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Optimal way to solve the below problem based on Data Structure

I was recently asked this in an interview.
Given below are the the candidates and the time at which they got a vote.
Q. Given a time, print the person winning till that time.
Cand. Time
A 4
B 10
C 15
C 18
C 21
B 35
B 40
B 42
E.g In the Qsn above, if we are asked to find the winner at time 20, answer would be C -> Since C has 2 votes.
Tried Solution
Have a Map<String, List> to store Map<Candidate, [Time, votes]>
We can iterate through the array & fetch only the times which are less than 20 (as per the question).
But I believe there will be a more optimum way to solve this type of problem.
Essentially store the given data in a proper Data Structure which will give us the result in optimum time.
Thanks

What I would do is as follows:
First, implement a naive solution, such as the one you thought of, or the one suggested by Pp88 above.
Then, immediately write a test for it, which shows(1) that it works.
Then, implement a more optimal solution, such as the one which follows.
Finally, re-use the previous test to show(1) that the more optimal solution also works.
A more optimal solution could be as follows:
Build a LinkedHashMap where the key is a time coordinate, and the value is one more map, in which the keys are the names of all candidates, and the values are the accumulated vote count of each candidate at that time coordinate.
Traverse this LinkedHashMap and create a new map, where the key is again a time coordinate, and the value is the name of the winning candidate at that time coordinate. Throw away the previous map.
Build an ArrayList containing all the keys in the map.
Once you have done all of the above, any query of the type "who is the winner at time X" can be answered by performing a binary search in the ArrayList to find the time coordinate which is closest to but does not exceed X, and then a look-up of the time coordinate in the map to find the name of the candidate who was winning at that moment.
Ignoring the overhead of preparing the data structures, the time complexity of each query is equal to the time complexity of binary search, which is O(log2 n). This is sub-linear, so it is better than O(N).
(1) At best, we can say that a test "shows that it works"; it does not prove anything. The most accurate way of putting it is that it "gives sufficient reason to believe that it works", but that's too long, so "it shows that it works" is a decent alternative.

This is an O(N) solution, I'm assuming you have in input 2 arrays one for candidates and one for times. Who will win if there are the same amount of votes is not clear so in this case the first wins.
public Optional<Character> findCandidateWithMaxVotes(Character[] candidates, int[] times, int timeLimit) {
Character cadidateWithMaxVotes = null;
int max = 0, count = 0;
Map<Character, Integer> numberOvVotesForCandidate = new HashMap<>();
for(int i = 0; i < times.length; i++) {
if(times[i] <= timeLimit) {
count = numberOvVotesForCandidate.merge(candidates[i], 1, Integer::sum);
if(max < count) {
max = count;
cadidateWithMaxVotes = candidates[i];
}
}
}
return Optional.ofNullable(cadidateWithMaxVotes);
}

It's a bit unclear should we answer just one question ("who is the winner at time 20") or we are
supposed to preprocess the data provided and then answer several queries ("who is the winner at time 20", "who is winner at time 8" etc.).
The first problem is easy:
// Nobody is leading before elections with 0 votes
String leader = null;
int leaderVotes = 0;
HashMap<string, Integer> ballots = new HashMap<string, Integer>();
for (int i = 0; i < candidates.length; ++i) {
// too late, don't count this vote
if (times[i] > givenTime)
continue;
// number of votes
int current = ballots.containsKey(candidates[i])
? ballots.get(candidates[i]) + 1
: 1;
ballots.set(candidates[i], current);
// do we have a leader change?
//TODO: add tie breaking logic here
if (current > leaderVotes) {
leaderVotes = current;
leader = candidates[i];
}
}
// at givenTime we have leader with leaderVotes
The second problem is trickier:
We sort the votes by time
Scan them as we do in the first problem
On every leader change we add a record into (time, leader) list
Having all these done we have a sorted list which is ready for binary search: for given time we are looking for the latest record which is not later than time.

How to prevent genetic algorithm from converging on local minima?

