Only Intervals Preserve the Invertibility of Arithmetic Operations

Only Intervals Preserve the Invertibility of Arithmetic Operations Olga Kosheleva 1 and Vladik Kreinovich 2 1 Department of Electrical and Computer Engineering 2 Department of Computer Science University of Texas at El Paso El Paso, TX 79968, USA E-mails: olga@ece.utep.edu, vladik@cs.utep.edu Abstract In standard arithmetic, if we, e.g., accidentally added a wrong number y to the preliminary result x, we can undo this operation by subtracting y from the result x + y. A similar possibility to invert (undo) addition holds for intervals. In this paper, we show that if we add a single non-interval set, we lose invertibility. Thus, invertibility requirement leads to a new characterization of the class of all intervals. 1 Formulation of the Problem Why Invertible? Many computer operations, including addition x x + y, are invertible in the sense that if we accidentally added the wrong number y, we can always reconstruct the original value x by simply subtracting y from the result of the addition: (x + y) y = x. Case of Partial Knowledge In many real life situations, our knowledge about the actual values x and y is incomplete. This means that we do not know the exact values of x and y, we only know the sets X and Y of possible values of x and y. The most common situation is when the sets X and Y are intervals. 1

If the only information that we have about the physical quantity x is that x X, and the only information that we have about the physical quantity y is that y Y, then the only information that we have about the sum x + y of these quantities is that x + y X + Y, where the sum of two sets is defined element-wise: X + Y = {x + y x X and y Y }. (1) Invertibility in Case of Partial Knowledge It would be nice to preserve the same invertibility property for set operations: namely, if we accidentally added the wrong set Y, we must be able to reconstruct the original set X from the sum X + Y and Y. For Intervals, Addition is Invertible If both sets X and Y are intervals, i.e., if X = [x, x + ] and Y = [y, y + ], then addition is invertible: indeed, in this case, the sum Z = X + Y is an interval [z, z + ] with the bounds z = x + y and z + = x + + y +. Therefore, if we know Z and Y, we can reconstruct X as x = z y and x + = z + y + (for a more general discussion of invertibility of interval computations, see [1, 2]). Not Only Intervals Partial knowledge is often represented by sets that are more general than intervals. For example, if we know that x 2 [1, 4], then the set of possible values of x is not an interval, but a union of two intervals [ 2, 1] [1, 2]. The Main Problem A natural question is: is invertibility preserved if we allow sets different from intervals? Our answer is: no. Hence, intervals are uniquely determined by the requirement that invertible operations stay invertible. The corresponding two results will be presented in the next section. 2

2 Definitions and the Main Result 2.1 First Result: If We Add a Non-Interval to the Class of All Intervals, Addition Stops Being Invertible Definition 1. numbers. Let C be a class of compact subsets of the set IR of all real For every two sets X, Y C, we define their sum X + Y by formula (1). We say that addition is invertible on C if for every three sets X, X, Y C, X + Y = X + Y implies that X = X. In other words, if we know Y and X + Y, then we can uniquely reconstruct X. This property corresponds to the inversion in the sense in which this word is used in computer science: namely, to the possibility of inverting (undoing) addition X + Y X. The reader should be warned that in algebra, the terminology is slightly different: what we have called invertibility is called cancelation property, and the word invertibility is used only when this undoing X +Y X is performed by applying the same operation, in this case, addition: X + Y (X + Y ) + Z = X for an appropriate Z. For addition of intervals, this algebraic invertibility is a much stronger requirement than the invertibility in computer science sense: On one hand, as we have seen (in Section 1), interval addition X + Y is invertible in our (computer science) sense, i.e., if we know X + Y and Y, we can reconstruct X. On the other hand, interval addition is not invertible in the algebraic sense: Indeed, if the interval Y is non-degenerate (i.e., has positive width), then the sum X + Y is wider than X; hence, whatever interval Z we add to X + Y, the resulting sum (X + Y ) + Z is at least as wide as X + Y and hence, wider than the desired interval X. So, (X + Y ) + Z X. PROPOSITION 1. Addition is invertible on the class of all (one-dimensional) intervals. If a class C contains all intervals and at least one non-interval set C, then addition is not invertible on C. Proof. In the previous section, we have already shown that addition is invertible on the set of all intervals. Let us now assume that C contains all intervals, and also a non-interval compact set C. Let us show that in this case, addition is not invertible. Indeed, since the set C is compact, it has finite infimum c and supremum c +. Let us take X = C and X = Y = [c, c + ]. Then, X + Y = [2c, 2c + ]. Let us show that the sum X + Y is also equal to the interval [2c, 2c + ]: 3

