Alternate recursion function definition syntax in Mathematica - recursion

Suppose I wish to define a recursive function theta whose functionality should be apparent below.
The following definition will work.
theta[0] = 0;
theta[i_ ] := theta[i-1] + 1
However, this will not work.
theta[0] = 0;
theta[i_ + 1] := theta[i] + 1
My question is, is it possible to make something like the second definition work, where I can define the function based on the i+1 term instead of the i term?
I understand that they are mathematically equivalent, but I am curious about whether Mathematica will permit something like the second syntax.

It is perfectly feasible to make your second definition work if you understand that default automatic simplifications are done, often before you can get control, and if you then use your definition with appropriate parameters that match your definition.
Example
In[1]:= theta[i_ + 1] := Sin[i]+1;
theta[a + 1]
Out[2]= 1+Sin[a]
but then you probably expect to use this as
In[3]:= theta[8]
Out[3]= theta[8]
and that fails because you defined a function that matches the sum of something and one, but gave it just an integer and you have no definition that matches that. Even this fails
In[4]:= theta[7 + 1]
Out[4]= theta[8]
because the default automatic rules turn the sum of two integers into an integer and you are back to the previous case.
It is sometimes said that Mathematica does "structural" matching, if two structure of two expressions match the Mathematica accepts this as a match. This is very different from the sort of matching that anyone with a bit of mathematical maturity would use. A decade or more ago someone wrote up an article in the Mathematica Journal showing that it would be possible to use a more mathematical version of matching within Mathematica. I think that was completely ignored and nothing more was ever done with that. It would be nice if someone with the skill needed could bring that code up to the current version of Mathematica, but I think this might be a substantial challenge.

There is always "a way". For example:
ClearAll[a];
a[i_] = a[i] /. First#RSolve[{a[i + 1] == a[i] + 1, a[0] == 0}, a[i], i]

Related

Define a new method with only a few changes

I want to write a version that accepts a supplementary argument. The difference with the initial version only resides in a few lines of codes, potentially within loops. A typical example is to user a vector of weight w.
One solution is to completely rewrite a new function
function f(Vector::a)
...
for x in a
...
s += x[i]
...
end
...
end
function f(a::Vector, w::Vector)
...
for x in a
...
s += x[i] * w[i]
...
end
...
end
This solution duplicates code and therefore makes the program harder to maintain.
I could split ... into different helper functions, which are called by both functions, but the resulting code would be hard to follow
Another solution is to write only one function and use a ? : structure for each line that should be changed
function f(a, w::Union(Nothing, Vector) = nothing)
....
for x in a
...
s += (w == nothing)? x[i] : x[i] * w[i]
...
end
....
end
This code requires to check a condition at every step in a loop, which does not sound efficient, compared to the first version.
I'm sure there is a better solution, maybe using macros. What would be a good way to deal with this?
There are lots of ways to do this sort of thing, ranging from optional arguments to custom types to metaprogramming with #eval'ed code generation (this would splice in the changes for each new method as you loop over a list of possibilities).
I think in this case I'd use a combination of the approaches suggested by #ColinTBowers and #GnimucKey.
It's fairly simple to define a custom array type that is all ones:
immutable Ones{N} <: AbstractArray{Int,N}
dims::NTuple{N, Int}
end
Base.size(O::Ones) = O.dims
Base.getindex(O::Ones, I::Int...) = (checkbounds(O, I...); 1)
I've chosen to use an Int as the element type since it tends to promote well. Now all you need is to be a bit more flexible in your argument list and you're good to go:
function f(a::Vector, w::AbstractVector=Ones(size(a))
…
This should have a lower overhead than either of the other proposed solutions; getindex should inline nicely as a bounds check and the number 1, there's no type instability, and you don't need to rewrite your algorithm. If you're sure that all your accesses are in-bounds, you could even remove the bounds checking as an additional optimization. Or on a recent 0.4, you could define and use Base.unsafe_getindex(O::Ones, I::Int...) = 1 (that won't quite work on 0.3 since it's not guaranteed to be defined for all AbstractArrays).
In this case, using Optional Arguments may play the trick.
Just make the w argument default to ones().
I've come up against this problem a few times. If you want to avoid the conditional if statement inside the loop, one possibility is to use multiple dispatch over some dummy types. For example:
abstract MyFuncTypes
type FuncWithNoWeight <: MyFuncTypes; end
evaluate(x::Vector, i::Int, ::FuncWithNoWeight) = x[i]
type FuncWithWeight{T} <: MyFuncTypes
w::Vector{T}
end
evaluate(x::Vector, i::Int, wT::FuncWithWeight) = x[i] * wT.w[i]
function f(a, w::MyFuncTypes=FuncWithNoWeight())
....
for x in a
...
s += evaluate(x, i, w)
...
end
....
end
I extend the evaluate method over FuncWithNoWeight and FuncWithWeight in order to get the appropriate behaviour. I also nest these types within an abstract type MyFuncTypes, which is the second input to f (with default value of FuncWithNoWeight). From here, multiple dispatch and Julia's type system takes care of the rest.
One neat thing about this approach is that if you decide later on you want to add a third type of behaviour inside the loop (not necessarily even weighting, pretty much any type of transformation will be possible), it is as simple as defining a new type, nesting it under MyFuncTypes, and extending the evaluate method to the new type.
UPDATE: As Matt B. has pointed out, the first version of my answer accidentally introduced type instability into the function with my solution. As a general rule I typically find that if Matt posts something it is worth paying close attention (hint, hint, check out his answer). I'm still learning a lot about Julia (and am answering questions on StackOverflow to facilitate that learning). I've updated my answer to remove the type instability pointed out by Matt.

