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Joined 2026-09-21 Message
Incoming links: Integer
Use dynamic programming in increasing order of the maximum permitted part and then of the remaining sum. Let Require . The outer loop invariant says all rows below equal . The inner loop invariant additionally says entries before column in the current row are correct. Its dependencies are the previous row and the earlier column , so no future entry is read. Each state is evaluated once, giving arithmetic operations and stored counts. A memoization table around part (a)'s recursion is another valid way to share states; its entries need a separate “computed” flag because zero is a legitimate answer. Growing count sizes affect bit complexity and can cause integer overflow in a fixed-width
Count be a sufficiently wide integer type and let table have dimensions (MaxN+1) by (MaxN+1):PROCEDURE CountPartitions(n: INTEGER): Count;
VAR i, j: INTEGER;
table: ARRAY MaxN+1, MaxN+1 OF Count;
BEGIN
table[0,0] := 1; j := 1;
WHILE j <= n DO table[0,j] := 0; j := j+1 END;
i := 1;
WHILE i <= n DO
table[i,0] := 1; j := 1;
WHILE j <= n DO
table[i,j] := table[i-1,j];
IF j >= i THEN
table[i,j] := table[i,j] + table[i,j-i]
END;
j := j+1
END;
i := i+1
END;
RETURN table[n,n]
END CountPartitions;Count.For a proposed loop invariant , initial condition , guard , body , and postcondition , establish these Hoare logic obligations:The first establishes before the loop; the second preserves it after each iteration; the third derives the desired result when the guard is false. Together they prove partial correctness. For total correctness, also prove algorithm termination, for example by a nonnegative integer loop variant that strictly decreases whenever the body executes. The loop invariant need not be the postcondition: it generally describes completed and unfinished work together.
Use the convention , as required by the paper's rotation example. Follow each permutation cycle once. Temporarily encode a visited value as , and restore the whole permutation before returning:Require
PROCEDURE Permute(VAR p: ARRAY OF INTEGER);
VAR s, j, next: INTEGER; saved: T;
BEGIN
s := 0;
WHILE s < N DO
IF p[s] >= 0 THEN
IF p[s] = s THEN
p[s] := -p[s]-1
ELSE
saved := a[s]; j := s;
next := p[j]; p[j] := -next-1;
WHILE next # s DO
a[j] := a[next]; j := next;
next := p[j]; p[j] := -next-1
END;
a[j] := saved
END
END;
s := s+1
END;
s := 0;
WHILE s < N DO p[s] := -p[s]-1; s := s+1 END
END Permute;LEN(p) >= N and that p[0..N) is the stipulated permutation. Its cells must be writable and distinct from the data array. The marking map is reversible and fits whenever the original valid indices and fit in the signed integer type.At each outer test, all marked indices belong to fully completed cycles; all unmarked cycles still contain their original data and permutation entries. Within a cycle the same shifting loop invariant as part (d) applies, with successor given by the original
p: all previously filled positions have their final values, and saved preserves the overwritten first source. A permutation has no path into a different cycle, so every new successor is unmarked until it returns to s; no encoded negative value is used as a source index. At outer-loop exit every index is marked and correct. The restoration loop has decoded precisely the indices below s.The procedure uses one extra value of type
T, time, and at most data reads plus data writes, with no data access for fixed points. It returns p unchanged, though it marks it temporarily. If p must remain read-only even during execution, test whether each index is its cycle's smallest member by traversing p before moving that cycle. That preserves constant space and the same optimal data accesses, but can require index-array accesses.Compare the two heads, emitting one copy of a tie and advancing both inputs in that case:
merge :: Ord a => [a] -> [a] -> [a]
merge [] ys = ys
merge xs [] = xs
merge xs@(x:xt) ys@(y:yt)
| x < y = x : merge xt ys
| x > y = y : merge xs yt
| otherwise = x : merge xt ytThe emitted head is the smallest remaining value, so induction preserves strict order and removes duplicates. The extra empty-input cases also make the definition useful on finite lists.
There is a mathematical qualification to the unrestricted infinite-input claim: the union must be enumerable in this order. If one input increases to one from below and the other starts at two, no finite output position can contain two. Locally finite sorted-stream merge excludes this obstruction, and applies to the positive integer streams needed below.
Tie the three scaled streams back to the same shared lazy list:
hamming :: [Integer]
hamming = 1 : merge (map (2*) hamming)
(merge (map (3*) hamming) (map (5*) hamming))The resulting values are precisely the Hamming numbers for nonnegative integers . Every noninitial value has a smaller predecessor obtained by removing a factor, so the recursive demand can be supplied from earlier cells. This is the productivity argument for a sorted multiplicative closure stream. The merge algorithm removes duplicate representations such as , retaining strict order and every generated value.
Choose the representative . Since is prime and does not divide , their greatest common divisor is one. The extended Euclidean algorithm gives the Bezout identityfor some integers . Reducing modulo gives , so the residue class of is a multiplicative inverse of . The argument uses the nonzero condition to exclude the only residue with greatest common divisor .
