A. Vogel Federal Republic of Germany THE TESTCOMPLEXITY OF FANOUT-FREE COMBINATIONAL NETWORKS FOR BRIDGING ·FAULTS
Introduction There are many investigations in the problem of finding minimal complete test" sets for single and multiple stuck-at faults. Because of the increasing importance of reliable hardware the problem of generating complete test-sets must be solved for all possible hardware-faults not only for stuck-at faults. Friedman /2/ and Mei /3/ pointed out the necessity to handle lrriQ.gj!l!LLaults.. Friedman described some phenomena of bridging faults whereas Mei investigated a class of bridging faults which do appear at the gate-inputs only. Since only extended ANO- and OR-gates are allowed they are logical equivalent to the stuck-at faults and are therefore detected by any complete test-set for stuck-at faults. In this paper we use the same bridging fault model as Mei did in /2/ but faults between any of two lines are allowed. For reasons of compatibility only one logical family is used within a net. Therefore either OR-bridges or AND-bridges are assumed to appear. We restrict ourselves to OR-bridges throughout the paper. Construction of complete test-sets and bounds on the test-complexity of fanoutfree nets are ~died for the class of combinational bridging faults, classes of bridging faults induced by appropriate neighbouring relations and the class of bridging and stuck-at faults.
Basic definitions Elementary units are "black boxes" with inputs and outputs. 0(9) is the number inputs and 1l£Ll is the number of outputs of the elementary unit ~. Let ~= {I, n, d, k, 0, 9, U, 1, Q} a set of elementary units with Q(I); = Q(n) Q(d) = Q(k) Q(9) Q(U) Q(O) 1 and
Q(.!) = Q(Q) 2 and Zed) 2 and Z(S) Z(O) = 2
=
1 and Z(I) Z(k) 1 Z(U) 2
=
Zen)
The set of nets over C is defined as follows:
=
Z(l)
=
Z(Q) =
Of
- 6 Definition 1:
i) Any g £ C is a net; ii) if 51 and 52 are nets then 51 x 52 is a net with 0(5 1 x 52) = 0(5 1) + 0(5 2 ) and 1(5 1 x 52} = 1(5 1) + 1(5 2 ); iii) if 51 and 52 are nets such that 0(5 1) = 1(5 2 ) then 51 0 52 is a net with 0(5 1 0 52) = 0(5 2) and 1(5 1 0 52) = 1(5 1 ); iv) the set
~
of nets over C is completely defined by i - iii).
To each net 5 £ ~ a di rected marked graph G = (V, E, g, m) is ass i gned: Vis the set of nodes, ·E the set of edges, g a function which maps each edge a £ E onto its ends. The function m : V -+ n, ... , 0(5)) un, ... , 1(5)) u C maps the i-th input node and the j-th output node to i and j respectively and all other nodes x £ V onto units of C such that indegree (x) = O(m(x). By nets we mean these special graphs and interpret the edges as leads of the net. For prio g simply gi (i = I, 2) is written. Let B = (0, I} and B~ the set of all functions f : Bn tion f(5)· £ B~gl is assigned as follows: Defintion 2:
-+
Bm. To all 5
£
5 a func-
i) f(I), f(n), f(Q), ft.!) £ B~ with f(l) = id, fin) = neg, f(Q) (x) = 0, f(.!-) (x) = 1; f(D) £ Bi with f(D)(x) = (x,x); f(k), f(d) £ B~ with f(k)(x,y) = x·y and f(d)(x,y) = x + y; .f(U), f(8) £ B~ with f(8)(x,y) =(max(x,y), max(x,y» and f(U)(x,y) = (min(x,y), min(x,y». ii) f(5 1 0 52) = f(5 1) 0 f(5 2} and f(5 1 x 52) = f(5 1) x f(5 2), where o.and x are the usual operations in ~ B~; iii) for 5 £ 5 and u £ E we define u £ B6(5} with
We now define bridging faults as special transformations of 5 50 £ ~ , executed on the graph.
