I am trying to create a recursive function call method that would print the Fibonacci until a specific location:
1 function f = fibonacci(n)
2 fprintf('The value is %d\n', n)
3 if (n==1)
4 f(1) = 1;
5 return;
6 elseif (n == 2)
7 f(2) = 2;
8 else
9 f(n) = fibonacci(n-1) + fibonacci(n-2);
10 end
11 end
As per my understanding the fibonacci function would be called recursively until value of argument n passed to it is 1. Then the function stack would rollback accordingly. So when I call this function from command:
>> fibonacci(4)
The value of n is 4, so line 9 would execute like:
9 f(4) = fibonacci(3) + fibonacci(2);
Now I believe that that first fibonacci(3) would be called - hence again for fibonacci(3)
9 if(3) = fibonacci(2) + fibonacci(1);
The ifs in line number 3 and 6 would take care.
But now how fibonacci(2) + fibonacci(1) statement would change to:
if(3) = 2 + 1;
I am receiving the below error and unable to debug further to resolve it:
>> fibonacci(4)
The value is 4
The value is 3
The value is 2
The value is 1
In an assignment A(I) = B, the number of elements in B and I must be the same.
Error in fibonacci (line 9)
f(n) = fibonacci(n-1) + fibonacci(n-2);
Error in fibonacci (line 9)
f(n) = fibonacci(n-1) + fibonacci(n-2);
Please provide some insight for the solution and with which parameter would fibonacci function be recursively called at line number 9 first and consequently.
Ex For n = 4
f(n) = fibonacci(3) + fibonacci(2);
So will MATLAB call fibonacci(3) or fibonacci(2) first?
Shouldn't the code be some thing like below:
1 function f = fibonacci(n)
2 fprintf('The valus is %d\n', n)
3 if (n==1)
4 f(1) = 1;
5 return f(1);
6 elseif (n == 2)
7 f(2) = 2;
8 return f(2);
9 else
10 f(n) = fibonacci(n-1) + fibonacci(n-2);
11 end
12 end
fibonacci(4)
Error: File: fibonacci.m Line: 5 Column: 12
Unexpected MATLAB expression.
Why return expression in a function is resulting in an error?
Try this:
function f = fibonacci(n)
if (n==1)
f= 1;
elseif (n == 2)
f = 2;
else
f = fibonacci(n-1) + fibonacci(n-2);
end
Note that this is also a recursion (that only evaluates each n once):
function f=fibonacci(n)
f=additive(n,1,2);
function a=additive(n,x0,x1)
if(n==1)
a=x0;
else
if(n==2)
a=x1;
else
a=additive(n-1,x1,x0+x1);
end
end
If you HAVE to use recursive approach, try this -
function out = fibonacci(n)
fprintf('The valus is %d\n', n)
if (n==1)
out = 1;
return;
elseif (n == 2)
out = 2;
return;
else
out = fibonacci(n-1) + fibonacci(n-2);
end
return;
Unlike C/C++, in MATLAB with 'return', one can't return a value, but only the control goes back to the calling function. The output to be returned to the calling function is to be stored in the output variable that is defined at the start of the function.
EDIT 1: For the entire fibonacci series and which assumes that the series starts from 1, use this -
N = 16; %// Number of fibonacci numbers needed
all_nums = zeros(1,N);
all_nums(1) = 1;
for k = 2:N
all_nums(k) = fibonacci(k-1);
end
Gives -
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987
Create a M-file for fibonacci function and write code as given below
function [ result ] = fibonacci( n )
if n==0|n==1
result = n;
else
result = fibonacci(n-2)+fibonacci(n-1);
end
end
Write following code in command window of matlab
for n = 0:10
fprintf('Fibonacci(%d)= %d\n', n, fibonacci(n));
end
Output :-
Fibonacci(0)= 0
Fibonacci(1)= 1
Fibonacci(2)= 1
Fibonacci(3)= 2
Fibonacci(4)= 3
Fibonacci(5)= 5
Fibonacci(6)= 8
Fibonacci(7)= 13
Fibonacci(8)= 21
Fibonacci(9)= 34
Fibonacci(10)= 55
Create a function, which returns Integer:
func fibonacci(number n : Int) -> Int
{
guard n > 1 else {return n}
return fibonacci(number: n-1) + fibonacci(number: n-2)
}
This will return the fibonacci output of n numbers, To print the series You can use this function like this in swift:
for _ in 0...10
{
print(fibonacci(number : 10))
}
It will print the series of 10 numbers.
