I am writing a program that will simulate how members of a certain timeshare will request their apartments. The apartments are only available for certain 'events' throughout the year, and the events each have different durations (counted in days). For each event, there are a number of apartments available, which are divided into groups according to cost (points per day).
We may also have the situation that some of the apartments are already rented out, so it is possible that certain events cannot admit a certain apartment type.
Now I want a member to request apartments according to this strategy:
Maximizing the number of total event days is first priority.
Maximizing the number of points spent is second priority (so the user requests the best possible apartment that he/she can afford and still have as many days as possible).
The total cost of the requested apartments cannot exceed the total amount of available points for that user.
Obviously, if there are no more apartments of a given type for a certain event, the user should not request that particular apartment/event combo.
I am wondering whether the problem is similar to the problem described here:
http://en.wikipedia.org/wiki/Hungarian_algorithm
in that it (I suppose) possible to frame this as a matrix problem where each entry has a cost associated with it.
However, the difference is that for my problem, I am allowed to use the same apartment for several events - it's not 'spent' once it has been used for one event. Also, the cost per entry is not really one-dimensional, since each event/apartment combo both has a number of days associated with it and a number of points - both of which should be maximized (but with priority given to the number of days).
As an example, let's say there are three apartment types, costing 75, 100, and 125 points per day, and three events, with a duration of 2, 10, and 4 days. Let's further say many of the apartments are taken, so the availability matrix looks like this:
cost
75 100 125
2 True False True
days 10 False False True
4 True False True
Let's also say the user has 1250 points available. The solution, in this case, would be that the user requests the 10-day event with the 125-point apartment and nothing else.
The brute-force way of doing this would perhaps be a recursive algorithm:
Let n be the number of events you are currently trying
Find all combinations of events and apartments, and calculate the combo that maximizes the number of days, then the number of points spent (this will both include all permutations of n events, but also the number of ways 3 apartment types can be assigned to n events).
Let n=n-1
This will quickly become overwhelming when the number of events goes up, I think, so I am wondering whether there are any algorithms that can solve this in a less expensive way?
If you have access to a library for http://en.wikipedia.org/wiki/Integer_programming you could try throwing this at it.
Even if you have only one choice of cost for each event, so that you are just trying to chose combinations of events that cover as many total days as possible without going over budget, I think this reduces to http://en.wikipedia.org/wiki/Knapsack_problem. This means that it is unlikely to be solved exactly by a worst case polynomial time algorithm such as the Hungarian algorithm.
Related
I have this problem:
You need to develop a meal regime based on the data entered by the user. We have a database with the meals and their prices (meals also have a mark whether they are breakfast, cause on breakfast we eat smth different from lunch and dinner most often). The input receives the amount of money (Currency is not important) and the number of days. At the output, we must get a meal regime for a given number of days. Conditions:
Final price does not differ from the given one by more than 3%.
meals mustn't repeat more than once every 5 days.
I found this not effective solution: We are looking for an average price per day = amount of money / number of days. Then, until we reach the given number of days, we iterate throught each breakfast, then lunch and dinner (3 for loops, 2 are nested) and if price is not too different, then we end the search and add this day to the result list. So the design now looks like this:
while(daysCounter < days){
for(){
for(){
for(){
}
}
It looks scary, although there is not a lot of data (number of meals is about 150). There are thoughts that it is possible to find a more effective solution. Also i think about dynamic programming, but so far there are no ideas how to implement it.
Dynamic programming won't work because a necessary part of your state is the meals from the last 5 days. And the number of possibilities for that are astronomical.
However it is likely that there are many solutions, not just a few. And that it is easy to find a working solution by being greedy. Also that an existing solution can be improved fairly easily.
So I'd solve it like this. Replace days with an array of meals. Take the total you want to spend. Divide it up among your meals, in proportion to the median price of the options for that meal. (Breakfast is generally cheaper than dinner.) And now add up that per meal cost to get running totals.
And now for each meal choose the meal you have not had in the last 5 days that brings the running total of what has been spent as close as possible to the ideal total. Choose all of your meals, one at a time.
This is the greedy approach. Normally, but not always, it will come fairly close to the target.
And now, to a maximum of n tries or until you have the target within 3%, pick a random meal, consider all options that are not eaten within the last or next 5 days, and randomly pick one (if any such options exist) that brings the overall amount spent closer to the target.
