Update: Previously, a system was introduced for detecting if an individual was stuck at a local optimum. After extensive testing, this system was shown to be fragile. This post has been updated to showcase a more robust system.

Previously our Evolutionary Algorithms had it pretty easy: there would be either one local optimum (like our Secret Message problem instance) or multiple, valid local optimums (like the 3-SAT problem instance). In the real world, we might not be so lucky.

Often, an Evolutionary Algorithm might encounter a local optimum within the search space, and it will not be so easy to escape — offspring generated will be in close proximity of the optimum, and the mutation will not be enough to start exploring other parts of the search space.

To add to the frustration, there might not enough time or patience to wait for the Evolutionary Algorithm to finish. We might have different criteria we are looking for, outside of just a fitness target.

We are going to tackle both of these issues.

Applying Termination Conditions

First, we will examine what criteria we want met before our Evolutionary Algorithm terminates. In general, there are six that are universal:

1. Date and Time. After a specified date and time, terminate.
2. Fitness Target. This is what we had before; terminate when any individual attains a certain fitness.
3. Number of Fitness Evaluations. Every generation, every individual's fitness is evaluated (in our case, every generation $\mu + \lambda$ fitnesses are evaluated). Terminate after a specified number of fitness evaluations.
4. Number of Generations. Just like the number of fitness evaluations, terminate after a specified generations.
5. No Change In Average Fitness. This is a bit tricky. After specify $N$ generations, we check every $N$ generations back to determine if the average fitness of a population has improved. We have to be careful in our programming; by preserving diversity, we almost always lose fitness.
6. No Change In Best Fitness. Just like No Change In Average Fitness, but instead of taking the average fitness, we take the best.

Later, we will see how Conditions 5 & 6 will come in handy to determining if we are stuck in a local optimum.

To make sure we are always given valid termination conditions, we will have a super class that all termination conditions will inherit from. From there, we will have a separate condition for each of the listed conditions above.

class _TerminationCondition:
    pass

class FitnessTarget(_TerminationCondition):
    """Terminate after an individual reaches a particular fitness."""

class DateTarget(_TerminationCondition):
    """Terminate after a particular date and time."""

class NoChangeInAverageFitness(_TerminationCondition):
    """Terminate after a there has been no change in the average fitness for a period of time."""

class NoChangeInBestFitness(_TerminationCondition):
    """Terminate after a there has been no change in the best fitness for a period of time."""

class NumberOfFitnessEvaluations(_TerminationCondition):
    """Terminate after a particular number of fitness evaluations."""

class NumberOfGenerations(_TerminationCondition):
    """Terminate after a particular number of generations."""

Now, we need something that will keep track of all these conditions, and tells us when we should terminate. And here's where we need to be careful.

First, we need to know when to terminate. We want to mix and match different conditions, depending on the use case. This begs the questions:

Should the Evolutionary Algorithm terminate when one condition has been met, or all of them?

Generally, it makes more sense to terminate when any of the conditions have been met, as apposed to all of them. Suppose the two termination conditions are date and target fitness. It does not make sense to keep going after the target fitness is reached, and (if in a time crunch) it does not make sense to keep going after a specified date.

Second, how should we define no change in average/best fitness? These values can be quite sinusoidal, so we want to be more conservative in our definition. One plausible solution is to take the average of the first quartile (the first 25% to ever enter the queue), and see if the there is a single individual with a better fitness in the second, third, or fourth quartile (the last 75% percent to enter the queue). This way, even if there were very dominant individuals in the beginning, a single, more dominant individual will continue the Evolutionary Algorithm.

From this, we have everything we might need to keep track of our terminating conditions.

class TerminationManager:
    def __init__(self, termination_conditions, fitness_selector):
        assert isinstance(termination_conditions, list)
        assert all(issubclass(type(condition), _TerminationCondition) for condition in termination_conditions), "Termination condition is not valid"

        self.termination_conditions = termination_conditions
        self.__fitness_selector = fitness_selector

        self.__best_fitnesses = []
        self.__average_fitnesses = []

        self.__number_of_fitness_evaluations = 0
        self.__number_of_generations = 0

