Related
Considering the classical staircase problem as "Davis has a number of staircases in his house and he likes to climb each staircase 1, 2, or 3 steps at a time. Being a very precocious child, he wonders how many ways there are to reach the top of the staircase."
My approach is to use memoization with recursion as
# TimeO(N), SpaceO(N), DP Bottom Up + Memoization
def stepPerms(n, memo = {}):
if n < 3:
return n
elif n == 3:
return 4
if n in memo:
return memo[n]
else:
memo[n] = stepPerms(n - 1, memo) + stepPerms(n - 2 ,memo) + stepPerms(n - 3 ,memo)
return memo[n]
The question that comes to my mind is that, is this solution bottom-up or top-down. My way of approaching it is that since we go all the way down to calculate the upper N values (imagine the recursion tree). I consider this bottom-up. Is this correct?
Recoursion strategies are as a general rule topdown approaches, whether they have memory or not. The underlaying algorithm design is dynamic programming, which traditionally built in a bottom-up fashion.
I noticed that you wrote your code in python, and python is generally not happy about deep recoursion (small amounts are okay, but performance quickly takes a hit and there is a maximum recousion depth of 1000 - unless it was changed since I read that).
If we make a bottom-up dynamic programmin version, we can get rid of this recousion, and we can also recognise that we only need constant amount of space, since we are only really interested in the last 3 values:
def stepPerms(n):
if n < 1: return n
memo = [1,2,4]
if n <= 3: return memo[n-1]
for i in range(3,n):
memo[i % 3] = sum(memo)
return memo[n-1]
Notice how much simpler the logic is, appart from the i is one less than the value, since the positions are starts a 0 instead of the count of 1.
In the top-down approach, the complex module is divided into submodules. So it is top down approach. On the other hand, bottom-up approach begins with elementary modules and then combine them further.
And bottom up approach of this solution will be:
memo{}
for i in range(0,3):
memo[i]=i
memo[3]=4
for i in range(4,n+1):
memo[i]=memo[i-1]+memo[i-2]+memo[i-3]
"1. The program will calculate the values using repetitive execution (loops)."
Recursion is repetitive execution but, i do not think it's a loop, do you think if I used recursion it would follow the guideline above?
No. In fact, it looks like the assignment is specifically asking for the "opposite" of recursion, iteration.
Loops are fundamentally about iteration, which is different to recursion. The main difference is that an iteration uses a constant amount of space, whereas recursion uses more space the deeper the recursion goes. For example, here are iterative and recursive procedures to compute the sum of the numbers from 1 to n
def sum_iter(n):
x = 0
for i in range(1,n+1):
x += i
return x
def sum_recursive(n):
if n == 0:
return 0
else:
return n + sum_recursive(n-1)
If you run these with a very large argument, you will run out of stack space (a "stack overflow") on the recursive version, but the iterative version will work fine.
There is a special kind of recursion called tail recursion, which is where the function doesn't have to do anything with the value from a recursive call except pass it to the caller. In this case you don't need to keep track of the stack - you can just jump straight to the top. This is called tail call optimization. A tail recursive function to calculate the sum of the integers 1 to n looks like
def sum_tailrec(n):
def helper(s,i):
if i == 0:
return s
else:
return helper(s+i, i-1)
return helper(0, n)
In this case people often refer to the function helper as an iterative recursion, because (with tail call optimization) it is only using a constant amount of space.
This is all a bit moot, because Python doesn't have tail call optimization, but some languages do.
This logic programming is really making a lap dance on my imperative programming skills. This is homework, so please just don't drop me the answer. This is what I have:
fibo(N,1) :-
N < 2,
!.
fibo(N,R) :-
N1 is N-1,
N2 is N-2,
fibo(N1,R1),
fibo(N2,R2),
R is R1+R2.
I'm suppose to make another function that looks like this; fib(N,Value,LastValue).
N is the n'th number, and value is the return value. I don't understand how I can rewrite this using accumulation. And since it counts backwards I don't see how it can "know" a last value before it calculates anything. :s Any input is appreciated.
I could post here the solution, but since that this is homework, it would be counter-productive. Instead, here's a lead:
The problem with the version of Fibonacci that you listed is that it is inefficient. Each call to fibo/2 causes another two calls, but some of these calls calculate the values of the same Fibonacci numbers. For example, in pseudo-code:
(a) fibo(4) -> fibo(3), fibo(2)
(b) fibo(3) -> fibo(2), fibo(1)
(c) fibo(2) -> fibo(1), fibo(0) % called from (a)
(d) fibo(2) -> fibo(1), fibo(0) % called from (b), redundant
To overcome this deficiency, you were asked to rephrase Fibonacci in terms of returning not just the last value, but the last two values, so that each call to fib/3 will cause only a single recursive call (hence calculate the Fibonacci series in linear time). You'll need to change the base cases to:
fib(1,1,0).
fib(2,1,1).
I'll leave the recursive case to you.
For the impatient
Here is the recursive case as well:
fib(N, Val, Last) :-
N > 2,
N1 is N - 1,
fib(N1, Last, Last1), % single call with two output arguments,
% instead of two calls with one output argument
Val is Last + Last1.
See the related discussion:
Generalizing Fibonacci sequence with SICStus Prolog
and consider ony's very good solution using finite domain constraints from there.
Perhaps using tail recursion is a good option
edit:
Instead of breaking fib(6) into fib(5)+fib(4) you might try something like fib(6) = fib(6,0,0) the first parameter is the count of steps, when it reaches 0 you stop, the second parameter the last value you calculated, and the third parameter is the value to calculate which is equal to the sum of current second and third parameters (with the exception of the first step, in which 0 + 0 will be 1)
So to calculate you set the second parameter at each call and acumulate in the third so fib(6,0,0) => fib(5,0,1) => fib(4,1,1) => fib(3,1,2) => fib(2,2,3) => fib(1,3,5) => fib(0,5,8) then you return 8
In that method you dont actually have to save in the stack the adress return, avoiding stack overflow
Remember that there is another way to calculate the Fibonacci sequence: starting from the base case and moving up.
Right now, to calculate fib(n), you add fib(n-1) and fib(n-2). Instead, flip that around and calculate fib(0) and fib(1) based on the definition of the Fibonacci sequence, and build up from that.
You almost already have it. Just rewrite:
fibo(N, Value) :-
N1 is N-1, N2 is N-2,
fibo(N1, LastValue),fibo(N2, SecondToLastValue),
Value is LastValue + SecondToLastValue.
in terms of
fibo2(N, Value, LastValue):- ...
I don't understand how I can rewrite
this using accumulation
Just don't, this is not needed (although it's possible to do so).
Is there a performance hit if we use a loop instead of recursion or vice versa in algorithms where both can serve the same purpose? Eg: Check if the given string is a palindrome.
I have seen many programmers using recursion as a means to show off when a simple iteration algorithm can fit the bill.
Does the compiler play a vital role in deciding what to use?
Loops may achieve a performance gain for your program. Recursion may achieve a performance gain for your programmer. Choose which is more important in your situation!
It is possible that recursion will be more expensive, depending on if the recursive function is tail recursive (the last line is recursive call). Tail recursion should be recognized by the compiler and optimized to its iterative counterpart (while maintaining the concise, clear implementation you have in your code).
I would write the algorithm in the way that makes the most sense and is the clearest for the poor sucker (be it yourself or someone else) that has to maintain the code in a few months or years. If you run into performance issues, then profile your code, and then and only then look into optimizing by moving over to an iterative implementation. You may want to look into memoization and dynamic programming.
Comparing recursion to iteration is like comparing a phillips head screwdriver to a flat head screwdriver. For the most part you could remove any phillips head screw with a flat head, but it would just be easier if you used the screwdriver designed for that screw right?
Some algorithms just lend themselves to recursion because of the way they are designed (Fibonacci sequences, traversing a tree like structure, etc.). Recursion makes the algorithm more succinct and easier to understand (therefore shareable and reusable).
Also, some recursive algorithms use "Lazy Evaluation" which makes them more efficient than their iterative brothers. This means that they only do the expensive calculations at the time they are needed rather than each time the loop runs.
That should be enough to get you started. I'll dig up some articles and examples for you too.
