17 replies to this topic

[Barcallica]

Spoiler for

if we assume they can count inside at same rhythm problem would be too easy so i assume not

[plainglazed]

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.

yes indeed, bushindo. well done.

Spoiler for Alternatively

all 8 logicians could perform your step 3) above. Then every digit of each logician will appear exactly once on each of the seven other logicians step 3). And each logician can already see six of those seven digits.

for example:
logician(1) step 3)-->d8(a8)+d7(a7)+…+d2(a2) and logician(8) can see d7(a7)+d6(a6)+…+d2(a2) so knowing the parity of logician(1)'s step 3), logician(8) can easily figure her eighth digit, etc.

[bushindo]

yes indeed, bushindo. well done.

Spoiler for Alternatively

all 8 logicians could perform your step 3) above. Then every digit of each logician will appear exactly once on each of the seven other logicians step 3). And each logician can already see six of those seven digits.

for example:
logician(1) step 3)-->d8(a8)+d7(a7)+…+d2(a2) and logician(8) can see d7(a7)+d6(a6)+…+d2(a2) so knowing the parity of logician(1)'s step 3), logician(8) can easily figure her eighth digit, etc.

That alternative solution is a lot more compact and elegant. Thanks for sharing this problem.

[superprismatic]

I think the solution still works if the word "unique" is removed from the problem statement. N'est pas?

[phil1882]

hmm, not convinced its as simple as bushindo/plain makes it out to be.

Spoiler for

let's say everyone has an even binary parity.
1 2 3 4 5 6 7 8
3 6 12 15 24 30 48 60

0000011
0000110
0001100
0001111
0011000
0011110
0110000
0111100
does no one raise their hand? or even if some poeple raise, thare are various ways to get all even binary parities. how would everyone know thier number?

[superprismatic]

Spoiler for The way I see it

The way I look at it, the i^th person raises her hand if and only if
d₁(a_{1+(i mod 8)}) ⊕ d₂(a_{1+(i+1 mod 8)}) ⊕ d₃(a_{1+(i+2 mod 8)}) ⊕ d₄(a_{1+(i+3 mod 8)})
⊕ d₅(a_{1+(i+4 mod 8)}) ⊕ d₆(a_{1+(i+5 mod 8)}) ⊕ d₇(a_{1+(i+6 mod 8)}) = 1
(where ⊕ is addition modulo 2). Using knowledge of the numbers on the heads
you can see, as well as which hands are raised, each person can determine her
number.

[plainglazed]

yes sp, thanks for the explanation. and

Spoiler for some visual aid using Phil1882's numbers

where h_i is the sum mod(2) determining if logician i raises a hand or not (or raises a left or right mittened paw)

so to figure say a8 -
to find the seventh digit logician 8 can see six digits of h1 (colored buff) whose sum mod(2)=0=h1 so the seventh digit of a8=0
to find the sixth digit logician 8 can see six digits of h2 (colored yellow) whose sum mod(2)=1 but h2=0 so the sixth digit of a8=1
etc.

I think the solution still works if the word "unique" is removed from the problem statement. N'est pas?

et @ superprismatique - mais bien sur

Spoiler for also

zero works as well as integers up to 127 but thought limiting to unique two digit integers might better obscure the path to the solution

[WitchOfSecrets]

An alternative solution:

Spoiler for Cheeeeeeap.

Before being placed in a circle, all agree on a march to hum inside their heads, and synchronize their humming. Then they go silent when placed in a circle, but keep humming and counting lines in their heads. This is MUCH more accurate than counting in words.

Each person raises a hand on the line # of the person opposite them.