Given the algorithm:

$B(n)$

$\{$if $n1$

st$1$;

return $B(n-2)$;

else return $c$; $\}$

The recurrence representing its running time function, for $n1,$ is

$T(n)=T(n-2)+c,$ where $c$ is a constant

$T(n)=T(\frac{n}{2})+c,$ where $c$ is a constant

$T(n)=T(\frac{n}{2})+c,$ where $c$ is a constant

$T(n)=T(n-1)+c,$ where $c$ is a constant