4 of 7 added 18 characters in body; added 1 character in body

Given a 128-trit state space, is a 3-symbol block cellular automaton secure against preimage attacks?

The following implementation, here for illustration:

void hash(int* bits, int bitsLen, int* out, int outLen){
  const int tt[9] = {1,2,0,0,0,1,1,2,2};
  int st[128], i, k, j, x, y, a, b, o = 0;
  for (i=0; i<128; ++i)
    st[i] = i%3;
  for (k=0; k < bitsLen + outLen; ++k){
    for (j=0; j<32; ++j){
      if (k < bitsLen)
        st[0] = bits[k];
      for (i=j%2; i<128; i+=2){
        x = i%128;
        y = (i+1)%128;
        a = st[x];
        b = st[y];
        st[x] = tt[a*3+b];
        st[y] = tt[b*3+a];
    if (k >= bitsLen)
      out[o++] = (st[0] + st[1]*3 + st[2]*9 + st[3]*27) % 2;

Uses a 3-symbol block cellular automata, with transition rules encoded as the 3x3 table defined by tt and using a 128-trit (about 202 bits) state space, to materialize an one-way function. It employs k interactions, where each interaction advances the automata by 32 steps, inserting one of the input bits on the first cell of the automata at each step. Under those conditions, is this function secure against preimage attacks? Here is a test run.