Riddler: find the fake coins

[TheOriginalHappyGoat]

The 538 Riddler this week is a good one. The easy version is this:

You have nine gold coins, but one isn’t pure. One has been minted with a cheap alloy, and is known to be heavier than the others. You have a simple balance scale. How do you determine the impure coin with only two weighings?​

If you think about it, you should get that one without too much trouble. But the hard version is a lot trickier:

You have 12 gold coins — or so you think! One is fake and is known to have a different weight than the others. It could be heavier or lighter; you only know it’s wrong. Using the same simple balance scale, how do you determine the incorrect coin, and whether it is heavier or lighter, in only three weighings?​

I thought I had it, but as I was typing the answer, I realized that I needed four weighings to be 100% sure. I still haven't figured it out, and I've been bouncing it around the ol' noggin for a while.

Any takers?

 

[toastedbread]

The 538 Riddler this week is a good one. The easy version is this:

You have nine gold coins, but one isn’t pure. One has been minted with a cheap alloy, and is known to be heavier than the others. You have a simple balance scale. How do you determine the impure coin with only two weighings?​

If you think about it, you should get that one without too much trouble. But the hard version is a lot trickier:

You have 12 gold coins — or so you think! One is fake and is known to have a different weight than the others. It could be heavier or lighter; you only know it’s wrong. Using the same simple balance scale, how do you determine the incorrect coin, and whether it is heavier or lighter, in only three weighings?​

I thought I had it, but as I was typing the answer, I realized that I needed four weighings to be 100% sure. I still haven't figured it out, and I've been bouncing it around the ol' noggin for a while.

Any takers?

Weigh 3 and 3. If the piles have different weight you know the coin is in one of the 2 piles. Then weigh the other 2 piles. Whichever pile is a different weight take 2 of those 3 coins and weigh them. If weights are equal it's the 3rd coin. If weights are not equal it's whichever coin is lighter/heavier depending upon whether the initial pile was lighter or heavier.

 

[tt68]

For the 2nd one...put them into three piles of 4. Weight two of them, which will tell you which of the 3 piles has the heavy coin. From that pile pick 2 and weigh them. If they are the same weigh the other 2 to determine which is heavier.

 

[TheOriginalHappyGoat]

For the 2nd one...put them into three piles of 4. Weight two of them, which will tell you which of the 3 piles has the heavy coin. From that pile pick 2 and weigh them. If they are the same weigh the other 2 to determine which is heavier.

Doesn't work, because the 2nd puzzle specifically says you don't know if the fake coin is heavier or lighter.

 

[TheOriginalHappyGoat]

Weigh 3 and 3. If the piles have different weight you know the coin is in one of the 2 piles. Then weigh the other 2 piles. Whichever pile is a different weight take 2 of those 3 coins and weigh them. If weights are equal it's the 3rd coin. If weights are not equal it's whichever coin is lighter/heavier depending upon whether the initial pile was lighter or heavier.

Your first weighing doesn't determine if the heavier or lighter is the culprit. You need a fourth weighing for that.

 

[toastedbread]

Your first weighing doesn't determine if the heavier or lighter is the culprit. You need a fourth weighing for that.

My second one does. Weigh 3 and 3, then another 3 and 3. Then we know which pile of 3 is the odd one out and will know if it weighs more or less.

 

[TheOriginalHappyGoat]

My second one does. Weigh 3 and 3, then another 3 and 3. Then we know which pile of 3 is the odd one out and will know if it weighs more or less.

No, we don't. We only know the odd one out is one of two. But unless you compare it to one of the other stacks of 3, you don't know if it's the heavy one or the light one (because you can't draw any assumptions about how the first two groups compare to the second two groups - it's a simple balance scale).

Your solution was very similar to mine, but it failed for the same reason. I got it down to where I could determine light or heavy in 6 out of 7 cases, but in that 7th case, I still needed a fourth weighing.