    def should_terminate(self):
        for condition in self.termination_conditions:
            if isinstance(condition, FitnessTarget) and self.__fitness_should_terminate():
                return True
            elif isinstance(condition, DateTarget) and self.__date_should_terminate():
                return True
            elif isinstance(condition, NoChangeInAverageFitness) and self.__average_fitness_should_terminate():
                return True
            elif isinstance(condition, NoChangeInBestFitness) and self.__best_fitness_should_terminate():
                return True
            elif isinstance(condition, NumberOfFitnessEvaluations) and self.__fitness_evaluations_should_terminate():
                return True
            elif isinstance(condition, NumberOfGenerations) and self.__generations_should_terminate():
                return True

        return False

    def reset(self):
        """Reset the best fitnesses, average fitnesses, number of generations, and number of fitness evaluations."""
        self.__best_fitnesses = []
        self.__average_fitnesses = []

        self.__number_of_fitness_evaluations = 0
        self.__number_of_generations = 0

    def __fitness_should_terminate(self):
        """Determine if should terminate based on the max fitness."""

    def __date_should_terminate(self):
        """Determine if should terminate based on the date."""

    def __average_fitness_should_terminate(self):
        """Determine if should terminate based on the average fitness for the last N generations."""

    def __best_fitness_should_terminate(self):
        """Determine if should terminate based on the average fitness for the last N generations."""

    def __fitness_evaluations_should_terminate(self):
        """Determine if should terminate based on the number of fitness evaluations."""

    def __generations_should_terminate(self):
        """Determine if should terminate based on the number of generations."""

And the changes to our Evolutionary Algorithm are minimal, too.

class EA:
    ...
    
    def search(self, termination_conditions):
        generation = 1
        self.population = Population(self.μ, self.λ)
        
        fitness_getter = lambda: [individual.fitness for individual in self.population.individuals]  # noqa
        termination_manager = TerminationManager(termination_conditions, fitness_getter)

        while not termination_manager.should_terminate():
            offspring = Population.generate_offspring(self.population)
            self.population.individuals += offspring.individuals
            self.population = Population.survival_selection(self.population)

            print("Generation #{}: {}".format(generation, self.population.fittest.fitness))
            generation += 1

        print("Result: {}".format(self.population.fittest.genotype))
        return self.population.fittest

However, we can still do better.

Generations Into Epochs

Before, the Evolutionary Algorithm framework we put in place was strictly a generational model. One generation lead to the next, and there were no discontinuities. Now, let's make our generational model into an epochal one.

We define an epoch as anytime our Evolutionary Algorithm encounters a local optimum. Once the end of an epoch is reached, the EA is reset, and the previous epoch is saved. Upon approaching the end of the next epoch, reintroduce the last epoch into the population; by this, more of the search space is covered.

How can we determine if we are at a local optimum?

We can't.

That does not mean we cannot have a heuristic for it. When there is little to no change in average/best fitness for a prolonged period of time, that typically means a local optimum has been reached. How long is a prolonged period of time? That's undetermined; it is another parameter we have to account for.

Note, if the Evolutionary Algorithm keeps producing more fit individuals, but the average fitness remains the same, the algorithm will terminate. Likewise, if the best fitness remains the same, but the average fitness closely approaches the best, the EA will terminate. Therefore, we should determine if the best fitness and the average fitness has not changed; only then should we start a new epoch.

Luckily, we already have something that will manage the average/best fitness for us.

class EA:
    ...

    def search(self, termination_conditions):
        epochs, generation, total_generations = 1, 1, 1
        self.population = Population(self.μ, self.λ)

        previous_epoch = []
        fitness_getter = lambda: [individual.fitness for individual in self.population.individuals]  # noqa

        termination_manager = TerminationManager(termination_conditions, fitness_getter)
        epoch_manager_best_fitness = TerminationManager([NoChangeInBestFitness(250)], fitness_getter)
        epoch_manager_average_fitness = TerminationManager([NoChangeInAverageFitness(250)], fitness_getter)

        while not termination_manager.should_terminate():
            if epoch_manager_best_fitness.should_terminate() and epoch_manager_average_fitness.should_terminate():
                if len(previous_epoch) > 0:
                    epoch_manager_best_fitness.reset()
                    epoch_manager_average_fitness.reset()

                    self.population.individuals += previous_epoch
                    previous_epoch = []
                else:
                    epoch_manager_best_fitness.reset()
                    epoch_manager_average_fitness.reset()

                    previous_epoch = self.population.individuals
                    self.population = Population(self.μ, self.λ)

                    generation = 0
                    epochs += 1

            self.population = Population.survival_selection(self.population)

            offspring = Population.generate_offspring(self.population)
            self.population.individuals += offspring.individuals

            self.__log(total_generations, epochs, generation)

            total_generations += 1
            generation += 1

        print("Result: {}".format(self.population.fittest.genotype))
        return self.population.fittest


    def __log(self, total_generations, epochs, generation):
        """Log the process of the Evolutionary Algorithm."""
        ...

