1 / 13

0,1

". ¥. d[1]:=0;. i>1: d[. i]:=. ;. Æ. S:=. ;. Repeat. Î. È. u:=arg. min. {d[. i]|. i. NS}, S:=S. {u}. ". Î. i. NS: d[. i]:=. min. {d[. i],d[u]+c[. u,i]}. until. S=N. 11,1. 10,2. 5,1. 5. 4. 16. 2. 13. 5. 15. 8. 11. 0,1. 17,1. 26,4. 8. 7. 17. 1.

kedma
Download Presentation

0,1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. " ¥ d[1]:=0; i>1: d[ i]:= ; Æ S:= ; Repeat Î È u:=arg min {d[ i]| i N\S}, S:=S {u} " Î i N\S: d[ i]:= min {d[ i],d[u]+c[ u,i]} until S=N 11,1 10,2 5,1 5 4 16 2 13 5 15 8 11 0,1 17,1 26,4 8 7 17 1 5 9 7 18,4 3 7 13,4 14 -5 3 12 3 M,0 8 9 25,4 6 14 Shortest path with negative arc-costs allowed. Dijkstra? No! Better now Bellman-Ford:

  2. Search(s: node); input A(i)’s of a (di)graph (N,A) [1];node-labels pr(i)N; integer k(=0); output for j in component of s an s-j path begin  iN: pr(i):=0; pr(s):=s; LIST:=<s>; whileLIST do remove the first node from LIST, say i; k:=k+1; /* just counting the number of list-removals/scannings for all(i,j)A(i) with pr(j)=0do pr(j):=i; put j on LIST; end; iN: pr(i)=0; compNr:=0; for all sN with pr(s)=0 do compNr:=compNr+1,Search(s) ; [1] if (N,A) is undirected then for any edge (i,j): (i,j) A(i) and (j,i) A(j)

  3. reNumbering input A(i)’s of acyclic digraph (N,A);node-labels nr(i), indgr(i)Z; integer k(=0); output topological numbering s.t. nr(i)<nr(j) for (i,j) begin  iN: nr(i):=0; indgr(i):=|inA(I)|; LIST:=<s: with indegree(s)=0>; whileLIST do remove a node from LIST, say i; k:=k+1; nr(i):=k; /* here is a purpose withNr(k):=i for all(i,j)A(i)do indgr(j):=indgr(j)-1; if indgr(j)=0 then put j on LIST; end; DPshortestpaths s:=withNr[1]; d(s):=0; for k:=2 to ndo i:=withNr[k], (j,i) inA(i) : if d(j)+c(j,i)< d(i)thend(i):=d(j)+c(j,i) ; k:=0; reNumbering; ifk=nthen write('graph is acyclic');

  4. Search(s: node) input A(i)’s of a (di)graph G=(N,A);labels pr(i)N{0}; output for j in component of s an s-j path begin  iN: pr(i):=0; pr(s):=s; LIST:=<s>; whileLIST do remove a node from LIST, say i; for all(i,j)A(i) with pr(j)=0do pr(j):=i; put j on LIST; end;

  5. BreadthFirst(s: node); /* scan first nodes first: fifo list management input A(i)’s of a (di)graph G=(N,A); labels pr(i)N{0}, d(i) Z0; output for j in component of s an arc-minimals-j path with d(j) arcs begin  iN: pr(i):=0, d(i):=; pr(s):=s; LIST:=<s>; d(s):=0; whileLIST do remove the first node from LIST, say i; for all(i,j)A(i) with d(j)=do pr(j):=i, d(j):=d(i)+1; put jlast on LIST; end;

  6. Dijkstra(s: node); /* scan 'best' nodes first input A(i)’s of a (di)graph G=(N,A);labels pr(i)N{0}, d(i) Z0; output for j in component of s a minimal-costs-j path of cost d(j) begin  iN: pr(i):=0, d(i):=; pr(s):=s; LIST:=<s>; d(s):=0; whileLIST do remove the best node from LIST, say i with minimal d(i); for all(i,j)A(i) with d(j)>d(i)+c(i,j)do pr(j):=i, d(j):=d(i)+c(i,j) put j on LIST (if not yet in) end;

