EPT(ABGH versus CDIJ)
This time, look at the pairs?
ex: ABCDEF is heavier, ABGH is heavier. Then the odd one out is one of AB, measure.
DAMNIT there are always the lighter-lighter options which increases the number of required weighs, doesn't work.
The fake coin problem - Page 2
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goldrush
Canada709 Posts
(ABGH versus CDIJ) This time, look at the pairs? ex: ABCDEF is heavier, ABGH is heavier. Then the odd one out is one of AB, measure. DAMNIT there are always the lighter-lighter options which increases the number of required weighs, doesn't work. | ||
KwarK
United States43675 Posts
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 2b tells you one coin is heavier than the other. Which is the fake? | ||
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SpiritoftheTunA
United States20903 Posts
On April 24 2009 09:37 goldrush wrote: (ABCDEF) versus (GHIJKL) (ABGH versus CDIJ) This time, look at the pairs? ex: ABCDEF is heavier, ABGH is heavier. Then the odd one out is one of AB, measure. in your example, J could be lighter and could be the fake | ||
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Tyxiquale
Australia424 Posts
1. Weigh 3 balls against 3 balls. If they are the same, move to 2a, if they are different, 2b. You now have the information of 6 balls that are normal. 2a) You take 3 balls from one end off the scale, and grab 3 "unidentified" balls from your other pile, and weigh them. If different to 3 normal balls, then it must be in that group of 3. If they are the same, then the mystery ball must be in the final 3 balls. MOST IMPORTANTLY - you know whether the ball is heavier or lighter. 2b) You take 3 of these balls and weigh it against 3 identified balls. If they are the same, the mystery ball must be in the other 3 balls that you weighed in your first go. If not, they are in the pile you just measured. AND you know whether the ball is heavier or lighter. 3. Regardless of which method you have taken, you have now identified 3 balls, of which 1 of them is the mystery ball. AND most importantly, the information of whether it is heavier or lighter. As such, the final weigh would be 1 ball against another in your group of 3. The rest is trivial. That’s it folks. The answer is 3 regardless of whether it is heavier or lighter. Amen. | ||
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Crunchums
United States11144 Posts
I can get a worst case of 4 with the second one: + Show Spoiler + Add one coin to each side, then check if each side weighs the same. Repeat this until the two sides don't weigh the same or until you weigh 5 coins against 5 coins and they still weigh the same. Once that happens, set aside the two most recently added coins or the two remaining coins if there are only two left and remove all the coins from the scale. Pick one of the two coins you set aside and weigh it against another coin that's not the other coin you set aside. If these two coins weigh the same, the fake coin is the other coin you set aside. If they do not it is the coin that you had set aside that is now on the weight scale. (This algorithm will also work with the same efficiency if you instead start with all the coins on the scale and remove them two at a time, one from each side, until it changes from not the same weight to the same weight and not measure if there are two left.) # of Measurements: Maximum - 6 Minimum - 2 Average - 3.3333... Put 3 coins on side A and 3 coins on side B and measure. If the two sides are the same weight: Add one coin to each side and measure. If the two sides are still the same weight, repeat this step unless there are now 10 coins on the scale, in which case set aside the 2 remaining coins. If the two sides are no longer the same weight, set aside the two coins you just added to each side. If the two sides are not the same weight: Remove one coin from each side and measure. If the two sides are still not the same weight, repeat this step unless there are now 2 coins on the scale, in which case set those 2 coins aside. If the two sides are now the same weight, set aside the two coins you just removed from each side. Remove all coins from the scale. Pick one of the two coins you set aside and weigh it against another coin that's not the other coin you set aside. If these two coins weigh the same, the fake coin is the other coin you set aside. If they do not it is the coin that you had set aside that is now on the weight scale. # of Measurements: Maximum - 4 Minimum - 3 Average - complicated | ||
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SpiritoftheTunA
United States20903 Posts
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TheYango
United States47024 Posts
