I am going through an algorithms and datastructures textbook and came accross this question:
1-28. Write a function to perform integer division without using
either the / or * operators. Find a fast way to do it.
How can we come up with a fast way to do it?
I like this solution: https://stackoverflow.com/a/34506599/1008519, but I find it somewhat hard to reason about (especially the |-part). This solution makes a little more sense in my head:
var divide = function (dividend, divisor) {
// Handle 0 divisor
if (divisor === 0) {
return NaN;
}
// Handle negative numbers
var isNegative = false;
if (dividend < 0) {
// Change sign
dividend = ~dividend+1;
isNegative = !isNegative;
}
if (divisor < 0) {
// Change sign
divisor = ~divisor+1;
isNegative = !isNegative;
}
/**
* Main algorithm
*/
var result = 1;
var denominator = divisor;
// Double denominator value with bitwise shift until bigger than dividend
while (dividend > denominator) {
denominator <<= 1;
result <<= 1;
}
// Subtract divisor value until denominator is smaller than dividend
while (denominator > dividend) {
denominator -= divisor;
result -= 1;
}
// If one of dividend or divisor was negative, change sign of result
if (isNegative) {
result = ~result+1;
}
return result;
}
We initialize our result to 1 (since we are going to double our denominator until it is bigger than the dividend)
Double our denominator (with bitwise shifts) until it is bigger than the dividend
Since we know our denominator is bigger than our dividend, we can minus our divisor until it is less than our dividend
Return result since denominator is now as close to the result as possible using the divisor
Here are some test runs:
console.log(divide(-16, 3)); // -5
console.log(divide(16, 3)); // 5
console.log(divide(16, 33)); // 0
console.log(divide(16, 0)); // NaN
console.log(divide(384, 15)); // 25
Here is a gist of the solution: https://gist.github.com/mlunoe/e34f14cff4d5c57dd90a5626266c4130
Typically, when an algorithms textbook says fast they mean in terms of computational complexity. That is, the number of operations per bit of input. In general, they don't care about constants, so if you have an input of n bits, whether it takes two operations per bit or a hundred operations per bit, we say the algorithm takes O(n) time. This is because if we have an algorithm that runs in O(n^2) time (polynomial... in this case, square time) and we imagine a O(n) algorithm that does 100 operations per bit compared to our algorithm which may do 1 operation per bit, once the input size is 100 bits, the polynomial algorithm starts to run really slow really quickly (compared to our other algorithm). Essentially, you can imagine two lines, y=100x and y=x^2. Your teacher probably made you do an exercise in Algebra (maybe it was calculus?) where you have to say which one is bigger as x approaches infinity. This is actually a key concept in divergence/convergence in calculus if you have gotten there already in mathematics. Regardless, with a little algebra, you can imagine our graphs intersecting at x=100, and y=x^2 being larger for all points where x is greater than 100.
As far as most textbooks are concerned, O(nlgn) or better is considered "fast". One example of a really bad algorithm to solve this problem would be the following:
crappyMultiplicationAlg(int a, int b)
int product = 0
for (b>0)
product = product + a
b = b-1
return product
This algorithm basically uses "b" as a counter and just keeps adding "a" to some variable for each time b counts down. To calculate how "fast" the algorithm is (in terms of algorithmic complexity) we count how many runs different components will take. In this case, we only have a for loop and some initialization (which is negligible in this case, ignore it). How many times does the for loop run? You may be saying "Hey, guy! It only runs 'b' times! That may not even be half the input. Thats way better than O(n) time!"
The trick here, is that we are concerned with the size of the input in terms of storage... and we all (should) know that to store an n bit integer, we need lgn bits. In other words, if we have x bits, we can store any (unsigned) number up to (2^x)-1. As a result, if we are using a standard 4 byte integer, that number could be up to 2^32 - 1 which is a number well into the billions, if my memory serves me right. If you dont trust me, run this algorithm with a number like 10,000,000 and see how long it takes. Still not convinced? Use a long to use a number like 1,000,000,000.
Since you didn't ask for help with the algorithm, Ill leave it for you as a homework exercise (not trying to be a jerk, I am a total geek and love algorithm problems). If you need help with it, feel free to ask! I already typed up some hints by accident since I didnt read your question properly at first.
EDIT: I accidentally did a crappy multiplication algorithm. An example of a really terrible division algorithm (i cheated) would be:
AbsolutelyTerribleDivisionAlg(int a, int b)
int quotient = 0
while crappyMultiplicationAlg(int b, int quotient) < a
quotient = quotient + 1
return quotient
This algorithm is bad for a whole bunch of reasons, not the least of which is the use of my crappy multiplication algorithm (which will be called more than once even on a relatively "tame" run). Even if we were allowed to use the * operator though, this is still a really bad algorithm, largely due to the same mechanism used in my awful mult alg.
PS There may be a fence-post error or two in my two algs... i posted them more for conceptual clarity than correctness. No matter how accurate they are at doing multiplication or division, though, never use them. They will give your laptop herpes and then cause it to burn up in a sulfur-y implosion of sadness.
I don't know what you mean by fast...and this seems like a basic question to test your thought process.
