Related
I'm trying to understand the semicolon functionality.
I have this code:
del(X,[X|Rest],Rest).
del(X,[Y|Tail],[Y|Rest]) :-
del(X,Tail,Rest).
permutation([],[]).
permutation(L,[X|P]) :- del(X,L,L1), permutation(L1,P).
It's the simple predicate to show all permutations of given list.
I used the built-in graphical debugger in SWI-Prolog because I wanted to understand how it works and I understand for the first case which returns the list given in argument. Here is the diagram which I made for better understanding.
But I don't get it for the another solution. When I press the semicolon it doesn't start in the place where it ended instead it's starting with some deep recursion where L=[] (like in step 9). I don't get it, didn't the recursion end earlier? It had to go out of the recursions to return the answer and after semicolon it's again deep in recursion.
Could someone clarify that to me? Thanks in advance.
One analogy that I find useful in demystifying Prolog is that Backtracking is like Nested Loops, and when the innermost loop's variables' values are all found, the looping is suspended, the vars' values are reported, and then the looping is resumed.
As an example, let's write down simple generate-and-test program to find all pairs of natural numbers above 0 that sum up to a prime number. Let's assume is_prime/1 is already given to us.
We write this in Prolog as
above(0, N), between(1, N, M), Sum is M+N, is_prime(Sum).
We write this in an imperative pseudocode as
for N from 1 step 1:
for M from 1 step 1 until N:
Sum := M+N
if is_prime(Sum):
report_to_user_and_ask(Sum)
Now when report_to_user_and_ask is called, it prints Sum out and asks the user whether to abort or to continue. The loops are not exited, on the contrary, they are just suspended. Thus all the loop variables values that got us this far -- and there may be more tests up the loops chain that sometimes succeed and sometimes fail -- are preserved, i.e. the computation state is preserved, and the computation is ready to be resumed from that point, if the user presses ;.
I first saw this in Peter Norvig's AI book's implementation of Prolog in Common Lisp. He used mapping (Common Lisp's mapcan which is concatMap in Haskell or flatMap in many other languages) as a looping construct though, and it took me years to see that nested loops is what it is really all about.
Goals conjunction is expressed as the nesting of the loops; goals disjunction is expressed as the alternatives to loop through.
Further twist is that the nested loops' structure isn't fixed from the outset. It is fluid, the nested loops of a given loop can be created depending on the current state of that loop, i.e. depending on the current alternative being explored there; the loops are written as we go. In (most of the) languages where such dynamic creation of nested loops is impossible, it can be encoded with nested recursion / function invocation / inside the loops. (Here's one example, with some pseudocode.)
If we keep all such loops (created for each of the alternatives) in memory even after they are finished with, what we get is the AND-OR tree (mentioned in the other answer) thus being created while the search space is being explored and the solutions are found.
(non-coincidentally this fluidity is also the essence of "monad"; nondeterminism is modeled by the list monad; and the essential operation of the list monad is the flatMap operation which we saw above. With fluid structure of loops it is "Monad"; with fixed structure it is "Applicative Functor"; simple loops with no structure (no nesting at all): simply "Functor" (the concepts used in Haskell and the like). Also helps to demystify those.)
So, the proper slogan could be Backtracking is like Nested Loops, either fixed, known from the outset, or dynamically-created as we go. It's a bit longer though. :)
Here's also a Prolog example, which "as if creates the code to be run first (N nested loops for a given value of N), and then runs it." (There's even a whole dedicated tag for it on SO, too, it turns out, recursive-backtracking.)
And here's one in Scheme ("creates nested loops with the solution being accessible in the innermost loop's body"), and a C++ example ("create n nested loops at run-time, in effect enumerating the binary encoding of 2n, and print the sums out from the innermost loop").
There is a big difference between recursion in functional/imperative programming languages and Prolog (and it really became clear to me only in the last 2 weeks or so):
In functional/imperative programming, you recurse down a call chain, then come back up, unwinding the stack, then output the result. It's over.
In Prolog, you recurse down an AND-OR tree (really, alternating AND and OR nodes), selecting a predicate to call on an OR node (the "choicepoint"), from left to right, and calling every predicate in turn on an AND node, also from left to right. An acceptable tree has exactly one predicate returning TRUE under each OR node, and all predicates returning TRUE under each AND node. Once an acceptable tree has been constructed, by the very search procedure, we are (i.e. the "search cursor" is) on a rightmost bottommost node .
Success in constructing an acceptable tree also means a solution to the query entered at the Prolog Toplevel (the REPL) has been found: The variable values are output, but the tree is kept (unless there are no choicepoints).
And this is also important: all variables are global in the sense that if a variable X as been passed all the way down the call chain from predicate to predicate to the rightmost bottommost node, then constrained at the last possible moment by unifying it with 2 for example, X = 2, then the Prolog Toplevel is aware of that without further ado: nothing needs to be passed up the call chain.
If you now press ;, search doesn't restart at the top of the tree, but at the bottom, i.e. at the current cursor position: the nearest parent OR node is asked for more solutions. This may result in much search until a new acceptable tree has been constructed, we are at a new rightmost bottommost node. The new variable values are output and you may again enter ;.
This process cycles until no acceptable tree can be constructed any longer, upon which false is output.
Note that having this AND-OR as an inspectable and modifiable data structure at runtime allows some magical tricks to be deployed.
There is bound to be a lot of power in debugging tools which record this tree to help the user who gets the dreaded sphynxian false from a Prolog program that is supposed to work. There are now Time Traveling Debuggers for functional and imperative languages, after all...
