GRAPH ALGORITHMS
Week 12 (19 Dec- 23 Dec 2016)
C. Croitoru
[email protected]
FII
December 17, 2016
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OUTLINE
Polynomial-Time Reductions for Graph Problems
I
I
I
I
Maximum Stables
Vertex Colorings
Hamiltonian Problems
Traveling Salesman Problem
Approaching NP-hard Graph Problems
I
I
Metric TSP: Christofides Algorithm
Vertex Coloring: Greedy-Color Algorithm
Exercises for the next week Seminar (9 Jan - 14 Jan 2017)
Homework 3 – Presentation
2 / 35
Polynomial-Time Reductions for Graph Problems
Hamiltonian Problems
Remember: Let G = (V (G ), E (G )) be a (di)graph.
A circuit C of G is a Hamiltonian circuit if V (C ) = V (G ). On open path P of
G is a Hamiltonian path if V (P) = V (G ). A Hamiltonian (di)graph is a
(di)graph which has a Hamiltonian circuit. A traceable (di)graph is a (di)graph
which has a Hamiltonian path.
Theorem 3 (Nash-Williams 1969) The following problems are polynomial
equivalent:
CH: Given a graph G . Is G Hamiltonian?
TR: Given a graph G . Is G traceable?
DCH: Given a digraph G . Is G Hamiltonian?
DTR: Given a digraph G . Is G traceable?
BCH: Given a bipartite graph G . Is G Hamiltonian?
Remark. P1 and P2 are polynomial equivalent if P1 ≤P P2 and P2 ≤P P1 .
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Polynomial-Time Reductions for Graph Problems
Hamiltonian Problems – Proof of Theorem 3
CH ≤P TR
Let G be a graph and v0 ∈ V (G ). We construct in polynomial time a
graph H such that G is Hamiltonian if and only if H is traceable.
Let V (H) = V (G ) ∪ {x, y , z} and
E (H) = E (G ) ∪ {xv0 , yz} ∪ {wy | w ∈ V (G )
∧
wv0 ∈ E (G )}:
x
vo
vo
y
z
N (v o
)
G
G
N (vo)
G
H
Then, H is traceable if and only if it has a Hamiltonian path, P, with extremities
x şi z (which have degree 1 in H). P exists in H if and only if in G there is a
Hamiltonian Path with an extremity in v0 and the other a neighbor of v0 , that is,
if and only if G is Hamiltonian.
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Polynomial-Time Reductions for Graph Problems
Hamiltonian Problems – Proof of Theorem 3
TR ≤P CH
Let G be a graph. Consider H = G + K1 . Then, H is Hamiltonian if and
only if G has a Hamiltonian path.
G
G+K1
The equivalence of the problems DCH and DTR can be proved in a similar way.
CH ≤P DCH
Let G be a graph. Let D be the digraph obtained from G by replacing each
edge with a symmetric pair of arcs. Clearly any circuit in G gives rise to a
circuit in D and conversely, any circuit in D gives rise to a circuit in G .
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Polynomial-Time Reductions for Graph Problems
Hamiltonian Problems – Proof of Theorem 3
DCH ≤P CH
Let D = (V (D), E (D)) be a digraph. Each vertex v ∈ V (D) is replaced by an
undirected graph P3 (v ) with extremities av and bv :
P3 (v ) = ({av , cv , dv , bv }, {av cv , cv dv , dv bv }).
Each arc vw ∈ E (D) is replaced by the (undirected) edge bv aw . Let G be the
graph obtained (in polynomial time) in this way:
v
aw
bv
w
av
b
w
bx
x
ax
bb
b
a
D
a
b
ba
aa
G
Each circuit C of D corresponds to a circuit in G and conversely, each circuit in
G corresponds to a circuit of D. It follows that D is Hamiltonian if and only if
G is Hamiltonian.
