoked function expression (or self executing blah blah blah), to which a cache is initiated with whatever argument was passed.
Not quite: fibonnaci is assigned the return value of an IIFE. There's a difference. Inside the IIFE, we se a return function(n) statement. The IIFE is, as it's name suggests, invoked immediatly. the function is created, executed and, once returned, is not being referenced (explicitly) anywhere. The function is returned, is assigned to the variable fibonacci.
This IIFE does create an object literal, called cache. This object resides in the scope of the IIFE, but because of JS's scope scanning(this answer links to others... all of them together explain how JS resolves names to their values), this object is still accessible to the returned function, now assigned to fibonaci.
If the argument was already in the cache, we just return it and live our lives in everlasting peace. If the argument is 1 or less, that is also the end of the function, everlasting peace ensues once more. But [...]
Well, now cache is not created over and over again on each function call (the IIFE is only called once, and that's where cache is created). If the returned function (fibonnaci) changes it, that change to the object will persist in memory. Closure vars, for that is what cache is can be used to hold state between function calls. All the other things you say (n <= 1) is standard recursive function stuff... it's the condition that prevents infinite recursion.
But if neither of these conditions exist, then the function returns this statement that makes me feel as if I am just a monkey in a human suit.
Well, this is actually the fun part. This is where the actual magic happens.
As I've explained, fibonnaci does not reference the IIFE, but rather, it references teh returned function:
function(n)
{
var cached = cache[n];
if (cached)
{
return cached;
}
if (n <= 1)
{
return n;
}
return (cache[n] = (fibonacci(n-2) + fibonnaci(n-1)));
}
This means, that every occurance of fibonacci can be replaced with the function body. When calling fibonnaci(10), the last (return) statement should be read as:
return (cahce[n] = fibonacci(8) + fibonnaci(9));
Now, as yousee, fibonacci(8) and fibonnaci(9) are called, in the return value. These expression can be written in full, too:
return (cache[10] = (fibonnaci(6) + fibonacci(7)) + (fibonacci(7) + fibonacci(8)));
//which is, actually:
return (cache[10 = ( retrun (cache[6] = fibonacci(4) + fibonacci(5))
//since fibonacci(6) cached both fibonacci(5) & fibonacci(6)
+ return (cache[7] = cache[5] + cache[6])
+ return cache[7] + return (cache[8] = cache[6] + cache[7]
That's how this cache function actually ties in...
We can repeat this until we call fibonnacci(1) or fibonacci(0), because in that case n<=1, and will be returned without any more recursive calls.
Also note that, in writing fibonnaci(9) in full, this actually breaks down into fibonacci(7) + fibonacci(8), both these calls have been made before, and that's why there results were cached. If you don't cache the results of each call, you'll waste time calculating the same thing twice...
BTW: this code is very much "condesed", and works because the specs say that an assignment expression (cache[n] = ...) evaluates to the assigned value, you're returning cache[n], essentially.
Great questions. Thinking in recursive terms is tricky. If you try to grasp all the process you probably will fail miserably. I remembered to be frustrated as you are for not understanding the recursive solution to the Hanoi towers problem. I tried to trace on paper the sequence of steps but it was not of any help to understand the magic of what was going on.
What it worked for me was to think that the recursive function is a kind of "oracle" that knows the return value of the function fib(i) for any value of i < n. If the oracle knows fib(n-1) and fib(n-2), then it just has to give the instructions to calculate fib(n) from these known values. As a consequence we can say that the oracle, this function, also knows the value of fib(n).
A warning: there is a tricky part in all recursive functions, we need at least one not recursive value known at start the process, otherwise we will have an infinite recursion. In fibonacci example, these values are: fib(0) = 0 and fib(1) = 1
Your example is a bit more complex than that, as is using memoization, a technique to store the values of fib(n) in a cache to avoid recalculating them. In this case, this cache is an sparse array (array with "holes") that stores in its position i the value of fib(i) the first time it is calculated. This is a way to avoid repeating a costly computation the next time the same value of fib(i) is requested.
Aswering your questions:
fib(n-2) doesn't need to know the value of fib(n) to be calculated, what it needs are the values of fib(n-3) and fib(n-4). The only thing to do is to invoke them in ordert to ask the "oracle" their values.
It depends, there are linear recursive functions and tree shaped recursive functions, depending of how many other values they use. In this case you have a tree shaped invocation order. (Actually memoization makes it more complex and I would represent it as as a directed acyclical graph instead of a tree, but it is not relevant for the discussion).
Keep thinking on it, one day you will have an "aha" moment and then recursive functions will become evident to you.
