Complexity of some Path Problems in DAGs and Linear Orders

S. Burc kel Complexit y of some Path Problem s in DA Gs and Linear Orders. Complexit y of some P ath Problems in D A Gs and Linear Orders. Abstract. W e in v estigate here the com putational compl exit y of three natural problems in directed ac yclic graphs. W e prov e their NP Completenes s and conside r their restrictions to li near orders. Mathematics Subje cts Classific ation : . 1. Introduc tion. In this pap er, w e precise and extend the r esults of [1] a bo ut the complexity of path pr oblems in directed g raphs. In pa rticular we fo cus ourselves on the case of D AG s and very particular ones : linear order s. 2. Problems. Problem 1 : Null W e igh ted P ath. input : a DA G G = ( V , E ) endow ed with a w eight function w : E 7→ Z and tw o vertices s, t . question : is there a directed path π from s to t such that X e ∈ π w ( e ) = 0 Prop ositio n (1.) The Null W eighted P ath Decision Pro blem is NP- Complete, even on linear orders. Pro of. Obviously , it is in NP . T o prov e that it is NP hard, we r educe the SUBSET SUM P roblem to it. Let A = { a 1 , . . . , a n } b e a subset of Z n . One can constr uc t in p olynomia l time a DA G G = ( V , E ), a weigh t function and t wo vertices s, t that admits a path from s to t of n ull weigh t if and only if there exists a non e mpt y set S ⊆ A such that X a ∈ S a = 0 The DA G G = ( V , E ) has n + 2 ordered vertices V = [ s = 0 , 1 , 2 , . . . , n, n + 1 = t ] and the arcs ar e all the pa irs e ij = ( i, j ) with i < j . Hence, G is a linear order. The weigh ts are : w ( e ij ) = ( a j for 0 ≤ i < j ≤ n 0 for 1 ≤ i ≤ n and j = t +1 for i = 0 and j = t 1 S. Burc kel Complexit y of some Path Problems in DA Gs and Linear Orders. The following example shows this construction for A = { 4 , 2 , − 5 } (arcs are oriented from left to right) : 0 0 0 t +1 s +4 +2 −5 +2 −5 −5 Assume ther e ex ists a non e mpt y subs et of A S = { a i 1 , . . . , a i k } suc h that a i 1 + . . . + a i k = 0. Then, one can a ssume that i 1 < i 2 < . . . < i k . Hence, π = [ s, a i 1 , . . . , a i k , t ] is a n ull weighted path from s to t . F or the conv erse, observe that every path fro m s to t (excepted the direct arc ( s, t ) of weight +1) co n tains at least one ar c with an ele men t of A . Hence, a nu ll w eight ed path from s to t des crib es a non empt y s ubset of A with null sum.  Now we consider a similar problem on D A Gs with po sitive weigh ts. Problem 2 : K W eigh ted P ath. input : a DA G G = ( V , E ) endow e d with a weigh t function w : E 7→ N , tw o vertices s, t and an integer K ≥ 0 question : is there a directed path π from s to t such that X e ∈ π w ( e ) = K Prop ositio n (2.) The K W eig hted Path Decisio n P roblem is NP-Co mplete, even on linear orders. Pro of. Obviously , it is in NP . T o prov e that it is NP hard, we reduce ag ain the SUBSET SUM Pro blem to it. Let A = { a 1 , . . . , a n } be a subset of Z n . One can constr uct in p o lynomial time a DA G G = ( V , E ), a weight function, 2 S. Burc kel Complexit y of some Path Problems in DA Gs and Linear Orders. an integer K and tw o vertices s, t that admits a pa th from s to t of weight K if and only if there ex ists a non e mpty set S ⊆ A such that X a ∈ S a = 0 Let P = − m in ( { 0 } ∪ A ) and fix K = ( n + 1) P . The constructio n is similar as pre v iously . How ev er, one trans lates the weigh t of e a ch arc ( i , j ) by ( j − i ) P . The DA G G = ( V , E ) has n + 2 o rdered vertices V = [ s = 0 , 1 , 2 , . . . , n, n + 1 = t ] and the arcs are all the pairs e ij = ( i, j ) with i < j . Hence G is a line a r order. The weigh ts are : w ( e ij ) =    ( j − i ) P + a j for 0 ≤ i < j ≤ n ( j − i ) P for 1 ≤ i ≤ n and j = t ( j − i ) P + 1 for i = 0 and j = t F or A = { 4 , 2 , − 5 } one obtains with P = − min ( { 0 , 4 , 2 , − 5 } ) = + 5 the follow- ing linear order DA G : P−5 P+2 3P 2P 4P+1 s t 2P−5 P+4 3P−5 2P+2 P Assume ther e ex ists a non e mpt y subs et of A S = { a i 1 , . . . , a i k } suc h that a i 1 + . . . + a i k = 0. Then, one can a ssume that i 1 < i 2 < . . . < i k . Hence, π = [ s, a i 1 , . . . , a i k , t ] is a path fro m s to t of weight K = ( n + 1) P . F or the conv erse, observe that every path fro m s to t (excepted the direct arc ( s, t ) of weigh t ( n + 1) P + 1) contains at least one arc that contributes to an element of A . Hence, a path of weigh t ( n + 1) P describ es a non empty subset of A with n ull s um.  Now, w e consider DA Gs without weights. W e just count the leng th of the paths, i.e, the n um ber of a rcs in them. 3 S. Burc kel Complexit y of some Path Problems in DA Gs and Linear Orders. Problem : P ath of Length K . input : a DA G G = ( V , E ), tw o vertices s, t and an integer K > 0. question : is there a directed path from s to t of le ngth K . This pro blem is solv able in determinis tic po lynomial time since one can as sume that K < | V | since G is a D AG . Hence, o ne can efficiently compute the matrix M K where M is the adjacency matrix of G . How ever, when one adds a supplementary condition, the problem beco mes NP- Complete. Problem 3 : Irreducible P ath of Length K . input : a DA G G = ( V , E ), tw o vertices s, t and an integer K > 0. question : is there an irre ducible directed path fro m s to t of length K . Here irr e ducible p ath means a se quence of vertices [ s = x 0 , x 1 , . . . , x K = t ] such that ( x i , x j ) ∈ E if and only if j = i + 1. Of course, in linear o rders, this problem is trivial since : for K ≥ 2 , the answer is necessarily NO (b y tr ansitivity). for K = 1 , that co rresp onds to s < t . F or mor e general DA Gs, this problem is no n trivial anymore. Prop ositio n (3.) The Irreducible Path of Length K Decision Problem is NP-Complete. Pro of. Obviously again, it is in NP . F or the NP Hardness, we reduce now the CNF SA T Problem to it. Let Φ b e a CNF fo r mula C 1 ∧ C 2 ∧ C 3 . . . ∧ C k where each clause C i is a disjunction of litterals on different v ariables. Assume that the total num ber of litterals in Φ is N . W e construct a DA G G = ( V , E ) with N + 2 vertices : one vertex per litteral in each c lause C i and tw o supplementary vertices s, t . The a rcs a re the pairs : ( s, x ) for x ∈ C 1 ( y , t ) for y ∈ C k ( x, y ) for x ∈ C i and y ∈ C j with j = i + 1 a nd x 6 = y ( x, y ) for x ∈ C i and y ∈ C j with j > i + 1 a nd x = y F or Φ = ( a ∨ b ∨ c ) ∧ ( d ∨ a ) ∧ ( a ∨ d ∨ c ), one obtains the following DA G : 4 S. Burc kel Complexit y of some Path Problems in DA Gs and Linear Orders. 2 1 C 3 C C t a s b c a d a d c The question is to find an irre ducible pa th of length K = ( k + 1). This is equiv alent to the satisfiability of Φ : one has to find at least one littera l assig ne d to ”true” in each cla use (hence a path of length k + 1 from s to t ) and with no contradictory assignments like x = tr ue and x = tr ue . F or co nsecutive clauses, such an arc ( x, x ) do es no t exist. Otherwise, this arc ( x, x ) would make the path not irreducible.  [1] S. Basagni,D. B r uschi, F. Ra v asio , On the difficult y of finding walks of le ng th k , Theoretical Informatics and Applications 31 (5) (1997) 429– 435. Serge Burck el. INRIA-LORIA, Campus Scien tifique BP 239 54506 V ando euvr e-l` es-Nancy Cedex sergeburckel@orange.fr This w ork has been partially supp or ted by the CNRS-ANR pro ject ”graphs decomp ositions and algorithms” . 5