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Incoming links: Time complexity
A divide-and-conquer algorithm divides an input into smaller instances, solves those instances recursively, and combines their answers. Directly solvable base cases terminate the recursion. The total time complexity includes both the recursive work and the work of dividing and combining; balanced sizes alone do not determine the running time.
Maintain one array index per input and a candidate value. Cycle through the arrays, counting consecutive agreements with that candidate; when a larger head is encountered, start counting agreements again. The loop invariant has three parts. All indices satisfy ; no discarded entry is common to every input; and every original common value is at least
p is an auxiliary array of indices, not a copy of any input.PROCEDURE commonMany(as, n, m);
VAR p: ARRAY m OF INTEGER;
j, candidate, agree, x: INTEGER;
BEGIN
FOR j := 0 TO m-1 DO
p[j] := 0;
IF n[j] = 0 THEN RETURN None END
END;
candidate := as[0][0];
j := 0; agree := 0;
WHILE agree < m DO
IF p[j] = n[j] THEN RETURN None END;
x := as[j][p[j]];
IF x < candidate THEN
p[j] := p[j] + 1
ELSIF x = candidate THEN
agree := agree + 1;
j := (j + 1) MOD m
ELSE
candidate := x;
agree := 1;
j := (j + 1) MOD m
END
END;
RETURN Some(candidate)
END commonMany;candidate. Additionally, the agree cyclically consecutive arrays immediately preceding index j have heads equal to candidate. These facts are true initially, with an empty agreement run.A head below
candidate is not globally common, by the lower-bound clause, so advancing its pointer is safe. A head above candidate gives a new lower bound: every globally common value lies in this array's unprocessed suffix and so is at least its head. Only this array is yet certified to agree with the new candidate. An equal head extends the run by one. An exhausted suffix proves absence; a run of agreements covers every array exactly once and proves that candidate is common, and the lower-bound clause proves minimality. The exhaustion test precedes each dereference, including after the last entry of a suffix is discarded.For algorithm termination and time complexity, charge comparisons to the pair
(j,p[j]). A fixed entry is discarded at most once, can raise the candidate at most once, and can be counted as an agreement at most once: before the same array is revisited, either the algorithm has finished or the candidate has strictly increased. Thus there are at most three comparisons per input entry, besides the initialization. The result is the smallest common value or None in time and auxiliary space. For , the first head immediately gives the answer.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.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.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.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]Perform a destructive binary search tree flattening by converting both child subtrees and splicing their endpoints around the original root:Save the child references before changing their interpretation. The input is a finite binary search tree satisfying part (b), and no external client may simultaneously rely on the old tree links. By structural induction, the two recursive calls return disjoint, ordered doubly linked lists containing exactly their original entries. The binary search tree bounds place every left word before
def flatten(t: Tree): (Tree, Tree) = {
if (t == null) (null, null)
else {
val oldLeft = t.left
val oldRight = t.right
val (ls, le) = flatten(oldLeft)
val (rs, re) = flatten(oldRight)
t.left = le
if (le != null) le.right = t
t.right = rs
if (rs != null) rs.left = t
val start = if (ls == null) t else ls
val end = if (re == null) t else re
(start, end)
}
}t.word and every right word after it. Splicing the left end, t, and the right start therefore preserves strict order and reciprocal links. A missing left list makes t.left null and t the start; a missing right list similarly makes t the end. The resulting outside links are null as required by part (i).The recursive loop variant analogue is the strictly smaller number of entries in each child subtree. Every entry is processed once and incurs a constant number of link writes, so time complexity is and recursion uses auxiliary space. No new
Tree entry is created; returning endpoint tuples and using a call stack are permitted auxiliary storage. The old binary search tree representation is consumed by this operation.Use pattern matching and an optional value to distinguish absence from a present result:The representation precondition is that equal keys always have equal associated values, with a stable equality relation on keys. This permits harmless repeated identical pairs; it excludes conflicting outputs for one input.
