LOWER BOUNDS ON COMMUNICATION ... - Semantic Scholar
LOWER BOUNDS ON C O M M U N I C A T I O N pavol Duris* Slovak Academy
Zvi Galil* Columbia University Tel-Aviv University
of S c i e n c e
Abstract: We prove the following four results on c o m m u n i c a t i o n complexity: i) For every k ~ 2, the language L k
p r o t o c o l requires exchanging n(n~/k41og3n) bits if any (k-l)-round protocol is used. 2) For every k ~ 1 and for infinitely many n ~ i, there exists a c o l l e c t i o n
0. I n t r o d u c t i o n . Suppose that a language L ~ [0, i}* must be recognized by two distant computers. Each computer receives half of the input bits, and the computation proceeds using some protocol for communication between the two computers. The minimum number of bits that has to be e x c h a n g ed in order to successfully recognize
{0, i} 2n that can be recog-
nized by exchanging O(k log n)**bits using a k - r o u n d protocol, and any (k-l)n round protocol recognizing L k requires exchanging O(n/k) bits. 3)
Given a set L c
{0, i} 2n,
there
L n [0, I} 2n, minimized over all Partitions of the input bits into two equal parts, and c o n s i d e r e d as a function of n, is called the c o m m u n i c a t i o n complexity Of L. This model was suggested by Papadimitriou and Sipser [i]. They motivated it by pointing out its relation to lower bound proofs in VLSI [2,3,4]. A closely related model, where the p a r t i t i o n of the input is fixed was studied in [5,6]. Both versions w e r e also studied in [7,8]. We now review the model [I] : A protocol on 2n inputs is a pair D = (~,~), where n (a) ~ is a p a r t i t i o n of {1,2 ..... 2n} into two equal sets S I and Sii (this cor-
is a set ~ {0, i} 8n such that any (kround) protocol recognizing ~ can be t r a n s f o r m e d to a (k-round) fixed p a r t f t i o n p r o t o c o l recognizing L with the same c o m m u n i c a t i o n complexity, and vice versa. 4) For every integer function f, I ~ f ( n ) % n, there are languages recognized by a one round d e t e r m i n i s t i c protocol exchanging f(n) bits, but not by any nondete~uinistic protocol exchanging f(n)-I bits. The first two results show in an incomparable way an exponential gap between (k-l)-round and k-round protocols, * R e s e a r c h supported by National S c i e n c e F o u n d a t i o n Grant MCS-8303139' **All logarithms in the paper are of base 2.
responds to the p a r t i t i o n of the input into the two halves for the two computers); and (b) ~ is a function from
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~J 1984 ACM 0 - 8 9 7 9 1 - 1 3 3 - 4 / 8 4 / 0 0 4 / 0 0 8 1
Georg Schnitger Penn State U n i v e r s i t y
settling a conjecture by P a p a d i m i t r i o u and Sipser. The third result shows that as long as we are interested in existence proofs, a fixed partition of the input is not a restriction. The fourth result extends a result by P a p a d i m i t r i o u and Sipser who showed that for every integer function f, 1 ~ f(n) ~ n, there is a language accepted by a deterministic protocol exchanging f(n) bits but not by any deterministic p r o t o c o l exchanging f(n) - 1 bits.
of encodings of directed graphs of out degree one that contain a path of length k+l from the first vertex to the last vertex and can be recognized by exchanging O(k log n) bits using a simple k - r o u n d
of sets L~ ~
COMPLEXITY
(0,i} n × [0, i,$}* t O (0,i}* U [accept, reject]. Intuitively, the first argument of is the local part of the input, while the second argument is the "log" of all previous messages, with $ serving as the delimiter b e t w e e n messages. The result of is the next message. For a given string
$00.75 81
c E [0, i,S]*,
the function
~
We also consider n o n d e t e r m i n i s t i c p r o tocols and the c o r r e s p o n d i n g class NCOMM(f(n)). In n o n d e t e r m i n i s t i c protocols is a "nondeterministic function"; i.e. it may h a v e several values (and t h e r e f o r e it is not a function). The d e f i n i t i o n s above apply if w h e n e v e r we w r i t e ~(x,c) we m e a n a p o s s i b l e value of ~(x,c). In [i], P a p a d i m i t r i o u and Sipser gave two open problems. The first is r e l a t e d to their main r e s u l t in which they showed a language L e N C O M M ( i o g n) - COMM(cn), for some c > 0; i.e. an exponential gap b e t w e e n det e r m i n i s t i c and n o n d e t e r m i n i s t i c protocols. However, L ~ NCOMM(cn) and they asked w h e t h e r there is a language in N C O M M ( i o g n) N c o - N C O M M ( l o g n) - COMM(cn) (i.e., w h e t h e r a language such that it and its c o m p l e m e n t are easy n o n d e t e r m i n i s t i c a l l y but exponentially h a r d e r deterministically). Recently, Aho Ullman and Y a n n a k a k i s [8] a n s w e r e d the q u e s t i o n affirmatively. P a p a d i m i t r i o u and Sipser d e f i n e d the n o t i o n of k - r o u n d protocols in w h i c h u p to k messages are exchanged. They d e n o t e d COMMk(f(n)) and NCOMMk(f(n)) the cortes u
has the pro-
perty that for every two y,y' E [0, i] n, ~(y,c) is not a proper prefix of ~(y',c) and if ~(y,c) E [accept,reject] then ~(y',c) = ~(y,c). (This second part is m i s t a k e n l y missing in [I].) ThiS p r e f i x - f r e e n e s s prop e r t y assures that the exchanged messages are self-delimitinq, and that no extra "end of transmission" symbol is required. A computation
of D n on input xE[0,1] 2n
is a string c = Cl$C2$...SCk$Ck+l, k ~ 0, Cl, .... c k E [0,13",
where
Ck+ 1 E [accept,
reject], and such that, for each integer ~, 0 ~ ~ < k, we have: (i) if L is odd, then C~+ 1 = ~ ( x i , c l $ c 2 S . . . S c ) , w h e r e x I is the input if i
x r e s t r i c t e d to the set Si; and is even, then c . =
(2)
~ ( x i i , c l S c 2 S , . . . , S c _ £) . ~ + l In other words, in a c o m p u t a t i o n the two c o m p u t e r s take turns computing the next m e s s a g e to be sent, by c o n s u l t i n g the local input and all previous exchanges ( a n d using, w i t h o u t loss of generality, the same function ~). Obviously, this process is c o m p l e t e l y deterministic. The lenqth of a computation c is the total length of all messages in c (ignoring $'s and the final accept/reject). Let L ~ [0, l] 2n be a language, and D n be a d e t e r m i n i s t i c protocol. W e say D re2
p o n d i n g classes of languages w h e n we restrict ourselves to k - r o u n d protocols. They d e f i n e d the languages L k = [W0Wl.. . W 2 m _ l l W i ~ [0,1] m and zJ0'''''Jk+llWji Jk+l = 2m-i]"
n--
L if, for each x E [0, i] n, the c o m p u t a t i o n of D on input x is always n finite, and ends w i t h accept iff x E L. Let f be a function from integers to integers. W e say that L is r e c o q n i z a b l e w i t h i n communication f, L E COMM(f(n)), if there is a protocol D recognizing L such that for n all x E [0,i] 2n the c o m p u t a t i o n of D n on x has length at most f(n). Let L c [0, i]* be a language, ~ = a sequence of d e t e r m i n i s t i c p r o t o c o l s cognizes
directed
= Ji+l w h e r e J0 = 0 and
A m e m b e r of L k encodes
graph of o u t d e g r e e
1 having
a a path
Of length k + 1 from v e r t e x 0 to v e r t e x 2m-l. It is easily seen that L 2 E COMM2(2 log n) and in fact L k E C O M M k ( k
log n).
They show-
ed that L2 ~ C O M M I ( ~ / ( 2
log n)),
thus ex-
h i b i t i n g an e x p o n e n t i a l gap b e t w e e n oneand two-round protocols. The second open p r o b l e m in their paper was w h e t h e r a similar gap exists b e t w e e n k- and (k-l)-round protocols. They c o n j e c t u r e d that indeed this is the case and that L k is the w i t n e s s to this fact.
and f a f u n c t i o n from i n t e g e r s to integers. W e say that d recognizes L (L is r e c o q n i z a b l e w i t h i n c o m m u n i c a t i o n f, L ~ COMM(f(n))) if Dn recognizes L n ~ L N [0, l]2n (if L n E COMM(f(n))) for
C o n j e c t u r e [i] : For every k > i, there is an L such that
all
n. The p r e f i x f r e e n e s s p r o p e r t y is motivated in [i]. W e need it only for our last r e s u l t w h e r e we w a n t to pin down exactly the c o m m u n i c a t i o n complexity. In other cases we augment the messages w i t h an endmarker. W e do not change the d e f i n i t i o n of the length of the message. Even if we c o u n t e d it in the first three results we w o u l d at most double the c o m m u n i c a t i o n complexity.
Lk ~ C O ~ k _ l ( n l / ~ ) . Our first two results show that indeed there is an e x p o n e n t i a l gap b e t w e e n k- and (k-l)-round protocols. T h e o r e m l, settles the c o n j e c t u r e above in almost the s t r o n g e s t sense. Theorem
i:
For every k ~
COMMk(nl/2/(36k4
82
log 3 n)).
i, Lk+ 1
Remark: E COM~
and c o n s i d e r Lk n [0,i] 2n,
One can easily show that Lk+ 1 (nl/2/(log n) 1/2).
