THE PROBLEM THE PROBLEM (1) THE PROBLEM (2) THE MINIMUM ENERGY BROADCAST PROBLEM THE MINIMUM SPANNING TREE

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1 THE MINIMUM ENERGY BROADCAST PROBLEM I.E. THE MINIMUM SPANNING TREE PROBLEM Prof. Tiziaa Calamoeri Network Algorithms A.y. 05/ THE PROBLEM THE PROBLEM () THE PROBLEM () What does it meas sufficietly close?! A wireless ad-hoc etwork cosists of a set S of (fixed) radio statios joit by wireless coectios.! We assume that statios are located o the Euclidea plae (oly partially realistic hp).! Nodes have omidirectioal ateas: each trasmissio is listeed by all the eighborhood (atural broadcast)!! Two statios commuicate either directly (sigle-hop) -if they are sufficietly close- or through itermediate odes (multi-hop).! A trasmissio rage is assiged to every statio: a rage assigmet r : S R determies a directed commuicatio graph G=(S,E), where edge (i, j) E iff dist(i, j) r(i) (dist(i, j)= euclidea distace betwee i ad j).! I other words, (i, j) E iff j belogs to the disk cetered at i ad havig radius r(i). 3

2 THE PROBLEM (3)! For reasos coected with eergy savig, each statio ca dyamically modulate its ow trasmissio power.! I fact, the trasmissio radius of a statio depeds o the eergy power supplied to the statio.! The geeral aim is to save eergy as much as possible. THE PROBLEM ()! I particular, the power P s required by a statio s to trasmit data to aother statio t must satisfy: P s dist(s,t) α where is the distace-power gradiet. Usually (it depeds o the evorimet) I the empty space =.! Hece, i order to have a commuicatio from s to t, the power P s must be proportioal to dist(s,t) 5 THE PROBLEM (5)! Statios of a ad hoc etwork cooperate i order to provide specific etwork coectivity properties by adaptig their trasmissio rages ad, at the same time, they try to save eergy.! THE PROBLEM ()! Accordig to the required property, differet problems are proposed.! For example:! The trasmissio graph is required to be strogly coected. I such a case, the problem is NP-hard ad there is a -approximate alg. i dim. [Kirousis, Kraakis, Krizac, Pelc 0]; there exists a r> s.t. the problem is ot r-approximable.! The trasmissio graph is required to have diameter at most h. Not trivial approximate results are ot kow.! Give a source ode s, the trasmissio graph is required to iclude a spaig tree rooted at s.

3 THE PROBLEM () I this latter case:! A Broadcast Rage Assigmet (for short Broadcast) is a rage assigmet that yields a commuicatio graph G cotaiig a directed spaig tree rooted at a give source statio s.! A fudametal problem i the desig of ad-hoc wireless etworks is the Miimum-Eergy Broadcast problem (for short Mi Broadcast), that cosists i fidig a broadcast of miimal overall eergy. THE PROBLEM () Th. Mi Broadcast is ot approximable withi ay costat factor. Proof. Recall the MiSetCover problem: give a collectio C of subsets of a fiite set S, fid a subset C of C with mi cardiality, s.t. each elemet i S belogs to at least oe elemet of C. Example: S={,,3,,5} C={{,}, {,,3}, {3}, {3,,5}} C ={{,,3},{3,,5}} 0 THE PROBLEM () Proof (ctd). Note. MiSetCover is ot approximable withi c log for some costat c>0, where = S. Give a istace x of MiSetCover it is possible to costruct a istace y of MiBroadcast s.t. there exists a solutio for x of cardiality k iff there exists a solutio for y of cost k+. So, if MiBroadcast is approximable withig a costat the eve MiSetCover is. Cotradictio. THE PROBLEM (0) Proof (ctd). Reductio: x=(s,c) istace of MiSetCover with: S={s, s,, s } ad C={C, C,, C m }. We costruct y=(g,w,s) of MiBroadcast. Nodes of G: {s} U {V C } U {V S } Edges of G:{(s, v ic ), i m}u{(v ic, v js ), i m, s.t. s j i C i } s v C v i C v m C v j S s.t. s j is i C i V C V S

