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Fix: Ensure all skills are tracked as files, not submodules

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2026-01-14 18:48:48 +01:00
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21
skills/loki-mode/benchmarks/results/humaneval-loki-solutions/0.py Normal file
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# HumanEval/0
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
from typing import List
def has_close_elements(numbers: List[float], threshold: float) -> bool:
""" Check if in given list of numbers, are any two numbers closer to each other than
given threshold.
>>> has_close_elements([1.0, 2.0, 3.0], 0.5)
False
>>> has_close_elements([1.0, 2.8, 3.0, 4.0, 5.0, 2.0], 0.3)
True
"""
for i in range(len(numbers)):
for j in range(i + 1, len(numbers)):
if abs(numbers[i] - numbers[j]) < threshold:
return True
return False

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/1.py Normal file
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# HumanEval/1
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
from typing import List
def separate_paren_groups(paren_string: str) -> List[str]:
""" Input to this function is a string containing multiple groups of nested parentheses. Your goal is to
separate those group into separate strings and return the list of those.
Separate groups are balanced (each open brace is properly closed) and not nested within each other
Ignore any spaces in the input string.
>>> separate_paren_groups('( ) (( )) (( )( ))')
['()', '(())', '(()())']
"""
result = []
current_group = []
depth = 0
for char in paren_string:
if char == ' ':
continue
if char == '(':
depth += 1
current_group.append(char)
elif char == ')':
depth -= 1
current_group.append(char)
if depth == 0:
result.append(''.join(current_group))
current_group = []
return result

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/10.py Normal file
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# HumanEval/10
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def is_palindrome(string: str) -> bool:
""" Test if given string is a palindrome """
return string == string[::-1]
def make_palindrome(string: str) -> str:
""" Find the shortest palindrome that begins with a supplied string.
Algorithm idea is simple:
- Find the longest postfix of supplied string that is a palindrome.
- Append to the end of the string reverse of a string prefix that comes before the palindromic suffix.
>>> make_palindrome('')
''
>>> make_palindrome('cat')
'catac'
>>> make_palindrome('cata')
'catac'
"""
if not string:
return ''
for i in range(len(string)):
if is_palindrome(string[i:]):
return string + string[:i][::-1]
return string

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/100.py Normal file
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# HumanEval/100
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def make_a_pile(n):
"""
Given a positive integer n, you have to make a pile of n levels of stones.
The first level has n stones.
The number of stones in the next level is:
- the next odd number if n is odd.
- the next even number if n is even.
Return the number of stones in each level in a list, where element at index
i represents the number of stones in the level (i+1).
Examples:
>>> make_a_pile(3)
[3, 5, 7]
"""
result = [n]
for _ in range(n - 1):
result.append(result[-1] + 2)
return result

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/101.py Normal file
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# HumanEval/101
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def words_string(s):
"""
You will be given a string of words separated by commas or spaces. Your task is
to split the string into words and return an array of the words.
For example:
words_string("Hi, my name is John") == ["Hi", "my", "name", "is", "John"]
words_string("One, two, three, four, five, six") == ["One", "two", "three", "four", "five", "six"]
"""
if not s:
return []
return s.replace(",", " ").split()

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/102.py Normal file
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# HumanEval/102
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def choose_num(x, y):
"""This function takes two positive numbers x and y and returns the
biggest even integer number that is in the range [x, y] inclusive. If
there's no such number, then the function should return -1.
For example:
choose_num(12, 15) = 14
choose_num(13, 12) = -1
"""
if x > y:
return -1
if y % 2 == 0:
return y
if y - 1 >= x:
return y - 1
return -1

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/103.py Normal file
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# HumanEval/103
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def rounded_avg(n, m):
"""You are given two positive integers n and m, and your task is to compute the
average of the integers from n through m (including n and m).
Round the answer to the nearest integer and convert that to binary.
If n is greater than m, return -1.
Example:
rounded_avg(1, 5) => "0b11"
rounded_avg(7, 5) => -1
rounded_avg(10, 20) => "0b1111"
rounded_avg(20, 33) => "0b11010"
"""
if n > m:
return -1
count = m - n + 1
total = count * (n + m) // 2
avg = round(total / count)
return bin(avg)

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/104.py Normal file
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# HumanEval/104
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def unique_digits(x):
"""Given a list of positive integers x. return a sorted list of all
elements that hasn't any even digit.
Note: Returned list should be sorted in increasing order.
For example:
>>> unique_digits([15, 33, 1422, 1])
[1, 15, 33]
>>> unique_digits([152, 323, 1422, 10])
[]
"""
even_digits = set("02468")
result = []
for num in x:
if not any(d in even_digits for d in str(num)):
result.append(num)
return sorted(result)

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/105.py Normal file
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# HumanEval/105
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def by_length(arr):
"""
Given an array of integers, sort the integers that are between 1 and 9 inclusive,
reverse the resulting array, and then replace each digit by its corresponding name from
"One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine".
For example:
arr = [2, 1, 1, 4, 5, 8, 2, 3]
-> sort arr -> [1, 1, 2, 2, 3, 4, 5, 8]
-> reverse arr -> [8, 5, 4, 3, 2, 2, 1, 1]
return ["Eight", "Five", "Four", "Three", "Two", "Two", "One", "One"]
If the array is empty, return an empty array:
arr = []
return []
If the array has any strange number ignore it:
arr = [1, -1 , 55]
-> sort arr -> [-1, 1, 55]
-> reverse arr -> [55, 1, -1]
return = ['One']
"""
names = ["One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"]
filtered = [x for x in arr if 1 <= x <= 9]
filtered.sort()
filtered.reverse()
return [names[x - 1] for x in filtered]

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/106.py Normal file
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# HumanEval/106
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def f(n):
""" Implement the function f that takes n as a parameter,
and returns a list of size n, such that the value of the element at index i is the factorial of i if i is even
or the sum of numbers from 1 to i otherwise.
i starts from 1.
the factorial of i is the multiplication of the numbers from 1 to i (1 * 2 * ... * i).
Example:
f(5) == [1, 2, 6, 24, 15]
"""
result = []
for i in range(1, n + 1):
if i % 2 == 0:
# Even i: compute factorial
factorial = 1
for j in range(1, i + 1):
factorial *= j
result.append(factorial)
else:
# Odd i: compute sum from 1 to i
result.append(i * (i + 1) // 2)
return result

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/107.py Normal file
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# HumanEval/107
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def even_odd_palindrome(n):
"""
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Input: 3
Output: (1, 2)
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Input: 12
Output: (4, 6)
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
"""
even_count = 0
odd_count = 0
for i in range(1, n + 1):
s = str(i)
if s == s[::-1]:
if i % 2 == 0:
even_count += 1
else:
odd_count += 1
return (even_count, odd_count)

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/108.py Normal file
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# HumanEval/108
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def count_nums(arr):
"""
Write a function count_nums which takes an array of integers and returns
the number of elements which has a sum of digits > 0.
If a number is negative, then its first signed digit will be negative:
e.g. -123 has signed digits -1, 2, and 3.
>>> count_nums([]) == 0
>>> count_nums([-1, 11, -11]) == 1
>>> count_nums([1, 1, 2]) == 3
"""
count = 0
for num in arr:
if num == 0:
continue
is_negative = num < 0
num = abs(num)
digits = [int(d) for d in str(num)]
if is_negative:
digits[0] = -digits[0]
if sum(digits) > 0:
count += 1
return count

