Can someone explain me this kind of recursion? - recursion

I want to understand recursion.
I understand stupid example with math but i m not sure to know the essence of it.
I have 1 example that i don t understand how it works:
TREE-ROOT-INSERT(x, z)
if x = NIL
return z
if z.key < x.key
x.left = TREE-ROOT-INSERT(x.left, z)
return RIGHT-ROTATE(x)
else
x.right = TREE-ROOT-INSERT(x.right, z)
return LEFT-ROTATE(x)
I know what this code does:
First insert a node in a BST and then rotate each time so the new node became the root.
But in my mind analysing the code i suppose that it insert the node where it has to go and then JUST 1 TIME it rotates the tree.
How is it possible that the tree is rotated every time?

You need to maintain your place in the recursive call for each level of the tree. When you hit return RIGHT-ROTATE (or left) for the first time, you're not completely done; you take the tree that is the result of the ROTATE function, and place it in the code where the recursive TREE-ROOT-INSERT call was one level higher in the stack. You then rotate again, and return the current tree one level higher up in the stack, until you've hit the original root of the tree.

What is important for understanding recursion is to think of the recursive function as an abstract black box. In other words, when reading or reasoning about recursive function, you should focus on the current iteration, treat the invocation of the recursive function as atomic (something you cannot explore into) assuming it can do what it is supposed to do, and see how its result can be used to solve the current iteration.
You already know the contract of your TREE-ROOT-INSERT(x, z):
insert z into a binary search tree rooted at x, transform the tree so that z will become the new root.
let's look at this snippet:
if z.key < x.key
x.left = TREE-ROOT-INSERT(x.left, z)
return RIGHT-ROTATE(x)
This says z is less than x so it goes to the left sub-tree (because it is BST). TREE-ROOT-INSERT is invoked again, but we won't follow into it. Instead we just assume it can do what it is meant to do: it will insert z to the tree rooted at x.left, and make z the new root. Then you will get a tree of the bellow structure:
x
/ \
z ...
/ \
... ...
Again, you don't know how exactly calling TREE-ROOT-INSERT(x.left, z) will get you the z-rooted sub-tree. At this moment you don't care, because the real important part is what follows: how to make this entire tree rooted at z? The answer is the RIGHT-ROTATE(x)
But in my mind analysing the code i suppose that it insert the node where it has to go and then JUST 1 TIME it rotates the tree.
How is it possible that the tree is rotated every time?
If I understand you correctly, you are still thinking how to solve the problem in a non-recursive way. It is true that you can insert z into the BST rooted at x using the standard BST insertion procedure. That will put z in the correct position. However to bring z to the root from that position, you need more than 1 rotation.
In the recursive version, rotation is required to bring z to the root after you get a z-rooted sub-tree. But to get the z-rooted sub-tree from the original x.left rooted sub-tree, you need a rotation as well. Rotation is called many times, but on different sub-trees.

Related

Find depth difference between two nodes in a tree without going all the way up to the root

I Came across a question I couldn't figure out.
Assume A binary tree, where each node has a pointer to its parent, and two children.
We receive input for r, p,q ( root, and two nodes from the tree) and we want to find the depth difference between p and q, Without Traversing all the way up to the root.
Limitations and Assumptions:
No memory usage (meaning no arrays, variables are ok)
Assume the nodes Lowest Common Ancestor isn't the root (and for simplicity let's assume the LCA is far from the root in Big-O notation Terms.
Just to clarify -
Basically, the question is how do we figure out the depth difference without going all the way up to the root, and without memory.
Thanks!
Basically, the question is how do we figure out the depth difference without going all the way up to the root...
It isn't saying that you cannot traverse all the way up to the root. It just says that the final outcome may not be the root.
First, define a function to determine the depth of a node (pseudo code):
function depthOf(r, p):
depth = 0
temp = p
while curr is not r:
depth++
temp = temp.parent
return depth
Simply call this function for both nodes, and take the difference.
As an additional feature, you can use this information to find the Lowest Common Ancestor:
Use the depth difference to walk up from the deepest node of the two up until you are at the same depth as the other. From then on walk upwards from both nodes in tandem, until you arrive at the same node from both sides:
function least(r, p, q)
diff = depthOf(r, p) - depthOf(r, q) # The answer
# In order to find least common ancestor, continue as follows:
while diff > 0: # p is deeper than q
p = p.parent
diff--
while diff < 0: # q is deeper than p
q = q.parent
diff++
while p is not q:
p = p.parent
q = q.parent
return p

