Lecture 8: Graph-theoretic problems (again)

Transcription

1 COMP36111: Advned Algorithms I Leture 8: Grph-theoreti prolems (gin) In Prtt-Hrtmnn Room KB2.38: emil: iprtt@s.mn..uk

2 Reding for this leture: Sipser: Chpter 7.

3 A grph is pir G = (V, E), where V is finite set nd E set of unordered pirs of (distint) elements of V. Grphs re typilly displyed you guessed it grphilly: v 0 v 1 v 2 v 3 v 4 v 5 v 6 v 7 Grphs n e enoded s strings over finite lphets. For instne: (n; (u 1, v 1 ),... (u m, v m )), where the u i nd v i re integers in the rnge [0, n 1], enodes the ovious grph over the verties {0,... n 1}.

4 Let G = (V, E) e grph. A k-olouring of G is funtion f : V {0, 1,..., k 1} suh tht, for ny edge (u, v) E, f (u) f (v). We sy tht G is k-olourle if there exists k-olouring for G. Of the two grphs elow, one is 3-olourle nd the other not. Cn you tell whih?

5 Let G = (V, E) e grph. A k-olouring of G is funtion f : V {0, 1,..., k 1} suh tht, for ny edge (u, v) E, f (u) f (v). We sy tht G is k-olourle if there exists k-olouring for G. Of the two grphs elow, one is 3-olourle nd the other not. Cn you tell whih?

6 Let G = (V, E) e grph. A k-olouring of G is funtion f : V {0, 1,..., k 1} suh tht, for ny edge (u, v) E, f (u) f (v). We sy tht G is k-olourle if there exists k-olouring for G. Of the two grphs elow, one is 3-olourle nd the other not. Cn you tell whih?

7 Let G = (V, E) e grph. A k-olouring of G is funtion f : V {0, 1,..., k 1} suh tht, for ny edge (u, v) E, f (u) f (v). We sy tht G is k-olourle if there exists k-olouring for G. Of the two grphs elow, one is 3-olourle nd the other not. Cn you tell whih?

8 Let G = (V, E) e grph. A k-olouring of G is funtion f : V {0, 1,..., k 1} suh tht, for ny edge (u, v) E, f (u) f (v). We sy tht G is k-olourle if there exists k-olouring for G. Of the two grphs elow, one is 3-olourle nd the other not. Cn you tell whih?

9 Let G = (V, E) e grph. A k-olouring of G is funtion f : V {0, 1,..., k 1} suh tht, for ny edge (u, v) E, f (u) f (v). We sy tht G is k-olourle if there exists k-olouring for G. Of the two grphs elow, one is 3-olourle nd the other not. Cn you tell whih?

10 Let G = (V, E) e grph. A k-olouring of G is funtion f : V {0, 1,..., k 1} suh tht, for ny edge (u, v) E, f (u) f (v). We sy tht G is k-olourle if there exists k-olouring for G. Of the two grphs elow, one is 3-olourle nd the other not. Cn you tell whih?

11 This gives us nturl prolem: k-colourability Given: A grph G. Return: Yes if G is k-olourle, nd No otherwise. We re now going to show tht 3-olourility is NPTime-hrd. We proeed y redution from 3-St: from set of 3-literl luses Γ, we ompute (in logrithmi spe) grph G Γ nd show tht Γ is stisfile iff G Γ is 3-olourle. We uild the grph using series of gdgets...

12 Here is the first gdget: The green nodes re internl to the gdget: the grey nodes will interfe to other nodes in the grph. It is ovious tht, in ny olouring, nd will hve the sme olour. Conversely, if nd hve the sme olour, we n lwys extend to three-olouring of the whole gdget.

13 Here is the first gdget: The green nodes re internl to the gdget: the grey nodes will interfe to other nodes in the grph. It is ovious tht, in ny olouring, nd will hve the sme olour. Conversely, if nd hve the sme olour, we n lwys extend to three-olouring of the whole gdget.

14 Here is the first gdget: The green nodes re internl to the gdget: the grey nodes will interfe to other nodes in the grph. It is ovious tht, in ny olouring, nd will hve the sme olour. Conversely, if nd hve the sme olour, we n lwys extend to three-olouring of the whole gdget.

15 Here is the seond gdget: The green nodes re internl to the gdget: the grey nodes will interfe to other nodes in the grph. It is ovious tht, in ny olouring, will hve the sme olour s extly one of nd. Moreover, if hs the sme olour s extly one of nd, we n extend to three-olouring.

16 Here is the seond gdget: The green nodes re internl to the gdget: the grey nodes will interfe to other nodes in the grph. It is ovious tht, in ny olouring, will hve the sme olour s extly one of nd. Moreover, if hs the sme olour s extly one of nd, we n extend to three-olouring.

17 Here is the seond gdget: The green nodes re internl to the gdget: the grey nodes will interfe to other nodes in the grph. It is ovious tht, in ny olouring, will hve the sme olour s extly one of nd. Moreover, if hs the sme olour s extly one of nd, we n extend to three-olouring.

18 Here is the third gdget: The green nodes re internl to the gdget: the grey nodes will interfe to other nodes in the grph.

