Premaster Course Algorithms 1 Chapter 6: Shortest Paths. Christian Scheideler SS 2018
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1 Premster Course Algorithms Chpter 6: Shortest Pths Christin Scheieler SS 8
2 Bsic Grph Algorithms Overview: Shortest pths in DAGs Dijkstr s lgorithm Bellmn-For lgorithm Johnson s metho SS 8 Chpter 6
3 Shortest Pths t s Centrl question: Determine fstest wy to get from s to t. SS 8 Chpter 6 3
4 Shortest Pths Shortest Pth Prolem: irecte/unirecte grph G=(V,E) ege costs c:e R SSSP (single source shortest pth): fin shortest pths from source noe to ll other noes APSP (ll pirs shortest pth): fin shortest pths etween ll pirs of noes SS 8 Chpter 6 4
5 Shortest Pths s δ(s,v): istnce etween s n v no pth from s to v δ(s,v) = - pth of ritrrily low cost from s to v min{ c(p) p is pth from s to v} SS 8 Chpter 6 5
6 Shortest Pths s When is the istnce -? If there is negtive cycle: s C v c(c)< SS 8 Chpter 6 6
7 Shortest Pths Negtive cycle necessry n sufficient for istnce of -. Negtive cycle sufficient: pth q pth p s C v c(c)< Cost for i-fol trversl of C: c(p) + i c(c) + c(q) For i this expression pproches -. SS 8 Chpter 6 7
8 Shortest Pths Negtive cycle necessry n sufficient for istnce of -. Negtive cycle necessry: l: miniml cost of simple pth from s to v suppose there is non-simple pth p from s to v with cost c(r)<l p non-simple: continuously remove cycle C till we re left with simple pth p since c(p) < l ut c(p ) l ue to the efinition of l, there must e cycle C with c(c)< SS 8 Chpter 6 8
9 Shortest Pths in DAGs Directe cyclic grph (DAG): DAG with ege costs : reth-first serch ( : eges of BFS-tree) s 4 SS 8 Chpter 6 9
10 Shortest Pths in DAGs Directe cyclic grph (DAG): DAG with ritrry ege costs: reth-first serch oes not work ny more! s SS 8 Chpter
11 Shortest Pths in DAGs Directe cyclic grph (DAG): Correct istnces: s SS 8 Chpter 6
12 Shortest Pths in DAGs Ie: use the fct tht noes in DAGs cn e topologiclly sorte (i.e., ll eges stisfy <) s SS 8 Chpter 6
13 Shortest Pths in DAGs Initilly, set the istnce [s] to n the istnces [v] of ll other noes v to SS 8 Chpter 6 3
14 Shortest Pths in DAGs Next, consier the noes u in the orer of the topologicl sorting n perform istnce upte for u. s SS 8 Chpter 6 4
15 Shortest Pths in DAGs Distnce upte for noe u: for ll eges (u,v) E, set [v] to min{[v],[u]+c(u,v)}. s SS 8 Chpter 6 5
16 Shortest Pths in DAGs Summry: First, compute topologicl sort of the noes (cn e one vi DFS). Then, upte the istnces in the orer of the topologicl sort of the noes. Runtime: O( V + E ). Why oes tht work? SS 8 Chpter 6 6
17 Shortest Pths in DAGs Consier shortest pth from s to v. This hs topologicl sort with t i <t i+ for ll i. t t t 3 t 4 t 5 s 3 4 c c c 3 c 4 v A visit in topologicl orer therefore results in the correct istnces ( i = j i c i ). SS 8 Chpter 6 7
18 Shortest Pths in DAGs Consier shortest pth from s to v. This hs topologicl sort with t i <t i+ for ll i. t t t 3 t 4 t 5 s 3 4 c c c 3 c 4 v Remrk: no noe i long the shortest pth to v cn hve istnce < i to s since otherwise there is shorter pth to v. SS 8 Chpter 6 8
19 Generl Strtegy: Shortest Pths Initilly, set (s):= n (v):= for ll other noes Visit noes in n orer tht ensures tht t lest one shortest pth from s to every v is visite in the orer of its noes For every visite v, upte istnces to noes w with (v,w) E, i.e., (w):= min{(w), (v)+c(v,w)} SS 8 Chpter 6 9
