sorry for limited code, as i have quite no idea how to do it, and parts of the code are not a code, just an explanation what i need. The base is:
arrayList<double> resultTopTen = new arrayList<double();
arrayList<double> conditions = new arrayList<double(); // this arrayList can be of a very large size milion+, gets filled by different code
double result = 0;
for (int i = 0, i < conditions.size(), i++){ //multithread this
loopResult = conditions.get(i) + 5;
if (result.size() < 10){
resultTopTen.add(loopResult);
}
else{
//this part i don't know, if this loopResult belongs to the TOP 10 loopResults so far, just by size, replace the smallest one with current, so that i will get updated resultTopTen in this point of loop.
}
}
loopResult = conditions.get(i) + 5; part is just an example, calculation is different, in fact it is not even double, so it is not possible simply to sort conditions and go from there.
for (int i = 0, i < conditions.size(), i++) part means i have to iterate through input condition list, and execute the calculation and get result for every condition in conditionlist, Don't have to be in order at all.
The multithreading part is the thing i have really no idea how to do, but as the conditions arrayList is really large, i would like to calculate it somehow in parallel, as if i do it just as it is in the code in a simple loop in 1 thread, i wont get my computing resources utilized fully. The trick here is how to split the conditions, and then collect result. For simplicity if i would like to do it in 2 threads, i would split conditions in half, make 1 thread do the same loop for 1st half and second for second, i would get 2 resultTopTen, which i can put together afterwards, But much better would be to split the thing in to as many threads as system resources provide(for example until cpu ut <90%, ram <90%). Is that possible?
Use parallel stream of Java 8.
static class TopN<T> {
final TreeSet<T> max;
final int size;
TopN(int size, Comparator<T> comparator) {
this.max = new TreeSet<>(comparator);
this.size = size;
}
void add(T n) {
max.add(n);
if (max.size() > size)
max.remove(max.last());
}
void combine(TopN<T> o) {
for (T e : o.max)
add(e);
}
}
public static void main(String[] args) {
List<Double> conditions = new ArrayList<>();
// add elements to conditions
TopN<Double> maxN = conditions.parallelStream()
.map(d -> d + 5) // some calculation
.collect(() -> new TopN<Double>(10, (a, b) -> Double.compare(a, b)),
TopN::add, TopN::combine);
System.out.println(maxN.max);
}
Class TopN holds top n items of T.
This code prints minimum top 10 in conditions (add 5 to each element).
Let me simplify your question, from what I understand, please confirm or add:
Requirement: You want to find top10 results from list called conditions.
Procedure: You want multiple threads to process your logic of finding the top10 results and accumulate the results to give top10.
Please also share the logic you want to implement to get top10 elements or it is just a descending order of list and it's top 10 elements.
Related
If have a workflow that removes elements of a List by a certain criteria. However certain items are skipped? Why is this happening?
List<Integer> listWithAge = new ArrayList<>();
int randomNumber = 100;
for (int i = 0; i < randomNumber; i++) {
listWithAge.add(i);
}
// this is my loop
for (int i = 0; i < listWithAge.size(); i++) {
System.out.println(i);
if ((listWithAge.get(i) % 3) == 2) listWithAge.remove(i);
}
Above code is my loop. I replaced my condition with something simpler. If I run this code my second loop only runs for 67 turns instead of 100.
It is problematic to iterate over a list and remove elements while iterating over it.
If you think about how the computer has to reconcile it, it makes sense...
Here's a thought experiment for you to go through.
If you have a list that is size 10 and you want to remove elements 1, 5, and 9 then you would think maybe the following would work:
List<String> listOfThings = ...some list with 10 things in it...;
list.remove(0);
list.remove(4);
list.remove(8);
However, after the first remove command, the list is only size 9.. Then after the second command, it's size has become 8. At this point, it hardly even makes sense to do list.remove(8) anymore because you're looking at an 8-element list and the largest index is 7.
You can also see now that the 2nd command didn't even remove the element now that you wanted.
If you want to keep this style of "remove as I go" syntax, the more appropriate way is to use Iterators. Here's an SO that talks about it and shows you the syntax you would need (see the question). It's easy to read up on elsewhere too.
How Iterator's remove method actually remove an object
Skipping a value would be the result of your list getting out of sync with your loop index because the list is reduced in size. This causes you to hop over some locations since the reduction in size affects future locations that have not been reached.
So the first thing you could do is simply correct the synchronization by decrementing i when you remove a value from the list. This will keep index at the same spot as the list shifts "left" caused by the removal.
for (int i = 0; i < listWithAge.size(); i++) {
if ((listWithAge.get(i) % 3) == 2) listWithAge.remove(i--);
}
The other option is to loop thru the list backwards.
for (int i = listWithAge.size()-1; i >= 0; i--) {
if ((listWithAge.get(i) % 3) == 2) {
listWithAge.remove(i);
}
}
This way, no values should be skipped since the removing of the element does affect the loop index's future positions relative to the changing size of the list.
But the best way would be to use an iterator as has already been mentioned by
Atmas
As a side note, I recommend you always use blocks {} even for single statements as I did above in the if block. It will save you some serious debugging time in the future when you decide you need to add additional statements and then wonder why things are no longer working.
And deleting like this from a list is very expensive, especially for large lists. I would suggest that if you don't have duplicate values, you use a Set. Otherwise, instead of deleting matching values, add the non-matching to a second list.