I'm trying to understand the theorem behind "call-by-need." I do understand the definition, but I'm a bit confused. I would like to see a simple example which shows how call-by-need works.
After reading some previous threads, I found out that Haskell uses this kind of evaluation. Are there any other programming languages which support this feature?
I read about the call-by-name of Scala, and I do understand that call-by-name and call-by-need are similar but different by the fact that call-by-need will keep the evaluated value. But I really would love to see a real-life example (it does not have to be in Haskell), which shows call-by-need.
The function
say_hello numbers = putStrLn "Hello!"
ignores its numbers argument. Under call-by-value semantics, even though an argument is ignored, the parameter at the function call site may need to be evaluated, perhaps because of side effects that the rest of the program depends on.
In Haskell, we might call say_hello as
say_hello [1..]
where [1..] is the infinite list of naturals. Under call-by-value semantics, the CPU would run off trying to build an infinite list and never get to the say_hello at all!
Haskell merely outputs
$ runghc cbn.hs
Hello!
For less dramatic examples, the first ten natural numbers are
ghci> take 10 [1..]
[1,2,3,4,5,6,7,8,9,10]
The first ten odds are
ghci> take 10 $ filter odd [1..]
[1,3,5,7,9,11,13,15,17,19]
Under call-by-need semantics, each value — even a conceptually infinite one as in the examples above — is evaluated only to the extent required and no more.
update: A simple example, as asked for:
ff 0 = 1
ff 1 = 1
ff n = go (ff (n-1))
where
go x = x + x
Under call-by-name, each invocation of go evaluates ff (n-1) twice, each for each appearance of x in its definition (because + is strict in both arguments, i.e. demands the values of the both of them).
Under call-by-need, go's argument is evaluated at most once. Specifically, here, x's value is found out only once, and reused for the second appearance of x in the expression x + x. If it weren't needed, x wouldn't be evaluated at all, just as with call-by-name.
Under call-by-value, go's argument is always evaluated exactly once, prior to entering the function's body, even if it isn't used anywhere in the function's body.
Here's my understanding of it, in the context of Haskell.
According to Wikipedia, "call by need is a memoized variant of call by name where, if the function argument is evaluated, that value is stored for subsequent uses."
Call by name:
take 10 . filter even $ [1..]
With one consumer the produced value disappears after being produced so it might as well be call-by-name.
Call by need:
import qualified Data.List.Ordered as O
h = 1 : map (2*) h <> map (3*) h <> map (5*) h
where
(<>) = O.union
The difference is, here the h list is reused by several consumers, at different tempos, so it is essential that the produced values are remembered. In a call-by-name language there'd be much replication of computational effort here because the computational expression for h would be substituted at each of its occurrences, causing separate calculation for each. In a call-by-need--capable language like Haskell the results of computing the elements of h are shared between each reference to h.
Another example is, most any data defined by fix is only possible under call-by-need. With call-by-value the most we can have is the Y combinator.
See: Sharing vs. non-sharing fixed-point combinator and its linked entries and comments (among them, this, and its links, like Can fold be used to create infinite lists?).
I learning more and more about Erlang language and have recently faced some problem. I read about foldl(Fun, Acc0, List) -> Acc1 function. I used learnyousomeerlang.com tutorial and there was an example (example is about Reverse Polish Notation Calculator in Erlang):
%function that deletes all whitspaces and also execute
rpn(L) when is_list(L) ->
[Res] = lists:foldl(fun rpn/2, [], string:tokens(L," ")),
Res.
%function that converts string to integer or floating poitn value
read(N) ->
case string:to_float(N) of
%returning {error, no_float} where there is no float avaiable
{error,no_float} -> list_to_integer(N);
{F,_} -> F
end.
%rpn managing all actions
rpn("+",[N1,N2|S]) -> [N2+N1|S];
rpn("-", [N1,N2|S]) -> [N2-N1|S];
rpn("*", [N1,N2|S]) -> [N2*N1|S];
rpn("/", [N1,N2|S]) -> [N2/N1|S];
rpn("^", [N1,N2|S]) -> [math:pow(N2,N1)|S];
rpn("ln", [N|S]) -> [math:log(N)|S];
rpn("log10", [N|S]) -> [math:log10(N)|S];
rpn(X, Stack) -> [read(X) | Stack].
As far as I understand lists:foldl executes rpn/2 on every element on list. But this is as far as I can understand this function. I read the documentation but it does not help me a lot. Can someone explain me how lists:foldl works?
Let's say we want to add a list of numbers together:
1 + 2 + 3 + 4.
This is a pretty normal way to write it. But I wrote "add a list of numbers together", not "write numbers with pluses between them". There is something fundamentally different between the way I expressed the operation in prose and the mathematical notation I used. We do this because we know it is an equivalent notation for addition (because it is commutative), and in our heads it reduces immediately to:
3 + 7.
and then
10.
So what's the big deal? The problem is that we have no way of understanding the idea of summation from this example. What if instead I had written "Start with 0, then take one element from the list at a time and add it to the starting value as a running sum"? This is actually what summation is about, and it's not arbitrarily deciding which two things to add first until the equation is reduced.
sum(List) -> sum(List, 0).
sum([], A) -> A;
sum([H|T], A) -> sum(T, H + A).
If you're with me so far, then you're ready to understand folds.