A simple function can be use a counter and keep subtracting the divisor from the dividend till it becomes 0. This is O(n) process.
int divide(int n, int d){
int c = 0;
while(1){
n -= d;
if(n >= 0)
c++;
else
break;
}
return c;
}
Another way can be using shift operator, which should do it in log(n) steps.
int divide(int n, int d){
if(d <= 0)
return -1;
int k = d;
int i, c, index=1;
c = 0;
while(n > d){
d <<= 1;
index <<= 1;
}
while(1){
if(k > n)
return c;
if(n >= d){
c |= index;
n -= d;
}
index >>= 1;
d >>= 1;
}
return c;
}
This is just like integer division as we do in High-School Mathematics.
PS: If you need a better explanation, I will. Just post that in comments.
EDIT: edited the code wrt Erobrere's comment.
The simplest way to perform a division is by successive subtractions: subtract b from a as long as a remains positive. The quotient is the number of subtractions performed.
This can be pretty slow, as you will perform q subtractions and tests.
With a=28 and b=3,
28-3-3-3-3-3-3-3-3-3=1
the quotient is 9 and the remainder 1.
The next idea that comes to mind is to subtract several times b in a single go. We can try with 2b or 4b or 8b... as these numbers are easy to compute with additions. We can go as for as possible as long as the multiple of b does not exceed a.
In the example, 2³.3 is the largest multiple which is possible
28>=2³.3
So we subtract 8 times 3 in a single go, getting
28-2³.3=4
Now we continue to reduce the remainder with the lower multiples, 2², 2 and 1, when possible
4-2².3<0
4-2.3 <0
4-1.3 =1
Then our quotient is 2³+1=9 and the remainder 1.
As you can check, every multiple of b is tried once only, and the total number of attempts equals the number of doublings required to reach a. This number is just the number of bits required to write q, which is much smaller than q itself.
This is not the fastest solution, but I think it's readable enough and works:
def weird_div(dividend, divisor):
if divisor == 0:
return None
dend = abs(dividend)
dsor = abs(divisor)
result = 0
# This is the core algorithm, the rest is just for ensuring it works with negatives and 0
while dend >= dsor:
dend -= dsor
result += 1
# Let's handle negative numbers too
if (dividend < 0 and divisor > 0) or (dividend > 0 and divisor < 0):
return -result
else:
return result
# Let's test it:
print("49 divided by 7 is {}".format(weird_div(49,7)))
print("100 divided by 7 is {} (Discards the remainder) ".format(weird_div(100,7)))
print("-49 divided by 7 is {}".format(weird_div(-49,7)))
print("49 divided by -7 is {}".format(weird_div(49,-7)))
print("-49 divided by -7 is {}".format(weird_div(-49,-7)))
print("0 divided by 7 is {}".format(weird_div(0,7)))
print("49 divided by 0 is {}".format(weird_div(49,0)))
It prints the following results:
49 divided by 7 is 7
100 divided by 7 is 14 (Discards the remainder)
-49 divided by 7 is -7
49 divided by -7 is -7
-49 divided by -7 is 7
0 divided by 7 is 0
49 divided by 0 is None
unsigned bitdiv (unsigned a, unsigned d)
{
unsigned res,c;
for (c=d; c <= a; c <<=1) {;}
for (res=0;(c>>=1) >= d; ) {
res <<= 1;
if ( a >= c) { res++; a -= c; }
}
return res;
}
The pseudo code:
count = 0
while (dividend >= divisor)
dividend -= divisor
count++
//Get count, your answer
I have written this code to check if a number is prime (for numbers upto 10^9+7)
Is this a good method ??
What will be the time complexity for this ??
What I have done is that I have made a unordered_set which stores the prime numbers upto sqrt(n).
When checking if a number is prime or not if first check if its is less than the max number in the table.
If it is less it is searched in the table so the complexity should be O(1) in this case.
If it is more the number is put through a divisibility test with the numbers from the set of number containing the prime numbers.
#include<iostream>
#include<set>
#include<math.h>
#include<unordered_set>
#define sqrt10e9 31623
using namespace std;
unordered_set<long long> primeSet = { 2, 3 }; //used for fast lookups
void genrate_prime_set(long range) //this generates prime number upto sqrt(10^9+7)
{
bool flag;
set<long long> tempPrimeSet = { 2, 3 }; //a temporay set is used for genration
set<long long>::iterator j;
for (int i = 3; i <= range; i = i + 2)
{
//cout << i << " ";
flag = true;
for (j = tempPrimeSet.begin(); *j * *j <= i; ++j)
{
if (i % (*j) == 0)
{
flag = false;
break;
}
}
if (flag)
{
primeSet.insert(i);
tempPrimeSet.insert(i);
}
}
}
bool is_prime(long long i,unordered_set<long long> primeSet)
{
bool flag = true;
if(i <= sqrt10e9) //if number exist in the lookup table
return primeSet.count(i);
//if it doesn't iterate through the table
for (unordered_set<long long>::iterator j = primeSet.begin(); j != primeSet.end(); ++j)
{
if (*j * *j <= i && i % (*j) == 0)
{
flag = false;
break;
}
}
return flag;
}
int main()
{
//long long testCases, a, b, kiwiCount;
bool primeFlag = true;
//unordered_set<int> primeNum;
genrate_prime_set(sqrt10e9);
cout << primeSet.size()<<"\n";
cout << is_prime(9999991,primeSet);
return 0;
}
This doesn't strike me as a particularly efficient way to do the job at hand.
