Appologies I don't know fortran I'll try my best to show you how to speed it up in javascript and my best at a fortran solution var memo = []; function fib(n) { if (memo[n-1]) { //check to see if you already calculated the answer return memo[n-1]; } memo[n-1] =...

Your problem is that you only return the total if n == 0, but your main function actually calls it with n > 0. Make your function always return total + 1, not only when n is zero (in other words just remove the else), and it should work.

c++,algorithm,recursion,fibonacci

long fibo(int N, bool print) { long value = 0; if(1 == N) value = 1; if(1 < N) value = fibo(N-1, print) + fibo(N-2, false); if(print) std::cout << N << " => " << value << std::endl; return value; } int main(){ fibo(5, true); return 0; } What you...

c#,output,windows-forms-designer,fibonacci,labels

lblOUT.Text += Convert.ToString(num2) + Environment.NewLine; That should be it...

The reason its returning 0: result = addFibNumbers(num, oldnum);//num=1,oldNum=0 //function while(num < 4000000) { //num is 1, so it enters while if (num % 2 == 0) {// 1 % 2 == 1, so skip this if return total;// this ends the function, returning total=0 as nothing was changed I...

The sums of the first ten and 100 fibonacchi number would be 88 and 573147844013817084100, respectively: >>> cache = {} >>> def fib(n): if n == 0: return 0 if n == 1: return 1 if not n in cache: cache[n] = fib(n - 1) + fib(n - 2) return...

exception,clojure,fibonacci,lazy-sequences

You are right, it is a lighttable bug. You should log it here The following work in lighttable without issue (->> [0 1] (iterate (fn [[a b]] [b (+ a b)])) (take 10) (my-map first)) and (take 10 (->> [0 1M] (iterate (fn [[a b]] [b (+ a b)])) (my-map...

arrays,algorithm,search,binary-search,fibonacci

Because if we have a Fibonacci sequence : 0,1,1,2,3,5,8,13,... F(k - 1) + F(k - 3) is what falls in the middle between F(k - 1) and F(k), and that's what we want because it's divide and conquer.

Let me first make the program easier to understand and also more general by using true arithmetic relations between integers instead of low-level arithmetic: :- use_module(library(clpfd)). fib(0, 0). fib(1, 1). fib(V, VF) :- V #> 1, B #= V - 1, C #= V - 2, VF #= BF +...

javascript,algorithm,big-o,fibonacci

If you don't have an Array then you save on memory and .push calls function fib(n) { var a = 0, b = 1, c; if (n < 3) return 1; while (--n) c = a + b, a = b, b = c; return c; } ...

It means that MATLAB can't find your function - the directory where you saved the file fibfun.m should either be the current directory or defined in your MATLAB path.

Using a loop you could store the values in an array that could stop immediately one key after finding the selected number in the previous keys value. function getFib($n) { $fib = array($n+1); // array to num + 1 $fib[0] = 0; $fib[1] = 1; // set initial array keys...

For your reference, here is the syntax fixed version of your code: mortalRabbits xs 0 = xs mortalRabbits xs n = mortalRabbits xs' (n-1) where xs' = updateRabbits xs updateRabbits (x:xs) = case x of 0 -> [1] ++ updateRabbits xs 1 -> [2, 0] ++ updateRabbits xs 2 ->...

Obviously memoisation is only useful if you do it precisely once, and then call to that multiple times. Whereas in your approach you keep memoising the function over and over again. That's not the idea! fib' n = memoize fib n is the right start, yet won't work as desired...

One way would be to keep track of all coordinates at all times and increment depending of the orientation of the following square to place. I think your whole structure is ok, but there are some info lacking, mainly keeping bottom and right coordinates. Something like this seems to work...

javascript,loops,for-loop,while-loop,fibonacci

Here's a working example with recursive call: function myFunction(getLucas) { var x = document.f1.n1.value; if (getLucas) { alert(lucas(x)); } else { alert(fib(x)); } } function fib(n) { if (n < 2) { return n; } else { return fib(n - 1) + fib(n - 2); } } function lucas(n) {...

python,runtime-error,fibonacci

enumerate() always starts counting at 0; it cannot know that you sliced your list: for index, element in enumerate(myarr[3:]): Here index starts at 0, so you are trying to index from the end when you use index - 1 and index - 2; these translate to myarr[-1] and myarr[-2]. Those...

