Riddles

[Arminius]

No I don't want to play with a recursive solution until you acknowledge that my inductive solution is sound.

It isn't "sound" because

Both of them know that both of them don't have 9. So it is not necessary for one to know that the other knows.

...that isn't true.

The entire game is figuring out what the other person must know. That is why it is in the category of "Perfect Logicians", else one couldn't be certain of what the other might deduce.

Exactly.

[James S Saint]

Like at the first no I know I don't have a 9, then at the second no you know you don't have a 9 and I know that you know.

They both know that they don't have a 9 from the start. But neither knows that the other knows that because one might have a 15 and thus the other would have a 12 or a 9. Each can see a 12, but they can't know they don't have a 15 and thus the other person thinking that he might have a 9 or a 12.

If the person saying "no" suspects that he might have a 9, the reasoning that he said "no" because he realizes that he must have either 12 or 15, doesn't hold. When the 15 is discounted, he would still wonder if he had 12 or 9.

[phoneutria]

Any of them would immediately say yes if they could see a 9. This is obvious to both.
If both of them say no on the 1st round, then none have 9.
This is evident.

[James S Saint]

Any of them would immediately say yes if they could see a 9. This is obvious to both.

?? How so?

If you see a 9, how do you know whether you have a 18 (for 27) or a 15 (for 24)?

If both of them say no on the 1st round, then none have 9.
This is evident.

Even after the first round, 9 is still an option.

[Arminius]

untab

Okay, I will give you the next step.

In the beginning A knows that (1) a = 12 or a = 15, and (2) B knows that b = 12 or b = 15.

Okay. But A does not know that B (2) knows, and B does not know that A (1) knows. So the statement above is not suited for the recursive conclusion.

But both A and B know all of the following statements and that each of them knows that the other one knows them:

(3) a = 24 - b or a = 27 - b and (4) b = 24 - a or b = 27 - a.

Now, from the first "no" of A and from (4) follows (5) b < 24, because if b >= 24, then A would be able to conclude a. This is the motor for the recursive conclusion.

Now, from the first "no" of B and from (3) and (5) follows (6) a > 3.

And so on.

Next step:

A: "No" => b < 21.
B: "No" => a > 6.

And so on.
__________

By this I have given you almost the whole solution. (Now, hurry up, because the others are coming soon.)

Good luck

Next step:

A: "No" => b < 18.
B: "No" => a > 9.

And so on.

[Arminius]

The whole solution (with the solution process):

untab

In the beginning A knows that (1) a = 12 or a = 15, and (2) B knows that b = 12 or b = 15.

Okay. But A does not know that B (2) knows, and B does not know that A (1) knows. So the statement above is not suited for the recursive conclusion.

But both A and B know all of the following statements and that each of them knows that the other one knows them:

(3) a = 24 - b or a = 27 - b and (4) b = 24 - a or b = 27 - a.

Now, from the first "No" of A and from (4) follows (5) b < 24, because if b >= 24, then A would be able to conclude a. This is the motor for the recursive conclusion.

Now, from the first "No" of B and from (3) and (5) follows (6) a > 3.

And so on.

A: "No" => b < 21.
B: "No" => a > 6.
A: "No" => b < 18.
B: "No" => a > 9.
A: "No" => b < 15.
B: "Yes". Because together with the information of (2) there remains only one possibility.

Now add the „No“s!
___________________

The game ends after 7 „no“s.

[Arminius]

I remind you of the riddle I posted on 14 January 2016:

Six people in two groups.

There are six people A, B, C, D, E, F which are in each case either in group 1 or group 2. The following statements are given:

1. Both A and B are in 1.
2. F is in 2, and if E is in 2, then C is also in 2.
3. D is in 1, and if F is in 2, then A is also in 2.
4. A and E are both in 2.
5. D is in 2, and E is in 1, and if C is in 2, then B is in 1.
6. D and B are both in 2.
7. The statements 1-6 are wrong.

Who is in which group?

[Arminius]

Who is depicted here?

Easy_Riddle.jpg (102.93 KiB) Viewed 4964 times

[James S Saint]

The whole solution (with the solution process):

untab

In the beginning A knows that (1) a = 12 or a = 15, and (2) B knows that b = 12 or b = 15.

Okay. But A does not know that B (2) knows, and B does not know that A (1) knows. So the statement above is not suited for the recursive conclusion.

But both A and B know all of the following statements and that each of them knows that the other one knows them:

(3) a = 24 - b or a = 27 - b and (4) b = 24 - a or b = 27 - a.

