Let’s define a (P = (Q, \Sigma, \Gamma, \delta, q_0, Z, F)) where:
This PDA accepts the language (L = a^i b^j c^k \mid j = i + k). The key insight was to split the (b)'s into two parts: the first (i) (b)'s match (a)'s, the next (k) (b)'s match (c)'s. This is a classic exercise in stack-based computation, demonstrating how a PDA can count two independent sequences ((a)'s and (c)'s) using a single stack by offsetting them against (b)'s. pda for a-ib-jc-k where j i k
| State | Input | Stack top | New state | Stack action | |-------|-------|-----------|-----------|--------------| | (q_0) | (a) | (Z_0) | (q_0) | push (X) | | (q_0) | (a) | (X) | (q_0) | push (X) | | (q_0) | (\varepsilon) | (Z_0) | (q_3) | pop | # empty string case (optional) | | (q_0) | (c) | (Z_0) | (q_1) | push (X) | | (q_0) | (c) | (X) | (q_1) | push (X) | | (q_0) | (b) | (Z_0) | (q_2) | pop | # no a or c, but j=0, i=k=0 handled later | Let’s define a (P = (Q, \Sigma, \Gamma,
Better: Use nondeterminism to decide when to stop popping A’s and start pushing B’s for extra b’s. | State | Input | Stack top |
Rewrite the language: (a^i b^j c^k, j = i + k) → (a^i b^i+k c^k). This is essentially seeing the string as (a^i b^i) followed by (b^k c^k).
If the stack is empty exactly when the input string ends, the condition is satisfied. Formal Transition Definition
To make sure I’ve captured exactly what you need for this logic, let me know: Do you need a formal state transition table to accompany the story? Should I explain the Pumping Lemma