4 THE PROBLEM () THE PROBLEM () Proof (ctd). Fially, defie w(e)= for ay edge e. Let C be a solutio for x. A sol. for y assigs to s ad to all odes of V C i C. The resultig trasmissio graph cotais a spaig tree rooted at s because each elemet i S is cotaied i at least oe elemet of C. The cost of such a sol. is C +. Proof (ctd). Coversely, assume that r is a feasible sol. for y, (w.l.o.g. r(v) is either 0 or if v is i V C : other values would be meaigless) ad r(v)=0 if v is i V S. We derive a sol. C for x selectig all subsets C i s.t. r(v ic )=. It holds that C =cost(r)-. " 3 THE PROBLEM (5) Note We proved that Mi Broadcast is ot approximable withi a costat factor, but we have dealt with the geeral problem. There are some special cases (e.g. the Euclidea bidimesioal oe) that are particularly iterestig ad that behave better! I the followig, we restrict to the special case of Euclidea plae 5 THE PROBLEM ()! Collaboratig i order to miimize the overall eergy is crucial: S S 3 S! S eeds to commuicate with S! let =! Cost of S!S = dist(s, S )! Cost of S!S 3!S = dist(s, S 3 ) +dist(s 3, S )! Whe agle S S 3 S is obtuse: dist(s, S ) > dist(s, S 3 ) +dist(s 3, S )

5 THE PROBLEM () THE PROBLEM ()! I the Euclidea case, a rage assigmet r ca be represeted by the correspodet family D = {D,..., D l } of disks, ad the overall eergy is defied as: cost(d) = where r i is the radius of D i. l i= r i α! Cosider the complete ad weighted graph G ( ) where the weight of each arc e=(u,v) is dist(u,v).! The broadcast problem is strictly related with the miimum spaig tree o G ( ), i view of some importat properties THE PROBLEM () The set of coectios used to perform a broadcast from s: caot geerate a cycle, because odes do ot eed to be iformed twice tree miimizes the overall eergy log arcs waste more eergy tha short oes. THE PROBLEM (0)! Nevertheless, the Miimum Broadcast problem is ot the same as the Mi Spaig Tree problem: The eergy wasted by each ode u is max (u,v) T { dist(u,v) } α (i.e. ot all the arcs appear with their cotributio) Leaves waste o eergy 0

6 THE PROBLEM ()! The Miimum Broadcast problem is NP-hard i its geeral versio ad it is either approximable withi (- ) where is the maximum degree of T ad is a arbitrary costat! Nothig is kow about the hardess of the geometric versio. THE PROBLEM ()! A approx algorithm is based o the computatio of the MST:! compute the MST of the complete graph iduced by S,! Assig a directio to arcs (from s to the leaves)! Assig to each ode i a radius equal to the legth of the logest arc outgoig from i! Easy to implemet # deep aalysis of the approx ratio.! [Clemeti+al. 0] the first costat approx ratio (about 0)! [Ambüehl 05] the best (tight) kow approx ratio () MINIMUM SPANNING TREE ()! Obs. : If the weights are positive, the a MST is i fact a miimum-cost subgraph coectig all odes.! Proof: A subgraph cotaiig cycles ecessarily has a higher total weight. " 3 THE MINIMUM SPANNING TREE PROBLEM (RECUP)! Obs. : There may be several miimum spaig trees of the same weight havig a miimum umber of edges.! I particular, if all the edge weights of a give graph are the same, the every spaig tree of that graph is miimum.

7 MINIMUM SPANNING TREE ()! Obs. 3: If each edge has a distict weight, the there is a uique MST.! This is true i may realistic situatios, where it's ulikely that ay two coectios have exactly the same cost! Proof: Assume by cotradictio that MST T is ot uique. So, there is aother MST with equal weight, say T. 5 MINIMUM SPANNING TREE (3) (proof ctd)! Let e be a edge that is i T but ot i T. As T is a MST, {e } U T cotais a cycle C ad there is at least oe edge e i T that is ot i T ad lies o C.! If the weight of e is less tha that of e : replacig e with e i T yields tree {e } U T \ {e } which has a smaller weight compared to T. Cotradictio, as we assumed T is a MST but it is ot.! If the weight of e is larger tha that of e : a similar argumet ivolvig tree {e } U T \ {e } also leads to a cotradictio.! We coclude that the assumptio that there is a further MST was false. " MINIMUM SPANNING TREE ()! Obs. : For ay cycle C i the graph, if the weight of a edge e of C is larger tha the weights of all other edges of C, the this edge caot belog to a MST.! Proof: Assumig the cotrary, i.e. that e belogs to a MST T, the deletig e will break T ito two subtrees with the two edpoits of e i differet subtrees. The remaider of C recoects the subtrees, i particular there is a edge f of C with edpoits i differet subtrees, i.e., it recoects the subtrees ito a tree T with weight less tha that of T, because the weight of f is less tha the weight of e. " MINIMUM SPANNING TREE (5)! Obs. 5: If the edge of a graph with the miimum cost e is uique, the this edge is icluded i ay MST.! Proof: If e was ot icluded i the MST, removig ay of the (larger cost) edges i the cycle formed after addig e to the MST, would yield a spaig tree of smaller weight. "