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/109.py Normal file
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# HumanEval/109
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def move_one_ball(arr):
"""We have an array 'arr' of N integers arr[1], arr[2], ..., arr[N].The
numbers in the array will be randomly ordered. Your task is to determine if
it is possible to get an array sorted in non-decreasing order by performing
the following operation on the given array:
You are allowed to perform right shift operation any number of times.
One right shift operation means shifting all elements of the array by one
position in the right direction. The last element of the array will be moved to
the starting position in the array i.e. 0th index.
If it is possible to obtain the sorted array by performing the above operation
then return True else return False.
If the given array is empty then return True.
Note: The given list is guaranteed to have unique elements.
For Example:
move_one_ball([3, 4, 5, 1, 2])==>True
Explanation: By performin 2 right shift operations, non-decreasing order can
be achieved for the given array.
move_one_ball([3, 5, 4, 1, 2])==>False
Explanation:It is not possible to get non-decreasing order for the given
array by performing any number of right shift operations.
"""
if len(arr) <= 1:
return True
# Count the number of break points where arr[i] > arr[i+1]
break_count = 0
n = len(arr)
for i in range(n - 1):
if arr[i] > arr[i + 1]:
break_count += 1
# If already sorted, return True
if break_count == 0:
return True
# If exactly one break point and the wrap-around is valid
# (last element <= first element), it can be rotated to sorted order
if break_count == 1 and arr[-1] <= arr[0]:
return True
return False

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/11.py Normal file
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# HumanEval/11
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
from typing import List
def string_xor(a: str, b: str) -> str:
""" Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
>>> string_xor('010', '110')
'100'
"""
result = []
for char_a, char_b in zip(a, b):
if char_a == char_b:
result.append('0')
else:
result.append('1')
return ''.join(result)

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/110.py Normal file
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# HumanEval/110
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def exchange(lst1, lst2):
"""In this problem, you will implement a function that takes two lists of numbers,
and determines whether it is possible to perform an exchange of elements
between them to make lst1 a list of only even numbers.
There is no limit on the number of exchanged elements between lst1 and lst2.
If it is possible to exchange elements between the lst1 and lst2 to make
all the elements of lst1 to be even, return "YES".
Otherwise, return "NO".
For example:
exchange([1, 2, 3, 4], [1, 2, 3, 4]) => "YES"
exchange([1, 2, 3, 4], [1, 5, 3, 4]) => "NO"
It is assumed that the input lists will be non-empty.
"""
odds_in_lst1 = sum(1 for x in lst1 if x % 2 != 0)
evens_in_lst2 = sum(1 for x in lst2 if x % 2 == 0)
if evens_in_lst2 >= odds_in_lst1:
return "YES"
else:
return "NO"

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/111.py Normal file
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# HumanEval/111
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def histogram(test):
"""Given a string representing a space separated lowercase letters, return a dictionary
of the letter with the most repetition and containing the corresponding count.
If several letters have the same occurrence, return all of them.
Example:
histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}
histogram('a b b a') == {'a': 2, 'b': 2}
histogram('a b c a b') == {'a': 2, 'b': 2}
histogram('b b b b a') == {'b': 4}
histogram('') == {}
"""
if not test or not test.strip():
return {}
letters = test.split()
counts = {}
for letter in letters:
if letter:
counts[letter] = counts.get(letter, 0) + 1
if not counts:
return {}
max_count = max(counts.values())
return {letter: count for letter, count in counts.items() if count == max_count}

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/112.py Normal file
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# HumanEval/112
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def reverse_delete(s,c):
"""Task
We are given two strings s and c, you have to deleted all the characters in s that are equal to any character in c
then check if the result string is palindrome.
A string is called palindrome if it reads the same backward as forward.
You should return a tuple containing the result string and True/False for the check.
Example
For s = "abcde", c = "ae", the result should be ('bcd',False)
For s = "abcdef", c = "b" the result should be ('acdef',False)
For s = "abcdedcba", c = "ab", the result should be ('cdedc',True)
"""
chars_to_delete = set(c)
result = ''.join(char for char in s if char not in chars_to_delete)
is_palindrome = result == result[::-1]
return (result, is_palindrome)

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/113.py Normal file
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# HumanEval/113
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def odd_count(lst):
"""Given a list of strings, where each string consists of only digits, return a list.
Each element i of the output should be "the number of odd elements in the
string i of the input." where all the i's should be replaced by the number
of odd digits in the i'th string of the input.
>>> odd_count(['1234567'])
["the number of odd elements 4n the str4ng 4 of the 4nput."]
>>> odd_count(['3',"11111111"])
["the number of odd elements 1n the str1ng 1 of the 1nput.",
"the number of odd elements 8n the str8ng 8 of the 8nput."]
"""
result = []
for s in lst:
odd_digit_count = sum(1 for d in s if int(d) % 2 == 1)
count_str = str(odd_digit_count)
template = "the number of odd elements in the string i of the input."
formatted = template.replace('i', count_str)
result.append(formatted)
return result

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/114.py Normal file
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# HumanEval/114
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def minSubArraySum(nums):
"""
Given an array of integers nums, find the minimum sum of any non-empty sub-array
of nums.
Example
minSubArraySum([2, 3, 4, 1, 2, 4]) == 1
minSubArraySum([-1, -2, -3]) == -6
"""
if not nums:
return 0
min_ending_here = nums[0]
min_so_far = nums[0]
for i in range(1, len(nums)):
min_ending_here = min(nums[i], min_ending_here + nums[i])
min_so_far = min(min_so_far, min_ending_here)
return min_so_far

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/115.py Normal file
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# HumanEval/115
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def max_fill(grid, capacity):
import math
"""
You are given a rectangular grid of wells. Each row represents a single well,
and each 1 in a row represents a single unit of water.
Each well has a corresponding bucket that can be used to extract water from it,
and all buckets have the same capacity.
Your task is to use the buckets to empty the wells.
Output the number of times you need to lower the buckets.
Example 1:
Input:
grid : [[0,0,1,0], [0,1,0,0], [1,1,1,1]]
bucket_capacity : 1
Output: 6
Example 2:
Input:
grid : [[0,0,1,1], [0,0,0,0], [1,1,1,1], [0,1,1,1]]
bucket_capacity : 2
Output: 5
Example 3:
Input:
grid : [[0,0,0], [0,0,0]]
bucket_capacity : 5
Output: 0
Constraints:
* all wells have the same length
* 1 <= grid.length <= 10^2
* 1 <= grid[:,1].length <= 10^2
* grid[i][j] -> 0 | 1
* 1 <= capacity <= 10
"""
return sum(math.ceil(sum(row) / capacity) for row in grid if sum(row) > 0)

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/116.py Normal file
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# HumanEval/116
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def sort_array(arr):
"""
In this Kata, you have to sort an array of non-negative integers according to
number of ones in their binary representation in ascending order.
For similar number of ones, sort based on decimal value.
It must be implemented like this:
>>> sort_array([1, 5, 2, 3, 4]) == [1, 2, 3, 4, 5]
>>> sort_array([-2, -3, -4, -5, -6]) == [-6, -5, -4, -3, -2]
>>> sort_array([1, 0, 2, 3, 4]) [0, 1, 2, 3, 4]
"""
return sorted(arr, key=lambda x: (bin(x).count('1'), x))

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/117.py Normal file
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# HumanEval/117
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def select_words(s, n):
"""Given a string s and a natural number n, you have been tasked to implement
a function that returns a list of all words from string s that contain exactly
n consonants, in order these words appear in the string s.
If the string s is empty then the function should return an empty list.
Note: you may assume the input string contains only letters and spaces.
Examples:
select_words("Mary had a little lamb", 4) ==> ["little"]
select_words("Mary had a little lamb", 3) ==> ["Mary", "lamb"]
select_words("simple white space", 2) ==> []
select_words("Hello world", 4) ==> ["world"]
select_words("Uncle sam", 3) ==> ["Uncle"]
"""
if not s:
return []
vowels = "aeiouAEIOU"
result = []
for word in s.split():
consonant_count = sum(1 for char in word if char.isalpha() and char not in vowels)
if consonant_count == n:
result.append(word)
return result

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/118.py Normal file
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# HumanEval/118
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def get_closest_vowel(word):
"""You are given a word. Your task is to find the closest vowel that stands between
two consonants from the right side of the word (case sensitive).
Vowels in the beginning and ending doesn't count. Return empty string if you didn't
find any vowel met the above condition.
You may assume that the given string contains English letter only.
Example:
get_closest_vowel("yogurt") ==> "u"
get_closest_vowel("FULL") ==> "U"
get_closest_vowel("quick") ==> ""
get_closest_vowel("ab") ==> ""
"""
if len(word) < 3:
return ""
vowels = set("aeiouAEIOU")
for i in range(len(word) - 2, 0, -1):
if word[i] in vowels:
if word[i - 1] not in vowels and word[i + 1] not in vowels:
return word[i]
return ""