The p r o o f of T h e o r e m 1 is c o m b i n a t o r ial. W e h a v e found a w a y to "force" our intuition. For m a n y open p r o b l e m s in comp u t a t i o n a l c o m p l e x i t y the s o l u t i o n is int u i t i v e l y clear: intuitively, P ~ NP bec a u s e we m u s t c h e c k all a s s i g n m e n t s w h e n solving SATISFIABILITY. U n f o r t u n a t e l y , we can rarely t r a n s f o r m such an i n t u i t i o n into a proof. In our case, our i n t u i t i o n tells us that if the two computers h a v e the w r o n g vertices, they m u s t e x c h a n g e k + 1 internal v e r t i c e s in order to check w h e t h e r a p a t h of length k + 2 exists. So, if only k rounds are allowed, the c o m p u t e r that is s u p p o s e d to m a k e the d e c i s i o n w i l l be "one v e r t e x behind". The other c o m p u t e r w i l l h a v e to send h i m a long list of values not k n o w i n g w h a t is the (k+l)-st v e r t e x on the path. Of course, our computers do not nec e s s a r i l y get the w r o n g v e r t i c e s neither do they always e x c h a n g e vertices. W h a t is worse, the input is p a r t i t i o n e d a r b i t r a r i l y and each c o m p u t e r may get only p a r t of the bits of the various edges. W e found a way a r o u n d it. By r e s t r i c t i n g a t t e n t i o n to a s u b s e t of the inputs w h i c h is large e n o u g h w e w e r e able to find graphs w i t h k + 2 layers such that, indeed, the two c o m p u t e r s h a v e the w r o n g vertices. Starting with this subset of inputs we fix a c e r t a i n p a t h by adding one v e r t e x at a time. Each time we further r e s t r i c t the inputs to c o n t a i n this initial path, say of length i, and to h a v e the same i m e s s a g e s exchanged. Not a l l o w i n g long enough messages, w e are still left w i t h a large number of such inputs, so after k m e s s a g e s the r e m a i n i n g set has both: inputs in Lk+ 1 and not in Lk+l,
L ~ [0, i] 2n d e s c r i b e d by 2 n × 2 n matrices. Once a p a r t i t i o n ~ of the input bits is g i v e n the set is fully d e s c r i b e d by a 0 - 1 m a t r i x M(L,~) w i t h 2 n rows and columns corr e s p o n d i n g to the p o s s i b l e bit strings seen by I and II. The proof of T h e o r e m 2 considers a fixe d p a r t i t i o n of the input. This is justified by T h e o r e m 3 stated below. Once a partition n is fixed we can assume w i t h o u t loss of g e n e r a l i t y that ~ = ~0 the natural p a r t i t i o n that gives I the first h a l f of the input. W e call the m a t r i x M ~ M ( L , ~ 0) the m a t r i x that corresponds to L and refer to L as the language that c o r r e s p o n d s to M. The m a t r i x r e p r e s e n t a t i o n is due to Yao [6]. The c o m p u t a t i o n can be v i e w e d as follows: two computers called ROW and C O L U M N h a v e to r e c o g n i z e L. Each c o m p u t e r has one h a l f of the input. (ROW knows the row in the m a t r i x and C O L U M N k n o w s the column.) They a l t e r n a t e sending m e s s a g e s (each one of them can start). B o t h computers k n o w the m a t r i x of L. At any stage, each i E [ROW,COLUMN] knows the subset S. of inputs the other may still have. W h e n l o n e of them sends a m e s s a g e the other one, j, obtains i n f o r m a t i o n that enables h i m to m a k e S. 3 smaller. In fact the p o s s i b l e messages j receives imply a p a r t i t i o n of S.. The com] p u t a t i o n terminates w h e n one of them, say ROW, has SRO W such that all the entries in
b e c a u s e s o m e initial paths h a v e the
c o m p l e t i n g edge and some don't. All of this is a c h i e v e d by an i n t e r e s t i n g i n d u c t i v e argument. This c o n t r a d i c t i o n proves the theorem. W e g i v e n another p r o o f for the exponential gap b e t w e e n k- and (k-l)-round complexity.
the row ROW has and the columns of S R W O
S C O L U M N should h a v e the same entries
(zeros
or ones), b e c a u s e all c o r r e s p o n d i n g input pairs h a v e the same communication. W e c o n s t r u c t the languages L~ in-
- COM~_l(n/4k).
d u c t i v e l y by c o n s t r u c t i n g the c o r r e s p o n d i n g n n m a t r i c e s M k. The m a t r i c e s ~ are d e r i v e d
T h e o r e m 1 and 2 are incomparable. On the one h a n d the gap of T h e o r e m 2 is wider. AS w e r e m a r k e d above, there is no such a large gap for L k of T h e o r e m i. A l s o if w e k
are
the same (0 or I). Note, that the submatrix c o r r e s p o n d i n g to the final S R O W and
T h e o r e m 2: For all k ~ 1 and for i n f i n i t e l y many n w i t h k ~ n/(96 log n) there exists n c [0, l}2n such that L k n E COM~(k Lk log n)
take in T h e o r e m 1
then T h e o r e m 2
is m e a n i n g f u l for a w i d e r range of k. On the other hand, w h i l e the languages of Theorem 1 are simple and constructive, those of T h e o r e m 2 are nonconstructive. The p r o o f of T h e o r e m 2 is an e x i s t e n c e proof. In addition, the two proofs are entirely different. W e suggest as an o p e n problem, to p r o v e a w i d e gap (from logn to ckn ) for a c o n s t r u c t i v e language. The proof of T h e o r e m 2 considers sets
from simple matrices. The latter are obtained by r e p e a t i n g b - a r y r e p r e s e n t a t i o n of the numbers 1,2,...,~times. (b and ~ are
to be a f u n c t i o n of n 83
carefully chosen parameters.) Then, all i's n in these matrices are replaced by ~i(Mk_2)
tocols, b e c a u s e Aho et al showed [8] that for fixed p a r t i t i o n there is only a polynomial (square) d i f f e r e n c e b e t w e e n determ i n i s t i c and n o n d e t e r m i n i s t i c protocols. The proof Of T h e o r e m 3 uses again prob a b i l i s t i c arguments. The m a t r i x M* = M(L*,r) is o b t a i n e d from M ~ M(L,~) by first d u p l i c a t i n g a large number of times the rows and columns of M and then by choosing two r a n d o m p e r m u t a t i o n s and p e r m u t i n g the rows and the columns of the r e s u l t i n g matrix. To establish (b) one observes that there are two such p e r m u t a t i o n s such that for any p a r t i t i o n of the input bits, the correspondihg m a t r i x contains a full copy of M or of MT° Note that the proof of T h e o r e m 3 introduces additional n o n c o n s t r u c t i v e n e s s to the languages of T h e o r e m 2. The second main result in [i] was showing that for any integer function f, 1 ~ f(n) ~ n, COMM(f(n)) - COMM(f(n)-i) ~ ~. Our last result is:
~i is a "random" permutation. The resulting m a t r i x is M~. In the p r o o f we define a m e a n i n g f u l p o r t i o n of M~. Let Pj be the claim that the s u b m a t r i ~ c o r r e s p o n d i n g to SRO W and S C O L U M N contains a m e a n i n g f u l p o r t i o n of n M.. Then we show inductively that if P. 3 3 holds, then after exchanging two messages Pj-2 holds. The proof makes use of the randomness of ~i" A n o t h e r way to look at it is that by some interesting counting arguments we show that there exist permutations n0,...,~b_l such that the above holds. The proof terminates w h e n we o b s e r v e that a m e a n i n g f u l p o r t i o n of M~ must c o n t a i n zeros and ones. R e c e n t l y [9] Yao c o n s i d e r e d p r o b a b i l istic p r o t o c o l s and p r o v e d an exponential gap b e t w e e n one- and t w o - r o u n d p r o b a b i l i s tic protocols. It is an interesting open p r o b l e m to p r o v e a result similar to Theorem 1 or 2 for such protocols. W h e n we fix a p e r m u t a t i o n n, we speak of a p r o t o c o l ~ = (~c,~r) (where c(r)
stands f o r
Let L 1 c
column
[0, i] 2n, L 2 c
(row))
for
T h e o r e m 4: For any integer function f, 1 ~ f(n) ~ n, C O M M i ( f ( n ) ) - N C O M M ( f ( n ) - i ~ .
protocol
(Li,~i) ~ (~c,~r)
functions
fl, f2:
for
(L,~).
[0,i] 2m w i t h parti-
(b) (L2,n2)
for
or
(~r(fl,),We(f2,
(L2,~2)
(L2,n2)~
T h e o r e m 3:
2:
NCOMM(f(n))-NCOMM(f(n)-i)/~.
The Proof of T h e o r e m
a contradiction. if
(Ll,n I)
For each language L ~
i.