£
5 into a net
- 7 Let S E
Definition 3:
~,
K de, U} . and n, v
El
i) The transformation S .. K(u,v)S
E. by rule x) is called
simple bridging fault of kind K. x) K(u,v) V = V u {w} K(u,v) E = (E -
{u,v} ) 0 {(g1(u),w),(g1(v),w),(w,g2( u) w,g2(v»}
fm(a) K(u,v)m(a)=lw
ii) i i i)
if
a '" w
otherwise
By 'iteration of rule x) multiple bridges are defined. K( S) : = ( K( u, v) S E ~
I (K, u, v) E {e, U} x E x E}
Illustration of a transformation by rule x
•
M-
Cl.
·b
b
> •c.
V
•el
Definition 4:
c.
Let S, S' E
~
with O(S) = O(S') = n
n i) d(S,S') : = (b E B I f(S)(b) '" f(S')(b)} n ii) bE B
detects
S'
iff b E d(S,S')
Two questions are looked for in the paper: 1.)
Finding complete test-sets of fan-out free circuits for some fault classes.
2.)
Finding bounds on the cardinality of minimal complete test-sets for tnese fault classes.
- 8 Definition 5:
Let 5 E
~
=
(5 £!
I Z(S)
= I}
and P(S)
c ~.
i) T c BQ(S) is complete test-set of 5 for P(S) iff for all d(S,S')4o" => d(S,S') n T • "
P(S) holds:
=(
ii) T(P,S)
T IT is complete test-set of 5 for PIS)}
= min ( ITI I'T E T(P,S)]
iii) C(P,S)
is called test-complexity.
We have some simple Propositions: Proposition 1: i) ii)
P(S) c P'(S) P(S)
c
P'(S)
= =
T(P',S) C(P,S)
c $
T(P,S) C(P',S)
Proposition 2: i) ii) 'ii)
((I,D)], {(D,l)} E T(e,k) = T(U,d) 2 T(e,d) = T(U,k) = !(B ) T(K,I) = T(K,n) = '1(B)
for K E {e,U}
Fanout-free nets and bridging faults
Definition 6:
The set B of fanout-free nets is completely defined by i -i i i ) . i) ii)
Lemma 1:
Let 5
I,n,d,k,EB; 5 E B => I 0 5 E Band noS E B;
-
= Go
-
(SI x 52) E!!. with Q(Si)
= ni
(i
= 1,2).
T E T(e, 5) => { (pr (b), ... ,pr (b)) I b £ D £ T(e, 51) n l 1 and {(pr
Proof:
n
+l(b), ... ,pr
1
n
(b))1 b £DET(e,S2)
2
Any complete test-set T(e,S) must detect all faults of 51 and 52' Therefore lemma 1. holds.
- 9 Lemma 1 gives the justification to build complete test-sets for S E B from complete test-sets of the nets SI' S2 E~, First of all we need some technical (j"fiititions. Definition 7:
Let S i) For
E
Band u
{v!' .... vl }
we define:
E.
E
= (v
EEl g2(v)
viu projects for b
E
= gl(u)J and 1 s is I BQ(S)
iff
= f(m(gl(u))) (v l (b) •...• vi (b) •.•.• vl (b)) • f(m(gl(u))) (v l (b) ..... vi (b) .....v l (b)) i i) ui ... ul E E* with g2(u i ) = gl (u i +l ) projects for u(b)
b
BQ(S)iff for all 1 s i S 1 ui u + projects for b. i 1
E
As in a net S E B there exists for all u E E a unique path to the output we say that the predicate S(u.b) is true"iff the path from u to the output of S projects for b. In the following we denote by q(G) the projection-value of G E {d.k} which is 0 for G = d and 1 for G = k. Definition 8:
For S p
Proposition 3:
E
E
B • C E B and Mc E we define pC(S.M)c'"(f-l(S)(c)):
pC(S~"')
For S
E
A
iff uEl''\
I\. : u(b) bEP
=
1 or S(U.b).
