Suppose now that a temporal graph on a (strongly) connected (di)graph G = (V, E) is given to a designer with the claim that it satisfies TC. In this scenario, the designer is allowed to only use the given set of edge availabilities, or a subset of them. If the given design is not minimal, they may wish to remove as many labels as possible, thus reducing the cost. Minimality of a design can be verified by running Algorithm 1 (cf. Section 2) for every s ∈ V.
A Partial Characterisation of Minimal Temporal Graphs
As mentioned earlier, if many edges have the same label, we can encounter trivial cases of minimal temporal graphs. To avoid such cases, we focus our attention here to the class of SLSE temporal graphs, in which every edge only becomes available at one moment in time and no two different edges become available at the same time. Are there minimal SLSE temporal graphs with non linear (in the size of the graph) cost? For example, any complete SLSE temporal graph satisfies TC. Are all these Θ(n
2) labels needed for TC, i.e., are there minimal temporal complete graphs? As we prove in Theorem 4, the answer is negative. However, we give below a minimal temporal graph on n vertices with non-linear in n cost, namely with O(nlogn) labels.
A Minimal Temporal Design of nlogn Cost
Definition 10 (Hypercube graph)
The k-hypercube graph, commonly denoted Q
k
, is a k-regular graph of 2k vertices and 2k − 1⋅k edges. The 1-hypercube is the graph of two vertices and one edge. Recursively, the n-hypercube is produced by taking two isomorphic copies of the (n − 1)-hypercube and adding edges between the corresponding vertices.
Definition 11 (Flat)
In geometry, a flat is a subset of the n-dimensional space that is congruent to a Euclidean space of lower dimension, e.g., the flats in the two-dimensional space are points and lines. In the n-dimensional space, there are flats of every dimension from 0, i.e., points, to n − 1, i.e., hyperplanes.
Theorem 3
There exists an infinite class of minimal temporal graphs on n vertices with Θ(n⋅ logn) edges and Θ(n⋅ logn) labels, such that different edges have different labels.
Proof
We present a minimal temporal graph on the hypercube graph of n vertices. Consider Protocol 2 for labelling the edges of G = Q
k
= (V, E). The temporal graph, G(L), that this labelling procedure produces on the hypercube is minimal. Indeed, first we will prove that the temporal graph produced by Protocol 2 satisfies TC on G = Q
k
.
Consider vertices u, v ∈ V and the steps described in Protocol 3 to reach v, starting from u, via temporal edges. The procedure described in Protocol 3 gives a journey from u to v, which is also unique. It suffices to consider the k-bit binary representation of the vertices of G. Notice that if the hamming distance of the labels of two vertices u, v ∈ V(G) is exactly m, then to reach v from u via a temporal path in the temporal graph on G, we need to move through vertices by consecutively swapping the bits in which u and v differ in the order of dimensions. This way, we maintain the strictly increasing order of the time labels we use and, swap by swap, we approach the destination. Note also that swapping only the bits in which u and v differ is the only way to not violate the increasing order of time labels we use: without loss of generality, suppose that the j
th bit of u is 1 and so is j
th bit of v. If, starting from u, we swap the j
th bit to 0, i.e., we use an edge, e, on the j
th dimension, then at a future step, we again need to swap the j
th bit back to 1 (otherwise, we never reach v). However, the two swaps cannot be consecutive, because then we would use edge e twice and we violate the increasing order of labels. So, we would need to move to a higher dimension after the first of the two swaps; but, then, we have used labels that are larger than all the labels of the j
th dimension, so using any edge of the j
th dimension would also violate the increasing order of labels.
Since our labelling gives a unique (u, v)-journey, for every u, v ∈ V, and since all labels assigned to the edges of E are used in the union of all those journeys, the deletion of any single label will violate TC. Therefore, G(L) is minimal. Finally, note that the temporal graph G(L) on the hypercube graph G = Q
k
has n = 2k vertices, \(\frac {1}{2}n\cdot \log {n}\) edges and \(\frac {1}{2}n\cdot \log {n}\) labels. □
A minimal temporal design of linear in n cost
In the previous section, we showed that there are graphs of non-linear cost (in the number of vertices) that are minimal. Here, we show that there are classes of minimal graphs whose cost is linear in the number of their vertices.
Indeed, as seen in Lemma 1 (Section 3), the star graph of n vertices needs at least Θ(n) labels to satisfy TC and, in fact, we present there a TC satisfying labelling of Θ(n) labels (cf. Fig. 1). Theorem 2(b) (Section 3) also gives a class of minimal temporal graphs of linear cost in the number of vertices. Therefore, we have the following Corollary:
Corollary 2
There exists an infinite class of minimal temporal graphs on n vertices with Θ(n) edges and Θ(n) labels.
SLSE Cliques of at least 4 vertices are not minimal
The complete graph on n vertices, K
n
, with an SLSE labelling L, i.e., a labelling that assigns a single label per edge, different labels to different edges, is an interesting case, since K
n
(L) always satisfies TC. However, it is not minimal as the theorem below shows.
Theorem 4
Let
\(n\in \mathbb {N},~n\geq 4\)
and denote by K
n
the complete graph on n vertices. There exists no minimal SLSE temporal graph on K
n
(L). In fact, we can remove (at least)
\(\lfloor \frac {n}{4} \rfloor \)
labels from any SLSE labelling on K
n
(L) without violating TC.
Proof
The proof is divided in two parts, as follows:
-
(a)
We first show that any SLSE labelling on the complete graph on 4 vertices produces a temporal graph that is not minimal, i.e., the theorem holds for K
4. Consider the six different labels a, b, c, d, x, y assigned by an SLSE labelling to the edges of K
4 as shown in Fig. 7.
Up to their renaming, there are three possible cases for the labels a, b, c, d. Counting all the cases of alternation, cycle, and entanglement (see below) would give us all possible 4! = 24 cases.
-
1.
(Alternation) a < b>d < c > a.
It is easy to see that in this case, both diagonals can be removed: v
1 can reach v
3 using labels a and then c; v
3 can reach v
1 using labels d and then b; v
2 can reach v
4 using labels a and then b; v
4 can reach v
2 using labels d and then c.
-
2.
(Cycle) a < b < d < c.
Here, diagonal x can be removed: v
2 can reach v
4 using labels a and then b; v
4 can reach v
2 using labels d and then c.
-
3.
(Entanglement) a < b < c < d.
This is a more complex case, for which we distinguish the following five sub-cases:
-
i)
x < b and y < c.
We can remove label a: v
1 can reach v
2 using labels y and then c; v
2 can reach v
1 using labels x and then b.
-
ii)
x < b and y > c.
We can remove label b: v
1 can reach v
4 using labels a, then c and then d; v
4 can reach v
1 using labels x, then c and then y (notice that x < b < c < y).
-
iii)
x > b and y > c.
We can remove label a: v
1 can reach v
2 using labels b and then x; v
2 can reach v
1 using labels c and then y.
-
iv)
x > b and b < y < c.
We can remove label x: v
2 can reach v
4 using labels a and then b; v
4 can reach v
2 using labels b, then y and then c.
-
v)
x > b and y < b.
We can remove label c: v
2 can reach v
3 using labels a, then b and then d; v
3 can reach v
2 using labels y, then b and then x.
Notice that the coverage of the above five cases is complete (cf. Fig. 8).
-
(b)
Now, consider the complete graph on n ≥ 4 vertices, K
n
= (V, E). Partition V arbitrarily into \(\lceil \frac {n}{4} \rceil \) subsets \(V_{1}, V_{2}, \ldots , V_{\lceil \frac {n}{4} \rceil }\), such that \(|V_{i}|=4, \forall i=1,2, \ldots , \lceil \frac {n}{4} \rceil -1 \) and \(|V_{\lceil \frac {n}{4} \rceil }| \leq 4\). In each 4-clique defined by \(V_{i},~i=1,2,\ldots , \lfloor \frac {n}{4} \rfloor \), we can remove a “redundant” label, as shown in (a). The resulting temporal graph on K
n
still preserves TC since for every ordered pair of vertices u, v ∈ V:
-
if u, v are in the same V
i
, \(i=1,2,\ldots , \lfloor \frac {n}{4} \rfloor \), then there is a (u, v)-journey that uses time edges within the 4-clique on V
i
, as proven in (a).
-
if u ∈ V
i
and v ∈ V
j
, i ≠ j, then there is a (u, v)-journey that uses the (direct) time edge on {u, v}.
□
Computing the Removal Profit is APX-hard
Note that it is straightforward to check in polynomial time whether a given L satisfies TC on a given (di)graph G, by just checking for every possible (ordered) pair (u, v) of vertices in G whether there is a (u, v)-journey in G(L). Recall that the removal profit is the largest number of labels that can be removed from a temporally connected graph without destroying TC. We now show that it is hard to approximate the value of the removal profit arbitrarily well for an arbitrary graph, i.e., there exists no PTASFootnote 4 for this problem, unless P=NP. In our hardness proof below, we consider undirected graphs.
We prove our hardness result by providing an approximation preserving polynomial reduction from a variant of the maximum satisfiability problem, namely from the monotone Max-XOR(3) problem. Consider a monotone XOR-boolean formula ϕ with variables x
1, x
2, …, x
n
, i.e., a boolean formula that is the conjunction of XOR-clauses of the form (x
i
⊕x
j
), where no variable is negated. The clause α = (x
i
⊕x
j
) is XOR-satisfied by a truth assignment τ if and only if x
i
≠x
j
in τ. The number of clauses of ϕ that are XOR-satisfied in τ is denoted by |τ(ϕ)|. If every variable x
i
appears in exactly r XOR-clauses in ϕ, then ϕ is called a monotone XOR(r) formula. The monotone Max-XOR(r) problem is, given a monotone XOR(r) formula ϕ, to compute a truth assignment τ of the variables x
1, x
2, …, x
n
that XOR-satisfies the largest possible number of clauses, i.e., an assignment τ such that |τ(ϕ)| is maximized. The monotone Max-XOR(3) problem essentially encodes the Max-Cut problem on 3 -regular (i.e., cubic) graphs, which is known to be APX-hard [4].