W e assume and derive
Let ~ = [Dn] be the cor-
r e s p o n d i n g k - r o u n d p r o t o c o l s that r e c o g n i z e Lk+ I. W i t h o u t loss of g e n e r a l i t y each com-
(Li,~i).
into
- C O M M ( f ( n ) - i ) / ~.
Lk+ 1 E C O ~ ( n l / 2 / ( 3 6 k 4 1 o g 3 n ) )
p u t a t i o n contains exactly k exchanged messages: by adding two bits we can record the fact w h e t h e r the input has b e e n accepted, r e j e c t e d or neither. This increases the c o m m u n i c a t i o n c o m p l e x i t y by a c o n s t a n t (2k). The proof consists Of three parts. We first define several constants and p r o v e a r e l a t i o n s h i p among them (Claim I). Next we define a subset of the inputs, S, corresp o n d i n g to c e r t a i n graphs. Then we p r o v e L e m m a 1 from w h i c h the t h e o r e m follows. W e c o n s i d e r inputs of length 2n = m2 m, n large enough as will be e x p l a i n e d below. W e choose the constants a, r, p (an integer), ~ and ~, t and s (an integer) in this order to satisfy
Remarks: The two cases in (a) mean that w e treat s y m m e t r i c a l l y ROW and COLUMN. If (Li,~i) ~ (L2,~2) , then any (Li,~i) comput a t i o n i m m e d i a t e l y translates c o m p u t a t i o n and v i c e v e r s a .
Corollary
i.
))
(L2,~2).
(Li,~i) ~ and
there exist
{0,i] n ~ {0, i] m such that
(~c(fl,)),~r(f2,)) is a p r o t o c o l
(L2,~2) , if for each (Li,~i)
I: COMM(f(n))
T h e o r e m 4 extends the r e s u l t in [i] (Corollary i) to n o n d e t e r m i n i s t i c protocols in the s t r o n g e s t way. There seems to be no way to change the direct proof of C o r o l l a r y 1 in order to prove theorem 4. The proof of T h e o r e m 4 is rather simple. The s t r u c t u r e of the paper is as follows: the p r o o f of T h e o r e m i, i = 1,2,3,4, appears in S e c t i o n i.
tions ~i,~2. (a)
Corollary
(L2,~2) [0, i] 2n
and p a r t i t i o n ~ there exists L* c {0, i] 8n such that (a) there exists 7 with (L,~)~(L*,7) (b) for all ~ (L,~) ~ (L*,c). In other words, w h e n e v e r lower and upper bounds are p r o v e d for a language L and a fixed p a r t i t i o n ~, then there exists another language L* such that these bounds h o l d ind e p e n d e n t l y of the partition. It is i n t e r e s t i n g to note that T h e o r e m 3 does not h o l d for n o n d e t e r m i n i s t i c pro-
(i)
84
n a = nl/2/(36k41og3n),
(2)
i i log n a=~ (r= 3 log log n+2 log 6k 2)"
(3)
p =
for each block b a set Xb ~ [0, i] m of p o s s i b l e inputs. (a) X 0 contains the u n i q u e number in B 1• (b) For b e B. and 1 ~ i ~ k we d e ~ fine X b as fol~ows. There are ~ m - ~ strings r e p r e s e n t i n g numbers in Bi+ I.
[r],
1 1 ~ = - - -2 2p ~ = log(3kp),
(4) (5)
(6) t = "-I2m=-~I/2, and (7) s = [~ T h e s e constan£s h a v e b e e n c h o s e n so that C l a i m i: Proof:
If n is large enough, By
(2) and ~log
1
(3), ~ p
T h e s e are p a r t i t i o n e d according to the fixed bits of b l o c k b into 2 m - ~ m subsets. One of these subsets has at
then s>kn a.
!es_2
1
- log n ~ 2r+2
least~m~-~]
log n + log
log n log log n + log k + ~ + 1 log n (by (2)-(5))
n ~-a >
(log n) 1/2k2~+i"
large enough, n
>
(log n) ~ S ~
t-I/2 =
Hence,
> 2 ~ + i k n a.
and
if n is
Thus
(log n) I/2 ~ m ~ - ~ ] / 2 - i / 2 2 2 m ~ - ~ - I - i / 2 > knM.D
Each input c o n s i s t s of 2 m blocks of length m w h i c h w i l l be i d e n t i f i e d w i t h the numbers O,l,...,2m-l. C o n s i d e r i n g the p r o t o c o l Dn: (n,~), each c o m p u t e r I, II sees a p a r t (possibly empty) of each b l o c k (according to ~). W e say that b l o c k i is free for one of the c o m p u t e r s if it sees at least ~m bits in it. (without loss of generality ~m is an integer.) N o t e that since 1 < ~, a b l o c k may be free for the two computers. Note also that there are at least 2m/(p+l) blocks free for each c o m p u t e r (because o t h e r w i s e the other c o m p u t e r w o u l d see m o r e than (2m-2m/(p+l)) (m-am) = m2 m-I = n bits). W e now i d e n t i f y k + 2 d i s j o i n t sets of blocks Bi, i = 0,1,...,k+l, B i
tain tains going tains
one v e r t e x (= block). L a y e r 0 conb l o c k O, and is c o n n e c t e d (by an outedge) to layer I. L a y e r k + 2 conb l o c k 2m-l. L a y e r i, 2 ~ i < k + I,
c o n t a i n s "~|2m-~| vertices. Each v e r t e x in layers l,...,k is c o n n e c t e d to one of [2 mu-~] s p e c i f i c v e r t i c e s in layers i + i. V e r t i c e s in layer k + i are c o n n e c t e d to one of two vertices, exactly one of w h i c h is the one in layer k + 2. Moreover, for 1 ~ i < k , i odd (even) II (I) has the entire i n f o r m a t i o n on layer i, b e c a u s e for each b l o c k in B i he has the free bits. I(II) has no i n f o r m a t i o n at all on layer i b e c a u s e all fixed bits in B i have the same v a l u e in X b. To each input x in S c o r r e s p o n d s a d i r e c t e d p a t h that starts at v e r t e x 0, goes through layers 1,2,...,k+l and either terminates in layer k+2, in w h i c h case x ¢ L or not, in w h i c h case x ~ L. F r o m now on we c o n s i d e r only inputs in S. For i = 1,2,...,k+l, let Pi be the possible input segments in the blocks of B i (the m a r k e d c o n c a t e n a t i o n of X b for b in Bi). A n element of S is r e p r e s e n t e d by an element of P1 × P2 ×'''× Pk+l" For conven-
{0,i,...,2m-2} that s a t i s f y (i) B 0 = ~0].
(ii)
choose from
tains l's (O's) in the fixed (free) bits of the block b. (d) If b ~ UiBi, X b = [0 m] (any fixed v a l u e w i l l do). W i t h the graph i n t e r p r e t a t i o n in mind, ignoring the blocks not in UiB i w e h a v e r e s t r i c t e d a t t e n t i o n to the following inputs. The p o s s i b l e d i r e c t e d graphs h a v e k + 3 layers (the first k + 2 c o r r e s p o n d to B0,...,Bk+i). Layers 0,i and k + 2 con-
since 2n = m2 m, 2 m~ - 2 ~ > n
We
one such s u b s e t [2 m~-~] elements to form X b. N o t e that the so called fixed bits h a v e fixed values in X b. (c) If b E Bk+l, ~ = [im,yb} , Yb con-
_ log(r+l) > 6k2 if n is log n log n 1 log p + large enough. So ~ - a ~ ~ p > log n 1 log log n + log 6k 2 1
strings.
rBlf = i.
(iii) 1Bii = ~ m - ~ ] for i = 2 ..... k+l. (iv) For i = l,...,k+l, i odd (even), the blocks in B i are free for II (I). A s i m p l e c o u n t i n g a r g u m e n t shows that this is indeed possible. W e say that the blocks in B~ w i t h odd (even) j b e l o n g to J II (I). C l e a r l y if a b l o c k belongs to I(II) then it is free for I(II). For each b l o c k that belongs to I (II) we c h o o s e ~m bits that I (II) sees, call them free bits, and call the other fixed bits. W e now d e s c r i b e a s u b s e t of the poss i b l e inputs S ~ X 0 - X l . . . X 2 m _ l , s p e c i f y i n g
ience we also include Pk+2 w h i c h is the set c o n t a i n i n g the empty string. W e d e s c r i b e b e l o w a process that chooses in turn values from Pi,P2,-... ter i stages,
Af-
values from Pi,...,Pi_i h a v e
already b e e n chosen and the value from Pi
85
is restricted 1 s [w i .... ,wi].
to one of
s
possible
1 s w i , . . . , w i are any
values
The input will be determined
once an element
of Pi × Pi+l x...
Pk+l is
chosen. If layer i belongs to computer I (say), then the input is determined once one of the s values of Pi as well as an element of Pi+2 × Pi+4"'" and an element of Pi+l × Pi+3 ×''"
are chosen.
IVll ~
U~=l{W ~] × W~ ~ Pi N Pi+2 ×''"
Induction i~k-l.