If you still want to trace the execution of this function maybe it would help to refactor your calculation to an equivalent way as it makes more evident the order of execution:
// var result = (cache[n] = fibonacci(n - 2) + fibonacci(n - 1));
var fib2 = fibonacci(n - 2);
var fib1 = fibonacci(n - 1);
var result = fib2 + fib1;
cache[n] = result;
I know the question is a bit old, and the answers are helpful. I was doing this exercise in GoLang, and thought how I would write in Javascript, and use this answer to refresh my mind as well. I see your code has a cache variable to store the value of the fib(n-2) + fib(n-1) iteration. If you're doing recursive, you don't need a variable to store the result, because each time the function is called it returns a number, and these numbers are added up to the first function call.
function fib(n) {
if (n<=1) return n;
return fib(n - 1) + fib(n - 2);
}
To see why you don't need the cache variable is go through each function call, and start calculate the values once n equal 1 or 0.
for example:
iteration 1)
fib(3) {
return fib(3-1) + f(3-2)
}
---------------------------
iteration 2)
fib(3-1) {
return fib(2 - 1) + fib(2-2)
}
iteration 3)
fib(3-2) {
return 1
}
---------------------------
iteration 4)
fib(2-1) {
return 1
}
iteration 5)
fib(2-2) {
return 0
}
----------------------
if you calculate it the return value backward from iteration 5)
5) 0
4) 1
3) 1
2) 1 <== 4) + 5) = 1 + 0
1) 2 <== 2) + 3) = 1 + 1
so fib(3) is 2
Related
This is a freecodecamp problem where the string argument should be multiplied by the num argument. I understood the other methods they provided, but I'm hung up on how this one works.
repeatStringNumTimes("abc", 3);
//should return "abcabcabc"
I am trying to figure out how the last part of this function (the else statement) inherently knows to multiply the parameters together even though there is no instruction to do so. The way I see it, all it says is: x + (x, y - 1) yet somehow it's still returning correctly.
What am I missing?
function repeatStringNumTimes(str, num) {
if(num < 0)
return "";
if(num === 1)
return str;
else
return str + repeatStringNumTimes(str, num - 1);
}
This is a form of computing called "recursion". It refers to functions that can refer to themselves, thus restarting their cycles until a certain condition is met. In this case, the function is recursively calling itself num times, which in this case yields a simple repetition of its commands.
I don't know if I chose the right title for this question, and maybe this is why I cannot also find an answer to this question.
While reading a javascript book, I found this example while talking about closures.
function multiplier(factor){
console.log('factor:'+factor);
return function(number){
console.log('number:'+number)
return number * factor;
};
}
var twice = multiplier(2);
console.log('twice:'+twice(5));
And in console I get this output:
factor:2
number:5
twice:10
I understand what a closure is meant to be, but I do not understand how the variable number, that by my knowledge I expected to be undefined, get the value 5.
My reasoning is the following:
When I call the function multiplier(2) the local variable factor is assigned the value 2, so the first output is correct.
But when it reaches the line return function(number){ it shall assign number undefined, since no value has been previously assigned to such a name.
So it shall crash at all, and not doing correctly the output I got.
May anyone help me understand why calling twice(5) it get the output
number: 5?
Thank you all, excuse me again If i did not post the question in the right way, feel free to modify anything to make this question more intelligible.
return function (number) { ... } returns a function. number is not a variable, it's a function parameter. The same way that factor in function multiplier(factor) is a parameter. It is neither undefined nor does it cause anything to crash either.
In essence, multiplier(2) returns this function:
function (number) {
console.log('number:' + number)
return number * 2;
}
...which you assign to twice, so twice is now the above function.
I think the key thing that you're missing here is that the returned function acts just like any other function. The function isn't entered until it's called.
return number * factor;
doesn't run until you call twice. It's just like how
console.log('factor:'+factor);
doesn't run until you call multiplier. The body of the function isn't entered until the function is called. number doesn't have a value until twice is called, but the code that uses number also doesn't run until twice is called.
But when it reaches the line return function(number){ it shall assign number undefined, since no value has been previously assigned to such a name.
Here's the misunderstanding. Remember: In Javascript, almost everything is an Object. Some will say that many things that you interact with regularly (strings, numbers, booleans (Notice how I put these types/primitives/words first letter in lowercase. I usually use uppercase for Classes and lowercase for primitives)) are primitives, not objects. This is true, but for the purpose of this thread let's consider (almost) everything is an Object.
Let's get back on this sentence you wrote:
when it reaches the line return function(number){ it shall assign number undefined
Here's the issue: when it reaches the line "return function(number){}", it actually returns a function, which is an object.