First, due to our choice of c + and c, we have X = C [c, c + ] = X, hence, X + Y X + Y = [2c, 2c + ]. Second, since C = X is a compact set, we have {c, c + } X, and therefore, for their sum X + Y (as defined by Definition 1) X + Y {c, c + } + Y = {c, c + } + [c, c + ] = [2c, c + c + ] [c + c +, 2c + ] = [2c, 2c + ]. So, X + Y [2c, 2c + ] and X + Y [2c, 2c + ]; hence, X + Y = [2c, 2c + ] = X + Y. However, X = C is not an interval, and X is; therefore, X X. Hence, this triple is the desired counterexample to invertibility. Q.E.D. 2.2 Second Result: If We Require Invertibility and Start with Arbitrary Sets, We End up with Intervals Definition 2. We say that compact sets X, X IR are equivalent if there exists a compact set Y for which X + Y = X + Y. If we use arbitrary sets, then we do not have invertibility. So, if we want invertibility, we must represent some sets X by their supersets I X. In particular, if the sets X and X are equivalent in the sense of the above definition, then we must represent the sets X and X by one and the same superset. The restriction to compact sets is justified by the fact that if we do not restrict ourselves to compact sets, then the resulting notion of equivalence will be trivial: for Y = IR, X + Y = X + Y = IR for all X and X and therefore, every two sets would be equivalent. PROPOSITION 2. Two compact sets X and X are equivalent iff inf X = inf X and sup X = sup X. Proof. Let us denote x = inf X, x = inf X, y = inf Y, x + = sup X, x + = sup X, and y + = sup Y. The infimum of X+Y is attained when x is the smallest, and y is the smallest, i.e., inf(x + Y ) = x + y. Similarly, inf(x + Y ) = x + y. Therefore, if X + Y = X + Y, then inf(x + Y ) = inf(x + Y ), x + y = x + y and hence, x = x. Likewise, we can prove that x + = x +. Let us now show that if x = x and x + = x +, then X + Y = X + Y for some Y. Indeed, as we have shown in the proof of Proposition 1, for Y = [x, x + ], we have X + Y = X + Y = [2x, 2x + ]. Q.E.D. Due to Proposition 2, if we want invertibility, then we must represent all equivalent sets by a single set. The class of all equivalent sets is the class of all sets X with given infimum and supremum. All these sets are contained in the interval [inf X, sup X], which can thus serve as the desired representative. So, if we want invertibility of addition, we get intervals. 4

3 Auxiliary Results In this section, we generalize the above results in two directions: to functions more general than addition and to sets in multi-dimensional space. 3.1 Invertible Operations beyond Addition There are other invertible operations besides addition: e.g., multiplication of positive numbers, exponentiation, etc. Like addition, these operations can also be extended from numbers to arbitrary sets of numbers (see, e.g., [3]), in particular, to intervals. For these operations, interval extensions are also invertible. Definition 3. Let A and B be connected subsets of IR. We say that a continuous function f : A B is invertible if f(a, b) = f(a, b) implies a = a, and f(a, b) = f(a, b ) implies b = b. By an interval extension of a function f, we mean a function that maps the intervals X and Y into a set f(x, Y ) = {f(x, y) x X and y Y }. We say that an interval extension f(x, Y ) is invertible if for every intervals X, X, Y, and Y, f(x, Y ) = f(x, Y ) implies X = X, and f(x, Y ) = f(x, Y ) implies Y = Y. PROPOSITION 3. If a continuous function f is invertible, then its interval extension is also invertible. Proof. Since f is invertible, for every a, the function b f(a, b) is continuous and 1-to-1; therefore, this function is strictly monotonic (i.e., strictly increasing or strictly decreasing). Let us show that either for every a the function f a (b) defined as f a (b) = f(a, b) is increasing, or for every a, this function is decreasing. If the set B consists of only one point b, then this statement is trivially true. So, let us consider the case when the set B contains at least two different points b < b. In this case, if for some a, the function f a (b) is increasing, then f(a, b ) = f a (b ) < f a (b ) = f(a, b ). Since f is continuous, we can conclude that f(a, b ) < f(a, b ) for all a from some open neighborhood of a. Therefore, for these a, the function f a (b) must also be strictly increasing. So, the set A in of all a A for which the function f a (b) is increasing is open. Similarly, the set A de of all a A for which the function f a (b) is decreasing is open. Therefore, 5

