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Incoming links: Loop invariant
Set . At each iteration with , compute the remainder , then setStop at the first index with and return . The positive second coordinates strictly decrease, so algorithm termination follows from the well-ordering principle. The equality of greatest common divisors established in (ii) is a loop invariant, and at termination it gives
Use an explicit stack of handles. Push the right subtree first, so that the left subtree is removed first by the LIFO stack.The loop invariant says that the already written prefix, followed by the preorder traversals of the pending subtrees in top-to-bottom stack order, equals the entire required output. Popping a leaf of a rooted tree transfers one marker to the prefix. Popping a fork transfers its marker and replaces its remaining work by left then right subtree work. Both preserve the loop invariant. The number of as-yet-unwritten vertices decreases by one per iteration, proving termination.The time complexity is linear in the output length. The maximum stack size is at most one plus the height of a binary tree.
PROCEDURE PreorderIter(t: Tree; VAR out: ARRAY OF INTEGER;
VAR count: INTEGER;
VAR stack: ARRAY OF Tree);
VAR top: INTEGER; p: Tree;
BEGIN
count := 0; top := 1; stack[0] := t;
WHILE top > 0 DO
DEC(top); p := stack[top];
IF p = 0 THEN
out[count] := 0; INC(count)
ELSE
out[count] := 1; INC(count);
stack[top] := nodes[p-1].right; INC(top);
stack[top] := nodes[p-1].left; INC(top)
END
END
END PreorderIter;Let be the final output length, the number of emitted markers, and the number of pending handles. Each pending subtree contributes at least one future marker, soStore output from the low end and the stack from the high end of the same array. A pop frees one cell before the next write. For a fork, the two pushed children each account for at least one of the remaining markers. Thus the two regions never overlap.The loop invariant includes
PROCEDURE PreorderShared(t: Tree; VAR out: ARRAY OF INTEGER;
VAR count: INTEGER);
VAR top, cap: INTEGER; p: Tree;
BEGIN
cap := LEN(out); count := 0;
top := cap-1; out[top] := t;
WHILE top < cap DO
p := out[top]; INC(top);
IF p = 0 THEN
out[count] := 0; INC(count)
ELSE
out[count] := 1; INC(count);
DEC(top); out[top] := nodes[p-1].right;
DEC(top); out[top] := nodes[p-1].left
END
END
END PreorderShared;count <= top, with pending handles in out[top..cap) and final markers in out[0..count). The preceding accounting proves every intermediate write also respects array bounds. This uses the integer handle representation chosen in part (b). Simply putting a pointer into a genuinely Boolean-typed cell would not be valid Oberon.Implement staircase matrix search from the top-right corner. In the following Oberon-style code, The loop invariant is that every occurrence not yet returned lies inInitially this is the whole array. If , every entry of row in the candidate rectangle is smaller than , so increasing discards no possible answer. If , every entry of column in that rectangle is greater than , so decreasing is safe. These conclusions follow from monotonicity even when repeated values occur.
& denotes short-circuit conjunction; all guarded accesses respect array bounds.PROCEDURE Staircase(x: INTEGER; VAR i, j: INTEGER): BOOLEAN;
VAR r, c: INTEGER;
BEGIN
r := 0; c := N-1;
WHILE (r < M) & (c >= 0) DO
IF f[r,c] = x THEN i := r; j := c; RETURN TRUE
ELSIF f[r,c] < x THEN INC(r)
ELSE DEC(c)
END
END;
RETURN FALSE
END Staircase;The loop variant decreases on every unsuccessful iteration. If a dimension becomes empty, the loop invariant proves absence. For positive dimensions the path inspects at most entries; an initially empty dimension causes no inspection. Thus
Assume The loop invariant says that the segment from
less is a lawful strict comparison, for example a strict weak order, and the two input linked lists are disjoint and sorted by it. Allocate a result header and move existing data linked list nodes; do not copy their values.PROCEDURE Merge(less: LESS; q, r: List; VAR p: List);
VAR result: List; last, a, b: Tail;
BEGIN
NEW(result); result.count := q.count + r.count;
result.tail := NIL; last := result;
a := q.tail; b := r.tail;
WHILE (a # NIL) & (b # NIL) DO
IF less(b,a) THEN
last.tail := b; b := b.tail
ELSE
last.tail := a; a := a.tail
END;
last := last.tail
END;
IF a # NIL THEN last.tail := a ELSE last.tail := b END;
q.tail := NIL; q.count := 0;
r.tail := NIL; r.count := 0;
p := result
END Merge;result.tail through last is a sorted prefix of the desired merge algorithm, containing exactly the nodes already removed from the two unread chains. Every element of this prefix is no greater than either unread head. Choosing the smaller unread head maintains this property; choosing from q on a tie also gives stable sorting.Allocate a fresh coefficient array, add matching polynomial coefficients, and normalize to remove any cancellation of leading terms.The loop invariant says that every completed coefficient equals the corresponding coefficient of , treating missing coefficients as zero.