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Polynomial-Time Reductions for Graph Problems
Hamiltonian Problems – Proof of Theorem 3
Note that if C is a circuit of G ,then this is generated by a circuit C 0 of D,
and lg(C ) = 3 · lg(C 0 ) + lg(C 0 ) = 4 · lg(C 0 ). It follows that any circuit of G
is even, therefore G is a bipartite graph.
Therefore the above proof (DCH ≤P CH) is in fact DCH ≤P BCH.
Since BCH ≤P CH is obvious, the Theorem 3 is completely proved.
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Polynomial-Time Reductions for Graph Problems
Hamiltonian Problems SM ≤P CH
Theorem 4.(Karp 1972) SM ≤P CH.
Proof. Let G = (V , E ) and j ∈ Z+ an instance of a problem SM. We will
construct in polynomial time (w.r.t. n = |V |) a graph H such that there is a
stable set S in G with |S| ≥ j if and only if H is Hamiltonian.
Let k = |V | − j. Suppose that k > 0 (to avoid trivial cases)
i) Let A = {a1 , a2 , . . . , ak } be a set of k distinct vertices.
ii) For each edge e = uv ∈ E (G ), consider the graph Ge0 = (Ve0 , Ee0 )
with Ve0 = {(w , e, i); w ∈ {u, v }, i = 1, 6}} and
Ee0 = {(w , e, i)(w , e, i + 1); w ∈ {u, v }, i = 1, 5}
v
(v,e,1)
(v,e,2)
(v,e,3)
(v,e,4)
(v,e,5)
(v,e,6)
u
(u,e,1)
(u,e,2)
(u,e,3)
(u,e,4)
(u,e,5)
(u,e,6)
e
8 / 35
Polynomial-Time Reductions for Graph Problems
Hamiltonian Problems SM ≤P CH
The graph has the property that, if it is an induced subgraph of a Hamiltonian
graph H, and no one of the vertices (w , e, i) with w ∈ {u, v } and i = 2, 5 has
another neighbor in H, then the only possibilities of traversing Ge0 by a
Haminltonian circuit are:
a
b
c
Ge0
Hence, if the Hamiltonian ”enters” in
by a vertex corresponding to u ((u, e, 1)
or (u, e, 6)) then it ”leaves” the graph Ge0 also by a vertex corresponding to u.
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Polynomial-Time Reductions for Graph Problems
Hamiltonian Problems SM ≤P CH
iii) For each vertex u ∈ V , consider (in an arbitrary order) the edges
incident in G with u: e1u = uv1 , e2u = uv2 , . . . , epu = uvp
u
, 1); i = 1, p − 1} and
(p = dG (u)). Let Eu00 = {(u, eiu , 6)(u, ei+1
000
u
u
Eu = {ai (u, e1 , 1), ai (u, ep , 6); i = 1, k}.
u
e
ep
v
2
u
(u,e ,6)
1
u
(u,e ,1)
2
e1
v
1
u
(u,e ,1)
1
a
1
2
vp
u
(u,e ,6)
2
a
u
(u,e p,1)
u
(u,e p,6)
k
The graph H has V (H) = A ∪ ∪e∈E Ve0 and E (H) = ∪e∈E Ee0 ∪ ∪u∈V (Eu00 ∪ Eu000 ).
Clearly, it can be constructed from G in polynomial (in n) time.
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Polynomial-Time Reductions for Graph Problems
Hamiltonian Problems SM ≤P CH
a2
Example:
1
3
2
4
1
3
2
4
G
3
1
a1
2
4
k=2
2
4
2
4
⇐ If H is Hamiltonian, then there is C a Hamiltonian circuit in H. Since
A is a stable set in H, A decomposes the circuit C in exactly k paths
internal disjoint: Dai1 ai2 , Dai2 a13 , . . . , Daik ai1 .
Let Daij aij+1 be such a path (j + 1 = 1 + (j(mod k)) ).
By the construction of H, it follows that the first vertex after aij on this
vi
vi
path will be (vij , e1 j , 1) or (vij , ep j , 6), where p = dG (vij ) vij ∈ V .
