Perpetually Fair Assignments Via Balanced Sequences of Permutations

P erp etually F air Assignmen ts Via Balanced Sequences of P erm utations T errence Adams, Erel Segal-Halevi F ebruary 26, 2026 Abstract There is a set of n indivisible items (or c hores), and a set of n pla yers. Eac h da y , a single item should be assigned to eac h pla yer. W e w ant to ensure that all play ers feel that they ha ve b een treated fairly , not only after the last da y , but after every single day . W e presen t t w o ’balance’ conditions on sequences of p erm utations. One condition can alwa ys b e satisfied, but is arguably to o weak; a second condition is strong, and can b e satisfied for all n ≤ 11, but cannot be satisfied for some larger v alues of n , including all n > 61. W e then relate the ’balance’ condition to the requiremen t that the cumulativ e assignmen t is pr op ortional up to one item (PROP1) , where prop ortionalit y holds in a strong ordinal sense — for every v aluations that are consisten t with the item ranking. W e presen t a third balance condition that implies ordinal PROP1. W e sho w that a sequence guaranteeing this balance condition exists for all n ≤ 12, but might not exist when n = 6 k for any k ≥ 19. Finally , w e present a fourth, w eaker balance condition on a sequence, that guaran tees ordinal prop ortionalit y up to tw o items (PROP2). Whether or not this condition can b e satisfied for all n remains an op en question. 1 In tro duction Some n house-mates need to divide the n daily house-chores among them. Eac h house- mate m ust do exactly one c hore per da y . What w ould be a fair w a y to assign the c hores? Ideally , a fair allo cation would giv e each housemate the same amount of work. But this migh t b e impossible, as some chores are easier and some are harder; in every single da y , some housemates must do the hard c hores. Ho wev er, if the same housemates do the same unw an ted chores da y after da y , they will rightfully complain that this r ep e ate d assignment is unfair. A similar problem may arise in different con texts. F or example, supp ose grandpa w ants to divide sweets to his children each week end. He has to decide who gets the sw eets first, who is second, and so on. Each w eek, the children who are last to get their sw eets might feel somewhat unhappy; but if the same children are last in the line w eek after week, they will naturally feel that the allo cation is unfair. Similar considerations are applicable when assigning seats in classro om to pupils — a task which ma y b e rep eated in differen t terms. 1 1 2 3 4 5 6 2 3 4 5 6 1 3 4 5 6 1 2 4 5 6 1 2 3 5 6 1 2 3 4 6 1 2 3 4 5 1 8 5 4 3 6 7 2 2 7 6 3 4 5 8 1 3 6 7 2 1 8 5 4 4 5 8 1 2 7 6 3 5 4 1 8 7 2 3 6 6 3 2 7 8 1 4 5 7 2 3 6 5 4 1 8 8 1 4 5 6 3 2 7 T able 1: Examples of p erm utation sequences (rep eated assignmen ts) of 6 items (left) and 8 items (righ t). Eac h column represents an assignment in a single day; eac h row represen ts the bundle of a single play er. In all these settings, the same assignmen t problem rep eats man y times. Whereas some inequalit y can b e tolerated in every single assignmen t, w e exp ect it to even tually “cancel out” as the same pro cess is repeated for long enough. W e call this problem F air R ep e ate d Assignment . Belo w w e define the problem formally . 2 Setting There is a set N of n players , N := { A 1 , . . . , A n } . There is a set M of n indivisible items , M := [ n ] = { 1 , . . . , n } . W e assume that 1 is the b est item (e.g. the easiest c hore, or the first place in the line), 2 is the second-b est item, etc. An assignment is a bijection b et ween M and N ; equiv alen tly , a p erm utation of M . A r ep e ate d assignment is a finite sequence of assignmen ts. W e say that eac h assignmen t is implemen ted in a different day . The multiset of items that a particular play er i has after a rep eated assignment is called a bund le . The bundle of i after da y t is denoted Z t i . The multiset of all items allo cated up to day t is denoted Z t N := ⊎ i ∈ N Z t i . W e often presen t a rep eated assignment in a table, in whic h each column con tains the assignmen t to all agents in a single day , and eac h row contains the items giv en to a single agent in all days. See T able 1 for examples. F or any bundle Z and any k ≤ | Z | , w e denote by Z [ k ] the k -th highest v alued item in Z , so that Z [1] is the b est item in Z , Z [2] the second-b est, etc. The matching b et ween the play ers and their bundles is called an al lo c ation . 3 F airness After n Da ys: Latin Square W e make no assumptions on ho w muc h exactly item 1 is b etter than item 2, 2 is b etter than 3, etc. Hence, we cannot assume e.g. that receiving items 4 and 5 is as fair as receiving items 1 and 8. Hence, after n days, a necessary condition for fairness is that e ach player r e c eives e ach item exactly onc e . This condition is satisfied in both examples in T able 1 , where each ro w is a p erm utation of [ n ]. As eac h column is also a permutation of [ n ], the en tire rep eated assignment table is a Latin square. There are man y w a ys to construct Latin squares, so this requirement is easy to satisfy . 2 Clearly , in few er than n da ys, it is imp ossible to b e exactly fair for ev eryone. F or example, if item 1 is v ery v aluable (a diamond) and the other items are muc h less v aluable (stones), then at any da y b efore the n -th day , some pla yers (those who did not get the diamond so far) will naturally feel that they did not get their fair share. Still, n days may b e to o long to w ait for fairness. W e would lik e the allo cation to b e “as fair as p ossible” after each da y — not only after day n . In the follo wing sections, we explore different w ay to define what this fairness means. 