Teaser #6

[Softus]

Didn't noseall fall down the same hole as me and Jackpot (and say as such in his post) ?

I didn't notice him saying as such.

Isn't his answer as follows (albeit written slightly differently)?

Split the coins into four equal piles of three each: A, B, C, D.

weigh A against B (weighing #1)

IF A & B don't balance
THEN
A or B contains the dissimilar coin (X)
weigh A against C (weighing #2)

IF A & C don't balance
THEN
A contains X
observe direction (T) of tipping of scales
ELSE
B contains X
ENDIF
ELSE
C or D contains X
weigh A against C (weighing #2)

IF A & C don't balance
THEN
C contains X
observe direction (T) of tipping of scales
ELSE
D contains X
ENDIF

take two coins (Y and Z) from the pile of three known to contain X, leaving one remaining coin (W)

weigh Y against Z (weighing #3)

IF Y and Z balance
THEN
W is X
ELSE
IF Y goes in direction T
THEN
Y is X
ELSE
ELSE
Z is X
ENDIF

 

[Softus]

Didn't noseall fall down the same hole as me and Jackpot (and say as such in his post) ?

I didn't notice him saying as such.

Isn't his answer as follows (albeit written slightly differently)?

[code:1]Split the coins into four equal piles of three each: A, B, C, D.

weigh A against B (weighing #1)

IF A & B don't balance
THEN
A or B contains the dissimilar coin (X)
weigh A against C (weighing #2)

IF A & C don't balance
THEN
A contains X
observe direction (T) of tipping of scales
ELSE
B contains X
ENDIF
ELSE
C or D contains X
weigh A against C (weighing #2)

IF A & C don't balance
THEN
C contains X
observe direction (T) of tipping of scales
ELSE
D contains X
ENDIF

take two coins (Y and Z) from the pile of three known to contain X, leaving one remaining coin (W)

weigh Y against Z (weighing #3)

IF Y and Z balance
THEN
W is X
ELSE
IF Y goes in direction T
THEN
Y is X
ELSE
ELSE
Z is X
ENDIF
[/code:1]

 

[johnny_t]

One of us is reading this wrong. Are you talking about the post that ends

'if they do balance then you have a pile of nine good coins and a pile of three with one dud.

DAMN!'



And that is the same hole. If you identify D as containing the dud, purely by working out that its not A, B or C, then you lose the vital info which is whether the fake is lighter or heavier, and it is only by knowing that that you can work out which of the final three is the fake.


Apologies if I'm misreading you.

 

[empip]

split the coins into four equal piles of three each.

weigh two of those piles putting three on each tray.

if they balance then you know the other six contain the duff coin.

if they don't balance you know the other six are good.

so now after one weighing, you have a two piles of six coins, one good and one bad.

take three coins from the good pile and weigh them with three coins from the duff pile.

if they don't balance, then you now know you have three coins which contain the fake, and, wether it is heavier or lighter.

weigh two of those coins to determine which is the fake.

if they do balance then you have a pile of nine good coins and a pile of three with one dud.

DAMN!

Groups A B C D holding 3 coins each

Weigh-1

if they balance then you know the other six contain the duff coin.

A v B result A = B therefore C or D holds the fake.

Weigh-2

take three coins from the good pile and weigh them with three coins from the duff pile.

A v C result A = C therefore D holds the fake.

if they don't balance, then you now know you have three coins which contain the fake, and, wether it is heavier or lighter.

We do not know wether D holds heavier or lighter weighted fake - yet -- because we have not weighed group D

Now have to find
i) the fake
and
ii) its weight difference, heavy or light.
Just one weighing remains and group D has not yet been weighed.

Group D comprises coins 10,11,12

Weigh-3

weigh two of those coins to determine which is the fake.

10 v 11 result 10 = 11 therefore 12 is odd but is it heavy or light in weight?
Worse
10 v 11 10 > 11 Now is 10 heavy or 11 light? ... not much use knowing 12 is neither heavy nor light weight at this stage.

..if they do balance then you have a pile of nine good coins and a pile of three with one dud.

DAMN..

I agree NA it is but part of a journey...

 

[empip]

My pattern matching for 10 coins in 3 weighings.
Maybe this helps to explain itself.
Simply pattern matching really.
Coin 10 set on one side and only weighed at w-3 if other weighings are equal.

The full weighing example is for 4,5 or 6 deemed lightweight following w-1 (weighing 1) and w-2 (weighing 2)

Went from 3 coins to 12 coins using this type of method.

