Retractions onto series-parallel posets

Discrete Mathematics 308 (2008) 2104–2114www.elsevier.com/locate/disc

Retractions onto series-parallel posets

Víctor Dalmaua, Andrei Krokhinb,∗, Benoit Larosec

aDepartment of Technology, University Pompeu Fabra, Passeig de la Circumval.lacio 8, Barcelona 08003, SpainbDepartment of Computer Science, University of Durham, Science Laboratories, South Road, Durham DH1 3LE, UK

cDepartment of Mathematics and Statistics, Concordia University, 1455 de Maisonneuve West, Montréal, Qué., Canada H3G 1M8

Received 13 October 2003; received in revised form 28 April 2006; accepted 4 August 2006Available online 6 May 2007

Abstract

The poset retraction problem for a poset P is whether a given poset Q containing P as a subposet admits a retraction onto P,that is, whether there is a homomorphism from Q onto P which fixes every element of P. We study this problem for finite series-parallel posets P. We present equivalent combinatorial, algebraic, and topological charaterisations of posets for which the problemis tractable, and, for such a poset P, we describe posets admitting a retraction onto P.© 2007 Elsevier B.V. All rights reserved.

Keywords: Poset retraction; Series-parallel; Posets; Complexity

1. Introduction

Throughout the paper, we consider only finite posets. For posets P, Q, etc., we shall denote their universes by P, Q,etc., and their partial orders by �P, �Q, etc., omitting the superscript if the order is clear from the context. A mappingh from poset Q to poset P is called monotone, or a homomorphism if, for any x�y in Q, we have h(x)�h(y) in P.In this case we write h : Q → P. If, in addition, P is a subposet of Q and h(x) = x for every x ∈ P then h is called aretraction from Q onto P.

The poset retraction problem for a poset P, denoted PoRet(P), is whether a given poset Q containing P as a subposetadmits a retraction onto P. The notion of retraction plays an important role in combinatorial theory (see, e.g., [3,6,7,15]).The poset retraction problem is also of interest in theoretical computer science, as the following examples show. It canbe considered a natural order-theory analogue of the much-studied graph H-coloring problem [5], or a special case ofthe constraint satisfaction problem (CSP), which is the problem of deciding whether there exists a homomorphism froma given relational structure to a fixed relational structure [4]. The CSP attracts much attention in artificial intelligence,combinatorics, and complexity theory, and it is known that a CSP for any fixed structure can be encoded as a posetretraction problem [4]. The poset retraction problem has also been studied in connection with type reconstruction[1,11,16].

In this paper, we consider the poset retraction problem in the case of series-parallel posets. This is an interesting classof posets from both the mathematical [17,21] and computer science [12–14] point of view. Recall that a linear sum of

∗ Corresponding author.E-mail addresses: [email protected] (V. Dalmau), [email protected] (A. Krokhin), [email protected] (B. Larose).

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V. Dalmau et al. / Discrete Mathematics 308 (2008) 2104–2114 2105

two (disjoint) posets P1 and P2 is a poset P1 + P2 with universe P1 ∪P2 and partial order �P1 ∪ �P2 ∪{(p1, p2)|p1 ∈P1, p2 ∈ P2}.

Definition 1. A poset is called series-parallel if it can be constructed from singletons by using disjoint union and linearsum.

Let k denote a k-antichain (that is, disjoint union of k singletons). A 4-crown is a poset isomorphic to 2 + 2. TheN-poset can be described as 2 + 2 with one comparability missing (its Hasse diagram looks like the letter “N”).Series-parallel posets can be characterized as N-free posets, that is, posets not containing the N-poset as an inducedsubposet [20].

2. Definitions and results

In this section we give all necessary definitions and state our results.

Definition 2. A poset P is said to admit an operation f : P n → P if, for any a, b ∈ P n such that a(i)�b(i) fori = 1, . . . , n, we have f (a)�f (b). In this case we also say that f is a polymorphism of P.

Definition 3. An n-ary operation f on P is called idempotent if f (x, . . . , x) = x for all x ∈ P .It is said to be Taylor if, in addition, it satisfies n(�2) identities of the form

f (xi1, . . . , xin) = f (yi1, . . . , yin), i = 1, . . . , n,

where xij , yij ∈ {x, y} for all i, j and xii �= yii for i = 1, . . . , n.An n-ary idempotent operation f is said to be totally symmetric idempotent (or TSI) if

f (x1, . . . , xn) = f (y1, . . . , yn)

for all sets of variables such that {x1, . . . , xn} = {y1, . . . , yn}.

Posets admitting Taylor and TSI operations were studied in [9,10,19]. Under the name ‘set functions’, TSI operationswere applied in the study of CSP [2].

Definition 4. A subset X of a poset P is called an idempotent subalgebra of P if there is a triple (Q, q0, g) where Q isa poset, q0 ∈ Q, and g is a partial map from Q to P with domain Y with the following property:

X = {h(q0)|h: Q → P and h|Y = g}.

Equivalently, X is definable by a first-order formula � with one free variable such that � contains only conjunction,existential quantification, the order predicate (of P), and constants (interpreted as elements of P).

The name “idempotent subalgebra” comes from another equivalent definition which goes as follows. A subset of Xof a poset P is called an idempotent subalgebra of P if it is preserved by all idempotent polymorphisms of P, that is,f (x1, . . . , xn) ∈ X for all n-ary idempotent polymorphisms of P and x1, . . . , xn ∈ X, n�1.

Of course, an idempotent subalgebra of P can be considered as a poset, with ordering induced by P.For basic definitions in algebraic topology, we refer to [18]. Recall that to each finite poset is associated naturally a

simplicial complex: the simplices are the totally ordered subsets of the poset (see [10] for details.) If P is a connectedposet, we shall denote the fundamental group of the associated complex by �(P); we shall say that P is simply connectedif this group is trivial. It is well known that �(2 + 2) is isomorphic to Z (see, e.g., [10]), so the 4-crown is not simplyconnected. It is also well known (and easy to show) that if a poset P is a retract of a poset Q then the group �(P) is aretract of �(Q).

Denote by X the class of all series-parallel posets P with the following property: if {a, a′, b, b′} is a 4-crown in P, aand a′ being the bottom elements, then at least one of the following conditions holds:

(1) there is e ∈ P such that a, a′, e, b, b′ form a subposet in P isomorphic to 2 + 1 + 2;

2106 V. Dalmau et al. / Discrete Mathematics 308 (2008) 2104–2114

(2) infP(a, a′) exists;(3) supP(b, b′) exists.

