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Joined 2026-09-21 Message
Incoming links: Recursion invariant
Array local maximum 2026-10-31
An entry of a finite array which strictly exceeds each existing immediate neighbour. At either endpoint only one neighbour is tested, and the unique entry of a singleton qualifies vacuously. For distinct entries, a divide-and-conquer algorithm finds such an entry in time. Maintain an index interval whose left endpoint exceeds the excluded entry immediately to its left, if any, and whose right endpoint exceeds the excluded entry immediately to its right, if any. If a midpoint has a larger internal neighbour, recurse towards it; the comparison supplies the new boundary inequality. Otherwise the midpoint exceeds its internal neighbours and the boundary inequalities cover any external neighbour. Halving the interval and preserving this recursion invariant proves correctness and the bound.
For parts (b)–(e), use array indices as tree handles. This also makes the shared storage in part (d) type-correct: both pending handles and output markers occupy integer cells. Zero denotes a leaf of a rooted tree; a positive handle denotes a fork whose children are in Initialize
nodes[t-1]. Only finite acyclic structures are admitted. An allocated fork's children are themselves valid handles.CONST MaxForks = 100000;
TYPE Tree = INTEGER;
Node = RECORD left, right: Tree END;
Arena = ARRAY MaxForks OF Node;
VAR nodes: Arena;
used: INTEGER;
PROCEDURE NewFork(left, right: Tree): Tree;
BEGIN
ASSERT(used < MaxForks);
nodes[used].left := left; nodes[used].right := right;
INC(used);
RETURN used
END NewFork;
PROCEDURE Preorder(t: Tree; VAR out: ARRAY OF INTEGER;
VAR count: INTEGER);
PROCEDURE Visit(p: Tree);
BEGIN
IF p = 0 THEN
out[count] := 0; INC(count)
ELSE
out[count] := 1; INC(count);
Visit(nodes[p-1].left); Visit(nodes[p-1].right)
END
END Visit;
BEGIN
count := 0; Visit(t)
END Preorder;used := 0 before constructing any trees; NewFork allocates within this arena. The fixed capacity is an implementation parameter and may be replaced by a dynamically allocated arena. The returned prefix out[0..count) encodes Boolean values by and . This encoding is part of the chosen representation, not a conversion of a pointer into a Boolean value.The recursion invariant is that
Visit(p) appends exactly the preorder traversal of the subtree represented by p. If the full binary tree has forks, it has leaves; henceEach occurrence is visited once, using linear time complexity and a call stack bounded by the height of a binary tree. The arena is the input representation, not an extra copy made by the traversal.Use half-open rectangles . Split the longer dimension at its middle. If rows are the longer dimension, binary search the middle row for its first entry at least . Equality returns immediately; otherwise part (e) leaves only an upper-right and a lower-left rectangle. A middle-column split uses the transposed argument.The lower-bound binary search invariant says that entries before
PROCEDURE SearchRect(x, r0, r1, c0, c1: INTEGER;
VAR i, j: INTEGER): BOOLEAN;
VAR mid, lo, hi, k: INTEGER;
BEGIN
IF (r0 >= r1) OR (c0 >= c1) THEN RETURN FALSE END;
IF r1-r0 >= c1-c0 THEN
mid := r0 + (r1-r0) DIV 2;
lo := c0; hi := c1;
WHILE lo < hi DO
k := lo + (hi-lo) DIV 2;
IF f[mid,k] < x THEN lo := k+1 ELSE hi := k END
END;
IF lo < c1 THEN
IF f[mid,lo] = x THEN i := mid; j := lo; RETURN TRUE END
END;
IF SearchRect(x, r0, mid, lo, c1, i, j) THEN
RETURN TRUE
END;
RETURN SearchRect(x, mid+1, r1, c0, lo, i, j)
ELSE
mid := c0 + (c1-c0) DIV 2;
lo := r0; hi := r1;
WHILE lo < hi DO
k := lo + (hi-lo) DIV 2;
IF f[k,mid] < x THEN lo := k+1 ELSE hi := k END
END;
IF lo < r1 THEN
IF f[lo,mid] = x THEN i := lo; j := mid; RETURN TRUE END
END;
IF SearchRect(x, r0, lo, mid+1, c1, i, j) THEN
RETURN TRUE
END;
RETURN SearchRect(x, lo, r1, c0, mid, i, j)
END
END SearchRect;
PROCEDURE DivideSearch(x: INTEGER; VAR i, j: INTEGER): BOOLEAN;
BEGIN
RETURN SearchRect(x, 0, M, 0, N, i, j)
END DivideSearch;lo are smaller than and entries from hi onward are at least , within the splitting row or column. At its end, either equality is found or the splitting line is strictly separated around . Endpoint cuts lo=c0, lo=c1, lo=r0, or lo=r1 are valid and merely make one candidate rectangle empty.The recursion invariant is that the current rectangle contains every still-possible answer assigned to that call. The discarded rectangles are excluded by monotonicity; both retained rectangles inherit monotonicity. Each nonempty child has a strictly smaller longest dimension after at most two levels, so the recursion terminates, including one-row and one-column cases. The concise construction is
Two complementary rectangles survive a middle-row search
. For a monotone matrix, the middle row's binary search separates entries smaller than from entries greater than when no equality is found. The upper-left and lower-right rectangles are excluded by monotonicity; only the blue and orange rectangles require further search algorithm. Details.
