PPT3

Quick function:

Write a function that checks to see if a number is even or not.

What TYPE does this return?

Boolean Values We now know functions can return:

Numbers (int, double)

Strings

True or False ? AKA Boolean Values

Yes, No

1, 0

We can use True and False as if they are a type)

E.g., return(True)

def ismultof3(x): if ((x%3) == 0): return(True) else: return(False)

When python executes the following statement, what is the result?(x%3)==0

def ismultof3(x): return((x%3) == 0)

def func2(x): if (ismultof3(x)): # Can we see why specifying what type

# is returned from a function is critical?!?

return(str(x) + " is a multiple of 3") else: return(str(x) + " is not a multiple of 3")

Returning to Boolean Values:

Quadratic Equation: x2 - 3x – 4 = 0

Is this true for 1? 2? 3? 4?

Can you write a function that returns the answer to this question? Hint: the function needs to return a boolean value.

What is the input?

How do you check for the output?

Function to represent this:#Name: eqcheck#Calculation: Determines if input value (x) will

solve#the problem:# x2 - 3x – 4 = 0#Input: x: a number#Output: a boolean value

def eqcheck(x): return (x**2 – 3*x – 4) == 0

What is returned?print(eqcheck(3))What is returned? print(eqcheck(4))

Remember?def Squr(par1):

return(par1 ** 2)

def dbl(par2):

return(par2 + par2)

def Func4(p1,p2):

return(dbl(Squr(p2)+ Squr(p2)+3))

print(Func4(2,4))

def Func5(p1):

return(dbl(dbl(dbl(p1))))

print(Func5(4))

What gets printed out?def f(n): if (n == 0): return (" ") else: print(“blug ") return f(n-1)f(3)

What about?def f(x): if (x == 0): return x else: return(x + f(x-1))

print(f(4))

def f2(x): if (x == 1): return (str(x)) else: return(str(x) + f2(x-1))

print(f2(4))

Recursion (one of 3 types of loops)

Recursion is when a function is defined in terms of itself (it calls itself). Def: A recursive definition is one that defines something in terms of itself

(that is, recursively)

Recursion is, in essence, making a function happen again and again without our having to call it (convenient, huh?)

It is one of 3 types of loops we will learn all do the same thing, just different syntax)

#This is recursion

def recurse():

return(recurse())

#This isn’t

def nonrecurse():

return(“nope”)

Try:def f(x):

return(x + f(x-1))

print(f(4))

def f2(x): if (x == 1): return x else: return(x + f2(x+1))

print(f2(4))

def f3(x): if (x == 1): return x else: return(x + f3(x-2))

print(f3(4))

How about:def f(x): if x == 100: return(“none") else: if (x**2 - 3 *x - 4) == 0: print(str(x) + " solves the equation ")

return(f(x+1))

print(f(-100))

Loop Essentials

We now have the basics:1. Must formulate a problem in terms of

itself. (Recursion: the function must call itself)

2. Must include a condition for stopping the recursion (base case)

3. Must make sure that we will always hit that stopping condition.

Sum numbers (1-5)def summing(x):

1. If we start by calling the function with 1, when should we stop?

if (x >= 5):

# must return something that does not make the loop happen again!!!!!

return(x) #why x?

Assume the smallest case possible: Summing one number.

No matter what the number is, if you are summing that number and that number only, you want to return that number.

Alt: if we start summing 1-5 by calling the function with 5, when should we stop?

if (x <= 1):

return(x)

Note: Sometimes we can have MORE THAN ONE stopping condition (you only have to make one of them happen)

Sum numbers (1-5)def summing(x):

1. Stopping Condition?

if (x >= 5):

return(x)

2. How do we make sure that we get to the stopping condition? Increase x every time we “loop”

E.g., summing(x+1)

Alternative: summing(x-1)

If we do these two things, we will likely avoid an infinite loop

Sum numbers 1-5 (cont.)def summing(x):

1. Stopping condition

if (x >= 5):

return (x)

2. Progressing towards stopping condition:

summing(x+1)

3. Finally, we need to write our function: Pretend there are only 2 cases – the last case (step 1, above) and

the case right before that. (e.g., summing 4-5)

We want to return(x + summing(x+1)), right?

