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comp-think / 2019-2020

The GitHub repository containing all the material related to the Computational Thinking and Programming course of the Digital Humanities and Digital Knowledge degree at the University of Bologna (a.a. 2019/2020).
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Lecture "Recursion", exercise 2 #26

Open essepuntato opened 5 years ago

essepuntato commented 5 years ago

Define a recursive function def fib(n) that implements the algorithm to find the nth Fibonacci number. In particular, if n is less than or equal to 0, then 0 is returned as a result. Otherwise, if n is equal to 1, then 1 is returned. Otherwise, return the sum of the same function called with n-1 and n-2 as input. Please accompany the function with the related test case.

ereuhl commented 5 years ago 
def fib(n):
    if n <= 0:
        return 0
    elif n == 1:
        return 1
    else:
        return fib(n-1) + fib(n-2)

def test_fib(n, expected):
    result = fib(n)
    if result == expected:
        return True
    else:
        return False

print(test_fib(0, 0)) # True
print(test_fib(1, 1)) # True
print(test_fib(2, 1)) # True
print(test_fib(3, 2)) # True
print(test_fib(4, 3)) # True
print(test_fib(5, 5)) # True
print(test_fib(6, 8)) # True
arcangelo7 commented 5 years ago 
def test_fib(n, expected):
    if fib(n) == expected:
        return True
    else:
        return False

def fib(n):
    if n <= 0:
        return n
    elif n == 1:
        return 1
    else:
        return fib(n-1) + fib(n-2)

print(test_fib(7, 13)) # True
Vince2992 commented 5 years ago

Ex2_11(A)

FrancescoFernicola commented 5 years ago 
def test_fib(n, expected):
    result = fib(n)
    if result == expected:
        return True
    else:
        return False

def fib(n):
    if n <= 0:
        return 0
    elif n == 1:
        return 1
    else:
        return fib(n-1) + fib(n-2)

print(test_fib(0, 0)) #True
print(test_fib(1, 1)) #True
print(test_fib(2, 1)) #True
print(test_fib(4, 3)) #True
print(test_fib(9, 34)) #True
arimuti commented 5 years ago

image

aschimmenti commented 5 years ago

fib1 fib2

elisasilvad commented 5 years ago

Excercise 8 2

essepuntato commented 4 years ago

Hi all,

please find attached my personal solution – also available online:

# Test case for the function
def test_fib(n, expected):
    result = fib(n)
    if expected == result:
        return True
    else:
        return False

# Code of the function
def fib(n):
    if n <= 0:
        return 0
    if n == 1:
        return 1
    else:
        return fib(n-1) + fib(n-2)

# Tests
print(test_fib(0, 0))
print(test_fib(1, 1))
print(test_fib(2, 1))
print(test_fib(7, 13))
print(test_fib(-15, 0))

@aschimmenti please look again at the definition of Fibonacci, because there is something wrong in your implementation. @elisasilvad please remember that all the tests must pass.

aschimmenti commented 4 years ago 
def fib(n):
    if n <= 0:
        return 0
    if n == 1:
        return 1
    else:
       return fib(n-1) + fib(n-2)

fibn = []
requested_number = input("Type the fibonacci number you want: ")
for i in range(int(requested_number)+1):
    fibn.append(fib(i))
    print(fibn)

def test_fib(n, expected):
    result = fib(n)
    if result == expected:
        return True
    else:
        return False
print(test_fib(n, expected))