PROCEDURE Add(VAR sum: Poly; a, b: Poly);
VAR result: Poly; d, cap, i: INTEGER;
BEGIN
d := a.degree;
IF b.degree > d THEN d := b.degree END;
cap := d+1; IF cap < 1 THEN cap := 1 END;
result.degree := d; NEW(result.coeff, cap);
result.coeff[0] := 0.0;
FOR i := 0 TO d DO
result.coeff[i] := 0.0;
IF i <= a.degree THEN
result.coeff[i] := result.coeff[i] + a.coeff[i]
END;
IF i <= b.degree THEN
result.coeff[i] := result.coeff[i] + b.coeff[i]
END
END;
Normalize(result); sum := result
END Add;Normalize changes no represented polynomial, and can reduce the result to the zero polynomial.Neither input coefficient array is changed. Output variables are separate from input record variables; the fresh result also remains safe when the two inputs share a coefficient array.Require a nonzero divisor. Polynomial long division repeatedly subtracts the multiple of that cancels the leading term of the current polynomial remainder. Use fresh arrays for both outputs.The loop invariant is the polynomial identity . Setting and replaces by and by , preserving this identity. The leading coefficient of that difference is exactly zero, so setting its stored value to zero and decreasing
PROCEDURE DivMod(VAR div, mod: Poly; a, b: Poly);
VAR q, r: Poly; cap, i, j, k, old: INTEGER; t: REAL;
BEGIN
ASSERT(b.degree >= 0);
cap := a.degree+1; IF cap < 1 THEN cap := 1 END;
r.degree := a.degree; NEW(r.coeff, cap);
r.coeff[0] := 0.0;
FOR i := 0 TO a.degree DO r.coeff[i] := a.coeff[i] END;
IF a.degree < b.degree THEN
Zero(q)
ELSE
q.degree := a.degree-b.degree;
NEW(q.coeff, q.degree+1);
FOR i := 0 TO q.degree DO q.coeff[i] := 0.0 END;
WHILE r.degree >= b.degree DO
old := r.degree; k := old-b.degree;
t := r.coeff[old]/b.coeff[b.degree];
q.coeff[k] := q.coeff[k]+t;
FOR j := 0 TO b.degree-1 DO
r.coeff[k+j] := r.coeff[k+j]-t*b.coeff[j]
END;
r.coeff[old] := 0.0;
DEC(r.degree); Normalize(r)
END;
Normalize(q)
END;
div := q; mod := r
END DivMod;degree implements its exact cancellation. Each iteration strictly decreases the degree of the polynomial remainder.At termination,The zero polynomial dividend, a lower-degree dividend, and a constant divisor are all covered. Division by the zero polynomial is undefined. The two output record variables must be distinct and separate from the input record variables; no input coefficient array is written. For , the work is .
Use destructive inorder traversal by right rotations. A right tree rotation preserves inorder traversal but moves the first unvisited node closer to the root:The outer loop invariant states that the labels already passed to
PROCEDURE InorderDestructive(t: Tree; p: Visit);
VAR q: Tree;
BEGIN
WHILE t # NIL DO
WHILE t.left # NIL DO
q := t.left;
t.left := q.right;
q.right := t;
t := q
END;
p(t.label);
t := t.right
END
END InorderDestructive;p, followed by the inorder traversal of t, equal the original inorder traversal. The remaining nodes form a binary tree disjoint from the nodes already visited. The inner loop invariant preserves the remaining inorder traversal: if the old root is , its left child is , and the three unchanged subtrees are , then both shapes have order .