There is a problem with the function above; it is too specific. It braids three ideas together without specifying any independently:
iteration
accumulation
addition
It is easy to miss the difference between iteration and accumulation because most of the time we never give this a second thought. Most languages accidentally encourage us to miss the difference, actually, by having the same storage location change its value each iteration of a similar function.
It is easy to miss the independence of addition merely because of the way it is written in this example because "+" looks like an "operation", not a function.
What if I had said "Start with 1, then take one element from the list at a time and multiply it by the running value"? We would still be doing the list processing in exactly the same way, but with two examples to compare it is pretty clear that multiplication and addition are the only difference between the two:
prod(List) -> prod(List, 1).
prod([], A) -> A;
prod([H|T], A) -> prod(T, H * A).
This is exactly the same flow of execution but for the inner operation and the starting value of the accumulator.
So let's make the addition and multiplication bits into functions, so we can pull that part of the pattern out:
add(A, B) -> A + B.
mult(A, B) -> A * B.
How could we write the list operation on its own? We need to pass a function in -- addition or multiplication -- and have it operate over the values. Also, we have to pay attention to the identity of the type and operation of things we are operating on or else we will screw up the magic that is value aggregation. "add(0, X)" always returns X, so this idea (0 + Foo) is the addition identity operation. In multiplication the identity operation is to multiply by 1. So we must start our accumulator at 0 for addition and 1 for multiplication (and for building lists an empty list, and so on). So we can't write the function with an accumulator value built-in, because it will only be correct for some type+operation pairs.
So this means to write a fold we need to have a list argument, a function to do things argument, and an accumulator argument, like so:
fold([], _, Accumulator) ->
Accumulator;
fold([H|T], Operation, Accumulator) ->
fold(T, Operation, Operation(H, Accumulator)).
With this definition we can now write sum/1 using this more general pattern:
fsum(List) -> fold(List, fun add/2, 0).
And prod/1 also:
fprod(List) -> fold(List, fun prod/2, 1).
And they are functionally identical to the one we wrote above, but the notation is more clear and we don't have to write a bunch of recursive details that tangle the idea of iteration with the idea of accumulation with the idea of some specific operation like multiplication or addition.
In the case of the RPN calculator the idea of aggregate list operations is combined with the concept of selective dispatch (picking an operation to perform based on what symbol is encountered/matched). The RPN example is relatively simple and small (you can fit all the code in your head at once, it's just a few lines), but until you get used to functional paradigms the process it manifests can make your head hurt. In functional programming a tiny amount of code can create an arbitrarily complex process of unpredictable (or even evolving!) behavior, based just on list operations and selective dispatch; this is very different from the conditional checks, input validation and procedural checking techniques used in other paradigms more common today. Analyzing such behavior is greatly assisted by single assignment and recursive notation, because each iteration is a conceptually independent slice of time which can be contemplated in isolation of the rest of the system. I'm talking a little ahead of the basic question, but this is a core idea you may wish to contemplate as you consider why we like to use operations like folds and recursive notations instead of procedural, multiple-assignment loops.
I hope this helped more than confused.
First, you have to remember haw works rpn. If you want to execute the following operation: 2 * (3 + 5), you will feed the function with the input: "3 5 + 2 *". This was useful at a time where you had 25 step to enter a program :o)
the first function called simply split this character list into element:
1> string:tokens("3 5 + 2 *"," ").
["3","5","+","2","*"]
2>
then it processes the lists:foldl/3. for each element of this list, rpn/2 is called with the head of the input list and the current accumulator, and return a new accumulator. lets go step by step:
Step head accumulator matched rpn/2 return value
1 "3" [] rpn(X, Stack) -> [read(X) | Stack]. [3]
2 "5" [3] rpn(X, Stack) -> [read(X) | Stack]. [5,3]
3 "+" [5,3] rpn("+", [N1,N2|S]) -> [N2+N1|S]; [8]
4 "2" [8] rpn(X, Stack) -> [read(X) | Stack]. [2,8]
5 "*" [2,8] rpn("*",[N1,N2|S]) -> [N2*N1|S]; [16]
At the end, lists:foldl/3 returns [16] which matches to [R], and though rpn/1 returns R = 16
I have been playing with an implementation of lookandsay (OEIS A005150) in J. I have made two versions, both very simple, using while. type control structures. One recurs, the other loops. Because I am compulsive, I started running comparative timing on the versions.
look and say is the sequence 1 11 21 1211 111221 that s, one one, two ones, etc.
For early elements of the list (up to around 20) the looping version wins, but only by a tiny amount. Timings around 30 cause the recursive version to win, by a large enough amount that the recursive version might be preferred if the stack space were adequate to support it. I looked at why, and I believe that it has to do with handling intermediate results. The 30th number in the sequence has 5808 digits. (32nd number, 9898 digits, 34th, 16774.)
When you are doing the problem with recursion, you can hold the intermediate results in the recursive call, and the unstacking at the end builds the results so that there is minimal handling of the results.
In the list version, you need a variable to hold the result. Every loop iteration causes you to need to add two elements to the result.
The problem, as I see it, is that I can't find any way in J to modify an extant array without completely reassigning it. So I am saying
try. o =. o,e,(0&{y) catch. o =. e,(0&{y) end.
to put an element into o where o might not have a value when we start. That may be notably slower than
o =. i.0
.
.
.
o =. (,o),e,(0&{y)
The point is that the result gets the wrong shape without the ravels, or so it seems. It is inheriting a shape from i.0 somehow.