Recall that infP and supP denote the greatest common lower bound and the least common upper bound in P, respectively.Usually we write simply inf and sup, as P is clear from the context, and we also call them meet and join, respectively.

Fix a poset P ∈ X. If P is a subposet of Q and A, B are two antichains in P, let QA,B denote the set of all elementsq ∈ Q such that both q �a for every a ∈ A and q �b for every b ∈ B. One of A, B may be empty, and in this case thecorresponding part of the condition disappears.

In order to describe posets Q that admit a retraction onto P, we need to introduce the following two conditionson Q:

(R1) For every pair (A, B) of antichains in P, QA,B is empty whenever PA,B is empty.(R2) For every non-empty and non-connected H ⊆ P that is of the form H = PA′,B ′ , and for every p1, p2 ∈ H that

belong to different connected components of H, there is no path (of comparable elements) in Q that connects p1and p2 and such that every element in this path belongs to some QA,B with PA,B ⊆ H .

Theorem 1. Let P ∈ X be connected. Then a poset Q containing P has a retraction onto P if and only if Q satisfiesconditions (R1) and (R2).

Theorem 2. Let P be a connected series-parallel poset. Then the following conditions are equivalent:

(1) P ∈ X;(2) P does not retract onto any of 1 + 2 + 2 + 2, 2 + 2 + 2 + 1, 1 + 2 + 2 + 2 + 2 + 1, 2 + 2, and 2 + 2 + 2;(3) P has a Taylor polymorphism;(4) P has a TSI polymorphism of every arity k�2;(5) every connected idempotent subalgebra of P is simply connected;(6) for every connected idempotent subalgebra Q of P, �(Q) does not retract onto Z.

Theorem 3. Let P be a series-parallel poset. If every connected component of P satisfies any of the conditions ofTheorem 2 then PoRet(P) is tractable. Otherwise it is NP-complete.

3. Proof of Theorem 1

For a poset P, denote by w(P) and w(P) the biggest and the smallest, respectively, elements in P that are comparablewith each element in P.

Lemma 1. Let P ∈ X be connected. Then w(P) and w(P) exist.

Proof. Since P is series-parallel, it follows that P is the sum of two posets. In particular, every maximal element inP is above any minimal one. If P has the greatest element then it is comparable with all other elements, so w(P) andw(P) exist.

Suppose P has at least two maximal elements and let V denote the set of all maximal common lower bounds of theset of all maximal elements of P. If V = {p} then p = w(P). Indeed, if there is a ∈ P incomparable with p then thereexists a maximal element of m such that a�m. Take any maximal element m′ such that a < m′. Then {a, p, m, m′}is an N-poset, a contradiction. Suppose |V |�2. If the elements of V have a unique greatest common lower bound,that is, inf V , then one can show that it is w(P). Assume inf V does not exist. If V1, V2 ⊆ V , V1 ∩ V2 �= ∅, andinf V1 and inf V2 would both exist then they would be comparable, because P is N-free. Moreover, we would haveinf(inf V1, inf V2) = inf(V1 ∪ V2). Since inf V does not exist, one can find vi, vj ∈ V such that inf(vi, vj ) does notexist. Consider W ={x ∈ P | x�vi, vj }. Note that this set contains all maximal elements of P. The poset W cannot beconnected, since otherwise any of its minimal elements would be below all maximal elements of P which is impossibleby the choice of V. Now pick two minimal elements wk, wl from different connected components of W. It is easy tosee that{vi, vj , wk, wl} form a 4-crown that contradicts the assumption that P ∈ X.

Since there exists an element in P comparable with all other elements, w(P ) exists as well. �


Lemma 2. Let P ∈ X and A, B ⊆ P be two antichains (one of which may be empty). Then every connected componentof the subposet PA,B belongs to X.

Proof. Follows immediately from definitions, as the offending 4-crown in PA,B would be such in P. �

Proof of Theorem 1. It is easy to see that any retraction from Q onto P must map QA,B to PA,B . If Q does not satisfy(R1) then Q has an element that cannot be mapped anywhere by a retraction, due to the rule above. If Q does not satisfy(R2) then, in Q, there is a path of comparable elements that are forced to be mapped to some disconnected H and thatconnect two elements from different connected components of H. Hence, if Q does not satisfy either (R1) or (R2) thenthere is no retraction from Q onto P.

Suppose now that Q satisfies both (R1) and (R2). We show that there is a retraction r: Q → P. In order to define rwe need some constructions and notation.

In the following we write � to denote �Q, this will cause no confusion. However sup and inf will always becalculated in P. Take q ∈ Q, and let Lq be the set of maximal elements of {a ∈ P | a�q}, and Uq the set of minimalelements of {b ∈ P | b�q}. Let Hq = PLq,Uq . Note that Hq is non-empty.

It is not hard to show that if Hq is not connected then there exists no set of the form PA,B such that Hq is properlycontained in PA,B and the elements from different connected components of Hq are still disconnected in PA,B . SinceQ satisfies (R2), there is at most one connected component Kq of Hq such that q is connected to an element of Kq inthe way described in (R2). For every disconnected Hq , let us fix an arbitrary element z(Hq) such that z(Hq) = w(K)

for some connected component K of Hq . Note that z(Hq) actually depends on the set Hq , not on q. If Uq and Lq areboth non-empty, denote by Xq the set of all x ∈ P such that x is minimal with the property that

(1) PLq,Uq∪{x} is non-empty;(2) Uq ∪ {x} is an antichain;(3) sup(x, u) does not exist for some u ∈ Uq .

Denote PLq,Uq∪Xq by H ′q . Throughout the proof we use the following property which can be easily derived from

definitions: for every x ∈ Xq , either PLq,Uq∪{x} = PLq,Uq or, for all u ∈ Uq , sup(x, u) does not exist. We need thefollowing three claims in order to define the retraction.

Claim 1. If Hq �= ∅ then H ′q �= ∅.

We show that every minimal element of Hq belongs to H ′q . Suppose, for contradiction, that there is a minimal element

a ∈ Hq and some x ∈ Xq such that a�x. Then {a, b, x, u} where b is some minimal element of Hq such that b�x,and u is an arbitrary element in Uq , form an N-poset.

Claim 2. If Hq is not connected then Hq = H ′q .