the set A is represented as the union of two disjoint open sets: A = A in A de. Since A is connected, one of these sets must be empty. Therefore, either the function b f a (b) is strictly increasing for all a, or the function b f a (b) is strictly decreasing for all a. Similarly, either the function a f(a, b) is strictly increasing for all b, or it is strictly decreasing for all b. Totally, we have four possible cases, depending on whether f is increasing or decreasing in a, and increasing or decreasing in b. We are now ready to show that the extension of f to intervals is invertible. We will show it for the case when f is strictly increasing in both a and b (the proof for the three other possible cases is similar). Indeed, in this case, due to monotonicity, f([x, x + ], [y, y + ]) = [f(x, y ), f(x +, y + )]. If we know f(x, y ) and y, then, due to the fact that x f(x, y ) is strictly increasing, we can uniquely reconstruct x. Similarly, we can uniquely reconstruct x +. So, the interval extension of f is truly invertible. Q.E.D. Without continuity, this result is not necessarily true. For example, let s(x) denote a function that swaps 0 and 1 and leaves all other values intact. Then, the function f(a, b) = s(a) + b is invertible. However, for intervals X = [ 0.5, 0.5], X = [ 0.5, 1], and Y = [0, 1], we have s(x) = [ 0.5, 0) (0, 0.5] {1}, and f(x, Y ) = s(x)+y = [ 0.5, 1) (0, 1.5] [1, 2] = [ 0.5, 2]; also, s(x ) = X, and f(x, Y ) = X + Y = [ 0.5, 1] + [0, 1] = [ 0.5, 2]. So, here, f(x, Y ) = f(x, Y ) but X X. 3.2 Sets in Multi-Dimensional Space: General Case Definition 1. Let d 1 be an integer, and let C be a class of compact subsets of the set IR d. We say that addition is invertible on C if for every three sets X, X, Y C, X + Y = X + Y implies that X = X. PROPOSITION 1. Addition is invertible on the class of all compact convex sets. If a class C contains all compact convex sets and at least one non-convex compact set C, then addition is not invertible on C. Proof. The fact that addition is invertible on compact convex sets easily follows from the known properties of these sets: indeed, it is known (see, e.g., [4], Section 13) that a compact convex set C is uniquely determined by its support function δ (x C) = sup{ x, x x C} (where a, b = a i b i is a scalar (dot) product), and that the support function of the sum of two sets is equal to the sum of their support functions: δ (x C 1 + C 2 ) = δ (x C 1 ) + δ (x C 2 ). Hence, if we know X + Y and Y, we can determine support functions for these sets, subtract these functions, get the support function for X, and reconstruct X. Let us now show that if we add a non-convex set C to the class of all convex compact sets, then we lose invertibility. Indeed, we will show that X + Y = 6