But even functions like } amend don't modify a list, they return a list that has a modification made to it, and if you want to save the list you need to assign it. As the size of the assigned list increases (as you walk the the number from the beginning to the end making the next number) the assignment seems to take more time and more time. This assignment is really the only thing I can see that would make element 32, 9898 digits, take less time in the recursive version while element 20 (408 digits) takes less time in the loopy version.
The recursive version builds the return with:
e,(0&{y),(,lookandsay e }. y)
The above line is both the return line from the function and the recursion, so the whole return vector gets built at once as the call gets to the end of the string and everything unstacks.
In APL I thought that one could say something on the order of:
a[1+rho a] <- new element
But when I try this in NARS2000 I find that it causes an index error. I don't have access to any other APL, I might be remembering this idiom from APL Plus, I doubt it worked this way in APL\360 or APL\1130. I might be misremembering it completely.
I can find no way to do that in J. It might be that there is no way to do that, but the next thought is to pre-allocate an array that could hold results, and to change individual entries. I see no way to do that either - that is, J does not seem to support the APL idiom:
a<- iota 5
a[3] <- -1
Is this one of those side effect things that is disallowed because of language purity?
Does the interpreter recognize a=. a,foo or some of its variants as a thing that it should fastpath to a[>:#a]=.foo internally?
This is the recursive version, just for the heck of it. I have tried a bunch of different versions and I believe that the longer the program, the slower, and generally, the more complex, the slower. Generally, the program can be chained so that if you want the nth number you can do lookandsay^: n ] y. I have tried a number of optimizations, but the problem I have is that I can't tell what environment I am sending my output into. If I could tell that I was sending it to the next iteration of the program I would send it as an array of digits rather than as a big number.
I also suspect that if I could figure out how to make a tacit version of the code, it would run faster, based on my finding that when I add something to the code that should make it shorter, it runs longer.
lookandsay=: 3 : 0
if. 0 = # ,y do. return. end. NB. return on empty argument
if. 1 ~: ##$ y do. NB. convert rank 0 argument to list of digits
y =. (10&#.^:_1) x: y
f =. 1
assert. 1 = ##$ y NB. the converted argument must be rank 1
else.
NB. yw =. y
f =. 0
end.
NB. e should be a count of the digits that match the leading digit.
e=.+/*./\y=0&{y
if. f do.
o=. e,(0&{y),(,lookandsay e }. y)
assert. e = 0&{ o
10&#. x: o
return.
else.
e,(0&{y),(,lookandsay e }. y)
return.
end.
)
I was interested in the characteristics of the numbers produced. I found that if you start with a 1, the numerals never get higher than 3. If you start with a numeral higher than 3, it will survive as a singleton, and you can also get a number into the generated numbers by starting with something like 888888888 which will generate a number with one 9 in it and a single 8 at the end of the number. But other than the singletons, no digit gets higher than 3.
Edit:
I did some more measuring. I had originally written the program to accept either a vector or a scalar, the idea being that internally I'd work with a vector. I had thought about passing a vector from one layer of code to the other, and I still might using a left argument to control code. With I pass the top level a vector the code runs enormously faster, so my guess is that most of the cpu is being eaten by converting very long numbers from vectors to digits. The recursive routine always passes down a vector when it recurs which might be why it is almost as fast as the loop.
That does not change my question.
I have an answer for this which I can't post for three hours. I will post it then, please don't do a ton of research to answer it.
assignments like
arr=. 'z' 15} arr
are executed in place. (See JWiki article for other supported in-place operations)
Interpreter determines that only small portion of arr is updated and does not create entire new list to reassign.
What happens in your case is not that array is being reassigned, but that it grows many times in small increments, causing memory allocation and reallocation.
If you preallocate (by assigning it some large chunk of data), then you can modify it with } without too much penalty.
After I asked this question, to be honest, I lost track of this web site.
Yes, the answer is that the language has no form that means "update in place, but if you use two forms
x =: x , most anything
or
x =: most anything } x
then the interpreter recognizes those as special and does update in place unless it can't. There are a number of other specials recognized by the interpreter, like:
199(1000&|#^)199
That combined operation is modular exponentiation. It never calculates the whole exponentiation, as
199(1000&|^)199
would - that just ends as _ without the #.
So it is worth reading the article on specials. I will mark someone else's answer up.
The link that sverre provided above ( http://www.jsoftware.com/jwiki/Essays/In-Place%20Operations ) shows the various operations that support modifying an existing array rather than creating a new one. They include:
myarray=: myarray,'blah'
If you are interested in a tacit version of the lookandsay sequence see this submission to RosettaCode:
las=: ,#((# , {.);.1~ 1 , 2 ~:/\ ])&.(10x&#.inv)#]^:(1+i.#[)
5 las 1
11 21 1211 111221 312211
Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
One of the topics that seems to come up regularly on mailing lists and online discussions is the merits (or lack thereof) of doing a Computer Science Degree. An argument that seems to come up time and again for the negative party is that they have been coding for some number of years and they have never used recursion.
So the question is:
What is recursion?
When would I use recursion?
Why don't people use recursion?
There are a number of good explanations of recursion in this thread, this answer is about why you shouldn't use it in most languages.* In the majority of major imperative language implementations (i.e. every major implementation of C, C++, Basic, Python, Ruby,Java, and C#) iteration is vastly preferable to recursion.
