Recursion revisited - can you help me?
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function v = reversal(v)
if length(v) > 1
v = [v(end) reversal(v(1:end-1))];
end
end
5 comentarios
Gerry Dumlao
el 28 de Jun. de 2021
Walter Roberson
el 28 de Jun. de 2021
Your second function is not recursive. Also it converts to double for no apparent reason.
Gerry Dumlao
el 28 de Jun. de 2021
Gerry Dumlao
el 28 de Jun. de 2021
Tests out okay
V = char(randi([33 126], 1, 31))
RV = reversal2(V)
isequal(RV, fliplr(V))
V = char(randi([33 126], 1, 32))
RV = reversal2(V)
isequal(RV, fliplr(V))
function v = reversal2(v)
if length(v) > 1
ii=round(length(v) /2 );
v = [reversal2(v(ii+1:end)) , reversal2(v(1:ii))];
end
end
Respuestas (2)
ghazal
el 2 de Jul. de 2022
I have problem and this is my code, anyone can help me?
function v=reversal(v)
if length(v)==1
ii=round(length(v)/2);
v=[reversal(v(ii+1:end)) , reversal(v(1:ii))];
end
end
3 comentarios
Walter Roberson
el 2 de Jul. de 2022
What should happen if length(v) is not 1? Currently the code returns v unchanged in that case.
If length(v)==1 is true, then you execute the body of the if, and take length(v) which we know must be 1 (otherwise the condition would have been false.) round(1/2) is 1, so ii will always be 1 in that case (no point in computing it.) So you would call reversal(vv(2:end)) which would be reversal([]) which would end up returning the [] unchanged. And you would call reversal(v(1:1)) which would pass the same v into reversal (v must be scalar or the if would not be true), so you would end up calling reversal() with exactly the same input... which is going to call reversal([]) and reversal(v(1)) again which is going to call reversal([]) and reversal(v(1)) again and...
ghazal
el 3 de Jul. de 2022
Thanks friend for your explanation actually I don't get where the problem is, but I changed my code to this and I get this Error!
Error:
Undefined function 'reversal' for input arguments of type 'double'.
Code:
function v = reversal2(v)
if length(v) > 1
ii=round(length(v) /2 );
v = [reversal2(v(ii+1:end)) , reversal2(v(1:ii))];
end
end
Walter Roberson
el 4 de Mzo. de 2024
You would have a problem running function reversal when the function is named reversal2
Jeevan
el 4 de Mzo. de 2024
0 votos
code run without output
1 comentario
Walter Roberson
el 4 de Mzo. de 2024
what code runs without output ?
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