I'm trying to do some operations that would cause a StackOverflow in Kotlin just now.
Knowing that, I remembered that Kotlin has support for tailrec functions, so I tried to do:
private tailrec fun Turn.debuffPhase(): List<Turn> {
val turns = listOf(this)
if (facts.debuff == 0 || knight.damage == 0) {
return turns
}
// Recursively find all possible thresholds of debuffing
return turns + debuff(debuffsForNextThreshold()).debuffPhase()
}
Upon my surprise that IDEA didn't recognize it as a tailrec, I tried to unmake it a extension function and make it a normal function:
private tailrec fun debuffPhase(turn: Turn): List<Turn> {
val turns = listOf(turn)
if (turn.facts.debuff == 0 || turn.knight.damage == 0) {
return turns
}
// Recursively find all possible thresholds of debuffing
val newTurn = turn.debuff(turn.debuffsForNextThreshold())
return turns + debuffPhase(newTurn)
}
Even so it isn't accepted. The important isn't that the last function call is to the same function? I know that the + is a sign to the List plus function, but should it make a difference? All the examples I see on the internet for tail call for another languages allow those kind of actions.
I tried to do that with Int too, that seemed to be something more commonly used than addition to lists, but had the same result:
private tailrec fun discoverBuffsNeeded(dragon: RPGChar): Int {
val buffedDragon = dragon.buff(buff)
if (dragon.turnsToKill(initKnight) < 1 + buffedDragon.turnsToKill(initKnight)) {
return 0
}
return 1 + discoverBuffsNeeded(buffedDragon)
}
Shouldn't all those implementations allow for tail call? I thought of some other ways to solve that(Like passing the list as a MutableList on the parameters too), but when possible I try to avoid sending collections to be changed inside the function and this seems a case that this should be possible.
PS: About the question program, I'm implementing a solution to this problem.
None of your examples are tail-recursive.
A tail call is the last call in a subroutine. A recursive call is a call of a subroutine to itself. A tail-recursive call is a tail call of a subroutine to itself.
In all of your examples, the tail call is to +, not to the subroutine. So, all of those are recursive (because they call themselves), and all of those have tail calls (because every subroutine always has a "last call"), but none of them is tail-recursive (because the recursive call isn't the last call).
Infix notation can sometimes obscure what the tail call is, it is easier to see when you write every operation in prefix form or as a method call:
return plus(turns, debuff(debuffsForNextThreshold()).debuffPhase())
// or
return turns.plus(debuff(debuffsForNextThreshold()).debuffPhase())
Now it becomes much easier to see that the call to debuffPhase is not in tail position, but rather it is the call to plus (i.e. +) which is in tail position. If Kotlin had general tail calls, then that call to plus would indeed be eliminated, but AFAIK, Kotlin only has tail-recursion (like Scala), so it won't.
Without giving away an answer to your puzzle, here's a non-tail-recursive function.
fun fac(n: Int): Int =
if (n <= 1) 1 else n * fac(n - 1)
It is not tail recursive because the recursive call is not in a tail position, as noted by Jörg's answer.
It can be transformed into a tail-recursive function using CPS,
tailrec fun fac2(n: Int, k: Int = 1): Int =
if (n <= 1) k else fac2(n - 1, n * k)
although a better interface would likely hide the continuation in a private helper function.
fun fac3(n: Int): Int {
tailrec fun fac_continue(n: Int, k: Int): Int =
if (n <= 1) k else fac_continue(n - 1, n * k)
return fac_continue(n, 1)
}
For me when making a recursive method. I always need to spend a lot of time to do it, because I will make some test cases and to see whether my recursive case works and to draw a stack diagram. However, when I ask other about it, they just say that I need to believe myself it will work. How am I suppose to believe that if you don't know what is going on in the recursive case?
You define what is going on in the recursive case, just as you define the rest of the method. Imagine someone else wrote a method to do what the one you are writing does; you wouldn't have a problem calling that, would you? The only difference is that you are that method's author, and it just happens to be the one being written.
