Q3 — Games: Strategy + Alpha-Beta

Three MIN children at the root: alpha-beta with running bounds kills C,E,F,I, and SSS* dismisses exactly the same five without a glance.

The Question

The figure shows a game tree with evaluation function values at the horizon nodes. The horizon nodes are labeled from A to I. Use these labels to enter a horizon node or a list of horizon nodes in short answers (textbox).

Tie-breaker: when several nodes carry the same best cost then select the deepest node, if tie persists then select the leftmost of the deepest nodes to break the tie.

Based on the above data, answer the given subquestions.

#QuestionOfficial answer
3.1Which of the following is a strategy for the MAX player?B
3.2List the horizon nodes in the best strategy for MAX. Enter the node labels in alG,H
3.3List the horizon nodes pruned by Alpha-Beta.C,E,F,I
3.4List the horizon nodes not processed (neither LIVE nor SOLVED) by SSS*.B,C,E,F,I

The Topic in 60 Seconds

Root has THREE MIN children; MAX picks the largest. With the course's aggressive bound-passing (a node's running value is passed to its children as their second window bound), cutoffs fire early and often.

Deep-dive: browse the Concepts section for Games, Alpha-Beta, SSS*.

Step 0 — Full minimax backup

NodeChildrenValue
X1 (MAX)B 80, C 608080
M1 (MIN)A 40, X1min⁡(40,80)=40\min(40,80) = 40
X2 (MAX)E 10, F 909090
M2 (MIN)D 30, X2min⁡(30,90)=30\min(30,90) = 30
X3 (MAX)H 70, I 507070
M3 (MIN)G 60, X3min⁡(60,70)=60\min(60,70) = 60
R (MAX)M1, M2, M3max⁡(40,30,60)=60\max(40,30,60) = \mathbf{60}

3.1 — Strategy → B

A strategy = one choice per MAX node (R, X1, X2, X3), leaves all under ONE root child. Option B is the listed set meeting the shape rules — e.g., committing R→M3 with one leaf per branch beneath. Options mixing leaves across different M-children are auto-invalid.

Answer: B

3.2 — Best strategy's leaves → G,H

Root value 60 flows through M3: opponent takes min⁡(60,70)=60\min(60,70)=60, touching exactly G, H.

Answer: G,H

3.3 — Alpha-Beta prunes → C,E,F,I

Left-to-right DFS passing RUNNING bounds down:

EventWindow logicCut
M1: A = 40 → M1 running min 40X1 called with β = 40
X1: B = 80 ≥ β=40 → MAX cutoffC never readC
X1 returns 80 → M1 = 40; Root α = 40M2 window α=40
M2: D = 30 ≤ α=40 → MIN cutoffX2 never openedE, F
M2 returns 30; Root α stays 40M3 window α=40
M3: G = 60 > 40 continue; running min 60X3 called with β = 60
X3: H = 70 ≥ β=60 → MAX cutoffI never readI
M3 = min(60, 70) = 60 → Root = 60 ✓

Pruned horizon nodes in order: C, E, F, I — exact key. Same root value as full minimax, four leaves saved.

3.4 — SSS*: not processed → B,C,E,F,I

SSS* resolves (node, bound) PROBLEM entries until the root is SOLVED:

  1. Establish M3 = 60 exactly: solve G (60), solve H (70) ⇒ M3 = 60.
  2. M1 side needs only a witness < 60: solve A (40) ⇒ M1 ≤ 40; X1's subtree becomes irrelevant.
  3. M2 side likewise: solve D (30) ⇒ M2 ≤ 30; E, F dismissed.

Processed/solved horizon nodes: A, D, G, H. Never processed (neither LIVE nor SOLVED): B, C, E, F, I — the exact key, mirroring alpha-beta's prune set plus B (which αβ evaluated but SSS*'s proof never needed).

Interactive trace

1 / 4
A=40 -> M1 running min 40 -> X1 gets beta=40. B=80 >= 40: MAX CUTOFF - C dies unread. M1 = 40. Root alpha = 40.
evaluatingpruned / not processedprincipal variationdone

Tricks & Tips

Answers: 3.1 B · 3.2 G,H · 3.3 C,E,F,I · 3.4 B,C,E,F,I