At inner-loop exit,
t.left = NIL, so the root is the first remaining node. Visiting it and replacing t by t.right preserves the outer loop invariant. The procedure changes child pointers and makes the old root reference unusable as the original binary tree. Assume p neither changes the binary tree nor depends on its old shape.Track the length The loop invariant is: ;
r of the trailing constant segment of an array, and the best length m anywhere in the processed prefix:i := 1; r := 1; m := 1;
WHILE i < n DO
IF a[i] = a[i-1] THEN r := r+1 ELSE r := 1 END;
IF r > m THEN m := r END;
i := i+1
END;r is the length of the longest constant suffix of ; and m is the length of the longest constant segment of an array within . Initially all three lengths are one, using .When the next entry equals its predecessor, the trailing constant segment extends by one; otherwise its new length is one. Every new plateau either was already in the old prefix or ends at the new entry, so taking the larger of the old
m and the new r preserves the loop invariant. The loop variant is , a nonnegative integer decreased by one. At termination , soThe auxiliary space is , and correctly returns one without entering the loop.Use binary search on an interval with unequal endpoint values. The array need not be sorted:The loop invariant isIt holds initially by the given endpoint inequality, which also implies . When , . If , replacing maintains unequal endpoints. Otherwise , so replacing maintains them. This does not require monotonicity or even just two possible values.
l := 0; r := n;
WHILE r-l > 1 DO
m := l + (r-l) DIV 2;
IF b[l] # b[m] THEN r := m ELSE l := m END
END;
k := l;The positive integer width is a loop variant, reduced to at most its old half rounded up. On exit , givingThere are at most iterations. The additive constant handles .
Return the exclusive endpoint of the maximal initial run:The loop invariant is and is nondecreasing. An unsuccessful adjacent-order test extends the run by one; a strict descent at prevents any extension across it. If , the whole segment is a run. The loop variant is .For a run of length , the running time is , including at most one failed extension test; an empty segment takes .
PROCEDURE Run(a, b: INTEGER): INTEGER;
VAR c: INTEGER;
BEGIN
IF a = b THEN RETURN b END;
c := a+1;
WHILE c < b DO
IF u[c-1] > u[c] THEN RETURN c END;
c := c+1
END;
RETURN c
END Run;Use a merge algorithm with one unread index in each run. Here Let denote the original two runs. The first loop invariant says and is the sorted merge of and ; moreover each emitted value is at most every unread value. Choosing the smaller head preserves that condition. Equal heads are taken from the first run, giving stable sorting within this merge. The loop variant is .
u is the module's array; the destination is disjoint from the source and lies within the array bounds:PROCEDURE Merge(a, b, c, d: INTEGER);
VAR i, j, k: INTEGER;
BEGIN
i := a; j := b; k := d;
WHILE (i < b) & (j < c) DO
IF u[i] <= u[j] THEN
u[k] := u[i]; i := i+1
ELSE
u[k] := u[j]; j := j+1
END;
k := k+1
END;
WHILE i < b DO
u[k] := u[i]; i := i+1; k := k+1
END;
WHILE j < c DO
u[k] := u[j]; j := j+1; k := k+1
END
END Merge;For each copying loop, the output remains the sorted merge of the consumed prefixes, at least one source run is exhausted, and only the indicated suffix remains. The respective loop variants are and . At exit both source runs are exhausted:The destination's multiset is the sum of the two source multisets.
Let be the original sequence and . The loop invariant isHere denotes the multiset of entries. The capacity condition uses the question's assumption of sufficient space. Removing two runs of combined length and appending their merge preserves the length and multiset. It also leaves all live values in one contiguous segment.
For a nonempty live sequence, let be its number of maximal initial runs obtained by repeatedly cutting at strict adjacent descents. The loop variant is . If
x = d, there is one run and the procedure returns a sorted permutation. Otherwise the iteration removes two runs and appends one; it cannot create a new descent inside that appended run. The appended run may join the last old run, further reducing , so decreases by at least one. There are at most merges.For , the loop is skipped and the empty segment is already sorted. No global stability claim is needed: moving a merged run behind other runs can change the order of equal-labelled occurrences.If the merge algorithm from part (b) is retained with a disjoint destination, use a circular buffer of cells. Represent the live segment by its start modulo and its length . The other cells are free; the next merge needs at most output cells. Appending its output and advancing the live start releases exactly as many obsolete source cells. Logical run boundaries and copying indices are reduced modulo . The final sorted sequence may wrap, in which case an array rotation makes it contiguous. ThusThis is also the worst-case requirement if all source cells must be retained until a whole disjoint merge has been written: the final two runs can have combined length .
