ALG 5.4. Lexicographical Search: Applied to Scheduling Theory. Professor John Reif

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1 Algorithms Professor John Reif ALG 54 input assume problem Scheduling Problems Set of n tasks in precidence relation on tasks Also, given m processors Unit time cost to process task using single processor Schedule execution of tasks in minimal time, using < m processors per unit time A task can not be executed before its predecessors Lexicographical Search: Applied to Scheduling Theory Reading Selection: "Algorithms for Minimal-length Schedules", R Sethi, Introduction Deterministic Scheduling Theory, Ch, pp 5-73, E Coffman, ed to Scheduling: using 3 processors

2 3 Dag = directed acyclic graph Level Strategy on Dags --Whenever a processor becomes available, assign it to the first unexecuted ready task at the highest level Levels l 4 l 3 4Theorem: Level Strategy Optimal for Forests of Trees Proof: Suppose not Consider a smallest task forest T with non-optimal level strategy S taking time D Then $ idle processor at time D- Hence, $ task q of level l executed at time D- So, no task of level l is executed before time D l Levels 3 4 l l 3 l 0 q l level l i = set of nodes such that $ path length i from node to terminal l Levels computed by reverse Breadth First Search of Dag in linear time! l 0

3 Derive a new task forest T from T: () Delete roots and children of roots 5 (ie delete l 0,l ) () Add a new root connected to all old grandchildren (ie level l ) 6 Scheduling Problem: T But level strategy S or T is length D-, a contradiction with minimality of T T' l k Task System is a Dag G (Note: If G has a cycle, then every task on the cycle is a predecessorof itself, so is not ever executable) l k- l Complexity of Scheduling with m Processors-- m= Processors: polynomial time m=3 Processors: Open! m>3 Processors: NP complete l new root l 0 root time D - D - D D - D - q q idle new root

4 7 Two Processor Scheduling 8 Task System G is Dag (Delete transitive edges in time n 8 ) Requirements of Scheduling Strategy: ()Consistent with level strategy: those tasks closer to root are executed later ()If successors(q) successors(q'), then q always executed before q' Lexicographic Ordering on Strings: ()For any non-empty string w, w>" " ()If w>w', then for any string u, uw>uw' compare them from left to right, choosing the string with first larger character Examples: 5>5 53>5

5 9 Lexicographical INPUT: dag G Search = ( V, E) Initialize: Let S be set of unassigned nodes with no unlabelled successors ( ) = ( ) ( ) For Each vœ S, let r v L Y L Y K i, where Y, K, Y i are successors of v with LY ( ) > LY ( ) > K > LY ( i ) 0 A Linear Time Algorithm for Lexicographical Search of dag G=(V,E) Idea: Do not compute sort of successor lists, instead keep list LEX of their suffix labels, from smallest to largest G H I J D E F 3 A B C l 3 l l Let x be an element of S with min r( x) with respect to Lexicographical order Lx ( ) k k k+ If Then Some vertex is still unlabeled, goto stop Number Q LEX - A= B= C=3 E=4 F=5 D=6 J=7 I=8 G=9 H=0 ABC BC C EFD FD DJI JIGH IGH GH H - - GF GFD G HG GH Example: After labeling E with 4, now label( H) =-4, label( G) =-4 fi LEX = HG

6 Algorithm: Lexicographic Search Input: dag G = ( V, E) Begin: Initialize: Q ( nodes with no successors) LEX ( ) "v ŒV, flag( v) 0 k 0 Loop: Choose and delete first v ŒQ Assign v number k k + for each immediate predecessor v of v do flag( v ) ; Add v' to LEX SUFFIX - LEX ( ); SUFFIX - Q ( ) For If Each v ŒLEX in increasing order such that flag( v ) = do delete ( v,v ) edge; flag( v ) 0 delete v from LEX if v has remaining departing edge: then add v to SUFFIX - LEX else add v to SUFFIX - Q od LEX LEX ( SUFFIX - LEX) ( ) SUFFIX - Q Q Q remaining predecessors of v delete all edges ( v,v ) entering v Q π f, then go to LOOP Application to Task Scheduling Input: Task dag G Compute LEXICOGRAPHIC Search of G Assign tasks by Level Strategy (scheduling highest numbered tasks earliest if they are ready) Theorem: LEXICOGRAPHIC Search yields an optimal scheduling for case of processors Proof: Nontrivial and interesting!

