Numbered Foreheads Concluded

[ThunderCloud]

Although I believe the logic to answer the original problem was present, it was distributed among several postings... It did not seem right to mark any single post as the best answer. Therefore, here is the bonus round.

If approached correctly, this version is not much harder than the original.

The puzzle:

Three perfect logicians had stickers placed on their foreheads so that none could see their own sticker but each could see one another's. They were told that each sticker has a single positive integer written on it (i.e.1, 2, 3, ...), and that the sum of the integers on all three stickers is either 1002 or 1003. They were then asked, in turn, to identify the number on their own sticker. Upon being asked, each logician would name their number if they were sure that they knew it, give up if they were sure that they would never know it, or otherwise 'pass' (or say "I don't know"). The question was repeated, again in turn, until EACH of the three logicians had either named their number or given up. All three stickers actually had the same number written on them. Who among the three logicians was able to deduce his number, and who among them gave up? (Furthermore, how did each answer?)

[ThunderCloud]

Hint:

This is not a tedious problem... not much harder than the original. There is a simple way to solve it; a pattern to be observed. An exhaustive description of how each logician answered in his turn would not be very long...

[ThunderCloud]

Still unanswered...

Can you find the complete sequence of responses? It is considerably shorter than one might initially think...

[harey]

This did not happen:

1a) Number One sees 334 and 335 and announces 334

1b) Number Two and Three announce 1003-(sum of numbers they see)

1c) Number One announces 334

So this must have happened:

2a) Number One sees 334 and 334, gives up and commits suicide

2b) Number Two and Three announce 334

I know a better and harder version, if I find it, I'll post it. Does it matter if it is in French?

[antel0pe]

Could they communicate beforehand so that they can deduce a plan?

[Rainman]

Let's start with what would happen if giving up hadn't been allowed. I will call the logicians A, B, and C. A goes first, then B, then C.

1) Distribution 334,334,334: All pass 333 times, then A passes, B wins, C wins.

2) Distribution 335,334,334: All pass 333 times, then A passes, B wins, C wins.

3) Distribution 335,333,334: All pass 333 times, then A wins, B wins, C wins.

4) Distribution 334,335,334: All pass 333 times, then A wins, B wins.

5) Distribution 334,335,333: All pass 333 times, then A wins, B wins.

6) Distribution 334,334,335: All pass 333 times, then A wins, B wins, C wins.

7) Distribution 334,333,335: All pass 333 times, then A wins, B wins, C wins.

Now, assume that A has 334 and B has 335. C can't distinguish between cases 4 and 5, and thus he will never be able to announce his number. So if giving up is allowed, then C would give up immediately if he saw that A had 334 and B had 335.

So here's what happens in the actual scenario: A passes, B passes, C passes, A passes. Now B announces 334, because C would have given up rather than passing if B had had 335. Now C announces 334, because if he had had 335 then B would have had no way of already having distinguished between cases 6 and 7. Finally A announces 334, as B wouldn't already have been able to distinguish between cases 2 and 3 if A had had 335.

Summary:

pass

pass

pass

pass

win

win

win

Edited November 13, 2013 by Rainman

[ThunderCloud]

This did not happen:

1a) Number One sees 334 and 335 and announces 334

1b) Number Two and Three announce 1003-(sum of numbers they see)

1c) Number One announces 334

So this must have happened:

2a) Number One sees 334 and 334, gives up and commits suicide

2b) Number Two and Three announce 334

I know a better and harder version, if I find it, I'll post it. Does it matter if it is in French?

I am not sure I follow your reasoning here -- care to elaborate further?

[ThunderCloud]

Could they communicate beforehand so that they can deduce a plan?

No. The logicians cannot agree upon a strategy in advance. However, you may assume each of them to be "perfect" in that they will deduce all that they logically can. You may further assume that each logician will assume the others to behave this way as well (i.e., that it is generally known that all three logicians are "perfect").

[harey]

We know the distributed numbers (334 334 334) are therefore what A, B and C see.

A sees 334 334. He thinks:
1a) if I have 335, B will see 335 334 and will announce 334

1b) if I have 334, B will see 334 334; not being able to choose between 334 335, he will pass

- > I will be able to know, so I pass

B sees 334 334

2a) not being able to choose between 334 335, he will pass (or even already give up, I am not very sure here)

C sees 334 334

3a) B did not see 334 335 (he would announce 344)

3b) -> B sees 334 334 -> I have 334

A knows he has 334 from 1b) or 3b) knows he has 334

C cannot know

Edited November 14, 2013 by harey

[ThunderCloud]

Let's start with what would happen if giving up hadn't been allowed. I will call the logicians A, B, and C. A goes first, then B, then C.

1) Distribution 334,334,334: All pass 333 times, then A passes, B wins, C wins.

2) Distribution 335,334,334: All pass 333 times, then A passes, B wins, C wins.

