I'm trying to devise the most efficient algorithm for a problem, but I'm having some difficulty. If someone could lend a hand, either by proposing an algorithm or classifying the problem so I can do some further research, I would be very appreciative.
The problem is as follows:
There are n (an integer) number of distinct red balls, each of which has its own number, and m number of distinct green balls, each of which has its own corresponding number as well. For example, if n = 3, then there are three red balls named Red Ball 1, Red Ball 2 and Red Ball 3.
There are also two boxes in which the balls can be placed.
Before the balls are placed in the boxes however, x number of people make predictions as to which balls will be placed in which box (either box 1 or box 2). Each person gets one prediction and for each prediction they can guess one ball to be in each box. The only condition is that the ball they guess in box 1 cannot be the same color as the ball they guess to be in box 2. An example prediction would be: "I think that Red Ball 2 will be in box 1 and Green Ball 3 will be in box 2"
After everyone has made their predictions, the balls will be placed in the boxes the maximize the number of predictions that are correct.
The code I must write will be prompted with n, m, and x as well as the predictions and then be asked to return the maximum number of predictions that are correct.
Once again, I am looking for either algorithmic help or help to identify the type of problem this is. I currently have a recursive algorithm running on (n^2), but I need something a little more efficient.
Thanks for your help! Cheers, Mates!
Related
I came across below problem
There are N houses in a row. Each house can be painted in either Red,
Green or Blue color. The cost of colouring each house in each of the
colours is different. Find the colour of each house such that no two
adjacent houses have the same colour and the total cost of colouring
all the houses is minimum.
Here is the complete question
The problem to me looks bit confusing, as objective is to minmize the cost and ensure that no adjacent house have same colour. In that case should not i select
the just two colours out of three whose cost is minimum.
Say here is the cost of colours
Red $100
Green $200
Blue $300
I have 5 row houses to paint
Here will be my Algo
Select Red and Green colour as their cost is minimum
calculate 5%2.
If 5%2 == 1 then start from last and select the colour of last house as Red($100). Now choose the alternate colour
If 5%2 == 0 then start from beginning and choose the alternate colour
I see Is "house coloring with three colors" NP? suggested the dynamic programming But i am not sure what is wrong with my approach and why dynamic programming is required here ?
This is (presumably) a homework problem. I think the intention is to teach you to implement a greedy algorithm:
Paint the first house using the cheapest color ("red").
Paint the next house using the cheapest color available.
The second will alternate between "red" and "green", but you don't have to choose the colors in advance.
If I were assigning this problem, the next problem would be "And solve this assuming you have only enough paint for one red house".
I have a problem which sounds like this:
We have a bag with balls in. There are R red balls, B blue balls and G green balls.
I need to find the minimum number of extractions from the bag such that i am sure that i will have at least K balls of same colour.
Anyone can help with any idea? Or tips, etc?
If K>max(R,G,B) then the problem has no solution. So let's assume K <= max(R,G,B).
If you don't have any control over which ball gets extracted, you'll need at most (i.e. this is a lower bound) min(R, (K-1))+min(G, (K-1))+min(B, (K-1))+1 extractions for obvious reasons: extract K-1 red balls (or all red balls if R<K), then extract K-1 green balls (or all green balls if G<K), and finally extract K-1 blue balls (or all blue balls if B<K). Now there is at least one ball remaining---because max(R,G,B)>=K by assumption---and when we extract it we have K balls of the same color. (This is clearly the worst case.)
You obviously need at least K extractions (this is the best case).
Since you tagged the question with probability, maybe you can only extract a ball chosen uniformly at random. In that case we can talk of the expected number of extractions needed before we end up with K balls of the same color. This is an interesting question.
You need to make sure the upper bond working... When I test the upper bond, it blows off the int range (I using java)
I'm having a hard time finding an admissible heuristic for the Hungarian Rings puzzle. I'm planing on using IDA* algorithm to solve and am writing the program in Visual Basic. All I am lacking is how to implement the actual solving of the puzzle. I've implemented both the left and right rings into their own arrays and have functions that rotate each ring clockwise and counterclockwise. I'm not asking for code, just somewhere to get started is all.
