I am new to programming and I am writing a simple code in Java that is using recursion. I want to show the product of two numbers (from start to End). The return of the method is the multiplication of the numbers from start to end. (For Example: If the numbers are 1 and 3 then I want the method to return 6.
I managed to do the recursion but I am not sure if the code is effective at all. Here is my code so far. Thanks
public class ÜbungsblattSieben {
public static void main(String[] args) {
System.out.println(multiplyFromStartToEnd(1, 3));
}
public static int multiplyFromStartToEnd(int start, int end) {
if (start == end) {
return end;
} else {
return start * multiplyFromStartToEnd(++start, end);
}
}
}
Your code is as effective as a recursive multiplication can be. Well done.
That said, here are a few notes:
You may write start + 1 instead of ++start. Generally easier to read and understand. Also you do not have to change the start variable itself, you just want to pass a bigger number to the method call, thats all.
You may also want to properly indent your code (just hit the auto-format key in your IDE).
I would also suggest to rename your method to multiplyFromTo, but thats a very subjective note.
All in all, your code would then look like:
public class ÜbungsblattSieben {
public static void main (String[] args) {
System.out.println(multiplyFromStartToEnd(1, 3));
}
public static int multiplyFromTo(int start, int end) {
if (start == end) {
return end;
} else {
return start * multiplyFromStartToEnd(start + 1, end);
}
}
}
For reference, here is how an iterative version could look like:
int result = 1;
for (int i = start; i <= end; i++) {
result *= i;
}
System.out.println(result);
Obviously, this is a lot faster than recursion.
I have a question about java collections such as Set or List. More generally objects that you can use in a for-each loop. Is there any requirement that the elements of them actually has to be stored somewhere in a data structure or can they be described only from some sort of requirement and calculated on the fly when you need them? It feels like this should be possible to be done, but I don't see any of the java standard collection classes doing anything like this. Am I breaking any sort of contract here?
The thing I'm thinking about using these for is mainly mathematics. Say for example I want to have a set representing all prime numbers under 1 000 000. It might not be a good idea to save these in memory but to instead have a method check if a particular number is in the collection or not.
I'm also not at all an expert at java streams, but I feel like these should be usable in java 8 streams since the objects have very minimal state (the objects in the collection doesn't even exist until you try to iterate over them or check if a particular object exists in the collection).
Is it possible to have Collections or Iterators with virtually infinitely many elements, for example "all numbers on form 6*k+1", "All primes above 10" or "All Vectors spanned by this basis"? One other thing I'm thinking about is combining two sets like the union of all primes below 1 000 000 and all integers on form 2^n-1 and list the mersenne primes below 1 000 000. I feel like it would be easier to reason about certain mathematical objects if it was done this way and the elements weren't created explicitly until they are actually needed. Maybe I'm wrong.
Here's two mockup classes I wrote to try to illustrate what I want to do. They don't act exactly as I would expect (see output) which make me think I am breaking some kind of contract here with the iterable interface or implementing it wrong. Feel free to point out what I'm doing wrong here if you see it or if this kind of code is even allowed under the collections framework.
import java.util.AbstractSet;
import java.util.Iterator;
public class PrimesBelow extends AbstractSet<Integer>{
int max;
int size;
public PrimesBelow(int max) {
this.max = max;
}
#Override
public Iterator<Integer> iterator() {
return new SetIterator<Integer>(this);
}
#Override
public int size() {
if(this.size == -1){
System.out.println("Calculating size");
size = calculateSize();
}else{
System.out.println("Accessing calculated size");
}
return size;
}
private int calculateSize() {
int c = 0;
for(Integer p: this)
c++;
return c;
}
public static void main(String[] args){
PrimesBelow primesBelow10 = new PrimesBelow(10);
for(int i: primesBelow10)
System.out.println(i);
System.out.println(primesBelow10);
}
}
.
import java.util.Iterator;
import java.util.NoSuchElementException;
public class SetIterator<T> implements Iterator<Integer> {
int max;
int current;
public SetIterator(PrimesBelow pb) {
this.max= pb.max;
current = 1;
}
#Override
public boolean hasNext() {
if(current < max) return true;
else return false;
}
#Override
public Integer next() {
while(hasNext()){
current++;
if(isPrime(current)){
System.out.println("returning "+current);
return current;
}
}
throw new NoSuchElementException();
}
private boolean isPrime(int a) {
if(a<2) return false;
for(int i = 2; i < a; i++) if((a%i)==0) return false;
return true;
}
}
Main function gives the output
returning 2
2
returning 3
3
returning 5
5
returning 7
7
Exception in thread "main" java.util.NoSuchElementException
at SetIterator.next(SetIterator.java:27)
at SetIterator.next(SetIterator.java:1)
at PrimesBelow.main(PrimesBelow.java:38)
edit: spotted an error in the next() method. Corrected it and changed the output to the new one.