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Polynomial-Time Reductions for Graph Problems
Hamiltonian Problems SM ≤P CH
After that, Daij aij+1 will enter the component Ge1 or Gep that will be leaved
also by a vertex corresponding to vij . If the next vertex is not aij+1 , it will
enter into the component corresponding to the next edge incident with vij ,
that will be leaved also by a vertex corresponding to vij .
Example.
a2
1
3
2
4
1
2
3
4
3
1
a1
2
4
2
4
2
4
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Polynomial-Time Reductions for Graph Problems
Hamiltonian Problems SM ≤P CH
It follows that each path Daij aij+1 corresponds to a unique vertex vij ∈ V ,
such that the first and the last edge of Daij aij+1 are
vi
vi
aij (vij , et j , x), aij+1 (vij , et 0 j , x 0 ), with t = 1 and t 0 = dG (u), x = 1, x 0 = 6
or t = dG (u), t 0 = 1, x = 6, x 0 = 1.
It follows that the vertices vij are distinct.
Let V ∗ = {vi1 , vi2 , . . . , vik }. Since C is a Hamiltonian circuit in H, it
follows that, ∀e ∈ E , there is a path Daij aij+1 traversing Ge0 , hence there is
v ∈ V ∗ incident cu e.
Therefore S = V − V ∗ is a stable set in G and |S| = j.
Hence, we proved that if H is Haminltonian, then in G there is a stable set
with j vertices, therefore the answer of SM for the instance G , j is yes.
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Polynomial-Time Reductions for Graph Problems
Hamiltonian Problems SM ≤P CH
⇒ Suppose that the answer of SM for the instance (G , j) is yes, hence
there is S0 stable set in G with |S0 | ≥ j. There is S ⊆ S0 stable set in G
with |S| = j. Let V ∗ = V − S = {v1 , v2 , . . . , vk }.
Consider in H for each e = uv ∈ E :
the two paths from case (c) in Ge0 depicted in Figure on slide 32, if u, v ∈ V ∗
the path from case (b) in Ge0 depicted in Figure on slide 32, if
u ∈ V ∗, v ∈
/ V∗
the path from case (a) in Ge0 depicted in Figure on slide 32, if
v ∈ V ∗, u ∈
/ V∗
To the union of thes all paths we add the edges ai (vi , e1vi , 1), (vi , e1vi , 6)(vi , e2vi , 1)
vi
, . . . , (vi , ep−1
, 6)(vi , epvi , 1), (vi , epvi , 6)ai+1 , (with p = dG (vi )), for i = 1, k.
It is not difficult to check that we obtain, in this way, a Hamiltonian circuit in G .
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Polynomial-Time Reductions for Graph Problems
Traveling Salesman Problem (TSP)
TSP Given G = (V , E ) a graph and d : E → R+ , an non-negative weight
function on its edges, find a Hamiltonian circuit, H0 , s.t. the sum of the weights
of the edges of H0 is minimum (over all Hamiltonian circuits of G ).
Let the graph G be a network consisting of a set V of cities together with a set E
of direct routes between cities and the weight function d giving, for each edge
uv ∈ E , d(uv )= the distance on the direct route between cities u and v . Fixing a
starting city v0 , the Hamiltonian circuit H0 represents the shortest way of visiting
all the cities exactly once (excepting v0 ) by a traveling salesman starting from v0
and returning to v0 .
Probably, the most studied NP-hard optimization problem!
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Polynomial-Time Reductions for Graph Problems
Traveling Salesman Problem (TSP)
We consider the following equivalent formulation of this problem.
TSP Given n ∈ Z+ (n ≥ 3) and d : E (Kn ) → R+ , find H0 , Hamiltonian
circuit in Kn , with d(H0 ) minimum over all Hamiltonian circuits of Kn ,
X
where
d(H0 ) =
d(e).
e∈E (H0 )
If the graph G , on which it is required to solve the Traveling Salesman
Problem, is not the complete graph Kn , then we can introduce the missing
edges with a big weight, M ∈ R+ , where M > |V |· maxe∈E d(e).