Now, from the first "No" of A and from (4) follows (5) b < 24, because if b >= 24, then A would be able to conclude a. This is the motor for the recursive conclusion.

Now, from the first "No" of B and from (3) and (5) follows (6) a > 3.

And so on.

A: "No" => b < 21.
B: "No" => a > 6.
A: "No" => b < 18.
B: "No" => a > 9.
A: "No" => b < 15.
B: "Yes". Because together with the information of (2) there remains only one possibility.

Now add the „No“s!
___________________

The game ends after 7 „no“s.

Incorrect (for several reasons).
Sorry, try again.

[Arminius]

My solution is absolutely correct.

Your "there-is-no-solution-solution" is incorrect.

I don't see how that one can end: "No's all the way down."

That is incorrect. Sorry. Try again.

[James S Saint]

My solution is absolutely correct.

"Famous last words."

You should know me well enough to at least accept a tiny bit of doubt if I am telling you that you are incorrect.
Would I say it without a reason?

Your "there-is-no-solution-solution" is incorrect.

I don't see how that one can end: "No's all the way down."

That is incorrect. Sorry. Try again.

My actual words were:

As it is worded, I don't see how that one can end: "No's all the way down."

First correction:
You must disqualify zero and all negative numbers when you word the riddle, else your count will be different.

Agreed?

[Arminius]

My solution is absolutely correct.

"Famous last words."

You should know me well enough to at least accept a tiny bit of doubt if I am telling you that you are incorrect.
Would I say it without a reason?

Your "there-is-no-solution-solution" is incorrect.

I don't see how that one can end: "No's all the way down."

That is incorrect. Sorry. Try again.

My actual words were:

As it is worded, I don't see how that one can end: "No's all the way down."

First correction:
You must disqualify zero and all negative numbers when you word the riddle, else your count will be different.

Agreed?

No.

The said riddle:

Perfect Logicians.

Players A and B both have got the number 12 written on her forehead. Everyone sees the number on the front of the other but does not know the own number. The game master tells them that the sum of their numbers is either 24 or 27 and that this numbers are positive integers (thus also no zero).

Then the game master asks repeatedly A and B alternately, if they can determine the number on her forehead.

A: "No".
B: "No".
A: "No".
B: "No".
A: "No".
....

After how many "no"s does the game end, if at all?

|=>

So I wrote in the said riddle: "this numbers are positive integers (thus also no zero)".

It seems that you have not read the said riddle.

[James S Saint]

So I wrote in the said riddle: " this numbers are positive integers (thus also no zero)"

Oops .. your right, I missed that (silly me). My apologies. One must VERIFY anything I say (age n all).
... so on to the next issue:

Second correction:
This is the issue with all such "perfect logician", "recursive" riddles.

Being so perfect, they both already know that the other knows this sort of game. Even without being perfect, both you and I know of this algorithmic method. And the whole issue is to be able to realize what the other person knows so that each member can depend upon the answers being given by the others.

As with the "Blued eyed puzzle" and all such similar algorithms, there must be a number with which to begin. You chose "24", as most people would. But perfect logicians are not "most people". They know to choose, from the many options, the starting point that would lead to the least number of rounds. The question is "how many no's are required?" They could have begun their count at 48 or at 100. That would be silly. Why would they? But then again, why would they start at 24?

In all of these scenarios, the place a more perfect logician would begin is the number that is the closest that both parties would necessarily not be able to resolve the puzzle by knowing. They both want for the first "no" to be informative, telling them of something they didn't already know. They both see a "12" and thus both know that the other knows that the only options for any party is either:

9
12
15

It is a waist to begin at 100 and count your way down when you already know that nothing is going to be resolved until you get close to those numbers. It is also silly to begin with 24 for the same reason. Both parties know that they could begin at any number higher than 15, but can't choose which number unless they privately begin the known algorithm at 24 (the lowest known sum) and simply count to themselves down to "18" (or merely add the difference of the sums to the 15). They both can deduce from the beginning that neither would be able to say "yes" if they began from the number 18. Thus that is where to begin.

a) They both already know that both already know that their number is <18.

b) And that means that after the first "no", they both know that their number is >9, eliminating one of the possibles.

c) Second "no", their number must be <15, eliminating a second possible, leaving only one possible number.

Puzzle resolved with more perfect logicians with only 2 "no"s.

But that isn't my last objection/"correction".

