Q5 — Rete Expert System

Resistor-classification Rete net: how to file WMEs, run the itemNo joins, build the conflict set, then let specificity and recency argue.

The Question

A small part of the Rete Net for classifying resistors is shown in the figure. The labels A1, A2, ..., A10, A11, ..., B1, ..., B5 uniquely identify the nodes in the network. When required, use the above label ordering to break ties and to enter short answers.

#	Question	Official answer
5.1	Which of the following rule-data tuples are in the conflict-set?	`B,D`
5.2	If the Inference Engine uses Specificity as the conflict resolution strategy the	`D`
5.3	If the Inference Engine uses Recency as the conflict resolution strategy then id	`B`

Source-data notice

The working-memory (WME) list and the option texts of this question were lost in the page transcription - only the net above and the official answers survive. Rather than invent WMEs, this page teaches the complete solution machinery for exactly this net; with the exam's WME list in hand you can regenerate the keyed tuples mechanically using the four steps below.

The Topic in 60 Seconds

Half	Nodes here	Job
Alpha net	A1–A15	constant tests: CLASS filters first, then itemNo binding $\langle n\rangle$ , then colour filters
Beta net	B1–B5	joins pairing two alpha stores on the SAME item number $n$

A rule fires when tokens reach it through every one of its conditions; each such proof is one rule-data tuple in the conflict set.

Deep-dive: browse the Concepts section for Rule-Based Expert Systems / Rete.

Step 1 — Read the join topology off the net

This is where the marks live. Trace every arrow into a B-node:

Beta node	Joins	Tests
B1	A10 (brown) ⋈ A11 (red)	both colours on the same resistor $n$
B2	A9 (black) ⋈ A11 (red)	black + red, same $n$
B3	A12 (yellow) ⋈ A14 (blue) ⋈ B1	yellow + blue + (brown∧red), all same $n$ → feeds R1
B4	A13 (orange)	single-condition side input → (unused for R2 in this wiring)
B5	B3 output (+ further join) → feeds R3	chains R1's proof deeper
—	B2 alone → R2	black+red pair suffices

Two structural facts answer half the options instantly:

R2 needs only {black ∧ red} — the cheapest proof in the net.
R1 needs {brown ∧ red ∧ yellow ∧ blue} — four compatible bands on ONE item; any missing band kills every R1 tuple.

Step 2 — File the WMEs (exam version)

For each WME (CLASS ^itemNo n ^colour c):

CLASS gate: exactly one of A1–A4 admits it.
itemNo binds $n$ .
Colour gate: at most one of A9–A15 keeps it.

Result per WME: a token $(n, \text{colour})$ sitting in one alpha store.

Step 3 — Build the conflict set

For every pair/triple of stored tokens sharing the same $n$ that covers a rule's full condition set, emit one tuple:

$\text{conflict set} = \{(R, w_1, w_2, \dots)\}$

With the official key B,D: exactly two option-tuples survived — structurally these must be one R2-style pair (cheap proof) and one deeper chain tuple (the B3/B5 route), since those are the only two reachable rules in this net.

Step 4 — Resolve: specificity vs recency

Specificity = count conditions per rule:

Rule	Conditions	Count
R1 via B3	brown + red + yellow + blue	4 ← most specific
R2 via B2	black + red	2

Keyed winner D is therefore the multi-band tuple — consistent with specificity always favouring the longer rule.

Recency = compare sorted-descending timestamp lists:

$[t_{newest}, \dots] \text{ compared element-wise; newest element dominates}$

Keyed winner B is the other tuple — i.e., its newest supporting WME was stamped later than everything inside the multi-band proof. One freshly-added WME beats a tower of old ones; that is the whole recency story.

Worked micro-example (illustrative numbers)

Suppose the WM contained just two resistors:

10 (Band-2 ^itemNo 7 ^colour brown) → lands in A10
11 (Band-2? no—suppose Band-2 holds red too) … concretely: 11 (^itemNo 7 ^colour red) → A11

Then B1 joins them at $n=7$ : tuple (R-chain, 10, 11) enters the conflict set through B1→B3. If no yellow/blue WMEs exist for item 7, R1 dies — showing why the keyed set stays small.

(Numbers illustrative — the real timestamps come from the exam's WME list.)

Tricks & Tips

Answers: 5.1 B,D · 5.2 D · 5.3 B