Under the question's literal wording, however, no running time requirement or disjoint-output restriction is imposed on the modified implementation. The absolute smallest array isTo achieve this, replace the disjoint-output merge by an in-place merge: compare the next unread values of the two adjacent runs; when the second is smaller, save it in one scalar, shift the intervening first-run block right by one, and insert it before that block. This is a sorted merge using no extra array. Its loop invariant is that the finished prefix is the sorted merge of the consumed portions and the two remaining runs stay sorted. Each step consumes one value, proving termination.
After merging the first two runs in place, rotate the merged prefix behind the untouched suffix. Three array reversals implement this rotation with one scalar: reverse the merged prefix, reverse the suffix, then reverse the whole segment. The resulting live sequence is exactly the one produced in part (c). Repeat the same run schedule until sorted. The shift and reversal loops respectively preserve the copied block and exchanged end pairs; their loop variants are the unshifted length and half the unreversed length. The modification can be substantially slower, but preserves the loop invariant and decreasing run count from part (d). Fewer than cells cannot even hold the specified input entries. Thus is the natural circular-workspace answer for the original merge primitive, while is attainable when minimizing storage over all implementations allowed by the stated time-free specification.
Keep For insertion, conceptually treat index
u[0..len) sorted, with the coupling invariantThe abstract list retains insertion order, while the concrete array retains its sorted multiset. Initially both are empty.PROCEDURE Add(x: INTEGER);
VAR i: INTEGER;
BEGIN
ASSERT(len < N);
i := len;
WHILE i > 0 DO
IF u[i-1] <= x THEN EXIT END;
u[i] := u[i-1];
i := i-1
END;
u[i] := x;
len := len+1
END Add;
PROCEDURE Get(k: INTEGER): INTEGER;
BEGIN
ASSERT((0 <= k) & (k < len));
RETURN u[k]
END Get;i as a hole. If is the old sorted prefix of length , the loop invariant is , and , with every shifted value strictly greater than . The loop variant is . The exit condition gives either no left neighbour or a left neighbour at most ; filling the hole establishes sortedness and adds exactly one occurrence, preserving the coupling invariant.When the new entry is smaller than every old entry, all old values must move in this contiguous sorted-prefix representation. Binary search can reduce comparisons to locate the insertion point, but cannot remove that linear movement cost. Appending a new largest value takes constant time in this implementation.The original page permits , but also indexes . The meaningful pivot-position contract is the stronger , with and . Implement this two-way array partition by reserving the last entry as pivot:The loop invariant isThe segment's multiset is preserved and entries outside are unchanged. The unclassified part is . A smaller entry is swapped into the left group; if , the displaced entry belongs to the greater-or-equal group. Otherwise the entry simply joins that group. The loop variant is .
PROCEDURE Partition(l, r: INTEGER): INTEGER;
VAR pivot, i, j, tmp: INTEGER;
BEGIN
ASSERT(l < r);
pivot := a[r-1];
i := l; j := l;
WHILE j < r-1 DO
IF a[j] < pivot THEN
tmp := a[i]; a[i] := a[j]; a[j] := tmp;
i := i+1
END;
j := j+1
END;
tmp := a[i]; a[i] := a[r-1]; a[r-1] := tmp;
RETURN i
END Partition;Store the current multiset in an unsorted prefix The loop invariant is , the complete array has its original multiset, every discarded entry on the left is at most every active entry, and every active entry is at most every discarded entry on the right. Thus the desired global order statistic is the local one of rank . The two-way array partition result either identifies it or excludes the pivot and the irrelevant side. The positive loop variant strictly decreases unless the procedure returns.
