Digitized by the Internet Archive 
 in 2013 
 
 http://archive.org/details/datadependencein837bane 
 
u> » ~r 
 
 Report No. UIUCDCS-R-76-837 
 
 73 J 
 
 NSF-0CA-MCS73-07980 A03-000025 
 
 ) 2- 
 
 DATA DEPENDENCE IN ORDINARY PROGRAMS 
 
 by 
 
 UTPAL BANERJEE 
 
 November 1976 
 
 DEPARTMENT OF COMPUTER SCIENCE 
 UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN 
 
 URBANA, ILLINOIS 
 
 The Library of the 
 
 University 01 Illinois 
 * Urbana-Charn w 
 
Report No. UIUCDCS-R-76-837 
 
 DATA DEPENDENCE IN ORDINARY PROGRAMS 
 
 by 
 
 UTPAL BANERJEE 
 
 November 1976 
 
 Department of Computer Science 
 
 University of Illinois at Urbana-Champaign 
 
 Urbana, Illinois 61801 
 
 This work was supported in part by the National Science Foundation under 
 Grant No. US NSF-MCS73-07980 A03 and was submitted in partial fulfillment 
 of the requirements for the degree of Master of Science in Computer 
 Science, November 1976. 
 
Ill 
 
 ACKNOWLEDGMENT 
 
 I would like to thank my thesis advisor, Professor David 
 
 Kuck, for his valuable advice, my wife, Aloka, for her infinite 
 
 patience, and my colleague, Michael Wolfe, for his great help in 
 debugging the PL/ 1 program in the thesis. 
 
IV 
 
 TABLE OF CONTENTS 
 
 Page 
 
 1. . INTRODUCTION 1 
 
 2. PRELIMINARIES 3 
 
 2.1 Definitions 3 
 
 2.2 Lemma 3 
 
 2.3 Lemma 4 
 
 2.4 Definitions 5 
 
 2.5 Definitions 5 
 
 3. THE MAIN RESULT 7 
 
 3.1 Theorem 10 
 
 3.2 Example 14 
 
 3.3 Example 18 
 
 3.4 Remarks 21 
 
 3.5 Extensions 24 
 
 4. THE LINEAR CASE 27 
 
 4.1 Theorem 27 
 
 4.2 Example : 30 
 
 4.3 Theorem 31 
 
 4.4 Example 35 
 
 4.5 Remarks 37 
 
 LIST OF REFERENCES 38 
 
 APPENDIX 39 
 
m 
 s no 
 
 e 
 
 1. INTRODUCTION 
 
 In recent times much effort has been spent in the 
 search for narallelism in existing nrop-rams. Here, the con- 
 cent of data dependence plays a very special role. T he ai 
 or this thesis is to study conditions under which there i 
 dependence between two assignment statements inside an TV_f re 
 loon in a language like Fortran. In our basic model, we deal 
 with array variables coming from a one-dimensional array and 
 let the subscripts be arbitrary polynomials In the index var- 
 iables. Generalization to multi-dimensional arrays noses no 
 additional problems. 
 
 In section 2 we establish our notation and e-ive some 
 useful definitions and lemmas. Our main result (Theorem 3.1) 
 alonr with some examples, remarks, and possible extensions is 
 riven in section 3. Finally, section * is devoted to a detailed 
 study of the very important special case where the subscripts 
 are linear functions. 
 
 So far, no work has been done in the non-linear case, 
 not even in the linear case with more than two index variables. 
 Cohagan [1] and Towle ([5], ch. 5) have considered the linear 
 case when the number of variables is restricted to one or two. 
 Towle rives a very large set of necessary conditions in the 
 two-variable case, and a complicated program based on that set. 
 Prom the nractical noint of view, it would be extremely difficult 
 to extend his oroP-ram to the linear case of three or more var- 
 iables. And It can not be extended to the non-linear case at all. 
 
2 
 The necessary conditions we have obtained are valid in thl general 
 
 non-linear case with an arbitrary number of index variable: . 
 They are also shown to be sufficient in a snecial linear case 
 (Theorem *J . 3 ) that occurs frequently in nractice. Five a 
 short and simnle nrorram (based on those conditions) that wor 
 for the reneral linear case with any number of variables, (See 
 the Anoendix.) And this oropxam can be easilv extended to the 
 non-linear case with onlv minor modifications. f "ore comments on 
 the results of Ooharan and ,p owle are civen at the end o r section >\ . 
 
 
2. PRELIMINARIES 
 
 2.1 Definitions 
 
 Let r denote a real number. 
 
 (1) We define Lr I to be the greatest integer not greater than 
 r. For example, [3 J = 3, 1.3. 9J = 3, and L-3.9J = -4. 
 
 (2) The positive nart r and the negative part r~" of r 
 
 are ^ivon by 
 
 r + = (!r| + r)/2 
 
 and 
 
 r" = (Irl - r)/2. 
 
 The basic nronerties of r and r~ are summarized 
 
 .1 n the followinp tv/o lemmas. 
 
 2 . 2 Lemma 
 
 For any real number r, 
 
 (1) r + _> 0, r~ _> 0; 
 
 (2) if r _> then r = r and r" = 0; 
 
 (3) if r <_ then r = -r~ and r + = 0; 
 
 (4) r = r + - r"; 
 
 (5) (-r) + = r*, (-r)" = r + ; 
 
 (6) (r + ) + = r + , (r + )~ = 0, (r~) + = r", (r - ) - = 
 
Proof. The proof follows directly from the deflnltior . 
 
 2.3 Lemma 
 
 Let r,s,z denote real numbers. Tf r i <_ z <_ s 
 
 then 
 
 + 
 -r s < rz < r s . 
 
 + 
 In oarticular, -r <_ r <_ r . 
 
 Proof. If r >_ 0, then 
 
 <_ rz <_ rs 
 and 
 
 r = r + , f" = 0, 
 
 so that 
 
 + 
 -r s < rz < r s. 
 
 and 
 
 so that 
 
 If r <_ 0, then 
 
 rs < rz < 
 
 + 
 
 r = -r , r = 0, 
 
 + 
 -r s < rz < r s. 
 
 Q . E . D 
 
 ^hese two lemmas will be frenuently used without 
 always beinn; exnlicitly mentioned. 
 
2.4 Definitions 
 
 Let y denote a vector (y,, Vp,,.., y ) where 
 
 y, , y ni .... y are all integer variables. 
 