It does not execute this function, it only declares it, and returns it, as an Object.
You could have wrote "return 666", it would have returned an Object. A Number.
Let's continue.
Your variable "twice" now contains a function. Guess which one. This one:
function(number){
console.log('number:'+number)
return number * factor;
}
Remember, you've declared it and returned it in only one statement:
"return function(number){...}"
Your "twice" variable is now equivalent to a named function you could've declared this way :
function twice(number){
console.log('number:'+number)
return number * factor;
}
Yes, functions are Objects, named functions are like named variables, and variables can be functions.
You can call it this way for example: twice(9), or this way: twice(5).
That's what you've done.
Now let's answer your question:
why calling twice(5) it get the output number: 5?
Because:
var twice = function(number){
console.log('number:'+number)
return number * factor;
}
And you've executed "twice(5);" which in turn executed console.log this way:
console.log('number:'+5);
As far as I've understood, in your "function multiplier()", you do not want to return a function but rather the result of this function itself. I advise you to read about IIFE (Immediately-invoked function expression).
With this you will be able, in only one statement, to:
- declare a function
- execute that function
- (and eventually return its result)
Have fun playing with Javascript. Javascript is great, only when you know what's going behind.
I was reading Eloquent JavaScript and I came across this example for the puzzle:
Consider this puzzle: By starting from
the number 1 and repeatedly either
adding 5 or multiplying by 3, an
infinite amount of new numbers can be
produced. How would you write a
function that, given a number, tries
to find a sequence of additions and
multiplications that produce that
number?
Here's the code for the solution:
function findSequence(goal) {
function find(start, history) {
if (start == goal)
return history;
else if (start > goal)
return null;
else
return find(start + 5, "(" + history + " + 5)") ||
find(start * 3, "(" + history + " * 3)");
}
return find(1, "1");
}
print(findSequence(24));
Could someone clear up how dod find get executed if it didn't have a value for the arguments start and goal? Also how did the recursion happen?
But find didn't get executed without a value for start and goal. It was first executed with the value 1 for start, and the only value for goal was 24.
Perhaps you're confused about the order of operations. There we see the declaration of a function, findSequence. During the declaration, no code is executed. The findSequence function only gets executed later, on the last line, where the result of executing the function gets printed out.
Within the declaration of findSequence, there's a declaration of another function, find. Once again, it doesn't get executed until later. The findSequence function has just one executable line of code, the one that calls find(1, "1"). Execution of that one line triggers the execution of find some number of times, recursively. The find function makes reference to goal; when the Javascript interpreter executes the code, goal always refers to the parameter of findSequence, and since in this example findSequence is only called once, goal always has the same value, 24.
You should be able to see where the recursion happened. If start was equal to goal, then the function stops; it returns the history of how it arrived at that number. If start is greater than goal, then it returns null, indicating that that path was not a path to the target number. If start is still less than goal, then the function tries calling itself with its start value plus 5. If that returns a non-null value, then that's what gets returned. Otherwise, it tries multiplying by 3 and returning that history value instead.
Note that although this code can return many numbers, it cannot return all numbers. If the goal is 2, for example, findSequence will return null because there is no way to start at 1 and get to 2 by adding 5 or multiplying by 3.
When find is called inside of findSequence, it has access to the goal variable that is set in findSequence's definition. A simple example of this is:
function outerFunction() {
var a = 2;
function innerFunction() {
alert(a);
}
innerFunction();
}
outerFunction();
The start variable is defined when it does:
return find(1, "1");
Effectively having an initial start variable of 1, goal variable of 24, and a history of "1" on the first pass.
EDIT: Per Rob's comment, closures aren't actually what's causing this here, as find() is not being executed outside of findSequence(), scoping is causing goal to be found.
If I'm understanding your question correctly: The final line of code is calling findSequence(), with a goal of 24. In findSequence() there's a function called find(), which is defined and then called in the return statement for findSequence, with start equaling 1, and history equaling 1.
I was reading Eloquent JavaScript and I came across this example for the puzzle:
Consider this puzzle: By starting from
the number 1 and repeatedly either
adding 5 or multiplying by 3, an
infinite amount of new numbers can be
produced. How would you write a
function that, given a number, tries
to find a sequence of additions and
multiplications that produce that
number?
Here's the code for the solution:
function findSequence(goal) {
function find(start, history) {
if (start == goal)
return history;
else if (start > goal)
return null;
else
return find(start + 5, "(" + history + " + 5)") ||
find(start * 3, "(" + history + " * 3)");
}
return find(1, "1");
}
print(findSequence(24));
Could someone clear up how dod find get executed if it didn't have a value for the arguments start and goal? Also how did the recursion happen?