class PartialFn[T, U](val data: List[(T, U)]) {
def get(x: T): Option[U] =
data.find { case (key, _) => key == x }.map(_._2)
def remove(x: T): PartialFn[T, U] =
new PartialFn[T, U](data.filterNot { case (key, _) => key == x })
def add(x: T, y: U): PartialFn[T, U] =
new PartialFn[T, U]((x, y) ::
data.filterNot { case (key, _) => key == x })
}get therefore returns Some(y) exactly when the represented partial function is defined at x. Removing filters out every occurrence of the key and leaves all other pairs unchanged. Adding first removes every old occurrence and prepends exactly the new pair, proving the partial function update postcondition even if the input contains duplicate identical pairs. The immutable linked list representation leaves the original mapping unchanged. Each operation has worst-case time complexity for pairs under constant-cost key equality.The second partial function accepts this one's outputs. Put this method inside the invariant class in part (b):The represented mapping is , since the paper explicitly applies
def compose[V](that: PartialFn[U, V]): PartialFn[T, V] = {
val pairs: List[(T, V)] = data.flatMap {
case (x, y) => that.get(y).map(z => (x, z)).toList
}
new PartialFn[T, V](pairs)
}that second despite the method name. A pair is emitted exactly when this mapping supplies a y and that.get(y) supplies a z. Hence its domain of a partial function is and its value there is . This is strict composition of partial functions: either missing stage makes the composite undefined. Repeated equal input pairs still produce the same output, preserving the representation invariant. With and pairs, direct association-list lookup gives worst-case time complexity, not a hidden constant-cost dictionary lookup.Represent the characteristic values of a bounded universe by one Boolean slot per possible member, a Boolean-array form of a bitset:The abstraction function is . The representation invariant is that
class BitMapSet(val MAXVALUE: Int) extends IntSet {
require(MAXVALUE >= 0)
private val bits = Array.fill(MAXVALUE)(false)
private var count = 0
def contains(elem: Int): Boolean =
0 <= elem && elem < MAXVALUE && bits(elem)
def add(elem: Int): Boolean = {
require(0 <= elem && elem < MAXVALUE, "Outside bitmap universe")
if (bits(elem)) false
else {
bits(elem) = true
count += 1
true
}
}
def size: Int = count
}bits.length==MAXVALUE, its entries are Boolean, and count equals the number of true entries. Initialization gives the empty set; changing a false entry to true adds exactly its index and increments the count once. A repeated addition preserves both. All three operations have time complexity after initialization; storage and explicit array initialization are .The constructor requires , and
add requires . contains remains total by returning false outside the representable universe; short-circuit evaluation protects the index. This finite representation cannot implement unrestricted addition of every Int while preserving the abstract contract. Its strengthened add precondition therefore makes it a restricted implementation, not a behaviorally substitutable implementation of the unrestricted trait for arbitrary clients. The sequence client below proves that it stays within the restricted universe.Let The loop invariant is , every entry of is less than , every entry of is at least , and the segment's multiset is unchanged. Entries outside are unchanged. When
a be the given enclosing array. A two-way array partition needs only one scan, and does not require the pivot to occur in the array:def partition(l: Int, r: Int, p: Int): Int = {
require(0 <= l && l <= r && r <= a.length)
var k = l
var j = l
while (j < r) {
if (a(j) < p) {
val t = a(k)
a(k) = a(j)
a(j) = t
k += 1
}
j += 1
}
k
}a(j)<p, swapping it into position k extends the low part; if k<j, the displaced entry was in the high part and remains there after j advances. If k==j, the self-swap is harmless. The other branch simply extends the high part. The loop variant is . At exit , so the returned has the required separation property, including empty segments and arbitrarily many duplicate entries. The time complexity is and auxiliary space complexity is .Keep a deficient interval of possible values and only the array segment containing occurrences from that interval. This is missing-value search by deficient value intervals:The precondition is that every original entry lies in . The loop invariant istogether with: every active entry lies in , and every original occurrence with a value in that interval is in the active segment. The whole array remains a permutation of its initial contents. Initially these assertions follow from the precondition and .
def missing(): Int = {
val n = a.length
var lo = 0L
var hi = n.toLong + 1L
var l = 0
var r = n
while (hi - lo > 1L) {
val p = lo + (hi - lo) / 2L
val k = partition(l, r, p.toInt)
if ((k - l).toLong < p - lo) {
hi = p
r = k
} else {
lo = p
l = k
}
}
lo.toInt
}After array partitioning, the low subinterval has occurrences and possible values. If it is deficient, retaining it preserves the loop invariant. Otherwise , while the original active count is strictly less than ; subtraction givesThus the high subinterval is deficient and retaining it also preserves the loop invariant. Discarded occurrences lie outside the new candidate interval. The loop variant is the positive integer , which strictly decreases because the midpoint is strictly internal. At exit the interval has one possible value and fewer than one occurrence, so
lo is absent from the entire original array. For the loop is skipped and zero is returned. The Long value bounds avoid overflowing ; every internal pivot is at most , so its checked mathematical range fits Int.For the time complexity, let be the interval width and its active occurrence count before iteration . The loop invariant gives , and binary search gives . Each two-way array partition costs . Until the width reaches one, the widths are bounded by , so the total scan work is bounded by a geometric series plus loop overhead: overall, hence for positive . Partitioning the full array again at every iteration would instead lose this bound. Auxiliary space complexity is .