 

[noseall]

when's bas gonna put us out of our misery.

i'm ready to self-harm here!

 

[empip]

when's bas gonna put us out of our misery.

i'm ready to self-harm here!

Thought you had already done so

There are more answers than one ...

 

[jackpot]

Here goes again.
Split the coins into 3 groups of 4 coins each A B C.
1. weigh A and B. If they balance then the fake coin is in the Group C

2. Now take 3 of the coins in group C and weight them against 3 of the coins that we know are real in group A OR B. If they balance, then the remaining coin is the fake one.

But if it didn't ballance then know that the fake coin is one of the 3 coins from group C, and we also know if it is lighter or heavier, because we know which side had the real coins.

3. Now weigh 2 of the coins against each other.... if they balance then the third coin is fake. if they differ we know which coin is fake, because we already know if it is lighter or heavier.

Now the confusing bit

If groups A and B didnt balance at the start then the the fake coin is in one of these groups. For this instance say group that the heavier group is group A, and the group thats lighter group B.

2, weigh two coins of group A and one coin of group B, against the other two coins of group A and one coin of group B......i.. If they balance then the fake coin must be one of the remaining 2

3. weigh these 2 and the lighter is the fake as group B is the lightest group.

If the second weigh did not balance then the fake coin is one of the two group A coins or the group B coin thats left

3ii. Take the two group A coins and weigh them. If they balance, then it is the group B coin that is fake and if they do not then the heavier one is the fake coin as this is from the heavy group A.

wOW its a teaser trying to explain this thing never mind solving it.
Is it right?

 

[hermes]

If the second weigh did not balance then the fake coin is one of the two group A coins or the group B coin thats left
?

But would there not be 4 group a coins and 2 group b coins left at this stage?

 

[empip]

I think JP means at

Now the confusing bit

:-

If A <> B at we-1
when A > B then grp A contains one heavy OR Grp B one lightwt coin
when B > A then grp B contains one heavy OR Grp A one lightwt coin

At we-2
If A > B at we-1
then ( 1 2 3 4 are equal weighted and one of 5 6 7 8 is a lightwt
OR
5 6 7 8 are equal and one of 1 2 3 4 is a heavywt )

testing the coins ... setting 7 and 8 (potential normal or lightwt) to one side.

(1 2 5) .v. (3 4 6)

If 1 2 5 > 3 4 6 then ( 1 or 2 is heavy OR 6 is lightwt)
..... at we-3 ( 1 ).v.( 2 ) if equal then 6 is odd and lightwt.
......Else scale deflection indicates which is heavy 1 or 2.

If 1 2 5 < 3 4 6 then ( 3 or 4 is heavy OR 5 is lightwt)
..... at we-3 ( 3 ).v.( 4 ) if equal then 5 is odd and lightwt.
......Else scale deflection indicates which is heavy 3 or 4.

If 1 2 5 = 3 4 6 then ( 7 or 8 is lightwt)
..... at we-3 ( 7 ).v.( 8 ) scale deflection indicates which is light 7 or 8.
=========
If B>A at first weighing then a similar approach gets the answer, bearing in mind B group now contains either all normal or one heavy coin and A group either all normal or one lightwt coin.

 

[hermes]

I think JP means at

Now the confusing bit

:-

If A <> B at we-1
when A > B then grp A contains one heavy OR Grp B one lightwt coin
when B > A then grp B contains one heavy OR Grp A one lightwt coin

At we-2
If A > B at we-1
then ( 1 2 3 4 are equal weighted and one of 5 6 7 8 is a lightwt
OR
5 6 7 8 are equal and one of 1 2 3 4 is a heavywt )

testing the coins ... setting 7 and 8 (potential normal or lightwt) to one side.

(1 2 5) .v. (3 4 6)

If 1 2 5 > 3 4 6 then ( 1 or 2 is heavy OR 6 is lightwt)
..... at we-3 ( 1 ).v.( 2 ) if equal then 6 is odd and lightwt.
......Else scale deflection indicates which is heavy 1 or 2.

If 1 2 5 < 3 4 6 then ( 3 or 4 is heavy OR 5 is lightwt)
..... at we-3 ( 3 ).v.( 4 ) if equal then 5 is odd and lightwt.
......Else scale deflection indicates which is heavy 3 or 4.

If 1 2 5 = 3 4 6 then ( 7 or 8 is lightwt)
..... at we-3 ( 7 ).v.( 8 ) scale deflection indicates which is light 7 or 8.
=========
If B>A at first weighing then a similar approach gets the answer, bearing in mind B group now contains either all normal or one heavy coin and A group either all normal or one lightwt coin.