Note that we are limited here only the symmetric TSP, a similar (asymmetric)
problem can be be considered for the case when G is a digraph.
In the study of the time complexity of this problem, we will take d(e) ∈ Z+ , for
each edge e.
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Polynomial-Time Reductions for Graph Problems
Traveling Salesman Problem (TSP)
The associated decision problem is
DTSP
Instance: n ∈ Z+ (n ≥ 3), d : E (Kn ) → Z+ şi B ∈ Z+
Question: Is there H0 , Hamiltonian circuit in Kn , s. t. d(H0 ) ≤ B ?
Theorem 5. CH ≤P DTSP.
Proof. Let G = (V , E ) (|V | = n) be an instance of the problem CH. We
construct in polynomial time an instance, d : E (Kn ) → Z+ and B ∈ Z+ , of the
problem DTSP such that there is a Hamiltonian circuit in Kn of total weight not
greater than B if and only if G is a Hamiltonian graph.Let
(
1 if vw ∈ E (G )
d(vw ) =
and B = n.
2 if vw ∈ E (G )
17 / 35
Polynomial-Time Reductions for Graph Problems
Traveling Salesman Problem (TSP)
Then, there is in Kn a Hamiltonian circuit of weight ≤ n if and only if
there is a Hamiltonian circuit C in Kn such that ∀e ∈ E d(e) = 1, that is,
if and only if G has a Hamiltonian circuit:
G
pondere 1
pondere 2
It follows that TSP is NP-hard problem.
A possible approach is to consider approximation algorithms A, which
construct in polynomial time, for each instance of TSP, a Hamiltonian
circuit HA of Kn , approximating the optimal solution Ho .
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Polynomial-Time Reductions for Graph Problems
Traveling Salesman Problem (TSP)
The quality of the approximation can be expressed using the following
ratios:
RA (n) =
sup
d:E (Kn )→R+
d(Ho )6=0
d(HA )
d(Ho )
RA = sup RA (n).
n≥3
Obviously, the approximation algorithm A is useful only if RA is finite.
Unfortunately, if the weight function d is arbitrary, the simple condition that RA is
finite is equal in difficulty as solving exactly TSP. More precisely, we have:
Theorem 6. If there is a polynomial time approximation algorithm A for
TSP s.t. RA < ∞, then the problem CH can be solved in polynomial time.
19 / 35
Polynomial-Time Reductions for Graph Problems
Traveling Salesman Problem (TSP)
Proof of Theorem 6. Let A be a polynomial time with RA < ∞. Hence
there is k ∈ Z+ such that RA ≤ k.
Let G = (V , E ) be an arbitrary graph, instance of CH. If n = |V |, then
we consider d : E (Kn ) → Z+ defined by
(
1
if uv ∈ E
d(uv ) =
kn if uv ∈
/ E.
Obviously, G is Hamiltonian if and only if Ho , the optimal solution of this
instance of TSP, satisfies d(H0 ) = n.
Apply A to solve approximately this instance of TSP.
If d(HA ) ≤ kn, then d(HA ) = n and HA = Ho .
If d(HA ) > kn, then d(Ho ) > n. Indeed, assuming that d(Ho ) = n,
A)
we have d(H
d(Ho ) ≤ k, therefore d(HA ) ≤ kd(Ho ) = kn, contradiction.
20 / 35
Polynomial-Time Reductions for Graph Problems
Traveling Salesman Problem (TSP)
It follows that G is Hamiltonian if and only if d(HA ) ≤ kn, and, since A
runs in polynomial time, it follows that CH can be solved in polynomial
time.
G
pondere 1
pondere 7k
Remark. The Theorem 6 can be formulated equivalently:
If P 6= NP, then there is no polynomial time approximation
algorithm A with RA < ∞.
21 / 35
Approaching NP-hard Graph Problems
Metric TSP: Christofides Algorithm
Theorem ( Christofides,1976) If in TSP the weight function d satisfies
∀v , w , u ∈ V (Kn ) distinct, d(vw ) ≤ d(vu) + d(uw ),
then there is a polynomial time approximation algorithm A with RA = 32 .
