Side-by-side comparison: base vs SFT vs DPO v2 vs DPO v3
Source: joshuasundance/mypo-training → alpaca-stripped-validation/alpaca-stripped-2026-04-23T005911Z/
Eval job: 69e96eba2aa1660eaffa8d00
Sample size: 30 stratified validation prompts (seed=42, 3 length buckets)
Decoding: batch_size=1, no left-padding, do_sample=False, max_new_tokens=384, single-prompt through tokenizer.apply_chat_template as a normal user turn
Scaffold handling: the Alpaca ### Instruction: / ### Input: / ### Output: wrapper is stripped from the dataset prompt; the model sees only the bare natural-language instruction
Aggregate results (n=30)
| subject | parses | black | ruff | mypy --strict | ann_cov |
|---|---|---|---|---|---|
| Qwen2.5-Coder-1.5B-Instruct (base) | 1.000 | 0.100 | 0.033 | 0.033 | 0.000 |
| mypo-dpo-v2 (misconfigured run) | 1.000 | 0.100 | 0.033 | 0.033 | 0.000 |
| mypo-sft | 1.000 | 0.967 | 0.633 | 0.767 | 0.970 |
| mypo-dpo-v3 | 0.967 | 0.933 | 0.600 | 0.733 | 0.943 |
Key gap: dpo_v3 − base = +0.700 on mypy-strict pass rate. This gap survives removing the Alpaca prompt scaffold, ruling out "the model just learned to respond to ### Output:".
Compare with the previous Alpaca-wrapped single-prompt run (single-prompt-2026-04-23T002137Z):
| subject | mypy (wrapped) | mypy (stripped) | delta |
|---|---|---|---|
| base | 0.000 | 0.033 | +0.03 |
| dpo-v2 | 0.000 | 0.033 | +0.03 |
| sft | 0.733 | 0.767 | +0.03 |
| dpo-v3 | 0.733 | 0.733 | 0.00 |
Effect is robust to prompt shape.
What changes between subjects
- Base and DPO v2 fail the same way: wrap code in
```pythonfences, no type annotations, verbose prose preamble. DPO v2 is a misconfigured training run that wiped the SFT adapter's behavior — it behaves like the base model. - SFT and DPO v3 both: emit bare Python (no markdown fence), open with
from typing import …, annotate parameters and return types, and satisfymypy --strict. At n=30 the two are statistically indistinguishable on this slice; earlier batched n=150 showed SFT 0.853 vs v3 0.920 mypy-strict, but left-padded batched decoding inflates scores by ~19 pts so treat that with care. - Trade-off on v3: picks up a small regression in
parses(one generation over-long) andblackvs SFT; otherwise matches.
Three concrete examples
The three examples below are from indices 0, 8, 22 of the stratified sample (seed=42). In each, base and dpo-v2 fail mypy --strict, sft and dpo-v3 pass. All four outputs come from the exact same prompt, same decoding config, same machine (A10G-Large).
Example 1: prompt Build a Python program that identifies valid Leap Years.
Prompt (bare instruction, no Alpaca scaffold):
Build a Python program that identifies valid Leap Years.
Qwen2.5-Coder-1.5B-Instruct (base) — parses=pass black=fail ruff=pass mypy-strict=fail ann_cov=0.00
Sure! A leap year is a year that is divisible by 4 but not by 100 unless it is also divisible by 400. Here's a simple Python program to check if a given year is a leap year:
```python
def is_leap_year(year):
# Check if the year is divisible by 4
if year % 4 == 0:
# If the year is divisible by 100, check if it is also divisible by 400
if year % 100 == 0:
return year % 400 == 0
else:
return True
else:
return False
# Test the function with some examples
print(is_leap_year(2000)) # Output: True
print(is_leap_year(1900)) # Output: False
print(is_leap_year(2020)) # Output: True
print(is_leap_year(2021)) # Output: False
```
In this program, we define a function `is_leap_year` that takes a year as input and returns `True` if it is a leap year, otherwise `False`. The function first checks if the year is divisible by 4. If it is, it then checks if the year is divisible by 100. If the year is divisible by both 4 and 100, it further checks if the year is divisible by 400. If all these conditions are met, the function returns `True`, indicating that the year is a leap year. Otherwise, it returns `False`.
We then test the function with some example years to verify its correctness.
mypo-dpo-v2 (broken run) — parses=pass black=fail ruff=pass mypy-strict=fail ann_cov=0.00
Sure! A leap year is a year that is divisible by 4 but not by 100 unless it is also divisible by 400. Here's a simple Python program to check if a given year is a leap year:
```python
def is_leap_year(year):
# Check if the year is divisible by 4
if year % 4 == 0:
# If the year is divisible by 100, check if it is also divisible by 400
if year % 100 == 0:
return year % 400 == 0
else:
return True
else:
return False
# Test the function with some examples
print(is_leap_year(2000)) # Output: True
print(is_leap_year(1900)) # Output: False
print(is_leap_year(2020)) # Output: True
print(is_leap_year(2021)) # Output: False
```
In this program, we define a function `is_leap_year` that takes a year as input and returns `True` if it is a leap year, otherwise `False`. The function first checks if the year is divisible by 4. If it is, it then checks if the year is divisible by 100. If the year is divisible by both 4 and 100, it must also be divisible by 400 to be a leap year. Otherwise, it is not a leap year. Finally, we test the function with some example years to verify its correctness.
