> For the complete documentation index, see [llms.txt](https://wenbo-notes.gitbook.io/cs1010-notes/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://wenbo-notes.gitbook.io/cs1010-notes/lec-tut-lab-exes/lecture/lec-10-more-recursion/diagnostic-quiz.md).

# Diagnostic Quiz

## Problems

### 1. Tower of Hanoi

> *Disk `i-1` is moved twice the number of times of Disk `i`*

It is **correct.** This can be reasoned using the recurrence relation $$T(n)=2T(n-1)+1$$ for the Tower of Hanoi problem.

### 5. Time Complexity for Nqueens

> *Consider the solution to N-Queens given in class. Suppose the running time of `nqueens` is* $$T(n)$$*. What is* $$T(1)$$*?*

This is an awesome question. $$T(1)$$ happens when we reach the **base case**, so here we need to think about what will happen in the **base case**.

The base case is:

{% code lineNumbers="true" %}

```c
if (row == n - 1) {
  if (!threaten_each_other_diagonally(queens, n - 1)) {
    cs1010_println_string(queens);
    return true;
  }
  return false;
}
```

{% endcode %}

It calls `threaten_each_other_diagonally` which runs in $$O(n)$$.

```c
bool threaten_each_other_diagonally(char queens[], size_t last_row) {
  for (size_t curr_row = 0; curr_row <= last_row; curr_row += 1) {
    if (has_a_queen_in_diagonal(queens, curr_row, last_row)) {
      return true;
    }
  }
  return false;
}
```

And print a string with length `n` takes $$O(n)$$. So, the overall running time for $$T(1)$$ is $$O(n^2)$$.