Treat a null reference as the empty binary tree and perform the visit before descending:By structural induction, the empty binary tree produces the empty output. A nonempty binary tree prints its root label, then the complete preorder traversal of the left subtree, then that of the right subtree. These are exactly the required output positions. The recursion terminates on a finite binary tree because each recursive call has strictly fewer vertices. For the three-node example, the output order is . The time complexity is linear in the number of vertices, with call stack depth bounded by the height of a binary tree plus a constant.
def preorderRecursive(t: Tree): Unit = {
if (t != null) {
println(t.datum)
preorderRecursive(t.left)
preorderRecursive(t.right)
}
}The number isA direct search over every -element subset must inspect candidates. Testing one candidate by generating its subset sums creates values; storing previously seen sums detects a collision. Generating them incrementally needs arithmetic operations, or for the straightforward independent calculation of each sum. Thus the simple overall search has cost on the scale of , apart from integer-arithmetic and data-structure costs.
The important time complexity distinction is that fixed gives polynomial growth in , of degree , but this can still become impractically large. If increases with , both the candidate count and the internal power set can grow exponentially. For example, near the binomial coefficient is already exponentially large. An exhaustive implementation can therefore encounter a combinatorial explosion; the fixed- estimate must not be described as exponential in .
Maintain the latest five temperature readings in a circular buffer, together with their running sum. The following C programming language code spells out the message passing interface rather than assuming a particular library:
send(dst,&m) is a blocking synchronous message passing send; receive(src,&m) receives from the specified sender or ANY and returns the sender's identity; sendrec(dst,&m) atomically sends the request and waits for that sender's reply in the same message object. Successful, reliable operations are assumed. Message contents are copied by the interface.enum { ANY = -1, TIMER = 1, TEMP = 2 };
enum { PULSE, PING, REQUEST, REPLY };
typedef struct {
int type;
int period_ms;
double value;
} Message;
extern void send(int dst, const Message *m);
extern int receive(int src, Message *m);
extern void sendrec(int dst, Message *m);
void smooth(void) {
double history[5] = {0};
double sum = 0;
int count = 0, next = 0;
Message m = { .type = PULSE, .period_ms = 1000 };
send(TIMER, &m); /* subscribe once */
for (;;) {
int sender = receive(count < 5 ? TIMER : ANY, &m);
if (sender == TIMER && m.type == PING) {
Message t = { .type = REQUEST };
sendrec(TEMP, &t); /* matching REPLY contains t.value */
if (count == 5)
sum -= history[next];
else
++count;
history[next] = t.value;
sum += t.value;
next = (next + 1) % 5;
} else if (count == 5 && m.type == REQUEST) {
Message answer = { .type = REPLY, .value = sum / 5.0 };
send(sender, &answer);
}
}
}Before the fifth reading, selective reception from
TIMER prevents premature client requests from being accepted. Such clients wait at their synchronous sends; they are not given a partial arithmetic mean. Thereafter the operating-system process receives both timer messages and client requests. Each accepted PING initiates one measurement; each accepted client request returns the arithmetic mean of the five most recently completed measurements, without initiating another measurement or changing the circular buffer.The loop invariant is that
count is the number of valid readings, capped at five, sum is their sum, and next is the next unused slot during startup or the oldest slot afterwards. Removing the overwritten reading and adding the new one preserves this loop invariant. Each update and reply uses time complexity and storage, since the window length is fixed. Under timely sensor replies, timer delivery and process scheduling, measurements are initiated once per second. The next parts explain failures of these timing assumptions; the code does not promise one-second sampling when the sensor itself cannot sustain it.A Haskell representation is:Every tree stores a value at its root, including internal nodes. Interpret the forest by root-left-right preorder traversal, concatenating the results in forest order. Valid values have correct cached sizes, perfect binary trees, nondecreasing sizes and only the permitted initial duplicate. The raw constructors can express invalid representations, so the operations below assume valid values and should form the public interface. Sizes are assumed to fit in
import Prelude hiding (null, head, tail)
data RTree a = Leaf a | Node a (RTree a) (RTree a)
newtype List a = List [(Int, RTree a)]Int; the stated time complexities use constant-time size arithmetic.