Exercise 1: Hand-Tracing MaxOfDivideConquer on [5, 12, 3, 8] — Possible Solution
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THE SPLIT
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list = [5, 12, 3, 8], length 4, mid = 4/2 = 2
Left half:  [5, 12]  (indices 0 to 1)
Right half: [3, 8]   (indices 2 to 3)

SOLVING THE LEFT HALF: MaxOfDivideConquer([5, 12])
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Length is 2, not 1, so split again: mid = 1
  Left:  [5]  -> base case, length 1, RETURN 5
  Right: [12] -> base case, length 1, RETURN 12
  Combine: left_max=5, right_max=12. Since 12 > 5, RETURN 12.

SOLVING THE RIGHT HALF: MaxOfDivideConquer([3, 8])
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Length is 2, split again: mid = 1
  Left:  [3] -> base case, length 1, RETURN 3
  Right: [8] -> base case, length 1, RETURN 8
  Combine: left_max=3, right_max=8. Since 8 > 3, RETURN 8.

FINAL COMBINE
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left_max (from [5,12]) = 12
right_max (from [3,8]) = 8
Since 12 > 8, RETURN 12.

FINAL ANSWER: 12

WHY THIS IS CORRECT
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12 is indeed the largest value in [5, 12, 3, 8] - confirmed by
inspection, and by the recursive process itself never discarding any
element without comparing it: every one of the four original values
was eventually part of some base-case return value, and every combine
step correctly kept the larger of its two inputs.

WHY THIS WORKS AS AN ANSWER
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The trace follows the recursion down to every base case explicitly
before combining back up, mirroring exactly the split-then-combine
structure this chapter's own worked example used, rather than jumping
straight to the final answer.