First prove for every nonnegative integer by mathematical induction. The case is immediate. If it holds at , the binomial theorem and the stated divisibility of the middle binomial coefficients giveThis establishes the induction. For an arbitrary integer , take a nonnegative representative of its residue class; then . When , part (b) supplies a multiplicative inverse modulo . Multiplying the congruence by that inverse proves Fermat's little theorem:
Use exponentiation by squaring, retaining the unprocessed exponent in Writing , the loop invariant is and . If the old exponent is , , multiplying the accumulator by and replacing the remaining power by preserves the invariant. When , the unused squaring may be omitted. The nonnegative integer
k:def eaBinary(n: Int): Double = {
require(n > 0)
var base = 1.0 + 1.0 / n.toDouble
var k = n
var result = 1.0
while (k > 0) {
if ((k & 1) != 0) result *= base
k = k / 2
if (k > 0) base *= base
}
result
}k is a loop variant, since for . At exit the remaining power is one, giving . There are iterations, at most one accumulator multiplication and one squaring per iteration, hence multiplications and constant space.First rearrange the ratios so that the intermediate powers remain moderate:This avoids separately forming enormous powers such as . Perform the two instances of exponentiation by squaring in one while loop, using coupled binary exponentiation:With and , the loop invariant is , , and . Each active branch preserves its own power identity by the parity argument in part (a)(ii), and an inactive branch changes nothing. The loop variant is the mathematical integer , which need not be computed in a fixed-width variable: it decreases whenever either exponent is positive. At exit , , proving the returned expression is . There are iterations, at most four loop multiplications each, and two final multiplications. Converting
def ec(n: Int): Double = {
require(n > 1)
val dn = n.toDouble
var x = 1.0 + 1.0 / dn
var y = 1.0 + 1.0 / (dn - 1.0)
var k = n
var l = n - 1
var p = 1.0
var q = 1.0
while (k > 0 || l > 0) {
if (k > 0) {
if ((k & 1) != 0) p *= x
k /= 2
if (k > 0) x *= x
}
if (l > 0) {
if ((l & 1) != 0) q *= y
l /= 2
if (l > 0) y *= y
}
}
(dn + 1.0) * p - dn * q
}n before adding one avoids integer overflow at Int.MaxValue.The rearrangement avoids floating-point overflow in the original powers, but subtraction still combines two quantities of order to obtain one of order one. Catastrophic cancellation can amplify their accumulated roundoff error. The mathematical approximation error therefore does not imply that the
Double implementation improves indefinitely as n grows.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.Take the abstract state to be a finite sequence of integers, initially empty. Let denote the entry state, the empty sequence, and concatenation. A Scala trait presents the interface; its comments give the mathematical preconditions and postconditions:The double-ended queue has no abstract capacity limit in this first specification. Both removals require nonemptiness; the query leaves the state unchanged, an instance of command-query separation.
trait DoubleEndedQueue {
// Pre: true. Post: result == s0.isEmpty and s == s0.
def isEmpty: Boolean
// Pre: true. Post: s == [x] ++ s0.
def addLeft(x: Int): Unit
// Pre: s0.nonEmpty. Post: result == s0.head, s == s0.tail.
def getLeft(): Int
// Pre: true. Post: s == s0 ++ [x].
def addRight(x: Int): Unit
// Pre: s0.nonEmpty. Post: result == s0.last, s == s0.init.
def getRight(): Int
}Programming language polymorphism means that one interface or program definition can be used with values of multiple types. In object-oriented programming, subtype polymorphism allows references with a common static type to denote instances of different concrete subclasses or implementers of a trait. An object method call uses dynamic dispatch to select the overriding implementation belonging to the receiver's runtime type. For example, different implementations of a playlist interface can respond to the same
insert call without the client testing their concrete classes.Parametric polymorphism uses type parameters, as in
Sequence[A], so the same definition works uniformly for songs, integers, or other elements. Ad hoc polymorphism includes method overloading, where different parameter signatures share a name and the static types determine overload selection. These mechanisms are distinct: selecting an overloaded signature is not the same step as dispatching an override of that selected signature. Class inheritance can supply implementations, but the reusable interface and legal subtyping relationship determine what a polymorphic client may do.Write the bit vector as , with the least significant bit. Its unsigned interpretation of a bit string is , where each . The range is the integers from through ; a leading one contributes positive weight , not a sign.
Not injective; its image of a function is exactly . For example, both and give the ceiling function value . Since , , so the value is never attained. For every positive integer ,Each such interval is nonempty and contained in , with the upper endpoint excluded when . Thus the answer concerning surjectivity depends on the convention for the natural numbers: onto if , and not onto if . The source does not settle that convention.
Pad every number with leading zeros to obtain exactly decimal digits. Its digits are a solution of in nonnegative integers. Conversely every such solution is a valid digit string, since no individual digit can exceed ; its digit sum also excludes the number zero. The stars and bars count therefore givesFor , , so . Thus , with an explicit Big O notation constant and threshold .
False. On every odd positive integer , . Given any proposed Big O notation constant and threshold , choose an odd . Then , contradicting the required bound. The behavior on even integers cannot repair this failure on an unbounded subsequence.
Write for a positive integer moment index. The moment isThis improper integral converges exactly when , namely . For those indices,At the divergence is logarithmic; for it is a power divergence. Thus, in the source's moment notation, the positive integers allowed are exactly . If , no positive integer moment is finite.