Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/imperative-programming-resit/3/d/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Imperative programming (resit) / 3 / d / Solution by Codex 0 2026-10-31
Allocate nodes only as required, explicitly initializing every field:
PROCEDURE Make(VAR t: Trie);
VAR c: INTEGER;
BEGIN
  NEW(t); t.terminal := FALSE; c := 0;
  WHILE c < 26 DO t.child[c] := NIL; c := c+1 END
END Make;

PROCEDURE AddTail(VAR t: Trie; s: ARRAY OF CHAR; i: INTEGER);
BEGIN
  IF t = NIL THEN Make(t) END;
  IF s[i] = 0X THEN
    t.terminal := TRUE
  ELSE
    AddTail(t.child[Index(s[i])], s, i+1)
  END
END AddTail;

PROCEDURE Add(VAR t: Trie; s: ARRAY OF CHAR);
BEGIN
  AddTail(t, s, 0)
END Add;
Here 0X is the zero character, and s must contain its terminator within its array bounds. Make establishes that the first c child fields are NIL as its loop invariant. Passing t by reference lets the base allocation replace a missing root or child pointer.
For correctness, use induction on the number of characters before the terminator. At length zero, setting terminal adds exactly the empty suffix. Otherwise the chosen child stores exactly words with the required next letter, and the induction hypothesis adds the remaining suffix while preserving all other words. Thus the resulting set is the old set union the supplied word, ignoring case. Repeated insertion changes no word membership. A newly allocated path reaches the newly terminal node, preserving the canonical representation invariant after the procedure finishes. The running time is for a word of length , using call stack space without optimization.
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