how to write a formula in c#?

how to write a formula like
v_r (t)=∑_(n=0)^(N-1)▒〖A_r (L_2-L_1 ) e^j(ω_c t-4π/λ (R+υt+L_(1+L_2 )/2 cos⁡〖(θ)sin⁡(ω_r t+2πn/N)))〗 ┤) sinc(4π/λ-L_(2-L_1 )/2 cos⁡(θ) sin⁡(ω_r t+2πn/N))〗
in c#?
You have to convert the formula to something the compiler recognizes.
To it's equivalent using the a combination of basic algebra and the Math class like so:
p = rho*R*T + (B_0*R*T-A_0-((C_0) / (T*T))+((E_0) / (Math.Pow(T, 4))))*rho*rho +
(b*R*T-a-((d) / (T)))*Math.Pow(rho, 3) +
alpha*(a+((d) / (t)))*Math.Pow(rho, 6) +
((c*Math.Pow(rho, 3)) / (T*T))*(1+gamma*rho*rho)*Math.Exp(-gamma*rho*rho);
Example taken from: Converting Math Equations in C#
Well, first you have to figure out what all those symbols mean. I see the sigma which usually indicates sum-of, with ∑_(n=0)^(N-1) probably translating to:
N-1
∑
n=0
This generally means the sum of the following expression where n varies from 0 to N-1. So I gather you'd need a loop there.
The expression to be calculated within that loop consists of a lof of trigonometric-type functions involving π, θ, sin and cos, and the little known sinc which I assume is a typo :-)
The bottom line is that you need to understand the current expression before you can think about converting it to another form like a C# program. Short of knowing where it came from, or a little bit of context, we probably can't help you that much though there's always a possibility that we have a savant/genius here that will recognise that formula off the top of their head.