Proof. Let A be the following algorithm:
1
Find T 0 the edge set of MST in Kn (the cost of each edge e is d(e)) (this
takes polynomial time using any MST algorithms).
2
Find M 0 a perfect matching in the subgraph induced in Kn by the set of
vertices of odd degrees in the MST T 0 (this takes polynomial time using
any maximum matching algorithm).
3
In the multi-graph obtained from < T 0 ∪ M 0 >Kn , by duplicating the edges
from T 0 ∩ M 0 (it is connected and all vertices have even degree) find a
closed Euler trail, (vi1 , vi2 , . . . , vi1 ). Eliminate the all multiple occurrences of
the internal vertices to obtain a Hamiltonian circuit HA in Kn with the edge
set HA = (vj1 vj2 , vj2 vj3 , . . . , vjn vj1 ) (both constructions need O(n2 ) time).
22 / 35
Approaching NP-hard Graph Problems
Metric TSP: Christofides Algorithm
Example:
o
Cuplajul M
Arborele T o
5
9
3
11
8
7
6
1
4
10
Graf Eulerian
2
Turul H
A
23 / 35
Approaching NP-hard Graph Problems
Metric TSP: Christofides Algorithm
HA is approximative solution of TSP given by Christofides. Let m = b n2 c and Ho
be the optmum solution. We prove (after Cornuejols & Nemhauser) that
∀n ≥ 3
d(HA ) ≤
3m − 1
d(H0 ).
2m
Let Ho = {v1 v2 , v2 v3 , . . . , vn v1 } (if necessary, we can rename the nodes).
Let A = {vi1 , vi2 , . . . , vi2k } the set of odd degree vertices in < T 0 >Kn ,
i1 < i2 < . . . < i2k . Let H = {vi1 vi2 , vi2 vi3 , . . . , vi2k−1 vi2k , vi2k vi1 } be the circuit
generated by A in Kn . Applying repeatedly the triangle inequality, we obtain
d(H) ≤ d(H0 ), (the weight each chord, d(vij vij+1 ), is upper bounded by the sum
of the weights of the edges on Ho joining the extremities of the chord vij vij+1 ).
Since H is an even circuit, it is the union of two perfect matchings in [A]Kn ,
M 1 ∪ M 2 . Suppose that d(M 1 ) ≤ d(M 2 ).
24 / 35
Approaching NP-hard Graph Problems
Metric TSP: Christofides Algorithm
o
H
M
1
M2
H
By the choosing of M 0 , we have d(M 0 ) ≤ d(M 1 ) ≤ 12 (d(M 1 ) + d(M 2 )) = 12 d(H)
≤ 12 d(Ho ). Let α ∈ R+ such that d(M 0 ) = αd(Ho ). Obviously, 0 < α ≤ 12 .
Decompose Ho into H 1 ∪ H 2 by taking into H i the edges of Ho connecting the
extremities of each chord in M i : (vij vij+1 ∈ M i ⇒ vij vij +1 , . . . , vij+1 −1 vij+1 ∈ H i ).
H2
H1
25 / 35
Approaching NP-hard Graph Problems
Metric TSP: Christofides Algorithm
By the triangle inequality, d(H i ) ≥ d(M i ) i = 1, 2.
At least one of H 1 or H 2 has at most m = b n2 c edges. Suppose that H 1 has this
property. Since d(H 1 ) ≥ d(M 1 ) ≥ d(M 0 ) = αd(Ho ), it follows that there is
α
e ∈ H 1 such that d(e) ≥ m
d(Ho ).
Let T be the spanning tree obtained from Ho by deleting an edge of maximum
α
d(H0 ).
weight. We have d(T ) = d(H0 ) − maxe∈H0 d(e) ≤ d(H0 ) − m
Since T 0 is MST in Kn , it follows that d(T 0 ) ≤ d(H0 )(1 −
α
m ).