[Arminius]

So I wrote in the said riddle: " this numbers are positive integers (thus also no zero)"

Oops .. your right, I missed that (silly me). My apologies. One must VERIFY anything I say (age n all).

Advantage for me. 1 : 0 (one to nothing).

First
Second correction:

It is your first correction, because your "first correction" was no correction but a mistake of you that was corrected by me. So it was my first correction. Therefore: Advantage for me. 1 : 0 (one to nothing).

This is the issue with all such "perfect logician", "recursive" riddles.

Being so perfect, they both already know that the other knows this sort of game. Even without being perfect, both you and I know of this algorithmic method. And the whole issue is to be able to realize what the other person knows so that each member can depend upon the answers being given by the others.

As with the "Blued eyed puzzle" and all such similar algorithms, there must be a number with which to begin. You chose "24", as most people would. But perfect logicians are not "most people". They know to choose, from the many options, the starting point that would lead to the least number of rounds. The question is "how many no's are required?" They could have begun their count at 48 or at 100. That would be silly. Why would they? But then again, why would they start at 24?

In all of these scenarios, the place a more perfect logician would begin is the number that is the closest that both parties would necessarily not be able to resolve the puzzle by knowing. They both want for the first "no" to be informative, telling them of something they didn't already know. They both see a "12" and thus both know that the other knows that the only options for any party is either:

9
12
15

It is a waist to begin at 100 and count your way down when you already know that nothing is going to be resolved until you get close to those numbers. It is also silly to begin with 24 for the same reason. Both parties know that they could begin at any number higher than 15, but can't choose which number unless they privately begin the known algorithm at 24 (the lowest known sum) and simply count to themselves down to "18" (or merely add the difference of the sums to the 15). They both can deduce from the beginning that neither would be able to say "yes" if they began from the number 18. Thus that is where to begin.

a) They both already know that both already know that their number is <18.

b) And that means that after the first "no", they both know that their number is >9, eliminating one of the possibles.

c) Second "no", their number must be <15, eliminating a second possible, leaving only one possible number.

Puzzle resolved with more perfect logicians with only 2 "no"s.

Okay. Now I can only say what you have said:

Oops .. your right, I missed that (silly me). My apologies. One must VERIFY anything I say (age n all).

The said riddle I posted about three month ago was a copy (the first copied riddle I posted). I did not question that there were too many "no"s in the text, because I did not question the whole copy, and that was my mistake. I am sorry. My apologies. The next riddle will be one of my own thoughts again, I can promise. You are right: One must VERIFY anything. But sometimes one is too busy or too lazy to do it anytime.

Advantage for you. 1 : 1 (one to one).

But that isn't my last objection/"correction".

No. But it was your first correction.

[James S Saint]

So I wrote in the said riddle: " this numbers are positive integers (thus also no zero)"

Oops .. your right, I missed that (silly me). My apologies. One must VERIFY anything I say (age n all).

Advantage for me. 1 : 0 (one to nothing).

Counting feathers are we.

Advantage for you. 1 : 1 (one to one).

Now see, if you were Carleas, you would argue this issue with me for the next 20 PAGES (as he did with the Blue eyed puzzle, another similar riddle, and the Stopped Clock Paradox). Some men just can't stand to lose a feather (those with too few).

Now let me "Re-Riddle" something from this riddle:

Re-Riddle:

Two men, John and Gerry, are walking along with a clearly visible number written on their foreheads.
John asks Gerry if he knows what number is on his own forehead.
Gerry honestly answers "No".
Gerry then asks the same of John.
John also honestly answers "No".
Then Gerry says, "Oh okay, my number must be __X__."
Immediately John replies, "Oh, then so is mine."

Now the question:
What must they have known about their numbers before John asked Gerry if he knew his own number?


.. and doesn't it seem eerie that such could actually happen.

[phoneutria]

Gah, James, you got it smartpants.

I was close though.

[Arminius]

Re-Riddle:

Two men, John and Gerry, are walking along with a clearly visible number written on their foreheads.
John asks Gerry if he knows what number is on his own forehead.
Gerry honestly answers "No".
Gerry then asks the same of John.
John also honestly answers "No".
Then Gerry says, "Oh okay, my number must be __X__."
Immediately John replies, "Oh, then so is mine."

Now the question:
What must they have known about their numbers before John asked Gerry if he knew his own number?

