[bushindo]

Suppose that you encounter 4 people, each of whom is either a truth teller (always tells the truth), a liar (always tell lies), a random answerer (always answer randomly with 'Yes' or 'No'), or an alternator. The alternator, upon being asked the first question, will randomly choose to be a truth teller, a liar, or a random answerer and then answer the question with that persona. Every time the alternator is asked a question thereafter, he'll switch his current persona in the following cycle: truthteller -> liar -> random answerer -> truth teller ..., etc. For instance, if the alternator acted as a truth teller for a particular question, he will act as a liar when he is asked the next question, and as a random answerer the question after that.

You know that in this group of 4, there are 1 random answerer, 1 truth teller, 1 liar, and 1 alternator, but you don't know which person is which. The puzzle is to identify the 1 random answerer with the following information:

1) You can only ask yes/no questions to 1 person at a time

2) You can ask the same question of different people, or address different questions to the same person.

3) You can not ask any question which is impossible for a truth teller/liar to answer with a yes/no (ie. asking a truth teller what a random answerer would say to a particular question, asking paradoxes, etc. )

4) Each person in the group of 4 knows what type of people the remaining 3 are.

Determine a strategy that is guaranteed to find the 1 random answerer in 8 questions or less.

Edited November 28, 2010 by bushindo

[araver]

Again, as seen before by asking "oblique" questions.

Assuming in the following that all questions are re-processed as "If I would ask you if P is true would you answer Yes?" so basically we have 2T (truth-tellers), 1 R(Random) and 1 A(Alternator).

Assume all 4 are standing in a line.

First question Q) "Is the random one of the last 2 (third or fourth) in the line?". Ask this question to each of the four ONCE. You spend 4 questions this way.

Depending on the number of YES/NO answers, there are 5 cases:

4YES -> Q is true. Alternator was either telling the truth or playing random (and guessing it, but that's not the point).

Note that because the random is either the third or the fourth (as Q is true), then the first two can be TT or TA.

Note that if asked the same question twice, the alternator must answer at least once as a truth-teller.

So ask the first two, TWICE each one, if the last one (4th) is Random.

These four answers are either 4 truths (if TT or TA and both times the alternator was saying the truth) or 3 truths and 1 unknown (if TA and one time alternator was randomly answering). Either way, at least 3 are truths so you know if last one is Random by majority vote of these answers. If majority is YES then Random is fourth. If majority is NO, then Random is third.

Finished in 8 questions this way.

3YES 1NO -> Q is true and two of the three Yes's are truth tellers.

Ask all 3 that said Yes to the first question if last (4th)is random.

Majority vote decides since there are at least two truth-tellers: pick last if YES, third if NO.

Finished in 7 questions this way.

2YES 2NO -> alternator was acting as random (and siding with random). This is the only piece of solid information in this case.

Pick 2 of the Yes's and 1 of the No's.

Either:

A) you picked 2 truth-tellers + 1 random/alternator in which case you have a majority.

B) you picked random+alternator+truth-teller. But as alternator acted like random at the previous question, he will act as a truth-teller for the next question you ask of him. Again you have guaranteed majority for the next question!

Ask the three of them the same question Q as before "Is the random one of the last 2 (third or fourth) in the line?.

This time you get a majority truth-tellers so you know what random and alternator answered to the first question. Moreover, you know what the truth-tellers answered to the first question (opposite than the random).

So you narrow down random to one of two and you have both the truth-tellers narrowed down.

To get the random from the two that answered falsely to the first question, ask a final (8th) question to any of the truth-tellers: Is X random (pointing at any of the two). Depending on his answer you get the random and the either is the alternator.

1YES 3NO -> Q is false. Two of the three No's are truth tellers.

Ask all 3 of them if first is random. Majority vote decides - pick first if YES, second if NO. Again, 7 questions.

4NO -> Q is false, so Random is one of the first two.

Ask the last 2, twice each one, if the first one is Random. As in the 4YES case, you get truth-answers in majority.

Thank you for the puzzle.