 1 d ' m 
 
 (1) For any positive integer L we define a vector y by 
 
 Li 
 
 y L 
 
 (2) A product term (or simply, a term ) in y 1s an expression 
 of the form 
 
 l l l 2 £ m 
 
 • 1 ' y 2 ,y m 
 
 where £ , L.,,,. I are non-negative integer constants. 
 1 d m 
 
 If for some r, £ > then we say that the term contains the 
 
 » r J 
 
 variable y . The degree of the term is the . non-negative in- 
 
 ' r ■ 
 
 teger (I. + l„ + ... + i ). (The term with degree is the 
 
 12 171 
 
 constant 1. ) 
 
 (3) A polynomial in y is a finite sum of the form 
 
 y a k P k (v")> where for each k, a R is an integer constant 
 K 
 
 and P. (7) a nroduct term in y. The degree of this noly- 
 
 K. 
 
 nomial is the largest of the degrees of all the product terms 
 P k (y) such that a k * 0. 
 
 2.5 Definitions 
 
 An assignment statement S is a statement of the form 
 
6 
 
 Z = R, where 7. Is a scalar or array var^ -1 
 
 formed arithmatic expression. Z is cai: e output variable 
 of S and is denoted by OUT(S). Any variable occurrlr 
 r is an input variable of S, and the set of all input var- 
 iables of S is denoted by IN(S). 
 
 If S and S« are two assignment statements such 
 that S is executed before r, ?i then we write Sj < ^ ? . 
 
 Finally, we have the followinr convention. Any sum 
 of the form *T ( ) where 1s the empty set, or of the 
 
 m 
 form V ( ) where M >m, is identically enual to 0. 
 
 I 
 
 k=M 
 
3. THE MAIN RESULT 
 
 We turn now to the problem of data dependence. Con- 
 sider a program segment of the following type: 
 
 DO 5 I 1 a 0, n 1 
 DO 5 Ip ■ 0, n 2 
 
 DO 5 I = 0, n 
 
 s 1 (T) 
 
 s 2 (T) 
 
 s N (D 
 
 5 CONTINUE, 
 where n>, n„,..., n are positive integers, T = (I-., I?'*'** * ^» 
 and S , S ,..., S„ are assignment statements. Take any two 
 
 statements S and S (1 < n, q < N). Let X(f(T)) denote 
 d q — — 
 
 a variable of S and X(g(T)) a variable of S , where X is 
 
 P <1 
 
 a one-dimensional array and f, g are integer valued functions 
 (such that the variables are well defined). 
 
 Three different types of dependence of X(g(I)) on 
 X(f(I)) may arise corresponding to the following three possib- 
 ilities : 
 
% 
 
 (1) x(f(T)) ■ OIJT(S ), X(pr(i)) e II 
 
 (2) X(f(T)) e IN(S ), X(r(D) = OUT<S q ); 
 
 (3) X(f(T)) = OUT (3 ), X(p,(I)) = OUT(S„). 
 
 For our purpose, there is no need to distinguish 
 
 between these three types. We simply assume that one of 
 
 (1), (2), (3) holds. We will say that the variable X(r(T)) 
 
 of statement S is dependent on the variable X(f(T)) of 
 
 statement S , and write 
 p' 
 
 X(f(T)) A X(f(T)), 
 if there exist two values i and J of such that 
 
 (1) S p (i) < S q (.1), 
 and 
 
 (ii) f(i) = fAT). 
 
 Our problem is to find conditions (necessary, sufficient, or 
 both) such that 
 
 X(f(T)) A X(g(T)). 
 
 In actual programs f(I) and fr(I) are almost 
 always nolynomials in I. We assume that this is indeed the 
 case. Let d denote the larger of the two dep-rees, and 
 {P, (I) k £ K) the set of all nroduct terms in X with de- 
 crees from 1 through d. (Here E is lust an index set. If 
 
9 
 
 the total number of these product terms Is NP, then we can take 
 
 E = (1, 2,..., NP}.) Let 
 
 f(I) = a + £ a k P k (I) 
 
 keE 
 and 
 
 „('d = b * £ b k P k (I) . 
 
 kEE 
 
 where a, 6, the a's, and the b's are all integer constants. 
 Now take any integer L such that 1 <_ L <_ m+1. A 
 polynomial in T can be expressed as a polynomial In T T 
 
 Li 
 
 (Definition 2. 1 \ (1)) without any change in degree. Thus we 
 
 can write 
 
 kcE 
 
 r L £*_, "Lk l k x *L', 
 
 where y t and the c's are integer constants. (Clearly, 
 
 Y m+1 = and c (m+1)k = b k , keE.) Also, let E L1 , E L2 , E L3 
 
 denote subsets of E such that E T , gives all product terms 
 
 in I (in the collection {P. (T)|k c E}) containing none of 
 
 ^t s I T ,!,.... I ; E T „ rrives all terms containing I T but 
 
 none of L ,,.... . I ; and E T _ gives all terms containing at 
 L+l' ' m' L3 
 
 least one of I T ,,,... , I . Then E T , , E TO , E T _ are pair- 
 
 L+l* ' m LI' L2* L3 
 
 wise disjoint (some of them may be empty) and 
 
 e l1 Ue l2 Ue L3 = e. 
 
10 
 
 Finally, let n denote the vector (r^, n ?i ..., 
 We have the following result. 
 
 3.1 Theorem 
 
 If X(f(T)) A X(g(T)), then 
 
 ( A ) (^ _ a ) is divisible by the gcd of all the a k ' s and the 
 b 's, k e E, if this p;cd is not zero; 
 
 and (B) 
 
 "vl Ca k - c Lk r P k (n L ) - £ (a" + c Lk ) + P^) 
 
 k e E L1 keE L2 
 
 " L (<< + c ^) p , Cn, ) 
 
 m< TO k Lk k l 
 
 <_ 3 - a 
 
 KU L1 L2 
 
 + y 
 
 >-L (a + c~ 
 
 k,E < a k + C Lk> W> 
 
 ''J 
 
 for some positive integer L such that 
 
 Tm+1 if p<q 
 
 L < < 
 
 ~ \ m if d >_ a 
 
 Proof Assume that X(f(I)) A X(p;(I)). Then there must 
 exist two values 
 
11 
 
 and 
 
 i « (i , i 2 ,..., i m ) 
 
 .1 vJ-ij J 2 * • • • * ^m 
 
 of I such that 
 
 (1) S p (i) < S (J), and 
 
 (2) f(i) = K (J). 
 
 Now (1) implies that there exists a positive integer 
 L such that 
 
 m if p > q; 
 
 W i r = J r , 1 < r < L * 1 (if L > 1); and 
 
 (5) 
 
 i L - ,1 L " - 1 (if L £ m )- 
 Usinp; the exoressions for f and p; we pet from (2) 
 
 that 
 
 (6) 3 - a = ^ a k P k (i) " I b k P k ( < 1} • 
 keE kcE 
 
 This implies condition (A) of the theorem. In order to derive 
 condition (B) we oroceed as follows. 
 
hynotl 
 
 I \ V T) " ^,, + I e Lk V 
 
 keF 
 
 
 12 
 
 Mence, (6) can be written as 
 
 (7) 3 
 
 or as 
 
 - « = -y l + 1 1\ v*> - c L k yv ] 
 
 keF 
 
 (8) 6 - a = -y L + £ [a k P k (D - C Lk P k (J L )] 
 
 keE 
 
 LI 
 
 + ^ [a k P k (i) " °T,k W 1 
 
 keF. 
 
