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Put , avoiding any dependence on whether the convention includes among the natural numbers. Under set inclusion, the results are
| Family | Maximum | Least upper bound in |
|---|---|---|
| None | None | |
| None |
The set union of all three-element subsets of is . An upper bound for under set inclusion would have to contain this entire infinite set, whereas members of are finite. Thus has no upper bound and hence no maximum or least upper bound.
For , every member is a subset of the finite set , and itself belongs to , making it both maximum and least upper bound. The union of the three-element subsets of is again , so is the least upper bound of . It is not a member of that family, and no three-element member contains all the others, so there is no maximum.
View the necklace as a cycle graph. Between the white bead and there are two edge-disjoint paths, containing and links. The beads remain connected if and only if at least one of these paths is uncut.
There are equally likely cut sets. Keeping the -link path uncut requires all three cuts to be among its complementary links, giving choices. Keeping the other path uncut gives choices. These two events are disjoint because three links are cut and they cannot all lie in both complementary paths. ThereforeUse for . The answer is unchanged by exchanging and , so either orientation around the necklace gives the same value. This is the connectivity calculation for random cuts of a cycle.
For each , surjectivity makes the fiber of a function nonempty. Choose one element of each such fiber of a function and define . Then for every , proving . For arbitrary sets, the simultaneous selection invokes the usual axiom of choice; this is the right-inverse characterization of the axiom of choice. For finite only finitely many choices are needed. If is empty, the unique empty function gives the same conclusion.
Let be the set of permutations fixing object . If a specified collection of objects is fixed, the remaining objects can be arranged in ways. There are such collections. Applying the inclusion-exclusion principle to exclude all fixed objects givesFor , the terms and cancel. Thus the number of derangements is , interpreting the remaining sum as empty when . The full inclusion-exclusion formula also gives .
The disjoint decomposition into sets uses only events in . Finite additivity and nonnegativity giveThus already follows from the two stated requirements. Equality is attained, for example, by choosing , since then . More generally it holds whenever . Neither normalization nor countable additivity is needed for this particular inequality.
Use coordinates with row and column in :
type Place = (Int,Int)
type Places = [Place]
type Board = (Places,Places,Places)
data Player = O | X deriving (Eq,Ord,Show)The components of
Board record the positions occupied by O, occupied by X, and still empty, respectively. On a nonterminal legal board, equal occupied counts mean O moves next; one extra O means X moves next.The type synonym does not enforce the representation invariant: coordinates are in range; each component has no duplicates; the three represented sets form a set partition of the nine positions; and the occupied counts differ as just described. Both players cannot have won. A winning
O board has one extra O, and a winning X board has equal counts. More fundamentally, a legal tic-tac-toe state has a legal alternating history that stops at the first win: satisfying a type alone never licenses continuing an already completed game.The empty occupancy sets leave all nine coordinates available:
empty :: Board
empty = ([],[],[(r,c) | r <- [0..2], c <- [0..2]])This board satisfies the representation invariant and gives
O the next move.Construct each possible winning line, then test whether one is contained in the player's occupancy set:
winningLines :: [Places]
winningLines =
[[(r,c) | c <- [0..2]] | r <- [0..2]] ++
[[(r,c) | r <- [0..2]] | c <- [0..2]] ++
[[(i,i) | i <- [0..2]], [(i,2-i) | i <- [0..2]]]
won :: Places -> Bool
won ps = any (\line -> all (`elem` ps) line) winningLinesThe list comprehension generates three rows, three columns and two diagonals. Each has exactly three coordinates; therefore
all holds precisely when every coordinate of that line is occupied by this player. Taking any over the eight lines proves the win test correct.Different move orders can lead to the same position, and the two occupancy lists also retain irrelevant insertion order. Thus this search tree repeatedly solves the same remaining game. For example, swapping the order of two
O moves while leaving the intervening X moves legal can reach the same occupied sets.Canonicalize each state by sorted occupancy sets, or by two bitboards, and apply memoization or a visited-state set. Expand each canonical position once, stopping at terminal states, and collect terminal positions in a set. There are at most cell assignments, many of which are illegal; this is much smaller than a tree of move permutations. Store positions, not histories, when the requested output is distinct final positions. Board rotations and reflections can reduce work further if handled as an additional equivalence: if the output must retain every oriented board, reconstruct those distinct orientations rather than silently returning one representative per symmetry class.
