Call Admission Control Algorithm for pre-stored VBR video streams

C. T ryf onas · D . Pa pamichail · A. Mehler · S. S. Skiena Call a dmission control algorithm f or pre-stor ed VBR vid eo str eams Abstract W e examine the problem of ac cepting a new request for a pre-stored VBR video strea m that has been smoothed using a ny of the smoothing algorithms found i n the lite rature. The output of these a lgorithms is a piecewise constant-rate schedule for a V ariable Bit-Rate (VBR) stream. The sche dule guarantees that the decoder b uf fer does not o verflo w or underflo w . The problem a ddressed i n this paper is the det ermination of the minimal time displacement of each ne w requested VBR stream so that it can b e accomodated by the network and/or the video server without overbook ing the committed tra f fic. W e prove that this call -admission control problem for multiple requested VBR streams is NP-complete and inapproximable within a constant factor, by reducing it from the V E RT E X C O L O R problem. W e a lso present a deterministic morp hology-sensiti ve algorithm that ca lculates t he mi nimal time displac ement of a VBR stream request. The complexity of the proposed al gorithm make it suitabl e for real -time dete rmination of t he ti me displaceme nt parameter during the call admission phase. Keyw ords V a riable Bit-Rate Stream, Call-Admission C ontrol, T ime Displacement, 3SUM hard, constant factor inapproximable 1 Introduction A signific ant portion of the forecasted network traffic is expected t o be multime dia (e.g. v oice and video) traffic. New services such as video-on-demand (V oD) and TV broadcasting are currently under massiv e de- ployment. One of the salient c haracterist ics of video traffic is t hat it usual ly exhibits high variability in its bandwidth demands in dif ferent t ime scales. The need to better understand t he bandwidth demands of video streams is essenti al for prop er resource provisioning of both the network resources and the resources of t he video se rvers when st ored video is transported. Pro per resource dimensioning has direct correla tion with the quality of the rec ov ered video on the decoder and, t herefore, a v ariety of te chniques ha v e been proposed i n the past. Significant w ork has bee n don e in the literature i n the a rea of stat istica l modeling of video traffic for re- source provisioning purposes, so that it can be ef fectiv ely transported ov er packet-switched networks [2, 3, 4, 9, 11]. In most c ases, the objective of t hese efforts is to b uild a general model that can be used for resource dimensioning for all the video traffic transported o ver the networ k. In some cases, the long -range depend ence (LRD) characteri stic o f video traffic is exploited to create a model of the traffic source [2, 8, 12]. These meth- ods, in general, c haracterize the traf fic source based on its statist ical properties, and pro vide v alue when the video stream is not kno wn a-priori. Howe ver , when dealing with pre-stored video, the resource dimensioning process can be made det erministic and any statist ical te chnique is of limi ted v alue since it does not ca pture the exact dy namics of the video strea m in the time do main. C. T ryfonas Kazeon Systems, Inc., 1161 San Antonio Road , Mountain V ie w , CA 94043, USA E-mail: tryfonas@kazeon.com D. Papamich ail Computer Science Department, Uni versity of Miami, Coral Gables, FL 33146, USA E-mail: dimitris@cs.sun ysb.e du A. Mehler · S. S. Skiena Computer Science Depar tment, SUNY at Stony Brook, Ston y Brook, NY 11794 , USA E-mail: { mehler | skiena } @cs.sun ysb.ed u 2 In video applications t hat transport stored video o ver a packet-switched network, the resource provision- ing process can ta ke advantage of the fact that video strea ms ca n be pre-processed offline. During the pre- processing of a video stream, a transmission schedule is typical ly computed to m inimize its rat e variability and, there fore, facili tate the resource provisioning and the call a dmission c ontrol process. The reduction in rate variability is done by work-ahead smoothing, i.e. sending more data to t he re ceiv er with respect to its playback tim e. Significant work ca n be found in the lit erature in t he area of wo rk-ahead video smoothing [6, 10, 13, 15]. The general idea behi nd most of these algorithms i s to maximize the time intervals (rate se gments) at which