Although considerably more complicated, this new Evolutionary Algorithm framework allows us to explore much more of a search space (without getting stuck).

Let's put it to the test.

A New 3-SAT Problem

We're going to take on a substantially harder 3-SAT instance: 1,000 clauses, 250 variables. To make it worse, the number of valid solutions is also lower. We will also include the following terminating conditions:

• Time of eight hours.
• Fitness of all clauses satisfied (100).
• A million generations.

So, how does our Evolutionary Algorithm fair?

Not well. After twenty epochs, and thousands of generations — we do not find a solution. Fear not; in subsequent posts, we will work on optimizing our Genetic Algorithm to handle much larger cases, more effectively.

We have been introduced to recombination operators before; however, that was merely an introduction. There are dozens of different Evolutionary Algorithm recombination operators for any established genotype; some are simple, some are complicated.

For a genotype representation that is a permutation (such as a vector^[1], bit-string, or hash-map^[2]), we have seen a possible recombination operator. Our 3-SAT solver uses a very popular recombination technique: uniform crossover.

Furthermore, we know a permutation is not the only, valid genotype for an individual: other possibilities can include an integer or a real-valued number.

Note, for simplicity, we will discuss recombination to form one offspring. This exact process can be applied to form a second child (generally with the parent's role reversed). Recombination can also be applied to more than two parents (depending on the operator). Again, for simplicity, we choose to omit it^[3].

First, let us start with permutations.

Permutation Crossover

In regard to premutation crossover, there are three, common operators:

1. Uniform Crossover
2. $N$ -Point Crossover
3. Davis Crossover

Uniform crossover we have seen before. We consider individual elements in the permutation, and choose one with a random, equal probability. For large enough genotypes, the offspring genotype should consist of 50% of the genotype from parent one, and 50% of the genotype from parent two.

$N$-Point crossover considers segments of a genotype, apposed to individual elements. This operator splits the genotype of Parent 1 and Parent 2 $N$ times (hence the name $N$-point), and creates a genotype by alternating segments from the two parents. For every $N$, there will be $N + 1$ segments. For 1-point crossover, the genotype should be split into two segments, and the offspring genotype should be composed of one segment from Parent 1, and one segment from Parent 2. For 2-point crossover, there will be three segments, and the offspring genotype will have two parts from Parent 1 and one part from Parent 2 (or two parts, Parent 1, one part, Parent 2).

Davis Crossover tries to preserve the ordering of the genotype in the offspring (apposed to the previous methods, where ordering was not considered). The premise is a bit complicated, but bear with me. Pick two random indices ($k_1$ and $k_2$), and copy the genetic material of Parent 1 from $k_1$ to $k_2$ into the offspring at $k_1$ to $k_2$. Put Parent 1 to the side, his role is finished. Start copying the genotype of Parent 2 starting at $k_1$ to $k_2$ at the beginning of the offspring. When $k_2$ is reached in the parent, start copying the beginning of Parent 2 into the genotype, and when $k_1$ is reached in the parent, skip to $k_2$. When $k_1$ is reached in the offspring, skip to $k_2$, and start copying until the end. If this seems a complicated (it very much is), reference the accompanying figure.

Those are considered the three, most popular choices for permutations. Now, let us look at integer crossover.

Integer Crossover

Integer crossover is actually quite an interesting case; integers can be recombined as permutations or real-valued numbers.

An integer is already a permutation, just not at first glance: binary. The individual bits in a binary string are analogous to elements in a vector, and the whole collection is a vector. Now it is a valid permutation. We can apply uniform crossover, $N$-point crossover, or Davis Crossover, just as we have seen.

An integer is also already a real-valued number, so we can treat it as such. Let's take a look at how to recombine it.

