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The fake coin problem

Blogs > Byo
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1 2 3 4 Next All
Byo
Profile Blog Joined July 2007
Canada209 Posts
April 23 2009 23:59 GMT
#1
here's a more concrete problem compared to the first entry. Though its still more logical than mathamatical.

You are given 12 identical coins, and one weight scale. The weight scale is only able to identify if side A is heavier or lighter than side B. You are able to put any number of coins on either side. Among the 12 identical coins is 1 fake coin, whose weight is different but whither it is heavier or not is unknown. The task is to use the scale the least number of times to determine which coin is fake. You will probably have to name the coins A-L for clarity.

vAltyR
Profile Blog Joined July 2008
United States581 Posts
April 24 2009 00:06 GMT
#2
+ Show Spoiler +
Should be seven times. Weigh two coins at a time, so that's six times. One of those will produce an imbalance. Take one of the coins from the imbalanced result and weigh it against a different coin. If there is an imbalance again, that's the fake coin. If it's balanced, the other coin from the imbalance is the fake.
내 호버크라프트는 장어로 가득 차 있어요
TheYango
Profile Joined September 2008
United States47024 Posts
Last Edited: 2009-04-24 00:09:56
April 24 2009 00:08 GMT
#3
+ Show Spoiler +
Put 6 coins on each side of the scale. One side will be lighter, and one will be heavier. Take all the coins off, and put 3 on each side from the set of 6 coins that were on the lighter side. If the piles weigh the same, then the fake is heavier than normal. If they're different, then the fake is lighter than normal.

If the fake is lighter than normal, take the lighter of the 2 piles you just weighed. Compare any 2 of those 3 on the scale. If they weigh the same, the third one is the fake. If they're different, the lighter one is the fake.

If you found out the fake is heavier than normal, then split the heavier pile from the original weighing into piles of 3 and compare them. Take the heavier of the 2, and weigh any 2 of those 3. If they weigh the same, the third one is the fake. If they're different, the heavier one is the fake.

Thats three weighings if the fake is lighter than normal, and four if the fake is heavier than normal.
Moderator
Dave[9]
Profile Blog Joined October 2003
United States2365 Posts
April 24 2009 00:10 GMT
#4
you otta balance two at a time, so if C1=C2 then they are identical and can be put aside, when Cn does not equal Cm where m=n+1 then you know you have the fake coin.

So at most it will take 7 times.
http://www.teamliquid.net/forum/viewmessage.php?topic_id=104154&currentpage=316#6317
SpiritoftheTunA
Profile Blog Joined August 2006
United States20903 Posts
Last Edited: 2009-04-24 00:16:29
April 24 2009 00:11 GMT
#5
+ Show Spoiler +
ABCDEF vs GHIJKL (1)

one will turn up heavier, say its the abc group

ABC vs DEF (2)

if one doesn't turn up heavier, then weigh other set

GHI vs JKL (3)

one of these will turn up lighter, weigh each of them 1 by 1 against a coin from the ABCDEF set, whichever turns up lighter is correct

J v A (4)

K v A (5)

L v A (6)


EDIT strikeout part was stupid but yeah up until then i had theyango's method

max by this method is 6
posting on liquid sites in current year
vAltyR
Profile Blog Joined July 2008
United States581 Posts
Last Edited: 2009-04-24 00:13:57
April 24 2009 00:12 GMT
#6
On April 24 2009 09:08 TheYango wrote:
+ Show Spoiler +
Put 6 coins on each side of the scale. One side will be lighter, and one will be heavier. Take all the coins off, and put 3 on each side from the set of 6 coins that were on the lighter side. If the piles weigh the same, then the fake is heavier than normal. If they're different, then the fake is lighter than normal.

If the fake is lighter than normal, take the lighter of the 2 piles you just weighed. Compare any 2 of those 3 on the scale. If they weigh the same, the third one is the fake. If they're different, the lighter one is the fake.

If you found out the fake is heavier than normal, then split the heavier pile from the original weighing into piles of 3 and compare them. Take the heavier of the 2, and weigh any 2 of those 3. If they weigh the same, the third one is the fake. If they're different, the heavier one is the fake.

Thats three weighings if the fake is lighter than normal, and four if the fake is heavier than normal.


Ouch. Owned. >_<;

On April 24 2009 09:11 SpiritoftheTunA wrote:
+ Show Spoiler +
ABCDEF vs GHIJKL (1)

one will turn up heavier, say its the abc group

ABC vs DEF (2)

if one doesn't turn up heavier, then weigh other set

GHI vs JKL (3)

one of these will turn up lighter, weigh each of them 1 by 1 against a coin from the ABCDEF set, whichever turns up lighter is correct

J v A (4)

K v A (5)