Well, as you see with your (now fixed) example, you can easily do it with Iterables/Iterators. Instead of having a backing collection, the example would've been nicer with just an Iterable that takes the max number you wish to calculate primes to. You just need to make sure that you handle the hasNext() method properly so you don't have to throw an exception unnecessarily from next().
Java 8 streams can be used easier to perform these kinds of things nowadays, but there's no reason you can't have a "virtual collection" that's just an Iterable. If you start implementing Collection it becomes harder, but even then it wouldn't be completely impossible, depending on the use cases: e.g. you could implement contains() that checks for primes, but you'd have to calculate it and it would be slow for large numbers.
A (somewhat convoluted) example of a semi-infinite set of odd numbers that is immutable and stores no values.
public class OddSet implements Set<Integer> {
public boolean contains(Integer o) {
return o % 2 == 1;
}
public int size() {
return Integer.MAX_VALUE;
}
public boolean add(Integer i) {
throw new OperationNotSupportedException();
}
public boolean equals(Object o) {
return o instanceof OddSet;
}
// etc. etc.
}
As DwB stated, this is not possible to do with Java's Collections API, as every element must be stored in memory. However, there is an alternative: this is precisely why Java's Stream API was implemented!
Streams allow you to iterate across an infinite amount of objects that are not stored in memory unless you explicitly collect them into a Collection.
From the documentation of IntStream#iterate:
Returns an infinite sequential ordered IntStream produced by iterative application of a function f to an initial element seed, producing a Stream consisting of seed, f(seed), f(f(seed)), etc.
The first element (position 0) in the IntStream will be the provided seed. For n > 0, the element at position n, will be the result of applying the function f to the element at position n - 1.
Here are some examples that you proposed in your question:
public class Test {
public static void main(String[] args) {
IntStream.iterate(1, k -> 6 * k + 1);
IntStream.iterate(10, i -> i + 1).filter(Test::isPrime);
IntStream.iterate(1, n -> 2 * n - 1).filter(i -> i < 1_000_000);
}
private boolean isPrime(int a) {
if (a < 2) {
return false;
}
for(int i = 2; i < a; i++) {
if ((a % i) == 0) {
return false;
}
return true;
}
}
}
I have a question about tail calls optimization, I need to know how this java code behaves:
private void doSomething(int v) {
inf f = someCalculation(v);
if (f < 0) doSomething(v/2);
else doSomething(v*2);
}
This code is a nonsense example but my question is, in such a case:
The first doSomething() call would be optimized?
The second doSomething() call would be optimized?
The if/else block affects in any way the optimization?
Thanks
EDIT:
Please provide an example on how you would do this if the language was not Java but something else that has TCO
Java 8 has no Tail Call Optimization whatsoever. No calls will be optimized (turned into iteration/goto statements).
The discussion over TCO for Java has a long history, though, with Guy Steele being one of its best-known proponents.
I recommend reading this post from the mlvm-dev mailing list for a recent review of the subject.
Try running the following code:
public static void main(String[] args) {
for (int i = 1; i > 0; i *= 2) { doSomething(i); }
}
private static void doSomething(int start) {
doSomething(start, start);
}
private static void doSomething(int i, int start) {
if (i == 0) { System.out.println("done from " + start); }
else { doSomething(i - 1, start); }
}
If the JVM can run it without stack overflow, then it should mean it can do tail recursion optimization (or a very good constant propagation).
Why I am not getting any Exception in the following code?
After running this code I am getting an infinite loop mentioning at test.fact(t.java:32)
No Compile-Time Error was found.
class test
{
int fact(int m) throws Exception
{
if (m==1)
{
return 1;
}
else
return (fact ((m-1)*m));
}
}
class main
{
public static void main(String ar[]) throws Exception
{
test t = new test();
System.out.println(t.fact(5));
}
}
while say for example i am using
return(a+b);
it executes successfully whats the problem with the recursion
to show an error???
You have a mistake in the return value expression of fact method.
It should be
return fact(m-1) * m;
Another way to calculate factorial using cycle (with no recursion):
int fact(int m) throws Exception
{
int f = 1;
for (int i = 0; i < m; f *= ++i);
return f;
}
return (fact ((m-1)*m));
returns
fact(20)
which returns
fact (380)
which returns
fact (379*380)
which ....
which never returns anything and should make a stack overflow (too much memory is used on the call stack).
return fact(m-1) * m;
should work.