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/119.py Normal file
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# HumanEval/119
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def match_parens(lst):
'''
You are given a list of two strings, both strings consist of open
parentheses '(' or close parentheses ')' only.
Your job is to check if it is possible to concatenate the two strings in
some order, that the resulting string will be good.
A string S is considered to be good if and only if all parentheses in S
are balanced. For example: the string '(())()' is good, while the string
'())' is not.
Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.
Examples:
match_parens(['()(', ')']) == 'Yes'
match_parens([')', ')']) == 'No'
'''
def is_balanced(s):
balance = 0
for char in s:
if char == '(':
balance += 1
else:
balance -= 1
if balance < 0:
return False
return balance == 0
# Try both concatenation orders
if is_balanced(lst[0] + lst[1]) or is_balanced(lst[1] + lst[0]):
return 'Yes'
return 'No'

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/12.py Normal file
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# HumanEval/12
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
from typing import List, Optional
def longest(strings: List[str]) -> Optional[str]:
""" Out of list of strings, return the longest one. Return the first one in case of multiple
strings of the same length. Return None in case the input list is empty.
>>> longest([])
>>> longest(['a', 'b', 'c'])
'a'
>>> longest(['a', 'bb', 'ccc'])
'ccc'
"""
if not strings:
return None
longest_str = strings[0]
for s in strings[1:]:
if len(s) > len(longest_str):
longest_str = s
return longest_str

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/120.py Normal file
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# HumanEval/120
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def maximum(arr, k):
"""
Given an array arr of integers and a positive integer k, return a sorted list
of length k with the maximum k numbers in arr.
Example 1:
Input: arr = [-3, -4, 5], k = 3
Output: [-4, -3, 5]
Example 2:
Input: arr = [4, -4, 4], k = 2
Output: [4, 4]
Example 3:
Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
Output: [2]
Note:
1. The length of the array will be in the range of [1, 1000].
2. The elements in the array will be in the range of [-1000, 1000].
3. 0 <= k <= len(arr)
"""
if k == 0:
return []
return sorted(arr)[-k:]

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# HumanEval/121
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def solution(lst):
"""Given a non-empty list of integers, return the sum of all of the odd elements that are in even positions.
Examples
solution([5, 8, 7, 1]) ==> 12
solution([3, 3, 3, 3, 3]) ==> 9
solution([30, 13, 24, 321]) ==>0
"""
return sum(x for x in lst[::2] if x % 2 == 1)

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# HumanEval/122
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def add_elements(arr, k):
"""
Given a non-empty array of integers arr and an integer k, return
the sum of the elements with at most two digits from the first k elements of arr.
Example:
Input: arr = [111,21,3,4000,5,6,7,8,9], k = 4
Output: 24 # sum of 21 + 3
Constraints:
1. 1 <= len(arr) <= 100
2. 1 <= k <= len(arr)
"""
total = 0
for element in arr[:k]:
if -99 <= element <= 99:
total += element
return total

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# HumanEval/123
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def get_odd_collatz(n):
"""
Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.
The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
as follows: start with any positive integer n. Then each term is obtained from the
previous term as follows: if the previous term is even, the next term is one half of
the previous term. If the previous term is odd, the next term is 3 times the previous
term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.
Note:
1. Collatz(1) is [1].
2. returned list sorted in increasing order.
For example:
get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
"""
odd_numbers = set()
while n >= 1:
if n % 2 == 1:
odd_numbers.add(n)
if n == 1:
break
if n % 2 == 0:
n = n // 2
else:
n = 3 * n + 1
return sorted(odd_numbers)

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# HumanEval/124
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def valid_date(date):
"""You have to write a function which validates a given date string and
returns True if the date is valid otherwise False.
The date is valid if all of the following rules are satisfied:
1. The date string is not empty.
2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
3. The months should not be less than 1 or higher than 12.
4. The date should be in the format: mm-dd-yyyy
for example:
valid_date('03-11-2000') => True
valid_date('15-01-2012') => False
valid_date('04-0-2040') => False
valid_date('06-04-2020') => True
valid_date('06/04/2020') => False
"""
if not date or not isinstance(date, str):
return False
parts = date.split('-')
if len(parts) != 3:
return False
mm, dd, yyyy = parts
if len(mm) != 2 or len(dd) != 2 or len(yyyy) != 4:
return False
try:
month = int(mm)
day = int(dd)
year = int(yyyy)
except ValueError:
return False
if month < 1 or month > 12:
return False
if month in [1, 3, 5, 7, 8, 10, 12]:
max_days = 31
elif month in [4, 6, 9, 11]:
max_days = 30
else:
max_days = 29
if day < 1 or day > max_days:
return False
return True

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# HumanEval/125
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def split_words(txt):
'''
Given a string of words, return a list of words split on whitespace, if no whitespaces exists in the text you
should split on commas ',' if no commas exists you should return the number of lower-case letters with odd order in the
alphabet, ord('a') = 0, ord('b') = 1, ... ord('z') = 25
Examples
split_words("Hello world!") ➞ ["Hello", "world!"]
split_words("Hello,world!") ➞ ["Hello", "world!"]
split_words("abcdef") == 3
'''
if ' ' in txt:
return txt.split()
elif ',' in txt:
return txt.split(',')
else:
count = 0
for char in txt:
if char.islower() and (ord(char) - ord('a')) % 2 == 1:
count += 1
return count

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# HumanEval/126
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def is_sorted(lst):
'''
Given a list of numbers, return whether or not they are sorted
in ascending order. If list has more than 1 duplicate of the same
number, return False. Assume no negative numbers and only integers.
Examples
is_sorted([5]) ➞ True
is_sorted([1, 2, 3, 4, 5]) ➞ True
is_sorted([1, 3, 2, 4, 5]) ➞ False
is_sorted([1, 2, 3, 4, 5, 6]) ➞ True
is_sorted([1, 2, 3, 4, 5, 6, 7]) ➞ True
is_sorted([1, 3, 2, 4, 5, 6, 7]) ➞ False
is_sorted([1, 2, 2, 3, 3, 4]) ➞ True
is_sorted([1, 2, 2, 2, 3, 4]) ➞ False
'''
# Check if sorted in ascending order
for i in range(len(lst) - 1):
if lst[i] > lst[i + 1]:
return False
# Check that no number appears more than twice
count = {}
for num in lst:
count[num] = count.get(num, 0) + 1
if count[num] > 2:
return False
return True

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# HumanEval/127
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def intersection(interval1, interval2):
"""You are given two intervals,
where each interval is a pair of integers. For example, interval = (start, end) = (1, 2).
The given intervals are closed which means that the interval (start, end)
includes both start and end.
For each given interval, it is assumed that its start is less or equal its end.
Your task is to determine whether the length of intersection of these two
intervals is a prime number.
Example, the intersection of the intervals (1, 3), (2, 4) is (2, 3)
which its length is 1, which not a prime number.
If the length of the intersection is a prime number, return "YES",
otherwise, return "NO".
If the two intervals don't intersect, return "NO".
[input/output] samples:
intersection((1, 2), (2, 3)) ==> "NO"
intersection((-1, 1), (0, 4)) ==> "NO"
intersection((-3, -1), (-5, 5)) ==> "YES"
"""
start = max(interval1[0], interval2[0])
end = min(interval1[1], interval2[1])
if start > end:
return "NO"
length = end - start
if length < 2:
return "NO"
for i in range(2, int(length ** 0.5) + 1):
if length % i == 0:
return "NO"
return "YES"

