Four liars

[BMAD]

You meet four quadruplets at a party with a unique character flaw: they are all liars. When answering questions, the youngest of them lies every other time. The second youngest lies twice for every three questions. The next oldest lies three times for every four questions. And the oldest lies four times for every five questions. Each person must maintain their rate of telling lies but does not have to tell them in a pattern....eg. The oldest must tell four lies and one truthful answer for every five questions, so if asked five questions they may respond with lie, lie, true, lie, lie..then when asked another five questions they may respond with true, lie, lie, lie, lie. What is the minimum amount of questions needed to work out the order of birth for the quadruplets if you can't repeat questions or ask the same person a question consecutively?

Edited April 30, 2013 by BMAD

[TimeSpaceLightForce]

for this solution, depending on how one interpret's 'minimum' amount of questions, i find the answer to exist in the range of

37 to 53

the problem is defending on how the brothers will maximize

the amount of questions, you'll find one solution because

they will not reply in random or individual preference but in

a coordinated plan..

1.Maximum to determine the 1/2: F T F T F..T =6 T F F T F..T=6

if 2/3 plays F T F T F..F T F F T F..F

2.Maximum to determine the 3/4 :T F F F F F F T F F F T F F F T F F F..T =20

if 4/5 plays T F F F F F F T F F F T F F F T F F F.. F

or 3/4 :T F F F F F T F F F T F F F F T F F F..T =20

if 4/5 plays T F F F F F T F F F T F F F F T F F F..F

but

3.Maximum to determine the 2/3: T F F F F T F F T F F ..T =12

if 3/4 plays T F F F F T F F T F F.. F

but 1/2 is determined in 3

and 4/5 is determined in 9 ... this will not maximize

[bonanova]

This is a challenging puzzle.
Here the the beginning of a brute-force approach.
Suppose we relax the rules and permit repeated questioning of one of the liars.
Suppose also that we ask a series of questions: Does 1=1? Does 2=2? and so forth.
We can then discern a sequence of false and truthful answers.

After a number of responses, it becomes possible to eliminate some of the 1/2 1/3 1/4 and 1/5 cases for some of the liars, and eventually we will know them all.

Since T answers are the least likely, we learn identities soonest if the first answer a liar gives us it truthful.

Even so, to distinguish 1/4 from 1/5, we might need to ask that liar as many as 32 questions.

If the first answer is F, we we need to ask more, perhaps 3 more.

Given that upper limit, we ask T/F questions in turn of three of the liars.

We can alternate liars being questioned until three are identified and then the fourth is known as well.

Worst case estimate for number of questions:

thirty-three questions asked of three liars, perhaps 100 questions in all.

That's an upper limit for a brute force approach.

Smarter questions might be asked, which would shorten the process.

For example if we have identified the 1/3 liar, he might be in a state where the truth value of his next answer is known. Then we could point to a brother and ask is he the 1/5 liar? And so forth.

[TimeSpaceLightForce]

Frontal Attack in 42 most
1:Y N N N N Y N N N N Y N N
2:Y N N N N Y N N Y N N N Y
3:Y N N N N Y N N Y N N Y
4:N Y N Y

[bonanova]

Frontal Attack in 42 most

1:Y N N N N Y N N N N Y N N

2:Y N N N N Y N N Y N N N Y

3:Y N N N N Y N N Y N N Y

4:N Y N Y

One of them gives this sequence of replies?

T F F F F F T F F F T F F F F T F F F . . . . . .

Is he the 1/4 or 1/5? And the sequence can be longer yet of it starts with F.

[TimeSpaceLightForce]

Frontal Attack in 42 most
1:Y N N N N Y N N N N Y N N
2:Y N N N N Y N N Y N N N Y
3:Y N N N N Y N N Y N N Y
4:N Y N Y

One of them gives this sequence of replies?

T F F F F F T F F F T F F F F T F F F . . . . . .

Is he the 1/4 or 1/5? And the sequence can be longer yet of it starts with F.

we expect T or F on the next ..

T F F F F F T F F F T F F F F T F F F..F 4/5

T F F F F F T F F F T F F F F T F F F..T 3/4

the 3 brood shall make their replies same as possible

we got 39 questions here plus 10 on the others ,so 49 questions ?

Edited April 30, 2013 by TimeSpaceLightForce

[TimeSpaceLightForce]

This is a challenging puzzle.

Here the the beginning of a brute-force approach.

Suppose we relax the rules and permit repeated questioning of one of the liars.

Suppose also that we ask a series of questions: Does 1=1? Does 2=2? and so forth.

We can then discern a sequence of false and truthful answers.

After a number of responses, it becomes possible to eliminate some of the 1/2 1/3 1/4 and 1/5 cases for some of the liars, and eventually we will know them all.

Since T answers are the least likely, we learn identities soonest if the first answer a liar gives us it truthful.

Even so, to distinguish 1/4 from 1/5, we might need to ask that liar as many as 32 questions.

If the first answer is F, we we need to ask more, perhaps 3 more.

Given that upper limit, we ask T/F questions in turn of three of the liars.

We can alternate liars being questioned until three are identified and then the fourth is known as well.

Worst case estimate for number of questions:

thirty-three questions asked of three liars, perhaps 100 questions in all.

That's an upper limit for a brute force approach.

Smarter questions might be asked, which would shorten the process.

For example if we have identified the 1/3 liar, he might be in a state where the truth value of his next answer is known. Then we could point to a brother and ask is he the 1/5 liar? And so forth.

