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Joined 2026-09-21 Message
Incoming links: Structural induction
Retain the first element, filter all its further occurrences out of the tail, and recurse:
removedup :: Eq a => [a] -> [a]
removedup [] = []
removedup (x:xs) = x : removedup (filter (/= x) xs)The boxed specification is . The Haskell type class
Eq is needed for comparing values. Structural induction on a finite list proves duplicate removal: filtering removes every further x, and the recursive result contains each remaining distinct value exactly once. In the worst case of all distinct values, successive filters give quadratic time complexity.The child results supplied by the tree fold are already their depth-first traversals. Prepend the current label to their list concatenation:
dftF :: RTree a -> [a]
dftF = foldRTree visit
where
visit :: a -> [[a]] -> [a]
visit x childLists = x : concat childListsThus . Substituting the equation for the tree fold gives exactly the recursive preorder traversal equation from part (i), proving equality by structural induction on a finite rose tree.
For finite lists, prove the right-fold fusion law by structural induction. The empty case uses the base-image premise:
f (foldr g a []) = f a = b = foldr h b []For the cons case, use the compatibility premise, then the induction hypothesis:
f (foldr g a (x:xs))
= f (g x (foldr g a xs))
= h x (f (foldr g a xs))
= h x (foldr h b xs)
= foldr h b (x:xs)The assumption that
f is a strict function addresses partial spines. Write for the bottom value. A right fold must inspect the outer list constructor, soThe same cons calculation therefore establishes equality for finite partial spines ending in , including partial heads, because compatibility was assumed for all arguments.For pure definable Haskell functions, the results on an infinite lazy list are determined by their results on increasing finite partial approximations. Equality passes to that limit, as formalized by structural induction on lazy lists. ThusFor finite total inputs alone, the empty and cons cases suffice. Strictness is needed to extend the claim to undefined spines; the finite-total proof by itself does not justify the infinite-input claim.
Take This is its standalone principal type; in the right fold,
h = (.):h :: (b -> c) -> (a -> b) -> a -> c
h f g x = f (g x)c specializes to the type of a function from the accumulator to itself.For fixed The second equality is the induction hypothesis. Function extensionality gives the required equality of composed functions. This is an instance of the right-fold fusion law.
f, g and e, prove equality pointwise by structural induction on the list. At [] both sides return e. For x:xs, assume the equality for xs and calculate:foldr f e (map g (x:xs))
= f (g x) (foldr f e (map g xs))
= f (g x) (foldr (h f g) e xs)
= h f g x (foldr (h f g) e xs)
= foldr (h f g) e (x:xs)For the full lazy Haskell interpretation, include a bottom value for an undefined list spine: both sides are bottom there. The same constructor argument covers every finite approximation with an undefined tail. The Scott continuity of the list operations then extends equality to their limits, including infinite lists. Ordinary induction on finite total lists alone proves only that restricted case.
Use an auxiliary parser that returns both a value and the unconsumed suffix. The arithmetic helper is:For the integer-valued expression language, this chooses integer division with A number consumes one token. An operator consumes the two successive complete expressions that follow it, so nested operators are handled by recursion. Structural induction on the expression proves that
applyOp :: Op -> Int -> Int -> Int
applyOp Add x y = x + y
applyOp Sub x y = x - y
applyOp Mul x y = x * y
applyOp Div x y = x `div` ydiv's rounding convention. The result is evaluated under the intended arithmetic model; bounded-Int overflow is separate from parsing.evalP :: Polish -> Int
evalP input =
case parse input of
(value, []) -> value
_ -> error "trailing tokens"
where
parse :: Polish -> (Int, Polish)
parse [] = error "missing operand"
parse (Number n : rest) = (n, rest)
parse (Operator op : rest) =
let (left, rest1) = parse rest
(right, rest2) = parse rest1
in (applyOp op left right, rest2)parse consumes exactly its Polish notation and returns its value. Keeping the suffix is essential: neither operand has a fixed token count. Division by zero need not be treated separately under the stated input convention.The principal type needs an equality type class constraint:The empty list is a prefix of every list, witnessed by the second list itself as the remaining suffix. A nonempty prefix cannot fit in the empty list. Otherwise the first elements must agree, and the remaining first list must be a prefix of the remaining second list. These cases prove correctness by structural induction on the candidate prefix. The
isPrefix :: Eq a => [a] -> [a] -> Bool
isPrefix [] _ = True
isPrefix (_:_) [] = False
isPrefix (x:xs) (y:ys) = x == y && isPrefix xs ysEq a constraint is required by x == y; no ordering constraint is needed.Choose a root. For each vertex , define as the maximum size in its descendant subtree when is included, and as that size when is excluded. For its children ,A leaf has . If is included, all its children must be excluded. If is excluded, either child state is allowed. Different child subtrees have no edges between them because the graph is a tree, so their contributions can be optimized independently. This proves the recurrence by structural induction.
