public class Factorial {
int factR(int n){
int result;
if(n==1)return 1;
result=factR(n-1)*n;
System.out.println("Recursion"+result);
return result;
}
I know that this method will have the output of
Recursion2
Recursion6
Recursion24
Recursion120
Recursive120
However, my question is how does java store the past values for the factorial? It also appears as if java decides to multiply the values from the bottom up. What is the process by which this occurs? It it due to how java stores memory in its stack?
http://www.programmerinterview.com/index.php/recursion/explanation-of-recursion/
The values are stored on Java's call stack. It's in reverse because of how this recursive function is defined. You're getting n, then multiplying it by the value from the same function for n-1 and so on, and so on, until it reaches 1 and just returns 1 at that level. So, for 5, it would be 5 * 4 * 3 * 2 * 1. Answer is the same regardless of the direction of multiplication.
You can see how this works by writing a program that will break the stack and give you a StackOverflowError. You cannot store infinite state on the call stack!
public class StackTest {
public static void main(String[] args) {
run(1);
}
private static void run(int index) {
System.out.println("Index: " + index);
run(++index);
}
}
It actually isn't storing 'past values' at all. It stores the state of the program in the stack, with a frame for each method call containing data such as the current line the program is on. But there is only one value for the variable result at any time, for the current method on top of the stack. That gets returned and used to compute result in the frame that called this, and so on backwards, hence the bottom up behaviour you see.
One way to make this less confusing is to take recursion out of the picture temporarily. Suppose Java did not support recursion, and methods were only allowed to call other, different methods. If you wanted to still take a similar approach, one crude way would be to copy paste the factR method into multiple distinct but similar methods, something like:
int fact1(int n){
int result;
if(n==1)return 1;
// Here's the difference: call the 'next' method
result=fact2(n-1)*n;
System.out.println("Recursion"+result);
return result;
}
Similarly define a fact2 which calls fact3 and so on, although eventually you have to stop defining new methods and just hope that the last one doesn't get called. This would be a horrible program but it should be very obvious how it works as there's nothing magical. With some thought you can realise that factR is essentially doing the same thing. Then you can see that Java doesn't 'decide' to multiply the values bottom up: the observed behaviour is the only logical possibility given your code.
well i am trying to understand you,
if someone call likewise then
factR(3) it's recursive process so obviously java uses Stack for maintaining work flow,
NOTE : please see below procedural task step by step and again note
where it get back after current task complete.
result=factR(2)*3 // again call where n=2
-> result=factR(1)*2 // again call where n=1
-> now n=1 so that it will return 1
-> result=1*2 // after return it will become 6
print "Recursion2" // print remaning stuff
return 2;
result=2*3 // after return it will become 6
print "Recursion3" // print remaning stuff
return 3
Related
This question already has answers here:
Understanding recursion [closed]
(20 answers)
Closed 5 years ago.
Before you get started, I have used google countless times in hopes of searching for a very brief and simple explanation of how recursion works when it has a return type. But I guess I'm not as bright as I thought since i still cant understand it quite well.
Take the following code snippet (in java) as an example
public static int recursion(int num)
{
int result;
if (num == 1)
result = 1;
else
result = recursion(num - 1) + num;
return result;
}
I grabbed this code from my professors lecture slide and he said this will return 1 + 2 + 3 + ... + num.
I just need someone to explain how the process works in the method that i provided. Maybe a step by step approach might help me understand how recursion works.
recursion(5) = recursion(4) + 5, let's figure out recursion(4) and come back to this later
recursion(4) = recursion(3) + 4, let's figure out recursion(3) and come back to this later
recursion(3) = recursion(2) + 3, ...
recursion(2) = recursion(1) + 2, ...
recursion(1) = 1, we know this!
recursion(2) = 1 + 2, now we can evaluate this
recursion(3) = (1+2) + 3, and now we can evaluate this
recursion(4) = (1+2+3) + 4, ...
recursion(5) = (1+2+3+4) + 5, the answer to our original question
Note: Without knowing recursion(1), we'd have gone to 0, -1, -2, and so on until forever. This known quantity is called the base case and it is a requirement for recursion.
