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Outward-rounded operations are inclusion-isotone and enclose DAG evaluations

Statement

For real intervals, complex rectangles or complex balls on which an operation is defined, enlarging either input enclosure cannot shrink the computed enclosure: for binary ∘∈{+,−,×,÷}, if X⊆X′ and Y⊆Y′, then the computed set enclosure of X∘Y is contained in that of X′∘Y′ whenever both are finite results and their domain restrictions are preserved. The square-root enclosure is inclusion-isotone on nonnegative intervals. Each finite ball operation in Outward-rounded interval, rectangle and ball operations contains the exact ball image. A split-domain, entire, domain-error, overflow, or inconclusive status makes no finite claim. For a finite expression DAG whose operation nodes all return finite enclosures, every exact node value from every admissible input assignment belongs to its computed node enclosure; in particular, the root enclosure contains the exact range. If any node returns a non-finite status, no finite DAG enclosure is asserted.

Facts & Assumptions

Given: The interval, rectangle, ball and operation definitions in Outward-rounded interval, rectangle and ball operations and the directed rounding lemma Directed dyadic rounding encloses with explicit underflow and overflow bounds.

[L1]

Outward-rounded interval, rectangle and ball operations: each finite real interval or rectangle operation uses its displayed endpoint formula and each ball operation uses its displayed origin-centered radius, followed by directed rounding; invalid domains return a non-finite status.

[L2]

Directed dyadic rounding encloses with explicit underflow and overflow bounds: every finite exact endpoint lies between its downward and upward dyadic roundings.

[L3]

Dyadic square-root bisection gives outward endpoints: every defined nonnegative square-root endpoint is bracketed by exact dyadic bisection, and its outward lower and upper endpoint maps are nondecreasing in the radicand.

[L4]

Real and complex modulus product and triangle inequalities: complex modulus is submultiplicative with equality on products and satisfies the triangle inequality; real absolute value satisfies the triangle inequality.

[L5]

Real intervals, complex rectangles and ball enclosures: for z=x+iy, ∣z∣=x2+y2; an expression DAG is a finite directed acyclic graph with input nodes, constant nodes, and operation nodes whose constants carry certified enclosing intervals when needed.

[L6]

Nonnegative real square roots exist uniquely: each nonnegative radicand has a unique nonnegative square root.

[L7]

Dedekind-complete ordered real field: ∣x∣=x for x≥0 and ∣x∣=−x for x<0, hence ∣x∣2=x2.

Proof

1.1givenL1L2L3algebra

For real addition and subtraction, every exact pair value lies between the endpoint formulas. For multiplication, fixing one variable makes the product affine in the other, so a minimum or maximum over a rectangle is attained at a corner. Reciprocal is monotone on either interval avoiding zero, and square root is monotone on [0,∞). Thus the exact ranges lie in the displayed pre-rounding intervals. Each computed lower endpoint is no greater than the exact lower endpoint and each upper endpoint is no smaller than the exact upper endpoint by [L2]; square-root endpoint bounds follow from [L3]. For complex rectangles, component addition and multiplication preserve the exact component formulas. If a rectangle excludes zero, at least one coordinate interval omits zero, so its exact squared interval has positive lower endpoint. The reciprocal rule retains the exact rational denominator and component quotient endpoints until final outward conversion, so underflow cannot introduce a zero divisor. Those converted components and subsequent multiplication enclose the reciprocal and quotient.

1.2givenL1L4L5L6L7algebra

For a ball B(c,r), the coordinate projection intervals in [L1] contain each real and imaginary component. By [L5]–[L6], ∣iℑz∣=∣ℑz∣; the complex triangle inequality [L4] therefore gives ∣z∣=∣ℜz+iℑz∣≤∣ℜz∣+∣ℑz∣≤MB. Also ∣z∣2=(ℜz)2+(ℑz)2≥(ℜz)2=∣ℜz∣2 by [L7]. If ∣z∣<∣ℜz∣, nonnegativity would imply ∣z∣2<∣ℜz∣2, a contradiction. Hence ∣z∣≥∣ℜz∣; the same argument gives ∣z∣≥∣ℑz∣. Every point of XB and YB has absolute value at least its corresponding d value, so ∣z∣≥mB. For z∈B1,w∈B2, [L4] gives ∣z±w∣≤MB1+MB2 and ∣zw∣≤MB1MB2. If mB2>0, then ∣w∣≥mB2 and ∣z/w∣≤MB1/mB2. Upward radius rounding in [L1] preserves these bounds, proving soundness of all finite ball results.

2.1step 1.1step 1.2L1L2L3algebra

Inclusion of balls implies inclusion of each coordinate projection interval. Therefore MB cannot decrease and mB cannot increase as an input ball enlarges. The ball addition, subtraction and multiplication radii in [L1] are consequently nondecreasing. If both divisor results are finite, the larger divisor ball has positive mB and the smaller one has at least as large an mB; the quotient radius therefore cannot decrease when either ball is enlarged. For intervals and rectangles, the exact endpoint hulls and their downward/upward roundings preserve inclusion, so their finite extensions are isotone as well; for square root, [L3] gives monotonicity of the computed endpoint maps. If a domain restriction is lost, the operation returns its stated non-finite status.

3.1L1L5step 1.1step 1.2step 2.1inductionconstruct∎

In any nonempty finite acyclic graph, some node has no predecessor: otherwise, repeatedly following predecessors would revisit a node by finiteness and create a directed cycle. Remove such a node and repeat to construct a finite topological order. By [L1], input nodes contain their assigned values and constants have certified enclosures. Induct along this order: if each predecessor of an operation node contains its exact value, steps 1.1, 1.2, and 2.1 show that the operation enclosure contains the node's exact value. The induction covers every node, including the root, for every admissible assignment.

Depends on

Used by

…and 27 more results.

Cited to discharge well-definedness by Outward-rounded interval, rectangle and ball operations.

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources