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Incoming links: Constant time
Maximum priority queue Created 2026-10-24 Updated 2026-10-31
A priority queue whose distinguished entry has largest key. A max-heap implements maximum inspection in constant time, and insertion, extraction of a maximum and increasing the key of a located entry in running time. An entry handle needs its current heap index to support that last bound; searching by an arbitrary key can take linear time. Multiple entries can have equal keys without requiring a particular tie order.
The accounting method assigns each operation a charge, its amortized cost, that can differ from its actual cost. Cheap operations save the excess charge as credits associated with the data structure or its entries. An expensive operation spends these previously saved credits. The essential invariant is that the credit balance never becomes negative.
If the initial balance is zero, operation has actual cost , charge , and leaves balance , thenThus a nonnegative credit balance makes the total charge an upper bound on the total actual cost for every operation sequence. In particular, a constant time charge proves a linear time total. The argument is a deterministic amortized analysis; it requires no probability distribution on the operations. A nonzero initial balance must be included as an additive term in the bound.
In binary long division, first double the divisor until it exceeds
x, then halve it successively. Keep p = 2^j and d = y*p, so no exponentiation is needed inside a loop:q := 0; r := x;
j := 0; p := 1; d := y;
WHILE d <= x DO
d := 2*d;
p := 2*p;
j := j+1
END;
WHILE j > 0 DO
d := d DIV 2;
p := p DIV 2;
j := j-1;
IF r >= d THEN
r := r-d;
q := q+p
END
ENDThe first loop invariant is , , , and . Every doubling preserves it. For integer inputs, after at most iterations when ,
d > x; if , this loop has no iterations. The unbounded increase of proves termination. Equivalently, one can use the decreasing nonnegative loop variant while the guard is true.At the top of the second loop the loop invariant isIt holds on entry because
d > x = r. Suppose the old d is and the old p is . After the three halving assignments, and the old remainder satisfies . If , the desired new bound already holds. Otherwise subtracting leaves and adding to q preserves because . The divisions by two are exact: before each halving, j > 0 makes both p and d even.The loop variant
j decreases by one each iteration. At exit , so d = y and the loop invariant gives the required remainder bound. The work per iteration is constant in the unit-cost model, and the two loops have the same number of iterations. For ,The case , including , takes constant time. These are arithmetic-operation counts with sufficient range for the doubled values; arbitrary-precision arithmetic incurs additional bit complexity. If the inputs were arbitrary positive real numbers, the uniform bound would depend on rather than only ; the stated is the conventional integer division interpretation.The representation invariant for the array stack is , with live entries at positions and the top at
s.n-1 when nonempty. The following Oberon procedures use the given data types:PROCEDURE makeEmpty(VAR s : stack);
BEGIN
s.n := 0
END makeEmpty;
PROCEDURE isEmpty(s : stack) : BOOLEAN;
BEGIN
RETURN s.n = 0
END isEmpty;
PROCEDURE push(VAR s : stack; t : node);
BEGIN
s.a[s.n] := t;
s.n := s.n+1
END push;
PROCEDURE pop(VAR s : stack) : node;
VAR t : node;
BEGIN
s.n := s.n-1;
t := s.a[s.n];
s.a[s.n] := NIL;
RETURN t
END pop;For
push, the supplied precondition s.n < N makes the old s.n a valid free index. For pop, the precondition s.n > 0 makes the decremented index the old top. Setting the vacated entry to NIL avoids retaining a reference to the removed binary tree; it does not deallocate that binary tree. makeEmpty need not clear inactive entries for the abstract stack behavior. None of these procedures has a loop. Each changes or inspects a constant number of fields, so its logical work is constant time; an implementation of the specified by-value isEmpty parameter may additionally copy the fixed-capacity record. The essential update rule isA doubly linked list supports backward traversal and removal of a known linked list node in constant time, because that node supplies its own predecessor. In a singly linked list, removing a known node generally requires finding its predecessor by a forward scan unless the caller already supplies it. Inserting before a known node has the same advantage.
