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Joined 2026-09-21 Message
Incoming links: Integer
Suppose with the same integer result type. The empty list forces . The two singleton calculations force and . But thencontrary to . No such
c,b exist. This is the information restriction in the fold factorisation criterion: equal folded tails must remain indistinguishable after the same head is attached. A larger, function-valued result type can retain the missing information, as the continuation representation of a left fold does below.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.
Accumulate the prefix by Horner's method in base two. The loop invariant, in exact arithmetic, is that
digit-'0' converts either permitted digit to its numerical value:var i = 0
var integerPart = 0.0
while (i < bin.length && bin(i) != '.') {
integerPart = 2.0 * integerPart + (bin(i) - '0')
i += 1
}i is the number of processed digits before the point andAppending a digit shifts the old positional value by one binary place and adds that digit, which proves preservation. The loop variant is . At exit, i is the binary-point index or the length if no point occurs, and integerPart is the value of the integer prefix. An empty prefix contributes zero. There is one multiplication per processed digit, with no separate powers or repeated divisions. Short-circuit evaluation prevents an out-of-range access when the point is absent.For
Double, this exact-arithmetic invariant describes the intended value; the operational value is the rounded Horner recurrence. Very long prefixes can lose precision or overflow, as discussed in part (g). A fixed-width return type cannot represent every finite binary digit string exactly.Extract low bits and fill the integer positions from right to left:At the start of an iteration, the loop invariant says that positions contain the correct low bits andThe bitwise AND extracts the remainder modulo two; the shift supplies its quotient. These two facts preserve the loop invariant. The loop variant is . The width established in part (c) ensures that
var remaining = int_part
var pos = digits - 1
while (pos >= 0) {
res(pos) = if ((remaining & 1) == 0) '0' else '1'
remaining = remaining >>> 1
pos -= 1
}remaining is zero at exit, so the first digits entries encode the whole integer part. For zero the single entry becomes '0'. There are no explicit multiplications or divisions, and no extra digit array is needed.Generate the fractional bits by repeated doubling, keeping only the unconverted residual:Let and . The loop invariant is , the first fractional digits have been written, and, in exact arithmetic,Doubling exposes the next digit, and subtracting that digit leaves the next residual in . This proves preservation. The loop variant is . The loop fills the whole suffix, even when the residual has become zero; subsequent bits are zero. There is one multiplication per output bit. With fractional positions, the result is truncated, with exact-arithmetic error . An integer has an all-zero suffix. No rounding-up step is performed.
res(digits) = '.'
var fraction = dec - int_part.toDouble
var out = digits + 1
while (out < res.length) {
fraction = 2.0 * fraction
if (fraction >= 1.0) {
res(out) = '1'
fraction -= 1.0
} else res(out) = '0'
out += 1
}Equivalence partitioning divides the input domain into classes expected to exercise the same relevant behavior, then chooses representatives and boundary cases from each class. It is a way to select tests, not a proof that every member of a class must behave identically in a faulty implementation.
Useful classes for this parser include integer-only inputs, a leading point, a trailing point, and digits on both sides; all-zero inputs and inputs containing ones; leading or trailing zeroes; one-digit and longer blocks; and large integer or tiny fractional magnitudes near floating-point rounding, floating-point overflow and floating-point underflow boundaries. For instance, compare
'1', '1.', '.1', '1.1', '0001.1000', and digit strings just beyond the exact integer precision of Double. Empty digit blocks have the zero convention used above. If a surrounding validator rejects an entirely empty representation or a bare point, test those as separate invalid-format classes. Unexpected digits and multiple points are likewise validator tests, since the conversion fragments assume well-formed input. Include representatives with independently calculated positional values, rather than relying only on inverse conversion.The first proposed identity can fail even without numerical error because binary fractions have nonunique textual representations. Leading zeroes, optional points and trailing fractional zeroes are normalized differently by a fixed-width output convention. It also fails if the input has more significant bits than
Double can preserve, more fractional places than the output buffer can hold, an integer part outside the assumed Int range, or a value that overflows or underflows during decoding. Array identity here must mean equal contents: a newly allocated Scala array is a different software object, so use sameElements rather than reference equality for a content test.The second identity fails when the stored fractional precision truncates a nonzero bit; with two fractional places, becomes zero. It can also fail outside the finite nonnegative input and integer-conversion assumptions. Within those assumptions, enough fractional positions to encode the finite
Double exactly can permit the round trip, but an arbitrary fixed small width does not guarantee it. Every finite Double has a finite binary expansion; saying that a decimal-looking Double necessarily has an infinite binary expansion would confuse the represented number with an exact decimal real.The source's introductory example is internally inconsistent: the displayed digit string has value , whereas its shortened character array encodes . The conversion rules above follow the actual input digits. Round-trip testing needs an independent value oracle as well as the two compositions, because two mutually consistent but incorrect routines can conceal each other's errors.