4 F airness After Ev ery Da y: T op-Balance A natural fairness requiremen t is to require that the top items are allo cated among the pla yers in a “balanced” w a y: in the first t wo days, each play er should get at least one item from the top half (the highest-v alued ⌈ n/ 2 ⌉ items); in the first three da ys, each pla yer should get at least one item from the top third (the highest-v alued ⌈ n/ 3 ⌉ items); and so on. This condition is succinctly formalized b elo w. Definition 4.1. A p erm utation sequence is called top-b alanc e d if after every da y t ∈ [ n ], ev ery play er i has at least one item from the top ⌈ n/t ⌉ . F ormally , Z t i [1] ≤ ⌈ n/t ⌉ . The top-balance condition is violated b y the sequences of T able 1 already in day 2 (for n = 6) or in day 3 (for n = 8). Are there other sequences that satisfy it? The answ er is yes. Prop osition 4.2. F or every p ositive inte ger n , ther e exists a top-b alanc e d se quenc e. Pr o of. F or each n , w e sho w how to define the assignmen t of items to pla yers and giv e a pro of b y induction in the n umber of da ys t . Let n b e a p ositiv e integer. Assign the n items arbitrarily on day t = 1. W e wan t to establish the following tw o claims inductiv ely in t . 1. The set of play ers assigned at least one of the top  n t  items in t days contains all pla yers, and 2. the play ers assigned at least one of the top  n t  − 1 items in t days are all unique. Both claims clearly hold for t = 1. Suppose claims (1) and (2) hold for t where 1 ≤ t < n . W e sho w how to progress to the next day t + 1. Consider t wo cases. If  n t +1  <  n t  , then we know by induction that the play ers assigned one of the top  n t +1  items after t days will b e unique. With priorit y , choose the play ers not in this collection and assign the top items on da y t + 1 to these play ers. Since ( t + 1) ⌈ n t +1 ⌉ ≥ n and we choose unique pla y ers with priority , all n pla yers will b e assigned one of the top ⌈ n t +1 ⌉ items in t + 1 da ys. Also, since ( t + 1)( ⌈ n t +1 ⌉ − 1) < n , the pla yers assigned one of the top ⌈ n t +1 ⌉ − 1 items in t + 1 days will b e unique. If ⌈ n t +1 ⌉ = ⌈ n t ⌉ , then b egin b y choosing pla yers who w ere not assigned items from the top ⌈ n t ⌉ − 1 items, and assign the top items to these play ers in da y t + 1. Since ⌈ n t +1 ⌉ = ⌈ n t ⌉ , the induction assumption implies that all pla yers w ere assigned at least one of the items from the top ⌈ n t +1 ⌉ items in t + 1 da ys. Also, since ( t + 1)( ⌈ n t +1 ⌉ − 1) < n , all pla yers assigned one of the top ⌈ n t +1 ⌉ − 1 in the first t + 1 days are unique. That concludes the pro of by induction, which yields our top-balanced sequence. 3 App endix B giv es a concrete example of this assignmen t algorithm for n = 15. 5 Stronger F airness After Ev ery Da y: Balance The top-balance condition considers only the top items for each pla yer. The allo cations of other items can b e arbitrary , and thus unfair. The condition can b e strengthened as follo ws: • In the first t wo days, eac h pla y er should receive at least one item from the top half (the highest-v alued ⌈ n/ 2 ⌉ items) — this condition is the same as in top-balance. • In the first three days, eac h play er should receive at least one item from the top third (the highest-v alued ⌈ n/ 3 ⌉ items), and at le ast two items fr om the top two-thir ds (the highest-value d ⌈ 2 n/ 3 ⌉ items) — the latter condition is new. • In general, in the first t days, for ev ery j ∈ { 1 , . . . , t } , eac h play er should receiv e at least j items from the highest-v alued ⌈ j n/t ⌉ items. All these conditions are formalized succinctly b elo w. Definition 5.1. A p erm utation sequence is called b alanc e d if after ev ery day t ∈ [ n ], the bundle Z t i of every agent i satisfies the following condition for every j ∈ [ t ]: Z t i [ j ] ≤ ⌈ j n/t ⌉ . (1) W e denote by Bal ( t, j ) the condition ( 1 ) for some particular t ∈ [ n ] and j ∈ [ t ]. So a rep eated assignmen t is balanced if it satisfies Bal ( t, j ) for all t ∈ [ n ] and j ∈ [ t ] (where t denotes the n umber of days and j denotes the num b er of items). Remark 5.2. (a) Bal ( t = n, j ) implies that, after n da ys, the j -b est highest item of eac h pla yer is item j ; this implies that each pla yer has exactly one copy of each item, whic h is the Latin-square condition of Section 3 . (b) Bal ( t, j = 1) is exactly the top-balance condition of Section 4 . (c) Bal ( t = 1 , · ) and Bal ( t, j = t ) hold trivially , so in fact w e only hav e to chec k t ∈ { 2 , . . . , n } and j ∈ { 1 , . . . , t − 1 } . Do balanced sequences exist for all n ? It is easy to find balanced sequences for n ∈ { 3 , 4 , 5 , 6 } ; the readers may w ant to try it themselves b efore lo oking at T able 2 . T o v erify that a p erm utation sequence is balanced, one should lo ok, for eac h t ∈ [ n ], only at the leftmost t columns; then one should chec k that each row satisfies Bal ( t, j ) for all j ∈ [ t − 1]. F or example, for n = 6 and t = 4, w e look only at the 4 leftmost columns of the righ tmost table; w e v erify that each row contains one n umber from { 1 , 2 } , t w o n umbers from { 1 , 2 , 3 } and three n umbers from { 1 , 2 , 3 , 4 , 5 } . There are balanced sequences for natural n umbers n ≤ 11. W e w ere able to hand- craft sequences for n ≤ 8. F or larger v alues of n , w e dev elop ed a C++ library that searc hes for a balanced sequence using a backtrac king algorithm. T ables 3 and 4 show t wo balanced sequences generated by the backtrac king algorithm. Tw elve (12) is the first natural n um b er that do es not admit a balanced sequence. Belo w is a result that sho ws there are infinitely man y natural n umbers with no balanced sequence, including n = 12. 4 1 2 3 2 3 1 3 1 2 1 4 3 2 2 3 4 1 3 2 1 4 4 1 2 3 1 3 5 4 2 2 4 1 5 3 3 5 2 1 4 4 1 3 2 5 5 2 4 3 1 1 4 5 3 2 6 2 5 3 6 1 4 3 6 1 5 4 2 4 1 6 2 3 5 5 2 4 1 6 3 6 3 2 4 5 1 T able 2: Balanced p ermutation sequences for n ∈ { 3 , 4 , 5 , 6 } . 