Nested if ... else (Python exercise)

You shouldn't use if-else as the 3 conditions are not exclusive.

For example, [3, 4, 1] should return in the 3rd condition but is also suitable in the 1st condition, so it returns nothing.

If you don't want to change your code a lot. You can use:

def minimal_three(a,b,c):
    if a < b:
        if a < c:
            return (a)
    if b < a:
        if b < c:
            return (b)
    if c < a:
        if c < b:
            return (c)
    return 'none'

For simple, you can try:

def minimal_three(a,b,c):
    return min(a, b, c)

why that code doesn't work:

def minimal_three(a,b,c):
    if a < b:
        if a < c:
            return (a)
        else:
            # what if a >= c and a < b ?
            return "i returned nothing"
    elif b < a:
        if b < c:
            return (b)
        else:
            # what if b >= c and a > b ?
            return "i returned nothing"
    elif c < a:
        if c < b:
            return (c)
        else:
            # what if b <= c and a < c ?
            return "i returned nothing"
    else:
        return 'none'

Alternative:

def min_of_two(a, b):
    if a > b:
        return b
    return a

def min_of_three(a, b, c):
    min_ab = min_of_two(a, b)
    min_abc = min_of_two(min_ab, c)
    return min_abc

def min_of_three_v2(a, b, c):
    min_ab = a
    if a > b:
        min_ab = b
    min_abc = min_ab
    if min_ab > c:
        min_abc = c
    return min_abc

def min_of_three_v3(a, b, c):
    min_abc = a
    if min_abc > b:
        min_abc = b
    if min_abc > c:
        min_abc = c
    return min_abc

if you really want to use nested if/else (this code is so long):

# if-elif-else is ok.
# nested if is hard to read
# if-elif-elif-elif-elif...-else is hard to read.
# hard to read == easy to have bugs, which is bad.

def min_abc_slower(a, b, c):
    if a > b:
        # a > b. This means min(a, b) == b
        if b > c:
            # b > c. This means min(c, min(a, b)) == c
            return c
        else:
            # b > c is False. This means b <= c.
            # So, min(c, min(a, b)) == b
            return b
    else:
        # a > b is False. This means a <= b.
        # So, min(a, b) = a
        if a > c:
            # a > c. This means min(c, min(a, b)) == c
            return c
        else:
            # a > c is False. This means a <= c
            # So, min(c, min(a, b)) == a
            return a

Use the power of if...elif...else over if...if...if or if...else...if...else. The way you have written your code, depreciates the power of elif. The correct way should be as this:

def minimal_three(a,b,c):
if a < b and a < c:
        return (a)
elif b < c:
        return (b)
else
        return (c)

This code will always return the min no matter what numbers you give.

Explaination:

In your code, the line if a < b already tells you the comparison between a and b that which one is greater. So checking for this condition again in the second if if b < a is useless. if a is not lesser than b, then obviously it is either greater than b or equal to b. So now you must just check if b is lesser than c to prove that b is smallest or even a or b both but the returned value is always minimum. I hope you get that.

Also I don't understand whyt do you want to return None. If you provide three numbers to a function to find the minimum number, it should always and always return a number. Tell me a case where you can expect a None.

Easily be done using min inbuilt function:

def min_value(a,b,c):
    return (min(a,min(b,c)))

also the number of steps in your code can be reduced with some tweaking.

It works - Finding two greatest in three int:

def Biggest(a, b, c):
    if a >= b >= c:
        print a, b, 'are the biggest two'
    elif b >= c >= a:
        print b, c, 'are the biggest two'
    else:
        print c, a, 'are the biggest two'

Find greatest in three integer:

def Biggest(a, y, z):
    Max = a
    if y > Max:
        Max = y    
    if z > Max:
        Max = z
        if y > z:
            Max = y
    return Max

You could probably get away with using a for loop to minimise lines of code! It would look something like this:

c = 3
b = 2
a = 1
list = [a,b,c]
def minimal_three(a,b,c):
    for y in list:
        n = y
    for x in list:
        if x < n:
           print(n)