(For a meal plan that is likely to vary more over a long period, I'd suggest trying simulated annealing. That will produce more interesting results, but it will also take longer.)
I am working on a scheduling problem for a team of volunteers. I have boiled my problem down to the following algorithmic problem:
I have a matrix with ~60 rows representing volunteers and ~14 columns representing days. Each entry is an integer in the range 0 to 3 inclusive representing how free the volunteer is on that day. I want to choose exactly 4 entries from each column (4 volunteers a day) such that (in order of importance)
A 0-entry is never chosen.
The workload is as spread out as possible (first give everyone one shift, then start giving out second shifts, etc. We can expect that most volunteers will only have one shift per 2-week period, and some may even have none.)
The sum over selected entries is maximised (volunteers get days that they prefer).
I want to output a decision matrix that has a 1 whenever a volunteer is chosen for a day, and 0 otherwise. I believe this is an instance of the nurse-rostering problem, so I'm not expecting a fast solution, but I just want to make a brute force algorithm that will work in a reasonable time for my ~60 person team. I'm just really not sure how to start tackling this problem. Is it suited for backtracking, or is there some way to calculate the best placement of each volunteer based on the distribution of his/her day-scores?
This question (last one) appeared in Benelux Algorithm Programming Contest-2007
http://www.cs.duke.edu/courses/cps149s/spring08/problems/bapc07/allprobs.pdf
Problem Statement in short:
A Company needs to figure out strategy when to - buy OR sell OR no-op on a given input so as to maximise profit. Input is in the form:
6
4 4 2
2 9 3
....
....
It means input is given for 6 days.
Day 1: You get 4 shares, each with price 4$ and at-max you can sell 2 of them
Day 2: You get 2 shares, each with price 9$ and at-max you can sell 3 of them
.
We need to output the maximum profit which can be achieved.
I m thinking about how to go for this problem. It seems to me that if we apply brute force, it will take too much time. If this can be converted to some DP problem like 0-1 Knapsack? Some help will be highly appreciated.
it can be solved by DP
suppose there are n days, and the total number of stock shares is m
let f[i][j] means, at the ith day, with j shares remaining, the maximum profit is f[i][j]
obviously, f[i][j]=maximum(f[i-1][j+k]+k*price_per_day[i]), 0<=k<=maximum_shares_sell_per_day[i]
it can be further optimized that, since f[i][...] only depends on f[i-1][...], a rolling array can be used here. hence u need only to define f[2][m] to save space.
total time complexity is O(n*m*maximum_shares_sell_per_day).
perhaps it can be further optimized to save time. any feedback is welcome
Your description does not quite match the last problem in the PDF - in the PDF you receive the number of shares specified in the first column (or are forced to buy them - since there is no decision to make it does not matter) and can only decide on how many shares to sell. Since it does not say otherwise I presume that short selling is not allowed (else ignore everything except the price and go make so much money on the derivatives market that you afford to both bribe the SEC or congress and retire :-)).
This looks like a dynamic program, where the state at each point in time is the total number of shares you have in hand. So at time n you have an array with one element for each possible number of shares you might have ended up with at that time, and in that element you have the maximum amount of money you can make up to then while ending up with that number of shares. From this you can work out the same information for time n+1. When you reach the end, then all your shares are worthless so the best answer is the one associated with the maximum amount of money.
We can't do better than selling the maximum amount of shares we can on the day with the highest price, so I was thinking: (this may be somewhat difficult to implement (efficiently))
It may be a good idea to calculate the total number of shares received so far for each day to improve the efficiency of the algorithm.
Process the days in decreasing order of price.
For a day, sell amount = min(daily sell limit, shares available) (for the max price day (the first processed day), shares available = shares received to date).
For all subsequent days, shares available -= sell amount. For preceding days, we binary search for (shares available - shares sold) and all entries between that and the day just processed = 0.
We might not need to physically set the values (at least not at every step), just calculate them on-the-fly from the history thus-far (I'm thinking interval tree or something similar).
I am currently working on writing an algorithm for my new site I plan to launch soon. The index page will display the "hottest" posts at the moment.