The choice of w~,
step:
of Pi"
The
It can still have
by P3 × P5 x...,
Let q =
so
(b) is immediate. Assume
the lemma holds
for
h P i + l l / ( 2 ( 2 n a ) i ~ . Coheir × Uj, uj e Pi+[" Uj ~ Pi+3 x
der V i = Uj[uj}
Pi+5 x .... U. is said to be large if 3 IUj I ~ ( IPi+3 I" IPi+5 I''" )/(2 (2 ha) i).
from
are allowed
Claim 2: For i < k - i, there are at least q large Uj's.
for I. Note that after stage i, all inputs that are still considered have the same ~ corresponding initial path go = 0, gl,...,g i.
represented
the second half of
The first two
for II and only values
elements
(IP211P41...)/2na.
all inputs
values are known to I, while the third is k n o w n to II. W h i l e fixing values in p~, j = 1,2,... we also restrict in a speclal way the possible continuations. After stage i only values from V i c Pi+l × Pi+3"" are allowed
s
messages sent from I to II in the first round imply a partition of the possible inputs for I. We choose a message c I with the largest corresponding part V I. So, by (i),
Proof : Otherwise, if there were only q' < q large Uj'S, then IVi ! ~ (q' Ipi+ 3 I" IP~+5 I''" )
j = l,...,s
+
(!Pi+ll-q')(!Pi+3
I IPi+5!..-)/2(2n~)i
1/8. But this contradicts Lemma 6. D
from every column block of < .
_Lemma 5: Let B be an u n d i s t i n g u i s h a b l e (l-'eo)-fragment of M~. A s s u m e that ROW sends messages of length < 6 log m w h i c h are answered by messages from COLUMN of
3. Proof of Theorem 3. Let M = M(L,n). W i t h o u t loss of generality ~ = ~0' the natural partition. M = (Cl,...,C2n), w h e r e c. e [0, I} 2n is the i-th column of
length < 6 log C L-I Then there exists k-2" one u n d i s t i n g u i s h a b l e (l-¢0-6-3/~)-fragment m/b of ~k-2 after these 2 rounds, p r o v i d e d
1
M. Let M 1 = (Cl,...,Cl,...,C2n,...,C2n), w h e r e each c. is repeated 23n times. T l M1 (r I ..... r2n) w h e r e __r' E [0,l} 24n is
.
I-~¢0-6-3/~ > 1/8.
=
•
m¢/2b2). at least
L e m m a 7: There is a m a t r i x M* equivalent to M2, such that the language L* corresponding to M* satisfies the following property: for each p a r t i t i o n 6, one of the matrices M, M T is equivalent to a submatrix of M(L*,6). W e consider M and M T in the lemma because we will allow either ROW or COLUMN to start the computation. Lemma 7 establishes part (b) of Theorem 3. Part (a) is immediate with ~ = ~0 because of the way M ~ was obtained from M. L e m m a 7 makes use of Claim 6 below.
2
But all column blocks of B 1 have .C~-i [ k_2 ] i-¢ 0 columns in each block.
So B 1 has one column block with i)
2)
b e-I/L inserted'matrices me/2b 2 ~ m ¢-3/~ rows each and
at least
having
1
the i-th row of M I. Let ~' = (r I ..... r 1 ..... r ..... r n ), where each r i ~, 2 n 2 is repeated 2 °~* times. W e say that two matrices A, B are equivalent if there are p e r m u t a t i o n matrices P,Q, such that B = PAQ. We prove below:
Proof: First p a r t i t i o n B according to the row messages. W e c o n c e n t r a t e on the largest class B 1 which must contain at i-¢0- 6 least m rows. A p p l y i n g L e m m a 3 we find: there exists one column block (= digit position) , Bi, w h e r e at least b ~-I/~ (¢ = i-¢N-6) inserted matrices (= digits) ~¢ h a v e m /2b rows each (= f r e q u e n c y ~ -
([e - --3.)-fragment
Lemma 6: Let e > 1/8. Every e-fragment of bL-t M1 contains zeros and ones.
Two A d d i t i o n a l Lemmas. A submatrix A of is said to be u n d i s t i n g u i s h a b l e after the i-th round, if for all rows and columns of A as p o s s i b l e inputs for ROW and C O L U M N the same first i messages are exchanged. A submatrix B of M~ is called a efraqment of M~ if B has at least m e rows ~-I e and there are at least (Ck_2) columns in B
Therefore we
[Ck-2 ~-I-] i-¢0 co lumns.
Next, p a r t i t i o n this column block according to the column messages. A g a i n we 89
Cl__._a!~ _6: Let G = [i .... ,24n] and let .... ,~p, p = ( ~ ) b e p different subsets zn of G of size 2 , and let £=GIU£2...£2 n
which correspond to the choices of ~ ~). G = ~.•UG2n is the p a r t i t i o n of G to classes of copies of rows (columns) of M. The conclusion of the claim implies that there is a p e r m u t a t i o n ~(~) such that for every possible choice of ~(~), if we permute the rows (columns) of M 2 by ~(7), we still have at least one copy of every row (column) of M among the rows (columns) of M 2 that correspond to ~(~) (~(i)). D
be a partition of G into disjoint sets of size 23n. Then, there is a p e r m u t a t i o n of £ such that for all i,j, 1 ~ i ~ p, I~ j~2 n (15)
~(Xi)
N Gj ~ ~.
Proof: Let Pi j be the number tions of £ tha£ violate (15). Pi,j (24n): =
=
i
4.
(24n - 22n)23n: (24n_23n): (24n): 23n
(24n-23n) "-" (24n-23n-22n+l) 24n
of permuta-
•..
~
Claim 7:
( 1 _ 2 - n ) 2 2n
24n Hence Z i . Pi, " < ( ): D ,3 3 4n 24n Let A be a 2 × matrix. Let i = [i 1 ..... i2n} and i = [Jl ..... J2n ] two between
of the form [xylx E {0,1~n+f(n)]. of them is in COMMi(f(n)). We consider
1 and 4n. Case
A[~,i] is the 22n × 22n submatrix of A that consists of all rows (columns) of A with numbers that are obtained by taking 4n bit strings, assigning bits il,...,i2n (Jl' .... J2n ) in all possible
ways
and all
•
!
of COLUMN
M(L*,6) [~',i']
= M*[~,~]
(column sets)
the following
Each one
two cases.
log n.
at most f(n)-i corend with accept. By
property p ~ 2 f(n)-I I For i = l,...,p, let X i (X ~I ) be the
freeness
set of inputs that computer I (II) sees and correspond to the computation c.. There is a one to one c o r r e s p o n d e n c e between L n and II Ui=iP X~i × X.i • This c o r r e s p o n d e n c e is determined by the p a r t i t i o n n. Therefore, the number of such L n is at most (2nn)~2f~n)-l((22n)2)p= P ~ 22n2 f(n)-l(2f(n)_l)~22n+l 22n(22n+l) 2f(n)-i
are defined
22n22n+f(n) 2 f(n)-I ~ p A 's, there m u s t be x I ~ and y in L n w i t h x E A and e A with m ~ m that c o r r e s p o n d to the same computation c.. But then also the z with l I xi II II z = and z = y m u s t be a c c e p t e d by c. w h i l e z ~ L n. l
yI
5.
References.
[i] C.H. P a p a d i m i t r i o u and M. Sipser, C o m m u n i c a t i o n complexity, Proc. 14th A n n u a l A C M Symp. o n Theory of Computing, 330-337, 1982. [2] C.D. Thompson, A r e a - t i m e c o m p l e x i t y for VLSI, proc. llth A n n u a l A C M Symp. on T h e o r y of Computing, 81-88, 1979. [3] R.J. L i p t o n and R. Sedgewick, L o w e r bounds for VLSI, Proc. 13th A n n u a l A C M Symp. on T h e o r y of Computing, 300-307, 1981. [4] A.C. Yao, The entropic limitations on VLSI computations, Proc. 13th A n n u a l A C M Symp. on T h e o r y of Computing, 308-311, 1981. [5] H. Abelson, L o w e r bounds on information t r a n s f e r in d i s t r i b u t e d computations, Proc. 19th A n n u a l IEEE Symp. F o u n d a t i o n s of C o m p u t e r Science, 151-158, 1978. [6] A.C. Yao, S o m e c o m p l e x i t y q u e s t i o n s r e l a t e d to d i s t r i b u t i v e computing, Proc. llth A n n u a l A C M Symp. on T h e o r y of Computing, 209-213, 1979. [7] K. M e h l h o r n and E.M. Schmidt, Las Vegas is b e t t e r than d e t e r m i n i s m in VLSI and d i s t r i b u t e d c o m p u t i n g (extended abstract), Proc. 14th A n n u a l A C M S y m p o s i u m on Theory of C o m p u t i n g ~ 330-337, 1982. [8] A.V. Aho, J.D. U l l m a n and M. Yannakakis, On notions of i n f o r m a t i o n t r a n s f e r in VLSI circuits, Proc. 14th A n n u a l Symp. on Theory of Computing, 133-139, 1983. [9] A.C. Yao, L o w e r bounds by p r o b a b i l istic arguments, Proc. 24th A n n u a l IEEE Symp. on F o u n d a t i o n s of C o m p u t e r Science, 420-428, 1983. [i0] J. Ja'Ja', V.K. P r a s a n n a K u m a r and J. Simon, I n f o r m a t i o n t r a n s f e r under d i f f e r e n t sets of protocols, to appear in S I A M J. on Computing.