Real-Valued Crossover

Real-Valued crossover is different than methods we have seen before. We could turn it into binary, but that would be a nightmare to deal with. However, we can exploit the arithmetic properties of real-valued numbers — with a weighted, arithmetic mean. For a child (of real value) $z$, we can generate it from Parent 1 $x$ and Parent 2 $y$ as such:

$$
z = \alpha \cdot x + (1 - \alpha) \cdot y
$$

Now, if we want to crossover a permutation of Parent 1 and Parent 2, we can do so for every element.

$$
z_i = \alpha \cdot x_i + (1 - \alpha) \cdot y_i
$$

This can be shown to have better performance than crossover methods discussed, but would entirely depend on use case.

Implementing Permutation Recombination

As always, we will now tackle implementing the permutation crossovers we've had before. None of them are incredibly complicated, except possibly $N$-point crossover.

class Individual
    ...

    @staticmethod
    def __uniform_crossover(parent_one, parent_two):
        new_genotype = SAT(Individual.cnf_filename)

        for variable in parent_one.genotype.variables:
            gene = choice([parent_one.genotype[variable], parent_two.genotype[variable]])
            new_genotype[variable] = gene

        individual = Individual()
        individual.genotype = new_genotype
        return individual

    @staticmethod
    def __n_point_crossover(parent_one, parent_two, n):
        new_genotype = SAT(Individual.cnf_filename)
        variables = sorted(parent_one.genotype.variables)
        splits = [(i * len(variables) // (n + 1)) for i in range(1, n + 2)]

        i = 0
        for index, split in enumerate(splits):
            for variable_index in range(i, split):
                gene = parent_one.genotype[variables[i]] if index % 2 == 0 else parent_two.genotype[variables[i]]
                new_genotype[variables[i]] = gene

                i += 1

        individual = Individual()
        individual.genotype = new_genotype

        return individual

    @staticmethod
    def __davis_crossover(parent_one, parent_two):
        new_genotype = SAT(Individual.cnf_filename)
        variables = sorted(parent_one.genotype.variables)
        split_one, split_two = sorted(sample(range(len(variables)), 2))

        for variable in variables[:split_one]:
            new_genotype[variable] = parent_two.genotype[variable]

        for variable in variables[split_one:split_two]:
            new_genotype[variable] = parent_one.genotype[variable]

        for variable in variables[split_two:]:
            new_genotype[variable] = parent_two.genotype[variable]

        individual = Individual()
        individual.genotype = new_genotype

        return individual

Recombination In General

By no means is recombination easy. It took evolution hundreds of thousands of years to formulate ours. The particular permutation operator to use entirely dependent on the context of the problem; and most of the time, it is not obvious by any stretch. Sometimes, there might not even be an established crossover operator for a particular genotype.

Sometimes, you might have to get a little creative.

1. List or array in programming terms. ↩︎
2. Dictionary or map in programming terms. ↩︎
3. View it as "an exercise left for the reader". ↩︎

Let's propose an Evolutionary Algorithm experiment; say we already have a framework in place (like the Secret Message framework we previously implemented). How difficult would it be to completely switch problem instances?

First, we need another problem instance. Our previous problem instance was pretty straightforward: it had one local optimum. Let's take on a problem with many local optimum, such as the 3-SAT problem.

The premise of 3-SAT is simple. From a global pool of variables ($x_1$, $x_2$, $\ldots$, $x_n$), we have a basic clause of three variables or-ed together (signified by $\vee$):

$$
x_p \vee x_q \vee x_r
$$

Then, and (signified by a $\wedge$) several clauses together:

$$\left(x_p \vee x_q \vee x_r\right) \wedge \left(x_s \vee x_t \vee x_u\right) \wedge \ldots \wedge \left(x_v \vee x_w \vee x_y\right)$$

The only stipulation is that any variable can be negated (signified by a $\neg$). So, supposing we want to negate $x_p$; $x_s$ and $x_u$; and $x_v$, $x_w$, and $x_y$; we can do the following:

$$
\left(\neg x_p \vee x_q \vee x_r\right) \wedge \left(\neg x_s \vee x_t \vee \neg x_u\right) \wedge \ldots \wedge \left(\neg x_v \vee \neg x_w \vee \neg x_y\right)
$$

Now, we simply have to assign all the variables such that all the clauses will evaluate to true. It may sound simple, but it belongs to the hardest classes of problems in computer science. There is no guaranteed algorithm to produce the right answer at this time.