To see why, walk through the steps that the above languages use to call a function:
space is carved out on the stack for the function's arguments and local variables
the function's arguments are copied into this new space
control jumps to the function
the function's code runs
the function's result is copied into a return value
the stack is rewound to its previous position
control jumps back to where the function was called
Doing all of these steps takes time, usually a little bit more than it takes to iterate through a loop. However, the real problem is in step #1. When many programs start, they allocate a single chunk of memory for their stack, and when they run out of that memory (often, but not always due to recursion), the program crashes due to a stack overflow.
So in these languages recursion is slower and it makes you vulnerable to crashing. There are still some arguments for using it though. In general, code written recursively is shorter and a bit more elegant, once you know how to read it.
There is a technique that language implementers can use called tail call optimization which can eliminate some classes of stack overflow. Put succinctly: if a function's return expression is simply the result of a function call, then you don't need to add a new level onto the stack, you can reuse the current one for the function being called. Regrettably, few imperative language-implementations have tail-call optimization built in.
* I love recursion. My favorite static language doesn't use loops at all, recursion is the only way to do something repeatedly. I just don't think that recursion is generally a good idea in languages that aren't tuned for it.
** By the way Mario, the typical name for your ArrangeString function is "join", and I'd be surprised if your language of choice doesn't already have an implementation of it.
Simple english example of recursion.
A child couldn't sleep, so her mother told her a story about a little frog,
who couldn't sleep, so the frog's mother told her a story about a little bear,
who couldn't sleep, so the bear's mother told her a story about a little weasel...
who fell asleep.
...and the little bear fell asleep;
...and the little frog fell asleep;
...and the child fell asleep.
In the most basic computer science sense, recursion is a function that calls itself. Say you have a linked list structure:
struct Node {
Node* next;
};
And you want to find out how long a linked list is you can do this with recursion:
int length(const Node* list) {
if (!list->next) {
return 1;
} else {
return 1 + length(list->next);
}
}
(This could of course be done with a for loop as well, but is useful as an illustration of the concept)
Whenever a function calls itself, creating a loop, then that's recursion. As with anything there are good uses and bad uses for recursion.
The most simple example is tail recursion where the very last line of the function is a call to itself:
int FloorByTen(int num)
{
if (num % 10 == 0)
return num;
else
return FloorByTen(num-1);
}
However, this is a lame, almost pointless example because it can easily be replaced by more efficient iteration. After all, recursion suffers from function call overhead, which in the example above could be substantial compared to the operation inside the function itself.
So the whole reason to do recursion rather than iteration should be to take advantage of the call stack to do some clever stuff. For example, if you call a function multiple times with different parameters inside the same loop then that's a way to accomplish branching. A classic example is the Sierpinski triangle.
You can draw one of those very simply with recursion, where the call stack branches in 3 directions:
private void BuildVertices(double x, double y, double len)
{
if (len > 0.002)
{
mesh.Positions.Add(new Point3D(x, y + len, -len));
mesh.Positions.Add(new Point3D(x - len, y - len, -len));
mesh.Positions.Add(new Point3D(x + len, y - len, -len));
len *= 0.5;
BuildVertices(x, y + len, len);
BuildVertices(x - len, y - len, len);
BuildVertices(x + len, y - len, len);
}
}
If you attempt to do the same thing with iteration I think you'll find it takes a lot more code to accomplish.
Other common use cases might include traversing hierarchies, e.g. website crawlers, directory comparisons, etc.
Conclusion
In practical terms, recursion makes the most sense whenever you need iterative branching.
Recursion is a method of solving problems based on the divide and conquer mentality.
The basic idea is that you take the original problem and divide it into smaller (more easily solved) instances of itself, solve those smaller instances (usually by using the same algorithm again) and then reassemble them into the final solution.
The canonical example is a routine to generate the Factorial of n. The Factorial of n is calculated by multiplying all of the numbers between 1 and n. An iterative solution in C# looks like this:
public int Fact(int n)
{
int fact = 1;
for( int i = 2; i <= n; i++)
{
fact = fact * i;
}
return fact;
}
There's nothing surprising about the iterative solution and it should make sense to anyone familiar with C#.
The recursive solution is found by recognising that the nth Factorial is n * Fact(n-1). Or to put it another way, if you know what a particular Factorial number is you can calculate the next one. Here is the recursive solution in C#:
public int FactRec(int n)
{
if( n < 2 )
{
return 1;
}
return n * FactRec( n - 1 );
}
The first part of this function is known as a Base Case (or sometimes Guard Clause) and is what prevents the algorithm from running forever. It just returns the value 1 whenever the function is called with a value of 1 or less. The second part is more interesting and is known as the Recursive Step. Here we call the same method with a slightly modified parameter (we decrement it by 1) and then multiply the result with our copy of n.
When first encountered this can be kind of confusing so it's instructive to examine how it works when run. Imagine that we call FactRec(5). We enter the routine, are not picked up by the base case and so we end up like this:
// In FactRec(5)
return 5 * FactRec( 5 - 1 );
// which is
return 5 * FactRec(4);
If we re-enter the method with the parameter 4 we are again not stopped by the guard clause and so we end up at:
// In FactRec(4)
return 4 * FactRec(3);
If we substitute this return value into the return value above we get
// In FactRec(5)
return 5 * (4 * FactRec(3));
This should give you a clue as to how the final solution is arrived at so we'll fast track and show each step on the way down:
return 5 * (4 * FactRec(3));
return 5 * (4 * (3 * FactRec(2)));
return 5 * (4 * (3 * (2 * FactRec(1))));
return 5 * (4 * (3 * (2 * (1))));
That final substitution happens when the base case is triggered. At this point we have a simple algrebraic formula to solve which equates directly to the definition of Factorials in the first place.