Both Uq and Lq contain more than one element. Take an arbitrary x ∈ Xq . If x is above one maximal element ofPLq,Uq then it is above all elements of PLq,Uq (or else one can easily find an N-poset), and then PLq,Uq = PLq,Uq∪{x}.Otherwise, by the choice of x, there is a in PLq,Uq∪{x}. Take a maximal element b in PLq,Uq such that a�b, and anyelement u ∈ Uq . One can check that {a, b, x, u} is an N-poset.

Claim 3. If Hq is connected then so is H ′q .

It is shown above that all minimal elements of PLq,Uq belong to PLq,Uq∪Xq . If there is only one minimal element thenit ensures connectivity. Otherwise, for every pair a, b of distinct minimal elements, fix their minimal common upperbound (in PLq,Uq ) va,b. If every va,b is below all x ∈ Xq then we get connectivity. Assume, on the contrary, that thereexist a, b and x such that va,b�x. Then {a, b, va,b, x} is a 4-crown, and there is no inf(a, b) (since a, b are minimal inPLq,Uq ). If there exists u = sup(x, va,b) then take any maximal element w in PLq,Uq such that sup(w, x) does not existand show that {x, va,b, u, w} is an N-poset. Indeed, w and u are incomparable. Obviously, u�w, and we have w�u,since otherwise u= sup(w, x). Thus, u= sup(x, va,b) does not exist, and we obtain a contradiction with P ∈ X. Claim3 is proved.


Now we are ready to define the mapping r. Let

r(q) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

w(P) if Lq = Uq = ∅,

w(Hq) if Lq �= ∅, Uq = ∅, and Hq is connected,

w(Hq) if Lq = ∅, Uq �= ∅, and Hq is connected,

w(H ′q) if Lq �= ∅, Uq �= ∅, and Hq is connected,

w(Kq) if Lq �= ∅, Hq is disconnected, and Kq is defined,

w(Kq) if Lq = ∅, Hq is disconnected, and Kq is defined,

z(Hq) if Hq is disconnected and Kq is not defined.

Using Lemmas 1 and 2 it can be verified that r is well defined, and that it is identity on P. It remains to show that r is ahomomorphism. Take q < q ′ in Q and prove that r(q)�r(q ′). We distinguish seven cases depending on Lq , Uq , Lq ′ ,and Uq ′ . Throughout the proof we use the following facts without explicit reference: (1) QA,B is non-empty wheneverPA,B is non-empty (since Q satisfies (R1)) and (2) if p1, p2 ∈ P and there is a path from p1 to p2 in QA,B then thereis a path from p1 to p2 in PA,B (this follows from the fact that Q satisfies (R2)). Moreover, in the second case p1 andp2 either are comparable or have a common upper or lower bound, since P is N-free.

Case 1: Lq = Uq = ∅ or Lq ′ = Uq ′ = ∅.Suppose that Lq = Uq = ∅, that is, r(q) = w(P). It follows that Uq ′ = ∅. Assume that Lq ′ �= ∅, since otherwise

trivially r(q)= r(q ′). If PLq′ ,∅ is not connected then it is easy to see that r(q) is below every element in PLq′ ,∅. If PLq′ ,∅is connected then r(q)�r(q ′) because r(q) is comparable with every element in P and r(q) > r(q ′) would contradictthe choice of r(q ′).

Case 2: Lq = Uq ′ = ∅.We have r(q)�w(P)�w(P)�r(q ′) (similarly to Case 1).

Case 3: Lq �= ∅, Uq = ∅.It follows that Lq ′ �= ∅ and Uq ′ = ∅.Subcase 3.1: PLq′ ,∅ is not connected.

We have PLq′ ,∅ ⊆ PLq,∅.Indeed, since q ′ ∈ QLq∪Lq′ ,∅, we know that PLq∪Lq′ ,∅ is non-empty. Then there exists a maximal element u ∈ PLq′ ,∅

such that u ∈ PLq,∅. Then every maximal element v from any connected component K of PLq′ ,∅ such that u /∈ K mustbelong to PLq,∅, or otherwise one can easily find an N-poset there. Now if there is x ∈ PLq′ ,∅ such that a�x for somea ∈ Lq then take any maximal element v from a connected component of PLq′ ,∅ not containing x, and take any elementb ∈ Lq ′ . It is easy to see that {a, b, x, v} is an N-poset.

If every element x ∈ PLq′ ,∅ is connected with exactly the same elements in PLq,∅ as it is in PLq′ ,∅ then PLq,∅=PLq′ ,∅.Indeed, otherwise one can choose a < u such that a /∈ PLq′ ,∅ but u ∈ PLq′ ,∅. Take any element v from a connectedcomponent of PLq′ ,∅ not containing u and any element b ∈ Lq ′ such that b�a. Then {a, b, u, v} form an N-poset, acontradiction. So, in this case r(q) = r(q ′) because q and q ′ are connected in QLq′ ,∅ = QLq,∅.

The argument from the previous paragraph also shows that if PLq,∅ strictly contains PLq′ ,∅ then there exist a andu as described above and, moreover, every such a must lie below each element from PLq′ ,∅ (or else we can find anN-poset). So we now assume that PLq′ ,∅ is contained in one connected component of PLq,∅. Now it is easy that if PLq,∅is connected then r(q)�r(q ′). If PLq,∅ has at least two components then every element of Lq ′ is above every elementof Lq , or else there is an N-poset (two incomparable elements from Lq ′ and Lq , and two maximal elements u, v suchthat u ∈ PLq′ ,∅ and v is from any other component of PLq,∅). Therefore, Lq ′ , PLq′ ,∅, r(q) and r(q ′) are all in the sameconnected component K of PLq,∅. Since r(q) is then below all elements of PLq′ ,∅, we have r(q)�r(q ′).

Subcase 3.2: PLq′ ,∅ is connected.Every maximal element of PLq′ ,∅ is a maximal element of PLq,∅ or otherwise one can easily find an N-poset.If PLq,∅ is connected then r(q)�r(q ′) because of the above inclusion. If PLq,∅ has at least two components then the

argument is the same as in the previous subcase.