Lemma 2
[4] The monotone Max-XOR(3) problem is APX-hard.
Now we provide our reduction from the monotone Max-XOR(3) problem to the problem of computing r(G, L). Let ϕ be an arbitrary monotone XOR(3) formula with n variables x
1, x
2, …, x
n
and m clauses. Since every variable x
i
appears in ϕ in exactly 3 clauses, it follows that \(m=\frac {3}{2}n\). We will construct from ϕ a graph G
ϕ
= (V
ϕ
, E
ϕ
) and a labelling L
ϕ
of G
ϕ
.
A very high-level description of the construction is as follows. G
ϕ
is composed of gadgets that represent the variables x
i
of the formula ϕ. Each variable x
i
is assigned a “source” vertex \(s^{x_{i}}\) and three “sink” vertices \(t_{1}^{x_{i}},t_{2}^{x_{i}},t_{3}^{x_{i}}\) in G
ϕ
; each of them corresponds to one of the three clauses of ϕ in which x
i
appears. The gadgets of G
ϕ
are connected in such a way that, if (x
i
⊕x
j
) is a clause of ϕ, then one of the sink vertices of x
i
coincides with one of the sink vertices of x
j
. Furthermore it turns out that, by the construction, a journey from each source vertex \(s^{x_{i}}\) to the corresponding sink vertices \(t_{1}^{x_{i}},t_{2}^{x_{i}},t_{3}^{x_{i}}\) represents a truth assignment of the variable x
i
. Moreover, the number of clauses of ϕ that can be satisfied by a truth assignment corresponds bijectively to the number of time-labels that can be removed from G
ϕ
without destroying TC. Thus an optimum solution of the monotone Max-XOR(3) problem on ϕ corresponds to an optimal removal profit in G
ϕ
.
Now we present the detailed construction of G
ϕ
from the formula ϕ. First we construct for every variable x
i
, where 1 ≤ i ≤ n, the gadget-graph G
ϕ, i
together with a labelling L
ϕ, i
of its edges, as illustrated in Fig. 9. In this figure, the labels of every edge in L
ϕ, i
are drawn next to the edge. We call the induced subgraph of G
ϕ, i
on the 4 vertices \( \{s^{x_{i}},u_{0}^{x_{i}},w_{0}^{x_{i}},v_{0}^{x_{i}}\}\) the base of G
ϕ, i
. Moreover, for every p ∈ {1, 2, 3}, we call the induced subgraph of G
ϕ, i
on the 4 vertices \( \{t_{p}^{x_{i}},u_{p}^{x_{i}},w_{p}^{x_{i}},v_{p}^{x_{i}}\}\) the p
th branch of G
ϕ, i
. Finally, we call the edges \(\{ u_{0}^{x_{i}} , w_{0}^{x_{i}} \}\) and \( \{ w_{0}^{x_{i}} , v_{0}^{x_{i}} \}\) the transition edges of the base of G
ϕ, i
and, for every p ∈ {1, 2, 3}, we call the edges \( \{ u_{p}^{x_{i}} , w_{p}^{x_{i}} \}\) and \( \{ w_{p}^{x_{i}} , v_{p}^{x_{i}} \}\) the transition edges of the pth branch of G
ϕ, i
. For every p ∈ {1, 2, 3} we associate the pth appearance of the variable x
i
with the pth branch of G
ϕ, i
.
We continue the construction of G
ϕ, i
and L
ϕ, i
as follows. First, we add an edge between any possible pair of vertices \( w_{p}^{x_{i}},w_{q}^{x_{j}}\), where p, q ∈ {0, 1, 2, 3} and i, j ∈ {1, 2, …, n}, and we assign to this new edge \( e= \{ w_{p}^{x_{i}} , w_{q}^{x_{j}} \} \) the unique label L
ϕ
(e) = {7}. The addition of the above described edges is not illustrated in Fig. 9. Note here that we add this edge \( \{ w_{p}^{x_{i}} , w_{q}^{x_{j}} \}\) also in the case where i = j (and p ≠ q).
Intuitively, the base of G
ϕ, i
(cf. Fig. 9) corresponds to the variable x
i
and, for every p ∈ {1, 2, 3}, the pth branch of G
ϕ, i
, together with the two edges \(\{u_{0}^{x_{i}},u_{p}^{x_{i}}\}\) and \(\{v_{0}^{x_{i}},v_{p}^{x_{i}}\}\), correspond to the clause of ϕ in which x
i
appears for the pth time in ϕ.
Consider now a clause α = (x
i
⊕x
j
) of ϕ. Assume that the variable x
i
(resp. x
j
) of α corresponds to the pth (resp. to the qth) appearance of x
i
(resp. of x
j
) in ϕ. Then we identify the vertices \({u_{p}^{x_{i}},v_{p}^{x_{i}},w_{p}^{x_{i}},t_{p}^{x_{i}}}\) of the pth branch of G
ϕ, i
with the vertices \( v_{q}^{x_{i}},u_{q}^{x_{i}},w_{q}^{x_{i}},t_{q}^{x_{i}}\) of the qth branch of G
ϕ, j
, respectively (cf. Fig. 10b). Now we add an edge between any possible pair of vertices \(t_{p}^{x_{i}},t_{q}^{x_{j}}\), i, j ∈ {1, 2, …, n}, and p, q ∈ {1, 2, 3}. We assign to this new edge \(e= \{ t_{p}^{x_{i}} , t_{q}^{x_{j}} \}\) the unique label L
ϕ
(e) = {7}.
Furthermore, for every i ∈ {1, 2, …, n} and every p ∈ {1, 2, 3} we define for simplicity of notation the temporal paths \( P_{i,p}=(s^{x_{i}},u_{0}^{x_{i}},u_{p}^{x_{i}},t_{p}^{x_{i}})\) and \( Q_{i,p}=(s^{x_{i}},v_{0}^{x_{i}},v_{p}^{x_{i}},t_{p}^{x_{i}})\).
The intuition behind the composition of the gadget-graphs G
ϕ, i
(cf. Fig. 10b) is the following. If variable x
i
is false in a truth assignment τ of ϕ, then all edges of the paths P
i,1, P
i,2, P
i,3 keep their labels as in L
ϕ
. Otherwise, if x
i
is true in τ, then all edges of the paths Q
i,1, Q
i,2, Q
i,3 keep their labels as in L
ϕ
. Furthermore, depending on the value of x
i
in the assignment τ, each of the transition edges \(\{u_{p}^{x_{i}},w_{p}^{x_{i}}\}\) and \(\{w_{p}^{x_{i}},v_{p}^{x_{i}}\}\), where p ∈ {1, 2, 3}, keeps exactly one of its two labels from L
ϕ
. Consider now a clause α = (x
i
⊕x
j
) of ϕ which corresponds to the pth branch of G
ϕ, i
and to the qth branch of G
ϕ, j
. Then the only case where both edges \(\{t_{p}^{x_{i}},u_{p}^{x_{i}}\}\) and \(\{t_{p}^{x_{i}},v_{p}^{x_{i}}\}\) keep their labels from L
ϕ
, is when the two variables x
i
, x
j
have equal truth value in the corresponding truth assignment τ of ϕ; that is, when the clause α = (x
i
⊕x
j
) is not XOR-satisfied by τ. Therefore, intuitively, by a careful counting of the labels it turns out that, if more clauses can be satisfied by a truth assignment τ, then a TC preserving sub-labelling L of L
ϕ
can be constructed which avoids more labels from L
ϕ
, and vice versa (cf. Theorem 5).
To finalize the construction of the graph G
ϕ
, we add a new vertex t
0 to ensure the existence of a temporal path between each pair of vertices of G
ϕ
, as follows. This new vertex t
0 is adjacent to vertex \(w_{0}^{x_{n}}\) and to all vertices in the set \(\{s^{x_{i}},t_{1}^{x_{i}},t_{2}^{x_{i}},t_{3}^{x_{i}},u_{p}^{x_{i}},v_{p}^{x_{i}}:1\leq i\leq n,\ 0\leq p\leq 3\} \). First we assign to the edge \(\{ t_{0} , w_{0}^{x_{n}} \}\) the unique label \( L_{\phi }(\{ t_{0} , w_{0}^{x_{n}} \} )=\{5\}\) . Furthermore, for every vertex \( t_{p}^{x_{i}}\), where 1 ≤ i ≤ n and 1 ≤ p ≤ 3, we assign to the edge \( \{ t_{0} , t_{p}^{x_{i}} \} \) the unique label \(L_{\phi }(\{ t_{0} , t_{p}^{x_{i}} \} )=\{5\}\). Finally, for each of the vertices \(z\in \{s^{x_{i}},u_{p}^{x_{i}},v_{p}^{x_{i}}:1\leq i\leq n,\ 0\leq p\leq 3\}\) we assign to the edge {t
0, z} the unique label L
ϕ
({t
0, z})={6}. The addition of the vertex t
0 and the labels of the (dashed) edges incident to t
0 are illustrated in Fig. 10a. Denote the vertex sets \(A= \{s^{x_{i}},u_{p}^{x_{i}},v_{p}^{x_{i}}:1\leq i\leq n,\ 0\leq p\leq 3\}\), \( B=\{w_{p}^{x_{i}}:1\leq i\leq n,\ 0\leq p\leq 3\}\) , and \(C=\{t_{p}^{x_{i}}:1 \leq i\leq n,\ 1\leq p\leq 3\}\). Note that V
ϕ
= A∪B∪C∪{t
0}. This completes the construction of the graph G
ϕ
and its labelling L
ϕ
.
For every i ∈ {1, 2, …, n} the graph G
ϕ, i
has 16 vertices. Furthermore, for every p ∈ {1, 2, 3}, the 4 vertices of the pth branch of G
ϕ, i
also belong to a branch of G
ϕ, j
, for some j ≠ i. Therefore, together with the vertex t
0, the graph G
ϕ
has in total 10n + 1 vertices. We now present the auxiliary Lemmas 3–5 which are necessary for the proof of Theorem 5.