A fixed array containing the distinct seen integers needs capacity at most , because at most one new distinct value is introduced per result entry. Its auxiliary space is therefore , rather than the bitmap's universe storage. A size field identifies the occupied prefix; insertion first checks membership, and appends only when absent.
An unsorted occupied prefix needs linear membership scans, giving total work by the same summation as in the original version. Keeping it sorted permits binary search membership in , but inserting a new value can shift entries, so the overall worst-case upper bound remains quadratic. Unlike a bitmap, the space cost depends on the number of stored values rather than their largest possible numerical magnitude. Compared with part (a)(ii), this extra set array also duplicates information already present in the output prefix; its principal benefit is a separate set interface, not an automatic asymptotic speedup.
The expression constructs a valid iterator object; it does not itself return the concatenated numeric or character values. The element types have a common supertype Successive calls return
AnyVal, and covariance lets Iterator[Int] and Iterator[Char] serve as the two inputs to ConcatIterator[AnyVal]. In the ordinary Scala 2 typing of this expression, the common element type is AnyVal. One can make the intended widening explicit:new ConcatIterator[AnyVal](
List(1, 2, 3).iterator,
List('a', 'b', 'c').iterator
)1, 2, 3, 'a', 'b', 'c', then exhaustion. Characters remain characters: viewing an existing iterator at a wider type does not convert its contents to integer character codes. No next() is invoked merely by constructing this adapter. The common-type argument uses subtype polymorphism and covariant type parameters, rather than requiring the two concrete input types to be equal.Use the remainder form of the Euclidean algorithm to compute the greatest common divisor, then divide both components of the fraction by it:The loop invariant is , , and . Initially it holds by symmetry of the greatest common divisor. For , the integer division identity , with , shows that and have exactly the same common integer divisors. Thus the update preserves the loop invariant. The loop variant is the nonnegative integer , which strictly decreases on each iteration. At exit , so . Dividing by preserves the rational number and makes the returned components coprime. In particular, zero is returned as . No multiplication is needed, and all remainders remain within the original nonnegative
def Reduce(num: Int, den: Int): (Int, Int) = {
require(0 <= num && num < den)
var a = den
var b = num
while (b != 0) {
val r = a % b
a = b
b = r
}
(num / a, den / a)
}Int range.For a positive remaining fraction , the condition is equivalent to . Since decreases as the positive integer increases, choose . The zero fraction needs no term. This is the greedy Egyptian fraction expansion.
The remaining fraction isThe defining inequalities give . Therefore the unreduced new numerator is a nonnegative integer strictly smaller than the old positive numerator. Passing to a reduced fraction can only reduce it further. That numerator is a loop variant, so the process terminates. The increasing denominators are not the termination argument; their size need not be bounded in advance.
There are entries in , which contains possible integers. If , then . Therefore some value in is missing, again by the pigeonhole principle. This is an existence conclusion in the chosen interval, not a claim that all missing values lie there; both sides can contain missing values when duplicates occur.