8 MINIMUM SPANNING TREE () MINIMUM SPANNING TREE ()! Obs. : For ay cut C i the graph, if the weight of a edge e of C is strictly smaller tha the weights of all other edges of C, the this edge belogs to all MSTs of the graph.! Proof: If e was ot icluded i the MST, addig e to the MST produces a cycle. Removig ay of the (larger cost) edges of the cut i the cycle, would yield a spaig tree of smaller weight. "! By similar argumets, if more tha oe edge is of miimum weight across a cut, the each such edge is cotaied i a miimum spaig tree. Three classical algorithms:! Kruskal [ 5]! Prim [ 5]! Boruvka [ ] 30 MINIMUM SPANNING TREE () MINIMUM SPANNING TREE ()! The three algorithms are all greedy algorithms ad based o the same structure:! Give a set of arcs A cotaiig some MST arcs, e is a safe arc w.r.t. A if A U {e} cotais oly MST arcs, too.! A=empty set While A is ot a MST fid a safe arc e w.r.t. A A=A U e difficult issue 3! A=empty set while A is ot a MST fid a safe arc e w.r.t. A A=A U e wheever:! A is acyclic! graph G A =(V, A) is a forest whose each coected compoet is either a ode or a tree! Each safe arc coects differet coected compoets of G A! the while loop is ru - times 3

9 KRUSKAL ALGORITHM ()! A=empty set While GA is ot a MST fid a safe arc e w.r.t. A A=A U {e} KRUSKAL ALGORITHM () Amog those coectig two differet coected compoets i GA, choose the oe with miimum weight 0 Implemetatio usig:! Nodes i a mi-priority queue w.r.t. key(v)=mi weight of a arc coectig v to a ode of the mai coected compoet; if it does ot exist! If the priority queue is a heap # Complexity: O(m log )! If the priority queue is a Fiboacci heap # Complexity: O(m+ log ) 35 [Ahuja, Magati & Orli 3] 0 33 A=empty set While GA is ot a MST fid a safe arc e w.r.t. A A=A U {e}! 0 [Johso 5, Cherito & Tarja ] Amog those coectig the mai coected compoet with a isolated ode, choose the oe with miimum weight Implemetatio usig:! Data structure Uio-Fid! The set of the arcs of G is sorted w.r.t. their weight! Time Complexity: O(m log ) PRIM ALGORITHM () PRIM ALGORITHM ()

10 BORUVKA ALGORITHM () (purpose: a efficiet electrical coverage of Moravia) Hipothesis: each arc has a distict weight! A=empty set While A is ot a MST for each coected compoet C i of G A fid a safe arc e i w.r.t. C i A=A U {e i } Amog those coectig C i to aother compoet, the oe with miimum weight Trick: hadle may arcs (exactly log of the # of coected compoets) durig the same loop 3 Impossible to itroduce cycles, thaks to the hipothesis! Complexity: O(m log ) BORUVKA ALGORITHM () OTHER ALGORITHMS () OTHER ALGORITHMS ()! [Friedma & Willard ] Liear time algorithm, but it assumes the edges are already sorted w.r.t. their weight. Not used i practice, as the asymptotic otatio hides a huge costat.! [Matsui 5] Liear time algorithm for plaar graphs (possible lesso)! [Frederickso 5, Eppstei ] Give a graph ad its MST, it is eve iterestig to fid a ew MST after that the origial graph has bee slightly modified. It ca be performed i average time O(log )! Oly O(+m) time is ecessary to verify whether a give spaig tree is miimum. 3 0

11 ANOTHER APPLICATION A telecommuicatio compay wats to lay cable to a ew eighborhood.! It is costraied to bury the cable oly alog certai paths (e.g. alog roads).! Model as a (ot geometrical) graph:!!!! odes: represet poits edges: represet those paths (edge) weight: cost of addig cable o that path. AGAIN ON MINIMUM ENERGY BROADCAST Note. some of those paths might be more expesive, because they are loger, or require the cable to be buried deeper Note. there is o requiremet for edge legths to obey ormal rules of geometry such as the triagle iequality.! A miimum spaig tree for that graph would be a subset of those paths that has o cycles but still coects to every house with the lowest total cost, thus would represet the least expesive path for layig the cable. HEURISTICS () I [Wieselthier, Nguye, Ephremides, 00]: three heuristics all based o the greedy techique:! SPT (spaig path tree): it rus Dijkstra algorithm to get the miimum path tree, the it directs the edges of the tree from the root to the leaves.! BAIP (Broadcast Average Icremetal Power): it is a modificatio of the Dijkstra algorithm based o the odes (i.e. a ew ode is added to the tree o the basis of its miimum average cost).! MST (mi spaig tree): it rus Prim algorithm to get a MST, the it directs the edges of the tree from the 3 root to the leaves. HEURISTICS () GREEDY IS NOT ALWAYS GOOD Greedy is ot always good [Wa, Caliescu, Li, Frieder 0]:! SPT: it rus Dijkstra algorithm to get the miimum path tree, the it directs the edges of the tree from the root to the leaves p3 p q3 ε ε q o q p qm pm (let =)! SPT outputs a tree with total eergy: +/(- )! If the root trasmits with radius the eergy is! Whe $0 SPT is far / from the optimal solutio.

