Better Adaptive Diagnosis of Hypercubes Seminars for the PH.D in Computer Science Arwa Zabian

HADA:Hypercube Adaptive Diagnosis Algorithm C.Feng , N.Bhuyan IHADA: Improvement HADA Hyp-DIAG: Hypercube Diagnosis E.Kranokis , A.Pelc

Some Interrogative Why Hypercube ? How the diagnosis algorithm perform when the number of faulty processors d diagnosted by it are less than the effective number of faulty processors t in the system ? There is an optimal Diagnosis Algorithm ? Yes We can implemented it?

Why Hypercube For Its regulare structure is hirarchical the connectivity is very limited.

Nakajima: proposed an adaptive diagnosis approach difference between adaptive and non adaptive approach: in the maner of definding the results in the scheduling in the time needed to identify t < n/2 faulty in the system when in the non adaptive needed at least t , but in the adaptive we can identify in a constant time the number of test to identify t < n/2 faulty : non adaptive required nt in the worst case adaptive n+t-1

HADA: Hypercube Adaptive Diagnosis Algorithm A diagnosis must be correct and complete. Model : G = (V,E) directed graph , n-hypercube contiene N= 2n V : processors E: the connection links Parametrs: Time Diagnosis Number of tests links

HADA Algorithm : Adaptive system-level diagnosis for hypercube multiprocessor Assumption - Each node can test only its neighbours - Any node cann’t be tester and tested at the same time - the status of processors is permanent during the diagnosis process

Algorithm :

Algorithm is based on divided -and-conquer mechanism n-hypercube is divided into subcubes ; each subcube is mapped into a ring using reflected Gray Code ,that can be represented by an array r; each ring recorsively divided into subring; Diagnosis each subring until finded a faulty - free ring or a ring that contien only one faulty unite;

The result obteined can be used to identify the status of the processors in the other direct subcube

Function Ring Diagnosis Each processor i test the processor [( i+1 )modN] in the same ring and if that test was not performed previously , if no one outcome is 1 then the ring is faulty-free processor i test[( i+N-1) modN] ,if the same faulty unit outcome by the first step is faulty by the second step then is tested correctly else the unit i is faulty

Function Subcube Diagnosis Call function RingDiagnosis (r1 ) Call function RingDiagnosis r2 - if the subcube faulty-free then observe the previous result and identify the faulty units - if the subcube diagnosed with one faulty units also observe the previous result to identify the remaining faulty units

Algorithm analysis Theorem 1: for an n-cube ,HADA is correct and complete provided that the number of faulty PEs is no more than n and n 3.

Theorem 2 The number of test link required in HADA is : - 2n if there is 0,1 faulty in the system - 3. 2n-1 if f1 = 0,1 1 < f2  n where f1 , f2 is a faulty units in the two direct (n-1) subcube on the n-cube respectively. - 3. 2n-1 +1 if f1 =1 1 < f2  n-1 the required diagnosis time is : 2 if f1 , f2 = 0 4 if f1 = 0 , f2 = 1 6 if f1 , f2 = 1 7 if f1 = 1 , 1< f2 <n-1

The number of test links for n.cubee is  2n log n The worst case 2n (log n  +2) tests and n+4 rouds round : is the time during an PE s applied a test to another PE iteration : is the time during which a single loop testing is performed in each iteration we use four round

Experiment on the n-cube: the experiment is performed for 6-cube

The diagnosis time

IHADA: Improvement HADA the diagnosis start from an arbitrary level m diagnosis is performed only in one direction bottom to up diagnosis m is choosed in base on the fault bound and the diagnosis cost Theorem 3: the diagnosis is correct by using IHADA to an n-cube provided that 2  m [( n+3)/2] and f< min (2m , 2m-1 + 2.( 2 n-m -1)) when m is the starting level of diagnosis and f the faulty bound.

Theorem4 : The proposed diagnosis algorithms (HADA,IHADA) for an n-cube is correct and complete iff: 2  m [( n+3)/2] and f< min (2m , 2m-1 + 2.( 2 n-m -1)) there is a faulty path between each of the Pes in V-Z and one in Z, where V the set of the PEs in the array Z the set of the identified fault-free Pes after the top down process terminate This theorem give a necesary and sufficient condizione for the complete and correcte diagnosis and guarantee that d < f <t

HYP-DIAG : Hypercube Diagnosis Model: indirected graph G = (U,E) T =(U,A) is a directed graph when the element of A is only the adjacent nodes of U that can be tested each other S = A 0,1 S is compatible with a faulty set F iff for each (u,v) F w(u,v) = 1

Algorithm HYP-DIAG For n > 9 The diagnosis is performed in four phases: 1- r =  log n +1 when 2r > n the diagnosis has two possibilty : - faulty-free ring - in each ring there is at least one faulty unit Lemma 1: there is at most one unguarded ring 2- identify a faulty-free ring and using its faulty nodes to identify all its neighbors rings, testing all this nodes required at most f+1 test where f il the maximum number of faulty unit in the system.

Lemma 2: in the unguarded ring ( if it exist ) there is at most one node x all of whose foreign neighbors are faulty. 3- diagnose all the units in the unguarded ring except x 4- diagnose x, there is two possibility: - we already identify n faulty units on the system that mean x is faulty-free. - we already identify n-1 faulty that mean x has an aleardy diagnosed fault-free neighbor , we can used to test x we use the procedure Guarded -Ring -Diag to diagnose all the node in a guarded ring

Lemma 3 : in a guarded ring the number of tests is at most f+1 if there is a f faults in the ring, and the number of test is one if there is one faulty in the ring. Lemma 4 : all nodes of the unguarded ring that have foreign faulty-free neighbors can be diagnosed using at most f+t tests, when f the number of faults a mong these nodes , if f=1 then we used one test. Theorem 5 : algorithm HYP-DIAG diadnose all node of an n-hypercube for 3  n  8, using at most 2n +3n/2 tests in the worst case .

There is an optimal diagnosis algorithm ? Yes is a HYP-DIAG HYP-DIAG uses at most 2+3n/2 tests in the worst case to diagnose n faulty in the n-hypercube. and have a lower bound of the number of tests to diagnose 2n processors with at most n faults is 2n +n-1 that mean is an optimal diagnosis algorithm . We can implemented it ? No because lemma 1 does not neccesarily hold