Meaning of $R=0$, $R_{ab}=0$. $R_{abcd}=0$ First let me state some definition
The Einstein tensor is given by
\begin{align}
  G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R
\end{align}
and note that
\begin{align}
G^{\mu}_{\phantom{\mu} \mu} = R^{\mu}_{\phantom{\mu}\mu} - \frac{1}{2} R g^{\mu}_{\phantom{\mu}\mu} = R-2R = -R
\end{align}
where we used $g_{\mu}^{\mu}=4$. With this we can express Ricci tensor in terms of Einstein tensor
\begin{align}
  R_{\mu\nu} = G_{\mu\nu} - \frac{1}{2} G g_{\mu\nu}
\end{align}
From this we can see $G_{\mu\nu}=0$ implies $R_{\mu\nu}=0$ and vice versa. ( for 4d)
What i confused is following 
suppose $R_{ab}=0$ 
Then $ R_{ab} = g^{cd} R_{acbd} =0$ since $g$ has inverse, we can deduce $R_{acbd}=0$ 
Is this results comes from 4d? 
Am i correct? 

cf. 
What i want to know is the meaning of $R=0$, $R_{ab}=0$. $R_{abcd}=0$. 
Firstly, i know $R_{abcd} =0 $ implies $R_{ab}=0$ and $R_{ab}=0$ implies $R=0$. 
Is inverse relation also true?
 A: Each of $R_{abcd} = 0$, $R_{ab} = 0$, and $R=0$ implies different things. $R_{abcd} =0$ is the strongest condition, and if this vanishes, then all contractions of it vanish, and hence:
$$R_{abcd} = 0\quad \Rightarrow \quad R_{ab} = 0, \quad R=0.$$
If $R_{abcd} =0$, the metric is either flat space or some identification of it. 
A weaker condition is $R_{ab} = 0$ (this condition is called Ricci flatness), which can be true even if $R_{abcd} \neq 0$. A good example is Schwarzschild, which has non-trivial curvature, but is a vacuum Einstein solution and so $R_{ab} =0$. Taking a trace yields
$$ R_{ab} = 0 \quad \Rightarrow \quad R = 0. $$
The weakest condition of the three is $R=0$, which can be true for $R_{abcd} \neq 0$, $R_{ab} \neq 0$. A good example of this is Reissner-Nordstrom in 4D: the black hole is not vacuum so $R_{ab} \propto T_{ab} \neq 0$ (there is some non-trivial stress tensor), but the trace of the stress tensor vanishes, so that $R=T=0$.
A: Surgical Commander's answer is perfect. I just want to add that Riemann tensor is a (1,3) rank tensor. When a tensor is zero, then since contraction is a sum of components in a basis, all the contractions that one can built from that tensor are zero! 