I liked the cycle-3 alternator very much. Made it different from any other problem I've seen so far.

I pondered for a little while if you can narrow down the alternator too at the same time. Probably not, but still pondering.

Edited November 28, 2010 by araver

[Guest]

araver, your research was conducted well and is mostly valid except for one thing that happens to negate the entire legitimacy of your argument....it is impossible (as far as I can tell) to distinguish between a truth teller and a random. Let's use the case of 3YES and 1NO as an example:

"3YES 1NO -> Q is true and two of the three Yes's are truth tellers.

Ask all 3 that said Yes to the first question if last (4th)is random.

Majority vote decides since there are at least two truth-tellers: pick last if YES, third if NO.

Finished in 7 questions this way."

Suppose the following:

Person A: Random

Person B: Liar

Person C: Truth

Person D: Alternator

Using your question sequence, the respondents' answers could look like this:

A: Yes, Yes

B: No, N/A

C: Yes, No

D: Yes, No

Switching the random's answer also leads to an inconclusive result:

A: Yes, No

B: No, N/A

C: Yes, No

D: Yes, No

[bushindo]

Again, as seen before by asking "oblique" questions.

Assuming in the following that all questions are re-processed as "If I would ask you if P is true would you answer Yes?" so basically we have 2T (truth-tellers), 1 R(Random) and 1 A(Alternator).

Assume all 4 are standing in a line.

First question Q) "Is the random one of the last 2 (third or fourth) in the line?". Ask this question to each of the four ONCE. You spend 4 questions this way.

Depending on the number of YES/NO answers, there are 5 cases:

4YES -> Q is true. Alternator was either telling the truth or playing random (and guessing it, but that's not the point).

Note that because the random is either the third or the fourth (as Q is true), then the first two can be TT or TA.

Note that if asked the same question twice, the alternator must answer at least once as a truth-teller.

So ask the first two, TWICE each one, if the last one (4th) is Random.

These four answers are either 4 truths (if TT or TA and both times the alternator was saying the truth) or 3 truths and 1 unknown (if TA and one time alternator was randomly answering). Either way, at least 3 are truths so you know if last one is Random by majority vote of these answers. If majority is YES then Random is fourth. If majority is NO, then Random is third.

Finished in 8 questions this way.

3YES 1NO -> Q is true and two of the three Yes's are truth tellers.

Ask all 3 that said Yes to the first question if last (4th)is random.

Majority vote decides since there are at least two truth-tellers: pick last if YES, third if NO.

Finished in 7 questions this way.

2YES 2NO -> alternator was acting as random (and siding with random). This is the only piece of solid information in this case.

Pick 2 of the Yes's and 1 of the No's.

Either:

A) you picked 2 truth-tellers + 1 random/alternator in which case you have a majority.

B) you picked random+alternator+truth-teller. But as alternator acted like random at the previous question, he will act as a truth-teller for the next question you ask of him. Again you have guaranteed majority for the next question!

Ask the three of them the same question Q as before "Is the random one of the last 2 (third or fourth) in the line?.

This time you get a majority truth-tellers so you know what random and alternator answered to the first question. Moreover, you know what the truth-tellers answered to the first question (opposite than the random).

So you narrow down random to one of two and you have both the truth-tellers narrowed down.

To get the random from the two that answered falsely to the first question, ask a final (8th) question to any of the truth-tellers: Is X random (pointing at any of the two). Depending on his answer you get the random and the either is the alternator.

1YES 3NO -> Q is false. Two of the three No's are truth tellers.

Ask all 3 of them if first is random. Majority vote decides - pick first if YES, second if NO. Again, 7 questions.

4NO -> Q is false, so Random is one of the first two.

Ask the last 2, twice each one, if the first one is Random. As in the 4YES case, you get truth-answers in majority.

Thank you for the puzzle.

I liked the cycle-3 alternator very much. Made it different from any other problem I've seen so far.

I pondered for a little while if you can narrow down the alternator too at the same time. Probably not, but still pondering.

