We are currently solving the following issue within our repository. Here is the issue text:
--- BEGIN ISSUE ---
Merge insertion sort doesn't work
```
>>> merge_insertion_sort([0, 1, 2, 3, 4])
[0, 2, 1, 3, 4]
```

Cc @ulwlu #2211 [`sorts/merge_insertion_sort.py`](../blob/master/sorts/merge_insertion_sort.py)

--- END ISSUE ---

Below are some code segments, each from a relevant file. One or more of these files may contain bugs.
--- BEGIN FILES ---
Path: `sorts/merge_insertion_sort.py`
Content:
```
1 """
2 This is a pure Python implementation of the merge-insertion sort algorithm
3 Source: https://en.wikipedia.org/wiki/Merge-insertion_sort
4 
5 For doctests run following command:
6 python3 -m doctest -v merge_insertion_sort.py
7 or
8 python -m doctest -v merge_insertion_sort.py
9 
10 For manual testing run:
11 python3 merge_insertion_sort.py
12 """
13 
14 from __future__ import annotations
15 
16 
17 def merge_insertion_sort(collection: list[int]) -> list[int]:
18     """Pure implementation of merge-insertion sort algorithm in Python
19 
20     :param collection: some mutable ordered collection with heterogeneous
21     comparable items inside
22     :return: the same collection ordered by ascending
23 
24     Examples:
25     >>> merge_insertion_sort([0, 5, 3, 2, 2])
26     [0, 2, 2, 3, 5]
27 
28     >>> merge_insertion_sort([99])
29     [99]
30 
31     >>> merge_insertion_sort([-2, -5, -45])
32     [-45, -5, -2]
33     """
34 
35     def binary_search_insertion(sorted_list, item):
36         left = 0
37         right = len(sorted_list) - 1
38         while left <= right:
39             middle = (left + right) // 2
40             if left == right:
41                 if sorted_list[middle] < item:
42                     left = middle + 1
43                 break
44             elif sorted_list[middle] < item:
45                 left = middle + 1
46             else:
47                 right = middle - 1
48         sorted_list.insert(left, item)
49         return sorted_list
50 
51     def sortlist_2d(list_2d):
52         def merge(left, right):
53             result = []
54             while left and right:
55                 if left[0][0] < right[0][0]:
56                     result.append(left.pop(0))
57                 else:
58                     result.append(right.pop(0))
59             return result + left + right
60 
61         length = len(list_2d)
62         if length <= 1:
63             return list_2d
64         middle = length // 2
65         return merge(sortlist_2d(list_2d[:middle]), sortlist_2d(list_2d[middle:]))
66 
67     if len(collection) <= 1:
68         return collection
69 
70     """
71     Group the items into two pairs, and leave one element if there is a last odd item.
72 
73     Example: [999, 100, 75, 40, 10000]
74                 -> [999, 100], [75, 40]. Leave 10000.
75     """
76     two_paired_list = []
77     has_last_odd_item = False
78     for i in range(0, len(collection), 2):
79         if i == len(collection) - 1:
80             has_last_odd_item = True
81         else:
82             """
83             Sort two-pairs in each groups.
84 
85             Example: [999, 100], [75, 40]
86                         -> [100, 999], [40, 75]
87             """
88             if collection[i] < collection[i + 1]:
89                 two_paired_list.append([collection[i], collection[i + 1]])
90             else:
91                 two_paired_list.append([collection[i + 1], collection[i]])
92 
93     """
94     Sort two_paired_list.
95 
96     Example: [100, 999], [40, 75]
97                 -> [40, 75], [100, 999]
98     """
99     sorted_list_2d = sortlist_2d(two_paired_list)
100 
101     """
102     40 < 100 is sure because it has already been sorted.
103     Generate the sorted_list of them so that you can avoid unnecessary comparison.
104 
105     Example:
106            group0 group1
107            40     100
108            75     999
109         ->
110            group0 group1
111            [40,   100]
112            75     999
113     """
114     result = [i[0] for i in sorted_list_2d]
115 
116     """
117     100 < 999 is sure because it has already been sorted.
118     Put 999 in last of the sorted_list so that you can avoid unnecessary comparison.
119 
120     Example:
121            group0 group1
122            [40,   100]
123            75     999
124         ->
125            group0 group1
126            [40,   100,   999]
127            75
128     """
129     result.append(sorted_list_2d[-1][1])
130 
131     """
132     Insert the last odd item left if there is.
133 
134     Example:
135            group0 group1
136            [40,   100,   999]
137            75
138         ->
139            group0 group1
140            [40,   100,   999,   10000]
141            75
142     """
143     if has_last_odd_item:
144         pivot = collection[-1]
145         result = binary_search_insertion(result, pivot)
146 
147     """
148     Insert the remaining items.
149     In this case, 40 < 75 is sure because it has already been sorted.
150     Therefore, you only need to insert 75 into [100, 999, 10000],
151     so that you can avoid unnecessary comparison.
152 
153     Example:
154            group0 group1
155            [40,   100,   999,   10000]
156             ^ You don't need to compare with this as 40 < 75 is already sure.
157            75
158         ->
159            [40,   75,    100,   999,   10000]
160     """
161     is_last_odd_item_inserted_before_this_index = False
162     for i in range(len(sorted_list_2d) - 1):
163         if result[i] == collection[-i]:
164             is_last_odd_item_inserted_before_this_index = True
165         pivot = sorted_list_2d[i][1]
166         # If last_odd_item is inserted before the item's index,
167         # you should forward index one more.
168         if is_last_odd_item_inserted_before_this_index:
169             result = result[: i + 2] + binary_search_insertion(result[i + 2 :], pivot)
170         else:
171             result = result[: i + 1] + binary_search_insertion(result[i + 1 :], pivot)
172 
173     return result
174 
175 
176 if __name__ == "__main__":
177     user_input = input("Enter numbers separated by a comma:\n").strip()
178     unsorted = [int(item) for item in user_input.split(",")]
179     print(merge_insertion_sort(unsorted))
180
```
--- END FILES ---