Evaluate the two states in postorder traversal. At root , the answer is . For reconstruction, pick an attaining root state; an included vertex forces every child to its excluded state, while an excluded vertex permits each child to choose its larger state. Record the included vertices during this second traversal, breaking ties consistently.
With adjacency lists, rooting, state evaluation and reconstruction each take time, with stored states and output. This is maximum independent set in a tree.
The source describes an independent set as a proper subset. For a tree with at least one edge, every independent set is automatically proper, so the algorithm meets that convention. If it is enforced literally for the single-vertex tree, return the empty set of size zero there; the usual non-strict subset convention would instead return its one vertex.
In executable Haskell, the algebraic data type declaration uses This is a preorder traversal of a full binary tree, including the leaves. Structural induction proves correctness: the leaf of a rooted tree case has precisely one output, and the fork case concatenates the correct preorder traversals of its two subtrees after its own marker.
data, rather than the printed type. A leaf of a rooted tree contributes False; a fork contributes True followed by its left and right subtrees in that order.data Tree = Leaf | Fork Tree Tree
preorder :: Tree -> [Bool]
preorder Leaf = [False]
preorder (Fork l r) = True : (preorder l ++ preorder r)An equivalent version avoids repeatedly copying the left list through list concatenation:Its strengthening is
preorder t = walk t []
where
walk Leaf rest = False : rest
walk (Fork l r) rest = True : walk l (walk r rest)walk t rest = preorder t ++ rest. Each vertex constructs one list cell, so this version has linear time complexity. The concise result isMaintain a cursor to the first unread marker. A fork consumes its own marker and then two complete encodings, in left-to-right order.Start with
PROCEDURE Decode(VAR code: ARRAY OF INTEGER;
VAR pos: INTEGER): Tree;
VAR marker: INTEGER; left, right: Tree;
BEGIN
marker := code[pos]; INC(pos);
IF marker = 0 THEN RETURN 0 END;
ASSERT(marker = 1);
left := Decode(code, pos);
right := Decode(code, pos);
RETURN NewFork(left, right)
END Decode;pos := 0; t := Decode(code, pos), with sufficient unused arena space. The valid-input precondition ensures that every read exists and every fork has two encoded subtrees. Evaluating the two recursive calls in separate statements fixes their order.Structural induction shows that decoding reconstructs the same binary tree shape and leaves the cursor just after its encoding. The tree handles need not coincide with the original handles: the abstract tree does. The result isThe time complexity is linear in the encoding length, with call stack space bounded by the height of a binary tree.
Use pattern matching on the two data constructors, concatenating the left subtree's inorder traversal, the root label, and the right subtree's inorder traversal:Here
data Tree = Leaf | Fork Tree Int Tree
inorder :: Tree -> [Int]
inorder Leaf = []
inorder (Fork l x r) = inorder l ++ [x] ++ inorder rLeaf represents the empty binary tree, rather than a labelled leaf. The printed declaration uses type; an executable Haskell algebraic data type requires data, as above. By structural induction, the two equations return precisely the prescribed inorder traversal:Give the visiting procedure a procedure type:Structural induction establishes algorithm correctness: an empty binary tree makes no calls; otherwise the first recursive call visits precisely the left subtree, then the root is visited once, then the second recursive call visits precisely the right subtree. Assume
TYPE Visit = PROCEDURE(x: INTEGER);
PROCEDURE Inorder(t: Tree; p: Visit);
BEGIN
IF t # NIL THEN
Inorder(t.left, p);
p(t.label);
Inorder(t.right, p)
END
END Inorder;p does not modify the binary tree. With constant-cost p, there is constant work per node and empty child occurrence:where is the height of a binary tree and storage is the call stack.Split the half-open array segment at its midpoint, using one entry for the root:The precondition is . Set . For the child sizes are and , differing by at most one. The same construction in both subtrees proves subtree-size balance at every node. Structural induction also gives the inorder traversal , followed by , followed by , hence precisely ; no sortedness assumption is needed.
