Expand the Shares Together: Envy-free Mechanisms with a Small Number of Cuts

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1 Nonme mnusript No. (will e inserted y the editor) Expnd the Shres Together: Envy-free Mehnisms with Smll Numer of uts Msoud Seddighin Mjid Frhdi Mohmmd Ghodsi Rez Alijni Ahmd S. Tjik Reeived: dte / Aepted: dte Astrt We study the prolem of fir division of heterogeneous resoure mong strtegi plyers. Given divisile heterogeneous ke, we wish to divide the ke mong n plyers to meet these onditions: (I) every plyer (wekly) prefers his lloted ke to ny other plyer s shre (suh notion is known s envy-freeness), (II) the llotion is dominnt strtegy-proof (truthful) (III) the numer of uts mde on the ke is smll. We provide methods for dividing the ke under different ssumptions on the vlution funtions of the plyers. First, we suppose tht the vlution funtion of every plyer is single intervl with speil property, nmely ordering property. For this se, we propose proess lled expnsion proess nd show tht it results in n envy-free nd truthful llotion tht uts the ke into extly n piees. Next, we remove the ordering restrition nd show tht for this se, n extended form of the expnsion proess, lled expnsion proess with unloking yields n envy-free llotion of the ke with t most 2(n 1) uts. Furthermore, we show tht in the expnsion proess with unloking, the plyers my misrepresent their vlutions to ern more shre. In ddition, we use more omplex form of the expnsion proess with unloking to otin A preliminry version of this pper is ppered in AAAI 2017 [1]. Msoud Seddighin Srif University of Tehnology, E-mil: mseddighin@e.shrif.edu Mjid Frhdi Shrif University of Tehnology, E-mil: mjid.frhdi94@gmil.om Mohmmd Ghodsi Shrif University of Tehnology, IPM - Institute for Reserh in Fundmentl Sienes, E-mil: ghodsi@shrif.edu Rez Alijni Duke university, E-mil: rezlijni71@gmil.om Ahmd S. Tjik University of Mihign - Ann Aror, E-mil: hmdshh.t@gmil.om

2 2 Msoud, Mjid, Mohmmd, Rez, Shh n envy-free nd truthful llotion tht uts the ke in t most 2(n 1) lotions. We lso, evlute our expnsion method in prtie. In the empiril results, we ompre the numer of uts mde y our method to the numer of uts in the optiml solution (n 1). The experiments revel tht the numer of uts mde y the expnsion nd unloking proess for envy-free division of the ke is very lose to the optiml solution. Finlly, we study pieewise onstnt nd pieewise uniform vlution funtions with m piees nd present the onditions, under whih generlized form of expnsion proess n llote the ke vi O(nm) uts. Keywords ke utting, envy-free, mehnism design, pproximtion, firness 1 Introdution The prolem of dividing ke mong set of individuls hs een widely studied in the pst 60 yers. The sujet ws first defined y Steinhus [17]. The desription of the prolem is s follows: given heterogeneous ke nd set of plyers, with potentilly different tendenies to different prts of the ke, how to ut the ke nd distriute it mong the plyers in fir mnner? Severl notions re defined for mesuring the firness of n llotion (see [16]). One of the most importnt notions is envy-freeness. An llotion of the ke is envy-free, if eh plyer (wekly) prefers his lloted shre to ny other plyers shre. Envy-free resoure llotion hs een vstly studied in the literture. For two plyers, the fmous method of ut nd hoose gurntees envy-freeness of the llotion. For three plyers, Selfridge nd onwy designed protool for finding n envy-free division of the ke. In their method, plyer my reeive more thn one piee (see [15] for detils). Brms nd Tylor generlized this method to n ritrry numer of plyers [9]. However, their method doesn t gurntee ny upper ound on the numer of uts. Reently, Aziz nd Mkenzie in [2], suggested ounded envy-free protool for ny numer of gents. In some settings, the numer of uts is importnt. In severl ppers (e.g. [18], [4], [19], [5]) the ke utting with minimum numer of uts hs een studied. In some ses, eh ut might hve n dditionl ost. As n exmple, suppose tht the ke models proessing time tht must e firly lloted mong set of tsks. Every tsk-swith imposes n overhed; minimizing the totl mount of overhed would e equivlent to minimizing the numer of uts on the ke. In ddition, plyers my not hve ny vlue for very smll piees mde y lrge numer of uts. In [10], this issue ws illustrted y the dvertisement exmple: think of the ke s time nd onsider the llotion of dvertising time. In suh setting, lrge numer of uts n yield so smll periods of time tht re not useful for dvertising. In n llotion with smll numer of uts, this issue is unlikely.

3 Envy-free Mehnisms with Smll Numer of uts 3 Stromquist, in [18], proved the existene of n envy-free division of the ke mong n plyers with n 1 uts whih is the minimum numer of uts required to divide ke mong n plyers. However, the proof is not onstrutive nd does not yield polynomil time lgorithm. He lso proved in [19] tht no finite protool n find n envy-free llotion with the minimum numer of uts for n 3. Deng, Qi, nd Seri in [12] showed tht the prolem of finding n envy-free llotion of the ke, with minimum numer of uts is PPAD-omplete. They lso proposed n FPTAS for the se of three plyers. In numer of reent ppers (e.g. [10, 7, 5, 14, 11, 3]) some restrited lsses of vlution funtions hve een studied. Pieewise onstnt nd pieewise uniform vlution funtions re two speil lsses of vlution funtions whih re very importnt in prtie. One of the importnt properties of these vlution funtions is tht they n e desried onisely. Kurokw, Li, nd Proi in [13] proved tht finding n envy-free llotion (in Roertson- We model) when the vlution funtions re pieewise-uniform is s hrd s solving the prolem without ny restrition on the vlution funtions. The lssi ke utting lgorithms ssume tht the gents re not strtegi nd honestly report their vlutions. However, this is not the se in mny rel life situtions. Reently, some studies onsidered the prolem from gme theoreti perspetive nd ttempt to find truthful mehnisms for dividing the ke. Similr to firness, there re different notions for the onept of truthfulness. Brms et. l. in [8], oserved wek notion of truthfulness: plyers don t risk telling lie if there exists senrio (for other plyers vlutions) in whih lying results in lower pyoff. As n exmple, they showed tht ut nd hoose protool is wekly truthful. My nd Nisn [14] designed truthful nd Preto-effiient mehnisms to divide the ke etween two plyers where eh plyer is interested in suset of the ke, uniformly. In [11], hen et l. onsidered strong notion of truthfulness (denoted y strtegy-proofness), in whih the plyers dominnt strtegies re to revel their true vlutions over the ke. They presented strtegy-proof mehnism for the se when the vlution funtions re pieewise uniform. They lso designed rndomized lgorithm tht is envy-free nd truthful in expettion for pieewise liner vlution funtions. However, their pproh for dividing the ke uses Ω(n 2 m) uts, where m is the numer of piees in eh vlution funtion. Aziz nd Ye [3] onsidered the prolem when vlution funtions re pieewise onstnt/uniform. Bsed on prmetri network flows, they designed n envy-free lgorithm tht is group strtegy-proof 1 for pieewise uniform vlutions. It is notle tht their method eomes equivlent to Mehnism 1 from [11], in the se of pieewise uniform vlutions. 1 Group strtegy-proof mens no group of plyers n misreport their vlutions, suh tht in the resulting llotion ll of them ern more pyoff

