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What gets printed out?

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def f(n):

if (n == 0):

return (" ")

else:

print(“blug ")

return f(n-1)

f(3)

deff(x):

if (x == 0):

return x

else:

return(x + f(x-1))

print(f(4))

deff2(x):

if (x == 1):

return (str(x))

else:

return(str(x) + f2(x-1))

print(f2(4))

- Recursion is when a function is defined in terms of itself (it calls itself).
- Def: A recursive definitionis 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?)
#This is recursion

def recurse():

recurse()

return(recurse())

#This isn’t

def nonrecurse():

return(“nope”)

def f3(x):

if (x == 1):

return x

else:

return(x + f3(x-2))

print(f3(4))

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 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))

- We now have the basics:
- Must formulate a problem in terms of itself. (the function must call itself)
- Must include a condition for stopping the recursion (base case)
- Must make sure that we will always hit that stopping condition.

Stack in memory

f(4) = 4 + f(3)?

f(3) = 3 + f(2)?

f(2) = 2 + f(1)?

f(1) = 1 + f(0)?

f(0) = 0

__________________________

f2(4) = “4” + f2(3)

f2(3) =“3” + f2(2)

f2(2) = “2” + f2(1)

f2(1) = "1"

def f(x):

if (x == 0):

return x

else:

return(x + f(x-1))

print(f(4))

deff2(x):

if (x == 1):

return (str(x))

else:

return(str(x) + f2( x-1))

print(f2(4))

def f7(a,b):

if (b <= 0):

return(a)

elif((b%3)== 0):

return(f7(a+1,b-1))

else:

return(f7(a,b-1))

print(f7(0,13))

def f5 (a):

if (a <= 0):

return(1)

elif ((a%2) ==0):

return (f5(a - 1))

else:

return (a*f5(a-1))

print(f5(6))

print(f5(5))

print(f5(-1))

def f6(x):

if (x <= 1):

return str(x)

else:

return(f6(x-1) + str(x) )

print(f6(5))

(Cool, but challenging)

def f4 (a, b):

if (b == 0):

return (a)

if (a < b):

return f4 (b, a)

else:

return f4 (b, a%b)

print(f4(27,12))

print(f4(25,50))

print(f4(15,20))

defnums(x,y,z):

if y == 1:

return z + x

else:

return(nums(x%y, y//10, z+x//y))

print(nums(1354,1000,0))

print(nums(254,100,0))

What did we just do?

- Write the base case (the stopping case) first!
- There can be more than one stopping condition

- Figure out how you’re going to get to the base case
- (e.g., write it as if it only needs to happen twice, once without the base case and once with, making sure the second case gets you to the first case).

- Write a recursive function that prints out every other number starting at 1 and ending at 10
- Write a recursive function that counts the number of numbers that is evenly divisible by 3 between x and y
- Write a recursive function that calculates x to the yth power, assuming we’ve only got multiplication (i.e., you can’t use **)
- 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)

defg(x):

if x == 10:

returnstr(x))

elif x > 10:

return()

else:

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

print(g(1))

defh(x):

if x > 15:

return(0)

elif x%2 == 0:

return (x + h(x + 1))

else:

return(h(x+1))

print(h(1))

defk(x,y):

if y == 0:

return(1)

else:

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

print(k(3,4))

print(k(2,4))

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))

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))