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# Numbered Foreheads Concluded

### #11

Posted 15 November 2013 - 01:07 AM

### #12

Posted 15 November 2013 - 02:25 AM

Spoiler for Comments

### #13

Posted 15 November 2013 - 09:20 AM

Spoiler for

**Edited by harey, 15 November 2013 - 09:29 AM.**

### #15

Posted 11 February 2014 - 09:48 AM

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?