### Remarks on Four-Dimensional Probabilistic Finite Automata

Makoto Sakamoto, Yasuo Uchida, Makoto Nagatomo, Tuo Zhang, Hikaru Susaki, Takao Ito, Tsunehiro Yoshinaga, Satoshi Ikeda, Masahiro Yokomichi, Hiroshi Furutani
2014 2014 7th International Conference on Signal Processing, Image Processing and Pattern Recognition
This paper deals with the study of four-dimensional automata. Recently, due to the advances in many application areas such as dynamic image processing, computer animation, augmented reality (AR), and so on, it is useful for analyzing computation of four-dimensional information processing (three-dimensional pattern processing with time axis) to explicate the properties of four-dimensional automata. From this point of view, we have investigated many properties of four-dimensional automata and
more » ... utational complexity. On the other hand, the class of sets accepted by probabilistic machines have been studied extensively. As far as we know, however, there is no results concerned with four-dimensional probabilistic machines. In this paper, we introduce four-dimensional probabilistic finite automata, and investigate some accepting powers of them. This section shows that the 4-PFA is incomparable with 4-AFA. We first give several preliminaries to get our desired results. Let M be a 4-PFA and Σ be the input Copyright ⓒ 2015 SERSC 79 alphabet of M. For each k, m, n (k≥ 1, m≥ k+1, 1 ≤n ≤m-1), an (m, n, k)-chunk over Σ is a four-dimensional object obtained from a four-dimensional tape in Σ (m, m, m, m) by cutting off the [(m-k+1, 1,1,1), (m, n, 1, 1)]-segment. We show the image of the first three-rectangular array of an (m, n, k)-chunk as shown in Figure 4 . Figure 4. The First Three-Dimensional Rectangular Array of an (m, n, k)chunk Especially, an (m, n, 1)-chunk over Σ is called an (m, n)-chunk over Σ. For an (m, n, Copyright ⓒ 2015 SERSC 81 states, so r ij =1 for all t  T r . For t  T r , let p* [t, R] denote the probability that Markov chain R is trapped in state t when started in state 1. We are now ready to prove our key lemma. Lemma 3.