I have written a java program to add elements in an array using Linear Recursion. The output obtained is not as expected. Can anyone point what is wrong with this program?
public class TestSum {
public int count = 0;
public int sum(int[] a){
count++;
if(a.length == count){
return a[count -1];
}
return sum(a) + a[count -1] ;
}
public static void main(String[] args) {
int[] a = {1,2,3};
int val = new TestSum().sum(a);
System.out.println(val);
}
}
I am expecting the output as 6 but obtained is 9. What is wrong?
Strangely if I change the order of addition i.e. return a[count -1] + sum(a); then it gives output as 6.
Generally, recursive programs that are not re-entrant (i.e. relying on external state) are suspicious. In your particular case count will change between invocations of sum, making the behavior hard to trace, and ultimately resulting in the error that you observe.
You should pass the index along with the array to make it work:
// The actual implementation passes the starting index
private static int sum(int[] a, int start){
if(a.length == start){
return 0;
}
return sum(a, start+1) + a[start];
}
// Make sure the method can be called with an array argument alone
public static int sum(int[] a) {
return sum(a, 0);
}
Unlike an implementation that increments the count external to the method, this implementation can be called concurrently on multiple threads without breaking.
Related
I have written a program to sort elements of an array based on the principle of quicksort. So what the program does is that it accepts an array, assumes the first element as the pivot and then compares it with rest of the elements of the array. If the element found greater then it will store at the last of another identical array(say b) and if the element is less than the smaller than it puts that element at the beginning of the array b. in this way the pivot will find its way to the middle of the array where the elements that are on the left-hand side are smaller and at the right-hand side are greater than the pivot. Then the elements of array b are copied to the main array and this whole function is called via recursion. This is the required code.
package sorting;
import java.util.*;
public class AshishSort_Splitting {
private static Scanner dogra;
public static void main(String[] args)
{
dogra=new Scanner(System.in);
System.out.print("Enter the number of elements ");
int n=dogra.nextInt();
int[] a=new int[n];
for(int i=n-1;i>=0;i--)
{
a[i]=i;
}
int start=0;
int end=n-1;
ashishSort(a,start,end);
for(int i=0;i<n;i++)
{
System.out.print(+a[i]+"\n");
}
}
static void ashishSort(int[]a,int start,int end)
{
int p;
if(start<end)
{
p=ashishPartion(a,start,end);
ashishSort(a,start,p-1);
ashishSort(a,p+1,end);
}
}
public static int ashishPartion(int[] a,int start,int end)
{
int n=start+end+1;
int[] b=new int[n];
int j=start;
int k=end;
int equal=a[start];
for(int i=start+1;i<=end;i++)
{
if(a[i]<equal)
{
b[j]=a[i];
j++;
}
else if(a[i]>equal)
{
b[k]=a[i];
k--;
}
}
b[j]=equal;
for(int l=0;l<=end;l++)
{
a[l]=b[l];
}
return j;
}
}
this code works fine when I enter the value of n up to 13930, but after that, it shows
Exception in thread "main" java.lang.StackOverflowError
at sorting.AshishSort_Splitting.ashishSort(AshishSort_Splitting.java:28)
at sorting.AshishSort_Splitting.ashishSort(AshishSort_Splitting.java:29)
I know the fact the error caused due to bad recursion but I tested my code multiple times and didn't find any better alternative. please help. thanks in advance.
EDIT: can someone suggest a way to overcome this.
I see perfrmance issues first. I see in your partition method:
int n = start+end+1
Right there, if the method was called on an int[1000] with start=900 and end=999, you are allocating an int[1900]... Not intended, I think...!
If you are really going to trash memory instead of an in-place partitioning,
assume
int n = end-start+1
instead for a much smaller allocation, and j and k indexes b[], they would be j=0 and k=n, and return start + j.
Second, your
else if(a[i]<equal)
is not necessary and causes a bug. A simple else suffice. If you don't replace the 0's in b[j..k] you'll be in trouble when you refill a[].