Variables to consider are:
Number of votes
How controversial the post is (# between 0-1)
Time since post
I have come up with two possible algorithms, the first and most simple is:
controversial * (numVotesThisHour / (numVotesTotal - numVotesThisHour)
Denom = numVotesTuisHour if numVotesTotal - numVotesThisHour == 0
Highest number is hottest
My other option is to use an algorithm similar to Reddit's (except that the score decreases as time goes by):
[controversial * log(x)] - (TimePassed / interval)
x = { numVotesTotal if numVotesTotal >= 10, 10 if numVotesTotal < 10
Highest number is hottest
The first algorithm would allow older posts to become "hot" again in the future while the second one wouldn't.
So my question is, which one of these two algorithms do you think is more effective? Which one do you think will display the truly "hot" topics at the moment? Can you think of any advantages or disadvantages to using one over the other? I just want to make sure I don't overlook anything so that I can ensure the content is as relevant as possible. Any feedback would be great! Thanks!
Am I missing something. In the first formula you have numVotesTotal in the denominator. So higher number of votes all time will mean it will never be so hot even if it is not so old.
For example if I have two posts - P1 and P2 (both equally controversial). Say P1 has numVotesTotal = 20, and P2 has numVotesTotal = 1000. Now in the last one hour P1 gets numVotesThisHour = 10 and P2 gets numVotesThisHour = 200.
According to the algorithm, P1 is more famous than P2. It doesn't make sense to me.
I think the first algorithm relies too heavily on instantaneous trend. Think of NASCAR, the current leader could be going 0 m.p.h. because he's at a pit stop. The second one uses the notion of average trend. I think both have their uses.
So for two posts with the same total votes and controversial rating, but where posts one receives 20 votes in the first hour and zero in the second, while the other receives 10 in each hour. The first post will be buried by the first algorithm but the second algorithm will rank them equally.
YMMV, but I think the 'hotness' is entirely dependent on the time frame, and not at all on the total votes unless your time frame is 'all time'. Also, it seems to me that the proportion of all votes in the relevant time frame, rather than the absolute number of them, is the important figure.
You might have several categories of hot:
Hottest this hour
Hottest this week
Hottest since your last visit
Hottest all time
So, 'Hottest in the last [whatever]' could be calculated like this:
votes_for_topic_in_timeframe / all_votes_in_timeframe
if you especially want a number between 0 and 1, (useful for comparing across categories) or, if you only want the ones in a specific timeframe, just take the votes_for_topic_in_timeframe values and sort into descending order.
If you don't want the user explicitly choosing the time frame, you may want to calculate all (say) four versions (or perhaps just the top 3), assign a multiplier to each category to give each category a relative importance, and calculate total values for each topic to take the top n. This has the advantage of potentially hiding from the user that no-one at all has voted in the last hour ;)
I'm trying to develop a rating system for an application I'm working on. Basically app allows you to rate an object from 1 to 5(represented by stars). But I of course know that keeping a rating count and adding the rating the number itself is not feasible.
So the first thing that came up in my mind was dividing the received rating by the total ratings given. Like if the object has received the rating 2 from a user and if the number of times that object has been rated is 100 maybe adding the 2/100. However I believe this method is not good enough since 1)A naive approach 2) In order for me to get the number of times that object has been rated I have to do a look up on db which might end up having time complexity O(n)
So I was wondering what alternative and possibly better ways to approach this problem?
You can keep in DB 2 additional values - number of times it was rated and total sum of all ratings. This way to update object's rating you need only to:
Add new rating to total sum.
Divide total sum by total times it was rated.
There are many approaches to this but before that check
If all feedback givers treated at equal or some have more weight than others (like panel review, etc)
If the objective is to provide only an average or any score band or such. Consider scenario like this website - showing total reputation score
And yes - if average is to be omputed, you need to have total and count of feedback and then have to compute it - that's plain maths. But if you need any other method, be prepared for more compute cycles. balance between database hits and compute cycle but that's next stage of design. First get your requirement and approach to solution in place.
I think you should keep separate counters for 1 stars, 2 stars, ... to calcuate the rating, you'd have to compute rating = (1*numOneStars+2*numTwoStars+3*numThreeStars+4*numFourStars+5*numFiveStars)/numOneStars+numTwoStars+numThreeStars+numFourStars+numFiveStars)
This way you can, like amazon also show how many ppl voted 1 stars and how many voted 5 stars...
Have you considered a vote up/down mechanism over numbers of stars? It doesn't directly solve your problem but it's worth noting that other sites such as YouTube, Facebook, StackOverflow etc all use +/- voting as it is often much more effective than star based ratings.