Use a programming closure whose private variable holds the next Fibonacci number. Use arbitrary-precision integers to enumerate all terms exactly:
def fibonacci(): Iter[BigInt] = {
  var following = BigInt(1)
  val step: BigInt => BigInt = (current: BigInt) => {
    val result = following
    following = current + following
    result
  }
  new FunIter[BigInt](BigInt(0), step)
}
At the cell containing , the private state immediately before computing its successor is following = F_(i+1). The successor returns that value and sets following to , preserving the recursion invariant. The resulting sequence starts
Each successor is computed once by the memoized implementation in part (c); interleaved copies see the same already generated successor instead of independently advancing shared programming closure state. Replacing BigInt by Int and the two initial values by 0,1 gives exactly the hinted Int => Int implementation, but eventually incurs integer overflow, so it does not enumerate all mathematical Fibonacci numbers.
An alternative avoiding mutable programming closure state represents the current and next Fibonacci numbers by a pair and uses the pure function with initial state ; a projecting iterator returns the first component. The closure construction above meets the suggested one-number interface directly.
Represent a multiset by a pure multiplicity function. Every operation returns a new function describing updated counts:
class ContraMSet[-T](f: T => Int) {
  def count(x: T): Int = f(x)

  def insert(x: T): ContraMSet[T] =
    new ContraMSet[T]((y: T) => f(y) + (if (y == x) 1 else 0))

  def delete(x: T): ContraMSet[T] =
    new ContraMSet[T]((y: T) => {
      val old = f(y)
      if (y == x && old > 0) old - 1 else old
    })

  def union[U <: T](that: ContraMSet[U]): ContraMSet[U] =
    new ContraMSet[U]((y: U) => this.count(y) + that.count(y))
}
An empty instance is new ContraMSet[T]((_: T) => 0). The constructor's precondition is that f is pure and gives a nonnegative integer count, with finite support of a multiset if only finite multisets are intended. Nonnegativity and the support condition are preserved by each operation. Integer overflow must be excluded for this fixed-width count interface.
A contravariant type parameter makes ContraMSet[T] a consumer of values of type T. Thus a counter able to answer queries for every animal can also answer queries restricted to cats. It does not expose or enumerate stored T values. The return type ContraMSet[T] is legal: its outer covariant result position and the class's contravariant argument position together place negatively.
For union, the argument itself is in a negative position, while its ContraMSet argument would reverse that sign. Using ContraMSet[T] directly as the parameter type would therefore put positively and fail the variance-position rule. Introduce : both operands can count every U, and their sum is a ContraMSet[U]. This is a restriction to a common query domain; operations do not promise to preserve counts outside the narrowed domain.[ref]
Pointwise evaluation gives
No old function is modified, so updates preserve immutable object. Constructing each wrapper takes constant work, but a later count may traverse a chain of previously created programming closures; its running time need not be constant in the history length. This functional representation trades enumeration and direct storage of occurrences for composable counting behavior.
Separate the maximal decimal prefix from the unconsumed suffix, rejecting an empty prefix:
import Data.Char (isDigit, digitToInt)

type Parse a = String -> Maybe (a,String)

