Lecture 6: Breadth-First Search

Transcription

1 Leture 6: Breth-First Serh Outline of this Leture The shortest pth prolem. The reth-first serh lgorithm. The running time of BFS. The orretness proof. Note: We introue BFS for unirete grphs, ut the sme lgorithm will lso work for irete grphs.

2 e Shortest Pths s Exmple: 3 simple pths from the soure s to : s,, s,,, s, e,, of length,, 3 respetively. So the shortest pth from s to is s,. The shortest pths from s to other verties re s,, s,, s,,, s,,, s, e. There re two shortest pths from s to.

3 The Shortest Pth Prolem e s Distne [v]: The length of shortest pth from s to v. For exmple [] =. We efine [s] = 0. The Prolem: Given grph G = (V, E) n soure vertex s V, fin the istnes [v] n shortest pth from s to eh other vertex in G. 3

4 Wht Does the Breth-First Serh Do Given grph G = (V, E), the BFS returns the istnes [v] from s to v; the preeessors pre[v], whih is use to erive shortest pth from s to every other vertex v. BFS is tully returning shortest pth tree in whih the unique pth from s to noe u is shortest pth from s to u in the originl grph. Remrks: In ition to the two rrys [v] n pre[v], the BFS lso uses nother uxiliry rry olor[v], whih hs three possile vlues: white (W, unisovere ), gry (G, isovere ut not proesse ), lk (B, isovere n proesse ). 4

5 The Ie of the BFS: Visit the verties s follows:. visit ll verties t istne. visit ll verties t istne 3. visit ll verties t istne 3 et. Initilly, s is me gry. The Breth-First Serh When gry vertex is visite, its olor is hnge to lk, n the olor of ll white neighors is hnge to gry. Gry verties re kept in queue Q. 5

6 The Breth-First Serh More etils. G given y its jeny list. Initiliztion, first prt: For eh vertex u V, olor[u] = W ; [u] = ; pre[u] = NIL; Initiliztion, seon prt: olor[s] = G, [s] = 0, Q = s. Min loop: if Q is nonempty, u = equeue(q) for eh v j[u], if (olor[v] == W ), o olor[v] = G [v] = [u] + pre[v] = u put v in Q olor[u] = B. 6

7 Exmple of the Breth-First Serh Prolem: Given the following unirete grph n soure vertex, fin the istne from s to eh vertex u V n the preeessor pre[u] long shortest pth y following the lgorithm esrie erlier. s f e 7

8 Initiliztion, first prt Exmple Continue vertex u s e f olor[u] W W W W W W W [u] pre[u] NIL NIL NIL NIL NIL NIL NIL s f e 8

9 Initiliztion, seon prt Exmple Continue vertex u s e f olor[u] G W W W W W W [u] 0 pre[u] NIL NIL NIL NIL NIL NIL NIL Q = s (Put s into Q (isovere) & mrk G, mening unproesse ) s 0 f e 9

10 While loop, first itertion Exmple Continue Dequeue s from Q. Fin j[s] =, e. Mrk n e G, mrk s B. Upte [], [e], pre[], pre[e]. Put, e in Q. vertex u s e f olor[u] B W G W W G W [u] 0 pre[u] NIL NIL s NIL NIL s NIL Q =, e s 0 f e 0

11 While loop, seon itertion Exmple Continue Dequeue from Q. Fin j[] = s,,, f. Mrk,, f G, mrk B. Upte [], [], [f], pre[], pre[], pre[f]. Put,, f in Q. vertex u s e f olor[u] B G B G W G G [u] 0 pre[u] NIL s NIL s Q = e,,, f s 0 f e

12 While loop, thir itertion Exmple Continue Dequeue e from Q. Fin j[e] = s,,, f. Mrk gry, mrk e B. Upte [], pre[]. Put in Q. vertex u s e f olor[u] B G B G G B G [u] 0 pre[u] NIL s e s Q =,, f, s 0 f e

13 While loop, forth itertion Exmple Continue Dequeue from Q. Fin j[] =, e. Mrk B. vertex u s e f olor[u] B B B G G B G [u] 0 pre[u] NIL s e s Q =, f, s 0 f e 3

14 While loop, fifth itertion Exmple Continue Dequeue from Q. Fin j[] =,. Mrk B. vertex u s e f olor[u] B B B B G B G [u] 0 pre[u] NIL s e s Q = f, s 0 f e 4

15 While loop, sixth itertion Exmple Continue Dequeue f from Q. Fin j[f] =, e. Mrk f B. vertex u s e f olor[u] B B B B G B B [u] 0 pre[u] NIL s e s Q = s 0 f e 5

16 While loop, seventh itertion Exmple Continue Dequeue from Q. Fin j[] =, e. Mrk B. vertex u s e f olor[u] B B B B B B B [u] 0 pre[u] NIL s e s Q = s 0 f e 6

