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Distributed Algorithms (22903)
The wait-free hierarchy and the universality of consensus Lecturer: Danny Hendler This presentation is based on the book “Distributed Computing” by Hagit attiya & Jennifer Welch
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Formally: the Consensus Object
Supports a single operation: decide Each process pi calls decide with some input vi from some domain. decide returns a value from the same domain. The following requirements must be met: Agreement: In any execution E, all decide operations must return the same value. Validity: The values returned by the operations must equal one of the inputs.
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Wait-free consensus can be solved easily by compare&swap
Comare&swap(b,old,new) atomically v read from b if (v = old) { b new return success } else return failure; How? Motorola 680x0 IBM 370 Sun SPARC 80X86 MIPS PowerPC DECAlpha
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Would this consensus algorithm from reads/writes work?
Initially decision=null Decide(v) ; code for pi, i=0,1 if (decision = null) decision=v return v else return decision
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A proof that wait-free consensus for 2 or more processes cannot be solved by registers.
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A FIFO queue Supports 2 operations: q.enqueue(x) – returns ack
q.dequeue – returns the first item in the queue or empty if the queue is empty.
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FIFO queue + registers can implement 2-process consensus
Initially Q=<0> and Prefer[i]=null, i=0,1 Decide(v) ; code for pi, i=0,1 Prefer[i]:=v qval=Q.deq() if (qval = 0) then return v else return Prefer[1-i] There is no wait-free implementation of a FIFO queue shared by 2 or more processes from registers
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A proof that wait-free consensus for 3 or more processes cannot be solved by FIFO queue (+ registers)
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The wait-free hierarchy
We say that object type X solves wait-free n-process consensus if there exists a wait-free consensus algorithm for n processes using only shared objects of type X and registers. The consensus number of object type X is n, denoted CN(X)=n, if n is the largest integer for which X solves wait-free n-process consensus. It is defined to be infinity if X solves consensus for every n. Lemma: If CN(X)=m and CN(Y)=n>m, then there is no wait-free implementation of Y from instances of X and registers in a system with more than m processes.
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The wait-free hierarchy (cont’d)
registers 1 FIFO queue, stack, test-and-set 2 … Compare-and-swap
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The universality of conensus
An object is universal if, together with registers, it can implement any other object in a wait-free manner. We will show that any object X with consensus number n is universal in a system with n or less processes An algorithm is lock-free if it guarantees that some operation terminates after some finite total number of steps performed by processes. The lock-freedom progress property is weaker than wait-freedom.
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Universal constructions
Given the sequential specification of any object, implement a linearizable wait-free concurrent version of it: A lock-free construction using CAS A lock-free construction using consensus A wait-free construction using consensus A bounded-memory wait-free construction using consensus
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A lock-free universal algorithm using CAS
Each operation is represented by a shared record of type opr. typedef opr structure { inv ;the operation invocation, including its parameters new-state ;the new state of the object, after applying the operation response ;The response of the operation } Head inv new-state response inv new-state response … inv new-state response
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A lock-free universal algorithm using CAS (cont’d)
Head anchor inv new-state response inv new-state response … inv new-state=init response Initially Head points to the anchor record. Head.newstate is initialized with the implemented object’s initial state. When inv occurs point:=new opr, point.inv:=inv repeat h:=Head point.new-state, point.response=apply(inv, h.new-state) until compare&swap(Head, h, point)=h return point.response
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A lock-free universal algorithm using consensus
Each operation is represented by a shared record of type opr. typedef opr structure { seq ;the operation’s sequential number (register) inv ;the operation invocation, including its parameters (register) new-state ;the new state of the object, after applying the operation (register) response ;The response of the operation, including its return value (register) after ;A pointer to the next record (consensus object) Head anchor seq seq=1 seq … inv new-state response after inv=null new-state=init response=null after inv new-state response after
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… A lock-free universal algorithm using consensus (cont’d) Head anchor
