Using a Genetic-esque Algorithm To Solve The 0/1 Knapsack Problem

11_genetic algorithm

Below is an example of a genetic algorithm that was coded in R as a solution to a specific 0/1 knapsack problem.  This algorithm is called a genetic algorithm because the methods it uses to maximize the value of our solution vectors are based on the types of things that occur within the reproduction of chromosomes. The basic idea behind the maximization effect in the genetic algorithm runs parallel to the basic ideas behind the reproductive mechanisms at play in survival of the fittest (SOF) type of evolutionary activity in populations.  In a population that is expanding along a survival of the fittest type of trajectory the members of the population that are the fittest for survival are also the members of the population that reproduce the most. These fit parents will give rise to offspring that contain some blend of the parent’s genetic material and through this blending of genetic material, there will be a general increase in the fitness of the offspring over the long run. That is, fit parents make fitter offspring and allowing fitter parents to reproduce more will produce a population that is increasingly fit.  The algorithm takes an initial collection of chromo- some (solutions) and uses them to create generation after generation of increasingly fit chromosomes through the action of fitness, selection, crossover, and mutation. The simulation requires ‘plyr’ and ‘Rcpp’ be installed.

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Let’s Make a Deal Simulation

Below is an R program that can be used to run Let’s Make a Deal simulations. The function, “LMAD(switch,stay)”, takes two arguments.  Switch informs the program how many simulations you would like to run where the strategy regarding the second door is to switch to the unchosen door.  Stay informs the program of the number of simulations to run where the strategy is to stay.

Let’s Make a Deal initially baffled people due to the somewhat unintuitive result that your chances of winning always improved by switching to the unchosen door.  The intuition behind this improvement in odds is obvious when looked at mathematically, but was illusive when the problem was considered in everyday terms.  The information that is obtained when discovering that one of the three doors does not contain the prize is somewhat invisible in the case of only three doors.  However, the reason for switching would be obvious if we imagine instead that there were 1,000 doors to initially choose from and then all but two were opened revealing nothing behind them.  This would make us rightly believe that there was a very good reason for the other door not being included in the 998 that were opened and shown to contain nothing.

#Here we set a seed for reproducibility.
	set.seed(10); trials=switch+stay

#Here we set up a vector called 'strategy' that assigns our switch or stay
	st=c(); for(i in 1:stay){st[i]=0}
	sw=c(); for(i in 1:switch){sw[i]=1}

#Vector 'x' is repeatedly sampled from in order to determine whether our
#first choice contained the winning prize.

#Here we repeatedly sampling the vector 'x' and place the sampled element
#into the vector 'u'.  This simulates the first guess in the game. The vector
#'u' stores whether or not we chose the door with the prize ('1') for game i.
	u=c(); for(i in 1:trials){u[i]=sample(x)[1]}

#Here we populate the vectors 'sw' and 'st' with the outcomes of the second
#part of the game.
	for(i in 1:trials){
	#If our initial guess contained the prize and our strategy was to switch
	#we would certainly lose because the two remaining doors are empty.
		if (u[i]==1 && strategy[i]==1){sw[i]=0; st[i]=0}
	#If our initial guess contained the prize and our strategy was to stay
	#we would certainly win because the choice we are staying with is the
	#winning choice.
		else if (u[i]==1 && strategy[i]==0){sw[i]=0; st[i]=1}
	#If our initial guess did not contain the prize and our strategy was to
	#switch then after Monty revealed the other losing door we would switch
	#to the remaining door, which is the winner.
		else if (u[i]==0 && strategy[i]==1){sw[i]=1; st[i]=0}	
	#If our initial guess did not contain the prize and our strategy was to
	#stay we would certainly lose because the choice we are staying with is a
	#losing choice.
		else if (u[i]==0 && strategy[i]==0){sw[i]=0; st[i]=0}

#The vector 'sw' contains 1's for all of the games we won under the switch
#strategy and the vector 'st' contains 1's for all of the games we won under
#the stay strategy.  Using these we can calculate our switch and stay win rates.
	switch_wins<-sum(sw)/switch; stay_wins<-sum(st)/stay

#Here we produce 95% lower and upper bounds on our percentage of wins.

#Here we output the results.

cat("Number of Games:",trials,"\n")
cat("Number of switches:",switch,"\n")
cat("Number of wins when switching",sum(sw),"\n")
cat("95% CI for wins when switching:",switch_CI,"\n")
cat("Number of stays:",stay,"\n")
cat("Number of wins when staying",sum(st),"\n")
cat("95% CI for wins when staying:",stay_CI)