Finally, your final copy is bogus, from [0 to end] is beyond the bounds of the invocation [start..end], AND most importantly, there is usually nothing of interest in b[nearby 0] with your b[] as it is. The zone of b[] (in your version) is [start..end] (in my suggested version it would be [0..n-1])
Here is my version, but it still has the O(n) stack problem that was mentioned in the comments.
public static int ashishPartion(int[] a, int start, int end) {
int n = end-start + 1;
int[] b = new int[n];
int bj = 0;
int bk = n-1;
int pivot = a[start];
for (int i = start + 1; i <= end; i++) {
if (a[i] < pivot) {
b[bj++] = a[i];
} else {
b[bk--] = a[i];
}
}
b[bj] = pivot;
System.arraycopy(b, 0, a, start, n);
return start+bj;
}
If you are free to choose a sorting algo, then a mergesort would be more uniform on performance, with logN stack depth. Easy to implement.
Otherwise, you will have to de-recurse your algo, using a manual stack and that is a nice homework that I won't do for you... LOL
Could anyone give me some clue about how could I Transform this code to recursion:
public class arrayExample {
public static void main (String[] args) {
int[] a = {2,2,2,2};
int[] b = {2,2,2,2};
int n = a.length;
int sum = 0;
for (int i = 0; i < n; i++) {
sum += a[i] * b[i];
}
System.out.println(sum);
}
}
So to do this do product with recursion.
You asked for a hint, so I'm not giving you the complete solution. When you want to process a list (or an array) recursively, the concept is nearly always:
public int recursiveFunction(List l, int carry) {
if (l.isEmpty()) {
return carry;
}
return recursiveFunction(l.subList(1, l.size()), operation(carry, l.get(0));
}
Where operation is whatever you want to do with your list. carry is used to provide an initial value (in the first call) and save the interim results.
You just have to change the code so it uses two arrays instead of one list and choose the correct operation.
Ok so hoping you have tried it before this is one possible way to code it.
public class ArrayExample {
public static void main (String[] args) {
int[] a = {2,2,2,2};
int[] b = {2,2,2,2};
int n = a.length;
int result = recurseSum(a, b, n-1);
System.out.println(result);
}
public static int recurseSum(int[] a, int[] b, int n){
if(n == 0)
return a[0]*b[0];
else{
return (a[n] * b[n]) + recurseSum(a,b,n-1);
}
}
}
This code is basically doing the same thing in the iteration.
The recursive call happens 4 times. When n hits 0, a[0]*b[0] is returned to the higher call. so basically from right to left it happens as follows:
a[3]*b[3] + a[2]*b[2] + a[1]*b[1] + a[0]*b[0]
One simple way to make a loop into a recursion is to answer these two questions:
What happens when the loop executes zero times?
If the loop has already executed n-1 times, how do I compute the result after the n-th iteration?
The answer to the first case produces your base case; the answer to the second question explains how to do the recursive invocation.
In your case, the answers are as follows:
When the loop executes zero times, the sum is zero.
When the loop executed n-1 times, add a[n] * b[n] to the previous result.
This can be translated into a recursive implementation
static int dotProduct(int[] a, int[] b, int n) {
... // your implementation here
}
For a classic binarySearch on an array of java Strings (say String[] a), which is the correct way of calling the search method? is it
binarySearch(a,key,0,a.length)
or
binarySearch(a,key,0,a.length-1)
I tried both for the below implementation,and both seems to work.. Is there a usecase where either of these calls can fail?