17 While loop, eigth itertion Sine Q is empty, stop. Exmple Continue vertex u s e f olor[u] B B B B B B B [u] 0 pre[u] NIL s e s Q = s 0 f e 7

18 Exmple Continue s 0 f e Question: How o you onstrut shortest pth from s to ny vertex y using the following tle? vertex u s e f olor[u] B B B B B B B [u] 0 pre[u] NIL s e s 8

19 The Breth-First Serh Algorithm for (eh vertex u V ) { olor[u] = W ; [u] = ; pre[u] = NIL; } olor[s] = G; [s] = 0; enqueue(q, s); if (Q is nonempty) { u = equeue(q); for (eh v j[u]) if (olor[v] == W ) { olor[v] = G; [v] = [u] + ; pre[v] = u; enqueue(c, v); } olor[u] = B; } 9

20 Anlysis of the Breth-First Serh Algorithm Let n = V n e = E. We ssume tht it tkes one time unit to test the olor of vertex, or to upte the olor of vertex, or to ompute [v] = [u] +, or to set pre[v] = u, or to enqueue, or to equeue. The following nlysis is vli for onnete grphs. The initiliztion requires 3n + 3 time units. Eh vertex u must e proesse n is proesse one. Wht is the totl mount of time for proessing u? For eh v j[u], if olor[v] = W, the inner loop tkes 4 time units. Otherwise the inner loop will not e rrie out. equeue[u] n set olor[u] = B tke time units. Hene the totl mount of time neee for proessing u is t most 5 eg(u) + Hene the totl mount of time neee for proessing ll the verties is t most (5 eg(u) + ) = 0e + n. u V Hene T (n, e) (3n + 3) + (0e + n) = 5n + 0e + 3 = O(n + e). 0

21 Anlysis of the Breth-First Serh Algorithm The nlysis n e improve: Eh vertex is olore G extly one. Therefore, the inner loop is exeute extly (n ) times. Hene T (n, e) = (3n + 3) + 4(n ) + u V (eg(u) + ) = (3n + 3) + (4n 4) + (e + n) = 9n + e. Compre to T (n, e) 5n + 0e + 3. Remrk: Note tht e n(n )/. Sine the grph is onnete, e n. If e = Θ(n), then T (n, e) = Θ(n).

22 Grphs tht re not onnete The BFS lgorithm lso works for grphs tht re not onnete. For suh grphs, only the verties v tht re in the sme omponent s s will get vlue [v]. In prtiulr, we n use the rry [ ] t the en of the omputtion to eie if the grph is onnete. Alterntively, we n use the rry olor[ ] or the rry pre[ ]. Explin why. How is the nlysis of the BFS lgorithm hnge if we o not ssume tht the grph is onnet?

23 We n tully moify BFS so tht it returns forest. More speifilly, if the originl grph is ompose of onnete omponents C, C,..., C k then BFS will return tree orresponing to eh C i. BF S(s) Strt BFS olor[s] = G; [s] = 0; enqueue(q, s); if (Q is nonempty) { u = equeue(q); for (eh v j[u]) if (olor[v] == W ) { olor[v] = G; [v] = [u] + ; pre[v] = u; enqueue(c, v); } olor[u] = B; } En BFS for (eh vertex u V ) Initilize { olor[u] = W ; [u] = ; pre[u] = NIL; } for (eh vertex u V ) Strt Connete Component if [u] ) BF S(u); 3

24 Corretness of the BFS Algorithm The orretness of the BFS lgorithm onsists of the following two prts.. Prove tht the BFS lgorithm outputs the orret istne [v].. Prove tht the pths otine y using the rry pre[v] re the shortest. Sine the pth onstrute with the rry pre[v] hs length extly [v], we nee to prove only the first prt! 4

25 Corretness of the BFS Algorithm Oservtions: Any vertex v in Q hs rel vlue [v]. For u, v Q t ny time, if [u] < [v] then u ws isovere erlier thn v n (will e proesse) erlier thn v. Proof: No proof is given here. You re enourge to ome up with your own proof. Theorem: The BFS lgorithm outputs the orret istne [v]. Proof: See next pge. 5

26 Corretness of the BFS Algorithm Proof: By inution on [v]. If [v] = 0, then v = s. onlusion is true. The Assume tht [v] is the orret istne for ll [v] < i. Consier the se [v] = i. If [v] were not the orret istne, then the true istne [v] < [v]. We then hve two pths: s t w () st... must e shortest pth y inution hypothesis s []=[v] =i <i () sw... must e shortest pth s it is supth of the shortest pth sw...v (3) istint from euse []<[], while oth [] n [] re true istne [] = [v] = i v [] = [v] < [] [v] < [v] = i Sine [] < [], shoul e proesse erlier thn, n shoul isover v. This is ontrry to tht isovere v. 6