91
Zvi Galil* Columbia University Tel-Aviv University
of S c i e n c e
Abstract: We prove the following four results on c o m m u n i c a t i o n complexity: i) For every k ~ 2, the language L k
p r o t o c o l requires exchanging n(n~/k41og3n) bits if any (k-l)-round protocol is used. 2) For every k ~ 1 and for infinitely many n ~ i, there exists a c o l l e c t i o n
0. I n t r o d u c t i o n . Suppose that a language L ~ [0, i}* must be recognized by two distant computers. Each computer receives half of the input bits, and the computation proceeds using some protocol for communication between the two computers. The minimum number of bits that has to be e x c h a n g ed in order to successfully recognize
{0, i} 2n that can be recog-
nized by exchanging O(k log n)**bits using a k - r o u n d protocol, and any (k-l)n round protocol recognizing L k requires exchanging O(n/k) bits. 3)
Given a set L c
{0, i} 2n,
there
L n [0, I} 2n, minimized over all Partitions of the input bits into two equal parts, and c o n s i d e r e d as a function of n, is called the c o m m u n i c a t i o n complexity Of L. This model was suggested by Papadimitriou and Sipser [i]. They motivated it by pointing out its relation to lower bound proofs in VLSI [2,3,4]. A closely related model, where the p a r t i t i o n of the input is fixed was studied in [5,6]. Both versions w e r e also studied in [7,8]. We now review the model [I] : A protocol on 2n inputs is a pair D = (~,~), where n (a) ~ is a p a r t i t i o n of {1,2 ..... 2n} into two equal sets S I and Sii (this cor-
is a set ~ {0, i} 8n such that any (kround) protocol recognizing ~ can be t r a n s f o r m e d to a (k-round) fixed p a r t f t i o n p r o t o c o l recognizing L with the same c o m m u n i c a t i o n complexity, and vice versa. 4) For every integer function f, I ~ f ( n ) % n, there are languages recognized by a one round d e t e r m i n i s t i c protocol exchanging f(n) bits, but not by any nondete~uinistic protocol exchanging f(n)-I bits. The first two results show in an incomparable way an exponential gap between (k-l)-round and k-round protocols, * R e s e a r c h supported by National S c i e n c e F o u n d a t i o n Grant MCS-8303139' **All logarithms in the paper are of base 2.
responds to the p a r t i t i o n of the input into the two halves for the two computers); and (b) ~ is a function from
Permission to copy without fee all or part of this material is granted provided that the copies are not made or distributed for direct commercial advantage, the ACM copyright notice and the title of the publication and its date appear, and notice is given that copying is by permission of the Association for Computing Machinery. To copy otherwise, or to republish, requires a fee a n d / o r specific permission.
~J 1984 ACM 0 - 8 9 7 9 1 - 1 3 3 - 4 / 8 4 / 0 0 4 / 0 0 8 1
Georg Schnitger Penn State U n i v e r s i t y
settling a conjecture by P a p a d i m i t r i o u and Sipser. The third result shows that as long as we are interested in existence proofs, a fixed partition of the input is not a restriction. The fourth result extends a result by P a p a d i m i t r i o u and Sipser who showed that for every integer function f, 1 ~ f(n) ~ n, there is a language accepted by a deterministic protocol exchanging f(n) bits but not by any deterministic p r o t o c o l exchanging f(n) - 1 bits.
of encodings of directed graphs of out degree one that contain a path of length k+l from the first vertex to the last vertex and can be recognized by exchanging O(k log n) bits using a simple k - r o u n d
of sets L~ ~
COMPLEXITY
(0,i} n × [0, i,$}* t O (0,i}* U [accept, reject]. Intuitively, the first argument of is the local part of the input, while the second argument is the "log" of all previous messages, with $ serving as the delimiter b e t w e e n messages. The result of is the next message. For a given string
$00.75 81
c E [0, i,S]*,
the function
~
We also consider n o n d e t e r m i n i s t i c p r o tocols and the c o r r e s p o n d i n g class NCOMM(f(n)). In n o n d e t e r m i n i s t i c protocols is a "nondeterministic function"; i.e. it may h a v e several values (and t h e r e f o r e it is not a function). The d e f i n i t i o n s above apply if w h e n e v e r we w r i t e ~(x,c) we m e a n a p o s s i b l e value of ~(x,c). In [i], P a p a d i m i t r i o u and Sipser gave two open problems. The first is r e l a t e d to their main r e s u l t in which they showed a language L e N C O M M ( i o g n) - COMM(cn), for some c > 0; i.e. an exponential gap b e t w e e n det e r m i n i s t i c and n o n d e t e r m i n i s t i c protocols. However, L ~ NCOMM(cn) and they asked w h e t h e r there is a language in N C O M M ( i o g n) N c o - N C O M M ( l o g n) - COMM(cn) (i.e., w h e t h e r a language such that it and its c o m p l e m e n t are easy n o n d e t e r m i n i s t i c a l l y but exponentially h a r d e r deterministically). Recently, Aho Ullman and Y a n n a k a k i s [8] a n s w e r e d the q u e s t i o n affirmatively. P a p a d i m i t r i o u and Sipser d e f i n e d the n o t i o n of k - r o u n d protocols in w h i c h u p to k messages are exchanged. They d e n o t e d COMMk(f(n)) and NCOMMk(f(n)) the cortes u
has the pro-
perty that for every two y,y' E [0, i] n, ~(y,c) is not a proper prefix of ~(y',c) and if ~(y,c) E [accept,reject] then ~(y',c) = ~(y,c). (This second part is m i s t a k e n l y missing in [I].) ThiS p r e f i x - f r e e n e s s prop e r t y assures that the exchanged messages are self-delimitinq, and that no extra "end of transmission" symbol is required. A computation
of D n on input xE[0,1] 2n
is a string c = Cl$C2$...SCk$Ck+l, k ~ 0, Cl, .... c k E [0,13",
where
Ck+ 1 E [accept,
reject], and such that, for each integer ~, 0 ~ ~ < k, we have: (i) if L is odd, then C~+ 1 = ~ ( x i , c l $ c 2 S . . . S c ) , w h e r e x I is the input if i
x r e s t r i c t e d to the set Si; and is even, then c . =
(2)
~ ( x i i , c l S c 2 S , . . . , S c _ £) . ~ + l In other words, in a c o m p u t a t i o n the two c o m p u t e r s take turns computing the next m e s s a g e to be sent, by c o n s u l t i n g the local input and all previous exchanges ( a n d using, w i t h o u t loss of generality, the same function ~). Obviously, this process is c o m p l e t e l y deterministic. The lenqth of a computation c is the total length of all messages in c (ignoring $'s and the final accept/reject). Let L ~ [0, l] 2n be a language, and D n be a d e t e r m i n i s t i c protocol. W e say D re2
p o n d i n g classes of languages w h e n we restrict ourselves to k - r o u n d protocols. They d e f i n e d the languages L k = [W0Wl.. . W 2 m _ l l W i ~ [0,1] m and zJ0'''''Jk+llWji Jk+l = 2m-i]"
n--
L if, for each x E [0, i] n, the c o m p u t a t i o n of D on input x is always n finite, and ends w i t h accept iff x E L. Let f be a function from integers to integers. W e say that L is r e c o q n i z a b l e w i t h i n communication f, L E COMM(f(n)), if there is a protocol D recognizing L such that for n all x E [0,i] 2n the c o m p u t a t i o n of D n on x has length at most f(n). Let L c [0, i]* be a language, ~ = a sequence of d e t e r m i n i s t i c p r o t o c o l s cognizes
directed
= Ji+l w h e r e J0 = 0 and
A m e m b e r of L k encodes
graph of o u t d e g r e e
1 having
a a path
Of length k + 1 from v e r t e x 0 to v e r t e x 2m-l. It is easily seen that L 2 E COMM2(2 log n) and in fact L k E C O M M k ( k
log n).
They show-
ed that L2 ~ C O M M I ( ~ / ( 2
log n)),
thus ex-
h i b i t i n g an e x p o n e n t i a l gap b e t w e e n oneand two-round protocols. The second open p r o b l e m in their paper was w h e t h e r a similar gap exists b e t w e e n k- and (k-l)-round protocols. They c o n j e c t u r e d that indeed this is the case and that L k is the w i t n e s s to this fact.
and f a f u n c t i o n from i n t e g e r s to integers. W e say that d recognizes L (L is r e c o q n i z a b l e w i t h i n c o m m u n i c a t i o n f, L ~ COMM(f(n))) if Dn recognizes L n ~ L N [0, l]2n (if L n E COMM(f(n))) for
C o n j e c t u r e [i] : For every k > i, there is an L such that
all
n. The p r e f i x f r e e n e s s p r o p e r t y is motivated in [i]. W e need it only for our last r e s u l t w h e r e we w a n t to pin down exactly the c o m m u n i c a t i o n complexity. In other cases we augment the messages w i t h an endmarker. W e do not change the d e f i n i t i o n of the length of the message. Even if we c o u n t e d it in the first three results we w o u l d at most double the c o m m u n i c a t i o n complexity.