 

[SuperHoosierFan]

The 538 Riddler this week is a good one. The easy version is this:

You have nine gold coins, but one isn’t pure. One has been minted with a cheap alloy, and is known to be heavier than the others. You have a simple balance scale. How do you determine the impure coin with only two weighings?​

If you think about it, you should get that one without too much trouble. But the hard version is a lot trickier:

You have 12 gold coins — or so you think! One is fake and is known to have a different weight than the others. It could be heavier or lighter; you only know it’s wrong. Using the same simple balance scale, how do you determine the incorrect coin, and whether it is heavier or lighter, in only three weighings?​

I thought I had it, but as I was typing the answer, I realized that I needed four weighings to be 100% sure. I still haven't figured it out, and I've been bouncing it around the ol' noggin for a while.

Any takers?

I'm thinking three groups of four. Weigh two of the groups. You could get lucky and they are even, meaning the fake is one of the four left. If they are uneven you know the fake is on the scale somewhere so switch one group of four and see if the scale is still balanced or not. This is as far as I'm thinking right now as I've had a four pack of HI-RES from SixPoint and they are 10.5%. Now I'm waiting to sober up and take the coins to a jeweler.

 

[TheOriginalHappyGoat]

@toastedbread the key is this. We need to get it down to where, for the last weighing, we have:

1. Only three coins left.
2. We know if the bad coin is heavy or light.

So we need to get to that point in two weighings. I feel like the trick will involve three groups of four, but I still can't make it happen.

 

[TheOriginalHappyGoat]

I'm thinking three groups of four. Weigh two of the groups. You could get lucky and they are even, meaning the fake is one of the four left. If they are uneven you know the fake is on the scale somewhere so switch one group of four and see if the scale is still balanced or not. This is as far as I'm thinking right now as I've had a four pack of HI-RES from SixPoint and they are 10.5%. Now I'm waiting to sober up and take the coins to a jeweler.

Yeah, I thought of that one, too, but you end up with 4 coins left and only one weighing. No way to narrow 4 to 1 with only one weighing left.

 

[SuperHoosierFan]

Yeah, I thought of that one, too, but you end up with 4 coins left and only one weighing. No way to narrow 4 to 1 with only one weighing left.

I'm too drunk to type it out, but if the scales were uneven with the initial 4-4 weighing, I could do it if I marked the coins and switched certain coins from side to side twice. I could do it in one switch if the scale was even and I knew it was one of the other four.

 

[toastedbread]

@toastedbread the key is this. We need to get it down to where, for the last weighing, we have:

1. Only three coins left.
2. We know if the bad coin is heavy or light.

So we need to get to that point in two weighings. I feel like the trick will involve three groups of four, but I still can't make it happen.

My guess is you start with 6 and 6 and somehow remove coins. Not sure though.

 

[TheOriginalHappyGoat]

I'm too drunk to type it out, but if the scales were uneven with the initial 4-4 weighing, I could do it if I marked the coins and switched certain coins from side to side twice. I could do it in one switch if the scale was even and I knew it was one of the other four.

I know where you're going, but if you switch coins side to side twice, you've run out of weighings, and still have four other coins you haven't touched.

There's got to be another solution to this.

 

[SuperHoosierFan]

I know where you're going, but if you switch coins side to side twice, you've run out of weighings, and still have four other coins you haven't touched.

There's got to be another solution to this.

The other four coins don't matter if the initial weighing is unbalanced. On the other hand, only the other four matter if the initial weighing is even.

 

[TheOriginalHappyGoat]

The other four coins don't matter if the initial weighing is unbalanced. On the other hand, only the other four matter if the initial weighing is even.

That's the problem. There has to be a solution that works in ALL cases.

The best I could do was starting with 4 groups of 3. Weigh A to B, and then A to C. Out of the seven possibilities, six of them tell you whether the bad coin is light or heavy, and you always end up with only 3 coins left. Unfortunately, that seventh case - in which both weighings are even - gives you three coins, but no information on whether the bad one is light or heavy. Since it doesn't work in all cases, it can't be the right answer.

 

[SuperHoosierFan]

That's the problem. There has to be a solution that works in ALL cases.

The best I could do was starting with 4 groups of 3. Weigh A to B, and then A to C. Out of the seven possibilities, six of them tell you whether the bad coin is light or heavy, and you always end up with only 3 coins left. Unfortunately, that seventh case - in which both weighings are even - gives you three coins, but no information on whether the bad one is light or heavy. Since it doesn't work in all cases, it can't be the right answer.