 L2 
 
 + > ^ \ (i) " C Lk W 3 
 
 :eE 
 
 L3 
 
 Now three cases arise. 
 
 Case (i) 
 
 1 1 ' 2'** 
 3 1 , J 2 » ' ' 
 
 Let k e E . Then P k U) is a function of 
 
 . , i T , alone and p . (T T ) is a function of 
 
 ' L-l k L 
 
 j T n alone. Therefore, (4) imnlies that 
 
 lj™" A. 
 
 P k (1) " W' 
 
lence, _ _ 
 
 a k P k (i) " C Lk W 
 
 (a k - C Lk } W 
 
 13 
 
 < (a k - c LR ) p k ( n L ) fe y Lemma 2.3, 
 
 since < P k (,1 L ) £ p k ^ n L ) - 
 
 Case (ii) 
 
 Let k e E-p. We can write 
 
 P k (T) . Q R (I) R k (I L ), 
 
 where (I) is a product term in I containing none of 
 
 j -t > It.u.i.j I , and R. (I T ) is a power of I T . 
 L' L+l m' k L r L 
 
 Since \(i) = Q k ( J~L ; ' We get 
 
 a k V T > 
 = a k Q k (i) R k (i L ) 
 
 = a k VV W 
 
 1 a k Q k (T L ) R k U L - 1) by Lemnma 2.3, 
 
 since < VT L > W * Vl 5 V 3 L" ^ 
 
 <k W 
 
14 
 
 Hence, a P (1) - c,^ P k (J L ) 
 
 i < a+ k - c Lk> W 
 
 i (a k " C LK ,+ W' slnce ° i W i v H ■ 
 
 case (iii) 
 
 Let k e E, ,. We have 
 
 a k p k (1) " c Lk \ { h> 
 
 i a k P k<"L> + °Zk P k (E L>- 
 
 since < P k (D, P k (J L ) £ V"^' 
 
 = (a k + <=Zk> W- 
 
 The second part of condition (B) now follows fron 
 
 (8) if we use the inequalities in all three cases. To ret the 
 first part, multiply (8) by -1 and oroceed as before. 
 This completes the proof of the theorem. 
 
 O.F..D 
 
 For any Riven polynomial functions f and g, we 
 
 must first find the constants y T . c T , and the sets E T , , 
 
 L' Lk LI' 
 
 E T p, E_ » before we can test for data dependence. Consider the 
 
 followinr examoles. 
 
 3.2 Examnle 
 
 Consider the following program: 
 
15 
 
 DO 5 I 1 - 0, n x 
 
 DO 5 I = o, n. 
 
 S d : X(a + a 2 l ? ) = R 
 
 S : X(3 + b 1 i l + b^i^) = r 2 
 
 5 CONTINUE. 
 
 Suppose that we want to test for dependence of the output 
 
 variable of S on the output variable of S . 
 
 n p 
 
 Here n = 2, and f(I) and p(T) are polynomials in 
 
 J of decrees 1 and 2 respectively. We can take 
 
 E = {1, 2, 3, 4, c ^} 
 
 and 
 
 P 1 (T) = l lf P 2 (D = i 2 , 
 
 P 3 CT) = ij > V X) = l 1 l 2 > V T) = x 2 
 
 Then a n = a = a,, = a,- = b = b n = b r = 0. There are three 
 13 4 b 2 3 5 
 
 oossible values of L: 1, 2, 3. 
 
 Case (i) 
 
 Since 
 
 Let L = 1. Then I = I = (^ - 1, I ) 
 
 b 1 I 1 + b 4 I x I 2 
 
 = b 1 + b 1 d 1 - 1) + b 4 i 2 .+ b i( (I 1 - di 2 , 
 
16 
 
 we F,et 
 
 Yt = b i » c n = b i» c io = h n> 
 
 1 1» 11 
 
 12 H 
 
 C 13 = °» C l>\ = b H» C l^ = °* 
 
 Also, E ia = 0, E 12 = (1, 3), E - {2, H, 5). 
 
 case (11) 
 
 Let L = 
 
 2. Then I. = I 2 
 
 = (I lf I 2 - 1). Since 
 
 b 1 T 1 + b,!^ 
 
 ■ < b i + V 1 ! + Vl (I 2 " 1) ' 
 
 we eet 
 
 Y 2 = 0, c 21 = b x + b i|} c 22 = 0, 
 
 c 23 ~ ' °24 ~ 4' C 2S 
 
 Als 
 
 o, E 21 = (1, 3}, E 22 = (2, 4, 5>, E 23 = 0, 
 
 Case (ill) 
 
 Let L = 3. Then I L = I = (I I ). V, T e have 
 
 Y 3 = °' C 31 = b l* ° 32 = °' 
 
 c 33 ~ °' °3^ ~ b ^ a °35 ~ ° 
 
 Also, E 31 = {1, 2, 3, 4, 5>, E 32 = , E 33 = 
 
17 
 
 Corresponding to the three values 1, 2, 3 of L, 
 the inequalities in condition (D) of Theorem 3.1 take the 
 following three forms: 
 
 (1) (L = 1) 
 
 b~ - b, n, - a" n^ - b^ n -\ n 2 
 
 < B - a 
 
 <_ - b + b~ n + a« rip + b^ run,,, 
 
 (2) (L = 2) 
 
 -(b 1 + b^) n 1 - a" (n - 1) - b 2| n 1 (n 2 - 1) 
 
 < 3 - a 
 
 <_ (b-j^ + b i4 )~n 1 + a ? (n 2 - 1) + b^ n.(n 2 - 1), 
 
 (3) (L = 3) 
 
 -b. n, - a ? n 2 - b^ n i n p 
 
 < 6 - a 
 
 < b, n, + a,, n~ + b7 ri.n~ 
 — 11 2. 2. 4 12 
 
 Now sunoose that the variable X(3 + b I, + b^I-Ip) 
 
 of statement S is dependent on the variable X(a + a I ) 
 q f 2 2 
 
 of statement S . Then, (3 - a) must be divisible by the n;cd 
 
oT a~, b , b i( . Tf p < q, then one of (1), ' , ' • 
 hold. Tf n > o, then one of (1) and (2) mm 
 
 3.3 Example 
 
 Consider the followinr nrofram serment: 
 
 DO 5 T 1 = 0, n 1 
 
 DO 5 I 2 - 0, n 2 
 
 S x : Z = X(p, - I* + 5lJ + 2I 1 I 2^ 
 
 S 2 : X(« + I 1 - 3I 2 ) = R 
 
 5 CONTINUE 
 
 Suppose that we want to test for dependence of the inout var- 
 iable of S. on the output variable of S . 
 1 2 
 
 Here, 
 
 m=2, p=2, q = 1 , 
 
 f(I) = a + I 1 - 3I 2 , 
 
 and «<T) = 3 - I* + 5 I? + 21^ . 
 
