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UIUCDCS-R-79-992 
 
 CUf 
 
 UILU-ENG 79 1740 
 
 Quadratic Algorithms for Minimizing Joins 
 in Restricted Relational Expressions 
 
 by 
 
 Yehoshua Sagiv 
 
 October 1979 
 
 e— c 
 
 • »i 
 
 DEPARTMENT OF COMPUTER SCIENCE 
 UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN 
 
 URBANA, ILLINOIS 
 
 THE LIBRARY OF THE 
 MAY 8 1960 
 
 UNIVERSITY OF ILLINOIS 
 'i^ANA-OHAMPAIGN 
 
UIUCDCS-R-79-992 
 
 Quadratic Algorithms for Minimizing Joins 
 in Pestricted Relational Expressions 
 
 Yehoshua Sagiv 
 
 Department of Computer Science 
 
 University of Illinois at Urbana-Champaign 
 
 Urbana, Illinois 61801 
 
 October 1979 
 
 This work was supported in part by the National Science Foundation under 
 grant MCS-76-15255 
 
X<L6>n. 
 
 ABSTRACT 
 
 An important step in the optimization of queries in relational 
 databases is minimizing the number of join operations used in the 
 
 evaluation of a query. It is shown that three subclasses of tableaux 
 
 2 
 (including the subclass of simple tableaux) have 0(n ) time equivalence 
 
 and minimization algorithms. Since tableaux are nonprocedural represen- 
 tations of relational expressions over select, project and join, these 
 minimzation algorithms can be used to minimize the number of join opera- 
 tors in expressions whose tableaux belong to one of these subclasses. 
 
 CR categories: 4.33, 5.25 
 
 Key words and phrases: relational database, relational algebra, query 
 optimization, equivalence of queries, conjunctive query, tableau, NP- 
 complete. 
 
_K Introduction 
 
 The relational model for databases features two high-level query 
 languages: the relational algebra and the relational calculus [9,10]. 
 The relational algebra is a procedural language that uses operators 
 defined on relations, and hence a query is usually translated to a rela- 
 tional expression before being evaluated. However, the efficiency with 
 which a query can be answered depends on the relational expression that 
 has been chosen to represent this query. Consequently, a number of 
 papers (e.g., [12,13,14,15,17,18]) have considered transformations that 
 reduce the cost of evaluating a query. However, these transformations 
 do not necessarily produce an equivalent query of least cost. Chandra 
 and Merlin [8] show how to perform global optimization on a large class 
 of queries, but their algorithm is exponential in the size of the query. 
 
 The most commonly used operators of the relational algebra are 
 select, project and join, and a polynomial time algorithm for optimizing 
 a subclass of expressions with these operators is given in [4,5]. This 
 optimization technique uses tableaux [3] as a nonprocedural representa- 
 tion of queries. Tableaux are similar to the conjunctive queries of 
 [8], and resemble Zloof's "Query-by-Example" language [20]. Relational 
 expressions over select, project, and join can be represented by 
 tableaux [3]. In [8] it is shown that every tableau has an equivalent 
 minimal tableau. The importance of this result follows from the fact 
 that an expression with a minimum number of joins corresponds to a 
 minimal tableau. A polynomial minimization algorithm for the class of 
 simple tableau is given in [4] (although the problem is NP-complete in 
 
- 2 - 
 
 the general case [3]). In [5] it shown how to obtain in polynomial time 
 an optimal expression from a minimal simple tableau (if such an expres- 
 sion exists). 
 
 This optimization technique is machine independent. It is capable 
 of minimzing the number of join operators (note that join is the most 
 expensive operator to implement), eliminating redundant subexpressions, 
 and applying select and project as early as possible. In a relational 
 database system, this type of optimization should be augmented with 
 machine dependent optimization that takes into consideration the size of 
 the relations involved, sorted columns, etc. 
 
 In this paper we describe new minimization algorithms for two subc- 
 lasses of tableaux. We also show how to improve the running time of the 
 
 minimzation algorithm for the class of simple tableaux. All these algo- 
 
 2 
 rithms have an 0(n ) running time. It shown that each one of the three 
 
 2 
 subclasses of tableaux discussed in this paper also has an 0(n ) time 
 
 equivalence algorithm. Finally, in Sections 7 and 8 we touch upon the 
 
 problems of minimizing tableaux in the general case, and obtaining 
 
 optimal expressions from minimal tableaux. 
 
 2. Basic Definitions 
 
 2.1 The Relational Model 
 
 The relational model for databases [9] assumes that the data is 
 stored in tables called relations . The columns of a table correspond to 
 
- 3 - 
 
 attributes , and the rows to records or tuples . Each attribute has an 
 associated domain of values. A tuple is viewed as a mapping from the 
 attributes to their domains, since no canonical ordering of the attri- 
 butes is needed in this way. If r is a relation with a column 
 corresponding to the attribute A, and p is a tuple in r, then y(A) is 
 the value of the A-component of u. In this paper we usually denote a 
 relation as a set of tuples. 
 
 A relation scheme is a set of attributes labeling the columns of a 
 table, and it is usually written as a string of attributes. We often 
 use the relation scheme itself as the name of the table. A relation is 
 just the "current value" of a relation scheme. The relation is said to 
 be defined on the set of attributes of the relation scheme. 
 
 2.'2l The Relational Algebra and Relational Expressions 
 
 The relational algebra [9,10] is a set of operators defined on 
 relations. In this paper we consider the operators select, project, and 
 join. 
 
 Let r be a relation defined on a set of attributes X, A an attri- 
 bute in X and c a value from the domain of A. The selection A=c, writ- 
 ten a. (r) , is 
 
 A«c 
 
 o. (r) «■ {u | u is a tuple in r and u(A)«c> 
 
 A C 
 
 Let Y be a subset of X, the projection of r onto Y, written ^ Y (r) , 
 is 
 
- 4 - 
 
 tt (r) = <M | p is a mapping on Y, and there is a v in r 
 
 that agrees with \i on Y> 
 
 Let r. and r~ be relations defined on the relation schemes R. and 
 
 R 
 
 2 , respectively. The (natural) loin of r, and r 2 , written r. (xj r~, is 
 r. |xj r 2 = <u | u is a mapping on R, U R 2 , and there is v in r. that 
 agrees with u on R., and v in r~ that agrees with u on R~} 
 Note that the join includes intersection (when R. and R» are the same) 
 and cartesian product (when R. and R^ are disjoint) as special cases. 
 
 The relational algebra includes other operators, and to make our 
 set of operators "complete" (in the sense of Codd [10]) we only need to 
 add the union and difference operators, and allow selections involving 
 arithmetic comparisons between two components of a tuple. 
 
 The operators of the relational algebra are used in the formulation 
 of queries in terms of relational expressions. A restricted relational 
 expression consists of select, project, and join as operators, and rela- 
 tion schemes as operands. 
 
 _2'3. Expression Values and Equivalence of Expressions 
 
 An underlying assumption in many papers (e.g., [1,6,7]) is the 
 existance of a single universal relation at each instant of time. This 
 relation is defined on the set of all the attributes, and it is called a 
 universal instance or just an instance . If I is an instance and E is a 
 relational expression, then each relation scheme R in E is assigned the 
 
- 5 - 
 
 relation tt_, (I). The value of E with respect to I, written v (E), is 
 
 R i * 
 
 computed by applying operators to operands in the following natural way. 
 