Lk ~ C O ~ k _ l ( n l / ~ ) . Our first two results show that indeed there is an e x p o n e n t i a l gap b e t w e e n k- and (k-l)-round protocols. T h e o r e m l, settles the c o n j e c t u r e above in almost the s t r o n g e s t sense. Theorem
i:
For every k ~
COMMk(nl/2/(36k4
82
log 3 n)).
i, Lk+ 1
Remark: E COM~
and c o n s i d e r Lk n [0,i] 2n,
One can easily show that Lk+ 1 (nl/2/(log n) 1/2).
The p r o o f of T h e o r e m 1 is c o m b i n a t o r ial. W e h a v e found a w a y to "force" our intuition. For m a n y open p r o b l e m s in comp u t a t i o n a l c o m p l e x i t y the s o l u t i o n is int u i t i v e l y clear: intuitively, P ~ NP bec a u s e we m u s t c h e c k all a s s i g n m e n t s w h e n solving SATISFIABILITY. U n f o r t u n a t e l y , we can rarely t r a n s f o r m such an i n t u i t i o n into a proof. In our case, our i n t u i t i o n tells us that if the two computers h a v e the w r o n g vertices, they m u s t e x c h a n g e k + 1 internal v e r t i c e s in order to check w h e t h e r a p a t h of length k + 2 exists. So, if only k rounds are allowed, the c o m p u t e r that is s u p p o s e d to m a k e the d e c i s i o n w i l l be "one v e r t e x behind". The other c o m p u t e r w i l l h a v e to send h i m a long list of values not k n o w i n g w h a t is the (k+l)-st v e r t e x on the path. Of course, our computers do not nec e s s a r i l y get the w r o n g v e r t i c e s neither do they always e x c h a n g e vertices. W h a t is worse, the input is p a r t i t i o n e d a r b i t r a r i l y and each c o m p u t e r may get only p a r t of the bits of the various edges. W e found a way a r o u n d it. By r e s t r i c t i n g a t t e n t i o n to a s u b s e t of the inputs w h i c h is large e n o u g h w e w e r e able to find graphs w i t h k + 2 layers such that, indeed, the two c o m p u t e r s h a v e the w r o n g vertices. Starting with this subset of inputs we fix a c e r t a i n p a t h by adding one v e r t e x at a time. Each time we further r e s t r i c t the inputs to c o n t a i n this initial path, say of length i, and to h a v e the same i m e s s a g e s exchanged. Not a l l o w i n g long enough messages, w e are still left w i t h a large number of such inputs, so after k m e s s a g e s the r e m a i n i n g set has both: inputs in Lk+ 1 and not in Lk+l,
L ~ [0, i] 2n d e s c r i b e d by 2 n × 2 n matrices. Once a p a r t i t i o n ~ of the input bits is g i v e n the set is fully d e s c r i b e d by a 0 - 1 m a t r i x M(L,~) w i t h 2 n rows and columns corr e s p o n d i n g to the p o s s i b l e bit strings seen by I and II. The proof of T h e o r e m 2 considers a fixe d p a r t i t i o n of the input. This is justified by T h e o r e m 3 stated below. Once a partition n is fixed we can assume w i t h o u t loss of g e n e r a l i t y that ~ = ~0 the natural p a r t i t i o n that gives I the first h a l f of the input. W e call the m a t r i x M ~ M ( L , ~ 0) the m a t r i x that corresponds to L and refer to L as the language that c o r r e s p o n d s to M. The m a t r i x r e p r e s e n t a t i o n is due to Yao [6]. The c o m p u t a t i o n can be v i e w e d as follows: two computers called ROW and C O L U M N h a v e to r e c o g n i z e L. Each c o m p u t e r has one h a l f of the input. (ROW knows the row in the m a t r i x and C O L U M N k n o w s the column.) They a l t e r n a t e sending m e s s a g e s (each one of them can start). B o t h computers k n o w the m a t r i x of L. At any stage, each i E [ROW,COLUMN] knows the subset S. of inputs the other may still have. W h e n l o n e of them sends a m e s s a g e the other one, j, obtains i n f o r m a t i o n that enables h i m to m a k e S. 3 smaller. In fact the p o s s i b l e messages j receives imply a p a r t i t i o n of S.. The com] p u t a t i o n terminates w h e n one of them, say ROW, has SRO W such that all the entries in
b e c a u s e s o m e initial paths h a v e the
c o m p l e t i n g edge and some don't. All of this is a c h i e v e d by an i n t e r e s t i n g i n d u c t i v e argument. This c o n t r a d i c t i o n proves the theorem. W e g i v e n another p r o o f for the exponential gap b e t w e e n k- and (k-l)-round complexity.
the row ROW has and the columns of S R W O
S C O L U M N should h a v e the same entries
(zeros
or ones), b e c a u s e all c o r r e s p o n d i n g input pairs h a v e the same communication. W e c o n s t r u c t the languages L~ in-
- COM~_l(n/4k).
d u c t i v e l y by c o n s t r u c t i n g the c o r r e s p o n d i n g n n m a t r i c e s M k. The m a t r i c e s ~ are d e r i v e d
T h e o r e m 1 and 2 are incomparable. On the one h a n d the gap of T h e o r e m 2 is wider. AS w e r e m a r k e d above, there is no such a large gap for L k of T h e o r e m i. A l s o if w e k
are
the same (0 or I). Note, that the submatrix c o r r e s p o n d i n g to the final S R O W and
T h e o r e m 2: For all k ~ 1 and for i n f i n i t e l y many n w i t h k ~ n/(96 log n) there exists n c [0, l}2n such that L k n E COM~(k Lk log n)
take in T h e o r e m 1
then T h e o r e m 2
is m e a n i n g f u l for a w i d e r range of k. On the other hand, w h i l e the languages of Theorem 1 are simple and constructive, those of T h e o r e m 2 are nonconstructive. The p r o o f of T h e o r e m 2 is an e x i s t e n c e proof. In addition, the two proofs are entirely different. W e suggest as an o p e n problem, to p r o v e a w i d e gap (from logn to ckn ) for a c o n s t r u c t i v e language. The proof of T h e o r e m 2 considers sets
from simple matrices. The latter are obtained by r e p e a t i n g b - a r y r e p r e s e n t a t i o n of the numbers 1,2,...,~times. (b and ~ are
to be a f u n c t i o n of n 83
carefully chosen parameters.) Then, all i's n in these matrices are replaced by ~i(Mk_2)
tocols, b e c a u s e Aho et al showed [8] that for fixed p a r t i t i o n there is only a polynomial (square) d i f f e r e n c e b e t w e e n determ i n i s t i c and n o n d e t e r m i n i s t i c protocols. The proof Of T h e o r e m 3 uses again prob a b i l i s t i c arguments. The m a t r i x M* = M(L*,r) is o b t a i n e d from M ~ M(L,~) by first d u p l i c a t i n g a large number of times the rows and columns of M and then by choosing two r a n d o m p e r m u t a t i o n s and p e r m u t i n g the rows and the columns of the r e s u l t i n g matrix. To establish (b) one observes that there are two such p e r m u t a t i o n s such that for any p a r t i t i o n of the input bits, the correspondihg m a t r i x contains a full copy of M or of MT° Note that the proof of T h e o r e m 3 introduces additional n o n c o n s t r u c t i v e n e s s to the languages of T h e o r e m 2. The second main result in [i] was showing that for any integer function f, 1 ~ f(n) ~ n, COMM(f(n)) - COMM(f(n)-i) ~ ~. Our last result is:
~i is a "random" permutation. The resulting m a t r i x is M~. In the p r o o f we define a m e a n i n g f u l p o r t i o n of M~. Let Pj be the claim that the s u b m a t r i ~ c o r r e s p o n d i n g to SRO W and S C O L U M N contains a m e a n i n g f u l p o r t i o n of n M.. Then we show inductively that if P. 3 3 holds, then after exchanging two messages Pj-2 holds. The proof makes use of the randomness of ~i" A n o t h e r way to look at it is that by some interesting counting arguments we show that there exist permutations n0,...,~b_l such that the above holds. The proof terminates w h e n we o b s e r v e that a m e a n i n g f u l p o r t i o n of M~ must c o n t a i n zeros and ones. R e c e n t l y [9] Yao c o n s i d e r e d p r o b a b i l istic p r o t o c o l s and p r o v e d an exponential gap b e t w e e n one- and t w o - r o u n d p r o b a b i l i s tic protocols. It is an interesting open p r o b l e m to p r o v e a result similar to Theorem 1 or 2 for such protocols. W h e n we fix a p e r m u t a t i o n n, we speak of a p r o t o c o l ~ = (~c,~r) (where c(r)
stands f o r
Let L 1 c
column
[0, i] 2n, L 2 c
(row))
for
T h e o r e m 4: For any integer function f, 1 ~ f(n) ~ n, C O M M i ( f ( n ) ) - N C O M M ( f ( n ) - i ~ .
protocol
(Li,~i) ~ (~c,~r)
functions
fl, f2:
for
(L,~).
[0,i] 2m w i t h parti-
(b) (L2,n2)
for
or
(~r(fl,),We(f2,
(L2,~2)
(L2,n2)~
T h e o r e m 3:
2:
NCOMM(f(n))-NCOMM(f(n)-i)/~.
The Proof of T h e o r e m
a contradiction. if
(Ll,n I)
For each language L ~
i.
