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;
}
Related
I am trying to add a random number to a set. If it is already in the set, then the loop should continue and try again.
Which is better practice,
do {
nextChosenInt = rand.nextInt(48) + 1;
addFailed = !chosenInts.add(nextChosenInt);
}
while (addFailed);
or
do {
nextChosenInt = rand.nextInt(48) + 1;
}
while (!chosenInts.add(nextChosenInt));
I would vote for the second option here, with a slight change.
You are utilising the fact that the .add method returns a boolean which is a beneficial, in that you don't need to have a redundant flag variable which, from the context of your code, serves no other purpose than just terminating the loop.
Personally however I would choose to extract:
!chosenInts.add(nextChosenInt)
into its own method which is more descriptive and readable. By doing this, you are allowing anyone to understand this condition with no prior knowledge of the Collections API.
According to Java documentation, "if this set already contains the element, the call leaves the set unchanged and returns false." Regardless of when you add to the set, if the set already contains the element, it would remain unchanged. Thus, it doesn't really make a difference in performance.
Neither since sets can't hold duplicates. Just do it until set has a desired number of elements. But you could also do something like this.
Random rand = new Random();
int next = 50;
for (int i = 0; i < 10; i++) {
next = rand.nextInt(48) + next+1;
set.add(next);
}
They are all unique the first time (but not entirely random). To add a number that is not in the set, the following is possible.
int size = set.size();
while (set.size() == size) {
set.add(rand.nextInt(48));
}
I am passing some parameters in the URL and then I add them in a list. My list has a limit of 5 elements. So if someone adds 6th element in the URL the list would simply ignore it. So I am trying to use a counter but the logic is not working as desired. I am using While loop to achieve this. So if list size is smaller than 5 set the agencyCds otherwise just return the list.
private List<IUiIntegrationDto> generateViewIntegrationReportData(ESignatureIntegrationConfig eSignConfig) throws Exception {
int counter = 1;
if(eSignConfig.getAdditionalAgencyCds() != null ) {
List<String> combinedAgencyCds = new ArrayList<String>();
for(String agencyCd : eSignConfig.getAgencyCd()) {
combinedAgencyCds.add(agencyCd);
}
StringTokenizer token = new StringTokenizer(eSignConfig.getAdditionalAgencyCds().toString(), StringConstants.COMMA);
while(token.hasMoreTokens()) {
combinedAgencyCds.add(token.nextToken());
}
while(combinedAgencyCds.size() < 5) {
counter = counter + 1;
eSignConfig.setAgencyCd(combinedAgencyCds);
}
// eSignConfig.setAgencyCd(combinedAgencyCds);
}
List<IUiIntegrationDto> intgList = getUiIntegrationManager().retrieveUiIntegrationReportData(eSignConfig.getAgencyCd(), eSignConfig.getCreatedDays(),
eSignConfig.getLob(), eSignConfig.getTransactionStatus(), eSignConfig.getAccounts(), eSignConfig.getSortKey(), eSignConfig.getSortOrder());
return intgList;
}
I am not completely sure about this logic if it is correct or if there is nay better approach.
Thanks
Try this instead of the last while in your code:
if(combinedAgencyCds.size() <= 5) {
eSignConfig.setAgencyCd(combinedAgencyCds);
} else {
eSignConfig.setAgencyCd(combinedAgencyCds.subList(0, 5));
}
The full combined list will then be used if it is less than 5 in size. Otherwise, only the first 5 elements are used.
Edit: Or even better:
eSignConfig.setAgencyCd(combinedAgencyCds.subList(0, Math.min(5, combinedAgencyCds.size())));
Ok so let's break down what your code is currently doing.
int counter = 1;
while(combinedAgencyCds.size() < 5) {
counter = counter + 1;
eSignConfig.setAgencyCd(combinedAgencyCds);
}
This snippet of code has a couple things wrong best I can tell. First, this loop has the possibility of running forever or not at all. Because combinedAgencyCds is never being manipulated, the size won't ever change and the logic being checked in the while loop never does anything. Second, there's a more efficient loop for doing this, assuming you don't need the counter variable outside of its usage in the while loop and that is using for loops.
Example syntax is as follows:
for (int i = 0; i < combinedAgencyCds.size(); i++) {
if (i < 5) {
// Do your logic here.
}
else {
break; // Or handle extra values however you want.
}
}
Notice there is no need for the explicit declaration for a counter variable as "i" counts for you.
Now in your actual logic in the loop, I'm not sure what the setAgencyCd method does, but if it simply sets a list variable in the eSignConfig like it appears to, repeating it over and over isn't going to do anything. From what I can see in your code, you are setting a variable with the same value 5 times. If you need any more explanation just let me know and I will be happy to revise the answer.
I have to do an Array List for an insertion sort and my teacher sent this back to me and gave me an F, but says I can make it up before Friday.
I do not understand why this isn't an A.L insertion sort.