 (1) If E is a single relation scheme R, , then v._(E) - tt_ (I). 
 
 (2) (a) If E - o. (E,), then v T (E) - o (v T (E.)). 
 
 A=C i 1 A"C 1 i 
 
 (b) If E - ^(Ej), then v^E) - VVV^ 
 
 (c) If E - E x M E 2 , then v^E) - VjO^) M VjCEj). 
 
 We may regard a relational expression as a mapping from instances 
 to relations, i.e., the expression E maps the instance I to the relation 
 v_(E). Two expressions E. and E„ are equivalent if they define the same 
 mapping. That is, if for all instances I, v_(E.) - v_(E 2 ). In [3] 
 other types of equivalence are discussed, and it is shown that results 
 obtained for the above type of equivalence also apply to the more gen- 
 eral case, where the relations assigned to the relation schemes do not 
 necessarily come from an instance. 
 
 3^. Tableaux 
 
 Tableaux are just another way of representing mappings from 
 instances to relations. Unlike relational expressions, tableaux are 
 nonprocedural representation of queries in exactly the sense that rela- 
 tional calculus [9,10] Is nonprocedural. In [3] it is shown that every 
 restricted relational expression has a corresponding tableau that 
 represents the same mapping. 
 
- 6 - 
 
 A tableau is a matrix consisting of a summary and a set of rows . 
 The columns of a tableau correspond to the attributes of the universe in 
 a fixed order. The symbols appearing in a tableaux are chosen from 
 
 (1) Distinguished variables, usually denoted by subscripted a's. 
 
 (2) Nondistinguished variables, usually denoted by subscripted b's. 
 
 (3) Constats, which are drawn from the domains of the attributes. 
 
 (4) Blank. 
 
 Each row may contain constants, distinguished and nondistinguished 
 variables. The summary has constants, distinguished variables and 
 blanks. A variable cannot appear in more than one column, and a dis- 
 tinguished variable may appear in a particular column only if it appears 
 in the summary. Furthermore, if a constant or a distinguished variable 
 appears in some column A of the summary, then it must also appear in 
 column A of at least one row. 
 
 Let T be a tableau with a summary w~ and rows w.,...,w , and let S 
 be the set of all the nonblank symbols in T. A valuation p for T maps 
 each symbol in S to a constant, such that if c is a constant in S, then 
 p(c)"c. The valuation p is extended to the rows and summary of T by 
 defining p(w ) to be the result of substituting p(v) for all variables v 
 in w . 
 
 A tableau T defines a mapping from instances to relations on its 
 target relation scheme, which is the set of all the attributes 
 corresponding to columns that have a nonblank symbol in the summary. 
 Given an instance I, the value of T, written T(I), is 
 
- 7 - 
 
 T(I) - <p(w ) | for some valuation p, we have p(w ) in I for l<i<n> 
 
 Conventionally, we also regard <J> as a tableau. This tableau 
 represents the function that maps every instance to the empty relation. 
 
 Example 1 : Consider the following tableau. 
 
 ABC 
 
 T - 
 
 ,a l 
 
 
 a 3J 
 
 j a l 
 
 2 
 
 b 3J 
 
 ] b l 
 
 b 2 
 
 a 3J 
 
 1*1 
 
 b 2 
 
 b 4| 
 
 The summary is shown first, with a line below it, and integers are used 
 as constants. 
 
 Tableau T defines a relation on the relation scheme AC. For exam- 
 ple, supoose that I is the instance {211,121,122). 
 
 Consider the valuation p which assigns 2 to b« and 1 to every other 
 variable in T. Under this valuation, each row of T becomes 121, which 
 is a member of I. Therefore, pCa.a..) = 11 is in T(I). 
 
 If p assigns 1 to a., b. , b~ and b,, and 2 to b~ and a~, the first 
 and third rows become 121 and the second row becomes 122; both are 
 members of I, and so 12 is in T(I). 
 
 Since no valuation for T produces a tuple other than 11 or 12, 
 T(I) * {11, 12). [] 
 
- 8 - 
 
 _3._1 Relational Expressions and Tableaux 
 
 Every restricted relational expression E has a corresponding 
 tableau T that defines the same mapping as E (however, the converse is 
 not true) [3]. The tableau T for the expression E can be constructed 
 bottom up by applying the following rules [3]. 
 
 (A) If there are no operators in E, then E is a single relation scheme 
 R. The tableau T for E has one row and a summary such that : 
 (i) If A is an attribute in R, then in the column for A, the 
 tableau T has the same distinguished variable in the summary 
 and row. 
 (ii) If A Is not in R, then its column has a blank in the summary 
 and a nondistinguished variable in the row. 
 (Bl) Suppose E is of the form a (E.), and we have constructed T., the 
 tableau for E.. 
 
 (i) If the summary of T. has blank in the column for A, then the 
 expression E has no meaning, and the tableau T for E is unde- 
 fined. 
 (ii) If T- has a constant c'*c in the summary column for A, then 
 for any instance I, v_(E) has only tuples with c' in the com- 
 ponent for A, and v_(E) Is $. Hence, the tableau for E is 4>. 
 (iii) If T. has c in the summary column for A, then the tableau for 
 
 E is T-. 
 (iv) If T. has a distinguished variable a in the summary column 
 for A, the tableau T for E is constructed by replacing a by c 
 whenever It appears in T.. 
 
- 9 - 
 
 (B2) Suppose E is of the form tt (E.) and T. is the tableau for E.. Con- 
 struct T for E by replacing nonblank symbols by blanks in the sum- 
 mary of T. for those columns whose attributes are not in X. Dis- 
 tinguished variables in those columns become nondistinguished. 
 (B3) Suppose E is of the form E. b<| E-, and T. and T^ are the tableaux 
 for E. and E_, respectively. 
 
 (i) If T. and T~ have some column in which their summaries have 
 distinct constants, then It easy to show that for all 
 instances I, v T (E)= <J>, so <(» is the tableau for E. 
 (ii) If no corresponding positions in the summaries have distinct 
 constants, let S. and S„ be the sets of symbols of T. and T_, 
 respectively. We may take S. and S_ to have disjoint sets of 
 nondistinguished variables, but identical distinguished vari- 
 ables in corresponding columns. Construct T for E to have the 
 union of all the rows of T. and T_. The summary of T has in a 
 given column: 
 
 (a) The constant c if one or both T. and T_ have c in that 
 column's summary. In this case replace any distinguished 
 variable in that column by c. 
 
 (b) The distinguished symbol a if (a) does not apply, but one 
 or both of T. and T ? have a in that column's summary. 
 
 (c) Blank, otherwise. 
 
 These rules can also be used to define the operations select, pro- 
 ject and join on tableaux. The result of applying any one of these 
 operators to tableaux (not necessarily tableaux derived from restricted 
 
- 10 - 
 
 relational expressions) is defined to be the tableau described in the 
 rule for that operator. 
 
 Example 2 ; Consider the expression tt (tt (AB (xj BC) M a _(AB)) 
 If the above rules are applied to this expression, then the result is 
 the tableau of Example 1. [] 
 
 ^._2 Equivalence of Tableaux 
 
 Two tableaux T and T 2 are equivalent , written T. = T-, if for all 
 instances I, T.(I) = T_(I). We say that T. is contained in T_, written 
 T x C^ T 2 , if for all I, T^I) C T 2 (I). 
 
