Exercise 1: Recursive Fibonacci and Its Real Call Count — Possible Solution
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THE FUNCTION
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    fun fib(n) {
        if (n <= 1) { return n; }
        return fib(n - 1) + fib(n - 2);
    }
    print fib(10);

RESULT
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    55   (correct: 0,1,1,2,3,5,8,13,21,34,55 -- the 10th Fibonacci number)

TOTAL RECURSIVE CALLS FOR fib(10)
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    177 total calls to fib() (measured by instrumenting a plain
    Python equivalent with a call counter, since the exercise's own
    Wisp interpreter doesn't expose a call counter directly, and
    the recursive shape is identical either way)

WHY THIS WORKS AS AN ANSWER
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177 is dramatically more than 10, and that's the whole point of this
exercise. fib(10) does NOT ask "what's the 10th Fibonacci number"
once -- it asks fib(9) and fib(8), each of which asks two more
smaller fib() calls, and so on. fib(8) gets computed TWICE: once as
part of fib(9)'s own call to fib(7)+fib(8)... no wait, more precisely:
fib(10) calls fib(9) and fib(8). fib(9) calls fib(8) and fib(7). So
fib(8) is computed once directly from fib(10) and once again from
fib(9) -- a second, entirely separate, entirely redundant recursive
descent down to n=0 and n=1 all over again.

This is the classic "naive recursion recomputes the same subproblem
exponentially many times" pattern -- the same shape Algorithms &
Complexity's own naive-Fibonacci chapter (algo1-9) measured directly:
an O(2^n) call count, not O(n). 177 calls for fib(10) is a small,
concrete instance of exactly that blowup; fib(30) would already be in
the millions of calls using this same naive shape, entirely because
of duplicated work, not because computing a single Fibonacci number
is inherently that expensive. Fixing it (memoization, or an iterative
version using a while loop instead of recursion) is out of this
chapter's own scope, but the call count above is the concrete number
that motivates why it would matter.