parseInt :: Parse Int
parseInt input =
  case span isDigit input of
    ([],_)  -> Nothing
    (ds,ys) -> Just (foldl (\n d -> 10*n + digitToInt d) 0 ds, ys)
The left fold uses the positional notation recurrence . This parses unsigned decimal operands as illustrated, leaving the first nondigit and everything after it untouched. As the required result type is Int, correct mathematical values require the parsed integer to fit that bounded type; arbitrary-length exact values would instead need Integer or an explicit integer overflow check.
Use two array indices, advancing only the smaller current entry. Here and in Questions 1–4, code is structured Oberon-style pseudocode: RETURN exits a procedure, Some(x) and None distinguish success and absence, and AND short-circuits from left to right. Arithmetic is exact where the source asks us to ignore integer overflow.
PROCEDURE common(a, N, b, M): INTEGER;
VAR i, j: INTEGER;
BEGIN
  i := 0; j := 0;
  WHILE a[i] # b[j] DO
    IF a[i] < b[j] THEN i := i + 1
    ELSE j := j + 1
    END
  END;
  RETURN a[i]
END common;
Let be the smallest common value, and let . Existence implies . A useful loop invariant is , , with no common value among the discarded prefixes and . Initially the prefixes are empty. If , then cannot equal any entry of because that sorted array is increasing; it cannot be a common value in the discarded prefix either. Moreover would imply , a contradiction. Thus and incrementing is safe. The other branch is symmetric.
Every guard and return therefore uses valid array indices, even without explicit bounds tests. The loop variant decreases on each iteration. At exit the equal entries form a common value no larger than , hence equal . There are at most increments and constant work per increment. The procedure returns the smallest common value in time and auxiliary space.
Evaluate the supplied function once at each positive integer, retaining its largest value so far. In particular do not initialize the maximum to zero: might be negative everywhere.
PROCEDURE record(f, emit);
VAR n, best, y: INTEGER;
BEGIN
  best := f(1);
  emit(1);
  n := 2;
  WHILE TRUE DO
    y := f(n);
    IF y > best THEN
      emit(n);
      best := y
    END;
    n := n + 1
  END
END record;
At the loop head the loop invariant is , best , and precisely the record values with argument less than have been emitted, in increasing argument order. The first argument is a record because its comparison set is empty. The strict test correctly excludes ties. Updating best only on a strict increase preserves the loop invariant.
Exactly one call of f is made per processed argument. For an arbitrary supplied function, an unexamined argument can have a value larger than all previous values, so skipping its evaluation cannot be correct in general. This assumes is a total, deterministic function; the requested infinite computation ignores integer overflow.
Keep both the exponent and its power. This Oberon procedure uses only the permitted arithmetic:
PROCEDURE IntLog(b, n: INTEGER): INTEGER;
VAR i, p: INTEGER;
BEGIN
  i := 0; p := 1;
  WHILE p*b <= n DO
    p := p*b; i := i+1
  END;
  RETURN i
END IntLog;
The loop invariant is and . It holds initially because . A successful guard permits the next power, and the two assignments preserve the relation between and . At exit, , which is exactly part (a). Each iteration increases , and only finitely many powers of are at most , proving algorithm termination. The returned value is . Its running time is arithmetic operations; integer overflow is deferred to part (e).
Use repeated squaring, including index zero:
b2[0] := b; top := 0;
WHILE b2[top]*b2[top] <= n DO
  b2[top+1] := b2[top]*b2[top];
  top := top+1
END;
The loop invariant is for , with for every stored entry except possibly the initial entry when . Squaring gives , proving preservation. The first failed guard identifies the first larger power outside the range. If , no power with needs storing.
Index zero is deliberately included. The original paper asks for indices but then claims that their products represent every allowable power. Taken without , that claim fails for odd exponents, since every available exponent would be even. The extra initialized entry makes the intended binary expansion argument valid without changing the requested entries.
Let be the largest representable positive integer. The worst case is , requiring indices through . Thus enough = 5 for signed 32-bit integers, or 6 for signed 64-bit integers, suffices for stored powers that fit; a larger convenient constant is harmless. This counts slots starting at zero and does not reserve a slot for an out-of-range sentinel. The multiplication in the guard can still suffer integer overflow, independently of this array capacity bound.
Let be the maximum positive machine integer. A positive product suffers integer overflow when . The vulnerable expressions are p*b in part (b), b2[top]*b2[top] in the table guard, and p*b2[k] in the descending search. Even a rejected candidate must be evaluated before its comparison with .
The table guard becomes dangerous as soon as its current entry exceeds : approximately for signed 32-bit integers, or for signed 64-bit integers. These are thresholds for the intermediate power, not for alone. For example, with 32-bit arithmetic, and reach the stored power and then try its overflowing square . If , even the initial square may overflow for a tiny . In part (b), the final test can overflow when the next power above exceeds ; at , this already happens for .
For positive operands, integer division gives the exact equivalence . Therefore replace the three guards respectively by
p <= n DIV b
b2[top] <= n DIV b2[top]
p <= n DIV b2[k]
and perform each multiplication only after its guard succeeds. Every accepted product is then at most . With these guarded products and sufficient table capacity, the algorithms work throughout , . The stored exponents and weights also fit in those machine ranges. Checking for overflow after multiplying would be too late.
Use dynamic programming in increasing order of the maximum permitted part and then of the remaining sum. Let Count be a sufficiently wide integer type and let table have dimensions (MaxN+1) by (MaxN+1):
PROCEDURE CountPartitions(n: INTEGER): Count;
VAR i, j: INTEGER;
    table: ARRAY MaxN+1, MaxN+1 OF Count;
BEGIN
  table[0,0] := 1; j := 1;
  WHILE j <= n DO table[0,j] := 0; j := j+1 END;
  i := 1;
  WHILE i <= n DO
    table[i,0] := 1; j := 1;
    WHILE j <= n DO
      table[i,j] := table[i-1,j];
      IF j >= i THEN
        table[i,j] := table[i,j] + table[i,j-i]
      END;
      j := j+1
    END;
    i := i+1
  END;
  RETURN table[n,n]
END CountPartitions;
Require . The outer loop invariant says all rows below equal . The inner loop invariant additionally says entries before column in the current row are correct. Its dependencies are the previous row and the earlier column , so no future entry is read. Each state is evaluated once, giving arithmetic operations and stored counts. A memoization table around part (a)'s recursion is another valid way to share states; its entries need a separate “computed” flag because zero is a legitimate answer. Growing count sizes affect bit complexity and can cause integer overflow in a fixed-width Count.
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:
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
}
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 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.
A second binary search locates the first entry strictly greater than 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))
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.
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.
The representation invariant says that root is either null or the root of a finite, acyclic binary search tree, with no software object shared between different child subtrees. Every reachable entry has a non-null word and a strictly positive frequency. For each entry , every word in its entire left subtree is strictly less than t.word, and every word in its entire right subtree is strictly greater. This is an inherited ordering condition on all descendants, rather than merely a comparison with immediate children. It implies that each distinct word has exactly one entry. The empty initial state satisfies it.
The abstraction function is the finite multiset assigning each stored word its positive count and assigning all other words zero. Frequencies and their increments must fit in Int; the implementation below explicitly rejects an increment at its representational limit instead of silently violating positivity through integer overflow.
Make defensive copies on both entry and exit, while keeping the supplied Dimension class unchanged:
class Frame(d: Dimension) {
  require(d != null)
  private val saved = new Dimension(d.width, d.height)
  private val savedArea = saved.width * saved.height

