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Distinguishing local variables with same name in a loop

I have found the following java code:
test:
for (int i = 0; i <= max; i++) {
int n = substring.length();
int j = i;
int k = 0;
while (n-- != 0) {
if (searchMe.charAt(j++) != substring.charAt(k++)) {
continue test;
}
}
foundIt = true;
break test;
}
System.out.println(foundIt ? "Found it" : "Didn't find it");
}
Inside the loop, the above code is creating 'n', 'j' and 'k' several times. How the program distinguishes between these variables of the same name?
I mean where they are stored in the memory to distinguish them?

With a bit of simplification:
Inside a { ... } block, int k = 0; creates a variable, and that variable exists up to the moment where you reach the end of the block, and there the variable gets destroyed. So, at any time during the program run, there's at most one n, j, or k in existence.
A bit closer to reality:
The compiler scans the whole method, finds the list of variables that might exist in parallel (i, n, j, k, and foundIt), and allocates enough places on the stack for these variables (5 places in your example). These stack places exist from the moment you enter your method until you return from it, but they are not used all the time, e.g. the k place only contains useful values from the time you execute int k = 0; to the end of the current loop iteration.

Java's local variables have a protection known as definite assignment this means that you can't read a value from them before you've assigned it a value.
They are also defined within a scope: you can only access the variable within a certain chunk of the program.
With the two of these things together, you don't need a separate variable for each iteration of the loop: you are guaranteed to assign a local variable a value before using it, so you are guaranteed to overwrite any value which was stored in it before, if any.
Variables are really just a helpful concept in the source code. Once compiled, the byte code doesn't have variable names: the compiler has simply determined that it can temporarily use a particular part of memory to store a value for a limited time. It will reuse this memory many times, but in ways that it guarantees do not overlap between usages.

Trouble with recursive method java

basically I have a brute force password guesser(I realize it's not very efficient) I have a process I want to make into a recursive method that i can pass a length integer and it will run with that amount of characters.
here is the code:
public static void generatePassword(int length)
{
// should be a recusive function learn how to do that
// 1 interval
for(int i =32;i<127;i++)// ascii table length
{
System.out.println((char)i);
}
// 2 interval
for(int z =32;z<127;z++)// ascii table length
{
for(int q =32;q<127;q++)// ascii table length
{
System.out.println((char)z+""+(char)q);
}
}
// 3 interval
for(int w =32;w<127;w++)// ascii table length
{
for(int o =32;o<127;o++)// ascii table length
{
for(int g =32;g<127;g++)// ascii table length
{
System.out.println((char)w+""+(char)o+""+(char)g);
}
}
}
}
the intervals return a string with that length example: 3rd interval will return every possible string combination with a length of 3. if anyone can help me automate this process and explain(i would like to learn rather then copy and paste) that would be great ! :)

A recursive method is a method that calls itself, it has a base-condition (also called stop condition) which prevents it from going into an infinite loop.
Lets take the first interval as an example:
for(int i = 32; i < 127; i++) { // ascii table length
System.out.println((char)i);
}
we can create a recursive method that'll do the same thing:
private void interval1(int i) {
if (i < 32 || i >= 127) return;
System.out.println((char)i);
interval1(i + 1);
}
in order to use it for our use-case, we should call this method with i=32: interval(32)
Hope this helps!

The function
Note that this will be EXTREMELY INEFFICIENT. This shouldn't ever be done in practice, since the number of String objects created is MIND-BOGGLINGLY HUGE (see bottom of answer)
public void recursivePrint(String prefix, int levels) {
if (levels <= 1) {
for (int i = 33; i < 127; ++i) {
System.out.println(prefix+(char)i);
}
} else {
for (int i = 33; i < 127; ++i) {
recursivePrint(prefix+(char)i, levels-1);
}
}
}
Then you call it with:
recursivePrint("", 5); // for printing all possible combinations of strings of length 5
The way it works
Each call to a function has it's own memory, and is stored seperately. When you first call the function, there is a String called prefix with a value of "", and an int called 'levels' which has a value of 5. Then, that function calls recursivePrint() with new values, so new memory is allocated, and the first call will wait until this new call has finished.
This new call has a String called prefix with a value of (char)34+"", and a levels with a value of 4. Note that these are completely separate instances of these variables to the first function call because remember: each function call has it's own memory (the first call is waiting for this one to finish). Now this second call makes another call to the recursivePrint() function, making more memory, and waiting until this new call finishes.
When we get to levels == 1, there is a prefix built up from previous calls, and all that remains is to use that prefix and print all the different combinations of that prefix with the last character changing.
Recursive methods are highly inefficient in general, and in this case especially.
Why you should never use it
This method is not just inefficient, though; it's infeasible for anything useful. Let's do a calculation: how many different possibilities are there for a string with 5 characters? Well there's 127-33=94 different characters you want to choose, then that means that you have 94 choices for each character. Then the total number of possibilities is 94^5 = 7.34*10^9 [that's not including the 5+ bytes to store each one] (to put that in perspective 4GB of RAM is around 4*10^9 bytes)

