Part 7: Table Constraints

We ask you not to publish your solutions on a public repository. The instructors interested to get the source-code of the solutions can contact us.

Theoretical questions

Table Constraint

The table constraint (also called extension constraint) specify the list of solutions (tuples) assignable to a vector of variables.

More precisely, given an array X containing n variables, and an array T of size m*n, this constraint holds:

\exists i: \forall\ j\ T_{i,j} = X_j

That is, each line of the table is a valid assignment to X.

Here is an example of a table, with five tuples and four variables :

Tuple index X[0] X[1] X[2] X[3]
1 0 1 2 3
2 0 0 3 2
3 2 1 0 3
4 3 2 1 2
5 3 0 1 1

In this particular example, the assignment X={2, 1, 0, 3} is then valid, but not X={4, 3, 3, 3} as there are no such line in the table.

Many algorithms exists to filter table constraints.

One of the fastest filtering algorithm nowadays is Compact Table (CT) [CT2016]. In this exercise you’ll implement a simple version of CT.

CT works in two steps:

  1. Compute the list of supported tuples. A tuple T[i] is supported if, for each element j of the tuple,
the domain of the variable X[j] contains the value T[i][j].
  1. Filter the domains. For each variable x[j] and value v in its
domain, the value v can be removed if it’s not used by any supported tuple.

Your task is to terminate the implementation in

TableCT maintains for each pair variable/value the set of tuples the pair maintains as an array of bitsets:

private BitSet[][] supports;

where supports[j][v] is the (bit)set of supported tuples for the assignment x[j]=v.


As an example, consider that variable x[0] has domain {0, 1, 3}. Here are some values for supports: supports[0][0] = {1, 2} supports[0][1] = {} supports[0][3] = {4,5}

We can infer two things from this example: first, value 1 does not support any tuples, so it can be removed safely from the domain of x[0]. Moreover, the tuples supported by x[0] is the union of the tuples supported by its values; we immediately see that tuple 3 is not supported by x[0] and can be discarded from further calculations.

If we push the example further, and we say that variable x[2] has domain {0, 1}, we immediately see that tuples 1 and 2 are not supported by variable x[2], and, as such, can be discarded. From this, we can infer that the value 0 can be removed from variable x[0] as they don’t support any tuple anymore.

Using bit sets

You may have assumed that the type of supports would have been List<Integer>[][] supportedByVarVal. This is not the solution used by CT.

CT uses the concept of bit sets. A bit set is an array-like data structure that stores bits. Each bit is accessible by its index. A bitset is in fact composed of an array of Long, that we call in this context a word. Each of these words store 64 bits from the bitset.

Using this structures is convenient for our goal:

  • Each supported tuple is encoded as a 1 in the bitset. 0 encodes unsupported tuples. In the traditional list/array representation, each supported tuple would have taken 32 bits to be represented.
  • Doing intersection and union of bit sets (and these are the main operation that will be made on supportedByVarVal) is very easy, thanks to the usage of bitwise operators included in all modern CPUs.

Java provides a default implementation of bit sets in the class BitSet, that we will use in this exercise. We encourage you to read its documentation before going on.

A basic implementation

You will implement a version of CT that makes no use of the reversible structure (therefore it is probably much less efficient that the real CT algo).

You have to implement the propagate() method of the class TableCT. All class variables have already been initialized for you.

You “simply” have to compute, for each call to propagate():

  • The tuples supported by each variable, which are the union of the tuples supported by the value in the domain of the variable
  • The intersection of the tuples supported by each variable is the set of globally supported tuples
  • You can now intersect the set of globally supported tuples with each variable/value pair in supports. If the value supports no tuple (i.e. the intersection is empty) then it can be removed.

Make sure you pass all the tests

[CT2016]Demeulenaere, J., Hartert, R., Lecoutre, C., Perez, G., Perron, L., Régin, J. C., & Schaus, P. (2016, September). Compact-table: Efficiently filtering table constraints with reversible sparse bit-sets. In International Conference on Principles and Practice of Constraint Programming (pp. 207-223). Springer.

Eternity Problem

Fill in all the gaps in order to solve the Eternity II problem.

Your task is to terminate the implementation in

  • Create the table
  • Model the problem using table constraints
  • Search for a feasible solution using branching combinators

Compact table algorithm for table constraints with short tuples


Of course you should get a strong inspiration from the implementation you did in a previous exercise.

Check that your implementation passes the tests