4 4 Msoud, Mjid, Mohmmd, Rez, Shh 1.1 Our work We investigte the prolem of finding envy-free nd truthful mehnisms with smll numer of uts. By truthful, we men tht the plyers must not gin from misreporting their vlution, regrdless of the tion of other plyers. By smll numer of uts, we men tht the numer of uts does not exeed O(nm), where m is the omplexity (i.e., numer of steps) of eh plyers (pieewise onstnt) vlution funtion. To the est of our knowledge, this is the first study tht ims to pproximte the numer of uts. The sis of our methods is simple nd elegnt proess lled expnsion proess. After desriing the proess, we strt with the se where eh plyer s vlution funtion is pieewise onstnt with only one step nd mintins speifi property tht we nme ordering property. For this se, we propose EFISM, whih is polynomil time, strtegy-proof nd envy-free llotion with n 1 uts (Theorem 1). Next, we omine the expnsion proess with nother proess lled unloking. Bsed on the expnsion proess with unloking, we propose EFS, whih is n envy-free llotion tht uts the ke into t most 2n 1 piees in polynomil time (Theorem 2). To the est of our knowledge, no previous work tries to pproximte the numer of uts in fir solution. Furthermore, using more omplex form of the expnsion proess with unloking, we propose EFGISM, whih is reursive, polynomil time, truthful, nd envy-free lgorithm tht uts the ke into t most 2n 1 piees (Theorem 3). We evlute the expnsion with unloking method in prtie. We ompre the numer of uts mde y our method to the numer of uts in the optiml solution (n 1). Interestingly, the numer of uts returned y our lgorithm is very lose to n 1, whih shows tht the lgorithm is lmost optiml (see Setion 8 for more detils). Finlly, we onsider the se where the vlution funtions re pieewise onstnt with m piees. When the numer of plyers is onstnt, we provide poly(m) time lgorithm for envy-free division of the ke with n 1 uts. Finlly, we onsider the se tht the plyers possess prtiulr property, nmely intersetion property nd show tht under this ssumption, modifition of the expnsion proess yields poly(m, n) time, envy-free lgorithm tht uts the ke in O(nm) lotions. 2 Model Desription nd Preliminries Throughout the pper, we use the term intervl for two purposes: vlution funtions nd the shres lloted to the plyers. For revity, denote the former type of intervls y I nd the ltter y I. Also, we Suppose tht for every vlution intervl I i, I i = [α i, β i ] nd for every shre intervl I i, I i = [ i, i ]. Given set N of n plyers nd ke. We represent the ke y the intervl [0, 1]. For every plyer p i N, vlution funtion ν i : [0, 1] R is given.

5 Envy-free Mehnisms with Smll Numer of uts 5 For eh p i N nd intervl I = [, ], we define V i (I) s ν i(x)dx. Our ssumption is tht the vlutions funtions re normlized, suh tht V i () = 1, for eh plyer p i. A piee of the ke, is set of mutully disjoint su-intervls of [0, 1]. For piee P, we define V i (P ) s I P V i(i). A vlution funtion ν is pieewise onstnt, if there exists set S ν = {I ν1, I ν2,..., I νk } of mutully disjoint intervls, suh tht for ny two points x, x in I νi, ν(x) = ν(x ) = r i nd for ny point x tht does not elong to ny intervl in S v, ν(x) = 0. To put it simply, pieewise onstnt funtion onsists of finite numer of intervls, suh tht ll the points in the sme intervl hve the sme vlue, nd for the points tht do not elong to ny intervl, the vlution is 0. We sy ν hs k steps, if S ν = k. A division of the ke mong set N of n plyers is set D = {P 1, P 2,..., P n } of piees, with eh piee P i = {I i,1, I i,2,..., I i, Pi } eing set of intervls with the following two properties: (I) every pir of intervls re mutully disjoint nd (II) no piee of the ke is left ehind: i,j I i,j =. The numer of uts in division D is ( i P i ) 1. A division D = {P 1, P 2,..., P n } is envy-free, if for every plyer p i nd piee P j D the inequlity V i (P i ) V i (P j ) holds. The mjority of this pper is foused on the se, where eh vlution funtion is single intervl. For this se, we suppose tht for every plyer p i N, S vi = {I i }, where I i = [α i, β i ]. Furthermore, denote y T the set of vlution intervls, i.e., T = {I 1, I 2,..., I n }. In this setting, the envy-free notion for division D n e interpreted s follows: for eh plyer p i nd k i we hve I i,j I i I k,j I i. j j For set of intervls X, we define DOM(X) s the miniml intervl tht inludes ll memers of X s su-intervls; e.g., in the se tht eh vlution funtion is single intervl, for set T T we hve DOM(T ) = [min Ij T α j, mx Ii T β i ]. Furthermore, we define the density of X, denoted y Φ(X) s λ(x)/ X where λ(x) is the totl length of DOM(X) tht is overed y t lest one intervl in X. We ll set X of intervls solid, if for every point x DOM(X), there exists n intervl I in X suh tht x I. For exmple, in Fig 1, the set T is solid. For solid set T, we hve: λ(t ) = DOM(T ) = mx I i T β i min I j T α j. Our ssumption is tht every piee of the ke is vlule for t lest one plyer. In Setion 8, we demonstrte tht slightly modified versions of our lgorithms n hndle the situtions where this ssumption does not hold. We end this setion with simple oservtion. Oservtion 21 If + we hve: +d = e f holds for positive rel numers,,, d, e nd f, e f d e f

6 6 Msoud, Mjid, Mohmmd, Rez, Shh T = {,,, d} d DOM(T ) Φ(T ) = DOM(T ) /4 Fig. 1: Domin nd density Aording to Oservtion 21, If T is not solid, then there exists suset T T whih is solid nd Φ(T ) Φ(T ). We use Oservtion 21 in the proof of Lemms 8 nd 9. 3 The Expnsion proess The min tool in our method for dividing the ke is proedure lled expnsion proess. The expnsion proess expnds some ssoited intervls to the plyers, inside their desired re (i.e., vlution intervls). We use exp(t ) to refer to the expnsion proess on set T of (vlution) intervls. We initite the expnsion proess for T y ssoiting zero-length intervl I i t the eginning of its orresponding I i T, i.e., I i = [ i = α i, i = α i ]. Denote y S, the set of these Intervls. We expnd the intervls in S onurrently, ll from the endpoint. The expnsion is performed in wy tht mintins two invrints:(i) The expnsion hs the sme speed for ll the intervls so s the lengths of the intervls remin the sme nd (II) I i lwys remins within I i. During the expnsion, the endpoint of n intervl I i my ollide with the strting point of nother intervl I j. In this se, I i pushes the strting point of I j forwrd during the expnsion. The push ontinues to the end of the proess. If I i pushes I j, we sy I i is stuk in I j. Note tht y the wy we initite the proess, the intervls remin sorted ording to their orresponding α i s. In the speil se of equl α i for two plyers, the one with smller β i omes first. Definition 1 During the expnsion, n intervl I i eomes loked, if the endpoint of I i rehes β i. Definition 2 A hin is sequene of intervls I σ1, I σ2,..., I σk, with the property tht for 1 i < k, I σi is stuk in I σi+1. A hin is loked, if I σk is loked. The size of hin is the numer of intervls in tht hin. By definition, single intervl is hin of size 1.