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# HumanEval/128
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def prod_signs(arr):
"""
You are given an array arr of integers and you need to return
sum of magnitudes of integers multiplied by product of all signs
of each number in the array, represented by 1, -1 or 0.
Note: return None for empty arr.
Example:
>>> prod_signs([1, 2, 2, -4]) == -9
>>> prod_signs([0, 1]) == 0
>>> prod_signs([]) == None
"""
if not arr:
return None
sign_product = 1
magnitude_sum = 0
for num in arr:
if num == 0:
sign_product = 0
elif num < 0:
sign_product *= -1
magnitude_sum += abs(num)
return sign_product * magnitude_sum

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# HumanEval/129
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
n = len(grid)
# Find the position of cell with value 1
row_1, col_1 = 0, 0
for i in range(n):
for j in range(n):
if grid[i][j] == 1:
row_1, col_1 = i, j
break
# Find the smallest neighbor of cell with value 1
min_neighbor = n * n + 1
directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
for dr, dc in directions:
nr, nc = row_1 + dr, col_1 + dc
if 0 <= nr < n and 0 <= nc < n:
min_neighbor = min(min_neighbor, grid[nr][nc])
# Build the result: alternating pattern of 1 and min_neighbor
result = []
for i in range(k):
if i % 2 == 0:
result.append(1)
else:
result.append(min_neighbor)
return result

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# HumanEval/13
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def greatest_common_divisor(a: int, b: int) -> int:
""" Return a greatest common divisor of two integers a and b
>>> greatest_common_divisor(3, 5)
1
>>> greatest_common_divisor(25, 15)
5
"""
a = abs(a)
b = abs(b)
while b:
a, b = b, a % b
return a

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# HumanEval/130
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def tri(n):
"""Everyone knows Fibonacci sequence, it was studied deeply by mathematicians in
the last couple centuries. However, what people don't know is Tribonacci sequence.
Tribonacci sequence is defined by the recurrence:
tri(1) = 3
tri(n) = 1 + n / 2, if n is even.
tri(n) = tri(n - 1) + tri(n - 2) + tri(n + 1), if n is odd.
For example:
tri(2) = 1 + (2 / 2) = 2
tri(4) = 3
tri(3) = tri(2) + tri(1) + tri(4)
= 2 + 3 + 3 = 8
You are given a non-negative integer number n, you have to a return a list of the
first n + 1 numbers of the Tribonacci sequence.
Examples:
tri(3) = [1, 3, 2, 8]
"""
if n == 0:
return [1]
result = [1, 3]
for i in range(2, n + 1):
if i % 2 == 0:
result.append(1 + i // 2)
else:
next_even = 1 + (i + 1) // 2
result.append(result[i - 1] + result[i - 2] + next_even)
return result

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# HumanEval/131
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def digits(n):
"""Given a positive integer n, return the product of the odd digits.
Return 0 if all digits are even.
For example:
digits(1) == 1
digits(4) == 0
digits(235) == 15
"""
product = 1
has_odd = False
while n > 0:
digit = n % 10
if digit % 2 == 1:
product *= digit
has_odd = True
n //= 10
return product if has_odd else 0

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# HumanEval/132
# Loki Mode Multi-Agent Solution
# Attempts: 2
# Passed: True
def is_nested(string):
'''
Create a function that takes a string as input which contains only square brackets.
The function should return True if and only if there is a valid subsequence of brackets
where at least one bracket in the subsequence is nested.
is_nested('[[]]') ➞ True
is_nested('[]]]]]]][[[[[]') ➞ False
is_nested('[][]') ➞ False
is_nested('[]') ➞ False
is_nested('[[][]]') ➞ True
is_nested('[[]][[') ➞ True
'''
depth = 0
max_depth = 0
for char in string:
if char == '[':
depth += 1
max_depth = max(max_depth, depth)
elif char == ']':
if depth > 0:
depth -= 1
if depth == 0 and max_depth >= 2:
return True
if depth == 0:
max_depth = 0
return False

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# HumanEval/133
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
import math
def sum_squares(lst):
"""You are given a list of numbers.
You need to return the sum of squared numbers in the given list,
round each element in the list to the upper int(Ceiling) first.
Examples:
For lst = [1,2,3] the output should be 14
For lst = [1,4,9] the output should be 98
For lst = [1,3,5,7] the output should be 84
For lst = [1.4,4.2,0] the output should be 29
For lst = [-2.4,1,1] the output should be 6
"""
return sum(math.ceil(x)**2 for x in lst)

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# HumanEval/134
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def check_if_last_char_is_a_letter(txt):
'''
Create a function that returns True if the last character
of a given string is an alphabetical character and is not
a part of a word, and False otherwise.
Note: "word" is a group of characters separated by space.
Examples:
check_if_last_char_is_a_letter("apple pie") ➞ False
check_if_last_char_is_a_letter("apple pi e") ➞ True
check_if_last_char_is_a_letter("apple pi e ") ➞ False
check_if_last_char_is_a_letter("") ➞ False
'''
if not txt or not txt[-1].isalpha():
return False
if len(txt) == 1:
return True
return txt[-2] == ' '

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# HumanEval/135
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def can_arrange(arr):
"""Create a function which returns the largest index of an element which
is not greater than or equal to the element immediately preceding it. If
no such element exists then return -1. The given array will not contain
duplicate values.
Examples:
can_arrange([1,2,4,3,5]) = 3
can_arrange([1,2,3]) = -1
"""
result = -1
for i in range(1, len(arr)):
if arr[i] < arr[i - 1]:
result = i
return result

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# HumanEval/136
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def largest_smallest_integers(lst):
'''
Create a function that returns a tuple (a, b), where 'a' is
the largest of negative integers, and 'b' is the smallest
of positive integers in a list.
If there is no negative or positive integers, return them as None.
Examples:
largest_smallest_integers([2, 4, 1, 3, 5, 7]) == (None, 1)
largest_smallest_integers([]) == (None, None)
largest_smallest_integers([0]) == (None, None)
'''
negatives = [x for x in lst if x < 0]
positives = [x for x in lst if x > 0]
largest_negative = max(negatives) if negatives else None
smallest_positive = min(positives) if positives else None
return (largest_negative, smallest_positive)

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# HumanEval/137
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def compare_one(a, b):
"""
Create a function that takes integers, floats, or strings representing
real numbers, and returns the larger variable in its given variable type.
Return None if the values are equal.
Note: If a real number is represented as a string, the floating point might be . or ,
compare_one(1, 2.5) ➞ 2.5
compare_one(1, "2,3") ➞ "2,3"
compare_one("5,1", "6") ➞ "6"
compare_one("1", 1) ➞ None
"""
def to_float(x):
if isinstance(x, str):
return float(x.replace(",", "."))
return float(x)
val_a = to_float(a)
val_b = to_float(b)
if val_a == val_b:
return None
elif val_a > val_b:
return a
else:
return b

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# HumanEval/138
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def is_equal_to_sum_even(n):
"""Evaluate whether the given number n can be written as the sum of exactly 4 positive even numbers
Example
is_equal_to_sum_even(4) == False
is_equal_to_sum_even(6) == False
is_equal_to_sum_even(8) == True
"""
return n >= 8 and n % 2 == 0

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# HumanEval/139
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def special_factorial(n):
"""The Brazilian factorial is defined as:
brazilian_factorial(n) = n! * (n-1)! * (n-2)! * ... * 1!
where n > 0
For example:
>>> special_factorial(4)
288
The function will receive an integer as input and should return the special
factorial of this integer.
"""
result = 1
current_factorial = 1
for k in range(1, n + 1):
current_factorial *= k
result *= current_factorial
return result

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# HumanEval/14
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
from typing import List
def all_prefixes(string: str) -> List[str]:
""" Return list of all prefixes from shortest to longest of the input string
>>> all_prefixes('abc')
['a', 'ab', 'abc']
"""
return [string[:i] for i in range(1, len(string) + 1)]

