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jek_recluse [69]
3 years ago
15

Again, suppose that n and r denote integer values where r > 0 and n ≥ 0. What are two "simple" values, say lowEnough and tooH

igh, such that lowEnough ≤ ⌊n√r⌋⌊nr⌋ < tooHigh. Explain based on your answer to the previous question.
Suppose you would like to know ⌊47226−−−−−√5⌋⌊472265⌋. Explain how you could find ⌊47226−−−−−√5⌋⌊472265⌋ using a guess and verify method. Note: Explain your answer by relating this problem to the number guessing game described earlier. Think about what would be an appropriate question to use in place of "Is secret_number < guess_number?" and think about good choices for the starting values of lowEnough and tooHigh.
Computers and Technology
1 answer:
enyata [817]3 years ago
6 0

Answer:

Answer explained

Explanation:

From the previous question we know that while searching for n^(1/r) we don't have to look for guesses less than 0 and greater than n. Because for less than 0 it will be an imaginary number and for rth root of a non negative number can never be greater than itself. Hence lowEnough = 0 and tooHigh = n.

we need to find 5th root of 47226. The computation of root is costlier than computing power of a number. Therefore, we will look for a number whose 5th power is 47226. lowEnough = 0 and tooHigh = 47226 + 1. Question that should be asked on each step would be "Is 5th power of number < 47227?" we will stop when we find a number whose 5th power is 47226.

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Suppose we have two String objects and treat the characters in each string from beginning to end in the following way: With one
Simora [160]

Answer:

If all the character pairs match after processing both strings, one string in stack and the other in queue, then this means one string is the reverse of the other.                            

Explanation:

Lets take an example of two strings abc and cba which are reverse of each  other.

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string2 = cba

Now push the characters of string1 in stack. Stack is a LIFO (last in first out) data structure which means the character pushed in the last in stack is popped first.

Push abc each character on a stack in the following order.

c

b

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Now add each character of string2 in queue. Queue is a FIFO (first in first out) data structure which means the character inserted first is removed first.

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First c is added to queue then b and then a.

Now lets pop one character from the stack and remove one character from queue and compare each pair of characters of both the strings to each other.

First from stack c is popped as per LIFO and c is removed from queue as per FIFO. Then these two characters are compared. They both match

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