{
  "translations": {
    "en": {
      "union": {
        "type": "select",
        "description": "It is easy to construct a deterministic PDA for $\\{a^nb^n: n \\ge 1\\}$, and also to construct a deterministic PDA for $\\{a^nb^{2n}: n \\ge 1\\}$.",
        "question": "Can we connect these two PDAs together into one PDA to accept the language that is their union?",
        "answer": "Yes",
        "choices": ["Yes", "No"],
        "correctFeedback": ["Similar to unioning two NFAs, we can union the two PDAs by creating a new start state and a new final state, and connect them to the existing PDAs using $\\lambda$ transitions. Then a string is accepted if and only if at least one PDA accepts."]
      },
      "DPDA": {
        "type": "select",
        "description": "Now we will show that $L$ is not a DCFL. Assume that there is a deterministic PDA $M$ such that $L=L(M)$. We will construct a PDA that recognizes a language that is not a CFL and thus derive a contradiction.",
        "question": "What are the conditions for a PDA to be DPDA?",
        "answer": ["$\\delta(q, a, b)$ contains at most one element.", "If $\\delta(q, \\lambda, b)$ is not empty, then $\\delta(q, c, b)$ must be empty for every $c \\in \\Sigma$."],
        "choices": ["$\\delta(q, a, b)$ contains at most one element.", "If $\\delta(q, \\lambda, b)$ is not empty, then $\\delta(q, c, b)$ must be empty for every $c \\in \\Sigma$."]
      },
      "NFADFA": {
        "type": "select",
        "description": "We found that a language like $L = \\{a^nb^n: n \\ge 1\\} \\cup \\{a^nb^{2n}: n \\ge 1\\}$ is a CFL and not aDCFL.",
        "question": "For NFAs and DFAs, which is more powerful?",
        "answer": "Both have the same power",
        "choices": ["Both have the same power", "DFA", "NFA"],
        "correctFeedback": ["Any DFA is an NFA, and any NFA can be converted to a DFA."]
      },
      "CFLDCFL": {
        "type": "select",
        "description": "We found that a language like $L = \\{a^nb^n: n \\ge 1\\} \\cup \\{a^nb^{2n}: n \\ge 1\\}$ is a CFL and not aDCFL.",
        "question": "What about CFL vs. DCFL? Do they have the same power?",
        "answer": "No",
        "choices": ["No", "Yes"]
      },
      "which": {
        "type": "select",
        "description": "We found that a language like $L = \\{a^nb^n: n \\ge 1\\} \\cup \\{a^nb^{2n}: n \\ge 1\\}$ is a CFL and not aDCFL.",
        "question": "Which is more powerful?",
        "answer": "CFL, as it can accept more languages than DCFL",
        "choices": ["CFL, as it can accept more languages than DCFL", "DCFL, as it can accept more languages than CFL"]
      }
    }
  }
}