## Turn 1

The user wants me to fix the `median` function in `stats.py`.
The current implementation:
```python
def median(values):
    values = sorted(values)
    mid = len(values) // 2
    return values[mid]
```

Let's look at the failing tests:
1. `assert median([3, 1, 2]) == 2`
   Sorted: `[1, 2, 3]`. Length is 3. `mid = 3 // 2 = 1`. `values[1] = 2`. This passes.
2. `assert median([10, 1, 2, 3]) == 2.5`
   Sorted: `[1, 2, 3, 10]`. Length is 4. `mid = 4 // 2 = 2`. `values[2] = 3`. This fails, expected 2.5.
   For an even length list, the median is the average of the two middle elements.
   So, `return (values[mid - 1] + values[mid]) / 2`
3. `assert median([]) is None`
   Sorted: `[]`. Length is 0. `mid = 0 // 2 = 0`. `values[0]` raises `IndexError`. This fails.
   Need to handle empty list: `if not values: return None`

Let's write the corrected function:
```python
def median(values):
    if not values:
        return None
    values = sorted(values)
    mid = len(values) // 2
    if len(values) % 2 == 0:
        return (values[mid - 1] + values[mid]) / 2
    return values[mid]
```

Wait, the prompt says "Return only a unified diff."
Let's construct the unified diff.

Original:
```
def median(values):
    values = sorted(values)
    mid = len(values) // 2
    return values[mid]
```

New:
```
def median(values):
    if not values:
        return None
    values = sorted(values)
    mid = len(values) // 2
    if len(values) % 2 == 0:
        return (values[mid - 1] + values[mid]) / 2
    return values[mid]
```

Let's format the diff:
```diff
--- stats.py
+++ stats.py
@@ -1,4 +1,8 @@
 def median(values):
+    if not values:
+        return None
     values = sorted(values)
     mid = len(values) // 2
-    return values[mid]
+    if len(values) % 2 == 0:
+        return (values[mid - 1] + values[mid]) / 2
+    return values[mid]
```

Wait, is there a simpler way to write the even case?
`return (values[mid - 1] + values[mid]) / 2`
Yes, that's correct.

Let's double check the tests.
`median([3, 1, 2])` -> sorted `[1, 2, 3]`, len 3, odd. `mid = 1`. `values[1] = 2`. Correct.
`median([10, 1, 2, 3])` -> sorted `[1, 2, 3, 10]`, len 4, even. `mid = 2`. `values[1] = 2`, `values[2] = 3`. `(2 + 3) / 2 = 2.5`. Correct.
`median([])` -> empty, returns `None`. Correct.

Let's check the diff format.
```diff
--- stats.py
+++ stats.py
@@ -1,4 +1,8 @@
 def median(values):
+    if not values:
+        return None
     values = sorted(values)
     mid = len(values) // 2
-    return values[mid]
+    if len(values) % 2 == 0:
+        return (values[mid - 1] + values[mid]) / 2
+    return values[mid]
```

Is there any edge case? What if `values` has 1 element?
`median([5])` -> sorted `[5]`, len 1, odd. `mid = 0`. `values[0] = 5`. Correct.

What about negative numbers?
`median([-1, -2, -3])` -> sorted `[-3, -2, -1]`, len 3, odd. `mid = 1`. `values[1] = -2`. Correct.

The diff looks solid.