This algorithm searches both left AND right subtrees regardless of the comparison result, making it an exhaustive search that examines every element. For 15 elements, it will make exactly 15 comparisons (one per element). Unlike binary search, it doesn't eliminate half the search space. Choice B assumes binary search behavior, C suggests partial optimization that doesn't exist, D miscounts the total comparisons.

This ternary search makes at most ⌊log₃(n)⌋ + 1 recursive calls. For 81 elements: first call examines 81 elements, second examines 27, third examines 9, fourth examines 3, fifth examines 1. If not found, there's no sixth call since the base case is reached. Maximum is 5 calls. Choice A miscounts by not including the final unsuccessful case, C includes the initial non-recursive call, D uses the wrong calculation.

When you encounter binary search questions, focus on how the loop condition affects when the search terminates, especially for unsuccessful searches.

Let's trace through both implementations searching for a non-existent value in an 8-element array. Both start with low = 0 and high = 7.

Implementation A (while (low <= high)):

• Iteration 1: mid = 3, adjust bounds
• Iteration 2: mid = 1 or 5, adjust bounds
• Iteration 3: mid = 0, 2, 6, or 7, adjust bounds
• Iteration 4: Now low > high, so loop terminates This performs 4 iterations.

Implementation B (while (low < high)):

• Iteration 1: mid = 3, adjust bounds
• Iteration 2: mid = 1 or 5, adjust bounds
• Iteration 3: mid = 0, 2, 6, or 7, adjust bounds
• Now low == high, so loop terminates (since condition is low < high) This performs 3 iterations.

The key difference is that Implementation A continues when low == high (one final check), while Implementation B stops as soon as the pointers meet.

Answer choice A is wrong because they don't both perform 3 iterations. Choice B incorrectly states both perform 4 iterations. Choice C reverses the counts—Implementation A does more work, not less.

Study tip: Remember that <= allows one extra iteration compared to < because it includes the equality case. This extra iteration often matters in unsuccessful searches where you need to exhaust all possibilities.

When analyzing hybrid search algorithms, you need to carefully trace through both phases to count the maximum comparisons. This type of question tests your understanding of how different search strategies combine and your ability to analyze worst-case scenarios.

Let's break down this hybrid approach. In the linear phase, the algorithm checks every 4th element of the 64-element array. This means examining positions 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, and 64 — that's 16 comparisons total. The worst case occurs when the target isn't found in any of these positions, requiring all 16 comparisons to identify which 4-element block might contain the target.

For the binary phase, you're searching within a 4-element block. Binary search on 4 elements requires at most 2 comparisons (since $$\log_2(4) = 2$$). You can verify this: compare with the middle element, then with one remaining element based on the result.

Answer A incorrectly calculates 3 comparisons for the binary phase — this would be correct for 5-8 elements, not 4. Answer B uses 15 linear comparisons, suggesting it missed one of the 16 positions that need checking. Answer D claims 17 linear comparisons (impossible since there are only 16 fourth-positions) and 3 binary comparisons (wrong for 4 elements).

Answer C correctly identifies 18 total comparisons: 16 + 2.

Study tip: For algorithm analysis questions, always trace through each phase systematically and remember that binary search on $$n$$ elements requires $$\lceil \log_2(n) \rceil$$ comparisons maximum.

When rounding up, the algorithm creates an infinite loop. For example, with low=0, high=1, standard binary search uses mid=0, but this version uses mid=1. If the target isn't found, low becomes 1, high remains 1, and mid keeps being calculated as 1, creating an infinite loop. Choice A ignores the termination problem, C incorrectly suggests faster performance, D assumes bounds violations that don't necessarily occur.

When you encounter search algorithm problems, you need to carefully trace through each algorithm step-by-step to count the actual comparisons made.

For binary search on this 12-element array, you start at the middle position. With indices 0-11, the middle is position 5 (value 41). Since 47 > 41, you search the right half. The new middle of positions 6-11 is position 8 (value 59). Since 47 < 59, you search positions 6-7. The middle is position 6 (value 47) - found! This takes exactly 3 comparisons at positions 5, 8, and 6.

For linear search, you check each position sequentially starting from index 0: position 0 (12), position 1 (18), position 2 (23), position 3 (29), position 4 (34), position 5 (41), and finally position 6 (47). That's 7 comparisons total.

Answer A incorrectly assumes zero-indexing affects the comparison count - it doesn't change that you still need 7 checks to reach the target. Answer B gives the wrong binary search sequence; the correct path is 5→8→6, not 6→9→7→8. Answer D wrongly suggests linear search needs 8 comparisons, but once you find the target at position 6 (the 7th comparison), you stop - there's no "unsuccessful seventh check."

Answer C correctly identifies both claims as accurate: binary search uses 3 comparisons (positions 5, 8, 6 in that order), and linear search uses 7 comparisons (positions 0 through 6).

Remember to always trace through algorithms step-by-step rather than making assumptions about how many steps they "should" take.

Both algorithms have O(1) best case when the target is found immediately (first position for standard, first or last position for bidirectional). However, bidirectional search has better average performance since it can find elements in the second half of the array faster. Choice B incorrectly states different best cases, C ignores the average performance difference, D incorrectly assigns logarithmic complexity to linear search.