PROCEDURE MakeTree(VAR u: ARRAY OF INTEGER;
a, b: INTEGER): Tree;
VAR m: INTEGER; t: Tree;
BEGIN
IF a = b THEN RETURN NIL END;
m := a + (b-a) DIV 2;
NEW(t);
t.label := u[m];
t.left := MakeTree(u, a, m);
t.right := MakeTree(u, m+1, b);
RETURN t
END MakeTree;Exactly records are allocated. Each nonempty recursive call makes two child calls, so there are calls including empty segments. Under constant-cost allocation and reference parameter passing,For positive this is proportional to ; the empty case takes constant time. The call stack has depth.
Store one entry per occurrence in an immutable list. Use lower type bounds for arguments that might widen the element type, and remove only the first matching occurrence:A covariant type parameter cannot appear directly as an ordinary object method argument type. The fresh with permits insertion, deletion queries and count queries without imposing such a negative occurrence of .
class CoMSet[+T](c: List[T]) {
private val elems: List[T] = c
def insert[U >: T](x: U): CoMSet[U] =
new CoMSet[U](x :: elems)
def delete[U >: T](x: U): CoMSet[T] = {
def removeOne(xs: List[T]): List[T] = xs match {
case Nil => Nil
case h :: tail =>
if (h == x) tail else h :: removeOne(tail)
}
new CoMSet[T](removeOne(elems))
}
def count[U >: T](x: U): Int =
elems.count(y => y == x)
def union[U >: T](that: CoMSet[U]): CoMSet[U] =
new CoMSet[U](elems ::: that.elems)
}delete can retain result type CoMSet[T] because it introduces no new elements; widening its result to CoMSet[U] would also be valid. union must widen because the other multiset may contain the broader type. The class-private immutable elems is accessible on another instance of the same software class; it is not private[this].[ref]Scala 2.11 basic declarations and definitions specification. Type Parameters, Variance Annotations; variable and method parameter positions
Let be a multiset multiplicity. The implementations satisfyThe brackets denote an indicator function. Structural induction on the list proves
removeOne removes precisely one equal entry if present, otherwise none. filterNot would incorrectly remove every occurrence. List concatenation adds multiplicities, implementing the paper's additive additive multiset union, rather than the maximum-multiplicity convention sometimes used for that name.All methods preserve the original immutable object.
insert takes constant time through structural sharing; count and delete take time, and union takes time to copy the receiver's list spine. Equality is ordinary stable value equality; finite representable counts are assumed.Use structural induction on the finite list
xs, with the assertion quantified over every accumulator e and every list ys. The equations of the left fold are foldl f e [] = e and foldl f e (x:xs) = foldl f (f e x) xs.For The third equality uses the induction hypothesis with accumulator
xs = [], both sides reduce to foldl f e ys. For the constructor case, equational reasoning givesfoldl f e ((x:xs) ++ ys)
= foldl f e (x : (xs ++ ys))
= foldl f (f e x) (xs ++ ys)
= foldl f (foldl f (f e x) xs) ys
= foldl f (foldl f e (x:xs)) ysf e x. Hence the stated identity holds. No inspection of ys was required: in Haskell this equality also permits infinite or partially defined ys, with both expressions denoting the same possibly undefined result.Again use structural induction on the finite list. At The penultimate equality uses the induction hypothesis. Thus
[], both the right fold and the left fold return e. For x:xs, use reverse (x:xs) = reverse xs ++ [x], part (a), and the definition of flip:foldl (flip f) e (reverse (x:xs))
= foldl (flip f) e (reverse xs ++ [x])
= foldl (flip f) (foldl (flip f) e (reverse xs)) [x]
= flip f (foldl (flip f) e (reverse xs)) x
= f x (foldl (flip f) e (reverse xs))
= f x (foldr f e xs)
= foldr f e (x:xs)foldr f e xs = foldl (flip f) e (reverse xs). Finiteness matters: reversing an infinite list does not produce its first element, whereas some right folds do.Compute either the common depth of the leaves of a rooted tree, or failure when the depths differ. A tree fold can return a A leaf has depth zero. An internal vertex has equally deep leaves exactly when every child rose tree is perfect and their returned depths agree; adding one accounts for the edge to each child. This proves the stated meaning of the result by structural induction. In particular, unary branching is allowed, and a single leaf is perfect.