Using the triangle inequality, we have
d(HA ) ≤ d(T 0 ) + d(M 0 ) ≤ d(H0 )(1 −
Since α ≤ 21 , we obtain d(HA ) ≤
α
α(m − 1)
) + αd(H0 ) = (1 +
)d(H0 ).
m
m
3m−1
2m d(H0 )
pentru n ≥ 3.
26 / 35
Approaching NP-hard Graph Problems
Vertex Coloring: Greedy-Color Algorithm
Let G = (V , E ) be a graph with V = {1, 2, . . . , n} and let π be a permutation of
V . We construct a vertex-coloring c : {1, 2, . . . , n} −→ {1, 2, . . . , χ(G , π)}.
c(π1 ) ← 1; χ(G , π) ← 1; S1 ← {π1 };
for i ← 2 to n do
{ j ← 0;
repeat
j ← j + 1;
v ← first vertex (in the order given by π), from Sj s.t. πi v ∈ E (G );
if ∃ v then first(πi , j) ← v
else { first(πi , j) ← 0; c(πi ) ← j; Sj ← Sj ∪ {πi } }
until first(πi , j) = 0 or j = χ(G , π);
if first(πi , j) 6= 0 then {c(πi ) ← j + 1;Sj+1 ← {πi }; χ(G , π) ← j + 1;}
}
Note that the number of colors used by algorithm is not greater than 1 + ∆(G ).
It follows χ(G ) ≤ 1 + ∆(G ).
27 / 35
Approaching NP-hard Graph Problems
Vertex Coloring: Greedy-Color Algorithm
χ(G , π), the number of colors returned by the Greedy-Color Algorithm, can be
arbitrarily larger than χ(G ). For example, let G be the graph obtained from the
complete bipartite graph Kn,n , with vertex set {1, 2, . . . , n} ∪ {10 , 20 , . . . , n0 }, by
removing the edges 110 , 220 , . . . , nn0 . If π gives the ordering 1, 10 , 2, 20 , . . . , n, n0 ,
the Greedy-Color Algorithm returns c(1) = c(10 ) = 1, c(2) = c(20 ) = 2, . . . ,
c(n) = c(n0 ) = n. Therefore, χ(G , π) = n, while χ(G ) = 2.
1
1’
2
2’
3
3’
4
4’
5
5’
On the other hand, for any graph G , there is a permutation π of its vertices such
1, 1’, 2, 2’, 3, 3’, 4, 4’, 5, 5’
that χ(G , π) = χ(G ).pi:Indeed,
let S1 , S2 , . . . , Sχ(G ) , the color classes of an
optimal coloring, such that each Si is a maximal (w.r.t. inclusion) stable set in
G − ∪i−1
j=1 Sj . Let π be a permutation which induces an ordering s.t. vertices
occurs in non-decreasing order of their colors. Then, χ(G , π) = χ(G ).
28 / 35
Approaching NP-hard Graph Problems
Vertex Coloring: Greedy-Color Algorithm
A sufficient condition for the correctness of the Greedy-Color Algorithm:
Theorem. If, for every edge and every j < min{c(v ), c(w )} such that
first(v , j) < first(w , j) in the ordering given by π, we have vfirst(w , j) ∈ E , then
χ(G , π) = χ(G ).
The above condition can be tested in O(m) as a last step in the algorithm. If it is
fulfilled, then we have a certificate for the optimality of the number of color found.
first (j ,v)
v
w
first (j ,w)
Sj
S
c (v)
S c (w)
The above theorem follows by proving that if the condition stated holds, then
χ(G , π) = ω(G )(≤ χ(G )). On the other hand, it is well known that there are
graphs G for which χ(G ) − ω(G ) is arbitrarily large; for such a graph G no
permutation π satisfies the condition in the theorem.
29 / 35
Exercises for Next week (9 Jan - 14 Jan) Seminar
După sărbători, voi crea (pe pagina cursului) un director ”Examen” ı̂n care
voi publica subiectele de la examenul final de anul trecut.
