**References**:

- Z. Pawlak, "Rough Sets: Theoretical Aspects of Reasoning about Data," in
*Theory and Decision Library: Series D, vol. 9*, W. Leinfellner and G. Eberlein eds, Kluwer Academic Publishers, 1991. - Xiaohua Hu and Nick Cercone, "Learning in Relational Databases: A Rough Set Approach," in
*Computational Intelligence*, v.11, no.2 (1995), pp. 323-338.

__Terminology from Rough Set Theory__

__Information Systems__

Formally, an information system is a triple *S* = (*U*, *A*, *V*) where
*U* = {*x _{1},...,x_{n}*} is a (finite) set of objects (called the universe),
and

*A*is a set of attributes for which each object of the universe has a value (from the domain set

*V*). The set of attributes is partitioned into the condition attributes (C), and the decision attributes (D). In the context of machine learning, there is typically only one decision attribute, which takes a value of true/false to indicate positive examples and negative examples.

__Indiscernibility__

A pair of objects *x _{i}*,

*x*are indiscernible over a set of attributes

_{j}*B*iff they have the same value for every attribute of

*B*. Obviously, objects that are indiscernible over

*B*will also be indiscernible over any subset of

*B*. However, objects that were indiscernible over

*B*may become discernible over a superset of

*B*. In the context of machine learning, where we want to create simple rules, we will be interested in eliminating "redundant" attributes, which have little or no effect on the discernibility of objects.

__Approximation Spaces__

Indiscernibility over a set of attributes *B* defines an
equivalence relation on the universe of objects. The **lower approximation** of a target set *X*
(denoted *B _{*}(X)*) is the union of all the equivalence classes that are fully contained in

*X*. The

**upper approximation**of

*X*(denoted

*B*) is the union of all equivalence classes that intersect

^{*}(X)*X*. Clearly,

*B*.

_{*}(X) ⊆ X ⊆ B^{*}(X)Consider the information system presented in the following table:

Condition Attributes |
Decision Attributes |
||||

Name | Hair | Height | Weight | Lotion | Result |

Sarah | blonde | average | light | no | sunburned (positive) |

Dana | blonde | tall | average | yes | none (negative) |

Alex | brown | short | average | yes | none |

Annie | blonde | short | average | no | sunburned |

Emily | red | average | heavy | no | sunburned |

Pete | brown | tall | heavy | no | none |

John | brown | average | heavy | no | none |

Katie | blonde | short | light | yes | none |

In this case, the objects of the universe are people (Sarah, Dana, Alex, Annie, Emily, Pete, John, Katie).
Suppose we wish to describe the set *X* of people who did get a sunburn (*X* = {Sarah, Annie, Emily}).
When using the set of attributes *B = C* = {Hair, Height, Weight, Lotion}, we find that each equivalence class
has exactly one element, so we can define *X* precisely as
*B _{*}(X) = X = B^{*}(X) = [Sarah] ∪ [Annie] ∪ [Emily]*.
If, however, we consider only the attribute Lotion, then there are only two equivalence classes:

- [Sarah] = {Sarah, Annie, Emily, Pete, John}
- [Dana] = {Dana, Alex, Katie}

Observe that we now cannot define *X* precisely, because [Sarah] contains *X*. We can, however,
make the observation that [Dana] is disjoint from *X*, so we have the following:

*∅ = B _{*}(X) ⊂ X ⊂ B^{*}(X) = [Sarah] ⊂ U*.

In terms of rules, the above indicates that

**not**wearing lotion may (but not necessarily) lead to sunburn, and that wearing lotion prevents sunburn.

__Reduct of Attributes__

In order to simplify the learned rules, we will be interested in eliminating condition attributes.
First, we will look for **absolute reducts**, which are subsets *B* of the set *C* of
condition attributes such that *B* preserves the indiscernibility equivalence classes of *C*.
In the "sunburn" example above, we observe that taking *B* = {Height, Weight, Lotion} (*i.e.*
eliminating Hair) leaves every element of the universe distinct. Other absolute reducts include the following:

- {Hair, Height, Lotion}
- {Hair, Height, Weight}

Consider, however, that for our purposes, absolute reducts are too strict. That is, we do not care if we make
formerly discernible objects to become indiscernible, as long as they have the same value for the decision attribute(s).
For example, eliminating the attribute Height makes Pete and John indiscernible, but since neither of them got a sunburn,
we have not lost any ability to predict a sunburn. Thus, *B* = {Hair, Weight, Lotion} is a **relative reduct of C
with respect to D**. Notice that Height is in every absolute reduct, but it need not be in a relative reduct. Next,
we will discuss these concepts formally.

__Positive Regions and Dependency__

- Let
*S = (U, A, V)*be an information system, with*A = C ∪ D*and*B ⊆ C*. - Let
*IND(B)*denote the set of equivalence classes of*U*with respect to*B*. - Let
*IND(D)*denote the set of equivalence classes of*U*with respect to*D*.