It's instructive to note that every call into the method results in either a base case being triggered or a call to the same method where the parameters are closer to a base case (often called a recursive call). If this is not the case then the method will run forever.
Recursion is solving a problem with a function that calls itself. A good example of this is a factorial function. Factorial is a math problem where factorial of 5, for example, is 5 * 4 * 3 * 2 * 1. This function solves this in C# for positive integers (not tested - there may be a bug).
public int Factorial(int n)
{
if (n <= 1)
return 1;
return n * Factorial(n - 1);
}
Recursion refers to a method which solves a problem by solving a smaller version of the problem and then using that result plus some other computation to formulate the answer to the original problem. Often times, in the process of solving the smaller version, the method will solve a yet smaller version of the problem, and so on, until it reaches a "base case" which is trivial to solve.
For instance, to calculate a factorial for the number X, one can represent it as X times the factorial of X-1. Thus, the method "recurses" to find the factorial of X-1, and then multiplies whatever it got by X to give a final answer. Of course, to find the factorial of X-1, it'll first calculate the factorial of X-2, and so on. The base case would be when X is 0 or 1, in which case it knows to return 1 since 0! = 1! = 1.
Consider an old, well known problem:
In mathematics, the greatest common divisor (gcd) … of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder.
The definition of gcd is surprisingly simple:
where mod is the modulo operator (that is, the remainder after integer division).
In English, this definition says the greatest common divisor of any number and zero is that number, and the greatest common divisor of two numbers m and n is the greatest common divisor of n and the remainder after dividing m by n.
If you'd like to know why this works, see the Wikipedia article on the Euclidean algorithm.
Let's compute gcd(10, 8) as an example. Each step is equal to the one just before it:
gcd(10, 8)
gcd(10, 10 mod 8)
gcd(8, 2)
gcd(8, 8 mod 2)
gcd(2, 0)
2
In the first step, 8 does not equal zero, so the second part of the definition applies. 10 mod 8 = 2 because 8 goes into 10 once with a remainder of 2. At step 3, the second part applies again, but this time 8 mod 2 = 0 because 2 divides 8 with no remainder. At step 5, the second argument is 0, so the answer is 2.
Did you notice that gcd appears on both the left and right sides of the equals sign? A mathematician would say this definition is recursive because the expression you're defining recurs inside its definition.
Recursive definitions tend to be elegant. For example, a recursive definition for the sum of a list is
sum l =
if empty(l)
return 0
else
return head(l) + sum(tail(l))
where head is the first element in a list and tail is the rest of the list. Note that sum recurs inside its definition at the end.
Maybe you'd prefer the maximum value in a list instead:
max l =
if empty(l)
error
elsif length(l) = 1
return head(l)
else
tailmax = max(tail(l))
if head(l) > tailmax
return head(l)
else
return tailmax
You might define multiplication of non-negative integers recursively to turn it into a series of additions:
a * b =
if b = 0
return 0
else
return a + (a * (b - 1))
If that bit about transforming multiplication into a series of additions doesn't make sense, try expanding a few simple examples to see how it works.
Merge sort has a lovely recursive definition:
sort(l) =
if empty(l) or length(l) = 1
return l
else
(left,right) = split l
return merge(sort(left), sort(right))
Recursive definitions are all around if you know what to look for. Notice how all of these definitions have very simple base cases, e.g., gcd(m, 0) = m. The recursive cases whittle away at the problem to get down to the easy answers.
With this understanding, you can now appreciate the other algorithms in Wikipedia's article on recursion!
A function that calls itself
When a function can be (easily) decomposed into a simple operation plus the same function on some smaller portion of the problem. I should say, rather, that this makes it a good candidate for recursion.
They do!
The canonical example is the factorial which looks like:
int fact(int a)
{
if(a==1)
return 1;
return a*fact(a-1);
}
In general, recursion isn't necessarily fast (function call overhead tends to be high because recursive functions tend to be small, see above) and can suffer from some problems (stack overflow anyone?). Some say they tend to be hard to get 'right' in non-trivial cases but I don't really buy into that. In some situations, recursion makes the most sense and is the most elegant and clear way to write a particular function. It should be noted that some languages favor recursive solutions and optimize them much more (LISP comes to mind).
A recursive function is one which calls itself. The most common reason I've found to use it is traversing a tree structure. For example, if I have a TreeView with checkboxes (think installation of a new program, "choose features to install" page), I might want a "check all" button which would be something like this (pseudocode):
function cmdCheckAllClick {
checkRecursively(TreeView1.RootNode);
}
function checkRecursively(Node n) {
n.Checked = True;
foreach ( n.Children as child ) {
checkRecursively(child);
}
}
So you can see that the checkRecursively first checks the node which it is passed, then calls itself for each of that node's children.
You do need to be a bit careful with recursion. If you get into an infinite recursive loop, you will get a Stack Overflow exception :)
I can't think of a reason why people shouldn't use it, when appropriate. It is useful in some circumstances, and not in others.
I think that because it's an interesting technique, some coders perhaps end up using it more often than they should, without real justification. This has given recursion a bad name in some circles.
Recursion is an expression directly or indirectly referencing itself.
