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Joined 2026-09-21 Message
Incoming links: While loop
Coupled binary exponentiation 2026-10-31
To compute two nonnegative integer powers in one while loop, maintain a separate accumulator, current base and remaining exponent for each exponentiation by squaring. The paired loop invariant is and . For each positive remaining exponent, transfer its odd factor to its accumulator and halve the exponent while squaring its base. A zero remaining exponent is left unchanged. The mathematical integer is a loop variant: at least one positive term strictly decreases in every iteration. Both powers are therefore complete when the loop stops. The iteration count is , with constant space and a constant number of multiplications per iteration. This is an exact-arithmetic identity; floating-point rounding introduces a separate error in a numerical implementation.
Hoare logic Created 2026-10-18 Updated 2026-11-05
Hoare logic proves properties of programs through Hoare triples . A precondition constrains initial states and a postcondition constrains returned states. The basic rules establish partial correctness; termination requires a further argument.[ref]
For a program expression whose evaluation is defined and has no side effects, the assignment rule isThe substitution replaces the free occurrences of in the desired postcondition. For example, obtaining after
x := x+1 requires beforehand, so suffices. Integer overflow must be excluded or handled by the stated arithmetic model.Sequential composition joins a first postcondition to the next precondition. For a while loop, proving establishesHere is the loop invariant. For
while x > 0 do x := x-1, starting with an exact nonnegative integer, is preserved and the false exit guard yields . The decreasing nonnegative integer is also a loop variant, establishing total correctness for this example.[ref]Implement the three consumers directly with while loops. These methods consume the receiver; For
exists stops immediately after the first witness:trait RichIter[T] extends Iter[T] {
def foreach(f: T => Unit): Unit = {
while (hasNext) f(next())
}
def exists(p: T => Boolean): Boolean = {
while (hasNext) {
if (p(next())) return true
}
false
}
def foldLeft[U](x: U)(f: (U, T) => U): U = {
var result = x
while (hasNext) result = f(result, next())
result
}
}foreach, the loop invariant is that f has been called once, in order, on exactly the consumed prefix. For exists, provided it has not returned, every consumed entry fails the predicate; a returned true has a witness, while exhaustion proves there is none. For foldLeft, after consumed entries the loop invariant iswith result = x at . Thus initialization, one further application, and exhaustion prove the required left fold. On finite inputs, the remaining length is a loop variant for each loop. Callbacks are assumed not to mutate this iterator independently.With constant-cost callbacks and iterator operations, full consumption costs ; exists may return earlier. Empty inputs respectively do nothing, return false, and return x. Infinite inputs need not terminate, except that exists terminates upon reaching a witness.First rearrange the ratios so that the intermediate powers remain moderate:This avoids separately forming enormous powers such as . Perform the two instances of exponentiation by squaring in one while loop, using coupled binary exponentiation:With and , the loop invariant is , , and . Each active branch preserves its own power identity by the parity argument in part (a)(ii), and an inactive branch changes nothing. The loop variant is the mathematical integer , which need not be computed in a fixed-width variable: it decreases whenever either exponent is positive. At exit , , proving the returned expression is . There are iterations, at most four loop multiplications each, and two final multiplications. Converting
def ec(n: Int): Double = {
require(n > 1)
val dn = n.toDouble
var x = 1.0 + 1.0 / dn
var y = 1.0 + 1.0 / (dn - 1.0)
var k = n
var l = n - 1
var p = 1.0
var q = 1.0
while (k > 0 || l > 0) {
if (k > 0) {
if ((k & 1) != 0) p *= x
k /= 2
if (k > 0) x *= x
}
if (l > 0) {
if ((l & 1) != 0) q *= y
l /= 2
if (l > 0) y *= y
}
}
(dn + 1.0) * p - dn * q
}n before adding one avoids integer overflow at Int.MaxValue.The rearrangement avoids floating-point overflow in the original powers, but subtraction still combines two quantities of order to obtain one of order one. Catastrophic cancellation can amplify their accumulated roundoff error. The mathematical approximation error therefore does not imply that the
Double implementation improves indefinitely as n grows.Use Horner's method, accumulating the high-degree coefficients before the low-degree ones:
def eval(x: Double): Double = {
var i = a.length
var value = 0.0
while (i > 0) {
i -= 1
value = a(i) + x * value
}
value
}At each while loop head the loop invariant isThe initial sum is empty. Decreasing
i and replacing the accumulator by a(i) + x * value extends that sum by one coefficient with the required powers shifted. The loop variant is i; on exit it is zero, so the accumulator equals . The code uses exactly multiplications and time, with constant auxiliary space, and also handles the empty coefficient array as the zero polynomial. Initializing from the last coefficient instead would save one multiplication for nonempty arrays.The invariant is an exact-arithmetic specification. With
Double, each arithmetic operation rounds; consequently the code evaluates the Horner expression in floating-point arithmetic, rather than guaranteeing the exact real value for arbitrary coefficients.First find a sign bracket by exponential search, then bisect that bracket without evaluating its endpoints again:
def floorRoot(): Int = {
require(eval(0.0) < 0.0)
var lo = 0
var hi = 1
while (eval(hi.toDouble) <= 0.0) {
lo = hi
require(hi <= Int.MaxValue / 2, "doubling bound exceeds Int range")
hi *= 2
}
while (hi - lo > 1) {
val mid = lo + (hi - lo) / 2
if (eval(mid.toDouble) <= 0.0) lo = mid
else hi = mid
}
lo
}For the first while loop the loop invariant is , with
hi a positive power of two, and either or . Every previously tested power of two has nonpositive value. Let be the least nonnegative integer for which . It exists by the positive leading coefficient. If the current upper endpoint is , a mathematical loop variant is while the guard holds; the next doubling decreases it by one. The first while loop's postcondition is . The second while loop has exactly the invariant, variant and adjacent-integer postcondition established in part c.There are evaluations at powers of two. If , the final bracket has width and takes exactly midpoint evaluations. Including the initial sign check, the total is evaluations for , and two for . Thus evaluation cost is for the first positive dyadic bound ; it is not a claim about the first positive root of a polynomial when the polynomial is nonmonotone.
