On the choice number of complete multipartite graphs with part size four
OON THE CHOICE NUMBER OF COMPLETE MULTIPARTITEGRAPHS WITH PART SIZE FOUR
H. A. KIERSTEAD, ANDREW SALMON, AND RAN WANG
Abstract.
Let ch( G ) denote the choice number of a graph G , and let K s ∗ k be thecomplete k -partite graph with s vertices in each part. Erdős, Rubin, and Taylor showedthat ch( K ∗ k ) = k , and suggested the problem of determining the choice number of K s ∗ k . The first author established ch( K ∗ k ) = (cid:6) k − (cid:7) . Here we prove ch( K ∗ k ) = (cid:6) k − (cid:7) . Introduction
Let G = ( V, E ) be a graph. A list assignment L for G is a function L : V → N , where N is the set of natural numbers and N is the power set of N . If | L ( v ) | = k for all vertices v ∈ V , then L is a k -list assignment for G . A n L -coloring f from a list assignment L is a function f : V → N such that f ( v ) ∈ L ( v ) for all vertices v ∈ V and f ( x ) (cid:54) = f ( y ) whenever xy ∈ E . G is L -colorable if there exists an L -coloring of G ; it is k -choosable if itis L -choosable for all k -list assignments L . The list chromatic number or choi ce number of G , denoted ch( G ) , is the smallest integer k such that G is k -choosable. The generallist coloring problem may consider list assignments with uneven list sizes.The study of list coloring was initiated by Vizing [13] and by Erdős, Rubin and Taylor[2]. It is a generalization of two well studied areas of combinatorics—graph coloring andtransversal theory. Restricting the list assignment to a constant function, yields ordinarygraph coloring; restricting the graph to a clique yields the problem of finding a systemof distinct representatives (SDR) for the family of lists. Both restrictions play a role inthis paper. Given the general nature of this parameter, it is hardly surprising that thereare not many graphs whose exact choice number is known. However, there are someamazingly elegant results that add to the subject’s charm. For example, Thomassen [12]proved that planar graphs have choice number at most , Voight [14] proved that thisis tight, and Galvin [3] proved that line graphs of bipartite graphs have choice numberequal to their clique number.Erdős et al. [2] suggested determining the choice number of uniform complete multi-partite graphs. More generally, let K ∗ k , ∗ k ... denote the complete multipartite graphwith k i parts of size i , where zero terms in the subscript are deleted. Since K ∗ k is aclique and K s ∗ is an independent set, these cases are trivial. Alon [1] proved the generalbounds c k log s ≤ ch( K s ∗ k ) ≤ c k log s for some constants c , c > . This was tightenedby Gazit and Krivelevich [4]. Theorem 1 (Gazit and Krivelevich [4]) . ch( K s ∗ k ) = (1 + o (1)) log s log(1+1 /k ) . The next well-known example provides the best lower bounds for small values of s . Mathematics Subject Classification.
Key words and phrases.
List coloring, choice number, choosable, on-line choice number.The first author thanks Institut Mittag-Leffler (Djursholm, Sweden) for the hospitality and creativeenvironment.Research of the first author is supported in part by NSA grant H98230-12-1-0212. a r X i v : . [ m a t h . C O ] J u l xample 2. ch( K s ∗ k ) ≥ (cid:100) s − k − s +2 s (cid:101) : Let G = K s ∗ k have parts { X , . . . , X k } with X i = { v i, , . . . , v i,s } . We will construct an ( l − -list assignment L from which G cannotbe colored. Equitably partition C := [2 k − into s parts C , . . . , C s . Define a listassignment L for G by L ( v i,j ) = C (cid:114) C j . Then each list has size at least k − − (cid:24) k − s (cid:25) = (cid:22) ks − s − k + 1 s (cid:23) = (cid:24) s − k − s + 2 s (cid:25) = l − . Consider any color α ∈ C . Then α ∈ C i for some i ∈ [ s ] . So α / ∈ L ( x i,j ) for every j ∈ [ k ] .Thus any L -coloring of G uses at least two colors for every part X j . Since vertices indistinct parts are adjacent, they require distinct colors. As there are k parts this wouldrequire k > | C | colors, which is impossible.Restricting the question of Erdös et al., we ask for those integers s such that:(1.1) ( ∀ k ∈ Z + ) (cid:20) ch( K s ∗ k ) = l ( s, k ) := (cid:24) s − k − s + 2 s (cid:25)(cid:21) . The first two cases s = 2 and s = 3 have been solved: Theorem 3 (Erdős, Rubin and Taylor [2]) . All positive integers k satisfy ch( K ∗ k ) = k . Theorem 4 (Kierstead [5]) . All positive integers k satisfy ch( K ∗ k ) = (cid:100) k − (cid:101) . Recently, Kozik, Micek, and Zhu [6] gave a very different proof of Theorem 4. Thefollowing more general result appears in [8].
Theorem 5 (Ohba [8]) . ch( K ∗ k , ∗ k ) = max { k, (cid:100) n + k − (cid:101)} , where k = k + k and n = k + 3 k . The next example shows that the largest s satisfying (1.1) is at most . Example 6. If k is even then ch( K ∗ k ) ≥ l := 2 k : Let G = K s ∗ k have parts { X , . . . , X k } with X i = { v i, , . . . , v i,s } . We will construct an ( l − -list assignment L from which G cannot be colored. Equitably partition C := [3 k − into parts C , . . . , C , and fix abijection f : [15] → (cid:0) [6]2 (cid:1) . Define a list assignment L for G by L ( v i,j ) = C (cid:114) (cid:91) { C h : h ∈ f ( i ) } . Then each list has size at least k − − (cid:24) k − (cid:25) = 2 k − l − . Consider any two colors α, β ∈ C . Then α, β ∈ (cid:83) { C h : h ∈ f ( i ) } for some i ∈ [15] . So α, β / ∈ L ( x i,j ) for every j ∈ [ k ] . Thus any L -coloring of G uses at least three colors forevery part X j . Since k > | C | , this is impossible.Yang [15] proved (cid:100) k (cid:101) ≤ ch( K ∗ k ) ≤ (cid:100) k (cid:101) , and Noel et al. [7] improved the upperbound to (cid:100) k − (cid:101) . The main result of this paper is that (1.1) holds for s = 4 . To provethis theorem we first extract a simple proof of Theorem 4 from [7], and then elaborateon it. Theorem 7. ch( K ∗ k ) = l (4 , k ) := (cid:100) k − (cid:101) . Some of the recent development of list coloring of complete multipartite graphs has beenmotivated by paintability, or on-line choosability. Introduced by Schauz [11], paintability is a coloring game played between two players Alice and Bob on a graph G = ( V, E ) anda function f : V → N . Let V i denote the vertex set at the start of round i ; so V = V . At ound i , Alice selects a nonempty set of vertices A i ⊆ V i , and Bob selects an independentset B i ⊆ A i . Then B i is deleted from the graph so that V i +1 = V i (cid:114) B i , and the roundsare continued until V n = ∅ . Alice’s goal is to present some vertex v more than f ( v ) times,while Bob’s goal is to choose every vertex before it has