Link 1: Haskel vs PHP (Recursion vs Iteration)
Here is an example where the programmer had to process a large data set using PHP. He shows how easy it would have been to deal with in Haskel using recursion, but since PHP had no easy way to accomplish the same method, he was forced to use iteration to get the result.
http://blog.webspecies.co.uk/2011-05-31/lazy-evaluation-with-php.html
Link 2: Mastering Recursion
Most of recursion's bad reputation comes from the high costs and inefficiency in imperative languages. The author of this article talks about how to optimize recursive algorithms to make them faster and more efficient. He also goes over how to convert a traditional loop into a recursive function and the benefits of using tail-end recursion. His closing words really summed up some of my key points I think:
"recursive programming gives the programmer a better way of organizing
code in a way that is both maintainable and logically consistent."
https://developer.ibm.com/articles/l-recurs/
Link 3: Is recursion ever faster than looping? (Answer)
Here is a link to an answer for a stackoverflow question that is similar to yours. The author points out that a lot of the benchmarks associated with either recursing or looping are very language specific. Imperative languages are typically faster using a loop and slower with recursion and vice-versa for functional languages. I guess the main point to take from this link is that it is very difficult to answer the question in a language agnostic / situation blind sense.
Is recursion ever faster than looping?
Recursion is more costly in memory, as each recursive call generally requires a memory address to be pushed to the stack - so that later the program could return to that point.
Still, there are many cases in which recursion is a lot more natural and readable than loops - like when working with trees. In these cases I would recommend sticking to recursion.
Typically, one would expect the performance penalty to lie in the other direction. Recursive calls can lead to the construction of extra stack frames; the penalty for this varies. Also, in some languages like Python (more correctly, in some implementations of some languages...), you can run into stack limits rather easily for tasks you might specify recursively, such as finding the maximum value in a tree data structure. In these cases, you really want to stick with loops.
Writing good recursive functions can reduce the performance penalty somewhat, assuming you have a compiler that optimizes tail recursions, etc. (Also double check to make sure that the function really is tail recursive---it's one of those things that many people make mistakes on.)
Apart from "edge" cases (high performance computing, very large recursion depth, etc.), it's preferable to adopt the approach that most clearly expresses your intent, is well-designed, and is maintainable. Optimize only after identifying a need.
Recursion is better than iteration for problems that can be broken down into multiple, smaller pieces.
For example, to make a recursive Fibonnaci algorithm, you break down fib(n) into fib(n-1) and fib(n-2) and compute both parts. Iteration only allows you to repeat a single function over and over again.
However, Fibonacci is actually a broken example and I think iteration is actually more efficient. Notice that fib(n) = fib(n-1) + fib(n-2) and fib(n-1) = fib(n-2) + fib(n-3). fib(n-1) gets calculated twice!
A better example is a recursive algorithm for a tree. The problem of analyzing the parent node can be broken down into multiple smaller problems of analyzing each child node. Unlike the Fibonacci example, the smaller problems are independent of each other.
So yeah - recursion is better than iteration for problems that can be broken down into multiple, smaller, independent, similar problems.
Your performance deteriorates when using recursion because calling a method, in any language, implies a lot of preparation: the calling code posts a return address, call parameters, some other context information such as processor registers might be saved somewhere, and at return time the called method posts a return value which is then retrieved by the caller, and any context information that was previously saved will be restored. the performance diff between an iterative and a recursive approach lies in the time these operations take.
From an implementation point of view, you really start noticing the difference when the time it takes to handle the calling context is comparable to the time it takes for your method to execute. If your recursive method takes longer to execute then the calling context management part, go the recursive way as the code is generally more readable and easy to understand and you won't notice the performance loss. Otherwise go iterative for efficiency reasons.
I believe tail recursion in java is not currently optimized. The details are sprinkled throughout this discussion on LtU and the associated links. It may be a feature in the upcoming version 7, but apparently it presents certain difficulties when combined with Stack Inspection since certain frames would be missing. Stack Inspection has been used to implement their fine-grained security model since Java 2.
http://lambda-the-ultimate.org/node/1333
There are many cases where it gives a much more elegant solution over the iterative method, the common example being traversal of a binary tree, so it isn't necessarily more difficult to maintain. In general, iterative versions are usually a bit faster (and during optimization may well replace a recursive version), but recursive versions are simpler to comprehend and implement correctly.
Recursion is very useful is some situations. For example consider the code for finding the factorial
int factorial ( int input )
{
int x, fact = 1;
for ( x = input; x > 1; x--)
fact *= x;
return fact;
}
Now consider it by using the recursive function
int factorial ( int input )
{
if (input == 0)
{
return 1;
}
return input * factorial(input - 1);
}
By observing these two, we can see that recursion is easy to understand.
But if it is not used with care it can be so much error prone too.
Suppose if we miss if (input == 0), then the code will be executed for some time and ends with usually a stack overflow.
In many cases recursion is faster because of caching, which improves performance. For example, here is an iterative version of merge sort using the traditional merge routine. It will run slower than the recursive implementation because of caching improved performances.
Iterative implementation
public static void sort(Comparable[] a)
{
int N = a.length;
aux = new Comparable[N];
for (int sz = 1; sz < N; sz = sz+sz)
for (int lo = 0; lo < N-sz; lo += sz+sz)
merge(a, lo, lo+sz-1, Math.min(lo+sz+sz-1, N-1));
}
Recursive implementation
private static void sort(Comparable[] a, Comparable[] aux, int lo, int hi)
{
if (hi <= lo) return;
int mid = lo + (hi - lo) / 2;
sort(a, aux, lo, mid);
sort(a, aux, mid+1, hi);
merge(a, aux, lo, mid, hi);
}
PS - this is what was told by Professor Kevin Wayne (Princeton University) on the course on algorithms presented on Coursera.
Using recursion, you're incurring the cost of a function call with each "iteration", whereas with a loop, the only thing you usually pay is an increment/decrement. So, if the code for the loop isn't much more complicated than the code for the recursive solution, loop will usually be superior to recursion.
Recursion and iteration depends on the business logic that you want to implement, though in most of the cases it can be used interchangeably. Most developers go for recursion because it is easier to understand.
It depends on the language. In Java you should use loops. Functional languages optimize recursion.
Recursion has a disadvantage that the algorithm that you write using recursion has O(n) space complexity.
While iterative aproach have a space complexity of O(1).This is the advantange of using iteration over recursion.
Then why do we use recursion?
See below.
Sometimes it is easier to write an algorithm using recursion while it's slightly tougher to write the same algorithm using iteration.In this case if you opt to follow the iteration approach you would have to handle stack yourself.
If you're just iterating over a list, then sure, iterate away.
A couple of other answers have mentioned (depth-first) tree traversal. It really is such a great example, because it's a very common thing to do to a very common data structure. Recursion is extremely intuitive for this problem.
Check out the "find" methods here:
http://penguin.ewu.edu/cscd300/Topic/BSTintro/index.html
Recursion is more simple (and thus - more fundamental) than any possible definition of an iteration. You can define a Turing-complete system with only a pair of combinators (yes, even a recursion itself is a derivative notion in such a system). Lambda calculus is an equally powerful fundamental system, featuring recursive functions. But if you want to define an iteration properly, you'd need much more primitives to start with.
As for the code - no, recursive code is in fact much easier to understand and to maintain than a purely iterative one, since most data structures are recursive. Of course, in order to get it right one would need a language with a support for high order functions and closures, at least - to get all the standard combinators and iterators in a neat way. In C++, of course, complicated recursive solutions can look a bit ugly, unless you're a hardcore user of FC++ and alike.
I would think in (non tail) recursion there would be a performance hit for allocating a new stack etc every time the function is called (dependent on language of course).
it depends on "recursion depth".
it depends on how much the function call overhead will influence the total execution time.
For example, calculating the classical factorial in a recursive way is very inefficient due to:
- risk of data overflowing
- risk of stack overflowing
- function call overhead occupy 80% of execution time
while developing a min-max algorithm for position analysis in the game of chess that will analyze subsequent N moves can be implemented in recursion over the "analysis depth" (as I'm doing ^_^)
Recursion? Where do I start, wiki will tell you “it’s the process of repeating items in a self-similar way"
Back in day when I was doing C, C++ recursion was a god send, stuff like "Tail recursion". You'll also find many sorting algorithms use recursion. Quick sort example: http://alienryderflex.com/quicksort/
Recursion is like any other algorithm useful for a specific problem. Perhaps you mightn't find a use straight away or often but there will be problem you’ll be glad it’s available.
