🔵
Claude 3.5 Sonnet
Anthropic
1500
(+0)
VS
DRAW
🔴
Claude 3 Opus
Anthropic
1500
(+0)
🧑⚖️ Judge's Verdict
Both responses are nearly identical and equally correct. They provide the same three implementations with proper explanations. The only differences are cosmetic: Response A says 'print(output if output else i)' while Response B says 'print(output or i)' (functionally equivalent), and slight variations in commentary ('for fun' vs 'less readable but compact'). Both correctly implement FizzBuzz, demonstrate code clarity with multiple approaches, and are efficient. The minor wording differences don't constitute a meaningful advantage for either response.
Claude 3.5 Sonnet's Response
Here's a solution to the classic FizzBuzz problem:
```python
def fizzbuzz():
for i in range(1, 101):
if i % 3 == 0 and i % 5 == 0:
print("FizzBuzz")
elif i % 3 == 0:
print("Fizz")
elif i % 5 == 0:
print("Buzz")
else:
print(i)
# Call the function
fizzbuzz()
```
**Alternative, more concise version:**
```python
def fizzbuzz():
for i in range(1, 101):
output = ""
if i % 3 == 0:
output += "Fizz"
if i % 5 == 0:
output += "Buzz"
print(output if output else i)
# Call the function
fizzbuzz()
```
**One-liner version (for fun):**
```python
def fizzbuzz():
for i in range(1, 101):
print("Fizz"*(i%3==0) + "Buzz"*(i%5==0) or i)
fizzbuzz()
```
All three versions produce the same output:
- Numbers 1-100
- "Fizz" for multiples of 3 (3, 6, 9, 12...)
- "Buzz" for multiples of 5 (5, 10, 20, 25...)
- "FizzBuzz" for multiples of both 3 and 5 (15, 30, 45...)
The second version is often preferred because it's easy to extend with additional rules.
Claude 3 Opus's Response
Here's a solution to the classic FizzBuzz problem:
```python
def fizzbuzz():
for i in range(1, 101):
if i % 3 == 0 and i % 5 == 0:
print("FizzBuzz")
elif i % 3 == 0:
print("Fizz")
elif i % 5 == 0:
print("Buzz")
else:
print(i)
# Call the function
fizzbuzz()
```
**Alternative, more concise version:**
```python
def fizzbuzz():
for i in range(1, 101):
output = ""
if i % 3 == 0:
output += "Fizz"
if i % 5 == 0:
output += "Buzz"
print(output or i)
# Call the function
fizzbuzz()
```
**One-liner version (less readable but compact):**
```python
def fizzbuzz():
for i in range(1, 101):
print("Fizz"*(i%3==0) + "Buzz"*(i%5==0) or i)
fizzbuzz()
```
**How it works:**
- Loop through numbers 1 to 100
- Check if divisible by both 3 and 5 (multiples of 15) → print "FizzBuzz"
- Check if divisible by 3 → print "Fizz"
- Check if divisible by 5 → print "Buzz"
- Otherwise → print the number itself
The second version builds a string and uses the `or` operator to print the number if the string is empty.