# simple implementation of old-school square root algorithm RBH2023# (my parent showed me this, not sure what it is called)# input: positive integer n# output: q = floor(sqrt(n)), and remainder r = q*qdef basehundred (n):L = []while n > 0:L.append(n% 100)n = n // 100L.reverse()return Ldef ossqrt(n):q, r = 0,0 # quotient, remainderL = basehundred (n)for dd in L: #dd next digit-pair from nS = 20*q #LHS multiplierq, r = 10*q, 100*r + dddigit = 0while ((s+digit+1)*(digit+1)<= r):digit += 1assert ((0 <= digit) and (digit <= 9))print(r, '=', s+digit, '*', digit, '+ ',end='')q, r = q + digit, r - (s+digit)*digitprint (r)assert (q*q <= n and ((q+1)*(q+1)) > n)return q, rn = 741177q, r = ossqrt(n)print('\n', n,'= ',q,' * ',q,' + '‚r‚ sep='') Use the old-school square root algorithm from class: by hand, find the largestinteger t ≤ 762309. Show your work. Check your answer:

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Use the old-school square root algorithm from class: by hand, find the largest integer t ≤ √√762309 . Show your work. Check your answer:

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

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Concept explainers

Eight puzzle problem

The N-puzzle problem contains N tiles (N+1 including empty tiles) where the value of N can be 8, 15, 24, etc. The puzzle is divided into the square root of (N+1) rows and the square root of (N+1) columns. These problems have an initial state or configuration and a goal state or configuration. The puzzle can be solved by shifting the tiles one by one in the empty space. The objective is to place the numbers on the tiles to match the final state using the empty space. The other name for the N-puzzle problem is the sliding puzzle.

Question

# simple implementation of old-school square root algorithm RBH2023
# (my parent showed me this, not sure what it is called)
# input: positive integer n
# output: q = floor(sqrt(n)), and remainder r = q*q
def basehundred (n):
L = []
while n > 0:
L.append(n% 100)
n = n // 100
L.reverse()
return L
def ossqrt(n):
q, r = 0,0 # quotient, remainder
L = basehundred (n)
for dd in L: #dd next digit-pair from n
S = 20*q #LHS multiplier
q, r = 10*q, 100*r + dd
digit = 0
while ((s+digit+1)*(digit+1)
<= r):
digit += 1
assert ((0 <= digit) and (digit <= 9))
print(r, '=', s+digit, '*', digit, '+ ',end='')
q, r = q + digit, r - (s+digit)*digit
print (r)
assert (q*q <= n and ((q+1)*(q+1)) > n)
return q, r
n = 741177
q, r = ossqrt(n)
print('\n', n,'
= ',q,' * ',q,' + '‚r‚ sep='')

Use the old-school square root algorithm from class: by hand, find the largest
integer t ≤ 762309. Show your work. Check your answer:

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