 We can take 
 
 E = {1, 2,..., 91, 
 
 and 
 
19 
 
 P 1 (I) = I 1 , P 2 (I) = I 2 , 
 
 P 3 (I) = I*, P 2] (D = 1^2, P 5 (D = l\ 
 
 - T 3 
 
 T^ = T 3 
 
 P 6 (I) = Ij>, P 7 (I) = IJI 2 , Pg(l) = I 1 I|, P g (l) = I|. 
 
 Then, 
 
 a, - 1, a g - -3, 
 
 a 3 a i4 a 5 ^f, " a 7 ~ a 3 ~ °s 
 
 a,, = a = a., = a r . = a,- = 
 
 b 3 = ~ lj b 6 = 5 » b 8 = 2 ' 
 
 b l = b 2 = b 4 = b 5 = b 7 = b 9 = °' 
 
 Since p > q, there are only two possible values of L: 1, 2 
 
 Case (i) 
 
 Let L = 1. Then I L - I. = (I- L - 1, I 2 ). 
 
 Since 
 
 - I? + 51? + 
 
 !I,I 
 
 1 "~1 --1 
 
 = - [(I-l - 1) + l] 2 + 5[(I 1 - 1) + I] 3 + 2[(I 1 - 1) + 1]I 2 
 
 = i\ + 13(I 1 - 1) + 14(I 1 - l) 2 + 2I 2 + 5(I X - I) 3 
 
 + 2(1.. - 1)1* 
 
 we n;et 
 
 Tn = ^> On = 13, c, Q = 14, 
 
 "11 
 J l6 
 
 = 2, c, A = 5, c lR = 2, 
 
 13 
 
 18 
 
 c 12 c l4 
 
 c 17 c 19 
 
 0. 
 

 Also, K,, = 0, I 
 
 ! 12 = (1, 3, 6), E 13 = {2, fc, 5, y, B, 
 
 Case (ii) 
 
 since 
 
 Let L = 2. Then T. = T. = (I_, I- -1). 
 
 L C id 
 
 2 1 2 
 
 = " J l + 5I 1 + 2I 1 [(I 2 " 1} + 1] 
 
 .2 
 *1 
 
 = 21, - i, + 'a^ip - D + 51J + 2i 1 d 2 
 
 i) 2 , 
 
 we ret 
 
 Yp > ^21 ^2^ — 
 c oh = ^» G o^ = 5, c o 2, 
 
 2^ 
 
 26 
 
 28 
 
 c 22 c 25 C 27 °29 °* 
 
 A 
 
 lso, E = (1, 3, 6), E 22 = {2, A, 5, 7, 8, 9>, E 23 = 0. 
 
 Corresponding to the two values of L, the inequalities 
 in condition (R) of Theorem 3.1 take the followinp two forms: 
 
 (1) 
 
 (L = 1) 
 
 (2) 
 
 2 c 3 
 n l " 5n l 
 
 2 
 2n,n 
 
 - 3n^ < 6 - a < -4, 
 
 (L = 2) 
 
 3 + n 1 - 9n ? + 9n 2 - 5n? - 2n_ L n 2 - 3n^ < 
 
 - a <_ n,, 
 
 If the input variable of S. is dependent on the out- 
 put variable of S ? , then by Theorem 3.1 either (1) or (2) must 
 hold. 
 
21 
 
 3.^1 Remarks 
 
 (1) The conditions of Theorem 3.1 are not sufficient 
 for data dependence in general. We show this by an example. 
 
 Let 
 
 m = 1, n 1 = 2, 
 
 f(I) = 2I 1 , g(T)- H - 3I X . 
 
 Since d 
 Then 
 
 1 we can take E = {1} and P,(I) 
 
 a = 0, 
 
 a x = 2, (5-4, b 1 = 
 
 = T. 
 
 - 3. 
 
 Let n > q. Then L = 1. Clearly, 
 
 y 1 = -3, c n = -3, 
 
 E 1n 0, E 10 (!}, E 13 = 0. 
 
 11 
 
 12 
 
 Condition (R) of Theorem 3.1 becomes 
 
 x 
 
 or, 
 
 (a l + C ll )+(n l " 1) - 3 ~ a - " Y l + (a l " 
 
 C ll^ ^ n l~ "^ 
 
 3-0 < ^ < 3 + 5 
 
 which is true. Condition (A) is also satisfied. However, two 
 values i, , ,j of I can not be found such that 
 
 and 
 
 < 1 1 < ^ < 2 
 
 21., = H - 33 ± 
 
22 
 
 Therefore, the variable '/.( <\ - 31.) 
 
 dependent on the variable /.(?!) 
 
 (2) Suppose for a moment that 1 ,..., i and .},,..., ' 
 denote real variables, and define a real valued function 
 
 P(i lf .... i m , .1^..., J m ) = V Ea k P y (T) - b k %(!)]. 
 
 keE 
 
 For a riven value of L, v/e consider a subset - P the 
 
 Li 
 
 2n-dimensional Kuclidean space, defined by 
 
 < i , ,1 < n ( 1 < r < m) , 
 
 and the conditions ( ; , (5) in the proof o^ Theorem 3.1. 
 Since D, is connact and connected and F is continuous, 
 the image set F(D T ) is a finite closed interval. (T,ee 
 Dieudonne' [2], m heorems 3.17.10 and 3.19.8.) Condition (B) 
 of Theorem 3.1 is necessary and sufficient for (ft - a.) to 
 lie in the interval F(D T ), i.e. for the existence of a "real" 
 noint in D T at which the value of F is (B - a). 
 
 (3) Suppose that condition (B) of Theorem 3.1 is satisfied 
 for at least one value of L. Let L n be the largest value 
 of L for which it is satisfied, and suppose that L~ <_ m. 
 Then, there will be no data dependence if the index variables 
 I,, T ,..., I are kept fixed at any set of permissible 
 values . 
 
 ( l \) The whole idea behind Theorem 3.1 is to ret the sharpest 
 
23 
 
 possible bounds for (p - a) that can be directly obtained using 
 
 the nroperties of the different terms in f(T) and g(T). 
 Let us comnare the second inequality in condition (B) against 
 two other inequalities that can be derived much more easily. 
 From (6) in the Droof of Theorem 3.1, we get 
 
 - « < ]T Ca* + b k ) P k (n). 
 
 keK 
 
 Let m = 1, f(I) ■ a + I,, and g(T) = 8 + -I-. Then, this 
 
 will i^ive 3 - a < n. . 
 