For a more visual approach, please reference the figure below. The goals is to make every inner node green, by having at lease one, connected, outer node be green. Note the green nodes have to account for negation as well.

This sounds like a good problem candidate for an Evolutionary Algorithms^[1].

The SAT Problem

We can skip over the problem specific parts to worry more about the Evolutionary Algorithm aspect. Suppose we already have a well-defined SAT class that takes care of SAT-specific properties and methods, like so:

class SAT:
    def __init__(self, filename):
        """Create a SAT object that is read in from a CNF file."""
        ...

    @property
    def variables(self):
        """Get *all* the variables."""
        ...

    @property
    def total_clauses(self):
        """Set the total number of clauses."""
        ...

    @property
    def clauses_satisfied(self):
        """Get the number of satisfied clauses."""
        ...

    def __getitem__(self, key):
        """Get a particular variable (key)"""
        ...

    def __setitem__(self, key, value):
        """Set a variable (key) to value (True/False)"""
        ...

From this, we can create a new genotype for our Individual.

The New Genotype

The genotype structure was very similar to what we had before:

• The genotype is the SAT problem we defined above.
• Fitness is defined by a percentage of the total satisfied clauses.
• Mutation is uniform, choose a percentage $p$ of alleles and flip their value.
• Recombination is uniform, randomly assemble values from both parents.

Looking at the refactoring, not much has changed.

The New EA Framework

Now that we have updated our Individual, next thing to updated would be the Evolutionary Algorithm framework, including:

• The Population
• The EA Itself

Except, we don't have to.

That's the beauty of Evolutionary Algorithms, they are incredibly adaptable. By swapping out the Individual, the rest of the evolutionary algorithm should still work.

For our SAT problem, there were some parameters updated, to make the algorithm more efficient:

• The mutation rate has been reduced to 5%
• The tournament size has been reduced to 15 individuals (out of $\lambda = 100$).

The Result

So, let's try our Evolutionary Algorithm. Taking a SAT instance with 75 variables and 150 clauses, this makes the search space

$$
2^{75} \approx 3.77 \times 10^{22}
$$

Great, so roughly 1,000 times the grain of sand on Earth, easy. So, can our EA do it?

After roughly 100 iterations, yes. See the visualization below.

Marvelous, our EA managed to find a solution after only 100 iterations in a giant search space. And all we had to do was swap out one class.

The Source Code

All source code can be found here.

1. In reality, it's not a great candidate for an evolutionary algorithm. The gradient is sometimes murky, because flipping one variable's value can drastically decrease/increase the fitness function. Also, there are several great heuristics for solving the SAT problem. ↩︎

This is an extenuation of a blog post I wrote a couple months ago. You can find it here.

One of the big takeaways in my introduction to Evolutionary Algorithm was the sheer number of numerical parameters.

• $\mu$ And $\lambda$
• Mutation Rate
• $k$ in k-Tournament Selection

Not only this, but the sheer number of parameters:

• The genotype
• The mutator operator
• The survivor selection algorithm

And one might be wondering, what is the best operator for $x$ or $y$? Let’s look at an example.

Recall the problem from the previous discussion: We are going to consider a sample problem, a deciphering program. The premise of the problem is such:

There is a string of characters (without spaces) hidden away that, after set, is inaccessible.
There are two ways to retrieve data about the hidden message:

1: Get the length of the string.

2: Given a string, the problem will output how many characters match within the two strings.

Disregarding the other technical details, let us focus on the survivor selection. We used $k$ -tournament selection (with $k = 50$). But, let’s run a little experiment:

Run the Evolutionary Algorithm, with $k$ ranging from $5$ (basically the bare minimum) to $100$ ($\lambda$, the population size), and see how fast the algorithm terminates. Do this 1,000 times to get accurate results.

The result?

This makes sense. Our problem has one local optimum: the actual solution. So we do not need a lot of genetic diversity, we need aggressive selective pressure^[1] to reach the top quickly.

As $k$ gets closer to $\mu$, the average termination time decreases. What does this tell us? We picked the wrong survivor selection algorithm.

With $k = \mu$, we no longer have $k$ -tournament selection; we have truncation selection (where only the most fit individuals survive). And that's the interesting part about Evolutionary Algorithms: there are no objective, best parameters.

