A Comment on Budach's Mouse-in-an-Octant Problem

In the original formulation of the problem, the Budach mouse is a kind of finite automaton with a very simple program, a cycle of states, q0 d1 −→ q1 d2 −→ . . . qn−1 dn −→ q0, with d1, . . . , dn ∈ {N,E}. The mouse moves among integer points in the octant of the Cartesian plane delimited by the lines y = 0 and y = x. It begins on (1, 0), at state q0, and advances in steps which increase the state number (cyclically) and move the mouse, one unit at a time, in the direction indicated in the program (North or East). If the mouse hits the line y = x, its position is reset from (x, x) to (x, 0), without affecting the control state. Operation proceeds as usual. The computational problem is to decide, given the mouse’s program and a distinguished state number k, whether the mouse will ever visit the line y = x while at state qk. We may refer to this event as a stopping condition, giving the problem the flavour of a halting problem. Whether this problem is decidable is not known. In this note, we increase the power of the mouse by allowing several cycles, with states denoted qc,0, . . . , qc,nc−1 where c is the index of the cycle. A function δ maps states to cycles, and its use is that if the mouse hits the line y = x in state qc,i, not only is it returned to the x axis, but also its state is set to the initial state of the cycle δ(qc,i). We shall prove that mortality of this mouse is undecidable, by reducing from the problem: does a given 2CM halt, when started with null counters?