Consider recursive acronyms as a simple example:
GNU stands for GNU's Not Unix
PHP stands for PHP: Hypertext Preprocessor
YAML stands for YAML Ain't Markup Language
WINE stands for Wine Is Not an Emulator
VISA stands for Visa International Service Association
More examples on Wikipedia
Recursion works best with what I like to call "fractal problems", where you're dealing with a big thing that's made of smaller versions of that big thing, each of which is an even smaller version of the big thing, and so on. If you ever have to traverse or search through something like a tree or nested identical structures, you've got a problem that might be a good candidate for recursion.
People avoid recursion for a number of reasons:
Most people (myself included) cut their programming teeth on procedural or object-oriented programming as opposed to functional programming. To such people, the iterative approach (typically using loops) feels more natural.
Those of us who cut our programming teeth on procedural or object-oriented programming have often been told to avoid recursion because it's error prone.
We're often told that recursion is slow. Calling and returning from a routine repeatedly involves a lot of stack pushing and popping, which is slower than looping. I think some languages handle this better than others, and those languages are most likely not those where the dominant paradigm is procedural or object-oriented.
For at least a couple of programming languages I've used, I remember hearing recommendations not to use recursion if it gets beyond a certain depth because its stack isn't that deep.
A recursive statement is one in which you define the process of what to do next as a combination of the inputs and what you have already done.
For example, take factorial:
factorial(6) = 6*5*4*3*2*1
But it's easy to see factorial(6) also is:
6 * factorial(5) = 6*(5*4*3*2*1).
So generally:
factorial(n) = n*factorial(n-1)
Of course, the tricky thing about recursion is that if you want to define things in terms of what you have already done, there needs to be some place to start.
In this example, we just make a special case by defining factorial(1) = 1.
Now we see it from the bottom up:
factorial(6) = 6*factorial(5)
= 6*5*factorial(4)
= 6*5*4*factorial(3) = 6*5*4*3*factorial(2) = 6*5*4*3*2*factorial(1) = 6*5*4*3*2*1
Since we defined factorial(1) = 1, we reach the "bottom".
Generally speaking, recursive procedures have two parts:
1) The recursive part, which defines some procedure in terms of new inputs combined with what you've "already done" via the same procedure. (i.e. factorial(n) = n*factorial(n-1))
2) A base part, which makes sure that the process doesn't repeat forever by giving it some place to start (i.e. factorial(1) = 1)
It can be a bit confusing to get your head around at first, but just look at a bunch of examples and it should all come together. If you want a much deeper understanding of the concept, study mathematical induction. Also, be aware that some languages optimize for recursive calls while others do not. It's pretty easy to make insanely slow recursive functions if you're not careful, but there are also techniques to make them performant in most cases.
Hope this helps...
I like this definition:
In recursion, a routine solves a small part of a problem itself, divides the problem into smaller pieces, and then calls itself to solve each of the smaller pieces.
I also like Steve McConnells discussion of recursion in Code Complete where he criticises the examples used in Computer Science books on Recursion.
Don't use recursion for factorials or Fibonacci numbers
One problem with
computer-science textbooks is that
they present silly examples of
recursion. The typical examples are
computing a factorial or computing a
Fibonacci sequence. Recursion is a
powerful tool, and it's really dumb to
use it in either of those cases. If a
programmer who worked for me used
recursion to compute a factorial, I'd
hire someone else.
I thought this was a very interesting point to raise and may be a reason why recursion is often misunderstood.
EDIT:
This was not a dig at Dav's answer - I had not seen that reply when I posted this
1.)
A method is recursive if it can call itself; either directly:
void f() {
... f() ...
}
or indirectly:
void f() {
... g() ...
}
void g() {
... f() ...
}
2.) When to use recursion
Q: Does using recursion usually make your code faster?
A: No.
Q: Does using recursion usually use less memory?
A: No.
Q: Then why use recursion?
A: It sometimes makes your code much simpler!
3.) People use recursion only when it is very complex to write iterative code. For example, tree traversal techniques like preorder, postorder can be made both iterative and recursive. But usually we use recursive because of its simplicity.
Here's a simple example: how many elements in a set. (there are better ways to count things, but this is a nice simple recursive example.)
First, we need two rules:
if the set is empty, the count of items in the set is zero (duh!).
if the set is not empty, the count is one plus the number of items in the set after one item is removed.
Suppose you have a set like this: [x x x]. let's count how many items there are.
the set is [x x x] which is not empty, so we apply rule 2. the number of items is one plus the number of items in [x x] (i.e. we removed an item).
the set is [x x], so we apply rule 2 again: one + number of items in [x].
the set is [x], which still matches rule 2: one + number of items in [].
Now the set is [], which matches rule 1: the count is zero!
Now that we know the answer in step 4 (0), we can solve step 3 (1 + 0)
Likewise, now that we know the answer in step 3 (1), we can solve step 2 (1 + 1)
And finally now that we know the answer in step 2 (2), we can solve step 1 (1 + 2) and get the count of items in [x x x], which is 3. Hooray!
We can represent this as:
count of [x x x] = 1 + count of [x x]
= 1 + (1 + count of [x])
= 1 + (1 + (1 + count of []))
= 1 + (1 + (1 + 0)))
= 1 + (1 + (1))
= 1 + (2)
= 3
When applying a recursive solution, you usually have at least 2 rules:
the basis, the simple case which states what happens when you have "used up" all of your data. This is usually some variation of "if you are out of data to process, your answer is X"
the recursive rule, which states what happens if you still have data. This is usually some kind of rule that says "do something to make your data set smaller, and reapply your rules to the smaller data set."
