Exercise 1: Bellman-Ford on S-T-U-V — Possible Solution
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GIVEN
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Edges: S->T(6), S->U(7), T->U(-3), U->V(2)
Source: S. Vertices: S,T,U,V (V=4, so 3 rounds maximum).

ROUND 1
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Start: dist = {S:0, T:inf, U:inf, V:inf}

Relax S->T: 0+6=6 < inf -> update dist[T]=6
Relax S->U: 0+7=7 < inf -> update dist[U]=7
Relax T->U: 6+(-3)=3 < 7 (U's current value) -> update dist[U]=3
Relax U->V: 3+2=5 < inf -> update dist[V]=5

After round 1: dist = {S:0, T:6, U:3, V:5}

ROUND 2
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Relax S->T: 0+6=6, not < 6 (already equal) - no change
Relax S->U: 0+7=7, not < 3 - no change
Relax T->U: 6+(-3)=3, not < 3 (already equal) - no change
Relax U->V: 3+2=5, not < 5 (already equal) - no change

No changes this round - the algorithm has already converged after
just one round. (Round 3 would also produce no changes, so it can
be skipped once a round makes zero updates.)

RESULT
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Final distances from S: {S:0, T:6, U:3, V:5}

U was updated twice within round 1 itself - first to 7 via the
direct edge S->U, then immediately corrected to 3 once T->U was
relaxed later in that same round (since S->T happens to be listed
before T->U in the edge order here). It converged in a single
round only because the edges happened to be listed in a
convenient order - a different order could have taken an extra
round to reach the same final answer, exactly as this chapter's
own warn-box describes.

WHY THIS WORKS AS AN ANSWER
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Every edge relaxation across each round is shown explicitly,
including the within-round correction to U, and the trace
correctly recognizes convergence (a round producing zero updates)
as the signal that no further rounds are needed, rather than
mechanically running all V-1 rounds regardless.