what is the lowelink mean of Tarjan's algorithm

I was reading the description of Tarjan's algorithm for finding the strongly connected components in a driected graph.
But I find it hard to understand these codes snippet:
if (w.index is undefined) then
// Successor w has not yet been visited; recurse on it
strongconnect(w)
v.lowlink := min(v.lowlink, w.lowlink)
else if (w is in S) then
// Successor w is in stack S and hence in the current SCC
v.lowlink := min(v.lowlink, w.index)
end if
the fourth and the seventh lines are different, this make me confused.
And in my opinion,the seveth line could write as the same way with the fourth line
v.lowkink := min(v.lowlink, w.index)
I test this in my program and it works fine, and for me, it's better to understand bcz verdex v cloud reach hight up root, but i couldn't prove itT_T.
I wrote a program that enumerated all graphs of size 4, then run each version (with either min(v.lowlink, w.index) or min(v.lowlink, w.lowlink) if w is in S) and compared the results. Both were exactly identical in all cases, even though w.lowlink and w.index were often different.
The reason why we can use w.index is this: consider where on the stack S relative to the current node v the other node w is.
If it's earlier on the stack then it has a smaller index than the current node (because it was visited earlier, duh), so the current node is not the "head" of its connected component and that would be reflected in v.lowlink <= w.index < v.index anyway. And it's not like w.lowlink has any particular meaning at this point either, it's in the progress of being computed and doesn't necessarily have its final value yet.
Now, if w is later in the stack than v, then the crucial property that the algorithm depends on is that then w is a descendant of v, not some sibling/cousing node still left there from an earlier recursive call. Or, as it is usually stated in a complete proof, strongly connected components never span several unconnected branches of our search tree (forest). Because since it's an SCC, there must be a path from w to v, and since we are enumerating stuff in a depth-first order, we must have visited v using that path from w before we have finished processing w, so w should be earlier in the stack than v!
And if w is a descendant of v then we already got its actual lowlink value the first time we visited it and are not interested in it any more.
On a side note, it's trivial to get rid of the lowlink property on nodes and make strongconnect return it directly. Then we wouldn't be tempted to check it instead of w.index in the second case =)

Creating a BST from an array

I need to create a binary search tree in the following (strange) way:
I am given an array (A[n]). A[1] becomes the root of the tree.
Then, I insert A[1]+A[2] to the left subtree (subtree1, used below) of the root and also insert A[1]-A[2] to the right subtree (subtree2) of the root.
I insert A[1]+A[2]+A[3] to the left subtree of subtree1 (subtree3) and A[1]+A[2]-A[3] to the right subtree of subtree1 (subtree4).
Then, I insert A[1]-A[2]+A[3] to the left subtree of subtree2 (subtree5) and A[1]-A[2]-A[3] to the right subtree of subtree2 (subtree6).
I repeat for subtree3, subtree4, subtree5, subtree6 until I reach the end of the array.
So, basically, the first element of the array becomes the root of the tree and then I move down: Every left subtree has for value the sum of its parent plus the next element of the array and every right subtree has for value the difference of its parent and of the next element in the array.
I understand I need to use the concept of recursion but in a modified way. Typing my problem here and trying to explain it to someone else apart from my brain actually made me form it in a way that gave me some ideas to try but I can see the problem I am dealing with being a usual problem so maybe you could give me some pointers on how to use recursion to build the tree.
Looking around at other questions and the discussions I understand there is a policy against asking whole solutions so I wanted to make it clear that I am not asking for the solution but for guidance to it. If someone would like to have a look I can show you what I've already done.
The way to do recursion is to always assume you already have a working function in hand. So let's see [using Java syntax]...
Tree buildTree(int currentSum, int[] array, int index, boolean sign);
Suppose that works. Then would do u need to do to build a tree at index i?
// current value to look at at this level
int curValue = array[index];
// depending on sign, it may be negative
if (!sign) {
curValue *= -1;
}
// add it to the running total
int nodeValue = currentSum + curValue;
Node nd = new Node(nodeValue);
nd.left = buildTree(nodeValue, array, index + 1, true);
nd.right = buildTree(nodeValue, array, index + 1, false);
That's basically it. You need to take care of the edge cases: of index = array.length, creation of the very first node, and the like

Simple Recursion?