7 Envy-free Mehnisms with Smll Numer of uts 7 Id I I I Id I I I A B d d Id I I I Id I I I D d d Fig. 2: An exmple of expnsion proess: (A): the intervls re single points, (B): I d strts pushing I, (): I strts pushing I, nd (D): I eomes loked nd the expnsion proess termintes The expnsion ends when n intervl eomes loked. The termintion ondition ensures tht the seond invrint is lwys preserved. In Figure 2, you n see detiled exmple of the expnsion proess. Definition 3 The expnsion proess for T is perfet, if the ssoited intervls over the entire DOM(T ). If the proess termintes due to loked intervl efore entirely overing DOM(T ), the proess is imperfet. Note tht if n expnsion proess on T ends perfetly, then for every ssoited intervl I i, we hve I i = Φ(T ). Oservtion 31 During the expnsion proess, every intervl I i is either eing pushed y nother intervl, or its strting point is still on α i. 3.1 Exeuting Expnsion Proess in Polytime Despite the ft tht we desried the expnsion proess ontinuously, it n e effiiently implemented vi sweeping the events. As sid, the expnsion proess strts with lloting zero-length intervl [ i = α i, i = α i ] to p i. We expnd the shres until one of these events our: 1. A hin eomes loked. 2. Two onseutive hins i nd i+1 get merged, i.e., the end-point of the lst shre intervl in i gets stuk in strting point of the first shre intervl of i+1. If the first event ours, we terminte the proess. For the seond event, we merge the hins i nd i+1 nd ontinue to expnd the shres. To expnd

8 8 Msoud, Mjid, Mohmmd, Rez, Shh the shres fter eh merging event, we should know the mximum length L tht we n expnd the shres, efore the next event oures. Let 1,..., m e the mximl hins in the urrent stte. For eh shre intervl I i, let i e its zero-sed index in the orresponding mximl hin, i.e., numer of intervls tht re pushing I i. Simultneous expnsion of ll the intervls y length l simply shifts eh intervl I i y i l, if no hin gets loked nor two mximl hins merge. We n find mximum l y onsidering ll possile events. L = min(min {β i i }, min i n i + 1 {gp( j, j+1 ) }) (1) j<m j where gp( j, j+1 ) denotes the size of urrently unlloted piee of the ke etween hins j nd j+1 nd j denotes the numer of intervls in j. After expnsion of the shres y size L, the proess either termintes y loked hin, or pir of hins get merged. onsidering the ft tht the merged hins will remin together during the expnsion proess, numer of merge events is ounded y m 1 < n. Thus, the expnsion proess n e simulted in poly(n). 4 EFISM: Speil Intervl Sheduling In this setion, we ssume tht the vlution funtion of eh plyer is single intervl. In ddition, we suppose tht the intervls hve the following property: i, j α i α j β i β j. (2) In other words, no intervl is su-intervl of nother (unless they strt or end in the sme ple). For exmple, when ll the vlution intervls hve the sme length, Eqution (2) is held. For this se, we present polynomil time, envy-free, nd truthful llotion with n 1 uts. We nme this lgorithm s EFISM. The ide of EFISM is repetedly expnding the intervls nd removing the loked hins. Let T e the vlution intervls orresponding to the plyers in N. We egin y lling exp(t ). As desried in Setion 3, the proedure termintes either perfetly or imperfetly. In the first se we re done. Otherwise, t lest one hin is loked. Let = I σ1, I σ2,..., I σk e loked hin with mximl size in S. Sine is mximl, no intervl is pushing I σ1. By Oservtion 31, σ1 is extly on α σ1. Let T e the set of vlution intervls orresponding to the intervls in. Lemm 1 DOM(T ) = [α σ1, β σk ]. Proof. By the struture of expnsion proedure, we know σ1 σ2... σk. Furthermore, y Eqution (2) we hve σ1 σ2... σk. By the ft

9 Envy-free Mehnisms with Smll Numer of uts 9 d A d B e f e f D e f e f f E Fig. 3: An exmple of expnsion nd removing proess: (A): Expnsion strts with single point intervls (B): I is loked nd the mximl hin I d, I gets removed (): Expnsion ontinues until I gets loked nd removed (D): I, I e gets loked nd removed (E): I f is loked nd removed. End of proess. tht I σk is loked nd regrding the definition of hin, we hve σ1 = α σ1 nd σk = β σk. Therefore, DOM( ) = [ min 1 j k σ j, mx 1 l k σ l ] = [α σ1, β σk ]. Now, we llote eh I σi to p σi. Lemm 2 sttes tht suh n llotion is envy-free for p σ1, p σ2,..., p σk. Lemm 2 For every intervl I σi nd I σj in, we hve V σi (I σi ) V σi (I σj ). Proof. By the seond invrint of the expnsion proess, we know tht I σi is entirely within I σi. Other intervls in hve the sme length s I σi nd hene, their vlue for plyer p σi n not e more thn I σi. Next, we remove plyers p σ1, p σ2,..., p σk from N. We lso remove DOM(T ) from. By removing DOM(T ), the ke is divided into two su-kes: the piee to the right of DOM(T ) nd the piee to the left of DOM(T ), respetively r nd l. Let N l nd N r e the set of plyers with their shre inside l nd r. Also, let T l nd T r e the sets of vlution intervls orresponding to N l nd N r. Now, we updte the vlution funtions of the plyers in l nd r. Speifilly, for every plyer p i N l with β i > α σ1, we hnge the vlue of β i to α σ1. Similrly, for every plyer p j N r with α j < β σk, we hnge α j to β σk. After removing the lloted prt of the ke long with its orresponding plyers nd updting the vlutions, we perform this expnsion nd removl

