Recursively reversing large vector efficiently

I am trying to write a code that will recursively compute reverse of given large vector . I have tried 2 codes one simple recursive code and other one iterative reversal code which flips 2 elements in one loop hence efficient

I want to replicate second code below in recursive way how can I approach that? Thank you for your time

% First code with simple recursion that flips end element one by one
function w=reversal_v2(v)
if length(v)==1
    w= v;
else
    w= [v(end),reversal_v2(v(1:end-1))];
end

% Second code with simple recursion that flips end element and first
% element together hence half the required time
b=1:1e4;                         %I/P vector
x=length(b);                     % To get length of vector
a=zeros(x);
for i=1:floor(length(b)/2)        %Loop uses simple for loop and temp to substitute the values
    tmp=b(i);
    a(i)=b(x+1-i);
    a(x+1-i)=tmp;
end

Answer 1

[per isakson]

Five ways to flip a row vector. The last one, reversal_v3(), answers your question. Recursion is by two order of magnitude slower than the for-loop.

v = 1:1e4;
tic,  v(end:-1:1); toc

Elapsed time is 0.000255 seconds.

tic, flip(v); toc

Elapsed time is 0.000254 seconds.

tic, reversal_v2(v); toc

Elapsed time is 0.274852 seconds.

tic, SecondCode(v); toc

Elapsed time is 0.001037 seconds.

tic, reversal_v3(v); toc

Elapsed time is 0.097940 seconds.

function w = reversal_v2(v)
    % First code with simple recursion that flips end element one by one
    if length(v)==1
        w = v;
    else
        w = [v(end),reversal_v2(v(1:end-1))];
    end
end
function a = SecondCode( b )
    % Second code with simple recursion that flips end element and first
    % element together hence half the required time
    % b=1:1e4;          % I/P vector
    x = length(b);      % To get length of vector
    a = zeros(1,x);     % Create row vector            
    % Loop uses simple for loop and temp to substitute the values
    for ii = 1:floor(length(b)/2)   
        tmp=b(ii);
        a(ii)=b(x+1-ii);
        a(x+1-ii)=tmp;
    end
end
function w = reversal_v3(v)
    % ... replicate second code below in recursive way ...
    if length(v) <= 2
        w = flip( v );
    else
        w = [ v(end), v(end-1), reversal_v3(v(1:end-2)) ];
    end
end

[Rayyan Ramrajkar]

Thank you for your elaborate codes and I got the logic need to be implemented.

Regarding for loop vs recursion, I understand that simple loop approach is far more uperior than Recursion, but Recusrsion is must have requirement