Continuous If-Then Statements Are Computable

In many practical situations, we must compute the value of an if-then expression f(x) defined as “if c(x) ≥ 0 then f+(x) else f−(x)”, where f+(x), f−(x), and c(x) are computable functions. The value f(x) cannot be computed directly, since in general, it is not possible to check whether a given real number c(x) is non-negative or non-positive. Similarly, it is not possible to compute the value f(x) if the if-then function is discontinuous, i.e., when f+(x0) 6= f−(x0) for some x0 for which c(x0) = 0. In this paper, we show that if the if-then expression is continuous, then we can effectively compute f(x). Practical need for if-then statements. In many practical situations, we have different models for describing a phenomenon: – a model f+(x) corresponding to the case when a certain constraint c(x) ≥ 0 is satisfied, and – a model f−(x) corresponding to the case when this constraint is not satisfied, i.e., when c(x) < 0 (usually, the second model is also applicable when c(x) ≤ 0). For example, in Newton’s gravitation theory, when we are interested in the gravitation force generated by a celestial body – i.e., approximately, a sphere of a certain radius R – we end up with two different formulas: – a formula f+(x) that describes the force outside the sphere, i.e., where c(x) def = ‖r‖ −R ≥ 0, and – a different formula f−(x) that describes the force inside the sphere, i.e., where c(x) = ‖r‖ −R ≤ 0. Towards a precise formulation of the computational problem. In such situations, we have the following problem: – we know how to compute the functions f+(x), f−(x), and c(x); – we want to be able to compute the corresponding “if-then” function f(x) def = if c(x) ≥ 0 then f+(x) else f−(x). In general, we say that a function f(x) is computable if there is an algorithm that, given an input x and a rational number ε > 0, produces a rational number r for which |f(x)− r| ≤ ε. In the above formulation, we assume that the function c(x) is computable for all possible values x from a given set X, and that: – the function f+(x) is computable for all values x ∈ X for which c(x) ≥ 0; and – the function f−(x) is computable for all values x ∈ X for which c(x) ≤ 0. Why this problem is non-trivial. The value f(x) cannot be computed directly, since in general, it is not possible to check whether a given real number c(x) is non-negative or non-positive; see, e.g., [2, 3]. Discontinuous if-then statements are not computable. It is known that every computable function is everywhere continuous; see, e.g., [3]. Thus, when the if-then function f(x) is not continuous, i.e., when f+(x0) 6= f−(x0) for some x0 for which c(x0) = 0, then the function f(x) is not computable. Our main result. In this paper, we show that in all other cases, i.e., when the if-then function f(x) is continuous, it is computable. Algorithm: main idea. The main idea behind our algorithm is that in reality, we have one of the three possible cases: – case of c(x) > 0, when f(x) = f+(x); – case of c(x) < 0, when f(x) = f−(x); and – case of c(x) = 0, when f(x) = f+(x) = f−(x). Let us analyze these three cases one by one. In the first case, let us compute c(x) with higher and higher accuracy ε = 2−k, k = 1, 2, . . . As soon as we reach the accuracy 2−k < c(x) 2 , for which c(x) > 2 · 2−k, we get an approximation rk for which |c(x)− rk| ≤ 2−k, i.e., for which rk > c(x)− 2−k ≥ 2 · 2−k − 2−k = 2−k and thus, rk > 2−k. Since we know that c(x) ≥ rk − 2−k, we thus conclude that c(x) > 0. Similarly, in the second case, if we compute c(x) with higher and higher accuracy 2−k, we will reach an accuracy 2−k < |c(x)| 2 , for which the corresponding approximate value rk satisfy the inequality rk < −2−k and thus, we can conclude that c(x) < 0. In the third case, since f+(x) = f−(x), if we compute f+(x) and f−(x) with accuracy ε > 0, then the resulting approximate values r+ and r− satisfy the inequalities |f(x) − r+| = |f+(x) − r+| ≤ ε and |f(x) − r−| = |f−(x) − r−| ≤ ε and therefore, |r+ − r−| ≤ |r+ − f(x)|+ |f(x)− r−| ≤ ε + ε = 2ε. Vice versa, if the inequality |r+ − r−| ≤ 2ε is satisfied (even if we know nothing about c(x)), then in reality, the value f(x) coincides wither with f+(x) or with f−(x). In the first subcase, when f(x) = f+(x), we have |f(x)− r+| = |f+(x)− r+| ≤ ε