Basically when there is a stack buildup for each item that is created beyond the last iteration. (Where num=1)
When n>1 the if statement kicks the iteration to the else which 'saves' the result in a stack and calls the same funtion again with n-1
what this effectively does is keep calling the same function until you hit your designated 'base case' which is n=1
Recursion is all about solving a problem by breaking it into a smaller problem. In your case, the question is "how do you sum the numbers from 1 to n", and the answer is "sum up all the numbers from 1 to n-1, and then add n to it". You've phrased the problem in terms of a smaller or simpler version of itself. This often involves separating out a "base case"—an irreducibly simple problem with a straightforward answer.
public static int recursion(int num)
{
int result;
if (num == 1)
result = 1; // Base case: the sum of the numbers from 1 to 1 is 1.
else
result =
// This is the sum of numers from 1 to n-1. The function calls itself.
recursion(num - 1)
// Now add the final number in the list, and return your result.
+ num;
return result;
}
You're defining the unsolved problem in terms of itself, which works because the solution always involves either the base case or a simpler version of the problem (which itself further involves either the base case or an even simpler version of the problem).
I'll close with one of my favorite jokes:
How do you explain recursion to a five-year-old?
You explain recursion to a four-year-old, and wait a year.
Going by the classic code example you posted. if you call your method like so with number passed in as 5:
recursion(5);
In layman terms just to understand, your function will create & call another copy of your function in the else block as below:
recursion(4);
and then
recursion(3);
recursion(2);
recursion(1);
as the number keeps decrementing.
Finally it will call the if part in the final copy of the method as num will satisfy num == 1. So from there it starts unwinding & returning each value to the previous call.
As each method call has its own stack to load method local variables on, there will be n number of stacks created for n calls. When the deepest call in recursion is made, then the stacks start unwinding. Hence recursion achieved
The most important thing however to note is that there is a base-most call in your code, which is done at 1 just because you have the check if (num == 1). Else it would be infinite recursion & of course a fatal & wrong program to write. The base-most call is from where its called as stack unwinding in recursion terms.
Example: Finding the factorial of a number is the most classic examples of recursion.
Performance: Do look into recursion vs iteration and recursion vs looping to see what are the performance impacts of recursion
I was just introduced to recursion and I was given the following lines of code:
public class RecursionThree
{
public void run(int x )
{
if(x<5)
run(x+1);
out.println(x);
}
public static void main(String args[] )
{
RecursionThree test = new RecursionThree ();
test.run(1);
}
}
and the output is supposed to be: 5 4 3 2 1. I get why it would print 5 (because 5<5 would equal false and it would print x, which is 5). However, I do not understand why it prints 4 3 2 1 too. Thanks for your help
How recursion works is you break down to the base case and build up backwards. In your case. Your base case was x>=5, the point at which it would stop expanding the recursive tree, you can think of the base case as the end of the tree, or leaf. After that it goes back up completing things that were to be done after run was called. SO in your case, each time after calling one, it printed out x.
When x=1, it calls run(2), after run(2) is resolved, it would go to the next line. After run(2), the print out is 5 4 3 2 after that it would print out x coming back to the original call of run(1), which would be 1. This is really great for traversing trees etc and a lot of other problems.
To picture it, when you call run(1)
run(1)
1<5
run(2)
2<5
run(3)
3<5
run(4)
4<5
run(5)
print(5)
print(4)
print(3)
print(2)
print(1)
As you can see it goes to the base case, and back up.
To get familiar with recursion more, you can do problems like, finding the largest int in an array with the method head being public int findLargest(int [] array, int someNumber) where you would use someNumber to whatever you think you need. Or reversing a string using recursion and one parameter.
For your x=4 case, run(5) was called. After that completed running, control was returned to the callee function, where x is presently 4. So, it executes the line after the if which prints the 4.
The same logic follows for x = 3, 2, 1.
This is the context of my program.
A function has 50% chance to do nothing, 50% to call itself twice.
What is the probability that the program will finish?
I wrote this piece of code, and it works great apparently. The answer which may not be obvious to everyone is that this program has 100% chance to finish. But there is a StackOverflowError (how convenient ;) ) when I run this program, occuring in Math.Random(). Could someone point to me where does it come from, and tell me if maybe my code is wrong?
static int bestDepth =0;
static int numberOfPrograms =0;
#Test
public void testProba(){
for(int i = 0; i <1000; i++){
long time = System.currentTimeMillis();
bestDepth = 0;
numberOfPrograms = 0;
loop(0);
LOGGER.info("Best depth:"+ bestDepth +" in "+(System.currentTimeMillis()-time)+"ms");
}
}
public boolean loop(int depth){
numberOfPrograms++;
if(depth> bestDepth){
bestDepth = depth;
}
if(proba()){
return true;
}
else{
return loop(depth + 1) && loop(depth + 1);
}
}
public boolean proba(){
return Math.random()>0.5;
}
.
java.lang.StackOverflowError
at java.util.Random.nextDouble(Random.java:394)
at java.lang.Math.random(Math.java:695)
.