How do we alleviate this? Trial and error. There is no telling when one parameter is going to perform better than another.

A after a couple of trial runs, and objectives in mind (average terminating fitness, best terminating fitness, time to termination), the answer might surprise (and delight) you.

1. How elitist the survivor selection algorithm is, picking the strongest individuals more often. ↩︎

Ever wondered how sorting algorithms looked in Swift? I did too — so I implemented some. Below you can find the following:

• Bubble Sort
• Selection Sort
• Insertion Sort
• Merge Sort
• Quicksort

As a preface: some of the algorithms use a self-implemented swap function — it is as follows:

extension Array {
    mutating func swap(i: Int, _ j: Int) {
        (self[i], self[j]) = (self[j], self[i])
    }
}

Bubble Sort

func bubbleSort(var unsorted: Array<Int>) -> Array<Int> {
    for i in 0..<unsorted.count - 1 {
        for j in 0..<unsorted.count - i - 1 {
            if unsorted[j] > unsorted[j + 1] {
                unsorted.swap(j, j + 1)
            }
        }
    }
    
    return unsorted
}

Selection Sort

func selectionSort(var unsorted: Array<Int>) -> Array<Int> {
    for i in 0..<unsorted.count - 1 {
        var minimumIndex = i
        for j in i + 1 ..< unsorted.count {
            if unsorted[j] < unsorted[minimumIndex] {
                minimumIndex = j
            }
            unsorted.swap(minimumIndex, i)
        }
    }
    
    return unsorted
}

Insertion Sort

func insertionSort(var unsorted: Array<Int>) -> Array<Int> {
    for j in 1...unsorted.count - 1 {
        let key = unsorted[j];
        var i = j - 1
        
        while i >= 0 && unsorted[i] > key {
            unsorted[i + 1] = unsorted[i]
            i--
        }
        
        unsorted[i + 1] = key
    }
    
    return unsorted
}

Merge Sort

func merge(leftArray left: Array<Int>, rightArray right: Array<Int>) -> Array<Int> {
    var result = [Int]()
    var leftIndex = 0, rightIndex = 0
    
    // Calculated variable to determine if the indexes have reached end of their respective arrays
    var indexesHaveReachedEndOfArrays: Bool { return leftIndex < left.count && rightIndex < right.count }
    
    while indexesHaveReachedEndOfArrays {
        if left[leftIndex] < right[rightIndex] {
            result.append(left[leftIndex])
            leftIndex++
        }
        else if left[leftIndex] > right[rightIndex] {
            result.append(right[rightIndex])
            rightIndex++
        }
        else {
            result.append(left[leftIndex])
            result.append(right[rightIndex])
            
            leftIndex++
            rightIndex++
        }
    }
    
    result += Array(left[leftIndex..<left.count])
    result += Array(right[rightIndex..<right.count])
    
    return result
}

func mergesort(unsorted: Array<Int>) -> Array<Int> {
    let size = unsorted.count
    if size < 2 { return unsorted }
    
    let split = Int(size / 2)
    
    let left = Array(unsorted[0 ..< split])
    let right = Array(unsorted[split ..< size])
    
    let sorted = merge(leftArray: mergesort(left), rightArray: mergesort(right))
    return sorted
}

Quick Sort

func quicksort(var unsorted: Array<Int>) -> Array<Int> {
    quicksort(&unsorted, firstIndex: 0, lastIndex: unsorted.count - 1)
    return unsorted
}

private func quicksort(inout unsorted: Array<Int>, firstIndex: Int, lastIndex: Int) -> Array<Int> {
    if firstIndex < lastIndex {
        let split = partition(&unsorted, firstIndex: firstIndex, lastIndex: lastIndex)
        
        quicksort(&unsorted, firstIndex: firstIndex, lastIndex: split - 1)
        quicksort(&unsorted, firstIndex: split + 1, lastIndex: lastIndex)
    }
    
    return unsorted
}

private func partition(inout unsorted: Array<Int>, firstIndex: Int, lastIndex: Int) -> Int {
    let pivot = unsorted[lastIndex]
    var wall = firstIndex
    
    for i in firstIndex ... lastIndex - 1 {
        if unsorted[i] <= pivot {
            unsorted.swap(wall, i)
            wall++
        }
    }
    
    unsorted.swap(wall, lastIndex)
    return wall
}