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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.

SHB
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  • Huh. The more I think about this, the more interesting edge cases I think of. Namely, r=3 (always solvable edge case) and r=2 (only sometimes solvable). That being said, I think it's still not incredibly difficult. – Mooing Duck Dec 09 '15 at 23:32
  • @MooingDuck Making an edit, r is always greater than 2. – SHB Dec 09 '15 at 23:37
  • @SHB This problem is identical to, "How many empty seats must there be in a minimal arrangement of tables?" That one might be easier to think through, because there are a lot of ways to arrange people at tables but few to force yourself to have many tables with empty spaces. – btilly Dec 09 '15 at 23:40
  • @btilly: That's true if `r` is large. For edge cases like `r==3`, they're _very_ different. – Mooing Duck Dec 09 '15 at 23:44
  • @MooingDuck, for r==3 each table will either have 0 empty chair or 1 empty chair. But i am not sure whether breaking groups into sizes of twos and threes is the best idea. – SHB Dec 09 '15 at 23:47
  • @MooingDuck No, they are completely equivalent. The number of people seated is `r` times the number of tables minus the number of empty seats. – btilly Dec 10 '15 at 17:07
  • @btilly: Upon review, I misread your original claim. I thought you were simply saying `sum(A)/r`. My bad. You're correct, what you actually said is identical. – Mooing Duck Dec 10 '15 at 20:18
  • Which programming contest is this question from? – tmyklebu Dec 11 '15 at 06:18
  • @tmyklebu ACM ICPC 2012-2013, NEERC, Southern Subregional Contest. Question A in http://codeforces.com/gym/100109/attachments/download/1283/20122013-acmicpc-neerc-southern-subregional-contest-en.pdf . – SHB Dec 12 '15 at 00:16

3 Answers3

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Does a Mixed Integer Programming model count?

enter image description here

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.
Erwin Kalvelagen
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  • I am not much familiar with integer programming, so cannot exactly comment whether this would work or not. According to me the intended solution for the question would not be making use of integer programming. – SHB Dec 10 '15 at 17:51
  • The formulation above is close but doesn't solve your exact problem, @SHB. I'm pretty sure you will have to resort to integer programming though, because you're asking specifically for "the minimum number of tables" subject to a couple of constraints. It's gonna be difficult to determine a deterministic algorithm that'll get you a good answer, let alone the optimal answer. – Koen Peters Dec 11 '15 at 14:03
  • @KoenPeters what did I miss in my formulation? – Erwin Kalvelagen Dec 11 '15 at 14:07
  • My bad, I didn't realise you were using _semi_-integer variables when I first looked at your formulation. – Koen Peters Dec 11 '15 at 15:20
  • @KoenPeters I added some sentences on this. It is a bit of an esoteric variable type. – Erwin Kalvelagen Dec 11 '15 at 15:50
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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:

  1. Sort the groups according decreasing size.

  2. For each group put it in the table that has least amount of space, but still can accommodate this group.

ElKamina
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  • This question was asked in some European ICPC regionals, so I am pretty sure that with the given constraints there should be a deterministic solution for this. – SHB Dec 10 '15 at 11:56
  • I don't see immediately how this is a knapsack problem. – Erwin Kalvelagen Dec 10 '15 at 13:43
  • @ErwinKalvelagen See bin-packing problem https://en.wikipedia.org/wiki/Bin_packing_problem – ElKamina Dec 10 '15 at 16:57
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    If we have a single table it would be a knapsack problem. But here we want to minimize the number of tables needed to seat all the persons. – Erwin Kalvelagen Dec 10 '15 at 17:30
  • @ErwinKalvelagen Let me give you a simple instance of the problem. Say you have n groups with total 2m people and capacity of each table is m. Asking if you can fit them in two tables (which is NP complete), is same as asking what is the minimum number of tables. If we could solve the original problem in polynomial time, we could solve the latter problem in polynomial time, which implies we could solve the NP complete problem in polynomial time which is impossible. – ElKamina Dec 11 '15 at 03:56
  • @ElKamina: "X can be solved as an instance of an NP-complete problem" does not mean that X itself is NP-complete. You can formulate this problem as a bin-packing problem or as a knapsack problem or however else you like; however, there is no requirement for solutions of this problem to make use of your formulation. – tmyklebu Dec 11 '15 at 06:28
  • What I am saying is problem X, which is NP complete, can be solved trivially if we solve another problem Y (which is OP's problem) in polynomial time. Anyways, if you have a polynomial solution to this problem I would be very interested. – ElKamina Dec 11 '15 at 08:26
  • @ElKamina: You did not describe how to solve the knapsack problem given a polynomial-time solution to this problem. – tmyklebu Dec 12 '15 at 03:43
  • @ElKamina: You haven't. Your second comment here is not such a transformation. You can split the groups in this problem, just not into singles. You also did not account for the bound of 1000 on the size of each group and on the size of the tables. – tmyklebu Dec 16 '15 at 03:31
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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.

tmyklebu
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  • This is a great way to deal with the pairing constraint, but doesn't take into account that the OP is looking for a way to get the _minimal_ number of tables. – Koen Peters Dec 11 '15 at 14:06
  • @KoenPeters: Whoops. I obviously didn't read the statement carefully. I read it as a partitioning problem, not a packing problem. – tmyklebu Dec 12 '15 at 03:42