Transitive feedback arc set (TFAS): NP-complete?

Some time ago, I posted a reference request for graph problems where we want to find a 2-partition of the edges where both sets fulfill a property not related to their cardinality. I was trying to prove that the following problem is NP-hard:

Given a tournament $G = (V,E)$, is there a feedback arc set $F \subseteq E$ in $G$ that defines a transitive relation?

I do have a construction for an attempt at a proof, but it seems that that is going to run into a dead end, so I thought I might ask here to see whether I'm missing something obvious. In order to not confine your creativity to lines of thought similar to the ones I used, I won't post my attempt here.

Is this problem NP-hard? If so, how to prove it?

• perfect, thanks! (I deleted the comment because I wrote G=(E,V) instead of the standard G=(V,E) :-) – Marzio De Biasi Jan 20 '14 at 16:13
• If I understand correctly, this is equivalent to asking whether the edges in a tournament can be partitioned into two DAGs, one of which is transitively closed. – dspyz Jan 22 '14 at 2:51
• re dspyz's comment, there are not that many problems on DAGs that can be studied due to the complexity of them. there arent even that many theorems at all on DAGs it would seem. trees are a little more accessible. your problem (while apparently interesting as reflected in votes) seems to mix a lot of unusual elements together & not fit into any particular category. – vzn Jan 23 '14 at 3:59
• @IgorShinkar the arcs of any digraph can be partitioned trivially into two DAGs: order the vertices arbitrarily; one DAG is the forward edges, the other DAG is the backward edges. – Sasho Nikolov Jan 28 '14 at 0:00
• @SashoNikolov of course! – Igor Shinkar Jan 28 '14 at 7:08

To add a little context, here's a construction for a graph that doesn't have a transitive feedback arc set. For this construction, I'll use the following gadget graph:

This tournament has the following properties (which I checked using a program, I didn't prove it formally):

• if (2,7) is not in a given TFAS, then (1,3) is
• if (5,1) is in a given TFAS, then so is (3,6)
• if (7,3) is in a given TFAS, then (5,1) is not

or slightly abusing predicate logic notation:

• $\neg (2,7) \rightarrow (1,3)$
• $(5,1) \rightarrow (3,6)$
• $(7,3) \rightarrow \neg (5,1)$

You'll notice that for each implication, the two edges are pairwise disjoint, so the following construction works:

I hope you can make out the idea of the graph: using the implication properties of the tournament above, we can construct a graph in which each transitive feedback arc set both includes and doesn't include the edge $A$, i.e. a contradiction, which means the graph doesn't have a transitive feedback arc set. Any completion of that graph cannot have one either since the same contradiction will remain in any completion. I left out a large number of vertices, all of which can be derived from substituting the tournament above for the implications.

• I'm sorry, I don't follow. Is there any reason you can't just post a list of the edges so I can run it through an ASP solver and try to verify it? According to clingo, your gadget graph has 8 different TFAS's. Here's the smallest one: tfas(edge(5,0)) tfas(edge(6,0)) tfas(edge(7,0)) tfas(edge(6,2)) tfas(edge(7,3)) tfas(edge(1,2)) tfas(edge(1,3)) tfas(edge(7,5)) – dspyz Jan 30 '14 at 4:13
• I just noticed you mentioned the edge (6,3) in the gadget graph, but the image you provided has the edge (3,6) – dspyz Jan 30 '14 at 5:20
• I figured it out, see my updated answer: cstheory.stackexchange.com/a/20778/13643 – dspyz Jan 30 '14 at 6:32
• @dspyz I thought the construction was clearer than just a list of the edges, since if my reasoning isn't wrong, all that would be required to verify is whether the tournament above the construction actually has those implication properties. Thanks for pointing out the mistake about edge (3,6)! I also got 8 TFAS for that gadget, the one you listed being one of them. – G. Bach Jan 30 '14 at 14:02
• I'm sorry. I constructed the graph wrong. I fixed it and clingo now reports no TFAS. – dspyz Feb 1 '14 at 19:10