Please first localize the bug based on the issue statement, and then generate a patch according to the `git diff` format fenced by three backticks.

Here is an example:

```diff
diff --git a/examples/server_async.py b/examples/server_async.py
--- a/examples/server_async.py
+++ b/examples/server_async.py
@@ -313,4 +313,4 @@
 
 
 if __name__ == "__main__":
-    asyncio.run(run_async_server("."), debug=True)
+    asyncio.run(run_async_server(), debug=True)
diff --git a/examples/server_sync.py b/examples/server_sync.py
--- a/examples/server_sync.py
+++ b/examples/server_sync.py
@@ -313,5 +313,5 @@
 
 
 if __name__ == "__main__":
-    server = run_sync_server(".")
+    server = run_sync_server()
     server.shutdown()

```

diff --git a/sorts/merge_insertion_sort.py b/sorts/merge_insertion_sort.py
--- a/sorts/merge_insertion_sort.py
+++ b/sorts/merge_insertion_sort.py
@@ -30,6 +30,12 @@
 
     >>> merge_insertion_sort([-2, -5, -45])
     [-45, -5, -2]
+
+    Testing with all permutations on range(0,5):
+    >>> import itertools
+    >>> permutations = list(itertools.permutations([0, 1, 2, 3, 4]))
+    >>> all(merge_insertion_sort(p) == [0, 1, 2, 3, 4] for p in permutations)
+    True
     """
 
     def binary_search_insertion(sorted_list, item):
@@ -160,7 +166,7 @@
     """
     is_last_odd_item_inserted_before_this_index = False
     for i in range(len(sorted_list_2d) - 1):
-        if result[i] == collection[-i]:
+        if result[i] == collection[-1] and has_last_odd_item:
             is_last_odd_item_inserted_before_this_index = True
         pivot = sorted_list_2d[i][1]
         # If last_odd_item is inserted before the item's index,