Consideration fib( 4) = fib( 2) = 1 * fib(2) fib( 6) = fib( 4) + fib( 2) = 2 * fib(2) fib( 8) = fib( 6) + fib( 4) = 3 * fib(2) fib(10) = fib( 8) + fib( 6) = 5 * fib(2) fib(12) = fib(10) + fib(...

Since your fibonacci function is taking an input there isn't exactly a need to pass a parameter. But in the case of your error n isn't defined in the global scope. I would just get rid of the n parameter. Also, just replace stopNumber with n. def fibonacci(): a=0 b=1...

Here's and example. Problems like the Fibonacci are better dealt with recursion. I see that with your do while loops you are trying to use some sort of recursion but it's not really working int fibonacci(int x) { if (x == 0) return 0; if (x == 1) return 1;...

haskell,recursion,fibonacci,reduction

The order of the equations in the definition does matter. The part fib n = fib (n-1) + fib (n-2) gets applied only when the previous lines do not apply. That is, only when n is not 0 nor 1. Because of this, the step fib 3 = fib (3-1)...

c++,recursion,iteration,fibonacci,timing

For each input N, you need to calculate Fibb(N) six times, and display the minimum and maximum of those times. You display one minimum per N, and one maximum per N. Does it help to imagine that there are 100 N, and each one is calculated 6 times? You'd expect...

This can be proven by induction, but as previous posters have shown, it's tricky to get formally correct when referring to K directly in the proof. Here's my suggestion: Let P(n) be the property we want to show: P(n) holds iff K(n) yields 1 for n = 1, and 0...

java,algorithm,dynamic-programming,fibonacci

You're calling getFib() recursively, and instantiating a new dictionary with each call. Make the dictionary a class-level variable.

Sure, just use some form of loop. For example, if you want to make a list of the first 11 Fibonacci numbers after x: fiblist = [x] for _ in range(10): z = x + y fiblist.append(z) x, y = z, x print(fiblist) (or use a loop instead of the...

haskell,memory-leaks,sequence,fibonacci,space-leak

Well that is a stack overflow, but you also have a space leak, which is easier to explain in a few words. When you perform the indexing u !! n, u looks like 1 : 2 : 1 : ... : 11 : <go thunk> : ... : <go thunk>...

recursion,f#,fibonacci,lazy-sequences

Yes, it's called "Tail Call Optimization" See here: http://blogs.msdn.com/b/chrsmith/archive/2008/08/07/understanding-tail-recursion.aspx Also Seq is lazy so its 500th member will not be evaluated until you will not have to access it in your program like: let elm = Seq.nth 500 fibSeq ...

#Fibonacci number calculator function def fib(n): if n==0: return 1 elif n==1: return 1 else: return fib(n-1)+fib(n-2) Should work for finding n. fibonacci number....

In your loop, you are using fib_seq(x) and it should be fib_seq(i) Also, if you want to reduce time a bit more, you can use memoization technique def fib_seq(n): if n == 0: return n elif n == 1: return n else: return fib_seq(n-1) + fib_seq(n-2) def memoize(fn, arg): memo...

It looks like the bottom piece of code simply calls the fib function on x=0, x=1 ... x=9 and stores it's return value at the end of the array. When times is invoked with an iteration variable (x), it begins at 0 and increments on each iteration through the loop....

The Fibonacci numbers will grow large quite fast. At the 46th number, you start getting negative numbers, e.g. -1323752223. This is because the numbers have grown so large that it overflows the int datatype. You can use long[] arrays, but that will only postpone the problem. You'll starting getting negative...