12 HEURISTICS (3) GREEDY IS NOT ALWAYS GOOD HEURISTICS () GREEDY IS NOT ALWAYS GOOD! BAIP (Broadcast Average Icremetal Power): it is a modificatio of the Dijkstra algorithm based o the odes: a ew ode is added to the tree o the basis of the mi average cost=eergy icreasig/# of added odes! It has bee desiged to solve the problems of SPT 5 (let =): 3 ε! The mi trasmissio power of the source to reach k receivig odes is k =k ad thus the average power efficiecy is k/k=! O the other had, the mi trasmissio power of the source to reach all receivig odes is ( - ) =- ad thus the average power efficiecy is (- )/=- / HEURISTICS (5) GREEDY IS NOT ALWAYS GOOD! BAIP will let the source to trasmit at power - to reach all odes i a sigle step.! However, the opt. routig is a path cosistig of all odes from left to right. Its mi power is: ( i i ) + ( ε ) < ( i i ) = i= i= 3 ε ( i + i ) (( i i )( i + i )) ( i i ) = = ( i + i ) i= ( i + i ) i= (i (i )) = = =+ ( i + i ) ( i + i ) ( i + i ) i= i= i= HEURISTICS () 3 GREEDY IS NOT ALWAYS GOOD (computatio of the performace ratio of BAIP ctd) + + i= i + (i ) + i i ) i + (i ) i= + i + (i ) =+ + i 3 (i ) i= Substitutig i=j+: + j + j = j = Thus the approx ratio of BAIP is at least: ε (ε 0) l( ) + 5 l( ) + 5 = l + o() i= i= ε j + l( ) + 5 (l( ) +) =

13 HEURISTICS () GREEDY IS NOT ALWAYS GOOD MST: it rus Prim algorithm to get a MST, the it directs the edges of the tree from the root to the leaves p p 5 +ε +ε p 3 +ε +ε o +ε p p! Path op p is the uique MST, ad its total eergy is.! O the other had, the opt. routig is the star cetered at o, whose eergy is (+ ).! The approx. ratio coverges to, as goes to 0. HEURISTICS ()! We have just show a lower boud o the approximatio ratio of MST.! This ratio is a costat ad a upper boud is.! The proof ivolves complicated geometric argumets, ad therefore we oly sketch some of them:! 50 p HEURISTICS ()! Ay pair of edges do ot cross each other The blue edge is ecessarily shorter tha at least oe of the two crossig edges HEURISTICS (0) (properties of the geometric MST ctd)! The agles betwee ay two edges icidet to a commo ode is at least π/3 The blue edge is ecessarily shorter tha at least oe of the two orage edges 5 5

14 HEURISTICS () (properties of the geometric MST - ctd)! The lue determied by each edge does ot cotai ay other odes. The lue through poits p ad p is the itersectio of the two ope disks of radius dist(p,p ) cetered at p ad p, respectively, hece a iteral ode would create a cycle 53 HEURISTICS () (properties of the geometric MST ctd)! Let p p be ay edge. The the two edpoits of ay other edge are either both outside the ope disk D(p, dist(p, p )) or both outside the ope disk D(p, dist(p, p )) The red edges are added before tha the blue edge because they are shorter. The blue edge would create a cycle. 5 HEURISTICS (3) HEURISTICS ()! Obs. The proof i [Wa, Caliescu, Li, Frieder 0] cotais a small flaw that ca be solved, arrivig to a approximatio ratio of,5 [Klasig, Navarra, Papadopoulos, Perees 0]! Idipedetly, a approximatio ratio of 0 has bee stated i [Clemeti, Crescezi, Pea, Rossi, Vocca 0]! For realistic istaces, experimets suggest that the tight approximatio ratio is ot but [Flammii, Navarra, Perees 0] -> possible lesso! Approx. ratio improved to, [Flammii, Klasig, Navarra, Perees 0]! Approx. ratio improved to,33 [Navarra 05]! Optimal boud [Ambüehl 05] 55 5