been presented f ( v ) + 1 times.We say that G is on-line f -choosable if player B has a strategy such that any vertex v ∈ V is in at most f ( v ) sets A i , and on-line k choosable if G is on-line f -choosable when f ( v ) = k for all v ∈ V . The on-line choice number , denoted ch OL ( G ) , is the least k suchthat G is on-line k -choosable.This game formulation hides the on-line nature of the problem. Another way of thinkingabout it is that Alice has secretly assigned lists of colors to all the vertices. At round i she reveals all vertices whose list contains color i , and Bob colors an independent set ofthem with color i . In this formulation it is clear that ch( G ) ≤ ch OL ( G ) .Surprisingly, Schauz [11] proved that many results on choice number, including Brooks’theorem, Thomassen’s theorem, and the Bondy-Boppana kernel lemma carry over to on-line choice number. It is unknown whether ch OL ( G ) − ch( G ) is bounded by a constant.Indeed, no graphs are known for which ch OL ( G ) − ch( G ) ≥ . It is known that ch( K , , ) = 3 < OL ( K , , ) . The explicit value of ch( K ∗ k ) provided by Theorem 7 may be useful for establishinglarger gaps. In Section 4 we show that ch( K ∗ ) < ch OL ( K ∗ ) .2. Set-up
Fix s, k ∈ Z + . Let G = ( V, E ) = K s ∗ k , and P be the partition of V into k independent s -sets. Let l = l ( k, s ) = (cid:100) ( s − k − s +2 s (cid:101) , and consider any l -list assignment L for G . Put C ∗ = (cid:83) x ∈ V L ( x ) . Let L ¬ α be the result of deleting α from every list of L .We may write x . . . x t for the subpart S = { x , . . . , x t } ⊆ X ∈ P ; when we use thisnotation we implicitly assume the x i are distinct. Also set S = X (cid:114) S . For a set ofverties S ⊆ V let L ( S ) = { L ( x ) : x ∈ S } , L ( S ) = (cid:84) L ( S ) , W ( S ) = (cid:83) L ( S ) , and l ( S ) = | L ( S ) | . The operation of replacing the vertices in S by a new vertex v S withthe same neighborhood as S is called merging. The new vertex v S is said to be merged ;vertices that are not merged are called original. When merging a set S we also create alist L ( v S ) = L ( S ) .For a color α ∈ C ∗ , let | X, α | = |{ x ∈ X : α ∈ L ( x ) }| be the number of times α appearsin the lists of vertices of X , N i ( X ) = { α ∈ C ∗ : | X, α | = i } be the set of colors that appearexactly i times in the lists of vertices in X , n i ( X ) = | N i ( X ) | , and N ( X ) = N ( X ) ∪ N ( X ) .Let σ i ( X ) = (cid:80) { l ( I ) : I ⊆ X ∧ | I | = i } and µ i ( X ) = max { l ( I ) : I ⊆ X ∧ | I | = i } .For a set S and element x we use the notation S + x = S ∪ { x } and S − x = S (cid:114) { x } .The following lemma was proved independently by Kierstead [5], and by Reed andSudakov [9], [10], and named by Rabern. Lemma 8 (Small Pot Lemma) . If ch( G ) > r then there exists a list assignment L suchthat G has no L -coloring, all lists have size r , and their union has size less than | V ( G ) | . If s does not satisfy (1.1) then there is a minimal counterexample k with ch( K s,k ) >l ( s, k ) . By the Small Pot Lemma, this is witnessed by a list assignment L with | (cid:83) { L ( x ) : x ∈ V ( G ) }| < | V | . We always assume L has this property. Lemma 9.
Every part X of G satisfies L ( X ) = ∅ .Proof. Otherwise there exists a list assignment L , a color α , and a part X such that α ∈ L ( X ) . Color each vertex in X with α , set G (cid:48) = G − X , and put L (cid:48) = L ¬ α . Then L (cid:48) witnesses that k − is a smaller counterexample, a contradiction. (cid:3) y Lemma 9, n s ( X ) = 0 for each part X ∈ P . So by the Small Pot Lemma, | W ( X ) | = (cid:80) s − i =1 n i ( X ) < sk . Also (cid:80) s − i =1 i n i ( X ) = sl is the number of occurrences of colors in thelists of vertices of X. Thus s − (cid:88) i =2 ( i − n i ( X ) ≥ sl − | W ( X ) | ≥ s ( l − k ) + 1 . (2.1)Now we warm-up by giving a short proof extracted from [7] of Theorem 4. Proof of Theorem 4.
Let s = 3 , l = l (3 , k ) , and assume G is a counterexample with k minimal. Then k > . By Lemma 9, n ( X ) = 0 for all X ∈ P . We obtain a contradictionby L -coloring G . First we use the following steps to partition V into sets of vertices thatwill receive the same color. Then we merge each set I into a single vertex v I , and assign v I the set of colors in L ( I ) . Finally we apply Hall’s Theorem to chose a system of distinctrepresentatives (SDR) for these new lists; this induces an L -coloring of G . Step 1.
Partition P into a set R of l − k reserved parts together with a set U = P (cid:114) R of k − l unreserved parts. Step 2.
Choose U ⊆ U maximum subject to |U | ≤ µ ( X ) for all X ∈ U , and subjectto this, ν = (cid:80) X ∈U µ ( X ) is maximum. Set u = |U | . For each X ∈ U choose a pair I X ⊆ X with l ( I X ) ≥ u maximum. Put U = U (cid:114) U and u = |U | . So(2.2) if u < k − l then µ ( X ) ≤ u for all X ∈ U , since otherwise we could increase ν by adding X to U , and deleting one part Y ∈ U with µ ( Y ) = u , if such a part Y exists. Step 3.
Using (2.1), each part X ∈ P satisfies n ( X ) ≥ l − k ) + 1 ≥ (cid:24) k − (cid:25) + 1 ≥ k − k. Form an SDR f for { L ( v I X ) : X ∈ U } ∪ { N ( X ) : X ∈ R} by greedily choosing represen-tatives for the first family and then for the second family. For each X ∈ R choose a pair I X ⊆ X so that f ( x ) ∈ L ( I X ) . Step 4.
For each X ∈ U ∪ R , merge I X to a new vertex v I X , let z X ∈ X (cid:114) I X , and set X (cid:48) = { v I X , z X } . If X ∈ U , set X (cid:48) = X . This yields a graph G (cid:48) with parts P (cid:48) = { X (cid:48) : X ∈ P} , and list assignment L .Next we use Hall’s Theorem to prove that { L ( x ) : x ∈ V ( G (cid:48) ) } has an SDR. For this itsuffices to prove:(2.3) | S | ≤ (cid:12)(cid:12)(cid:12)(cid:91) { L ( x ) : x ∈ S } (cid:12)(cid:12)(cid:12) for every S ⊆ V ( G (cid:48) ) . To prove (2.3), let S ⊆ V ( G (cid:48) ) be arbitrary, and set W = W ( S ) := (cid:83) { L ( x ) : x ∈ S } . Weconsider several cases in order, always assuming all previous cases fail. Case 1:
There exists X ∈ P with | S ∩ X (cid:48) | = 3 . Then | S | ≤ k + u , X (cid:48) = X ∈ U and u ≥ . Thus u ≤ k − l − u < k − l , and so by (2.2), u ≥ µ ( X ) ≥ σ ( X ) / . Usinginclusion-exclusion, and Lemma 9, | W | ≥ | W ( X ) | ≥ σ ( X ) − σ ( X ) + σ ( X ) ≥ l − u = 3 l − k − l − u ) ≥ l − k ) + 3 u ≥ (2 k −
2) + (2 + u ) ≥ k + u ≥ | S | . Case 2:
There is X ∈ U with | S ∩ X (cid:48) | = 2 . Then X = X (cid:48) and | S | ≤ k. Since u = 2 k − l − u < k − l , (2.2) yields | W | ≥ | W ( S ∩ X ) | ≥ l − l ( S ∩ X ) ≥ l − u ≥ l − (2 k − l − u ) ≥ l + 1 − k = 2 k ≥ | S | . U U U R R R u u u u r r r k − l l − k Figure 3.1.
