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Incoming links: Loop invariant
trait Ordered[T] {
def <(that: T): Boolean
}
class MinMSet[T <: Ordered[T]](contents: List[T])
extends MSet[T](contents) {
def min: T = {
if (data.isEmpty) throw new NoSuchElementException("MinMSet.min")
var best = data.head
var rest = data.tail
while (rest.nonEmpty) {
if (rest.head < best) best = rest.head
rest = rest.tail
}
best
}
}The upper type bound
T <: Ordered[T] guarantees the method signature needed for rest.head < best. Assuming that comparison implements the intended total ordering, the loop invariant says best is a minimum of the inspected prefix; the remaining suffix length decreases. The method returns a minimum in comparisons and auxiliary space, without deleting it. The bound supplies a comparison operation, not a proof of its order laws.Before testing candidate , every earlier integer has been printed exactly when it belongs to the spectrum, and every positive satisfies . During the inner loop that exclusion remains true: an excluded candidate is passed over only after checking its square.
At termination of the inner loop, , so is the least positive possible witness. If its square is already at least , every larger witness is too large as well; otherwise it satisfies both required inequalities and is printed. Advancing preserves the earlier-witness exclusion because the new lower bound only increases. These loop invariants establish both completeness and absence of extra outputs.
Add
bestStart = 0 initially and, whenever a strictly longer run is found, recordbest = run
bestStart = i-run+1Then is a plateau of the returned maximum length. Keeping the old answer on equal lengths selects the earliest maximum. The loop invariant from part (a) gains the assertion that the recorded interval realizes
best; time and auxiliary-space bounds are unchanged. For empty input use the empty interval at zero.Source paper: released examination archive, member 2017/T/a10098w1.pdf. The printed cover dates this examination to Wednesday 13 June 2018, Trinity Term; the archive directory does not determine its year. All eight questions and every printed option are covered. Examples use Scala 2 syntax. Arithmetic loop invariants describe exact arithmetic unless a floating-point arithmetic qualification is stated.
Use exponentiation by squaring, retaining the unprocessed exponent in Writing , the loop invariant is and . If the old exponent is , , multiplying the accumulator by and replacing the remaining power by preserves the invariant. When , the unused squaring may be omitted. The nonnegative integer
k:def eaBinary(n: Int): Double = {
require(n > 0)
var base = 1.0 + 1.0 / n.toDouble
var k = n
var result = 1.0
while (k > 0) {
if ((k & 1) != 0) result *= base
k = k / 2
if (k > 0) base *= base
}
result
}k is a loop variant, since for . At exit the remaining power is one, giving . There are iterations, at most one accumulator multiplication and one squaring per iteration, hence multiplications and constant space.Keep a running product of the fixed base. In particular, The loop invariant is and . Initially the empty product is one. Multiplication by
1.0/n uses floating-point arithmetic, avoiding the truncation that integer division would introduce:def eaLinear(n: Int): Double = {
require(n > 0)
val base = 1.0 + 1.0 / n.toDouble
var k = 0
var result = 1.0
while (k < n) {
result *= base
k += 1
}
result
}base and incrementing k preserve the invariant. The loop variant decreases by one. On exit , so the result is , with exactly multiplications and time complexity. With Double, the operational result is the rounded recurrence, rather than an exact real power; this distinction applies to the next two power computations too.Generate each reciprocal factorial from the preceding one, instead of recomputing a factorial for every summand:The loop invariant is , , and . Initially and . The division produces , which is then added to the previous partial sum. The loop variant is . Consequently the returned value is , using exactly divisions and additions, no multiplications, and time complexity. The slightly weaker precondition also permits the useful zero-term boundary case. For
def eb(n: Int): Double = {
require(n >= 0)
var i = 0
var term = 1.0
var sum = 1.0
while (i < n) {
i += 1
term /= i.toDouble
sum += term
}
sum
}Double, small positive terms eventually cease to change sum because of floating-point rounding.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.Return the lower-bound search boundary, permitting both zero and the array length:The precondition includes nondecreasing order. The loop invariant isInitially both excluded ranges are empty. If , sortedness justifies excluding every position through
def lowerBound(a: Array[Int], x: Int): Int = {
require(a != null) // a is nondecreasing
var lo = 0
var hi = a.length
while (lo < hi) {
val mid = lo + (hi - lo) / 2
if (a(mid) < x) lo = mid + 1
else hi = mid
}
lo
}mid; otherwise it justifies excluding every position from mid onwards. The loop variant is a nonnegative integer that strictly decreases. When the old width is positive, mid is a valid array index. The midpoint expression avoids adding two potentially large indices. At exit , so lo satisfies the required partition, including when . Each new width is at most half the old width, rounded down; this gives time complexity and constant space, conventionally for nontrivial sizes.A second binary search locates the first entry strictly greater than For upper-bound search, the loop invariant is , , and ; its loop variant is again . The same midpoint and sortedness arguments prove preservation, termination and time complexity. The first binary search in part (b) already has its own loop invariant and loop variant.