class BS{
public static int binarySearch(String[] a,String key){
return binarySearch(a,key,0,a.length);
//return binarySearch(a,key,0,a.length-1);
}
public static int binarySearch(String[] a,String key,int lo,int hi) {
if(lo > hi){
return -1;
}
int mid = lo + (hi - lo)/2;
if(less(key,a[mid])){
return binarySearch(a,key,lo,mid-1);
}
else if(less(a[mid],key)){
return binarySearch(a,key,mid+1,hi);
}
else{
return mid;
}
}
private static boolean less(String x,String y){
return x.compareTo(y) < 0;
}
public static void main(String[] args) {
String[] a = {"D","E","F","M","K","I"};
Arrays.sort(a);
System.out.println(Arrays.toString(a));
int x = binarySearch(a,"M");
System.out.println("found at :"+x);
}
}
Consider the case where
a = [ "foo" ]
and you search key "zoo" with binarySearch(a,key,0,a.length);
The code will search for it in interval[0,1], see it should be right than that,
next recursion searches interval [1,1], causing an indexing of a[1] at line
if(less(key,a[mid])){
resulting in a array out of bounds error.
The second solution will work fine.
I think the second approach will be safe.
Consider this case - you have an array of 9 elements and the key is situated at the last index (8-th element). Then you might have a method call like this if you follow the first approach -
binarySearch(a, key, 9, 9);
Now, in that method execution, the integer division in the following line will result in 9 -
int mid = 9 + (9 - 9)/2;
and you will be indexing your array with 9 in the next line -
if( less(key,a[mid]) ) { // You'll face ArrayIndexOutOfBoundException
....
}
which will be invalid and cause ArrayIndexOutOfBoundException.
The second approach however will be just fine.
I am having a lot of trouble with this basic recursion problem in java; any pointers would be great.
"Write a static recursive method to print out the nth term of the
geometric sequence: 2, 6, 18, 54."
From what I can gather, somewhere in the code I should be recursively multiplying something by 3, but I'm struggling to figure out how to do this. I know I need a termination statement, but when does that occur? Do I need a helper method?
A Recursive Function is a function whose implementation references itself. Below is some funny example:
public class Inception {
public void dream() {
boolean enoughDreaming = false;
//Some code logic below to check if it's high time to stop dreaming recursively
...
...
if(!enoughDreaming) {
dream(); //Dream inside a Dream
}
}
}
And the solution for your problem:
public class GeometricSequence {
public static void main(String[] args) {
//Below method parameters - 5 = n, 1 = count (counter), res = result (Nth number in the GP.
System.out.println(findNthNumber(5, 1, 2));
}
public static int findNthNumber(int n, int count, int res) {
return ((count == n)) ? res : findNthNumber(n, count+1, res *3);
}
}
EDIT:
The above class uses "int", which is good only for small numbers (because of Integer Overflow problem). The below class is better for all types/numbers:
public class GeometricSequence {
public static void main(String[] args) {
//Below method parameters - 5 = n, 1 = count (counter), res = result (Nth number in the GP.
System.out.println(findNthNumber(2000, 1, new BigInteger("2")));
}
public static BigInteger findNthNumber(int n, int count, BigInteger res) {
return ((count == n)) ? res : findNthNumber(n, count+1, res.multiply(new BigInteger("3")));
}
}
This is the simplest example of recursion.
You need a method declaration.
You need to check if the end has been reached.
Otherwise you need to call the method again with an operation which makes the difference between one term and the next.
Yes, you need a termination condition - basically when you've taken as many steps as you need. So consider how you want to transition from one call to another:
How are you going to propagate the results so far?
What extra state do you need to keep track of how many more steps you need to take?
What are you going to return from the method?
Here's a C# example (I know your doing Java but it's pretty similar)
public static void Recursive(int counter, int iterations, int value, int multiplier)
{
if (counter < iterations)
{
Console.WriteLine(value);
counter++;
Recursive(counter, iterations, (value * multiplier), multiplier);
}
}
So when you run the function you enter the parameters
"counter" will always be 0 when you first call it
"iterations" is the value of n
"value" is your starting value, in your case 2
"multiplier" is how much you want to multiply by each iteration, in your case 3
Every time it runs it will check to see if counter is less than iterations. If it is more, the value is printed, the counter is incremented, the value is multiplied by the multiplier and you add the same parameters back in to the function.