 But the second inequality in condition (B) gives 
 
 - a <_ - 1 for L = 1 
 
 and 
 
 B - a < for L = 2. 
 
 From equation (7) in the nroof, we get 
 
 8 " n i " Y L + I (a k + c Zk> P k (n L ). 
 keE 
 
 The second inequality in condition (B) is clearly better than 
 this. Because, for any two real numbers a and c we have 
 
 (a - c) < (a -c) <a + c , 
 
 and these relations may be strict. 
 
 One could give a similar discussion about the first 
 ineouality in condtlon (B). 
 
24 
 
 'U [ j Kx tons Ions 
 
 (1) Let I denote any one of the indie I., I,, 
 ... I . For simplicity, we have assumed that I runs thro, 
 an ordered set of values of the type (0, 1,..., n}, n 
 is a positive integer . However, one frequently has a statement 
 like 
 
 "DO 5 I = M , n x , e" 
 
 so that I runs throurh the values {N., M_ + e,..., .'.'}. 
 (Here, H~, H , and e(/ 0) are integers and e divides 
 I! - M- . ) In this case we can introduce a new index variable 
 I f by 
 
 T f = (I - !I )/e. 
 
 Then the above DO statement mav be replaced bv 
 
 "DO 5 i' = 0, n", 
 
 Where n = (N, - U~)/e if every occurrence of I inside 
 
 i 
 
 the loop is replaced by Ii + el . 
 
 (2) So far, we have worked only with one-dimensional arrays. 
 Suppose now that X is multi-dimensional. Let ND denote the 
 number of dimensions of X, and ^B^.» LB, the upper and 
 
 lower bounds, respectively, of the k dimension of X. Then 
 
 i 
 X can be mapped onto a one-dimensional array X (in a one 
 
 to one fashion) by 
 
 X(£ -,..., H,m ) *> X (P(£-,..., *,»Tr\ ) ) 
 
25 
 
 where a is defined below: 
 
 aU,,..., *, JD ) = £\l\ ~ LB k ) 
 
 k=l 
 with 
 
 '(UB k+1 - LB k+1 + 1) ... (UB ND - LB ND + 1) if k .< ND - 1 
 
 if K = ND. 
 
 Thus Theoren 3.1 can be applied to multi dimensional arrays 
 after they have been transformed this way. 
 
 (3) Theorem 3.1 can be extended so that the statements 
 
 S and S are not required to be within the same loon. 
 
 n q 
 
 Let us illustrate this bv the following examnle: 
 
 DO 5 I 1 = 0, n. 
 
 DO 5 I t = 0, n 1 
 
 S 
 
 CONTINUE 
 
 DO 7 I t+1 = 0, n t+ 
 
 S q 
 
 DO 7 I = 0. n 
 
 7 CONTINUE 
 
26 
 
 Here, f Is a function of I,,..., I alone and r '-a 
 
 function of I^.t,.... I alone. Equation (6) In the proof 
 t + 1 ' m 
 
 Theorem 3.1 holds, where 1 and j are now any two arl Ltn 
 values of I. We ret the condition 
 
 -X ( <\ + b k> p k ( "' i b - ° - T u t. + b k> p k (n_) 
 
 keE keE 
 
 which replaces condition (R). 
 
27 
 
 4. THE LINEAR CASE 
 
 In this section we assume that both f(T) and p 5 (T) 
 are polynomials in I of degree 1. This is the most important 
 special case that one encounters in practice. Our first 
 theorem here is the linear version of Theorem 3.1. 
 
 4.1 Theorem 
 
 Let 
 
 m 
 
 f(T) = « + y a k I k 
 
 and 
 
 m 
 
 „(i) - u ♦ y b k i k 
 
 where «, 3, the a's, and the b f s are all integer constants 
 
 If X(f(I)) A X(g(T)), then 
 (A) (8 - a) is divisible by the ";cd of all the a. ' s 
 
 and the b. 's, 1 < k < m if this o;cd is not zero: 
 
 k ' — — > ' 
 
 and (B) 
 L-l 
 
 m 
 
 -b L . y (a k - b k ) n R - (a" + b L ) (n L - 1)- 
 
 (a!~ + b, )n, 
 k k k 
 
 y 
 
 l 
 
 k=?L+l 
 
 < 8 - a 
 
 L-l 
 
 - - b L + 
 
 fa k " b k )+n k + (a lt " b L ) + (n L 
 
 - 1 
 
 ) + £ (a* ♦ bj)n, 
 
 = 1 
 
 k=L+l 
 
 for some oositive integer L such that 
 
L < 
 
 m + 1 i r D < fl 
 
 2fl 
 
 n 
 
 if p > q . 
 
 (For L = rn + 1, we define a, = b. = n, = 0.) 
 
 Proof We will use the notation of section 3. Since d ■ 1, 
 
 We can take 
 
 and 
 
 Case (i) 
 
 'hen 
 
 = { 1 , 2 , . . . , rn } 
 
 P k (I) = T k 
 
 , 1 < k < m, 
 
 
 Let 1 < L < m. 
 
 Ej. - {1, 2,..., L - 1}, 
 
 F, L2 = (L), 
 
 E T _ = {L + 1, L + 2, . . . , m}. 
 
 (If L = 1 then E T , * 0f. IP L = m then E T - = #. ) 
 
 We have 
 
 I 
 
 keF 
 m 
 
 I 
 
 L-l 
 
 b k P k (I) 
 
 b, I, 
 k k 
 
 m 
 
 Zb, I, . + b T I T + V 
 k k| l l ; 
 
 K=l k=Wl 
 
 b. I. 
 k k 
 
- \ + £ 
 
 m 
 
 k 
 
 k=l 
 
 b, I. + b T (I r - 1) + V b. L 
 
 k k L L / k k 
 
 k=l+L 
 
 29 
 
 = b L + y \ W 
 
 keE 
 
 'herefore, 
 
 and 
 
 \ ' \ 
 
 c T1 = b, , k e E 
 
 'Lk k 
 
 .*> 
 
 case (ii) 
 
 m hen 
 
 Let L « m + 1. 
 
 E L1 = E and E L2 = E L3 = 0. 
 
 .Since I T = I, we <ret 
 
 L 
 
 I b k P k (I) = I b k W' 
 keE keE 
 
 so that 
 
 and 
 
 y l ■ ° " \ 
 
 c T . = b , . k e E 
 
 Lk k ' 
 
 rp he conditions (A) and (B) of the present theorem 
 now follow from the similar conditions of Theorem 3.1. 
 
 Q . E . D . 
 
 Theorem 4.1 is easier to anply than Theorem 3.1 as 
 we see below. 
 