a tra nsmission rate fo r the video stream is used without ca using under/overflo w of the re ceiv er bu ffer . The algorithms differ in the selec tion of the starti ng point of these rate segments. The output of these algo- rithms is a piecewise constant-rate schedule for the smoothed video stream. The schedule guarante es that the decoder buf fer does not overflo w or underflo w . Due to the fact that computing the smoothing schedule is not a trivial process and c annot be performed online, the pre-computed smoothed schedules of a video fil e for v a rious decoder profiles can be stored along with the file i tself in the video serv e rs, so that they can be used at the time of the corresponding video request to guarantee a deterministi c quality at the decoder . The problem addressed by this paper i s that of acc epting a new request for a pre-stored VBR video stream that has been pre-smoothed using any of the smoothing a lgorithms. Si nce the re quest c an come at any parti cular point in ti me, the problem is relat ed to the accomodation of the new request prov ided that t he en velope of the dynamics of the committe d traffic, a nd t herefore, t he env el ope of the av aila ble bandwidth in t he channel, do e s not introduce overallocation at any time interval (see Fig. 1). The goal is to displace the pre-computed smoothed schedule of the n ew request into the fu t ure to a void overallocation. More specifically , we want to find the mi nimum time displac ement of the new schedu le so that t he channel can accomodate the new request. T he problem des cribed is a n optimizat ion problem that can be extended in several ways. For example, given a set of requests, find the displace ment points of the associat ed schedules so that the overall schedule is the smoothest. In this paper we prese nt two algorithms that solve the p roblem of computing the minimum dispacement of a new request, also referred as the T W O S T R E A M S C H E D U L I N G problem ( 2-SS ): (i) a si mple algorithm with O ( n 2 l o gn ) complex ity , and (ii) a morphology-sensiti ve algorithm wit h lo wer computational complexity . The morpho l ogy-sensitiv e algorithm makes specific observ ations about the smoothed schedule so that certain peaks can be skipped by the algorithm to speed-up the final calculation considerably , depending on the input. Then we prese nt a lower bound on the complexity of the 2-SS problem, which i s sho wn to bel ong in the 3SUM -hard problem group. W e also dem onstrate that the problem of c omputing t he minimum displacement of multiple new requests ( also referred to as the M U L T I P L E S T R E A M S C H E D U L I N G problem or m-S S ) is NP- complete, and c annot be po lynomially a pproximated within a constant factor . This is prov e n by reducing the S T R I N G P A C K problem to m-SS . S T R I N G P AC K was introduced and sho wn NP-complete in [14]. T o obtai n the appro xi mability results, we further reduce V E RT E X C O L O R [16] to S T R I N G P AC K . This reduction yields new hardness of approximability bounds for S T R I N G P AC K , thus impro vi ng pre vious result s. The rest of this paper is or ganized as foll o ws: In Section 2, we present the formal definit ion of the prob- lem for c all admi ssion of two VBR streams (or e qui v a lently the admission control of a new request over the en velope of av ailable bandwidth in a channel). W e also propose two algorithms that are ef ficient for the 2-SS problem. In Section 3, we extend the problem to multiple streams. W e prove that the problem of admit- ting multiple streams i s NP-complete b y reducing the S T R I N G P AC K problem t o it. Fi nally , in Section 4 we conclude the paper with a summary of this work. 2 Admission Contr ol of a new r eque st 2.1 Formal definition The input of the T W O S T R E A M S C H E D U L I N G ( 2-SS ) problem is t wo ortholinea r traf fic en velopes (streams) S 1 and S 2 of total length L 1 and L 2 respectively and the channel bandwidth B . A stream env elope S k can be described by an ordered set of triple ts r i = ( h k i , s k i , e k i ) , i = 1 · · · n , with h k i being the h e ight v alue (bandwidth demand of vi deo) and s k i and e k i the starting and ending time points of the i th peak respect iv el y , of a t otal of n non-ov erlapping peaks in the strea m. Let l k i = e k i − s k i be the le ngth of the i th peak. For the 2-SS problem, S 1 consists of n such triplets and S 2 of m = O ( n ) t riplets. W e consider S 1 being requested and transmit ted a t time point 0, so being fixed at that positi on. This allows us to subtract its content allocat ion from the total bandwidth, creating a rev erse en velope, as in Fig. 1. 