Use a programming closure whose private variable holds the next Fibonacci number. Use arbitrary-precision integers to enumerate all terms exactly:At the cell containing , the private state immediately before computing its successor is
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)
}following = F_(i+1). The successor returns that value and sets following to , preserving the recursion invariant. The resulting sequence startsEach 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.
Maintain an inclusive interval with these recursion invariants: if , then ; if , then . The original full interval satisfies both conditions vacuously.In the left-recursion branch, the new right endpoint satisfies and the old left boundary is retained. The right-recursion branch is symmetric. Thus both recursion invariants persist.
LocalMaximum(A):
require length(A) >= 1
return Peak(0, length(A)-1)
Peak(lo, hi):
m = floor((lo+hi)/2)
if m > lo and A[m-1] > A[m]:
return Peak(lo, m-1)
if m < hi and A[m+1] > A[m]:
return Peak(m+1, hi)
return mIf neither branch is taken, exceeds its neighbours within the current interval because all entries are distinct. If is an endpoint of that interval and there is a neighbour outside, the boundary recursion invariant supplies the missing strict inequality. Hence the returned index is an array local maximum with respect to the original array, not just the restricted interval. Every branch shortens the interval, so termination is guaranteed; this also establishes existence, including the singleton case.
Each call performs at most two comparisons, and a recursive interval has at most half as many elements, up to rounding. Thus andThe conventional bound is for , with constant work for a singleton. Distinctness is essential to this proof of a strict local maximum.
Write parts in nonincreasing order. The array The recursion invariant is that
part stores the current prefix, and Emit(i,r,len) completes it using parts at most with remaining sum . With Out as the Oberon output module:VAR part: ARRAY MaxParts OF INTEGER;
PROCEDURE Emit(i, r, len: INTEGER);
VAR j: INTEGER;
BEGIN
IF r = 0 THEN
j := 0;
WHILE j < len DO
IF j > 0 THEN Out.Char("+") END;
Out.Int(part[j], 0); j := j+1
END;
Out.Ln
ELSIF i > 0 THEN
IF i <= r THEN
part[len] := i;
Emit(i, r-i, len+1)
END;
Emit(i-1, r, len)
END
END Emit;
PROCEDURE PrintPartitions(n: INTEGER);
BEGIN
Emit(n, n, 0)
END PrintPartitions;part[0..len) is nonincreasing, has sum , and every further part must be at most and at most the last existing part. The first branch includes the next part ; the second excludes it. These alternatives partition all completions, so each integer partition is printed exactly once. A recursive call reduces ; the printing loop's loop invariant is that the first entries have been printed with exactly the necessary separators.For , no partition has more than parts, and the all-ones partition has exactly . Thus the fixed array requires to handle every partition. When a write occurs, and therefore
len < n, so this bound prevents an out-of-range write. Zero prints one empty line, representing the empty integer partition. Available call stack space imposes a further practical limit.Perform a depth-first traversal, retaining the path's letters in a buffer. A terminal node must be printed before its children, so a word precedes its proper extensions in lexicographic order.The recursion invariant is that
VAR word: ARRAY N OF CHAR;
PROCEDURE Visit(t: Trie; depth: INTEGER);
VAR j, c: INTEGER;
BEGIN
IF t # NIL THEN
IF t.terminal THEN
j := 0;
WHILE j < depth DO
Out.Char(word[j]); j := j+1
END;
Out.Ln
END;
c := 0;
WHILE c < 26 DO
IF t.child[c] # NIL THEN
word[depth] := CHR(ORD("a")+c);
Visit(t.child[c], depth+1)
END;
c := c+1
END
END
END Visit;
PROCEDURE PrintTrie(t: Trie);
BEGIN
Visit(t, 0)
END PrintTrie;word[0..depth) spells the path to t. The child loop's loop invariant says children with indices below c have been completely printed. Each edge appends its lowercase label; each terminal flag prints exactly one represented string. All words therefore appear once, in lowercase. The output loop prints precisely the prefix of length depth; no terminating zero needs storing in this buffer.An
N-character buffer suffices for words of length at most N. At depth N there can be no child, so word[N] is never accessed. A terminal root prints an empty line; a NIL root prints nothing. The running time is linear in allocated nodes plus the number of printed characters, since the alphabet size is fixed at 26.Let be the number of leaves; the total number of nodes is . List concatenation copies its left operand. For a tree whose left subtree contains leaves at each successive internal node, these copied lengths are . Their sum is , so the worst-case running time is .