The partition P (cid:48) of K ∗ k . Case 3:
There is X ∈ U with | S ∩ X (cid:48) | = 2 . As | S | ≤ k − u = l + u and L ( v I X z X ) = L ( X ) = ∅ , | W | ≥ | W ( S ∩ X (cid:48) ) | ≥ l ( v I X ) + l ( z X ) − l ( v I X z X ) ≥ u + l ≥ | S | . Case 4: S has an original vertex. Then | S | ≤ l ≤ | W | . Case 5:
All vertices of S have been merged. Then | S | ≤ | f ( S ) | ≤ | W | . (cid:3) The main theorem
In this section we prove our main result, Theorem 7. The case when k is odd is con-siderably more technical. Casual or first time readers may wish to avoid these additionaldetails; the proof is organized so that this is possible. In particular, in the even case Step11 and Lemmas 13 and 14 are not involved. We often use the partition k = (2 k − l )+( l − k ) of the integer k , and note that k − l = l − k + b , where b = k mod 2 . Proof of Theorem 7.
Our set-up is the same as in the proof of Theorem 4. Let s = 4 , l = l (4 , k ) , and G = K ∗ k . The theorem is trivial if k = 1 . Let k > be a minimalcounterexample, and let L be an l -list assignment for G with | W ( V ) | ≤ k − and L ( X ) = ∅ for all parts X ∈ P . Again we partition V into sets of vertices that will receivethe same color, and then find an SDR for the induced list assignment that in turn inducesan L -coloring of G . See Figure 3.1. Step 1.
Partition V as P = U ∪ R , where |R| = l − k , |U | = 2 k − l , R = R ∪ R ∪ R and U = U ∪ U ∪ U ∪ U as follows. Step 2.
Choose U ⊆ P maximum subject to |U | ≤ k − l and for every X ∈ U thereis a pair I X ⊆ X with l ( I X ) , l ( I X ) ≥ k . Put U ⊆ U , and let u := |U | . Then:(3.1) If u < k − l then ( ∀ X ∈ P (cid:114) U )( ∀ I ⊆ X )[ | I | = 2 → min { l ( I ) , l ( I ) } ≤ k − . Step 3.
Choose U ⊆ P (cid:114) U maximum subject to |U | ≤ k − l − u and |U | ≤ µ ( X ) for all X ∈ U ; subject to this let ν = (cid:80) X ∈U µ ( X ) be maximum. Put U ⊆ U , and let u = |U | . If U (cid:54) = ∅ then let ˙ Z ∈ U ; else ˙ Z = ∅ . For each X ∈ U choose a triple I X ⊆ X with l ( I X ) ≥ u maximum. Since ν cannot be increased:(3.2) If u + u < k − l then ( ∀ X ∈ U ∪ U ∪ R )[ µ ( X ) ≤ u ] . tep 4. Choose R ⊆ P (cid:114) ( U ∪ U ) maximum subject to |R | ≤ l − k and for all X ∈ R there exists I X ⊆ X with | I X | = 3 such that there is an SDR f of L ( M ) , where M := { v I X : X ∈ U ∪ R } ; let C = ran( f ) . Put R ⊆ R , and let r := |R | . Then:(3.3) If r < l − k then ( ∀ X ∈ U ∪ U ∪ R ∪ R )[ N ( X ) ⊆ C ] . Step 5.
Choose U ⊆ P (cid:114) ( U ∪ U ∪ R ) maximum subject to |U | ≤ k − l − u − u and l − k + u + |U | ≤ µ ( X ) for all X ∈ U ; subject to this let ν = (cid:80) X ∈U µ ( X ) bemaximum. Put U ⊆ U , and u = |U | . Since ν cannot be increased:(3.4) If u + u + u < k − l then ( ∀ X ∈ U ∪ R ∪ R )[ µ ( X ) ≤ l − k + u + u ] . For all X ∈ U choose a pair I X = xy ⊆ X with l ( I X ) ≥ l − k + u + u maximum;subject to this choose I X so that ∆ ( I X ) := l ( I X ) − l ( I X ) is maximum. Set ∆ ( I X ) :=2 u − l ( xyz ) − l ( xyw ) , where zw = I X . Using r ≤ l − k , extend f to an SDR f of L ( M ) , where M := M ∪ { v I X : X ∈ U } ; set C = ran( f ) . Step 6.
Choose R ⊆ P (cid:114) ( U ∪ U ∪ U ∪ R ) maximum subject to |R | ≤ l − k − r and σ ( X ) − σ ( X ) ≥ l − k ) + 2 u + 2 u + u + r + |R | for all X ∈ R ; subject tothis let (cid:80) X ∈R σ ( X ) − σ ( X ) be maximum. Put R ⊆ R , and set r = |R | . Then:If r + r < l − k then ( ∀ X ∈ U ∪ R )[ σ ( X ) − σ ( X ) ≤ l − k ) + 2 u + 2 u + u + r + r ] . (3.5) Step 7.
Choose R ⊆ P (cid:114) ( U ∪ U ∪ U ∪ R ∪ R ) with |R | = l − k − r − r , and set R = R ∪ R ∪ R . Let r = |R | . For I ⊆ X , put L (cid:48) ( I ) = L ( I ) (cid:114) C and l (cid:48) ( I ) = | L (cid:48) ( I ) | .Using Lemma 11, for all X ∈ R there exists a pair I X ⊆ X with l (cid:48) ( I X ) ≤ l (cid:48) ( I X ) suchthat f can be extended to an SDR f of L ( M ) , where M := M ∪ { v I X : X ∈ R } . Let C = ran( f ) . Step 8.
Put U = P (cid:114) R , U := U (cid:114) ( U ∪ U ∪ U ) , and u := |U | . Step 9.
Using Lemma 12, choose a pair I X ⊆ X for all X ∈ R so that L ( M ) has anSDR f extending f , where M := M ∪ { v I X , v I X : X ∈ U ∪ R } . Step 10.
Let G (cid:48) := ( V (cid:48) , E (cid:48) ) be the graph obtained from G by merging each I X with X ∈ U ∪ U ∪ U ∪ R ∪ R ∪ R and each I X with X ∈ U ∪ R . For a part X , let X (cid:48) be the corresponding part in G (cid:48) , and set P (cid:48) = { X (cid:48) : X ∈ P} . Step 11.