In C++ if the recursive function is a templated one, then the compiler has more chance to optimize it, as all the type deduction and function instantiations will occur in compile time. Modern compilers can also inline the function if possible. So if one uses optimization flags like -O3 or -O2 in g++, then recursions may have the chance to be faster than iterations. In iterative codes, the compiler gets less chance to optimize it, as it is already in the more or less optimal state (if written well enough).
In my case, I was trying to implement matrix exponentiation by squaring using Armadillo matrix objects, in both recursive and iterative way. The algorithm can be found here... https://en.wikipedia.org/wiki/Exponentiation_by_squaring.
My functions were templated and I have calculated 1,000,000 12x12 matrices raised to the power 10. I got the following result:
iterative + optimisation flag -O3 -> 2.79.. sec
recursive + optimisation flag -O3 -> 1.32.. sec
iterative + No-optimisation flag -> 2.83.. sec
recursive + No-optimisation flag -> 4.15.. sec
These results have been obtained using gcc-4.8 with c++11 flag (-std=c++11) and Armadillo 6.1 with Intel mkl. Intel compiler also shows similar results.
Mike is correct. Tail recursion is not optimized out by the Java compiler or the JVM. You will always get a stack overflow with something like this:
int count(int i) {
return i >= 100000000 ? i : count(i+1);
}
You have to keep in mind that utilizing too deep recursion you will run into Stack Overflow, depending on allowed stack size. To prevent this make sure to provide some base case which ends you recursion.
Using just Chrome 45.0.2454.85 m, recursion seems to be a nice amount faster.
Here is the code:
(function recursionVsForLoop(global) {
"use strict";
// Perf test
function perfTest() {}
perfTest.prototype.do = function(ns, fn) {
console.time(ns);
fn();
console.timeEnd(ns);
};
// Recursion method
(function recur() {
var count = 0;
global.recurFn = function recurFn(fn, cycles) {
fn();
count = count + 1;
if (count !== cycles) recurFn(fn, cycles);
};
})();
// Looped method
function loopFn(fn, cycles) {
for (var i = 0; i < cycles; i++) {
fn();
}
}
// Tests
var curTest = new perfTest(),
testsToRun = 100;
curTest.do('recursion', function() {
recurFn(function() {
console.log('a recur run.');
}, testsToRun);
});
curTest.do('loop', function() {
loopFn(function() {
console.log('a loop run.');
}, testsToRun);
});
})(window);
RESULTS
// 100 runs using standard for loop
100x for loop run.
Time to complete: 7.683ms
// 100 runs using functional recursive approach w/ tail recursion
100x recursion run.
Time to complete: 4.841ms
In the screenshot below, recursion wins again by a bigger margin when run at 300 cycles per test
If the iterations are atomic and orders of magnitude more expensive than pushing a new stack frame and creating a new thread and you have multiple cores and your runtime environment can use all of them, then a recursive approach could yield a huge performance boost when combined with multithreading. If the average number of iterations is not predictable then it might be a good idea to use a thread pool which will control thread allocation and prevent your process from creating too many threads and hogging the system.
For example, in some languages, there are recursive multithreaded merge sort implementations.
But again, multithreading can be used with looping rather than recursion, so how well this combination will work depends on more factors including the OS and its thread allocation mechanism.
I found another differences between those approaches.
It looks simple and unimportant, but it has a very important role while you prepare for interviews and this subject arises, so look closely.
In short:
1) iterative post-order traversal is not easy - that makes DFT more complex
2) cycles check easier with recursion
Details:
In the recursive case, it is easy to create pre and post traversals:
Imagine a pretty standard question: "print all tasks that should be executed to execute the task 5, when tasks depend on other tasks"
Example:
//key-task, value-list of tasks the key task depends on
//"adjacency map":
Map<Integer, List<Integer>> tasksMap = new HashMap<>();
tasksMap.put(0, new ArrayList<>());
tasksMap.put(1, new ArrayList<>());
List<Integer> t2 = new ArrayList<>();
t2.add(0);
t2.add(1);
tasksMap.put(2, t2);
List<Integer> t3 = new ArrayList<>();
t3.add(2);
t3.add(10);
tasksMap.put(3, t3);
List<Integer> t4 = new ArrayList<>();
t4.add(3);
tasksMap.put(4, t4);
List<Integer> t5 = new ArrayList<>();
t5.add(3);
tasksMap.put(5, t5);
tasksMap.put(6, new ArrayList<>());
tasksMap.put(7, new ArrayList<>());
List<Integer> t8 = new ArrayList<>();
t8.add(5);
tasksMap.put(8, t8);
List<Integer> t9 = new ArrayList<>();
t9.add(4);
tasksMap.put(9, t9);
tasksMap.put(10, new ArrayList<>());
//task to analyze:
int task = 5;
List<Integer> res11 = getTasksInOrderDftReqPostOrder(tasksMap, task);
System.out.println(res11);**//note, no reverse required**
List<Integer> res12 = getTasksInOrderDftReqPreOrder(tasksMap, task);
Collections.reverse(res12);//note reverse!
System.out.println(res12);
private static List<Integer> getTasksInOrderDftReqPreOrder(Map<Integer, List<Integer>> tasksMap, int task) {
List<Integer> result = new ArrayList<>();
Set<Integer> visited = new HashSet<>();
reqPreOrder(tasksMap,task,result, visited);
return result;
}
private static void reqPreOrder(Map<Integer, List<Integer>> tasksMap, int task, List<Integer> result, Set<Integer> visited) {
if(!visited.contains(task)) {
visited.add(task);
result.add(task);//pre order!
List<Integer> children = tasksMap.get(task);
if (children != null && children.size() > 0) {
for (Integer child : children) {
reqPreOrder(tasksMap,child,result, visited);
}
}
}
}
private static List<Integer> getTasksInOrderDftReqPostOrder(Map<Integer, List<Integer>> tasksMap, int task) {
List<Integer> result = new ArrayList<>();
Set<Integer> visited = new HashSet<>();
reqPostOrder(tasksMap,task,result, visited);
return result;
}
private static void reqPostOrder(Map<Integer, List<Integer>> tasksMap, int task, List<Integer> result, Set<Integer> visited) {
if(!visited.contains(task)) {
visited.add(task);
List<Integer> children = tasksMap.get(task);
if (children != null && children.size() > 0) {
for (Integer child : children) {
reqPostOrder(tasksMap,child,result, visited);
}
}
result.add(task);//post order!
}
}
Note that the recursive post-order-traversal does not require a subsequent reversal of the result. Children printed first and your task in the question printed last. Everything is fine. You can do a recursive pre-order-traversal (also shown above) and that one will require a reversal of the result list.
Not that simple with iterative approach! In iterative (one stack) approach you can only do a pre-ordering-traversal, so you obliged to reverse the result array at the end:
List<Integer> res1 = getTasksInOrderDftStack(tasksMap, task);
Collections.reverse(res1);//note reverse!
System.out.println(res1);
private static List<Integer> getTasksInOrderDftStack(Map<Integer, List<Integer>> tasksMap, int task) {
List<Integer> result = new ArrayList<>();
Set<Integer> visited = new HashSet<>();
Stack<Integer> st = new Stack<>();
st.add(task);
visited.add(task);
while(!st.isEmpty()){
Integer node = st.pop();
List<Integer> children = tasksMap.get(node);
result.add(node);
if(children!=null && children.size() > 0){
for(Integer child:children){
if(!visited.contains(child)){
st.add(child);
visited.add(child);
}
}
}
//If you put it here - it does not matter - it is anyway a pre-order
//result.add(node);
}
return result;
}
Looks simple, no?
But it is a trap in some interviews.