3 Fig. 1 Example of accomodating a new video call request throu gh time displacement. Basicall y , stream S 1 corresponds to the committed t raffic. The second stream can be displa ced by a positive time interv al T to its right, resulti ng in d elayed transmiss ion. W e assume that the en velopes are rigid and none of the peaks can be altered either in length or height. The order of the peaks is fixed. Let S 2 be displaced by T ti me un its. An intersection (t ime o verlap) of the r 1 i = ( h 1 i , s 1 i , e 1 i ) peak t riplet from S 1 with the r 2 j = ( h 2 j , s 2 j , e 2 j ) peak t riplet from S 2 occurs when l 1 i + l 2 j > B a nd ∃ t ∈ [ s 1 i , e 1 i ] : t ∈ [ s 2 j + T , e 2 j + T ] . The set of all time points such that r 1 i intersect s r 2 j defines a time interv al (referred to from no w on as inte rsection interv al ) t i j of length h 1 i + h 2 j , starting at time point T 1 = s 1 i − e 2 j and ending at T 2 = e 1 i − s 2 j . Intersection parameters are depicted graphically in Fig. 2. The second stream cannot be displac ed by any v alue corresponding to this inte rsection interv al , or there will occur a band width overallocation. The output of the 2-SS algorithm will be the minimum displacement of S 2 , such t hat there is no bandwidth ov era llocati on. The second strea m can be shifted only by a displacement that does not f all into a ny intersecti on interval t i j , for 1 < i < n and 1 < j < m . So, the ou t put of the al gorithm c ould be described as the minimum displacement that does not fall into an intersection interval. 2.2 A morphology sensitive algorithm In this s ection we describe a n algorithm to solve the 2-SS overallocation problem. The a lgorithm processes all segments, in ord e r to ca lculate their intersection interval. It could be the case though that many peaks will not be as high a s to i ntersect. By sorting the peaks by height (bandwidth demand), one c an actually calc ulate the intersect ion intervals only fo r the ones that ac tually intersect and n ot consider the rest. Let P be t he number of peak pairs, where the first peak i s selecte d fro m env elope S 1 and the second from en velope S 2 , that hav e sum of heights greater than the bandwidth B and thus define an inte rsection interv al . The algorithm then goes as follo ws: 1. Sort the pea k information (triplets) of both en velopes according to height. 4 Total Bandwidth Time Bandwidth 0 T 1 2 e s 1 2 r r 1 2 i j i j T e 1 i s 2 j t i j S S 1 2 (reversed) Fig. 2 Intersection interval parameter di splay . 2. Ite rate through sorted peaks in S 1 and c alculat e their intersec tion interval with all peaks from the sorted li st of S 2 that c ause bandwidth o veralloca tion. Stop when the height of the ne xt peak in S 1 does not intersect the highest peak of S 2 . 3. Sort al l interv a ls according to their starting point. 4. Ite rate through sorted intervals, merg ing them i nto an aggregate interval, until an inte rv al that does not intersect the aggreg ate interval is discovered, or we run out of interv al s. 5. Output the e nd point of the aggreg ate interv a l as the solution. For the correctness of the algorithm we can ar gue that by i terati ng throug h all intersecting peaks of both streams, we ha ve discov e red all possible ti me interv al s where the second stream cannot be shifte d. The first “gap” bet ween the aggregate inte rv a l a nd the currentl y examined inte rv a l will provide the minimum dis- placement, since any position in the aggregate inte rv a l defines a forbidden displacement, belong ing to some pre viously examined i ntersecti on interval. The start of the “gap” descri bed abov e c annot belong to any in- terval, since, if such an interval existed, its start would occur before the end of the aggregate inte rv al and, as such, before the currently examined i nterval, which means it would have already bee n incl uded in t he aggregate interv al . A visual representat ion of the procedure can be seen in Fig. 3. Sorting the peaks by height takes O ( n log ( n )) time , t he iterat ion t hrough sorted p e aks tak e s O ( P ) ti me and sorting al