But find didn't get executed without a value for start and goal. It was first executed with the value 1 for start, and the only value for goal was 24.
Perhaps you're confused about the order of operations. There we see the declaration of a function, findSequence. During the declaration, no code is executed. The findSequence function only gets executed later, on the last line, where the result of executing the function gets printed out.
Within the declaration of findSequence, there's a declaration of another function, find. Once again, it doesn't get executed until later. The findSequence function has just one executable line of code, the one that calls find(1, "1"). Execution of that one line triggers the execution of find some number of times, recursively. The find function makes reference to goal; when the Javascript interpreter executes the code, goal always refers to the parameter of findSequence, and since in this example findSequence is only called once, goal always has the same value, 24.
You should be able to see where the recursion happened. If start was equal to goal, then the function stops; it returns the history of how it arrived at that number. If start is greater than goal, then it returns null, indicating that that path was not a path to the target number. If start is still less than goal, then the function tries calling itself with its start value plus 5. If that returns a non-null value, then that's what gets returned. Otherwise, it tries multiplying by 3 and returning that history value instead.
Note that although this code can return many numbers, it cannot return all numbers. If the goal is 2, for example, findSequence will return null because there is no way to start at 1 and get to 2 by adding 5 or multiplying by 3.
When find is called inside of findSequence, it has access to the goal variable that is set in findSequence's definition. A simple example of this is:
function outerFunction() {
var a = 2;
function innerFunction() {
alert(a);
}
innerFunction();
}
outerFunction();
The start variable is defined when it does:
return find(1, "1");
Effectively having an initial start variable of 1, goal variable of 24, and a history of "1" on the first pass.
EDIT: Per Rob's comment, closures aren't actually what's causing this here, as find() is not being executed outside of findSequence(), scoping is causing goal to be found.
If I'm understanding your question correctly: The final line of code is calling findSequence(), with a goal of 24. In findSequence() there's a function called find(), which is defined and then called in the return statement for findSequence, with start equaling 1, and history equaling 1.
This question already has answers here:
How do JavaScript closures work?
(86 answers)
Closed 8 years ago.
I'm a javascript noob trying to wrap my head around the closure exercise below.
Now, I know the result is 122. Can anyone walk me through this step by step (what gets passed to what), so I can understand how closures work?
var hidden = mystery(3);
var jumble = mystery3(hidden);
var result = jumble(2);
function mystery ( input ){
var secret = 4;
input+=2;
function mystery2 ( multiplier ) {
multiplier *= input;
return secret * multiplier;
}
return mystery2;
}
function mystery3 ( param ){
function mystery4 ( bonus ){
return param(6) + bonus;
}
return mystery4;
}
In order to understand this you must know what is the difference between a function call and a reference to a function. As well as how scopes work in javascript.
Assuming you do know these things, let's get explaining.
So you first have a variable hidden that is being assigned a value of mystery(3). So immediately look at the function mystery and see what it returns. it returns a reference to an inner function mystery2. So now hidden holds a reference, meaning that it has no actual numeric value. Following you have a second variable declaration
var jumble = mystery3(hidden);. Now in order to know what jumble holds you need to look at the function mystery3 and the value it returns. It, again, returns a reference to an inner function mystery4. So now the two variables you have contain references to inner functions of the closures mystery and mystery3.
Now let's have a look at var result = jumble(2). Executing jumble(2) is an actual function call to the function that jumble holds a reference to, which is mystery4. When mystery4 runs you see it requires a parameter bonus, which will be 2 given from the line var result = jumble(2). It returns param(6) + bonus. bonus is 2, ok, but what is param(6)? That is the value given to jumble: hidden, which was a reference to mystery2, remember? So running param(6) will execute mystery2 with a parameter 6
And so, tracing back the functions may have turned out a little confusing, but let's follow that with actual values to make it a little clearer ( if that's even a word ).
Executing var result = jumble(2) will give us a return value of param(6) + 2 to get param(6) we go into mystery2 with multiplier = 6, where we set multiplier = 6 * input. Our input is equal to 3+2=5, so multiplier becomes 6*5=30 and as a return value we multiply that by 4 which is our var secret. By the end of the execution of mystery2 we have a value of 120, which is returned to our param(6) in mystery4. And if you remember that our bonus was 2, 120+2=122 Voila!
I get the feeling I didn't do a very good job at explaining this simply, but that's probably the best I could do. Hope that helped!