Case 4: Uq ′ �= ∅, Lq ′ = ∅.Dual to Case 3.From now on, we assume that Uq �= ∅ and Lq ′ �= ∅, and at least one of Lq , Uq ′ is non-empty.Obviously, if u� l for some u ∈ Uq and l ∈ Lq ′ then r(q)�r(q ′). Assume that l1≮u1, l1 < u2 for some l1 ∈ Lq ′

and u1, u2 ∈ Uq . Then l1≮r(q). If r(q) is incomparable with l1 then {r(q), l1, u1, u2} is an N-poset, a contradiction.So r(q)� l1 �r(q ′). The argument is similar if there are l1, l2 ∈ Lq ′ and u ∈ Uq such that l1 < u and l2≮u.

In what follows we will always assume that either l < u for all l ∈ Lq ′ , u ∈ Uq , or all such l and u are incomparable.Case 5: Lq = ∅ and Uq ′ �= ∅.Subcase 5.1: Hq is disconnected, and l < u for all l ∈ Lq ′ , u ∈ Uq .Suppose there are l′1, l′2 ∈ Lq ′ belonging to two different connected components of Hq . If r(q ′)�u for some

element u ∈ Uq then r(q)�r(q ′). If not then, since Hq is disconnected, there is u ∈ Uq such that r(q ′) and u areincomparable. Let l′1 not belong to the connected component of Hq containing r(q). Then r(q)�r(q ′), since otherwise{r(q), l′1, u, r(q ′)} is an N-poset.

So let Lq ′ be contained in one connected component of Hq . Then, for every a ∈ Hq ′ , either a ∈ Hq , or a > b forevery b ∈ Hq , since otherwise {u, a, l, x} is an N-poset, where l ∈ Lq ′ is such that l < a, u ∈ Uq is such that u and aare incomparable, and x is a suitable maximal element of Hq .

Assume first that H ′q ′ ⊆ Hq . Then, for any l ∈ Lq ′ , there is a path in Q, from q to l, such that every element

on this path belongs to some QA,B with PA,B ⊆ Hq . Therefore, both r(q) and r(q ′) belong to the same connectedcomponent Kq of Hq , and r(q) = w(Kq). Then r(q) and r(q ′) are comparable. Moreover, by the definition of Uq ,r(q) ∈ Hq ′ . If r(q) ∈ H ′

q ′ then r(q)�r(q ′), as r(q) and r(q ′) are both comparable with all elements of H ′q ′ . As-

sume that r(q) > r(q ′). Then r(q)≮x for some x ∈ Xq ′ . Let A = {a|a < r(q) and r(q ′)�a}. This set is non-empty,since otherwise r(q) could not be above r(q ′). Let A′ be the set of all maximal elements of A. If some a ∈ A′satisfies a�x for some x ∈ Xq then {a, r(q ′), r(q), x} is an N-poset, a contradiction. If there exists inf(r(q ′), a)

for all a ∈ A′ then there exists c = inf(A′ ∪ {r(q ′)}), and c would contradict the choice of r(q) because it iseasy to show that c is comparable with all elements in Hq . So inf(r(q ′), a) does not exist for some a ∈ A′. Then{a, r(q ′), r(q), x} is a crown contradicting P ∈ X because the existence of sup(r(q), x) would imply the exis-tence of sup(x, u) for every u ∈ Uq which, in turn, would contradict the definition of x ∈ Xq . So if Hq ′ ⊆ Hq

then r(q)�r(q ′).Assume now that there is a ∈ H ′

q ′ such that a > b for every b ∈ Hq . (Note that then Hq ′ is connected or every a ∈ Hq ′

is above every b ∈ Hq .) Consider minimal such elements. If there is only one such element a then r(q)�a�r(q ′). Ifthere are more than two such elements a1, . . . , an then every pair (ai, aj ) must have a supremum (or else {ai, aj , y, z},where y, z are from different connected components of Hq , forms a crown that contradicts P ∈ X). Then there existsc = sup({a1, . . . , an}). It is easy to check that c is comparable with all elements in H ′

q ′ . Hence r(q)�c�r(q ′).Subcase 5.2: Hq is disconnected, and l and u are incomparable for all l ∈ Lq ′ , u ∈ Uq .If there are l ∈ Lq ′ and a ∈ Hq such that l > a then l > b for every b from any connected component of Hq not

containing a, or otherwise {u, l, a, b} is an N-poset for any u ∈ Uq . So in this case r(q) < l�r(q ′). We may assumefor the rest of this subcase that a�l for all l ∈ Lq ′ and a ∈ Hq .

One can easily show that u�u′ for every u ∈ Uq and u′ ∈ Uq ′ ∪ Xq ′ (or else one can find an N-poset).For every l ∈ Lq ′ there exists bl ∈ Hq such that l < bl and u < bl for some u ∈ Uq , since there is a path from l to u

in Q∅,Uq′ . Then bl > u for each u ∈ Uq , or otherwise one can easily find an N-poset.Take arbitrary u1, u2 ∈ Uq . Since P ∈ X, there exists u = sup(u1, u2). We know that u�u′ for every u ∈ Uq and

u′ ∈ Uq ′ ∪Xq ′ . If u > l for some l ∈ Lq ′ then one can take bl =u, and, in fact u= sup Uq . We show that u is comparablewith all elements of H ′

q ′ . Indeed, if u is incomparable with some a ∈ H ′q ′ then, for each ui ∈ Uq either {u, a, ui, l} is an

N-poset or a > ui . In the latter case, a must be comparable with u = sup Uq . Now we have r(q) < u�r(q ′) regardlesswhether u ∈ H ′

q ′ or not.

We may assume now that l�u for each l ∈ Lq ′ .Assume that H ′

q ′ is disconnected. If there is x ∈ Hq ′ such that u�x then there exists c ∈ Hq ′ such that c > x andc > u, and then one can always find a maximal element y ∈ Hq ′ such that {x, u, c, y} is an N-poset. If u�x for allx ∈ Hq ′ then we have r(q)�r(q ′).

Assume that H ′q ′ is connected. We know that r(q ′)�u. Assume that r(q ′) and u are incomparable and derive a

contradiction. Since r(q ′) and u are connected by a path in Q∅,Uq′ , they have a common bound in P∅,Uq′ . It is easy


to show that if they have a common lower bound then one can prove as in the first paragraph of this subcase thatr(q)�r(q ′). So assume that r(q ′) and u have no common lower bound, and there is b ∈ Hq ′ such that u < b andr(q ′) < b. Let B be the set of all minimal b with this property. It is not hard to verify that there cannot be only one suchminimal element. Any b ∈ B must belong to H ′

q ′ . Indeed, if b�x for some x ∈ Xq ′ then {b, x, u, r(q ′)} form a crowncontradicting the assumption that P ∈ X. If sup(bi, bj ) exists for any pair bi, bj ∈ B then there exists sup B. It is easyto check this element belongs to H ′

q ′ , and it is comparable with every element of this set. This is a contradiction withthe choice of r(q ′). So sup(bi, bj ) does not exist for some bi, bj ∈ B. Then {u, r(q ′), bi, bj } is a crown contradictingthe assumption that P ∈ X. Thus we have r(q)�u�r(q ′).

