17 replies to this topic

[plainglazed]

Each of eight logicians has a unique positive integer less than 100 stapled to their foreheads. They stand in a circle facing one another so every logician can see everyone's number, save for their own. Each logician must discern their number and raise their hand. After all logicians have raised a hand, they must then all declare their number in unison. What strategy (within the assumed spirit of these types of puzzles) might they employ such that all are correct?

EDIT

[Time Out]

Am guessing that they can agree on a system of communication beforehand

Spoiler for ?Show of Hands

Each logician communicates with the other directly opposite by raising his hand in a precise way. Fingers represent binary values. Left hand, right hand, palm out or in represent four meanings. Perhaps add from 0, add from 33, add from 66 & subtract from 100. In this way each is told his number. I'll keep my fingers crossed.

I like the image of the "stapled to head"

[plainglazed]

I thought for sure I mentioned in the OP that they were attending the Arctic Circle of Logicians annual Fresh Air/Fresh Ideas Seminar and were all wearing mittens. Apparantly not, so your answer does indeed fit within the constraints of the OP, Time Out. But for grins, let's do assume the logicians are all wearing plain white mittens.

[bushindo]

I thought for sure I mentioned in the OP that they were attending the Arctic Circle of Logicians annual Fresh Air/Fresh Ideas Seminar and were all wearing mittens. Apparantly not, so your answer does indeed fit within the constraints of the OP, Time Out. But for grins, let's do assume the logicians are all wearing plain white mittens.

When you say 'numbers less than 100', do you mean 'integers less than 100 but larger than 0'? Also, can a number be repeated?

Edited by bushindo, 14 May 2012 - 03:54 PM.

[mewminator]

Spoiler for A possible way of doing it

(assumming numbers cannot be repeated) each logician will look at the person with the smallest number on his head, the person with the smallest number on his head will have 7 people looking at him, so he will know that his number is smaller than the rest, that's all I could think of now.

or I could go for the simpler way

Spoiler for simply

(assuming all numbers are integers) blinking system, each person will blink the number of the person standing directly in front of him, to make things easier make 2 sets of blinks, one for tens and on for units, zeros can be winks, if no winking is allowed, then go for normal blinking, though it's not too practical to blink 100 times

[plainglazed]

hey bushindo - right you are, rewrote this this a.m. and forgot that bit. hadn't saved my .txt draft and Microsoft was kind enough to update and reboot my machine last night. thanks and sorry

and sorry, too - no blinking in morse code, looking in the reflection of the irises to your right, hoof stomping, shouting the number across from you in pig latin, etc., etc...

[Molly Mae]

Once a hand is raised, can it be put down? Or would that qualify as illegal communication?

[plainglazed]

hey Mo Ma - no, once a hand is raised that is the only bit of info the rest of the group can directly discern from that logician.

[bushindo]

Each of eight logicians has a unique positive integer less than 100 stapled to their foreheads. They stand in a circle facing one another so every logician can see everyone's number, save for their own. Each logician must discern their number and raise their hand. After all logicians have raised a hand, they must then all declare their number in unison. What strategy (within the assumed spirit of these types of puzzles) might they employ such that all are correct?

EDIT

Here's an attempt.

Spoiler for Solution

Let the logicians be indexed by numbers going from 1 to 8. Let a_j represent the positive integer of logician j. Let d_i( b ) represent the i-th digit from the right in the binary expansion of b. For instance, if b = 99 (which is 1100011 in binary), then d₂( b) = 1, d₃( b ) = 0, d₄(b) = 0, and so on.

The general strategy is as follows.

1) For each logician i, where 1<= i <=7, look around and sum up the remaining hats. Let's call this sum T_i. The logician i is then supposed to raise his hand if d_i( T_i) = 1. Otherwise the logician should not raise his hand.
2) After logician 1 to 7 have done the previous procedure, then logician 8 should be able to figure out his number.
3) Logician 8 then calculates the sum ( d₁( a₁) + d₂( a₂) + ... + d₇(a₇) ). He then should raise his hand if this sum is odd. The remaining 7 logicians should be able to figure out his number as well.

Edited by bushindo, 14 May 2012 - 05:35 PM.

[bushindo]

I just want to come back and say that this is a really well designed puzzle. I didn't explain the rationale behind the solution before, and so here it is

Spoiler for rationale

The rationale of the solution is that we want to use the logicians 1-7 to convey (in binary digits) to the 8th logician his number. We need 7 binary digits to describe a number between 1 and 100, hence the 7 logicians.

The procedure that we did above will also allow each of logician 1 to 7 to determine six of his seven binary digits. We then simply used the 8th logician to convey this information back to the previous 6.

Edited by bushindo, 06 July 2012 - 07:35 PM.