I'm not thinking "light" or "heavy", just "different" with the initial weighing. With three groups of four coins you will eliminate a minimum of four coins from the get go. You could get lucky and eliminate eight immediately. You just need to know if the "different" coin is on the scale or the table and the initial weighing will tell you. Sunuvabich, I'm sobering up

 

[TheOriginalHappyGoat]

I'm not thinking "light" or "heavy", just "different" with the initial weighing. With three groups of four coins you will eliminate a minimum of four coins from the get go. You could get lucky and eliminate eight immediately. You just need to know if the "different" coin is on the scale or the table and the initial weighing will tell you. Sunuvabich, I'm sobering up

I know, but it's still not good enough, because if you don't get lucky - if you only eliminate four coins - then you end up in a situation where you need a fourth weighing to be sure.

There has to be an answer that works in ALL situations.

Luckily we have until next Friday morning to figure it out.

 

[SuperHoosierFan]

I know, but it's still not good enough, because if you don't get lucky - if you only eliminate four coins - then you end up in a situation where you need a fourth weighing to be sure.

There has to be an answer that works in ALL situations.

Luckily we have until next Friday morning to figure it out.

Now that I think about it more, you'd need the four on the table almost as a control group if the initial weighing is unbalanced.

The initial weighing has to be three groups of four. I'm not budging on that one.

 

[TheOriginalHappyGoat]

Now that I think about it more, you'd need the four on the table almost as a control group if the initial weighing is unbalanced.

The initial weighing has to be three groups of four. I'm not budging on that one.

I've been over every iteration of that, and I can't make it work. Something's missing. My gut tells me you're on the right track, but there has to be a trick with the second weighing we haven't thought of.

 

[SuperHoosierFan]

I've been over every iteration of that, and I can't make it work. Something's missing. My gut tells me you're on the right track, but there has to be a trick with the second weighing we haven't thought of.

I've got the easy scenario figured out. You have to distinguish the coins somehow first. This is IU, so color four coins red, four white, and leave four plain.

Weigh the red vs white. If they balance you know they are all real coins so you also know one of the four plain coins if fake. Weigh two known good coins vs two potential fake. If they balance, just weigh a known good one vs either of the other not on the scale. If the scale moves, just pick the opposite of the known good one. If the scale remains balanced you know it's the one you didn't weigh.

It's when the initial weighing is unbalanced is where it gets tricky.

 

[TheOriginalHappyGoat]

I've got the easy scenario figured out. You have to distinguish the coins somehow first. This is IU, so color four coins red, four white, and leave four plain.

Weigh the red vs white. If they balance you know they are all real coins so you also know one of the four plain coins if fake. Weigh two known good coins vs two potential fake. If they balance, just weigh a known good one vs either of the other not on the scale. If the scale moves, just pick the opposite of the known good one. If the scale remains balanced you know it's the one you didn't weigh.

It's when the initial weighing is unbalanced is where it gets tricky.

No, you're overcomplicating that. The first one is easy:

Spoiler: Answer inside

Split the coins into three groups of three. Pick two at random, and weigh them.

If they are equal, the bad coin is in the group you didn't weigh. If they are unequal, the bad coin is in the heavier group. Either weigh, with one weighing, you've narrowed it down to three coins.

Take two of those three coins, and weigh them. If they are equal, the bad coin is the last coin left over. If they are unequal, the bad coin is the heavier of the two on the scales.

Easy. Two weighings, and you found the coin. The second scenario is trickier because you can't assume the bad coin is heavier. That's what messes it all up.

 

[SuperHoosierFan]

No, you're overcomplicating that. The first one is easy:

Spoiler: Answer inside

Split the coins into three groups of three. Pick two at random, and weigh them.

If they are equal, the bad coin is in the group you didn't weigh. If they are unequal, the bad coin is in the heavier group. Either weigh, with one weighing, you've narrowed it down to three coins.

Take two of those three coins, and weigh them. If they are equal, the bad coin is the last coin left over. If they are unequal, the bad coin is the heavier of the two on the scales.

Easy. Two weighings, and you found the coin. The second scenario is trickier because you can't assume the bad coin is heavier. That's what messes it all up.

I thought you didn't know if the fake was heavy or light. This scenario would seem to imply you already know the fake is heavy.

 

[TheOriginalHappyGoat]

I thought you didn't know if the fake was heavy or light. This scenario would seem to imply you already know the fake is heavy.