Define the **positive region of B in IND(D)** as

*POS _{B}(D)* = ∪{

*B*:

_{*}(X)*X*∈

*IND(D)*}.

That is, *POS _{B}(D)* includes all objects that can be sorted into classes of

*IND(D)*without error, based solely on the classification information in

*IND(B)*.

Furthermore, we say that the set of attributes *D* **depends in degree k** on the subset

*R*of

*C*if

*k(R, D) = card(POS _{R}(D)) ÷ card(POS_{C}(D))*

Clearly, *k*(*R*, *D*) ≤ *k(C, D)* ≤ 1. Also, the following are equivalent to each other:

*k(R, D)*=*k(C, D)*.*R*is a reduct of*C*with respect to*D*.*POS*_{R}(D) = POS_{C}(D).

Also, since we have a numerical description of dependency, we can speak of near-perfect reducts (*i.e.* *k(R, D)* ≈ *k(C, D)*),
so that we may discard even more attributes if they have little effect on the dependency.

__Computing the Best Reduct__

In general, the problem of finding all reducts is NP-hard. Therefore, we will consider a greedy algorithm for finding the "best" reduct.

Now, it is known that there is a (possibly empty) subset of attributes, called the **core**, which is common
to all reducts. Furthermore, it can be shown that *CO = CORE(C, D) = {a ∈ C : POS _{C}(D) ≠ POS_{C-{a}}(D)}*. So, we test every attribute to
see if it belongs to the core, then we pass the core to the following algorithm:

- REDU = CO.
- AR = C - REDU.
- While k(REDU, D) < k(C, D) -
*margin*, do the following:- Find attribute a ∈ AR such that k(REDU ∪ {a}, D) is maximized.
- Set REDU = REDU ∪ {a}; AR = AR - {a}.

- Output REDU.

This algorithm is greedy because it hopes to minimize the number of attributes by choosing the attribute that creates the greatest increase in k at each step.

__Example__

Consider the "sunburn" data above. First, let us compute the core. Every object is discernible,
so *POS _{C}(D) = U*.

Attribute (a) | IND(C - {a}) | POS_{C - {a}}(D) | In Core? |

Hair | {Sarah}, {Dana}, {Alex}, {Annie}, {Emily, John}, {Pete}, {Katie} | U - {Emily, John} | Yes |

Height | {Sarah}, {Dana}, {Alex}, {Annie}, {Emily}, {Pete, John}, {Katie} | U | No |

Weight | {Sarah}, {Dana}, {Alex}, {Annie}, {Emily}, {Pete}, {John}, {Katie} | U | No |

Lotion | {Sarah}, {Dana}, {Alex}, {Annie}, {Emily}, {Pete}, {John}, {Katie} | U | No |

So, CORE(C, D) = CO = {Hair}, and POS_{CO}(D) = {Alex, Emily, Pete, John}, and k(CO, D) = 4/8 = 0.5.
Now we proceed with the algorithm.

Step | REDU | AR | a ∈ AR | k(REDU ∪ {a}, D) | Choice |

1 | {Hair} | {Height, Weight, Lotion} | Height | 0.5 | Lotion |

Weight | 0.5 | ||||

Lotion | 1.0 |

The algorithm terminates after one iteration, with the answer REDU = {Hair, Lotion}. The reduced table is as follows:

C | D | ||

Vote | Hair | Lotion | Sunburn |

2 | Blonde | No | Yes |

2 | Blonde | Yes | No |

1 | Brown | Yes | No |

2 | Brown | No | No |

1 | Red | No | Yes |

__Learning Rules__

Before generating rules, we want to combine similar tuples. Tuples are similar if they differ in only one attribute (call it
*a*). Ideally, the similar tuples, taken together, would have all possible values for *a* (known as saturation), but
this is not necessary. In the example, notice that Brown/Yes/No and Brown/No/No differ only in the Lotion attribute, and that
Lotion is saturated. Therefore, we can combine these into one tuple (Brown/?/No with vote = 3). Similarly, if we assume that
Blonde, Brown, and Red are the only hair colours, then we can combine Blonde/No/Yes with Red/No/Yes to get ¬Brown/No/Yes.
So the simplified table is as follows:

C | D | ||

Vote | Hair | Lotion | Sunburn |

3 | ¬Brown | No | Yes |

2 | Blonde | Yes | No |

3 | Brown | ? | No |

So, finally, we are ready to read the rules out of the table:

- (Hair ≠ Brown ∧ Lotion = No) ⇒ Sunburn = Yes
- (Hair = Blonde ∧ Lotion = Yes) ∨ (Hair = Brown) ⇒ Sunburn = No

Dealing with noisy data.

Comparison with Decision Trees.