As pointed out in the comments, your approach is a tad overcomplicated. In particular, you don't need to use recursive calls, or even the reverse function, in order to generate the Fibonacci sequence. A linear-time implementation In addition to your own answer, here is a textbook one-liner, which uses memoization:...

java,dictionary,hashmap,fibonacci

No, it does not. You can read in the JavaDocs, there it clearly states that put returns the previous value associated with key, or null if there was no mapping for key. V put(K key, V value) Associates the specified value with the specified key in this map (optional operation)....

You have a typo, fibFromSafe calls your original fibFrom.

java,eclipse,recursion,fibonacci,nosuchelementexception

Do not call scanner.close(); When you do that, you close() System.in! Then when you attempt to construct your new Scanner(System.in); it doesn't work (because System.in is closed)....

Search Google for getline and awk and you'll mostly find reasons to avoid getline completely! Often it's a sign you're not really doing things the "awk" way. Find an awk tutorial and work through the basics and I'm sure you'll see quickly why your attempt using getlines is not getting...

You set H = 1 as the first statement in your while loop; so every time you enter the for loop, H = 1 and you'll only get the Fibonacci number for n=0 You need to set H = 1 outside the while loop: def Frange(x): A = [0] H...

You need to declare the biggest as a global using global statement (especially when there's an assignment to the variable). Otherwise, it is treated as a local variable. cache = {} biggest = 1 def fib(n): global biggest # <----- .... ...

Your code isn't wrong, it's just too slow. In order to solve Project Euler problems, not only does your code have to be correct, but your algorithm must be efficient. Your fibonacci computation is extremely expensive - that is, recursively trying to attain the next fibonacci number runs in O(2^n)...

assembly,recursion,stack,masm,fibonacci

Your formula is wrong. The first two values of a Fibonacci sequence are both 1, so fib(1) and fib(2) have to return 1 (not 'n - 1'). Edit: ok, ok... there can be another start value, so your formula can be right ;-) At the end of the procedure you...

Do you mean that you want to calculate the sum of all the odd numbers? Firstly you haven't declared and initialised febCount so you need to fix that. In this case febCount would be the same as Fibcnt, but you could use feb.length instead. public class Fibonacci { public static...

As everyone mentioned here, this is basically recursion. Just to get a feel of how this program works, I have made the recursion tree with initial fibIndex as 5. 5 5 calls 4 and 3. / \ 4 3 4 calls 3 and 2. 3 calls 2 and 1. /...

This can be solved using Dynamic programming. Lets call the function F(N) = probability to reach 0 using only 2 and 3 when the starting number is N F(N) = 0.2*F(N-2) + 0.3*F(N-3) Base case: F(0) = 1 and F(k)= 0 where k< 0 So the DP code would be...

I think he explained very clearly in the book. You should notice that when an element is added to the heap, there is no work done O(1) , the new element is just simply attached to the root, and the heap is only reorganized until an element is removed.( Look...

c#,.net,stack-overflow,biginteger,fibonacci

You can use BigInteger without recursion: public static int FibHugesUntil(BigInteger huge1, BigInteger huge2, int reqDigits) { int number = 1; while (huge2.ToString().Length < reqDigits) { var huge3 = huge1 + huge2; huge1 = huge2; huge2 = huge3; number++; } return number; } static void Main(string[] args) { Console.WriteLine(FibHugesUntil(BigInteger.Zero, BigInteger.One, 1000));...

php,recursion,fibonacci,memoization

Well, Edd Mann shows an excellent way to implement a memoize function in php in His post Here is the example code (actually taken from Edd Mann's post): $memoize = function($func) { return function() use ($func) { static $cache = []; $args = func_get_args(); $key = md5(serialize($args)); if ( !...