I highly recommend you reading again the basics, and examples ( here, for example - http://www.toves.org/books/java/ch18-recurex/index.html)
Try writing the recursion tree yoursels, in order to understand what happens.
I'm pretty new to the idea of recursion and this is actually my first attempt at writing a recursive method.
I tried to implement a recursive function Max that passes an array, along with a variable that holds the array's size in order to print the largest element.
It works, but it just doesn't feel right!
I have also noticed that I seem to use the static modifier much more than my classmates in general...
Can anybody please provide any general tips as well as feedback as to how I can improve my code?
public class RecursiveTry{
static int[] n = new int[] {1,2,4,3,3,32,100};
static int current = 0;
static int maxValue = 0;
static int SIZE = n.length;
public static void main(String[] args){
System.out.println(Max(n, SIZE));
}
public static int Max(int[] n, int SIZE) {
if(current <= SIZE - 1){
if (maxValue <= n[current]) {
maxValue = n[current];
current++;
Max(n, SIZE);
}
else {
current++;
Max(n, SIZE);
}
}
return maxValue;
}
}
Your use of static variables for holding state outside the function will be a source of difficulty.
An example of a recursive implementation of a max() function in pseudocode might be:
function Max(data, size) {
assert(size > 0)
if (size == 1) {
return data[0]
}
maxtail = Max(data[1..size], size-1)
if (data[0] > maxtail) {
return data[0]
} else {
return maxtail
}
}
The key here is the recursive call to Max(), where you pass everything except the first element, and one less than the size. The general idea is this function says "the maximum value in this data is either the first element, or the maximum of the values in the rest of the array, whichever is larger".
This implementation requires no static data outside the function definition.
One of the hallmarks of recursive implementations is a so-called "termination condition" which prevents the recursion from going on forever (or, until you get a stack overflow). In the above case, the test for size == 1 is the termination condition.
Making your function dependent on static variables is not a good idea. Here is possible implementation of recursive Max function:
int Max(int[] array, int currentPos, int maxValue) {
// Ouch!
if (currentPos < 0) {
raise some error
}
// We reached the end of the array, return latest maxValue
if (currentPos >= array.length) {
return maxValue;
}
// Is current value greater then latest maxValue ?
int currentValue = array[currentPos];
if (currentValue > maxValue) {
// currentValue is a new maxValue
return Max(array, currentPos + 1, currentValue);
} else {
// maxValue is still a max value
return Max(array, currentPos + 1, maxValue);
}
}
...
int[] array = new int[] {...};
int currentPos = 0;
int maxValue = array[currentPos] or minimum int value;
maxValue = Max(array, currentPos, maxValue);
A "max" function is the wrong type of thing to write a recursive function for -- and the fact you're using static values for "current" and "maxValue" makes your function not really a recursive function.
Why not do something a little more amenable to a recursive algorithm, like factorial?
"not-homework"?
Anyway. First things first. The
static int[] n = new int[] {1,2,4,3,3,32,100};
static int SIZE = n.length;
have nothing to do with the parameters of Max() with which they share their names. Move these over to main and lose the "static" specifiers. They are used only once, when calling the first instance of Max() from inside main(). Their scope shouldn't extend beyond main().
There is no reason for all invocations of Max() to share a single "current" index. "current" should be local to Max(). But then how would successive recurrences of Max() know what value of "current" to use? (Hint: Max() is already passing other Max()'s lower down the line some data. Add "current" to this data.)
The same thing goes for maxValue, though the situation here is a bit more complex. Not only do you need to pass a current "maxValue" down the line, but when the recursion finishes, you have to pass it back up all the way to the first Max() function, which will return it to main(). You may need to look at some other examples of recursion and spend some time with this one.
Finally, Max() itself is static. Once you've eliminated the need to refer to external data (the static variables) however; it doesn't really matter. It just means that you can call Max() without having to instantiate an object.
As others have observed, there is no need for recursion to implement a Max function, but it can be instructive to use a familiar algorithm to experiment with a new concept. So, here is the simplified code, with an explanation below:
public class RecursiveTry
{
public static void main(String[] args)
{
System.out.println(Max(new int[] {1,2,4,3,3,32,100}, 0, 0));
}
public static int Max(int[] n, int current, int maxValue)
{
if(current < n.Length)
{
if (maxValue <= n[current] || current == 0))
{
return Max(n, current+1, n[current]);
}
return Max(n, current+1, maxValue);
}
return maxValue;
}
}
all of the static state is gone as unnecessary; instead everything is passed on the stack. the internal logic of the Max function is streamlined, and we recurse in two different ways just for fun
Here's a Java version for you.
public class Recursion {
public static void main(String[] args) {
int[] data = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
System.out.println("Max: " + max(0, data));
}
public static int max(int i, int[] arr) {
if(i == arr.length-1) {
return arr[i];
}
int memo = max(i+1, arr);
if(arr[i] > memo) {
return arr[i];
}
return memo;
}
}
The recurrence relation is that the maximum element of an array is either the first element, or the maximum of the rest of the array. The stop condition is reached when you reach the end of the array. Note the use of memoization to reduce the recursive calls (roughly) in half.