~. C E B and M c E we have pC(S,;M) '"
0.
In the following definition we define for S E ~ a set H(S) which contains complete test-sets. The elements of H(S) are in general not minimal but guarantee some properties which will be used in" the proof for theorem 1. ;Definition 9:
For S
E
B we define H(S) c 'Il(BQ(S)) inductively:
i) H(d): =(((O.O), (0,1), (1.0)}}. H(k): = ({(l.l). (0.1). (l,O))}, H(I) = H(n) : = ({(D). (1)l) ii) H(n
0
S) : = H(S) ;
"ii) Let S = G 0 (SI x S2)
E ~"and
W= hI x P2 utI x h2 as -1 well as W= hI x t 2 U PI x h2 with t i E f (Si) (q ). hi E H(Si) and Pi E pq (Si' Ei ) for i 1,2 and q = q (G). -
- 10Then h
H(S) with
E
if h =
{
Propos iti on 4:
;
E
~
the following three properties hold:
i\*"
ii)
u
/\H(S) A b-E h /\ 1\E hE H(S)
E
E
h
1\ c B
iii )
:s; IWI
otherwi se.
For S i)
u(b)
E
=0
and S(u,b)
u E
E
( u Proof:
IWI
j f-1(S)
The proof goes
(c) '"
0
=
3b
E
1 f- (S)(C) u h : u(b)
by induction over the construction of S
E
1)
B.
See /1/.
/\B
Theorem 1: Proof:
S
E
(by induction)
:
H(S)
c
T(9,S)
For d, k, I, n follows the assertion from Proposition
2. Let S = G 0(~1 x S2) E~, such that for SI' S2 the assertion holds. Without loss of generality let h = hI x P2 u t 1 x h2 E H(S). Because of the preassumption H( Si) C T(9, Si)' the set hI x b2 u b 1 x h2 with b E f- 1 (Si) (q(G)) is a complete test-set of all faults i 9(u,v) S E 9(S) such that u, v E E for i = 1,2 ; therefore h detects i all 9(u,v) S E 9(S) with u, v E E . Let 9(u,v) S £ 9(S) such that i u E El' v E E and d(S,9(u,v)S) "'~. For b = (b , b ) E d(S,9(u,v)S) 2 1 2 we assume that v(S)(b) = 1. Two cases are to distinguish: a) v(S2)(t) = 1 for some t will detect 9(u,v)S. b) v(S2)jp
2 t
E
hI
E
= f(~).
E
P2. Because of Proposition 4ii) hI x P2
hI x P2 will detect 9(u,v) S if there is
f- 1 (Sl)(q(G)) such that u(Sl)(t) = 1. (Proposition 4iii);
- 11 -
Otherwise u(SI)
I hI
n f- 1(SI)(Q(G)( f(Q) such that
u(t 1) = o. Since v(S)(b) = v(S2)(b 2 ) = 1 there is t £ h2 with v(S2)(t) = 1; hence t 1 x t £ t 1 x h n d(S,9(u,v) S). 2 Proposition 4ii) The case that u(b) = 1 for b
£
d(S,9(u,v) S) is proofed similarly.
To estimate the cardinality of h £ H(S) we have to estimate the cardinality of the sets being used for the construction.
Proposition 5:
For S E ~ and c E B there is pE pC(S,E) such that Ipl ~ 2.
From Proposition 5 follows Lemma 2: min Ihl hEH(S) ii)
min (2 min Ihl + minlhl, minlhl + 2 min Ihl hEH(SI) hEH(S2) hEH(Sl) heH(S2)
For S = Go (S' min hEH(S)
Proof:
~
Ihi
~
S
I)
£
B
min IhI + 1 hEH(S' )
The proof follows from Definition 9 and Proposition 5 for ii). i) follows as for some p E pC(I,E) we have p E H(I) and Ipl = 1.
From Lemma 2 we get the following relation between the test-comple~ity C(9,S) of nets S E ~ and the number of inputs of S.