I suspect the stack and the amount of function in it is limited, but I don't really see the problem here.
Any advice or clue are obviously welcome.
Fabien
EDIT: Thanks for your answers, I ran it with java -Xss4m and it worked great.
Whenever a function is called or a non-static variable is created, the stack is used to place and reserve space for it.
Now, it seems that you are recursively calling the loop function. This places the arguments in the stack, along with the code segment and the return address. This means that a lot of information is being placed on the stack.
However, the stack is limited. The CPU has built-in mechanics that protect against issues where data is pushed into the stack, and eventually override the code itself (as the stack grows down). This is called a General Protection Fault. When that general protection fault happens, the OS notifies the currently running task. Thus, originating the Stackoverflow.
This seems to be happening in Math.random().
In order to handle your problem, I suggest you to increase the stack size using the -Xss option of Java.
As you said, the loop function recursively calls itself. Now, tail recursive calls can be rewritten to loops by the compiler, and not occupy any stack space (this is called the tail call optimization, TCO). Unfortunately, java compiler does not do that. And also your loop is not tail-recursive. Your options here are:
Increase the stack size, as suggested by the other answers. Note that this will just defer the problem further in time: no matter how large your stack is, its size is still finite. You just need a longer chain of recursive calls to break out of the space limit.
Rewrite the function in terms of loops
Use a language, which has a compiler that performs TCO
You will still need to rewrite the function to be tail-recursive
Or rewrite it with trampolines (only minor changes are needed). A good paper, explaining trampolines and generalizing them further is called "Stackless Scala with Free Monads".
To illustrate the point in 3.2, here's how the rewritten function would look like:
def loop(depth: Int): Trampoline[Boolean] = {
numberOfPrograms = numberOfPrograms + 1
if(depth > bestDepth) {
bestDepth = depth
}
if(proba()) done(true)
else for {
r1 <- loop(depth + 1)
r2 <- loop(depth + 1)
} yield r1 && r2
}
And initial call would be loop(0).run.
Increasing the stack-size is a nice temporary fix. However, as proved by this post, though the loop() function is guaranteed to return eventually, the average stack-depth required by loop() is infinite. Thus, no matter how much you increase the stack by, your program will eventually run out of memory and crash.
There is nothing we can do to prevent this for certain; we always need to encode the stack in memory somehow, and we'll never have infinite memory. However, there is a way to reduce the amount of memory you're using by about 2 orders of magnitude. This should give your program a significantly higher chance of returning, rather than crashing.
We can do this by noticing that, at each layer in the stack, there's really only one piece of information we need to run your program: the piece that tells us if we need to call loop() again or not after returning. Thus, we can emulate the recursion using a stack of bits. Each emulated stack-frame will require only one bit of memory (right now it requires 64-96 times that, depending on whether you're running in 32- or 64-bit).
The code would look something like this (though I don't have a Java compiler right now so I can't test it):
static int bestDepth = 0;
static int numLoopCalls = 0;
public void emulateLoop() {
//Our fake stack. We'll push a 1 when this point on the stack needs a second call to loop() made yet, a 0 if it doesn't
BitSet fakeStack = new BitSet();
long currentDepth = 0;
numLoopCalls = 0;
while(currentDepth >= 0)
{
numLoopCalls++;
if(proba()) {
//"return" from the current function, going up the callstack until we hit a point that we need to "call loop()"" a second time
fakeStack.clear(currentDepth);
while(!fakeStack.get(currentDepth))
{
currentDepth--;
if(currentDepth < 0)
{
return;
}
}
//At this point, we've hit a point where loop() needs to be called a second time.
//Mark it as called, and call it
fakeStack.clear(currentDepth);
currentDepth++;
}
else {
//Need to call loop() twice, so we push a 1 and continue the while-loop
fakeStack.set(currentDepth);
currentDepth++;
if(currentDepth > bestDepth)
{
bestDepth = currentDepth;
}
}
}
}
This will probably be slightly slower, but it will use about 1/100th the memory. Note that the BitSet is stored on the heap, so there is no longer any need to increase the stack-size to run this. If anything, you'll want to increase the heap-size.