W e assume and derive
Let ~ = [Dn] be the cor-
r e s p o n d i n g k - r o u n d p r o t o c o l s that r e c o g n i z e Lk+ I. W i t h o u t loss of g e n e r a l i t y each com-
(Li,~i).
into
- C O M M ( f ( n ) - i ) / ~.
Lk+ 1 E C O ~ ( n l / 2 / ( 3 6 k 4 1 o g 3 n ) )
p u t a t i o n contains exactly k exchanged messages: by adding two bits we can record the fact w h e t h e r the input has b e e n accepted, r e j e c t e d or neither. This increases the c o m m u n i c a t i o n c o m p l e x i t y by a c o n s t a n t (2k). The proof consists Of three parts. We first define several constants and p r o v e a r e l a t i o n s h i p among them (Claim I). Next we define a subset of the inputs, S, corresp o n d i n g to c e r t a i n graphs. Then we p r o v e L e m m a 1 from w h i c h the t h e o r e m follows. W e c o n s i d e r inputs of length 2n = m2 m, n large enough as will be e x p l a i n e d below. W e choose the constants a, r, p (an integer), ~ and ~, t and s (an integer) in this order to satisfy
Remarks: The two cases in (a) mean that w e treat s y m m e t r i c a l l y ROW and COLUMN. If (Li,~i) ~ (L2,~2) , then any (Li,~i) comput a t i o n i m m e d i a t e l y translates c o m p u t a t i o n and v i c e v e r s a .
Corollary
i.
))
(L2,~2).
(Li,~i) ~ and
there exist
{0,i] n ~ {0, i] m such that
(~c(fl,)),~r(f2,)) is a p r o t o c o l
(L2,~2) , if for each (Li,~i)
I: COMM(f(n))
T h e o r e m 4 extends the r e s u l t in [i] (Corollary i) to n o n d e t e r m i n i s t i c protocols in the s t r o n g e s t way. There seems to be no way to change the direct proof of C o r o l l a r y 1 in order to prove theorem 4. The proof of T h e o r e m 4 is rather simple. The s t r u c t u r e of the paper is as follows: the p r o o f of T h e o r e m i, i = 1,2,3,4, appears in S e c t i o n i.
tions ~i,~2. (a)
Corollary
(L2,~2) [0, i] 2n
and p a r t i t i o n ~ there exists L* c {0, i] 8n such that (a) there exists 7 with (L,~)~(L*,7) (b) for all ~ (L,~) ~ (L*,c). In other words, w h e n e v e r lower and upper bounds are p r o v e d for a language L and a fixed p a r t i t i o n ~, then there exists another language L* such that these bounds h o l d ind e p e n d e n t l y of the partition. It is i n t e r e s t i n g to note that T h e o r e m 3 does not h o l d for n o n d e t e r m i n i s t i c pro-
(i)
84
n a = nl/2/(36k41og3n),
(2)
i i log n a=~ (r= 3 log log n+2 log 6k 2)"
(3)
p =
for each block b a set Xb ~ [0, i] m of p o s s i b l e inputs. (a) X 0 contains the u n i q u e number in B 1• (b) For b e B. and 1 ~ i ~ k we d e ~ fine X b as fol~ows. There are ~ m - ~ strings r e p r e s e n t i n g numbers in Bi+ I.
[r],
1 1 ~ = - - -2 2p ~ = log(3kp),
(4) (5)
(6) t = "-I2m=-~I/2, and (7) s = [~ T h e s e constan£s h a v e b e e n c h o s e n so that C l a i m i: Proof:
If n is large enough, By
(2) and ~log
1
(3), ~ p
T h e s e are p a r t i t i o n e d according to the fixed bits of b l o c k b into 2 m - ~ m subsets. One of these subsets has at
then s>kn a.
!es_2
1
- log n ~ 2r+2
least~m~-~]
log n + log
log n log log n + log k + ~ + 1 log n (by (2)-(5))
n ~-a >
(log n) 1/2k2~+i"
large enough, n
>
(log n) ~ S ~
t-I/2 =
Hence,
> 2 ~ + i k n a.
and
if n is
Thus
(log n) I/2 ~ m ~ - ~ ] / 2 - i / 2 2 2 m ~ - ~ - I - i / 2 > knM.D
Each input c o n s i s t s of 2 m blocks of length m w h i c h w i l l be i d e n t i f i e d w i t h the numbers O,l,...,2m-l. C o n s i d e r i n g the p r o t o c o l Dn: (n,~), each c o m p u t e r I, II sees a p a r t (possibly empty) of each b l o c k (according to ~). W e say that b l o c k i is free for one of the c o m p u t e r s if it sees at least ~m bits in it. (without loss of generality ~m is an integer.) N o t e that since 1 < ~, a b l o c k may be free for the two computers. Note also that there are at least 2m/(p+l) blocks free for each c o m p u t e r (because o t h e r w i s e the other c o m p u t e r w o u l d see m o r e than (2m-2m/(p+l)) (m-am) = m2 m-I = n bits). W e now i d e n t i f y k + 2 d i s j o i n t sets of blocks Bi, i = 0,1,...,k+l, B i
tain tains going tains
one v e r t e x (= block). L a y e r 0 conb l o c k O, and is c o n n e c t e d (by an outedge) to layer I. L a y e r k + 2 conb l o c k 2m-l. L a y e r i, 2 ~ i < k + I,
c o n t a i n s "~|2m-~| vertices. Each v e r t e x in layers l,...,k is c o n n e c t e d to one of [2 mu-~] s p e c i f i c v e r t i c e s in layers i + i. V e r t i c e s in layer k + i are c o n n e c t e d to one of two vertices, exactly one of w h i c h is the one in layer k + 2. Moreover, for 1 ~ i < k , i odd (even) II (I) has the entire i n f o r m a t i o n on layer i, b e c a u s e for each b l o c k in B i he has the free bits. I(II) has no i n f o r m a t i o n at all on layer i b e c a u s e all fixed bits in B i have the same v a l u e in X b. To each input x in S c o r r e s p o n d s a d i r e c t e d p a t h that starts at v e r t e x 0, goes through layers 1,2,...,k+l and either terminates in layer k+2, in w h i c h case x ¢ L or not, in w h i c h case x ~ L. F r o m now on we c o n s i d e r only inputs in S. For i = 1,2,...,k+l, let Pi be the possible input segments in the blocks of B i (the m a r k e d c o n c a t e n a t i o n of X b for b in Bi). A n element of S is r e p r e s e n t e d by an element of P1 × P2 ×'''× Pk+l" For conven-
{0,i,...,2m-2} that s a t i s f y (i) B 0 = ~0].
(ii)
choose from
tains l's (O's) in the fixed (free) bits of the block b. (d) If b ~ UiBi, X b = [0 m] (any fixed v a l u e w i l l do). W i t h the graph i n t e r p r e t a t i o n in mind, ignoring the blocks not in UiB i w e h a v e r e s t r i c t e d a t t e n t i o n to the following inputs. The p o s s i b l e d i r e c t e d graphs h a v e k + 3 layers (the first k + 2 c o r r e s p o n d to B0,...,Bk+i). Layers 0,i and k + 2 con-
since 2n = m2 m, 2 m~ - 2 ~ > n
We
one such s u b s e t [2 m~-~] elements to form X b. N o t e that the so called fixed bits h a v e fixed values in X b. (c) If b E Bk+l, ~ = [im,yb} , Yb con-
_ log(r+l) > 6k2 if n is log n log n 1 log p + large enough. So ~ - a ~ ~ p > log n 1 log log n + log 6k 2 1
strings.
rBlf = i.
(iii) 1Bii = ~ m - ~ ] for i = 2 ..... k+l. (iv) For i = l,...,k+l, i odd (even), the blocks in B i are free for II (I). A s i m p l e c o u n t i n g a r g u m e n t shows that this is indeed possible. W e say that the blocks in B~ w i t h odd (even) j b e l o n g to J II (I). C l e a r l y if a b l o c k belongs to I(II) then it is free for I(II). For each b l o c k that belongs to I (II) we c h o o s e ~m bits that I (II) sees, call them free bits, and call the other fixed bits. W e now d e s c r i b e a s u b s e t of the poss i b l e inputs S ~ X 0 - X l . . . X 2 m _ l , s p e c i f y i n g
ience we also include Pk+2 w h i c h is the set c o n t a i n i n g the empty string. W e d e s c r i b e b e l o w a process that chooses in turn values from Pi,P2,-... ter i stages,
Af-
values from Pi,...,Pi_i h a v e
already b e e n chosen and the value from Pi
85
is restricted 1 s [w i .... ,wi].
to one of
s
possible
1 s w i , . . . , w i are any
values
The input will be determined
once an element
of Pi × Pi+l x...
Pk+l is
chosen. If layer i belongs to computer I (say), then the input is determined once one of the s values of Pi as well as an element of Pi+2 × Pi+4"'" and an element of Pi+l × Pi+3 ×''"
are chosen.
IVll ~
U~=l{W ~] × W~ ~ Pi N Pi+2 ×''"
Induction i~k-l.
