Class Notes: Conditionals

Boolean Type:

• booleans: True and False

Comparison operators:

Equality

x = 10
y = 3
y == x

False

4 == 4.0

True

3.0 == 3.5

False

"Disequality" (not equal)

4.0 != 4

False

3 != 7

True

Less than

4 < 6

True

4.0 < 3.9

False

5 < 6 < 7

True

Less than or equal

32 <= 32

True

32 <= 32.8

True

7.5 <= 4

False

Greater than

5 > 6

False

8 > 4

True

Greater than or equal

4.0 >= 4

True

4.5 >= 3

True

4.5 >= 9

False

Boolean operators:

Just like we have operators with numeric types, we also have operators with booleans

And

a	b	a and b
True	True	True
True	False	False
False	True	False
False	False	False

True and False

False

x = 5
print(0 < x and x != 5)
print(x <= 10 and x > 6)
print(x < 5 and x <= 4)
print(x <= 5 and x >= 3)

False
False
False
True

Or

a	b	a or b
True	True	True
True	False	True
False	True	True
False	False	False

True or False

True

x = 5
print(0 < x or x != 5)
print(x <= 10 or x > 6)
print(x < 5 or x <= 4)
print(x <= 5 or x >= 3)

True
True
False
True

4 < 0 or 3 < 0

False

Not

a	not a
True	False
False	True

x = 5
not x < 0

True

not x >= 4

False

If conditions

if-else

def myAbs(x):
    if x < 0:
        x = -x
        # ... other commands may appear here, for example, another assignment or another if!
    else:
        x = x
        # else case cannot be empty
    return x

myAbs(-3)

3

if (without else)

If an else case "does nothing", we can get rid of it.

def myAbs(x):
    if x < 0:
        x = -x
    return x

myAbs(-32)

32

Exercise 1: sign function

Write the function that implements the signum function: $$ \text{sgn }x := \begin{cases} -1 & \text{if } x < 0 \\ 0 & \text{if } x = 0 \\ 1 & \text{if } x > 0 \end{cases}$$

When we need multiple conditional expressions after the if condition or between the if and else conditions, we can use the if-elif statements:

if-elif

def sign(n):
    s = 0
    if n < 0:
        s = -1
    elif n > 0:
            s = 1
    #else:  # we can omit the else here. why?
    #    s = 0
        
    return s
        
sign(-43)

-1

Act like a computer

What's the output of the following function for x = 3?

def myFunction(x):
    if x > 4:
        x = x + 1
    elif x < 3.5 and x >= 2:
        x = x * 2
    elif x > 0:
        x = 1
    else:
        x = -1
    x = x + 4
    return x

The answer is 10

print(myFunction(3))

10

Exercise 2: middle number

Implement the function middle(a, b, c) that takes three different numbers as input, and returns the middle among them.

Let's enumerate all possible cases

$ b < a < c$

$ c < a < b$

$ a < b < c$

$ c < b < a$

$ a < c < b$

$ b < c < a$

Solution only with if statements

def middle(a, b, c):
    if b < a < c:
        return a
    if c < a < b:
        return a
    if a < b < c:
        return b
    if c < b < a:
        return b
    if a < c < b:
        return c
    if b < c < a:
        return c
        
        
def testMiddle():
    assert(middle(1,2,3) == 2)
    assert(middle(3,2,1) == 2)
    assert(middle(1,0,3) == 1)
    assert(middle(1,200,100) == 100)
    assert(middle(7, 100, 1)==7)
    print("Passed...")
testMiddle()

Passed...

Another solution using elif statements

# elif instead of if
def middle(a, b, c):
    if b < a < c:
        return a
    elif c < a < b:
        return a
    elif a < b < c:
        return b
    elif c < b < a:
        return b
    elif a < c < b:
        return c
    elif b < c < a:
        return c
        
        
def testMiddle():
    assert(middle(1,2,3) == 2)
    assert(middle(3,2,1) == 2)
    assert(middle(1,0,3) == 1)
    assert(middle(1,200,100) == 100)
    assert(middle(7, 100, 1)==7)
    print("Passed...")
testMiddle()

Passed...

Another solution using elif statements and else at the end

# else at the end
def middle(a, b, c):
    if b < a < c:
        return a
    elif c < a < b:
        return a
    elif a < b < c:
        return b
    elif c < b < a:
        return b
    elif a < c < b:
        return c
    else:
        return c
        
        
def testMiddle():
    assert(middle(1,2,3) == 2)
    assert(middle(3,2,1) == 2)
    assert(middle(1,0,3) == 1)
    assert(middle(1,200,100) == 100)
    assert(middle(7, 100, 1)==7)
    print("Passed...")
testMiddle()

Passed...

Another equivalent solution with fewer checks

# more compact
def middle(a, b, c):
    if b < a < c:
        return a
    elif c < a < b:
        return a
    elif a < b < c:
        return b
    elif c < b < a:
        return b
    else:
        return c
        
        
def testMiddle():
    assert(middle(1,2,3) == 2)
    assert(middle(3,2,1) == 2)
    assert(middle(1,0,3) == 1)
    assert(middle(1,200,100) == 100)
    assert(middle(7, 100, 1)==7)
    print("Passed...")
testMiddle()

Passed...

Another (more compact) solution

# even more compact
def middle(a, b, c):
    if b < a < c or c < a < b:
        return a
    elif a < b < c or c < b < a:
        return b
    else:
        return c
        
        
def testMiddle():
    assert(middle(1,2,3) == 2)
    assert(middle(3,2,1) == 2)
    assert(middle(1,0,3) == 1)
    assert(middle(1,200,100) == 100)
    assert(middle(7, 100, 1)==7)
    print("Passed...")
testMiddle()

Passed...

All previous solutions are equivalent and correct