10 10 Msoud, Mjid, Mohmmd, Rez, Shh Algorithm 1 EFISM lgorithm funtion EFISM( = [, ], N, T ) if then exp(t ) Let = I σ1, I σ2,..., I σk e mximl loked hin for 1 i k do Allote I σi to p σi l = [, α σ1 ] r = [β σk, ] for every p k N do if k < σ1 then β k = min(β k, α σ1 ) Add p k to N l Add I k to T l else if k > σk then α k = mx(α k, β σk ) Add p k to N r Add I k to T r EFISM( l, N l, T l ) EFISM( r, N r, T r) Expnsion proess on T seprtely for oth T l nd T r. The proess ontinues until ll the plyers re removed. In Algorithm 1, you n find pseudoode for EFISM. In ddition, In Figure 3 you n find detiled exmple of EFISM. Theorem 1 EFISM is envy-free, truthful, nd uts the ke in extly n 1 lotions. Proof. Envy-freeness: We prove y indution on the numer of plyers, n. For n = 1, the lim holds trivilly. For n > 1, onsider the first expnsion proess. If the proess ws perfet, y Lemm 2, the llotion is envy-free. Otherwise, If the proess ws imperfet, let e mximl loked hin nd let N e the plyers orresponding to the intervls in. By Lemm 2, none of the plyers in N envy eh other. Furthermore, sine the entire DOM( ) is lloted, the plyers in N do not envy the plyers in N l nd N r. By indution hypothesis, the llotion of r nd l is envy free for N r nd N l, respetively. Sine l ( r ) hs no vlue for the plyers in N r (N l ), the plyers in N l nd N r do not envy eh other. Finlly, regrding the fts tht the expnsion proess for T l nd T r expnds the ssoited intervls t lest to the length of intervls in, the plyers in N l N r do not envy the plyers in N. Numer of uts: During the lgorithm, one piee of the ke is ssoited to eh plyer, whih mens tht the totl numer of uts on the ke is n 1. Note tht no piee of is left ehind. Truthfulness: For n ritrry plyer p i whose true vlution intervl is I i, we show his utility nnot e inresed y lying. However, first of ll, we must lerly determine wht do we men y lying?. Note tht the EFISM lgorithm is proposed for the se tht the vlution intervls stisfy the ordering property, i.e., stisfy Eqution (2). Regrding this,

11 Envy-free Mehnisms with Smll Numer of uts 11 p i n hnge his vlution in wy tht this property no longer is stisfied. Here, we show tht p i n not ern more pyoff even y misreporting his vlution in wy tht the ordering property no longer stisfies. Note tht this type of misreporting n hrm other properties of the lgorithm; For exmple, Figure 4 shows n exmple tht removing the ordering property results in n llotion tht is no longer envy-free. Let L e the length of lloted shre to p i in se he tells the truth nd let e the loked hin tht inludes I i. onsidering the expnsion proess s ontinuous proess tht the shres grow with the sme speed, let t e the (reltive) time, when eomes loked. Furthermore, let T e the set of vlution intervls relted to the shre intervls in. Aording to the struture of the expnsion proess, Φ(T ) = L, nd the size of every shre intervl in is extly L. Let α i, β i e the strting point nd the ending points of the intervl tht p i sttes. We wnt to show tht no mtter wht the vlue of α i nd β i re, there would e no senrio in whih p i hieves more thn L from DOM(T ). First, note tht for every suset T of T, Φ(T ) L, whih mens if p i tells the truth, no intervl in T eomes loked during the expnsion, efore time t. Thus, if p i lies in wy tht loked hin forms efore time t, I i would e one of the intervls in. In other words, p i n not fore other intervls in T to lok erlier, without prtiipting in the loked hin. On the other hnd, if p i stisfies in ny time erlier thn t, his shre would e less thn L. Hene, the shre of ll the plyers in remins t lest L, even if p i lies. Moreover, DOM(T ) = L T whih mens tht p i n not gin n intervl with size more thn L T L ( T 1) = L from DOM(T ), independent of wht he sumits s his vlution intervl. Regrding the ft tht if he reports truthfully I i = L, he n not gin more pyoff y lying. Remrk tht removing the ordering property desried in the eginning of this setion my result in n inpproprite llotion. For exmple, onsider the input desried in Figure 4. Oviously, running EFISM on this input does not yield n envy-free llotion; here p envies p. In ddition, the llotion does not llote the entire ke, euse piee etween I nd I is left over. To overome these issues, in the next setion we introdue more generl form of the expnsion proess 5 Expnsion Proess with Unloking In this setion, we introdue more generl form of the expnsion proess. The si ide is the ft tht during the expnsion proess, there might e some ses tht loked hin eomes unloked y re-permuting some of its intervls, wihtout violting the expnsion invrints.

12 12 Msoud, Mjid, Mohmmd, Rez, Shh I I I Fig. 4: EFISM for intervls without ordering property Definition 4 Let = I σ1, I σ2,..., I σk e mximl loked hin. A permuttion I δ1, I δ2,..., I δr of the intervls in is sid to e -unloking, if the following onditions re held: (I) : All the intervls of the permuttion re memers of the loked hin, i.e., i, I δi, nd the lst intervl of the permuttion is the loked intervl, i.e., δ r = σ k. (II) : For every j < r, the shre ssoited to plyer p δj is totlly within the vlution intervl of plyer p δj+1 (with its endpoint stritly less thn the endpoint of the vlution intervl), i.e., 1 j r 1, δj α δj+1 nd δj < β δj+1. (III) : The shre ssoited to plyer p δr is within the vlution intervl of plyer p δ1 (with its endpoint stritly less thn the endpoint of the vlution intervl), i.e., α δ1 δr nd β δ1 > δr. The intuition ehind the definition of unloking permuttion is s follows: let I δ1, I δ2,..., I δr e -unloking permuttion, where = I σ1, I σ2,..., I σk. Then, we n hnge the order of the intervls in y pling I δj in the lotion of I δj 1 for 1 < j r nd pling I δ1 in the lotion of I δr. By the definition of unloking permuttion, fter suh opertions I δr (I σk ) is no longer loked. Thus, I σk is not rrier for the expnsion proess nd the expnsion n e ontinued. It is worthwhile to mention tht there my e multiple loked intervls in moment. In suh se, we seprtely try to unlok eh intervl. For set T of vlution intervls, we use U-exp(T ) to refer to the expnsion proess with unloking for T. See Figure 5 for n exmple of this proess. Definition 5 A loked hin = I σ1, I σ2,..., I σk dmits no unloking permuttion. is strongly loked, if Definition 6 An expnsion proess with unloking U-exp(.) is strongly loked, if t lest one of its hins is strongly loked. Definition 7 A permuttion grph for loked hin = I σ1, I σ2,..., I σk is direted grph G V, E. For every intervl I σi, there is vertex v σi in V. The edges in E re in two types E l nd E r, i.e., E = E l E r. The edges in E l nd E r re determined s follows: (I) For eh I σi nd I σj, the edge (v σi, v σj ) is in E l, if i > j nd α σi σj.(ii) For eh I σi nd I σj, edge (v σi, v σj ) is in E r, if i < j nd β σi > σj.