In the first one you do. Don't mix up the two scenarios. That's the key difference that makes the first one easy and the second one difficult.

 

[SuperHoosierFan]

In the first one you do. Don't mix up the two scenarios. That's the key difference that makes the first one easy and the second one difficult.

Oh, well mine is for if you don't know.

Even if I knew the fake was heavy or light, I'd still weigh four vs four and do it in three tries.

It's when you don't know if the fake is heavy or light and the initial weighing is uneven that I'm having trouble with.

 

[TheOriginalHappyGoat]

Oh, well mine is for if you don't know.

Even if I knew the fake was heavy or light, I'd still weigh four vs four and do it in three tries.

It's when you don't know if the fake is heavy or light and the initial weighing is uneven that I'm having trouble with.

In the second puzzle, yes, if you know heavy or light, it's easy to do in three. That's the point. That's what makes it such a difficult puzzle. Not knowing going in, I can't guarantee it in less than four weighings. Like I said, I can get it in three 6 out of 7 times, but to answer the puzzle, you need it to be 100%. That's why it's such a tricky puzzle.

Again, we still have a week to figure it out.

 

[Marvin the Martian]

All weightings are of 4 coins. The trick is each coin never shares the side of the scale with the same coin. So if you weigh 1,2,3, and 4 on the left side to start, 1 can never be on the same side as 2, 3, 4 again.

Left. Right
1234. 5678
159a 627b
16a0. 5b49

Coins being weighed once are 3, 8, 0. If they are off, their side will be lighter/heavier. The two weightings without the will be equal.

1, 5, 6 are weighed 3 times. Their side will always be off and either heavier or lighter.

All other coins are weighed twice. In the weighing without them, if they are off the sides will be equal in the weighting they are not in.

Note, I didn't double check the distribution above to make sure it is perfect. It may be I have to slightly rearrange the coins. But the concept is accurate, three coins are weighed 3 times, 3 once, and 6 twice. The counselor weighed twice cannot share/oppose sides with the same coin twice unless it is a 3 measure coin.

 

[TheOriginalHappyGoat]

All weightings are of 4 coins. The trick is each coin never shares the side of the scale with the same coin. So if you weigh 1,2,3, and 4 on the left side to start, 1 can never be on the same side as 2, 3, 4 again.

Left. Right
1234. 5678
159a 627b
16a0. 5b49

Coins being weighed once are 3, 8, 0. If they are off, their side will be lighter/heavier. The two weightings without the will be equal.

1, 5, 6 are weighed 3 times. Their side will always be off and either heavier or lighter.

All other coins are weighed twice. In the weighing without them, if they are off the sides will be equal in the weighting they are not in.

Note, I didn't double check the distribution above to make sure it is perfect. It may be I have to slightly rearrange the coins. But the concept is accurate, three coins are weighed 3 times, 3 once, and 6 twice. The counselor weighed twice cannot share/oppose sides with the same coin twice unless it is a 3 measure coin.

I think you have it, but you're right that your distribution needs changed. Two coins only weighed once can't go at the same time. In your case, if either 3 or 8 is the bad coin, you won't know which one.

But I'd bet money you've got the concept figured out.

 

[Marvin the Martian]

I think you have it, but you're right that your distribution needs changed. Two coins only weighed once can't go at the same time. In your case, if either 3 or 8 is the bad coin, you won't know which one.

But I'd bet money you've got the concept figured out.

I had the distribution figured out when a friend and I worked it out. He has the paper we got to work. But you are right why the above attempt fails. I imagine someone figured it out quickly, we sure didn't. I do believe the first puzzle's solution makes the second puzzle harder.

 

[TheOriginalHappyGoat]

I had the distribution figured out when a friend and I worked it out. He has the paper we got to work. But you are right why the above attempt fails. I imagine someone figured it out quickly, we sure didn't. I do believe the first puzzle's solution makes the second puzzle harder.

Because it serves as a little misdirection?

 

[Marvin the Martian]

Because it serves as a little misdirection?