For proper lazy lists, the generic condition is a finite first list, or an empty second list. A finite first list reaches its terminating constructor, after which list concatenation exposes the second one. An infinite first list never yields control to the second; it therefore cannot deliver occurrences originating there unless there are none. This concerns input occurrences, rather than an accidental overlap of the two inputs' value sets.
If , return its sole index without a query. Otherwise first evaluate for until two different outputs are found. Part i says these outputs are precisely , so two values with must exist. This uses at most calls to the function-evaluation oracle.
Maintain a candidate index set , initially all indices. Split into two nearly equal pieces and choose one, . Query the input that is on and everywhere else. Its output equals exactly when , and otherwise equals . Retain the indicated half and repeat until one index remains.
The loop invariant is , and the witness outputs keep each test unambiguous. There are at most splitting queries. The total is at most evaluations, which is for , and in fact a stronger additive bound. Only oracle calls are being counted; explicitly constructing each query vector can require ordinary work. This is identification of a unique essential variable.
Let be the final
A set produced by the algorithm. It contains the initial A set, and every vertex outside has at most half its neighbours in : otherwise its final key would still be positive. Thus is closed under the activation rule.Now consider any legal sequence, including sequences which change some
A preferences back to B. We prove by induction that its current A set is always a subset of . This is true initially. A switch to B preserves the inclusion. If a vertex outside were to switch to A, its A neighbours would all lie in , so their number could not exceed half its degree. Such a switch is forbidden. A switch inside also preserves the inclusion.Consequently every attainable final
A set has size at most . The algorithm supplies a legal sequence attaining itself, so its sequence is optimal. This closure dominance for asynchronous threshold switching proves more than a local-greedy claim: temporary losses of A vertices cannot unlock an additional vertex outside the activation closure.For nonnegative integer amount , let be the minimum number of coins making amount . Set and compute, in increasing order of ,An empty candidate set has minimum . Here denomination ensures that every nonnegative amount is attainable. Removing one coin from an optimal nonempty solution gives a smaller amount, and adding any allowed last coin gives a feasible solution; these two directions prove the recurrence by induction on .
Return whether . This solves the at-most- decision without needing a separate dimension for . The table for the minimum coin change recurrence takes running time and space complexity. The bound is pseudopolynomial time: the amount, rather than its binary encoding length, appears in it. Amount zero needs zero coins.
Use the multiplicative score exactly as defined in the paper. Let be the largest score of a matching length- route ending at . Initialize and mark all other entries unreachable. For each stage, clear the new layer and inspect matching edges from reachable entries:An empty candidate set is marked unreachable, rather than given score zero: a genuine route may have score zero. A Boolean reachability table or a sentinel with a guard before multiplication distinguishes the cases. Store the maximizing predecessor edge at each reachable layer entry, retaining one on a tie.
Multiplication by a nonnegative edge factor preserves order, so replacing a prefix by a larger-scoring matching prefix cannot worsen its extension. Induction proves this maximum-product edge-labelled walk recurrence correct. Choose a reachable final entry of greatest score and follow its layer-indexed predecessors backwards to reconstruct the answer. If no final entry is reachable, report that no matching route exists, even though all numeric scores might otherwise look like zero.
The running time is , and rolling scores have space complexity , while retaining all reconstruction pointers gives space complexity . For the vertex-simple interpretation, retain the visited set in each state as in part i, prohibit revisits and maximize the same products over those states. This exact exponential-state variant preserves both the zero-score distinction and predecessor reconstruction. If the starting vertex is unrestricted rather than inherited from part i, initialize for every vertex. The calculation needs no independence assumption beyond the paper's definition of a product score; repeated physical edge-activation events would require a separate probabilistic model.
The printed formula for the edge label has mismatched variable names. The type-correct reading is . Interpret the requested route as a directed walk of consecutive labelled edges, allowing repeated vertices; the distinction from a vertex-simple directed path is discussed below.