mypo-sft — parses=pass black=pass ruff=pass mypy-strict=pass ann_cov=1.00
def is_leap(year: int) -> bool:
if (year % 4 == 0 and year % 100 != 0) or year % 400 == 0:
return True
else:
return False
# Test the function
print(is_leap(2000)) # Output: True
mypo-dpo-v3 — parses=pass black=pass ruff=fail mypy-strict=pass ann_cov=1.00
from typing import Union
def is_leap(year: int) -> bool:
if year % 4 == 0 and (year % 100 != 0 or year % 400 == 0):
return True
else:
return False
# Test the function
print(is_leap(2000)) # Output: True
Example 9: prompt Develop a Python class to represent a point on 2D-space.
Prompt (bare instruction, no Alpaca scaffold):
Develop a Python class to represent a point on 2D-space.
Qwen2.5-Coder-1.5B-Instruct (base) — parses=pass black=fail ruff=pass mypy-strict=fail ann_cov=0.00
```python
class Point:
def __init__(self, x=0, y=0):
# Initialize the point with coordinates (x, y)
self.x = x
self.y = y
def get_x(self):
# Return the x-coordinate of the point
return self.x
def set_x(self, value):
# Set the x-coordinate of the point
self.x = value
def get_y(self):
# Return the y-coordinate of the point
return self.y
def set_y(self, value):
# Set the y-coordinate of the point
self.y = value
def distance_to_origin(self):
# Calculate and return the Euclidean distance from the point to the origin (0, 0)
return (self.x**2 + self.y**2)**0.5
def move(self, dx, dy):
# Move the point by the given displacement in both x and y directions
self.x += dx
self.y += dy
def __str__(self):
# Return a string representation of the point in the format "(x, y)"
return f"({self.x}, {self.y})"
```
This Python class `Point` represents a point in 2D space with methods to initialize, retrieve, modify, calculate distances, and perform translations. The `__str__` method provides a readable string representation of the point.
mypo-dpo-v2 (broken run) — parses=pass black=fail ruff=pass mypy-strict=fail ann_cov=0.00
```python
class Point:
def __init__(self, x=0, y=0):
# Initialize the point with coordinates (x, y)
self.x = x
self.y = y
def get_x(self):
# Return the x-coordinate of the point
return self.x
def set_x(self, value):
# Set the x-coordinate of the point
self.x = value
def get_y(self):
# Return the y-coordinate of the point
return self.y
def set_y(self, value):
# Set the y-coordinate of the point
self.y = value
def distance_to_origin(self):
# Calculate the Euclidean distance from the origin (0, 0)
return (self.x**2 + self.y**2)**0.5
def move(self, dx, dy):
# Move the point by dx units in the x-direction and dy units in the y-direction
self.x += dx
self.y += dy
def __str__(self):
# Return a string representation of the point
return f"({self.x}, {self.y})"
```
This Python class `Point` represents a point in 2D space with attributes for its x and y coordinates. It includes methods to get and set these coordinates, calculate the distance from the origin, and move the point by specified amounts in the x and y directions. The `__str__` method provides a readable string representation of the point.
mypo-sft — parses=pass black=pass ruff=pass mypy-strict=pass ann_cov=0.50
class Point:
def __init__(self, x: float, y: float):
self.x = x
self.y = y
def get_x(self) -> float:
return self.x
def get_y(self) -> float:
return self.y
mypo-dpo-v3 — parses=pass black=pass ruff=pass mypy-strict=pass ann_cov=0.67
from typing import Tuple
class Point:
def __init__(self, x: float, y: float) -> None:
self.x = x
self.y = y
def get_coordinates(self) -> Tuple[float, float]:
return (self.x, self.y)
Example 23: prompt Create a Python program that reads in two strings and outputs the longest common
Prompt (bare instruction, no Alpaca scaffold):
Create a Python program that reads in two strings and outputs the longest common substring.
Input:
str1 = "GeeksforGeeks"
str2 = "Practice"
Qwen2.5-Coder-1.5B-Instruct (base) — parses=pass black=fail ruff=pass mypy-strict=fail ann_cov=0.00
Here is a simple Python program to find the longest common substring between two strings:
```python
def longest_common_substring(str1, str2):
m = len(str1)
n = len(str2)
# Create a 2D array to store lengths of longest common suffixes
L = [[0 for x in range(n+1)] for y in range(m+1)]
# Build the table in bottom up manner
for i in range(1, m+1):
for j in range(1, n+1):
if str1[i-1] == str2[j-1]:
L[i][j] = L[i-1][j-1] + 1
else:
L[i][j] = 0
# Find the length of the longest common substring
max_length = 0
end_index = 0
for i in range(1, m+1):
for j in range(1, n+1):
if L[i][j] > max_length:
max_length = L[i][j]
end_index = i - 1
# Return the longest common substring
return str1[end_index-max_length+1:end_index+1]
# Test the function
str1 = "GeeksforGeeks"
str2 = "Practice"
print(longest_common_substring(str1, str2))
```
This program uses dynamic programming to build a 2D array `L` where `L[i][j]` represents the length of the longest common suffix of `str1[0..i-1]` and `str2[0..j-1]`. The algorithm then finds the maximum value in this array and returns the corresponding substring from `str1`.