Here is the 2 ring arrays:
Dim leftRing(19) As Integer
' leftRing(16) is bottom intersection and leftRing(19) is top intersection
Dim rightRing(19) As Integer
' rightRing(4) is top intersection and rightRing(19) is bottom intersection
In the arrays, I store the following as the values for each color:
Red value = 1 Yellow = 2 Blue = 3 and Black = 4
I suggest counting "errors" in each ring separately - how many balls need to be replaced to make the ring solved (1 9-color, 1 10-color, one lone ball from a 9-color). At most two balls can be fixed using a rotation, then another rotation is needed to fix another two. Compute the distance of each ring individually = 2n-1 where n is half the amount of bad positions and take the larger of them. You can iterate over all twenty positions when looking for one that has the least amount of errors, but I suppose there's a better way to compute this metric (apart from simple pruning).
Update:
The discussion with Gareth Reed points to the following heuristic:
For each ring separately, count:
the number of color changes. The target amount is three color changes per ring, and at most four color changes may be eliminated at a time. Credits go to Gareth for this metric.
the count of different colors, neglecting their position. There should be: 10 balls of one 10-color, 9 balls of one 9-color and one ball of the other 9-color. At most 2 colors can be changed at a time.
The second heuristic can be split into three parts:
there should be 10 10-balls and 10 9-balls. Balls over ten need to be replaced.
there should be only one color of 10-balls. Balls of the minor color need to be replaced.
there should be only one ball of a 9-color. Other balls of the color need to be replaced. If all are the same color, and 9-color is not deficient, one additional ball need to be replaced.
Take the larger of both estimates. Note that you will need to alternate the rings, so 2n-1 moves are actually needed for n replacements. If both estimates are equal, or the larger one is for the latest moved ring, add an additional one. One of the rings will not be improved by the first move.
Prune all moves that rotate the same ring twice (assuming a move metric that allows large rotations). These have already been explored.
This should avoid all large local minima.
I'm trying to come up with a reasonable algorithm for this problem:
Let's say you have a bunch of balls. Each ball has at least one color, but can also be multicolored. Each ball has a weight and a value associated with it. There are also a bunch of boxes which are each only one color. Each box has a maximum number of balls it can hold. The goal is to maximize the sum of the value in the boxes while staying under some total weight, W, and the only rule is:
In order to place a ball in a box, it has to at least have the box's color on it
(For example, you can put a blue and green ball into a blue box or a green box, but not into a red box.)
I've dome some research and this seems similar to the knapsack problem and also similar to being solvable by the Hungarian algorithm, but I can't quite seem to reduce it to either problem.
I'm just curious is there's some kind of dynamic programming algorithm for this type of problem to make it solvable in polynomial time, or if it's just the traveling salesman problem in disguise. Would it help if I knew there were at most X colors? Any help is greatly appreciated. I could also formalize the problem a bit with variable names if it would help. Thanks!
Here's a simple example:
Maximum weight: 5
Balls:
1 red ball - (value = 5, weight = 1)
1 blue ball - (value = 3, weight = 1)
1 green/red/blue ball - (value = 2, weight = 4)
1 green/blue ball - (value = 4, weight = 1)
1 red/blue ball - (value = 1, weight = 1)
Boxes:
1 red (holds 1 ball)
1 blue (holds 2 balls)
1 green (holds 1 ball)
Optimal Solution:
red ball in red box
blue ball and red/blue ball in blue box
green/blue ball in green box
Total value: 13 (5 + 3 + 1 + 4)
Total weight: 4 (1 + 1 + 1 + 1)
Note: even though the green/red/blue ball was more valuable than the red/blue ball, it's weight would have put us over the limit.
Edit:
One clarifying point: balls with the same color combination will not necessarily have the same weights and values. For example, you could have a red ball with value 3 and weight 1 and another red ball with value 2 and weight 5.