I am trying to build a 4 x 4 sudoku solver by using the genetic algorithm. I have some issues with values converging to local minima. I am using a ranked approach and removing the bottom two ranked answer possibilities and replacing them with a crossover between the two highest ranked answer possibilities. For additional help avoiding local mininma, I am also using mutation. If an answer is not determined within a specific amount of generation, my population is filled with completely new and random state values. However, my algorithm seems to get stuck in local minima. As a fitness function, I am using:
(Total Amount of Open Squares * 7 (possible violations at each square; row, column, and box)) - total Violations
population is an ArrayList of integer arrays in which each array is a possible end state for sudoku based on the input. Fitness is determined for each array in the population.
Would someone be able to assist me in determining why my algorithm converges on local minima or perhaps recommend a technique to use to avoid local minima. Any help is greatly appreciated.
Fitness Function:
public int[] fitnessFunction(ArrayList<int[]> population)
{
int emptySpaces = this.blankData.size();
int maxError = emptySpaces*7;
int[] fitness = new int[populationSize];
for(int i=0; i<population.size();i++)
{
int[] temp = population.get(i);
int value = evaluationFunc(temp);
fitness[i] = maxError - value;
System.out.println("Fitness(i)" + fitness[i]);
}
return fitness;
}
Crossover Function:
public void crossover(ArrayList<int[]> population, int indexWeakest, int indexStrong, int indexSecStrong, int indexSecWeak)
{
int[] tempWeak = new int[16];
int[] tempStrong = new int[16];
int[] tempSecStrong = new int[16];
int[] tempSecWeak = new int[16];
tempStrong = population.get(indexStrong);
tempSecStrong = population.get(indexSecStrong);
tempWeak = population.get(indexWeakest);
tempSecWeak = population.get(indexSecWeak);
population.remove(indexWeakest);
population.remove(indexSecWeak);
int crossoverSite = random.nextInt(14)+1;
for(int i=0;i<tempWeak.length;i++)
{
if(i<crossoverSite)
{
tempWeak[i] = tempStrong[i];
tempSecWeak[i] = tempSecStrong[i];
}
else
{
tempWeak[i] = tempSecStrong[i];
tempSecWeak[i] = tempStrong[i];
}
}
mutation(tempWeak);
mutation(tempSecWeak);
population.add(tempWeak);
population.add(tempSecWeak);
for(int j=0; j<tempWeak.length;j++)
{
System.out.print(tempWeak[j] + ", ");
}
for(int j=0; j<tempWeak.length;j++)
{
System.out.print(tempSecWeak[j] + ", ");
}
}
Mutation Function:
public void mutation(int[] mutate)
{
if(this.blankData.size() > 2)
{
Blank blank = this.blankData.get(0);
int x = blank.getPosition();
Blank blank2 = this.blankData.get(1);
int y = blank2.getPosition();
Blank blank3 = this.blankData.get(2);
int z = blank3.getPosition();
int rando = random.nextInt(4) + 1;
if(rando == 2)
{
int rando2 = random.nextInt(4) + 1;
mutate[x] = rando2;
}
if(rando == 3)
{
int rando2 = random.nextInt(4) + 1;
mutate[y] = rando2;
}
if(rando==4)
{
int rando3 = random.nextInt(4) + 1;
mutate[z] = rando3;
}
}

The reason you see rapid convergence is that your methodology for "mating" is not very good. You are always producing two offspring from "mating" of the top two scoring individuals. Imagine what happens when one of the new offspring is the same as your top individual (by chance, no crossover and no mutation, or at least none that have an effect on the fitness). Once this occurs, the top two individuals are identical which eliminates the effectiveness of crossover.
A more typical approach is to replace EVERY individual on every generation. There are lots of possible variations here, but you might do a random choice of two parents weighted fitness.
Regarding population size: I don't know how hard of a problem sudoku is given your genetic representation and fitness function, but I suggest that you think about millions of individuals, not dozens.
If you are working on really hard problems, genetic algorithms are much more effective when you place your population on a 2-D grid and choosing "parents" for each point in the grid from the nearby individuals. You will get local convergence, but each locality will have converged on different solutions; you get a huge amount of variation produced from the borders between the locally-converged areas of the grid.
Another technique you might think about is running to convergence from random populations many times and store the top individual from each run. After you build up a bunch of different local minima genomes, build a new random population from those top individuals.

I think the Sudoku is a permutation problem. therefore i suggest you to use random permutation numbers for initializing population and use the crossover method which Compatible to permutation problems.

How to train data correctly using libsvm?