It means the following: with the recursive approach, you can implement Depth First Traversal and then select what order you need pre or post(simply by changing the location of the "print", in our case of the "adding to the result list"). With the iterative (one stack) approach you can easily do only pre-order traversal and so in the situation when children need be printed first(pretty much all situations when you need start print from the bottom nodes, going upwards) - you are in the trouble. If you have that trouble you can reverse later, but it will be an addition to your algorithm. And if an interviewer is looking at his watch it may be a problem for you. There are complex ways to do an iterative post-order traversal, they exist, but they are not simple. Example:https://www.geeksforgeeks.org/iterative-postorder-traversal-using-stack/
Thus, the bottom line: I would use recursion during interviews, it is simpler to manage and to explain. You have an easy way to go from pre to post-order traversal in any urgent case. With iterative you are not that flexible.
I would use recursion and then tell: "Ok, but iterative can provide me more direct control on used memory, I can easily measure the stack size and disallow some dangerous overflow.."
Another plus of recursion - it is simpler to avoid / notice cycles in a graph.
Example (preudocode):
dft(n){
mark(n)
for(child: n.children){
if(marked(child))
explode - cycle found!!!
dft(child)
}
unmark(n)
}
It may be fun to write it as recursion, or as a practice.
However, if the code is to be used in production, you need to consider the possibility of stack overflow.
Tail recursion optimization can eliminate stack overflow, but do you want to go through the trouble of making it so, and you need to know you can count on it having the optimization in your environment.
Every time the algorithm recurses, how much is the data size or n reduced by?
If you are reducing the size of data or n by half every time you recurse, then in general you don't need to worry about stack overflow. Say, if it needs to be 4,000 level deep or 10,000 level deep for the program to stack overflow, then your data size need to be roughly 24000 for your program to stack overflow. To put that into perspective, a biggest storage device recently can hold 261 bytes, and if you have 261 of such devices, you are only dealing with 2122 data size. If you are looking at all the atoms in the universe, it is estimated that it may be less than 284. If you need to deal with all the data in the universe and their states for every millisecond since the birth of the universe estimated to be 14 billion years ago, it may only be 2153. So if your program can handle 24000 units of data or n, you can handle all data in the universe and the program will not stack overflow. If you don't need to deal with numbers that are as big as 24000 (a 4000-bit integer), then in general you don't need to worry about stack overflow.
However, if you reduce the size of data or n by a constant amount every time you recurse, then you can run into stack overflow when n becomes merely 20000. That is, the program runs well when n is 1000, and you think the program is good, and then the program stack overflows when some time in the future, when n is 5000 or 20000.
So if you have a possibility of stack overflow, try to make it an iterative solution.
As far as I know, Perl does not optimize tail-recursive calls, but you can fake it.
sub f{
my($l,$r) = #_;
if( $l >= $r ){
return $l;
} else {
# return f( $l+1, $r );
#_ = ( $l+1, $r );
goto &f;
}
}
When first called it will allocate space on the stack. Then it will change its arguments, and restart the subroutine, without adding anything more to the stack. It will therefore pretend that it never called its self, changing it into an iterative process.
Note that there is no "my #_;" or "local #_;", if you did it would no longer work.
"Is there a performance hit if we use a loop instead of
recursion or vice versa in algorithms where both can serve the same purpose?"
Usually yes if you are writing in a imperative language iteration will run faster than recursion, the performance hit is minimized in problems where the iterative solution requires manipulating Stacks and popping items off of a stack due to the recursive nature of the problem. There are a lot of times where the recursive implementation is much easier to read because the code is much shorter,
so you do want to consider maintainability. Especailly in cases where the problem has a recursive nature. So take for example:
The recursive implementation of Tower of Hanoi:
def TowerOfHanoi(n , source, destination, auxiliary):
if n==1:
print ("Move disk 1 from source",source,"to destination",destination)
return
TowerOfHanoi(n-1, source, auxiliary, destination)
print ("Move disk",n,"from source",source,"to destination",destination)
TowerOfHanoi(n-1, auxiliary, destination, source)
Fairly short and pretty easy to read. Compare this with its Counterpart iterative TowerOfHanoi:
# Python3 program for iterative Tower of Hanoi
import sys
# A structure to represent a stack
class Stack:
# Constructor to set the data of
# the newly created tree node
def __init__(self, capacity):
self.capacity = capacity
self.top = -1
self.array = [0]*capacity
# function to create a stack of given capacity.
def createStack(capacity):
stack = Stack(capacity)
return stack
# Stack is full when top is equal to the last index
def isFull(stack):
return (stack.top == (stack.capacity - 1))
# Stack is empty when top is equal to -1
def isEmpty(stack):
return (stack.top == -1)
# Function to add an item to stack.
# It increases top by 1
def push(stack, item):
if(isFull(stack)):
return
stack.top+=1
stack.array[stack.top] = item
# Function to remove an item from stack.
# It decreases top by 1
def Pop(stack):
if(isEmpty(stack)):
return -sys.maxsize
Top = stack.top
stack.top-=1
return stack.array[Top]
# Function to implement legal
# movement between two poles
def moveDisksBetweenTwoPoles(src, dest, s, d):
pole1TopDisk = Pop(src)
pole2TopDisk = Pop(dest)
# When pole 1 is empty
if (pole1TopDisk == -sys.maxsize):
push(src, pole2TopDisk)
moveDisk(d, s, pole2TopDisk)
# When pole2 pole is empty
else if (pole2TopDisk == -sys.maxsize):
push(dest, pole1TopDisk)
moveDisk(s, d, pole1TopDisk)
# When top disk of pole1 > top disk of pole2
else if (pole1TopDisk > pole2TopDisk):
push(src, pole1TopDisk)
push(src, pole2TopDisk)
moveDisk(d, s, pole2TopDisk)
# When top disk of pole1 < top disk of pole2
else:
push(dest, pole2TopDisk)
push(dest, pole1TopDisk)
moveDisk(s, d, pole1TopDisk)
# Function to show the movement of disks
def moveDisk(fromPeg, toPeg, disk):
print("Move the disk", disk, "from '", fromPeg, "' to '", toPeg, "'")
# Function to implement TOH puzzle
def tohIterative(num_of_disks, src, aux, dest):
s, d, a = 'S', 'D', 'A'
# If number of disks is even, then interchange
# destination pole and auxiliary pole
if (num_of_disks % 2 == 0):
temp = d
d = a
a = temp
total_num_of_moves = int(pow(2, num_of_disks) - 1)
# Larger disks will be pushed first
for i in range(num_of_disks, 0, -1):
push(src, i)
for i in range(1, total_num_of_moves + 1):
if (i % 3 == 1):
moveDisksBetweenTwoPoles(src, dest, s, d)
else if (i % 3 == 2):
moveDisksBetweenTwoPoles(src, aux, s, a)
else if (i % 3 == 0):
moveDisksBetweenTwoPoles(aux, dest, a, d)
# Input: number of disks
num_of_disks = 3
# Create three stacks of size 'num_of_disks'
# to hold the disks
src = createStack(num_of_disks)
dest = createStack(num_of_disks)
aux = createStack(num_of_disks)
tohIterative(num_of_disks, src, aux, dest)
Now the first one is way easier to read because suprise suprise shorter code is usually easier to understand than code that is 10 times longer. Sometimes you want to ask yourself is the extra performance gain really worth it? The amount of hours wasted debugging the code. Is the iterative TowerOfHanoi faster than the Recursive TowerOfHanoi? Probably, but not by a big margin. Would I like to program Recursive problems like TowerOfHanoi using iteration? Hell no. Next we have another recursive function the Ackermann function:
Using recursion:
if m == 0:
# BASE CASE
return n + 1
elif m > 0 and n == 0:
# RECURSIVE CASE
return ackermann(m - 1, 1)
elif m > 0 and n > 0:
# RECURSIVE CASE
return ackermann(m - 1, ackermann(m, n - 1))
Using Iteration:
callStack = [{'m': 2, 'n': 3, 'indentation': 0, 'instrPtr': 'start'}]
returnValue = None
while len(callStack) != 0:
m = callStack[-1]['m']
n = callStack[-1]['n']
indentation = callStack[-1]['indentation']
instrPtr = callStack[-1]['instrPtr']
if instrPtr == 'start':
print('%sackermann(%s, %s)' % (' ' * indentation, m, n))
if m == 0:
# BASE CASE
returnValue = n + 1
callStack.pop()
continue
elif m > 0 and n == 0:
# RECURSIVE CASE
callStack[-1]['instrPtr'] = 'after first recursive case'
callStack.append({'m': m - 1, 'n': 1, 'indentation': indentation + 1, 'instrPtr': 'start'})
continue
elif m > 0 and n > 0:
# RECURSIVE CASE
callStack[-1]['instrPtr'] = 'after second recursive case, inner call'
callStack.append({'m': m, 'n': n - 1, 'indentation': indentation + 1, 'instrPtr': 'start'})
continue
elif instrPtr == 'after first recursive case':
returnValue = returnValue
callStack.pop()
continue
elif instrPtr == 'after second recursive case, inner call':
callStack[-1]['innerCallResult'] = returnValue
callStack[-1]['instrPtr'] = 'after second recursive case, outer call'
callStack.append({'m': m - 1, 'n': returnValue, 'indentation': indentation + 1, 'instrPtr': 'start'})
continue
elif instrPtr == 'after second recursive case, outer call':
returnValue = returnValue
callStack.pop()
continue
print(returnValue)
And once again I will argue that the recursive implementation is much easier to understand. So my conclusion is use recursion if the problem by nature is recursive and requires manipulating items in a stack.
