What working memory does this recogniser carry between input symbols?
- The full input history in a separate growing list
- Its current state, which can summarise earlier input
- An unlimited count held separately from the current state
Which list defines the deterministic string recogniser used here?
- States, alphabet, transitions, start state and accepting states
- States, alphabet, transitions, stored input history and an unlimited stack
- States, alphabet, stored input history, accepting states and a clock speed
A state diagram and a transition table can describe the same recogniser.
Circle one: True False
A binary recogniser starts at A; only C accepts. On 0, every state goes to B. On 1: A goes to A, B to C, C to A. Does it accept 010?
- No; its final state is B
- Yes; visiting C once is enough
- Yes; containing 01 is enough
A binary recogniser starts at A; only C accepts. On 0, every state goes to B. On 1: A goes to A, B to C, C to A. Run 01101. Where do you finish?
- At B, a non-accepting state
- At A, a non-accepting state
- At C, the accepting state
A binary recogniser uses B exactly for strings ending in 0 and C exactly for strings ending in 01. Reading a symbol appends it. From C, reading 0 sends it to B.
Circle one: True False
Start at A. On 0, every state goes to B. On 1: A goes to A, B to C, C to A. Both 0 and 110 finish in B. What shared fact does B record?
- The complete strings have equal lengths and identical symbols
- The current string ends in 0
- The complete strings contain equal numbers of zeros and ones
Why can no finite-state recogniser accept exactly the balanced-parentheses strings at every nesting depth?
- Every closing parenthesis must be rejected, even when matching openings came before it
- Every nesting depth needs a different input alphabet, so one alphabet cannot cover all depths
- Two opening depths share a state, so an identical closing suffix cannot distinguish them