l intersect ion interv al s takes O ( Pl o gP ) time. The merging iterati on tak e s at most O ( P ) time . So the total complexity is O (( P + n ) l ogn ) . It should be noted t hat in the worst c ase senario where the number of intersecti ng pea ks between the two streams is O ( n 2 ) , the assymptotic complexity becomes O ( n 2 l o gn ) , dominated by sorting the intersecti on intervals’ starting points. W e c an further improv e the running t ime of this al gorithm by excluding intersec- tion interval calculation for peak pairs that result in negati ve second stream displacement, although such an optimizati on does not result in any assymptotic gain. 5 } gap intervals intersection ith interval aggregate interval after i−1 merging steps Fig. 3 Merging intersection interv als into an aggregate interv al at the i − 1 step. The start of the gap defines the S 2 stream displacement where no bandwidth ov erallocation occurs. 2.3 2-SS scheduling is 3SUM hard In this sec tion we will prov e that 2-SS i s 3SUM -hard, a cla ss of problems introduced in [7 ]. The 3SUM problem is to decide whether there exist integers a , b , c in a set of n integers, such that a + b + c = 0, which is currently considered to hav e complexity Θ ( n 2 ) . The notion of 3SUM -hardness (or n 2 -hardness) i s formally introduced in [1, 7], the notati on of which we follow . In brief, we will mention t hat a problem is considered 3SUM -hard i f a ny instance of t he 3SUM problem can be reduced to some instance (with a comparable size ) of the other prob lem i n o ( n 2 ) time , where n is the size of the input. For o ur proof, we will need the follow ing definition: Definition 1 Gi ven two pro bl ems PR 1 and PR 2 we say that PR 1 is f(n)-solvable using PR 2 if ev e ry instance of PR 1 of size n can be solved by using a constant number of instance s of PR 2 (of size O ( n ) ) and O ( f ( n )) additional time. W e denote this by PR 1 ≪ f ( n ) PR 2 T o prov e that 2-SS is 3SUM -hard, it will be suf fici ent to sho w t hat another 3S UM -hard pr oble m i s o ( n 2 ) - solv able using 2-SS . For that pu rpose, we will use the following 3SUM -hard prob l em: Problem: S C P (Segments Containing Points): Given a se t P of n real numbers a nd a se t Q of m = O ( n ) pairwise-disjoint i nterv als of real numbers, i s there a rea l number tra nslation u such t hat P + u ⊆ Q ? P + u here indicates the set of intervals in P translat ed by u . S C P was sho wn 3SUM -hard in [1]. W e wil l no w prove the follo wing: Theor em 1 S CP ≪ nl ogn 2 − S S Pr oof Gi ven an i nstance of the S C P prob lem, we construct two streams in t he following way: The m intervals in set Q and n real numbers in set P are sorte d and Stream S 1 is constructed to ha ve p eaks of height 0 on t hese intervals and p eaks of height 1 in between. T he length L 1 of S 1 is determined by the start of the first interv al s 1 and end of the last i nterv al e m in the sorte d list and st arts at time point 0, with every segment displaced i n time by subtracting s 1 from each of it s coord inates. Stream S 2 is constructed with peaks of length ε with ε → 0 of height 1 at l ocations de fined by the sorted numbers of set P , w ith peaks of height 0 in the interv al s in between. The length L 2 of S 2 is agai n determined from the smallest and largest elem ents of P ( p 1 and p n respectively) and original displacement T of 0 is achiev ed by subtracting p 1 from all peak se gment coordinates. W e set the channel bandwidth B = 1. The c onstruct can be seen in Fig. 4. W e will no w argue t hat the inst ance of S C P has a soluti on if and only if the corresponding instance of 2-SS has a dis placement soluti on less than L 1 − L 2 (if L 2 > L 1 there is no solution). From the construction it is ob vious that a peak of S 2 of height 1 can fit under a 0 height peak of stream S 1 only if the correspond ing number in P falls in the corresponding interval of Q . If for a certain displacement T of S 2 we hav e T < L 1 − L 2 and there is no ov era llocati on of bandwidth, then all peaks of S 2 of height 1 fit un der 0-h eight peaks of stream S 1 , which would imply that ∃ u = T + s 1 − p 1 : P + u ⊆ Q . Also, by t he same arguments, if t here ∃ u : P + u ⊆ Q , then S 2 displaced by T = u + p 1 − s 1 will result in scheduling the two streams with no ov erall ocation. Based on this result, we c an conclude that ou r morphology -se nsitiv e algorithm for schedu ling tw o streams is within a log factor from optimality . 