Lemma 3
The labelling L
ϕ
assigns
\(\frac {17}{4}n^{2}+28n+1\)
labels to the edges of G
ϕ
.
Proof
The vertex t
0 has in total 3 incident edges (to vertices \( s^{x_{i}},u_{0}^{x_{i}},v_{0}^{x_{i}}\)) to every base of a variable x
i
of ϕ, 3 incident edges (to vertices \( t_{p}^{x_{i}},u_{p}^{x_{i}},v_{p}^{x_{i}}\), where 1 ≤ p ≤ 3) to every clause (x
i
⊕x
j
) of ϕ (i.e., to one branch of x
i
and one branch of x
j
), and one incident edge to vertex \(w_{0}^{x_{n}}\). That is, t
0 has in total \(3n+3m+1=3n+3\cdot \frac {3}{2}n+1=\frac {15}{2} n+1\) incident edges, each of them having one label in L
ϕ
.
Furthermore there exist in total \(\frac {m(m-1)}{2} \) edges among the vertices \(\{t_{p}^{x_{i}}:1\leq i\leq n,\ 1\leq p\leq 3\}\) , as well as \( \frac {(n+m)(n+m-1)}{2}\) edges among the vertices \(\{w_{p}^{x_{i}}:1\leq i\leq n,\ 0\leq p\leq 3\}\) , each of them having one label in L
ϕ
. Therefore, since \(m=\frac {3}{2}n\) , L
ϕ
assigns in total \( \frac {17}{4}n^{2}- 2n\) labels for these edges.
Moreover, the labelling L
ϕ
assigns to every variable x
i
of ϕ in total 12 labels, i.e., two labels for each of the transition edges \( \{ u_{0}^{x_{i}} , w_{0}^{x_{i}} \} , \ \{ w_{0}^{x_{i}} , v_{0}^{x_{i}} \}\) and one label for each of the edges \(\{ \{ s^{x_{i}} , u_{0}^{x_{i}} \} ,\ \{ s^{x_{i}} , v_{0}^{x_{i}} \} ,\ \{ u_{0}^{x_{i}} , u_{p}^{x_{i}} \} , \ \{ v_{0}^{x_{i}} , v_{p}^{x_{i}} \} :1\leq p\leq 3\}\).
Finally, L
ϕ
assigns to every clause (x
i
⊕x
j
) of ϕ in total 7 labels, i.e., two labels for each of the transition edges \( \{ u_{p}^{x_{i}} , w_{p}^{x_{i}} \},\ \{ w_{p}^{x_{i}} , v_{p}^{x_{i}} \} \) and one label for each of the edges \( \{u_{p}^{x_{i}} , t_{p}^{x_{i}} \} ,\ \{ v_{p}^{x_{i}} , t_{p}^{x_{i}} \},\ \{ t_{p}^{x_{i}} , w_{p}^{x_{i}} \} \), where x
i
is associated with the pth branch of G
ϕ, i
. That is, L
ϕ
assigns in total \(7m=\frac {21}{2}n\) labels for all clauses of ϕ.
Summarizing, the labelling L
ϕ
assigns to the edges of the graph G
ϕ
a total of \(\left (\frac {15}{2}n+1\right ) +\left (\frac {17}{4}n^{2}- 2n\right ) +12n+ \frac {21}{2}n=\frac {17}{4}n^{2}+28n+1\) labels. □
Lemma 4
The labelling L
ϕ
satisfies TC on G
ϕ
.
Proof
We will prove that there exists a temporal path in L
ϕ
between any pair of vertices of V
ϕ
= A∪B∪C∪{t
0}.
For any two vertices b, b
′ ∈ B there exists a temporal path from b to b
′ and from b
′ to b, due to the edge {b, b
′} with label 7. Similarly, for any two vertices c, c
′ ∈ C there exists a temporal path from c to c
′ and from c
′ to c, due to the edge {c, c
′} with label 7. Let a
1, a
2 ∈ A. There exists a temporal path from a
1 to a
2 as follows: start from a
1, follow P
i, p
(or Q
i, p
) upwards until \( t_{p}^{x_{i}}\) with greatest label 4, then go to t
0 with label 5, and finally from t
0 to a
2 with label 6. In the special case where a
1 and a
2 lie on the same path P
i, p
(resp. Q
i, p
) and a
1 appears before a
2 in P
i, p
(resp. Q
i, p
), there exists clearly a temporal path from a
1 to a
2 along P
i, p
(resp. Q
i, p
).
Let a ∈ A and b ∈ B. Note that \(b=w_{p}^{x_{i}}\) for some i ∈ {1, 2, …, n} and some p ∈ {0, 1, 2, 3}. There exists the temporal path from b to a as follows. First follow the edge \( \{ w_{p}^{x_{i}} , u_{p}^{x_{i}} \} \) (with label 1), then follow upwards the path P
i, p
until one of the vertices \( \{t_{1}^{x_{i}},t_{2}^{x_{i}},t_{3}^{x_{i}}\}\) (with maximum label 4), then go to t
0 with label 5 and finally reach a with label 6. Furthermore there exists the temporal path from a to b as follows. Assume first that \(a=s^{x_{i}}\), for some i ∈ {1, 2, …, n}. If \( b=w_{0}^{x_{i}}\) then there exists the temporal path on the edges \( \{ s^{x_{i}} , u_{0}^{x_{i}} \} \) (with label 1) and \( \{ u_{0}^{x_{i}} , w_{0}^{x_{i}} \}\) (with label 2). If \(b\neq w_{0}^{x_{i}}\) then there exists the temporal path from \(s^{x_{i}}\) to \(w_{0}^{x_{i}}\) (with maximum label 2), followed by the edge \(\{ w_{0}^{x_{i}} , b \} \) (with label 7). Assume now that \(a\neq s^{x_{i}}\), for every i ∈ {1, 2, …, n}. That is, \(a=u_{p}^{x_{i}}\) or \(a=v_{p}^{x_{i}}\), for some i ∈ {1, 2, …, n} and some p ∈ {0, 1, 2, 3}. If \(b=w_{p}^{x_{i}}\) then there exists the temporal path from a to b on the edge {a, b} (with label 1). If \(b\neq w_{p}^{x_{i}}\) then there exists the temporal path from a to b through the edges \( \{ a , w_{p}^{x_{i}} \} \) (with label 1) and \( \{ w_{p}^{x_{i}} , b \} \) (with label 7). That is, there exists a temporal path in L
ϕ
between any a ∈ A and any b ∈ B.
Let b ∈ B, i.e., \(b=w_{p}^{x_{i}}\) for some i ∈ {1, 2, …, n} and some p ∈ {0, 1, 2, 3}. Then there exists a temporal path from b to every vertex c ∈ C as follows. If p = 0 then start with the edge \( \{ w_{0}^{x_{i}} , u_{0}^{x_{i}} \} \) (of label 1), continue upwards with a temporal path (of maximum label 4) until \(t_{1}^{x_{i}}\in C\) and then continue to any other vertex c ∈ C with the edge \( \{ t_{1}^{x_{i}} , c \} \) (of label 7). If p ∈ {1, 2, 3} then reach \(t_{p}^{x_{i}}\in C\) with the edge \( \{ w_{p}^{x_{i}} , t_{p}^{x_{i}} \} \) (of label 1) and continue to any other vertex c ∈ C with the edge \( \{ t_{p}^{x_{i}} , c \} \) (of label 7). That is, there exists a temporal path from any b ∈ B to any vertex of the set C. Now let c ∈ C, i.e., \(c=t_{p}^{x_{i}}\) for some i ∈ {1, 2, …, n} and some p ∈ {1, 2, 3}. Then there exists a temporal path from c to every vertex b ∈ B as follows. First reach the vertex \(w_{p}^{x_{i}}\in B\) with the edge \( \{ t_{p}^{x_{i}} , w_{p}^{x_{i}} \} \) (of label 1) and then continue to any other vertex c ∈ C with the edge \( \{ w_{p}^{x_{i}} , c \} \) (of label 7). That is, there exists a temporal path in L
ϕ
between any b ∈ B and any c ∈ C.
Let a ∈ A, i.e., \(a\in \{s^{x_{i}},u_{p}^{x_{i}},v_{p}^{x_{i}}\}\) for some i ∈ {1, 2, …, n} and some p ∈ {0, 1, 2, 3}. Then there exists at least one path from a upwards to a vertex \(c\in \{t_{1}^{x_{i}},t_{2}^{x_{i}},t_{3}^{x_{i}}\}\) (with maximum label 4). Once we have (temporally) reached c from a, we can (temporally) continue to any other c
′ ∈ C through the edge {c, c
′} (of label 7 ). That is, there exists a temporal path from any a ∈ A to any vertex of C. Now let c ∈ C, i.e., \(c=t_{p}^{x_{i}}\) for some i ∈ {1, 2, …, n} and some p ∈ {1, 2, 3}. Then there exists a temporal path from c to every vertex a ∈ A as follows. First reach the vertex t
0 with the edge \( \{ t_{p}^{x_{i}} , t_{0} \}\) (of label 5) and then continue to any vertex a ∈ A with the edge {t
0, a} (of label 6). That is, there exists a temporal path in L
ϕ
between any a ∈ A and any c ∈ C.
Finally, there exists a temporal path between t
0 and every vertex of \( A\cup C\cup \{w_{0}^{x_{n}}\}\), since t
0 is a neighbour with all these vertices. Let b ∈ B, i.e., \(b=w_{p}^{x_{i}}\) for some i ∈ {1, 2, …, n} and some p ∈ {0, 1, 2, 3}. Then there exists a temporal path from \( w_{p}^{x_{i}}\) to t
0 with the edges \( \{ w_{p}^{x_{i}} , u_{p}^{x_{i}} \} \) (with label 1) and \( \{ u_{p}^{x_{i}} , t_{0} \} \) (with label 6). On the other hand, there exists a temporal path from t
0 to every vertex \( b=w_{p}^{x_{i}}\in B\), as follows. First reach the vertex \(w_{0}^{x_{n}}\) with the edge \( \{ t_{0} , w_{0}^{x_{n}} \} \) (of label 5) and then, if \(b\neq w_{0}^{x_{n}}\), continue with the edge \( \{ w_{0}^{x_{n}} , b \} \) (of label 7). That is, there exists a temporal path in L
ϕ
between t
0 and any vertex in A∪B∪C.