Thanks for working on the puzzle. This answer is entirely satisfactory and I'd consider the OP solved. The note you made at the end made me ponder about it a bit too,

The random answerer in fact can be narrowed down in 7 questions instead of 8. For instance, take the case of 4 YES in the above example. Having narrowed the random answerer to the last two people in line, we can pick a person from the first two in line and ask him the same question twice, "Is the random answerer the 4th person in this line?"

a) If the chosen person is a truth teller, we'll get two identical true answers, which allows us to single out the random answerer in 4 + 2 questions.

b) If the chosen person is a alternator, he will either give us two identical true answers, or one YES and one NO. In the first case, we know where the random answerer is in 4 + 2 = 6 questions; in the latter case, we will know that he is the alternator, and pick the remaining person from the first two (a truth teller) and get the random answerer's position in 1 question (4 + 2 + 1 = 7 total).

I'm not able to show that it is possible to get both the random answerer and the alternator in 7 questions. But it should be straightforward to get the positions of both in 8 questions.

araver, your research was conducted well and is mostly valid except for one thing that happens to negate the entire legitimacy of your argument....it is impossible (as far as I can tell) to distinguish between a truth teller and a random. Let's use the case of 3YES and 1NO as an example:

"3YES 1NO -> Q is true and two of the three Yes's are truth tellers.

Ask all 3 that said Yes to the first question if last (4th)is random.

Majority vote decides since there are at least two truth-tellers: pick last if YES, third if NO.

Finished in 7 questions this way."

Suppose the following:

Person A: Random

Person B: Liar

Person C: Truth

Person D: Alternator

Using your question sequence, the respondents' answers could look like this:

A: Yes, Yes

B: No, N/A

C: Yes, No

D: Yes, No

Switching the random's answer also leads to an inconclusive result:

A: Yes, No

B: No, N/A

C: Yes, No

D: Yes, No

Araver's post notes that we could phrase the question in such a way that liars and truth tellers will give us the same answer. That should remove the difficulty you mentioned above.

Edited November 29, 2010 by bushindo

[araver]

Thanks for working on the puzzle. This answer is entirely satisfactory and I'd consider the OP solved. The note you made at the end made me ponder about it a bit too,

The random answerer in fact can be narrowed down in 7 questions instead of 8. For instance, take the case of 4 YES in the above example. Having narrowed the random answerer to the last two people in line, we can pick a person from the first two in line and ask him the same question twice, "Is the random answerer the 4th person in this line?"

a) If the chosen person is a truth teller, we'll get two identical true answers, which allows us to single out the random answerer in 4 + 2 questions.

b) If the chosen person is a alternator, he will either give us two identical true answers, or one YES and one NO. In the first case, we know where the random answerer is in 4 + 2 = 6 questions; in the latter case, we will know that he is the alternator, and pick the remaining person from the first two (a truth teller) and get the random answerer's position in 1 question (4 + 2 + 1 = 7 total).

I'm not able to show that it is possible to get both the random answerer and the alternator in 7 questions. But it should be straightforward to get the positions of both in 8 questions.

Ah, brilliant!

I admit I was a little lazy and didn't bother to optimize or check in detail. I thought one case was unbreakable in less than 8 and didn't bother to check the others.

Now, after a little more pondering, I'll call your claim and raise you

that both the random and the alternator can be found in at most 7

I'm borrowing a trick from your sleeve (in part3 puzzle) that 4 YES's means we could have stopped before the fourth YES. I'm also borrowing one of my tricks, asking about the next-in-line's alignment to narrow the cases faster.

And I'm possibly finding a new trick on how to detect flipping answers (see case 2YES/2NO below).

Again, as in my previous solution(s), rephrase all questions as oblique-questions to make the liar sound like a truth-teller. Also, replace any "Is X truth-teller?" from the following with "Is X a liar or a truth-teller?".

Round 1

Q: "Is Random one of the last 2 (third or fourth) in the line?".

Ask question Q to first three of them. After 3 of these questions you get:

- either 2YES/1NO or 2NO/1YES in which case it's undecided and proceed with asking the fourth man the same question.