x:def upperBound(a: Array[Int], x: Int): Int = {
require(a != null) // a is nondecreasing
var lo = 0
var hi = a.length
while (lo < hi) {
val mid = lo + (hi - lo) / 2
if (a(mid) <= x) lo = mid + 1
else hi = mid
}
lo
}
def equalRange(a: Array[Int], x: Int): (Int, Int) =
(lowerBound(a, x), upperBound(a, x))Let the two boundaries be . There cannot be a value greater than before the first value at least , so . Every entry between them is simultaneously at least and at most , hence equals . Entries before are smaller and entries at or after are larger, so the interval cannot be extended at either end. This proves the equal-range search result is maximal. If
x is absent, i == j, both equal to its insertion position; this can be zero, , or an interior gap. Two logarithmic searches still have logarithmic total time complexity. Computing an upper boundary as lowerBound(a, x+1) would introduce an unnecessary integer overflow bug at the largest integer.Black-box testing chooses inputs and expected outputs from the public specification without using the implementation's control structure. Equivalence class testing partitions the input space into classes expected to exercise similar specified behavior, then chooses representatives; boundary cases test where that behavior changes. For equal-range search, use these concrete representatives:
- Empty array:
([], 4)gives(0,0). - One entry, with the target below, equal, and above it:
([4],3),([4],4),([4],5)give(0,0),(0,1),(1,1). - An absent interior value:
([1,3,5],4)gives(2,2). - One interior match:
([1,3,5],3)gives(1,2). - A repeated interior block:
([1,3,3,3,5],3)gives(1,4). - A repeated block touching the left or right boundary:
([3,3,5],3)and([1,3,3],3)give(0,2)and(1,3). - Every entry equal:
([3,3,3],3)gives(0,3); targets two and four give(0,0)and(3,3). - Extreme keys:
([Int.MinValue,0,Int.MaxValue],Int.MaxValue)gives(2,3), and searching forInt.MinValuegives(0,1).
For each result independently check , the strict inequalities outside the returned interval, and equality inside it. Unsorted inputs violate the precondition; they are not ordinary members of the specified valid-input classes. Software testing supplies evidence about implementations, while the loop invariant proof explains correctness for every valid input.