I want to use SVM (Support vector machine) in my program, but I could not get the true result.
I want to know that how we must train data for SVM.
What I am doing:
Think that we have 5 document (the numbers are just an example), 3 of them is on first category and others (2 of them) are on second category, I merge the categories to each other (it means that the 3 doc that are in the first category will merge in one document), after that I made a train array like this:
double[][] train = new double[cat1.getDocument().getAttributes().size() + cat2.getDocument().getAttributes().size()][];
and I will fill the array like this:
int i = 0;
Iterator<String> iteraitor = cat1.getDocument().getAttributes().keySet().iterator();
Iterator<String> iteraitor2 = cat2.getDocument().getAttributes().keySet().iterator();
while (i < train.length) {
if (i < cat2.getDocument().getAttributes().size()) {
while (iteraitor2.hasNext()) {
String key = (String) iteraitor2.next();
Long value = cat2.getDocument().getAttributes().get(key);
double[] vals = { 0, value };
train[i] = vals;
i++;
System.out.println(vals[0] + "," + vals[1]);
}
} else {
while (iteraitor.hasNext()) {
String key = (String) iteraitor.next();
Long value = cat1.getDocument().getAttributes().get(key);
double[] vals = { 1, value };
train[i] = vals;
i++;
System.out.println(vals[0] + "," + vals[1]);
}
i++;
}
so I will continue like this to get the model :
svm_problem prob = new svm_problem();
int dataCount = train.length;
prob.y = new double[dataCount];
prob.l = dataCount;
prob.x = new svm_node[dataCount][];
for (int k = 0; k < dataCount; k++) {
double[] features = train[k];
prob.x[k] = new svm_node[features.length - 1];
for (int j = 1; j < features.length; j++) {
svm_node node = new svm_node();
node.index = j;
node.value = features[j];
prob.x[k][j - 1] = node;
}
prob.y[k] = features[0];
}
svm_parameter param = new svm_parameter();
param.probability = 1;
param.gamma = 0.5;
param.nu = 0.5;
param.C = 1;
param.svm_type = svm_parameter.C_SVC;
param.kernel_type = svm_parameter.LINEAR;
param.cache_size = 20000;
param.eps = 0.001;
svm_model model = svm.svm_train(prob, param);
Is this way correct? if not please help me to make it true.
these two answers are true : answer one , answer two,

Even without examining the code one can find conceptual errors:
think that we have 5 document , 3 of them is on first category and others( 2 of them) are on second category , i merge the categories to each other (it means that the 3 doc that are in the first category will merge in one document ) ,after that i made a train array like this
So:
training on the 5 documents won't give any reasonable effects, with any machine learning model... these are statistical models,there is no reasonable statistics in 5 points in R^n, where n~10,000
You do not merge anything. Such approach can work for Naive Bayes, which do not really treat documents as "whole" but rather - as probabilistic dependencies between features and classes. In SVM each document should be separate point in the R^n space, where n can be number of distinct words (for bag of words/set of words representation).

A problem might be that you do not terminate each set of features in a training example with an index of -1 which you should according to the read me...
I.e. if you have one example with two features i think you should do:
Index[0]: 0
Value[0]: 22
Index[1]: 1
Value[1]: 53
Index[2]: -1
Good luck!

Using SVMs to classify text is a common task. You can check out research papers by Joachims [1] regarding SVM text classification.
Basically you have to:
Tokenize your documents
Remove stopwords
Apply stemming technique
Apply feature selection technique (see [2])
Transform your documents using features achieved in 4.) (simple would be binary (0: feature is absent, 1: feature is present) or other measures like TFC)
Train your SVM and be happy :)
[1] T. Joachims: Text Categorization with Support Vector Machines: Learning with Many Relevant Features; Springer: Heidelberg, Germany, 1998, doi:10.1007/BFb0026683.
[2] Y. Yang, J. O. Pedersen: A Comparative Study on Feature Selection in Text Categorization. International Conference on Machine Learning, 1997, 412-420.