Here is your method implemented using recursion:
public static void generatePassword(int length, String s) {
if (length == 0) {
System.out.println(s);
return;
}
for (int i = 32; i < 127; i++) {
String tmp = s+((char) i);
generatePassword(length - 1, tmp);
}
}
All you have to do is to pass length and initial String (ie "") to it.
At if statement there is checked, if recursion should be stopped (when length of generated password is equals to expected).
At for-loop there is new character added to actual String and the method is invoked with shorter length and a new String as an argument.
Hope it helps.

Recursive binary search -- Java

I've searched extensively online but all solutions I've found use two parameters to keep track of the size of the area being used. This would be easy if I was allowed to do that, but I'm not. As you can see below, the code lacks a stop value, because I have no idea how to retain the original information.
This is the code on Wikipedia, you can see they use imin and imax for tracker variables: http://en.wikipedia.org/wiki/Binary_search_algorithm#Recursive
My (very incorrect) code is below. The mid variable doesn't mean anything, because I don't know how to set low and high correctly if I'm not allowed to have any extra arguments in the function.
public static int findRecursiveB( String s, char c)
{
int low = 0;
int high = s.length()-1;
int mid = (low+high)/2;
if (s.charAt(mid) < c) {
return findRecursiveB(s.substring(low, mid), c);
}
else if (s.charAt(mid) >= c) {
return findRecursiveB(s.substring(mid+1, high), c);
}
else return mid;
}

One crucial point here is what does the original String s contain? For this to work, it has to be a sorted String, meaning that the characters in the String must be in order. Specifically, it looks like you've written code that expects s to be sorted in reverse order. Otherwise, unless I'm totally missing something, your code does exactly what it is supposed to: no need to pass extra params because you are passing in the substring on each recursive call.
Otherwise, good job.

Problem with recursive backtracking

Hey guys, recently posted up about a problem with my algorithm.
Finding the numbers from a set which give the minimum amount of waste
Ive amended the code slightly, so it now backtracks to an extent, however the output is still flawed. Ive debugged this considerablychecking all the variable values and cant seem to find out the issue.
Again advice as opposed to an outright solution would be of great help. I think there is only a couple of problems with my code, but i cant work out where.
//from previous post:
Basically a set is passed to this method below, and a length of a bar is also passed in. The solution should output the numbers from the set which give the minimum amount of waste if certain numbers from the set were removed from the bar length. So, bar length 10, set includes 6,1,4, so the solution is 6 and 4, and the wastage is 0. Im having some trouble with the conditions to backtrack though the set. Ive also tried to use a wastage "global" variable to help with the backtracking aspect but to no avail.
SetInt is a manually made set implementation, which can add, remove, check if the set is empty and return the minimum value from the set.
/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
package recursivebacktracking;
/**
*
* #author User
*/
public class RecBack {
int WASTAGE = 10;
int BESTWASTAGE;
int BARLENGTH = 10;
public void work()
{
int[] nums = {6,1,2,5};
//Order Numbers
SetInt ORDERS = new SetInt(nums.length);
SetInt BESTSET = new SetInt(nums.length);
SetInt SOLUTION = new SetInt(nums.length);
//Set Declarration
for (int item : nums)ORDERS.add(item);
//Populate Set
SetInt result = tryCutting(ORDERS, SOLUTION, BARLENGTH, WASTAGE);
result.printNumbers();
}
public SetInt tryCutting(SetInt possibleOrders, SetInt solution, int lengthleft, int waste)
{
for (int i = 0; i < possibleOrders.numberInSet(); i++) // the repeat
{
int a = possibleOrders.min(); //select next candidate
System.out.println(a);
if (a <= lengthleft) //if accecptable
{
solution.add(a); //record candidate
lengthleft -= a;
WASTAGE = lengthleft;
possibleOrders.remove(a); //remove from original set
if (!possibleOrders.isEmpty()) //solution not complete
{
System.out.println("this time");
tryCutting(possibleOrders, solution, lengthleft, waste);//try recursive call
BESTWASTAGE = WASTAGE;
if ( BESTWASTAGE <= WASTAGE )//if not successfull
{
lengthleft += a;
solution.remove(a);
System.out.println("never happens");
}
} //solution not complete
}
} //for loop
return solution;
}
}