  7. Complexiteitstheorie: YES\NO recognition problems TSP>=k Instance: netwerk (N,E,c) met c:EZ+ ; getal k Question:Is er een tour met lengte < k? • Stel • algoritme A lost op: yes-no versies TSP<=k in complexiteit tA(n) • er is algoritme B voor TSPoptin complexiteit tB(n)= log(nC) · tA(n) • algoritme B*/ past bisectie toe met A • U:=nC; */ C=max {cij| (i,j) in E} • L:=0; */U is boven- en L is ondergrens op z* • Repeat • k:=(U+L)/2; • if algoritme A op TSP<=k returns YES then U:=k • else L:=k ; • /* opnieuw geldt: U is boven- en L is ondergrens op z* • en het interval [L,U] is gehalveerd ! • Until |U-L|< 1;

  8. Voorbeelden YES\NO recognition problems Graph-Connectedness Instance: graph G=(V, E) Question:Is G connected? Satisfiability (informeel) Instance:Boolean expression Question: Kan deze expressie de waarde ‘true’ aannemen (is it satisfiable) Langste Pad Instance: netwerk (N,E,c) met c:EZ+ ; s,t N; getal k Question:Is er een s-t pad P met lengte > k? LP Instance: (c,A,b) resp. n-vector, mxn matrix en m-vector;getal k Question:Is max{cx| xRn met Ax<=b} > k? ILP Zn Graph-Isomorfisme (informeel) Instance: graphs G1=(V1, E1) en G2=(V1, E1) Question:Zijn G1 en G2 hetzelfde?

  9. Graph-Connectedness Instance: graph G=(V, E) Question:Is G connected? MST Instance: graph G=(V, E,c); number k; Question:Is there a tree T spanning V of total weight <=k? Steiner tree Instance: graph G=(V, E,c); subset of nodes K; number k; Question:Is there a tree spanning K of length <=k? UFL Instance: clients i=1,…,m with unit demand, facilities j=1,…n with one-time fixed charge opening costs fj; costs cij for delivering the unitof i from j; number k; Question: Can we satisfy all demandsat a total cost <=k?

  10. DEFINITIONS thetime complexity(function)tAof algorithm A forproblemP is tA(n)= max {#operaties of A on instance I | over IP with |I|=n}P is polynomial solvable - notation P- if algorithm A with tA(n)of O(nq) for someqZ+P is non-deterministicallypolynomial solvable-notation P- if allIP with answer YEShave a ‘-certificate’, checkable in polytime EXAMPLES  (if P is polynomial solvable then P is in ) TSP: Instance:network (N,E,c) with number kR Question: tour oflength <=k? a -certificate is a set of n edges; polytime checkable: tour of length <= k coTSP ?! Instance:netwerk (N,E,c) with number kR Question:Are all tours of length > k? no -certificate?!

  11. 5 4 16 2 13 5 15 8 11 8 7 10 1 5 9 7 3 7 14 2 3 12 3 8 9 6 14 TSP>=k Instance: netwerk (N,E,c) met c:EZ+ ; getal k Question:Is er een tour met lengte < k? TSP Instance: netwerk (N,E,c) met c:EZ+ ; getal k Question:Is er een tour met lengte < k? ShortestPath Instance: netwerk (N,E,c) met c:EZ+ ; s,t N ; getal k Question:Is er een s-t path van lengte < k?

  12. Transformatie-afbeelding van ITSP naar I':=(I)Kortste Pad - 1) Neem (N',E')=(N,E) met extra kopie n' van n ( dezelfde kanten als n); 2) c' verlaagt alle kantkosten met M 3) s=n en t=n'' 4 +c -M -M 2 ij +16 4) k'=-nM+k 9'' 7 9 1 5 -M +3 Kortste 9 - 9'' pad bevat n kanten - als mogelijk - 3 8 tourlengte voor lengte - nM + 6 I heeft TSP tour <= 63 desda I' heeft kortste 9-9'' pad <= -9M+63

  13. algoritme in O(|I|qr) time voor I in P: doe eerst AP/* geeft I' in SAT van size O(| |r) doe dan A alg. A O(| |q) yes/no P(I)SAT AP O(| |r) IP probleem P1 is reduceerbaar tot P2 als  afbeelding :P1  P2zo dat: I yes-instantie van P1 (I) yes-instantie van P2 als ook:  polytime-algoritme, zeg A,met inputs IP1 en outputs (I)P2 dan P1 is 'polynomial‑reducable' tot P2 , notatie: P1 P2 TheoremCook (1971): SAT is -compleet, i.e.: each problemP of  is polynomial reducable to SAT Corollary If SAT can be solved by some polynomial algorithm A, then all problems P are polynomial solvable (=)Proof:

More Related