I encountered an Interview Question:
There is an event in the auditorium and Given capacity of the auditorium is NxM.
Every group of person booked ticket and all the tickets are booked, Now you have to assign seat number to all of them such that minimum number of group split.
So basically a 2-D array is given and we have some groups of certain size(different groups may be of different size).Array needs to be completely filled with minimum number of groups split.
One Brute force Recursive approach I found is :Place first group ,then second group and so on.Permute this arrangement to find the arrangement with minimum split.
One efficient solution I found was using subset sum problem.
https://en.wikipedia.org/wiki/Subset_sum_problem
I could not understand how subset sum problem can be used to solve this problem.
Please suggest how can I approach this problem.I am not looking for code,just psuedo-code or algorithm will suffice.
Firstly, I'm assuming that "group-split" means that some part of the group is in one row and remaining is in another. If the number of seats in a row are N, and given a set which contains the size of the different groups, you need to find a subset that will sum to N. If such a subset is found, that row will be filled without breaking any groups. If there is no such subset found, then you will need to break at least 1 group. Then there can be multiple strategies here.
1) You can pick a group that will be split across 2 rows. This group can be the largest of the remaining, or the smallest or can be picked at random. Once this group is decided, you have 2 rows with less than N empty seats that need to be filled recursively.
2) The strategy can be to find a subset that sums to 2*N - if found, 1 group will be split. If not found, then find a subset that sums to 3*N with 2 group-splits and so on. The maximum number of group-splits will be M-1 for M rows.
Continue 1) or 2) to fill M number of rows in the theatre.
Related
Given a list K of groups of items, each group possessing some fixed number of items, what would be a time efficient algorithm to generate all (full) permutations of items from said groups?
Example: If K = {A:3, B:2, C:4}, then these are three valid permutations:
AAABBCCCC
ABACCBCCA
CBACCBCAA
There seems to be some relation to Gray code, but with the added weights I'm not sure how to extend it. Trying to do better than monotonously increasing the overall permutation value and resetting on each iteration.
You can have a look at "13.2 Permutations of a multiset"
here https://jjj.de/fxt/fxtbook.pdf
I have an array of integers. Lets denote it by A. There is another array W containing weights associated with each entry of A.
In other words, associated with each entry A[i] is a certain weight entry W[i]. (Note that the weights are out of a limited set S_w = {w1,w2,w3,w4} so only few possible values)
The problem statement is as follows: Pick a random number of entries out of A such that when summed together, give you the highest value (SUM_A) under the constraint that the sum of their respective weights (SUM_W) doesn't exceed a threshold, W_threshold.
One possibility is brute force: Compute all permutations of A and for each permutation, select first n entries such that their sum weight SUM_W doesn't exceed W_threshold. Finally, the permutation that gives the maximum SUM_A shall be selected. But the complexity of this scheme is very high due to permutation computation step since the length of A is not constrained.
One other (sub-optimal) possibility is to sort A in descending order and then select first n entries such that their SUM_W doesn't exceed W_threshold. This has lower complexity but the result would be suboptimal.
Could someone give me tips if their already exists an algorithm to resolve the above stated problem? Or if anyone has ideas better than the ones I described above. Many thanks
There are n (n < 1000) groups of friends, with the size of the group being characterized by an array A[] (2 <= A[i] < 1000). Tables are present such that they can accommodate r(r>2) people at a time. What is the minimum number of tables needed for seating everyone, subject to the constraint that for every person there should be another person from his/her group sitting at his/her table.
The approach I was thinking was to break every group into sizes of twos and threes and try to solve this problem, but there are many ways of dividing a number n into groups of twos and threes and not all of them may be optimal.
Does a Mixed Integer Programming model count?
Some notes on this formulation:
I used random data to form the groups.
x(i,j) is the number of people of group i sitting at table j.
x(i,j) is a semi-integer variable, that is: it is an integer variable with values zero or between LO and UP. Not all MIP solvers offer semi-continuous and semi-integer variables but it may come handy. Here I use it to enforce that at least 2 persons from the same group need to sit at a table. If a solver does not offer these type of variables, we can formulate this construct using additional binary variables as well.
y(j) is a binary variable (0 or 1) indicating if a table is used.
the capacity equation is somewhat smart: if a table is not used (y(j)=0) its capacity is reduced to zero.
the option optcr=0 indicates we want to solve to optimality. For large, difficult problems we may want to stop say at 5%.
the order equation makes sure we start filling tables from table 1. This also reduces the symmetry of the problem and may speed up solution times.
the above model (with 200 groups and 200 potentially used tables) generates a MIP problem with 600 equations (rows) and 40k variables (columns). There are 37k integer variables. With a good MIP solver we find the proven optimal solution (with 150 tables used) in less than a minute.
Notice this is certainly not a knapsack problem (as suggested in another answer -- a knapsack problem has just one constraint) but it resembles a bin-packing problem.
