Two Illustrative Examples
5.2 The Second Example
The treeT2 is shown in Figure 5.3. Note that the edges inT2 are shorter edges in G0. We demostrate the algorithm twice with two processing sequence,(v, a, c, d, e, f, x, g, b, u, w)and (f, x, d, e, g, b, u, w, c, a, v). The 2nd processed node is a. Its unprocessed neighbor isc. Now the algorithm starts determining ifv ∈ SUCC(a). Becauseα+b_time(a) = 2+0 = 2≤ 17 = tandα+w(a, v)+b_time(v) = 2+3+0 = 5 ≤ 17 = t, the algorithm adds v to SU CC(a) and b_time(a) is 5 now. Because P RED∗(a) = ∅, it is obviously pred(a) = ∅. Because pred(a) =∅ and α + w(c, a) + b_time(a) = 2 + 2 + 5 = 9 ≤ 17 = t, the algorithm adds ato SU CC∗(c).
The 3rd processed node isc. Its unprocessed neighbor is w. Now the algorithm starts determining ifa ∈ SUCC(c). Becauseα + b_time(c) = 2+0 = 2 ≤ 17 = tandα+w(c, a)+b_time(a) = 2+2+5 = 9 ≤ 17 = t, the algorithm adds a to SU CC(c) and b_time(c) is 9 now. Because P RED∗(c) = ∅, it is obviouslypred(c) = ∅. Because pred(c) = ∅ and α + w(w, a) + b_time(a) = 2 + 4 + 9 = 15 ≤ 17 = t, the algorithm adds ato SU CC∗(w).
The 4th processed node is d. Its unprocessed neighbor is e. Because SU CC∗(d) = ∅ and P RED∗(d) = ∅, it is obviously SU CC(d) = ∅, b_time(d) = 0, pred(d) = ∅ and spare(d) = 0. Because pred(d) = ∅ and α + w(e, d) + b_time(d) = 2 + 5 + 0 = 7 ≤ 17 = t, the algorithm addsd to SU CC∗(e).
The 5th processed node ise. Its unprocessed neighbor is w. Now the algorithm starts determining ifd ∈ SUCC(e). Becauseα+b_time(e) = 2+0 = 2≤ 17 = tandα+w(e, d)+b_time(d) = 2+5+0 = 7 ≤ 17 = t, the algorithm adds d to SU CC(e) and b_time(e) is 7 now. Because P RED∗(e) = ∅, it is obviously pred(e) = ∅. Because pred(e) = ∅ and α + w(w, e) + b_time(e) = 2 + 2 + 7 = 11 ≤ 17 = t, the algorithm adds eto SU CC∗(w).
The 6th processed node is f. Its unprocessed neighbor is x. Because SU CC∗(f ) = ∅ and P RED∗(f ) = ∅, it is obviously SU CC(f ) = ∅, b_time(f ) = 0, pred(f ) = ∅ and spare(f ) = 0. Because pred(f ) = ∅ and α + w(x, f ) + b_time(f ) = 2 + 4 + 0 = 6 ≤ 17 = t, the algorithm addsf to SU CC∗(x).
The 7th processed node isx. Its unprocessed neighbor isw. Now the algorithm starts determining ifd ∈ SUCC(x). Becauseα+b_time(x) = 2+0 = 2≤ 17 = tandα+w(x, f )+b_time(f ) = 2+4+0 = 6 ≤ 17 = t, the algorithm adds f to SU CC(x) and b_time(x) is 6 now. Because P RED∗(x) = ∅, it is obviously pred(x) = ∅. Becausepred(x) = ∅and α + w(w, x) + b_time(x) = 2 + 3 + 6 = 11 ≤ 17 = t, the algorithm adds xto SU CC∗(w).
The 8th processed node is g. Its unprocessed neighbor is b. Because SU CC∗(g) = ∅ and P RED∗(g) = ∅, it is obviously SU CC(g) = ∅, b_time(g) = 0, pred(g) = ∅ and spare(g) = 0. Because pred(g) = ∅ and α + w(b, g) + b_time(g) = 2 + 3 + 0 = 5 ≤ 17 = t, the algorithm addsg to SU CC∗(b).