For example: I am writing the following method:
// Sort array a[i..j-1] in ascending order
method sort_array( a, i, j ) {
..
}
The base case is easy:
if ( i >= j-1 ) // there is at most one element to be sorted
return; // a[i..j-1] is already sorted
Now, when that isn't true, I could do the following:
else {
k = index_of_max( a, i, j );
swap( a, j-1, k );
At this point, I know that a[j-1] has the correct value, so I just need to sort what comes before it -- fortunately, I have a method to do just that:
sort_array( a, i, j-1 );
}
No leap of faith is required; I know that recursive call will work because I wrote the method to do just that.
I got this question in an interview. So, this seems to me a messed up Fibonacci seq. sum generator and this gives a stackoverflow. Because if(n==0) should be if(n<3) (exit condition is wrong). What should be the precise answer to this question? What was expected as an answer?
foo (int n) {
if (n == 0) return 1;
return foo(n-1) + foo(n-2);
}
UPDATE
What does this recursive function do?
What do you infer when you see these 3 lines of code. No debugging.
Yes, it's just a recursive Fibonacci method with an incorrect base case (although I don't think I'd use n < 3 either... it depends on how you define it).
As for "what was expected as an answer" - that would depend on the question. Were you asked exactly the question in the title of your post? If so, the answer is "it recurses forever until it blows up the stack when you pass it any value other than 0" - with an explanation, of course. (It will never end because either n-1 isn't 0, or n-2 isn't 0, or both.)
EDIT: The above answers the first question. To answer "What do you infer when you see these 3 lines of code" - I would infer that the developer in question has never run the code with a value other than 0, or doesn't care about whether or not the code works.
The problem with the code posted is that if we evaluate foo(1) we need to find foo(0) and foo (-1), foo(-1) then needs to find foo(-2) and foo(-3) and so on. This will keep putting more calls to foo() until there is no more space in the memory resulting in a stack overflow. How many calls to foo are made will depend on the size of the call stack, which will be implementation specific.
When I see these lines of code I immediately get the impression that whoever wrote it hasn't thought about all the possible inputs that could be fed to the function.
To make a recursive Fibonacci function that doesn't fail for foo(1) or a negative input we get:
foo (int n) {
if( n < 0 ) return 0;
if (n == 0) return 1;
return foo(n-1) + foo(n-2);
}
Edit: perhaps the return for a negative number should be something else, as the fibonacci sequence isn't implicitly defined for negative indexes.
However if we use the extension that fib(-n) = (-1)^(n+1) fib(n) we could get the following code:
int fib(int n) {
if( n == -1){
return 1;
}else if ( n < 0 ){
return ( (-1)^(-n+1 ) * fib(-n) );
}else if (n == 0){
return 1;
}else{
return fib(n-1) + fib(n-2);
}
}
suppose, you call foo ( 1 ), it will have two sub calls of foo(0) and foo(-1). As you can see, once you get to foo(-1), n will decrease indefinitely and will never reach a terminating condition.
The precise answer would be:
foo (int n) {
if (n <= 1) return 1; // assuming 0th and 1st fib is 1
return foo(n-1) + foo(n-2);
}
Its a recursive fibonacci function with 0th element being 1.
Ignoring the incorrect termination condition, the code is the "naive" recursive implementation of the Fibonacci function. It has very poor big-O time complexity. It would work fine for small n, but if n is greater than say 50 (I'm just guessing that number), then it would take "forever" to run.
A theoretical question here about the base or halting case in a recursive method, what's its standards?
I mean, is it normal not to have body in it, just a return statement?
Is it always like the following:
if (input operation value)
return sth;
Do you have different thoughts about it?
The pattern for recursive functions is that they look something like this:
f( value )
if ( test value )
return value
else
return f( simplify value )
I don't think you can say much more than that about general cases.
The base case is to terminate the loop (avoid becoming an infinite recursion). There's no standard in the base case, any input that is simple enough to be solved exactly can be chosen as one.