X + Y for X = C, X = conv(x) (a convex hull of X), and Y = d conv(x). Namely, we will show that X + Y = X + Y = (d + 1) conv(x). It is clear that X + Y = (d + 1) conv(x), and that X + Y X + Y = (d + 1) conv(x). So, it is sufficient to prove that every point z (d + 1) conv(x) belongs to the sum X + Y. Indeed, from z (d + 1) conv(x) it follows that z/(d + 1) conv(x). Since X is a subset of a d dimensional space, the point z can be represented as a convex combination of d + 1 points from X: z = α k x k for some α k 0, αk = 1, x k X (strictly speaking, we need at most d + 1 points x k ; however, if less than d + 1 points are sufficient, then we can add a few other points with α k = 0). Since the sum of (d + 1) values α k is equal to 1, at least one of these values α k is 1/(d + 1). Let us denote one of such points by α k0. Let us take β k = α k for k k 0 and β k = α k0 1/(d + 1). Then, z d + 1 = α k x k = 1 d + 1 x k 0 + β k x k. Here, β k 0, and β k = 1 1/(d + 1) = d/(d + 1). Hence, if we multiply both sides of the displayed equality by d + 1, we can conclude that z = x k0 + d ( γ k x k ), where γ k = (d + 1)β k /d are non-negative numbers with the sum equal to 1. Here, x k0 X, and d ( γ k x k ) d X = Y. Hence, z X + Y. Q.E.D. Definition 2. Let d 1 be an integer. We say that compact sets X, X IR d are equivalent if there exists a compact set Y IR d for which X + Y = X + Y. PROPOSITION 2. Two compact sets X, X IR d are equivalent (in the sense of Definition 2 ) iff conv(x) = conv(x ). Proof. If X + Y = X + Y for some X, X, and Y, then the support functions coincide: δ (x X + Y ) = δ (x X + Y ); from δ (x X + Y ) = δ (x X) + δ (x Y ) = δ (x X + Y ) = δ (x X ) + δ (x Y ), we conclude that δ (x X) = δ (x X ) and hence, that conv(x) = conv(x ). Vice versa, if conv(x) = conv(x ), then, as we have proved in the proof of Proposition 1, there exists a set Y (namely, Y = d conv(x)) for which X + Y = X + Y (= (d + 1) conv(x)). Q.E.D. So, if we want invertibility, we must restrict ourselves to convex sets only. 3.3 Sets in Multi-Dimensional Space: Families of Neighborhoods Before we formulate the corresponding results, let us first explain why it is natural to describe families of neighborhoods. 7

Why Neighborhoods Let us assume that we do not have a complete knowledge about the values x 1,..., x d of d physical quantities. This uncertainty means that the set X IR d of the possible values of x = (x 1,..., x d ) IR d consists of more than one point. According to the previous section, if we want to keep addition invertible, then it is reasonable to use compact convex sets to describe uncertainty. So, the set X must be convex and compact. Typically, sets that describe real-life uncertainty contain an open subset. In many cases, the user wants an estimate; in this case, we produce an element x Int(X). In topological terms, we can say that the set X is a neighborhood of the estimate x. To describe how good this estimate is, we must describe the set of possible values of errors x = x x. Since X is a neighborhood of x, the resulting set is a neighborhood of 0. So, to describe possible uncertainties, we must describe possible convex compact neighborhoods of 0. Why Linearly Ordered Families of Neighborhoods It is often desirable to be able to compare the uncertainties corresponding to different measurement, or to different knowledge bases. If a neighborhood is contained in the neighborhood (, ), then the uncertainty described by is smaller than the uncertainty described by. If we consider arbitrary convex neighborhoods, however, then some pairs will be incompatible: e.g., neither of the two multi-interval neighborhoods [ 0.1, 0.1] [ 1, 1] and [ 1, 1] [ 0.1, 0.1] is contained in the other one. Therefore, if we want to be able to compare arbitrary two types of uncertainty, we must restrict ourselves only to a family F of neighborhoods that is linearly ordered with respect to the relation, i.e., for which for every, F, either, or, or =. The property that we can have only one of the three possibilities is called trichotomy. The trichotomy property of neighborhoods was first formulated in a slightly stronger sense in [1, 2]. Namely, in these papers, it is required that for every and, at least one of the equations = + X and = + Y has a solution. If we look for solutions X and Y that are also neighborhoods of 0, then from = + X, we can conclude that, and from = + Y, we can conclude that. In other words, trichotomy in this algebraic sense really implies trichotomy in the sense of the relation. A Family of Neighborhoods Must Be an Additive Semigroup The reason why we restricted ourselves to convex neighborhoods is that we wanted computations to be invertible relative to addition. In other words, we 8