The downside of recursion is that it starts filling up your stack which will eventually cause a stack overflow if your recursion is too deep. If you want to ensure that the test ends you can increase your stack size using the answers given in the follow Stackoverflow thread:
How to increase to Java stack size?
I have the below recursive function to compute factorial of a number. The program works fine except when I remove the if condition. Can someone explain why?
This is the code that works fine --
public static long factUsingRecursion(int number) {
if (number == 1) {
return 1;
} else {
return number * factUsingRecursion(number - 1);
}
}
Without the if condition (Code that throws the error),
public static long factUsingRecursion(int number) {
return number * factUsingRecursion(number - 1);
}
I get the stack overflow error.
Exception in thread "main" java.lang.StackOverflowError
at birst.FactorialUsingRecursion.factUsingRecursion(FactorialUsingRecursion.java:10)
Request experts to please advise me why this is the case?
In recursion, there must always be a base case that stops the recursion. Without the if, you have no base case and nothing stops it. Eventually too many method calls are on the stack and a StackOverflowError results.
This line causing number variable to be decreased by 1
return number * factUsingRecursion(number - 1);
and it will handle all values of number except when it is 1
so this line of code is a break condition
if (number == 1) {
return 1;
}
and it prevent you to avoid stackoverflow exception
Recursion requires a base case. Without it, it will continue calling the function over and over and never stop. The if statement is the base case, which terminates the recursion. That is why if you remove it, you get a StackOverflowError.
Imagine what happens when you call:
factUsingRecursion(3);
With the if:
3*factUsingRecursion(2)
3*2*factUsingRecursion(1)
3*2*1
Without the if:
3*factUsingRecursion(2)
3*2*factUsingRecursion(1)
3*2*1*factUsingRecursion(0)
3*2*1*0*factUsingRecursion(-1)
3*2*1*0*-1*factUsingRecursion(-2)
3*2*1*0*-1*-2*factUsingRecursion(-3)
...
And so on... It will not stop until you encounter the StackOverflow error
It loses one of the things that makes a recursive function recursive in that it has no exit condition.
All recursive solutions must satisfy three rules or properties:
A recursive solution must contain a base case.
A recursive solution must contain a recursive case.
A recursive solution must make progress toward the base case.
From: Data Structures and Algorithms Using Python
The program will no longer work when you remove the if condition because you will just be left with return number * factUsingRecursion(number - 1); and the factUsingRecursion(number - 1) here would have the same return calling return number * factUsingRecursion(number - 1);. Your function constantly calls itself, never able to evaluate to anything. By setting the condition, you function is able to evaluate to a definitive value at some point in the recursive chain, and the first call can evaluate.
For every integer i, you are calling the function with i -1. Integersa are infinite, so you would never stop calling the function. eg: -1000 would call -1001 and this would keep going as long as JVM has some space in it's stack.
Ok I'm really confused about something about recursion in Java. Say I have the following code:
static int findShortestString(String[] paths, int lo, int hi) {
if(lo==hi)
return lo;
int minindex=findShortestString(paths,lo+1, hi);
if(safeStringLength(paths[lo])<safeStringLength(paths[minindex]))
return lo;
return minindex;
Now the question is not really about the code itself, but just about how recursion works. minindex is being set equal to a recursive call. So the first time the function run and tries to set minindex to something, it does so, and then the function calls itself. But when does the if statement run then? Will it only run when minindex finally actually holds a real value? I just cant wrap my head around this. If minindex causes the function to recurse and recurse, then when will the if statement ever be checked? When lo==hi? I dont get it:(
minindex is not assigned until findShortestString returns, which won't happen until lo == hi.
Each time the method calls itself, it narrows the difference between lo and hi by 1, so eventually they'll be equal* and that value will be returned.
An example, with paths = ["p1", "path2", "longpath3"]:
lo = 0, hi = 2
lo != hi -> call findShortestString(paths, 1, 2)
lo = 1, hi = 2
lo != hi -> call findShortestString(paths, 2, 2)
lo = 2, hi = 2
lo == hi -> return lo (=2)
lo = 1, hi = 2, minindex = 2
length of "path2" < length of "longpath3" -> return lo (= 1)
lo = 0, hi = 2, minindex = 1
length of "p1" < length of "path2" -> return lo (= 0)
I've tried to illustrate the variable values at each level of recursion using increasing amounts of indentation. At the beginning of each recursive call, the previous values of lo, hi and minindex are saved off (in a structure called a "stack") and the new values used instead. When each invocation of the method returns, the previously saved values are "popped" off the stack for use, and minindex assigned from the previous return value.