Although it probably won't make a big difference in the end, the efficient way to generate all the primes up to some specific limit is clearly to use a sieve--the sieve of Eratosthenes is simple and fast. There are a couple of modifications that can be faster, but for the small size you're dealing with, they're probably not worthwhile.
These normally produce their output in a more effective format than you're currently using as well. In particular, you typically just dedicate one bit to each possible prime (i.e., each odd number) and end up with it zeroed if the number is composite, and one if it's prime (you can, of course, reverse the sense if you prefer).
Since you only need one bit for each odd number from 3 to 31623, this requires only about 16 K bits, or about 2K bytes--a truly minuscule amount of memory by modern standards (especially: little enough to fit in L1 cache quite easily).
Since the bits are stored in order, it's also trivial to compute and test by the factors up to the square root of the number you're testing instead of testing against all the numbers in the table (including those greater than the square root of the number you're testing, which is obviously a waste of time). This also optimizes access to the memory in case some of it's not in the cache (i.e., you can access all the data in order, making life as easy as possible for the hardware prefetcher).
If you wanted to optimize further, I'd consider just using the sieve to find all primes up to 109+7, and look up inputs. Whether this is a win will depend (heavily) upon the number of queries you can expect to receive. A quick check shows that a simple implementation of the Sieve of Eratosthenes can find all primes up to 109 in about 17 seconds. After that, each query is (of course) essentially instantaneous (i.e., the cost of a single memory read). This does require around 120 megabytes of memory for the result of the sieve, which would once have been a major consideration, but (except on fairly limited systems) normally wouldn't be any more.
The very short answer: do research on the subject, starting with the term "Miller-Rabin"
The short answer is no:
Looking for factors of a number is a poor way to check for primality
Exhaustively searching through primes is a poor way to look for factors
Especially if you search through every prime, rather than just the ones less than or equal to the square root of the number
Doing a primality test on each number of them is a poor way to generate a list of primes
Also, you should take in primeSet by reference rather than copy, if it really needs to be a parameter.
Note: testing small primes to see if they divide a number is a useful first step of a primality test, but should generally only be used for the smallest primes before switching to a better method
No, it's not a very good way to determine if a number is prime. Here is pseudocode for a simple primality test that is sufficient for numbers in your range; I'll leave it to you to translate to C++:
function isPrime(n)
d := 2
while d * d <= n
if n % d == 0
return False
d := d + 1
return True
This works by trying every potential divisor up to the square root of the input number n; if no divisor has been found, then the input number could not be composite, meaning of the form n = p × q, because one of the two divisors p or q must be less than the square root of n while the other is greater than the square root of n.
There are better ways to determine primality; for instance, after initially checking if the number is even (and hence prime only if n = 2), it is only necessary to test odd potential divisors, halving the amount of work necessary. If you have a list of primes up to the square root of n, you can use that list as trial divisors and make the process even faster. And there are other techniques for larger n.
But that should be enough to get you started. When you are ready for more, come back here and ask more questions.
I can only suggest a way to use a library function in Java to check the primality of a number. As for the other questions, I do not have any answers.
The java.math.BigInteger.isProbablePrime(int certainty) returns true if this BigInteger is probably prime, false if it's definitely composite. If certainty is ≤ 0, true is returned. You should try and use it in your code. So try rewriting it in Java
Parameters
certainty - a measure of the uncertainty that the caller is willing to tolerate: if the call returns true the probability that this BigInteger is prime exceeds (1 - 1/2^certainty). The execution time of this method is proportional to the value of this parameter.
Return Value
This method returns true if this BigInteger is probably prime, false if it's definitely composite.
Example
The following example shows the usage of math.BigInteger.isProbablePrime() method
import java.math.*;
public class BigIntegerDemo {
public static void main(String[] args) {
// create 3 BigInteger objects
BigInteger bi1, bi2, bi3;
// create 3 Boolean objects
Boolean b1, b2, b3;
// assign values to bi1, bi2
bi1 = new BigInteger("7");
bi2 = new BigInteger("9");
// perform isProbablePrime on bi1, bi2
b1 = bi1.isProbablePrime(1);
b2 = bi2.isProbablePrime(1);
b3 = bi2.isProbablePrime(-1);
String str1 = bi1+ " is prime with certainity 1 is " +b1;
String str2 = bi2+ " is prime with certainity 1 is " +b2;
String str3 = bi2+ " is prime with certainity -1 is " +b3;
// print b1, b2, b3 values
System.out.println( str1 );
System.out.println( str2 );
System.out.println( str3 );
}
}
Output
7 is prime with certainity 1 is true
9 is prime with certainity 1 is false
9 is prime with certainity -1 is true
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