 V 
 

 H.2 Kxamole 
 
 Let 
 
 m • 3, 
 
 and 
 
 Then 
 
 f(I) - o + HI 1 - Bl 2 , 
 r(T) = B - 6I 1 + 4l 2 + 161 
 
 a i = 4 » a 2 = ~ 8, a 3 = °» 
 b - - -6, b 2 = ^, b = 16. 
 
 The red of the a's and the b's is 2. Corresoondinp to the 
 possible values 1, 2, 3, ^ of L, the inequalities in con- 
 dition (B) of Theorem *1 . 1 take the following r or r ns : 
 
 (1) (L = 1) 
 
 6 - 12n 2 - l6n < 3 - a <_ -4 + 10n 1 , 
 
 (2) (L = 2) 
 
 8 - 12n 2 - l6n < B - a <_ -*i + lOn , 
 
 (3) (L = 3) 
 
 -12 n 2 - l6n- <_ B - a <_ -16 + lOn.. , 
 
 CO (L = 1|) 
 
 -12n 2 - l6n„ 1 B - a <_ lOn 
 
31 
 
 Suppose now that X(f(I)) A X(r,(T)). Then by 
 
 Theorem '1.1, (8 - a) must be divisible by 2. If p < q, then 
 at least one of (1) - (4) must hold. If p _> q, then at least 
 one of (1) - (3) must hold. 
 
 A computer program that tests for data dependence 
 usinr the conditions of Theorem 4.1 is riven in the Appendix. 
 
 The example in Remark 3.4 (1) shows that the conditions 
 of Theorem 4.1 are not sufficient in the peneral linear case. 
 Fortunately, they are sufficient in a special linear case that 
 occurs frequently in actual nroprams. This is shown by Theorem 4.3 
 
 4.3 Theorem 
 
 Let 
 
 m 
 
 f(I) = a + t £ u k I k 
 k=l 
 m 
 and r(D = 3 + tV v k I k , 
 
 k=l 
 
 where ct, 8, t are integer constants, t > 0, and 
 u, , v, e {-1, 0, 1 • }, 1 _< k _< m. 
 Then, X(f(D) 4X(g(!)) if and only if 
 
 (A) t divides ( 8 - a) , 
 
 and (B) 
 
 L-l m 
 
 Y (u k " v k r \ " {u l + v r/ (n L - 1} - ^ (u k + v k )n i 
 
 k=l k=L+l 
 
u 
 
 < (3 - a)/t 
 
 L-l 
 
 m 
 
 < -v, 
 
 + I (Ul 
 
 " V\ + K " v l/ (n L- 1} + 
 
 + 7 " 
 
 k=l 
 
 I 
 
 for some positive integer L such that 
 
 L < 
 
 m+1 if n < n 
 
 m 
 
 if p > a 
 
 (For L = n + 1, We define u T = v T = n, = 0.) 
 ' L L Jj 
 
 Proof Since the 'only if nart follows directly fron ^heorem 
 4.1, we need prove only the 'if part. Assume that the conditions 
 
 (A) and (13) hold. 
 
 Let (0 - cx)/t + v, > 0. (The negative case can be 
 handled in an exactly similar way.) The second inequality in 
 
 (B) imnlies that 
 
 <_ (3 - ot)/t + V! 
 
 L-l 
 
 m 
 
 1 £ (u k- V +n k + (U L - V L ) + (n L " 1} + X ^ + ^ )rV 
 
 k=l 
 
 k = L+l 
 
 Therefore, we can write 
 
 m 
 
 (1) (P - a)/t + v L = 
 
 I v 
 
 k=l 
 
 where 
 
 if 
 
 (2) < h R <_ 
 
 (K k " v k )+n k 
 
 (u - v T ) (n T _ i) 
 
 Cu * + V k )n k 
 
 1 < k <_ L - 1 
 L - -w if k = L 
 
 if L + 1 < k < m. 
 
33 
 
 Now, in order to show that X(f(I)) A X(g(I)), 
 
 it will suffice if we can find two values i = (i n i ) 
 
 1' ■ rn 
 
 and J = (J , . . . , ,1 ) of T such that 
 
 (3) S (i) < S (j) 
 v p o q 
 
 and 
 
 ('0 f(i) 
 
 g(T). 
 
 Substituting for f and g in (4) we get 
 
 m 
 
 m 
 
 a + t r u k i k = B + t y v k J k 
 
 m 
 
 or 
 
 Z [ - 
 
 M 
 
 i _ v, ,1, ] = (B - a)/t. 
 k k k' k 
 
 k^ 
 
 k=l 
 
 By virtue of (1), this is equivalent to 
 
 m 
 
 (5) 
 
 I [ Vk " Vk ] + ^L 1 ! " V L ( ' 1 L " 1)]= I 
 
 V 
 
 k=l 
 k^L 
 
 k=l 
 
 We will find i and J such that (5) and (3) are 
 both satisfied. This will be done by picking suitable solutions 
 to the individual equations 
 
 (6) 
 
 Vk - V k = h k ; 1 - k - m ' k * L; 
 
 u T i- v, (j - 1) = h T (if L < m). 
 
 We give below a table that lists these solutions. For any given 
 
 k, detailed values of u, and v are considered only when 
 
 k r 
 
 h. j* 0. If h, = 0, then we always take the trivial solution. 
 Remember that the range of h is given by (2). 
 

 Range of k 
 
 U k 
 
 v k 
 
 
 
 Ranpe of h 
 k 
 
 Equation 
 
 Solution 
 
 
 he 
 
 •>k 
 
 l<k<L-l 
 
 
 
 h k -o 
 
 Vk-Vk" 
 
 
 
 
 
 
 
 
 -1 
 
 0<h k <n k 
 
 ^k ■ h k 
 
 
 
 h k 
 
 (if L>1) 
 
 1 
 
 
 
 0<h k <n k 
 
 l k ■ h k 
 
 h k 
 
 n 
 k 
 
 
 1 
 
 -1 
 
 0<h k <2n k 
 
 W h k 
 
 Lh k /2j 
 
 h k -Lh k /2j 
 
 k-L 
 
 
 
 h L = 
 
 Vl- v l ( Jl" 1h 
 
 
 
 1 
 
 
 -1 
 
 -1 
 
 °£ h L £ n L -l 
 
 "W^L 
 
 
 
 V 1 
 
 (if L<m) 
 
 
 
 -1 
 
 0<h. <n T -1 
 — u— l 
 
 h- 1 ' h L 
 
 
 
 V 1 
 
 
 1 
 
 
 
 0£h L <n L -l 
 
 h mh L 
 
 h L 
 
 n L 
 
 
 1 
 
 -1 
 
 0<h L £2n.-2 
 
 W^L 
 
 Lh L /2j 
 
 h L +l-Lh L /2j 
 
 L+l<k<m 
 
 
 
 v° 
 
 "kWk" 
 
 
 
 
 
