Problem 34 - How to approach to the Problem

Arman, hello!

And I tried to use sympy solver however its not working

Not sure what you may mean by "not working", this may work... but...

Using "sympy" for this task is what we call "overengineering" :)

The problem has no trick of any kind, just "implement it". You see it is called "Binary Search" so it is supposed to be solved by binary search.

The problem statement has a link to some explanatory article... You want Math function example part from this article. Hope this helps!

Well, it seems you've put a couple of strange things into your algorithm. Speaking strictly it is not the algorithm you want :)

This is confusing for me:

num1, num2 = 0, 100
for i in range(10**4):

we need not that many iterations. moreover it may lead to precision loss (could be seen on some values if reducing amount of iterations).

Now also curious thing:

if abs(func(num1)) > abs(func(num2)):

Why "abs"? and why testing ends? - this definitely is not the correct conditional, though definitely it may work sometimes. :)

Just take func value in the middle between num1 and num2 - your condition should rely only upon this single value.

I thought that increasing the range increases the accuracy

Well, as this is "binary" search and the range is reduced twice on every step, we get great accuracy just after few dozen steps. However, if you continue, due to finite representation of numbers, num1 becomes equal to num2 and when you sum them and divide by two, it can become slightly reduced for some values...

After 30 iterations we have num2-num1 = 100 / (2^30) i.e. about 1e-7...

Now about algorithm - what you do is more like somewhat broken mix of Newton's algorithm with Binary search :)

See, we know that function is monotonous, so it is:

• negative at func(num1)
• positive at func(num2)

Let us use value mid = (num1+num2)/2 - the middle coordinate of the range. Let us check func(mid);

• if it is greater than zero, then function have crossed zero somewhere in the left half, between num1 and mid. Hence we should reduce range to the left half, i.e. num2 = mid...
• if on contrary it is less than zero, then function crosses zero somewhere between mid and num2, so we shoul choose other half, i.e. num1 = mid...

Hello! I have problem with this task. I think my code (below) is right but my answers is different from yours a little bit. I suppose this is because of some calculation algorithms in my programm and computer and I don't know how to fix it unfortunately.

Could you please check my code? If it's right please help me to pass this problem. Thank you in advance.

import math

filename = 'binary_search.txt'
with open(filename) as f:
    nums = f.readlines()
    for i in range(len(nums)):
        nums[i] = nums[i].rstrip()
        nums[i] = nums[i].split()
    for n in nums:
        for i in range(len(n)):
            n[i] = float(n[i])

for n in nums:
    y = 0
    z = 100
    while True:
        x = (z + y) / 2
        a = n[0] * x + n[1] * math.sqrt(x**3) - n[2] * math.exp(-x/50) - n[3]
        if a < 0:
            y = x
        else:
            z = x
        if len(str(x)) >= 15:
            print(x, end=' ')
            break

my answers = 73.590087890625 41.900634765625 your answers = 73.595368554162 41.899174957955

Hi Friend!

I think your solution is correct but doesn't calculate for enough iterations so the answer is not very precise

if len(str(x)) >= 15:

that is most queer ending condition I ever seen :) Just change it to something more reasonable and most surely you'll get right answers!