I ran a short clingo program which reported no graph without a TFAS, but there was a bug. I fixed it and now it verifies there's no graph without a TFAS for n=8 or less. For n=9, it finds this one:

is_edge(edge(2,3)) is_edge(edge(1,4)) is_edge(edge(2,4)) is_edge(edge(3,5)) is_edge(edge(4,5)) is_edge(edge(1,6)) is_edge(edge(2,6)) is_edge(edge(3,6)) is_edge(edge(5,6)) is_edge(edge(1,7)) is_edge(edge(4,7)) is_edge(edge(5,7)) is_edge(edge(6,7)) is_edge(edge(1,8)) is_edge(edge(3,8)) is_edge(edge(4,8)) is_edge(edge(5,9)) is_edge(edge(6,9)) is_edge(edge(7,9)) is_edge(edge(2,1)) is_edge(edge(3,1)) is_edge(edge(4,3)) is_edge(edge(5,1)) is_edge(edge(5,2)) is_edge(edge(6,4)) is_edge(edge(7,2)) is_edge(edge(7,3)) is_edge(edge(8,2)) is_edge(edge(8,5)) is_edge(edge(8,6)) is_edge(edge(8,7)) is_edge(edge(9,1)) is_edge(edge(9,2)) is_edge(edge(9,3)) is_edge(edge(9,4)) is_edge(edge(9,8))


Here's the (fixed) encoding

% tfas.asp
#show is_edge/1.
vertex(1..n).

opp_edges(edge(A,B),edge(B,A)) :- vertex(A), vertex(B), A < B.
possible_edge(E1;E2) :- opp_edges(E1,E2).

{is_edge(E1); is_edge(E2)} = 1 :- opp_edges(E1, E2).
ntfas(E) :- possible_edge(E), not is_edge(E).
ntfas(edge(X, X)) :- vertex(X).

tfas(E) | fs(E) :- is_edge(E).
ntfas(E) :- fs(E).

broken :- ntfas(edge(A,C)), tfas(edge(A, B)), tfas(edge(B,C)).

reachable(X, Y) :- fs(edge(X, Y)), is_edge(edge(X, Y)).
reachable(X, Z) :- reachable(X, Y), fs(edge(Y, Z)), is_edge(edge(Y, Z)).
broken :- reachable(X, X).

tfas(E) :- broken, possible_edge(E).
fs(E) :- broken, possible_edge(E).
:- not broken.


Run it with clingo -c n=7 tfas.asp (Using clingo 4.2.1)

(the n=7 indicates graphs of exactly 7 vertices)

It should return satisfiable if and only if there exists a graph with no TFAS on 7 vertices.

Ok, I figured out what graph @G.Bach was describing and coded it up in clingo (see the clingo description below. It starts with a description of the gadget graph and proceeds to describe how to join copies of it together to get the full 34-vertex tournament graph G.Bach is describing. I've attached the grounded graph description as well).

I then proceeded to run clingo on that graph and it claimed to have found a TFAS with 241 edges. But I made a mistake in the graph encoding. I fixed the mistake and clingo now reports unsatisfiable (ie there is no TFAS).

Here's the program for finding TFAS's on a graph

{tfas(E)} :- is_edge(E).
:- not tfas(edge(A,C)), tfas(edge(A, B)), tfas(edge(B,C)).

reachable(X, Y) :- not tfas(edge(X, Y)), is_edge(edge(X, Y)).
reachable(X, Z) :- reachable(X, Y), not tfas(edge(Y, Z)), is_edge(edge(Y, Z)).
:- reachable(X, X).

tfas_count(N) :- N = #count{tfas(E) : tfas(E)}.

#show tfas/1.
#show tfas_count/1.


Here's the (updated) program for generating G.Bach's graph. I added indicators at the end to check that the graph is a well-formed tournament graph:

gadget_vertex(0..7).

special_edge(a;b;c;d;e).

forces(a,b).
forces(b,c).
forcesn(c,a).
nforces(a,d).
forces(d,e).
forces(e,a).

relates(A,B) :- forces(A,B).
relates(A,B) :- nforces(A,B).
relates(A,B) :- forcesn(A,B).

is_se_pair(se_pair(A,B)) :- relates(A,B).

matches(from(A), v(5, se_pair(A,B))) :- forces(A,B).
matches(to(A), v(1, se_pair(A,B))) :- forces(A,B).
matches(from(B), v(3, se_pair(A,B))) :- forces(A,B).
matches(to(B), v(6, se_pair(A,B))) :- forces(A,B).

matches(from(A), v(2, se_pair(A,B))) :- nforces(A,B).
matches(to(A), v(7, se_pair(A,B))) :- nforces(A,B).
matches(from(B), v(1, se_pair(A,B))) :- nforces(A,B).
matches(to(B), v(3, se_pair(A,B))) :- nforces(A,B).

matches(from(A), v(7, se_pair(A,B))) :- forcesn(A,B).
matches(to(A), v(3, se_pair(A,B))) :- forcesn(A,B).
matches(from(B), v(5, se_pair(A,B))) :- forcesn(A,B).
matches(to(B), v(1, se_pair(A,B))) :- forcesn(A,B).

same_vertex(V, V) :- vertex_name(V).
same_vertex(M, N; N, M) :- matches(X, M), matches(X, N).

already_found(v(Y,N2)) :- vertex_name(v(X,N1)), same_vertex(v(X,N1),v(Y,N2)), N1 < N2.

is_edge(edge(A,B)) :- vertex(A), vertex(B), A < B, not from_gadget_edge(edge(B,A)).

vertex_count(VN) :- VN = #count{vertex(V) : vertex(V)}.
edge_count(EN) :- EN = #count{is_edge(E) : is_edge(E)}.

#show vertex_count/1.
#show edge_count/1.

bidirectional :- is_edge(edge(A,B)), is_edge(edge(B,A)).
phantom_vertex :- is_edge(edge(A,B)), not vertex(A).
phantom_vertex :- is_edge(edge(A,B)), not vertex(B).
incomplete :- vertex(A), vertex(B), not is_edge(edge(A,B)), not is_edge(edge(B,A)), A != B.

#show bidirectional/0.
#show phantom_vertex/0.
#show incomplete/0.

• I'm positive there is a tournament on 18 vertices that doesn't have a TFAS. – G. Bach Jan 29 '14 at 19:07
• Can you please give it as an example? Just attach a file with the edges listed – dspyz Jan 29 '14 at 19:08
• How do I attach a file? It may take a few hours, I don't have the tournament to hand right now. I also miscalculated, it should have 34 vertices. It's probably easier to verify if I give the building blocks of the tournament. – G. Bach Jan 29 '14 at 19:25
• Upload to any file host and link to it (see meta.stackexchange.com/a/4643/185877), or if it has a regular structure, just describe it (give the building blocks) – dspyz Jan 29 '14 at 19:29
• cant follow your code yet but youre asserting that you are validating for finite n≤20? last sentence which asserts it is true for all $n$... if so quite remarkable(!) but not a "solution" to the general question (which would require a proof of all size cases). ps thx for working/posting an empirical observation to this site which in general is rare ... @G.Bach it would be great to see an image of the graph rather than just edge/vertex lists if possible – vzn Jan 30 '14 at 2:07

SWAG conjecture [something better than nothing?]:

Given a tournament $G = (V,E)$, there exists a feedback arc set $F \subseteq E$ in $G$ that defines a transitive relation. hence the problem is $O(1)$

notes: shootdown counterexamples welcome! none seem to be given so far. even better would be some observations of patterns of edge orientations related to particular graph classes. or some more motivation or tieing it into some existing literature. offered in the style of Proofs and refutations (Lakatos)... also, since it seems such an offbeat problem that doesnt [yet?] relate to much, suggest studying it empirically....

• I did run a program to check whether this holds and found that there are tournaments that do not have a transitive feedback arc set. I'll post one tomorrow, I won't get around to it today. – G. Bach Jan 28 '14 at 17:09
• @vzn can you prove the conjecture for a random tournament? – Igor Shinkar Jan 28 '14 at 19:53
• Counterexample with only 5 vertices: a->b, a->c, b->c, d->a, b->d, c->d, e->a, e->b, c->e, d->e. For any four vertices, the induced graph contains a cycle, so a transitive DAG can contain at most 3 edges among 3 vertices of the graph. There are only 5 possibilities (all other triplets are cycles): abc,eab,dea,bcd,cde It's easy to check that in each of the five cases there is a cycle among the other 7 edges – dspyz Jan 29 '14 at 6:31
• Yeah, nvr mind, it's not a counterexample – dspyz Jan 29 '14 at 6:39
• @dspyz I ran a brute force check on all tournaments on up to 8 vertices. All of them have transitive feedback arc sets, but there are some you can use to build a tournament that doesn't. – G. Bach Jan 29 '14 at 15:16