13 Envy-free Mehnisms with Smll Numer of uts 13 I I I I I I A B I I I Fig. 5: An exmple of Expnsion proess with Unloking See Figure 6 for n exmple of permuttion grph. A trivil neessry nd suffiient ondition for hin to e strongly loked is tht G ontins no yle inluding v σk. However, regrding the speil struture of G, we n define more restrited neessry nd suffiient ondition for strongly loked sitution. Definition 8 A direted irle in G is one-wy, if it ontins extly one edge from E r. Note tht no yle in G n ontin only the edges from one of E l or E r. In Lemm 3, we use one-wy yles to give neessry nd suffiient ondition for hin to e strongly loked. Lemm 3 A hin = I σ1, I σ2,..., I σk dmits no one-wy yle ontining v σk. is strongly loked, if nd only if G Proof. To prove the non-trivil diretion it suffies to show the existene of one-wy yle for ny non-strongly loked hin. Let = v δ1=σ k, v δ2,..., v δr=σ k e the shortest yle in G tht inludes v σk. We lim tht ontins extly one edge from E r (tht is the edge (v δr 1, v σk )). As ontrdition, let (v δi, v δi+1 ) e the first edge in E r tht ppers in. By definition of E r, we hve δ i+1 > δ i. Furthermore, y definition of E l, we hve: δ 1 > δ 2 >... > δ i < δ i+1. Hene, δ j > δ i+1 > δ j+1 for some 1 j i 1. Note tht sine v δj hs left edge to v δj+1, it hs n edge to v δi+1 s well (see the struture of the expnsion grph in Figure 6). Thus, = v σk =δ 1, v δ2,..., v δj, v δi+1,..., v δr=σ k is shorter yle tht inludes v σk, whih is ontrdition.

14 14 Msoud, Mjid, Mohmmd, Rez, Shh I I I Id Ie d e Fig. 6: An exmple of permuttion grph. Here the loked hin I, I, I, I d, I e n e unloked y repermuting I, I, I, I d, I e to I, I e, I, I, I d 5.1 Expnsion Proess With Unloking n e Exeuted in Polytime In this setion, we extend the simultion desried in 3.1 for loked sitution. Similrly, in eh itertion, we expnd ll the shres y size L s determined in Eqution (1). The differene is: when enountering loked hin, regrding Lemm 3, we serh for n unloking one-wy yle (in polynomil time) nd re-permute the shres through it. The proess termintes when hin eomes strongly loked. As mentioned in Setion 3.1, the numer of expnsions tht join two mximl hins is ounded y n. Thus, it only remins to ound the numer of lternting expnsions nd unloking proesses etween mximl hin merging stges. onsider n unloking proedure, tht unloks hin loked y shre I σi tht is positioned t σi (s defined in Setion 3.1). We lel this unloking y the pir (σ i, σi ). The point is tht no other unloking in this mximl hin will e leled y pir (σ i, σi ). We prove this y showing tht fter the unloking proess, I σi goes to new ple nd never returns k to σi. This is due to the ft tht fter the unloking nd its susequent expnsion, the endpoint of σi rosses β σi nd hene, I σi n never return k to ple σi y ny unloking re-permuttion. Sine no two unloking proedures n hve the sme lel, the totl numer of unloking proedures is upper ounded y O(n 2 ). Thus, one n simulte the Expnsion proess with Unloking in polytime. 6 Generl Intervl Sheduling In this setion, we ssume tht the vlution funtion of eh plyer is n intervl, without ny restrition on the strting nd ending points of the intervls. For this se, we suggest n envy-free nd truthful llotion tht

15 Envy-free Mehnisms with Smll Numer of uts 15 e d x e d Fig. 7: The ke nd the intervls,,, d nd e efore nd fter shrinking intervl x. uses less thn 2n uts. Our lgorithm for finding proper llotion is sed on the expnsion proess with unloking. Generlly speking, we itertively run U-exp(.) on the remining plyers shres. This proess llotes the entire ke or stops in strongly loked sitution. We prove some desirle properties for this sitution nd leverge these properties to llote piee of the ke to the plyers in the strongly loked hin. Next, we remove the stisfied plyers nd shrink the lloted piee (s defined in Definition 9) nd solve the prolem reursively for the remining plyers nd the remining prt of the ke. Definition 9 Let e ke nd I = [I s, I e ] e n intervl. By the term shrinking I, we men removing I from nd gluing the piees to the left nd right of I together. More formlly, every vlution intervl [α i, β i ] turns into [f(α i ), f(β i )], where x x < I s f(x) = I s I s x I e x I e + I s I e < x (see Figure 7). As wrm-up, we ignore the truthfulness property nd show tht the expnsion proess with unloking yields n envy-free llotion with 2(n 1) uts. 6.1 EFS: Envy-free llotion with 2(n 1) uts In this setion, we propose EFS, whih is polynomil time envy-free lgorithm with 2(n 1) uts. Our lgorithm is s follows: in the eginning, we run U-

16 16 Msoud, Mjid, Mohmmd, Rez, Shh exp(t ). The proess either ends perfetly nd the desired llotion is found, or strongly loked hin ppers. For the se tht the proess ends imperfetly, let = I σ1, I σ2,..., I σk e mximl strongly loked hin. Now, onsider G. By Lemm 3, G ontins no one-wy yle. Let l e the minimum index, suh tht there is direted pth from v σk to v σl using the edges in E l. Regrding the speil struture of G, Lemm 4 holds. This lemm is used to prove the envy-freeness of EFS. Lemm 4 There is direted pth from v σk to every vertex v σl with l > l, using the edges in E l. Proof. Let P = v σk =δ 1, v δ2,..., v δr=σ l e the pth from v σk to v σl using the edges in E l. By definition of the left edge, σ k = δ 1 > δ 2 >... > δ r = σ l. Note tht if σ l = δ j for some j, then we re done. Otherwise, sine l > l, δ j < σ l < δ j+1 for some 1 j < l. By the struture of the grph, v δj hs left edge to v σl. Thus, v σk =δ 1, v δ2,..., v δj, v σl is the desired pth etween v σk nd v σl. Bsed on Lemm 4 nd the ft tht G ontins no one-wy yle, there is no edge from v σl to v σk in E r for ny l l, whih mens: l l β σl σk. (3) On the other hnd, there is no pth from v σk to v σl for l < l, tht is: l l α σl > σl 1. (4) We now llote every intervl I σl to p σl for l l k, remove plyers p σl, p σl+1,..., p σk from N, nd shrink the intervl [ σl, σk ]. Next, we ontinue the expnsion with unloking proess for the remining plyers nd the remining prt of the ke. The itertion etween expnsion proess with unloking nd lloting the ke in strongly loked sitution goes on, until the entire ke is lloted. Theorem 2 EFS is envy-free nd uts the ke in t most 2(n 1) lotions. Proof. Envy-freeness: Let p i nd p j e two ritrry plyers. We wnt to show tht p i does not envy p j. To show this, let r i nd r j e the steps, tht piee of the ke is lloted p i nd p j, respetively. There re three possiilities: (I)r i < r j, (II)r i = r j nd (III)r i > r j. For r i < r j, regrding Inequlity 4, I j I i would e less tht I i, whih mens p i does not envy p j. If r i = r j, onsidering Lemm 2, p i does not envy p j. Finlly, for the se r i > r j, y the ft tht the totl size of the shre lloted to p i is stritly lrger thn the one lloted to p j, p i does not envy p j. Numer of uts: We use indution to prove tht the numer uts is t most 2(n 1). For n = 1, the expnsion proess llotes single piee to the