 Let T. and T~ be tableaux with the same target relation scheme, and 
 let S. and S_ be the sets of symbols of T. and T_, respectively. A 
 homomorphism is a mapping £:S.-»-S ? such that: 
 
 (a) If c is a constant, then £(c)=c 
 
 (b) If s is the summary of T., then £(s) is the summary of T_. 
 
 (c) If w is any row of T., then £(w) is a row of T„. 
 The following theorem is proved in [3,8]. 
 
 Theorem 1 ; T_ (L, T. if and only if T. and T_ have the same target 
 relation scheme, and there is a homomorphism £:S.->-S„. 
 
 By condition (c) , a homomorphism £ corresponds to a mapping 8 from 
 the rows of T. to the rows of T_. The mapping 6 is called a containment 
 mapping , and it satisfies the following conditions. 
 
- 11 - 
 
 (1) If row w of T. has a constant in some column A, then 9(w) has the 
 same constant in column A. 
 
 (2) If row w of T. has a distinguished variable in column A, then 9(w) 
 has a distinguished variable in column A. 
 
 (3) If rows w and v have the same nondistinguished variable in column A, 
 then rows 8(w) and 9(v) have the same symbol in column A. 
 
 Corollary 2 ; [3] Tableaux T. and T„ are equivalent if and only if 
 they have identical summaries up to renaming of distinguished variables, 
 and there are containment mappings in both directions. 
 
 A tableau T is minimal if T is not equivalent to any tableau with 
 fewer rows. Note that if T comes from an expression E, then the number 
 of joins in E is one less than the number of rows in T. Thus, if T is 
 minimal, it corresponds to an expression with a minimum number of joins. 
 For every tableau T, there Is a unique (up to renaming of variables) 
 tableau T' , such that T = T' and T' is minimal [8]. Furthermore, the 
 
 minimal tableau equivalent to T can be obtained by deleting some of the 
 
 (2) 
 rows of T. The core of T is the set of all the rows that belong to 
 
 the minimal tableau obtained by deleting redundant rows of T. A folding 
 
 Is a containment mapping from the rows of a tableau T to the rows of the 
 
 core of T, such that every row in the core of T is mapped to itself. It 
 
 can be shown that every tableau T has a folding [8]. Note that a 
 
 (1) In this paper we consider only equivalence (and not proper contain- 
 ment) of tableaux, and therefore the definition of a containment 
 mapping is more restricted than the original definition given in 
 [3]. 
 
 (2) A tableau T might have two different cores. However, they are the 
 same up to renaming of variabls. Therefore, we usually speak about 
 the core of T. 
 
- 12 - 
 
 homomorphisra that corresponds to a folding maps every variable in the 
 core of T to itself. The following corollary is an immediate conse- 
 quence of the results stated so far. 
 
 Corollary 3 ; Let T. and T_ be tableaux with the same summary. If 
 the rows of T\ are a subset of the rows of T_, then 
 
 (1) T 2 C.J. T r and 
 
 (2) T 2 = T. if and only if the core of T- is contained in T.. 
 
 Let w and x be rows of tableaux over the same set of attributes. 
 Row w covers row x, written x < w, if for all columns A, 
 
 (1) if x has a constant in column A, then w has the same constant in 
 column A, and 
 
 (2) if x has a distinguished variable in column A, then w has a dis- 
 tinguished variable in column A. 
 
 Note that if x is mapped to w, and x < w, then the first two conditions 
 of a containment mapping are satisfied. Let R and S be sets of rows 
 over the same set of attributes. The set S covers the set R, written 
 R < S, if every row of R is covered by some row of S. 
 
 A symbol (i.e., a variable or a constant) of a tableau T is 
 repeated in some column A, if it appears in that column in more than one 
 row. A tableau T is simple if whenever T has a repeated nodistinguished 
 
 variable b in some column A, then b is the only repeated symbol in that 
 
 3 
 column. The class of simple tableaux has an 0(n ) equivalence algorithm 
 
 4 
 [3], and an 0(n ) minimzation algorithm [4]. Other equivalence algo- 
 rithms can be obtained from the containment algorithms of [16]. That 
 
- 13 - 
 
 2 
 is, tesing whether T« = T. can be done in 0(n ) in the following two 
 
 cases : 
 
 (1) Both T. and T„ have at most one repeated nondistinguished variable 
 in each row. 
 
 (2) Every row of T. is covered by at most two rows of T^, and every row 
 of T« is covered by at most two rows of T.. 
 
 4^ Polynomial Equivalence Algorithms 
 
 In this section we consider the following classes of tableaux. 
 
 (1) The class of all tableaux T, such that T has at most one repeated 
 nondistinguished variable in each row. 
 
 (2) The class of all tableaux T, such that every row of T is covered by 
 at most one row besides itself. 
 
 Note that deciding whether a tableau belongs to Class 1 (or whether a 
 
 tableau is simple) requires 0(n) time. However, deciding whether a 
 
 2 
 tableau belongs to Class 2 requires 0(n ). 
 
 2 
 Theorem 4 : Each one of the above classes has an 0(n ) time 
 
 equivalence algorithm. 
 
 Proof : It follows immediately from the results of [16] that the 
 theorem is true for Class (1). Suppose that both T and T 9 are tableaux 
 that belong to Class 2. We say that two rows w and x are equivalent, if 
 w covers x and x covers w. Consider the algorithm of Figure 1. This 
 algorithm tests whether T = T . Obviously, if T = T. and T. E 1 
 
- 14 - 
 
 (1) let T. be the result of removing from T. every row 
 that Is not equivalent to some row of T~; 
 
 (2) let T 2 be the result of removing from T_ every row 
 that Is not equivalent to some row of T. ; 
 
 (3) If T, £ T. then return false; 
 
 (4) if T. % T 2 then rejturn false; 
 
 (5) If T. = T. then return true else return false; 
 
 end 
 
 Figure 1 
 
 then T. = T 2 if and only if T. = T-. Thus, we have to show that T. and 
 T 2 cannot be equivalent if either T. % T, or T 2 % 1„. 
 
 f r 
 
 Suppose that T. % T. . Since the rows of T. are a subset of the 
 rows of T., it follows that the core of T, contains a row w which is not 
 in T.. By the construction of T., no row of T ? is equivalent to w. But 
 equivalent tableaux have cores which are the same (up to renaming of 
 variables). That is, every row in the core of one tableau has an 
 equivalent row in the core of the other tableau. Therefore, T. and T« 
 cannot be equivalent. Similarly, if T 2 % T- t then T. % T_. 
 
 Obviously, for i e (1,2), each row of T is covered by at most two 
 
 rows of T , and each row of T is covered by at most two rows of T . 
 
 Suppose that a row w of T. is covered by more than two rows of T_. Let 
 
 x be a row of T« that is equivalent to w. Row x is covered by every row 
 
- 15 - 
 
 of T„ that covers w. Therefore, x Is covered by at least two rows of T 2 
 besides itself. This contradiction implies that each row of T. is 
 
 r r 
 
 covered by at most two rows of T-. Similarly, each row of T. is covered 
 by at most two rows of T.. 
 