 
 1 
 
 1 
 
 0£h k <n k 
 
 1 k--'k- h k 
 
 h k 
 
 
 
 (if L<m) 
 
 -1 
 
 -1 
 
 0£h k <n k 
 
 -V^k-^t 
 
 
 
 h k 
 
 
 
 
 -1 
 
 0<h k <n k 
 
 Jk- h k 
 
 
 
 h k 
 
 
 1 
 
 
 
 0£h k <n k 
 
 V h k 
 
 h k 
 
 
 
 
 1 
 
 -1 
 
 0<h k <2n k 
 
 Vrt 
 
 Lh k /2j 
 
 h k -Lh k /2j 
 
 Thus two values T = (i, ,..., i ) and J ■ (J-,,..., J„) 
 
 l m l m 
 
 of T can be found such that (6) is satisfied and 
 
 < \ < J k < \» 1 < k < L - 1 (if L > 1); 
 < i L < J L - 1 < n L - 1, (if L < m); 
 
 < 1 k» J k - n k» L + x I k i m (if L K m )« 
 
35 
 
 This means that i and j satisfy (3) and (4). Therefore, 
 
 X(f(T)) A X(g(T)). 
 
 Q.E.D 
 
 Let us illustrate the application of Theorem 4.3 by 
 an example. 
 
 4.4 Example 
 
 Consider the following; program segment: 
 
 DO 5 I 1 = 0, 10 
 DO 5 I ? = 0, 5 
 
 DO 5 I 3 = 0, 9 
 
 S x : X(13 + 6l 1 - 61 ) = R 1 
 
 Spt X(l + 61 - 6l 2 + 61 ') = R 2 
 
 5 CONTINUE. 
 
 And suppose that we want to test for dependence of the output 
 variable of S ? on the output variable of S, . 
 
 Here we have 
 
 m = 3, n i = 10 > n 2 = 5, n„ = 9, 
 
 p = 1, q = 2, 
 
 f(T) = 13 + 6I 1 - 6I 3 
 
 g(I) = 1 + 6l 1 - 6lp + 61-,. 
 
36 
 f(T) and g(T) have the forms needed for 
 
 a - 13, p - 1, t = 6, 
 
 u 1 =1, u 2 = 0, u^ = -1, 
 
 v x =1, v 2 = -1, v^ = 1. 
 
 t divides (fl - a). And for L = 1, the inequalities in condition 
 (13) of Theorem J l . 3 become 
 
 " V l " ^ U l + V l^ ^ n l 
 
 1) - (u~ + v 2 )n 2 - (u~ + v^)n^ 
 
 < (3 - «)/t 
 
 <_ - v 1 + (u 1 - v^) (n 1 
 
 1) + (u 2 + v~)n 2 + (u_ + v ^)ru 
 
 or, -28 < -2 < k 
 
 which is true. Therefore, by Theorem H . 3 the outnut variable 
 X(l + 61 - 6lp + 61-,) of statement S ? is denendent on 
 the outnut variable X(13 + 61, - 6l_) of statement S, . 
 Indeed, taking 
 
 i = (0, 0, 0) 
 
 and 
 
 .1 = (1, 0, 1) , 
 
 we see that 
 
 S, (i) < S (.1) 
 1 o 2 
 
 and 
 
 f(D = g(J). 
 
37 
 
 4 . 5 Remarks 
 
 We g-ive now some comments on the results of Cohap;an 
 [1] and Towle ([51, ch. 5). 
 
 Theorem 1 of Cohafran can be derived from Theorem 4.1. 
 His Theorem. 2 and Lemmas 2.1, 2.2 follow immediately from well 
 known results in the theory of Diophantine Equations. (See 
 Mordell [4], ch. 5, Sec. 1.) 
 
 Now consider Towle. Parts (1) and (3) of his Theorems 
 5.3 and 5.4 can be derived from our Theorem 4.1. However, 
 part (2) of each of these two theorems is false. (Take n, = 10, 
 p = 20, q = 4, a, = 1, b 2 = 2. Since 20 + = 4 + 2*8, 
 there is data dependence. But (d - q)/(b - a, ) = 16 > n ) . 
 His theorems 5.5 and 5.6 are contained in the nroof of our 
 Theorem 3.1. Towle uses a theorem of I 7 an([3], Th . 10) on 
 the inconsistency of a system of linear equations to derive 
 a set of necessary conditions for data dependence. These con- 
 ditions should be equivalent to the inequalities in (B) of 
 Theorem 4.1, since both sets are (separately) necessary and 
 sufficient for the existence of real solutions. However, 
 these conditions are rather imnractical. Because, they are 
 difficult to derive, and even in the two variable case with 
 L = 1, there are 29 of these conditions for the sinple pair 
 of inequalities in (B) of Theorem 4.1. (See Towle [5], 
 Appendix B. ) 
 
38 
 LIST OF RRPERENC 
 
 [1] Coha^an, W.L., "Vector Ontimization for the A.': r ;", 
 
 Proceedings of the Seventh Annual Princeton Conference 
 on Information Sciences and Systems, 1973, nn. 169-174. 
 
 [2] Dieudonne', J., "Foundations of Modern Analysis", 
 Academic Press, .'Jew York, 1969. 
 
 [3] Fan, K., "On Systems of Linear Inequalities", Anna. 
 
 of Mathematics Studies, NO. 38, nn. 99-156, Princeton 
 University Press, 1956. 
 
 [4] Mordell, L.J., "Dionhantine Equations", Academic Press, 
 New York, 1969. 
 
 [5] Towle, R., "Control and Data Dependence for Program 
 
 Transformations", Ph.D. Thesis, Denartment of Comnuter 
 Science, University of Illinois at Urbana-Chamnaipn, 
 Report 788, March - 1976. 
 
39 
 
 APPENDIX 
 
 Here we reive a PL/I program that tests for data 
 dependence usinp: the conditions of Theorem ^.1. 
 
 The innut to the program consists of the values of 
 W, ALPHA, A, BETA, B, N, P, Q where 
 
 M = m; 
 
 ALPHA = a ; 
 
 A is an array of size M with A(K) = a,, 
 
 1 < k < M; 
 BETA = 3; 
 B is an array of size M + 1 with B(K) = b. , 
 
 1 < k < M, B(M + 1) = 0; 
 N is an array of size M with M(K) = n, , 
 
 1 < k < M; 
 
 P - p, Q - q. 
 
 The gcd of the set (A(l),..., A(M), B(l),..., B(M)} 
 is comnuted and condition (A) of Theorem '4.1 is tested. If 
 
 it holds then the program goes on to test condition (B). ((A) 
 
 i i it 
 
 is referred to as the GCD Test , and (B) as the Minimax Test .) 
 