6 I R Q P u T S S 1 2 s e p p 1 1 n m Bandwidth Time Fig. 4 2SS construct. S 1 was translated in height for better vie wing. 3 Scheduling multiple streams W e now extend the 2-SS problem to M U LT I P L E S T R E A M S C H E D U L I N G ( m-SS ), where the input would consist of multiple VBR streams that we want to schedule for transmission over a fix ed bandwidth channel. Althoug h we could set dif fe rent object iv es for optimizat ion, we wil l sele ct minimi zing the displac ement of the la st stream bei ng transmi tted. For streams of the same siz e this is e qui v a lent wi th mi nimizing the total length of trasmission, starti ng from the t ime point of t he first stream t ransmission and e nding when the la st stream has been transferred ov er t he channel. 3.1 Multi-stream scheduling is NP-complete T o demonstrat e m-SS is NP-complet e, we will reduce the S T R I N G P A C K problem to it. T he S T R I N G P AC K problem appeared in [14] and was pro ved hard by reduction from 3 - P A RT I T I O N . The S T R I N G P A C K i s defined as foll o ws: Giv en a set of m st rings of length n , over the bi nary al phabet Σ = { 0 , 1 } , find a minim um le ngth l packing (ali gnment) of the strings, such that no column has more than one ‘1’. An example of the inpu t and the output of the problem are shown in Fig. 5. 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 1 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 1 1 Input Output l Fig. 5 S T R I N G P AC K example with m = 4 , n = 6 and l = 8 The reduction is straightforward. W e will transform the input binary st rings i nto streams with peaks of height 1 for e ach ’1’ encountered in the s tring and peaks of hei ght 0 for e ach ’0’ appearing in the st ring, as sho wn i n Fig. 6 1 0 0 1 1 1 0 1 0 0 1 Fig. 6 String and equi valent stream transformation 7 W e let the total av aila ble b andwidth B = 1, such that no peaks from any stream can ov erl ap. This adheres to the requirement of t he S T R I N G P AC K problem not having a ny column wi th more than one ’1’. Si nce the input t o the S T R I N G P A C K problem i s a se t of stri ngs with equal length n , m inimizing t he total length of t he outputed alignment is equiv alent to minimi zing the displaceme nt of the last string. Thus, t he output of t he m-SS on the transformed strings prov i des that exact minimum. So we ha ve the fo l lowing: Theor em 2 S T R I N G P AC K ≤ p 2-SS Pr oof Gi ven an instance of S T R I N G P AC K , create a 2- SS instance by transforming the binary st rings to equi v- alent streams as described abov e. The mi nimum displa cement of the last strea m t o be transmitted, added to the length n of the strings, provides the minimum length of the m strings’ packing. The result that m-SS is NP-complete follows from the observation that giv en a string packing, i t can be verified i n time O ( mn ) (thus polynomial in the input length) that it consti tutes a v a lid solution, where no column in the p a cking has more than one ’1’, and that the length o f t he packing is less than a specified length k , which would be an input of the decision version of the prob l em. 3.2 M U LT I P L E S T R E A M S C H E D U L I N G i s polyn omially inapproximable withi n a constant V E RT E X C O L O R is a we ll known problem[16], defined as follows: Given a graph G = ( V , E ) , col or t he vertices of V with the minimum number of colors such tha t for eac h edge ( i , j ) ∈ E , vertices i and j have different colors. It h as been sh o wn that V E RT E X C O L O R is inapproximable w ithin | V | 1 − ε for any ε > 0, unless Z p p = N P [5]. B y reducing V E RT E X C O L O R to S T R I N G P AC K a nd w ith the reduction of the latter t o m-S S , sho wn in the pre vious secti on, we will demonstrate that any constant approximation of m-SS is NP-hard. 3.2.1 V ertex Co l or W e now show that V E RT E X C O L O R reduces to S T R I N G P A C K ; and that t his reduction also yie lds a polynomial approximation reduction. Consider a graph G = ( V , E ) , and its v e rtex-edge incide nce matrix. As a running example, we will use the graph giv en in (Fig. 7) whose inci dence matrix is sho wn below . v 1 v 2 v 3 v 4      e 1 e 2 e 3 e 4 e 5 e 6 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 1 1      2 1 3 4 Fig. 7 A Graph on 4 vertices. It is clea r that