Returning the smallest integer of an increasing function

Question

Write a function lowest_integer() which takes 2 input arguments:

• a function f representing an increasing function (),
• a number fmin,

and returns an integer nmin such that >0 is the smallest integer that satisfies ()>.

To test the function we use code like this where we define f(n) then print(lowest_integer(f, fmin)) and get the expected output:

def f(n):
    return 2*n
print(lowest_integer(f, 10))
## 6

def f(n):
    return n**2 + 6*n - 3
print(lowest_integer(f, 500))
## 20

My attempt at the code:

def lowest_integer(f, fmin):
    f = fmin
    nmin = n + 1
    return nmin

I'm having trouble figuring out how to define the function f(n) within the lowest_integer() function

Answer 1

You can loop through the integers starting at 1, until you find one that satisfies your condition, in that case, return it:

def lowest_integer(f, fmin):
    nmin = 1
    while True:
        if f(nmin) > fmin:
            return nmin
        nmin += 1

def f(n):
    return 2*n
print(lowest_integer(f, 10))

def f(n):
    return n**2 + 6*n - 3
print(lowest_integer(f, 500))

Output:

6
20

This approach, of course, is a naive one, and will be slow if the function f is slow and fmin is very big, but in general it does good.

A faster approach would be using a library like sympy or something, that have functions that use more sophisticated methods and techniques that are suited to these kinds of problems.

Answer 2

You can do it like that :

def lowest_integer(f, fmin):
    nmin=1
    while f(nmin)<=fmin :
        nmin += 1
    return nmin 

You begin with nmin = 1 and you add 1 to nmin while f(nmin)<=fmin. To do that, you can use a while loop.

By doing f(nmin), you call the function f and calculate its value for the entry nmin.

Answer 3

def f(n):
    return 2*n
def lowest_integer(f, fmin):
    for findmin in range(fmin):
        if f(findmin) > fmin:
            return findmin
    return -1
print(lowest_integer(f, 10))