- 3 YES's or 3NO's in which case proceed to Round 2 without asking the fourth question.

Depending on the number of YES/NO answers, there are 5 cases for Round 1:

- 2 YES/2NO - 4 questions spent

- 3 YES/1 NO or 3 NO/1 YES - 4 questions spent

- 3 YES or 3 NO - 3 questions spent!

2YES/2NO - you can get both Random and Alternator in at most 7 questions.

Round 1 - You must spend your first 4 questions till you get 2YES 2NO, since up until the last of the 4, you could still get a majority.

You only know that the alternator was acting as random (and siding with random) so there's no definite way to point a person that's not random.

Good news is that Alternator will tell the truth the next 2 times you ask him a question.

For Round 2:

- If you take 1 of the 4 you could be looking at either case (T, A, R). Very uncertain.

- If you take 2 from same side you get TT or AR. Since Alternator will act as a truth-teller this time, you get in fact either TT or TR. Which is good enough.

So, pick one side, take the two that answered YES to the first question.

Ask them again the same question as in Round 1. You're looking at either 2YES, 2NO, or 1 YES/1 NO.

- If you get 1 YES / 1 NO, then this is the Random+Alternator side since one of them changed his answer from Round 1. The other side is the truth-tellers' side.

So you can finish in 7 by asking a truth-teller which of the two asked in Round 2 is random. The other one is the alternator.

- If you get 2 YES, then the answer to the question is surely YES (either TT or AR - one of them is definitely telling the truth this time). But since the answer is YES and the question is the same, you've selected the two truth-tellers in Round 2, since they can't be the NO guys in Round 1 (the fact that it's the same question is crucial here!). So you can still finish in 7 by asking one of the truth-tellers if one of the other two is random. The other one on his side in Round 1 is the alternator.

- If you get 2 NO's, then both have changed their opinion since Round 1 (since you picked the YES side). So this is the Random+Alternator side. Again finish in 7 by asking one of the truth-tellers (the other side) if one of the guys in Round 2 is Random. The other one in Round 2 is the Alternator, so you get both.

3YES 1NO - Finish in 7 finding out both the alternator and the random:

From Round 1 - Q is true and two of the three YES's are truth tellers and the other is either random (in which case alternator was random in Round 1) or alternator (either randomly guessing or truth-teller).

You know that Random is either 3rd or 4th in line (since Q is true). Avoid him by asking only the first 2 in line in Round 1.

Now, the first two answered either 2YES's or YES/NO in Round 1:

- If the first two answered 1 YES/1 NO in Round 1, then the NO guy is the alternator lying in Round 1. You can finish in 5 by asking the other whether random is 4th.

- If the first two answered 2 YES's in Round 1 - they are either 2 truth-tellers or one truth-teller and alternator (either randomly guessing or truth-teller, cannot know for sure).

Ask these two if the Alternator is one of the last two. At least one of them tells the truth.

---If you get 2 YES's, then both Random and Alternator are 3rd and 4th which makes the first two truth-tellers. Finish in 7 asking a truth-teller if Random is 4th.

---If you get 2 NO's then Alternator is one of the first two (the answer to the question in Round 2 says so)

---If you get 1 YES and 1 NO, then Alternator is one of the first two (since truth-tellers don't disagree).

In both cases in which the Alternator is one of the first two, we get that the other YES guy in Round 1 (who we didn't ask in Round 2) is a truth-teller and the NO guy in Round 1 the Random. Finish in 7 by asking the truth-teller if Alternator is 1st.

3NO 1YES is symmetrical to 3YES 1NO (replace lying with minority vote above, and ask the 3rd and 4th instead of 1st and 2nd to avoid Random)

3YES - Finish in 7 finding out both the alternator and the random:

Rememember Round 1 spent only 3 questions since we stopped before asking the last.

Q is true after round 1 so Random is either 3rd or 4th.

Then the first two can be TT or TA.

Ask the first if the second is a truth-teller.

-If he answers YES then either he's the alternator and the second is the truth-teller or he's a truth-teller and the second is the truth-teller.