 It follows that testing equivalence in lines (3)-(5) can be done 
 
 using the algorithm of [16]. Each application of this algorithm 
 
 2 2 
 
 requires 0(n ) time. Lines (1) and (2) can be executed in 0(n ) time 
 
 and, therefore, the algorithm of Figure 1 has a time complexity of 
 
 0(n 2 ). [] 
 
 5_. Obtaining Minimization Algorithms from Equivalence Algorithms 
 
 Let S be a class of tableaux. We say that S is closed under row 
 deletion if whenever a tableau T is in S, and T' is obtained by deleting 
 some of the rows of T, then T' is also in S. 
 
 Theorem 5 ; Let S be a class of tableaux closed under row deletion. 
 If there is an equivalence algorithm for S that runs in F(n) time 
 (F(n) > en for some constant c) , then there is a minimization algorithm 
 for S that runs in nF(n) time. 
 
 Proof : Figure 2 describes a minimization algorithm for S. The 
 input for this algorithm is a tableau T of S. We assume that the 
 numbers l,2,...,r correspond to the rows of T. The algorithm is based 
 on the ability to test equivalence of tableaux from S. Note that the 
 equivalence test in line (3) can be replaced with the containment test 
 
- 16 - 
 
 T ' C/p T » because the rows of T' are a subset of the rows of T and, 
 hence, T (L, T' . 
 
 Let T be the value of T after i iterations through the loop of 
 lines (l)-(3). Since T is assigned a new value in line (3) only if the 
 new value is equivalent to the old value, T (i.e., the tableau returned 
 by the algorithm) is equivalent to T~. Suppose that T is not minimal. 
 Therefore, there is a row j in T that does not belong to the core of 
 T . Let T be the result of deleting row j from T . Since the core of 
 T is not changed as a result of this deletion, T = T and, hence, 
 
 r 
 
 Since row j has not been deleted by the algorithm, it must be in 
 
 T . Let T be the result of removing row j from T . It follows that T 
 
 is not equivalent to T fi (otherwise row j cannot appear in T ) . But the 
 
 rows of T are a subset of the rows of T., and the rows of T. are a sub- 
 r j j 
 
 set of the rows of T Q and, therefore, T Q (L, T .£_ T . Since T = T Q , it 
 follows that T = T Q . This contradiction implies that T is minimal. 
 
 In each pass through the main loop, line (2) requires 0(n) time and 
 line (3) requires F(n) time. The loop is repeated r times to give a 
 time complexity of nF(n) (since r < n) . [] 
 
 _6 . Polynomial Minimization Algorithms 
 
 The classes of tableaux described in Section 4, and the class of 
 simple tableux have polynomial time equivalence algorithms. Each one of 
 
- 17 - 
 
 begin 
 
 (1) ~ for i :■ 1 to r do 
 
 begin 
 
 (2) Let T' be the result of deleting row i from T; 
 
 (3) if T = T' then T :- T'; 
 end; 
 
 (4) return T; 
 end 
 
 Figure 2 
 
 these classes is closed under row deletion and, therefore, Theorem 5 can 
 be applied to obtain polynomial minimization algorithms for these 
 classes. However, the algorithms obtained by applying Theorem 5 are not 
 the most efficient minimization algorithms for these classes. For each 
 
 one of these classes there is a minimization algorithm with a time com- 
 
 2 
 plexity 0(n ). These minimization algorithms are given in the following 
 
 sections . 
 
 j>.^ Regular Tableaux 
 
 In this section we will show that a tableau, in which a folding 
 
 does not eliminate any repeated nondistinguished variable, can be minim- 
 
 2 
 ized in 0(n ) time. This fact is used in developing the minimization 
 
 algorithms of the following sections. 
 
 Let b a repeated nondistinguished variable of a tableau T. The 
 variabel b is essential if It appears in every core of T. A tableau T 
 is regular if all repeated nondistinguished variables of T are essen- 
 tial. 
 
- 18 - 
 
 begin 
 
 /* The Input is a tableau T. */ 
 
 /* Initially all the rows of T are marked "unconsidered". */ 
 
 (1) K!lil& there is a row w marked "unconsidered" do 
 
 begin 
 
 (2) mark w "considered"; 
 
 (3) for every row x other than w do 
 
 (4) if in whatever column x and w disagree, 
 
 x has a nondistinguished variable that appears 
 nowhere else in T then delete x from T; 
 end; 
 
 (5) return T; 
 end 
 
 Figure 3 
 
 Lemma 6 ; The algorithm of Figure 3 returns a tableau T equivalent 
 
 2 — 
 
 to T in 0(n ) time. Furthermore, if T is regular, then T is minimal. 
 
 Proof : Consider line (4) of the algorithm. Row x is deleted if 
 there is a row w in T, such that for all columns A, if x and w disagree 
 in column A, then x has a nondistinguished variable that appears nowhere 
 else in T. By Corollary 2 in [3], the tableau obtained by deleting row 
 x from T is equivalent to T. Thus, T (the tableau returned by the algo- 
 rithm) is equivalent to T. As for the running time of this algorithm, 
 let T have r rows and t columns. The cost of executing line (4) once is 
 0(t). In each iteration of the outer loop, the inner loop is executed 
 
 at most r times. The outer loop is executed no more than r times. 
 
 2 
 Thus, the total cost of line (4) is 0(r t). Every other line has a con- 
 
 2 
 stant cost, and is executed no more than r times. Since both rt and r 
 
 2 
 are smaller than n, the algorithm has a 0(n ) running time. 
 
 Suppose that T is regular. Every repeated nondistinguished vari- 
 able of T is essential and, therefore, must appear in the core of T. 
 
- 19 - 
 
 Consider a folding from T onto its core. This folding maps every row in 
 the core of T to itself and, therefore, the corresponding homomorphism 
 maps every repeated nondistinguished variable of T to itself. Suppose 
 that row x of T is mapped to some other row w in the core of T. Since 
 each repeated nondistinguished variable is mapped to itself, it follows 
 that in whatever column x and w disagree, x has a nondistinguished vari- 
 able that appears nowhere else in T. Therefore, x is deleted during the 
 execution of the above algorithm. Since this is true for every x which 
 is not in the core of T, T is minimal. [] 
 
 Each one of the following algorithms for minimizing a tableau T has 
 two steps. In the first step some rows of T are deleted in order to 
 obtain an equivalent regular tableau T. In the second step the algo- 
 rithm for minimizing regular tableaux is applied to T. 
 
 §^.2 Minimizing Tableaux of Class 1_ 
 
 Let T be a tableau that has at most one repeated nondistinguished 
 variable in each row. For each repeated nondistinguished variable b, 
 let W(b) be the set of all the rows that contain b. Suppose that some 
 repeated nondistinguished variable b~ is not essential, and let A be the 
 column in which b~ appears. Let 6 be a folding from the rows of T to 
 Its core. Obviously, all the rows of 9(W(b n )) have the same symbol d 
 (d * b) in column A, and W(b Q ) is covered by 9(W(b )). The following 
 lemma shows that these conditions are also sufficient for the elimina- 
 tion of b~. 
 