  def area: Int = savedArea
  def dimension: Dimension = new Dimension(saved.width, saved.height)
  def isPortrait: Boolean = saved.height >= saved.width
}
The private stored dimensions have no aliases exposed to the caller and are never changed. Mutating the constructor argument affects only the caller's object; mutating a returned Dimension affects only a fresh copy. Consequently the stored area is calculated once and stays equal to the stored dimensions' product. This defensive copying is sufficient because Dimension contains only primitive fields; a more deeply nested mutable structure could require a deep copy or an immutable representation.
The multiplication above deliberately retains the supplied class's Int arithmetic and accessor type. If an exact mathematical area outside that range is required, compute once with saved.width.toLong * saved.height.toLong and expose Long instead; avoiding aliases and avoiding integer overflow are separate concerns.
Source paper: released examination archive, member 2018/L/a10098w1.pdf. The cover identifies the Long Vacation sitting of Thursday 5 September 2019. That printed identity takes precedence over the archive member's directory name.
The implementations use Scala 2 syntax. Fixed-width results are checked when arbitrary-precision arithmetic is needed internally; the mathematical contracts do not make integer overflow harmless. All eight questions are covered.
An Egyptian fraction need not be proper: several positive unit fractions may sum to at least one. Accordingly, accumulate a general nonnegative reduced fraction, rather than calling the proper-input-only Reduce on every intermediate value. First define an exact helper:
def toRatExact(e: Array[Int]): (BigInt, BigInt) = {
  var n = BigInt(0)
  var d = BigInt(1)
  var i = 0
  while (i < e.length) {
    require(e(i) > 0)
    val x = BigInt(e(i))
    val g = d.gcd(x)
    val nextN = n * (x / g) + d / g
    val nextD = d * (x / g)
    val h = nextN.gcd(nextD)
    n = nextN / h
    d = nextD / h
    i += 1
  }
  (n, d)
}