1 10 7 4 6 3 9 2 8 5 2 9 6 3 8 5 10 4 1 7 3 8 2 6 10 4 7 5 9 1 4 6 10 2 7 1 8 3 5 9 5 7 3 10 2 9 6 1 4 8 6 4 8 1 9 2 5 10 7 3 7 3 9 5 1 10 4 8 6 2 8 5 1 9 4 7 3 6 2 10 9 2 5 8 3 6 1 7 10 4 10 1 4 7 5 8 2 9 3 6 T able 3: Balanced sequence for n = 10. 1 11 8 6 5 4 10 3 9 7 2 2 10 7 5 4 11 8 1 6 9 3 3 9 6 11 2 8 5 7 1 4 10 4 8 11 3 7 2 9 6 5 10 1 5 7 4 2 11 10 3 9 8 1 6 6 5 3 10 9 1 7 11 4 2 8 7 4 10 1 8 5 11 2 3 6 9 8 3 9 4 1 6 2 10 7 5 11 9 1 5 7 10 3 6 8 2 11 4 10 6 2 9 3 7 1 4 11 8 5 11 2 1 8 6 9 4 5 10 3 7 T able 4: Balanced sequence for n = 11. Prop osition 5.3. Ther e is no b alanc e d se quenc e of length n = 6 k for k ≥ 2 . Pr o of. Let A i represen t the play er who go es i th in the first round. Let B i represen t the pla yer who go es i th in the second round; C i the play er who go es i th in the third round and D i the pla yer who go es i th in the fourth round. W e will show that balance must break do wn within the fourth round. Notice that for k ≥ 2,  6 k 4  < 2 k . Here is the list of implications that preven t balance: 1. { A i : 3 k + 1 ≤ i ≤ 6 k } = { B i : 1 ≤ i ≤ 3 k } b y Bal (2 , 1). Each pla yer that app ears in the second half in the first round must app ear in the first half in the second round, since  6 k 2  = 3 k . 2. { A i : 1 ≤ i ≤ 3 k } = { B i : 3 k + 1 ≤ i ≤ 6 k } . This follo ws by ( 1 ). 5 3. { C i : 1 ≤ i ≤ 2 k } = { A i : 2 k + 1 ≤ i ≤ 3 k } ∪ { B i : 2 k + 1 ≤ i ≤ 3 k } by Bal (3 , 1). Since  6 k 3  = 2 k , eac h pla yer must go at least once in the top 2 k p ositions after the first 3 rounds. But, { A i : 2 k + 1 ≤ i ≤ 3 k } ∪ { B i : 2 k + 1 ≤ i ≤ 3 k } are 2 k distinct pla y ers that ha ve not gone in the top 2 k p ositions after the first 2 rounds. 4. { C 2 k } ⊂ { D i : 1 ≤ i ≤ 3 k } b y Bal (4 , 1). Let z :=  6 k 4  . Since z < 2 k , then pla yer C 2 k did not get any of the top z items in an y of the first three days. Thus, C 2 k ∈ { D i : 1 ≤ i ≤ z } (Moreov er, { C i : z + 1 ≤ i ≤ 2 k } ⊆ { D i : 1 ≤ i ≤ z } ). 5. { C i : 3 k + 1 ≤ i ≤ 6 k } ⊆ { D i : 1 ≤ i ≤ 3 k } b y Bal (4 , 2). Each play er is in the first 3 k p ositions in the first or second round, but not b oth rounds. This follows from ( 1 ) and ( 2 ). Since 2(6 k ) 4 = 3 k , then each play er go es in the top 3 k p ositions at least t wice in the first 4 rounds. But, each of C 3 k +1 , . . . , C 6 k ha ve gone once in the first 3 rounds. But there isn’t room for all of C 2 k , C 3 k +1 , . . . , C 6 k to go in the top 3 k p ositions in the fourth round. I.e., |{ C 2 k , C 3 k +1 , C 3 k +2 , . . . , C 6 k }| = |{ D 1 , D 2 , . . . , D 3 k }| + 1 whic h leads to a contradiction. Prop osition 5.3 can b e extended to n = 6 k + j for j = 1 , 2 , 3 , 4 , 5. The table b elow sho ws a lo w er b ound on the v alue of k for eac h j whic h is guaranteed to prev en t balance for n . A pro of is giv en in App endix A . T able 5 shows there are at most finitely many n 6 k 6 k + 1 6 k + 2 6 k + 3 6 k + 4 6 k + 5 b ounds on k k ≥ 2 k ≥ 11 k ≥ 3 k ≥ 8 k ≥ 4 k ≥ 9 b ounds on n n ≥ 12 n ≥ 67 n ≥ 20 n ≥ 51 n ≥ 28 n ≥ 59 T able 5: V alues of n = 6 k + j with no balanced sequence v alues n which admit a balanced sequence. In particular, the following theorem follows from T able 5 . Theorem 5.4. Ther e exist at most finitely many n such that ther e is a b alanc e d se quenc e of length n . Mor e over, ther e is no b alanc e d se quenc e for n > 61 . Pr o of. Since ev ery natural n umber can b e expressed uniquely in the form n = 6 k + j for j = 0 , 1 , 2 , 3 , 4 , 5, then we can list the natural num b ers that fall under the b ounds in T able 5 . It is straightforw ard to c hec k that n = 61 is the largest natural n um b er which p ossibly has a balanced sequence. See the App endix A for a pro of of T able 5 and this theorem. 6 V aluations and Prop ortionalit y As the concept of a balanced sequence is to o strong, we would like to relax it. T o do so (and also to link our balance concepts to the literature on fair allo cation), w e assume 6 that eac h item j ∈ [ n ] has a value v j ∈ R , suc h that v 1 > v 2 > · · · > v n . 1 The total v alue of any bundle (multiset of items) Z is denoted v ( Z ) := P k ∈ Z v k . A natural fairness concept is that each pla yer should receive a bundle of items that is worth at least 1 /n of the total v alue, that is, v ( Z t i ) ≥ v ( Z t N ) /n . 2 This concept is prominen t b oth in law and in the literature on fair division ev er since its inception ( Steinhaus , 1948 ). No wada ys, it is called pr op ortionality ( Brams and T aylor , 1996 ; Rob ertson and W ebb , 1998 ). W e require a strong v arian t of prop ortionality , whic h holds not only for sp ecific v aluations but for any v aluations that satisfy the condition v 1 > · · · > v n . W e call this v arian t or dinal pr op ortionality , as it is based only on the order among items rather than on their exact v alues. 3 Belo w, whenever we write “prop ortional”, we mean “ordinally-prop ortional”. Giv en the Latin square constraint (Section 3 ), the allocation after day n is alw a ys prop ortional. This is b ecause the bundle of ev ery pla y er i satisfies v ( Z n i ) = P j ∈ M v j and the total v alue of all items is v ( Z n N ) = n · P j ∈ M v j . Ho wev er, b efore day n it is imp ossible to guarantee prop ortionality . As explained ab ov e, if v 1 is v ery high and the other item v alues are very lo w, then at any day before the n -th day , some pla y ers (those who did not get item 1 so far) will hav e a less-than-prop ortional v alue. A natural relaxation of prop ortionality for indivisible items is pr op ortionality up to c items , abbreviated PROP c ( Conitzer et al. , 2017 ; Chakrab orty et al. , 2021 ; Gourv ` es et al. , 2021 ; Bu et al. , 2023 ). It means that, for each play er