the graph can be colored with 2 colors. v 4 gets one color , and { v 1 , v 2 , v 3 } get a nother color . Also, the ro ws of t he incidence matrix corresponding to a color group can a ll be packed wi th no offset. For example, putting together the ro ws for { v 1 , v 2 , v 3 } gives 8 v 1 v 2 v 3 Sum      e 1 e 2 e 3 e 4 e 5 e 6 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 1 1      But if we t ry to pac k stri ngs from a djacent vertices, we will a lways get a collision (X); since a djacent vertices ha ve an edge in common. For example v 1 v 4 Sum    e 1 e 2 e 3 e 4 e 5 e 6 0 0 1 0 0 0 0 0 1 0 1 1 0 0 X 0 1 1    So if we pac k rows of t he i ncidence ma trix, the vertices the rows correspond t o must all be non-adjacent (i.e. can have the same color in a coloring). But i f we can color a graph wi th c colors, t hen we would be able to pack the rows of the incidence matrix into c groups. This is a good st art, but the S T R I N G P A C K problem has no way of enforcing gr oups . The strings are allowed to ov e rlap an arbit rary amount. For inst ance, with the example matrix, it may gi ve the follo wing as a solution: v 1 v 2 v 3 v 4    0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 1 1    In order to c omplete the reduction, we need to fla nk the inci dence mat rix wi th special strings t hat will force any feas ible solution to group t he strings in the manner we desire. That is, we want strings to ov erlap completely , or not at all. That way , if S T R I N G P A C K gi ves a solution with c groups, we kno w there must exist a c coloring of G . It turns out tha t we ca n not construct strings t hat ov erlap complete ly , or not a t all. W e c an make strings that overlap c ompletely , or by some small, bounded, a mount. T his turns out to be suf ficient. Consider what ov erl aps we want to allow and dissall o w . In the following, a n ’x’ represents a region of the fla nking region, and the 0,1 a re t he rows of our incidence ma trix. T he following 2 types of ali gnments should be a llowed by the flanking regions. Complete ov erla p of incidence matrix ro ws.  x x . . . x 00100 0 x x . . . x x x . . . x 00000 1 x x . . . x  Zero ov erl ap of incidence matrix ro ws.  x x . . . x 001000 x x . . . x x x . . . x 000001 x x . . . x  But we do not want to al low these following types of alignments, as they interfere with our ’grouping’ of the incidence matrix ro ws. Interference of flanking region with inci dence matrix ro ws.  x x . . . x 0 0 1 0 0 0 x x . . . x x x . . . x 0 0 0 0 0 1 x x . . . x  Partial o verlap of incidence matrix ro ws.  x x . . . x 0 0 1 0 0 0 x x . . . x x x . . . x 0 0 0 0 0 1 x x . . . x  The solution to our example would then look like 9 v 1 v 2 v 3 v 4    x x . . . x 001000 x x . . . x x x . . . x 000010 x x . . . x x x . . . x 000001 x x . . . x x x . . . x 001011 x x . . . x    And we c ould recover t he number of colors from the number o f groups is the string alignment (t he number of groups is recovered from the span of t he solution). Because these fla nking st rings force grouping, we call them self-aligning strings. Now we procede to desc ribe what t hese flanking regions (se lf-aligning s trings) look lik e. T o moti vate the pr oc ess, we present an example. In the following set o f strings, it i s obvious that the first string can not be shifted by any amoun t to the right and n ot cause an y collisions. The first four charact ers will always collide w ith the other strings. 1 2 3 4    1111 1000 1000 1000 1000 1000 0100 0100 0100 0100 0100 0100 0010 0010 0010 0010 0010 0010 0001 0001 0001 0001 0001 0001    Thus, a first a ttempt at t he se lf-aligning strings would be n consecutiv e 1’ s foll o wed by repea ted ident ity matrices . 1 2 3 4    1111 0000 0000 0000 1000 1000 1000 1000 1000 . . . 0000 1111 0000 0000 0100 0100 0100 0100 0100 . . . 0000 0000 1111 0000 0010 0010 0010 0010 0010 . . . 0000 0000 0000 1111 0001 0001 0001 0001 0001 . . .    The strings are g rouped for clarity . In the first 4 bl ocks, each ro w gets a sequence of 4 consecuti ve 1’ s. The rest of the blocks are ident ity matrice s. In the region wi th indentity ma trix, any submatrix of 4 c onsecutiv e columns i s a permuta tion mat rix (ie e ach row has a 1 in i t). Thus, once the 4-consecutive 1’ s are shifted into this region, they will always collide with e very string. Ho wever , we see t hese are not sel f-aligning st rings, since the consecutive 1 blocks must be shifted by as much as 16 places before they are in the indentity matrix reg ion of the other strings. For e xampl e 1 2 3 4    1111 0000 0000 0000 1000 1000 1000 1000 1000 . . . 