assumed addition to be one of the basic operations. In real-life data processing, addition is indeed one of the most important operations: it corresponds, e.g., to the case when we bring two physical systems together: then, the total charge, the total mass, and many other physical characteristics simply add up. So, if we know that x is described by an estimate x and an error neighborhood, and that x is described by an estimate x and an error neighborhood, then it is natural to consider the sum x + x of these two sets of variables. This sum is naturally described by an estimate x + x and an error set +. It is therefore natural to choose the family F in such a way that if the sets and belong to F, then their sum + should also belong to F. In mathematical terms, a set that is closed under addition is called an additive semigroup; so, the above requirement on F means that the family F is an additive semigroup w.r.t. addition on IR d. Now, we are ready for the formal definitions. Definitions and the Main Result of This Subsection Definition 4. A convex compact set IR d is called a neighborhood of 0 (or simply neighborhood, for short) if 0 Int( ). A family of neighborhoods F is called linearly ordered if for every, F, either, or. A family of neighborhoods F is called an additive semigroup if for every, F, we have + F. To characterize additive semigroups of neighborhoods, we need to recall the notion of similarity: Definition 5. Neighborhoods and are called similar if = λ for some real number λ > 0. PROPOSITION 4. If F is a linearly ordered additive semigroup of neighborhoods, then all neighborhoods from F are similar to each other. Proof. Since neighborhoods are convex compact sets, we can describe them by support functions. Since each F is a neighborhood of 0, each support function δ (x ) is > 0 for x 0. In terms of support functions, = λ iff δ (x ) = λδ (x ). Therefore, the statement that we want to prove means that for every two support functions δ (x ) and δ (x ), their ratio δ (x ) δ (x ) 9

is a constant (if this is indeed a constant, then we can take this constant as the desired λ). We will prove this statement by reduction to a contradiction. Indeed, if this ratio is not a constant, then δ (x ) δ (x ) δ (y ) δ (y ) for some x and y. W.l.o.g., we can assume that δ (x ) δ (x ) < δ (y ) δ (y ). In between two non-equal positive real numbers, there always is a positive rational number p/q; therefore, there exist natural numbers p and q such that and δ (x ) δ (x ) < p q p q < δ (y ) δ (y ). If we multiply both sides of both inequalities by the common denominator, we will conclude that q δ (x ) < p δ (x ) and q δ (y ) > p δ (y ). Since and are convex sets, we have q δ (x ) = δ (x q ) and p δ (x ) = δ (x p ). Since F is an additive semigroup, the sets q = +... + (q times) and p = +... + (p times) also belong to the family F. For these two sets, we thus have δ (x q ) < δ (x p ) (2) and δ (y q ) > δ (y p ). (3) But the neighborhoods from the family F are linearly ordered; therefore: either q p, or p q. In both cases, we will get a contradiction: In the first case, we will have δ (y q ) δ (y p ), which contradicts to (3). In the second case, we will have δ (x q ) δ (x p ), which contradicts to (2). 10

These contradictions show that our assumption is impossible and therefore, for every two neighborhoods, the ratio of their support function is indeed a constant. Hence, every two neighborhoods are similar. Q.E.D. In other words, if F is a linearly ordered semigroup of neighborhoods, then all neighborhoods F have the form λ 0, where 0 is one of these neighborhoods, for a real λ > 0. In particular, if we additionally require that the neighborhoods be symmetric w.r.t. 0 (i.e., that = ), then the set 0 defines a norm x = min{λ x/λ 0 }, and neighborhoods from F become closed balls of different radii in this norm. This result explains why such families are often used to describe uncertainty: when 0 is a unit sphere, we get balls in Euclidean sense; when 0 is a unit cube, we get cubic boxes that are balls w.r.t. the metric x = max x i, etc. Acknowledgments. This work was partially supported by NSF Grant No. EEC-9322370, by NASA Grant No. NAG 9-757, and by the German Science Foundation. The authors are very thankful to Svetoslav Markov and to the anonymous referees for fruitful discussions, suggestions, and encouragement. References [1] S. M. Markov, On directed interval arithmetic and its applications, Journal of Universal Computer Science, 1995, Vol. 1, No. 7, pp. 510 521 (electronic journal; available at www server: http://hgiicm.tu-graz.ac.at). [2] S. M. Markov, On the foundations of interval arithmetic, In: G. Alefeld and A. Frommer (eds.), Scientific Computing and Validated Numerics (Proceedings of SCAN-95), Akademie-Verlag, Berlin, 1996 (to appear). [3] G. Matheron, Random sets and integral geometry, Wiley, New York, 1975. [4] R. T. Rockafellar, Convex Analysis, Princeton University Press, Princeton, NJ, 1970. 11