20 Shortest Pths Next step: compute shortest pths for ritrry grphs with positive ege costs. Prolem: visiting noes of shortest pth in the right orer s 3 4 v Solution: visit noes in the orer of their istnce to s. SS 8 Chpter 6
21 Shortest Pths Assumption: ll ege costs re positive. (Actully, Dijkstr lso works if ege costs re non-negtive.) Dijkstr s lgorithm:. Let S e the set of lrey processe noes.. Initilly, S:={s} n [s]:= 3. while V S o 4. choose noe v V\S with t lest one ege from S n [v]:= min u S:(u,v) E [u] + c(u,v) eing miniml mong ll v V\S 5. v to S SS 8 Chpter 6
22 Shortest Pths Assumption: ll ege weights re positive. Dijkstr s lgorithm:. Let S e the set of lrey processe noes.. Initilly, S:={s} n [s]:= How to o tht efficiently?? 3. while V S o 4. choose noe v V\S with t lest one ege from S n [v]:= min u S:(u,v) E [u] + c(u,v) eing miniml mong ll v V\S 5. v to S SS 8 Chpter 6
23 Shortest Pths Assumption: ll ege weights re positive. Dijkstr s lgorithm:. Let S e the set of lrey processe noes.. Initilly, S:={s} n [s]:= 3. while V S o Use hep for tht! 4. choose noe v V\S with t lest one ege from S n [v]:= min u S:(u,v) E [u] + c(u,v) eing miniml mong ll v V\S 5. v to S SS 8 Chpter 6 3
24 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o Here, ege cost of e: w(e) 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u s 7 c DK(Q,v,): DecreseKey opertion Reuces vlue [v] of v in Q to n performs hepifyup to repir hep property from position of v. SS 8 Chpter 6 4
25 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u s c 5 SS 8 Chpter 6 5
26 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u s c 5 Q: s c SS 8 Chpter 6 6
27 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u s c 5 Q: s c SS 8 Chpter 6 7
28 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u s c 5 Q: c SS 8 Chpter 6 8
29 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u s c 5 Q: c SS 8 Chpter 6 9
30 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u s c 5 Q: c SS 8 Chpter 6 3
31 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u s c 5 Q: c SS 8 Chpter 6 3
32 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u s c 5 Q: c SS 8 Chpter 6 3
33 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u s c 5 7 Q: c SS 8 Chpter 6 33
34 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u s c 5 7 Q: c SS 8 Chpter 6 34
35 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u s c 5 7 Q: c SS 8 Chpter 6 35
36 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u s c 5 7 Q: c SS 8 Chpter 6 36
37 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c 5 7 Q: c SS 8 Chpter 6 37
38 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c 5 7 Q: c SS 8 Chpter 6 38
39 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c 5 4 Q: c SS 8 Chpter 6 39
40 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c 5 4 Q: c SS 8 Chpter 6 4
41 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c 5 4 Q: c SS 8 Chpter 6 4
42 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c 5 4 Q: c SS 8 Chpter 6 4
43 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c 5 4 Q: c SS 8 Chpter 6 43
44 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c 5 4 Q: c SS 8 Chpter 6 44
45 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c 5 4 Q: SS 8 Chpter 6 45