*unless lo > hi to begin with, I guess...
Here's a play by play of the execution:
You call findShortestString() yourself
if lo doesn't not equal hi things continue. Otherwise they stop here and the function returns.
Once you call findShortestString() again, everything in this instance of the function completely stops and will not resume until the computer has a value to give minindex (aka the function returns.) We start over in a new instance of the function at the top. The only code executed until one of the functions return is the code BEFORE the method call. This could be compared to a while loop.
We only get beyond that line once one of the function instances has lo==hi and returns.
Control switches to the function instance before that, which assigns the returned lo value to minindex.
If (safeStringLength(paths[lo])<safeStringLength(paths[minindex])) then we return lo. Else, we return minindex. Either way, this function instance is complete and control returns to the one before it.
Each function called is now only executing the code AFTER the method call, as the method will not get called again. We are unwinding the stack of calls. All of the returns will now be from the last 2 statements, as the code at the top does not get executed again. Note how only one function instance returns with the top part of the code, terminating the while loop. All the rest terminate with the return statements in the part of the function after the recursive call.
Eventually the last function returns and you go back to the code you called the function from originally.
Here's a more readable version of what the code is actually doing:
In the code before the recursive call, all that happens is the creation of a chain of calls until lo==hi. Each time the function is called with lo being 1 greater. Here's a sample stack of calls:
findShortestString(2,5);
findShortestString(3,5);
findShortestString(4,5);
findShortestString(5,5);
When they unwind, each function instance compares the string lengths of the strings at the indexes lo and the index the previous index with the shortest string.
compare strings at indexes 2 and 5
if the string at 2 is smaller, compare the strings at indexes 2 and 4.
Otherwise, compare the strings with indexes at 3 and 5.
If lo>hi at the beginning, the code will continue to run until lo overflows an integer and becomes negative, then until lo finally gets all the way up to hi, or 4,94,967,296 - (original lo - original hi). In other words, in will take a long time. To fix this, add a check at the beginning of the method that throws an exception if lo>hi.
The code could be better rewritten as this:
static int findShortestString(String[] paths, int lo, int hi) {
int indexWithShortestString=lo;
for( int i=lo; i<=hi-1; i++) {
//assumption: lo and hi are both valid indexes of paths
if (paths[i+1].length < paths[i].length)
indexWithShortestString=i+1;
}
}
Think of a stack. Every time the recursive method is called a new "frame" is put on top of the stack. The frame contains its own "slots" for each variable, independent and distinct from those in the frames below.
Eventually, a new frame will be created where the value of lo and hi are equal, and the method will return without pushing another frame. (This is called the "base case".) When that return occurs, that frame is popped off the stack, and the frame that was just below it continues its execution at the second if statement. Eventually that frame is also popped off and the same happens to the frame just below, and so on, until execution returns to the original caller.
Each time findShortestString calls itself, minindex will eventually be assigned, then the value is used in the if statement. It is always set the index of the shortest string at a higher index than lo.
So if there is a call stack with 10 levels of findShortestString, minindex is assigned 9 times (the first call is from another function).
This is a really confusing recursive function. But you generally have it correct. Every call to findShortestString() will push the function onto the stack. It will keep doing this until lo==hi. At that point, the stack is unwound and corresponding recursive calls will be assigned to their corresponding ints.
In this function, it seems that you'll only ever be returning lo. Because either (safeStringLength(paths[lo])<safeStringLength(paths[minindex]) will be true and you'll return lo. Or lo==hi will be true and you'll return lo
In order for the statement
int minindex=findShortestString(paths,lo+1, hi);
to evaluate, the method call findShortestString(paths,lo+1, hi) must return a value. Thus the following if statement will not happen until this method call returns a value. However, this method might call itself again, and you get a nesting effect.
Basically an execution of a function ends when a return statement is called. Everything after a return statement which is called no longer matters (or "exists").
Hence, the local variable minindex will only exist in an execution of a findShortestString function when the first if-statement is false.
Treat each execution of a findShortestString function independently, whether they are called recursively or from somewhere else in the code. i.e. different execution of a findShortestString function may return at different paths and have their own values and local variables. Depending on the input values, they may return at line 3, 6 or 7.
minindenx only exists in an execution that can run line 4, and it is assigned findShortestString(paths,lo+1, hi) which is guaranteed have a value, if the code is correct, otherwise you will get an infinite recursion, resulting in a stack overflow (pun unintended).