def toRat(e: Array[Int]): (Int, Int) = {
  val result = toRatExact(e)
  require(result._1.isValidInt && result._2.isValidInt,
          "Reduced result does not fit (Int, Int)")
  (result._1.toInt, result._2.toInt)
}
The loop invariant is , , , , and
Writing , the fraction for the extended prefix has numerator and denominator . These are the nextN and nextD in the code. Reducing them preserves the value and restores coprimality. The loop variant is . At exit the helper returns the exact sum in lowest terms, including for the empty array.
There are three distinct ways to control integer overflow: cancel the common denominator factor before multiplication, reduce after each addition, and use arbitrary-precision arithmetic when the remaining values still exceed fixed-width bounds. Cancellation reduces intermediate size but cannot guarantee that an exact answer fits (Int, Int). For example, the sum is in lowest terms, whose denominator exceeds Int.MaxValue. No implementation returning that fixed type can represent every permitted input exactly. The wrapper therefore checks this unavoidable result-type limitation instead of silently truncating; toRatExact supplies the unrestricted mathematical result.[ref]
Use a strict record test; tying the previous record is not enough:
def emitHighlyComposite(): Unit = {
  var candidate = 1
  var best = 0
  while (true) {
    val count = divisors(candidate)
    if (count > best) {
      emit(candidate)
      best = count
    }
    candidate += 1
  }
}
At the start of an iteration, best is the maximum divisor count among the positive integers below candidate, taking the maximum of the empty family as zero. The already emitted values are exactly the highly composite numbers in that range, in increasing order. The next count exceeds best precisely when this candidate is a new strict record. Updating the record and advancing the candidate preserves both assertions. The program deliberately has no termination requirement; integer overflow in this unbounded enumeration is ignored as directed by the source.
Keep total in the callee-saved register r4, and the first returned sum in the callee-saved register r5. Suppose the surrounding function has made an aligned stack frame with the local seven-halfword array at the current sp, occupying fourteen bytes in a sixteen-byte slot. One possible frame setup is push {r4,r5,r6,lr} followed by sub sp,sp,#16; the earlier code initializes total, regular, and special before this fragment.
    ldr  r0, =regular       @ assembler literal load of the global array address
    movs r1, #7
    bl   sum16
    movs r5, r0             @ first sum must survive the next call

    add  r0, sp, #0         @ current frame's local special array
    movs r1, #7
    bl   sum16
    muls r0, r5, r0         @ second sum times first sum
    adds r4, r4, r0         @ update total
The AAPCS32 permits each call to overwrite r0–r3, r12, and condition flags, but requires preservation of r4–r8, r10, r11, and sp. Thus both r4 and r5 remain valid even if sum16 is replaced by another conforming implementation. bl changes the link processor register lr, so the enclosing function saves its own return address in its prologue. Its epilogue can use add sp,sp,#16 and pop {r4,r5,r6,pc}. The frame uses thirty-two bytes in total, preserving the required eight-byte stack pointer alignment at both calls.[ref]
The two calls may be evaluated in this order. Assume the product and final addition fit the signed 32-bit result expected by the source program; unlike each individual seven-element sum, their product need not fit. The assembly language operations retain their low 32 bits if this assumption fails, whereas signed integer overflow in the C programming language has different semantics.
Use r0 for the argument and result. With 32-bit integers and pointers, the three fields have byte offsets 0, 4 and 8. Assume an acyclic finite binary tree, no concurrent mutation, a valid aligned call stack, and no signed integer overflow in any evaluated sum. These are representation and arithmetic assumptions, rather than properties of arbitrary C programming language implementations.
.syntax unified
.cpu cortex-m0
.thumb
.global sum_rec
.thumb_func
sum_rec:
    cmp  r0, #0
    beq  .Lempty
    push {r4, r5, r6, lr}    @ 16-byte frame; r6 is alignment padding
    movs r4, r0              @ retain this invocation's node
    ldr  r5, [r4, #0]        @ running value begins with its label
    ldr  r0, [r4, #4]
    bl   sum_rec             @ r0 = left subtree sum
    adds r5, r5, r0
    ldr  r0, [r4, #8]
    bl   sum_rec             @ r0 = right subtree sum
    adds r0, r0, r5
    pop  {r4, r5, r6, pc}    @ restore caller state and return
.Lempty:
    movs r0, #0
    bx   lr
The node and partial sum are held in callee-saved registers r4,r5. Each recursive invocation saves its caller's values before replacing them, so a child returns with the parent's values intact. Saving lr preserves the return address across the two bl instructions. Caller-saved registers may be overwritten by a child; no parent value is left live only in them. Pushing four words maintains eight-byte stack alignment at recursive calls.[ref]
Correctness follows by structural induction. An empty subtree returns zero. For a nonempty subtree, assume both recursive calls return their respective label sums. The final addition then combines the node's label with exactly those two sums, so r0 returns the sum of all labels. The instruction forms and field offsets fit the short load and arithmetic encodings.[ref]
Use column heights and update the cached count by the number of removed cells:
class Chomps(N: Int) {
  require(N > 0 && N.toLong * N.toLong <= Int.MaxValue)
  private val height = Array.fill(N)(N)
  private var remaining = (N.toLong * N.toLong).toInt