7 3 5 Levels l 7 4 Convention Assign task to processor before processor Let S be scheduled by LEXICOGRAPHIC Search 4 l 6 Follow Sets: F k, F, k-, F 0 D i * * The F i are predecessors of F i- 3 l 5 DominatingTasks: D k, D k+,,d 0 F i - The D i "dominate " the F i- l 4 Jumpahead Tasks: J k, J k+,, J 0 The J i "jump ahead " of F i l 3 -- Both D i and J i are executed at the same time unit l D i F i- D i- J i J i - 3 time processor processor l Define: J i as (possibly idle) task scheduled as late as possible st number ( J i ) < number ( D i - ) time ( J i ) < time ( D i - ) Define: D i as task executed at the same time as J i ; idle Define: F i Ïtasks scheduled strictly before = { Di}» Ì Ó Di and after Di+ ( if Di+ exists)

8 F 3 5 D 3 5 F F 3 4 D Lemma : The D i nodes are always (not necessarily immediate) predecessors of the F i - nodes D J 0 J F 0 3 D 0 7 J 3 Proof: By definition, number ( J i ) < number ( D i - ), and J i is the latest such task with time ( J i ) < time ( D i - ) Thus, number ( J i ) < number ( D i - ) number v for any v ŒFi- - Di- So each element of F i - is numbered larger than J i, but Scheduling attempts to label the highest numbered tasks first, if possible (ie if they are ready) So, no element of F i - could be ready at the time D i is assigned Thus, D i dominates all the F :" v Œ F, $ path from D i to v i- i- ( ) processor F 3 5 F 4 3 F F 0 3 QED processor J 3 J J 0 idle task

9 Lemma3: Proof : All the F i nodes are predecessors of the F i- nodes By Lemma, D i precedes all F i- By definition of F i, number( F i ) > number( D i ) (else F i node is set to J i +, which is not in F i ) By our scheduling algorithm, D i has a higher number than any task executed at a later time Thus, the elements of F i - are the highest numbered tasks executed after D i Then it follows that if Lexicographic Search gives r( D i ) = ( LY ( ), LY ( ),, LY ( j )), then $ prefix Y,Y,,Y j ŒF i - the highest numbered successors of D i which are in F i- (but with no predecessors in F i- ), and thus all the F i nodes are predecessors of nodes Y,Y,,Y j in F i- Q E D 8 Claim: Lexicographic Search Ordering yields optimal scheduling for two processors Proof : By Lemma 3, each v ŒF i must preceed all v ŒF i - in time Since F is always odd, thus, t i = F i + is time required to schedule the tasks of F i Optimal scheduling requires time u Ât i i= which is exactly what Lexographic Search Ordering Algorithm achieves! Q E D

10 9 Given: List L of task ordering 0 L = ( ) ListScheduling : Whenever a processor becomes evaluable, scan the list from left to right Assign to the processor the first unexecuted ready task AlgorithmforListScheduling : Build up the schedule by constructing schedules for increasingly longer prefixes of L = ( n,n-, n-,,,) L = (,) L 6 = (,,3,) schedule L = ( n) schedule L = ( n, n- ) idle idle idle L 8 = (,,0) schedule L n = ( n,n-,,,)= L L 9 = (,,9) 9 arrows give st time $ free processor

11 Ê Almost LinearTimeO Ánlogn * ˆ Ë AlgorithmforListScheduling t'œ SET() t iff t' is the st time at, or after time t s t Begin $ a free processor initialize : forall times t, do SET()= t {} t for all tasks x with no predecessors do ready time( x) for each task x in decreasing order do begin t FIND( ready time( x) ) assign x to time t ift has no more free processors for thenmerge SETt () with SETt ( + ) each immediate successor Y to x st all predecessors of Y are visited do ready timey ( ) t + end Output ready time ( ),,ready time ( ) Task Systemwithm 3 processors If t= time used by Lex schedule t 0 = optimal schedule time t t0 - m