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# HumanEval/140
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
import re
def fix_spaces(text):
"""
Given a string text, replace all spaces in it with underscores,
and if a string has more than 2 consecutive spaces,
then replace all consecutive spaces with -
fix_spaces("Example") == "Example"
fix_spaces("Example 1") == "Example_1"
fix_spaces(" Example 2") == "_Example_2"
fix_spaces(" Example 3") == "_Example-3"
"""
def replace_spaces(match):
spaces = match.group(0)
if len(spaces) > 2:
return '-'
else:
return '_' * len(spaces)
return re.sub(r' +', replace_spaces, text)

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# HumanEval/141
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def file_name_check(file_name):
"""Create a function which takes a string representing a file's name, and returns
'Yes' if the the file's name is valid, and returns 'No' otherwise.
A file's name is considered to be valid if and only if all the following conditions
are met:
- There should not be more than three digits ('0'-'9') in the file's name.
- The file's name contains exactly one dot '.'
- The substring before the dot should not be empty, and it starts with a letter from
the latin alphapet ('a'-'z' and 'A'-'Z').
- The substring after the dot should be one of these: ['txt', 'exe', 'dll']
Examples:
file_name_check("example.txt") # => 'Yes'
file_name_check("1example.dll") # => 'No' (the name should start with a latin alphapet letter)
"""
# Check for exactly one dot
if file_name.count('.') != 1:
return 'No'
# Split by dot
parts = file_name.split('.')
prefix = parts[0]
suffix = parts[1]
# Check prefix is non-empty and starts with a letter
if len(prefix) == 0 or not prefix[0].isalpha():
return 'No'
# Check suffix is valid
if suffix not in ['txt', 'exe', 'dll']:
return 'No'
# Count digits in entire filename
digit_count = sum(1 for c in file_name if c.isdigit())
if digit_count > 3:
return 'No'
return 'Yes'

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# HumanEval/142
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def sum_squares(lst):
""""
This function will take a list of integers. For all entries in the list, the function shall square the integer entry if its index is a
multiple of 3 and will cube the integer entry if its index is a multiple of 4 and not a multiple of 3. The function will not
change the entries in the list whose indexes are not a multiple of 3 or 4. The function shall then return the sum of all entries.
Examples:
For lst = [1,2,3] the output should be 6
For lst = [] the output should be 0
For lst = [-1,-5,2,-1,-5] the output should be -126
"""
result = 0
for i in range(len(lst)):
if i % 3 == 0:
result += lst[i] ** 2
elif i % 4 == 0:
result += lst[i] ** 3
else:
result += lst[i]
return result

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# HumanEval/143
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def words_in_sentence(sentence):
"""
You are given a string representing a sentence,
the sentence contains some words separated by a space,
and you have to return a string that contains the words from the original sentence,
whose lengths are prime numbers,
the order of the words in the new string should be the same as the original one.
Example 1:
Input: sentence = "This is a test"
Output: "is"
Example 2:
Input: sentence = "lets go for swimming"
Output: "go for"
Constraints:
* 1 <= len(sentence) <= 100
* sentence contains only letters
"""
def is_prime(n):
if n < 2:
return False
if n == 2:
return True
if n % 2 == 0:
return False
for i in range(3, int(n ** 0.5) + 1, 2):
if n % i == 0:
return False
return True
words = sentence.split()
prime_words = [word for word in words if is_prime(len(word))]
return " ".join(prime_words)

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# HumanEval/144
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def simplify(x, n):
"""Your task is to implement a function that will simplify the expression
x * n. The function returns True if x * n evaluates to a whole number and False
otherwise. Both x and n, are string representation of a fraction, and have the following format,
<numerator>/<denominator> where both numerator and denominator are positive whole numbers.
You can assume that x, and n are valid fractions, and do not have zero as denominator.
simplify("1/5", "5/1") = True
simplify("1/6", "2/1") = False
simplify("7/10", "10/2") = False
"""
x_num, x_den = map(int, x.split('/'))
n_num, n_den = map(int, n.split('/'))
numerator = x_num * n_num
denominator = x_den * n_den
return numerator % denominator == 0

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# HumanEval/145
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def order_by_points(nums):
"""
Write a function which sorts the given list of integers
in ascending order according to the sum of their digits.
Note: if there are several items with similar sum of their digits,
order them based on their index in original list.
For example:
>>> order_by_points([1, 11, -1, -11, -12]) == [-1, -11, 1, -12, 11]
>>> order_by_points([]) == []
"""
def digit_sum(n):
s = str(n)
if s[0] == '-':
return -int(s[1]) + sum(int(d) for d in s[2:])
else:
return sum(int(d) for d in s)
return sorted(nums, key=digit_sum)

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# HumanEval/146
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def specialFilter(nums):
"""Write a function that takes an array of numbers as input and returns
the number of elements in the array that are greater than 10 and both
first and last digits of a number are odd (1, 3, 5, 7, 9).
For example:
specialFilter([15, -73, 14, -15]) => 1
specialFilter([33, -2, -3, 45, 21, 109]) => 2
"""
odd_digits = {'1', '3', '5', '7', '9'}
count = 0
for num in nums:
if num > 10:
s = str(num)
if s[0] in odd_digits and s[-1] in odd_digits:
count += 1
return count

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# HumanEval/147
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def get_max_triples(n):
"""
You are given a positive integer n. You have to create an integer array a of length n.
For each i (1 ≤ i ≤ n), the value of a[i] = i * i - i + 1.
Return the number of triples (a[i], a[j], a[k]) of a where i < j < k,
and a[i] + a[j] + a[k] is a multiple of 3.
Example :
Input: n = 5
Output: 1
Explanation:
a = [1, 3, 7, 13, 21]
The only valid triple is (1, 7, 13).
"""
if n < 3:
return 0
a = [i * i - i + 1 for i in range(1, n + 1)]
count = 0
for i in range(n):
for j in range(i + 1, n):
for k in range(j + 1, n):
if (a[i] + a[j] + a[k]) % 3 == 0:
count += 1
return count

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# HumanEval/148
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def bf(planet1, planet2):
'''
There are eight planets in our solar system: the closerst to the Sun
is Mercury, the next one is Venus, then Earth, Mars, Jupiter, Saturn,
Uranus, Neptune.
Write a function that takes two planet names as strings planet1 and planet2.
The function should return a tuple containing all planets whose orbits are
located between the orbit of planet1 and the orbit of planet2, sorted by
the proximity to the sun.
The function should return an empty tuple if planet1 or planet2
are not correct planet names.
Examples
bf("Jupiter", "Neptune") ==> ("Saturn", "Uranus")
bf("Earth", "Mercury") ==> ("Venus")
bf("Mercury", "Uranus") ==> ("Venus", "Earth", "Mars", "Jupiter", "Saturn")
'''
planets = ("Mercury", "Venus", "Earth", "Mars", "Jupiter", "Saturn", "Uranus", "Neptune")
if planet1 not in planets or planet2 not in planets:
return ()
idx1 = planets.index(planet1)
idx2 = planets.index(planet2)
if idx1 > idx2:
idx1, idx2 = idx2, idx1
return planets[idx1 + 1:idx2]

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# HumanEval/149
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def sorted_list_sum(lst):
"""Write a function that accepts a list of strings as a parameter,
deletes the strings that have odd lengths from it,
and returns the resulted list with a sorted order,
The list is always a list of strings and never an array of numbers,
and it may contain duplicates.
The order of the list should be ascending by length of each word, and you
should return the list sorted by that rule.
If two words have the same length, sort the list alphabetically.
The function should return a list of strings in sorted order.
You may assume that all words will have the same length.
For example:
assert list_sort(["aa", "a", "aaa"]) => ["aa"]
assert list_sort(["ab", "a", "aaa", "cd"]) => ["ab", "cd"]
"""
filtered = [s for s in lst if len(s) % 2 == 0]
return sorted(filtered, key=lambda x: (len(x), x))

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# HumanEval/15
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def string_sequence(n: int) -> str:
""" Return a string containing space-delimited numbers starting from 0 upto n inclusive.
>>> string_sequence(0)
'0'
>>> string_sequence(5)
'0 1 2 3 4 5'
"""
return ' '.join(str(i) for i in range(n + 1))