The cost is another pointer per node, more link assignments on every update, and a stronger consistency invariant to maintain. HenceThis does not make locating an arbitrary data value faster: a search still takes linear time in the worst case.
A list header is a permanent dummy linked list node, distinct from the ordinary elements. It gives a stable reference to the linked list and can store metadata such as its length. In a circular linked list, following successor links eventually returns to the starting point; with a list header, the last ordinary element links to the header, and the empty structure has the header linked to itself.
A singly linked list stores a successor reference per linked list node. A doubly linked list additionally stores a predecessor reference. The latter supports traversal in both directions and deletion of a known linked list node in constant time, without first searching for its predecessor. It costs another reference per linked list node and additional updates whose mutual consistency must be maintained. A list header makes insertion and deletion at an endpoint use the same local splicing rules as an interior change; the empty case does not require replacing an external head reference.
The runtime precondition check rules out negative dimensions:Both links point to the fresh list header, so all reciprocal-link conditions hold and there are no ordinary entries to violate ordering or nonzero-value requirements. The abstraction function therefore yields the zero vector of dimension
def makeZeroVector(size: Int): Node = {
require(size >= 0)
val h = new Node(0.0, size)
h.prev = h
h.next = h
h
}size. Construction uses constant time and constant space.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.A fixed-size array cannot implement unbounded successful insertion. Either specify a bounded double-ended queue with an extra not-full insertion precondition, or retain the unbounded interface and specify a possible full-capacity failure. Silently overwriting an endpoint would change the stated behavior. Here use the bounded contract, with a fixed positive capacity
MAX and abstract state . Both additions require ; the other postconditions remain unchanged. The class declaration in part (c) is interpreted as implementing this explicitly revised contract, rather than as a behavioral subtype of the unrestricted contract in part (a).The concrete state is an array
a of length MAX, a first-position index head, and an occupied-entry count size. Its representation invariant isThe abstraction function is for . Unoccupied slots are unconstrained. Storing size distinguishes empty and full states even though their two circular endpoint positions coincide. MAX=1 is valid. Using a circular buffer avoids shifting entries and makes every endpoint operation constant time.The following class uses the bounded insertion contract from part (b). For
slot performs mathematical circular indexing with a wider intermediate, so even large legal indices do not overflow before reduction:class ArrayDoubleEndedQueue(val MAX: Int) extends DoubleEndedQueue {
require(MAX > 0)
private val a = new Array[Int](MAX)
private var head = 0
private var size = 0
private def slot(offset: Int): Int =
((head.toLong + offset.toLong) % MAX.toLong).toInt
private def before(i: Int): Int = if (i == 0) MAX - 1 else i - 1
private def after(i: Int): Int = if (i == MAX - 1) 0 else i + 1
def isEmpty: Boolean = size == 0
def addLeft(x: Int): Unit = {
require(size < MAX)
head = before(head)
a(head) = x
size += 1
}
def addRight(x: Int): Unit = {
require(size < MAX)
a(slot(size)) = x
size += 1
}
def getLeft(): Int = {
require(size > 0)
val x = a(head)
head = after(head)
size -= 1
x
}
def getRight(): Int = {
require(size > 0)
val x = a(slot(size - 1))
size -= 1
x
}
}addLeft, moving the first position back one slot makes the old position the new logical index one; inserting at the new position prepends x. For addRight, the old size is the first unoccupied logical index, so writing there appends x. The removals read the appropriate endpoint and exclude that slot from the occupied interval. Every operation preserves the representation invariant and realizes the abstraction function postcondition. Old contents in unused slots cannot be observed through this interface. Each operation performs a fixed number of accesses and arithmetic operations, hence has constant time and constant space per call, apart from the class's backing storage.Version snapshots simplify correctness: a complicated command restores the complete prior state, so a bespoke inverse cannot accidentally omit an auxiliary field. Redo restores the recorded result without re-executing behavior that might depend on time, randomness or external input. Restoring an immutable root takes constant time, though refreshing the display can cost more.