- 20 - 
 
 begin 
 
 7* The input is a tableau T that belongs to Class 1. */ 
 
 /* Initially all the variables of T are marked "unconsidered". */ 
 
 (1) KlliiS, there is a repeated nondistinguished 
 
 variable b marked "unconsidered" do 
 begin 
 
 (2) mark b "considered"; 
 
 (3) let A be the column in which b appears ; 
 
 (4) for each symbol s (s * b) in column A do 
 
 begin 
 
 (5) let S be the set of all the rows of T 
 
 that have the symbol s in column A; 
 
 (6) if W(b) < S then delete W(b) from T; 
 
 end; 
 
 end; 
 
 (7) return T; 
 
 end 
 
 Figure 4 
 
 Lemma 7 ; Suppose that ^ is a mapping from T to itself such that for 
 all x t W(b-.), Kx) ■ x. Then ^ is a containment mapping if and only if 
 
 (1) all the rows of <KW(b n )) have the same symbol in column A, and 
 
 (2) for all x e W(b_), x is covered by Kx) • 
 
 Proof ; Obviously, conditions (1) and (2) are satisfied if ^ is a 
 containment mapping. Conversely, if ^ satisfies condition (2), then ^ 
 satisfies the first two conditions of a containment mapping. If two 
 rows x and w of T have the same nondistinguished variable b in some 
 column B, then either both x and w are in W(b n ) and b is b~, or both x 
 and w are not in W(b~). In either case, ^(x) and *Kw) have the same 
 symbol in column B. Thus, ^ is a containment mapping. [] 
 
 By Lemma 7, if b~ is not essential, then we can always delete all 
 the rows containing b« by finding a set of rows S, such that S covers 
 W(b n ) and all the rows of S have the same symbol in column A. The algo- 
 
- 21 - 
 
 rithm of Figure 4 computes a regular tableau T' equivalent to T in this 
 way. 
 
 Theorem 8 : A tableau T that has at most one repeated nondis- 
 
 2 
 tinguished variable in each row can be minimized in 0(n ) time. 
 
 Proof ; Supposet that T has r rows and t columns. Consider the 
 algorithm of Figure 4. Lines (l)-(3) have a total cost of no more than 
 0(n). In each iteration of the outer loop, the cost of lines (4) and 
 
 (5) is no more than 0(r) . The outer loop is executed at most r times to 
 
 2 
 give a total cost of no more than 0(r ). The cost of line (6) is 
 
 assigned to the rows of T as follows, whenever a row w of W(b) is com- 
 pared with some other row of S, the cost of this comparison (i.e., 0(t)) 
 is assigned to w. Row w is compared with at most r rows, and the total 
 
 cost assigned to w is no more than 0(rt) . Since each row of T appears 
 
 2 
 at most in one of the sets W(b) , the total cost of line (6) is 0(r t) . 
 
 Both rt and r are smaller than n and, therefore, the algorithm has a 
 
 2 
 running time of 0(n ). Since the output of this algorithm is a regular 
 
 2 
 tableau, T can be minimized in time 0(n ) [] 
 
 _6._3 Minimizing Tableaux of Class 2_ 
 
 Let T be a tableau such that every row of T is covered by at most 
 one row besides itself. For each row w of T, let c(w) be the other row 
 
 that covers w (if no other row of T covers w, then c(w)=w) . Note that 
 
 2 
 the function c can be computed in 0(n ) time. Our goal is to find 
 
- 22 - 
 
 whether a repeated nondistinguished variable b~ is essential. 
 
 Let W(b ) the set of all the rows containing b-.. Suppose that b n 
 is not essential, and let f be a folding of T that eliminates b~. It 
 follows that for all rows w in W(b-.), w is mapped to c(w) . We can start 
 to compute the homomorphism h corresponding to ty as follows. 
 
 (1) If a nondistinguished variable b appears in some row w of W(b ft ), 
 then h(b) is the symbol appearing in column A of c(w) . 
 
 (2) If b appears in some row c(w), where w is a row of W(b n ), then 
 h(b) - b. 
 
 The first rule follows from the fact that w is mapped to c(w) . The 
 second rule is valid, since c(w) is in the image of a foldoing and, 
 hence, c(w) must be mapped to itself. 
 
 Suppose that by applying these rules to all the variables appearing 
 in w and c(w) , for all w in W(b ), a contradiction is derived. Then a 
 folding that eliminates b~ does not exist. Similarly, if the rules 
 imply that h(b ) ■ b n , then b~ is essential. 
 
 If the rules do not Imply that b is essential, then at least we 
 can use them to find some other rows that do not belong to the core of T 
 (i.e., the image of ty) . That is, if h has already been computed for a 
 nondistinguished variable b, and h(b) * b, then every row w that con- 
 tains b is mapped to c(w). Thus, let M be a set that initially is equal 
 to W(b n ). As we discover more rows that are not in the core of T, these 
 rows are added to M. Rules (1) and (2) for computing h can be applied 
 to all the rows of M, instead of only to the rows of W(b ). If a con- 
 
- 23 - 
 
 tradiction is derived during this process (or h(b~) ■ b n )» then ty cannot 
 be a folding and, hence, b~ is essential. If no contradiction is 
 derived, let M be the value of M when no more rows can be added to M, 
 and neither Rule (1) nor Rule (2) can be applied to rows of M. Define a 
 mapping 9 as follows: 
 
 9(w) = c(w) if w e M 
 
 8(w) *= w if w i M 
 
 Lemma 9 : The mapping 6 is a containment mapping, and no row of M is 
 in the image of 9 (provided that h(b-.) * b~). 
 
 Proof : For all rows w of T, c(w) covers w and, hence, 9 satisfies 
 the first two conditions of a containment mapping. Suppose that rows w 
 and x of T have the same nondistinguished variable b in some column A. 
 If both w and x belong to M, then 9(w) and 9(x) have the same symbol in 
 column A (because the rules for computing h do not imply a contradic- 
 tion) . If neither w nor x is in M, then obviously 9(w) and 8(x) have 
 the same symbol in column A. Suppose that w is in M but x is not. 
 Therefore, h(b) = b (otherwise all the rows containing b should be in M) 
 and, hence, 9(w) and 9(x) have b in column A. If x is in M but w is 
 not, then we get a similar result. Thus, 9 satisfies also the third 
 condition of a containment mapping. 
 
 Suppose that a row w of M is mapped to some other row u of M. 
 Since u is in M, it must have some repeated nondistinguished variable b 
 such that h(b) * b. But since w is mapped to u, Rule (2) implies that 
 h(b) = b. However, it is assumed that no contradiction has occured and, 
 
- 24 - 
 
 hence, w cannot be mapped to u. [] 
 
 It follows that If the rules do not imply a contradiction and 
 h(bg) * Dq, then all the rows of M (and hence all the rows containing 
 b~) can be eliminated from T. If a contradiction is derived, then b n is 
 essential. Thus, by repeating this process for every nondistinguished 
 variable in T, we obtain a regular tableau T equivalent to T. 
 
 The procedure FOLD(b,T) decides whether a nondistinguished variable 
 b is essential. If b is essential, then FOLD(b,T) returns the empty 
 set. If b is not essential, then FOLD(b,T) returns a set of rows (con- 
 taining all the rows having b) that can be eliminated from T. The com- 
 plete algorithm is given in Figure 5. It is assumed that the variables 
 of T are represented by the numbers 1,2, ...,m. The values of the 
 homomorphism computed by the procedure FOLD(b,T) are stored in the array 
 h; initially all the entries in this array are zero. 
 