L v A (6)

max by this method is 6


+ Show Spoiler +
Actually, max is 4. Read TheYango's post to find out why.
내 호버크라프트는 장어로 가득 차 있어요
FirstProbe
Profile Joined June 2004
1206 Posts
Last Edited: 2009-04-24 00:16:00
April 24 2009 00:13 GMT
#7
Break it up so that you're weighing 3 coins against 3 coins.
If both sets weigh the same, then repeat the process. This will leave you with one set of coins weighing more than the other.
Take one of those sets in which there is a discrepancy, and weigh it against another set. If it weighs the same, then the other set is the one with the fake coin. If they do not weight the same, then the fake coin is in the set that you have re-tested.
Following that, take the 3 coins that weighed in lighter, and weigh one against another. You'll either have your answer there, or have concluded that the coin not weighed is the lightest.
This should take you 4 attempts.
edit: or 3 attempts if you are more fortunate.
Byo
Profile Blog Joined July 2007
Canada209 Posts
April 24 2009 00:15 GMT
#8
actually, optimum method can find the result in 3 weighs regardless of heavier or lighter....
goldrush
Profile Blog Joined June 2004
Canada709 Posts
April 24 2009 00:17 GMT
#9
+ Show Spoiler +


Split it up into two groups of six. Weight it. One group will be heavier than the other. (A-F) (G-L) (#1)

Take one group of six. Split it into two groups of three. If they are identical weight, then use the other group of six for the next step. If they are not of identical weight, then use this group for the next step. (#2) Say that (A-C) (D-F) are of identical weight, and (G-I) and (J-L) are not.

Weight (G-I) against (A-C). (#3) If they are equal weight, then use the group (J-L) for the next step. If not, then use the group (G-I) for the next step.

Choose two random coins (say G and H if G-I was used from the previous step) from that group and weigh them from each other. (#4) If they are equal, then the one that remains (I) is the odd one out. If not, then weigh one of them (say, G) against coin A. (#5) If they are equal, then the other one (H) is the odd one out. If they are not, then G is the odd one out.

Steps 4-5 can probably be optimized though. =/


Deleted User 31060
Profile Blog Joined September 2007
3788 Posts
April 24 2009 00:17 GMT
#10
Compare ABCD with EFGH. If same, compare IJ with AB. If same, compare A with K. If same, the coin is L (if different, coin is K)

If IJ is different than AB, compare A with I. If same, the coin is J (if different, coin is I)

If ABCD is different with EFGH, compare ABCD with IJKL. If ABCD and IJKL are the same, compare AB with EF. If same, compare A with G. If same, the coin is H (if different, coin is G)

If EF is different than AB, compare A with E. If same, the coin is F (if different, coin is E)

If ABCD and IJKL are different, compare AB with EF. If same, compare A with E. If same, the coin is B (if different, coin is A)

If EF is different than AB, compare C with E. If same, the coin is D (if different, coin is C)



I think that takes care of all possibilities in 4 tries
Peaked at C- on ICCUP and proud of it! @Sunyveil
Bebop Berserker
Profile Joined April 2009
United States246 Posts
April 24 2009 00:19 GMT
#11

On April 24 2009 09:08 TheYango wrote:
+ Show Spoiler +
Put 6 coins on each side of the scale. One side will be lighter, and one will be heavier. Take all the coins off, and put 3 on each side from the set of 6 coins that were on the lighter side. If the piles weigh the same, then the fake is heavier than normal. If they're different, then the fake is lighter than normal.

If the fake is lighter than normal, take the lighter of the 2 piles you just weighed. Compare any 2 of those 3 on the scale. If they weigh the same, the third one is the fake. If they're different, the lighter one is the fake.

If you found out the fake is heavier than normal, then split the heavier pile from the original weighing into piles of 3 and compare them. Take the heavier of the 2, and weigh any 2 of those 3. If they weigh the same, the third one is the fake. If they're different, the heavier one is the fake.

Thats three weighings if the fake is lighter than normal, and four if the fake is heavier than normal.


Nice. Mine came out to be four/four but this seems right. I was just being stupid.(go figure)
Whatever happens, happens.
TheYango
Profile Joined September 2008
United States47024 Posts
Last Edited: 2009-04-24 00:25:50
April 24 2009 00:22 GMT
#12
On April 24 2009 09:15 Byo wrote:
actually, optimum method can find the result in 3 weighs regardless of heavier or lighter....

How about this:
+ Show Spoiler +

Coins A-L.
Weigh ABCD v EFGH and ABCD v IJKL.
If ABCD is the same as either, the other 4 contain the fake and you know if the fake is lighter/heavier by whether that pile is lighter/heavier (e.g. if ABCD = EFGH and ABCD > IJKL, then IJKL has the fake and the fake is lighter). If ABCD is different, then ABCD contains the fake, and you know if the fake is lighter/heaver by whether ABCD is lighter/heaver than the other two.

Call the pile with the fake now PQRS.
Weigh PQ v RS, choose the one you know has the fake from whether the fake is heavier/lighter as determined above, and weigh those. The fake is lighter/heavier as determined above.