The 9th processed node isb. Its unprocessed neighbor is u. Now the algorithm starts determining ifv ∈ SUCC(b). Becauseα + b_time(b) = 2+0 = 2≤ 17 = tandα+w(a, v)+b_time(g) = 2+3+0 = 5 ≤ 17 = t, the algorithm adds g to SU CC(b) and b_time(b) is 5 now. Because P RED∗(b) = ∅, it is obviously pred(b) = ∅. Because pred(b) = ∅ and α + w(u, b) + b_time(b) = 2 + 4 + 5 = 11 ≤ 17 = t, the algorithm addsb to SU CC∗(u).
The 10th processed node isu. Its unprocessed neighbor isw. Now the algorithm starts determining ifa ∈ SUCC(u). Becauseα+b_time(u) = 2+0 = 2 ≤ 17 = tandα+w(c, a)+b_time(b) = 2+4+5 = 11 ≤ 17 = t, the algorithm adds b to SU CC(u) and b_time(u) is 9 now. Because P RED∗(u) = ∅, it is obviously pred(u) = ∅. Becausepred(u) = ∅and
α + w(w, u) + b_time(b) = 2 + 2 + 11 = 15 ≤ 17 = t, the algorithm addsb to SU CC∗(w).
The last processed node is w. The algorithm reorders SU CC∗(w) = {c, e, x, u}first. Because w(w, c) + b_time(c) = 4 + 9 = 13,w(w, e) + b_time(e) = 2 + 7 = 9 = 13− 2.0α, w(w, x) + b_time(x) = 3 + 6 = 9− 2.0α and w(w, u) + b_time(u) = 2 + 11 = 13, the algorithm gives u1 = c, list0 = {c, u}, list1 = ∅, list2 = {e, x}, list3 = ∅, list4 = ∅. Therefore, the reordered sequence of SU CC∗(v)is (c, u, e, x).
Now the algorithm starts determining SU CC(w). Because
α + b_time(w) = 2 + 0 = 2 ≤ 17 = t and α + w(w, x) + b_time(x) = 2 + 3 + 6 = 11 ≤ 17 = t, the algorithm adds x to SU CC(w) and b_time(w)is 11 now. Because α + b_time(w) = 2 + 11 = 13 ≤ 17 = t and α + w(w, e) + b_time(e) = 2 + 2 + 7 = 11 ≤ 17 = t, the algorithm addsetoSU CC(w)andb_time(w)is 13 now. Becauseα+b_time(w) = 2 + 13 = 15 ≤ 17 = t and α + w(w, u) + b_time(u) = 2 + 2 + 11 = 15 ≤ 17 = t, the algorithm adds u to SU CC(w) and b_time(w) is 15 now. Becauseα + b_time(c) = 2 + 15 = 17 ≤ 17 = tandα + w(w, c) + b_time(c) = 2 + 4 + 9 = 15 ≤ 17 = t, the algorithm addsctoSU CC(w) and b_time(w) is 17 now. All successor candidates ofw are successors ofw. BecauseP RED∗(w) = ∅, it is obviouslypred(w) = ∅. The result is that one broadcast center is enough. The location of the center given by the algorithm isw.
Now we run the algorithm again, but with an another processing se-quence,(f, x, d, e, g, b, u, w, c, a, v). The 1st processed node isf. Its
un-processed neighbor is x. Because SU CC∗(f ) = ∅ and P RED∗(f ) =
∅, it is obviously SU CC(f ) = ∅, b_time(f ) = 0, pred(f ) = ∅ and spare(f ) = 0. Because pred(f ) = ∅ and α + w(x, f ) + b_time(f ) = 2 + 4 + 0 = 6 ≤ 17 = t, the algorithm addsf to SU CC∗(x).