Set u = ˙ r = ¨ u . If k is odd ( b = 1 ) then we merge one more pair of verticesunder any of the following special circumstances:(a) there exists X ∈ U with | W ( X ) | < | G (cid:48) | . Fix such an X = ˙ X . By Lemma 13, r = 0 and there is a pair ˙ I ⊆ ˙ X such that (i) f can be extended to an SDR f of L ( M ) , where M := M + v ˙ I ; (ii) | W ( { v ˙ I , v } ) | ≥ k − , and if equality holds then | W ( { v ˙ I , v } ∪ ˙ Z (cid:48) ) ∪ C | ≥ k for both v ∈ ˙ I ; and (iii) W ( ˙ I + v ˙ I ) ≥ | G (cid:48) | − . Merge ˙ I and set ˙ u = 1 .(b) u = r = 0 and there is Y ∈ R with | W ( Y ) | ≤ k − − u − r . Then (a) failssince r ≥ . Fix such a Y = ˙ Y . As u = 0 = r , M = M . Since r (cid:54) = 0 , (a) isnot executed. By Lemma 11, f can be chosen so that it is an SDR of L ( M ) , where M := M + v I ˙ Y . Merge I ˙ Y and set ˙ r = 1 .(c) condition (a) fails and there exist X ∈ U and xyz ⊆ X with | W ( xyz ∪ ˙ Z (cid:48) ) | ≤ k + u − < | W ( X ) | . Fix such an X = xyzw = ¨ X . By Lemma 14 there is a pair ¨ I ⊆ xyz such that (i) f can be extended to an SDR f of L ( M ) , where M := M + v ¨ I ; (ii) | W ( { v ¨ I , v } ) | ≥ k or v ∈ xyz (cid:114) ¨ I and | W ( { v ¨ I , w } ) | ≥ k − ; and (iii) | W ( ¨ I + v ¨ I ) | ≥ k + u . Merge I ¨ X := ¨ I and set ¨ u = 1 . Step 12.
Recall that G (cid:48) is the graph obtained after the first ten steps. Let H be thefinal graph obtained by this merging procedure. (If b = 0 , and possibly otherwise, H = G (cid:48) ). Also let M be the final set of merged vertices, f be the final SDR of L ( M ) ,and C = ran( f ) .Our next task is to state and prove the four lemmas on which the algorithm is based.We will need the following easy claim. Claim 10.
Let P , P , P be the three partitions of a -set X into pairs. For all I ∈P , I ∈ P , I ∈ P there exists v ∈ X such that either (i) v ∈ I ∩ I ∩ I or (ii) v / ∈ I ∪ I ∪ I . Lemma 11.
There is a family I = { I X : X ∈ R } such that I X ⊆ X , | I X | = 2 , l (cid:48) ( I X ) ≥ l (cid:48) ( I X ) , and L ( M ∪ { v I X : X ∈ R } ) has an SDR f extending f .Furthermore, if u = 0 = r and there is ˙ Y ∈ R with | W ( ˙ Y ) | ≤ k − − u − r , then I ˙ Y can be chosen so that there is an SDR f of L ( M ) extending f , where M = M + v I ˙ Y .Proof. Consider any X ∈ R , and let A ( X ) = N ( X ) (cid:114) C be the set of colors availablefor coloring a pair of vertices from X . Then L (cid:48) ( I ) = L ( I ) ∩ A ( X ) for all pairs I ⊆ X .For each color α ∈ A , set I ( α ) = { x ∈ X : α ∈ L ( x ) } . As A ( X ) ⊆ N ( X ) , | I ( α ) | = 2 .Let B ( X ) = { α ∈ A ( X ) : l (cid:48) ( I ( α )) ≥ l (cid:48) ( I ( α )) } . For the first part, it suffices to show that B = { B ( Z ) : Z ∈ R } has an SDR g : for each X ∈ R set I X = I ( α ) , and f ( v I X ) = α ,where α = g ( B ( X )) .By (3.3), N ( X ) ⊆ C ⊆ C ; so n ( X ) ≤ u + r . By (2.1)(3.6) n ( X ) + 2 n ( X ) ≥ l − | W ( X ) | ≥ l − k ) + 1 ≥ k − . Thus | A ( X ) | = n ( X ) + n ( X ) − | C | ≥ n ( X ) + 2 n ( X ) − n ( X ) − | C | (3.7) ≥ k − − (2 u + u + 2 r ) ≥ r − . If α ∈ A ( X ) (cid:114) B ( X ) then A ( X ) ∩ L ( I ( α )) ⊆ B ( X ) . So | B ( X ) | ≥ (cid:100)| A ( X ) | / (cid:101) ≥ r . Hence B has an SDR g .Now suppose ˙ Y is defined in Step 11(b). Then b = 1 , u = r = 0 , and | W ( ˙ Y ) | ≤ k − − u − r . As b = 1 , k is odd; so k ≥ . If r ≥ then fix Z ∈ R − ˙ Y . A partition Q = { I , I } of ˙ Y into pairs is bad if l (cid:48) ( I ) = 0 or l (cid:48) ( I ) = 0 ; else it is good. It is weak if r ≥ , L (cid:48) ( I ) ∪ L (cid:48) ( I ) ⊆ B ( Z ) and | B ( Z ) | = r ; else it is strong.For the second part, it suffices to show that ˙ Y has a good, strong partition: If { ˙ I, ˙ I } is a good, strong partition then choose α, β ∈ L (cid:48) ( I ) ∪ L (cid:48) ( I ) with | B ( Z ) − α | ≥ r and α ∈ L (cid:48) ( I ) iff β ∈ L (cid:48) ( I ) . Then α and β are the representatives for L (cid:48) ( I ) and L (cid:48) ( I ) , or vice versa . We are done if r = 1 . If r ≥ then continue by greedily choosing anSDR of B − B ( ˙ Y ) − B ( Z ) + ( B ( Z ) − α ) + L (cid:48) ( I ) + L (cid:48) ( I ) by picking representatives for B − B ( ˙ Y ) − B ( Z ) , and finally picking a representative for B ( Z ) − α .Using the first half of (3.6), n ( ˙ Y ) + 2 n ( ˙ Y ) ≥ l − | W ( ˙ Y ) | ≥ l + u + r . So by (3.7), | A ( ˙ Y ) | ≥ l + u + r − (2 u + u + 2 r ) ≥ l − u − u − r ≥ k + r − . irst suppose for a contradiction that ˙ Y has no good partition. For each partition P of X into pairs, choose I ∈ P with L ( I ) ∩ A ( ˙ Y ) = ∅ . Using Claim 10, there exists w ∈ ˙ Y such that either (i) L ( wx ) ∩ A ( ˙ Y ) = ∅ for all x ∈ ˙ Y (cid:114) w or (ii) L ( xy ) ∩ A ( ˙ Y ) = ∅ for all xy ⊆ ˙ Y (cid:114) w . If (i) holds then L ( w ) ∩ A ( ˙ Y ) = ∅ . This yields the contradiction l + 2 k + r − ≤ l ( w ) + | A ( ˙ Y ) | ≤ | W ( ˙ Y ) | ≤ k − − u − r < l + 2 k − . If (ii) holds then A ( ˙ Y ) ⊆ L ( w ) , and so l < | A ( ˙ Y ) | ≤ l ( w ) , another contradiction.So ˙ Y has a good partition (say) Q = { xy, zw } . Suppose Q is weak. Then r ≥ and | A | ≥ k − , where A := A ( ˙ Y ) (cid:114) B ( Z ) ⊆ A ( ˙ Y ) (cid:114) ( L (cid:48) ( xy ) ∪ L (cid:48) ( zw )) . The former implies ≤ r ≤ l − k ≤ k − l ; so (*) l ≤ k − . If the other two partitions of ˙ Y are both bad thenthere is v ∈ ˙ Y with A ⊆ L ( v ) . So k − ≤ | A | ≤ l contradicting (*). Say Q = { xw, yz } is good. If Q is weak then A ⊆ A ( ˙ Y ) (cid:114) ( L (cid:48) ( xy ) ∪ L (cid:48) ( zw ) ∪ L (cid:48) ( xw ) ∪ L (cid:48) ( yz )) . Then | L (cid:48) ( xz ) ∪ L (cid:48) ( yw ) | ≥ k − . So Q = { xz, yw } is strong. By (*), l (cid:48) ( xz ) , l (cid:48) ( yw ) ≤ l < k − .Thus l (cid:48) ( xz ) , l (cid:48) ( yw ) ≥ , and so Q is also good. (cid:3) Lemma 12.