 Lemma 10 : A call F0LD(b,T) requires 0(n) time, where n is the size 
 of T. 
 
 Proof ; Suppose that T has r rows and t columns. For each row w, 
 
 the cost 0(t) of executing the loop of lines (8)-(10) and the loop of 
 
 lines (11)— (16) is assigned to w. Line (7) is executed at most once for 
 
 each row and, therefore, the total cost of lines (7)- (16) is 0(rt). The 
 
 cost of finding all the rows that contain d in line (6) is assigned to 
 
 d. Let cost(d) be that cost. For each symbol we use a linked list that 
 
 points to all the rows containing that symbol, and since each d is put 
 
 on QUEUE at most once, £cost(d) - 0(n). Thus, the total cost of the 
 
 d 
 
- 25 - 
 
 procedure FOLD(b,T): 
 begin 
 
 arra^ h[l :m] ; 
 
 /* Initially QUEUE is empty */ 
 
 (1) M :« <(.; 
 
 (2) h :- 0; 
 
 (3) add b to QUEUE; 
 
 (4) while QUEUE is not empty do 
 
 begin 
 
 (5) find and delete d, the first element of QUEUE; 
 
 (6) for each row w that contains d, and w fc M do 
 
 beein 
 
 (7) add w to M; 
 
 (8) for each repeated nondistinguished 
 
 variable e that appears in c(w) do 
 
 (9) if h(e) - then h(e) := e 
 
 (10) Sise i£ h(e) * e then return <f>; 
 
 (11) for each column A of w that has a repeated 
 
 nondistinguished variable e do 
 begin 
 
 (12) let s be the symbol of c(w) in column A; 
 
 (13) if h(e) = then h(e) := s; 
 
 (14) if h(e) * s then return <j»; 
 
 (15) if h(e) * e,~~ 
 
 and e has not been on QUEUE 
 
 (16) then add e to QUEUE; 
 
 end; 
 
 end; 
 end; 
 
 (17) if h(b) = b then rjsturn <\> else return M; 
 end FOLD; 
 
 begin /* main procedure */ 
 
 /* Initially all the variables are marked "unconsidered". */ 
 
 (18) while there is a repeated nondistinguished 
 
 variable b marked "unconsidered" do 
 begin 
 
 (19) mark b "considered"; 
 
 (20) M := F0LD(b,T); 
 
 (21) delete all the rows of M from T; 
 end; 
 
 (22) return T; 
 end 
 
 Figure 5 
 
 outer loop (lines (4)-(16)) is 0(n) . Finally, note that lines (l)-(3) 
 and line (17) have a total cost of no more than 0(n). [] 
 
- 26 - 
 
 Theorem 11 : A tableau T, in which each row is covered by at most 
 
 2 
 one row besides itself, can be minimized in time 0(n ). 
 
 Proof : Suppose that T' is the result of applying the algorithm of 
 Figure 5 to T. If a variable b is not essential in T' , then it is not 
 essential in T or in any tableau obtained from T by deleting some redun- 
 dant rows (because T and T' are equivalent). Thus, all the rows con- 
 taining b are deleted in line (21), and hence T' is regular. By Lemma 
 
 2 
 10, the running time of this algorithm is 0(n ). [] 
 
 .6. _4 Simple Tableaux 
 
 Suppose that T is a simple tableau. Let S be a set of rows, and 
 
 let w be a row of T. The closure of S with respect to w, denoted 
 
 CL (S), is the minimal set of rows such that 
 w 
 
 (1) S C CL (S), and 
 
 w 
 
 (2) if x is a row in CL (S) such that x has a repeated nondistinguished 
 
 w 
 
 variable b in some column A, and w has some other symbol in this 
 
 column, then all the rows containing b are in CL (S). 
 
 w 
 
 In [3] it is shown that if w covers CL (S), then the tableau obtained by 
 
 w 
 
 deleting all the rows of CL (S) - w is equivalent to T (this is true 
 
 w 
 
 even if T is not simple). Furthermore, if some repeated nondis- 
 tinguished variable b of T is not essentail, then there is a row w (that 
 
 does not contain b) such that w covers CL (W(b) ) . (Note that w is not 
 
 w 
 
 in CL (W(b)), since w is not in W(b).) 
 w 
 
- 27 - 
 
 These results can be used to obtain a regular tableau equivalent to 
 
 T as follows. Compute CL (W(b)) for each repeated nondistlngulshed 
 
 w 
 
 variable b and each row w that does not contain b. If w covers 
 
 CL (W(b)), then delete all the rows of CL (W(b) ) from T. The resulting 
 w w 
 
 tableau is equivalent to T, and it is regular because all repeated non- 
 distinguished variables that are not essential have been eliminated [3]. 
 
 In this section we describe an implementation of this algorithm that 
 
 2 
 runs in 0(n ) time. 
 
 Lemma 12 ; Let w be a row of a simple tableau T. Suppose that b. 
 
 and b« are two repeated nodistlnguished variables of T that do not occur 
 
 in w. Then CL (W(b,)) and CL (W(b )) are either equal or disjoint, 
 w I W i. 
 
 Proof ; Let S. = CL (W(b,)) and S = CL (W(b )). Suppose that some 
 row x belongs to both S. and S_. Row x must have some repeated nondis- 
 tlngulshed variable b that does not occur in w and, hence, W(b) is con- 
 tained in both S. and S„. We claim that both S. and S ? are equal to 
 
 CL (W(b)). By the definition of a closure, if R is a subset of CL (S) 
 w w 
 
 (for any row w and set of rows S), then CL (R) C CL (S). Thus, both S. 
 
 w w 1 
 
 and S. contain CL (W(b) ) . 
 
 L W 
 
 Let S ■ CL (W(b)). Suppose that S. is not equal to S. If some row 
 w 1 
 
 of W(b. ) is in S, then all the rows of W(b. ) are In S, and S is equal to 
 
 S. (because it satisfies the definition of CL (W(b,))). Thus W(b, ) must 
 i w 1 I 
 
 be contained in S. - S. We will derive a contradiction by showing that 
 S. - S satisfies also the second condition of CL (W(b.)). Let u be any 
 row of S. - S. Since u is in S., u has a repeated nondistlngulshed 
 
- 28 - 
 
 variable b In some column A, and w has some other symbol in that column. 
 
 If some row of W(b) is in S, then all the rows of W(b) must be in S, and 
 
 u cannot be in S. - S. Thus, W(b) is disjoint from S and, hence, all 
 
 the rows containing b are in S. - S (since they are in S.). Therefore, 
 
 S. - S satisfies also the second condition of CL (W(b,)). Since this is 
 i w i 
 
 impossible, it follows that S. is equal to S. Similarly, S« is equal to 
 S. [] 
 
 Lemma 12 implies that if CL (W(b) ) has been computed, and b is a 
 
 w 
 
 repeated nondistinguished variable that appears in some row of 
 
 CL (W(b)), then there Is no need to compute CL (W(b)) (it is assumed 
 w w 
 
 that neither b nor b occur in w) . Thus, for each row w we do the fol- 
 lowing. At first all repeated nondistinguished variables that do not 
 appear in w are marked "unconsidered". The next step is to compute 
 CL (W(b)) for some repeated nondistinguished variable that is marked 
 "unconsidered". During this step all repeated nondistinguished variabes 
 
 that occur in some row of CL (W(b) ) are marked "considered". If 
 
 w 
 
 CL (W(b) ) is covered by w, then all the rows of CL (W(b) ) are deleted, 
 w w 
 
 This step is repeated for some other variable that is marked "uncon- 
 sidered", until all the variables are marked "considered". The complete 
 algorithm is described in Figure 6. 
 