A difference list representation avoids copying prefixes:
flattenFast :: Tree a -> [a]
flattenFast t = go t []
where
go (One x) acc = x:acc
go (Two l r) acc = go l (go r acc)The recursion invariant is
go t acc = flatten t ++ acc. The leaf case is immediate; the internal-node case uses associativity of list concatenation. Each node is visited once and each leaf contributes one output constructor, giving running time when the result is fully consumed. This is linear flattening of a binary tree and does not assume balance.Assume the subtrees rooted at the children in a rooted tree of already obey the max-heap condition. Compare with each present child and choose an index of the largest of these at most three keys. If that index is , stop. Otherwise exchange the two keys, continue at the chosen child, and repeat. This is sift down, with the inequality reversed from a minimum-oriented binary heap.
The exchange repairs the old root position because its new key dominates both child roots; only the subtree receiving the displaced key can still be defective. This recursion invariant proves correctness. Each exchange descends one level and costs constant work, so the worst-case running time is for subtree height , hence . A small key at the root of a suitably ordered heap must descend to a leaf, so the logarithmic worst-case bound is attained.
Change the right fold's result type to an accumulator-transforming function type:Thus
foldlAsFoldr :: (b -> a -> b) -> b -> [a] -> b
foldlAsFoldr s n xs = foldr c id xs n
where
c x k a = k (s a x)c x k = \a -> k (s a x) and k = id in the question's notation. Here c :: a -> (b -> b) -> (b -> b), and the fold's base value is the identity function, not the initial accumulator n.The recursion invariant is At an undefined spine, applying the undefined folded function also gives the bottom value. The finite-approximation argument then covers infinite lazy lists. This is a continuation representation of a left fold: the result retains the operation still to be performed, avoiding the information loss in part (d).
foldr c id xs a = foldl s a xs for every a. For [], both sides equal a. For a cons:foldr c id (x:xs) a
= c x (foldr c id xs) a
= foldr c id xs (s a x)
= foldl s (s a x) xs
= foldl s a (x:xs).To preserve the actual printed function, including its handling of nonpositive exponents and wrapping
Int bases, use a tail-recursive accumulator with the same multiplication order:def anotherfuncRec(x: Int, n: Long): Double = {
require(x > 0)
@scala.annotation.tailrec
def loop(y: Int, i: Long, result: Double): Double = {
if (i <= 0) result
else {
val nextResult = if (i % 2 == 1) result * y else result
loop(y * y, i / 2, nextResult)
}
}
loop(x, n, 1.0)
}Each recursive call corresponds to one source iteration and carries exactly its next program state. Hence it reproduces the printed numerical behavior, rather than silently repairing it. For positive
n, both routines use steps; for nonpositive n, both take constant time. The recursive call is in tail position, so Scala's tail-call optimization avoids a growing call stack; the annotation asks the compiler to check that condition.If the intended exact power is required instead, change the base and accumulator parameters of this same recurrence to
BigInt and require n >= 0, as in part (d). Its mathematical loop invariant becomes a recursion invariant, with the same logarithmic number of multiplication steps. Recursion invariant 2026-10-24
A relationship between a recursive procedure's arguments, accumulated state and original input that holds at every reachable recursive call. For accumulator-based exponentiation by squaring, is a recursion invariant under exact arithmetic. The base case converts it into the desired result, while the recursive step and a decreasing argument justify preservation and termination.