Answer a Loaded Question (with Lojban!) [closed]

Question

A loaded question is something like 'Have you stopped swearing?' in which both yes and no answers mean that I have swore, meaning it is impossible to answer the question if I haven't swore.

However, the constructed language Lojban has a word na'i, which means there is something wrong with what you just said.

Your input should be a list of newline-seperated strings containing the following tokens:

1. start followed by any integer (e.g. -1, 0, 1)
2. stop + integer
3. haveYouStopped + integer

You can assume that each stop n is followed by the corresponding start n.

Then for each haveYouStopped n in the code:

• Print Yes, I have. + a newline if the last stop n occurs before the last start n
• Print No, I have not. + a newline if the last stop n occures after the last start n, or at least one start n but no stop n is found in the previous line
• Print .i na'i + a newline if no start n occurs in the previous lines.

The last newline is optional.

Examples:

start 1
stop 1
haveYouStopped 1

Output: Yes, I have.

start 1
haveYouStopped 1

Output: No, I haven't.

haveYouStopped 1

Output: .i na'i

All code is counted in UTF-8 bytes. Codepoints outside the UTF-8 range are not allowed. Use any language, shortest code wins. Also, no external sources are allowed.

Answer 1

Scala 3, 255 bytes

255 bytes. It can be golfed much more.

---

Golfed version. Attempt This Online!

def p(t:String,d:Map[String,String]=Map[String,String]()):Unit={t.split("\n").filterNot(_.isEmpty).foreach{l=>val Array(c,v)=l.split(" ");if(c.contains("s"))d(v)=if(c.contains("p"))"Yes, I have."else"No, I haven't."else println(d.getOrElse(v,".i na'i"))}}

Ungolfed version. Attempt This Online!

object Main {
  import scala.io.Source
  import scala.collection.mutable

  def processText(text: String, dict: mutable.Map[String, String] = mutable.Map[String, String]()): Unit = {
    val lines = text.split("\n").filterNot(_.isEmpty)
    for (line <- lines) {
      val Array(command, value) = line.split(" ")
      if (command.contains("s")) {
        dict(value) = if (command.contains("p")) "Yes, I have." else "No, I haven't."
      }
      else {
        println(dict.getOrElse(value, ".i na'i"))
      }
    }
  }

  def main(args: Array[String]): Unit = {
    val text =
      """
        |start 1
        |haveYouStopped 1
      """.stripMargin
    processText(text)
  }
}

Answer 2

05AB1E, (68^† bytes) 107 bytes^UTF-8

^{† Since the challenge states the answers should be counted in UTF-8 bytes for some reason.. -_-}

|ηʒθ'hÅ?}ε€#R.¡θ}н€н€θDgi".i na'i"ëJ…dptQi“Yes, I€¡.“ë“No, I€¡€–.“]»

Try it online.

Explanation:

|                           # Get all inputs as a list of strings
 η                          # Pop and push its prefixes
  ʒ     }                   # Filter this list of prefixes, keeping those where:
   θ                        #  The last string of the prefix
      Å?                    #  ends with
    'h                     '#  an "h"
ε                           # Map over each prefix ending with "haveYouStopped"
 €#                         #  Split each inner string by spaces
   R                        #  Reverse this list of pairs
    .¡ }                    #  Then group it by:
      θ                     #   The last item of the pair (the numbers)
        н                   #  Keep the first group (of the reversed list)
 €н                         #  Keep the strings of each inner pair
   €θ                       #  Keep the last letter of each string
     Dg                     #  Duplicate, pop and push the amount of characters in this list
       i                    #  If this length is 1 (thus only "haveYouStopped"):
        ".i na'i"           #   Push string ".i na'i"
       ë                    #  Else:
        J                   #   Join the list of 2 or 3 characters to a string
         …dptQi             #   If it's equal to "dpt" (thus "haveYouStopped,stop,start"):
               “Yes, I€¡.“  #    Push dictionary string "Yes, I have."
              ë             #   Else:
               “No, I€¡€–.“ #    Push dictionary string "No, I have not."
]                           # Close both if-else statements and map
 »                          # Join the list of strings by newlines
                            # (after which it is output implicitly with additional trailing newline)

See this 05AB1E tip of mine (section How to use the dictionary?) to understand why “Yes, I€¡.“ is "Yes, I have." and “No, I€¡€–.“ is "No, I have not.".

Answer 3

Retina, 92 75 74 bytes

L$ms`d( \S+)$(?<=([pt])\1$.+?)?
$2.i na'i
p.+
YesI.
t.+
NoIn't.
I
, I have

Try it online! Link includes test cases. Explanation:

L$ms`d( \S+)$(?<=([pt])\1$.+?)?
$2.i na'i

For each haveYouStopped line, look for a previous start or stop line with that signed integer, and if so prefix the output with the final t or p, otherwise just output .i na'i.

p.*
YesI.
t.*
NoIn't.
I
, I have

If a start or stop was found then decode it to the desired output string.

Edit: Saved 17 bytes and corrected the Yes/No order thanks to @tsh.

Since this question counts in UTF-8, I didn't post a Charcoal answer, but had it been allowed it would have been 59 SBCS bytes:

≔⦃⦄θＷＳ¿№ιs§≔θΣι⪫⪪⎇№ιp¦Yes,.¦No,n't.¦,¦, I have⟦∨§θΣι.i na'i

Attempt This Online! Link is to verbose version of code. Takes input as a list of newline-terminated strings. Explanation:

≔⦃⦄θ

Start with all questions being impossible.

ＷＳ

Loop through the input strings.

¿№ιs

Does this string resolve a question?

§≔θΣι⪫⪪⎇№ιp¦Yes,.¦No,n't.¦,¦, I have

If so then save the appropriate answer.

⟦∨§θΣι.i na'i

Otherwise, output the saved answer, or the impossible answer if one wasn't saved.

Answer 4

Python, 160 158 bytes

def d(t,i={}):
 for l in t.strip().split("\n"):
  c,v=l.split()
  if"s"in c:i[v]="Yes, I have."if"p"in c else"No, I haven't."
  else:print(i.get(v,".i na'i"))

-2 thanks to UndoneStudios

(Assumes that you are referring to taking input as a newline-delimited string)

Ungolfed version

def decode(text):
    data = {}
    for line in text.strip().split("\n"):
        command, value = line.split()
        if "s" in command:
            data[value] = "Yes, I have." if "p" in command else "No, I haven't."
        else:
            print(data.get(v, ".i na'i"))