I'm going to answer your question by designing a Haskell data structure by "induction", which is a sort of "dual" to recursion. And then I will show how this duality leads to nice things.
We introduce a type for a simple tree:
data Tree a = Branch (Tree a) (Tree a)
| Leaf a
deriving (Eq)
We can read this definition as saying "A tree is a Branch (which contains two trees) or is a leaf (which contains a data value)". So the leaf is a sort of minimal case. If a tree isn't a leaf, then it must be a compound tree containing two trees. These are the only cases.
Let's make a tree:
example :: Tree Int
example = Branch (Leaf 1)
(Branch (Leaf 2)
(Leaf 3))
Now, let's suppose we want to add 1 to each value in the tree. We can do this by calling:
addOne :: Tree Int -> Tree Int
addOne (Branch a b) = Branch (addOne a) (addOne b)
addOne (Leaf a) = Leaf (a + 1)
First, notice that this is in fact a recursive definition. It takes the data constructors Branch and Leaf as cases (and since Leaf is minimal and these are the only possible cases), we are sure that the function will terminate.
What would it take to write addOne in an iterative style? What will looping into an arbitrary number of branches look like?
Also, this kind of recursion can often be factored out, in terms of a "functor". We can make Trees into Functors by defining:
instance Functor Tree where fmap f (Leaf a) = Leaf (f a)
fmap f (Branch a b) = Branch (fmap f a) (fmap f b)
and defining:
addOne' = fmap (+1)
We can factor out other recursion schemes, such as the catamorphism (or fold) for an algebraic data type. Using a catamorphism, we can write:
addOne'' = cata go where
go (Leaf a) = Leaf (a + 1)
go (Branch a b) = Branch a b
Whilst starting to learn lisp, I've come across the term tail-recursive. What does it mean exactly?
Consider a simple function that adds the first N natural numbers. (e.g. sum(5) = 0 + 1 + 2 + 3 + 4 + 5 = 15).
Here is a simple JavaScript implementation that uses recursion:
function recsum(x) {
if (x === 0) {
return 0;
} else {
return x + recsum(x - 1);
}
}
If you called recsum(5), this is what the JavaScript interpreter would evaluate:
recsum(5)
5 + recsum(4)
5 + (4 + recsum(3))
5 + (4 + (3 + recsum(2)))
5 + (4 + (3 + (2 + recsum(1))))
5 + (4 + (3 + (2 + (1 + recsum(0)))))
5 + (4 + (3 + (2 + (1 + 0))))
5 + (4 + (3 + (2 + 1)))
5 + (4 + (3 + 3))
5 + (4 + 6)
5 + 10
15
Note how every recursive call has to complete before the JavaScript interpreter begins to actually do the work of calculating the sum.
Here's a tail-recursive version of the same function:
function tailrecsum(x, running_total = 0) {
if (x === 0) {
return running_total;
} else {
return tailrecsum(x - 1, running_total + x);
}
}
Here's the sequence of events that would occur if you called tailrecsum(5), (which would effectively be tailrecsum(5, 0), because of the default second argument).
tailrecsum(5, 0)
tailrecsum(4, 5)
tailrecsum(3, 9)
tailrecsum(2, 12)
tailrecsum(1, 14)
tailrecsum(0, 15)
15
In the tail-recursive case, with each evaluation of the recursive call, the running_total is updated.
Note: The original answer used examples from Python. These have been changed to JavaScript, since Python interpreters don't support tail call optimization. However, while tail call optimization is part of the ECMAScript 2015 spec, most JavaScript interpreters don't support it.
In traditional recursion, the typical model is that you perform your recursive calls first, and then you take the return value of the recursive call and calculate the result. In this manner, you don't get the result of your calculation until you have returned from every recursive call.
In tail recursion, you perform your calculations first, and then you execute the recursive call, passing the results of your current step to the next recursive step. This results in the last statement being in the form of (return (recursive-function params)). Basically, the return value of any given recursive step is the same as the return value of the next recursive call.
The consequence of this is that once you are ready to perform your next recursive step, you don't need the current stack frame any more. This allows for some optimization. In fact, with an appropriately written compiler, you should never have a stack overflow snicker with a tail recursive call. Simply reuse the current stack frame for the next recursive step. I'm pretty sure Lisp does this.
An important point is that tail recursion is essentially equivalent to looping. It's not just a matter of compiler optimization, but a fundamental fact about expressiveness. This goes both ways: you can take any loop of the form
while(E) { S }; return Q
where E and Q are expressions and S is a sequence of statements, and turn it into a tail recursive function
f() = if E then { S; return f() } else { return Q }
Of course, E, S, and Q have to be defined to compute some interesting value over some variables. For example, the looping function
sum(n) {
int i = 1, k = 0;
while( i <= n ) {
k += i;
++i;
}
return k;
}
is equivalent to the tail-recursive function(s)
sum_aux(n,i,k) {
if( i <= n ) {
return sum_aux(n,i+1,k+i);
} else {
return k;
}
}
sum(n) {
return sum_aux(n,1,0);
}
(This "wrapping" of the tail-recursive function with a function with fewer parameters is a common functional idiom.)
This excerpt from the book Programming in Lua shows how to make a proper tail recursion (in Lua, but should apply to Lisp too) and why it's better.
A tail call [tail recursion] is a kind of goto dressed
as a call. A tail call happens when a
function calls another as its last
action, so it has nothing else to do.
For instance, in the following code,
the call to g is a tail call:
function f (x)
return g(x)
end
After f calls g, it has nothing else
to do. In such situations, the program
does not need to return to the calling
function when the called function
ends. Therefore, after the tail call,
the program does not need to keep any
information about the calling function
in the stack. ...
Because a proper tail call uses no
stack space, there is no limit on the
number of "nested" tail calls that a
program can make. For instance, we can
call the following function with any
number as argument; it will never
overflow the stack:
function foo (n)
if n > 0 then return foo(n - 1) end
end
... As I said earlier, a tail call is a
kind of goto. As such, a quite useful
application of proper tail calls in
Lua is for programming state machines.
Such applications can represent each
state by a function; to change state
is to go to (or to call) a specific
function. As an example, let us
consider a simple maze game. The maze
has several rooms, each with up to
four doors: north, south, east, and
west. At each step, the user enters a
movement direction. If there is a door
in that direction, the user goes to
the corresponding room; otherwise, the
program prints a warning. The goal is
to go from an initial room to a final
room.
This game is a typical state machine,
where the current room is the state.
We can implement such maze with one
function for each room. We use tail
calls to move from one room to
another. A small maze with four rooms
could look like this:
function room1 ()
local move = io.read()
if move == "south" then return room3()
elseif move == "east" then return room2()
else print("invalid move")
return room1() -- stay in the same room
end
end
function room2 ()
local move = io.read()
if move == "south" then return room4()
elseif move == "west" then return room1()
else print("invalid move")
return room2()
end
end
function room3 ()
local move = io.read()
if move == "north" then return room1()
elseif move == "east" then return room4()
else print("invalid move")
return room3()
end
end
function room4 ()
print("congratulations!")
end
So you see, when you make a recursive call like:
function x(n)
if n==0 then return 0
n= n-2
return x(n) + 1
end
This is not tail recursive because you still have things to do (add 1) in that function after the recursive call is made. If you input a very high number it will probably cause a stack overflow.