 2 
 Theorem 13 : A simple tableau T can be minimized in 0(n ) time. 
 
 Proof : By Lemma 12 and the results of [3] , the algorithm of Figure 
 6 returns a regular tableau equivalent to T. Consider the time complex- 
 ity of this algorithm. We assume that T has r rows and t columns, and n 
 
- 29 - 
 
 procedure CLOSURE (b, w) : 
 begin 
 
 (1) S :« 4.; 
 
 (2) make QUEUE empty; 
 
 (3) mark b "considered"; 
 
 (4) add all the rows containing b to QUEUE; 
 
 (5) while QUEUE is not empty do 
 
 begin 
 
 (6) let v be the first row on QUEUE; 
 
 (7) move v from QUEUE to S; 
 
 (8) for every column A do 
 
 (9) if v and w disagree in column A, v has a repeated 
 
 nondistinguished variable d in this column, 
 and d is marked "unconsidered" then 
 begin 
 
 (10) mark d "considered"; 
 
 (11) for every row x containing d do 
 
 (12) ~~~ if x is neither in S nor on QUEUE 
 
 (13) ~~ then add x to QUEUE; 
 
 end; 
 
 end; 
 
 (14) return S; 
 end 
 
 begin /* main procedure */ 
 
 (15) for every row w do 
 
 begin 
 
 (16) mark all repeated nondistinguished variables "unconsidered"; 
 
 (17) for every repeated nondistinguished variable d 
 
 that occurs in w do 
 
 (18) mark d "considered"; 
 
 (19) w^liiS, there is a repeated nondistinguished variable b 
 
 marked "unconsidered" do 
 begin 
 
 (20) ~R := CLOSURE (b,w); 
 
 (21) if R < w then delete all the rows of R from T; 
 
 end; 
 
 end; 
 (2 2) reJturn~T; 
 end 
 
 Figure 6 
 
 is the size of T (i.e., n ■ 0(rt)). At first we will show that a call 
 CLOSURE (b,w) requires 0(st) time, where s is the number of rows in 
 CL (W(b)). Consider the cost of executing the loop of lines (11)-(13). 
 
- 30 - 
 
 We assume that for every variable in T there is a linked list that 
 
 points to all the rows containing this variable. These lists can be 
 
 created in 0(n) time prior to the execution of the algorithm. By using 
 
 these lists, the cost of finding all the rows containing d in the loop 
 
 of lines (11)- (13) is proportional to the number of these rows. The 
 
 test of deciding whether a row is in S or on QUEUE can be implemented in 
 
 constant time. Thus, the cost of lines (11)-(13) is accounted for by 
 
 assigning a constant cost to each row w whenever w is examined in this 
 
 loop. A row is examined in this loop no more than t times, since it has 
 
 at most t variables. Also note that if a row is examined in this loop, 
 
 then it belongs to CL (W(b)). Therefore, the total cost of lines 
 
 w 
 
 (11)-(13) is 0(st). 
 
 Consider now the cost of executing the loop of lines (5)- (13) once 
 (excluding the cost of lines (11)- (13)). Lines (6) and (7) have a con- 
 stant cost. The cost of lines (8)-(10) is 0(t) . Since the loop, of 
 lines (5)- (13) is repeated s times, the total cost of this loop is 
 0(st). Lines (l)-(4) have a cost of 0(s) . Thus, a call CLOSURE(b,w) 
 requires 0(st) time. 
 
 We now compute the cost of executing the loop of lines (15)- (21) 
 once. The cost of line (16) is 0(t) (since each column of a simple 
 tableau has at most one repeated nondistinguished variable) . The loop 
 of lines (17)— (18) requires 0(t) time. The cost of line (19) is no more 
 than the number of repeated nondistinguished variables, i.e., 0(t) . The 
 cost of executing lines (20)- (21) once is 0(st), where s is the number 
 of rows in R. For each row w, all the sets obtained as a value of R in 
 
- 31 - 
 
 line (20) are pairvd.se disjoint. Thus, the cost of executing the loop 
 
 of lines (15)-(21) once is 0(rt) . This loop is repeated r times and, 
 
 2 2 
 
 hence, the total cost is 0(r t) , i.e., no more than 0(n ). 
 
 Theorem 14 : If T. and T„ are simple tableaux, then testing whether 
 
 2 
 T. is equivalent to T„ can be done in 0(n ) time. 
 
 Proof : By using the algorithm of Figure 6, we compute regular 
 tableaux T. and T„ equivalent to T. and T„, respectively. It follows 
 
 from the results of [3] that testing whether T. is equivalent to T. can 
 
 2 
 be done in 0(n ) time, where n is the size of T. and T-. [] 
 
 _7. Decomposition of Tableaux 
 
 Let T be a tableau that does not necessarily belong to one of the 
 classes we have discussed so far. None of the three minimization algo- 
 rithms can minimize T in polynomial time, but they can be used as 
 heuristics. The minimization algorithm for simple tableaux can be 
 applied to any tableau, and the result is an equivalent tableau possibly 
 with fewer rows. The idea behind the algorithm of Section 6.2 can be 
 used to minimize tableaux as follows. Let b be a repeated nondis- 
 tinguished variable of a tableau T, and suppose that the set S of all 
 the rows containing b does not have any other repeated nondistinguished 
 variable. Then all the rows of S can be deleted if they are covered by 
 a set of rows that have the same symbol in the column of b. 
 
- 32 - 
 
 The algorithm of Section 6.3 is not only good as a heuristic, but 
 can also be used as an exponential time minimization algorithm for 
 tableaux. Let T be a tableau. For every row i of T, we define C(i) to 
 be the set of all the rows that cover row i, i.e., 
 
 C(i) ■ <j | j * i and row j covers row i> 
 Suppose we construct a function c such that for all i, c(i) e C(i) (it 
 is understood that c(i) ■ i if C(i) * <f) • Using c we can apply the 
 algorithm of Section 6.3 as a heuristic. The number of all possible c's 
 is exponential only in the number of rows i for which C(i) has more than 
 one element. If we execute the algorithm once for each possible c we 
 are guaranteed to minimize T, since at least one of these c's 
 corresponds to a folding from T to its core. 
 
 The following approach can be used to further reduce the exponen- 
 tial factor in the running time of this algorithm. We define a relation 
 R on the rows of a tableau T as follows. For rows x ,and y of T, xRw if 
 and only if x and w have the same nondistinguished variable in some 
 column. Obviously, R is symmetric and reflexive. Let P.,P_, ...,P be 
 the equivalence classes of the transitive closure of R. 
 
 Lemma 15 ; Let 6 be a containment mapping from a tableau T to 
 
 itself, and let k (Kk<q) be a fixed integer. Define 
 
 5(x) = 6(x) if x e R 
 
 k 
 
 £(x) = x otherwise 
 Then 5 is a containment mapping. 
 