--Either way, the 2nd is a truth-teller, so ask him if the first is the alternator:

---If he says YES, you got the alternator. Ask the truth-teller if the 4th is Random and you get Random too in 3+1+1+1=6 questions

---If he says NO, you got the two truth-tellers as 1st and 2nd. Ask any of them if the 4th is Random and you get Random(either 4th or 3rd) and the Alternator is the other guy(either 3rd or 4th) in 3+1+1+1=6 questions.

-If he answers NO then either he's the alternator and the second is the truth-teller or he's a truth-teller and the second is the alternator. You've spent 3+1=4 questions so far.

Note that if you ask an Alternator something twice (with oblique-questions that liars and truth-tellers answer the same), he's bound to act like a truth-teller at least once.

So, ask the first TWICE if the 4th is Random (6 questions so far)

--If he gives you 2YES's or 2NO's, then you know Random is the 4th or the 3rd. Now, the other guy (3rd or 4th) is definitely a truth-teller (since we already narrowed it down that the alternator is among the first two). Ask him if Alternator is first and finish in 7 questions with both Random and Alternator found.

--If he gives you 1 YES and 1 NO, then he's the Alternator. Ask the second (who is surely a truth-teller)if Random is 4th and you finish in 7 questions again.

3NO - I will discuss this case because one may think the symmetry to the 3YES case is less clear since we stopped after 3 questions. But it is symmetrical since we did not use any information about who answered what in the first round, just the answer itself was important.

Rememember Round 1 spent only 3 questions since we stopped before asking the last.

Q is false after Round 1 so Random is either 1st or 2nd.

Then the last two can be TT or TA.

Ask the 4th if the 3rd is a truth-teller.

-If he answers YES then either he's the alternator and the 3rd is the truth-teller or he's a truth-teller and the 3rd is the truth-teller.

--Either way, the third is a truth-teller, so ask him if the fourth is the alternator:

---If he says YES, you got the alternator. Ask the truth-teller if the 1st is Random and you get Random too in 3+1+1+1=6 questions

---If he says NO, you got the two truth-tellers as 3rd and 4th. Ask any of them if the 1st is Random and you get botth Random(either 1st or 2nd) and the Alternator is the other guy(either 2nd or 1st) in 3+1+1+1=6 questions.

-If he answers NO then either he's the alternator and the 3rd is the truth-teller or he's a truth-teller and the 3rd is the alternator. You've spent 3+1=4 questions so far.

Note that if you ask an Alternator something TWICE (with oblique-questions that liars and truth-tellers answer the same), he's bound to act like a truth-teller at least ONCE.

So, ask the 4th TWICE if the 1st is Random (6 questions so far)

--If he gives you 2YES's (or 2NO's), then you know Random is the 1st (or the 2nd). Now, the other guy (2nd or 1st) is definitely a truth-teller (since we already narrowed it down that the alternator is among the last two). Ask him if Alternator is 4th and finish in 7 questions with both Random and Alternator found.

--If he gives you 1 YES and 1 NO, then he's the Alternator. Ask the 3rd (who is surely a truth-teller) if Random is 1st and you finish in 7 questions again.

And, it follows trivially

that after getting random and alternator in 7 questions, one can get the liar from the truth teller, thus completing the arrangement in at most 8 questions. By asking another oblique question one of the identified "truth-tellers" after 7 questions: "If I would ask you if X is a liar would you answer YES?" where X is the other "truth-teller".

Edited November 29, 2010 by araver

[bushindo]

Ah, brilliant!

I admit I was a little lazy and didn't bother to optimize or check in detail. I thought one case was unbreakable in less than 8 and didn't bother to check the others.

Now, after a little more pondering, I'll call your claim and raise you

that both the random and the alternator can be found in at most 7

I'm borrowing a trick from your sleeve (in part3 puzzle) that 4 YES's means we could have stopped before the fourth YES. I'm also borrowing one of my tricks, asking about the next-in-line's alignment to narrow the cases faster.