If , the upper side contains the whole permitted range and hence a missing value. If , the condition is impossible. Outside the internal-pivot range this conclusion therefore supplies no improved localization.
For an integer pivot , exactly entries lie in . That interval contains possible integers. If , the pigeonhole principle implies that some value in is missing. Counting occurrences rather than distinct entries only strengthens this argument: occurrences cannot cover more than distinct values.
For completeness, if the lower side contains the entire permitted value range, which already has a missing value. If , the condition is impossible. Thus the general conclusion is a missing permitted value below ; an internal pivot gives useful localization.
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 .
Search for the predecessor of the matching linked list node, or for the last linked list node if there is no match. This helper makes deletion uniform:The search loop invariant says that
class LinkedBook extends Book {
private val header = new Node("", "", null)
// Pre: name != null.
// Post: result is header or a real node.
// All real nodes through result have names different from name.
// result.next is null or has name equal to name.
// No state is changed.
private def findPrev(name: String): Node = {
require(name != null)
var p = header
while (p.next != null && p.next.name != name) {
p = p.next
}
p
}
def store(name: String, number: String): Unit = {
require(name != null && number != null)
val p = findPrev(name)
if (p.next == null)
p.next = new Node(name, number, null)
else
p.next.number = number
}
def recall(name: String): String = {
require(name != null)
val p = findPrev(name)
require(p.next != null, "Name is not in the book")
p.next.number
}
def isInBook(name: String): Boolean = {
require(name != null)
findPrev(name).next != null
}
def delete(name: String): Boolean = {
require(name != null)
val p = findPrev(name)
if (p.next == null) false
else {
p.next = p.next.next
true
}
}
}p is the header or a reachable real linked list node, and every real linked list node from the start through p has a name different from the requested name. The linked list has not been changed. Short-circuit evaluation protects p.next.name when the end is reached. The loop variant is the number of real linked list nodes strictly after p; finiteness and acyclicity make it a nonnegative decreasing integer. Consequently the helper's stated postcondition holds.For
store, a found name changes only its number; an absent name adds exactly one fresh binding at the end, maintaining uniqueness. For delete, unlinking the found linked list node removes exactly the requested binding; absence returns false without mutation. Both query operations preserve the abstraction function. Non-null preconditions are checked at entry; recall additionally checks membership using the one completed search, avoiding a redundant whole-list membership scan. The helper is the only additional operation and its complete behavioral contract is stated above.Keep the same player until an occupied, in-range cell is supplied:The outer loop invariant is that
val board = new Chomps(6)
var player = 1
while (board.Count() > 0) {
board.Display()
var x = -1
var y = -1
while (!board.Legal(x, y)) {
println("Player " + player + ": enter x, then y")
try {
x = scala.io.StdIn.readInt()
y = scala.io.StdIn.readInt()
} catch {
case _: NumberFormatException =>
x = -1
y = -1
}
if (!board.Legal(x, y)) println("Choose an occupied cell")
}
board.Move(x, y)
if (board.Count() == 0)
println("Player " + player + " loses; player " + (3 - player) + " wins")
else player = 3 - player
}player is the next player's number, either one or two, and the board is the result of the accepted moves. The inner loop invariant says that the board and player have not changed; rejected inputs do not consume a turn. Assuming the input source continues and eventually supplies a legal coordinate pair, the inner loop finishes. No program can guarantee that a user stops entering illegal moves, so that interaction assumption is necessary.Every accepted move removes at least its selected token, so the outer loop variant is