Consider the simpler problem of summing the first 100 positive integers: sum [x | x <- [1,2..], x <= 100] This doesn't work either. As a human, you know that once x <= 100 returns False, it will never return True again, because x is getting larger. But Haskell doesn't...

clojure,fibonacci,integer-overflow

The default integer type in Clojure is long. If you want to specify that an integer literal should be considered a clojure.lang.BigInt just add an N right after the number. (defn fib-pair [[a b]] [b (+ a b)]) (nth (map first (iterate fib-pair [1N 1N])) 500) ;= 225591516161936330872512695036072072046011324913758190588638866418474627738686883405015987052796968498626N You can...

There is a typo in line 14. You had just written "Fibonaccie" but must have been Fibonacci.. Btw, just note that you dont have to write elseif condition. You can do it with only one else using or statement. However to make your code more robust you should consider n...

The following code works to take user input from an Entry, send it to a function, and take the output of that function and update the text of the Label: import Tkinter as Tk class App(object): def __init__(self): self.root = Tk.Tk() self.root.wm_title("Fibonacci Calculator") #self.root.wm_iconbitmap("@icon2.xbm") self.label = Tk.Label(self.root, text="Set the digit...

there is always a value for x. So shouldn't select always fall on case 'c <- x' ? No, because this channel is unbuffered, the send will block until someone can receive from it. Read about channels on Effective Go: Receivers always block until there is data to receive....

An implementation could be: from itertools import count def fast_fib_generator(): F = [1, 1] yield 1 yield 1 for k in count(1): F.append(F[k] ** 2 + F[k - 1] ** 2) yield F[-1] F.append(F[k] * (2 * F[k + 1] - F[k])) yield F[-1] for x in fast_fib_generator(): print x...

The difference between a,b=b,a+b and a=b b=a+b is that in the second one, a is assigned value of b, and then b is assigned the sum of a and b, which means it is twice its original value. Consider: a = 1 b = 2 a,b = b,a+b This gives...

SO isn't a debugging service, as I said, learn to use a debugger yourself. There are a few problems in your code: AH not being zero after keyin The user's input must be 2 digit (doesn't work for 1 digit) Where your comment says loop until counter = 0 you...

c++,dynamic-programming,fibonacci

In C++, the two solutions at your disposal that would save you time are the dynamic programming approach and the memoization approach. Dynamic Programming We just build a table from [1..n] and fill it in: int fib(int n) { if (n <= 1) return 1; std::vector<int> table(n + 1); table[0]...

You can use a List to keep track of all numbers in the sequence, and then print whichever number you want. Try this: number = 1 last = 0 before_last = 0 def Fibonacci(number, last, before_last): Question = raw_input('Would you like a List or the Number: ') l=[1] X =...

Just move it all to a function and handle special cases. def fib(n): #TODO: handle invalid case(negative or non-int) and return if(n == 0): print(0) return if(n == 1): print(1) return a, b = 0, 1 i = 2 #this will go up to the n-th number while i<=n: a,b...

I love APL's symbols too, as well as its array programming power. Other array languages may be more powerful, such as J, but they lack the beauty of APL's symbols and explicit syntax. I just tried the example you link to in GNU APL and it works all right: ↑0...

python,generator,fibonacci,yield,def

The message means that generators do not support indexing, so iterable[i] fails. Instead, use the next() function to get the next item from the iterator. def take(n, iterable): x = [] if n > 0 itr = iter(iterable) # Convert the iterable to an iterator for _ in range(n): #...

So, most of your reasoning is correct. In particular, you described correctly how each new element of the list is evaluated in terms of older ones. You are also correct that to fully evaluate fibs would require repeating the steps you outlined and would, in fact, loop forever. The key...

With modulus, long long int a = fib(k + 1); long long int b = fib(k); a >= b is not necessary true, (2 * a >= b neither). so return (b * ((2 * a - b) % MOD)) % MOD; may return negative number....

The code from intel document "int x = cilk_spawn fib(n-1);" asks for permission to run it in another threat, while the next line of "int y = fib(n-2);" will be run on the same threat as the main program. The wiki code asks for a new threat also for computing...