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;
}

Java and Increasing the Efficiency of Genetic Algorithms

I was wondering if I could get some advice on increasing the overall efficiency of a program that implements a genetic algorithm. Yes this is an assignment question, but I have already completed the assignment on my own and am simply looking for a way to get it to perform better
Problem Description
My program at the moment reads a given chain made of the types of constituents, h or p. (For example: hphpphhphpphphhpphph) For each H and P it generated a random move (Up, Down, Left, Right) and adds the move to an arrayList contained in the "Chromosome" Object. At the start the program is generating 19 moves for 10,000 Chromosomes
SecureRandom sec = new SecureRandom();
byte[] sbuf = sec.generateSeed(8);
ByteBuffer bb = ByteBuffer.wrap(sbuf);
Random numberGen = new Random(bb.getLong());
int numberMoves = chromosoneData.length();
moveList = new ArrayList(numberMoves);
for (int a = 0; a < numberMoves; a++) {
int randomMove = numberGen.nextInt(4);
char typeChro = chromosoneData.charAt(a);
if (randomMove == 0) {
moveList.add(Move.Down);
} else if (randomMove == 1) {
moveList.add(Move.Up);
} else if (randomMove == 2) {
moveList.add(Move.Left);
} else if (randomMove == 3) {
moveList.add(Move.Right);
}
}
After this comes the selection of chromosomes from the Population to crossover. My crossover function selections the first chromosome at random from the fittest 20% of the population and the other at random from outside of the top 20%. The chosen chromosomes are then crossed and a mutation function is called. I believe the area in which I am taking the biggest hit is calculating the fitness of each Chromosome. Currently my fitness function creates a 2d Array to act as a grid, places the moves in order from the move list generated by the function shown above, and then loops through the array to do the fitness calculation. (I.E. found and H at location [2,1] is Cord [1,1] [3,1] [2,0] or [2,2] also an H and if an H is found it just increments the count of bonds found)
After the calculation is complete the least fit chromosome is removed from my population and the new one is added and then the array list of chromosomes is sorted. Rinse and repeat until target solution is found
If you guys want to see more of my code to prove I actually did the work before asking for help just let me know (dont want to post to much so other students cant just copy pasta my stuff)
As suggested in the comments I have ran the profiler on my application (have never used it before, only a first year CS student) and my initial guess on where i am having issues was somewhat incorrect. It seems from what the profiler is telling me is that the big hotspots are:
When comparing the new chromosome to the others in the population to determine its position. I am doing this by implementing Comparable:
public int compareTo(Chromosome other) {
if(this.fitness >= other.fitness)
return 1;
if(this.fitness ==other.fitness )
return 0;
else
return -1;
}
The other area of issue described is in my actual evolution function, consuming about 40% of the CPU time. A codesample from said method below
double topPercentile = highestValue;
topPercentile = topPercentile * .20;
topPercentile = Math.ceil(topPercentile);
randomOne = numberGen.nextInt((int) topPercentile);
//Lower Bount for random two so it comes from outside of top 20%
int randomTwo = numberGen.nextInt(highestValue - (int) topPercentile);
randomTwo = randomTwo + 25;
//System.out.println("Selecting First: " + randomOne + " Selecting Second: " + randomTwo);
Chromosome firstChrom = (Chromosome) populationList.get(randomOne);
Chromosome secondChrom = (Chromosome) populationList.get(randomTwo);
//System.out.println("Selected 2 Chromosones Crossing Over");
Chromosome resultantChromosome = firstChrom.crossOver(secondChrom);
populationList.add(resultantChromosome);
Collections.sort(populationList);
populationList.remove(highestValue);
Chromosome bestResult = (Chromosome) populationList.get(0);
The other main preformance hit is the inital population seeding which is performed by the first code sample in the post

I believe the area in which I am taking the biggest hit is calculating the fitness of each Chromosome
If you are not sure then I assume you have not run a profiler on the program yet.
If you want to improve the performance, profiling is the first thing you should do.

Instead of repeatedly sorting your population, use a collection that maintains its contents already sorted. (e.g. TreeSet)

If your fitness measure is consistent across generations (i.e. not dependent on other members of the population) then I hope at least that you are storing that in the Chromosome object so you only calculate it once for each member of the population. With that in place you'd only be calculating fitness on the newly generated/assembled chromosome each iteration. Without more information on how fitness if calculated it's difficult to be able to offer any optimisations in that area.

Your random number generator seed doesn't need to be cryptographically strong.
Random numberGen = new Random();

A minor speedup when seeding your population is to remove all the testing and branching:
static Move[] moves = {Move.Down, Move.Up, Move.Left, Move.Right};
...
moveList.add(moves[randomMove]);