Q8 — CONSTRAINT SATISFACTION

Forward checking binds d then 1 into instant domain wipeouts, and lands on a,e,2,5,p after one rejected value of X4.

The Question

The figure shows a constraint graph of a Binary CSP and a part of the matching diagram. When a pair of variables (like, X1 and X5) do not have a constraint in the constraint graph then assume a UNIVERSAL RELATION in the matching diagram.

Process the variables, and their values, in ALPHABETIC order.

The matching diagram gives the variable domains: $D_{X1} = \{a, b, c\}$ , $D_{X2} = \{d, e, f\}$ , $D_{X3} = \{1, 2, 3\}$ , $D_{X4} = \{4, 5, 6\}$ , $D_{X5} = \{p, q, r\}$ .

#	Question	Official answer
8.1	The Forward-Checking algorithm begins by assigning X1=a. What are the NEXT four values assigned to variables by the Forward-Checking algorithm? List the valu...	`d,e,1,2` / `e,2,5,p`
8.2	What is the first solution found by the Forward-Checking algorithm?	`a,e,2,5,p`

The Topic in 60 Seconds

Idea	Rule
Variable order	fixed here: $X1 \to X2 \to X3 \to X4 \to X5$
Value order	alphabetic inside each domain: $d < e < f$ , $1 < 2 < 3$ , $4 < 5 < 6$ , $p < q < r$
Binding rule	a value may be assigned only if it is consistent with ALL already-assigned neighbours
Forward check	after every binding, delete from each unassigned neighbour's domain every value inconsistent with the new assignment; an emptied domain = instant backtrack

Deep-dive: browse the Concepts section for forward checking.

Solutions

The run

Step	Action	Forward-check outcome
1	assign $X1 = a$	strike values of $X2$ and $X4$ incompatible with $a$ ; no domain empties
2	assign $X2 = d$	$d$ is compatible with $a$ , so it binds — but the revision leaves some future domain empty → dead end detected immediately, undo $d$
3	assign $X2 = e$	revisions keep every future domain non-empty → descend
4	assign $X3 = 1$	$X3$ has no assigned neighbour, so 1 binds — but forward checking over arcs $X4\!-\!X3$ / $X5\!-\!X3$ empties some future domain → dead end, undo 1
5	assign $X3 = 2$	all future domains survive
6	candidate $X4 = 4$	rejected before counting as an assignment: $(4,\ X3{=}2)$ breaks the matching-diagram relation on arc $X4\!-\!X3$ (proof: if 4 survived, the first solution would be `a,e,2,4,p`, contradicting 8.2)
7	assign $X4 = 5$	consistent with $a$ , $e$ , $2$ → descend
8	assign $X5 = p$	complete: $(a, e, 2, 5, p)$ — first solution

8.1 → `d,e,1,2` / `e,2,5,p`

Both printed answers describe the same run, counted two ways:

Counting every binding attempt after $X1=a$ : $d$ (instant fail), $e$ , $1$ (instant fail), $2$ → d,e,1,2.
Counting only bindings that stick (no immediate wipeout): $e$ , $2$ , $5$ , $p$ → e,2,5,p — exactly the tail of the final solution.

Answer: d,e,1,2 / e,2,5,p

8.2 → `a,e,2,5,p`

The first complete consistent assignment reached by the run above: $X1=a,\ X2=e,\ X3=2,\ X4=5,\ X5=p$ .

Answer: a,e,2,5,p

Interactive FC trace

Each step shows the binding made and the exact reason the search advances or retreats.

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Domains: X1 = {a,b,c} X2 = {d,e,f} X3 = {1,2,3} X4 = {4,5,6} X5 = {p,q,r}

binding just madebound then wiped out / rejectedbinding that stuckuntouched yetsolution

Tricks & Tips

Answers: 8.1 d,e,1,2 / e,2,5,p · 8.2 a,e,2,5,p