  def Count(): Int = remaining

  def Legal(x: Int, y: Int): Boolean =
    0 <= x && x < N && 0 <= y && y < height(x)

  def Move(x: Int, y: Int): Unit = {
    require(Legal(x, y), "illegal move")
    var i = x
    while (i < N && height(i) > y) {
      remaining -= height(i) - y
      height(i) = y
      i += 1
    }
  }

  def Display(): Unit = {
    require(N < 10, "single-digit display labels required")
    var y = N - 1
    while (y >= 0) {
      print(y + "  ")
      var x = 0
      while (x < N) {
        print(if (y < height(x)) "* " else "  ")
        x += 1
      }
      println()
      y -= 1
    }
    print("   ")
    var x = 0
    while (x < N) { print(x + " "); x += 1 }
    println()
  }
}
The strengthened representation invariant includes . The move loop's processed columns have their final truncated heights, unprocessed columns retain their old heights, and each subtraction removes exactly the corresponding height reduction. Because heights are nonincreasing, once a column is no higher than y, neither is any later column; stopping there is valid. A legal move removes at least one token and leaves a order ideal.
The constructor computes in Long and checks representability before converting, avoiding integer overflow in the cached Int count. Legal rejects out-of-range coordinates before indexing. The methods meet the bounds in part (a); Display additionally checks its stated single-digit-label assumption.
Search only the suffix strictly after each candidate first index. This enforces distinct indices even when the target is twice an existing value. Compute the complement in Long to prevent integer overflow: subtracting two Int values can require 33 signed bits.
def twoSum(a: Array[Int], target: Int): (Int, Int) = {
  def lowerBound(from: Int, wanted: Long): Int = {
    var lo = from
    var hi = a.length
    while (lo < hi) {
      val mid = lo + (hi - lo) / 2
      if (a(mid).toLong < wanted) lo = mid + 1
      else hi = mid
    }
    lo
  }
  var i = 0
  while (i < a.length - 1) {
    val wanted = target.toLong - a(i).toLong
    val j = lowerBound(i + 1, wanted)
    if (j < a.length && a(j).toLong == wanted) return (i, j)
    i += 1
  }
  (0, 0)
}
The outer loop invariant says that no solution has a smaller first index than i, when a pair is written in increasing index order. A failed suffix binary search proves there is no partner for the current i, extending the excluded prefix by one. A successful return has and in exact integer arithmetic. At exhaustion, every possible ordered pair has been excluded, so the sentinel (0,0) means no solution. The inner loop invariant is the standard lower-bound partition, restricted to the suffix starting at i + 1.
There are at most suffix searches, each taking running time, so the total is , customarily written for . Auxiliary space is . Empty and singleton arrays immediately return the sentinel. This is distinct-index two-sum by suffix lower bounds; a faster two-pointer technique is also possible on a sorted array, but the code here directly meets the requested search-based bound.