3) Distribution 335,333,334: All pass 333 times, then A wins, B wins, C wins.

4) Distribution 334,335,334: All pass 333 times, then A wins, B wins.

5) Distribution 334,335,333: All pass 333 times, then A wins, B wins.

6) Distribution 334,334,335: All pass 333 times, then A wins, B wins, C wins.

7) Distribution 334,333,335: All pass 333 times, then A wins, B wins, C wins.

Now, assume that A has 334 and B has 335. C can't distinguish between cases 4 and 5, and thus he will never be able to announce his number. So if giving up is allowed, then C would give up immediately if he saw that A had 334 and B had 335.

So here's what happens in the actual scenario: A passes, B passes, C passes, A passes. Now B announces 334, because C would have given up rather than passing if B had had 335. Now C announces 334, because if he had had 335 then B would have had no way of already having distinguished between cases 6 and 7. Finally A announces 334, as B wouldn't already have been able to distinguish between cases 2 and 3 if A had had 335.

Summary:

pass

pass

pass

pass

win

win

win

I think you are on the right track, and very close.

[harey]

We know the distributed numbers (334 334 334) are therefore what A, B and C see.

A sees B=334 C=334, he thinks:

- I can have 334 or 335

- whether I pass or give up, B will know I have not seen B=335 C=334 -> B will announce 334

- C will know I did not see B=334 C=335 -> C will announce 334

-> I will not learn anything new, so I give up

B sees A=334 C=334, he thinks:

A did not see B=335 C=334 (he would announce 334) -> B=334

C sees A=334 C=334, he thinks:

A did not see B=334 C=335 (he would announce 334) -> C=334

[ThunderCloud]

Quoting, emphasis added:

A sees B=334 C=334, he thinks:

- I can have 334 or 335

- whether I pass or give up, B will know I have not seen B=335 C=334 -> B will announce 334

***

Actually A cannot make the second deduction. Hypothetically, had A in fact seen B=335 and C=334, he would have known only that his number was either A=334 or A=333. So he would still have to pass (or give up) the first time he is asked, regardless. His response is therefore not very immediately helpful to B...

...

[harey]

ThunderCloud

Hypothetically, had A in fact seen B=335 and C=334

All three stickers actually had the same number written on them.

We know that A sees B=334 and c=334 - A CANNOT see B=335 and C=334.

(Not to confound with: "As this information in not available to him, he ASSUMES he has 334 or 335.")

So my first answer was correct. In the 2nd try, I got lost and could not correct quickly enough.

Edited November 15, 2013 by harey

[ThunderCloud]

ThunderCloud

Hypothetically, had A in fact seen B=335 and C=334

All three stickers actually had the same number written on them.

We know that A sees B=334 and c=334 - A CANNOT see B=335 and C=334.

(Not to confound with: "As this information in not available to him, he ASSUMES he has 334 or 335.")

So my first answer was correct. In the 2nd try, I got lost and could not correct quickly enough.

B does not know that all three stickers have 334 on them. Therefore, B does not know whether A sees B=334 or B=335. When A responds by passing, this still does not tell B what A has seen. B thinks:

1. if A saw B=334, then A must have been unable to decide if his number was A=334 or A=335.

2. if A saw B=335, then A must have been unable to decide if his number was A=334 or A=333.

Either way, B cannot call out his number yet; he doesn't know it.

[harey]

There can be a solution in one pass disregarding the answers of C. The numbers are a little bit high for practical purposes, so I will illustrate it on a sum of 12 or 13, everyone having 4.

Notation:
"!n" means "I do not see the number n"
"<-(xy)" means "because otherwise I would know the combination is (a=x b=y)"

Both A and B see c=4, they both know a+b=8 or a+b=9.
They also know that the other one knows that.
By saying "I do not know", they in fact tell:

1 A: !8 <- (18)
B: !8 <- (81)
!1 <- (17); (18) excluded in 1A
2 A: !1 <- (71); (81) excluded in 1B
!7 <- (27); (17) excluded in 1B
B: !7 <- (72); (71) excluded in 2A
!2 <- (26); (27) excluded in 2A
3 A: !2 <- (62); (72) excluded in 2B
!6 <- (36); (26) excluded in 2B
B: !6 <- (63); (62) excluded in 3A
!3 <- (35); (36) excluded in 3A
4 A: !3 <- (53); (63) excluded in 3B
!5 <- (45); (35) excluded in 3B
--> B knows A sees 4.
--> A realises he never will get more information

A already knows that b=4, he can imagine this dialogue (the answers of B will not change whether a=4 or a=5), so his very first answer is:
1 A: I will never be able to tell.

B thinks:
if b=4, this makes sense
if b=5, A would have announced he would be able to tell.

From the answers of A and B, C deduces c=4 (if necessary, I will explain).

The remaining problem is whether the answers of C are redundant or whether they lead to another distribution.

Comments?