0000 1111 0000 0000 0100 0100 0100 0100 0100 . . . 0000 0000 1111 0000 0010 0010 0010 0010 0010 . . . 0000 0000 0000 1111 0001 0001 0001 0001 0001 . . .    T o pre vent shifts of 1 to 15 we can add the following types of strings to the end o f the a bov e stri ngs. 1 2 3 4    1000 0000 0000 0000 0111 1111 1111 1111 0000 0000 0000 0000 0000 0000 0000 0000    This matrix prevents t he first row from shifti ng a n amount 1 to 15 with the second ro w . W e concatenat e strings l ike these for ev ery pair of ro ws (4 2 = 16). The final idea in this constructi on is that there is no li mit on the number of i dentity mat rices we incl uded. Thus we can make these st rings as long as we need, until the allowed ov e rlap i s a smal l enough fraction (for example, add n 100 identity ma trices). For a more preci se explanation of self-aligning strings, consult the Appendix. Theor em 3 V E RT E X C O L O R ≤ p S T R I N G P A C K Pr oof Gi ven an instance of V E RT E X C O L O R , c reate a S T R I N G P AC K insta nce wit h the v e rtex-edge incidence matrix flanked by self-aligning strings. The number of groups in the solution to S T R I N G P A C K is t he numb er of colors in an optimal coloring. Theor em 4 S T R I N G P AC K is har d to appr oximate (No constant factor appr oximation). 10 Pr oof W e ca n approximate V E RT E X C O L O R with S T R I N G P AC K . The approximation depends on t he length of the flanking regions. In the next secti on, we construct flanking strings of size O ( n 5 ) . T hus the total siz e of the S T R I N G P A C K inst ance is O ( n 6 ) . Since n is the number of vertices, the s ize of V E RT E X C O L O R problems a re O ( m ) = O ( n 2 ) . So if w e have an f ( n ) approximation t o S T R I N G P AC K , we ge t an f ( n 6 ) = f ( m 3 ) approximation to V E RT E X C O L O R . Since V E RT E X C O L O R is not constant factor approximable, S T R I N G P A C K is not. 4 Conclusions In this paper, we examined the problem of acce pting new video requests for pre-st ored VBR video stream s that hav e been pre-smoothed using any of the smoothing algorithms found in the lit erature. W e prov ed that this problem is an NP-complete problem by reducing it to the S T R I N G P A C K problem. W e al so presente d two optimiz ation algorithms t hat ca n be used to c ompute t he minimum t ime displace - ment of a new request to av oid resource ov erbooking. The morphology -sensitive a lgorithm is capable of computing the time di splacement in O ( P + n ) l ogn time complexity , where P is the number of peak pairs when the first peak is selecte d from the schedule of t he new request, and the second from t he c urrent traffic en velope, and n corresponds to the number of peaks in the schedule of the new video request. This work c an be extended in sev era l ways. In particular, when the cost to the e nd-user i s a variable that needs t o be considered, and t he c ost i s a function of the time displacement, the problem can be transformed into one that finds the minimal displaceme nt at the minima lly acceptable cost fo r the end-user . Other opti mizati on objectiv es c ould be a nalyzed, when gi ven a set of requests, the requirement is to find the displaceme nt p oi nts of the associated schedules that produces smoothest combined schedule. References 1. G. Barequet and S. Har-Peled. Polygon-con tainment and translational min-Hausdorf f-distance between segment sets are 3SUM-hard. Int. J. Comput. Geom. , 11:465–47 4, 2001. 2. E. Casilari, A. Reyes Lecuona, A. Daz Estrella, and F . Sandov al. Classification and comparison of modelling strategies for vbr video traf fic. In Procee dings of International Teletraf fic Congr ess (ITC-16) ’99 , June 1999. 3. K. M. Elsayed and H. G Perros. On the effecti ve bandwidth of arbitrary on/off sources. In Proceedings of the Sixth IFIP WG6.3 Confer ence on P erformance of Computer Networks , pages 257–271, October 1995. 4. A. I. Elwalid and D. Mitra. Ef fecti ve bandwidth of general marko vian traffic sources and admission control of high speed networks. In Pr oceedings of IEEE INFOCOM ’93 , volume 1, pages 256–265, March 1993. 5. U. Feig e and J. Kilian. Zero knowledge and the chromatic number. J. Comput. System Sci. , 57:187–1 99, 1998. 