The choice of w~,
step:
of Pi"
The
It can still have
by P3 × P5 x...,
Let q =
so
(b) is immediate. Assume
the lemma holds
for
h P i + l l / ( 2 ( 2 n a ) i ~ . Coheir × Uj, uj e Pi+[" Uj ~ Pi+3 x
der V i = Uj[uj}
Pi+5 x .... U. is said to be large if 3 IUj I ~ ( IPi+3 I" IPi+5 I''" )/(2 (2 ha) i).
from
are allowed
Claim 2: For i < k - i, there are at least q large Uj's.
for I. Note that after stage i, all inputs that are still considered have the same ~ corresponding initial path go = 0, gl,...,g i.
represented
the second half of
The first two
for II and only values
elements
(IP211P41...)/2na.
all inputs
values are known to I, while the third is k n o w n to II. W h i l e fixing values in p~, j = 1,2,... we also restrict in a speclal way the possible continuations. After stage i only values from V i c Pi+l × Pi+3"" are allowed
s
messages sent from I to II in the first round imply a partition of the possible inputs for I. We choose a message c I with the largest corresponding part V I. So, by (i),
Proof : Otherwise, if there were only q' < q large Uj'S, then IVi ! ~ (q' Ipi+ 3 I" IP~+5 I''" )
j = l,...,s
+
(!Pi+ll-q')(!Pi+3
I IPi+5!..-)/2(2n~)i
1/8. But this contradicts Lemma 6. D
from every column block of < .
_Lemma 5: Let B be an u n d i s t i n g u i s h a b l e (l-'eo)-fragment of M~. A s s u m e that ROW sends messages of length < 6 log m w h i c h are answered by messages from COLUMN of
3. Proof of Theorem 3. Let M = M(L,n). W i t h o u t loss of generality ~ = ~0' the natural partition. M = (Cl,...,C2n), w h e r e c. e [0, I} 2n is the i-th column of
length < 6 log C L-I Then there exists k-2" one u n d i s t i n g u i s h a b l e (l-¢0-6-3/~)-fragment m/b of ~k-2 after these 2 rounds, p r o v i d e d
1
M. Let M 1 = (Cl,...,Cl,...,C2n,...,C2n), w h e r e each c. is repeated 23n times. T l M1 (r I ..... r2n) w h e r e __r' E [0,l} 24n is
.
I-~¢0-6-3/~ > 1/8.
=
•
m¢/2b2). at least
L e m m a 7: There is a m a t r i x M* equivalent to M2, such that the language L* corresponding to M* satisfies the following property: for each p a r t i t i o n 6, one of the matrices M, M T is equivalent to a submatrix of M(L*,6). W e consider M and M T in the lemma because we will allow either ROW or COLUMN to start the computation. Lemma 7 establishes part (b) of Theorem 3. Part (a) is immediate with ~ = ~0 because of the way M ~ was obtained from M. L e m m a 7 makes use of Claim 6 below.
2
But all column blocks of B 1 have .C~-i [ k_2 ] i-¢ 0 columns in each block.
So B 1 has one column block with i)
2)
b e-I/L inserted'matrices me/2b 2 ~ m ¢-3/~ rows each and
at least
having
1
the i-th row of M I. Let ~' = (r I ..... r 1 ..... r ..... r n ), where each r i ~, 2 n 2 is repeated 2 °~* times. W e say that two matrices A, B are equivalent if there are p e r m u t a t i o n matrices P,Q, such that B = PAQ. We prove below:
Proof: First p a r t i t i o n B according to the row messages. W e c o n c e n t r a t e on the largest class B 1 which must contain at i-¢0- 6 least m rows. A p p l y i n g L e m m a 3 we find: there exists one column block (= digit position) , Bi, w h e r e at least b ~-I/~ (¢ = i-¢N-6) inserted matrices (= digits) ~¢ h a v e m /2b rows each (= f r e q u e n c y ~ -
([e - --3.)-fragment
Lemma 6: Let e > 1/8. Every e-fragment of bL-t M1 contains zeros and ones.
Two A d d i t i o n a l Lemmas. A submatrix A of is said to be u n d i s t i n g u i s h a b l e after the i-th round, if for all rows and columns of A as p o s s i b l e inputs for ROW and C O L U M N the same first i messages are exchanged. A submatrix B of M~ is called a efraqment of M~ if B has at least m e rows ~-I e and there are at least (Ck_2) columns in B
Therefore we
[Ck-2 ~-I-] i-¢0 co lumns.
Next, p a r t i t i o n this column block according to the column messages. A g a i n we 89
Cl__._a!~ _6: Let G = [i .... ,24n] and let .... ,~p, p = ( ~ ) b e p different subsets zn of G of size 2 , and let £=GIU£2...£2 n
which correspond to the choices of ~ ~). G = ~.•UG2n is the p a r t i t i o n of G to classes of copies of rows (columns) of M. The conclusion of the claim implies that there is a p e r m u t a t i o n ~(~) such that for every possible choice of ~(~), if we permute the rows (columns) of M 2 by ~(7), we still have at least one copy of every row (column) of M among the rows (columns) of M 2 that correspond to ~(~) (~(i)). D
be a partition of G into disjoint sets of size 23n. Then, there is a p e r m u t a t i o n of £ such that for all i,j, 1 ~ i ~ p, I~ j~2 n (15)
~(Xi)
N Gj ~ ~.
Proof: Let Pi j be the number tions of £ tha£ violate (15). Pi,j (24n): =
=
i
4.
(24n - 22n)23n: (24n_23n): (24n): 23n
(24n-23n) "-" (24n-23n-22n+l) 24n
of permuta-
•..
~
Claim 7:
( 1 _ 2 - n ) 2 2n
24n Hence Z i . Pi, " < ( ): D ,3 3 4n 24n Let A be a 2 × matrix. Let i = [i 1 ..... i2n} and i = [Jl ..... J2n ] two between
of the form [xylx E {0,1~n+f(n)]. of them is in COMMi(f(n)). We consider
1 and 4n. Case
A[~,i] is the 22n × 22n submatrix of A that consists of all rows (columns) of A with numbers that are obtained by taking 4n bit strings, assigning bits il,...,i2n (Jl' .... J2n ) in all possible
ways
and all
•
!
of COLUMN
M(L*,6) [~',i']
= M*[~,~]
(column sets)
the following
Each one
two cases.
log n.
at most f(n)-i corend with accept. By
property p ~ 2 f(n)-I I For i = l,...,p, let X i (X ~I ) be the
freeness
set of inputs that computer I (II) sees and correspond to the computation c.. There is a one to one c o r r e s p o n d e n c e between L n and II Ui=iP X~i × X.i • This c o r r e s p o n d e n c e is determined by the p a r t i t i o n n. Therefore, the number of such L n is at most (2nn)~2f~n)-l((22n)2)p= P ~ 22n2 f(n)-l(2f(n)_l)~22n+l 22n(22n+l) 2f(n)-i
are defined
22n22n+f(n) 2 f(n)-I ~ p A 's, there m u s t be x I ~ and y in L n w i t h x E A and e A with m ~ m that c o r r e s p o n d to the same computation c.. But then also the z with l I xi II II z = and z = y m u s t be a c c e p t e d by c. w h i l e z ~ L n. l
yI
5.
References.
[i] C.H. P a p a d i m i t r i o u and M. Sipser, C o m m u n i c a t i o n complexity, Proc. 14th A n n u a l A C M Symp. o n Theory of Computing, 330-337, 1982. [2] C.D. Thompson, A r e a - t i m e c o m p l e x i t y for VLSI, proc. llth A n n u a l A C M Symp. on T h e o r y of Computing, 81-88, 1979. [3] R.J. L i p t o n and R. Sedgewick, L o w e r bounds for VLSI, Proc. 13th A n n u a l A C M Symp. on T h e o r y of Computing, 300-307, 1981. [4] A.C. Yao, The entropic limitations on VLSI computations, Proc. 13th A n n u a l A C M Symp. on T h e o r y of Computing, 308-311, 1981. [5] H. Abelson, L o w e r bounds on information t r a n s f e r in d i s t r i b u t e d computations, Proc. 19th A n n u a l IEEE Symp. F o u n d a t i o n s of C o m p u t e r Science, 151-158, 1978. [6] A.C. Yao, S o m e c o m p l e x i t y q u e s t i o n s r e l a t e d to d i s t r i b u t i v e computing, Proc. llth A n n u a l A C M Symp. on T h e o r y of Computing, 209-213, 1979. [7] K. M e h l h o r n and E.M. Schmidt, Las Vegas is b e t t e r than d e t e r m i n i s m in VLSI and d i s t r i b u t e d c o m p u t i n g (extended abstract), Proc. 14th A n n u a l A C M S y m p o s i u m on Theory of C o m p u t i n g ~ 330-337, 1982. [8] A.V. Aho, J.D. U l l m a n and M. Yannakakis, On notions of i n f o r m a t i o n t r a n s f e r in VLSI circuits, Proc. 14th A n n u a l Symp. on Theory of Computing, 133-139, 1983. [9] A.C. Yao, L o w e r bounds by p r o b a b i l istic arguments, Proc. 24th A n n u a l IEEE Symp. on F o u n d a t i o n s of C o m p u t e r Science, 420-428, 1983. [i0] J. Ja'Ja', V.K. P r a s a n n a K u m a r and J. Simon, I n f o r m a t i o n t r a n s f e r under d i f f e r e n t sets of protocols, to appear in S I A M J. on Computing.
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