And I'm possibly finding a new trick on how to detect flipping answers (see case 2YES/2NO below).

Again, as in my previous solution(s), rephrase all questions as oblique-questions to make the liar sound like a truth-teller. Also, replace any "Is X truth-teller?" from the following with "Is X a liar or a truth-teller?".

Round 1

Q: "Is Random one of the last 2 (third or fourth) in the line?".

Ask question Q to first three of them. After 3 of these questions you get:

- either 2YES/1NO or 2NO/1YES in which case it's undecided and proceed with asking the fourth man the same question.

- 3 YES's or 3NO's in which case proceed to Round 2 without asking the fourth question.

Depending on the number of YES/NO answers, there are 5 cases for Round 1:

- 2 YES/2NO - 4 questions spent

- 3 YES/1 NO or 3 NO/1 YES - 4 questions spent

- 3 YES or 3 NO - 3 questions spent!

2YES/2NO - you can get both Random and Alternator in at most 7 questions.

Round 1 - You must spend your first 4 questions till you get 2YES 2NO, since up until the last of the 4, you could still get a majority.

You only know that the alternator was acting as random (and siding with random) so there's no definite way to point a person that's not random.

Good news is that Alternator will tell the truth the next 2 times you ask him a question.

For Round 2:

- If you take 1 of the 4 you could be looking at either case (T, A, R). Very uncertain.

- If you take 2 from same side you get TT or AR. Since Alternator will act as a truth-teller this time, you get in fact either TT or TR. Which is good enough.

So, pick one side, take the two that answered YES to the first question.

Ask them again the same question as in Round 1. You're looking at either 2YES, 2NO, or 1 YES/1 NO.

- If you get 1 YES / 1 NO, then this is the Random+Alternator side since one of them changed his answer from Round 1. The other side is the truth-tellers' side.

So you can finish in 7 by asking a truth-teller which of the two asked in Round 2 is random. The other one is the alternator.

- If you get 2 YES, then the answer to the question is surely YES (either TT or AR - one of them is definitely telling the truth this time). But since the answer is YES and the question is the same, you've selected the two truth-tellers in Round 2, since they can't be the NO guys in Round 1 (the fact that it's the same question is crucial here!). So you can still finish in 7 by asking one of the truth-tellers if one of the other two is random. The other one on his side in Round 1 is the alternator.

- If you get 2 NO's, then both have changed their opinion since Round 1 (since you picked the YES side). So this is the Random+Alternator side. Again finish in 7 by asking one of the truth-tellers (the other side) if one of the guys in Round 2 is Random. The other one in Round 2 is the Alternator, so you get both.

3YES 1NO - Finish in 7 finding out both the alternator and the random:

From Round 1 - Q is true and two of the three YES's are truth tellers and the other is either random (in which case alternator was random in Round 1) or alternator (either randomly guessing or truth-teller).

You know that Random is either 3rd or 4th in line (since Q is true). Avoid him by asking only the first 2 in line in Round 1.

Now, the first two answered either 2YES's or YES/NO in Round 1:

- If the first two answered 1 YES/1 NO in Round 1, then the NO guy is the alternator lying in Round 1. You can finish in 5 by asking the other whether random is 4th.

- If the first two answered 2 YES's in Round 1 - they are either 2 truth-tellers or one truth-teller and alternator (either randomly guessing or truth-teller, cannot know for sure).

Ask these two if the Alternator is one of the last two. At least one of them tells the truth.

---If you get 2 YES's, then both Random and Alternator are 3rd and 4th which makes the first two truth-tellers. Finish in 7 asking a truth-teller if Random is 4th.

---If you get 2 NO's then Alternator is one of the first two (the answer to the question in Round 2 says so)

---If you get 1 YES and 1 NO, then Alternator is one of the first two (since truth-tellers don't disagree).

In both cases in which the Alternator is one of the first two, we get that the other YES guy in Round 1 (who we didn't ask in Round 2) is a truth-teller and the NO guy in Round 1 the Random. Finish in 7 by asking the truth-teller if Alternator is 1st.