Assume The search loop invariant says that the structure is unchanged and every ordinary linked list node passed has index smaller than
vector already satisfies the full representation invariant in part (b). Check the local bounds and value preconditions, then find the first index not smaller than the requested one:def setElement(vector: Node, value: Double, index: Int): Unit = {
require(vector != null)
require(0 <= index && index < vector.index)
require(value != 0.0 && !java.lang.Double.isNaN(value) &&
!java.lang.Double.isInfinite(value))
var p = vector.next
while ((p ne vector) && p.index < index) p = p.next
if ((p ne vector) && p.index == index) p.value = value
else {
val left = p.prev
val fresh = new Node(value, index, left, p)
left.next = fresh
p.prev = fresh
}
}index; p is the next candidate. The number of unpassed ordinary entries is a loop variant. The short-circuit evaluation prevents testing the header index as an ordinary entry. If an equal index exists, changing only its nonzero value preserves the representation invariant. Otherwise left and p delimit the unique insertion position: initializing both new links before redirecting their neighbors preserves reciprocal links and strict index order. Exactly the requested coordinate changes. With stored entries, the search costs and the splice costs constant time; checking the whole input structure is a caller obligation, rather than an unmentioned scan inside each call.An already-zero entry needs no new linked list node:The full input representation invariant is a precondition. The search has the same loop invariant and loop variant as in part (d). At exit, either
def zeroElement(vector: Node, index: Int): Unit = {
require(vector != null)
require(0 <= index && index < vector.index)
var p = vector.next
while ((p ne vector) && p.index < index) p = p.next
if ((p ne vector) && p.index == index) {
val left = p.prev
val right = p.next
left.next = right
right.prev = left
}
}p is the only stored entry with that index or none exists. In the first case, bypassing it in both directions removes exactly that coordinate and preserves the order of all remaining entries. This also works when its two neighbors are the same list header, leaving a self-linked empty structure. The header itself is never removed. The requested coordinate is zero and all others are unchanged, in time complexity.Append entries in increasing index order, keeping a tail reference so that repeated searches do not make construction quadratic:The loop invariant is , a valid circular representation of the nonzero entries in
def makeVector(elems: Array[Double]): Node = {
require(elems != null)
val h = makeZeroVector(elems.length)
var tail = h
var i = 0
while (i < elems.length) {
val v = elems(i)
require(!java.lang.Double.isNaN(v) &&
!java.lang.Double.isInfinite(v))
if (v != 0.0) {
val fresh = new Node(v, i, tail, h)
tail.next = fresh
h.prev = fresh
tail = fresh
}
i += 1
}
h
}elems[0..i), a fixed header dimension elems.length, and tail eq h.prev. The next nonzero index is greater than every stored index; appending the fresh linked list node preserves reciprocal links, including tail.next eq h. The loop variant is elems.length-i. At exit the abstraction function yields exactly the input array as a vector. There is one pass and constant work per entry: time complexity for , including emptiness, and output storage for nonzero entries. Both signs of floating-point zero are represented by absence.Use the sorted sparse dot product merge, processing a coordinate only when it appears in both supports of a vector:Both structures must satisfy the representation invariant; equal dimension is checked explicitly. The loop invariant says that
def dotProduct(vec1: Node, vec2: Node): Double = {
require(vec1 != null && vec2 != null)
require(vec1.index == vec2.index)
var p = vec1.next
var q = vec2.next
var result = 0.0
while ((p ne vec1) && (q ne vec2)) {
if (p.index < q.index) p = p.next
else if (q.index < p.index) q = q.next
else {
result += p.value * q.value
p = p.next
q = q.next
}
}
result
}result contains exactly the products for all matching indices already passed, and that every skipped unequal index has zero in the other vector. The latter follows from sortedness: if p.index < q.index, that index cannot appear at q or later, and all earlier entries of the other vector have already been settled. Equality contributes its product exactly once. The total number of remaining ordinary entries is a loop variant; every iteration advances at least one cursor. Once either cursor reaches its header, all remaining entries pair with zeros. Thus the exact-arithmetic result is . With stored entries, time complexity is and auxiliary space is constant. Floating-point products and additions may round or overflow even when input coordinates are finite; no exact real-arithmetic accuracy claim is made for Double.Place this object method inside the concrete class. Its precondition is the representation invariant; its postcondition is that the abstract data type state is unchanged and its left-to-right contents have been printed in bracket notation followed by a newline:Assume ordinary sequential execution and successful output. The loop invariant is , unchanged representation, and output consisting of