Summarizing, there exists a temporal path between any pair of vertices of V
ϕ
= A∪B∪C∪{t
0}, i.e., the labelling L
ϕ
satisfies TC on G
ϕ
. □
Lemma 5
Let L ⊆ L
ϕ
be a labelling of the graph G
ϕ
. If L satisfies TC on G
ϕ
, then L contains:
-
(a)
at least one label for every transition edge
\( \{ u_{p}^{x_{i}} , w_{p}^{x_{i}} \} \)
and
\( \{ w_{p}^{x_{i}} , v_{p}^{x_{i}} \}\), where i ∈ {1, 2, …, n} and p ∈ {0, 1, 2, 3},
-
(b)
the label of each edge
\( \{ t_{p}^{x_{i}} , w_{p}^{x_{i}} \} \), where i ∈ {1, 2,…, n} and p ∈ {1, 2, 3},
-
(c)
the labels of all edges of G
ϕ
among the vertices
\( \{t_{p}^{x_{i}}:1\leq i\leq n,\ 1\leq p\leq 3\}\),
-
(d)
the labels of all edges among the vertices
\(\{w_{p}^{x_{i}}:1\leq i\leq n,\ 0\leq p\leq 3\}\),
-
(e)
the label of each edge incident to t
0, and
-
(f)
the labels of all edges of the path P
i,p
or the labels of all edges of the path Q
i,p
, where i ∈ {1, 2, …, n} and p ∈ {1, 2, 3}.
Proof
-
(a)
First assume that L does not keep any time label on the transition edge \( \{ u_{p}^{x_{i}} , w_{p}^{x_{i}} \}\) (resp. \( \{ w_{p}^{x_{i}} , v_{p}^{x_{i}} \} \)), where i ∈ {1, 2, …, n} and p ∈ {0, 1, 2, 3}. Then there does not exist in L any temporal path from \(u_{p}^{x_{i}}\) (resp. \(v_{p}^{x_{i}}\)) to \(w_{p}^{x_{i}}\), even if L maintains all other edge labels from L
ϕ
. This is a contradiction. Therefore L keeps at least one label on the transition edge \( \{ u_{p}^{x_{i}} , w_{p}^{x_{i}} \}\) (resp. \( \{ w_{p}^{x_{i}} , v_{p}^{x_{i}} \} \)).
-
(b)
Now assume that L does not contain the label of some edge \( \{ t_{p}^{x_{i}} , w_{p}^{x_{i}} \} \), where i ∈ {1, 2, …, n} and p ∈ {1, 2, 3}. Then there does not exist in L any temporal path from \( t_{p}^{x_{i}}\) to any vertex \(w_{q}^{x_{j}}\in B\), even if L maintains all other edge labels from L
ϕ
. This is a contradiction to the assumption that L satisfies TC on G
ϕ
. Therefore L contains the label of each edge \( \{ t_{p}^{x_{i}} , w_{p}^{x_{i}} \} \), where i ∈ {1, 2, …, n} and p ∈ {1, 2, 3}.
-
(c)
Consider two vertices \(t_{p}^{x_{i}}\neq t_{q}^{x_{j}}\), 1 ≤ i < j ≤ n, 1 ≤ p, q ≤ 3. If L does not contain the label of the edge \( \{ t_{p}^{x_{i}} , t_{q}^{x_{j}} \} \), then there does not exist in L any temporal path from \(t_{p}^{x_{i}}\) to \(t_{q}^{x_{j}}\), which is a contradiction. Therefore L contains the labels of all edges of G
ϕ
among the vertices \(\{t_{p}^{x_{i}}:1\leq i\leq n,\ 1\leq p\leq 3\}\).
-
(d)
Assume that L does not contain the label of the edge \( \{ w_{p}^{x_{i}} , w_{q}^{x_{j}} \} \), for some i, j ∈ {1, 2, …, n} and p, q ∈ {0, 1, 2, 3}. Then there does not exist in L any temporal path from \(w_{p}^{x_{i}}\) to \(w_{q}^{x_{j}}\), which is a contradiction. Therefore L contains the labels of all edges among the vertices \( \{w_{p}^{x_{i}}:1\leq i\leq n,\ 0\leq p\leq 3\}\).
-
(e)
We now prove that L contains the label of each edge incident to t
0. Recall that the neighbours of t
0 in G
ϕ
are exactly the vertices of the set \(A\cup C\cup \{w_{0}^{x_{n}}\}\). Assume L does not have the label of the edge \(e= \{ t_{0} , w_{0}^{x_{n}} \} \). Then there exists no temporal path in L from t
0 to any vertex \( w_{p}^{x_{i}}\in B\), even if L maintains all other edge labels from L
ϕ
. This is a contradiction to the assumption that L satisfies TC on G
ϕ
. Now assume that there exists a vertex \(a\in A=\{s^{x_{i}},u_{p}^{x_{i}},v_{p}^{x_{i}}:1\leq i\leq n,\ 0\leq p\leq 3\}\) such that L does not have the label of the edge e = {t
0, a}. Then there does not exist in L any temporal path from vertex t
0 to vertex a, which is again a contradiction. Finally assume that there exists a vertex \(t_{p}^{x_{i}}\in C\), such that L does not have the label of the edge \(e= \{ t_{0} , t_{p}^{x_{i}} \} \). Then there does not exist in L any temporal path from vertex \(u_{p}^{x_{i}}\) to vertex \( s^{x_{i}}\), which is a contradiction. Therefore L contains the label of each edge incident to t
0.
-
(f)
Assume that L misses from L
ϕ
at least one label of the path P
i, p
and at least one label of the path Q
i, p
, for some i ∈ {1, 2, …, n} and p ∈ {1, 2, 3}. Then there does not exist any temporal path from \(s^{x_{i}}\) to \(t_{p}^{x_{i}}\), which is a contradiction. Therefore L contains the labels of all edges of the path P
i, p
or the labels of all edges of the path Q
i, p
, where i ∈ {1, 2, …, n} and p ∈ {1, 2, 3}.
□
We are now ready to provide the proof of Theorem 5.
Theorem 5
There exists a truth assignment τ of ϕ with |τ(ϕ)|≥k if and only if there exists a TC satisfying labelling L⊆L
ϕ
of G
ϕ
such that |L
ϕ
∖L|≥9n+k.
Proof
(⇒) Assume that there is a truth assignment τ that XOR-satisfies k clauses of ϕ. We construct a labelling L of G
ϕ
by removing 9n + k labels from L
ϕ
, as follows. First we keep in L all labels of L
ϕ
on the edges incident to t
0. Furthermore we keep in L the label {7} of all the edges \( \{ t_{p}^{x_{i}} , t_{q}^{x_{j}} \} \) and the label {7} of all the edges \(w_{p}^{x_{i}}w_{q}^{x_{j}}\). Moreover we keep in L the label {1} of all the edges \( \{ t_{p}^{x_{i}} , w_{p}^{x_{i}} \} \). Let now i = 1, 2, …, n. If x
i
=0 in τ, we keep in L the labels of the edges of the paths P
i,1, P
i,2, P
i,3, as well as the label 1 of the edge \( \{ v_{0}^{x_{i}} , w_{0}^{x_{i}} \} \) and the label 2 of the edge \( \{ w_{0}^{x_{i}} , u_{0}^{x_{i}} \} \). Otherwise, if x
i
=1 in τ, we keep in L the labels of the edges of the paths Q
i,1, Q
i,2, Q
i,3, as well as the label 1 of the edge \( \{ u_{0}^{x_{i}} , w_{0}^{x_{i}} \} \) and the label 2 of the edge \( \{ w_{0}^{x_{i}} , v_{0}^{x_{i}} \} \).
We now continue the labelling L as follows. Consider an arbitrary clause α = (x
i
⊕x
j
) of ϕ. Assume that the variable x
i
(resp. x
j
) of the clause α corresponds to the pth (resp. to the qth) appearance of variable x
i
(resp. x
j
) in ϕ. Then, by the construction of G
ϕ
, the pth branch of G
ϕ, i
coincides with the qth branch of G
ϕ, j
, i.e., \( u_{p}^{x_{i}}=v_{q}^{x_{j}}\), \(v_{p}^{x_{i}}=u_{q}^{x_{j}}\), \( w_{p}^{x_{i}}=w_{q}^{x_{j}}\), and \(t_{p}^{x_{i}}=t_{q}^{x_{j}}\) (cf. Fig. 10b). Let α be XOR-satisfied in τ, i.e., \(x_{i}=\overline {x_{j}}\). If \(x_{i}=\overline {x_{j}}=0\) (i.e., x
i
=0 and x
j
=1) then we keep in L the label 1 of the edge \( \{ v_{p}^{x_{i}} , w_{p}^{x_{i}} \} \) and the label 2 of the edge \( \{ w_{p}^{x_{i}} , u_{p}^{x_{i}} \} \) , cf. Fig. 11a. In the symmetric case, where \(x_{i}=\overline {x_{j}}=1\) (i.e., x
i
=1 and x
j
=0), we keep in L the label 1 of the edge \( \{ u_{p}^{x_{i}} , w_{p}^{x_{i}} \} \) and the label 2 of the edge \( \{ w_{p}^{x_{i}} , v_{p}^{x_{i}} \} \). Let now α be XOR-unsatisfied in τ , i.e., x
i
= x
j
. Then, in both cases where x
i
= x
j
=0 and x
i
= x
j
=1, we keep in L the label 1 of both edges \( \{ v_{p}^{x_{i}} , w_{p}^{x_{i}} \} \) and \( \{ w_{p}^{x_{i}} , u_{p}^{x_{i}} \} \), cf. Fig 11b. This finalizes the labelling L of G
ϕ
. It is easy to check that L satisfies TC on G
ϕ
.