I'm new to programming and have had a hard time understanding recursion. There's a problem I've been working on but can't figure out. I really just don't understand how they are solvable.
"Define a procedure plus that takes two non-negative integers and returns their sum. The only procedures (other than recursive calls to plus) that you may use are: zero?, sub1, and add1.
I know that this is a built in functions in scheme, so I know they're possible to solve, I just don't understand how. Recursion is so tricky!
Any help would be greatly appreciated!
=] Thanks
I'm working in Petite Chez Scheme (with the SWL editor)
Recursion is a very important concept in software development. I don't know (petit chez) scheme so I will approach this from a general angle.
The concept of a recursive function is to repeat the same task over and over again until you reach some limiting boundary. Taking your first question, you have two numbers and you need to add them together. However, you only have the ability to add 1 to a number or subtract 1 from a number. You also have the literal value zero.
So. consider you numbers as two buckets. They each have 10 stones in them. You want to "add" those two buckets together. You are only permitted to move one stone at a time (i.e. you can't grab a handful or tip one bucket into the other).
Lets say you want to move everything from the left bucket into the right bucket, one stone at a time. What are you going to have to do?
First, you have to take 1 stone from the left bucket, i.e. you are using sub1 to remove one stone from the bucket. You then add that same stone to the right bucket, i.e. you add1 to the right bucket.
Now you could do this in a loop, but you don't know how many stones there will be in any given solution. What you really want to do is say "Take one stone from the left bucket, put it in the right bucket and repeat until there are no stones in the left bucket.' This case of there being no stones in the left bucket is called you "Base Case". It's the point at which you say OK, I'm done now.
A pseudocode example of this situation would be (using your plus, add1, sub1 and zero):
plus(leftBucket, rightBucket)
{
if(leftBucket == zero) // check if the left bucket is empty yet
{
// the left bucket is empty, we've moved all the stones
return rightBucket; // the right bucket must be full
}
else
{
// we still have stones in the left bucket, remove 1,
// put it in the right bucket, repeat.
return plus(sub1(leftBucket), add1(rightBucket));
}
}
If you still need more help, let me know, I can run through other examples but this looks like it's probably a homework problem for you and recursion is incredibly important to understand so I don't want to just give you all the answers.
Recursion is simply a function that calls itself. The most common, easily understood examples of recursion is walking a data structure that looks like a tree.
How would you visit each branch of a tree? You would start at the trunk and call visit(branch), passing the trunk of the tree as the first branch. Visit() calls itself for each branch of each branch, and so on.
public void visit(Branch branch)
{
// do something with this branch here
// visit the branches of this branch
foreach(var subbranch in branch.branches)
{
visit(subbranch)
}
}
Recursion is closely related to induction - first you solve (or prove) a base case, and then you assume your solution is correct for some value n, and use that to solve (or prove) it for n + 1.
So the first step here is to look at the first problem. What would be a good base case for adding two numbers together?
Alright, so we have our base case: when one of the numbers is zero.
For simplicity's sake, we'll assume that the second number is zero, just to make things a little easier.
So we know that (+ n 0) is equal to n. So now for our recursive step, we want to take an arbitrary call (+ x y), and turn that into a call which is closer to our ideal (+ n 0). That way we'll have made some progress and will eventually solve our problem.
So how are we going to do this?
(+ x y) is of course equivalent to (+ (add1 x) (sub1 y)) - which takes us closer to our base case of (zero? y).
This gives us our final solution:
(define (+ x y)
(if (zero? y)
(x)
(+ (add1 x) (sub1 y))
))
(you can, of course, swap the order of the arguments and it will still be equivalent).
A similar mechanism can be used to solve the other two problems.

What is recursion and when should I use it?

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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.

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