The storage comparison depends on the representation. A change-specific record for these elementary edits retains only a few indices and song references. Copying a length- array for each of edits can require storage and copying per edit. A persistent data structure with structural sharing can preserve many versions much more cheaply; balanced path copying can make small edits allocate new entries. Retaining versions also keeps their reachable data alive. Snapshot references are only constant-sized roots, not a guarantee of constant total retained memory. Neither scheme automatically reverses external effects such as playing audio or writing files. The desired undo boundary must be specified.
The empty skew-binary random-access list contains no trees:Both construction of
nil :: List a
nil = List []
null :: List a -> Bool
null (List []) = True
null _ = Falsenil and testing null take constant time.Repeated constant time insertion constructs the representation:The right fold preserves order because
fromList :: [a] -> List a
fromList = foldr cons nilcons prepends its argument. On a finite length- list, there are calls of cons; completely constructing the result takes time and constructors. This claim concerns finite inputs, not productivity of conversion from an infinite lazy list.Repeated root removal gives the conventional list:The representation interpretation satisfies
toList :: List a -> [a]
toList rs
| null rs = []
| otherwise = head rs : toList (tail rs)toList (cons x rs) = x : toList rs; alternatively, structural induction on the number of stored elements proves that each step emits exactly the next preorder traversal value. The output has cons cells, and each step takes constant time, so fully consuming it costs time. No repeated list concatenation is needed.Skip whole trees using their cached sizes, then descend through one tree:After skipping a size- tree, the desired index decreases by . Inside a node, position zero is the root, positions belong to the left child and the following positions to the right child. This proves correctness by structural induction on the search.
index :: List a -> Int -> a
index (List ts) i
| i < 0 = error "index: negative index"
| otherwise = seek ts i
where
seek [] _ = error "index: out of bounds"
seek ((w,t):us) j
| j < w = at w t j
| otherwise = seek us (j-w)
at _ (Leaf x) 0 = x
at w (Node x l r) j
| j == 0 = x
| j <= h = at h l (j-1)
| otherwise = at h r (j-1-h)
where h = w `div` 2
at _ _ _ = error "index: invalid representation"There are at most forest entries by the supplied bound, and each tree has logarithmic height. Consequently a valid query takes time, with logarithmic worst cases. On a conventional linked list,
xs !! i follows links and costs , hence in the worst case. The first element remains a constant time query in either representation.The following shared step consumes exactly one input element per recursive call of The helper's input consists of nonempty runs. If its first run starts with
group1:prependRun :: Eq a => a -> [[a]] -> [[a]]
prependRun x xss = (x:same):rest
where
~(same,rest) = case xss of
[] -> ([],[])
run@(y:_):runs
| x == y -> (run,runs)
| otherwise -> ([],xss)
group1 :: Eq a => [a] -> [[a]]
group1 [] = []
group1 (x:xs) = prependRun x (group1 xs)x, prepend x to that run; otherwise create a new singleton run. Structural induction proves that these are the maximal equal runs in order. The irrefutable pair binding ~(same,rest) preserves lazy evaluation: the first output group exposes its initial x without inspecting the recursively grouped tail. Thus even an infinite constant input can expose the first group and each finite prefix of that group. Finite inputs take linear time when equality tests take constant time. Skew-binary random-access list 2026-10-25
Represent a finite list by a forest of value-bearing perfect binary trees of sizes , in nondecreasing order, using the greedy largest sizes and allowing a repeated size only in the first two positions. Concatenated preorder traversals recover the list. Cached tree sizes guide indexing.
Front insertion merges the first two equal-size trees beneath a new root, or adds a singleton if they differ. Removing the first root either discards a singleton or splits its two children into the initial equal-size pair. Both operations inspect and allocate only a bounded number of constructors, taking worst-case constant time and sharing the remaining representation. A positional query skips whole cached tree sizes and then descends through one tree, taking time. Conversion in either direction takes linear time on finite input.
The duplicate-size restriction follows directly from greediness. If two size- trees had any preceding nonempty portion, there would be at least elements available, enough for the next larger perfect binary tree. Three equal-size trees would likewise suffice for that larger tree. Both contradict the greedy representation.