Using regular recursion, each recursive call pushes another entry onto the call stack. When the recursion is completed, the app then has to pop each entry off all the way back down.
With tail recursion, depending on language the compiler may be able to collapse the stack down to one entry, so you save stack space...A large recursive query can actually cause a stack overflow.
Basically Tail recursions are able to be optimized into iteration.
The jargon file has this to say about the definition of tail recursion:
tail recursion /n./
If you aren't sick of it already, see tail recursion.
Instead of explaining it with words, here's an example. This is a Scheme version of the factorial function:
(define (factorial x)
(if (= x 0) 1
(* x (factorial (- x 1)))))
Here is a version of factorial that is tail-recursive:
(define factorial
(letrec ((fact (lambda (x accum)
(if (= x 0) accum
(fact (- x 1) (* accum x))))))
(lambda (x)
(fact x 1))))
You will notice in the first version that the recursive call to fact is fed into the multiplication expression, and therefore the state has to be saved on the stack when making the recursive call. In the tail-recursive version there is no other S-expression waiting for the value of the recursive call, and since there is no further work to do, the state doesn't have to be saved on the stack. As a rule, Scheme tail-recursive functions use constant stack space.
Tail recursion refers to the recursive call being last in the last logic instruction in the recursive algorithm.
Typically in recursion, you have a base-case which is what stops the recursive calls and begins popping the call stack. To use a classic example, though more C-ish than Lisp, the factorial function illustrates tail recursion. The recursive call occurs after checking the base-case condition.
factorial(x, fac=1) {
if (x == 1)
return fac;
else
return factorial(x-1, x*fac);
}
The initial call to factorial would be factorial(n) where fac=1 (default value) and n is the number for which the factorial is to be calculated.
It means that rather than needing to push the instruction pointer on the stack, you can simply jump to the top of a recursive function and continue execution. This allows for functions to recurse indefinitely without overflowing the stack.
I wrote a blog post on the subject, which has graphical examples of what the stack frames look like.
The best way for me to understand tail call recursion is a special case of recursion where the last call(or the tail call) is the function itself.
Comparing the examples provided in Python:
def recsum(x):
if x == 1:
return x
else:
return x + recsum(x - 1)
^RECURSION
def tailrecsum(x, running_total=0):
if x == 0:
return running_total
else:
return tailrecsum(x - 1, running_total + x)
^TAIL RECURSION
As you can see in the general recursive version, the final call in the code block is x + recsum(x - 1). So after calling the recsum method, there is another operation which is x + ...
However, in the tail recursive version, the final call(or the tail call) in the code block is tailrecsum(x - 1, running_total + x) which means the last call is made to the method itself and no operation after that.
This point is important because tail recursion as seen here is not making the memory grow because when the underlying VM sees a function calling itself in a tail position (the last expression to be evaluated in a function), it eliminates the current stack frame, which is known as Tail Call Optimization(TCO).
EDIT
NB. Do bear in mind that the example above is written in Python whose runtime does not support TCO. This is just an example to explain the point. TCO is supported in languages like Scheme, Haskell etc
Here is a quick code snippet comparing two functions. The first is traditional recursion for finding the factorial of a given number. The second uses tail recursion.
Very simple and intuitive to understand.
An easy way to tell if a recursive function is a tail recursive is if it returns a concrete value in the base case. Meaning that it doesn't return 1 or true or anything like that. It will more than likely return some variant of one of the method parameters.
Another way is to tell is if the recursive call is free of any addition, arithmetic, modification, etc... Meaning its nothing but a pure recursive call.
public static int factorial(int mynumber) {
if (mynumber == 1) {
return 1;
} else {
return mynumber * factorial(--mynumber);
}
}
public static int tail_factorial(int mynumber, int sofar) {
if (mynumber == 1) {
return sofar;
} else {
return tail_factorial(--mynumber, sofar * mynumber);
}
}
The recursive function is a function which calls by itself
It allows programmers to write efficient programs using a minimal amount of code.
The downside is that they can cause infinite loops and other unexpected results if not written properly.
I will explain both Simple Recursive function and Tail Recursive function
In order to write a Simple recursive function
The first point to consider is when should you decide on coming out
of the loop which is the if loop
The second is what process to do if we are our own function
From the given example:
public static int fact(int n){
if(n <=1)
return 1;
else
return n * fact(n-1);
}
From the above example
if(n <=1)
return 1;
Is the deciding factor when to exit the loop
else
return n * fact(n-1);
Is the actual processing to be done
Let me the break the task one by one for easy understanding.
Let us see what happens internally if I run fact(4)
Substituting n=4
public static int fact(4){
if(4 <=1)
return 1;
else
return 4 * fact(4-1);
}
If loop fails so it goes to else loop
so it returns 4 * fact(3)
In stack memory, we have 4 * fact(3)
Substituting n=3
public static int fact(3){
if(3 <=1)
return 1;
else
return 3 * fact(3-1);
}
If loop fails so it goes to else loop
so it returns 3 * fact(2)
Remember we called ```4 * fact(3)``
The output for fact(3) = 3 * fact(2)
So far the stack has 4 * fact(3) = 4 * 3 * fact(2)
In stack memory, we have 4 * 3 * fact(2)
Substituting n=2
public static int fact(2){
if(2 <=1)
return 1;
else
return 2 * fact(2-1);
}
If loop fails so it goes to else loop
so it returns 2 * fact(1)
Remember we called 4 * 3 * fact(2)
The output for fact(2) = 2 * fact(1)
So far the stack has 4 * 3 * fact(2) = 4 * 3 * 2 * fact(1)
In stack memory, we have 4 * 3 * 2 * fact(1)
Substituting n=1
public static int fact(1){
if(1 <=1)
return 1;
else
return 1 * fact(1-1);
}
If loop is true
so it returns 1
Remember we called 4 * 3 * 2 * fact(1)
The output for fact(1) = 1
So far the stack has 4 * 3 * 2 * fact(1) = 4 * 3 * 2 * 1
Finally, the result of fact(4) = 4 * 3 * 2 * 1 = 24
The Tail Recursion would be
public static int fact(x, running_total=1) {
if (x==1) {
return running_total;
} else {
return fact(x-1, running_total*x);
}
}
Substituting n=4
public static int fact(4, running_total=1) {
if (x==1) {
return running_total;
} else {
return fact(4-1, running_total*4);
}
}
If loop fails so it goes to else loop
so it returns fact(3, 4)
In stack memory, we have fact(3, 4)
Substituting n=3
public static int fact(3, running_total=4) {
if (x==1) {
return running_total;
} else {
return fact(3-1, 4*3);
}
}
If loop fails so it goes to else loop
so it returns fact(2, 12)
In stack memory, we have fact(2, 12)
Substituting n=2
public static int fact(2, running_total=12) {
if (x==1) {
return running_total;
} else {
return fact(2-1, 12*2);
}
}
If loop fails so it goes to else loop
so it returns fact(1, 24)
In stack memory, we have fact(1, 24)
Substituting n=1
public static int fact(1, running_total=24) {
if (x==1) {
return running_total;
} else {
return fact(1-1, 24*1);
}
}
If loop is true
so it returns running_total
The output for running_total = 24
Finally, the result of fact(4,1) = 24
In Java, here's a possible tail recursive implementation of the Fibonacci function:
public int tailRecursive(final int n) {
if (n <= 2)
return 1;
return tailRecursiveAux(n, 1, 1);
}
private int tailRecursiveAux(int n, int iter, int acc) {
if (iter == n)
return acc;
return tailRecursiveAux(n, ++iter, acc + iter);
}
Contrast this with the standard recursive implementation:
public int recursive(final int n) {
if (n <= 2)
return 1;
return recursive(n - 1) + recursive(n - 2);
}
I'm not a Lisp programmer, but I think this will help.
Basically it's a style of programming such that the recursive call is the last thing you do.
Here is a Common Lisp example that does factorials using tail-recursion. Due to the stack-less nature, one could perform insanely large factorial computations ...
