Problem 12. Fibonacci sequence

easy. classical iteration problem. solving this problem is easy, but iteration is hard to master.

hahahahah

Classic!

Is there a shorter way?

是的。
Also this solutions will fail fib(0)

I broke Cody server with this.

function f = fib(n)
a = [0 1];
indx = n-1;
if n==0
f=a(1);
elseif n==1
f=a(2);
else
while indx <= n
a(indx+1)=fib(n-1)+fib(n-2);
indx = indx+1;
end
f=a(n);

end

我的大小是-1不应该赢吗？

When you want to calculate fib(1), it calls fib(0) + fib(-1). There is no fib(-1) so you never leave the while loop.

function f = fib(n)

c =[1 2];
for i=1:n
c（i+2）= c（i）+c（i+1）
end
f = c（：，i）
end
could anybody please give me hint. I am struck here
f(1) fails. All other cases passed

将C = [1 2]更改为C = [1 1]

i wrote a code just like i would in c++. The code is huge by size but the answers are right however cody seems to disagree.

nice task

nice

为什么fibonacci功能在这里不起作用？

Can someone explain about the solution size please..

good job

这太容易了

The Test Suite should include a check using random integer inputs to prevent simple indexing to generate a solution.

我认为我的代码没有任何错误.......

很容易

很容易

.

i had to add a line of code because you didnt expect the number to be a float rather then an integer so it counted my last test as wrong all because it had .000 at the end of each one.

I tried this one in MATLAB and it worked but here it keeps displaying assertion failed for every number different from 1

因为您使用的是浮点计算，所以结果可能不准确。如果您将结果围绕到最近的整数，则应起作用。

fibonacii系列OS OS该系列，其中每个数字是先前数字的总和

Can anyone suggest how to reduce the size of the code ?

是的，有两种方法。您可以将F1初始化为矢量[1，1]。如果使用start i = 1：n，您也不需要if else语句，然后使用f1（i+2）= ...

Not sure what went wrong...

Put this at the end (between the two 'end's)
f = f（n）;

:)

我什至没有考虑走递归路线。现在似乎很明显。

我正在使用平方纤维契数的属性。

对齐在这里是一个问题:)

Yay recursion

我承认这有点作弊。

我不知道怎么作弊。你能不能告诉我？

Testsuite seems to have problem for value n=1

I don't understand why the size of this code is so huge, can anyone help ?

Hi all,
I have used regular expressions in other languages before but not in MATLAB. Can someone explain the solution to me? I have looked up the format of regexp in the help file, but the use of the regexp in the solution does not seem to fit the description in help.

当我尝试在八度中运行代码时，它会出现错误，因为它无法识别“无效调用Regexp”错误。

Can someone please help!

Nice !

problem in the test suite my Solution was perfect, please update the test suite

我不确定您是如何得出的。给出的示例说FIB（5）应为5。您的代码返回15。

这被打破了：我急于在第一个版本中修复错字，我不小心用1个错误的2替换了错误的2。

我不明白为什么这不正确。

I got the equation from 1) and the calculations are correct when I use Matlab at home (see 2)).

1) https://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form
2）https://en.wikipedia.org/wiki/fibonacci_number#list_of_fibonacci_numbers

如果您返回地板（F），这将起作用。

Thank you =)

这是一个很好的答案。比'圆（1.61803^n /2.236）更好。如果n> 24，后者是错误的。

编程中的递归........................

不是我的工作；以前的领导者这样做了，我只是使用“ Ans Trick”将大小减少了2。

这并不是真正的解决方案，因为对于大N，它最终会产生不准确的结果。

True. It can be improved a bit though by using more decimals for phi.

Even though there are smaller solutions, I prefer this one.

好东西 ：）

哇！你是怎么想到的？

我们终于开始递归生成斐波那契

i want to know how does the second line work , i don't understand the reslut ,can u help???,

A clever idea, knowing that the determinant of the dramadah matrix obeys a Fibonacci series

该解决方案在n = 23之后不起作用。

It's still one hell of a hack! :)