Grupele care au seminar luni vor discuta in prima săptămână din ianuarie
probleme din aceste subiecte (nu toate, sunt 30 in total) iar in a doua
săptămână, problemele din temă.
Celelalte grupe vor schimba ordinea celor două săptămâni.
30 / 35
Homework 3 – Presentation
Problema 1.
Problema 1. Fie G = (V , E ) un graf cu V = {1, 2, . . . , n} şi ω(G ) = 2.
Construim graful M(G ) făcând reuniunea disjunctă a lui G cu graful K1,n (a cărui
bipartiţie a mulţimii vârfurilor este ({0}, {10 , 20 , . . . , n0 }) şi adăugând toate
muchiile ∪ij∈E (G ) {ij 0 , j 0 i}. Demonstraţi că ω(M(G )) = 2 şi că
χ(M(G )) = χ(G ) + 1. Deduceţi că, pentru orice p ∈ N∗ , există grafuri K3 -free şi
cu numărul cromatic p.
(1+2+1= 4 puncte)
31 / 35
Homework 3 – Presentation
Problema 2.
Digraful G = (V , E ) se numeşte ciclabil dacă există k ≥ 1 circuite ı̂n G ,
C1 , . . . , Ck , ale căror mulţimi de vârfuri, V (C1 ), . . . , V (Ck ), formează o
partiţie a lui V (G ). Demonstraţi că se poate testa ı̂n timp polinomial (ı̂n
raport cu |V | + |E |) dacă un digraf G = (V , E ) dat este ciclabil, rezolvând
o problemă de flux de valoare maximă ı̂ntr-o reţea de fluxuri convenabil
construită.
(2+2=4 puncte)
32 / 35
Homework 3 – Presentation
Problema 3.
a) Considerăm următoarea problemă de decizie:
Bip[3-1]
Instance : G = (S, T ; E ), graf bipartit, cu dG (s) = 3 , ∀s ∈ S .
Question : Există S 0 ⊆ S a.ı̂. ı̂n H = [S 0 ∪ T ]G avem dH (t) = 1, ∀t ∈ T ?
Demonstraţi că 3SAT ≤P Bip[3-1].
b) Fie R = (G , c, s, t) o reţea de fluxuri cu funcţia de capacitate luând
valori ı̂ntregi, c : E (G ) → Z≥0 . Pe fiecare arc este dată şi o margine
inferioară u : E (G ) → Z≥0 , u(e) ≤ c(e), ∀e ∈ E (G ). Un flux x ı̂n R se
numeşte special dacă x(e) ∈ {0} ∪ {u(e), u(e) + 1, . . . , c(e)}, ∀e ∈ E (G ).
Fie problema de decizie:
Spec-Flow
Instance : R, reţea de fluxuri cu margini inferioare (date ı̂ntregi), v ∈ Z≥0 .
Question : Există ı̂n R un flux special de valoare v ?
Demonstraţi că Bip[3-1] ≤P Spec-Flow.
((1+1+1)+(1+1+1)=6 puncte)
33 / 35
Homework 3 – Presentation
Soluţiile se primesc
luni 9 ianuarie ı̂ntre orele 12 şi 14, la cabinetul C-402
Precizări
Este ı̂ncurajată asocierea ı̂n echipe formate din 2 studenţi care să
rezolve ı̂n comun tema.
Depistarea unor soluţii copiate ı̂ntre echipe diferite conduce la
anularea punctajelor tuturor acestor echipe.
Nu e nevoie să se rescrie enunţul problemelor. Nu uitaţi să vă treceţi
numele şi grupa din care faceţi parte, la ı̂nceputul lucrarii.
Este ı̂ncurajată redactarea latex a soluţiilor.
Nu se primesc soluţii prin e-mail.
Economisiţi hârtia!! (temele ce depăşesc 8 pagini nu sunt binevenite)
34 / 35
La Mulţi Ani!
35 / 35