List<Integer> listWithAge = new ArrayList<>();
int randomNumber = 100;
for (int i = 0; i < randomNumber; i++) {
listWithAge.add(i);
}
// this is my loop
List<Integer> itemsToBeDeleted = new ArrayList<>();
for (int i = 0; i < listWithAge.size(); i++) {
System.out.println(i);
if ((listWithAge.get(i) % 3) == 2) {
itemsToBeDeleted.add(i);
}
//delete all outside the loop
//deleting inside the loop messes the indexing of the array
listWithAge.removeAll(itemsToBeDeleted);
I want to make a program that each semester, creates a schedule depending of a list of classes I input, so if I add various classes, depending of the time and days that these classes will happen then the program would be able to match what classes can be added without intercepting each other. I want to know if there is an algorithms or way of comparing the class schedule in order to determine what classes can I have from the list of courses.
The only way of doing these that I can imagine is with many if statements or adding an starting course and then having an array that tracks the hours, starting each position at 0 and then each time an hour is occupied I change the array position to 1. Then when adding a course I check what positions are different to 1 and try to add the class.
I want to find a more optimal solution to this problem than the ones I can imagine.
This is a planning problem. Planning problem are very difficult to solve in a efficient way: performance issues arise very quickly in this kind of problem.
If you just want to solve the problem, you should check an existing planning problem solver like OptaPlanner: it's open source, so you can try to understand how it work and there is a blog with interresting thoughts about planning problem.
If you're wanting a method where, given a group of classes with their times and a chosen class, output the available classes, you can use simple iteration to do the job. Suppose you have a 5 hour slot with 4 classes you could represent each class with a single array:
int[][] times = {
{0,1,1,0,0},
{1,1,0,0,0},
{0,1,0,0,0},
{0,0,0,1,1}
};
So if you were to choose the class taking up the last 2hrs then the remaining options would be:
{0,1,1,0,0},
{1,1,0,0,0},
{0,1,0,0,0}
Given this representation you could do something like:
import java.util.*;
public class C {
static int[][] available(int[] c,int[][] times){
ArrayList<Integer> index = new ArrayList<>();
ArrayList<Integer> result = new ArrayList<>();
for(int i=0;i<c.length;i++)
if(c[i]==1) index.add(i);
for(int i = 0; i < times.length; i++){
if(!times[i].equals(c)) {
for (int j = 0; j < times[0].length; j++) {
if(times[i][j]==1){
if(index.contains(j)) break;
}
if(j==times[0].length-1) result.add(i);
}
}
}
int[][] r = new int[result.size()][c.length];
for(int i=0;i<result.size();i++){
r[i] = times[result.get(i)];
}
return r;
}
public static void main(String[] args) {
int[][] times = {
{0,1,1,0,0},
{1,1,0,0,0},
{0,1,0,0,0},
{0,0,0,1,1}
};
int[] c = {0,0,0,1,1};
available(c,times);
System.out.println(Arrays.deepToString(available(c,times)));
}
}
This is a chunk of code in Java, I'm trying to output random numbers from the tasks array, and to make sure none of the outputs are repeated, I put them through some other loops (say you have the sixth, randomly-chosen task "task[5]"; it goes through the for loop that will check it against every "tCheck" element, and while task[5] equals one of the tCheck elements, it will keep trying to find another option before going back to the start of the checking forloop... The tCheck[i] elements are changed at the end of each overall loop of output to the new random number settled on for the task element).
THE PROBLEM is that, despite supposedly checking each new random task against all tCheck elements, sometimes (not always) there are repeated tasks output (meaning, instead of putting out say 2,3,6,1,8,7,5,4, it will output something like 2,3,2,1,8,7,5,4, where "2" is repeated... NOT always in the same place, meaning it can sometimes end up like this, too, where "4" is repeated: 3,1,4,5,4,6,7,8)
int num = console.nextInt();
String[] tasks = {"1","2","3","4","5","6","7","8"};
String[] tCheck = {"","","","","","","",""};
for(int i = 0; i<= (num-1); i++){
int tNum = rand.nextInt(8);
for(int j = 0; j <=7; j++){
if(tasks[tNum].equals(tCheck[j])){
while(tasks[tNum].equals(tCheck[j])){
tNum = rand.nextInt(8);
}
j = 0;
}
}
tCheck[i] = tasks[tNum];
System.out.println(tasks[tNum]+" & "+tCheck[i]);
}
None of the other chunks of code affect this part (other than setting up Random int's, Scanners, so on; those are all done correctly). I just want it to print out each number randomly and only once. to never have any repeats. How do I make it do that?
Thanks in advance.
Firstly, don't use arrays. Use collections - they are way more programmer friendly.
Secondly, use the JDK's API to implement this idea:
randomise the order of your elements
then iterate over them linearly
In code:
List<String> tasks = Arrays.asList("1","2","3","4","5","6","7","8");
Collections.shuffle(tasks);
tasks.forEach(System.out::println);
Job done.
you can check if a certain value is inside your array with this approach.
for(int i = 0; i<= (num-1); i++){
int tNum = rand.nextInt(8);
boolean exist = Arrays.asList(tasks).contains(tNum);
while(!exist){
//your code
int tNum = rand.nextInt(8);
exist = Arrays.asList(tasks).contains(tNum);
}
}
if you are using an arraylist then you can check it with contains method since you are using an array we have to get the list from the array using asList() and then use the contains method. with the help of the while loop it will keep generating random numbers untill it generates a non duplicate value.
I used to created something similar using an ArrayList
public class Main {
public static void main(String[] args) {
String[] array = { "a", "b", "c", "d", "e" };
List<String> l = new ArrayList<String>(Arrays.asList(array));
Random r = new Random();
while(!l.isEmpty()){
String s = l.remove(r.nextInt(l.size()));
System.out.println(s);
}
}
}
I remove a random position in the list until it's empty. I don't use any check of content. I believe that is kind of effective (Even if I create a list)
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()
}
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;
}