(defun ! (n &optional (product 1))
(if (zerop n) product
(! (1- n) (* product n))))
And then for fun you could try (format nil "~R" (! 25))
A tail recursive function is a recursive function where the last operation it does before returning is make the recursive function call. That is, the return value of the recursive function call is immediately returned. For example, your code would look like this:
def recursiveFunction(some_params):
# some code here
return recursiveFunction(some_args)
# no code after the return statement
Compilers and interpreters that implement tail call optimization or tail call elimination can optimize recursive code to prevent stack overflows. If your compiler or interpreter doesn't implement tail call optimization (such as the CPython interpreter) there is no additional benefit to writing your code this way.
For example, this is a standard recursive factorial function in Python:
def factorial(number):
if number == 1:
# BASE CASE
return 1
else:
# RECURSIVE CASE
# Note that `number *` happens *after* the recursive call.
# This means that this is *not* tail call recursion.
return number * factorial(number - 1)
And this is a tail call recursive version of the factorial function:
def factorial(number, accumulator=1):
if number == 0:
# BASE CASE
return accumulator
else:
# RECURSIVE CASE
# There's no code after the recursive call.
# This is tail call recursion:
return factorial(number - 1, number * accumulator)
print(factorial(5))
(Note that even though this is Python code, the CPython interpreter doesn't do tail call optimization, so arranging your code like this confers no runtime benefit.)
You may have to make your code a bit more unreadable to make use of tail call optimization, as shown in the factorial example. (For example, the base case is now a bit unintuitive, and the accumulator parameter is effectively used as a sort of global variable.)
But the benefit of tail call optimization is that it prevents stack overflow errors. (I'll note that you can get this same benefit by using an iterative algorithm instead of a recursive one.)
Stack overflows are caused when the call stack has had too many frame objects pushed onto. A frame object is pushed onto the call stack when a function is called, and popped off the call stack when the function returns. Frame objects contain info such as local variables and what line of code to return to when the function returns.
If your recursive function makes too many recursive calls without returning, the call stack can exceed its frame object limit. (The number varies by platform; in Python it is 1000 frame objects by default.) This causes a stack overflow error. (Hey, that's where the name of this website comes from!)
However, if the last thing your recursive function does is make the recursive call and return its return value, then there's no reason it needs to keep the current frame object needs to stay on the call stack. After all, if there's no code after the recursive function call, there's no reason to hang on to the current frame object's local variables. So we can get rid of the current frame object immediately rather than keep it on the call stack. The end result of this is that your call stack doesn't grow in size, and thus cannot stack overflow.
A compiler or interpreter must have tail call optimization as a feature for it to be able to recognize when tail call optimization can be applied. Even then, you may have rearrange the code in your recursive function to make use of tail call optimization, and it's up to you if this potential decrease in readability is worth the optimization.
In short, a tail recursion has the recursive call as the last statement in the function so that it doesn't have to wait for the recursive call.
So this is a tail recursion i.e. N(x - 1, p * x) is the last statement in the function where the compiler is clever to figure out that it can be optimised to a for-loop (factorial). The second parameter p carries the intermediate product value.
function N(x, p) {
return x == 1 ? p : N(x - 1, p * x);
}
This is the non-tail-recursive way of writing the above factorial function (although some C++ compilers may be able to optimise it anyway).
function N(x) {
return x == 1 ? 1 : x * N(x - 1);
}
but this is not:
function F(x) {
if (x == 1) return 0;
if (x == 2) return 1;
return F(x - 1) + F(x - 2);
}
I did write a long post titled "Understanding Tail Recursion – Visual Studio C++ – Assembly View"
A tail recursion is a recursive function where the function calls
itself at the end ("tail") of the function in which no computation is
done after the return of recursive call. Many compilers optimize to
change a recursive call to a tail recursive or an iterative call.
Consider the problem of computing factorial of a number.
A straightforward approach would be:
factorial(n):
if n==0 then 1
else n*factorial(n-1)
Suppose you call factorial(4). The recursion tree would be:
factorial(4)
/ \
4 factorial(3)
/ \
3 factorial(2)
/ \
2 factorial(1)
/ \
1 factorial(0)
\
1
The maximum recursion depth in the above case is O(n).
However, consider the following example:
factAux(m,n):
if n==0 then m;
else factAux(m*n,n-1);
factTail(n):
return factAux(1,n);
Recursion tree for factTail(4) would be:
factTail(4)
|
factAux(1,4)
|
factAux(4,3)
|
factAux(12,2)
|
factAux(24,1)
|
factAux(24,0)
|
24
Here also, maximum recursion depth is O(n) but none of the calls adds any extra variable to the stack. Hence the compiler can do away with a stack.
here is a Perl 5 version of the tailrecsum function mentioned earlier.
sub tail_rec_sum($;$){
my( $x,$running_total ) = (#_,0);
return $running_total unless $x;
#_ = ($x-1,$running_total+$x);
goto &tail_rec_sum; # throw away current stack frame
}
This is an excerpt from Structure and Interpretation of Computer Programs about tail recursion.
In contrasting iteration and recursion, we must be careful not to
confuse the notion of a recursive process with the notion of a
recursive procedure. When we describe a procedure as recursive, we are
referring to the syntactic fact that the procedure definition refers
(either directly or indirectly) to the procedure itself. But when we
describe a process as following a pattern that is, say, linearly
recursive, we are speaking about how the process evolves, not about
the syntax of how a procedure is written. It may seem disturbing that
we refer to a recursive procedure such as fact-iter as generating an
iterative process. However, the process really is iterative: Its state
is captured completely by its three state variables, and an
interpreter need keep track of only three variables in order to
execute the process.
One reason that the distinction between process and procedure may be
confusing is that most implementations of common languages (including Ada, Pascal, and
C) are designed in such a way that the interpretation of any recursive
procedure consumes an amount of memory that grows with the number of
procedure calls, even when the process described is, in principle,
iterative. As a consequence, these languages can describe iterative
processes only by resorting to special-purpose “looping constructs”
such as do, repeat, until, for, and while. The implementation of
Scheme does not share this defect. It
will execute an iterative process in constant space, even if the
iterative process is described by a recursive procedure. An
implementation with this property is called tail-recursive. With a
tail-recursive implementation, iteration can be expressed using the
ordinary procedure call mechanism, so that special iteration
constructs are useful only as syntactic sugar.
Tail recursion is the life you are living right now. You constantly recycle the same stack frame, over and over, because there's no reason or means to return to a "previous" frame. The past is over and done with so it can be discarded. You get one frame, forever moving into the future, until your process inevitably dies.
The analogy breaks down when you consider some processes might utilize additional frames but are still considered tail-recursive if the stack does not grow infinitely.
Tail Recursion is pretty fast as compared to normal recursion.
It is fast because the output of the ancestors call will not be written in stack to keep the track.
But in normal recursion all the ancestor calls output written in stack to keep the track.
To understand some of the core differences between tail-call recursion and non-tail-call recursion we can explore the .NET implementations of these techniques.
Here is an article with some examples in C#, F#, and C++\CLI: Adventures in Tail Recursion in C#, F#, and C++\CLI.
C# does not optimize for tail-call recursion whereas F# does.
The differences of principle involve loops vs. Lambda calculus. C# is designed with loops in mind whereas F# is built from the principles of Lambda calculus. For a very good (and free) book on the principles of Lambda calculus, see Structure and Interpretation of Computer Programs, by Abelson, Sussman, and Sussman.
Regarding tail calls in F#, for a very good introductory article, see Detailed Introduction to Tail Calls in F#. Finally, here is an article that covers the difference between non-tail recursion and tail-call recursion (in F#): Tail-recursion vs. non-tail recursion in F sharp.
If you want to read about some of the design differences of tail-call recursion between C# and F#, see Generating Tail-Call Opcode in C# and F#.
If you care enough to want to know what conditions prevent the C# compiler from performing tail-call optimizations, see this article: JIT CLR tail-call conditions.
Recursion means a function calling itself. For example:
(define (un-ended name)
(un-ended 'me)
(print "How can I get here?"))