def display: Unit = {
print("[")
var i = 0
while (i < size) {
if (i > 0) print(", ")
print(a(slot(i)))
i += 1
}
println("]")
}[ followed by precisely the first abstract values, separated by commas and spaces. The loop variant decreases. At exit the closing bracket completes the required display, including [] for emptiness. This takes output operations and constant space.Replace suspended recursive calls by an explicit stack for an inorder traversal. Save each original right child before overwriting that link:The method belongs to the concrete class owning
def flattenIterative(): (Tree, Tree) = {
val pending = new scala.collection.mutable.Stack[Tree]
var current = root
var start: Tree = null
var end: Tree = null
while (current != null || pending.nonEmpty) {
while (current != null) {
pending.push(current)
current = current.left
}
val t = pending.pop()
val nextSubtree = t.right
t.left = end
if (end == null) start = t else end.right = t
t.right = null
end = t
current = nextSubtree
}
root = null
(start, end)
}root. Clearing root records that ownership has transferred to the returned doubly linked list; callers retain both returned endpoints. The inner loop invariant says that the current subtree lies before all its pending ancestors in inorder traversal; pushing and descending left preserves this fact. Its loop variant is the number of entries along the remaining left-child path, a finite decreasing quantity.The outer loop invariant says that
start through end is precisely the already emitted sorted prefix, with reciprocal links and null outside endpoints. current and the pending ancestors describe exactly the remaining traversal in order; original links that are still needed have been retained. A pending ancestor has not yet had its right child overwritten. The next popped entry is therefore the next sorted entry. Saving nextSubtree preserves its future work before assigning t.right = null; linking it to end extends the prefix by exactly one entry. The number of original entries not yet emitted is an outer loop variant, since each full outer iteration emits one. At exit nothing remains, proving the same preservation and ordering postconditions as the recursive version.Each entry is pushed and popped once, giving time complexity and auxiliary space. No new
Tree entries are allocated. The historical mutable stack interface provides the push and pop operations used here; its later deprecation does not change their meaning.[ref]Use four working processor registers by packing the growing quotient into the vacated low bits of the shifting dividend. The following is A32 assembly language with conditional execution, an allowed instance of the question's general 32-bit Reduced instruction set computer. It is not restricted to the sixteen-bit encodings in the supplied Thumb instruction set table. On entry
r0 is the dividend and r1 the positive divisor; r2 becomes the high remainder machine word and r3 counts iterations. The quotient is returned in r0 and the ordinary remainder remains in r2.mov r2, #0
mov r3, #32
loop:
adds r0, r0, r0
adcs r2, r2, r2
cmp r2, r1
subcs r2, r2, r1
adc r0, r0, #0
subs r3, r3, #1
bne loop
bx lrHere
subcs means conditional sub when arithmetic carry is set, without the flag-setting suffix. adc also preserves flags. Consequently the arithmetic carry from cmp survives until it is inserted as the next quotient bit. The loop has seven instructions, one of them conditional, and no data-dependent branch. Its sole branch is the backwards loop branch, taken 31 times. It uses no separate quotient processor register, zero processor register, temporary comparison processor register or saved arithmetic carry processor register. A Thumb instruction set rendering using conditional execution would need an explicit conditional-execution encoding such as it cs; that is an additional encoded instruction, not an unexplained free operation.The packed restoring division loop invariant relates it to the displayed algorithm. Specifically, let , let retain its original mathematical value, and let be the number of completed iterations. Define the processed prefix , its quotient and remainder . The processor register loop invariant isThe two terms in the low machine word occupy disjoint bit positions. Its next outgoing arithmetic carry is therefore the next original dividend bit, not a quotient bit.