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# HumanEval/150
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def x_or_y(n, x, y):
"""A simple program which should return the value of x if n is
a prime number and should return the value of y otherwise.
Examples:
for x_or_y(7, 34, 12) == 34
for x_or_y(15, 8, 5) == 5
"""
if n <= 1:
return y
if n == 2:
return x
if n % 2 == 0:
return y
i = 3
while i * i <= n:
if n % i == 0:
return y
i += 2
return x

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# HumanEval/151
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def double_the_difference(lst):
'''
Given a list of numbers, return the sum of squares of the numbers
in the list that are odd. Ignore numbers that are negative or not integers.
double_the_difference([1, 3, 2, 0]) == 1 + 9 + 0 + 0 = 10
double_the_difference([-1, -2, 0]) == 0
double_the_difference([9, -2]) == 81
double_the_difference([0]) == 0
If the input list is empty, return 0.
'''
total = 0
for n in lst:
if isinstance(n, int) and not isinstance(n, bool) and n > 0 and n % 2 != 0:
total += n * n
return total

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# HumanEval/152
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def compare(game,guess):
"""I think we all remember that feeling when the result of some long-awaited
event is finally known. The feelings and thoughts you have at that moment are
definitely worth noting down and comparing.
Your task is to determine if a person correctly guessed the results of a number of matches.
You are given two arrays of scores and guesses of equal length, where each index shows a match.
Return an array of the same length denoting how far off each guess was. If they have guessed correctly,
the value is 0, and if not, the value is the absolute difference between the guess and the score.
example:
compare([1,2,3,4,5,1],[1,2,3,4,2,-2]) -> [0,0,0,0,3,3]
compare([0,5,0,0,0,4],[4,1,1,0,0,-2]) -> [4,4,1,0,0,6]
"""
return [abs(g - s) for g, s in zip(game, guess)]

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# HumanEval/153
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def Strongest_Extension(class_name, extensions):
"""You will be given the name of a class (a string) and a list of extensions.
The extensions are to be used to load additional classes to the class. The
strength of the extension is as follows: Let CAP be the number of the uppercase
letters in the extension's name, and let SM be the number of lowercase letters
in the extension's name, the strength is given by the fraction CAP - SM.
You should find the strongest extension and return a string in this
format: ClassName.StrongestExtensionName.
If there are two or more extensions with the same strength, you should
choose the one that comes first in the list.
For example, if you are given "Slices" as the class and a list of the
extensions: ['SErviNGSliCes', 'Cheese', 'StuFfed'] then you should
return 'Slices.SErviNGSliCes' since 'SErviNGSliCes' is the strongest extension
(its strength is -1).
Example:
for Strongest_Extension('my_class', ['AA', 'Be', 'CC']) == 'my_class.AA'
"""
strongest_ext = extensions[0]
max_strength = sum(1 for c in strongest_ext if c.isupper()) - sum(1 for c in strongest_ext if c.islower())
for ext in extensions[1:]:
strength = sum(1 for c in ext if c.isupper()) - sum(1 for c in ext if c.islower())
if strength > max_strength:
max_strength = strength
strongest_ext = ext
return f"{class_name}.{strongest_ext}"

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# HumanEval/154
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def cycpattern_check(a , b):
"""You are given 2 words. You need to return True if the second word or any of its rotations is a substring in the first word
cycpattern_check("abcd","abd") => False
cycpattern_check("hello","ell") => True
cycpattern_check("whassup","psus") => False
cycpattern_check("abab","baa") => True
cycpattern_check("efef","eeff") => False
cycpattern_check("himenss","simen") => True
"""
if len(b) == 0:
return True
if len(b) > len(a):
return False
for i in range(len(b)):
rotation = b[i:] + b[:i]
if rotation in a:
return True
return False

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# HumanEval/155
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def even_odd_count(num):
"""Given an integer. return a tuple that has the number of even and odd digits respectively.
Example:
even_odd_count(-12) ==> (1, 1)
even_odd_count(123) ==> (1, 2)
"""
even_count = 0
odd_count = 0
for digit in str(abs(num)):
if int(digit) % 2 == 0:
even_count += 1
else:
odd_count += 1
return (even_count, odd_count)

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# HumanEval/156
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def int_to_mini_roman(number):
"""
Given a positive integer, obtain its roman numeral equivalent as a string,
and return it in lowercase.
Restrictions: 1 <= num <= 1000
Examples:
>>> int_to_mini_roman(19) == 'xix'
>>> int_to_mini_roman(152) == 'clii'
>>> int_to_mini_roman(426) == 'cdxxvi'
"""
value_symbol_pairs = [
(1000, 'm'),
(900, 'cm'),
(500, 'd'),
(400, 'cd'),
(100, 'c'),
(90, 'xc'),
(50, 'l'),
(40, 'xl'),
(10, 'x'),
(9, 'ix'),
(5, 'v'),
(4, 'iv'),
(1, 'i')
]
result = ''
for value, symbol in value_symbol_pairs:
while number >= value:
result += symbol
number -= value
return result

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# HumanEval/157
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def right_angle_triangle(a, b, c):
'''
Given the lengths of the three sides of a triangle. Return True if the three
sides form a right-angled triangle, False otherwise.
A right-angled triangle is a triangle in which one angle is right angle or
90 degree.
Example:
right_angle_triangle(3, 4, 5) == True
right_angle_triangle(1, 2, 3) == False
'''
epsilon = 1e-9
a_sq = a * a
b_sq = b * b
c_sq = c * c
if abs(a_sq + b_sq - c_sq) < epsilon:
return True
if abs(a_sq + c_sq - b_sq) < epsilon:
return True
if abs(b_sq + c_sq - a_sq) < epsilon:
return True
return False

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# HumanEval/158
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def find_max(words):
"""Write a function that accepts a list of strings.
The list contains different words. Return the word with maximum number
of unique characters. If multiple strings have maximum number of unique
characters, return the one which comes first in lexicographical order.
find_max(["name", "of", "string"]) == "string"
find_max(["name", "enam", "game"]) == "enam"
find_max(["aaaaaaa", "bb" ,"cc"]) == ""aaaaaaa"
"""
return sorted(words, key=lambda w: (-len(set(w)), w))[0]

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# HumanEval/159
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def eat(number, need, remaining):
"""
You're a hungry rabbit, and you already have eaten a certain number of carrots,
but now you need to eat more carrots to complete the day's meals.
you should return an array of [ total number of eaten carrots after your meals,
the number of carrots left after your meals ]
if there are not enough remaining carrots, you will eat all remaining carrots, but will still be hungry.
Example:
* eat(5, 6, 10) -> [11, 4]
* eat(4, 8, 9) -> [12, 1]
* eat(1, 10, 10) -> [11, 0]
* eat(2, 11, 5) -> [7, 0]
Variables:
@number : integer
the number of carrots that you have eaten.
@need : integer
the number of carrots that you need to eat.
@remaining : integer
the number of remaining carrots thet exist in stock
Constrain:
* 0 <= number <= 1000
* 0 <= need <= 1000
* 0 <= remaining <= 1000
Have fun :)
"""
actual_eaten = min(need, remaining)
return [number + actual_eaten, remaining - actual_eaten]

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# HumanEval/16
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def count_distinct_characters(string: str) -> int:
""" Given a string, find out how many distinct characters (regardless of case) does it consist of
>>> count_distinct_characters('xyzXYZ')
3
>>> count_distinct_characters('Jerry')
4
"""
return len(set(string.lower()))

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# HumanEval/160
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def do_algebra(operator, operand):
"""
Given two lists operator, and operand. The first list has basic algebra operations, and
the second list is a list of integers. Use the two given lists to build the algebric
expression and return the evaluation of this expression.
The basic algebra operations:
Addition ( + )
Subtraction ( - )
Multiplication ( * )
Floor division ( // )
Exponentiation ( ** )
Example:
operator['+', '*', '-']
array = [2, 3, 4, 5]
result = 2 + 3 * 4 - 5
=> result = 9
Note:
The length of operator list is equal to the length of operand list minus one.
Operand is a list of of non-negative integers.
Operator list has at least one operator, and operand list has at least two operands.
"""
expression = str(operand[0])
for i, op in enumerate(operator):
expression += op + str(operand[i + 1])
return eval(expression)

