There will be N lines of outputs, each line being the lowest starting water level that would get to within the acceptable margins for the sweet spot evaporation rate s. Absolute or relative error must not exceed 10-6. Print "Sweet spot cannot be reached! Those cheeky developers!" (without the quotation marks) if the sweet spot is impossible to be reached. Sample Input 0 5 0.7119 -0.1206 2.9930 4.0577 0.9219 0.1963 2.7140 4.3201 0.2692 -0.0695 2.9132 1.6116 0.3601 -0.1771 0.6213 1.1870 0.3492 -0.3865 3.8889 2.5649 Sample Output 0 0.697784503037 0.887908531876 0.921929270250 0.645118392190 0.599166042326 Explanation 0 On the first day, a starting water level of 0.6977845, with d= -0.1206, k = 2.9930, and a = 4.0577 will result in an evaporation rate within 10-6 absolute error of 0.7119. The same is true for the other days. Sample Input 1 7 0.2772 0.2253 4.2550 1.9385 0.2807 -0.2682 1.5963 1.6372 0.9055 0.0522 2.5857 2.2600 0.3196 -0.2520 2.7628 0.9047 0.6853 -0.2366 3.6572 2.5567 0.7148 -0.1846 0.9737 4.1721 0.9411 -0.1370 2.0860 2.6794 Sample Output 1 Sweet spot cannot be reached! Those cheeky developers! 0.793247141838 0.387402764833 0.340837109300 0.353158012405 0.817511673620 Sweet spot cannot be reached! Those cheeky developers!

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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Question

Based on the infomration in the attached screenshots I need help in coding the solution in python.

Additional Explanation

Explanation 1

On the first day, it is impossible to reach an evaporation rate within 10^-16 of 0.2772 given d = 0.2253,k=4.2550 and a = 1.9385.

The Python Code so far-

import math

"""
Parameters:
s : float - target rate as described in problem
d : float - constant in formula described in problem
k : float - constant in formula described in problem
a : float - constant in formula described in problem

Return:
A STRING containing the answer
"""

def rate(x, d, k, a):
return 1-(1/(1 + math.exp(-k * (x - 0.5))))**a + d

def solve(s, d, k, a):
# binary search for the lowest starting water level
l, r = 0, 1
while r - l >= 1e-12:
mid = (l + r) / 2
if rate(mid, d, k, a) > s:
l = mid
else:
r = mid
return l if abs(rate(l, d, k, a) - s) <= 1e-12 else None

def main():
n = int(input().strip())
for i in range(n):
s, d, k, a = map(float, input().strip().split())
result = solve(s, d, k, a)
if result is None:
print("Sweet spot cannot be reached! Those cheeky developers!")
else:
print("{:.12f}".format(result))

if __name__ == '__main__':
main()

While this works for some test cases, its incorrect for a bunch as well. I need help in fixing any possible errors in this existing code while also trying to improve it (if possible) to make it more faster(less time complexity)

There will be N lines of outputs, each line being the lowest starting water level that would get to within the
acceptable margins for the sweet spot evaporation rate s. Absolute or relative error must not exceed 10-6.
Print "Sweet spot cannot be reached! Those cheeky developers!" (without the quotation marks) if the sweet
spot is impossible to be reached.
Sample Input 0
5
0.7119 -0.1206 2.9930 4.0577
0.9219 0.1963 2.7140 4.3201
0.2692 -0.0695 2.9132 1.6116
0.3601 -0.1771 0.6213 1.1870
0.3492 -0.3865 3.8889 2.5649
Sample Output 0
0.697784503037
0.887908531876
0.921929270250
0.645118392190
0.599166042326
Explanation 0
On the first day, a starting water level of 0.6977845, with d= -0.1206, k = 2.9930, and a = 4.0577 will
result in an evaporation rate within 10-6 absolute error of 0.7119. The same is true for the other days.
Sample Input 1
7
0.2772 0.2253 4.2550 1.9385
0.2807 -0.2682 1.5963 1.6372
0.9055 0.0522 2.5857 2.2600
0.3196 -0.2520 2.7628 0.9047
0.6853 -0.2366 3.6572 2.5567
0.7148 -0.1846 0.9737 4.1721
0.9411 -0.1370 2.0860 2.6794
Sample Output 1
Sweet spot cannot be reached! Those cheeky developers!
0.793247141838
0.387402764833
0.340837109300
0.353158012405
0.817511673620
Sweet spot cannot be reached! Those cheeky developers!

To simulate rice paddy maintenance, a game company decided to use the following function to calculate the
evaporation rate of water:
rate = 1
-)a + d
Where:
1
1+e-k(x-0.5)
• x-how full the rice paddy is at the start of the day, the value of which ranges from 0.0 (0% water level) to
1.0 (100% water level)
d, k, and a - variables that control the rate of evaporation.
When a rice paddy is filled with water, the evaporation rate is set for that day, and water would slowly
evaporate following this rate.
Upon releasing the game, players discovered that if a sweet spot evaporation rates, with an error margin of
0.1 is used, the water level would be perfect every time and would result in the best quality crop. However,
because players were not yet aware of factors d, k and a, they were unable to figure out how to properly set
their starting water level to exploit this sweet spot.
After data mining through the game, however, you managed to uncover factors d, k, and a, and you found out
that these three factors were arbitrarily set at the start of each day. Moreover, there are indeed days where,
no matter what you set your starting water level to, the sweet spot cannot be reached (darn those cheeky
game developers!). Your task is to write a program that, given the values of d, k, and a, would determine the
starting water level that would result in an evaporation rate of s for that particular day. On days where s
cannot be reached, indicate that "Sweet spot cannot be reached! Those cheeky developers!".
Input Format
The first line of the input contains an integer N, indicating the number of days you would need to figure out
the starting water level for.
N lines follow, each containing four space-separated real numbers that correspond to the value of s, d, k, and
a. s is the sweet spot evaporation rate. d, k, and a are described in the problem statement.
Constraints
0≤N≤5.10³
0 ≤s≤1
-0.5 ≤ d ≤ 0.5
0 ≤ a, k≤ 5

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