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An explanation of a recursive function named 'mystery' along with an example of its computation. The function takes two integer arguments, 'm' and 'n'. If 'n' equals 1, the function returns 'm'. Otherwise, it returns the sum of 'm' and the result of the function called with arguments 'm' and 'n-1'. The document also includes a table showing the values of 'm' and 'n' for a specific example, as well as the value of 'mystery(m,n)' for that pair.
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Figure 10.2 from Hanley and Koffman (modified) int mystery(int m, int n) { int ans; if (n == 1) /* simple case / ans = m; else / recursive step */ ans = m + mystery(m, n-1); return (ans); } compute mystery(5, 4); Notice m stays the same. Recursion is on n, which is 4 in our example Make a table m n mystery(m,n)
