a) Place each value of the one-dimensional array into a row of the bucket array, based on the value’s “ones” (rightmost) digit. For example, 97 is placed in row 7, 3 is placed in row 3, and 100 is placed in row 0. This procedure is called a distribution pass. b) Loop through the bucket array row by row, and copy the values back to the original array. This procedure is called a gathering pass. The new order of the preceding values in the one-dimensional array is 100, 3 and 97. c) Repeat this process for each subsequent digit position (tens, hundreds, thousands, etc.). On the second (tens digit) pass, 100 is placed in row 0, 3 is placed in row 0 (because 3 has no tens digit), and 97 is placed in row 9. After the gathering pass, the order of the values in the one-dimensional array is 100, 3 and 97. On the third (hundreds digit) pass, 100 is placed in row 1, 3 is placed in row 0, and 97 is placed in row 0 (after the 3). After this last gathering pass, the original array is in sorted order. The two-dimensional array of buckets is 10 times the length of the integer array being sorted. This sorting technique provides better performance than a selection and insertion sorts but requires much more memory—the selection and insertion sorts require space for only one additional element of data. This comparison is an example of a space/time trade-off: The bucket sort uses more memory than the selection and insertion sorts, but performs better. This version of the bucket sort requires copying all the data back to the original array on each pass. Another possibility is to create a second two-dimensional bucket array and repeatedly swap the data between the two bucket arrays. complete disctribute_element() and collect_element() functions without modifying rest of the code. Code 11.7 # Exercise 11.17 import math import numpy as np def bucket_sort(data): """Sort an array's values into ascending order using bucket sort.""" # Determine largest number of digits in numbers to sort totalDigits = int(math.log10(max(data)) + 1) # bucket array where numbers will be placed pail = [[0] * len(data) for i in range(10)] # go through all digit places and sort each number # according to digit place value for pass_number in range(1, totalDigits + 1): distribute_elements(data, pail, pass_number) # distribution pass collect_elements(data, pail) # gathering pass if pass_number != totalDigits: empty_bucket(pail) # set size of buckets to 0 def distribute_elements(data, pail, digit): """Distribute elements into buckets based on specified digit""" """ COMPLETE CODE HERE """ def collect_elements(data, pails): """Return elements to original array""" """ COMPLETE CODE HERE """ def empty_bucket(pails): """Set size of all buckets to zero""" for i in range(10): pails[i][0] = 0 # set size of bucket to 0 # MAIN data = np.random.randint(100, size=20) print(f'Unsorted array: {data}') bucket_sort(data) # sort array print(f'Sorted array: {data}')
11.17 (Bucket Sort) Use Python. Show the whole code input and output when done.
A bucket sort begins with a one-dimensional array of positive integers to be sorted and a two-dimensional array of integers with rows indexed from 0 to 9 and columns indexed from 0 to n – 1, where n is the number of values to be sorted.
Each row of the two-dimensional array is referred to as a bucket. Write a class named BucketSort containing a function called sort that operates as follows:
- a) Place each value of the one-dimensional array into a row of the bucket array, based on the value’s “ones” (rightmost) digit. For example, 97 is placed in row 7, 3 is placed in row 3, and 100 is placed in row 0. This procedure is called a distribution pass.
- b) Loop through the bucket array row by row, and copy the values back to the original array. This procedure is called a gathering pass. The new order of the preceding values in the one-dimensional array is 100, 3 and 97.
- c) Repeat this process for each subsequent digit position (tens, hundreds, thousands, etc.). On the second (tens digit) pass, 100 is placed in row 0, 3 is placed in row 0 (because 3 has no tens digit), and 97 is placed in row 9. After the gathering pass, the order of the values in the one-dimensional array is 100, 3 and 97. On the third (hundreds digit) pass, 100 is placed in row 1, 3 is placed in row 0, and 97 is placed in row 0 (after the 3). After this last gathering pass, the original array is in sorted order.
The two-dimensional array of buckets is 10 times the length of the integer array being sorted. This sorting technique provides better performance than a selection and insertion sorts but requires much more memory—the selection and insertion sorts require space for only one additional element of data. This comparison is an example of a space/time trade-off: The bucket sort uses more memory than the selection and insertion sorts, but performs better. This version of the bucket sort requires copying all the data back to the original array on each pass. Another possibility is to create a second two-dimensional bucket array and repeatedly swap the data between the two bucket arrays.
complete disctribute_element() and collect_element() functions without modifying rest of the code.
Code 11.7
# Exercise 11.17
import math
import numpy as np
def bucket_sort(data):
"""Sort an array's values into ascending order using bucket sort."""
# Determine largest number of digits in numbers to sort
totalDigits = int(math.log10(max(data)) + 1)
# bucket array where numbers will be placed
pail = [[0] * len(data) for i in range(10)]
# go through all digit places and sort each number
# according to digit place value
for pass_number in range(1, totalDigits + 1):
distribute_elements(data, pail, pass_number) # distribution pass
collect_elements(data, pail) # gathering pass
if pass_number != totalDigits:
empty_bucket(pail) # set size of buckets to 0
def distribute_elements(data, pail, digit):
"""Distribute elements into buckets based on specified digit"""
""" COMPLETE CODE HERE """
def collect_elements(data, pails):
"""Return elements to original array"""
""" COMPLETE CODE HERE """
def empty_bucket(pails):
"""Set size of all buckets to zero"""
for i in range(10):
pails[i][0] = 0 # set size of bucket to 0
# MAIN
data = np.random.randint(100, size=20)
print(f'Unsorted array: {data}')
bucket_sort(data) # sort array
print(f'Sorted array: {data}')
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