3NO 1YES is symmetrical to 3YES 1NO (replace lying with minority vote above, and ask the 3rd and 4th instead of 1st and 2nd to avoid Random)

3YES - Finish in 7 finding out both the alternator and the random:

Rememember Round 1 spent only 3 questions since we stopped before asking the last.

Q is true after round 1 so Random is either 3rd or 4th.

Then the first two can be TT or TA.

Ask the first if the second is a truth-teller.

-If he answers YES then either he's the alternator and the second is the truth-teller or he's a truth-teller and the second is the truth-teller.

--Either way, the 2nd is a truth-teller, so ask him if the first is the alternator:

---If he says YES, you got the alternator. Ask the truth-teller if the 4th is Random and you get Random too in 3+1+1+1=6 questions

---If he says NO, you got the two truth-tellers as 1st and 2nd. Ask any of them if the 4th is Random and you get Random(either 4th or 3rd) and the Alternator is the other guy(either 3rd or 4th) in 3+1+1+1=6 questions.

-If he answers NO then either he's the alternator and the second is the truth-teller or he's a truth-teller and the second is the alternator. You've spent 3+1=4 questions so far.

Note that if you ask an Alternator something twice (with oblique-questions that liars and truth-tellers answer the same), he's bound to act like a truth-teller at least once.

So, ask the first TWICE if the 4th is Random (6 questions so far)

--If he gives you 2YES's or 2NO's, then you know Random is the 4th or the 3rd. Now, the other guy (3rd or 4th) is definitely a truth-teller (since we already narrowed it down that the alternator is among the first two). Ask him if Alternator is first and finish in 7 questions with both Random and Alternator found.

--If he gives you 1 YES and 1 NO, then he's the Alternator. Ask the second (who is surely a truth-teller)if Random is 4th and you finish in 7 questions again.

3NO - I will discuss this case because one may think the symmetry to the 3YES case is less clear since we stopped after 3 questions. But it is symmetrical since we did not use any information about who answered what in the first round, just the answer itself was important.

Rememember Round 1 spent only 3 questions since we stopped before asking the last.

Q is false after Round 1 so Random is either 1st or 2nd.

Then the last two can be TT or TA.

Ask the 4th if the 3rd is a truth-teller.

-If he answers YES then either he's the alternator and the 3rd is the truth-teller or he's a truth-teller and the 3rd is the truth-teller.

--Either way, the third is a truth-teller, so ask him if the fourth is the alternator:

---If he says YES, you got the alternator. Ask the truth-teller if the 1st is Random and you get Random too in 3+1+1+1=6 questions

---If he says NO, you got the two truth-tellers as 3rd and 4th. Ask any of them if the 1st is Random and you get botth Random(either 1st or 2nd) and the Alternator is the other guy(either 2nd or 1st) in 3+1+1+1=6 questions.

-If he answers NO then either he's the alternator and the 3rd is the truth-teller or he's a truth-teller and the 3rd is the alternator. You've spent 3+1=4 questions so far.

Note that if you ask an Alternator something TWICE (with oblique-questions that liars and truth-tellers answer the same), he's bound to act like a truth-teller at least ONCE.

So, ask the 4th TWICE if the 1st is Random (6 questions so far)

--If he gives you 2YES's (or 2NO's), then you know Random is the 1st (or the 2nd). Now, the other guy (2nd or 1st) is definitely a truth-teller (since we already narrowed it down that the alternator is among the last two). Ask him if Alternator is 4th and finish in 7 questions with both Random and Alternator found.

--If he gives you 1 YES and 1 NO, then he's the Alternator. Ask the 3rd (who is surely a truth-teller) if Random is 1st and you finish in 7 questions again.

And, it follows trivially

that after getting random and alternator in 7 questions, one can get the liar from the truth teller, thus completing the arrangement in at most 8 questions. By asking another oblique question one of the identified "truth-tellers" after 7 questions: "If I would ask you if X is a liar would you answer YES?" where X is the other "truth-teller".

Nice work! Once again you made a nice contribution to this topic. Thanks!