46 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c 5 4 Q: SS 8 Chpter 6 46
47 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c 5 4 Q: SS 8 Chpter 6 47
48 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c 5 4 Q: SS 8 Chpter 6 48
49 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c 5 4 Q: SS 8 Chpter 6 49
50 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c 5 4 Q: SS 8 Chpter 6 5
51 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c 5 4 Q: SS 8 Chpter 6 5
52 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c Q: SS 8 Chpter 6 5
53 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c Q: SS 8 Chpter 6 53
54 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c Q: SS 8 Chpter 6 54
55 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c Q: SS 8 Chpter 6 55
56 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c Q: SS 8 Chpter 6 56
57 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q BuilHep(V) 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c Q: SS 8 Chpter 6 57
58 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q V 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c Runtime: Q BuilHep(V) (sets up hep): O( V ) SS 8 Chpter 6 58
59 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q V 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c Runtime: eletemin un DecreseKey: O(log V ) SS 8 Chpter 6 59
60 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q V 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c Runtime: eletemin: V times, DecreseKey: E times SS 8 Chpter 6 6
61 Shortest Pths Dijkstr(G,w,s). for ech vertex v V o [v] ; π[v] nil. [s] ; S ; Q V 3. while Q o 4. u eletemin(q) 5. S S {u} 6. for ech vertex v Aj[u] o 7. if [v]>[u]+w(u,v) then DK(Q,v,[u]+w(u,v)); π[v] u 4 s c Runtime: ltogether, O(( V + E ) log V ) SS 8 Chpter 6 6
62 Shortest Pths in Aritrry Grphs Generl Strtegy: Initilly, set (s):= n (v):= for ll other noes Visit noes in n orer tht ensures tht t lest one shortest pth from s to every v is visite in the orer of its noes For every visite v, upte istnces to noes w with (v,w) E, i.e., (w):= min{(w), (v)+c(v,w)} SS 8 Chpter 6 6
63 Bellmn-For Algorithm Consier grphs with ritrry ege costs. Prolem: visit noes long shortest pth from s to v in the right orer s 3 4 v Dijkstr s lgorithm cnnot e use in this cse ny more. SS 8 Chpter 6 63
64 Bellmn-For Algorithm Exmple: v - s Noe v hs wrong istnce vlue! SS 8 Chpter 6 64
65 Bellmn-For Algorithm Strtegy: visit (n-)-times ll noes in the grph n upte istnces. Then ll shortest pths hve een consiere. s 3 4 Run 3 4 v SS 8 Chpter 6 65
66 Bellmn-For Algorithm Prolem: etection of negtive cycles s Conclusion: in negtive cycle, istnce of t lest one noe keeps ecresing in ech roun, strting with roun <n SS 8 Chpter 6 66
67 Bellmn-For Algorithm Lemm : No ecrese of istnce in roun (i.e., [v]+c(v,w) [w] for ll w): Done ecuse [w]=δ(s,w) for ll w Decrese of istnce even in n-th roun (i.e., [v]+c(v,w)<[w] for some w): There re negtive cycles for ll of these noes, so noe w hs istnce δ(s,w)=-. If this is true for w, then lso for ll noes rechle from w. Proof: exercise SS 8 Chpter 6 67
68 Bellmn-For Algorithm Proceure BellmnFor(s: NoeI) :=<,, >: NoeArry of R {-, } prent:=<,, >: NoeArry of NoeI [s]:=; prent[s]:=s for i:= to n- o // upte istnces for n- rouns forll e=(v,w) E o if [w] > [v]+c(e) then // etter istnce? [w]:=[v]+c(e); prent[w]:=v forll e=(v,w) E o // still etter in n-th roun? if [w] > [v]+c(e) then infect(w) Proceure infect(v) // set - -istnce strting with v if [v]>- then [v]:=- forll (v,w) E o infect(w) SS 8 Chpter 6 68