Yes

 

[i'vegotwinners]

puzzle 2

3 stacks of 4 coins each.

weigh stacks 1 & 2 against each other.

if the same, then stack 3 has the ringer, and you have 2 more weighs and only 4 coins.

if stacks 1 & 2 come out unequal,

remove 2 coins from stack 1, and 1 coin from stack 2, and set those 3 coins aside.

stack 1 now has 2 coins, stack 2 now has 3 coins. (5 coins total).

move 2 coins from stack 2 to stack 1, and 1 original coin from stack 1 to stack 2. (you've flipped 3 coins, 2 coins are still in their original stack).

stack 1 now has 3 coins, and stack 2 has 2 coins.

now take 1 coin from stack 3, (which you now know isn't the ringer, as none from stack 3 are the ringer) and put it in stack 2.

you now have 2 stacks of 3 coins each.

if the 2 stacks are now balanced, then the ringer is one of the 3 you just removed and set aside.

if the stacks are still unbalanced, but the imbalance is flipped from one stack to the other, then it's one of the 3 coins you flipped from one stack to the other.

if the 2 stacks are still unbalanced, and the imbalance remains in favor of your original measurement, then it's one of the 2 coins still in their original stack.

regardless, you've now narrowed it down to the 3 coins you removed and set aside, the 3 coins you flipped from one stack to the other, or the 2 coins still in their original stack.

if down to 3 coins, then weigh 2 of the 3 against each other.

if down to 2 coins, weigh one of them against 1 from stack 3.

bingo.

 

[TheOriginalHappyGoat]

puzzle 2

3 stacks of 4 coins each.

weigh stacks 1 & 2 against each other.

if the same, then stack 3 has the ringer, and you have 2 more weighs and only 4 coins.

if stacks 1 & 2 come out unequal,

remove 2 coins from stack 1, and 1 coin from stack 2, and set those 3 coins aside.

stack 1 now has 2 coins, stack 2 now has 3 coins. (5 coins total).

move 2 coins from stack 2 to stack 1, and 1 original coin from stack 1 to stack 2. (you've flipped 3 coins, 2 coins are still in their original stack).

stack 1 now has 3 coins, and stack 2 has 2 coins.

now take 1 coin from stack 3, (which you now know isn't the ringer, as none from stack 3 are the ringer) and put it in stack 2.

you now have 2 stacks of 3 coins each.

if the 2 stacks are now balanced, then the ringer is one of the 3 you just removed and set aside.

if the stacks are still unbalanced, but the imbalance is flipped from one stack to the other, then it's one of the 3 coins you flipped from one stack to the other.

if the 2 stacks are still unbalanced, and the imbalance remains in favor of your original measurement, then it's one of the 3 coins still in their original stack.

regardless, you've now narrowed it down to the 3 coins you removed and set aside, the 3 coins you flipped from one stack to the other, or the 3 coins still in their original stack.

then weigh 2 of the 3. bingo.

Nice try, but that doesn't work in 100% of cases. If you end up with the fake coin in the stack of 3 that haven't been weighed yet, you can't just weigh two coins and be done, because if they are uneven, you won't know if the culprit is the heavy or the light one.

I think Marvin's answer checks out.

 

[i'vegotwinners]

Nice try, but that doesn't work in 100% of cases. If you end up with the fake coin in the stack of 3 that haven't been weighed yet, you can't just weigh two coins and be done, because if they are uneven, you won't know if the culprit is the heavy or the light one.

I think Marvin's answer checks out.

got it

 

[i'vegotwinners]

as opposed to saying remove 2 coins from stack 1, i should have said remove 2 coins from the heavy stack, and one coin from the light stack, and set them aside.

if it gets down to those 3 you set aside, then in your final weigh, weigh 1 of the 2 you removed from the heavy side, and the one you removed from the lighter side, against 2 from stack 3.

if the 2 from stack 3 is heavier, then it's the one you removed from the original lighter stack.

if the 2 from stack 3 is lighter, then it's the one you just paired from the original heavy stack.

if the two are balanced, then it's the one you removed from the heavy side, but didn't include in the 3rd weigh.


if it's down to the three you flipped from one stack to the other, (say you flipped 2 from the heavy side to the light side, and one from the light side to the heavy), then take one you flipped from the heavy side and the one you flipped from the light side, and weigh it against two coins from stack 3. (which we'll now call the control stack). (same concept as we just did with the 3 set asides).

if it's heavier than the 2 from stack 3, then it's the one you just weighed of the 2 flipped from the original heavy stack.

if lighter, then it's the one you flipped from the original light stack.

if equal, then it's the one you flipped from the original heavy stack, that didn't just get weighed against the 2 from stack 3.

back to BINGO.

don't think you can get through all of Marv's 16 needed combinations in 3 weighs.