Can someone help me fix this so it hits his criteria?
Thanks.
HE SAID:
After checking your first insertion sort you all did it incorrectly. I specifically said to shift the numbers and move the number into its proper place and NOT SWAP THE NUMBER INTO PLACE. In the assignment in MySA I said if you do this you will get a 0 for the assignment.
import java.util.ArrayList;
public class AListINSSORT {
private static void insertionSort(ArrayList<Integer> arr) {
insertionSort();
}
private static void insertionSort() {
ArrayList<Integer> swap = new ArrayList<Integer>();
swap.add(1);
swap.add(2);
swap.add(3);
swap.add(4);
swap.add(5);
int prior = 0;
int latter = 0;
for (int i = 2; i <= latter; i++)
{
for (int k = i; k > prior && (swap.get(k - 1) < swap.get(k - 2)); k--)
{
Integer temp = swap.get(k - 2);
swap.set(k - 2, swap.get(k - 1));
swap.set(k - 1, temp);
}
}
System.out.println(swap);
}
}
First of all, it seems your teacher asked you to use a LinkedList instead of an ArrayList. There is quite a difference between them.
Secondly, and maybe more to the point. In your inner loop you are saving a temp variable and swapping the elements at position k - 2 and k - 1 with each other. From the commentary this is not what your teacher intended. Since he wants you to solve the problem with element insertion, I recommend you look at the following method definition of LinkedList.add(int i, E e): https://docs.oracle.com/javase/7/docs/api/java/util/LinkedList.html#add(int,%20E).
This should point you in the right direction.
As far as I see, your code does nothing at all.
The condition of the outer for loop
for (int i = 2; i <= latter; i++)
is not fulfilled.
As you start with i = 2 and as latter = 0, it never holds i <= latter.
Thus, you never run through the outer for loop and finally just give back the input values.
If you add the input values to swap in a different order (not already ordered), you will see that your code does not re-order them.
There's a lot of stuff wrong here.
Firstly, your method:
private static void insertionSort(ArrayList<Integer> arr) {
insertionSort();
}
takes an ArrayList and completely ignores it. This should presumably be the List which requires sorting.
Then in insertionSort() you create a new ArrayList, insert some numbers already in order, and then attempt something which looks nothing like insertion sort, but slightly more like bubble sort.
So, when you call insertionSort(List) it won't actually do anything to the list at all, all the work in insertionSort() happens to a completely different List!
Since on SO we don't generally do people's homework for them, I suggest looking at the nice little animated diagram on this page
What you should have then is something like:
public void insertionSort(LinkedList<Integer> numbers) {
//do stuff with numbers, using get() and add()
}
I have an ArrayList, which contains game objects sorted by their 'Z' (float) position from lower to higher. I'm not sure if ArrayList is the best choice for it but I have come up with such a solution to find an index of insertion in a complexity faster than linear (worst case):
GameObject go = new GameObject();
int index = 0;
int start = 0, end = displayList.size(); // displayList is the ArrayList
while(end - start > 0)
{
index = (start + end) / 2;
if(go.depthZ >= displayList.get(index).depthZ)
start = index + 1;
else if(go.depthZ < displayList.get(index).depthZ)
end = index - 1;
}
while(index > 0 && go.depthZ < displayList.get(index).depthZ)
index--;
while(index < displayList.size() && go.depthZ >= displayList.get(index).depthZ)
index++;
The catch is that the element has to be inserted in a specific place in the chain of elements with equal value of depthZ - at the end of this chain. That's why I need 2 additional while loops after the binary search which I assume aren't too expensive becouse binary search gives me some approximation of this place.
Still I'm wondering if there's some better solution or some known algorithms for such problem which I haven't heard of? Maybe using different data structure than ArrayList? At the moment I ignore the worst case insertion O(n) (inserting at the begining or middle) becouse using a normal List I wouldn't be able to find an index to insert using method above.
You should try to use balanced search tree (red-black tree for example) instead of array. First you can try to use TreeMap witch uses a red-black tree inside to see if it's satisfy your requirements. Possible implementation:
Map<Float, List<Object>> map = new TreeMap<Float, List<Object>>(){
#Override
public List<Object> get(Object key) {
List<Object> list = super.get(key);
if (list == null) {
list = new ArrayList<Object>();
put((Float) key, list);
}
return list;
}
};
Example of usage:
map.get(0.5f).add("hello");
map.get(0.5f).add("world");
map.get(0.6f).add("!");
System.out.println(map);
One way to do it would to do a halving search, where the first search is half way thru your list (list.size()/2), then for the next one you can do half of that, and so on. With this exponential method, instead of having to do 4096 searches when you have 4096 objects, you only need 12 searches
sorry for the complete disregard for technical terms, I am not the best at terms :P
Unless I overlook something, your approach is essentially correct (but there's an error, see below), in the sense that your first while tries to compute the insert-index such that it will be placed after all lower OR EQUAL Z: there's correctly an equal sign in your first test (updating "start" if it yields TRUE).