A recursive solution: Seq(1) is the first element of the sequence .... Seq(n-th)
public static void main(String args[]) throws Exception {
int x = Seq(3); //x-> 18
}
public static int Seq(int n){
return SeqRec(n);
}
private static int SeqRec(int n){
if(n == 1)
return 2;
else return SeqRec(n - 1) * 3;
}
Non-Recursive solution:
public static int Non_RecSeq(int n){
int res = 2;
for(int i = 1; i < n; i ++)
res *= 3;
return res;
}
public static void main(String args[]) throws Exception {
int x = Non_RecSeq(3); //x-> 18
}
I am doing this assignment and I am having trouble writing this method recursively.
I have this way to do it which is effective but not recursive:
public static <T extends Comparable< ? super T>> T getLargest(T [] a, int low,
int high)
{
if(low>high)
throw new IllegalArgumentException();
return Collections.max(Arrays.asList(Arrays.copyOfRange(a, low, high)));
So from there I went to this one, which kind of extends it but is not recursive either:
T[] arrCopy = (T[]) new Object[high-low];
for(int i=low;i<high;i++){
if(a[i].compareTo(a[i-1])>0)
arrCopy[i]=a[i];
else
arrCopy[i]=a[i+1];
}
return arrCopy[0];
And I've been working on it for hours and can't seem a way to make it recursive and make it work.
Any help and ideas are greatly appreciated!
Well, here is a template for turning a for-loop into a tail-recursive method:
//iterative version
public Object getIteratively(Object[] a) {
Object retVal = null;
for (int i = 0; i < a.length; a++ ) {
//do something
}
return retVal;
}
//recursive version
public Object getRecursively(Object[] a) {
doGetRecursively(a, 0, null);
}
private Object doGetRecursively(Object[] a, int i, Object retVal) {
if ( i == a.length ) {
return retVal;
}
//do something
return doGetRecursively(a, i+1, retVal);
}
Why you would ever want to do this in a non-functional language is beyond me though.
In this case //do something would be the same in both cases, e.g.:
if ( a[i].compareTo(retVal) > 0 ) {
retVal = a[i];
}
First, your method signature is incorrect. You do not need a 'low'. You should take an array/list as input and return the largest element. You may find however that you want a secondary method that requires extra arguments.
When approaching recursion and you're stuck, it's often best to identify your base case(s) first, then deal with your recursive case(s) next.
Your base case is the simplest case in which you know the answer. In this problem, you know what the largest element is right away if the size of your list is 1 - you just return the only element. You may want to think about the case where your list is empty as well.
Your recursive case then, is whenever your list has size greater than 1. In your recursive case, you want to try and 'break a piece off' and then send the rest to a recursive call. In this case, you can look at the first element in the list, and compare it to the result you get from a recursive call on the rest of the list.
This would be the right answer:
T tmp = a[low];
for(int i=0;i<=high;i++){
if(a[i].compareTo(tmp)>0){
tmp = a[i];
getLargest(a,i,high);
}
}
return tmp;
Okay before this gets out of hand, here's a simple iterative and the equivalent recursive solution to this - implemented with ints though so you have to change it a bit ;)
public static int getLargest(int[] vals) {
int max = vals[0];
for (int i = 1; i < vals.length; i++) {
max = Math.max(max, vals[i]);
}
return max;
}
public static int getLargestRec(int[] vals) {
return getLargestRec(vals, 0, vals.length);
}
private static int getLargestRec(int[] vals, int low, int high) {
if (low + 1 == high) {
return vals[low];
}
int middle = low + (high - low) / 2;
int left = getLargestRec(vals, low, middle);
int right = getLargestRec(vals, middle, high);
return Math.max(left, right);
}
public static void main(String[] args) {
int[] vals = {5, 23, 32, -5, 4, 6};
System.out.println(getLargestRec(vals));
System.out.println(getLargest(vals));
}
Note that as usual for recursive problems the lower bound is inclusive and the higher bound is exclusive. Also we could implement this differently as well, but the usual divide & conquer approach is rather useful and lends itself nicely to parallelization with a fork framework, so that's fine. (And yes for an empty array both versions will fail)