The Weighted All-Pairs-Shortest-Path-Length Problem on Two ...

<strong>The</strong> 25th Workshop on Combinatorial Mathematics and Computation <strong>The</strong>ory
v = lt(G(q)) and v = rt(G(q)); and let v =
i
lt(G(f)) and
v j
be performed.
k
= rt(G(f)). <strong>The</strong> following codes will
recursively compute L[s, t], and L[t, s], for ,
v t
∈ V(G(q));
recursively compute L[s, t], and L[t, s], for ,
v t
∈ V(G(f));
L[i, j] = L[i, k] + L[k, j];
L[j, i] = L[j, k] + L[k, i];
for each pair of vertices v ≠ v such that v ∈
V(G(q)) – { v , v } and v ∈ V(G(f)) – { v ,
v j
}
i
k
/* If one of V(G(q)) – { v , v } and V(G(f)) –
{ v , v } is empty, then we just skip this loop. */
k
j
L[a, b] = L[a, k] + L[k, b];
L[b, a] = L[b, k] + L[k, a];
endfor
vi
∈ V(G(f)) – { vk
, v
j
}
v v
for each pair of vertices and v such that
a
i
b
k
b
b
k
v s
v s
a
k
vb
/* If V(G(f)) – {
k
,
j
} is empty, then we just
skip this loop. */
L[i, b] = L[i, k] + L[k, b];
L[b, i] = L[b, k] + L[k, i];
endfor
for each pair of vertices v and v such that
∈ V(G(q)) – { vi
, vk
}
/* If V(G(q)) – { vi
, vk
} is empty, then we just
skip this loop. */
L[a, j] = L[a, k] + L[k, j];
L[j, a] = L[j, k] + L[k, a];
endfor
a
j
va
Case 3. z is a P-node
Let q and f be the left child and the right child of z,
respectively, i.e., G(z) = G(q) + G(f). In this case,
lt(G(q)) = lt(G(f)) = vi
and rt(G(q)) = lt(G(f)) = v
j
.
<strong>The</strong> following codes will be performed.
recursively compute L[s, t], and L[t, s], for ,
v t
∈ V(G(q));
L
1
[i, j] = L[i, j];
[j, i] = L[j, i];
L 1
recursively compute L[s, t], and L[t, s], for ,
v t
∈ V(G(f));
L
2
[i, j] = L[i, j];
L 2
[j, i] = L[j, i];
v s
v s
L[i, j] = min{ L1
[i, j], L2
[i, j]};
L[j, i] = min{ [j, i], L [j, i]};
L1
2
2
Lemma 5: Phase I can be done in O( n ) time.
Proof: To complete Phase I, each pair of vertices
v j
and is examined constant times and each node of
T(r) is also examined constant times. This implies
2
that Phase I can be done in O( n ) time.
3.2 Phase II
It is easy to see that LPNA(u), for each P-node u,
can be obtained by the post-order traversal on T and
the time-complexity is linear. Based on the
reasoning and the results of so far, Phase II can be
achieved by breadth-first traversal on T from the
root r. For each non-top P-node u, assume that f =
LPNA(u) and the terminals of G(u) and G(f) are
< vi
, v
j
> and < vs
, vt
>, respectively. <strong>The</strong>
following code will be performed.
if (u ∈ left_descendants(f))
OL[i, j] = L[i, s] + [s, t] + L[t, j];
L 2
L 2
OL[j, i] = L[j, t] + [t, s] + L[s, i];
else /* u ∈ right_descendants(f). */
OL[i, j] = L[i, s] + [s, t] + L[t, j];
OL[j, i] = L[j, t] +
endif
L 1
[t, s] + L[s, i];
L 1
<strong>The</strong> following lemma can be verified easily.
2
Lemma 6: Phase II can be done in O( n )-time.
3.3 Phase III
This phase will be achieved by depth-first
traversal on T from the root r. For each P-node u, let
q and f be the left child and the right child of u,
respectively, i.e., G(u) = G(q) + G(f). In this case,
lt(G(q)) = lt(G(f)) = vi
and rt(G(q)) = lt(G(f)) = v
j
.
<strong>The</strong> following codes will be performed.
for each pair { va
, vb
} ∈ pair_set(u)
docase
case va
, vb
∈ V(G(q))
L[a, b] = min{L[a, b],
L[a, i] + L 2
[i, j] + L[j, b],
L[a, j] + L 2
[j, i] + L[i, b],
L[a, i] + OL[i, j] + L[j, b],
L[a, j] + OL[j, i] + L[i, b]};
L[b, a] = min{L[a, b], L[b, i] + L 2
[i, j]
v i
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