____________ reference is used when you want to use the same calculation across multiple rows or columns.
2 answers:
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Answer:
A python code (Python recursion) was used for this given question
Explanation:
Solution
For this solution to the question, I am attaching code for these 2 files:
item.py
code.py
Source code for item.py:
class Item(object):
def __init__(self, name: str, weight: int, value: int) -> None:
self.name = name
self.weight = weight
self.value = value
def __lt__(self, other: "Item"):
if self.value == other.value:
if self.weight == other.weight:
return self.name < other.name
else:
return self.weight < other.weight
else:
return self.value < other.value
def __eq__(self, other: "Item") -> bool:
if is instance(other, Item):
return (self.name == other.name and
self.value == other.value and
self.weight == other.weight)
else:
return False
def __ne__(self, other: "Item") -> bool:
return not (self == other)
def __str__(self) -> str:
return f'A {self.name} worth {self.value} that weighs {self.weight}'
Source code for code.py:
#!/usr/bin/env python3
from typing import List
from typing import List, Generator
from item import Item
'''
Inductive definition of the function
fun3(0) is 5
fun3(1) is 7
fun3(2) is 11
func3(n) is fun3(n-1) + fun3(n-2) + fun3(n-3)
Solution 1: Straightforward but exponential
'''
def fun3_1(n: int) -> int:
result = None
if n == 0:
result = 5 # Base case
elif n == 1:
result = 7 # Base case
elif n == 2:
result = 11 # Base case
else:
result = fun3_1(n-1) + fun3_1(n-2) + fun3_1(n-3) # Recursive case
return result
''
Solution 2: New helper recursive function makes it linear
'''
def fun3(n: int) -> int:
''' Recursive core.
fun3(n) = _fun3(n-i, fun3(2+i), fun3(1+i), fun3(i))
'''
def fun3_helper_r(n: int, f_2: int, f_1: int, f_0: int):
result = None
if n == 0:
result = f_0 # Base case
elif n == 1:
result = f_1 # Base case
elif n == 2:
result = f_2 # Base case
else:
result = fun3_helper_r(n-1, f_2+f_1+f_0, f_2, f_1) # Recursive step
return result
return fun3_helper_r(n, 11, 7, 5)
''' binary_strings accepts a string of 0's, 1's, and X's and returns a generator that goes through all possible strings where the X's
could be either 0's or 1's. For example, with the string '0XX1',
the possible strings are '0001', '0011', '0101', and '0111'
'''
def binary_strings(string: str) -> Generator[str, None, None]:
def _binary_strings(string: str, binary_chars: List[str], idx: int):
if idx == len(string):
yield ''.join(binary_chars)
binary_chars = [' ']*len(string)
else:
char = string[idx]
if char != 'X':
binary_chars[idx]= char
yield from _binary_strings(string, binary_chars, idx+1)
else:
binary_chars[idx] = '0'
yield from _binary_strings(string, binary_chars, idx+1)
binary_chars[idx] = '1'
yield from _binary_strings(string, binary_chars, idx+1)
binary_chars = [' ']*len(string)
idx = 0
yield from _binary_strings(string, binary_chars, 0)
''' Recursive KnapSack: You are looking to rob a jewelry store. You have been staking it out for a couple of weeks now and have learned
the weights and values of every item in the store. You are looking to
get the biggest score you possibly can but you are only one person and
your backpack can only fit so much. Write a function that accepts a
list of items as well as the maximum capacity that your backpack can
hold and returns a list containing the most valuable items you can
take that still fit in your backpack. '''
def get_best_backpack(items: List[Item], max_capacity: int) -> List[Item]:
def get_best_r(took: List[Item], rest: List[Item], capacity: int) -> List[Item]:
if not rest or not capacity: # Base case
return took
else:
item = rest[0]
list1 = []
list1_val = 0
if item.weight <= capacity:
list1 = get_best_r(took+[item], rest[1:], capacity-item.weight)
list1_val = sum(x.value for x in list1)
list2 = get_best_r(took, rest[1:], capacity)
list2_val = sum(x.value for x in list2)
return list1 if list1_val > list2_val else list2
return get_best_r([], items, max_capacity)
Note: Kindly find an attached copy of the code outputs for python programming language below