Count(), a decreasing nonnegative integer. In a nonempty order ideal, is occupied. Removing it removes the entire board, whereas a move at any other cell leaves it. Therefore the player making the final move is exactly the loser; the code reports that player before changing the turn. At most legal moves occur.Programming language polymorphism lets one expression or operation work with values of different types or concrete representations. In object-oriented programming, subtype polymorphism lets a reference of an interface or superclass type designate instances of different implementing software classes. A call through that reference uses dynamic dispatch to select the receiver's overriding object method. For example, a routine taking
Book can invoke recall on a linked-list or an array-based implementation while depending only on the common abstract contract.Parametric polymorphism is another form: the type parameter in
FiniteDist[T] lets the same interface describe distributions over Booleans, integers or characters. A generic method introduces a further type parameter when its output type differs. Method overloading, in contrast, supplies several statically selected signatures with one name; it should not be confused with receiver-dependent dynamic dispatch. Correct substitutability also requires implementations to respect the common behavioral contract, not merely to share method names.The inclusive interval is nonempty exactly when . Reject with a failed precondition, rather than assigning a nonexistent normalization to an empty interval. Compute its width in The loop invariant says that and
Long before taking a reciprocal:class Uniform(l: Int, u: Int) extends FiniteDist[Int] {
require(l <= u, "Empty uniform interval")
private val width = u.toLong - l.toLong + 1L
private val mass = 1.0 / width.toDouble
def getProb(x: Int): Double =
if (l <= x && x <= u) mass else 0.0
def getDom: List[Int] = {
var x = l.toLong
var rev = List.empty[Int]
while (x <= u.toLong) {
rev = x.toInt :: rev
x += 1L
}
rev.reverse
}
}rev contains exactly the integers in reverse order. The loop variant is while the guard is true. Long handles both the full interval width and the increment past Int.MaxValue. At exit, reversing yields each support value once. All support values have mass , so their exact sum is one; Double approximates that rational number. Very large valid domains require correspondingly large lists and sufficient memory; that resource issue does not make the mathematical endpoint values invalid.Use the conventional integer calling convention:
$a0 holds the argument and $v0 holds the result. The function is recursive, so every call must preserve its own return address and original argument. The first recursive result must also survive the second call; a caller-saved register alone would not protect it.This implementation reserves a 32-byte stack frame, keeping the stack pointer eight-byte aligned. Offsets 0–15 are available as the outgoing argument area, 16 stores the original argument, 20 stores the first result, 24 is padding, and 28 stores the return address. It changes no callee-saved register. Assemble in classic MIPS delayed-branch mode with instruction reordering disabled, such as
.set noreorder.foo:
addiu $sp,$sp,-32 # allocate 32-byte aligned frame
sw $ra,28($sp) # save caller return address
sw $a0,16($sp) # save original n
beq $a0,$zero,foo_base
sll $zero,$zero,0
addiu $t0,$zero,1
beq $a0,$t0,foo_base
sll $zero,$zero,0
addiu $t0,$zero,2
beq $a0,$t0,foo_base
sll $zero,$zero,0
addiu $a0,$a0,-1
jal foo
sll $zero,$zero,0
sw $v0,20($sp) # preserve foo(n-1) across next call
lw $a0,16($sp) # recover original n
sll $zero,$zero,0
addiu $a0,$a0,-3
jal foo
sll $zero,$zero,0
lw $t0,20($sp)
sll $zero,$zero,0
addu $v0,$v0,$t0 # combine the two recursive results
beq $zero,$zero,foo_return
sll $zero,$zero,0
foo_base:
addu $v0,$a0,$zero # base case result is n
foo_return:
lw $ra,28($sp)
sll $zero,$zero,0
addiu $sp,$sp,32 # release the frame
jr $ra
sll $zero,$zero,0All displayed operation mnemonics denote single machine instructions. In particular,
sll $zero,$zero,0 is the actual harmless shift encoding often abbreviated by the pseudoinstruction nop; addu $v0,$a0,$zero performs a copy without the move pseudoinstruction. Each branch or jump has an explicit harmless branch delay slot, and harmless shifts also separate loads from their subsequent uses. The base cases return the original argument; other calls return the sum of the two recursive values. Every return restores the saved return address and exactly undoes the stack allocation.[ref]MIPS32 Architecture For Programmers Volume II by MIPS Technologies. Revision 2.00, ADDIU and ADDU, pp. 37–38; JAL and JR, pp. 139 and 144