17 Envy-free Mehnisms with Smll Numer of uts 17 I I I(ont) I I rel shre Fig. 8: Plyer n inrese his shre y misreporting plyer tht requires no ut. Now, suppose tht the proposition is true for every n < n. We prove it for n plyers. First, note tht if the proess U-exp(.) ends suessfully, the ke is divided vi n 1 uts. Thus, ssume tht the proess is in strongly loked sitution. Let = I σ1, I σ2,..., I σk e strongly loked hin nd let l e the minimum index suh tht there is pth from v σk to v σl in G. In the lgorithm we llote I σl, I σl+1,..., I σk to plyers p σl, p σl+1,..., p σk, respetively, nd shrink the lloted prt. Every plyer in {p σl, p σl+1,..., p σk } gets single piee of the ke ((k l) uts). Furthermore, y indution hypothesis, the remining prt of the ke requires t most 2(n k + l 2) uts. onsidering the uts in the eginning nd the ending points of the shres relted to, we hve 2(n k + l 2) + (k l) + 2 = 2(n 1) k + l 2(n 1) uts. 6.2 EFGISM: truthful, envy-free llotion with 2(n 1) uts It is worth mentioning tht EFS is not truthful. onsider the exmple in Figure 8. It n e oserved tht plyer n inrese his shre y misreporting α. In this setion, we try to resolve this issue. Our strtegy to del with this diffiulty is to run U-exp(.) only for speil suset of plyers in every step. Lemm 5 onstitutes the ore of our method. Lemm 5 Let T e set of intervls, with the property tht for every T T, Φ(T ) > Φ(T ) (we ll suh set s irreduile). Then we n divide DOM(T ) into t most 2 T 1 piees nd ssoite them to the intervls, suh tht:(i) totl length of the piees ssoited with ny intervl is extly Φ(T ). (II) the piees lloted to ny intervl is totlly within tht intervl. Proof. We use indution on T. For T = 1 the lim trivilly holds: we n ssoite DOM(T ) to the intervl in T tht needs no ut. Suppose tht the proposition is true for T < k. We prove it for T = k. onsider U-exp(T ). If U-exp(T ) ends perfetly, then we re done. Otherwise, let = I σ1, I σ2,..., I σk

18 18 Msoud, Mjid, Mohmmd, Rez, Shh e mximl strongly loked hin fter the proess. onsidering G, let l e the minimum index, suh tht there is direted pth from v σk to v σl using the edges in E l. In Lemm 6 we show l is stritly greter thn 1. This ft is lter used to rek the prolem into two stritly smller su-prolems nd solve eh su-prolem reursively. Lemm 6 Let = I σ1, I σ2,..., I σk e mximl strongly loked hin fter running U-exp(T ) nd let l e the minimum index, suh tht there is direted pth from v σk to v σl in G using the edges in E L. Then, we hve l > 1. Proof. By ontrdition, suppose l = 1. Regrding Eqution (3), for ll i we hve β σi σk. Now, we show tht for ll i, α σi σ1. Let nd x = min 1 i k α σ i j = rg min 1 i k α σ i. In the eginning of the expnsion proess, the strting point of I σj is on x. During the expnsion, there my e situtions tht I σj is repled y nother intervl I σj, ut y the definition of unloking permuttion, α σj = x. Furthermore, the strting point of the rer intervl in lwys remins on x, euse is mximl nd hene, no intervl pushes the intervl with the strting point on x. Therefore, in strongly loked sitution, α σ1 = x nd for everey i > 1, we hve α σi x. Thus, DOM(T ) = [ σ1, σk ], where T is the vlution intervls orresponding to the shre intervls in. Sine, [ σ1, σk ] is overed y, Φ(T ) = I σ1 = I σ2 = = I σk. On the other hnd, sine U-exp(T ) ws imperfet, I σk < Φ(T ), whih ontrdits the irreduilinty of T. By Lemm 4, we know tht Equtions (3) nd (4) re held for. Now, let x = β σk (k l + 1)Φ(T ). (5) In Lemm 7, we show tht the lotion of x in [0, 1] is some vlue etween σl 1 nd σl. This ft, llows us to rek DOM(T ) into two piees, oth of whih preserve the properties defined in Lemm 5. Lemm 7 Let x = β σk (k l + 1)Φ(T ), where l is the minimum index, suh tht there is direted pth from v σk to v σl. Then, we hve σl 1 < x < σl. Proof. Regrding Inequlity (3) nd the ft tht σk = β σk (note tht is strongly loked) we hve Furthermore, y Inequlity (4), l i k β σi β σk. l i k α σi > σl 1.

19 Envy-free Mehnisms with Smll Numer of uts 19 Now, let T = {p σl, p σl+1,..., p σk }. We hve: Φ(T ) = DOM(T ) T On the other hnd, regrding Eqution (5), Φ(T ) = β σ k x k l + 1. < β σ k σl 1 k l + 1. onsidering the ft tht T hs the minimum possile density, x > σl 1. Sine U-exp(T ) ws imperfet, we hve: I σ1 = I σ2 =... = I σk < Φ(T ). Hene, x = σk (k l + 1) Φ(T ) < σk (k l + 1) I σk < σl 1 = σl. Note tht sine the intervl I σl 1 is stuk in I σl, we hve σl 1 = σl. Now, we show tht the piee [x, β σk ] n e lloted to plyers p σl, p σl+1,..., p σk using 2(k l + 1) 2 uts. For this, onsider the vlution intervls T = {I σ l, I σ l+1,..., I σ k } suh tht: l i k I σ i = [mx(x, α σi ), β σi ] Note tht DOM(T ) = [x, β σk ] nd hene, Φ(T ) = β σ k x k l + 1 = σ k x k l + 1 Regrding Inequlity (5), Φ(T ) = Φ(T ). (6) Lemm 8 T is irreduile, i.e., for ll T T, we hve Φ(T ) > Φ(T ). Proof. We sy intervl I σ i is trimmed in T, if α σi < x. Note tht for every intervl I σ i tht is not trimmed, I σ i is extly I σi. Hene, if T ontins no trimmed intervl, regrding irreduiility of T, we hve Φ(T ) > Φ(T ) = Φ(T ). Thus, it only remins to prove the lim for the se tht T ontins trimmed intervl. By Oservtion 21, we n ssume tht DOM(T ) is solid. Let j e the mximum index, suh tht I σ j T. Sine I σj is entirely within I σj, β σj σj. Sine σ j = σj, DOM(T ) = [x, σj ]. On the other hnd, T ontins t most j l + 1 intervls. By Inequlity (6), Φ(T ) = σ k x k l + 1

