SKEDSOFT

Iteration 2 : u = b, the permanent label of b is 1.0 T becomes T–{b} there are three edges incident with b, i.e., bc, bd and be where c, d, e ∈ T.
L(c) = min {old L(c), L(b) w(bc)} = min {4.0, 1.0 2.0} = 3.0
L(d) = min {old L(d), L(b) w(bd)} = min {α, 1.0 6.0} = 7.0
L(e) = min {old L(e), L(b) w(be)} = min {α, 1.0 5.0} = 6.0

Thus minimum label is L(c) = 3.0.

Iteration 3 : u = c, the permanent label of e is 3.0, T becomes T–{c}. There is one edge incident
with c, i.e., c, e where e ∈ T.
L(c) = min {old L(e), L(c) w(ce)} = min {6.0, 3.0 1.0} = 4.0
Thus minimum label is L(c) = 4.0

Iteration 4 : u = e, the permanent label of e is 4.0, T becomes T–{e}. There are two edges incident with e, i.e., ed and ef where d, f ∈ T

L(d) = min {old L(d), L(e) w(ed)} = min {7.0, 4.0 3.0} = 7.0
L(f) = min {old L(f), L(e) w(ef)} = min {α, 4.0 7.0} = 11.0

Thus minimum label is L(d) = 7.0
Iteration 5 : u = d, the permanent label of d is 7.0. T becomes T–{d}. There is one edge incident with d, i.e., d, f where f ∈ T
L(f) = min {old L(f), L(d) w(df)} = min {11.0, 7.0 2.0} = 9.0

The minimum label is L(f) = 9.0
Since u = f, the only choice. Iteration stops. Thus the shortest distance between a and f is 9, and moreover, the shortest paths is {a, b, c, e, d, f}.