69 Bellmn-For Algorithm Runtime: O(n m) Improvements: Check in ech upte roun if we still hve [v]+c[v,w]<[w] for some (v,w) E. No: one! Visit in ech roun only those noes w with some ege (v,w) E where [v] hs ecrese in the previous roun. SS 8 Chpter 6 69
70 All Pirs Shortest Pths Assumption: grph with ritrry ege costs, ut no negtive cycles Nive Strtegy for grph with n noes: run n times Bellmn-For Algorithm (once for every noe s the source) Runtime: O(n m) SS 8 Chpter 6 7
71 All Pirs Shortest Pths Better Strtegy: Reuce n Bellmn-For pplictions to n Dijkstr pplictions Prolem: we nee non-negtive ege costs Solution: convert ege costs into nonnegtive ege costs without chnging the shortest pths (not so esy!) SS 8 Chpter 6 7
72 All Pirs Shortest Pths Counter exmple to itive increse y c: efore cost + everywhere v 3 v s s - : shortest pth SS 8 Chpter 6 7
73 Johnson s Metho Let φ:v R e function tht ssigns potentil to every noe. The reuce cost of e=(v,w) is: r(e) := c(e) + φ(v) - φ(w) Lemm : Let p n q e pths connecting the sme enpoints in G. Then for every potentil φ: r(p)<r(q) if n only if c(p)<c(q). SS 8 Chpter 6 73
74 Johnson s Metho Lemm : Let p n q e pths connecting the sme enpoints in G. Then for every potentil φ: r(p)<r(q) if n only if c(p)<c(q). Proof: Let p=(v,,v k ) e n ritrry pth n e i =(v i,v i+ ) for ll i. It hols: r(p) = i r(e i ) = i (φ(v i ) + c(e i ) - φ(v i+ )) = φ(v ) + c(p) - φ(v k ) SS 8 Chpter 6 74
75 Johnson s Metho Lemm 3: Suppose tht G hs no negtive cycles n tht ll noes cn e reche from s. Let φ(v)=δ(s,v) for ll v V. With this φ, r(e) for ll e. Proof: Accoring to our ssumption, δ(s,v) R for ll v We know: for every ege e=(v,w), δ(s,v)+c(e) δ(s,w) (otherwise, we hve contriction to the efinition of δ!) Therefore, r(e) = δ(s,v) + c(e) - δ(s,w) SS 8 Chpter 6 75
76 Johnson s Metho. Crete new noe s n new eges (s,v) for ll v in G with c(s,v)= (ll noes rechle!). Compute δ(s,v) using Bellmn-For n set φ(v):=δ(s,v) for ll v 3. Compute the reuce costs r(e) 4. Compute for ll noes v the istnces δ(v,w) using Dijkstr with the reuce costs on grph G without noe s 5. Compute the correct istnces δ(v,w) vi δ(v,w):=δ(v,w)+φ(w)-φ(v) SS 8 Chpter 6 76
77 Johnson s Metho Exmple: c - c - SS 8 Chpter 6 77
78 Johnson s Metho Step : crete new source s s c - SS 8 Chpter 6 78
79 Johnson s Metho Step : pply Bellmn-For to s φ()= φ()= s c φ(c)=- - φ()= SS 8 Chpter 6 79
80 Johnson s Metho Step 3: compute r(e)-vlues φ()= The reuce cost of e=(v,w) is: r(e) := φ(v) + c(e) - φ(w) φ()= c φ(c)=- φ()= SS 8 Chpter 6 8
81 Johnson s Metho Step 4: compute ll istnces δ(v,w) vi Dijkstr φ()= δ c φ()= 3 3 c 3 c φ(c)=- φ()= SS 8 Chpter 6 8
82 Johnson s Metho Step 5: compute correct istnces vi the formul δ(v,w)=δ(v,w)+φ(w)-φ(v) φ()= φ()= δ c 3 c c φ(c)=- - φ()= SS 8 Chpter 6 8
83 All Pirs Shortest Pths Runtime of Johnson s Metho: O(T Bellmn-For (n,m) + n T Dijkstr (n,m)) = O(n m + n(n log n + m)) = O(n m + n log n) when using Fioncci heps. Prolem with the runtime oun: m cn e quite lrge in the worst cse (up to ~n ) Nevertheless, est known oun: O(n m) SS 8 Chpter 6 83
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