Summarizing, the labelling L misses in total 6 labels of L
ϕ
for the edges \( \{ \{ s^{x_{i}} , u_{0}^{x_{i}} \} ,\ \{ s^{x_{i}} , v_{0}^{x_{i}} \} ,\ \{ u_{0}^{x_{i}} , w_{0}^{x_{i}} \} ,\ \{ w_{0}^{x_{i}} , v_{0}^{x_{i}} \} ,\ \{ u_{0}^{x_{i}} , u_{p}^{x_{i}} \} ,\ \{ v_{0}^{x_{i}} , v_{p}^{x_{i}} \} :1\leq p\leq 3,~ i = 1,2,{\ldots } ,n \}\). That is, L misses in total 6n labels of L
ϕ
for all variables x
1, x
2, …, x
n
. For each of the k XOR-satisfied clauses (x
i
⊕x
j
) of ϕ, the labelling L misses in total 3 labels of L
ϕ
for the edges \( \{ u_{p}^{x_{i}} , w_{p}^{x_{i}} \} ,\ \{ w_{p}^{x_{i}} , v_{p}^{x_{i}} \} ,\ \{ u_{p}^{x_{i}} , t_{p}^{x_{i}} \} ,\ \{ v_{p}^{x_{i}} , t_{p}^{x_{i}} \} ,\ \{ t_{p}^{x_{i}} , w_{p}^{x_{i}} \} \), where x
i
is associated with the pth branch of G
ϕ, i
. That is, L misses in total 3k labels of L
ϕ
for all XOR-satisfied clauses. Furthermore, for each of the m − k XOR-satisfied clauses (x
i
⊕x
j
) of ϕ, the labelling L misses in total 2 labels of L
ϕ
for the edges \( \{ u_{p}^{x_{i}} , w_{p}^{x_{i}} \} ,\ \{ w_{p}^{x_{i}} , v_{p}^{x_{i}} \} ,\ \{ u_{p}^{x_{i}} , t_{p}^{x_{i}} \} ,\ \{ v_{p}^{x_{i}} , t_{p}^{x_{i}} \} ,\ \{ t_{p}^{x_{i}} , w_{p}^{x_{i}} \} \), where x
i
is associated with the pth branch of G
ϕ, i
. That is, L misses in total 2(m − k) =3n − 2k labels of L
ϕ
for all XOR-satisfied clauses. All other labels of L
ϕ
remain in the labelling L ⊆ L
ϕ
. Therefore, L misses in total 6n+3k+3n − 2k = 9n + k labels from L
ϕ
.
(⇐) Assume that r(G
ϕ
, L
ϕ
) ≥ 9n + k and let L ⊆ L
ϕ
be a TC preserving labelling of G
ϕ
with |L
ϕ
∖L| = r(G
ϕ
, L
ϕ
) ≥ 9n + k, i.e., G
ϕ
(L) is minimal. Let i ∈ {1, 2, …, n}. For every p ∈ {1, 2, 3}, L contains by Lemma 5(f) the labels of all edges of the path P
i, p
or the labels of all edges of the path Q
i, p
. Therefore, there exist at least two indices p
1, p
2∈{1,2,3} such that L contains the labels of all edges of the paths \(P_{i,p_{1}},P_{i,p_{2}}\)
or the labels of all edges of the paths \(Q_{i,p_{1}},Q_{i,p_{2}}\). Without loss of generality let p
1=1 and p
2=2 and let L contain the labels of all edges of the paths P
i,1, P
i,2 (the other cases can be dealt with in the same way by symmetry). Assume that L also contains the labels of all edges of the path \( Q_{i,3}=(s^{x_{i}},v_{0}^{x_{i}},v_{3}^{x_{i}},t_{3}^{x_{i}})\). Then we can modify the labelling L to a labelling L
′ as follows. First remove from L the labels of the edges \( \{ s^{x_{i}} , v_{0}^{x_{i}} \} \) and \( \{ v_{0}^{x_{i}} , v_{3}^{x_{i}} \} \) and add instead the labels of the edges \( \{ u_{0}^{x_{i}} , u_{3}^{x_{i}} \} \) and \( \{ u_{3}^{x_{i}} , t_{3}^{x_{i}} \} \) (if they do not exist yet in L). Furthermore change the labels of the transition edges \( \{ v_{0}^{x_{i}} , w_{0}^{x_{i}} \} \) and \( \{ w_{0}^{x_{i}} , u_{0}^{x_{i}} \} \) to the labels 1 and 2, respectively. Note that in the resulting labelling L
′, both edges \( \{ u_{3}^{x_{i}} , t_{3}^{x_{i}} \} \) and \( \{ v_{3}^{x_{i}} , t_{3}^{x_{i}} \} \) are labelled. Furthermore L
′ ⊆ L
ϕ
and L
′ does not have more labels than L, and thus |L
ϕ
∖L
′|≥|L
ϕ
∖L| = r(G
ϕ
, L
ϕ
). Moreover, it is easy to check that L
′ still satisfies TC on G
ϕ
, as L satisfies TC as well. So, it must also be |L
ϕ
∖L
′| = r(G
ϕ
, L
ϕ
), i.e., G
ϕ
(L
′) is also minimal. Therefore, we may assume without loss of generality that for any minimal labelling L ⊆ L
ϕ
, L contains the labels of all edges of the paths P
i,1, P
i,2, P
i,3
or the labels of all edges of the paths Q
i,1, Q
i,2, Q
i,3.
From Lemma 5(a), L contains at least 2n + 2m labels on the edges of the form \(\{u_{p}^{x_{i}}, w_{p}^{x_{i}}\}\) or \(\{w_{p}^{x_{i}}, v_{p}^{x_{i}}\}\), since there are exactly 2n transition edges on the different bases of G
ϕ
and 2m transition edges on the different branches of G
ϕ
. From Lemma 5(b), L contains m additional labels, one for each branch, more specifically for the respective edge \(\{t_{p}^{x_{i}}, w_{p}^{x_{i}} \}\) of the branch. From Lemma 5(c), L contains \(\frac {m(m-1)}{2}\) extra labels among the vertices \(\{t_{p}^{x_{i}}: 1 \leq i \leq n, 1 \leq p \leq 3\}\). From Lemma 5(d), L also contains \(\frac {(n+m) (n+m-1)}{2}\) additional labels among the vertices \(\{w_{p}^{x_{i}}: 1\leq i \leq n, 0\leq p \leq 3 \}\) . From Lemma 5(e), L also contains \(\frac {15}{2}n+1\) labels on the edges incident to t
0. Finally, from Lemma 5(f), L contains at least 4n + m additional labels: for each G
ϕ, i
, L contains at least 4 labels, namely one label on the base edge \(\{s^{x_{i}}, u_{0}^{x_{i}}\}\) or on the base edge \(\{s^{x_{i}}, v_{0}^{x_{i}}\}\) and, for every p ∈ {1, 2, 3}, one label on the edge \(\{u_{0}^{x_{i}},u_{p}^{x_{i}}\}\) or on the edge \(\{v_{0}^{x_{i}},v_{p}^{x_{i}}\}\); also, for each branch of G
ϕ
, L contains at least 1 label, namely a label on the edge \(\{u_{p}^{x_{j}}, t_{p}^{x_{j}}\}\) or on the edge \(\{v_{p}^{x_{j}}, t_{p}^{x_{j}}\}\), for some p ∈ {1, 2, 3} and j ∈ {1, 2, …, m}.
Notice that all the labels of L mentioned above are on different edges, so no subset of labels has been accounted for more than once. Therefore, since \(m=\frac {3n}{2}\), L contains at least:
$$ c(L) \geq \frac{17}{4} n^{2} +17n +\frac{n}{2} +1 $$
(1)
labels.
Now we construct from the labelling L ⊆ L
ϕ
a truth assignment τ for the formula ϕ as follows. For every i ∈ {1, 2, …, n}, if L contains the labels of all edges of the paths P
i,1, P
i,2, P
i,3, then we define x
i
=0 in τ. Otherwise, if L contains the labels of all edges of the paths Q
i,1, Q
i,2, Q
i,3, then we define x
i
=1 in τ. We will prove that |τ(ϕ)|≥k, i.e., that τ XOR-satisfies at least k clauses of the formula ϕ.
Let α = (x
i
⊕x
j
), where i, j ∈ {1, 2, …, n}, be a clause of ϕ that is not XOR-satisfied by τ in ϕ. Let x
i
(resp. x
j
) be associated with the pth (resp. qth) branch of G
ϕ, i
(resp. of G
ϕ, j
). Since α is not XOR-satisfied, either x
i
= x
j
=0 or x
i
= x
j
=1 in τ. If x
i
= x
j
=0 in τ, it follows by the definition of the assignment τ that the labelling L contains the labels of all edges of the path P
i, p
and of the path P
j, q
. Therefore, the p
th branch of G
ϕ, i
, which is identified with the q
th branch of G
ϕ, j
, has both edges \(\{t_{p}^{x_{i}}, u_{p}^{x_{i}} \} \equiv \{t_{q}^{x_{j}}, v_{q}^{x_{j}} \}\) and \(\{t_{p}^{x_{i}}, v_{p}^{x_{i}} \} \equiv \{t_{q}^{x_{j}}, u_{q}^{x_{j}} \}\) labelled under L, with one label each. The same holds if x
i
= x
j
=1, where all edges of both paths Q
i, p
and
Q
j, q
are labelled. So, for all the branches of G
ϕ
that correspond to non-satisfied clauses of ϕ by the truth assignment τ, L contains an additional label (to the ones accounted for by using the result of Lemma 5(f)). The number of clauses that are not satisfied by τ in ϕ is exactly \(m-|\tau (\phi )|= \frac {3}{2} n - |\tau (\phi )|\).
Thus, it follows by (1), by adding the extra \(\frac {3}{2} n - |\tau (\phi )|\), that L contains in total at least:
$$\begin{array}{@{}rcl@{}} c(L) &\geq& \frac{17}{4} n^{2} +17n +\frac{n}{2} +1 + (\frac{3n}{2} - |\tau(\phi)|) \\ & = & \frac{17}{4} n^{2} +19n +1 - |\tau(\phi)| \end{array} $$
labels.