For each X ∈ R there is a pair I X ⊆ X such that { L ( I X ) : X ∈ P (cid:114) U } ∪{ L ( I X ) : X ∈ U ∪ R } has an SDR f that extends f .Proof. Each X ∈ U satisfies L ( I X ) , L ( I X ) ≥ k . Thus | L ( I X ) (cid:114) C | , | L ( I X ) (cid:114) C | ≥ k − u − u − r − r ≥ u . By Theorem 3, { L ( I X ) (cid:114) C , L ( I X ) (cid:114) C : X ∈ U } has an SDR,and so f can be extended to an SDR g for L ( M (cid:48) ) , where M (cid:48) := M ∪ { I X , I X : X ∈ U } .Let C g = ran( g ) . Then | C g | = 2 u + u + u + r + r . Consider any X = xyzw ∈ R .Let A ( X ) = N ( X ) (cid:114) C g . Again by Theorem 3 it suffices to show:(3.8) ( ∃ I X ⊆ X )[ | I X | = 2 ∧ | L ( I X ) ∩ A ( X ) | ≥ r ∧ | L ( I X ) ∩ A ( X ) | ≥ r ] . Observe σ ( X ) = n ( X ) + 3 n ( X ) and σ ( X ) = n ( X ) . So n ( X ) = n ( X ) + n ( X ) = σ ( X ) − σ ( X ) . By (3.3), N ( X ) ⊆ C g , and by (3.4) σ ( X ) ≤ u + r . So n ( X ) = σ ( X ) − σ ( X ) ≥ l − k ) + 2 u + 2 u + u + r + r − ( u + r ) ≥ l − k ) + 2 u + u + u + r and(3.9) | A ( X ) | = | N ( X ) (cid:114) C g | = | N ( X ) ∪ N ( X ) (cid:114) C g | = n ( X ) − | C g |≥ l − k ) + 2 u + u + u + r − (2 u + u + u + r + r ) ≥ l − k ) − r + r − r ≥ l − k ) + 2 r . (3.10)Suppose (3.8) fails. Then for each of the three partitions of X into pairs, there is apair uv with | L ( uv ) ∩ A ( X ) | ≤ r − . Using Claim 10, there exists v ∈ X such thateither (i) | L ( vw ) ∩ A ( X ) | ≤ r − for all w ∈ X − v or (ii) | L ( vw ) ∩ A ( X ) | ≤ r − forall w ∈ X − v .If (i) holds then | L ( v ) ∩ N ( X ) | ≤ | C g | + (cid:88) w ∈ X − v | L ( vw ) ∩ A ( X ) | ≤ | C g | + 3 r − . Since | L ( w ) ∩ N ( X ) | ≤ l for all w ∈ X − v , n ( X ) ≤ (cid:88) v ∈ X | L ( v ) ∩ N ( X ) | ≤ l + ( | C g | + 3 r − . Using | C g | = 2 u + u + u + r + r and (3.9) implies l − k ) + 4 u + 2 u + 2 u + 2 r ≤ l − u + u + u + r + r + 3 r − (3.11) l − k + (6 l − k + 3) + 2 u + u + u ≤ l + r + r + r ≤ l − k. ince l − k = − b , both b = 1 and u = u = u . Now, by (3.4), µ ( X ) ≤ l − k .So | L ( w ) ∩ N ( X ) | ≤ l − k ) for all w ∈ X . Strengthening the estimate in (3.11) yieldsthe contradiction: l − k ) + 2 r ≤ l − k ) + ( | C g | + 3 r − l − k ≤ r + r + r − < l − k. Thus (ii) holds. So | A ( X ) | ≤ l ( v ) + (cid:88) wx ⊆ X − v | L ( uv ) ∩ A ( X ) | ≤ l + 3( r − . (3.12)Using (3.10), (3.12) and l − k = − b , this yields the contradiction l − k ) + 2 r ≤ | A ( X ) | ≤ l + 3( r − l − k + 2 ≤ l − k + 3 ≤ r ≤ l − k. (cid:3) Lemma 13.
Suppose X = xyzw ∈ U and | W ( X ) | < | G (cid:48) | . Then b = 1 , u = 0 = r , u + u ≥ , and there exists a pair J ⊆ X such that:(1) L ( J ) (cid:42) C ; (2) | W ( { v J , v } ) | ≥ k − and if | W ( { v J , v } ) | = 2 k − then | W ( { v J , v }∪ ˙ Z ) ∪ C | ≥ k for both v ∈ J ;(3) | W ( J + v J ) | ≥ | G (cid:48) | − ; in particular | W ( X ) | ≥ | G (cid:48) | − .Proof. Now | G (cid:48) | = 3 k − u − u + u − r − r . Observe that(3.13) σ ( X ) − σ ( X ) ≥ l − k ) + 2 u + 2 u + u + r + r + 1 , since otherwise inclusion-exclusion yields the contradiction: | W ( X ) | = σ ( X ) − σ ( X ) + σ ( X ) ≥ l − l − k ) − u − u − u − r − r ≥ k + (2 k − l − u − u − u ) − u − u − r − r ≥ k − u − u + u − r − r = | G (cid:48) | > | W ( X ) | . By (3.13) and (3.5), r + r = l − k and r = 0 . Consider any pair I = xy ⊆ X . Then | W ( I + v I ) | ≥ l ( xy ) + l ( z ) + l ( w ) − l ( xyz ) − l ( xyw ) − l ( zw ) (3.14) ≥ l − u + ∆ ( I ) + ∆ ( I ) | G (cid:48) | − | W ( I + v I ) | ≤ b − u + ( u + u + u − l + k ) − ∆ ( I ) − ∆ ( I ) (3.15) ≤ b − u − u − ∆ ( I ) − ∆ ( I ) . (3.16)By (3.16), ∆ ( I ) + ∆ ( I ) ≤ . As ∆ ( I ) = − ∆ ( I ) and ∆ ( I ) , ∆ ( I ) ≥ , we couldchoose I with ∆ ( I ) + ∆ ( I ) ≥ . So b = 1 , u = 0 , u ≤ , and(3.17) ≤ | G | − | W ( I + v I ) | ≤ − u − ∆ ( I ) − ∆ ( I ) ≤ . Furthermore, using ∆ ( I ) = − ∆ ( I ) again,(3.18) ≤ u − σ ( X ) = ∆ ( I ) + ∆ ( I ) = ∆ ( I ) + ∆ ( I ) + ∆ ( I ) + ∆ ( I ) ≤ . By (3.13), r + r = l − k , σ ( X ) ≤ µ ( X ) , (3.4), and σ = 4 u − ∆ ( I ) − ∆ ( I ) , l − k ) + 2 u + u + σ ( X ) ≤ σ ( X ) ≤ l − k + u + u ) (3.19) u + 6( l − k + u ) − ∆ ( I ) − ∆ ( I ) ≤ σ ( X ) ≤ l − k + u + u ) . (3.20) y (3.19) u + u ≥ . So the first three assertions of the lemma have been proved. Itremains to find a pair J ⊆ X satisfying (1–3).First suppose u = 1 . By (3.17), ∆ ( I ) + ∆ ( I ) = 0 for all pairs I ⊆ X . So ∆ ( I ) ≤ and ∆ ( I ) ≤ . As ∆ ( I ) = − ∆ ( I ) , this implies ∆ ( I ) = 0 = ∆ ( I ) . So ∆ ( I ) = 0 =∆ ( I ) . By (3.20), there exists a pair I ⊆ X with l ( I ) = l − k + u + u . As ∆ ( I ) = 0 , l ( I ) = l − k + u + u . Thus | W ( { v I , v I } ) | = l ( I ) + l ( I ) = 2( l − k + u + u ) > l − k ) + u + u ≥ | C | . Pick J ∈ { I, I } such that L ( J ) (cid:42) C . Then (1) holds. For (2), let v (cid:48) ∈ J , and observe | W ( { v J , v (cid:48) } ) | = l ( J ) + l ( v (cid:48) ) − l ( J + v (cid:48) ) ≥ l − k + u + u − u = 2 k. Thus (2) holds. As u = 1 , (3.17) implies (3).Otherwise u = 0 . Then u ≥ , and so ˙ Z is defined in Step 3. Put C := C ∪ W ( ˙ Z (cid:48) ) .By Step 3, | C | ≥ | W ( ˙ Z (cid:48) ) | ≥ l + u . Call a vertex x ∈ X bad if | L ( x ) ∪ C | ≤ k − ;otherwise x is good . If x is bad then | C (cid:114) L ( x ) | ≤ k − − l ≤ l − k . If another vertex y is also bad, then using (3.4) and (3.17), l − k + u ≥ l ( xy ) ≥ | L ( xy ) ∩ C | ≥ | C | − | C (cid:114) L ( x ) | − | C (cid:114) L ( y ) |≥ l + u − l − k ) ≥ l − k + u + 1 , a contradiction. So at most one vertex of X is bad.Call a pair I ⊆ X bad if L ( I ) ⊆ C ; otherwise I is good . Note that if I is good then I satisfies (1). By (3.18), (3.20), and u = 0 , l − k + u ) − ≤ σ ≤ l − k + u ) ; and soby (3.4), every pair I ⊆ X satisfies l − k + u − ≤ l ( I ) ≤ l − k + u . If the upper bound is sharp then call I normal ; otherwise call I abnormal . Then there isat most one abnormal pair. If I is normal then l ( I ) ≤ l ( I ) ; so ∆ ( I ) ≥ .By (3.2), every triple T ⊆ X satisfies l ( T ) ≤ u . If equality holds then call T normal ;otherwise call T abnormal ; if | L ( T ) ∩ C | ≤ u − then call T very abnormal . Supposetwo pairs I, J ⊆ T are both bad. At least one, say I , is normal. Then l − k ) + u ≥ | C | ≥ | L ( I ) ∪ L ( J ) | ≥ l − k + u + l ( J ) − l ( I ∪ J ) (3.21) l ( I ∪ J ) ≥ l ( J ) − l + k = (cid:40) u if J is normal u − if J is abnormal . So an abnormal triple contains at most one bad, normal pair, and a very abnormaltriple contains at most one bad pair. A pair I contained in an abnormal triple satisfies ∆ ( I ) ≥ .Let J be a good, normal pair contained in a abnormal triple T with w ∈ X (cid:114) T . Then ∆ ( J ) + ∆ ( J ) ≥ . So J satisfies (3) by (3.17). Also, | W ( v J , v ) | = l ( J ) + l ( v ) − l ( J + v ) ≥ (cid:40) l − k + u − ( u −
1) = 2 k if v ∈ T (cid:114) J l − k + u − u = 2 k − if v = w . So (*) J satisfies (2), provided | W ( v j , w ) ∪ C | ≥ k . In particular, (2) holds if w is good.By (3.20) and (3.18), ≤ ∆ ( I ) + ∆ ( I ) ≤ . As σ = 4 u − ∆ ( I ) − ∆ ( I ) , we have u − ≤ σ ( X ) = 4 u − . In the first case there is one abnormal triple. In the secondcase, either there is a very abnormal triple or there are two abnormal triples.First suppose there are two abnormal triples. Choose an abnormal triple T so that ifthere is a bad vertex then it is in T . As T contains three pairs and at most one is bad nd at most one is abnormal, T contains a good, normal pair J . Say J = yz , T = xyz ,and w ∈ X (cid:114) T . Then w is good, and thus J satisfies (2) by (*).Otherwise, let T = xyz be the only abnormal triple and w ∈ X (cid:114) T . There is at mostone abnormal pair, and only if T is very abnormal. So T contains at most one bad pair.Now suppose T has two good, normal pairs xy and yz . By (*), some J ∈ P := { xy, yz } satisfies (2), unless C ⊆ L ( J ) ∪ L ( w ) for both J ∈ P . Then, using u = u = r = 0 , l + u = | C | ≤ | L ( xy ) ∪ L ( w ) | + | L ( yz ) ∪ L ( w ) | − | L ( xy ) ∪ L ( yz ) ∪ L ( w ) | . As T is abnormal, and both xy and yz are normal, | L ( xy ) ∪ L ( yz ) ∪ L ( w ) | = l ( xy ) + l ( yz ) + l ( w ) − l ( xyw ) − l ( yzw ) − l ( xyz ) ≥ l − k + 2 u − (3 u −
1) = k + l − u . Combining the last two expressions yields the contradiction, l + u ≤ | C | ≤ k − − ( k + l − u ) = 3 k − − l + u − l + u − . Otherwise, T is very abnormal, and (say) both xz is bad and J = yz is normal. As T contains at most one bad pair, yz is also good. Since xz is bad, xz ⊆ C . Now | C (cid:114) W ( { v J , w } ) | ≥ | L ( xz ) (cid:114) ( L ( w ) ∪ L ( J )) | ≥ l − k + u − ( u − ≥ , and (2) holds by (*), since ∅ (cid:54) = L ( xz ) (cid:114) ( L ( w ) ∪ L ( J )) ⊆ C implies | W ( { v J , w } ∪ C ) | ≥ | W ( { v J , v } ) | + 1 ≥ k. (cid:3) Lemma 14.