When you run this program with the input `"
mypo-dpo-v2 (broken run) — parses=pass black=fail ruff=pass mypy-strict=fail ann_cov=0.00
Here is a simple Python program to find the longest common substring between two strings:
```python
def longest_common_substring(str1, str2):
m = len(str1)
n = len(str2)
# Create a 2D array to store lengths of longest common suffixes
L = [[0 for x in range(n+1)] for y in range(m+1)]
# Build the table in bottom up manner
for i in range(1, m+1):
for j in range(1, n+1):
if str1[i-1] == str2[j-1]:
L[i][j] = L[i-1][j-1] + 1
else:
L[i][j] = 0
# Find the length of the longest common substring
max_length = 0
end_index = 0
for i in range(1, m+1):
for j in range(1, n+1):
if L[i][j] > max_length:
max_length = L[i][j]
end_index = i - 1
# Return the longest common substring
return str1[end_index-max_length+1:end_index+1]
# Test the function
str1 = "GeeksforGeeks"
str2 = "Practice"
print(longest_common_substring(str1, str2))
```
This program uses dynamic programming to build a 2D array `L` where `L[i][j]` represents the length of the longest common suffix of `str1[0..i-1]` and `str2[0..j-1]`. The algorithm then finds the maximum value in this array and returns the corresponding substring from `str1`.
When you run this program with the input `"
mypo-sft — parses=pass black=pass ruff=pass mypy-strict=pass ann_cov=1.00
def longest_common_substring(str1: str, str2: str) -> str:
m = len(str1)
n = len(str2)
# Create a table to store lengths of longest common suffixes
dp = [[0 for _ in range(n + 1)] for _ in range(m + 1)]
res_len = 0
res_end = 0
for i in range(1, m + 1):
for j in range(1, n + 1):
if str1[i - 1] == str2[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
if dp[i][j] > res_len:
res_len = dp[i][j]
res_end = i
else:
dp[i][j] = 0
return str1[res_end - res_len : res_end]
# Driver code
str1 = "GeeksforGeeks"
str2 = "Practice"
print(longest_common_substring(str1, str2))
mypo-dpo-v3 — parses=pass black=pass ruff=pass mypy-strict=pass ann_cov=1.00
from typing import List
def longest_common_substring(str1: str, str2: str) -> str:
m = len(str1)
n = len(str2)
# Create a table to store lengths of longest common suffixes
dp: List[List[int]] = [[0 for _ in range(n + 1)] for _ in range(m + 1)]
ans = ""
max_length = 0
# Fill dp[][] in bottom up manner
for i in range(1, m + 1):
for j in range(1, n + 1):
if str1[i - 1] == str2[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
# Update answer if current substring is longer
if dp[i][j] > max_length:
start_index = i - dp[i][j]
end_index = i
max_length = dp[i][j]
else:
dp[i][j] = 0
# Extract the longest common substring from the dp array
for i in range(start_index, end_index):
ans += str1[i - 1]
return ans
# Driver code
str1 = "GeeksforGeeks"
str2 = "Practice"
print(longest_common_substring(str1, str2))
Reproducing this
Scripts in the mypo-training repo:
- Dataset prefix inspection (
dataset-inspection/2026-04-22/):hf jobs uv run --flavor cpu-upgrade --secrets HF_TOKEN .copilot_tmp/dataset_prefix_inspection.py - Alpaca-stripped validation (this file):
hf jobs uv run --flavor a10g-large --timeout 60m --secrets HF_TOKEN .copilot_tmp/alpaca_stripped_validation.py --sample-size 30 - Prior single-prompt (Alpaca-wrapped) validation:
hf jobs uv run --flavor a10g-large --timeout 60m --secrets HF_TOKEN .copilot_tmp/single_prompt_validation.py --sample-size 30
Honest caveats
- n=30 is small. 95 % Wilson interval for a 0.733 pass rate with n=30 is roughly [0.55, 0.86].
- The dataset is heavily biased: 99.8 % of
chosentraining rows have a type annotation vs 2.8 % ofrejected(dataset-inspection/2026-04-22/summary.json). So the improvement shown is specifically on "produce type-annotated Python for Alpaca-style toy prompts". It has NOT been tested on HumanEval+, MBPP, or any real codebase task. ruffpass rates around 0.60 for SFT/v3 mean both trained models still emit code that would trigger lint warnings (unused imports, etc.). "Passes mypy --strict" is the narrow strong claim; "lint-clean" is not.- DPO v2 is kept public as a worked example of a failed training run and is not recommended for use.