6. W . Feng. 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In Pro ceedings of A CM SIGMETRICS ’96 , volume 1, 1 996. 16. S.S. Skiena. The Algorithm Design Manual . Springer , 1998. 11 APPENDIX Self-Aligning Strings W e now gi ve a more precise account of Self-Aligning strings. W e call a set S = { s 1 . . . s n } of strings ( n , k , L ) -aligning if the following properties hold. 1. | S | = n 2. | s i | = L 3. { s i ( 0 ) , s j ( 0 ) } is feasible 4. { s i ( 0 ) , s j ( r ) } is not feasible for 1 ≤ r ≤ L − k Fig. A-1 Self aligning strings will either ov erlap completely (left) or overlap by some small, limited amount (right) For a giv en ( n , k , L ) , there may or may not exists a set of self-aligning strings. W e want to show a set exists that will make the reduction in the pre vious section work. That is, we need to be able to construct them in polynomial time (it is clear from construction that it takes O ( nL ) time to construct), and also we need certain constraints on n , k , and L . The following two constraints are suf ficient: Flank Flank Row for v i > k Fig. A-2 The prefix copy of a j needs to over lap at least k with the suffix copy of a i First, we want the ‘grouping’ ef fect. Thus our S T R I N G - P A C K strings should only be able to overlap by at most k (or equi valently only allow shifts of at least 2 L +  n 2  − k ) . Since our S T R I N G - P AC K strings contain self-aligning strings as sub- strings; it is obvious that shifts of 1 to L − k are not allowed. Also, once we shift by  n 2  + k + 1, the prefix flanker of the shifted string overl aps the suf fix flanker of the other string. Thus shifts of  n 2  + k + 1 to 2 L +  n 2  − k are not allowed (Figure A-2). T o make these 2 ranges ov erlap, we need L − k ≥  n 2  + k + 1 The second constraint is to be able to recov er the number of groups from the span of the solution. Since two strings will ov erlap completely only if their corresponding vertices are non-adjacent, we can recover a coloring by grouping strings that ov erlap completely . Say the answer to S T R I N G - P AC K has C groups of strings that overlap completely . Since (from above) each can ov erlap at most k , this means the span of the solution is in the range C ∗ ( | s | − k ) + k ≤ span ≤ C ∗ | s | If the answer had C − 1 groups, the range would be ( C − 1 ) ∗ ( | s | − k ) + k ≤ span ≤ ( C − 1 ) ∗ | s | T o be able to distinguish the number of completely overlapping groups from the span, we w ould need C ∗ ( | s | − k ) + k > ( C − 1 ) ∗ | s | That is, the smallest span from C groups is larger than the larges t span from C − 1 groups. This yields | s | + k > k ∗ C Since C ≤ n , this inequality is achiev ed if | s | + k > kn 2 L +  n 2  + k > kn L > 1 2 k ( n − 1 ) −  n 2  12 These constraints are easy to achie ve with the outlined construction. T o be precise, our self-aligning strings are the ro ws of [ R 1 R 2 . . . R n I l P 1 , 1 . . . P n , n ] Where R i is the n x n matrix 1 2 . . . i . . . n          0000 . . . 0 0000 . . . 0 . . . 1111 . . . 1 . . . 0000 . . . 0          and P i , j is the n x n 2 matrix 1 2 . . . i . . . . . . j . . . n                     0000 . . . 0 0000 . . . 0 . . . 1000 . . . 0 . . . 0000 . . . 0 . . . 0111 . . . 1 . . . 0000 . . . 0                     The R i gi ve each ro w n consecutiv e 1 s . Once a string is shifted right by n 2 , its R i will all lie in the I region of the other strings (That is, it will have n consecuti ve 1’ s overlappi ng the I matrices). Since in this I region, a string has a 1 every n positions, this shift is not feasible (the n 1s in the shifted string must conflict with a 1 in the other strings). Thus shifts of n 2 through l n are not feasible ( n 2 ensures that an R i is completely in the I regios; l n ensures an R i isnt shifted past the I region). The P i , j eliminate shifts of 1 through n 2 − 1. This is done explicitly , as can be seen in the construction of the P i , j Thus shifts of 1 through l n are not feasible. So the maximum overlap, k , is bounded by k ≤ L − l n = ( n ∗ | R i | + l ∗ | I | + n 2 | P i , j | ) − l n = n 2 + l n + n 4 − l n k ≤ n 2 + n 4 This is a good result since k is fixed for any size l . All we want is L > max ( 2 k +  n 2  , nk 2 ) , which we get for large enough n by setting l = n 4 which gets L = n 2 + l n + n 4 = O ( n 5 ) ; a polynomial length string, as desired.