i , if w e add to i ’s bundle some c items not allo cated to i yet, then i ’s v alue will b e at least 1 /n of the total v alue. Definition 6.1. Given c ≥ 1, a rep eated assignmen t is called p erp etual ly PROP c if after every single da y , the cumulativ e allo cation satisfies ordinal-PR OP c . The sequence for n = 6 at T able 1 fails this condition already after the second day . This is b ecause the total v alue allo cated up to this time is 2 · ( v 1 + v 2 + v 3 + v 4 + v 5 + v 6 ), whereas the bundle of pla yer 5 is { 5 , 6 } . Even if w e add the most v aluable item 1, the resulting bundle v alue is v 1 + v 5 + v 6 , which migh t b e smaller than 1 / 6 of the total v alue (e.g. if v 5 = v 6 = 0 and v 1 = · · · = v 4 = 1). Hence, the rep eated assignment at T able 1 is not ordinally-PR OP1 after day 2, and hence it is not p erp etually-PROP1. Belo w we relate balance to p erp etual fairness. Lemma 6.2. A b alanc e d p ermutation se quenc e gener ates a p erp etual ly-PROP1 r ep e ate d assignment. Pr o of. W e fix a da y t , and prov e that the cumulativ e allo cation up to day t satisfies ordinal-PR OP1. So far, nt items ha ve b een allo cated, so | Z t N | = nt . Note that the multiset Z t N con tains exactly t clones of every item in 1 , . . . , n . 1 The literature on fair division studies a more general setting, in whic h different play ers ma y hav e differen t rankings and differen t v aluations ov er the items. It should be easy to extend the positive result in this paper to the more general setting. 2 The phrase “at least” comes from the more general setting (fo otnote 1 ). In the case of a common v aluation, prop ortionality implies that eac h pla yer receives exactly 1 /n of the total v alue. 3 Ordinal fairness is also kno wn as ne c essary fairness or sto chastic-dominanc e fairness (sd-fairness) . Ordinal fairness in item allo cation has b een studied e.g. by Pruhs and W o eginger ( 2012 ); Aziz et al. ( 2015 ); Bouveret et al. ( 2010 ); Brams et al. ( 2022 , 2026 ) 7 Fix an agent i . Let us add to i ’s bundle Z t i a copy of the highest-v alued item (1). Although it is not strictly needed for the pro of, let us also r emove the least-v aluable item from i ’s bundle, so that i ’s new bundle still has t items ov erall. Denote this new bundle C i . T o show PROP1, it is sufficien t to show that the v ( C i ) ≥ v ( Z t N ) /n . Construct a new m ultiset D i , containing n clones of each item in C i ; no w, it is sufficien t to sho w that v ( D i ) ≥ v ( Z t N ). W e do this b y showing a bijection f : D i → Z t N , suc h that x ≤ f ( x ) for all x ∈ D i ; hence v ( x ) ≥ v ( f ( x )) for all x ∈ D i . W e first show the bijection for the sp ecial case that n/t is an integer. • W e first map the n clones of item 1 in D i (the added item). W e map them to the t clones of items 1 , 2 , . . . , n/t in Z t N . • Next, we map the n clones of the b est item held b y i , denoted Z t i [1]. By Bal ( t, j = 1), Z t i [1] ≤ ⌈ n/t ⌉ = n/t . Hence, w e can map it to the t clones of items n/t + 1 , 2 , . . . , 2 n/t in Z t N . • Similarly , by Bal ( t, j ) for all j ∈ [ t − 1], Z t i [ j ] ≤ j n/t ; hence, we can map the n clones of Z t i [ j ] to the the t clones of items j n/t + 1 , 2 , . . . , ( j + 1) n/t in Z t N . When n/t is not an integer, the mapping works as follows: • The n clones of item 1 in D i are mapped to the t clones of items 1 , 2 , . . . , ⌊ n/t ⌋ in Z t N , as w ell as to some n % t clones of item ⌈ n/t ⌉ (here % denotes the in teger remainder op eration). • The n clones of Z t i [1] in D i , which is at most ⌈ n/t ⌉ by Bal ( t, j = 1), are mapp ed to the remaining t − n % t clones of item ⌈ n/t ⌉ in Z t N , as well as to the t clones of ⌈ n/t ⌉ + 1, and some low er-v alued items. Overall, the 2 n clones of 1 and Z t i [1] in D i are mapp ed to the t clones of items 1 , . . . , ⌊ 2 n/t ⌋ as well as to some clones of ⌈ 2 n/t ⌉ in Z t N . • Similarly , for all j ∈ [ t − 1], the ( j − 1) n clones of items 1 , Z t i [1] , . . . , Z t i [ j − 1] in D i are mapp ed to the t clones of items 1 , . . . , ⌊ j n/t ⌋ , as well as to some clones of ⌈ j n/t ⌉ in Z t N . By Bal ( t, j ) Z t i [ j ] ≤ ⌈ j n/t ⌉ , so the clones of Z t i [ j ] can b e mapp ed to the remaining clones of ⌈ j n/t ⌉ and to low er-v alued items. F or illustration, T able 6 sho ws the items in Z t N (left), and the items 1 and ⌈ j n/t ⌉ for j ∈ [ t − 1] in D i (righ t) for n = 10 , t = 6. One can see that each item in the D i table is at least as go o d as the item at the same lo cation in the Z t N table, as required. Lemma 6.2 provides a PROP1 solution for an y n for whic h a balanced sequence exists. How ev er, b y Theorem 5.4 there are only finitely many suc h n . Hence, in the next section w e study a weak er balance notion. 7 Ordinal PR OP1: W eak Balance W e note that the balance condition ( 1 ) is slightly stronger than required to guarantee PR OP1. W e can w eak en it to use the condition we call WeakBal ( t, j ): In the first t 8 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 8 8 8 8 8 8 9 9 9 9 9 9 10 10 10 10 10 10 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 7 7 7 7 7 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 T able 6: Example for the pro of of Lemma 6.2 . Here n = 10 , t = 6. 