For example, this is perfectly valid:
int factorial (int n) {
if (n <= 5) {
// Not just a return statement
int x = 1;
while (n > 0) {
x *= n;
-- n;
}
return x;
} else {
return n * factorial(n-1);
}
}
In some cases, your base case is
return literal
In some cases, your base case is not simply "return a literal".
There cannot be a "standard" -- it depends on your function.
The "Syracuse Function" http://en.wikipedia.org/wiki/Collatz_conjecture for example,
doesn't have a trivial base case or a trivial literal value.
"Do you have different thoughts about it??" Isn't really a sensible question.
The recursion has to terminate, that's all. A trivial tail recursion may have a "base case" that returns a literal, or it may be a calculation. A more complex recursion may not have a trivial "base case".
It depends entirely on the particular recursive function. In general, it can't be an empty return; statement, though - for any recursive function that returns a value, the base case should also return a value of that type, since func(base) is also a perfectly valid call. For example, a recursive factorial function would return a 1 as the base value.
It seems quite a few mainstream languages support function literals these days. They are also called anonymous functions, but I don't care if they have a name. The important thing is that a function literal is an expression which yields a function which hasn't already been defined elsewhere, so for example in C, &printf doesn't count.
EDIT to add: if you have a genuine function literal expression <exp>, you should be able to pass it to a function f(<exp>) or immediately apply it to an argument, ie. <exp>(5).
I'm curious which languages let you write function literals which are recursive. Wikipedia's "anonymous recursion" article doesn't give any programming examples.
Let's use the recursive factorial function as the example.
Here are the ones I know:
JavaScript / ECMAScript can do it with callee:
function(n){if (n<2) {return 1;} else {return n * arguments.callee(n-1);}}
it's easy in languages with letrec, eg Haskell (which calls it let):
let fac x = if x<2 then 1 else fac (x-1) * x in fac
and there are equivalents in Lisp and Scheme. Note that the binding of fac is local to the expression, so the whole expression is in fact an anonymous function.
Are there any others?
Most languages support it through use of the Y combinator. Here's an example in Python (from the cookbook):
# Define Y combinator...come on Gudio, put it in functools!
Y = lambda g: (lambda f: g(lambda arg: f(f)(arg))) (lambda f: g(lambda arg: f(f)(arg)))
# Define anonymous recursive factorial function
fac = Y(lambda f: lambda n: (1 if n<2 else n*f(n-1)))
assert fac(7) == 5040
C#
Reading Wes Dyer's blog, you will see that #Jon Skeet's answer is not totally correct. I am no genius on languages but there is a difference between a recursive anonymous function and the "fib function really just invokes the delegate that the local variable fib references" to quote from the blog.
The actual C# answer would look something like this:
delegate Func<A, R> Recursive<A, R>(Recursive<A, R> r);
static Func<A, R> Y<A, R>(Func<Func<A, R>, Func<A, R>> f)
{
Recursive<A, R> rec = r => a => f(r(r))(a);
return rec(rec);
}
static void Main(string[] args)
{
Func<int,int> fib = Y<int,int>(f => n => n > 1 ? f(n - 1) + f(n - 2) : n);
Func<int, int> fact = Y<int, int>(f => n => n > 1 ? n * f(n - 1) : 1);
Console.WriteLine(fib(6)); // displays 8
Console.WriteLine(fact(6));
Console.ReadLine();
}
You can do it in Perl:
my $factorial = do {
my $fac;
$fac = sub {
my $n = shift;
if ($n < 2) { 1 } else { $n * $fac->($n-1) }
};
};
print $factorial->(4);
The do block isn't strictly necessary; I included it to emphasize that the result is a true anonymous function.
Well, apart from Common Lisp (labels) and Scheme (letrec) which you've already mentioned, JavaScript also allows you to name an anonymous function:
var foo = {"bar": function baz() {return baz() + 1;}};
which can be handier than using callee. (This is different from function in top-level; the latter would cause the name to appear in global scope too, whereas in the former case, the name appears only in the scope of the function itself.)