adds shifts the low machine word; adcs doubles the high remainder and incorporates that dividend bit. cmp decides whether this trial remainder is at least the divisor; subcs performs the conditional subtraction, and adc appends its decision to the shifted quotient. This is exactly the doubling, compare/subtract and quotient-bit update, with the counter maintained separately. Before processing bit , the trial remainder is at most , so the high-word addition does not require a hidden 33rd remainder bit for this 32-bit dividend problem. After 32 iterations the residual dividend portion is zero, giving the stated outputs.There is a printed source error in the auxiliary bound. At the initial state, choosing valid inputs makes the printed strict inequality false. The algorithm instead preserves for its unpacked 64-bit variable, together with its printed equality loop invariant. Doubling and subtracting when necessary preserves that corrected bound. At termination the equality becomes ; since is then divisible by , it gives . This source error does not invalidate the algorithm or the packed implementation. The figure is mathematical pseudocode: its
2^32 denotes exponentiation, rather than C's XOR operator.The displayed A32 routine assembled successfully with Clang and was disassembled to check the conditional and non-flag-setting forms. Its carry flag model was checked exhaustively at an eight-bit machine word size and against boundary and randomized 32-bit divisions. No native ARM architecture family execution is claimed.
Maintain a counter and independent timed finite-state machine for each detector–light pair, while sharing the interrupt scan and timer infrastructure. Blocking for one channel's flash would unnecessarily delay every other channel.
eye_handler():
for i in all_channels:
if EYE_DETECTED[i] != 0:
EYE_DETECTED[i] = 0
pending[i] = pending[i] + 1
main_event_loop():
initialise each LIGHT[i] = 0
initialise pending[i] = 0, state[i] = IDLE
enable detector interrupts
forever:
now = monotonic_timer_snapshot()
for i in all_channels:
service_channel(i, now)
arrange the next wake-up for new work or the earliest deadlineservice_channel is the three-state sequencer from (a), with its own started[i] and an interrupt-safe try_take_event(i). Shared interrupt demultiplexing uses separate if tests, not an else if chain: simultaneous pending detectors must all be acknowledged and counted. If a new detector becomes pending during the scan, the controller must retrigger while a flag remains set or the handler must rescan appropriately; the exact interrupt controller semantics must be checked rather than assuming a single scan can never miss an edge.One hardware timer can schedule the earliest active channel deadline. On its expiry, service every due channel and rearm for the next deadline; a periodic tick with adequate resolution is a simpler alternative. Detectors and the timer can then interleave without serial one-second waits. The handler/main-program counter accesses need the same short critical section protection as in (a). Keep timing state in one execution context, or protect it too if both a timer handler and the main program modify it.
The loop invariants are per channel: the pending count equals recorded but unstarted events,
ON means that channel's light is on, and GAP prevents a new flash until one second after its preceding light-off transition. Different channels may flash simultaneously. The event latch rearming limitation, finite backlog capacity and timer-resolution qualifications apply independently to each channel. These cannot be solved by replacing several event counters with one global Boolean variable.Use a logical right shift to discard one inspected bit at a time. The following Thumb instruction set subroutine accepts the word in
r0, returns its population count in r0, preserves r4, and otherwise uses caller-saved processor registers.popcount_loop:
push {r4}
movs r1, #0
movs r2, #1
movs r3, #32
pop_loop:
movs r4, r0
ands r4, r4, r2
adds r1, r1, r4
lsrs r0, r0, #1
subs r3, r3, #1
bne pop_loop
movs r0, r1
pop {r4}
bx lrands extracts the current least significant bit; adds includes it in the count. lsrs fills the newly vacant high position with zero. subs immediately precedes bne, so the branch tests the zero flag from decrementing the loop counter, not a flag left by the shift.After iterations, the accumulator contains the number of one bits in the original word's lowest positions, and the remaining word is the original unsigned interpretation shifted right by . This loop invariant holds initially and is preserved by each iteration. At every position has been inspected, proving the result. The given example therefore returns 10, and the all-zero and all-one words return 0 and 32. The accumulator cannot overflow: it never exceeds 32.