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 8 8 8 8 8 8 9 9 9 9 9 9 10 10 10 10 10 10 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 4 4 4 4 4 4 4 4 4 4 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 T able 7: Example for the pro of of Lemma 7.1 . Here n = 10 , t = 6. days, e ach player gets at le ast j items fr om the top ⌊ j n/t + 1 ⌋ . Equiv alently: Z t i [ j ] ≤ ⌊ j n/t + 1 ⌋ . (2) Define a weakly-balanced sequence as a sequence satisfying WeakBal ( t, j ) for all t ∈ [ n ] and j ∈ [ t ]. Note that WeakBal ( t, j ) is w eaker than Bal ( t, j ) only when j n/t is an in teger; otherwise they are equiv alent. Lemma 7.1. A we akly-b alanc e d p ermutation se quenc e guar ante es that, after every day, the cumulative al lo c ation is or dinal ly-PR OP1. Pr o of. The pro of follo ws a similar structure to that of Lemma 6.2 . See T able 7 for an illustration for t = 6 , n = 10. F or these num b ers, WeakBal ( t, j ) is w eaker than Bal ( t, j ) only for j = 3 (there are 6-s in the rightmost table, in place of the 5-s) . Unfortunately , the imp ossibility result for a balanced sequence extends to weakly- balanced sequences, though for larger n : Prop osition 7.2. Ther e is no we akly-b alanc e d se quenc e of length n = 6 k for any k ≥ 19 . 9 Pr o of. The follo wing list of implications lead to a contradiction. 1. { A i : 3 k + 2 ≤ i ≤ 6 k } ⊆ { B i : 1 ≤ i ≤ 3 k + 1 } by WeakBal (2 , 1), since  6 k 2 + 1  = 3 k + 1. 2. |{ A i : 1 ≤ i ≤ 3 k + 1 } ∩ { B i : 1 ≤ i ≤ 3 k + 1 }| ≤ 2, as b y ( 1 ), there are already 3 k − 1 agents who m ust b e contained in a set of 3 k + 1 agents, so there is room for at most tw o other agen ts in that set. 3. Define an auxiliary set Z 1 := { A i : 2 k + 2 ≤ i ≤ 3 k + 1 } ∩ { B i : 3 k + 2 ≤ i ≤ 6 k } . Then | Z 1 | ≥ k − 2 b y ( 2 ). Similarly , define the set Z 2 := { B i : 2 k + 2 ≤ i ≤ 3 k + 1 } ∩ { A i : 3 k + 2 ≤ i ≤ 6 k } . Then | Z 2 | ≥ k − 2 b y analogous considerations. 4. { C i : 1 ≤ i ≤ 2 k + 1 } ⊇ Z 1 ∪ Z 2 b y WeakBal (3 , 1). Since  6 k 3 + 1  = 2 k + 1, eac h play er m ust receive at least one of the top 2 k + 1 items during the first three rounds. But none of the pla yers in Z 1 ∪ Z 2 receiv ed one of the top 2 k + 1 items during the first tw o rounds. Note that Z 1 ∩ Z 2 = ∅ , so | Z 1 ∪ Z 2 | ≥ 2 k − 4 b y ( 3 ). 5. Let z :=  6 k 4 + 1  and note that z < 2 k − 4 for k ≥ 11. Define an auxiliary set Z 3 := ( Z 1 ∪ Z 2 ) ∩ { C i : z + 1 ≤ i ≤ 2 k + 1 } . Then: { D i : 1 ≤ i ≤ z } ⊇ Z 3 b y WeakBal (4 , 1), since all pla yers m ust receive one of their top z items at least once in the first three days, and the pla yers in Z 1 ∪ Z 2 did not get any of these items in the first tw o da ys. Note that | Z 3 | ≥ 2 k − 4 − z by ( 4 ), as Z 3 = ( Z 1 ∪ Z 2 ) \ { C i : 1 ≤ i ≤ z } . 6. |{ D i : 1 ≤ i ≤ 3 k + 1 } ∩ { C i : 3 k + 2 ≤ i ≤ 6 k }| ≥ 3 k − 3 by WeakBal (4 , 2). Since 2(6 k ) 4 + 1 = 3 k + 1, eac h play er must get one of the top 3 k + 1 items at least t wice in the first four rounds. So the set { D i : 1 ≤ i ≤ 3 k + 1 } m ust con tain all pla yers who got a top 3 k + 1 item at most once in previous rounds. By ( 2 ), at most t wo play ers got a top 3 k + 1 item twice in the first tw o days. So, from among the 3 k − 1 pla yers in { C i : 3 k + 2 ≤ i ≤ 6 k } , at least 3 k − 3 pla y ers m ust b e in { D i : 1 ≤ i ≤ 3 k + 1 } . Com bining ( 5 ) and ( 6 ) w e see that { D i : 1 ≤ i ≤ 3 k + 1 } m ust con tain at least 2 k − 4 − z pla yers from { C i : z + 1 , 2 k + 1 ≤ i ≤} and at least 3 k − 3 play ers from { C i : 3 k + 2 ≤ i ≤ 6 k } , which m ust b e different pla yers. All in all, { D i : 1 ≤ i ≤ 3 k + 1 } m ust contain at least 5 k − 7 − z play ers. Note that 5 k − 7 − z = 5 k − 8 − ⌊ 6 k / 4 ⌋ ≈ 3 . 5 k − 8, which is larger than 3 k + 1 when k is sufficien tly large. Sp ecifically , when k ≥ 19 w e get a contradiction. W e pro ved in Section 4 that there is no balanced sequence for n = 12. How ev er, a w eakly-balanced sequence for n = 12 do es exist; see T able 8 for an example. This raises the question of whether there is any reasonable weak er v arian t of the balance condition ( 2 ), that can b e guaran teed for all n ? 10 1 12 9 7 5 11 4 8 3 10 6 2 2 11 8 6 4 12 7 5 1 9 10 3 3 10 7 5 12 9 2 6 11 4 8 1 4 9 12 3 8 7 1 11 6 5 2 10 5 8 11 4 3 2 12 10 7 6 1 9 6 7 5 2 11 10 3 12 9 1 4 8 7 6 4 12 2 8 11 1 10 3 9 5 8 5 10 1 9 4 6 3 12 2 7 11 9 4 6 11 1 5 10 2 8 12 3 7 10 3 2 9 7 1 8 4 5 11 12 6 11 2 1 10 6 3 9 7 4 8 5 12 12 1 3 8 10 6 5 9 2 7 11 4 T able 8: W eakly-balanced sequence for n = 12 8 Ordinal PR OP2: Op en Question F or Ordinal PR OP2 (prop ortionality up to t w o items), it is sufficien t to satisfy the follo wing condition for all agents i ∈ N , after every da y t ∈ [ n ], for every j ∈ [ t ]: Z t i [ j ] ≤ ⌈ ( j + 1) n/t ⌉ . (3) In fact, ev en the following sligh tly weak er condition is sufficient: Z t i [ j ] ≤ ⌊ ( j + 1) n/t + 1 ⌋ . (4) Condition ( 4 ) is non-trivial only for t ≥ 3 and j ≤ n − 2, for example: • In the first three days, eac h play er gets at least one item from the top ⌊ 2 n/ 3 + 1 ⌋ ; • In the first four da ys, each play er gets at least one item from the top ⌊ 2 n/ 4 + 1 ⌋ and at least tw o items from the top ⌊ 3 n/ 4 + 1 ⌋ ; and so on. Op en Question 1. Is there, for every n ≥ 1, a rep eated assignmen t that satisfies condition ( 3 ), or at least condition ( 4 ), for all t ∈ [ n ] and j ∈ [ t ]? More generally , what is the strongest upp er b ound on Z t i [ j ] that can b e guaranteed for every p ositive n ? 9 Related W ork 9.1 F air division The literature on fair division is very large, with new pap ers app earing every sev eral w eeks. See Amanatidis et al. ( 2022 ); Aziz et al. ( 2022 ); Nguy en and Rothe ( 2023 ); Lang and Rothe ( 2024 ) for some recent surveys. Most works on fair allo cation consider a one-round division, but in recen t y ears there is a growing interest in multiple-round di- vision. Belo w we presen t sev eral settings that are related to ours, ordered b y increasing difficult y . 