Sodium Polysulfide production from NaOH and Sulfur
If you calculate the oxidation numbers for everything before and after, you will see that the oxidation number of the oxygen atoms will always be $-2$. However, sulfur $\ce{S},\ (0)$ is being both reduced to $\ce{S4^2-},\ (\frac{1}{2})$ and oxidized to $\ce{SO3},\ (+6)$. Sulfur is both the oxidant and reductant in what might be called a disproportionation. Sodium hydroxide is thus acting as both a source of oxygen atoms in the formation of $\ce{SO3}$ and probably as the electrolyte if this is occurring in aqueous solution.


  Oxidation numbers:


Element    Reactant     Product
S           0            -1/2 in polysulfide
S           0            +6 in sulfur trioxide
Na         +1            +1
O          -2            -2 in sulfur trioxide and water
H          +1            +1


The half reactions (unbalanced) are:


Oxidation: $\ce{S -&gt; SO3}$
Reduction: $\ce{S -&gt; S4^2-}$


You might get some sodium hydrogen sulfate forming through the followin;
$$\ce{NaOH + SO3 &lt;=&gt; NaHSO4}$$