Recall now that we have already shown in Lemma 3 that L
ϕ
has a total of \(\frac {17}{4} n^{2} +28n +1\) labels. Therefore, we have:
$$|L_{\phi }\setminus L | = c(L_{\phi}) - c(L) \leq 9n+|\tau (\phi )|.$$
However, by our initial assumption:
$$|L_{\phi }\setminus L | =r(G_{\phi}, L_{\phi}) \geq 9n+k.$$
Therefore 9n + k ≤ |L
ϕ
∖L| ≤ 9n + |τ(ϕ)|, and thus |τ(ϕ)| ≥ k, i.e., the truth assignment τ satisfies at least k clauses of ϕ. This completes the proof of the theorem. □
The next corollary follows immediately by Theorem 5.
Corollary 3
Let OPT
mon-Max-XOR(3)
(ϕ) the greatest number of clauses that can be simultaneously XOR-satisfied by a truth assignment of ϕ. Then r(G
ϕ
,L
ϕ
)=9n+OPT
mon-Max-XOR(3)
(ϕ).
Proof
Let τ be a truth assignment that satisfies k = OPT
mon-Max-XOR(3)(ϕ) clauses of ϕ. Then there exists by Theorem 5 a TC satisfying labelling L ⊆ L
ϕ
of G
ϕ
such that |L
ϕ
∖L| ≥ 9n + k. Thus, since r(G
ϕ
, L
ϕ
) ≥ |L
ϕ
∖L|, it follows that r(G
ϕ
, L
ϕ
) ≥ 9n + OPT
mon-Max-XOR(3)(ϕ). Conversely, let L ⊆ L
ϕ
be a labelling of G
ϕ
such that |L
ϕ
∖L| = r(G
ϕ
, L
ϕ
). Then there exists by Theorem 5 a truth assignment τ that satisfies at least r(G
ϕ
, L
ϕ
)−9n clauses of ϕ . Thus OPT mon-Max-XOR(3)(ϕ) ≥ r(G
ϕ
, L
ϕ
)−9n, which completes the proof. □
Using Theorem 5 and Corollary 3, we are now ready to prove the main theorem of this section.
Theorem 6
The problem of computing r(G,L) on an undirected temporally connected graph G(L) is APX-hard.
Proof
Denote by OPT mon-Max-XOR(3)(ϕ) the greatest number of clauses that can be simultaneously XOR-satisfied by a truth assignment of ϕ. The proof is done by an L-reduction [31] from the monotone Max-XOR(3) problem, i.e. by an approximation preserving reduction which linearly preserves approximability features. For such a reduction, it suffices to provide a polynomial-time computable function g and two constants α, β>0 such that:
-
r(G
ϕ
, L
ϕ
) ≤ α⋅OPT
mon-Max-XOR(3)(ϕ), for any monotone XOR(3) formula ϕ, and
-
for any TC satisfying labelling L ⊆ L
ϕ
of G
ϕ
, g(L) is a truth assignment for ϕ and OPT mon-Max-XOR(3)(ϕ)−|g(L)| ≤ β⋅(r(G
ϕ
, L
ϕ
)−|L
ϕ
∖L|), where |g(L)| is the number of clauses of ϕ that are satisfied by g(L).
We will prove the first condition for α = 13. Note that a random truth assignment XOR-satisfies each clause of ϕ with probability \(\frac {1}{2} \), and thus there exists an assignment τ that XOR-satisfies at least \( \frac {m}{2}\) clauses of ϕ. Therefore OPT\(_{\text {mon-Max-XOR}(3)}(\phi )\geq \frac {m}{2}=\frac {3}{4}n\), and thus \(n\leq \frac {4}{3}\textit {OPT}_{\text { mon-Max-XOR}(3)}(\phi )\). Now Corollary 3 implies that:
$$\begin{array}{@{}rcl@{}} r(G_{\phi },L_{\phi }) &=&9n+\text{OPT}_{\text{mon-Max-XOR}(3)}(\phi ) \\ &\leq &9\cdot \frac{4}{3}\text{OPT}_{\text{mon-Max-XOR}(3)}(\phi )+\text{OPT} _{\text{mon-Max-XOR}(3)}(\phi ) \\ &=&13\cdot \text{OPT}_{\text{mon-Max-XOR}(3)}(\phi ) \end{array} $$
(2)
To prove the second condition for β = 1, consider an arbitrary labelling L ⊆ L
ϕ
of G
ϕ
. As described in the (⇐ )-part of the proof of Theorem 5, we construct in polynomial time a truth assignment g(L) = τ that satisfies at least |L
ϕ
∖L|−9n clauses of ϕ, i.e. |g(L)|=|τ(ϕ)|≥|L
ϕ
∖L|−9n. Then:
$$\begin{array}{@{}rcl@{}} OPT_{\text{mon-Max-XOR}(3)}(\phi )-|g(L)| &\leq &OPT_{\text{mon-Max-XOR} (3)}(\phi )-|L_{\phi }\setminus L|+9n \\ &=&r(G_{\phi },L_{\phi })-9n-|L_{\phi }\setminus L|+9n \\ &=&r(G_{\phi },L_{\phi })-|L_{\phi }\setminus L| \end{array} $$
(3)
This completes the proof of the Theorem. □
Note
In fact, we have also shown (Theorem 5) that the problem of computing the removal profit is NP-hard in the strong sense, since all numbers used in the reduction are constant integers.
Open Problem
Is there a polynomial-time constant factor approximation algorithm to compute r(G, L)?
Temporally connected random labellings have high removal profit
In this section, we show that dense graphs with random labels have the property TC and have a very high removal profit asymptotically almost surely. More specifically, we consider the complete graph and the Erdös-Renyi model of random graphs, G
n, p
and we examine whether we can delete labels from such temporal graphs and continue preserving TC.
The (single-labelled) model of temporal graphs that we consider here is that of uniform random temporal graphs [1].
Definition 12
[1] A uniform random temporal graph is a graph G on n vertices, \(n \in \mathbb {N}\), each edge of which receives exactly one label uniformly at random from a set \(\{1,2,\ldots , \alpha \},~\alpha \in \mathbb {N}\) and the selection of the label of an edge is independent from the selection of the label of any other edge.
High removal profit in the complete graph
Theorem 7
In the uniform random temporal graph where the underlying graph G is the complete graph (clique) of n vertices and α≥4, we can delete all but Θ(nlogn) labels without violating TC, with probability at least
\(1-\frac {1}{n^{2}}\).
Proof
First, note that any set {1, 2, …, α} of α consecutive natural numbers can be partitioned into 4 disjoint almost equal subsets of consecutive numbers, A
1, A
2, A
3, A
4. Indeed, let α = 4k + v, where \(k\in \mathbb {N}\) and v ∈ {1,2,3,4}.
For v = 0, we use A
1 = {1, …, k}, A
2 = {k + 1,…,2k}, A
3 = {2k + 1,…,3k}, A
4 = {3k + 1,…,4k}.
For v = 1, we use A
1 = {1, …, k}, A
2 = {k + 1,…,2k}, A
3 = {2k + 1,…,3k}, A
4 = {3k + 1,…,4k + 1}.
For v = 2, we use A
1 = {1, …, k}, A
2 = {k + 1,…,2k}, A
3 = {2k + 1,…,3k + 1},A
4 = {3k + 2,…,4k + 2}.
For v = 3, we use A
1 = {1, …, k}, A
2 = {k + 1,…,2k + 1},A
3 = {2k + 2,…,3k + 2},A
4 = {3k+3,…,4k+3}.
In any of the above four cases, each particular edge of the clique K
n
receives a single random label l, with:
$$Pr[l \in A_{i}] \geq \frac{k}{4k+3},~ \forall i=1,2,3,4$$
Since k ≥ 1 (because α ≥ 4), we have \(\frac {k}{4k+3} \geq \frac {1}{7}\). So, we get the following Lemma:
Lemma 6
For each particular edge e of K
n
and for the label l that it receives, it holds that
\(Prob[l \in A_{i}] \geq \frac {k}{4k+3},~ \forall i=1,2,3,4\).
Now, colour green(g), yellow(y), blue(b) and red(r) the edges that are assigned a label in A
1, A
2, A
3 and A
4 respectively.
Definition 13
A temporal router (cf. Fig. 12) of a clique G = K
n
= (V, E) is a subgraph R = (V
R
, E
R
) of G, with 2γ log n + 1 vertices, γ being a constant such that \(\gamma \geq 4 \cdot \frac {1}{\log _{2}{\frac {2500}{2499}}}\), with the following properties (all logarithms are with base 2 here):
-
a)
V
R
is the union of a particular vertex v
0 (called the centre of R) and two equisized vertex sets V
i
n
and V
o
u
t
, each of γlogn vertices, and
-
b)
R is the induced subgraph of G formed from V
R
(so it is a clique itself).
Note that R has \(|E_{R}|=2 \gamma \log {n} + \frac {(2 \gamma \log {n})\cdot (2 \gamma \log {n}-1)}{2}\) edges.
Let w, w
′ be any two vertices of the clique that are not in V
R
. We consider the edges connecting w to V
i
n
and the edges connecting w
′ to V
o
u
t
; using those edges and the edges of R, there are γlogn edge-disjoint paths of length 4 (each) connecting w and w
′. Let us call those paths special paths and note that every such path uses edges of the form {w, v
i
n
}, {v
i
n
, v
0}, {v
0, v
o
u
t
}, {v
o
u
t
, w
′}, where v
i
n
∈ V
i
n
and v
o
u
t
∈ V
o
u
t
(cf. Fig. 13).
Each special path P = (w, v
i
n
, v
0, v
o
u
t
, w
′) connecting w and w
′ becomes a (w, w
′)-journey if the label l
1 of {w, v
i
n
} is in A
1, the label l
2 of {v
i
n
, v
0} is in A
2, the label l
3 of {v
0, v
o
u
t
} is in A
3, and the label l
4 of {v
o
u
t
, w
′} is in A
4. Then, the probability that P is a journey is at least \(\left (\frac {1}{7} \right )^{4}\), due to independence of the labels’ selection.