 Mote that condition (B) can be written as 
 -PSI(M,L, -A, -B, N) < BETA - ALPHA + B(L)< PSI(M, L, A, B, M), 
 
 where 
 
 L-l 
 PSI(M, L, A, B, M) + y-j (A(K) - B(K)) N(K) 
 
 + ((A(L)) - B(L)) + (M(L) - 1) + £ ((A(K)) + + (-B(K) ) + )N(K) . 
 
 k=L+l 
 
Since PS] Ls always non negative 
 of the two inequalities in the condition, d< 
 of [BETA - ALPHA + B(L)]. The number of tines the fimctJ I 
 Is comoutod (and then used in a comparison) ls at mo^t N ♦ 1, 
 The similarity between the conditions o? Theor' .1 
 and ^ . 1 shows that this prop-ram can be extended to the non lin' 
 case without too much difficulty. The only orobl<- that we 
 must know the sets BL , , ^ T p> Ej ^ and the interers y. , c. . 
 and incornorate then into the program. 
 
PROCEDURE OPTIONS (MAIN); 41 
 
 PSI ; PROCEDURE (M.L.A,B,N) RETURNS (FIXED BIN (31)): 
 
 PLUS : PROCEDURE (R) RETURNS (FIXED BIN (31)); 
 DCL R FIXED EIN (31) ; 
 
 IF R>0 THEN R ETURN (R) ; 
 
 It SB F.ETUKN (0) ; 
 
 END PLUS^; 
 
 DCL_XM. L. A(*), B(*). N(»), K, SUM) FIXED BIN(31); 
 3TJH = ~^~ 
 
 DO K = \ TO L-1 ; 
 
 SUM = SUM + PLUS (A (K) - B (K) ) *N (K) ; 
 
 rl^L<=M THEN SUM = SUM ♦ PLUS (PLUS (A (L) ) - B (L) ) * (N (L)- 1) ; 
 
 DO K = L+1 TO H; 
 
 SUM = SUM ♦ (PLUS (A (K) ) ♦ PLUS (-B (K) ) ) *N (K) ; 
 
 END: „ 
 
 PUTSKIP LI3T~f • PSI=', SUM) ; 
 
 RETURN (SUM) ; 
 END PSI; 
 
 HCL (M, L r ALPHA, BETA, P, Q) FIXED BIN131) ; 
 
 DCL \K. DIF, GCD, X, 3, UB) FIXED BIN(31); 
 DCL DEPENDENT BIT (1) ; 
 _9JL ENDFILE ( SYSIN) S TOP; 
 
 DO V'HILE (' 1'B) ; 
 GET LIST (M) ; 
 
 BEGIN; 
 
 DCL (A(M), B(M*1), N(H)J FIXED BIN(JI); 
 
 GET LIST (ALPHA, A, BETA- B, N, P, Q) ; 
 PUT SKIP(3) DATA (M, ALPHA, A, BETA, B, N, P, Q) ; 
 
 GCD = AM); /* COMPUTE THE GCD OF ELEMENTS OF A S B */ 
 
 DO K = 1 TO M-1 ; 
 X = AJK+1) ; 
 DO WHILE (X -.= 0) ; 
 
 R = MOD(GCD, X) ; 
 GCD = X; 
 
 X = F; 
 
 EWj 
 
 END; 
 
 DO K = 1 TO M; 
 
 DO WHILE (X -i= 0) ; 
 
 R = MOD (GCD, X) ; 
 GCD = X; 
 X = R: 
 
 END; 
 END; 
 PUT SKIP DATA (GCD) ; /* GCD HAS BEEN COMPUTED */ 
 
 D I F = BETA - ALP H A ;' 
 
 IF GCD -.= S MOD (DIF, GCD) -»= THEN 
 
 PUT SKIP LIST (•NOT DEPENDENT BY THE GCD TEST 1 ); 
 ELSE DO: 
 
 IF P < C THEN UB = M+1; ELSE OB = ft; 
 
 DEPENDENT = • O'B; 
 DO L = 1 TO UB WHILE (-DEPENDENT) ; 
 K = DIF ♦ B(L)j 
 
 I'F (K>0"6"K <= PSI (M,L,A,B, N) ) 
 
 | (K<0 S K >= -PSI (M,L,-A,-3,N) ) | (K=0) THEN CO; 
 DEPENDENT = • 1 • B; 
 PU T SKIP LI S T ('PROBABLY DE PEN DENT' .' 1=' . L) ; 
 
 I END; 
 
 IIP DDT; 
 
 END; 
 END * 
 IF -DEPENDENT THEN PUT SKIP LIST 
 
 ('NOT DEPENDENT BY THE MNIMAX TEST'J_; 
 
 "END; ■" 
 
 END; 
 
BIBLIOGRAPHIC DATA 
 SHEET 
 
 1. Report No. 
 
 UIUCDCS-R-76-837 
 
 itle and Subtitle 
 
 DATA DEPENDENCE IN ORDINARY PROGRAMS 
 
 3. Recipient's Accession No. 
 
 5. Report Date 
 
 November 1976 
 
 6. 
 
 Author(s) 
 
 Utpal Baner.iee 
 
 8- Performing Organization Rept. 
 
 No - UIUCDCS-R-76-837 
 
 '. Performing Organization Name and Address 
 
 University of Illinois at Urbana-Champaign 
 Department of Computer Science 
 Urbana, Illinois 61801 
 
 10. Project/Task/Work Unit No. 
 
 11. Contract/Grant No. 
 
 2. Sponsoring Organization Name and Address 
 
 National Science Foundation 
 Washington, D. C. 
 
 13. Type of Report & Period 
 Covered 
 
 Master's Thesis 
 
 14. 
 
 5. Supplementary Notes 
 
 6. Abstracts 
 
 Certain conditions have been obtained under which there is no dependence between 
 two assignment statements inside an IF-free loop in a language like FORTRAN. The 
 number of index variables is arbitrary, and the array subscripts are assumed to be 
 general polynomials in the index variables. 
 
 7. Key Words and Document Analysis. 17a. Descriptors 
 
 Assignment statements 
 Data dependence 
 Input variables 
 Output variables 
 
 'b. Identifiers/Open-Ended Terms 
 
 fc. COSATI Field/Group 
 
 |J. Availability Statement 
 
 Release Unlimited 
 
 19. Security Class (This 
 Report) 
 
 UNCI ASSIFIED 
 curitv Class (Thi 
 
 20. Security Class (This 
 Page 
 
 UNCLASSIFIED 
 
 21. No. of Pages 
 
 45 
 
 22. Price 
 
 3RM NTIS-3S ( 10-70) 
 
 USCOMM-DC 403Z9-P7 1 
 
• - 
 
 ^ 
 
 % 
 
M 
 
UNWERal 
 51(1 MIltRno I 
 Intirm! rtport / 
 
 XINO»*-Cmt!*H» 
 
 3 0112 088403149