Replacing functions with Table Lookups

I've been watching this MSDN video with Brian Beckman and I'd like to better understand something he says:
Every imperitive programmer goes through this phase of learning that
functions can be replaced with table lookups
Now, I'm a C# programmer who never went to university, so perhaps somewhere along the line I missed out on something everyone else learned to understand.
What does Brian mean by:
functions can be replaced with table lookups
Are there practical examples of this being done and does it apply to all functions? He gives the example of the sin function, which I can make sense of, but how do I make sense of this in more general terms?
Brian just showed that the functions are data too. Functions in general are just a mapping of one set to another: y = f(x) is mapping of set {x} to set {y}: f:X->Y. The tables are mappings as well: [x1, x2, ..., xn] -> [y1, y2, ..., yn].
If function operates on finite set (this is the case in programming) then it's can be replaced with a table which represents that mapping. As Brian mentioned, every imperative programmer goes through this phase of understanding that the functions can be replaced with the table lookups just for performance reason.
But it doesn't mean that all functions easily can or should be replaced with the tables. It only means that you theoretically can do that for every function. So the conclusion would be that the functions are data because tables are (in the context of programming of course).
There is a lovely trick in Mathematica that creates a table as a side-effect of evaluating function-calls-as-rewrite-rules. Consider the classic slow-fibonacci
fib[1] = 1
fib[2] = 1
fib[n_] := fib[n-1] + fib[n-2]
The first two lines create table entries for the inputs 1 and 2. This is exactly the same as saying
fibTable = {};
fibTable[1] = 1;
fibTable[2] = 1;
in JavaScript. The third line of Mathematica says "please install a rewrite rule that will replace any occurrence of fib[n_], after substituting the pattern variable n_ with the actual argument of the occurrence, with fib[n-1] + fib[n-2]." The rewriter will iterate this procedure, and eventually produce the value of fib[n] after an exponential number of rewrites. This is just like the recursive function-call form that we get in JavaScript with
function fib(n) {
var result = fibTable[n] || ( fib(n-1) + fib(n-2) );
return result;
}
Notice it checks the table first for the two values we have explicitly stored before making the recursive calls. The Mathematica evaluator does this check automatically, because the order of presentation of the rules is important -- Mathematica checks the more specific rules first and the more general rules later. That's why Mathematica has two assignment forms, = and :=: the former is for specific rules whose right-hand sides can be evaluated at the time the rule is defined; the latter is for general rules whose right-hand sides must be evaluated when the rule is applied.
Now, in Mathematica, if we say
fib[4]
it gets rewritten to
fib[3] + fib[2]
then to
fib[2] + fib[1] + 1
then to
1 + 1 + 1
and finally to 3, which does not change on the next rewrite. You can imagine that if we say fib[35], we will generate enormous expressions, fill up memory, and melt the CPU. But the trick is to replace the final rewrite rule with the following:
fib[n_] := fib[n] = fib[n-1] + fib[n-2]
This says "please replace every occurrence of fib[n_] with an expression that will install a new specific rule for the value of fib[n] and also produce the value." This one runs much faster because it expands the rule-base -- the table of values! -- at run time.
We can do likewise in JavaScript
function fib(n) {
var result = fibTable[n] || ( fib(n-1) + fib(n-2) );
fibTable[n] = result;
return result;
}
This runs MUCH faster than the prior definition of fib.
This is called "automemoization" [sic -- not "memorization" but "memoization" as in creating a memo for yourself].
Of course, in the real world, you must manage the sizes of the tables that get created. To inspect the tables in Mathematica, do
DownValues[fib]
To inspect them in JavaScript, do just
fibTable
in a REPL such as that supported by Node.JS.
In the context of functional programming, there is the concept of referential transparency. A function that is referentially transparent can be replaced with its value for any given argument (or set of arguments), without changing the behaviour of the program.
Referential Transparency
For example, consider a function F that takes 1 argument, n. F is referentially transparent, so F(n) can be replaced with the value of F evaluated at n. It makes no difference to the program.
In C#, this would look like:
public class Square
{
public static int apply(int n)
{
return n * n;
}
public static void Main()
{
//Should print 4
Console.WriteLine(Square.apply(2));
}
}
(I'm not very familiar with C#, coming from a Java background, so you'll have to forgive me if this example isn't quite syntactically correct).
It's obvious here that the function apply cannot have any other value than 4 when called with an argument of 2, since it's just returning the square of its argument. The value of the function only depends on its argument, n; in other words, referential transparency.
I ask you, then, what the difference is between Console.WriteLine(Square.apply(2)) and Console.WriteLine(4). The answer is, there's no difference at all, for all intents are purposes. We could go through the entire program, replacing all instances of Square.apply(n) with the value returned by Square.apply(n), and the results would be the exact same.
So what did Brian Beckman mean with his statement about replacing function calls with a table lookup? He was referring to this property of referentially transparent functions. If Square.apply(2) can be replaced with 4 with no impact on program behaviour, then why not just cache the values when the first call is made, and put it in a table indexed by the arguments to the function. A lookup table for values of Square.apply(n) would look somewhat like this:
n: 0 1 2 3 4 5 ...
Square.apply(n): 0 1 4 9 16 25 ...
And for any call to Square.apply(n), instead of calling the function, we can simply find the cached value for n in the table, and replace the function call with this value. It's fairly obvious that this will most likely bring about a large speed increase in the program.