seq seq=1 seq … inv new-state response after inv=null new-state=init response=null after inv new-state response after Initially all Head entries points to the anchor record. When inv occurs point:=new opr, point.inv:=inv for j=0 to n-1 ; find a record with the maximum sequenece number if Head[j].seq > Head[i].seq then Head[i]=Head[j] repeat win:=decide(Head[i].after,point) ; try to thread your operation win.seq:=Head[i].seq+1 < win.new-state, win.response > :=apply(win.inv, Head[i].new-state) Head[i]=win ; point to the following record until win=point return point.response
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inv new-state response after
A wait-free universal algorithm using consensus Each operation is represented by a shared record of type opr. typedef opr structure { seq ;the operation’s sequential number (register) inv ;the operation invocation, including its parameters (register) new-state ;the new state of the object, after applying the operation (register) response ;The response of the operation, including its return value (register) after ;A pointer to the next record (consensus object) We add a helping mechanism Announce inv new-state response after seq When performing operation with sequence number j, try to help process (j mod n)
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A wait-free universal algorithm using consensus (cont’d)
Initially all Head and Announce entries point to the anchor record. When inv occurs Announce[i]:=new opr, Announce[i].inv:=inv,Announce[i].seq:=0 for j=0 to n-1 ; find a record with the maximum sequenece number if Head[j].seq > Head[i].seq then Head[i]=Head[j] while Announce[i].seq=0 do priority:=Head[i].seq+1 mod n ; ID of process with priority if Announce[priority].seq=0 ; If help is needed then point:=Announce[priority] ; help the other process else point:=Announce[i] ; perform own operation win:=decide(Head[i].after, point) < win.new-state,win.reponse > :=apply(win.inv,Head[i].new-state) win.seq:=Head[i].seq+1 Head[i]=win return Announce[i].reponse
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A proof that the universal algorithm using consensus is wait-free
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What is the number of records needed by the algorithm?
A bounded-memory wait-free universal algorithm using consensus What is the number of records needed by the algorithm? Unbounded! The following algorithm uses a bounded # of records Each process allocates records from its private pool A record is recycled once we’re sure it will not be referenced anymore We don’t need this mechanism if we use a language with a GC (such as Java)
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A bounded-memory wait-free universal algorithm using consensus (cont’d)
When can we recycle record #k? No process trying to thread record (k+n+1) or higher will write record k. After all the processes that thread records k…k+n terminate, record k can be freed. When process p finishes threading record m it releases records m-1…m-n. After record k is released by the operations threading records k+1…k+n – it can be recycled.
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A bounded-memory wait-free universal algorithm using consensus: data structures
Each operation is represented by a shared record of type opr. typedef opr structure { seq ;the operation’s sequential number (register) inv ;the operation invocation, including its parameters (register) new-state ;the new state of the object, after applying the operation (register) response ;The response of the operation, including its return value (register) after ;A pointer to the next record (consensus object) before ;A pointer to the previous record released[1..n] initially true Head anchor inv new-state response before after seq inv new-state response before after seq … inv new-state response before after seq
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A bounded-memory wait-free universal algorithm using consensus (cont’d)
Initially all Head and Announce entries point to the anchor record. When inv occurs point:=a free record from private pool, point.inv:=inv,point.seq:=0 for r:=1 to n do point.released[r]:=false, Announce[i]:=point for j=0 to n-1 ; find a record with the maximum sequenece number if Head[j].seq > Head[i].seq then Head[i]=Head[j] while Announce[i].seq=0 do priority:=Head[i].seq+1 mod n ; ID of process with priority if Announce[priority].seq=0 ; If help is needed then point:=Announce[priority] ; help the other process else point:=Announce[i] ; perform own operation win:=decide(Head[i].after, point) < win.new-state,win.reponse > :=apply(win.inv,Head[i].new-state) win.before:=Head[i] win.seq:=Head[i].seq+1 Head[i]=win temp:=Announce[i].before for r:=1 to n do if temp<> anchor then before-temp:=temp.before, temp.released[r]:=true, temp:= before-temp return Announce[i].response
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How many records are required by the algorithm?
Each incomplete operation may waste n distinct records There may be up to n incomplete operations At any point in time, up to n2 non-recycable records All non-recycable records may belong to same process! Each pool should have O(n2) records, O(n3) total records needed
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