20 20 Msoud, Mjid, Mohmmd, Rez, Shh nd Thus, we hve Φ(T ) σ j x j l + 1. Φ(T ) = σ k x k l + 1 = ( σ j x) + ( σk σj ) (j l + 1) + (k j) (7) Note tht ( σk σj ) (k j) = I σ1 = I σ2 =... = I σk < Φ(T ) = Φ(T ). (8) omining Inequlities (7) nd (8) together with Oservtion 21 yields: ( σj x) (j l + 1) > Φ(T ) Φ(T ) > Φ(T ). Lemm 8 sttes tht the set of intervls in T dmit the properties desried in Lemm 5. Furthermore, regrding Lemm 6, T is strit suset of T. By indution hypothesis, we know tht one n ut DOM(T ) into t most 2(k l + 1) 2 piees nd llote them to plyers p σl, p σl+1,..., p σk suh tht oth the properties in Lemm 5 re stisfied. Denote y N T, the plyers with vlutions in T. We shrink DOM(T ) nd remove the plyers p σl, p σl+1,..., p σk from N T. Lemm 9 ssures tht the onditions in Lemm 5 re held for the remining ke nd remining plyers. Lemm 9 Let T e the intervls relted to the plyers in N T = N T \ {p σl, p σl+1,..., p σk } fter shrinking DOM(T ). Then, T is irreduile nd Φ(T ) = Φ(T ). Proof. We hve Φ(T ) = Φ(T ), whih mens: Thus, DOM(T ) T DOM(T ) DOM(T ) T T = DOM(T ) T. = Φ(T ) = Φ(T ), tht is Φ(T ) = Φ(T ). ll n intervl I i shrinked, if I i DOM(T ). onsider set ˆT T of intervls. If ˆT ontins no shrinked intervl, then Φ( ˆT ) hs the sme vlue s efore shrinking DOM(T ), whih y irreduiility, we hve: Φ( ˆT ) > Φ(T ) = Φ(T ).

21 Envy-free Mehnisms with Smll Numer of uts 21 On the other hnd, ssume tht ˆT ontins t lest one shrinked intervl. Let ˆT e the set of intervls in ˆT T efore shrinking T. Sine ˆT hs shrinked intervl, ˆT is solid. By irreduiility, we know: Φ( ˆT ) > Φ(T ) = Φ(T ) = Φ(T ). On the other hnd, DOM( ˆT ) = DOM( ˆT ) DOM(T ). Thus, we hve: Φ( ˆT ) = DOM( ˆT ) ˆT = DOM( ˆT ) DOM(T ) ˆT T Regrding Oservtion 21, we hve Φ( ˆT ) > Φ(T ) = Φ(T ). Aording to Lemm 9, we n use indution hypothesis to show tht T n e lloted to the plyers in N T vi 2(l 1) 2 uts. Totl numer of uts would e 2(l 1) 2 + 2(k l + 1) 2 = 2k 4 uts plus two uts on x nd β σk tht results in 2k 2 uts. Bsed on lemm 5, we introdue EFGISM s follows: mong ll susets of N, we find suset whose orresponding intervls hve the minimum density (nd the set with minimum size, if there were multiple options). Let N e this suset nd let T e the intervls orresponding to the plyers in N. In Lemm 10, we show tht T (nd onsequently N) n e found in polynomil time. Lemm 10 Let N e suset whose orresponding intervls hve the minimum density nd let T e the set shre intervls whih hve the minimum density.then, T n e found in polynomil time. Proof. There re n 2 different possile hoies for the strting point nd the ending points of DOM(T ). By fixing these points, in order to minimize the density of T, we must dd ll the intervls tht re within the seleted domin to T. Thus, the set T n e found y minimizing over ll possile hoies of domin. Sine T hs the minimum possile density, T is irreduile. Hene, we n llote to every plyer in N, piee from DOM(T ) with the properties defined in Lemm 5. Afterwrds, we remove the plyers in N from N nd shrink DOM(T ) from. Next, we reursively llote the remining piee of the ke to remining plyers using EFGISM. In Algorithm 2 you n find pseudoode for EFGISM.

22 22 Msoud, Mjid, Mohmmd, Rez, Shh Algorithm 2 EFGISM lgorithm funtion EFGISM(N, T, ) if then T = rg min T T Φ(T ) N = plyers with intervl in T Allote(N, DOM(T )) By Lemm 5 Shrink(, DOM(T )) T is lso updted EFGISM(N \ N, T, ) Theorem 3 EFGISM is envy-free, truthful, nd uses t most 2(n 1) uts. Proof. Envy-freeness nd truthfulness: We redit the proof for envyfreeness nd truthfulness to [11]. In [11], the following sttement is proved (restted for the se of intervls): Let A e the lgorithm tht in eh step finds set T of intervls with minimum Φ(T ) nd llotes it to the gents orresponding to the intervls in T, suh tht every gent gets shre of size Φ(T ) tht is totlly within his vlution intervl; then A is envy-free nd truthful. Regrding the ft tht our lgorithm hs the sme struture s stted ove, the lgorithm is envy-free nd truthful. Numer of uts: We use indution to prove tht the lgorithm uts the ke in t most 2(n 2) lotions. For n = 1, the lgorithm trivilly llotes the entire ke to the plyer whih needs no ut (2 (1 1)). Now, onsider the first step of the lgorithm, when set T of intervls re seleted. By Lemm 5, we ut T into t most 2(T 1) points. Furthermore, we shrink DOM(T ) nd solve the prolem reursively for the remining prt. By indution hypothesis, the reursive prt needs t most 2(n T 1) uts. In ddition, two uts re needed in the eginning nd ending point of DOM(T ). Thus, totl numer of uts would e: 2(T 1) + 2(n T 1) + 2 = 2(n 1). 7 Pieewise onstnt Funtions In this setion, we study more generl se of the prolem in whih the vlution funtions of the plyers re pieewise onstnt. Denote y m the mximum numer of intervls tht every vlution funtion n hve; tht is, for every plyer p i, S i m. Here, we ssume tht for every p i, S i = m. This is without loss of generlity, sine one n rek n intervl into severl su-intervls. Thus, for every plyer p i, we suppose S i = {I i,1, I i,2,..., I i,m }.