The 2ndprocessed node isx. Its unprocessed neighbor isw. Now the algorithm starts determining ifd ∈ SUCC(x). Becauseα+b_time(x) = 2+0 = 2≤ 17 = tandα+w(x, f )+b_time(f ) = 2+4+0 = 6 ≤ 17 = t, the algorithm adds f to SU CC(x) and b_time(x) is 6 now. Because P RED∗(x) = ∅, it is obviously pred(x) = ∅. Becausepred(x) = ∅and α + w(w, x) + b_time(x) = 2 + 3 + 6 = 11 ≤ 17 = t, the algorithm adds xto SU CC∗(w).
The 3rd processed node isd. Its unprocessed neighbor is e. Because SU CC∗(d) = ∅ and P RED∗(d) = ∅, it is obviously SU CC(d) = ∅, b_time(d) = 0, pred(d) = ∅ and spare(d) = 0. Because pred(d) = ∅ and α + w(e, d) + b_time(d) = 2 + 5 + 0 = 7 ≤ 17 = t, the algorithm addsd to SU CC∗(e).
The 4th processed node ise. Its unprocessed neighbor is w. Now the algorithm starts determining ifd ∈ SUCC(e). Becauseα+b_time(e) = 2+0 = 2≤ 17 = tandα+w(e, d)+b_time(d) = 2+5+0 = 7 ≤ 17 = t, the algorithm adds d to SU CC(e) and b_time(e) is 7 now. Because P RED∗(e) = ∅, it is obviously pred(e) = ∅. Because pred(e) = ∅ and α + w(w, e) + b_time(e) = 2 + 2 + 7 = 11 ≤ 17 = t, the algorithm adds eto SU CC∗(w).
The 5th processed node is g. Its unprocessed neighbor is b. Because
SU CC∗(g) = ∅ and P RED∗(g) = ∅, it is obviously SU CC(g) = ∅, b_time(g) = 0, pred(g) = ∅ and spare(g) = 0. Because pred(g) = ∅ and α + w(b, g) + b_time(g) = 2 + 3 + 0 = 5 ≤ 17 = t, the algorithm addsg to SU CC∗(b).
The 6th processed node isb. Its unprocessed neighbor is u. Now the algorithm starts determining ifv ∈ SUCC(b). Becauseα + b_time(b) = 2+0 = 2≤ 17 = tandα+w(a, v)+b_time(g) = 2+3+0 = 5 ≤ 17 = t, the algorithm adds g to SU CC(b) and b_time(b) is 5 now. Because P RED∗(b) = ∅, it is obviously pred(b) = ∅. Because pred(b) = ∅ and α + w(u, b) + b_time(b) = 2 + 4 + 5 = 11 ≤ 17 = t, the algorithm addsb to SU CC∗(u).
The 7th processed node isu. Its unprocessed neighbor isw. Now the algorithm starts determining ifa ∈ SUCC(u). Becauseα+b_time(u) = 2+0 = 2 ≤ 17 = tandα+w(c, a)+b_time(b) = 2+4+5 = 11 ≤ 17 = t, the algorithm adds b to SU CC(u) and b_time(u) is 9 now. Because P RED∗(u) = ∅, it is obviously pred(u) = ∅. Becausepred(u) = ∅and α + w(w, u) + b_time(b) = 2 + 2 + 11 = 15 ≤ 17 = t, the algorithm addsb to SU CC∗(w).
The 8th processed node is w. Its unprocessed neighbor is c. The algorithm reorders SU CC∗(w) = {x, e, u} first. Because w(w, x) + b_time(x) = 3 + 6 = 9− 2.0α, w(w, e) + b_time(e) = 2 + 7 = 9 = 13− 2.0α and w(w, u) + b_time(u) = 2 + 11 = 13, the algorithm gives u1 = u, list0 = {u}, list1 = ∅, list2 = {x, e},list3 = ∅. Therefore, the reordered sequence ofSU CC∗(v) is (u, x, e).