11 Rep eated fair assignmen t. This setting is most closely related to ours, as it inv olv es a single assignment problem (of exactly one item to each agent) that is rep eated ev ery da y . It w as recen tly studied by Micheel and Wilczynski ( 2024 ). They studied three fairness notions: (a) Mirr or e d envy : if some agent envies another agent a certain num b er of times, then the latter envied them the same num b er of times. They study the run- time complexit y of deciding whether a mirror-envy-free and P areto-efficien t allocation exists. (b) Equal tr e atment of e quals (ETE) : agen ts with the same preferences should get exactly the same final bu ndle. They prov e that an ETE and Pareto-efficien t allo cation alw ays exists when the size of ev ery equiv alence class of agen ts divides the n um b er of future rounds. They also study the run-time complexity of deciding whether such an allo cation exists in general. (c) Minimizing the numb er of times an agent is envious : If there are T future rounds, then there is alw a ys an allo cation in whic h the maxim um n umber of times eac h agen t is en vious is at most ⌈ T / 2 ⌉ , and this b ound is tight. But deciding whether there is an allo cation with at most T / 3 envy rounds is NP-hard. Caragiannis and Narang ( 2024 ) study a generalization of rep eated fair assignmen t, in whic h the v alue of an agen t for an item ma y dep end on the num b er of rounds the item was previously used by the same agent. In this setting, ev en obtaining fairness at the last day b ecomes challenging, as the round-robin tec hnique might not work. They study the run-time complexity of computing an allo cation that maximizes the sum of v alues (“utilitarian”), and present some sp ecial cases in which it is p ossible to ensure EF1 (envy-free up-to one item) at the last day . Rep eated fair allo cation. This setting is more general than ours, as it allo ws the n umber of items allocated each day to b e different than the num b er of agents, and do not require that each agent receives exactly one item p er da y ( Igarashi et al. , 2024 ). As their setting is more general than ours, their p ositiv e results are more restricted. In particular: (a) When n ≥ 3 and the num b er of da ys is fixed in adv ance (as in our setting), they only guarantee last-day fairness, and only when the n um b er of da ys is a multiple of n ; there are no guarantees on fairness after eac h day . (b) When n = 2 and the n umber of days is fixed in adv ance and is even, there are rep eated allo cations that are en vy-free after the last day , and in addition, the allo cation in eac h individual da y satisfies a w eak v ariant of EF1. (c) When the n umber of rounds can b e c hosen b y the algorithm, there are sequences that are simultaneously last-da y en vy-free and P areto-efficient, and individual-day PROP1 The result is obtained by establishing a connection with the probabilistic and divisible settings. The question of b ounding the n umber of rounds required for such an allo cation remains op en. Co okson et al. ( 2025 ) strengthen their results to or dinal fairness , as in our setting. They sho w polynomial-time algorithms that guaran tee the following: (a) F or n = 2 agen ts: ordinal EF1 after each da y , as well as in each individual da y; (b) F or n ≥ 3 agen ts: ordinal EF1 at eac h individual da y , as well as ordinal PR OP1 after the last da y . They also prov e that, for n ≥ 3, ordinal-EF1 after each day cannot b e guaran teed ev en when all agents hav e identical preferences. In contrast to these w orks, we fo cus on the imp ortan t sp ecial case of one item p er agen t p er da y , and [hop e to] get stronger p ositive results: ordinal PROP c for some small c , after each day , for any n . 12 T emp oral fair allo cation. This setting is ev en more general, as it allo ws the n umber and v alues of items to be different in eac h da y . The guaran tees are naturally weak er; see He et al. ( 2019 ); Elkind et al. ( 2024 ); Co okson et al. ( 2025 ). Still, the fact that the same agents participate in each daily allo cation is adv an tageous, as it allo ws us to comp ensate agen ts who got a lo w v alue in early days, by giving them a higher v alue in later days. One-shot allo cation. The most w ell-studied setting in fair division inv olv es a one- shot allo cation (a single da y). F or this setting, there are several algorithms that attain an allo cation that is envy-fr e e up to one item (EF1) — a condition stronger than PROP1 ( Lipton et al. , 2004 ; Budish , 2011 ; Caragiannis et al. , 2019 ). These algorithms cannot b e applied in our setting, as they might allocate more than one item to the same play er in the same day . One-shot assignmen t. Here, eac h agen t must receive a single item. In this setting ev ery allocation is trivially PROP1 (and EF1), whic h mak es these conditions unin ter- esting. F airness is usually attained either b y randomization; see Basteck ( 2018 ) and Segal-Halevi ( 2022 ) resp ectively for some recen t surv eys. In our r ep e ate d assignment setting, we exp ect to attain fair outcomes even without randomization or money . One-shot allo cation with cardinalit y constrain ts. Here, items are partitioned in to pre-sp ecified categories, and each agent can receiv e at most a fixed num b er of items (called the “capacit y”) of eac h category ( Bisw as and Barman , 2018 ; Dror et al. , 2023 ; Shoshan et al. , 2023 ). The rep eated assignment setting can b e reduced to this setting: define, for each day , a category containing all items allo cated at that day , with a capacity of 1. There are algorithms that guarantee EF1 (hence also PR OP1), but only for sp ecific v aluations. F or example, the algorithm of Bisw as and Barman ( 2018 ), translated to our setting, works as follo ws: (1) Assign the items in day 1 arbitrarily . (2) In each day t ≥ 2, order the pla yers by ascending v alue, from the play er with the smallest to the one with the largest total v alue; give item 1 to the first play er in the order, item 2 to the second play er, etc. Supp ose that, after some da y t , some pla yer i has a smaller v alue than some other pla yer j , i.e., v ( Z t i ) < v ( Z t j ). Let t ′ b e the maxim um t ′ < t in whic h v ( Z t ′ i ) ≥ v ( Z t ′ j ) held. Then at da y t ′ + 1, pla yer j received an item with a higher v alue than i , and at days t ′ + 2 , . . . , t , pla yer i received an item with an equal-or-higher v alue than j . Hence, if the item of day t ′ + 1 is remov ed from j ’s bundle, then i do es not envy . Hence, the allo cation is EF1. In con trast, our aim is to guarantee or dinal appr oximate fairness — a condition that holds sim ultaneously for all v aluations consistent with the ranking. Another adv antage of this approac h is that our algorithms do not need to know the v aluations at all; they only need to know the ranking (which item is b est, which is second-b est, etc.). Pic king sequences The output of our algorithms can also be understoo d as a picking se quenc e — an order by which eac h agent pic ks a fav orite item from the remaining items. Pic king sequences hav e a very lo w informational requirements, as pla yers do not hav e to reveal their v aluations — they only hav e to pick the b est remaining item when their turn arrives ( Bouveret and Lang , 2011 ; Brams and Kaplan , 2004 ; Segal-Halevi , 2020 ). 13 Online fair allocation. This is the hardest setting studied, as it assumes that the items and their v aluations are not known in advanc e (in contrast to temp oral fair divi- sion). Each day , a new item arriv es, and m ust be allo cated immediately and irrev o cably to one of the agents. As the future is not kno wn, the algorithm might make allo cations that are bad in hindsight. This translates to m uc h w eak er fairness guarantees. It is not p ossible to guarantee PR OP c for any constan t c , ev en in the last round; such a guaran tee is p ossible only if c grows at least lik e √ t . See Benade et al. ( 2018 ); Benad` e et al. ( 2024 ); Kahana and Hazon ( 2023 ); He et al. ( 2019 ); Neoh et al. ( 2025 ); Zeng and Psomas ( 2020 ); Jiang et al. ( 2019 ) for some relev an t pap ers. Latin squares The n -b y- n matrices formed b y balanced sequences (e.g. T able 2 ) are L atin squar es , as they contain each num b er in [ n ] exactly once in eac h ro w and eac h column. Recen tly , Kaw ase et al. ( 2025 ) studied an allo cation problem with “Latin Square” constrain ts. Their goal is not to attain prop ortionalit y , but rather to maximize the so cial w elfare. Note that the requiremen ts for a balanced sequence are stronger than the requirements for a Latin square. A In tegers for whic h No Balanced Sequence Exists W e prov e the b ounds presented in T able 5 . The tec hnique used here is similar to that in the pro of Prop osition 5.3 . W e also use the same notation A i , B i , C i , D i as in that pro of. A.1 Case n = 6 k + 1 The following formulas hold: l 6 k + 1 2 m = 3 k + 1 , l 6 k + 1 3 m = 2 k + 1 and l 6 k + 1 4 m =  (6 k + 4) / 4 if k is even (6 k + 2) / 4 if k is o dd W e ha ve the follo wing: { A 3 k +2 , . . . , A 6 k +1 } ⊂ { B 1 , . . . , B 3 k +1 } and { B 3 k +2 , . . . , B 6 k +1 } ⊂ { A 1 , . . . , A 3 k +1 } . Let A k = { A 3 k +2 , . . . , A 6 k +1 } ∩ { B 2 k +2 , . . . , B 3 k +1 } and B k = { B 3 k +2 , . . . , B 6 k +1 } ∩ { A 2 k +2 , . . . , A 3 k +1 } . Th us, A k and B k are disjoint and |A k | ≥ k − 1, |B k | ≥ k − 1. A.1.1 Ev en k Let a k = 6 k +4 4 and C k = { C a k +1 , . . . , C 2 k +1 } . Since A k ∪ B k ⊂ { C 1 , . . . , C 2 k +1 } , then    ( A k ∪ B k ) ∩ C k    ≥ 2 k + 1 − a k − 3 = 1 2 k − 3 . 14 W e know that ( A k ∪ B k ) ∩ C k ⊂ { D 1 , . . . , D a k } . Also,    { C 3 k +2 , . . . , C 6 k +1 } ∩ { D 1 , . . . , D 3 k +1 }    ≥ 3 k − 1 . There is a contradiction, if 3 k − 1 + 1 2 k − 3 > 3 k + 1. Hence, there is no solution, if k > 10, or equiv alen tly , k ≥ 12 and n ≥ 73. A.1.2 Odd k Using an argumen t similar to even k , we get a contradiction, if 3 k − 1 + 1 2 k − 5 2 > 3 k + 1 . Hence, there are no solutions if k ≥ 11 or n ≥ 67. A.2 Case n = 6 k + 2 The following formulas hold: l 6 k + 2 2 m = 3 k + 1 , l 6 k + 2 3 m = 2 k + 1 and l 6 k + 2 4 m =  (6 k + 4) / 4 if k is even (6 k + 2) / 4 if k is o dd W e ha ve the follo wing: { A 3 k +2 , . . . , A 6 k +2 } = { B 1 , . . . , B 3 k +1 } and { B 3 k +2 , . . . , B 6 k +2 } = { A 1 , . . . , A 3 k +1 } . Let A k = { A 3 k +2 , . . . , A 6 k +2 } ∩ { B 2 k +2 , . . . , B 3 k +1 } and B k = { B 3 k +2 , . . . , B 6 k +2 } ∩ { A 2 k +2 , . . . , A 3 k +1 } . Th us, A k and B k are disjoint and |A k | = k , |B k | = k . A.2.1 Ev en k Let a k = 6 k +4 4 and C k = { C a k +1 , . . . , C 2 k +1 } . Since A k ∪ B k ⊂ { C 1 , . . . , C 2 k +1 } , then    ( A k ∪ B k ) ∩ C k    ≥ 2 k + 1 − a k − 1 = 1 2 k − 1 . W e know that ( A k ∪ B k ) ∩ C k ⊂ { D 1 , . . . , D a k } . Ho w ever, { C 3 k +2 , . . . , C 6 k +2 } = { D 1 , . . . , D 3 k +1 } . There is a con tradiction, if 1 2 k − 1 > 0. Hence, there is no solution, if k > 2, or equiv alen tly , n ≥ 26. 15 A.2.2 Odd k Using an argumen t similar to even k , we get a contradiction, if 1 2 k − 1 2 > 0 . Hence, there are no solutions if k > 1 or n ≥ 20. A.3 Case n = 6 k + 3 The following formulas hold: l 6 k + 3 2 m = 3 k + 2 , l 6 k + 3 3 m = 2 k + 1 and l 6 k + 3 4 m =  (6 k + 4) / 4 if k is even (6 k + 6) / 4 if k is o dd W e ha ve the follo wing: { A 3 k +3 , . . . , A 6 k +3 } ⊂ { B 1 , . . . , B 3 k +2 } and { B 3 k +3 , . . . , B 6 k +3 } ⊂ { A 1 , . . . , A 3 k +2 } . Let A k = { A 3 k +3 , . . . , A 6 k +3 } ∩ { B 2 k +2 , . . . , B 3 k +2 } and B k = { B 3 k +3 , . . . , B 6 k +3 } ∩ { A 2 k +2 , . . . , A 3 k +2 } . Th us, A k and B k are disjoint and |A k | = k , |B k | = k . A.3.1 Ev en k Let a k = 6 k +4 4 and C k = { C a k +1 , . . . , C 2 k +1 } . Since A k ∪ B k ⊂ { C 1 , . . . , C 2 k +1 } , then    ( A k ∪ B k ) ∩ C k    ≥ 2 k + 1 − a k − 1 = 1 2 k − 1 . W e know that ( A k ∪ B k ) ∩ C k ⊂ { D 1 , . . . , D a k } . Ho w ever,    { C 3 k +3 , . . . , C 6 k +3 } ∩ { D 1 , . . . , D 3 k +2 }    ≥ 3 k. There is a contradiction, if 1 2 k − 1 > 2. Hence, there is no solution, if k > 6, or equiv alen tly , n ≥ 51. A.3.2 Odd k Using an argumen t similar to even k , we get a contradiction, if 1 2 k − 3 2 > 2 . Hence, there are no solutions if k > 7 or n ≥ 57. 