Subcase 5.3: Hq is connected, and l < u for all l ∈ Lq ′ , u ∈ Uq .In this case r(q) = w(Hq). We assume that r(q) ∈ Hq ′ , since otherwise r(q)� l�r(q ′) for some l ∈ Lq ′ .Assume that H ′

q ′ is connected, and r(q) is incomparable with some x ∈ Xq ′ , that is, r(q) /∈ H ′q ′ . Note that sup(x, r(q))

does not exist. If r(q) and r(q ′) are incomparable then, for any u′ ∈ Uq ′ , {r(q), r(q ′), u′, x} is an N-poset. Assume, forcontradiction that r(q) > r(q ′). Let A be the set of all a minimal with the following property: a < r(q) and a≯r(q ′). Ifinf(a, r(q ′)) exists for each a ∈ A then there exists sup(A ∪ {r(q ′)}), and this element is comparable with all elementsin Hq , a contradiction with the choice of r(q). So inf(a, r(q ′)) does not exist for some a ∈ A. Then {a, r(q ′), r(q), x}is a crown contradicting P ∈ X. Consequently, if H ′

q ′ is connected then r(q) ∈ H ′q ′ . Then, as r(q ′) = w(H ′

q ′),r(q) andr(q ′) are comparable. Assume, for contradiction, that r(q) > r(q ′). Let B be the set of all elements b minimal with thefollowing property: b < r(q) and b≯r(q ′). Similarly, let C be the set of all elements c ∈ H ′

q ′ minimal with the followingproperty: c > r(q ′) and c≮r(q). By definitions of r(q) and r(q ′), both B and C are non-empty. Arguing as above, onecan show that there are b ∈ B and c ∈ C such that neither inf(b, r(q ′)) nor sup(c, r(q)) exist. Then {b, r(q ′), c, r(q)}is a crown contradicting that P ∈ X.

Assume now that Hq ′ is disconnected. Note that Lq ′ has at least two elements. Let b be a minimal element fromthe connected component of Hq ′ containing r(q) and a a minimal element from any other connected component. InHq , consider a maximal (with respect to inclusion) antichain D such that Lq ′ ⊆ D. It is easy to show that, for everyd ∈ D, both d < a and d < b holds, or else P /∈X. The standard argument shows that there must be d1, d2 ∈ D suchthat inf(d1, d2) does not exist. By the choice of a and b, sup(a, b) does not exist either. Then {a, b, d1, d2} is a crowncontradicting that P ∈ X.

Subcase 5.4: Hq is connected, and l and u are incomparable for all l ∈ Lq ′ , u ∈ Uq .Assume that Hq ′ is disconnected. If r(q)� l for some l ∈ Lq ′ then r(q)�r(q ′).Assume now that r(q) is incomparable

with each l ∈ Lq ′ . Since, in Q∅,Uq′ , there is a path from r(q) to any such l, we see that r(q) and l have a commonbound in P∅,Uq′ . If, for some l ∈ Lq ′ , there is b such that b > l and b > r(q) then using disconnectedness of Hq ′ it iseasy to prove that r(q) < a for every a ∈ Hq ′ , in particular, r(q) < r(q ′). Assume now that, for each l ∈ Lq ′ , r(q) andl have only lower bounds in common. It follows, in particular, that sup(l, r(q)) exist for no l ∈ Lq ′ . Fix an arbitraryl ∈ Lq ′ and consider the set A consisting of all maximal common lower bounds of r(q) and l. Extend A to a maximalantichain D in Hq . Arguing as in the previous subcase, one can derive a contradiction with P ∈ X.

Assume now that H ′q ′ is connected. We know that r(q ′)�r(q). Assume that these elements are incomparable and

derive a contradiction. If r(q)≮x for some x ∈ Xq ′ then {r(q), r(q ′), x, u′} is an N-poset for every u′ ∈ Uq ′ . So wehave r(q) ∈ P∅,Uq′∪Xq′ . Arguing as above, one can prove that if inf(r(q), r(q ′)) exists then it must be r(q). Similarly, ifsup(r(q), r(q ′)) exists then it must be r(q ′). So by our assumption neither inf(r(q), r(q ′)) nor sup(r(q), r(q ′)) exists.We know that r(q) and r(q ′) have a common upper or lower bound in P∅,Uq′ , since they are connected by a path inQ∅,Uq′ . Assume that none of the common bounds belong to P∅,Uq′∪Xq′ . Then all their common bounds are upper. Takea minimal upper bound c ∈ P∅,Uq′ . There is x ∈ Xq ′ such that c and x are incomparable. Then, by construction ofXq ′ , sup(c, x) does not exist, and {r(q), r(q ′), x, c} form a crown contradicting P ∈ X. So r(q) and r(q ′) do havecommon bounds in P∅,Uq′∪Xq′ . Now depending on whether they have lower or upper bounds, one can find, using the“antichain” method above, a crown with the top or the bottom elements r(q) and r(q ′) such that this crown contradictsthe assumption P ∈ X.

Case 6: Lq �= ∅ and Uq ′ = ∅.Subcase 6.1: Hq ′ is disconnected, and l < u for all l ∈ Lq ′ , u ∈ Uq .The argument dual to the one from the first two paragraphs of Subcase 5.1 shows that it is enough to consider the

case when Uq is contained in one connected component of Hq ′ , and, for every a ∈ Hq , either a ∈ Hq ′ or a < b forevery b ∈ Hq ′ .