19 Suppose nd re differently oloured (lue nd red) Then n e given either of these olours: On the other hnd, if hs different olour, we get stuk:

20 Suppose nd re differently oloured (lue nd red) Then n e given either of these olours: On the other hnd, if hs different olour, we get stuk:

21 Suppose nd re differently oloured (lue nd red) Then n e given either of these olours: On the other hnd, if hs different olour, we get stuk:

22 Suppose nd re differently oloured (lue nd red) Then n e given either of these olours: On the other hnd, if hs different olour, we get stuk:

23 Suppose nd hve the sme olour (red) Then n e given this olour too: On the other hnd, if hs different olour (white), we get stuk:

24 Suppose nd hve the sme olour (red) Then n e given this olour too: On the other hnd, if hs different olour (white), we get stuk:

25 Suppose nd hve the sme olour (red) Then n e given this olour too: On the other hnd, if hs different olour (white), we get stuk:

26 Suppose nd hve the sme olour (red) Then n e given this olour too: On the other hnd, if hs different olour (white), we get stuk:

27 Thus, in this gdget, node hs the sme vlues s either node or node (or oth)

28 We n stik two of these gdgets together to mke fourth gdget, in whih hs to hve the sme olour s ny of, or d (with ll suh olourings possile) d d

29 Theorem The prolem 3-COLOURABILITY is NPTime-hrd. Proof. p1 p2 p3 p2 p4 p1 p1 p3 p2 p1 p1 p2 p2 p3 p3 p4 p4 Enoding of {( p 1 p 2 p 3 ), ( p 2 p 4 p 1 ), ( p 1 p 3 p 2 )}

30 Let G = (V, E) e onneted grph. A iruit of G is sequene of nodes v 0,... v n 1 suh tht, for ll i (0 i < n) (v i, v (i+1) mod n ) E. An Eulerin iruit of G is iruit of G in whih eh edge of G is trversed extly one. A Hmiltonin iruit of G is iruit of G in whih eh node is enountered extly one.

31 Extly one of the following grphs hs n Eulerin iruit:

32 Extly one of the following grphs hs n Eulerin iruit:

33 Extly one of the following grphs hs n Eulerin iruit:

34 Extly one of the following grphs hs n Eulerin iruit:

35 Extly one of the following grphs hs n Eulerin iruit:

36 Extly one of the following grphs hs n Eulerin iruit:

37 Extly one of the following grphs hs n Eulerin iruit:

38 Extly one of the following grphs hs n Eulerin iruit:

39 Extly one of the following grphs hs n Eulerin iruit:

40 Extly one of the following grphs hs n Eulerin iruit:

41 Extly one of the following grphs hs n Eulerin iruit:

42 Extly one of the following grphs hs n Eulerin iruit:

43 Extly one of the following grphs hs n Eulerin iruit:

44 Extly one of the following grphs hs n Hmiltonin iruit:

45 Extly one of the following grphs hs n Hmiltonin iruit:

46 We thus hve the prolems: EULERIAN-CIRCUIT Given: A Grph G Return: Yes if G hs n Eulerin iruit, nd No otherwise. HAMILTONIAN-CIRCUIT Given: A Grph G Return: Yes if G hs n Hmiltonin iruit, nd No otherwise.

47 Theorem (Euler) A onneted grph hs n Eulerin iruit if nd only if every node hs even degree. Proof. The only-if-diretion is ovious. For the if-diretion, let v 0,..., v m e the longest wlk in G in whih no edge is trversed more thn one. Sine every node hs even degree, v m = v 0. Suppose e E is not in this wlk. By onnetedness of G let e e (v i, u), where 0 i < m. Then v i,..., v m (= v 0 )v 1,..., v i u is longer wlk in G in whih no edge is trversed more thn one. Contrdition.

48 Corollry The prolem EULERIAN-CIRCUIT is in PTime. Proof. The ondition tht every node hs even degree n oviously e tested in time O(n 2 ).

49 By ontrst, HAMILTONIAN-CIRCUIT is NPTime-omplete. Agin, we use series of gdgets: The first gdget ensures tht ny Hmiltonin iruit trverses extly one of the edges (u, u ) or (v, v ), u u u u v v v v with oth possiilities relizle

50 The seond gdget ensures tht ny Hmiltonin iruit trverses t lest one of the lower three edges, with ll omintions possile.

51 Theorem HAMILTONIAN-CIRCUIT is NPTime-omplete. Proof. We use the ove gdgets to enode 3-SAT, long the following lines p 1 p 2 p 3 p 1 p 2 p 3 p 1 p 2 p 3

52 The prolem HAMILTONIAN-CIRCUIT is losely relted to the following prolem Imgine slesmn given the tsk of touring n ities, whih we identify s 1,..., n. Tke the distne etween the ities i nd j to e i,j. In wht order should the slesmn tour the ities so s to minimize his totl round trip? We ll this prolem the trvelling slesmn prolem (TSP). It hs mny prtil pplitions, nd een widely studied.

53 More formlly: TSP Given: Return: n n n symmetri mtrix { i,j } over N tour though this mtrix of miniml ost. This is not prolem in our tehnil sense (it does not hve Yes/No nswers); ut the following is: TSP-FEASIBILITY Given: n n n ost mtrix { i,j } over N, nd some k N Return: Yes, if there is tour of { i,j } with totl ost < k; nd No otherwise. In some sense, this is just s good....

54 Theorem TSP-FEASIBILITY is NPTime-omplete. Proof. Clerly, TSP-FEASIBILITY is in NP. Given G = (V, E), where V = {1,..., n}, set i,j to e 1 if (v i, v j ) E, nd 2 otherwise. Set k = n + 1. Thus we hve funtion f from instnes of HAMILTONIAN-CIRCUIT to instnes of TSP-FEASIBILITY, omputle in logrithmi spe. Suppose there is Hmiltonin iruit of G. Then tht is tour in { i,j } of length n; nd n < k. Suppose there is tour in { i,j } of length n (hene extly n). Then suessive nodes on the tour must e joined y n edge in G, so this tour is is Hmiltonin iruit in G. Hene, f is redution, nd so TSP-FEASIBILITY is NP-hrd.