Maybe Integer: Just h means that the rose tree is a perfect rose tree with all leaves at depth relative to its root; Nothing means it is not perfect.perfect :: RTree a -> Bool
perfect t =
case foldRTree step t of
Nothing -> False
Just _ -> True
where
step :: a -> [Maybe Integer] -> Maybe Integer
step _ [] = Just 0
step _ (Nothing : _) = Nothing
step _ (Just h : hs)
| all (== Just h) hs = Just (h + 1)
| otherwise = Nothingperfect returns precisely the required Boolean. Its time complexity is linear in the number of vertices, assuming unit-cost depth arithmetic; labels are never compared or evaluated.The constructors For example,
ZigType and ZagType record which of the two alternating positions contains the last element. The tail types swap the two type parameters, so the return type of lastZag swaps them back:lastZig :: Zig a b -> ZigOrZagType a b
lastZig Nil = error "lastZig: empty input"
lastZig (Cins x Nal) = ZigType x
lastZig (Cins _ zs) = lastZag zs
lastZag :: Zag a b -> ZigOrZagType b a
lastZag Nal = error "lastZag: empty input"
lastZag (Cans x Nil) = ZagType x
lastZag (Cans _ zs) = lastZig zslastZig passes a tail of type Zag b a to lastZag, whose result then has type ZigOrZagType a b, exactly as required. Conversely, lastZag passes a Zig b a tail to lastZig. A singleton returns its element in the appropriate constructor; a longer input discards one element and uses mutual recursion on the shorter tail. This proves correctness by structural induction on the finite alternating sequence. The three source examples give ZigType 2, ZagType "D", and ZigType True.Build columns by prepending the first row's entries to the columns obtained from the remaining rows:The singleton case turns each entry into a one-element column. In the recursive case,
transpose :: Matrix a -> Matrix a
transpose [] = []
transpose [row] = map (:[]) row
transpose (row : rows) = zipWith (:) row (transpose rows)transpose rows has exactly as many columns as row has entries, so zipWith (:) extends every column once. Structural induction on the number of rows proves that each output column contains the corresponding input entries in order. This uses list zip to implement the transpose, without requiring Num a.For one element, return it as both extrema without a comparison. Otherwise compare consecutive pairs once, put each smaller element in a candidate array and each larger element in a candidate array . If one element is unpaired, put it in both candidate arrays. This simultaneous minimum and maximum construction computes the minimum of and maximum of by balanced divide-and-conquer algorithms: split the candidate array, recursively find the desired extremum in each half, and compare the two returned values. These extremum routines use one comparison per merge and therefore comparisons on candidates.
Every element excluded from is at least its paired element, so a global minimum remains in ; likewise a global maximum remains in . The recursive extremum routines are correct by structural induction. For even , there are pair comparisons and two searches on candidates, givingFor odd , there are pair comparisons and two searches on candidates, givingTogether with , both cases giveFor powers of two, the equivalent direct divide-and-conquer algorithm returns the ordered pair of extrema for each half, then combines the two minima with one comparison and the two maxima with one comparison. Its base cases must include , rather than performing two unnecessary comparisons on a pair. The recurrence relation is for , whence . Arbitrary near-equal splitting without the initial pairing can exceed the desired comparison bound for other values of ; the pairing construction above handles every positive size.
For finite, defined lists, structural induction requires and for every head and tail . This proves for every finite list.
An infinite list cannot be reached by finitely many constructor steps. For structural induction on lazy lists, also check , where is the bottom value, and require an admissible predicate on lazy lists: if increasingly defined finite approximations all satisfy , their limit does too. Together with the empty-list and constructor cases this covers finite, partial and infinite lists. Merely asserting the two finite-list cases is insufficient; “the list is finite” itself satisfies those cases but fails on infinite lists.