Related
I just wanted to simply find the index of (row, col) that is a minimum point of a matrix A. I can use
[minval, imin] = min( A(:) )
and MATLAB built in function
[irow, icol] = ind2sub(imin);
But for efficiency reason, where matrix A is trigonal, i wanted to implement the following function
function [i1, i2] = myind2ind(ii, N);
k = 1;
for i = 1:N
for j = i+1:N
I(k, 1) = i; I(k, 2) = j;
k = k + 1;
end
end
i1 = I(ii, 1);
i2 = I(ii, 2);
this function returns 8 and 31 for the following input
[irow, icol] = myind2ind(212, 31); % irow=8, icol = 31
How can I implement myind2ind function more efficient way without using the internal "I"?
The I matrix can be generated by nchoosek.
For example if N = 5 we have:
N =5
I= nchoosek(1:N,2)
ans =
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5
so that
4 repeated 1 times
3 repeated 2 times
2 repeated 3 times
1 repeated 4 times
We can get the number of rows of I with the Gauss formula for triangular number
(N-1) * (N-1+1) /2 =
N * (N -1) / 2 =
10
Given jj = size(I,1) + 1 - ii as a row index I that begins from the end of I and using N * (N -1) / 2 we can formulate a quadratic equation:
N * (N -1) / 2 = jj
(N^2 -N)/2 =jj
So
N^2 -N - 2*jj = 0
Its root is:
r = (1+sqrt(8*jj))/2
r can be rounded and subtracted from N to get the first element (row number of triangular matrix) of the desired output.
R = N + 1 -floor(r);
For the column number we find the index of the first element idx_first of the current row R:
idx_first=(floor(r+1) .* floor(r)) /2;
The column number can be found by subtracting current linear index from the linear index of the first element of the current row and adding R to it.
Here is the implemented function:
function [R , C] = myind2ind(ii, N)
jj = N * (N - 1) / 2 + 1 - ii;
r = (1 + sqrt(8 * jj)) / 2;
R = N -floor(r);
idx_first = (floor(r + 1) .* floor(r)) / 2;
C = idx_first-jj + R + 1;
end
Need a little help understanding what is happening in this function particularly line 7 [Fnm1,Fnm2] = fibrecurmemo(N-1); I don't understand how a new variable can be declared here with in the array. an example of what is happening would be appreciated.
function [Fn,Fnm1] = fibrecurmemo(N)
% Computes the Fibonacci number, F(N), using a memoized recursion
if N <= 2
Fn = 1;
Fnm1 = 1;
else
[Fnm1,Fnm2] = fibrecurmemo(N-1);
Fn = Fnm1 + Fnm2;
end
end
Say we start with:
fibrecurmemo(3) %// N is 3
The else statements run (since N > 2):
[Fnm1,Fnm2] = fibrecurmemo(2); %//statement 1
Fn = Fnm1 + Fnm2; %//statement 2
Before statement 2 can run, fibrecurmemo(2) must first run.
The if statements in fibrecurmemo(2) run (since N <= 2):
Fn = 1;
Fnm1 = 1;
As a result, fibrecurmemo(2) returns 1, 1.
Contininuing from statement 1 above,
[1,1] = fibrecurmemo(2); %//statement 1
Fn = 1 + 1; %//statement 2
Finally,
[2, 1] = fibrecurmemo(3);
The function returns two values.
function [xFive,yFive] = addFive(x,y)
xFive = x + 5;
yFive = y + 5;
end
xx = (addFive(3,4))
xx will be equal to 8 in this example
the syntax for assignment for multiple return values is
[a,b,c,...] = someFunc();
where someFunc() has output of [a,b,c,...]
[aa,bb] = addFive(3,4);
cc = addFive(3,4);
if you do it this way you would get
aa == 8
bb == 9
cc == 8
in the case of cc instead of [aa,bb] Then you will just get the first return value.
i.e. you could do
x = fibrecurmemo(5)
[y,z] = fibrecurmemo(5)
in this case x == y
I have a function that I was wondering if it was possible to vectorize it and not have to use a for loop. The code is below.
a=1:2:8
for jj=1:length(a)
b(jj)=rtfib(a(jj)); %fibbonacci function
end
b
output below:
a =
1 3 5 7
>>>b =
1 3 8 21
I was trying to do it this way
t = 0:.01:10;
y = sin(t);
but doing the code below doesn't work any suggestions?
ps: I'm trying to keep the function rtfib because of it's speed and I need to use very large Fibonacci numbers. I'm using octave 3.8.1
a=1:2:8
b=rtfib(a)
Here's the rtfib code below as requested
function f = rtfib(n)
if (n == 0)
f = 0;
elseif (n==1)
f=1;
elseif (n == 2)
f = 2;
else
fOld = 2;
fOlder = 1;
for i = 3 : n
f = fOld + fOlder;
fOlder = fOld;
fOld = f;
end
end
end
You can see that your function rtfib actually computes every Fibonacci number up to n.