Then, of course, there's no need to worry anymore about its position among equals. However, your follow-up while destroys this nice situation: the test in the first follow-up while yields always TRUE (one time) and so you move back; and then you need the second follow-up while to undo that. So, you should remove BOTH follow-up whiles and you're done...
However, there's a little problem with your first while, such that it doesn't always exactly do what the purpose is. I guess that the faulty outcomes triggered you to implement the follow-up whiles to "repair" that.
Here's the issue in your while. Suppose you have a try-index (start+end)/2 that points to a larger Z, but the one just before it has value Z. You then get into your second test (elseif) and set "end" to the position where that Z-value resides. Finally you wind up with precisely that position.
The remedy is simple: in your elseif assignment, put "end = index" (without the -1). Final remark: the test in the elseif is unnecessary, just else is sufficient.
So, all in all you get
GameObject go = new GameObject();
int index = 0;
int start = 0, end = displayList.size(); // displayList is the ArrayList
while(end - start > 0)
{
index = (start + end) / 2;
if(go.depthZ >= displayList.get(index).depthZ)
start = index + 1;
else
end = index;
}
(I hope I haven't overlooked something trivial...)
Add 1 to the least significant byte of the key (with carry); binary search for that insert position; and insert it there.
Your binary search has to be so constructed as to end at the leftmost of a sequence of duplicates, but this is trivial given an understanding of the various Binary search algorithms.
I have to implement a Selection Sort in Java according to these parameters:
Implement a variation to the SelectionSort that locates both the smallest and largest elements while scanning the list and positions them at the beginning and the end of the list, respectively. On pass number one, elements x0,...,xn-1 are scanned; on pass number two, elements x1,...,xn-2 are scanned; and so on.
I am passing the method an array of size 32, and when I print the array it is not sorted. What's the matter with my code?
static void selectionSort() {
scramble();
int smallIndex = 0; //index of smallest to test
int largeIndex = array.length - 1; //index of largest to test
int small = 0; //smallest
int large; //largest
int smallLimit = 0; //starts from here
int largeLimit = array.length - 1; //ends here
int store; //temp stored here
int store2;
for(int i = 0; i < array.length/2; i++) { //TODO not working...
small = array[smallLimit];
large = array[largeLimit];
for(int j = smallLimit; j <= largeLimit; j++) {
if(array[j] < small) {
smallIndex = j;
small = array[j];
}
else if(array[j] > large) {
largeIndex = j;
large = array[j];
}
}
store = array[smallLimit];
store2 = array[smallIndex];
array[smallLimit] = store2;
array[smallIndex] = store;
store = array[largeLimit];
array[largeLimit] = array[largeIndex];
array[largeIndex] = store;
smallLimit++;
largeLimit--;
}
print();
}
Think about the extreme cases: what happens when the largest or smallest item is found at smallLimit or largeLimit. When that happens you have two problems:
largeIndex and smallIndex are not set. They maintain their values from a previous iteration.
Swapping the smallest item to its correct place moves the largest item. The second swap moves the smallest item where the largest should go, and the largest ends up in a random location. Step through the code on paper or in a debugger if you find this hard to visualize.
These problems are easy to fix. You could have avoided the problem following a few guidelines:
Use fewer moving parts. You can always get the value of small using smallIndex, if you just used smallIndex there would be no danger of different variables falling out of step.
Declare the variables in the smallest possible scope. If smallIndex was declared in the loop body and not outside the compiler would have told you there's a chance it was not set before the swap.
Destructive updates like the first swap here can always make a previous calculation obsolete. When you can't avoid this from happening, look for ways two updates can step on each other's toes.
Like #Joni, clearly pointed out, there is big caveat with swapping two elements twice during a traversal of the array. Since you have to implement the sorting algorithm in-place, you need to take into account the positions of the elements to be swapped as it happens in succession.
Another limiting case that you need to see is when there are just three elements left i.e. the last iteration of the for loop. This is how I would go about it:
store = array[smallLimit];
store2 = array[smallIndex];
array[smallLimit] = small;
array[smallIndex] = store;
smallLimit++;
//No problem with swapping the first two elements
store = array[largeLimit];
//However the first swap might cause the other elements to shift
//So we do this check
if((array[largeIndex] == large))
{array[largeLimit] = array[largeIndex];
array[largeIndex] = store;
largeLimit--;
}
//Just a limiting case, where amongst the last three elements, first swap happens.
//The smallest element is in place, just take care of the other two elements.
if(largeLimit - smallLimit == 1){
if(array[largeLimit] != large){
array[smallLimit] = array[largeLimit];
array[largeLimit] = large;
largeLimit--;
}
}
Working DEMO for the snippet mentioned above, building upon your code. Hope it gets you started in the right direction.