Instead of using backtracking, have you considered using a bitmask algorithm instead? I think it would make your algorithm much simpler.
Here's an outline of how you would do this:
Let N be number of elements in your set. So if the set is {6,1,2,5} then N would be 4. Let max_waste be the maximum waste we can eliminate (10 in your example).
int best = 0; // the best result so far
for (int mask = 1; mask <= (1<<N)-1; ++mask) {
// loop over each bit in the mask to see if it's set and add to the sum
int sm = 0;
for (int j = 0; j < N; ++j) {
if ( ((1<<j)&mask) != 0) {
// the bit is set, add this amount to the total
sm += your_set[j];
// possible optimization: if sm is greater than max waste, then break
// out of loop since there's no need to continue
}
}
// if sm <= max_waste, then see if this result produces a better one
// that our current best, and store accordingly
if (sm <= max_waste) {
best = max(max_waste - sm);
}
}
This algorithm is very similar to backtracking and has similar complexity, it just doesn't use recursion.
The bitmask basically is a binary representation where 1 indicates that we use the item in the set, and 0 means we don't. Since we are looping from 1 to (1<<N)-1, we are considering all possible subsets of the given items.
Note that running time of this algorithm increases very quickly as N gets larger, but with N <= around 20 it should be ok. The same limitation applies with backtracking, by the way. If you need faster performance, you'd need to consider another technique like dynamic programming.
For the backtracking, you just need to keep track of which element in the set you are on, and you either try to use the element or not use it. If you use it, you add it to your total, and if not, you proceeed to the next recursive call without increasing your total. Then, you decrement the total (if you incremented it), which is where the backtracking comes in.
It's very similar to the bitmask approach above, and I provided the bitmask solution to help give you a better understanding of how the backtracking algorithm would work.
EDIT
OK, I didn't realize you were required to use recursion.
Hint1
First, I think you can simplify your code considerably by just using a single recursive function and putting the logic in that function. There's no need to build all the sets ahead of time then process them (I'm not totally sure that's what you're doing but it seems that way from your code). You can just build the sets and then keep track of where you are in the set. When you get to the end of the set, see if your result is better.
Hint2
If you still need more hints, try to think of what your backtracking function should be doing. What are the terminating conditions? When we reach the terminating condition, what do we need to record (e.g. did we get a new best result, etc.)?
Hint3
Spoiler Alert
Below is a C++ implementation to give you some ideas, so stop reading here if you want to work on it some more by yourself.
int bestDiff = 999999999;
int N;
vector< int > cur_items;
int cur_tot = 0;
int items[] = {6,1,2,5};
vector< int > best_items;
int max_waste;
void go(int at) {
if (cur_tot > max_waste)
// we've exceeded max_waste, so no need to continue
return;
if (at == N) {
// we're at the end of the input, see if we got a better result and
// if so, record it
if (max_waste - cur_tot < bestDiff) {
bestDiff = max_waste - cur_tot;
best_items = cur_items;
}
return;
}
// use this item
cur_items.push_back(items[at]);
cur_tot += items[at];
go(at+1);
// here's the backtracking part
cur_tot -= items[at];
cur_items.pop_back();
// don't use this item
go(at+1);
}
int main() {
// 4 items in the set, so N is 4
N=4;
// maximum waste we can eliminiate is 10
max_waste = 10;
// call the backtracking algo
go(0);
// output the results
cout<<"bestDiff = "<<bestDiff<<endl;
cout<<"The items are:"<<endl;
for (int i = 0; i < best_items.size(); ++i) {
cout<<best_items[i]<<" ";
}
return 0;
}

How to detect an infinite loop in a recursive call?