Let be the set of vertices reachable from the specified start after reading exactly the first labels. Initialize and updateAfter stages, answer yes exactly when . A directed walk realizing a prefix decomposes into a walk realizing the shorter prefix and one correctly labelled final edge; conversely every such extension is valid. This proves the dynamic programming invariant by induction on . Store reachable vertices in Boolean arrays and scan the edges per stage. The running time is with rolling-table space complexity, in addition to graph storage. The empty label sequence is realized by staying at .
If “path” is instead required to mean vertex-simple, augment the state to , where is the set of visited vertices. Start with and extend a matching edge to only when , replacing by . This gives an exact dynamic programming algorithm for that interpretation too; reject immediately when . A straightforward bound is running time with rolling-table storage. The exponential distinction is substantive: all labels equal and include the Hamiltonian path problem from a fixed start. The efficient layered recurrence answers the consecutive-edge interpretation without pretending to enforce distinct vertices.
Let denote the entry state. A conventional meaning for the Boolean returned by Initially . In particular, re-adding an existing element returns false and does not increase the size. All operations have the stated empty precondition at the abstract level. The comment that
add is whether the call changes the set:// add(elem):
// S' = S0 union {elem}
// result == !S0.contains(elem)
// contains(elem):
// S' = S0
// result == S0.contains(elem)
// size:
// S' = S0
// result == cardinality(S0)add adds an element does not itself completely specify its Boolean result; the explicit changed/not-changed convention above completes that contract. As with any Int-valued size operation, exact cardinalities are understood to be representable by that result type.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.For result entries the largest required index is , so the bound in part (a)(iii) permits universe size . Check that the chosen universe can itself be expressed as an array length:The loop invariant is that the result prefix contains exactly and the abstraction function of
def recamanBitmap(a: Array[Int]): Unit = {
val n = a.length
if (n == 0) return
val capacity = n.toLong * (n.toLong - 1L) / 2L + 1L
require(capacity <= Int.MaxValue.toLong, "Bitmap universe too large")
val seen = new BitMapSet(capacity.toInt)
a(0) = 0
seen.add(0)
var i = 1
while (i < n) {
val candidate = a(i - 1).toLong - i.toLong
val next =
if (candidate > 0L && !seen.contains(candidate.toInt)) candidate
else a(i - 1).toLong + i.toLong
a(i) = next.toInt
seen.add(a(i))
i += 1
}
}seen is their set of values. The universe bound ensures that each positive candidate being queried and each actual next value is less than capacity; values are nonnegative. Thus the conversions and the restricted add precondition are justified, rather than merely hoped to succeed. At a repeated addition, add returning false is harmless: the set of previously seen values is still updated correctly. The loop variant is . At return every result entry agrees with the original recurrence.Give each solving session a shared visited set keyed by immutable snapshots of permutation contents, for example
game.toList, and insert the initial snapshot immediately. Have all successor objects refer to that same session's set. The refined flip first constructs the candidate state; if its key is present, return None. Otherwise add the key immediately and return Some(candidate). The search enqueues only Some results. Keys must compare contents rather than array identities, and a fresh solving session must start with a fresh visited set.Marking at generation time prevents the same state from being queued repeatedly, including the immediate cycle made by applying the same reversal twice. First discovery in breadth-first search already has minimum depth, so discarding later discoveries cannot remove a shorter solution. The graph now has at most admitted states. If only cycles along one path were to be rejected, a path-local set would suffice, but it would still repeat states reached by different paths; the shared session-wide set avoids both kinds of repetition. Public game fields must not be externally mutated while their states are being searched.
The abstract state is a finite multiset of ordered values, initially empty. Multiplicity matters: deleting the minimum removes one occurrence, not every equal occurrence. Assume a consistent total order on This abstract data type places no fixed-capacity bound on insertion. The only operation requiring a nonempty collection is
T.trait PriorityQueue[T] {
// Pre: true. Post: Q' is the empty multiset.
def clear(): Unit
// Pre: true. Post: Q' = Q with one additional occurrence of x.
def insert(x: T): Unit
// Pre: Q is nonempty.
// Post: result is a minimum of Q;
// Q' is Q with exactly one occurrence of result removed.
def delMin(): T
// Pre: true. Post: result iff Q is empty, and Q' = Q.
def isEmpty(): Boolean
}delMin. The representation must preserve duplicates even if it uses a data structure commonly associated with sets.