3 weighs every time.
EDIT: wow, I can't count. Will figure out in a bit.
Moderator
SpiritoftheTunA
Profile Blog Joined August 2006
United States20903 Posts
April 24 2009 00:23 GMT
#13
On April 24 2009 09:22 TheYango wrote:
Show nested quote +
On April 24 2009 09:15 Byo wrote:
actually, optimum method can find the result in 3 weighs regardless of heavier or lighter....

How about this:
+ Show Spoiler +

Coins A-L.
Weigh ABCD v EFGH and ABCD v IJKL.
If ABCD is the same as either, the other 4 contain the fake and you know if the fake is lighter/heavier by whether that pile is lighter/heavier (e.g. if ABCD = EFGH and ABCD > IJKL, then IJKL has the fake and the fake is lighter). If ABCD is different, then ABCD contains the fake, and you know if the fake is lighter/heaver by whether ABCD is lighter/heaver than the other two.

Call the pile with the fake now PQRS.
Weigh PQ v RS, choose the one you know has the fake from the fake weight determined above, and weigh those. The fake is lighter/heavier as determined above.

3 weighs every time.

i count 4 man
posting on liquid sites in current year
TheYango
Profile Joined September 2008
United States47024 Posts
April 24 2009 00:24 GMT
#14
On April 24 2009 09:23 SpiritoftheTunA wrote:
Show nested quote +
On April 24 2009 09:22 TheYango wrote:
On April 24 2009 09:15 Byo wrote:
actually, optimum method can find the result in 3 weighs regardless of heavier or lighter....

How about this:
+ Show Spoiler +

Coins A-L.
Weigh ABCD v EFGH and ABCD v IJKL.
If ABCD is the same as either, the other 4 contain the fake and you know if the fake is lighter/heavier by whether that pile is lighter/heavier (e.g. if ABCD = EFGH and ABCD > IJKL, then IJKL has the fake and the fake is lighter). If ABCD is different, then ABCD contains the fake, and you know if the fake is lighter/heaver by whether ABCD is lighter/heaver than the other two.

Call the pile with the fake now PQRS.
Weigh PQ v RS, choose the one you know has the fake from the fake weight determined above, and weigh those. The fake is lighter/heavier as determined above.

3 weighs every time.

i count 4 man

Wow I'm dumb.
Edit in a bit.
Moderator
SpiritoftheTunA
Profile Blog Joined August 2006
United States20903 Posts
April 24 2009 00:25 GMT
#15
you'd probably need to mix up the groups across weighings
posting on liquid sites in current year
goldrush
Profile Blog Joined June 2004
Canada709 Posts
Last Edited: 2009-04-24 00:34:21
April 24 2009 00:29 GMT
#16
Maybe this...

+ Show Spoiler +


(ABCDEF) versus (GHIJKL)
(ABCGHI) versus (DEFJKL)

Find which group of three is heavier or lighter.

Take two from the group of three, measure. If equal, odd one out. If not equal, you know whether the odd one is supposed to be lighter/heavier and that one is it.



edit: nope, nvm
TheYango
Profile Joined September 2008
United States47024 Posts
April 24 2009 00:31 GMT
#17
On April 24 2009 09:29 goldrush wrote:
Second try, I think I got it.

+ Show Spoiler +


(ABCDEF) versus (GHIJKL)
(ABCGHI) versus (DEFJKL)

Find which group of three is heavier or lighter.

Take two from the group of three, measure. If equal, odd one out. If not equal, you know whether the odd one is supposed to be lighter/heavier and that one is it.



edit: or not

Yeah, your first 2 weighings don't guarantee finding which group is heavier.
Moderator
OmgIRok
Profile Blog Joined June 2008
Taiwan2699 Posts
April 24 2009 00:33 GMT
#18


1a) weigh 5 coins on each scale, if there is an imbalance, then move on to step 2a
1b) same as 1a except they both weigh the same
2b) weigh the remaining 2 coins

2a) keep taking out 1 coin from both piles until the weight is the same,
3) weigh the 2 coins that you took out from the pile before it made it even weight
"Wanna join my [combo] clan?" "We play turret d competitively"
TheYango
Profile Joined September 2008
United States47024 Posts
Last Edited: 2009-04-24 00:41:36
April 24 2009 00:35 GMT
#19
On April 24 2009 09:33 OmgIRok wrote:
3) weigh the 2 coins that you took out from the pile before it made it even weight

How does this determine the fake? None of your previous steps ascertain whether the fake is lighter or heavier.

Edit: However, weighing one of those two with a third coin does. Still iffy on whether taking coins out constitutes a separate weighing, but this at least works.
Moderator
SpiritoftheTunA
Profile Blog Joined August 2006
United States20903 Posts
April 24 2009 00:36 GMT
#20
On April 24 2009 09:33 OmgIRok wrote:


1a) weigh 5 coins on each scale, if there is an imbalance, then move on to step 2a
1b) same as 1a except they both weigh the same
2b) weigh the remaining 2 coins

2a) keep taking out 1 coin from both piles until the weight is the same,
3) weigh the 2 coins that you took out from the pile before it made it even weight

yeah the existence of 2a means you can't guarantee < 3

try again mang
posting on liquid sites in current year
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