From step 1 we know summing (x+1) returns 5

And we know x holds 4.

So summing(4) returns 9

Now what if we summed from 3-5? Can we use what we just did?

Sum Numbers 1-5 (cont.) Put it all together:

def summing(x):

if (x >= 5):

return(x)

else:

return(x + summing(x+1))

print(summing(1))

Let’s try:

1. Write a recursive function that creates a string of every other number starting at 1 and ending at 10

2. Write a recursive function that sums every other number between two integers (you can assume the second integer is greater than the first integer)

3. Write a recursive function that counts the number of numbers that is evenly divisible by 3 between x and y

Problem 1:Write a recursive function that prints out every other number starting at 1 and ending at 10

def g(x):

if x == 10:

return(str(x))

elif x > 10:

return()

else:

return(str(x) + g(x+2))

print(g(1))

Problem 2: Write a recursive function that sums every other number between two integers

def g(x,y):

if x == y:

return(x)

elif x > y:

return()

else:

return(x + g(x+2))

print(g(3,12))

Problem 3: Write a recursive function that counts the number of numbers that is evenly divisible by 3 between x and y

def h(x,y):

if x >=y:

return(0)

elif x%3 == 0:

return (1 + h(x + 1,y))

else:

return(h(x+1,y))

print(h(3,28))

Same?

def h(x,y,z):

if x >=y:

return(z)

elif x%3 == 0:

return (h(x + 1,y,z+1))

else:

return(h(x+1,y,z))

print(h(3,28,0))

Let’s try:

1. Write a recursive function that calculates x to the yth power, assuming we’ve only got multiplication (i.e., you can’t use **)

2. Write a recursive function that determines whether a number is prime or not (and returns True if it is, and False if it isn’t)

Problem 3: Write a recursive function that finds x to the yth power, assuming we’ve only got multiplication (i.e., you can’t use **)

def k(x,y):

if y == 0:

return(1)

else:

return(x * k(x,y-1))

print(k(3,4))

print(k(2,4))

Same?

def k(x,y):

if y == 0:

return(x)

else:

return( k(x*x,y-1))

print(k(3,4))

print(k(2,4))

Now?

def k(x,y,z):

if y == 0:

return(z)

else:

return( k(x,y-1,z*x))

print(k(3,4,___))

print(k(2,4,___))

What does z need to start at?

Write a recursive function that determines whether a number is prime or not (and returns True if it is, and False if it isn’t)def f(x,y):

if (y>(x//2)):

return(True)

elif (x%y==0):

print(y)

return(False)

else:

return (f(x,y+1))

print(f(6,2))

print(f(137,2))

print(f(55,2))

print(f(29,2))

def f(x,y):

if (y>(x//2)):

return(True)

else:

return (x%y!=0 and f(x,y+1))

print(f(6,2))

print(f(137,2))

print(f(55,2))

print(f(29,2))

Lab cheat sheet Random numbers:

from random import *

#imports the random library, including all the functions to generate random numbers

And then

x=randrange(1,10)

#x now holds a number between 1 and 9 (doesn’t include 10)

Example:from random import *import turtle

def f(x): if (x == 5): return("Done") else:

x = randrange(0,10) print("the random number is " + str(x)) return(f(x))

print(f(10))

Turtle colors cheat sheet: Colors in turtle are represented with r g b values (red,green, and blue). Each

value should be between 0 and 1.

So .9,.1,.1 will be pretty red, whereas

.9,.1,.9 will be purple.

To use colors in turtle:from random import *import turtle

def f(x,r,g,b): if (x == 0): return("Done") else: r = randrange(0,100)/100 g = randrange(0,100)/100 b = randrange(0,100)/100 turtle.color(r,g,b) turtle.circle(10) turtle.penup() turtle.goto(randrange(0,100),randrange(0,100)) turtle.pendown()

return(f(x-1,r,g,b))print(f(10,.5,.5,.5))