low = 0 high = len(list) = n mid = (low + high)//2 If high = 0, return None Else if low < high (Hint: Use this comparison as base case) a. Compare sNum and middle element = list[mid] b. If sNum is equal/matches middle element, return mid + 1 (add 1 since index starts at 0) c. Else if x is greater than the middle element, then x can only lie in the right (greater) half sublist after the middle element i. low = mid +1 ii. mid = (low + high)//2 d. Else if x is less than the middle element, then x can only lie in the left (lesser) half sublist after the middle element i. high = mid – 1 ii. mid e. Go back to step 9 using new values for low, high, and mid Else, return None wwww w %3D (low + high)//2 Print “Element", x, “found:", and value returned (can save as a variable if use loop) if it is not None; if it is None, print “Not Found!"
low = 0 high = len(list) = n mid = (low + high)//2 If high = 0, return None Else if low < high (Hint: Use this comparison as base case) a. Compare sNum and middle element = list[mid] b. If sNum is equal/matches middle element, return mid + 1 (add 1 since index starts at 0) c. Else if x is greater than the middle element, then x can only lie in the right (greater) half sublist after the middle element i. low = mid +1 ii. mid = (low + high)//2 d. Else if x is less than the middle element, then x can only lie in the left (lesser) half sublist after the middle element i. high = mid – 1 ii. mid e. Go back to step 9 using new values for low, high, and mid Else, return None wwww w %3D (low + high)//2 Print “Element", x, “found:", and value returned (can save as a variable if use loop) if it is not None; if it is None, print “Not Found!"
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
Related questions
Question
Can the picture below be added to the following code.
import random
n=int(input("Enter the number of elements that will be in the list:"))
sNum=int(input("Enter number you want to search:"))
randomlist = []
for i in range(n):
n = random.randint(1,50) # we had choose number between 1 to 50
randomlist.append(n) #you can choose your own
#print(randomlist)
for i in range(len(randomlist)):
if(randomlist[i]==sNum):
print(i)
print("Number not found")
![low = 0
high = len(list) = n
mid = (low + high)//2
If high = 0, return None
Else if low < high (Hint: Use this comparison as base case)
a. Compare sNum and middle element = list[mid]
b. If sNum is equal/matches middle element, return mid + 1 (add 1 since index starts
at 0)
c. Else if x is greater than the middle element, then x can only lie in the right
(greater) half sublist after the middle element
i. low = mid + 1
ii. mid = (low + high)//2
d. Else if x is less than the middle element, then x can only lie in the left (lesser) half
sublist after the middle element
i. high = mid – 1
ii. mid = (low + high)//2
-
e. Go back to step 9 using new values for low, high, and mid
Else, return None
Print "Element", x, “found:", and value returned (can save as a variable if use loop) if it is
not None; if it is None, print “Not Found!"](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F72b83cc2-495e-4bc5-a028-f79dc88fc308%2F0f22bd2e-ca00-44dc-b9b0-0efa7ad379db%2Fcumhtw_processed.png&w=3840&q=75)
Transcribed Image Text:low = 0
high = len(list) = n
mid = (low + high)//2
If high = 0, return None
Else if low < high (Hint: Use this comparison as base case)
a. Compare sNum and middle element = list[mid]
b. If sNum is equal/matches middle element, return mid + 1 (add 1 since index starts
at 0)
c. Else if x is greater than the middle element, then x can only lie in the right
(greater) half sublist after the middle element
i. low = mid + 1
ii. mid = (low + high)//2
d. Else if x is less than the middle element, then x can only lie in the left (lesser) half
sublist after the middle element
i. high = mid – 1
ii. mid = (low + high)//2
-
e. Go back to step 9 using new values for low, high, and mid
Else, return None
Print "Element", x, “found:", and value returned (can save as a variable if use loop) if it is
not None; if it is None, print “Not Found!"
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