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# HumanEval/161
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def solve(s):
"""You are given a string s.
if s[i] is a letter, reverse its case from lower to upper or vise versa,
otherwise keep it as it is.
If the string contains no letters, reverse the string.
The function should return the resulted string.
Examples
solve("1234") = "4321"
solve("ab") = "AB"
solve("#a@C") = "#A@c"
"""
has_letter = any(c.isalpha() for c in s)
if not has_letter:
return s[::-1]
result = []
for c in s:
if c.isalpha():
result.append(c.swapcase())
else:
result.append(c)
return ''.join(result)

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# HumanEval/162
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def string_to_md5(text):
"""
Given a string 'text', return its md5 hash equivalent string.
If 'text' is an empty string, return None.
>>> string_to_md5('Hello world') == '3e25960a79dbc69b674cd4ec67a72c62'
"""
if text == '':
return None
import hashlib
return hashlib.md5(text.encode()).hexdigest()

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# HumanEval/163
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def generate_integers(a, b):
"""
Given two positive integers a and b, return the even digits between a
and b, in ascending order.
For example:
generate_integers(2, 8) => [2, 4, 6, 8]
generate_integers(8, 2) => [2, 4, 6, 8]
generate_integers(10, 14) => []
"""
low, high = min(a, b), max(a, b)
even_digits = [2, 4, 6, 8]
return [d for d in even_digits if low <= d <= high]

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# HumanEval/17
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
from typing import List
def parse_music(music_string: str) -> List[int]:
""" Input to this function is a string representing musical notes in a special ASCII format.
Your task is to parse this string and return list of integers corresponding to how many beats does each
not last.
Here is a legend:
'o' - whole note, lasts four beats
'o|' - half note, lasts two beats
'.|' - quater note, lasts one beat
>>> parse_music('o o| .| o| o| .| .| .| .| o o')
[4, 2, 1, 2, 2, 1, 1, 1, 1, 4, 4]
"""
if not music_string:
return []
note_beats = {'o': 4, 'o|': 2, '.|': 1}
tokens = music_string.split()
return [note_beats[token] for token in tokens if token]

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# HumanEval/18
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def how_many_times(string: str, substring: str) -> int:
""" Find how many times a given substring can be found in the original string. Count overlaping cases.
>>> how_many_times('', 'a')
0
>>> how_many_times('aaa', 'a')
3
>>> how_many_times('aaaa', 'aa')
3
"""
if not string or not substring:
return 0
count = 0
for i in range(len(string) - len(substring) + 1):
if string[i:i + len(substring)] == substring:
count += 1
return count

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# HumanEval/19
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
from typing import List
def sort_numbers(numbers: str) -> str:
""" Input is a space-delimited string of numberals from 'zero' to 'nine'.
Valid choices are 'zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight' and 'nine'.
Return the string with numbers sorted from smallest to largest
>>> sort_numbers('three one five')
'one three five'
"""
if not numbers or not numbers.strip():
return ''
word_to_num = {
'zero': 0,
'one': 1,
'two': 2,
'three': 3,
'four': 4,
'five': 5,
'six': 6,
'seven': 7,
'eight': 8,
'nine': 9
}
words = numbers.split()
sorted_words = sorted(words, key=lambda w: word_to_num[w])
return ' '.join(sorted_words)

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# HumanEval/2
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def truncate_number(number: float) -> float:
""" Given a positive floating point number, it can be decomposed into
and integer part (largest integer smaller than given number) and decimals
(leftover part always smaller than 1).
Return the decimal part of the number.
>>> truncate_number(3.5)
0.5
"""
return number - int(number)

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# HumanEval/20
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
from typing import List, Tuple
def find_closest_elements(numbers: List[float]) -> Tuple[float, float]:
""" From a supplied list of numbers (of length at least two) select and return two that are the closest to each
other and return them in order (smaller number, larger number).
>>> find_closest_elements([1.0, 2.0, 3.0, 4.0, 5.0, 2.2])
(2.0, 2.2)
>>> find_closest_elements([1.0, 2.0, 3.0, 4.0, 5.0, 2.0])
(2.0, 2.0)
"""
sorted_numbers = sorted(numbers)
min_diff = float('inf')
closest_pair = (sorted_numbers[0], sorted_numbers[1])
for i in range(len(sorted_numbers) - 1):
diff = sorted_numbers[i + 1] - sorted_numbers[i]
if diff < min_diff:
min_diff = diff
closest_pair = (sorted_numbers[i], sorted_numbers[i + 1])
return closest_pair

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# HumanEval/21
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
from typing import List
def rescale_to_unit(numbers: List[float]) -> List[float]:
""" Given list of numbers (of at least two elements), apply a linear transform to that list,
such that the smallest number will become 0 and the largest will become 1
>>> rescale_to_unit([1.0, 2.0, 3.0, 4.0, 5.0])
[0.0, 0.25, 0.5, 0.75, 1.0]
"""
min_val = min(numbers)
max_val = max(numbers)
range_val = max_val - min_val
return [(x - min_val) / range_val for x in numbers]

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# HumanEval/22
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
from typing import List, Any
def filter_integers(values: List[Any]) -> List[int]:
""" Filter given list of any python values only for integers
>>> filter_integers(['a', 3.14, 5])
[5]
>>> filter_integers([1, 2, 3, 'abc', {}, []])
[1, 2, 3]
"""
return [x for x in values if isinstance(x, int) and not isinstance(x, bool)]

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# HumanEval/23
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def strlen(string: str) -> int:
""" Return length of given string
>>> strlen('')
0
>>> strlen('abc')
3
"""
return len(string)

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# HumanEval/24
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def largest_divisor(n: int) -> int:
""" For a given number n, find the largest number that divides n evenly, smaller than n
>>> largest_divisor(15)
5
"""
for i in range(n // 2, 0, -1):
if n % i == 0:
return i
return 1

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# HumanEval/25
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
from typing import List
def factorize(n: int) -> List[int]:
""" Return list of prime factors of given integer in the order from smallest to largest.
Each of the factors should be listed number of times corresponding to how many times it appeares in factorization.
Input number should be equal to the product of all factors
>>> factorize(8)
[2, 2, 2]
>>> factorize(25)
[5, 5]
>>> factorize(70)
[2, 5, 7]
"""
factors = []
d = 2
while d * d <= n:
while n % d == 0:
factors.append(d)
n //= d
d += 1
if n > 1:
factors.append(n)
return factors

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# HumanEval/26
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
from typing import List
from collections import Counter
def remove_duplicates(numbers: List[int]) -> List[int]:
""" From a list of integers, remove all elements that occur more than once.
Keep order of elements left the same as in the input.
>>> remove_duplicates([1, 2, 3, 2, 4])
[1, 3, 4]
"""
counts = Counter(numbers)
return [num for num in numbers if counts[num] == 1]

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# HumanEval/27
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def flip_case(string: str) -> str:
""" For a given string, flip lowercase characters to uppercase and uppercase to lowercase.
>>> flip_case('Hello')
'hELLO'
"""
return string.swapcase()

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/28.py Normal file
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# HumanEval/28
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
from typing import List
def concatenate(strings: List[str]) -> str:
""" Concatenate list of strings into a single string
>>> concatenate([])
''
>>> concatenate(['a', 'b', 'c'])
'abc'
"""
return ''.join(strings)

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# HumanEval/29
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
from typing import List
def filter_by_prefix(strings: List[str], prefix: str) -> List[str]:
""" Filter an input list of strings only for ones that start with a given prefix.
>>> filter_by_prefix([], 'a')
[]
>>> filter_by_prefix(['abc', 'bcd', 'cde', 'array'], 'a')
['abc', 'array']
"""
return [s for s in strings if s.startswith(prefix)]