You are essentially writing an iterative version but using tail recursion for the looping. Also, by making so many variables static, you are essentially using global variables instead of objects. Here is an attempt at something closer to a typical recursive implementation. Of course, in real life if you were using a language like Java that doesn't optimize tail calls, you would implement a "Max" function using a loop.
public class RecursiveTry{
static int[] n;
public static void main(String[] args){
RecursiveTry t = new RecursiveTry(new int[] {1,2,4,3,3,32,100});
System.out.println(t.Max());
}
RecursiveTry(int[] arg) {
n = arg;
}
public int Max() {
return MaxHelper(0);
}
private int MaxHelper(int index) {
if(index == n.length-1) {
return n[index];
} else {
int maxrest = MaxHelper(index+1);
int current = n[index];
if(current > maxrest)
return current;
else
return maxrest;
}
}
}
In Scheme this can be written very concisely:
(define (max l)
(if (= (length l) 1)
(first l)
(local ([define maxRest (max (rest l))])
(if (> (first l) maxRest)
(first l)
maxRest))))
Granted, this uses linked lists and not arrays, which is why I didn't pass it a size element, but I feel this distills the problem to its essence. This is the pseudocode definition:
define max of a list as:
if the list has one element, return that element
otherwise, the max of the list will be the max between the first element and the max of the rest of the list
A nicer way of getting the max value of an array recursively would be to implement quicksort (which is a nice, recursive sorting algorithm), and then just return the first value.
Here is some Java code for quicksort.
Smallest codesize I could get:
public class RecursiveTry {
public static void main(String[] args) {
int[] x = new int[] {1,2,4,3,3,32,100};
System.out.println(Max(x, 0));
}
public static int Max(int[] arr, int currPos) {
if (arr.length == 0) return -1;
if (currPos == arr.length) return arr[0];
int len = Max (arr, currPos + 1);
if (len < arr[currPos]) return arr[currPos];
return len;
}
}
A few things:
1/ If the array is zero-size, it returns a max of -1 (you could have another marker value, say, -MAX_INT, or throw an exception). I've made the assumption for code clarity here to assume all values are zero or more. Otherwise I would have peppered the code with all sorts of unnecessary stuff (in regards to answering the question).
2/ Most recursions are 'cleaner' in my opinion if the terminating case is no-data rather than last-data, hence I return a value guaranteed to be less than or equal to the max when we've finished the array. Others may differ in their opinion but it wouldn't be the first or last time that they've been wrong :-).
3/ The recursive call just gets the max of the rest of the list and compares it to the current element, returning the maximum of the two.
4/ The 'ideal' solution would have been to pass a modified array on each recursive call so that you're only comparing the first element with the rest of the list, removing the need for currPos. But that would have been inefficient and would have bought down the wrath of SO.
5/ This may not necessarily be the best solution. It may be that by gray matter has been compromised from too much use of LISP with its CAR, CDR and those interminable parentheses.
First, let's take care of the static scope issue ... Your class is defining an object, but never actually instantiating one. Since main is statically scoped, the first thing to do is get an object, then execute it's methods like this:
public class RecursiveTry{
private int[] n = {1,2,4,3,3,32,100};
public static void main(String[] args){
RecursiveTry maxObject = new RecursiveTry();
System.out.println(maxObject.Max(maxObject.n, 0));
}
public int Max(int[] n, int start) {
if(start == n.length - 1) {
return n[start];
} else {
int maxRest = Max(n, start + 1);
if(n[start] > maxRest) {
return n[start];
}
return maxRest;
}
}
}
So now we have a RecursiveTry object named maxObject that does not require the static scope. I'm not sure that finding a maximum is effective using recursion as the number of iterations in the traditional looping method is roughly equivalent, but the amount of stack used is larger using recursion. But for this example, I'd pare it down a lot.
One of the advantages of recursion is that your state doesn't generally need to be persisted during the repeated tests like it does in iteration. Here, I've conceded to the use of a variable to hold the starting point, because it's less CPU intensive that passing a new int[] that contains all the items except for the first one.