That is a good idea, the alternating youngest brother will give out all the others

[TimeSpaceLightForce]

Frontal Attack in 42 most

1:Y N N N N Y N N N N Y N N

2:Y N N N N Y N N Y N N N Y

3:Y N N N N Y N N Y N N Y

4:N Y N Y

Spilling the beans 27

=========================================

1: Y N N N N Y

2: Y N N N N Y N

3: Y N N N N Y N

4: N Y N Y 1=2nd? 1=3rd? 2=2nd?

4=1st! Y N Y

1=4th! 2=3rd!

3=2nd!

[TimeSpaceLightForce]

with question skips but not consecutive
1. ? ? ?
2. ? ? ?
3: N Y N Y N N ? ?
4: N Y N Y N 1=2nd? 1=3rd? 2=2nd?
N Y N
4=1st! 1=4th! 2=3rd!
3=2nd

Edited April 30, 2013 by TimeSpaceLightForce

[TimeSpaceLightForce]

with question skips but not consecutive questioning to same person..

1. N Y N N

2. N Y N N

3: N Y N Y N N Y ?

4: N Y N Y N 1=2nd? 1=3rd? 2=2nd?

N/T Y/F N/T

4=1st! 1=4th! 2=3rd!

3=2nd

[BMAD]

would be to focus on the least common multiple

[TimeSpaceLightForce]

...this is also a good puzzle! I guess what the brothers want is

to cover the youngest who lies alternately every other time
Solution 4:
1. F T F F
2. F T F F
3: F T F T F F 1=3rd?N
3=2nd! 1=4th!
4: F T F T F 1=2nd?N ....4=1st! 2=3rd!

[bonanova]

You meet four quadruplets at a party with a unique character flaw: they are all liars. When answering questions, the youngest of them lies every other time. The second youngest lies twice for every three questions. The next oldest lies three times for every four questions. And the oldest lies four times for every five questions. Each person must maintain their rate of telling lies but does not have to tell them in a pattern....eg. The oldest must tell four lies and one truthful answer for every five questions, so if asked five questions they may respond with lie, lie, true, lie, lie..then when asked another five questions they may respond with true, lie, lie, lie, lie. What is the minimum amount of questions needed to work out the order of birth for the quadruplets if you can't repeat questions or ask the same person a question consecutively?

So the 4/5 liar could not respond like this? TLLLL LLLLT ..... .....

That is, he could never lie on 5 consequetive responses?

That makes it much easier.

[BMAD]

You meet four quadruplets at a party with a unique character flaw: they are all liars. When answering questions, the youngest of them lies every other time. The second youngest lies twice for every three questions. The next oldest lies three times for every four questions. And the oldest lies four times for every five questions. Each person must maintain their rate of telling lies but does not have to tell them in a pattern....eg. The oldest must tell four lies and one truthful answer for every five questions, so if asked five questions they may respond with lie, lie, true, lie, lie..then when asked another five questions they may respond with true, lie, lie, lie, lie. What is the minimum amount of questions needed to work out the order of birth for the quadruplets if you can't repeat questions or ask the same person a question consecutively?

So the 4/5 liar could not respond like this? TLLLL LLLLT ..... .....

That is, he could never lie on 5 consequetive responses?

That makes it much easier.

He could. What is meant in the op is that for every set of five questions the 4/5 liar will lie four times (with 4 lies at the end possibly). if asked another set of five questions the liar will again lie four out of five times (with 4 lies at the beginning this time). So:

if asked 5 questions

T L L L L (possibly)

next 5 questions

L L L L T (possibly)

so T L L L L L L L L T = 8L / 10total which reduces to 4/5

[bonanova]

You meet four quadruplets at a party with a unique character flaw: they are all liars. When answering questions, the youngest of them lies every other time. The second youngest lies twice for every three questions. The next oldest lies three times for every four questions. And the oldest lies four times for every five questions. Each person must maintain their rate of telling lies but does not have to tell them in a pattern....eg. The oldest must tell four lies and one truthful answer for every five questions, so if asked five questions they may respond with lie, lie, true, lie, lie..then when asked another five questions they may respond with true, lie, lie, lie, lie. What is the minimum amount of questions needed to work out the order of birth for the quadruplets if you can't repeat questions or ask the same person a question consecutively?

So the 4/5 liar could not respond like this? TLLLL LLLLT ..... .....

That is, he could never lie on 5 consequetive responses?

That makes it much easier.

He could. What is meant in the op is that for every set of five questions the 4/5 liar will lie four times (with 4 lies at the end possibly). if asked another set of five questions the liar will again lie four out of five times (with 4 lies at the beginning this time). So:

if asked 5 questions

T L L L L (possibly)

next 5 questions

L L L L T (possibly)

so T L L L L L L L L T = 8L / 10total which reduces to 4/5

So it's four lies for question 1-5. Four more lies for questions 6-10 and so forth.

Not quite what the OP says.

If it were four lies for every "five" questions then the lies, after the first truth,

would always come in groups of exactly four. It could not be T LLLLL LLLT

because questions 2-6 comprise "five" questions that do not contain a truth.

OK now it's hard again.

[TimeSpaceLightForce]

If no info about the other brothers is permitted..

and rotation of questions until knowing one's order.

the brothers will choose the longest uncertainty:

T F F F F F F T F F F T F F F T F F F.. F 4/5:20
T F F F F F F T F F F T F F F T F F F..T 3/4:19
or
T F F F F F T F F F T F F F F T F F F..F 4/5:20
T F F F F F T F F F T F F F F T F F F..T 3/4:19

plus
T F F T F ..F 2/3:6
T F F T F ..T 1/2:5
=50

..but i think the askings is not limited so the solution is lower

[BMAD]

Would it be easier to find the 1:1 lie:truth teller or the 4:1 lie:truth teller?

[BMAD]

for this solution, depending on how one interpret's 'minimum' amount of questions, i find the answer to exist in the range of

37 to 53