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# HumanEval/3
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
from typing import List
def below_zero(operations: List[int]) -> bool:
""" You're given a list of deposit and withdrawal operations on a bank account that starts with
zero balance. Your task is to detect if at any point the balance of account fallls below zero, and
at that point function should return True. Otherwise it should return False.
>>> below_zero([1, 2, 3])
False
>>> below_zero([1, 2, -4, 5])
True
"""
balance = 0
for operation in operations:
balance += operation
if balance < 0:
return True
return False

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/30.py Normal file
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# HumanEval/30
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def get_positive(l: list):
"""Return only positive numbers in the list.
>>> get_positive([-1, 2, -4, 5, 6])
[2, 5, 6]
>>> get_positive([5, 3, -5, 2, -3, 3, 9, 0, 123, 1, -10])
[5, 3, 2, 3, 9, 123, 1]
"""
return [x for x in l if x > 0]

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# HumanEval/31
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def is_prime(n):
"""Return true if a given number is prime, and false otherwise.
>>> is_prime(6)
False
>>> is_prime(101)
True
>>> is_prime(11)
True
>>> is_prime(13441)
True
>>> is_prime(61)
True
>>> is_prime(4)
False
>>> is_prime(1)
False
"""
if n <= 1:
return False
if n == 2:
return True
if n % 2 == 0:
return False
i = 3
while i * i <= n:
if n % i == 0:
return False
i += 2
return True

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# HumanEval/32
# Loki Mode Multi-Agent Solution
# Attempts: 3
# Passed: False
def find_zero(xs: list):
""" xs are coefficients of a polynomial.
find_zero find x such that poly(x) = 0.
find_zero returns only only zero point, even if there are many.
Moreover, find_zero only takes list xs having even number of coefficients
and largest non zero coefficient as it guarantees
a solution.
>>> round(find_zero([1, 2]), 2) # f(x) = 1 + 2x
-0.5
>>> round(find_zero([-6, 11, -6, 1]), 2) # (x - 1) * (x - 2) * (x - 3) = -6 + 11x - 6x^2 + x^3
1.0
"""
# Find initial bounds where polynomial changes sign
low, high = -1000.0, 1000.0
# Ensure we have opposite signs at bounds
while poly(xs, low) * poly(xs, high) > 0:
low *= 2
high *= 2
# Binary search (bisection method)
tolerance = 1e-10
while high - low > tolerance:
mid = (low + high) / 2
if poly(xs, mid) == 0:
return mid
if poly(xs, low) * poly(xs, mid) < 0:
high = mid
else:
low = mid
return (low + high) / 2

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# HumanEval/33
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def sort_third(l: list):
"""This function takes a list l and returns a list l' such that
l' is identical to l in the indicies that are not divisible by three, while its values at the indicies that are divisible by three are equal
to the values of the corresponding indicies of l, but sorted.
>>> sort_third([1, 2, 3])
[1, 2, 3]
>>> sort_third([5, 6, 3, 4, 8, 9, 2])
[2, 6, 3, 4, 8, 9, 5]
"""
third_indices_values = sorted([l[i] for i in range(0, len(l), 3)])
result = l[:]
for idx, val in enumerate(third_indices_values):
result[idx * 3] = val
return result

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# HumanEval/34
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def unique(l: list):
"""Return sorted unique elements in a list
>>> unique([5, 3, 5, 2, 3, 3, 9, 0, 123])
[0, 2, 3, 5, 9, 123]
"""
return sorted(set(l))

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/35.py Normal file
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# HumanEval/35
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def max_element(l: list):
"""Return maximum element in the list.
>>> max_element([1, 2, 3])
3
>>> max_element([5, 3, -5, 2, -3, 3, 9, 0, 123, 1, -10])
123
"""
return max(l)

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/36.py Normal file
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# HumanEval/36
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def fizz_buzz(n: int):
"""Return the number of times the digit 7 appears in integers less than n which are divisible by 11 or 13.
>>> fizz_buzz(50)
0
>>> fizz_buzz(78)
2
>>> fizz_buzz(79)
3
"""
count = 0
for i in range(n):
if i % 11 == 0 or i % 13 == 0:
count += str(i).count('7')
return count

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# HumanEval/37
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def sort_even(l: list):
"""This function takes a list l and returns a list l' such that
l' is identical to l in the odd indicies, while its values at the even indicies are equal
to the values of the even indicies of l, but sorted.
>>> sort_even([1, 2, 3])
[1, 2, 3]
>>> sort_even([5, 6, 3, 4])
[3, 6, 5, 4]
"""
even_elements = sorted(l[::2])
result = l[:]
for i, val in enumerate(even_elements):
result[i * 2] = val
return result

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# HumanEval/38
# Loki Mode Multi-Agent Solution
# Attempts: 2
# Passed: True
def encode_cyclic(s: str):
"""
returns encoded string by cycling groups of three characters.
"""
# split string to groups. Each of length 3.
groups = [s[(3 * i):min((3 * i + 3), len(s))] for i in range((len(s) + 2) // 3)]
# cycle elements in each group. Unless group has fewer elements than 3.
groups = [(group[1:] + group[0]) if len(group) == 3 else group for group in groups]
return "".join(groups)
def decode_cyclic(s: str):
"""
takes as input string encoded with encode_cyclic function. Returns decoded string.
"""
# split string to groups. Each of length 3.
groups = [s[(3 * i):min((3 * i + 3), len(s))] for i in range((len(s) + 2) // 3)]
# cycle elements in each group back to original. Unless group has fewer elements than 3.
groups = [(group[-1] + group[:-1]) if len(group) == 3 else group for group in groups]
return "".join(groups)

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# HumanEval/39
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def prime_fib(n: int):
"""
prime_fib returns n-th number that is a Fibonacci number and it's also prime.
>>> prime_fib(1)
2
>>> prime_fib(2)
3
>>> prime_fib(3)
5
>>> prime_fib(4)
13
>>> prime_fib(5)
89
"""
def is_prime(num):
if num < 2:
return False
if num == 2:
return True
if num % 2 == 0:
return False
for i in range(3, int(num ** 0.5) + 1, 2):
if num % i == 0:
return False
return True
count = 0
fib_prev, fib_curr = 0, 1
while count < n:
fib_prev, fib_curr = fib_curr, fib_prev + fib_curr
if is_prime(fib_curr):
count += 1
return fib_curr

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skills/loki-mode/benchmarks/results/humaneval-loki-solutions/4.py Normal file
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# HumanEval/4
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
from typing import List
def mean_absolute_deviation(numbers: List[float]) -> float:
""" For a given list of input numbers, calculate Mean Absolute Deviation
around the mean of this dataset.
Mean Absolute Deviation is the average absolute difference between each
element and a centerpoint (mean in this case):
MAD = average | x - x_mean |
>>> mean_absolute_deviation([1.0, 2.0, 3.0, 4.0])
1.0
"""
if not numbers:
return 0.0
mean = sum(numbers) / len(numbers)
absolute_deviations = [abs(x - mean) for x in numbers]
return sum(absolute_deviations) / len(absolute_deviations)

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# HumanEval/40
# Loki Mode Multi-Agent Solution
# Attempts: 1
# Passed: True
def triples_sum_to_zero(l: list):
"""
triples_sum_to_zero takes a list of integers as an input.
it returns True if there are three distinct elements in the list that
sum to zero, and False otherwise.
>>> triples_sum_to_zero([1, 3, 5, 0])
False
>>> triples_sum_to_zero([1, 3, -2, 1])
True
>>> triples_sum_to_zero([1, 2, 3, 7])
False
>>> triples_sum_to_zero([2, 4, -5, 3, 9, 7])
True
>>> triples_sum_to_zero([1])
False
"""
n = len(l)
if n < 3:
return False
sorted_list = sorted(l)
for i in range(n - 2):
target = -sorted_list[i]
left = i + 1
right = n - 1
while left < right:
current_sum = sorted_list[left] + sorted_list[right]
if current_sum == target:
return True
elif current_sum < target:
left += 1
else:
right -= 1
return False

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