On April 24 2009 09:55 Tyxiquale wrote: 2a) You take 3 balls from one end off the scale, and grab 3 "unidentified" balls from your other pile, and weigh them. If different to 3 normal balls, then it must be in that group of 3. If they are the same, then the mystery ball must be in the final 3 balls. MOST IMPORTANTLY - you know whether the ball is heavier or lighter. Uhh... HOW? ABC = DEF ABC = GHI How does that tell you whether the fake is lighter or heavier? | ||
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Byo
Canada210 Posts
in the optimum solution, there is a case where the fake coin is not weigh'd | ||
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SpiritoftheTunA
United States20903 Posts
On April 24 2009 09:58 TheYango wrote: Uhh... HOW? ABC = DEF ABC = GHI How does that tell you whether the fake is lighter or heavier? that flaw too what a joke lol | ||
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Tyxiquale
Australia424 Posts
On April 24 2009 09:58 SpiritoftheTunA wrote: tyxiquale, use ABC DEF GHI JKL notation please, i think there's a flaw in your 2b but it's too ambiguous for me to evaluate Sorry for not using that notation, cos I cbf.. but there is no flaw. re-read it again. Read between the lines of what I skipped cos I thought it was too obvious. | ||
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SpiritoftheTunA
United States20903 Posts
On April 24 2009 10:00 Tyxiquale wrote: Sorry for not using that notation, cos I cbf.. but there is no flaw. re-read it again. Read between the lines of what I skipped cos I thought it was too obvious. your signature is especially ironic, even more so since you used "their" instead of "they're" | ||
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TheYango
United States47024 Posts
On April 24 2009 10:00 Tyxiquale wrote: Sorry for not using that notation, cos I cbf.. but there is no flaw. re-read it again. Read between the lines of what I skipped cos I thought it was too obvious. Apparently not, because Byo would have recognized your answer as correct if it were. -_- You still haven't addressed the situation I raised. | ||
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ghostWriter
United States3302 Posts
On April 24 2009 10:01 SpiritoftheTunA wrote: your signature is especially ironic, even more so since you used "their" instead of "they're" I'm pretty sure it was done intentionally so that some high school student with a misguided sense of superiority can try to call him out on it. | ||
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sigma_x
Australia285 Posts
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TheYango
United States47024 Posts
On April 24 2009 10:05 sigma_x wrote: tyx is correct. The final weight of 1 ball against any of the other (two in your group) determines the solution. Ok, hold on. I'll carry out an example using his procedure. I follow it as well as i can. Balls are labeled A-L. Suppose you weigh ABC against DEF. They are the same. Step 2a says you then take GHI and weigh it against ABC. They are also the same. This says that the fake is in JKL. However, you don't know whether the fake is heavier or lighter because you've weighed 9 things that are all the same. Step 3 says weigh J with K. They are different. You still don't know which is fake without making another measurement. | ||
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Crunchums
United States11144 Posts
edit: nevermind he's just got another way to get 4 | ||
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TheYango
United States47024 Posts
On April 24 2009 10:09 Crunchums wrote: tyx is correct he just didn't explain very well Then explain it better, or tell me how the situation I raised is resolvable. I followed his steps, and in a worst-case scenario, it requires one more measurement. For God's sake, people, if someone doesn't understand, then explain WHY they're wrong, instead of just saying "tyx is correct, read more carefully." | ||
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SpiritoftheTunA
United States20903 Posts
On April 24 2009 10:04 ghostWriter wrote: I'm pretty sure it was done intentionally so that some high school student with a misguided sense of superiority can try to call him out on it. it's not much of a trap like affect/effect would be why are you defending this clown | ||
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Byo
Canada210 Posts
if weight 1 is same, and 2a is same....... and in 3, the weight are different, then which ball is the fake one? Edit: hint: + Show Spoiler + its significant whither A is > B or A is < B | ||
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Gliche
United States811 Posts
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