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.
Related
Currently learning JS.
Can't figure out at what point in this function "prev1" parameter somehow modified during recursion?
If "console.log(prev1)" on each iteration "prev1" is actually modifies, although in none of the code below seems to change this parameter.
Please, help me understand this thing.
function fibonacci(n, prev1, prev2){
//console.log(prev1);
var current = prev1 + prev2;
var fibonacci_string = current + " ";
if(n > 1)
fibonacci_string += fibonacci(n - 1, current, prev1);
return fibonacci_string;
}
console.log(fibonacci(10, 1, 0));
console.log(prev1):
1
1
2
3
5
8
13
21
34
55
If you look at the recursive call:
fibonacci(n - 1, current, prev1)
// calls: ⬇️ ⬇️ ⬇️
function fibonacci(n, prev1, prev2)
you can see that prev1 gets the value of the previous current.
prev1 is an argument to the function fibonacci.
When the line
fibonacci_string += fibonacci(n - 1, current, prev1);
is called current takes place of prev1 and is provided as argument to the fibonacci function. prev1 is not an actual variable, its just a placeholder for the value that is going to be received by the function. If in place of prev1 you give value as 5 inside the function prev1 will have a value of 5 and treated accordingly, in the above code, current is taking place of prev1 when it is called recursively and how value of current is changed you can see that.
As you can see in your if statement there is another call of fibonacci() with 2nd parameter current which is the sum of prev1 and prev2. So basically in this if statement when you call fibonacci() it enters in the same function again so the code goes to the definition of function which is function fibonacci(n, prev1, prev2) where prev1 stands for the 2nd passed parameter from wherever you called this function: in this case the 2nd parameter you passed was current.
It's a recursive function, it each iteration the parameters have different values:
In the first iteration prev1=1, prev2=0
Second iteration: prev1=prev1+prev2=1, prev2=prev1=1
Third iteration: prev1=prev1+prev2=2, prev2=prev1=1
Fourh iteration: prev1=prev1+prev2=3, prev2=prev1=2
Fifth iteration: prev1=prev1+prev2=5, prev2=prev1=3
prev1 is a parameter of the recursive function. Each time the function is called, new execution context is created with it's own set of identifiers. Identifiers registration implies binding of values passed to the function call (arguments) with function parameters. In the provided example the invocation happens at:
fibonacci(n - 1, current, prev1);
prev1 is the second parameter of fibonacci function therefore it is bound to the second argument passed to an invocation (value of current variable).
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'm reading "Eloquent JavaScript". Chapter 3 introduces "Closure" concept and gives you a couple of examples. One of these is next one:
function multiplier(factor) {
return function(number) {
return number * factor;
};
}
var twice = multiplier(2);
console.log(twice(5));
// → 10
I think I understood the concept. If first I execute console.log(twice), since variable number is undefined, what I get is [Function]. What I don't understand is how twice(5) works. Why local variable number is initialized with value 5?
Also, why if I execute console.log(multiplier(2,5)) I don't get 10 as a result?
Thanks.
Because multiplier returns a function, so twice is equal to the returned function, NOT the multiplier function.
However, when multiplier is called the factor variable is passed and used within the returned function.
So to make it easier to understand, consider that twice is basically:
var twice = function(number) {
return number * 2;
};
Where factor has been replaced by the value you passed in when calling multiplier(2).
I think I understood the concept. If first I execute console.log(twice), since variable number is undefined, what I get is [Function].
When you use console.log(twice) you are not actually calling the function twice, you are simply logging the value of it. So the output of [Function] is not because number is undefined, it is because you are outputting the actual function rather than the result of it.
Also, why if I execute console.log(multiplier(2,5)) I don't get 10 as a result?
Here you are calling the multiplier by providing 2 arguments, though you have only defined the function to accept one parameter (factor). In javascript, this won't cause an error, but you will just get the first value mapped in factor (factor = 2).
Note: There are ways to still access all the supplied arguments even if you don't have parameters defined for them (here's an example)
Something that would be possible to get a result of 10 which might be of interest is to use the following code:
var result = multiplier(2)(5); // result = 10
multiplier is a function which returns anonymous function which accepts an argument(number)
var twice = multiplier(2);
Basically is :-
var twice = function(number) {
return number * 2;
};
If you execute
console.log(multiplier(2,5))
you call the function giving two parameters, whereas
function multiplier(factor) {}
only takes one parameter.
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