{"golden_diff": "diff --git a/sorts/merge_insertion_sort.py b/sorts/merge_insertion_sort.py\n--- a/sorts/merge_insertion_sort.py\n+++ b/sorts/merge_insertion_sort.py\n@@ -30,6 +30,12 @@\n \n     >>> merge_insertion_sort([-2, -5, -45])\n     [-45, -5, -2]\n+\n+    Testing with all permutations on range(0,5):\n+    >>> import itertools\n+    >>> permutations = list(itertools.permutations([0, 1, 2, 3, 4]))\n+    >>> all(merge_insertion_sort(p) == [0, 1, 2, 3, 4] for p in permutations)\n+    True\n     \"\"\"\n \n     def binary_search_insertion(sorted_list, item):\n@@ -160,7 +166,7 @@\n     \"\"\"\n     is_last_odd_item_inserted_before_this_index = False\n     for i in range(len(sorted_list_2d) - 1):\n-        if result[i] == collection[-i]:\n+        if result[i] == collection[-1] and has_last_odd_item:\n             is_last_odd_item_inserted_before_this_index = True\n         pivot = sorted_list_2d[i][1]\n         # If last_odd_item is inserted before the item's index,\n", "issue": "Merge insertion sort doesn't work\n```\r\n>>> merge_insertion_sort([0, 1, 2, 3, 4])\r\n[0, 2, 1, 3, 4]\r\n```\r\n\r\nCc @ulwlu #2211 [`sorts/merge_insertion_sort.py`](../blob/master/sorts/merge_insertion_sort.py)\n", "before_files": [{"content": "\"\"\"\nThis is a pure Python implementation of the merge-insertion sort algorithm\nSource: https://en.wikipedia.org/wiki/Merge-insertion_sort\n\nFor doctests run following command:\npython3 -m doctest -v merge_insertion_sort.py\nor\npython -m doctest -v merge_insertion_sort.py\n\nFor manual testing run:\npython3 merge_insertion_sort.py\n\"\"\"\n\nfrom __future__ import annotations\n\n\ndef merge_insertion_sort(collection: list[int]) -> list[int]:\n    \"\"\"Pure implementation of merge-insertion sort algorithm in Python\n\n    :param collection: some mutable ordered collection with heterogeneous\n    comparable items inside\n    :return: the same collection ordered by ascending\n\n    Examples:\n    >>> merge_insertion_sort([0, 5, 3, 2, 2])\n    [0, 2, 2, 3, 5]\n\n    >>> merge_insertion_sort([99])\n    [99]\n\n    >>> merge_insertion_sort([-2, -5, -45])\n    [-45, -5, -2]\n    \"\"\"\n\n    def binary_search_insertion(sorted_list, item):\n        left = 0\n        right = len(sorted_list) - 1\n        while left <= right:\n            middle = (left + right) // 2\n            if left == right:\n                if sorted_list[middle] < item:\n                    left = middle + 1\n                break\n            elif sorted_list[middle] < item:\n                left = middle + 1\n            else:\n                right = middle - 1\n        sorted_list.insert(left, item)\n        return sorted_list\n\n    def sortlist_2d(list_2d):\n        def merge(left, right):\n            result = []\n            while left and right:\n                if left[0][0] < right[0][0]:\n                    result.append(left.pop(0))\n                else:\n                    result.append(right.pop(0))\n            return result + left + right\n\n        length = len(list_2d)\n        if length <= 1:\n            return list_2d\n        middle = length // 2\n        return merge(sortlist_2d(list_2d[:middle]), sortlist_2d(list_2d[middle:]))\n\n    if len(collection) <= 1:\n        return collection\n\n    \"\"\"\n    Group the items into two pairs, and leave one element if there is a last odd item.\n\n    Example: [999, 100, 75, 40, 10000]\n                -> [999, 100], [75, 40]. Leave 10000.\n    \"\"\"\n    two_paired_list = []\n    has_last_odd_item = False\n    for i in range(0, len(collection), 2):\n        if i == len(collection) - 1:\n            has_last_odd_item = True\n        else:\n            \"\"\"\n            Sort two-pairs in each groups.\n\n            Example: [999, 100], [75, 40]\n                        -> [100, 999], [40, 75]\n            \"\"\"\n            if collection[i] < collection[i + 1]:\n                two_paired_list.append([collection[i], collection[i + 1]])\n            else:\n                two_paired_list.append([collection[i + 1], collection[i]])\n\n    \"\"\"\n    Sort two_paired_list.