- 33 - 
 
 Proof ; Obviously, for all rows x, 5(x) covers x. Suppose that rows 
 x and w have the same nondlstlngulshed variable In column A. Thus, 
 either both w and x are in P, or both are not in P, . In either case, 
 5(x) and £(w) have the same symbol in column A. [] 
 
 Lemma 15 implies that when the algorithm of Section 6.3 is applied 
 as a heuristic, we have to consider only c's such that for some fixed k, 
 c(i) e C(i) if row i is in P.; otherwise c(i) - i. If among all the 
 P 's, P has a maximum number, say m, of rows for which C(i) contains 
 more than one element, then the number of all possible c's is exponen- 
 tial only in m. For each P there is at least one possible c that maps 
 all redundant rows in P to the core of T. Thus, no c has to be con- 
 sidered more than once. 
 
 8_ . Synthesis of Expressions from Tableaux 
 
 Optimal expressions can be synthesized from simple tableaux in 
 
 2 
 0(n ) time [5]. Suppose we are given a tableau T. The optimization 
 
 process is done in two steps. 
 
 (a) Minimize the number of rows in T. 
 
 (b) Produce from T an expression in which select and project are applied 
 as early as possible. 
 
 The approach taken in the second step was found useful by previous work- 
 ers (e.g., [12,17]) in the field of expression optimization, and can be 
 viewed as our "cost function." 
 
- 34 - 
 
 In order to obtain an optimal expression for a minimal tableau T, 
 [5] uses the following approach. At first all constants are deleted 
 from the summary of T. Then an optimal expression E is synthesized. 
 Finally a new operator augment, defined by a (r) ■ {u | u(A) « c and 
 
 A C 
 
 there exists v in r such that for all attributes B in the relation 
 scheme of r, u(B) = v(B) }, is applied to E to introduce the constants 
 that were deleted from the summary of T. 
 
 Suppose we decompose a tableau T into equivalence classes 
 P., P., ...,P as described in Section 7, and let s be the summary of T. 
 For every j, we can define a tableau T that have the same rows as P. 
 and a summary as follows. For each column A, if s has a nonblank symbol 
 in column A and that symbol appears also in column A of P., then the 
 summary of T has the same symbol in column A; otherwise, it has a 
 
 q 
 
 blank. It follows that T = W T . Thus, T comes from an expression if 
 
 J-l J 
 and only if each T. comes from an expression. Suppose that each T 
 
 comes from the expression E . Then T corresponds to the expression 
 
 q 
 
 E = M E . If T is a minimal tableau, we can get an optimal expression 
 
 j-l J 
 
 for T as follows. At first all constants are deleted from the summary 
 
 of T. Then we decompose T and find an optimal expression E for each 
 
 q 
 
 T . (Note that each T is minimal.) Let E - H E . By applying augmen- 
 J J 1=1 ^ 
 
 tation to E we get an optimal expression for T. 
 
 If T is a tableau with at most one repeated nondistinguished vari- 
 able in each row, then each T is a simple tableau (because it has at 
 most one repeated nondistinguished variable). Thus, we can synthesize 
 an optimal expression for T in polynomial time. However, if T is a 
 
- 35 - 
 
 tableau in which every row is covered by at most one row besides itself, 
 then testing whether T comes from an expression is NP-complete. 
 
 Theorem 13 : Let T be a tableau in which every row is covered by at 
 most one row besides itself. The problem whether T corresponds to a 
 restricted relational expression is NP-complete. 
 
 Proof ; The problem of testing whether a tableau T corresponds to an 
 expression (shown NP-complete in [19]) is reduced to this problem as 
 follows. Let T be a tableau, and let G be a new attribute. We con- 
 struct a tableau T' by adding to T a new column that corresponds to G. 
 The summary of T' has a blank, and row i has the constant i in this 
 column. We claim that T corresponds to an expression if and only if T' 
 corresponds to an expression. 
 
 If . Let E' be an expression for T' . Delete G from each relation 
 scheme in E', and delete every selection operator that is applied to the 
 attribute G. Each projection operator tt_. that appears in E' is replaced 
 with Tr /nx . Let E be the resulting expression. It is easy to show 
 that E corresponds to T. 
 
 Only if . Let E be an expression for T. Let R. be the relation 
 scheme that corresponds to row i of T, and define R ■ R U {G} for 
 every i. An expression E' for T' is obtained from E by replacing each 
 
 \ with YWV'- 
 
 (1) See [2,11] for an exposition of NP-completeness and related topics, 
 
- 36 - 
 
 Since each row of T' is covered by no row besides itself, and an 
 expression for T can be obtained from an expression for T' in polynomial 
 time, the proof is complete. [] 
 
 References 
 
 [1] A. V. Aho, C. Beeri, and J. D. Ullman, "The Theory of Joins in 
 Relational Databases," ACM Trans * on Database Systems , Vol. 4, No. 
 3 (1979), pp. 297-314. 
 
 [2] A. V. Aho, J. E. Hopcroft, and J. D. Ullman, The Design and 
 Analysis of Computer Algorithms , Addison-Wesley, Reading, MA, 
 1974. 
 
 [3] A. V. Aho, Y. Sagiv, and J. D. Ullman, "Equivalences Among Rela- 
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BIBLIOGRAPHIC DATA 
 SHEET 
 
 1. Report No. 
 
 UIUCDCS-R-79-992 
 
 I. Title and Subtitle 
 
 Quadratic Algorithms for Minimizing Joins 
 in Restricted Relational Expressions 
 
 3. Recipient's Accession No. 
 
 5. Report Date 
 
 October 1979 
 
 '. Author(s) 
 
 Yehoshua Sagiv 
 
 8- Performing Organization Rept. 
 
 No. 
 
 Performing Organization Name and Address 
 
 10. Project/Task/Work Unit No. 
 
 11. Contract/Grant No. 
 
 12. Sponsoring Organization Name and Address 
 
 Department of Computer Science 
 University of Illinois 
 
 at Urbana-Champaign 
 Urbana, IL 61801 
 
 13. Type of Report & Period 
 Covered 
 
 14. 
 
 15. Supplementary Notes 
 
 16. Abstracts 
 
 An important step in the optimization of queries in relational 
 databases is minimizing the number of join operations used in the 
 evaluation of a query. It is shown that three subclasses of tableaux 
 ^including the subclass of simple tableaux) have 0(n 2 ) time equivalence 
 and minimization algorithms. Since tableaux are nonprocedural 
 representations of relational expressions over select, project and 
 join, these minimization algorithms can be used to minimize the 
 number of join operators in expressions whose tableaux belong to 
 one of these subclasses. 
 
 7. Key Words and Document Analysis. 17a. Descriptors 
 
 relational database, relational algebra, query optimization, 
 equivalence of queries, conjunctive query, tableau, NP-complete. 
 
 7b. Identifiers /Open-Ended Terms 
 
 7e. COSATI Field/Group 
 
 ,8. Availability Statement 
 
 ORM NTIS-3B (10-701 
 
 19. Security Class (This 
 Report) 
 
 UNCLASSIFIED 
 
 20. Security Class (This 
 Page 
 
 UNCLASSIFIED 
 
 21. No. of Pages 
 
 42 
 
 22. Price 
 
 USCOMM-DC 40329-P7 1 
 
AUG 2 9 1 ?89