Forward and Backward Reasoning

Forward and Backward Reasoning

• In forward reasoning, reasoning proceeds forward, beginning with factor, chaining through rules and finally establishing the goal.

• When the left side of a sequence of rules is instantiated first and the rules are executed from left to right the process is called forward chaining/reasoning. This is also known as data-driven search, since, input data are used to guide the direction of the inference process. For example, one can chain forward to show that when a student is encouraged, is healthy, and has goals, the student will succeed.

ENCOURAGED (student) → MOTIVATED (students)

MOTIVATED (student) and HEALTHY (student) → WORKHARD (student)

WORKHARD (student) and HASGOALS (student) → EXCELL (student)

EXCELL (student) → SUCCEED (student)

• On the other hand, when the right side of the rule is instantiated first, the left-hand conditions become subgoals. These subgoals may in turn cause sub-subgoals to be established, and so on until facts are found to match the lowest subgoal conditions. When this form of inference takes place, it is said that backward chaining is performed. This form of inference is also known as goal-driven inference since an initial goal establishes the backward direction of the inferring.

For example, in MYCIN the initial goal in a consultation is "Does the patient have a certain disease?" This causes subgoals to be established such as "are certain bacteria present in the patient?" Determining if certain bacteria are present may require such things as tests on cultures taken from the patient. This process of setting up subgoals to confirm a goal continues until all the subgoals are eventually satisfied or fail. If satisfied, the backward chain is established thereby confirming the main goal.

• Some systems use both forward and backward chaining/reasoning, depending on the type of problem and the information available. Likewise rules may be tested or exhaustively or selectively, depending on the control structure.

Solved Example

Example 7.5.1 Consider an incandescent bulb manufacturing unit. Here machines M₁, M₂ and M₃ make 30 %, 30 % and 40% of the total bulbs of their, output, let's assume that 2 %, 3 % and 4 % are defective. A bulb is drawn at random and is found defective. What is the probability that the bulb is made by machine M₁ or M₂ or M₃.

Solution:

• Let E₁, E₂ and E₃ be the events that a bulb selected at random is made by machine M₁, M₂ and M₃.

• Let Q denote that it is defective.

Prob (E₁) = 0.3

Prob (E₂) = 0.3 and Prob (E₃) = 0.4 (given data),

These represent the prior probabilities.

• Probability of drawing a defective bulb made by M₁ = Prob (Q/E₁) = 0.02

• Probability of drawing a defective bulb made by M₂ = Prob (Q/E₂) = 0.03

• Probability of drawing a defective bulb made by M₃ = Prob (Q/E₃) = 0.04

•These values are the posterior probabilities

Therefore,

Prob (E₁/Q) = Prob (E₁/) * Prob (Q/E₁)/ Σ³_i=1 Prob (E_i) * Prob (Q/E_i)

= 0.3* 0.02/ (0.03* 0.2) + (0.03* 0.3) + (0.04 * 0.4)

= 0.1935

Similarly,

Prob (E₂/Q) = 0.3* 0.03/ (0.03* 0.2) + (0.03* 0.3) + (0.04* 0.4)

= 0.2903

Prob (E₃/Q) = (1-(Prob(E₁/Q) + Prob(E₂/Q)))

= (1-((0.1935) (0.2903)))