You should check the new element for being to big, not the last found element for being smaller. I won't go into more detail, because you said it's homework and probably don't need more.

java,recursion,stack,fibonacci

how many calls will be stored in stack for the recursive method below and why? With an input of 50, the max amount of stack frames at any given moment, including the initial call to fib, will be 50. (So there will be at most 49 recursive calls stored on...

java,dynamic-programming,fibonacci

Basically int[] fib is a cache in which the ith fibonacci number is stored. This is a great time saver. Otherwise the recursive fibonacci procedure would need to recalculate a lot of values. E.g. fib[8] = fibonacci(7) + fibonacci(6) But then: fib[7] = fibonacci(6) + fibonacci(5) As you can see,...

recursion,fortran,fibonacci,subroutine

Subroutines are not that nice for this particular problem. Nonrecursive solution would be more readable. You tried to use subroutine as a function. You cannot do that, they are very different. You must use them only in the call statement and only one at a time. If you want some...

Your problem is that k and l are local variables. As such they are lost every time the function exits, and re-start at zero and one respectively when is called again (yes, even when it's called from itself). Nick's code aims to store a single instance each of k and...

Time complexity of your algorithm is close to 2^(n) . Hence it is O( 2^n ) Here is an example of how your algorithm will solve if n = 6 You algorithm does T(n) = T(n-1) + T(n-2) + O(1) . Assume T(n-1) = O(2n-1), therefore T(n) = T(n-1) +...

Your code would throw ArrayIndexOutOfBoundsEception in some cases. Let's consider what you are doing : You are getting an input x from user Then you create array a with x elements, input by user Then you find the max element of a You create an array fib of max elements,...

python,loops,for-loop,fibonacci

for i in terms[2:n+1]: should be: for i in range(2, n+1): You're adding to terms, you don't want to iterate over its current contents....

c++,variables,global-variables,fibonacci

When using local variables, you have 'several' local variables a,b,c,d,k (one independent by scope). When using global variables, you share the variable with the different call to fib: so a = fib(k+1); b = fib(k); The second call will modify previous k, a (if k > 2) and so have...

As the OP is a very new Java Programmer, I thought it might be be helpful to give a small tutorial, as one might in beginners class. The other's that have responded have been correct, but everyone has to start somewhere. OK. The section you don't understand has several integer...

The problem was apparently with the casting mechanism at line remainder_array[index_counter] = (int) (Sequence[index_counter]%n); The cast rounded the integer down from a double, so the remainders turned out to be different in the two arrays. I couldn't use long since it doesn't have as much memory capacity. So instead I...

I am not sure exactly what you are asking, maybe you can clarify The time complexity of this algorithm is O(n). The loop will execute n times until i is greater than n. i starts at zero and increments by 1 every iteration of this loop until n. I hope...

If you just want to reverse the list at the end, use reverse/2: fib(N, F) :- fib(N, 0, [1], FRev), reverse(FRev, F). If you want to built the list in the reversed order, you would have to rethink the clauses. One solution is to use last/2 and append/3 to put...

The evaluation took longer than you expected because your function does not use memoization. See e.g. this question or that question for answers to how to define the fibonacci function in Haskell using memoization.

When n>20, you can first reach 20th floor, then go all the way up => stairs(20) you can also reach 19th floor, then go to 21st floor, from 21st floor, you have stairs(n-21) ways to floor n, so => stairs(19)*stairs(n-21) So sum it up is stairs(20) + stairs(19) * stairs(n-21)....

The problem is that a simple test of the values returned by your hochn() function shows that it yields the wrong answers. Your version is hochn0() in the code below; a fixed version is hochn1(). I've included test code for the layout too. Note that your printing of the heading...

java,c#,scala,haskell,fibonacci

The Scala API docs for Stream contains an example on how to do this in Scala: val fibs: Stream[BigInt] = BigInt(0) #:: BigInt(1) #:: fibs.zip(fibs.tail).map { n => n._1 + n._2 } Edit: To implement memoization in a language which doesn't have it built-in like Haskell, you would obviously need...