How to generate the LR(0) finite state machine for a grammar
There is no loop, an item like $S\to \bullet{+}SS$ in your example might occur in several item sets of the LR automaton.  Let us continue the construction of state 2:

[State 2]
S -&gt; * . S S
S -&gt; . + S S
S -&gt; . * S S
S -&gt; . s


It has transitions to itself on "$*$", to state 1 on "$+$", to some state 3 with item set $\{S\to s\bullet\}$ on "$s$", and to yet another state on "$S$":

[State 4]                 [State 5]
S -&gt; * S . S   --(S)--&gt;   S -&gt; * S S .
S -&gt; . + S S
S -&gt; . * S S
S -&gt; . s


State 4 has again a transition to state 1 on "$+$", state 2 on "$*$", and state 3 on "$s$".  The construction is not complete yet, since you are missing some outgoing transitions from states 1 and 0, but hopefully you can finish it yourself.