Lecture "Introduction to Computational Thinking", exercise 2

[essepuntato]

What is the result of applying the second natural language definition of the Fibonacci function in Section "Natural languages vs programming languages" using “7” as input?

[lelax]

127

[LudovicaErre]

According to the definition, the number 7 is not less or equal to 0 and it is not equal to 1. So, we have to return the sum between 7-1=6 and 7-2=5.
Considering that the F. sequence starts from 0 and 1 and that each number is the sum of the two preceding ones, we have F2=F1+F0=1, F3=F2+F1=2, F4=F3+F2=3. Continuing in this way we will obtain the results of 6 and 5, respectively 5 and 8, so the result is 13.

[federicabonifazi]

Implementing the function of Fibonacci with the concept of recursion we know that if "n" is greater than 1 the function returns the sum of the same function with “n-1” as input and still the same function with “n-2” as input, so:

[martasoricetti]

The result is 13 and I gained it following recursively the instructions that say that F(n) is the sum of F (n - 1) and F (n - 2), where n in this case is 7.
F(7) = F(6) + F(5)
F(6) = F(5) + F(4)
F(5) = F(4) + F(3)
F(4) = F(3) + F(2)
F(3) = F(2) + F(1)
F(2) = F(1) + F(0)
So F(2)=1; F(3)=2; F(4)=3; F(5)=5; F(6)=8; F(7)=13

[OrsolaMBorrini]

The result is 13.
Being 7 (our input n) different (and bigger) than both 0 and 1, we have to sum two fibonacci functions. The first with input (n-1) and the second with input (n-2).
Therefore, we will have F(6)+F(5).
We apply the same instructions recursively and at the end we go back summing together all the results. We end up with F(7) = F(6)+F(5) = 13.

[Pablo751]

n = 7 (function input)

a = 1
b = 1

Loop while():

Loop 1:
c = a + b = 1 + 1 = 2

Is n less or equal to 3? NO

a = b = 1
b = c = 2
n = n - 1 = 6

Loop 2:
c = a + b = 1 + 2 = 3
Is n less or equal to 3? NO

a = b = 2
b = c = 3
n = n - 1 = 5

Loop 3:
c = a + b = 2 + 3 = 5
Is n less or equal to 3? NO

a = b = 3
b = c = 5
n = n - 1 = 4

Loop 4:
c = a + b = 3 + 5 = 8
Is n less or equal to 3? NO

a = b = 5
b = c = 8
n = n - 1 = 3

Loop 5:
c = a + b = 5 + 8 = 13

Is n less or equal to 3? NO

a = b = 8
b = c = 13
n = n - 1 = 2

[olgagolgan]

The function recalls itself 2 times every time that we call it with an input different from either 0 or 1, in other words to find the result we have to apply the function again and again to every number starting from "n" until we get 1 or 0 as a result in every step.
At the first iteration "n" is 7, reaching the third condition of the function, we have to recall the function using 6 and 5 as input and sum the final results, 8 and 5, obtained by recalling the function two times for every intermediate result. We get 13 as the final result.

[Marethyu6]

It's "13", but my first take was "11". ... a dumb error, that demonstrated me remembering a bit more math definitions would have helped avoiding me the trial/error step, since the first result I got didn't match casting out nines with the Fibonacci rules.
It happen because I misinterpreted the natural language and was applying the "(n-1)+(n-2)" rule not recursively on the sequence i.e. ignoring it was a function; so just applying the rules partially, with the assumption that "the sequence" was the one of the natural numbers.
What I got then was the sequence of odd numbers, and for "n=7": (n-1)+(n-2) = (nx2)-3 = 13-3 = 11 .

[CarmenSantaniello]

Applying those instructions, it leads us to the concept of recursion. If 'n' is less or equal 0, it returns 0 and if 'n' is equal 1, it returns 1. But taking '7' as input, it returns the sum of the function respectively with "7-1" and "7-2".

Then, according to the recursion concept, it will be:
F(0) = 0 F(1) = 1
F2 = F(1) + F(0) = 1
F3 = F(2) + F(1) = 2
F4 = F(3) + F(2) = 3
F5 = F(4) + F(3) = 5
F6 = F(5) + F(4) = 8
F7 = F(6) + F(5) = 13

The final result is 13.

[ManueleVeggi]

Since 7>1, I have to consider the "else" statement. As I tried to show below, the correct number is 13 (i.e. the sum of the yellow cells).

[angstigone]

after realizing that n isn't either equal or less to zero and neither equal to one, I had to obtain the sum of the F(n-1) and F(n-2), which was thirteen.

[francescabudel]

(a,b) = 1
(n) = 7
(c) = (a+b) = 2
n > 3 --> a=b; b=c

F(n) = a+b+c
F(n>3) = (a+b) + (a+b) + (a+b) + (a+b)
F(7 - 1) = 4(a+b) - 1
F(7 - 1) = 6 + 7 = 13

If we apply the concept of recursion:

n > 1 —> n = 1 | n= (n-1) + (n-2)

F2 = F(1) + F(0) = 1
F3 = F(2) + F(1) = 2
F4 = F(3) + F(2) = 3
F5 = F(4) + F(3) = 5
F6 = F(5) + F(4) = 8
F7 = F(6) + F(5) = 13

[mirna-regolo]

n = 7
i.e.: n > 0; n > 1; n > 2

a = 1;
b = 1;
c = a + b = 2;
n > 3

a = b = 1;
b = c = 2;
c = a + b = 3;
n = n - 1 = 6;
n > 3

a = b = 2;
b = c = 3;
c = a + b = 5;
n = n - 1 = 5;
n > 3

a = b = 3;
b = c = 5;
c = a + b = 8;
n = n - 1 = 4;
n > 3

a = b = 5;
b = c = 8;
c = a + b = 13;
n = n - 1 = 3;
n = 3

a = b = 8;
b = c = 13;
c = a + b = 21;
n = n - 1 = 2;
n < 3

[MaddaGh]

Being n=7, therefore n>3. According to the instructions, one must "repeat the following
operations indefinitely until a value is returned. Set the variable
“c” as the sum of “a” plus “b”. Then assign the value of “b” to “a” and the
value of “c” to “b”, and finally decrease the value of “n” by 1
before repeating".
Knowing that
a=1
b=1
c= a+b,
the operations should unfold as follows:
(n=7) a+b=c --> 1+1=2
(n=7-1=6) 1+2=3
(n=6-1=5) 2+3=5
(n=5-1=4) 3+5=8
(n=4-1=3) 5+8=13
At this point n=3 so we can apply the following instructions: If “n” is less than or equal to “3”
then return “c”.
Therefore the result should be c=13

(I had to read the C programming language to understand that though, lol. Natural language instructions were very confusing...)