Q5 — Rete Expert System

Mushroom net with negated colour tests and a three-way beta chain: the join topology alone decides which tuples may exist, specificity picks the deepest.

The Question

A part of the Rete Net that classifies mushrooms (as edible or poisonous) is shown in the figure. The labels A1, A2, ..., A10, A16, ..., B1, B2, B3, R1, …, R4 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?	`A,B,C,D`
5.2	If the Inference Engine uses Specificity as the conflict resolution strategy the	`C,D`
5.3	If the Inference Engine uses Recency as the conflict resolution strategy then wh	`A`

Source-data notice

The working-memory list and option texts were lost in transcription — only the net and official answers survive. The complete machinery for THIS net follows; with the exam's WMEs the keyed tuples regenerate mechanically in four steps.

Step 0 — Read the rule structure off the net

Follow every arrow into rules/betas:

Rule	Fed by	Full condition set (same $n$ throughout)
R1	A14	Mushroom + odour NONE + sp-colour ≠ GREEN — single-WME chain
R2	B1	Stalk + br-surface SCALY ⋈ root BULBOUS
R3	B2	Mushroom + odour NONE ⋈ (habitat-LEAVES + ar-colour ≠ BROWN)
R4	B3	Cap + WHITE ⋈ SMOOTH (+ RED via A16 hangs off A13's branch)

Structural facts that pre-answer options:

R1 needs no join — any mushroom passing odour-NONE with sp-colour present-and-not-GREEN activates it alone. Negated tests still demand the attribute EXIST.
R2 is a two-stalk join — both conditions on ONE specimen's stalk WMEs... note B1 joins two tokens from the SAME alpha parent pair (A10, A11), so it pairs distinct stalk facts of one $n$ .
B3 pairs cap-facts: WHITE and SMOOTH(+RED branch) on one cap.

Step 1–3 — File, join, resolve

Filing: each WME lands in exactly one CLASS store (A1/A2/A3), binds $n$ , then survives at most one colour/odour/habitat filter per branch.

Joining: every B-node intersects $n$ -sets:

$B1: n(\text{scaly}) \cap n(\text{bulbous}) \qquad B2: n(\text{odour-none}) \cap n(\text{leaves} \wedge \text{arc} \neq \text{brown}) \qquad B3: n(\text{white}) \cap n(\text{smooth})$

Each surviving $n$ emits one tuple per rule downstream.

Specificity counts patterns per rule — the deepest chains win:

$\underbrace{\{R3, R4\}}_{\text{join-backed, } \ge 3 \text{ tests}} \;>\; \underbrace{R2}_{2 \text{ stalk tests}} \;>\; \underbrace{R1}_{\text{single WME}}$

Keyed winners C,D = the join-backed tuples.

Recency compares sorted-descending timestamp lists, newest element first, longer-list-wins-on-prefix-ties. Keyed winner A = the tuple whose newest supporting WME was stamped latest — usually the last-inserted band of the winning combination.

Worked micro-example (illustrative numbers)

Suppose WM holds: 301 Mushroom ^specimen M9 ^odour NONE ^sp-colour WHITE, 302 Cap ^specimen C4 ^colour WHITE ^surface SMOOTH, 303 Stalk ^specimen S7 ^br-surface SCALY, 304 Stalk ^specimen S7 ^root BULBOUS:

Filing: 301→A4→A7→A14 ✓ (WHITE ≠ GREEN) → (R1, 301) fires solo.
303 & 304 share $n$ =S7 across A10/A11 → B1 joins → (R2, 303, 304).
No cap pair shares an $n$ → B3 starves; no leaves-mushroom → B2 starves.
Specificity ranks R2's tuple top (join-backed); recency compares [304,303] vs [301] → newest wins.

Tricks & Tips

Answers: 5.1 A,B,C,D · 5.2 C,D · 5.3 A