Derivative Calculator

I'm interested in building a derivative calculator. I've racked my brains over solving the problem, but I haven't found a right solution at all. May you have a hint how to start? Thanks
I'm sorry! I clearly want to make symbolic differentiation.
Let's say you have the function f(x) = x^3 + 2x^2 + x
I want to display the derivative, in this case f'(x) = 3x^2 + 4x + 1
I'd like to implement it in objective-c for the iPhone.
I assume that you're trying to find the exact derivative of a function. (Symbolic differentiation)
You need to parse the mathematical expression and store the individual operations in the function in a tree structure.
For example, x + sin²(x) would be stored as a + operation, applied to the expression x and a ^ (exponentiation) operation of sin(x) and 2.
You can then recursively differentiate the tree by applying the rules of differentiation to each node. For example, a + node would become the u' + v', and a * node would become uv' + vu'.
you need to remember your calculus. basically you need two things: table of derivatives of basic functions and rules of how to derivate compound expressions (like d(f + g)/dx = df/dx + dg/dx). Then take expressions parser and recursively go other the tree. (http://www.sosmath.com/tables/derivative/derivative.html)
Parse your string into an S-expression (even though this is usually taken in Lisp context, you can do an equivalent thing in pretty much any language), easiest with lex/yacc or equivalent, then write a recursive "derive" function. In OCaml-ish dialect, something like this:
let rec derive var = function
| Const(_) -> Const(0)
| Var(x) -> if x = var then Const(1) else Deriv(Var(x), Var(var))
| Add(x, y) -> Add(derive var x, derive var y)
| Mul(a, b) -> Add(Mul(a, derive var b), Mul(derive var a, b))
...
(If you don't know OCaml syntax - derive is two-parameter recursive function, with first parameter the variable name, and the second being mathched in successive lines; for example, if this parameter is a structure of form Add(x, y), return the structure Add built from two fields, with values of derived x and derived y; and similarly for other cases of what derive might receive as a parameter; _ in the first pattern means "match anything")
After this you might have some clean-up function to tidy up the resultant expression (reducing fractions etc.) but this gets complicated, and is not necessary for derivation itself (i.e. what you get without it is still a correct answer).
When your transformation of the s-exp is done, reconvert the resultant s-exp into string form, again with a recursive function
SLaks already described the procedure for symbolic differentiation. I'd just like to add a few things:
Symbolic math is mostly parsing and tree transformations. ANTLR is a great tool for both. I'd suggest starting with this great book Language implementation patterns
There are open-source programs that do what you want (e.g. Maxima). Dissecting such a program might be interesting, too (but it's probably easier to understand what's going on if you tried to write it yourself, first)
Probably, you also want some kind of simplification for the output. For example, just applying the basic derivative rules to the expression 2 * x would yield 2 + 0*x. This can also be done by tree processing (e.g. by transforming 0 * [...] to 0 and [...] + 0 to [...] and so on)
For what kinds of operations are you wanting to compute a derivative? If you allow trigonometric functions like sine, cosine and tangent, these are probably best stored in a table while others like polynomials may be much easier to do. Are you allowing for functions to have multiple inputs,e.g. f(x,y) rather than just f(x)?
Polynomials in a single variable would be my suggestion and then consider adding in trigonometric, logarithmic, exponential and other advanced functions to compute derivatives which may be harder to do.
Symbolic differentiation over common functions (+, -, *, /, ^, sin, cos, etc.) ignoring regions where the function or its derivative is undefined is easy. What's difficult, perhaps counterintuitively, is simplifying the result afterward.
To do the differentiation, store the operations in a tree (or even just in Polish notation) and make a table of the derivative of each of the elementary operations. Then repeatedly apply the chain rule and the elementary derivatives, together with setting the derivative of a constant to 0. This is fast and easy to implement.

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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