If a solution is possible, then they have to know a certain number range, thus the upper limit and the lower limit of the range of their numbers, and they have to know another aspect, for example: a possible sum, a possible difference, a possible product, a. possible quotient of their two numbers (for instance: Gerry only knows a sum of the two numbers of a certain number range, whereas John only knows a product of the two numbers of the same certain number range).

If a solution is not possible, then they have to know how to order two beers.

Bier_und_Alt-Bier.jpg (108.73 KiB) Viewed 4778 times

.. and doesn't it seem eerie that such could actually happen.

In any case, ordering two beers could actually happen. Couldn't it?

[James S Saint]

If a solution is possible, then they have to know a certain number range, thus the upper limit and the lower limit of the range of their numbers, and they have to know another aspect, for example: a possible sum, a possible difference, a possible product, a. possible quotient of their two numbers (for instance: Gerry only knows a sum of the two numbers of a certain number range, whereas John only knows a product of the two numbers of the same certain number range).

A little vague.

If a solution is not possible, then they have to know how to order two beers.

file.jpg (108.73 KiB) Viewed 4743 times

.. and doesn't it seem eerie that such could actually happen.

In any case, ordering two beers could actually happen. Couldn't it?

I didn't know Mithus was a beer drinker.
{{hmmm... how come his is bigger than mine}}

[phoneutria]

Three perfect logicians walk into a bar.
The bartender says "do you guys know what you're having?
The first one says: "No"
The second one says: "No"
The third one says: "Yes"
...

[James S Saint]

Three perfect logicians walk into a bar.
The bartender says "do you guys know what you're having?
The first one says: "No"
The second one says: "No"
The third one says: "Yes"
...

[Arminius]

If a solution is possible, then they have to know a certain number range, thus the upper limit and the lower limit of the range of their numbers, and they have to know another aspect, for example: a possible sum, a possible difference, a possible product, a. possible quotient of their two numbers (for instance: Gerry only knows a sum of the two numbers of a certain number range, whereas John only knows a product of the two numbers of the same certain number range).

A little vague.

No.

John and Gerry, who are walking along with a clearly visible number written on their foreheads, have to know a certain number range, thus the upper limit and the lower limit of the range of their numbers, and they have to know another aspect, for example: a possible sum, a possible difference, a possible product, a. possible quotient of their two numbers (for instance: Gerry only knows a sum of the two numbers of a certain number range, whereas John only knows a product of the two numbers of the same certain number range). So they know enough, even more than enough (!), in order to solve the riddle.

The primises in the riddle "Perfect Logicians" are enough too. Again:

Perfect Logicians.

Players A and B both have got the number 12 written on her forehead. Everyone sees the number on the front of the other but does not know the own number. The game master tells them that the sum of their numbers is either 24 or 27 and that this numbers are positive integers (thus also no zero).

Then the game master asks repeatedly A and B alternately, if they can determine the number on her forehead.

A and B know enough in order to solve the riddle.

If it is said that two humans see a number, then we can surely assume that this humans are capable of seeing and reading, and of knowing what they see and read. That is common sense.

[Arminius]

untab

Who is depicted here?

The following solution is false:

False_Solution.jpg (170.08 KiB) Viewed 4639 times

[James S Saint]

Six people in two groups.

There are six people A, B, C, D, E, F which are in each case either in group 1 or group 2. The following statements are given:

1. Both A and B are in 1.
2. F is in 2, and if E is in 2, then C is also in 2.
3. D is in 1 and if F is in 2, then A is also in 2.
4. A and E are both in 2.
5. D is in 2 and E is in 1, and if C is in 2, then B is in 1.
6. D and B are both in 2.
7. The statements 1-6 are wrong.

Who is in which group?

Solution:

untab

1) A, C, D
2) B, E, F

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

Who is depicted here?

Looks an awful lot like:

untab

[Arminius]

Six people in two groups.

There are six people A, B, C, D, E, F which are in each case either in group 1 or group 2. The following statements are given:

1. Both A and B are in 1.
2. F is in 2, and if E is in 2, then C is also in 2.
3. D is in 1 and if F is in 2, then A is also in 2.
4. A and E are both in 2.
5. D is in 2 and E is in 1, and if C is in 2, then B is in 1.
6. D and B are both in 2.
7. The statements 1-6 are wrong.

Who is in which group?

Solution:

untab

1) A, C, D
2) B, E, F

That is right.

What about the solution process?

Who is depicted here?

Looks an awful lot like:

untab

That is false.

Try again.

[Moreno]

It looks like HIlary crossed with a gnome.