If we translate the above to pseudocode, we get:
numberOfItems(set)
if set is empty
return 0
else
remove 1 item from set
return 1 + numberOfItems(set)
There's a lot more useful examples (traversing a tree, for example) which I'm sure other people will cover.
Well, that's a pretty decent definition you have. And wikipedia has a good definition too. So I'll add another (probably worse) definition for you.
When people refer to "recursion", they're usually talking about a function they've written which calls itself repeatedly until it is done with its work. Recursion can be helpful when traversing hierarchies in data structures.
An example: A recursive definition of a staircase is:
A staircase consists of:
- a single step and a staircase (recursion)
- or only a single step (termination)
To recurse on a solved problem: do nothing, you're done.
To recurse on an open problem: do the next step, then recurse on the rest.
In plain English:
Assume you can do 3 things:
Take one apple
Write down tally marks
Count tally marks
You have a lot of apples in front of you on a table and you want to know how many apples there are.
start
Is the table empty?
yes: Count the tally marks and cheer like it's your birthday!
no: Take 1 apple and put it aside
Write down a tally mark
goto start
The process of repeating the same thing till you are done is called recursion.
I hope this is the "plain english" answer you are looking for!
A recursive function is a function that contains a call to itself. A recursive struct is a struct that contains an instance of itself. You can combine the two as a recursive class. The key part of a recursive item is that it contains an instance/call of itself.
Consider two mirrors facing each other. We've seen the neat infinity effect they make. Each reflection is an instance of a mirror, which is contained within another instance of a mirror, etc. The mirror containing a reflection of itself is recursion.
A binary search tree is a good programming example of recursion. The structure is recursive with each Node containing 2 instances of a Node. Functions to work on a binary search tree are also recursive.
This is an old question, but I want to add an answer from logistical point of view (i.e not from algorithm correctness point of view or performance point of view).
I use Java for work, and Java doesn't support nested function. As such, if I want to do recursion, I might have to define an external function (which exists only because my code bumps against Java's bureaucratic rule), or I might have to refactor the code altogether (which I really hate to do).
Thus, I often avoid recursion, and use stack operation instead, because recursion itself is essentially a stack operation.
You want to use it anytime you have a tree structure. It is very useful in reading XML.
Recursion as it applies to programming is basically calling a function from inside its own definition (inside itself), with different parameters so as to accomplish a task.
"If I have a hammer, make everything look like a nail."
Recursion is a problem-solving strategy for huge problems, where at every step just, "turn 2 small things into one bigger thing," each time with the same hammer.
Example
Suppose your desk is covered with a disorganized mess of 1024 papers. How do you make one neat, clean stack of papers from the mess, using recursion?
Divide: Spread all the sheets out, so you have just one sheet in each "stack".
Conquer:
Go around, putting each sheet on top of one other sheet. You now have stacks of 2.
Go around, putting each 2-stack on top of another 2-stack. You now have stacks of 4.
Go around, putting each 4-stack on top of another 4-stack. You now have stacks of 8.
... on and on ...
You now have one huge stack of 1024 sheets!
Notice that this is pretty intuitive, aside from counting everything (which isn't strictly necessary). You might not go all the way down to 1-sheet stacks, in reality, but you could and it would still work. The important part is the hammer: With your arms, you can always put one stack on top of the other to make a bigger stack, and it doesn't matter (within reason) how big either stack is.
Recursion is the process where a method call iself to be able to perform a certain task. It reduces redundency of code. Most recurssive functions or methods must have a condifiton to break the recussive call i.e. stop it from calling itself if a condition is met - this prevents the creating of an infinite loop. Not all functions are suited to be used recursively.
hey, sorry if my opinion agrees with someone, I'm just trying to explain recursion in plain english.
suppose you have three managers - Jack, John and Morgan.
Jack manages 2 programmers, John - 3, and Morgan - 5.
you are going to give every manager 300$ and want to know what would it cost.
The answer is obvious - but what if 2 of Morgan-s employees are also managers?
HERE comes the recursion.
you start from the top of the hierarchy. the summery cost is 0$.
you start with Jack,
Then check if he has any managers as employees. if you find any of them are, check if they have any managers as employees and so on. Add 300$ to the summery cost every time you find a manager.
when you are finished with Jack, go to John, his employees and then to Morgan.
You'll never know, how much cycles will you go before getting an answer, though you know how many managers you have and how many Budget can you spend.
Recursion is a tree, with branches and leaves, called parents and children respectively.
When you use a recursion algorithm, you more or less consciously are building a tree from the data.
In plain English, recursion means to repeat someting again and again.
In programming one example is of calling the function within itself .
Look on the following example of calculating factorial of a number:
public int fact(int n)
{
if (n==0) return 1;
else return n*fact(n-1)
}
Any algorithm exhibits structural recursion on a datatype if basically consists of a switch-statement with a case for each case of the datatype.
for example, when you are working on a type
tree = null
| leaf(value:integer)
| node(left: tree, right:tree)
a structural recursive algorithm would have the form
function computeSomething(x : tree) =
if x is null: base case
if x is leaf: do something with x.value
if x is node: do something with x.left,
do something with x.right,
combine the results
this is really the most obvious way to write any algorith that works on a data structure.
now, when you look at the integers (well, the natural numbers) as defined using the Peano axioms
integer = 0 | succ(integer)
you see that a structural recursive algorithm on integers looks like this
function computeSomething(x : integer) =
if x is 0 : base case
if x is succ(prev) : do something with prev
the too-well-known factorial function is about the most trivial example of
this form.
function call itself or use its own definition.