Theorem 2: Proof:
S E B => C(9,S) S nld3
Because of Proposition 2 the assertion holds for S £ ~ with Q(S) For all SE B with Q(S) 2: 2 we show by induction that Ihl ~ 3ld for some h E H(S).
1.
- 12 -.
The assertion holds for all 5 E B with 0(5) = 2 from Definition 9, Let 5 = G 0(5 1 x 52) E ~, such that the assertion is true for 51' 52' Let O(5 i ) = ni for i = 1,2 and let !Hm) = (5 £ ~ 10(5) = m} . From Lemma 2 and the preassumption we get: max (minlhl) 5EB(n) hEH(5)
= ma~·
n1'n 2 2:2 nI +n 2=n
(min(2 min Ih I + min Ih I hEH(5 I ) h£H(5 2 ) + 2 ;nin Ihl )))
hEH(5 2 )
,s
max ,
2 S nIS n
nI +n 2=n
2
where we assume n1 ,s n2 without loss of generality. The function f = It x It -. Il,f(x,y) - 2'3 ldx + 3 ldy with x + Y = z and x ,s y has a maximum for x = y =~. Therefore we get for x, y, Z E 1R+ with x + y = z: ld z 1dZ max 2.3 ldx + 3 ldy = 3.3 ~ = 3 ld2 ,3 ~ = 3 ldz x,sy x +Y = z
From theorem follows the assertion with 1093n 1 109 33 31dn = 3'0932 = n1093Z = n'og3 2 = nld3
·Neighbouring relations and bridging faults, 5ince in practice bridging faults occur between neighboured lines only on the board, we introduce some neighbouring relations on E, We denote by Ei the set· of all edges lying on the path from the i-th input of a net 5 to the output of 5,
- 13 Definition 10: a) R(S) c ll(ExE) for SE.!! is defined inductively: ({(u,v) I UE E1(G),v E E2 (G)}}
i) R(G) R( I)
i i) for S
R( n) : = =
~
for G E {d,k}
;
G0
r i E R(Si) and 1 s ji
~
Q(Si) for i
= 1,2.
b) For SE.!! and r E R(S) define: i) r (8,S)
{8 (u,v) S E 8 (S)
i i) R (8,S)
(u,v) E r}
(r (8,S) IrE R(S)}
For all r. (8,S) E R(8,S) we construct complete sets similar to those of Definition 9. Since for S E Band Ei(S) with i ~ i s Q(S) there exists PE pC(S,E i ) with Ipl ~ I-we get the following result which must be given without proof. Theorem 3:
For SE Band r'E R(S)
C(r(8;S), S) s Q(S) + 1.
For the pro·of see /1/ . Complete test-sets of bridging and stuck-at faults. There ,is no triv·ial relation between bridging and stuck-at faults such that. every complete test-set for stuck-at faults is also complete test-set for How.::ver the follo~Jing theorem holds: bridging faults or vice versa. Theorem 4: For S E ~ let L(S) be the set of multiple stuck-at faults of S. Then we have: H(S)
c
T (L,Sj
- 14 For the proof of theorem 4 see /1/ or /4/. Conclusion We have presented complete test - sets of fanout - free circuits which detect all multiple stuck - at faults as well as all single bridging faults. For the test-complexity of fanout free circuits we have derived an upper bound which is polynomial in the number of the circuit inputs. References: /1/
A. Vogel:
"KurzschluBfehler in Schaltkreisen" to appear in "Berichte der Abteilung Informatik der Universitat llortmund", BRD,(1977).
/2/
K.C.Y. Mei: "Bridging and stuck-at faults", IEEE-TC, Vol. C 23, No. 7 (1974), pp. 720 - 727.
/3/
A.D. Friedman: "Diagnosis of short-circuit faults in combinatorial circuits", IEEE-TC, Vol. C-23, No. 7, (1974), pp. 746-752.
/4/
W. Coy:
"Die Testko~lexitat von Schaltkreisen", Dissertation Darmstadt, (1975).