Suppose b = 1 and X = xyzw ∈ U . If | W ( xyz ) | ≤ k + u − < | W ( X ) | then u = 0 and there exists a pair J ⊆ X such that:(1) L ( J ) (cid:42) C ; (2) | W ( { v J , v } ) | ≥ k for v ∈ xyz (cid:114) J and | W ( { v J , w } ) | ≥ k − u ; and(3) | W ( J + v J ) | ≥ k + u .Proof. Consider a pair vv (cid:48) ⊆ xyz . Then k + u − ≥ | W ( xyz ) | ≥ | W ( vv (cid:48) ) | ≥ l ( v ) + l ( v (cid:48) ) − l ( vv (cid:48) ) ≥ l − ( l − k + u + u ) ≥ k − − k + u + u ≥ k + u + u − ≥ k. So u = 0 , l ( vv (cid:48) ) = l − k + u + u , and W ( xyz ) = W ( vv (cid:48) ) . Since vv (cid:48) is arbitrary, everycolor in W ( xyz ) appears in at least two of the lists L ( x ) , L ( y ) , L ( z ) . So W ( { v J , v } ) = W ( xyz ) and | W ( { v J , v } ) | ≥ k for every pair J ⊆ xyz and vertex v ∈ xyz (cid:114) J . As | C | < k ≤ | W ( xyz ) | , there is a pair J ⊆ xyz with L ( J ) (cid:42) C . Furthermore, | W ( { v J , w } ) | ≥ l ( J ) + l ( w ) − l ( J + w ) ≥ l − k + u + u + l − u = 2 k − u . Finally, as W ( { v J , v } ) = W ( xyz ) for v ∈ xyz (cid:114) J , | W ( J + v J ) | = | W ( { v J , v } ) ∪ W ( w ) | = | W ( xyzw ) | ≥ k + u . (cid:3) Lemma 15. G (cid:48) is L -choosable. roof. First observe that if k is even then b = ˙ u = ¨ u = ˙ r = 0 and H = G (cid:48) . In this casethe following argument is much simpler.Using Hall’s Theorem it suffices to show | S | ≤ | W | := (cid:12)(cid:12)(cid:83) x ∈ S L ( x ) (cid:12)(cid:12) for every S ⊆ V ( H ) .Suppose for a contradiction that | S | > | W | for some S ⊆ V ( H ) . We consider severalcases. Each case assumes the previous cases fail. Case 1:
There is X ∈ U with | S ∩ X | = 4 . Then | W | < | S | ≤ | G (cid:48) | . By Lemma 13, b = 1 and | G (cid:48) | − | W ( X ) | < | S | = | G (cid:48) | . So Step 11(a) is executed, and S = V ( G (cid:48) ) . Inparticular, ˙ X ⊆ S . Thus | S | ≤ | H | = | G (cid:48) | − ≤ | W ( J + v J ) | ≤ | W ( ˙ X ) | ≤ | W | . Case 2:
There exists Z = xyzw ∈ U with | S ∩ Z (cid:48) | = 3 . Now | S | ≤ k − u − u − r − r − ˙ r ,since Case 1 fails. Say I Z = xy . By Step 4, ∆ ( xy ) ≥ and l ( xyz ) + l ( xyw ) =2 u − ∆ ( xy ) . By Step 3, l ( xyz ) + l ( xyw ) ≤ u + r . So | W | ≥ | W ( Z (cid:48) ) | ≥ l ( xy ) + l ( z ) + l ( w ) − l ( xyz ) − l ( xyw ) − l ( zw ) (3.22) = 2 l + ∆ ( xy ) − u + ∆ ( xy ) = 3 k − b + ∆ ( xy ) − u + ∆ ( xy ) ≥ k − b + ∆ ( xy ) − u − r ≥ | S | − b. As | S | > | W | equality holds throughout. Thus b = 1 , u = r = ˙ r = ∆ ( xy ) = 0 , r ≤ u ,and (*) Y (cid:48) ⊆ S for all Y ∈ R . If u = 0 then k = l − k + u + u ≤ l ( xy ) = l ( xy ) + ∆ ( xy ) = l ( xy ) . By (3.1) this contradicts u = 0 . So u ≥ , u + u ≥ , and r + r ≤ u + 0 = 2 k − l − u − u ≤ l − k − . Thus r ≥ . Say Y ∈ R . By (*), Y (cid:48) ⊆ S ; by (3.22), | W ( Y (cid:48) ) | ≤ | W | = 3 k − − u − r .So, using b = 1 and u = 0 = r , Step 11(b) is executed, and ˙ r = 1 , a contradiction. Case 3:
There exists X = wxyz ∈ R with | S ∩ X (cid:48) | = 3 . Say I X = xy . Now | S | ≤ k − u − u − u − r − r − ˙ r . By Step 7, l (cid:48) ( xy ) ≥ l (cid:48) ( wz ) . By (3.3), N ( X ) ⊆ C ⊆ C .So l (cid:48) ( xyz ) = 0 = l (cid:48) ( xyw ) . Set t = | C ∩ W | . Then t ≤ u + u + r . So | W | = | W (cid:114) C | + | C ∩ W | ≥ l (cid:48) ( xy ) + l (cid:48) ( z ) + l (cid:48) ( w ) − l (cid:48) ( xyz ) − l (cid:48) ( xyw ) − l (cid:48) ( zw ) + t ≥ l (cid:48) ( xy ) + l ( z ) − t + l ( w ) − t − l (cid:48) ( zw ) + t ≥ k − b − ( u + u + r ) ≥ | S | − b. Thus b = 1 , r = u = ˙ r , and | W ( X ) | ≤ | W | ≤ k − − u − r . So Step 11(b) isexecuted, and ˙ r = 1 , a contradiction. Case 4:
There exists X ∈ U with | S ∩ X (cid:48) | = 3 . As the previous cases fail, | S | ≤ k + u .Let xy ⊆ S ∩ X (cid:48) (cid:114) M . By (3.4), | W | ≥ l ( x ) + l ( y ) − l ( xy ) ≥ k − b − ( l − k + u + u ) ≥ k + (2 k − l ) − ( u + u ) − b ≥ k + u + u − b ≥ | S | − b. So b = 1 , u = 0 , | W | = 2 k + u − , and | S | = 2 k + u . Thus S has exactly two verticesin every class of P (cid:48) (cid:114) U (cid:48) and exactly three vertices in every class of U (cid:48) . In particular, ˙ Z (cid:48) ⊆ S . If ˙ u = 1 , then ˙ X (cid:48) ⊆ S and | W ( ˙ X (cid:48) ) | ≥ | G (cid:48) | − ≥ k + u ≥ | S | by Lemma 13;else | W ( X ) | ≥ k + u . If ¨ u = 1 then ¨ X (cid:48) ⊆ S and | W | ≥ | W ( ¨ X (cid:48) ) ≥ | S | by Lemma 14;else X = X (cid:48) . As Step 11(c) is not executed, | W | ≥ | W (( S ∩ X ) ∪ ˙ Z ) | ≥ k + u ≥ | S | . ase 5: There exists X ∈ U with | S ∩ X (cid:48) | = 2 . Say S ∩ X (cid:48) = { v I , v I } . As the previouscases fail, | S | ≤ k . Now | W | ≥ L ( v I ) + L ( v I ) ≥ k ≥ | S | . Case 6:
There exists X ∈ U with | S ∩ X (cid:48) | = 2 . Say S ∩ X (cid:48) = vv (cid:48) . As the previous casesfail, | S | ≤ k − u . If v, v (cid:48) / ∈ M then I X = vv (cid:48) . By (3.1), l ( I X ) ≤ k − . So | W ( vv (cid:48) ) | ≥ l ( v ) + l ( v (cid:48) ) − l ( vv (cid:48) ) ≥ l − ( k − ≥ k ≥ | S | . Otherwise v = v xy , where I X = xy, and v (cid:48) = z / ∈ M . Then | W ( vv (cid:48) ) | ≥ l ( v xy ) + l ( z ) − l ( xy + z ) ≥ l − k + u + u + l − u ≥ k − b + u ≥ k ≥ | S | . Case 7:
There exists X ∈ U with | S ∩ X (cid:48) | = 2 . Say S ∩ X (cid:48) = vv (cid:48) . If possible, choose X so that S ∩ X (cid:48) ∩ M = ∅ . As the previous cases fail, | S | ≤ k − u − u . If v, v (cid:48) / ∈ M then | W ( vv (cid:48) ) | = l ( v ) + l ( v (cid:48) ) − l ( vv (cid:48) ) ≥ l − ( l − k + u + u ) ≥ k − b + u + u ≥ k ≥ | S | . (3.23)Else b = 1 , and (say) v ∈ M . By Step 11, v = v ˙ I or v = v ¨ I , and u = 0 .If v = v ˙ I then Step 11(a) was executed. So (i) r = 0 , (ii) | W ( vv (cid:48) ) | ≥ k − , and (iii)if | W ( vv (cid:48) ) | = 2 k − then u ≥ and | W ( vv (cid:48) ∪ ˙ Z (cid:48) ) ∪ C | ≥ k . Since k ≥ | S | > | W | ≥ | W ( vv (cid:48) ) | ≥ k − , | S | = 2 k . Thus S contains exactly two vertices of each part Y (cid:48) ∈ P (cid:48) . In particular, ˙ Z (cid:48) ⊆ S . The choice of X implies u = 0 and u = 1 ; thus u = l − k ≥ . Since u = 0 = r , M ⊆ S . So | W | ≥ | W ( vv (cid:48) ∪ ˙ Z (cid:48) ) ∪ C | ≥ k , a contradiction.Otherwise x = v ¨ I . Then Step 11(c) was executed. So there is a part ¨ X = xyzw ∈ U with ¨ I = xy such that | W ( xyz ∪ ˙ Z ) | ≤ k + u − < | W ( ¨ X ) | , | W ( { v xy , w } ) | ≥ k − u , and | W ( { v xy , z } ) | ≥ k . So we are done, unless v (cid:48) = w and k ≥ | S | > | W ( { v xy , w } ) | ≥ k − u . Thus u = 0 and | S | = 2 k . So S contains exactly two vertices of each class Y (cid:48) ∈ P (cid:48) . Inparticular, ˙ Z (cid:48) ⊆ S . As | W ( ¨ X ) | > | W ( xyz ∪ ˙ Z ) | , we have | L ( w ) (cid:114) W ( xyz ∪ ˙ Z (cid:48) ) | ≥ . So | W | ≥ | W ( { v xy , w } ∪ ˙ Z (cid:48) ) | ≥ W ( ˙ Z (cid:48) ) + 1 = l + u + 1 = 2 l − k + 1 = 2 k. Case 8:
There exists X = xyzw ∈ U with | S ∩ X (cid:48) | = 2 . Say S ∩ X (cid:48) = { v I , w } . As theprevious cases fail, | S | ≤ k − u − u − u = l + u . Since L ( xyz ) ∩ L ( w ) = ∅ , we have | W | ≥ | W ( X (cid:48) ) | ≥ l ( xyz ) + l ( w ) ≥ u + l ≥ | S | . Case 9:
Otherwise. As the previous cases fail, | S | ≤ u + u + u + u + 2 |R| = l. As L ( M ) has an SDR, there is a vertex x ∈ S (cid:114) M . Thus | W | ≥ l ( x ) = l ≥ | S | . (cid:3) We are done. (cid:3) Figure 4.1.
Strategy for Alice demonstrating ch OL ( K ∗ ) ≥ .4. On-line Choosability
By Theorem 7, ch( K ∗ ) = 4 . Using a computer we have checked that ch OL ( K ∗ ) = 5 ,but do not have a readable argument to verify the upper bound. Here we prove the lowerbound. Theorem 16. ch OL ( K ∗ ) ≥ .Proof. Figure 4.1 describes a strategy for Alice. The top left matrix depicts the initialgame position, and Alice’s first move. The positions in the matrix correspond to thevertices of K ∗ arranged so that vertices in the same part correspond to positions in thesame column. The order of vertices within a column is irrelevant, as is the order of thecolumns. The numbers represent the size of the list of each corresponding vertex. Thesequence of numbers represents a function f . The shaded positions represent the verticesthat Alice presents on here first move.As play progresses Bob chooses certain vertices presented by Alice and passes overothers. When a vertex is chosen its position is removed from the next matrix (and thepositions in its column of the remaining vertices and the order of the columns may berearranged). When he passes over a vertex its list size is decreased by one (and itsposition in its column and the order of the columns may change). The arrows betweenthe matrices point to the possible new game positions that arise from Bob’s choice, notcounting equivalent positions and omitting clearly inferior positions for Bob. In particularwe assume Bob always chooses a maximal independent set.For example, after Bob’s first move there is only one possible game position, providedBob chooses a maximal independent set. It is shown in the second column of the first row,along with Alice’s second move. Now Bob has two possible responses that are pointedto by two arrows. Also consider the matrix in the third row and third column. Thereare three nonequivalent responses for Bob, but choosing the offered vertex in the second olumn of the matrix results in a position that is inferior to choosing the offered vertexin the first column. So this option is not shown.Eventually, Alice forces one of five positions ( G, f ) such that G is not f -choosable, andBob, being a gentleman, resigns. (cid:3) References
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School of Mathematical and Statistical Sciences, Arizona State University, Tempe,AZ 85287, USA
E-mail address : [email protected] E-mail address : [email protected] School of Mathematical and Statistical Sciences, Arizona State University, Tempe,AZ 85287, USA
E-mail address : [email protected]@asu.edu