It is same problem as knapsack problem which is NP complete (see https://en.wikipedia.org/wiki/Bin_packing_problem ). So finding optimal solution is pretty hard.
A heuristic that works most of the time:
Sort the groups according decreasing size.
For each group put it in the table that has least amount of space, but still can accommodate this group.
Your approach is workable. If a solution exists for a given number of tables, then a solution exists where you've split every group into some number of twos and some number of threes. First, split a three off of every group of odd size. You're left with a bunch of groups of even size. Next, split twos off of every group whose size isn't divisible by six. And forget that it's one bigger group; split it into a bunch of groups of six.
At this point, you have split all of your groups into some number of twos, some number of threes, and some number of sixes. Give each table of odd size one three, splitting sixes as necessary; now all tables have even size. All remaining sixes can now be split into twos and seated arbitrarily.
I have this homework assignment that I think I managed to solve, but am not entirely sure as I cannot prove my solution. I would like comments on what I did, its correctness and whether or not there's a better solution.
The problem is as follows: we have N groups of people, where group ihas g[i]people in it. We want to put these people on two rows of S seats each, such that: each group can only be put on a single row, in a contiguous sequence, OR if the group has an even number of members, we can split them in two and put them on two rows, but with the condition that they must form a rectangle (so they must have the same indices on both rows). What is the minimum number of seats S needed so that nobody is standing up?
Example: groups are 4 11. Minimum S is 11. We put all 4 in one row, and the 11 on the second row. Another: groups are 6 2. We split the 6 on two rows, and also the two. Minimum is therefore 4 seats.
This is what I'm thinking:
Calculate T = (sum of all groups + 1) / 2. Store the group numbers in an array, but split all the even values x in two values of x / 2 each. So 4 5 becomes 2 2 5. Now run subset sum on this vector, and find the minimum value higher than or equal to T that can be formed. That value is the minimum number of seats per row needed.
Example: 4 11 => 2 2 11 => T = (15 + 1) / 2 = 8. Minimum we can form from 2 2 11 that's >= 8 is 11, so that's the answer.
This seems to work, at least I can't find any counter example. I don't have a proof though. Intuitively, it seems to always be possible to arrange the people under the required conditions with the number of seats supplied by this algorithm.
Any hints are appreciated.
I think your solution is correct. The minimum number of seats per row in an optimal distribution would be your T (which is mathematically obvious).
Splitting even numbers is also correct, since they have two possible arrangements; by logically putting all the "rectangular" groups of people on one end of the seat rows you can also guarantee that they will always form a proper rectangle, so that this condition is met as well.
So the question boils down to finding a sum equal or as close as possible to T (e.g. partition problem).
Minor nit: I'm not sure if the proposed solution above works in the edge case where each group has 0 members, because your numerator in T = SUM ALL + 1 / 2 is always positive, so there will never be a subset sum that is greater than or equal to T.
To get around this, maybe a modulus operation might work here. We know that we need at least n seats in a row if n is the maximal odd term, so maybe the equation should have a max(n * (n % 2)) term in it. It will come out to max(odd) or 0. Since the maximal odd term is always added to S, I think this is safe (stated boldly without proof...).
Then we want to know if we should split the even terms or not. Here's where the subset sum approach might work, but with T simply equal to SUM ALL / 2.
say i have a list of 100 numbers, i want to split them into 5 groups which have the sum within each group closest to the mean of the numbers.
the simplest solution is to sort the hundred numbers and take the max number and keep adding the smallest numbers until the sum goes beyond the avg.
obviously that is not going to bring the best results. i guess we could use BFS or DFS or some other search algo. like A* to get the best result.
does anyone have a simple solution out there? pseudo code is good enough. thanks!
This sounds like a variation on the knapsack problem and if I'm interpreting you correctly, it may be the multiple knapsack problem. Can't you come up with an easy problem? :)
An efficient algorithm (solution) that can be used is a variation of the Best Fit of bin packing algorithm. However, we would have to be applying a variation in which we have specifically 5 different groups of numbers that we want rather than looking for using the smallest number of groups.
The algorithm begins by finding the mean of the list of all 100 numbers. This mean will be used as the max capacity for all 5 of the groups (bins) we are trying to fit numbers into. We then find the largest number in our list of 100 numbers that doesn’t exceed the max capacity of our group and assign it to our first group. (We can find this in log(n) time because we can use a self-balancing binary search tree). We keep track of the how filled our current group is. We then find the next largest number that fits into our current group until we have either reached max capacity or there are no other numbers that will allow this group to reach max capacity. In either of these cases we move on to the next group and repeat our algorithm with the numbers left in our list of numbers. Once we move on from a group we must also keep track of the current sum of that group. We continue until we have reached the highest capacity for all 5 groups. If there are any numbers left we place them in the groups that have the lowest total sum (because we kept track of these sums as we went).
This is in fact a greedy algorithm with a Θ(nlogn) running time due to the nature of bin packing.