16 A.4 Case n = 6 k + 4 The following formulas hold: l 6 k + 4 2 m = 3 k + 2 , l 6 k + 4 3 m = 2 k + 2 and l 6 k + 4 4 m =  (6 k + 4) / 4 if k is even (6 k + 6) / 4 if k is o dd W e ha ve the follo wing: { A 3 k +3 , . . . , A 6 k +4 } = { B 1 , . . . , B 3 k +2 } and { B 3 k +3 , . . . , B 6 k +4 } = { A 1 , . . . , A 3 k +2 } . Let A k = { A 3 k +3 , . . . , A 6 k +4 } ∩ { B 2 k +3 , . . . , B 3 k +2 } and B k = { B 3 k +3 , . . . , B 6 k +4 } ∩ { A 2 k +3 , . . . , A 3 k +2 } . Th us, A k and B k are disjoint and |A k | = k , |B k | = k . A.4.1 Ev en k Let a k = 6 k +4 4 and C k = { C a k +1 , . . . , C 2 k +2 } . Since A k ∪ B k ⊂ { C 1 , . . . , C 2 k +2 } , then    ( A k ∪ B k ) ∩ C k    ≥ 2 k + 2 − a k − 2 = 1 2 k − 1 . W e know that ( A k ∪ B k ) ∩ C k ⊂ { D 1 , . . . , D a k } . Ho w ever, { C 3 k +3 , . . . , C 6 k +4 } = { D 1 , . . . , D 3 k +2 } . There is a con tradiction, if 1 2 k − 1 > 0. Hence, there is no solution, if k > 2, or equiv alen tly , n ≥ 28. A.4.2 Odd k Using an argumen t similar to even k , we get a contradiction, if 1 2 k − 3 2 > 0 . Hence, there are no solutions if k > 3 or n ≥ 34. A.5 Case n = 6 k + 5 The following formulas hold: l 6 k + 5 2 m = 3 k + 3 , l 6 k + 5 3 m = 2 k + 2 and l 6 k + 5 4 m =  (6 k + 8) / 4 if k is even (6 k + 6) / 4 if k is o dd 17 W e ha ve the follo wing: { A 3 k +4 , . . . , A 6 k +5 } ⊂ { B 1 , . . . , B 3 k +3 } and { B 3 k +4 , . . . , B 6 k +5 } ⊂ { A 1 , . . . , A 3 k +3 } . Let A k = { A 3 k +4 , . . . , A 6 k +5 } ∩ { B 2 k +3 , . . . , B 3 k +3 } and B k = { B 3 k +4 , . . . , B 6 k +5 } ∩ { A 2 k +3 , . . . , A 3 k +3 } . Th us, A k and B k are disjoint and |A k | = k , |B k | = k . A.5.1 Ev en k Let a k = 6 k +8 4 and C k = { C a k +1 , . . . , C 2 k +2 } . Since A k ∪ B k ⊂ { C 1 , . . . , C 2 k +2 } , then    ( A k ∪ B k ) ∩ C k    ≥ 2 k + 2 − a k − 2 = 1 2 k − 2 . W e know that ( A k ∪ B k ) ∩ C k ⊂ { D 1 , . . . , D a k } . Ho w ever,    { C 3 k +4 , . . . , C 6 k +5 } ∩ { D 1 , . . . , D 3 k +3 }    ≥ 3 k + 1 . There is a con tradiction, if 1 2 k − 2 > 2. Hence, there is no solution, if k > 8, or equiv alen tly , n ≥ 65. A.5.2 Odd k Using an argumen t similar to even k , we get a contradiction, if 1 2 k − 3 2 > 2 . Hence, there are no solutions if k > 7 or n ≥ 59. B Example of T op-Balanced Sequence Algorithm Here, we demonstrate a top-balanced sequence for n = 15 play ers. This example is sufficien tly complex to show how this pro cedure w orks for any p ositive integer n . F or day 1, assign items 1 through 15 to pla y ers 1 through 15 in increasing order. F or da y 2, assign items 1 through 7 to play ers 9 through 15. The remaining items can b e assigned arbitrarily to the remaining pla y ers for da y 2. F or da y t = 3, w e ha v e ⌈ 15 3 ⌉ = 5. Notice that for the first 2 da ys, there are 10 unique play ers who were assigned one of the items 1 through 5. There are 5 pla yers not in this group. In particular, these are pla yers 6 , 7 , 8 , 14 , 15. Assign these pla yers items 1 through 5. Assign the remaining items arbitrarily to the remaining play ers for da y 3. F or day t = 4, we hav e ⌈ 15 4 ⌉ = 4. F or the first 3 days, the play ers assigned one of the items 1 through 4 are 12 unique play ers. F or day 4, c ho ose the 3 pla yers who are not in this group of 12 pla yers, and assign items 1 through 3 to these play ers. These will b e pla y ers 5 , 13 , 15. Assign the remaining items arbitrarily for day 4. F or day t = 5, we ha ve ⌈ 15 5 ⌉ = 3. A k ey p oin t is that the group of play ers assigned one of items 1 through 3 on the first 4 days are unique. There are 3 play ers outside this 18 group. Given the previous assignmen ts, these will b e play ers 4 , 12 , 14. Assign items 1 through 3 to these 3 play ers on day 5. Assign the remaining items to the remaining pla yers arbitrarily . F or da y t = 6, w e hav e ⌈ 15 6 ⌉ = 3. Every play er has already b een assigned one of items 1 through 3. Ho wev er, we still need some care in assigning items on day 6. Cho ose 2 play ers that w ere not assigned items 1 or 2, and assign them items 1 and 2 on day 6. These could b e play ers 3 and 11. Assign the remaining items arbitrarily . F or day t = 7, we still hav e ⌈ 15 7 ⌉ = 3. There are at least 2 play ers who were not assigned items 1 or 2 on the first 6 da ys. Assign items 1 and 2 to these pla yers on day 7. These could b e play ers 8 and 15. On day t = 8, there is one play er not previously assigned items 1 or 2. Based on previous assignmen ts, this would b e play er 14. Assign item 1 to this play er. The remaining items ma y b e assigned arbitrarily on day 8. Ev ery pla yer has been assigned either item 1 or item 2 after 8 days. In particular, 8 different play ers were assigned item 1. Arbitrarily order the remaining 7 pla yers and assign item 1 to each of the 7 play ers on each of the remaining days. The remaining 7 pla yers could b e ordered as 2 , 10 , 7 , 13 , 12 , 8 , 14 . This pro cedure can b e applied to an arbitrary p ositive integer n . The k ey prop ert y is that at each stage all play ers are assigned one of the top ⌈ n t ⌉ items in the first t days. If n t is an integer, then these play ers will b e unique. Mo ving to the next da y , we pick the pla yers that ha ve not b een assigned any items from 1 through ⌈ n t +1 ⌉ and assign those items to these play ers on day t + 1. As so on as those play ers hav e b een exhausted, we fo cus on the play ers that ha ve not b een assigned an y items from 1 through ⌈ n t +2 ⌉ and so on. This pro cedure will pro duce a top-balanced sequence. 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