Since all special paths that connect w and w
′ are edge-disjoint, the probability that none of them is a (w, w
′)-journey is:
$$\begin{array}{@{}rcl@{}} Pr[\text{no special path is a }(w,w^{\prime}) \text{-journey}] = \left(1- \frac{1}{7^{4}} \right)^{\gamma \log{n}} &<& \left(1- \frac{1}{2500} \right)^{\gamma \log{n}} \\ &=& n^{-\gamma \log{\frac{2500}{2499}}}. \end{array} $$
Therefore, we have:
Lemma 7
For any two particular vertices w,w
′
of V∖V
R
, the probability that there is a special path P from w to w
′
that is a (w,w
′
)-journey is at least
\(1-n^{-\gamma \log _{2}{\frac {2500}{2499}}}\).
Now, we consider only the edges and labels of R and, for each w ∈ V∖V
R
, we consider only the edges connecting w to each vertex of R; the sparsified graph G
′= (V, E
′) has, thus, \(|E^{\prime }|= \frac {(2\gamma \log {n} +1) \cdot 2 \gamma \log {n}}{2} +\left (n-(2\gamma \log {n}+1) \right ) \cdot (2\gamma \log {n} +1)={\Theta }(n\log {n}+ \log ^{2}{n}) = {\Theta }(n\log {n}) \) edges. We will show that we need only consider the edges (and labels) of G
′ to maintain TC in G, i.e., that G
′ itself is temporally connected, with probability at least \(1-\frac {1}{n^{2}}\).
Consider any pair, w, w
′, of vertices of the uniform random temporal graph on K
n
and a temporal router R. Also, consider the graph G
′ as described above, with the labelling implied by the uniform random labelling on the clique. If w, w
′ ∈ V
R
, then they are directly connected via a labelled edge in G
′ and thus a journey exists both ways between them. If w ∈ V
R
and w ∈ V∖V
R
, then again there is a direct labelled edge in G
′ connecting w and w
′, so there is a journey between them either way.
It remains to examine the existence in G
′ of journeys between pairs of vertices w, w
′, w ≠ w
′, none of which is in V
R
; there are at most n
2 such pairs of vertices. Under the random labelling on G, let \(\mathcal {E}_{1}\) be the event that there exists a pair w, w
′ ∈ V∖V
R
such that there is no (w, w
′)-journey via a special path through R. Also, let \(\mathcal {E}_{2}\) be the event that for a specific pair w, w
′ ∈ V∖V
R
, there is no (w, w
′)-journey via a special path through R. Then,
$$Pr[\mathcal{E}_{1}] \leq n^{2} Pr[\mathcal{E}_{2}] \text{ (by the Union Bound).}$$
So, we have:
$$Pr[G^{\prime} \text{ is not temporally connected}] \leq n^{2} n^{-\gamma \log_{2}{\frac{2500}{2499}}}. $$
Note that Lemma 7 gives an upper bound on the probability of the event \(\mathcal {E}_{2}\). Set γ to be \(\gamma \geq 4 \cdot \frac {1}{\log _{2}{\frac {2500}{2499}}}\). Then, we have:
$$Pr[G^{\prime} \text{ is not temporally connected}] \leq n^{-2} . $$
□
High removal profit in dense random Erdös-Renyi graphs
In this section, we consider the underlying graph G = (V, E) to be an instance of the Erdös-Renyi graph model, G
n, p
, with n ≥ 14 and \(p\geq 7 \left (\frac {\gamma \ln {n}}{n} \right )^{\frac {1}{7}},~\gamma \geq 24\).
Definition 14 (Erdos-Renyi graphs̈)
An instance of G
n, p
is formed when for every pair of vertices u, v among a total number of n vertices, the edge {u, v} is chosen to exist with probability p independently of any other edge.
We will also use the Multiplicative Chernoff bound, as described below:
Fact Chernoff Bound [28]
Suppose X
1, …, X
n
are independent random variables taking values in {0,1}. Let X denote their sum and let μ = E[X] denote the sum’s expected value. Then, for 0 < δ < 1:
$$\begin{array}{@{}rcl@{}} &&Pr[X > (1+\delta) \mu] \leq e^{- \frac{\delta^{2} \mu}{3}}, \text{ and}\\ &&Pr[X \leq (1-\delta) \mu] \leq e^{- \frac{\delta^{2} \mu}{2}}. \end{array} $$
Notice that G
n, p
is almost surely connected for any \(p \geq 2 \frac {\ln {n}}{n}\) [8]. As in the previous section, we consider here a uniform random temporal graph on G, i.e., we consider each edge of G to receive exactly one label uniformly at random from a set {1, 2, …, α}, with α ≥ 4. The selection of the label of an edge is independent of the selection of the label of any other edge. Also, the label selection process is independent of the process of selection of edges in G
n, p
. As in Theorem 7, we consider partitioning {1, 2, …, α} into four consecutive subsets, A
1, A
2, A
3, A
4, of consecutive positive integers, where each subset is of size either \(\lfloor \frac {\alpha }{4} \rfloor \) or \(\lfloor \frac {\alpha }{4} \rfloor +1\); such a partition is always possible. Now colour green(g), yellow(y), blue(b) and red(r) the edges that are assigned a label in A
1, A
2, A
3 and A
4, respectively. As in Lemma 6, we have:
Lemma 8
For each particular edge of G and for the label l that it receives, it holds that
\(Prob[l \in A_{i}] \geq \frac {1}{7},~ \forall i=1,2,3,4\).
In such instances of G
n, p
, we cannot assume the existence of cliques such as the clique of the temporal router used in the previous section. Indeed, even for very dense instances of G
n, p
, with \(p=\frac {1}{2}\), the largest clique is at most of size 2lnn [8].
In order to “sparsify” labelled instances G of G
n, p
, by removing labels without violating TC, we need to guarantee the existence of much sparser routing subsets of G.
Definition 15
Given two vertices v
1, v
2 of G
n, p
, a temporal router, R(v
1, v
2), in an instance
I
of
G
n, p
is a subgraph of I that has vertices v
1, v
2 and additional vertices a
1, …, a
k
and b
1, …, b
k
so that:
-
v
1 connects directly to each a
i
, b
i
, i = 1, …, k,
-
v
2 connects directly to each a
i
, b
i
, i = 1, …, k,
-
each pair a
i
, b
i
is directly connected, i = 1, …, k,
-
each edge {a
i
, b
i
} receives a green label, i = 1, …, k,
-
each edge {a
i
, v
1} receives a yellow label, i = 1, …, k,
-
each edge {v
1, b
i
} receives a blue label, i = 1, …, k,
-
each edge {v
2, a
i
} receives a blue label, i = 1, …, k,
-
each edge {b
i
, v
2} receives a yellow label, i = 1, …, k.
Figures 14 and 15 show a temporal router R(v
1, v
2) for k = 1 and k = 2 respectively.
Note 1
A temporal router R(v
1, v
2) in an instance I of G
n, p
is temporally connected since:
-
any a
i
can reach any b
j
, via a journey through v
1, i.e., (a
i
, v
1, b
j
) is a journey,
-
any b
i
can reach any a
j
, via a journey through v
2, i.e., (b
i
, v
2, a
j
) is a journey,
-
any a
i
can reach any a
j
≠ a
i
, via a journey through b
i
and then v
2, i.e., (a
i
, b
i
, v
2, a
j
) is a journey,
-
any b
i
can reach any b
j
≠ b
i
, via a journey through a
i
and then v
1, i.e., (b
i
, a
i
, v
1, b
j
) is a journey,
-
v
1 can reach v
2, via any a
i
, i.e., (v
1, a
i
, v
2) is a journey,
-
v
2 can reach v
1, via any b
i
, i.e., (v
2, b
i
, v
1) is a journey, and
-
all other (temporal) connections are direct.
Definition 16
We denote by R
i
and call it the i
th
theta subgraph of
R(v
1, v
2) the labelled subgraph of R(v
1, v
2) induced by the vertices v
1, v
2, a
i
, and b
i
, for some i = 1, …, k (cf. Fig. 16).
Note that the following Lemma holds:
Lemma 9
Let I be an instance of G
n, p
with a uniform random labelling from the set {1, 2, …, α}, α ≥ 4. Fix two vertices v
1
,v
2
and 2k vertices a
i
,b
i
, i = 1, …, k, in I. Then, for each particular i = 1, …, k, the probability that the subgraph induced by the vertices v
1
,v
2
, a
i
, b
i
is a theta subgraph is:
$$Pr[R_{i} \text{ exists in } I] \geq \left(\frac{p}{7} \right)^{5}$$
Proof
Each edge of R
i
is realized in G
n, p
with probability p and receives the correct type of label (green, yellow, blue, or red) with probability at least \(\frac {1}{7}\). Note also that the edges of different theta subgraphs R
i
and R
j
, i ≠ j, are disjoint. Thus, in G
n, p
, the random experiments of each of the theta subgraphs R
i
appearing are independent from each other and each succeeds with probability at least \(\left (\frac {p}{7} \right )^{5}\). □
Now, consider the set of vertices V∖{v
1, v
2} and partition it into two almost equal sets V
1 and V
2; note that \(|V_{i}| \geq \lfloor \frac {n}{2} \rfloor -2 \geq \frac {n}{3} = n^{\prime },~i=1,2\) (since n ≥ 14). Consider a pairing of n
′ vertices of V
1 to n
′ vertices of V
2 and let the n
′ different pairs be the possible pairs of vertices a
i
, b
i
in a theta subgraph R
i
. By Lemma 9 and since the random experiments are independent, the number of appearances of R
i
is at least the number of successes in a Bernoulli distribution of n
′ trials, with success probability \(\left (\frac {p}{7} \right )^{5}\) per trial. Therefore, by the Chernoff bound, we have the following Lemma:
Lemma 10
Consider an instance I of a G
n,p
that has been labelled uniformly at random and fix two vertices v
1
,v
2
in I. The probability that there is a temporal router R(v
1
,v
2
) consisting of at least
\(k= \frac {n^{\prime }}{2} \left (\frac {p}{7} \right )^{5} = \frac {n}{6} \left (\frac {p}{7} \right )^{5} \)
theta subgraphs R
i
is at least
\(1- e^{-\frac {1}{8} n^{\prime } \left (\frac {p}{7} \right )^{5}} = 1- e^{-\frac {n}{24} \left (\frac {p}{7} \right )^{5}} \).
Corollary 4
Note that, again by the Chernoff bound, k asymptotically almost surely does not exceed
\(\frac {3}{2}n^{\prime } \left (\frac {p}{7} \right )^{5} \)
, since
\(Pr[k> \left (1+\frac {1}{2} \right ) E(k)] \leq e^{-\frac {n}{36} \left (\frac {p}{7} \right )^{5}}\).
We now condition on the event, \(\mathcal {E}_{1}\), that the instance I of a labelled G
n, p
has a temporal router R(v
1, v
2) of at least \(k=\frac {n}{6} \left (\frac {p}{7} \right )^{5}\) theta subgraphs. By Lemma 10, we know that:
$$Pr[ \bar{\mathcal{E}_{1}}] \leq e^{-\frac{n}{24} \left(\frac{p}{7} \right)^{5}}.$$
Given that R(v
1, v
2) exists, any vertex u that is not in R(v
1, v
2) can reach any vertex u
′ that is also not in R(v
1, v
2)through
R(v
1, v
2) via a journey, if u connects to some a
i
directly with a green edge, and b
i
connects to u
′ directly with a red edge. Then, (u, a
i
, v
1, b
i
, u
′) is a journey.
The probability of the edge {u, a
i
} being green and the edge {b
i
, u
′} being red, for any i, is \(\left (\frac {p}{7} \right )^{2}\) and it is independent of edge experiments inside R(v
1, v
2). So, we have:
Lemma 11
Condition on the event
\(\mathcal {E}_{1}\)
of the existence of R(v
1
,v
2
) in G
n,p
with at least
\(k= \frac {n}{6} \left (\frac {p}{7} \right )^{5}\)
theta subgraphs. Let u,u
′
be any two (different) vertices of G that are not in R(v
1
,v
2
). Then:
$$Pr[\text{there exists a } (u,u^{\prime}) \text{-journey through } R(v_{1},v_{2})] \geq 1- \left(1- (\frac{p}{7})^{2} \right)^{k} $$
Proof
For the vertices u, u
′ as described above and any one of the k possible journeys of the form (u, a
i
, v
1, b
i
, u
′), the probability that such a journey fails (i.e., is not realized) is at most \(1-\left (\frac {p}{7} \right )^{2}\). Therefore, given \(\mathcal {E}_{1}\), we have:
$$Pr[\text{there exists no } (u,u^{\prime}) \text{-journey through } R(v_{1},v_{2})] \leq \left(1- (\frac{p}{7})^{2} \right)^{k} $$
□
Let \(\mathcal {E}_{2}\) be the event that given k pairs of vertices a
i
, b
i
in a possible R(v
1, v
2), each vertex pair u, u
′, with u ≠ u
′ and u, u
′∉V(R(v
1, v
2)), satisfies the following: there is at least one pair of vertices a
i
, b
i
such that u connects to a
i
with a green edge and
u
′ connects to b
i
with a red edge.
Notice that \(\bar {\mathcal {E}_{2}}\) is the event that there is a pair of vertices u, u
′ that are not in R(v
1, v
2) that fails to connect as described above. Since the number of possible pairs of vertices u, u
′ is less than n
2, we have:
$$\begin{array}{@{}rcl@{}} Pr[\bar{\mathcal{E}_{2}}] & \leq & n^{2} \left(1- (\frac{p}{7})^{2} \right)^{k} \\ & \leq & n^{2} e^{-k (\frac{p}{7})^{2}} \\ & = & e^{-k (\frac{p}{7})^{2} +2 \ln{n}} \end{array} $$
Now, condition on \(\mathcal {E}_{1}\) and on \(\mathcal {E}_{2}\) (given \(\mathcal {E}_{1}\)). Then, for each vertex u∉V(R(v
1, v
2)), keep one of its green edges (to some a
i
) and one of its red edges (to the corresponding b
i
), since by \(\mathcal {E}_{2}\), those exist. Then, remove all edges of I except for the edges of R(v
1, v
2) and the two edges we keep for every vertex that is not in R(v
1, v
2). Notice that the resulting labelled subgraph of I is temporally connected, since:
-
a)
R(v
1, v
2) is temporally connected itself, by construction,
-
b)
any u∉V(R(v
1, v
2)) has a journey via R(v
1, v
2) to any other u
′ ∈ V in the graph,
-
c)
any a
i
or b
j
can reach any u∉V(R(v
1, v
2)) via a journey through v
1(using first a green edge, if we start from a b
j
vertex, and then using a yellow, a blue and a red edge to reach u), and
-
d)
v
1 and v
2 can reach any u∉V(R(v
1, v
2)) via a journey through some vertex b
i
(using first a blue -or yellow, respectively- edge to b
i
, and then a red edge to u).
The temporally connected instance I of G
n, p
after the removal of redundant edges as described above has a number of labelled edges (i.e., time-edges) that is at most 2n+Θ(k). Since \(k=\frac {n}{6} \left (\frac {p}{7}\right )^{5}\), I has at most Θ(n + n
p
5) labels after the removal of the redundant edges.
Recall that we require \(p \geq 7 \left (\frac {\gamma \ln {n}}{n} \right )^{\frac {1}{7}},~\gamma \geq 24\). Therefore, we get the following:
Theorem 8
Consider a G
n,p
, with
\(p \geq 7 \left (\frac {\gamma \ln {n}}{n} \right )^{\frac {1}{7}}\)
, for some γ≥24, labelled uniformly at random. Then, any instance I of G
n,p
needs only Θ(n+np
5
) time-edges to be temporally connected, with probability at least
\(1-2e^{-\frac {\gamma }{24} \ln {n}}\).
Proof
Any instance I of G
n, p
becomes temporally connected by using at most Θ(n + n
p
5) edges (and, thus, labels) as described above, with probability at least:
$$\begin{array}{@{}rcl@{}} Pr[\mathcal{E}_{1}] \cdot Pr[\mathcal{E}_{2} | \mathcal{E}_{1}] & \geq & \left(1- e^{-\frac{n}{24} \left(\frac{p}{7} \right)^{5}} \right) \cdot \left(1- e^{-k (\frac{p}{7})^{2} +2 \ln{n}} \right). \end{array} $$
(4)
Since \(p \geq 7 \left (\frac {\gamma \ln {n}}{n} \right )^{\frac {1}{7}}\) and \(k = \frac {n}{6} \left (\frac {p}{7}\right )^{5}\), we have:
$$\begin{array}{@{}rcl@{}} k\left(\frac{p}{7}\right)^{2} & = & \frac{n}{6} \cdot \left(\frac{p}{7}\right)^{7} \\ & \geq & \frac{n}{6} \cdot \frac{\gamma \ln{n} }{n} \\ &= & \frac{\gamma \ln{n}}{6}. \end{array} $$
(5)
Therefore, from relations (4) and (5), we have:
$$\begin{array}{@{}rcl@{}} Pr[\mathcal{E}_{1}] \cdot Pr[\mathcal{E}_{2} | \mathcal{E}_{1}] &\geq & \left(1- e^{-\frac{n}{24} \left(\frac{p}{7} \right)^{5}} \right) \cdot \left(1- e^{- \frac{\gamma \ln{n}}{6} +2 \ln{n}} \right)\\ & \geq & \left(1- e^{-\frac{n}{24} \left(\frac{p}{7} \right)^{7}} \right) \cdot \left(1- e^{- \frac{\gamma \ln{n}}{6} +2 \ln{n}} \right) \end{array} $$
(6)
Again, since \(p \geq 7 \left (\frac {\gamma \ln {n}}{n} \right )^{\frac {1}{7}}\), we have that: \(\left (\frac {p}{7} \right )^{7} \geq \frac {\gamma \ln {n}}{n}\), so relation (6) becomes:
$$\begin{array}{@{}rcl@{}} Pr[\mathcal{E}_{1}] \cdot Pr[\mathcal{E}_{2} | \mathcal{E}_{1}] & \geq & \left(1- e^{-\frac{n}{24} \frac{\gamma \ln{n}}{n} } \right) \cdot \left(1- e^{- \frac{\gamma \ln{n}}{6} +2 \ln{n}} \right)\\ & \geq & 1- e^{- \frac{\gamma \ln{n}}{6} + 2\ln{n} } - e^{-\frac{n}{24} \cdot \frac{\gamma \ln{n}}{n} } \\ & = & 1 - e^{- \ln{n} \left(\frac{\gamma}{6} -2 \right) } - e^{- \frac{\gamma}{24} \cdot \ln{n} } \\ & = & 1- n^{-\left(\frac{\gamma}{6} -2 \right) } - n^{-\frac{\gamma}{24} }. \end{array} $$
(7)
Now, since γ ≥ 24, we have that \(\frac {\gamma }{6} -2 \geq \frac {\gamma }{24}\). Therefore, from relation (7), we get:
$$\begin{array}{@{}rcl@{}} Pr[\mathcal{E}_{1}] \cdot Pr[\mathcal{E}_{2} | \mathcal{E}_{1}] & \geq & 1- 2n^{-\frac{\gamma}{24} }\\ & = & 1 - 2e^{-\frac{\gamma}{24} \ln{n}}. \end{array} $$
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Note that for the sparsest possible G
n, p
here, i.e., for \(p= 7 \left (\frac {\gamma \ln {n}}{n} \right )^{\frac {1}{7}} \) , we need only \({\Theta }(n+n^{\frac {2}{7}} (\ln {n})^{\frac {5}{7}}) = {\Theta }(n)\) edges (and, thus, labels) to satisfy TC, with probability at least \(1 - 2e^{-\frac {\gamma }{24} \ln {n}},~ \gamma \geq 24\).