a[0..len); Add appends exactly as in part (a). The coupling invariant is equality of length and multiset with abstract xs, rather than equality of insertion order. Get applies quickselect:PROCEDURE Get(k: INTEGER): INTEGER;
VAR l, r, m: INTEGER;
BEGIN
ASSERT((0 <= k) & (k < len));
l := 0; r := len;
LOOP
m := Partition(l, r);
IF k = m THEN RETURN a[m]
ELSIF k < m THEN r := m
ELSE l := m+1
END
END
END Get;Balanced pivots give , whose geometric series is . More generally, choosing a uniformly random pivot before each call, by swapping a random active entry into
a[r-1], gives expected running time on distinct keys: with constant probability, the pivot is in the middle half and the retained side has size at most . Expected work over these shrinking stages is a convergent geometric series.The particular last-entry pivot in the displayed code has no unconditional expected guarantee without an input distribution. Its worst case repeatedly discards one entry. Duplicate keys require further care: on an all-equal segment, the strict partition always returns , so selecting the last rank costs even with randomized pivot positions. A three-way array partitioning that groups all keys equal to the pivot restores the usual randomized linear expectation with duplicates; alternatively the stated expectation assumes distinct keys. This is a qualification to the bound, not to the algorithm correctness of the displayed code.The first matching entry must be retained, and copying an already-started prefix-dropping iterator must not restart the search. Use a private initial buffer together with the underlying iterator snapshot:Before preparation, the iterator still represents the original remaining sequence. During the search, the loop invariant says that every discarded entry fails the predicate, and
class DropUntil[T] private (
p: T => Boolean, iter: Iter[T],
saved: Option[T], alreadyStarted: Boolean) extends Iter[T] {
def this(p: T => Boolean, iter: Iter[T]) =
this(p, iter, None, false)
private var pending: Option[T] = saved
private var started: Boolean = alreadyStarted
private def prepare(): Unit = {
if (!started) {
var found = false
while (!found && iter.hasNext) {
val x = iter.next()
if (p(x)) {
pending = Some(x)
found = true
}
}
started = true
}
}
def hasNext: Boolean = {
prepare()
pending.nonEmpty || iter.hasNext
}
def next(): T = {
prepare()
pending match {
case Some(x) => pending = None; x
case None =>
if (!iter.hasNext) throw new NoSuchElementException("DropUntil.next")
iter.next()
}
}
def copy(): Iter[T] =
new DropUntil[T](p, iter.copy(), pending, started)
}pending is either empty or contains the first satisfying entry just consumed. For a finite input, the number of unread entries is a loop variant. After preparation, the abstract remaining sequence is pending.toList followed by the underlying iterator's remaining sequence. Repeated hasNext calls do not discard or test more entries after the first match.The iterator snapshot copies both the cached entry and the started flag. In particular, copying after the matching entry has been returned continues with the remaining suffix, including later entries that fail
p. Restarting new DropUntil(p, iter.copy()) at that point would wrongly drop another prefix.As usual, assume a deterministic predicate without shared mutable effects. On an infinite input having no match, the first demand cannot terminate: no finite inspection establishes absence. Independence here concerns cursor state; arbitrary mutation of shared element objects is outside the iterator contract.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.Use unsigned inputs as in the preceding parts. Store a 16-bit accumulator , a 16-bit shifted multiplicand , an eight-bit multiplier , and a three-bit iteration binary counter . The sequential circuit has one 16-bit adder, a 16-bit multiplexer selecting or zero from , fixed one-place bit shifts, and load/hold selection for the digital registers. The input values are captured when
start is accepted, so later input changes cannot alter the operation.
Use three control states with the following synchronous transition table. All assignments in a row use the old digital register values and occur together at the active clock edge.
| State | Condition | Register updates | Next state | IDLE | start = 0 | Hold all digital registers | IDLE | IDLE | start = 1 | , , , | RUN | RUN | , , , | RUN | RUN | , , ; hold | DONE | DONE | start = 1 | Hold all digital registers | DONE | DONE | start = 0 | Hold all digital registers | IDLE |
|---|
Reset selects IDLE and clears the digital registers. Set and
done = 1 exactly in DONE. A start pulse in IDLE launches one multiplication; further pulses during RUN are ignored. Holding start high cannot repeatedly launch operations: DONE waits for it to go low before rearming. The product remains stored while waiting and through the following IDLE state. There are eight RUN clock edges after the loading edge; the last addition is captured before done becomes high.For correctness, let denote the captured values. After iterations, for , the loop invariant isso . Writing , the next accumulation adds , the left bit shift doubles , and the logical right shift changes to . Hence the invariant is preserved. At , , giving . The largest product is , so the accumulator fits. Even the final shifted multiplicand is , so it also fits.