In Perl 6:
my $f = -> $n { if ($n <= 1) {1} else {$n * &?BLOCK($n - 1)} }
$f(42); # ==> 1405006117752879898543142606244511569936384000000000
F# has "let rec"
You've mixed up some terminology here, function literals don't have to be anonymous.
In javascript the difference depends on whether the function is written as a statement or an expression. There's some discussion about the distinction in the answers to this question.
Lets say you are passing your example to a function:
foo(function(n){if (n<2) {return 1;} else {return n * arguments.callee(n-1);}});
This could also be written:
foo(function fac(n){if (n<2) {return 1;} else {return n * fac(n-1);}});
In both cases it's a function literal. But note that in the second example the name is not added to the surrounding scope - which can be confusing. But this isn't widely used as some javascript implementations don't support this or have a buggy implementation. I've also read that it's slower.
Anonymous recursion is something different again, it's when a function recurses without having a reference to itself, the Y Combinator has already been mentioned. In most languages, it isn't necessary as better methods are available. Here's a link to a javascript implementation.
In C# you need to declare a variable to hold the delegate, and assign null to it to make sure it's definitely assigned, then you can call it from within a lambda expression which you assign to it:
Func<int, int> fac = null;
fac = n => n < 2 ? 1 : n * fac(n-1);
Console.WriteLine(fac(7));
I think I heard rumours that the C# team was considering changing the rules on definite assignment to make the separate declaration/initialization unnecessary, but I wouldn't swear to it.
One important question for each of these languages / runtime environments is whether they support tail calls. In C#, as far as I'm aware the MS compiler doesn't use the tail. IL opcode, but the JIT may optimise it anyway, in certain circumstances. Obviously this can very easily make the difference between a working program and stack overflow. (It would be nice to have more control over this and/or guarantees about when it will occur. Otherwise a program which works on one machine may fail on another in a hard-to-fathom manner.)
Edit: as FryHard pointed out, this is only pseudo-recursion. Simple enough to get the job done, but the Y-combinator is a purer approach. There's one other caveat with the code I posted above: if you change the value of fac, anything which tries to use the old value will start to fail, because the lambda expression has captured the fac variable itself. (Which it has to in order to work properly at all, of course...)
You can do this in Matlab using an anonymous function which uses the dbstack() introspection to get the function literal of itself and then evaluating it. (I admit this is cheating because dbstack should probably be considered extralinguistic, but it is available in all Matlabs.)
f = #(x) ~x || feval(str2func(getfield(dbstack, 'name')), x-1)
This is an anonymous function that counts down from x and then returns 1. It's not very useful because Matlab lacks the ?: operator and disallows if-blocks inside anonymous functions, so it's hard to construct the base case/recursive step form.
You can demonstrate that it is recursive by calling f(-1); it will count down to infinity and eventually throw a max recursion error.
>> f(-1)
??? Maximum recursion limit of 500 reached. Use set(0,'RecursionLimit',N)
to change the limit. Be aware that exceeding your available stack space can
crash MATLAB and/or your computer.
And you can invoke the anonymous function directly, without binding it to any variable, by passing it directly to feval.
>> feval(#(x) ~x || feval(str2func(getfield(dbstack, 'name')), x-1), -1)
??? Maximum recursion limit of 500 reached. Use set(0,'RecursionLimit',N)
to change the limit. Be aware that exceeding your available stack space can
crash MATLAB and/or your computer.
Error in ==> create#(x)~x||feval(str2func(getfield(dbstack,'name')),x-1)
To make something useful out of it, you can create a separate function which implements the recursive step logic, using "if" to protect the recursive case against evaluation.
function out = basecase_or_feval(cond, baseval, fcn, args, accumfcn)
%BASECASE_OR_FEVAL Return base case value, or evaluate next step
if cond
out = baseval;
else
out = feval(accumfcn, feval(fcn, args{:}));
end
Given that, here's factorial.
recursive_factorial = #(x) basecase_or_feval(x < 2,...