Sorry, you lost me on the first line!

 

[empip]

The real deal. from a W.McWorter.

[code:1]Number coins 1 to 12
Scale pan assignments.
left pan right pan
(9) first weighing 5,6,8,10 7,9,11,12
(3) second weighing 2,3,4,7 5,6,11,12
(1) third weighing 1,4,10,1 12,5,7,8

Multiplier.
Allow 0 for an equal balance situation.
Allow +1 for left pan down.
Allow -1 for right pan down.

then assign using base 3 as
weighing-1 3^2 x multiplier from above (0, -1 or 1 ) = 9 x multiplier.
weighing-2 3^1 x multiplier from above (0, -1 or 1 ) = 3 x multiplier.
weighing-3 3^0 x multiplier from above (0, -1 or 1 ) = 1 x multiplier.

NB.# coins 7,9,11,12 final sign to be reversed.

If coin 7 is light weight.
weighing-1 = left pan down = +1 x 9 = + 9
weighing-2 = right pan down = -1 x 3 = -3
weighing-3 = left pan down = +1 x 1 = +1 result = + 7 Nb #. -7 coin 7 is lightweight.

If coin 12 is heavy weight
weighing-1 = right pan down = -1 x 9 = - 9
weighing-2 = right pan down = -1 x 3 = -3
weighing-3 = balance = 0 x 1 = 0 result = -12 Nb #. +12 coin 12 is heavy

So if coin 8 is heavy.
weighing-1 = left pan down = 1 x 9 = 9
weighing-2 = balance = 0 x 3 = 0
weighing-3 = right pan down -1 x 1 =-1 result 9+0+(-1) = + 8 coin 8 is heavy.
[/code:1]

He may have whispered 'Bish-Bash-Bosh '

 

[ban-all-sheds]

OK - This is the one remaining piece of unfinished business I have here, so I felt I should clear it up before I go.

Here's the solution I have:

[code:1]W1: 1 2 3 4 v 5 6 7 8

If that balances W2: 8 9 v 10 11

If that balances W3: 12 v any coin

If 8 9 are heavier than 10 11, then as we know
from W1 that 8 is OK:

9 is H or 10 is L or 11 is L so
W3: 10 v 11

If 8 9 are lighter than 10 11, then as we know
from W1 that 8 is OK:

9 is L or 10 is H or 11 is H so
W3: 10 v 11


If 1 2 3 4 are lighter than 5 6 7 8 then either one of 1 2 3 4 is
light or one of 5 6 7 8 is heavy, so:

W2: 1 2 5 v 3 6 9 (9 is a known good coin from W1)

If 1 2 5 are lighter than 3 6 9, then as we know from W1
that a fake 5 has to be a heavy fake, and a fake 3 a light
one, we know that they must be OK, since we've swapped them
from one side to the other, and the right-hand scale pan is
still heavier.

So either 6 is H, or 1 or 2 are L, so W3: 1 v 2
which will find the light fake or confirm 6 as a heavy fake.

If 1 2 5 are heavier than 3 6 9, then because we've swapped
sides with 3 and 5, and the imbalace followed suit, then
either 3 is L or 5 is H, so
W3: either of those v a known good one.

If 1 2 5 v 3 6 9 balances then at W2 we must have removed
the fake coin, i.e. 4 or 7 or 8, and we know from W1 that
that means that 4 is L or 7 is H or 8 is H, so W3: 7 v 8.

If 1 2 3 4 are heavier than 5 6 7 8 then either one of 1 2 3 4 is
heavy or one of 5 6 7 8 is light, and hopefully you can see the
symmetry of what happens then after W2: 1 2 5 v 3 6 9.

[/code:1]


Anyway, in reverse order of finding a solution, the winners are:

3rd: noodlz for a method similar to mine.

2nd: empip

1st: Winner of the BAS Memorial Trophy, for not only being the first with a solution, but also for providing a supremely elegant one which does not require subsequent weighings to depend on the results of earlier ones:











bethrob99.

Cheers, everyone.

 

[noodlz]

Thank you BAS.

Looking at it again, I must admit that bethrob99's solution is mathematically superb...I was trying to get something like this, but making the mental leap from the first weighing to the others is a stroke of pure genius. Hats off to you

As for empip's....which one...the three times edited version or one of the others? (but I'm not bitter )