The mathematical argument uses unbounded integer endpoints and reliable signs. The guarded
Int implementation requires the successful dyadic bound to fit in Int; otherwise it explicitly fails rather than wrapping to a negative bound. The coefficient assumptions alone cannot guarantee that a requested integer result fits a fixed-width type. Floating-point overflow or unreliable signs likewise need a wider or certified evaluator for a fully general implementation.Scan the two suffixes until a mismatch or an endpoint:
def longestFrom(i: Int, j: Int): Int = {
require(0 <= i && i <= N && 0 <= j && j <= N)
var k = 0
while (k < N - i && k < N - j && a(i + k) == a(j + k)) {
k += 1
}
k
}The loop invariant is and equality of the two array segments of length . The loop variant is , which decreases whenever the while loop runs. Short-circuit evaluation checks both bounds before indexing. On exit either a string suffix has ended or the next characters differ, so no longer common string prefix is possible. Thus the returned value is the longest common prefix length. Overlap is allowed: the code only reads the array, so the two matching segments need not occupy disjoint storage.
Compare every unordered pair of distinct starting positions:
def longestRepeated(): (Int, Int, Int) = {
require(N >= 2)
var besti = 0
var bestj = 1
var bestk = 0
var i = 0
while (i < N - 1) {
var j = i + 1
while (j < N) {
val k = longestFrom(i, j)
if (k > bestk) {
besti = i; bestj = j; bestk = k
}
j += 1
}
i += 1
}
(besti, bestj, bestk)
}After each comparison, the stored triple attains the largest common-prefix length among pairs processed so far; its positions remain distinct. The nested while loops enumerate exactly pairs. Reversing the two positions does not change their common-prefix length, so no omitted ordered pair could improve the answer. Initialization also supplies a valid length-zero answer if no character repeats.
Write for the final maximal length. Each comparison takes time, including a possible first mismatch. Hence time is , at worst , and auxiliary space is . Writing merely would miss the case. The explicit precondition excludes arrays with fewer than two valid starting positions; the requested distinct-position triple has no answer there under this domain convention.
Reuse the verified longest-common-prefix scan, handling equality before scanning:
def before(i: Int, j: Int): Boolean = {
require(0 <= i && i < N && 0 <= j && j < N)
if (i == j) false
else {
val k = longestFrom(i, j)
if (j + k == N) false
else if (i + k == N) true
else a(i + k) < a(j + k)
}
}There is no additional while loop: the helper's loop invariant is that the first
k characters agree and both indices remain in bounds. If the second string suffix ends first, the first cannot precede it; if the first ends first, it is a proper string prefix and precedes the second. Otherwise the first mismatching characters decide lexicographic order. Distinct suffixes cannot both end after the same number of characters, since their lengths differ.The
i == j case ensures a strict comparator, returning false for a string suffix compared with itself, in constant time. For distinct suffixes its running time is , where is the longest repeated-substring length of the full array.Sort string suffix starting positions, then examine only adjacent entries:
def longestRepeated(): (Int, Int, Int) = {
require(N >= 2)
val suffixes = (0 until N).toArray.sortWith(before)
var besti = suffixes(0)
var bestj = suffixes(1)
var bestk = 0
var t = 1
while (t < N) {
val i = suffixes(t - 1)
val j = suffixes(t)
val k = longestFrom(i, j)
if (k > bestk) {
besti = i; bestj = j; bestk = k
}
t += 1
}
(besti, bestj, bestk)
}The while loop retains a best triple among adjacent pairs already visited. All string suffix starting positions are distinct, and the adjacent-suffix characterization proves that considering these pairs suffices. The two definitions named
longestRepeated in parts b and e are alternatives, not duplicate members of one software class.Assuming a comparison sort using comparator calls, sorting takes time, because each comparison scans characters. The final pass takes . Thus overall time is , at worst , with auxiliary storage for indices and ordinary sorting workspace. Merely storing indices does not make string suffix comparisons constant-time; claiming here would omit their cost. Faster suffix-array constructions require additional machinery beyond this comparator-based solution.
For a finite acyclic linked list define the node sequence, retaining software object identities, byThen scan without changing any links:
def findMin(first: Node): Node = {
require(first != null)
var best = first
var cursor = first.next
while (cursor != null) {
if (cursor.datum < best.datum) best = cursor
cursor = cursor.next
}
best
}At each while loop head, write . The loop invariant says is nonempty,
best belongs to , and its datum is no larger than that of every node in . Initially contains only first; comparing the next node extends the processed string prefix and preserves minimality. The loop variant strictly decreases. At exit the whole list is processed, so the returned reference identifies a minimum node. Ties retain the earliest minimum. Time is linear in the number of nodes and auxiliary space is constant.