Now the algorithm starts determining SU CC(w). Because
α + b_time(w) = 2 + 0 = 2 ≤ 17 = t and α + w(w, e) + b_time(e) = 2 + 2 + 7 = 11 ≤ 17 = t, the algorithm adds e to SU CC(w) and b_time(w)is 11 now. Because α + b_time(w) = 2 + 11 = 13 ≤ 17 = t andα + w(w, x) + b_time(x) = 2 + 3 + 6 = 11 ≤ 17 = t, the algorithm addsxtoSU CC(w)andb_time(w)is 13 now. Becauseα+b_time(w) = 2 + 13 = 15 ≤ 17 = t and α + w(w, u) + b_time(u) = 2 + 2 + 11 = 15 ≤ 17 = t, the algorithm adds u to SU CC(w) and b_time(w) is 15 now. All successor candidates ofw are successors of w.
Because P RED∗(w) = ∅, it is obviously pred(w) = ∅. Because α+w(c, w)+b_time(w) = 2+4+15 = 21 > 17 = t, the algorithm gives w cannot be a successor candidate of c. Now the algorithm determines spare(w). Becausearrive(w) + (3 + 1)α + w(w, e) + b_time(e) = 0 + 8+2+7 = 15≤ 17 = t, the algorithm givesecan delayαtime. Because arrive(w)+(2+1)α+w(w, x)+b_time(x) = 0+6+3+6 = 13 ≤ 17 = t, the algorithm givesxcan delayαtime. Becausearrive(w) + (1 + 1)α + w(w, u)+b_time(u) = 0+4+2+11 = 17 ≤ 17 = t, the algorithm gives ucan delayα time. All successors of w can delayα time, the algorithm tells us spare(v) = 0. Because spare(w) + α + w(w, c) = 0 + 2 + 4 = 6≤ 17 = t, the algorithm addsw to P RED∗(c).
The 9th processed node is c. Its unprocessed neighbor is a. Because SU CC∗(c) = ∅, it is obviously SU CC(c) = ∅ and b_time(c) = 0. There is only one predecessor candidate of c, w. Because spare(w) + α + w(w, c) + b_time(c) = 0 + 2 + 5 + 0 = 7 ≤ 17 = t, the algorithm
gives pred(c) = w, spare(c) = arrive(c) = 6 and that c cannot apply for being a successor candidate ofa. Becausespare(c) + α + w(c, a) = 6 + 2 + 2 = 10≤ 17 = t, the algorithm adds c to P RED∗(a).
The 10th processed node isa. Its unprocessed neighbor isv. Because SU CC∗(a) = ∅, it is obviously SU CC(a) = ∅ and b_time(a) = 0. There is only one predecessor candidate ofa,c. Becausespare(c) + α + w(c, a) + b_time(a) = 6 + 2 + 2 + 0 = 10 ≤ 17 = t, the algorithm gives pred(a) = c, spare(a) = arrive(a) = 10 and that a cannot apply for being a successor candidate ofv. Becausespare(a) + α + w(a, v) = 10 + 2 + 3 = 15 ≤ 17 = t, the algorithm addsa to P RED∗(v).
The last processed node isv. BecauseSU CC∗(v) = ∅, it is obviously SU CC(v) =∅andb_time(v) = 0. There is only one predecessor candi-date ofv,a. Becausespare(a)+α+w(a, v)+b_time(v) = 10+2+3+0 = 15 ≤ 17 = t, the algorithm gives pred(v) = a.
The results with the processing sequence (f, x, d, e, g, b, u, w, c, a, v) and(f, x, d, e, g, b, u, w, c, a, v)are the same. Both of them say one broad-cast center is enough and the location of the center isw.
Chapter 6 Conclusion
We made an improvement for finding the location of broadcast centers and the broadcast sequences on any treeT,OP T (T, t), such that broad-casting can be done within the time constraint t, which is introduced in [20]. We extended this problem from uniform telephone model to heterogeneous postal model and improved the time complexity toO(n). Although broadcast problem has several decades of history, There are still many open problems; many of them are shown in Table 6.1, where
”?” means open problems and ”×” means not defined.
Table 6.1: Broadcast problems in uniform telephone model and heterogeneous postal model
approximation ratio : O(loglog(n)log(n) ) ? time : O(|V ||E|) [8]
heuristic O(|E|) [10] ?
How to extend the results from uniform telephone model to hetero-geneous postal model or more complicated environment, from trees to more general structures, and so on, are researchable topics.
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