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Joined 2026-09-21 Message
Incoming links: Recursive call
A strict majority element in the whole array must be a strict majority element in at least one half: otherwise summing the two non-majority element frequency bounds would give at most half of all entries. This observation supplies at most two candidates, but they must still be verified in the whole subarray.
majority(A, left, right):
if left == right: return NoMajority
if right - left == 1: return Candidate(A[left])
middle = floor((left + right)/2)
first = majority(A, left, middle)
second = majority(A, middle, right)
for candidate in the two returned tagged results:
if candidate is Candidate(x):
count = 0
for i from left to right-1:
if A[i] == x: count = count + 1
if 2*count > right-left: return Candidate(x)
return NoMajorityNoMajority is a separate result tag, not a possible input element. The two-candidate loop does not require ordering, hashing or conversion of arbitrary elements to a set. It may count the same candidate twice without changing correctness or the bound.Induction proves the recursive candidates correct. If the whole subarray has a majority element, the observation ensures it reaches the verification step; if a candidate passes the strict frequency test, it really is a majority element. The worst-case recurrence has two balanced recursive calls and linear combination work, so the running time is , with auxiliary recursion depth when the array is passed by index ranges. A different cancellation-based method can achieve linear time, but is not needed for this divide-and-conquer construction.
Use the empty reference as the base case:Induction on the tree proves the order: the first recursive call prints exactly the left subtree, then the procedure prints the root, then the final call prints exactly the right subtree. A finite acyclic tree supplies decreasing proper subtrees, so the recursion terminates. The traversal is valid for arbitrary integer labels; sorted output would additionally require a binary search tree invariant.
def printInorder(root: Tree): Unit = {
if (root != null) {
printInorder(root.left)
println(root.datum)
printInorder(root.right)
}
}Split the index range at its middle:For , the child sizes are and , differing by at most one. Induction gives the same property at every descendant, so the result is nearly balanced. Its inorder traversal is the initial subrange, then , then the final subrange, reproducing exactly with every occurrence retained.
def MakeTree(u: Array[Int], a: Int, b: Int): Tree = {
require(0 <= a && a <= b && b <= u.length)
if (a == b) null
else {
val middle = a + (b - a) / 2
new Tree(u(middle),
MakeTree(u, a, middle),
MakeTree(u, middle + 1, b))
}
}There is one nonempty recursive call per constructed node and one constant-time base call per empty child position. Therefore construction takes time for , or including the empty case, and allocates exactly nodes. The balanced recursion has logarithmic stack depth. The input array need not be sorted: the required property concerns subtree sizes and inorder order, not a search-tree ordering.
To preserve the actual printed function, including its handling of nonpositive exponents and wrapping
Int bases, use a tail-recursive accumulator with the same multiplication order:def anotherfuncRec(x: Int, n: Long): Double = {
require(x > 0)
@scala.annotation.tailrec
def loop(y: Int, i: Long, result: Double): Double = {
if (i <= 0) result
else {
val nextResult = if (i % 2 == 1) result * y else result
loop(y * y, i / 2, nextResult)
}
}
loop(x, n, 1.0)
}Each recursive call corresponds to one source iteration and carries exactly its next program state. Hence it reproduces the printed numerical behavior, rather than silently repairing it. For positive
n, both routines use steps; for nonpositive n, both take constant time. The recursive call is in tail position, so Scala's tail-call optimization avoids a growing call stack; the annotation asks the compiler to check that condition.If the intended exact power is required instead, change the base and accumulator parameters of this same recurrence to
BigInt and require n >= 0, as in part (d). Its mathematical loop invariant becomes a recursion invariant, with the same logarithmic number of multiplication steps. Recursion (computer science) Created 2026-10-13 Updated 2026-10-18
Recursion solves a problem by invoking the same procedure on smaller instances, with base cases answered directly. For merge sort, an empty or one-entry array is already sorted, and every larger array is split into two shorter parts. The decreasing lengths justify termination; a self-call without a progress rule would not suffice.
A recursive procedure refers to itself, directly or through another procedure. A recursive call starts another invocation, with its own arguments and local variables; the caller can wait for its returned output. A base case supplies an answer without further recursive calls. To establish algorithm termination, also show that every recursive call progresses in a well-founded relation. “Smaller” means smaller in that chosen relation: argument values need not all decrease, and the arguments need not even be numbers.
Recursion is especially expressive when a data structure already decomposes into smaller structures. A binary tree has two child subtrees, so structural recursion can follow its shape. A short rule for each case can describe an entire traversal or computation. The distinction between a recursive procedure and the recursive process it generates helps connect that compact description to its resource use.[ref]
Structure and Interpretation of Computer Programs by Harold Abelson and others. Structure and Interpretation of Computer Programs, second edition, §1.2.1: procedure versus process; §1.2.2: closing discussion before Counting change, hierarchical data and straightforward definitions
Recursion invariant 2026-10-24
A relationship between a recursive procedure's arguments, accumulated state and original input that holds at every reachable recursive call. For accumulator-based exponentiation by squaring, is a recursion invariant under exact arithmetic. The base case converts it into the desired result, while the recursive step and a decreasing argument justify preservation and termination.
Well-founded progress in recursion 2026-10-18
A base case and a decreasing natural number measure work together. Check that every invocation satisfies its precondition, each base case finishes with the correct output, and every other invocation makes finitely many recursive calls with strictly smaller measure. All local work must also terminate. By well-founded induction, the children finish, so their caller finishes; if their answers are correct and the combination rule preserves the postcondition, the caller is correct too. This proves total correctness, not just correctness conditional on return.
Strict decrease without a lower bound is insufficient: subtracting one forever from a negative integer never reaches zero. Even a positive real-valued measure can decrease indefinitely, as shows. A well-founded relation rules out that descent. In structural recursion, a proper child subtree of a finite binary tree supplies progress without numeric arguments. The proof obligation is the relation between parent and child instances, together with terminating local work and complete base cases.