Edit 2:
We can assume integer weights and values if it helps us come up with a dynamic programming algorithm.
This is at least as complex as the Knapsack problem - consider a case where all balls are red and there is only one red box.
In the case when balls that have the same combination of colors must have the same weights and values consider a case when you have red/blue, red/green, etc. balls and only one red box.
If there is no bound on the number of boxes, then this problem is strongly NP-hard by reduction from 3-partition (set up n/3 boxes and make all the things rainbow-colored with value = weight).
If the number of boxes is constant, then there's a pseudo-polynomial time algorithm via dynamic programming, where each DP state consists of how full each box is.
The reduction from knapsack is as follows. Given your knapsack instance, you create an instance of the balls and bins problem: for each item of the knapsack instance you have a ball with the same weight and value as the item. Then you have a box representing the knapsack. The balls and the box are all blue. The capacity of the box is the limit given in the knapsack problem. Given a solution to your problem, we have a set of balls in the box whose total weight is at most the knapsack limit, and whose total value is maximised.
This problem is NP-complete, because it subsumes the knapsack problem.
That is, it's not just similar to the knapsack problem: if there is one bowl, all the balls have that bowl's color, and the maximum number of balls in the bowl is the total number of balls, then the problem is exactly the knapsack problem.
If an algorithm could solve this problem in polynomial time, it could solve any knapsack problem in polynomial time. But, since the knapsack problem is NP-complete, this problem is, too.
The best you can do in this situation is get an approximation of the optimal solution - the knapsack problem is not solvable in polynomial time itself. You may be able to still get good (although not guaranteed to be optimal) results in polynomial time if you can generate a good algorithm for it.
I'm trying to come up with a reasonable algorithm for this problem:
Let's say you have a bunch of balls. Each ball has at least one color, but can also be multicolored. Each ball also has a number on it. There are also a bunch of boxes which are each only one color. The goal is to maximize the sum of the numbers on the balls in the boxes, and the only rules are:
in order to place a ball in a box, it
has to at least have the box's color
on it
you can only put one ball in each
box.
For example, you can put a blue and green ball into a blue box or a green box, but not into a red box.
I've come up with a few optimizations that help a lot in terms of running time. For example, you can sort the balls in descending order of point value. Then as you go from highest number to lowest, if the ball only has one color, and there are no other higher-point balls that contain that color, you can put it in that box (and thus remove that box and that ball from the remaining combinations).
I'm just curious is there's some kind of dynamic algorithm for this type of problem, or if it's just the traveling salesman problem in disguise. Would it help if I knew there were at most X colors? Any help is greatly appreciated. Thanks!
Edit - here's a simple example:
Balls:
1 red ball - 5 points
1 blue ball - 3 points
1 green/red ball - 2 points
1 green/blue ball - 4 points
1 red/blue ball - 1 point
Boxes:
1 red
1 blue
1 green
Optimal Solution:
red ball in red box
blue ball in blue box
green/blue ball in green box
Total value: 12 points (5 + 3 + 4)
This is a special case of the maximum weight matching problem on a weighted bipartite graph. Construct a graph whose left vertices correspond to balls, whose right vertices correspond to boxes and with the edge joining a ball and a box having weight V where V is the number on the ball if the ball can be placed in the box, and 0 otherwise. Add extra boxes or balls joined to the other side with edges of weight zero until you have the same number of vertices on each side. The assignment you're looking for is determined by the set of edges of nonzero weight in the maximum (total) weight matching in the resulting graph.
The assignment algorithm can be solved in O(n^3) time, where n is here the maximum of the number of balls or boxes, using the Hungarian algorithm. (BTW, I should make the disclaimer that I only mention the Hungarian algorithm because it is the theoretical result I happen to be familiar with and it presumably answers the question in the title of whether the original problem is NP-hard. I have no idea whether it is the best algorithm to use in practice.)
Have you tried a greedy alg?
Sort by points/value and place in box if possible.
If there are any exceptions im missing id like to see them.