Before I talk about the theory behind RCC-8, I think an example is in order. If you can’t wait for my next write-up on RCC-8 go ahead and check out the wikipedia page. I’m also going to borrow the example from the page, but I’m going to show how the path-consistency algorithm works and how the composition table is used.

Consider two houses with a road connecting them. Each house is on its own property.  The first house (h1) may or may not be touching the boundary of its property (p1) , the second house (h2)  definitely does not touch the boundary. What can we say about the second property (p2)  in regards to the road (r)?

Let’s start with what we know:

• {DC}(h1, h2)
• {TPP, NTPP}(h1, p1)
• {DC, EC}(h1, p2)
• {EC}(h1, r)
• {DC, EC}(h2, p1)
• {NTPP}(h2, p2)
• {EC}(h2, r)
• {EC, DC, PO, EQ, TPP, TPPi, NTPP, NTPPi}(r, p1)
• {EC, DC, PO, EQ, TPP, TPPi, NTPP, NTPPi}(r, p2)
• {DC, EC}(p1, p2)

The composition table can be used to find more facts about our problem. R(a, b);S(b,c) can be used to find T(a,c) (we use ; to indicate composition). To use the composition table find the entry for each element of the Cartesian product and take the union of all those. I think an example will make it clearer. Let R(a,b) = {DC, EC}(h1, p2) and S(b,c) = {DC,EC}(p2, p1). The inverse of DC = DC and the inverse of EC = EC, so if {DC, EC}(p1, p2) is true, then so is {DC, EC}(p2, p1) (the inverse of TPP/NTPP are TPPi/NTPPi). T(a,c) = {?, ?}(h1, p1).  The elements of the Cartesian product of {DC, EC}x{DC, EC} are {{DCxDC}, {DCxEC}, {ECxDC}, {ECxEC}}. Looking up those combinations in the composition table and taking the union of all is {*}, where * is the universal relation.  In this case we didn’t learn anything more, actually we already know more about the relation between h1 and p1.

What we really want is to find more information about (r, p2).  For that we can try {EC}(r, h2) and {NTPP}(h2, p2). The result is {PO, TPP, NTPP}(r, p2). Now we have greatly reduced the possibly relations for (r, p2) from eight to three.  We can continue to use the composition table to find new information about the relations.  Try using the compostion table yourself and see what other information about these relations you can find.

The next part to answering our query is using past-consistency. Given three relations R(xy), S(y,z) and T(x,z) we can find new information by applying path-consistency = {(R; S) intersect T} (x,z). That is we use the composition table to find (R; S) and then the intersection with T. Let’s look at an example. R(x, y)= {EC}(r, h2), S(y,z) ={NTPP}(h2, p2) and T(x, z) = {PO, TPP, NTPP}(r, p2). {R;S} = {PO, TPP, NTPP} intersect {PO, TPP, NTPP} = {PO, TPP, NTPP}. If the intersection is the empty set, you’ve found an inconsistency.

Your overall approach to using the composition table and applying path-consistency should be similar to this:

1. Create all useful combinations of possible relations using the composition table. For example, make sure that there is a relations for(h1, h2), (h1, p1), (h1, p2), (h1, r) …  You don’t need to write down both directions i.e. (h1, h2) and (h2, h1), but only need to remember the inverses of the eight relations.
2. Create and list all possible pairs of relations: (h1, h2)(h2,p1), (h1, h2)(h2, p2), (h1,h2)(h2, r) … and use path consistency to learn new information. Here is an example, (h2, p2)(p2, p1) intersect (h2, p1) = NTPP; {DC, EC} intersect {DC, EC} = {DC} intersect {DC, EC} = {DC}. Now we know that (h2, p1) is not {DC, EC}, but is really only {DC}.
3. Make a note of all relations which have changed due to step 2.
4. Continue until you reach the end of the list or find an inconsistency. If you reach the end without an inconsistency, then go back through the list and check all path-consistencies involving of the relations marked in step 3.
5. Repeat steps 2 – 4 until all relations remain unchanged.

For more mathematical details check out these lecture slides form the University of Leeds.