1,...
str2func(getfield(dbstack, 'name')),...
{x-1},...
#(z)x*z);
And you can call it without binding.
>> feval( #(x) basecase_or_feval(x < 2, 1, str2func(getfield(dbstack, 'name')), {x-1}, #(z)x*z), 5)
ans =
120
It also seems Mathematica lets you define recursive functions using #0 to denote the function itself, as:
(expression[#0]) &
e.g. a factorial:
fac = Piecewise[{{1, #1 == 0}, {#1 * #0[#1 - 1], True}}] &;
This is in keeping with the notation #i to refer to the ith parameter, and the shell-scripting convention that a script is its own 0th parameter.
I think this may not be exactly what you're looking for, but in Lisp 'labels' can be used to dynamically declare functions that can be called recursively.
(labels ((factorial (x) ;define name and params
; body of function addrec
(if (= x 1)
(return 1)
(+ (factorial (- x 1))))) ;should not close out labels
;call factorial inside labels function
(factorial 5)) ;this would return 15 from labels
Delphi includes the anonymous functions with version 2009.
Example from http://blogs.codegear.com/davidi/2008/07/23/38915/
type
// method reference
TProc = reference to procedure(x: Integer);
procedure Call(const proc: TProc);
begin
proc(42);
end;
Use:
var
proc: TProc;
begin
// anonymous method
proc := procedure(a: Integer)
begin
Writeln(a);
end;
Call(proc);
readln
end.
Because I was curious, I actually tried to come up with a way to do this in MATLAB. It can be done, but it looks a little Rube-Goldberg-esque:
>> fact = #(val,branchFcns) val*branchFcns{(val <= 1)+1}(val-1,branchFcns);
>> returnOne = #(val,branchFcns) 1;
>> branchFcns = {fact returnOne};
>> fact(4,branchFcns)
ans =
24
>> fact(5,branchFcns)
ans =
120
Anonymous functions exist in C++0x with lambda, and they may be recursive, although I'm not sure about anonymously.
auto kek = [](){kek();}
'Tseems you've got the idea of anonymous functions wrong, it's not just about runtime creation, it's also about scope. Consider this Scheme macro:
(define-syntax lambdarec
(syntax-rules ()
((lambdarec (tag . params) . body)
((lambda ()
(define (tag . params) . body)
tag)))))
Such that:
(lambdarec (f n) (if (<= n 0) 1 (* n (f (- n 1)))))
Evaluates to a true anonymous recursive factorial function that can for instance be used like:
(let ;no letrec used
((factorial (lambdarec (f n) (if (<= n 0) 1 (* n (f (- n 1)))))))
(factorial 4)) ; ===> 24
However, the true reason that makes it anonymous is that if I do:
((lambdarec (f n) (if (<= n 0) 1 (* n (f (- n 1))))) 4)
The function is afterwards cleared from memory and has no scope, thus after this:
(f 4)
Will either signal an error, or will be bound to whatever f was bound to before.
In Haskell, an ad hoc way to achieve same would be:
\n -> let fac x = if x<2 then 1 else fac (x-1) * x
in fac n
The difference again being that this function has no scope, if I don't use it, with Haskell being Lazy the effect is the same as an empty line of code, it is truly literal as it has the same effect as the C code:
3;
A literal number. And even if I use it immediately afterwards it will go away. This is what literal functions are about, not creation at runtime per se.
Clojure can do it, as fn takes an optional name specifically for this purpose (the name doesn't escape the definition scope):
> (def fac (fn self [n] (if (< n 2) 1 (* n (self (dec n))))))
#'sandbox17083/fac
> (fac 5)
120
> self
java.lang.RuntimeException: Unable to resolve symbol: self in this context
If it happens to be tail recursion, then recur is a much more efficient method:
> (def fac (fn [n] (loop [count n result 1]
(if (zero? count)
result
(recur (dec count) (* result count))))))