23 Envy-free Mehnisms with Smll Numer of uts 23 This setion onsists of two prts. In the first prt, we show tht for onstnt numer of plyers, one n find n envy-free llotion with n 1 uts in time poly(m). Next, in the seond prt, we utilize the expnsion proess with unloking to devise poly(n, m) time envy-free lgorithm with O(nm) uts on the ke. Rell tht finding n envy-free llotion with n 1 uts for n plyers is PPAD-omplete even for the se of n = 3 [12]. In Theorem 4, we show tht for onstnt numer of plyers with pieewise onstnt vlution, the prolem n e solved in time poly(m). Theorem 4 An envy-free llotion with n 1 uts n e found for onstnt numer of plyers whose vlution funtions re pieewise onstnt with m steps in time poly(m). Proof. Firstly, note tht from [18] we know tht there exists n envy-free llotion with n 1 uts. In suh n llotion, there re n 1 utting points. Let n 1 1 e those utting points. In ddition, the vlution of eh plyer n e speified y 2m onstnt points (2 onstnt points for eh step) nd m onstnt vlues whih desrie the density of eh step. Therefore, there re t most 2mn onstnt points on the ke in wy tht eh plyer likes the ke etween two onseutive onstnt points uniformly. In other words, the density vlue of the ke etween two onseutive onstnt points is uniform vlue for eh of the plyers. Now, if we know the rnge of eh utting point (it n e etween whih of the two onseutive onstnt points), then we n write the vlue of the i th piee reted y the utting points ([ i 1, i ]) for eh plyer p j s liner omintion of the utting points. However, in order to stisfy envy-freeness, we lso need to know how the piees will e lloted to the plyers. If we know ll this informtion is given, then we n formulte the prolem s liner progrm (n(n 1) onstrints for envy-freeness, n 1 onstrints gurntees n 1 1, nd other onstrints fix the rnge of the utting points). Any fesile solution to the liner progrm is n envy-free llotion with n 1 uts. If we ouldn t find fesile solution for one liner progrm then we need to hek the next possiility of the rnge of the utting points nd llotion of the piees. In the worst se, we need to hek every possiility whih mens tht we need to solve n (2mn+n 1)! (2mn)! = O(m n ) different liner progrms. Finlly, we know tht suh n llotion exists nd t lest one of these liner progrms finds fesile solution. Hene, for onstnt n, y solving polynomil numer of different liner progrms, we n find n envy-free llotion. In the seond prt, we exploit the expnsion method with unloking to find proper llotion. Here, we ssume tht the vlution funtions hve speil property, nmely, intersetion property. Denote y R i,j,k the set of intervls in S k tht hve non-empty intersetion with I i,j. We suppose tht for every vlution intervl I i,j nd every plyer p k (k i), R i,j,k 1. For this se, we prove Theorem 5.

24 24 Msoud, Mjid, Mohmmd, Rez, Shh Theorem 5 Let N e set of plyers whose vlution funtions re pieewise onstnt with m steps. Assuming tht the intersetion property holds, there exists poly(m, n) time llotion lgorithm tht is envy-free nd uts the ke in O(nm) ples. Proof. onsider n instne of the prolem with nm plyers where the vlution funtion of plyer p i,j is I i,j. We run EFGISM for this instne. By the properties of EFGISM, we know tht the resulting llotion is envy-free nd uts the ke in t most 2(nm 1) lotions. Let P i,j e the set of intervls lloted to p i,j in EFGISM. We show tht the llotion tht llotes P i = 1 j m P i,j to plyer p i is lso envy-free. To prove envy-freeness, we use struturl property of the expnsion proess: y the first invrint of the expnsion proess, the finl solution llotes to every plyer p i,j set of piees tht re totlly within I i,j. In ddition, note tht for intervl I i,j, R i,j,k 1 holds for every plyer p k. We hve V i (P i ) = 1 j m V i(p i,j ) nd V i (P k ) = 1 j m V i(p k,j ). Furthermore, y the intersetion property, t most one vlution intervl of p k, sy I k,l hs non-empty intersetion with I i,j. By the envy-freeness of EFGISM, we know tht p i,j prefers his shre to the shre lloted to p k,l, tht is V i,j (P i,j ) V i,j (P k,l ). Regrding the ft tht I i,j I k,l = for ll l l, we hve V i,j (P i,j ) l V i,j(p k,l ). Thus, V i,j (P i,j ) V i,j (P k,l ) j j l V i (P i ) V i,j (P k,l ). (9) j The right hnd side of Eqution (9) is t lest V i (P k ). l 8 Expnsion Proess with Unloking in Prtie In ddition to otining worst-se gurntee on the numer of uts, we wish to evlute the ehvior of the expnsion proess with unloking method in prtie. Our experiments im t illustrting the performne of the expnsion nd unloking method from the spet of the numer of uts. As we show in this setion, the expnsion nd unloking method hieves signifint performne in prtie. In Setions 3.1, nd 5.1 we riefly disussed the implementtion detils of the expnsion nd unloking proesses. We implemented oth the proesses nd the odes re ville in e.shrif.edu/~m_frhdi/soure. As mentioned in Setion 2, our si ssumption is tht every piee of the ke is vlule to t lest one plyer. In Lemm 11 we show tht this ssumption is w.l.o.g sine every lgorithm for the former n e extended to the se tht the ke hs zero-vlued piees, without ny dditionl uts. In ft, Lemm 11 demonstrtes how to prevent dditionl uts on the ke when prts of it hve zero vlution to ll the plyers.

25 Envy-free Mehnisms with Smll Numer of uts 25 pn p1 p2 p3 pn 1 Fig. 9: The hthed piees re lloted to p n in EFGISM Lemm 11 The ssumption tht every piee of the ke is vlule to t lest one plyer is w.l.o.g. Proof. Let {Z 1,..., Z q } e the set of mximl intervls tht re zero vlued for ll the plyers. Shrink every Z i in into point z i to hieve new ke in whih our vlution ssumption holds. Now, utilizing the lgorithms proposed in this pper (or ny other llotion lgorithm), we divide mong the plyers. Next, for eh Z i, we llote it to the plyer tht reeived (left / right neighorhood) of z i. Z i will introdue single ut (t its right / left endpoint) in only if ontins ut t z i. Thus, the numer of uts is identil for nd. Finlly, onsidering the ft tht eh Z i hs zero vlution for ll the plyers, one n esily verify tht envy-freeness nd truthfulness of the mehnism re preserved. 8.1 Study Design To test the performne of the expnsion nd unloking proedure, we rn set of experiments on the EFS method. We generted ke utting instnes with etween 2 nd 500 plyers. For every 2 i 500, we generted 4 tests ontining i plyers. For every test, we drw the vlution intervl of every plyer uniformly from (0, 1). To generte rndom intervl, we seprtely smpled the end-points from U(0, 1). For every test, we lulted two ojetives: tot : totl numer of uts tht re mde on the ke y EFS. mx : the mximum numer of piees lloted to ny plyer y EFS. Note tht, despite the ft tht the verge numer of piees lloted to ny plyer in EFGISM is less thn 2n/n = 2, there re situtions in whih plyer my reeive Ω(n) piees. For exmple, onsider the instne in Figure 9. Sine in EFS the shre lloted to every plyer is totlly within his vlution intervl, the hthed prts of the ke will e lloted to p n, whih mens his shre inludes n piees of the ke. We ompre tot nd mx otined from every test to the vlues in the optiml solution. We lredy know tht in the optiml llotion, the numer of uts is n 1 nd every plyer reeives ontinuous piee of the ke. Therefore,