\n\n    Example: [100, 999], [40, 75]\n                -> [40, 75], [100, 999]\n    \"\"\"\n    sorted_list_2d = sortlist_2d(two_paired_list)\n\n    \"\"\"\n    40 < 100 is sure because it has already been sorted.\n    Generate the sorted_list of them so that you can avoid unnecessary comparison.\n\n    Example:\n           group0 group1\n           40     100\n           75     999\n        ->\n           group0 group1\n           [40,   100]\n           75     999\n    \"\"\"\n    result = [i[0] for i in sorted_list_2d]\n\n    \"\"\"\n    100 < 999 is sure because it has already been sorted.\n    Put 999 in last of the sorted_list so that you can avoid unnecessary comparison.\n\n    Example:\n           group0 group1\n           [40,   100]\n           75     999\n        ->\n           group0 group1\n           [40,   100,   999]\n           75\n    \"\"\"\n    result.append(sorted_list_2d[-1][1])\n\n    \"\"\"\n    Insert the last odd item left if there is.\n\n    Example:\n           group0 group1\n           [40,   100,   999]\n           75\n        ->\n           group0 group1\n           [40,   100,   999,   10000]\n           75\n    \"\"\"\n    if has_last_odd_item:\n        pivot = collection[-1]\n        result = binary_search_insertion(result, pivot)\n\n    \"\"\"\n    Insert the remaining items.\n    In this case, 40 < 75 is sure because it has already been sorted.\n    Therefore, you only need to insert 75 into [100, 999, 10000],\n    so that you can avoid unnecessary comparison.\n\n    Example:\n           group0 group1\n           [40,   100,   999,   10000]\n            ^ You don't need to compare with this as 40 < 75 is already sure.\n           75\n        ->\n           [40,   75,    100,   999,   10000]\n    \"\"\"\n    is_last_odd_item_inserted_before_this_index = False\n    for i in range(len(sorted_list_2d) - 1):\n        if result[i] == collection[-i]:\n            is_last_odd_item_inserted_before_this_index = True\n        pivot = sorted_list_2d[i][1]\n        # If last_odd_item is inserted before the item's index,\n        # you should forward index one more.\n        if is_last_odd_item_inserted_before_this_index:\n            result = result[: i + 2] + binary_search_insertion(result[i + 2 :], pivot)\n        else:\n            result = result[: i + 1] + binary_search_insertion(result[i + 1 :], pivot)\n\n    return result\n\n\nif __name__ == \"__main__\":\n    user_input = input(\"Enter numbers separated by a comma:\\n\").strip()\n    unsorted = [int(item) for item in user_input.split(\",\")]\n    print(merge_insertion_sort(unsorted))\n", "path": "sorts/merge_insertion_sort.py"}], "after_files": [{"content": "\"\"\"\nThis is a pure Python implementation of the merge-insertion sort algorithm\nSource: https://en.wikipedia.org/wiki/Merge-insertion_sort\n\nFor doctests run following command:\npython3 -m doctest -v merge_insertion_sort.py\nor\npython -m doctest -v merge_insertion_sort.py\n\nFor manual testing run:\npython3 merge_insertion_sort.py\n\"\"\"\n\nfrom __future__ import annotations\n\n\ndef merge_insertion_sort(collection: list[int]) -> list[int]:\n    \"\"\"Pure implementation of merge-insertion sort algorithm in Python\n\n    :param collection: some mutable ordered collection with heterogeneous\n    comparable items inside\n    :return: the same collection ordered by ascending\n\n    Examples:\n    >>> merge_insertion_sort([0, 5, 3, 2, 2])\n    [0, 2, 2, 3, 5]\n\n    >>> merge_insertion_sort([99])\n    [99]\n\n    >>> merge_insertion_sort([-2, -5, -45])\n    [-45, -5, -2]\n\n    Testing with all permutations on range(0,5):\n    >>> import itertools\n    >>> permutations = list(itertools.permutations([0, 1, 2, 3, 4]))\n    >>> all(merge_insertion_sort(p) == [0, 1, 2, 3, 4] for p in permutations)\n    True\n    \"\"\"\n\n    def binary_search_insertion(sorted_list, item):\n        left = 0\n        right = len(sorted_list) - 1\n        while left <= right:\n            middle = (left + right) // 2\n            if left == right:\n                if sorted_list[middle] < item:\n                    left = middle + 1\n                break\n            elif sorted_list[middle] < item:\n                left = middle + 1\n            else:\n                right = middle - 1\n        sorted_list.insert(left, item)\n        return sorted_list\n\n    def sortlist_2d(list_2d):\n        def merge(left, right):\n            result = []\n            while left and right:\n                if left[0][0] < right[0][0]:\n                    result.append(left.pop(0))\n                else:\n                    result.append(right.pop(0))\n            return result + left + right\n\n        length = len(list_2d)\n        if length <= 1:\n            return list_2d\n        middle = length // 2\n        return merge(sortlist_2d(list_2d[:middle]), sortlist_2d(list_2d[middle:]))\n\n    if len(collection) <= 1:\n        return collection\n\n    \"\"\"\n    Group the items into two pairs, and leave one element if there is a last odd item.\n\n    Example: [999, 100, 75, 40, 10000]\n                -> [999, 100], [75, 40]. Leave 10000.\n    \"\"\"\n    two_paired_list = []\n    has_last_odd_item = False\n    for i in range(0, len(collection), 2):\n        if i == len(collection) - 1:\n            has_last_odd_item = True\n        else:\n            \"\"\"\n            Sort two-pairs in each groups.\n\n            Example: [999, 100], [75, 40]\n                        -> [100, 999], [40, 75]\n            \"\"\"\n            if collection[i] < collection[i + 1]:\n                two_paired_list.append([collection[i], collection[i + 1]])\n            else:\n                two_paired_list.append([collection[i + 1], collection[i]])\n\n    \"\"\"\n    Sort two_paired_list.\n\n    Example: [100, 999], [40, 75]\n                -> [40, 75], [100, 999]\n    \"\"\"\n    sorted_list_2d = sortlist_2d(two_paired_list)\n\n    \"\"\"\n    40 < 100 is sure because it has already been sorted.\n    Generate the sorted_list of them so that you can avoid unnecessary comparison.\n\n    Example:\n           group0 group1\n           40     100\n           75     999\n        ->\n           group0 group1\n           [40,   100]\n           75     999\n    \"\"\"\n    result = [i[0] for i in sorted_list_2d]\n\n    \"\"\"\n    100 < 999 is sure because it has already been sorted.\n    Put 999 in last of the sorted_list so that you can avoid unnecessary comparison.\n\n    Example:\n           group0 group1\n           [40,   100]\n           75     999\n        ->\n           group0 group1\n           [40,   100,   999]\n           75\n    \"\"\"\n    result.append(sorted_list_2d[-1][1])\n\n    \"\"\"\n    Insert the last odd item left if there is.\n\n    Example:\n           group0 group1\n           [40,   100,   999]\n           75\n        ->\n           group0 group1\n           [40,   100,   999,   10000]\n           75\n    \"\"\"\n    if has_last_odd_item:\n        pivot = collection[-1]\n        result = binary_search_insertion(result, pivot)\n\n    \"\"\"\n    Insert the remaining items.\n    In this case, 40 < 75 is sure because it has already been sorted.\n    Therefore, you only need to insert 75 into [100, 999, 10000],\n    so that you can avoid unnecessary comparison.\n\n    Example:\n           group0 group1\n           [40,   100,   999,   10000]\n            ^ You don't need to compare with this as 40 < 75 is already sure.\n           75\n        ->\n           [40,   75,    100,   999,   10000]\n    \"\"\"\n    is_last_odd_item_inserted_before_this_index = False\n    for i in range(len(sorted_list_2d) - 1):\n        if result[i] == collection[-1] and has_last_odd_item:\n            is_last_odd_item_inserted_before_this_index = True\n        pivot = sorted_list_2d[i][1]\n        # If last_odd_item is inserted before the item's index,\n        # you should forward index one more.\n        if is_last_odd_item_inserted_before_this_index:\n            result = result[: i + 2] + binary_search_insertion(result[i + 2 :], pivot)\n        else:\n            result = result[: i + 1] + binary_search_insertion(result[i + 1 :], pivot)\n\n    return result\n\n\nif __name__ == \"__main__\":\n    user_input = input(\"Enter numbers separated by a comma:\\n\").strip()\n    unsorted = [int(item) for item in user_input.split(\",\")]\n    print(merge_insertion_sort(unsorted))\n", "path": "sorts/merge_insertion_sort.py"}]}