Assume first that H ′q ⊆ Hq ′ . Then, for any u ∈ Uq, u and q ′ are connected by a path in Hq ′ , and therefore r(q)

and r(q ′) are in the same connected component Kq ′ of Hq ′ , and r(q ′) = w(Kq ′). Then r(q) and r(q ′) are comparable.If r(q ′)�u for each u ∈ Uq then, since r(q ′) = w(Kq ′), we have r(q ′) > u for some u ∈ Uq , and r(q) < r(q ′). So letr(q ′) ∈ Hq . Then Hq is connected and r(q)=w(H ′

q). Assume, for contradiction, that r(q) > r(q ′). Let A be a maximal(with respect to inclusion) antichain in Kq ′ such that Uq ⊆ A, and let B = {b ∈ A | r(q)�b}. If A = B then it iseasy to show that r(q) is comparable with each element in Kq ′ , a contradiction with the choice of r(q ′). Assume thatA �= B. If, for every x ∈ A\B and every u ∈ Uq , there exists sup(x, u) then there exists c = sup(Uq ∪ (A\B)). If b�c

for every b ∈ B then it is easy to show that c is comparable with each element in Kq ′ , a contradiction with the choiceof r(q ′). If b�c for some b ∈ B then, for any x ∈ A\B, {x, r(q), b, c} is an N-poset. Hence there are x ∈ A\B andu ∈ Uq such that sup(x, u) does not exist. Then x ∈ Xq . Since r(q)�x, we obtain a contradiction with the choice ofr(q). Thus if H ′

q ⊆ Hq ′ then r(q)�r(q ′).Assume now that there is a ∈ H ′

q such that b < a for every every b ∈ Hq ′ . If all elements in H ′q are such then,

obviously, r(q) < r(q ′). If not then H ′q is connected and then r(q) = w(H ′

q). Note that there exist l ∈ Lq and l′ ∈ Lq ′such that l < l′. Take any element x from a connected component of Hq ′ not containing Uq . The element l ensures thatx ∈ Xq . Thus we have r(q)�x, and r(q) must be below all elements in Hq ′ , in particular, r(q) < r(q ′).

Subcase 6.2: Hq ′ is disconnected, and l and u are incomparable for all l ∈ Lq ′ , u ∈ Uq .Take arbitrary l and u. Since l and u are connected by a path in QLq,∅, they have a common bound in PLq,∅. Assume

that some l and u have a common upper bound b ∈ P�Lq ,∅ . Let l′≮b for some l′ ∈ Lq ′ . Then either b < a for every

a ∈ Hq ′ (and then r(q)�r(q ′)) or there is a from a suitable connected component of Hq ′ such that {l, l′, a, b} is anN-poset, a contradiction. Assume now that l′ < b for all l′ ∈ Lq ′ . Again, either u < a for every a ∈ Hq ′ (and thenr(q)�r(q ′)) or one can find an N-poset in P, a contradiction.

Assume for the rest of this subcase that no l ∈ Lq ′ and u ∈ Uq have a common upper bound. Fix l ∈ Lq ′ . For anyu ∈ Uq , there is a common lower bound du ∈ PLq,∅ of l and u. Choose du to be minimal in PLq,∅. If du≮u1 for someu1 ∈ Uq then {du1 , du, u1, l} is an N-poset. Hence du ∈ Hq , and, moreover, PLq,Hq∪{l} is non-empty. If Hq is connectedthen, by the definition of Xq , there is x ∈ Xq such that x� l, and then r(q)�x� l�r(q ′). If Hq is disconnected thena < l for every a ∈ Hq , and then r(q) < l < r(q ′), or else one can choose a ∈ Hq so that {a, du, u, l} is an N-poset.

Subcase 6.3: Hq ′ is connected, and l < u for all l ∈ Lq ′ , u ∈ Uq .In this case r(q ′) = w(Hq ′). We assume that r(q ′) ∈ Hq , since otherwise r(q)�u�r(q ′) for some u ∈ Uq .LetHq be connected. If r(q) and r(q ′) are incomparable then there isx ∈ Xq such that r(q ′)≮x, and {r(q), r(q ′), x, u}

is an N-poset for any u ∈ Uq . So r(q) and r(q ′) are comparable. Assume, for contradiction, that r(q) > r(q ′). Theargument similar to the one from the second paragraph of the previous subcase shows that this leads to a contradictionwith the choice of r(q ′)

Let Hq be disconnected. It is easy to see that r(q) and r(q ′) belong to the same connected component of Hq . Thenthe proof is similar to (the dual of) the last paragraph of Subcase 5.3.

Subcase 6.4: Hq ′ is connected, and l and u are incomparable for all l ∈ Lq ′ , u ∈ Uq .We have r(q ′) = w(Hq ′), and we may assume that r(q ′) is in comparable with u for each u ∈ Uq , since otherwise

we immediately have r(q) < r(q ′).Take l ∈ Lq ′ and u ∈ Uq . Since u and l are connected by a path in QLq,∅, they have a common bound in PLq,∅.Assume they have a common upper bound b. Then either a < l for every a ∈ Hq , and then r(q) < r(q ′), or u1 < b for

every u1 ∈ Uq . Similarly, either u < c for every c ∈ Hq ′ , and then r(q) < r(q ′), or l1 < b for every l1 ∈ Lq ′ . Assumethat b is above each element in Uq ∪ Lq ′ . Then b is comparable with r(q ′). If b�r(q ′) then, of course, r(q) < r(q ′).So let r(q ′) < b for every such b. Let B be the set of all minimal elements of {b|r(q ′) < b and u < b for all u ∈ Uq}.It is easy to check that every a such that r(q ′)�a is comparable with some element of B. If there exist sup(b1, b2)

for all b1, b2 in B then there exists sup B, and this element contradicts the choice of r(q ′). So sup(b1, b2) does notexist for some b1, b2 in B. Since P ∈ X, there is inf(r(q ′), u) for every u ∈ Uq . As in Subcase 6.2, we can choosedu � inf(r(q ′), u) so that du ∈ Hq and then derive that r(q) < r(q ′).

Case 7: Lq �= ∅ and Uq ′ �= ∅.Subcase 7.1: The inequality l < u holds for all l ∈ Lq ′ , u ∈ Uq .Assume, for contradiction, that r(q) and r(q ′) are incomparable. Then, since P is N-free, for every u ∈ Uq and

l′ ∈ Lq ′ , we have r(q) > l′ or r(q ′) < u. For the same reason, if r(q) > l′ for some l′ ∈ Lq ′ then r(q) > l′ for eachl′ ∈ Lq ′ , and if r(q ′) < u for some u ∈ Uq then r(q ′) < u for each u ∈ Uq . Hence, r(q) ∈ Hq ′ or r(q ′) ∈ Hq , or both.It is easy to verify that r(q) ∈ Hq ′ implies r(q) ∈ H ′

q ′ and r(q ′) ∈ Hq implies r(q ′) ∈ H ′q .