Tail-Recursion means the recursion that conclude the function:
(define (un-ended name)
(print "hello")
(un-ended 'me))
See, the last thing un-ended function (procedure, in Scheme jargon) does is to call itself. Another (more useful) example is:
(define (map lst op)
(define (helper done left)
(if (nil? left)
done
(helper (cons (op (car left))
done)
(cdr left))))
(reverse (helper '() lst)))
In the helper procedure, the LAST thing it does if the left is not nil is to call itself (AFTER cons something and cdr something). This is basically how you map a list.
The tail-recursion has a great advantage that the interpreter (or compiler, dependent on the language and vendor) can optimize it, and transform it into something equivalent to a while loop. As matter of fact, in Scheme tradition, most "for" and "while" loop is done in a tail-recursion manner (there is no for and while, as far as I know).
Tail Recursive Function is a recursive function in which recursive call is the last executed thing in the function.
Regular recursive function, we have stack and everytime we invoke a recursive function within that recursive function, adds another layer to our call stack. In normal recursion
space: O(n) tail recursion makes space complexity from
O(N)=>O(1)
Tail call optimization means that it is possible to call a function from another function without growing the call stack.
We should write tail recursion in recursive solutions. but certain languages do not actually support the tail recursion in their engine that compiles the language down. since ecma6, there has been tail recursion that was in the specification. BUt none of the engines that compile js have implemented tail recursion into it. you wont achieve O(1) in js, because the compiler itself does not know how to implement this tail recursion. As of January 1, 2020 Safari is the only browser that supports tail call optimization.
Haskell and Java have tail recursion optimization
Regular Recursive Factorial
function Factorial(x) {
//Base case x<=1
if (x <= 1) {
return 1;
} else {
// x is waiting for the return value of Factorial(x-1)
// the last thing we do is NOT applying the recursive call
// after recursive call we still have to multiply.
return x * Factorial(x - 1);
}
}
we have 4 calls in our call stack.
Factorial(4); // waiting in the memory for Factorial(3)
4 * Factorial(3); // waiting in the memory for Factorial(2)
4 * (3 * Factorial(2)); // waiting in the memory for Factorial(1)
4 * (3 * (2 * Factorial(1)));
4 * (3 * (2 * 1));
We are making 4 Factorial() calls, space is O(n)
this mmight cause Stackoverflow
Tail Recursive Factorial
function tailFactorial(x, totalSoFar = 1) {
//Base Case: x===0. In recursion there must be base case. Otherwise they will never stop
if (x === 0) {
return totalSoFar;
} else {
// there is nothing waiting for tailFactorial to complete. we are returning another instance of tailFactorial()
// we are not doing any additional computaion with what we get back from this recursive call
return tailFactorial(x - 1, totalSoFar * x);
}
}
We dont need to remember anything after we make our recursive call
There are two basic kinds of recursions: head recursion and tail recursion.
In head recursion, a function makes its recursive call and then
performs some more calculations, maybe using the result of the
recursive call, for example.
In a tail recursive function, all calculations happen first and
the recursive call is the last thing that happens.
Taken from this super awesome post.
Please consider reading it.
A function is tail recursive if each recursive case consists only of a call to the function itself, possibly with different arguments. Or, tail recursion is recursion with no pending work. Note that this is a programming-language independent concept.
Consider the function defined as:
g(a, b, n) = a * b^n
A possible tail-recursive formulation is:
g(a, b, n) | n is zero = a
| n is odd = g(a*b, b, n-1)
| otherwise = g(a, b*b, n/2)
If you examine each RHS of g(...) that involves a recursive case, you'll find that the whole body of the RHS is a call to g(...), and only that. This definition is tail recursive.
For comparison, a non-tail-recursive formulation might be:
g'(a, b, n) = a * f(b, n)
f(b, n) | n is zero = 1
| n is odd = f(b, n-1) * b
| otherwise = f(b, n/2) ^ 2
Each recursive case in f(...) has some pending work that needs to happen after the recursive call.
Note that when we went from g' to g, we made essential use of associativity
(and commutativity) of multiplication. This is not an accident, and most cases where you will need to transform recursion to tail-recursion will make use of such properties: if we want to eagerly do some work rather than leave it pending, we have to use something like associativity to prove that the answer will be the same.
Tail recursive calls can be implemented with a backwards jump, as opposed to using a stack for normal recursive calls. Note that detecting a tail call, or emitting a backwards jump is usually straightforward. However, it is often hard to rearrange the arguments such that the backwards jump is possible. Since this optimization is not free, language implementations can choose not to implement this optimization, or require opt-in by marking recursive calls with a 'tailcall' instruction and/or choosing a higher optimization setting.
Some languages (e.g. Scheme) do, however, require all implementations to optimize tail-recursive functions, maybe even all calls in tail position.
Backwards jumps are usually abstracted as a (while) loop in most imperative languages, and tail-recursion, when optimized to a backwards jump, is isomorphic to looping.
This question has a lot of great answers... but I cannot help but chime in with an alternative take on how to define "tail recursion", or at least "proper tail recursion." Namely: should one look at it as a property of a particular expression in a program? Or should one look at it as a property of an implementation of a programming language?
For more on the latter view, there is a classic paper by Will Clinger, "Proper Tail Recursion and Space Efficiency" (PLDI 1998), that defined "proper tail recursion" as a property of a programming language implementation. The definition is constructed to allow one to ignore implementation details (such as whether the call stack is actually represented via the runtime stack or via a heap-allocated linked list of frames).
To accomplish this, it uses asymptotic analysis: not of program execution time as one usually sees, but rather of program space usage. This way, the space usage of a heap-allocated linked list vs a runtime call stack ends up being asymptotically equivalent; so one gets to ignore that programming language implementation detail (a detail which certainly matters quite a bit in practice, but can muddy the waters quite a bit when one attempts to determine whether a given implementation is satisfying the requirement to be "property tail recursive")
The paper is worth careful study for a number of reasons:
It gives an inductive definition of the tail expressions and tail calls of a program. (Such a definition, and why such calls are important, seems to be the subject of most of the other answers given here.)
Here are those definitions, just to provide a flavor of the text:
Definition 1 The tail expressions of a program written in Core Scheme are defined inductively as follows.
The body of a lambda expression is a tail expression
If (if E0 E1 E2) is a tail expression, then both E1 and E2 are tail expressions.
Nothing else is a tail expression.
Definition 2 A tail call is a tail expression that is a procedure call.
(a tail recursive call, or as the paper says, "self-tail call" is a special case of a tail call where the procedure is invoked itself.)
It provides formal definitions for six different "machines" for evaluating Core Scheme, where each machine has the same observable behavior except for the asymptotic space complexity class that each is in.
For example, after giving definitions for machines with respectively, 1. stack-based memory management, 2. garbage collection but no tail calls, 3. garbage collection and tail calls, the paper continues onward with even more advanced storage management strategies, such as 4. "evlis tail recursion", where the environment does not need to be preserved across the evaluation of the last sub-expression argument in a tail call, 5. reducing the environment of a closure to just the free variables of that closure, and 6. so-called "safe-for-space" semantics as defined by Appel and Shao.
In order to prove that the machines actually belong to six distinct space complexity classes, the paper, for each pair of machines under comparison, provides concrete examples of programs that will expose asymptotic space blowup on one machine but not the other.
(Reading over my answer now, I'm not sure if I'm managed to actually capture the crucial points of the Clinger paper. But, alas, I cannot devote more time to developing this answer right now.)
Many people have already explained recursion here. I would like to cite a couple of thoughts about some advantages that recursion gives from the book “Concurrency in .NET, Modern patterns of concurrent and parallel programming” by Riccardo Terrell:
“Functional recursion is the natural way to iterate in FP because it
avoids mutation of state. During each iteration, a new value is passed
into the loop constructor instead to be updated (mutated). In
addition, a recursive function can be composed, making your program
more modular, as well as introducing opportunities to exploit
parallelization."
Here also are some interesting notes from the same book about tail recursion:
Tail-call recursion is a technique that converts a regular recursive
function into an optimized version that can handle large inputs
without any risks and side effects.
NOTE The primary reason for a tail call as an optimization is to
improve data locality, memory usage, and cache usage. By doing a tail
call, the callee uses the same stack space as the caller. This reduces
memory pressure. It marginally improves the cache because the same
memory is reused for subsequent callers and can stay in the cache,
rather than evicting an older cache line to make room for a new cache
line.