You can modify it so that is stores and returns all these number, so that you only have to call the function once with the maximum number you need:
function f = rtfib(n)
f=zeros(1,n+1);
if (n >= 0)
f(1) = 0;
end
if (n>=1)
f(2)=1;
end
if (n >= 2)
f(3) = 2;
end
if n>2
fOld = 2;
fOlder = 1;
for i = 3 : n
f(i+1) = fOld + fOlder;
fOlder = fOld;
fOld = f(i+1);
end
end
end
(It will return fibonnaci(n) in f(n+1), if you don't need the 0 you could change it so that it returns fibonnaci(n) in f(n) if you prefer)
Then you only need to call
>>f=rtfib(max(a));
>>b=f(a+1)
b =
1 3 8 21
If you don't want to store everything you could modify the function rtfib a little more, so that it takes the the array a as input, compute the Fibonacci numbers up to max(a) but only stores the one needed, and it would directly return b.
Both solutions will slow down rtfib itself but it will be a lot faster than calculating the Fibonacci numbers from 0 each time.
I am unable to get converging values using a Gauss-Seidel algorithm
Here is the code:
A = [12 3 -5 2
1 6 3 1
3 7 13 -1
-1 2 -1 7];
b = [2
-3
10
-11];
ep = 1e-8;
[m, n] = size(A);
[n, p] = size(b);
x = zeros(n, 1001);
x(:, 1) = []
for k=0:1000
ka = k + 1;
if ka == 1001
break;
end
xnew = zeros(n,1);
for i=1:n
sum = 0;
j = 1;
while j < i
s1 = s1 + A(i,j) * x(j, ka + 1);
j = j + 1;
end
j = i + 1;
while j <= n
sum = sum + A(i,j) * x(j, ka);
j = j + 1;
end
xnew(i) = (b(i) - sum) / A(i, i);
% if result is within error bounds exit loop
if norm(b - A * xnew, 2) < ep * norm(b, 2)
'ending'
break
end
end
x(:,ka + 1) = xnew;
end
I cannot get the A * xnew to converge on b what am I doing wrong?
I have tried running this changing the syntax several times, but I keep getting values that are way off.
Thanks!
Gabe
You have basically two problems with your code:
(1) You are using two different variables "sum" and "s1". I replaced it by mySum. By the way, dont use "sum", since there is a matlab function with this name.
(2) I think there is also a problem in the update of x;
I solved this problem and I also tried to improve your code:
(1) You dont need to save all "x"s;
(2) It is better to use a "while" than a for when you dont know how many iterations you need.
(3) It is good to use "clear all" and "close all" in general in order to keep your workspace. Sometimes old computations may generate errors. For instance, when you use matrices with different sizes and the same name.
(4) It is better to use dot/comma to separate the lines of the matrices
You still can improve this code:
(1) You can test if A is square and if it satisfies the conditions necessary to use this numerical method: to be positive definite or to be diagonally dominant.
clear all
close all
A = [12 3 -5 2;
1 6 3 1;
3 7 13 -1;
-1 2 -1 7];
b = [2;
-3;
10;
-11];
ep = 1e-8;
n = length(b); % Note this method only works for A(n,n)
xNew=zeros(n,1);
xOld=zeros(n,1);
leave=false;
while(~leave)
xOld=xNew;
for i=1:n
mySum = 0;
j = i + 1;
while j <= n
mySum = mySum + A(i,j) * xOld(j,1);
j = j + 1;
end
j = 1;
while j < i
mySum = mySum + A(i,j) * xNew(j,1);
j = j + 1;
end
mySum=b(i,1)-mySum;
xNew(i,1) = mySum / A(i, i);
end
if (norm(b - A * xNew, 2) < ep * norm(b, 2))
disp('ending');
leave=true;
end
xOld = xNew;
end
xNew
I wrote functions to convert 100,000 hex strings to values, but it takes 10 seconds to perform on the whole array. Does Matlab have a function to do this, so that it is faster, ... ie: less than 1 second for the array?
function x = hexstring2dec(s)
[m n] = size(s);
x = zeros(1, m);
for i = 1 : m
for j = n : -1 : 1
x(i) = x(i) + hexchar2dec(s(i, j)) * 16 ^ (n - j);
end
end
function x = hexchar2dec(c)
if c >= 48 && c <= 57
x = c - 48;
elseif c >= 65 && c <= 70
x = c - 55;
elseif c >= 97 && c <= 102
x = c - 87;
end
Try using hex2dec. It should be faster much faster than looping over each character.
shoelzer's answer is obviously the best.
However, if you want to do the conversion by yourself, then you might find this useful:
Assuming s is a char matrix: all hex numbers are of the same length (zero padded if necessary) and each row has a single number. Then
ds = double( upper(s) ); % convert to double
sel = ds >= double('A'); % select A-F
ds( sel ) = ds( sel ) - double('A') + 10; % convert to 10 - 15
ds(~sel) = ds(~sel) - double('0'); % convert 0-9
% do the sum through vector product
v = 16.^( (size(s,2)-1):-1:0 );
x = s * v(:);