If u�r(q ′) then, obviously, r(q)�r(q ′). Assume that r(q ′) is incomparable with some u ∈ Uq . Then r(q ′) isincomparable with each u ∈ Uq and we have r(q) ∈ H ′

q ′ . Since P is N-free, it follows that u < u′ for each u ∈ Uq ,u′ ∈ Uq ′ , that is Uq ⊆ Hq ′ , and, moreover, Uq ⊆ H ′

q ′ . Since, by definition, r(q ′) is comparable with all elements from itsconnected component of H ′

q ′ , we conclude that Hq ′ is disconnected, and the elements r(q) and r(q ′) belong to differentconnected components of H ′

q ′ . Moreover, sup(u, r(q ′)) does not exist for any u ∈ Uq . There exists a ∈ Hq such thata /∈ Hq ′ , since otherwise, by definition, r(q) and r(q ′) would belong to the same connected component of Hq ′ . Thenl′�a for some l′ ∈ Lq ′ . If a� l′ then a ∈ PLq,Uq∪{r(q ′)} and r(q)�x�r(q ′) for some x ∈ Xq , a contradiction with ourassumption that r(q) and r(q ′) are incomparable.If a and l′ incomparable then it is easy to show that l′ ∈ PLq,Uq∪{r(q ′)},so r(q)�r(q ′), a contradiction again. We conclude that r(q ′) cannot be incomparable with each u ∈ Uq , and sor(q ′) ∈ H ′

q . Similarly, one can show that r(q) ∈ H ′q ′ .

Then both Hq and Hq ′ are disconnected, and r(q) and r(q ′) belong to different connected components of both Hq

and Hq ′ , since otherwise r(q) and r(q ′) would be comparable. It follows that P�Lq′ ,Uq

is non-empty and disconnected,

and, of course, PLq′ ,Uq ⊆ Hq ∩ Hq ′ . As we have already noticed when defining Hq , such an inclusion impliesP�L

q′ ,Uq= Hq = Hq ′ . But then, by definition, r(q) and r(q ′) would belong to the same connected component of this

set, and so would be comparable. This is a contradiction with our assumption that r(q) and r(q ′) are incomparable.Assume now, for contradiction, that r(q) > r(q ′). In this case, we again have r(q ′) ∈ Hq and r(q) ∈ Hq ′ . If is easy

to show that if x′ ∈ Xq ′ then either x′ ∈ Xq or u < x′ for some u ∈ Uq . In both cases r(q) must be below x′, and sor(q) ∈ H ′

q ′ . We have r(q ′) ∈ H ′q , as r(q ′) < r(q). Now choose A and B as in the second paragraph of Subcase 6.1, and

repeat the argument from that paragraph to derive a contradiction.Subcase 7.2: Hq ′ is disconnected, and l and u are incomparable for all l ∈ Lq ′ , u ∈ Uq .Note that l < l′ for every l ∈ Lq , l′ ∈ Lq ′ , and u < u′ for every u ∈ Uq , u′ ∈ Uq ′ because the poset P is N-free.Take arbitrary l and u. Since l and u are connected by a path in QLq,Hq′ , they have a common bound in P�Lq ,H

q′ .

The rest of the proof for this subcase is identical to Subcase 6.2.Subcase 7.3: Hq ′ is disconnected, and l and u are incomparable for all l ∈ Lq ′ , u ∈ Uq .Similar to Subcase 6.4.The theorem is proved. �


(4) ⇒ (3): trivial.(3) ⇒ (5): this follows immediately from [10, Theorem 3.2].(5) ⇒ (6): trivial.(6) ⇒ (2): suppose that (2) does not hold. This means that P retracts onto a poset R which has a connected idempotent

subalgebra which is not simply connected: in fact, this subalgebra is a 4-crown in every case. Indeed, for R = 2 + 2,it is R itself; the four top elements of 2 + 2 + 2 do the trick, and similarly for 1 + 2 + 2 + 2. For 2 + 2 + 2 + 1,the bottom four elements will do, and finally the four middle elements do the job for 1 + 2 + 2 + 2 + 2 + 1. In eachcase, it is easy to check, by using Definition 4, that the selected 4-crown is indeed an idempotent subalgebra. Thus Pretracts onto a poset R which has an idempotent subalgebra isomorphic to a 4-crown. By results in [8] this implies thatthe 4-crown is a retract of an idempotent subalgebra of P, call it Q. It is easy to check that the connected componentsof an idempotent subalgebra are themselves idempotent subalgebras, so we may assume that Q is connected. But thenthe fundamental group of Q retracts onto the fundamental group of the 4-crown, which is Z, a contradiction.

(2) ⇒ (1): we prove the contrapositive by induction on the size of P. If P = P1 ∪ P2 then P1 or P2 contains theoffending crown, so we are done by induction. Otherwise we have that P = P1 + P2. Suppose first that P1 contains anoffending crown. Then, by induction hypothesis, P1 retracts onto one of the five posets in the list. Since any combinationof retractions of P1 and P2 is a retraction of P, it follows that it is enough to prove the result assuming that P1 actuallyis one of the five posets in the list. If P1 is neither 2 + 2 nor 1 + 2 + 2 + 2 then P retracts onto 2 + 2 + 2 + 1 or1 + 2 + 2 + 2 + 2 + 1 by retracting P2 to a point, so we are done. Now suppose that P1 = 1 + 2 + 2 + 2. Then theoffending crown must consist of the top four elements, which means that P2 must have at least 2 minimal elements; ifthese have an upper bound then P2 retracts onto 2 + 1, so P retracts onto 1 + 2 + 2 + 2 + 2 + 1; otherwise (that is, ifno pair of minimal elements have an upper bound), P2 retracts onto 2 and so P retracts onto 1 + 2 + 2 + 2 + 2 , which


retracts onto 1 + 2 + 2 + 2. If P2 contains an offending crown the proof is dual. So we may now suppose that P1 hastwo maximal elements with no meet and P2 has two minimal elements with no join.

Claim. Let Q be a series-parallel poset of height at least two, and let x, y be maximal elements in Q that have a lowerbound but no meet. Then Q retracts onto 1 + 2 + 2.