I have a function that is recursively calling itself, and i want to detect and terminate if goes into an infinite loop, i.e - getting called for the same problem again. What is the easiest way to do that?
EDIT: This is the function, and it will get called recursively with different values of x and y. i want to terminate if in a recursive call, the value of the pair (x,y) is repeated.
int fromPos(int [] arr, int x, int y)

One way is to pass a depth variable from one call to the next, incrementing it each time your function calls itself. Check that depth doesn't grow larger than some particular threshold. Example:
int fromPos(int [] arr, int x, int y)
{
return fromPos(arr, x, y, 0);
}
int fromPos(int [] arr, int x, int y, int depth)
{
assert(depth < 10000);
// Do stuff
if (condition)
return fromPos(arr, x+1, y+1, depth + 1);
else
return 0;
}

If the function is purely functional, i.e. it has no state or side effects, then you could keep a Set of the arguments (edit: seeing your edit, you would keep a Set of pairs of (x,y) ) that it has been called with, and every time just check if the current argument is in the set. That way, you can detect a cycle if you run into it pretty quickly. But if the argument space is big and it takes a long time to get to a repeat, you may run out of your memory before you detect a cycle. In general, of course, you can't do it because this is the halting problem.

You will need to find a work-around, because as you've asked it, there is no general solution. See the Halting problem for more info.

An easy way would be to implement one of the following:
Pass the previous value and the new value to the recursive call and make your first step a check to see if they're the same - this is possibly your recursive case.
Pass a variable to indicate the number of times the function has been called, and arbitrarily limit the number of times it can be called.

You can only detect the most trivial ones using program analysis. The best you can do is to add guards in your particular circumstance and pass a depth level context. It is nearly impossible to detect the general case and differentiate legitimate use of recursive algorithms.

You can either use overloading for a consistent signature (this is the better method), or you can use a static variable:
int someFunc(int foo)
{
static recursionDepth = 0;
recursionDepth++;
if (recursionDepth > 10000)
{
recurisonDepth = 0;
return -1;
}
if (foo < 1000)
someFunc(foo + 3);
recursionDepth = 0;
return foo;
}
John Kugelman's answer with overloading is better beacuse it's thread safe, while static variables are not.
Billy3

Looks like you might be working on a 2D array. If you've got an extra bit to spare in the values of the array, you can use it as a flag. Check it, and terminate the recursion if the flag has been set. Then set it before continuing on.
If you don't have a bit to spare in the values, you can always make it an array of objects instead.

If you want to keep your method signature, you could keep a couple of sets to record old values of x and y.
static Set<Integer> xs;
static Set<Integer> ys;//Initialize this!
static int n=0;//keeps the count function calls.
int fromPos(int [] arr, int x, int y){
int newX= getX(x);
int newY= getY(y);
n++;
if ((!xs.add(Integer.valueOf(newX)) && !ys.add(Integer.valueOf(newY))){
assert(n<threshold); //threshold defined elsewhere.
fromPos(arr,newx,newy);
}
}

IMHO Only loops can go into an infinite loop.
If your method has too many level of recursion the JVM will throw a StackOverflowError. You can trap this error with a try/catch block and do whatever you plan to do when this condition occurs.

A recursive function terminates in case a condition is fulfilled.
Examples:
The result of a function is 0 or is 1
The maximum number of calls is reached
The result is lower/greater than the input value
In your case the condition is ([x0,y0] == [xN,yN]) OR ([x1,y1] == [xN,yN]) OR ([xN-1,yN-1] == [xN,yN])
0, 1, ...N are the indexes of the pairs
Thus you need a container(vector, list, map) to store all previous pairs and compare them to the current pair.

First use mvn findbugs:gui to open a gui which point to the line where this error is present.
I also faced the same problem and I solved it by adding a boolean variable in the loop verification.
Code before ->
for (local = 0; local < heightOfDiv; local = local + 200) { // Line under Error
tileInfo = appender.append(tileInfo).append(local).toString();
while (true) {
try {
tileInfo = appender.append(tileInfo).append(getTheTextOfTheElement(getTheXpathOfTile(incr))).toString();
incr++;
} catch (Exception e) {
incr = 1;
tileInfo = appender.append(tileInfo).append("/n").toString();
}
}
To Solve this problem, I just added a boolean variable and set it to false in the catch block. Check it down
for (local = 0; local < heightOfDiv; local = local + 200) {
tileInfo = appender.append(tileInfo).append(local).toString();
boolean terminationStatus = true;
while (terminationStatus) {
try {
tileInfo = appender.append(tileInfo).append(getTheTextOfTheElement(getTheXpathOfTile(incr))).toString();
incr++;
} catch (Exception e) {
incr = 1;
tileInfo = appender.append(tileInfo).append("/n").toString();
terminationStatus = false;
}
}
This is how i Solved this problem.
Hope this will help. :)