Proof of Claim. We use induction on the size of Q. Clearly the result holds for small posets, and if Q = Q1 ∪ Q2 theresult follows immediately by induction. So suppose that Q = Q1 + Q2. Then {x, y} ⊆ Q2. Since x and y have noinfimum in Q they have no infimum in Q2. If {x, y} has a lower bound in Q2 then by induction it retracts onto 1+2 +2and so does Q, else Q2 retracts onto 2 because Q2 is N-free. Since x and y have no meet in Q it follows that Q1 hasat least two maximal elements, and these have a lower bound. Hence Q1 retracts onto 1 + 2 and so Q retracts onto1 + 2 + 2.

By the claim and its dual, P1 retracts onto 2 or 1 + 2 + 2, and P2 retracts onto 2 or 2 + 2 + 1. Combining the fourcases yields the desired result.

(1) ⇒ (4): we must show that every poset P ∈ X has TSI operations of every arity k�2. Let K = K(P) denotethe following poset, as defined in [10] (see also [9,19]): a subset C of a poset P is (weak) convex if for all c, c′ ∈ C,if c�x�c′ then x ∈ C. K is the poset of all non-empty convex sets in P ordered as follows: X�Y if and only if forevery x ∈ X there exists y ∈ Y such that x�y, and for every v ∈ Y there exists u ∈ X such that u�v. The embeddingof P into K sends p to {p}. By [19, Proposition 1, the last part of the proof of Theorem 3] (see also [4, remark (1), p.95]), P admits TSI operations of all arities if and only if it admits one of arity |P|, if and only if P is a retract of K.Hence, it suffices to prove that P is a retract of the poset K. Here we invoke Theorem 1.

(R1) Let (A, B) be a pair of antichains in P such that KA,B �= ∅; we must show that PA,B �= ∅. But it is immediate bythe definition of K and the embedding of P in K that if X ∈ KA,B then X ⊆ PA,B .

(R2) Let H =PA′,B ′be non-empty, let p1, p2 ∈ H be such that there exists a path from {p1} to {p2} in K such that everyelement in this path lies in some KA,B with PA,B ⊆ H . We must show that p1 and p2 are connected via a path inH itself. More precisely, consider a path

{p1} = X0, X1, . . . , Xn = {p2}

in K such that for each i=1, . . . , n−1, there are some antichains Ai, Bi in P such that Xi ∈ KAi,Biand PAi,Bi

⊆ H .As in the proof of property (R1) above, we have Xi ⊆ PAi,Bi

⊆ H for all i. By definition of comparability in K,we may find a sequence of elements x1 ∈ X1, x2 ∈ X2, . . . such that p1, x1, x2, . . . , xn−1, p2 is a path in H fromp1 to p2. �


For any poset P′, let Ext(P′) denote the following problem: given a poset Q and a partial mapping f from Q to P′,does f extend to a homomorphism from Q to P′. We will use the following result in our NP-completeness proof. Thisresult is more or less folklore, but, for completeness, we present it with a proof.

Proposition 1. For any finite poset P′, the problems PoRet(P′) and Ext(P′) are polynomial time equivalent.

Proof. Obviously, PoRet(P′) is a subproblem of Ext(P′), with f being the identity mapping defined on P′. We nowshow that Ext(P′) reduces to PoRet(P′) in polynomial time. Let (Q, f ) be an instance of Ext(P′). We may withoutloss of generality assume that Q ∩ P ′ = ∅. Let Q′ be a poset that is a disjoint union of Q and P′, and f ′ a partialmapping from Q′ to P′ that coincides with f on Q and is identity on P′. Obviously, the instances (Q, f ) and (Q′, f ′)are equivalent. Let �′ be the transitive closure of the following relation � on Q′:(x, y) ∈ � if and only if f ′(x) = f ′(y)

or x�y in Q′. It is easy to see that �′ can be computed from (Q, f ) in polynomial time. Moreover, if (p1, p2) ∈ �′ forsome p1, p2 ∈ P ′ such that p1�p2 in P′ then f does not extend to a homomorphism. Assume that this condition is notsatisfied. Then all elements of P ′ belong to different classes of the equivalence relation � = �′ ∩ (�′)−1. Now let Q′′ bea complete set of representatives of �-classes such that P ′ ⊆ Q′′, and let � be the partial order on Q′′ induced by �′.


Then the poset Q′′ = (Q′′, �) is an instance of PoRet(P′). It is straightforward to check that Q′′ retracts onto P′ if andonly f extends to a homomorphism. Since Q′′ can computed from (Q, f ) in polynomial time, the result follows. �Proof of Theorem 3. It is easy to see that if P is not connected then Q retracts onto P if and only if different connectedcomponents of P are not connected in Q and, for every connected component of P, the connected component of Qcontaining it retracts onto it. Since these conditions can be checked in polynomial time, we may from now on assumethat P is connected.

We begin with the NP-completeness part. Suppose that P /∈X. Theorem 2 implies that P has no Taylor polymorphism.Corollary 10 and Theorem 4 of [9] imply that if P has no Taylor polymorphism then Ext(P) is NP-complete. ByProposition 1, PoRet(P) is NP-complete.

Assume now that P satisfies the conditions of Theorem 2. In particular, P has TSI polymorphisms of every arityk�2. Then Proposition 1 above and Proposition 9 [9], imply that PoRet(P) is polynomial-time equivalent to a certainCSP (see [9]) whose tractability follows from [2] (or [4]). �

6. Conclusion

We have studied the poset retraction problem for series-parallel posets. We have obtained a full complexity classifi-cation for such problems and have characterized in several ways those posets for which the problem is tractable. It isan interesting research problem to find out whether some of the conditions of Theorem 2 (in particular, conditions (3)and (6)) determine tractable poset retraction problems for general posets.

Acknowledgments

Part of this work was done while the second author was visiting Concordia University, Montreal. Financial helpof NSERC Canada is gratefully acknowledged. Víctor Dalmau is partially supported by the MEC under the program“Ramon y Cajal”, Grants TIN 2006-15387-C03-03, TIN 2004-04343, the EU PASCAL Network of Excellence IST-2002-506778, and the MODNET Marie Curie Research Training Network MRTN-CT-2004-512234. Andrei Krokhin ispartially supported by the UK EPSRC Grants EP/C543831/1 and EP/C54384X/1. Benoit Larose is partially supportedby NSERC, FQRNT, and CRM.

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Documents

Retractions onto series-parallel posets