 

[TheOriginalHappyGoat]

as opposed to saying remove 2 coins from stack 1, i should have said remove 2 coins from the heavy stack, and one coin from the light stack, and set them aside.

if it gets down to those 3 you set aside, then in your final weigh, weigh 1 of the 2 you removed from the heavy side, and the one you removed from the lighter side, against 2 from stack 3.

if the 2 from stack 3 is heavier, then it's the one you removed from the original lighter stack.

if the 2 from stack 3 is lighter, then it's the one you just paired from the original heavy stack.

if the two are balanced, then it's the one you removed from the heavy side, but didn't include in the 3rd weigh.


if it's down to the three you flipped from one stack to the other, (say you flipped 2 from the heavy side to the light side, and one from the light side to the heavy), then take one you flipped from the heavy side and the one you flipped from the light side, and weigh it against two coins from stack 3. (which we'll now call the control stack). (same concept as we just did with the 3 set asides).

if it's heavier than the 2 from stack 3, then it's the one you just weighed of the 2 flipped from the original heavy stack.

if lighter, then it's the one you flipped from the original light stack.

if equal, then it's the one you flipped from the original heavy stack, that didn't just get weighed against the 2 from stack 3.

back to BINGO.

brain too fried to do the math, but i question if you can get through all of Marv's needed combinations in 3 weighs.

No, you're solution still doesn't help if the bad coin turns out to be one of the three you never weighed in the first two rounds. If that's the case, you can't find the culprit with only one more weighing.

Marv's checks out. As long as he can arrange the coins such that:
1. A maximum of three coins get weight only once, and never at the same time.
2. Any coin weighed twice must have unique partners and unique opponents in each weighing, unless the non-unique partner/opponent also has a third weighing.

He'll end up with a unique solution regardless of which coin is the culprit.

 

[i'vegotwinners]

No, you're solution still doesn't help if the bad coin turns out to be one of the three you never weighed in the first two rounds. If that's the case, you can't find the culprit with only one more weighing.

Marv's checks out. As long as he can arrange the coins such that:
1. A maximum of three coins get weight only once, and never at the same time.
2. Any coin weighed twice must have unique partners and unique opponents in each weighing, unless the non-unique partner/opponent also has a third weighing.

He'll end up with a unique solution regardless of which coin is the culprit.

no idea what 3 coins you're talking about.

if you're talking about the FOUR coins that didn't get weighed in the original weigh, (stack 3), guess i thought i didn't need to cover that in depth. (4 coins with 2 weighs left).


that said, if the ringer was one of the original four in stack 3, (which means all the coins in stacks 1 and 2 are legit), then take 3 coins of the 4 from stack 3, and weigh them against 3 from stacks 1 or 2. (stacks 1 and 2 now become control stacks).

if those 3 weigh the same as the 3 from the control group, then the 1 from stack 3 that i didn't just weigh again, is the ringer.

if the 3 from stack 3 is heavier than the control stack, then it's one of those 3, and the ringer is heavier than the others.

if lighter, then the ringer is lighter than the real coins.

i then take 2 of those 3, and weigh them against each other, already knowing if the ringer is heavier or lighter than the real coins.

if those 2 weigh equal, then it's the 3rd coin i didn't just weigh.

if one is heavier or lighter than the other, then i also know which is the ringer, since i now already know if the ringer is heavier or lighter than the real coins.





all my bases have been covered, problem solved.

 

[TheOriginalHappyGoat]

no idea what 3 coins you're talking about.

if you're talking about the FOUR coins that didn't get weighed in the original weigh, (stack 3), i have already covered that.

all my bases have been covered, problem solved.

Sorry, you're skipping a lot of steps with your "BINGO." I got confused. Either way, you never explained how you find the bad coin if you know it's in stack 3.

I'll have to think about your heavy-light flipping some more. That might work, but you need to show your work for how to resolve the problem in cases where the bad coin ends up in stack 3. (EDIT: Actually, I think I have that part worked out, but I'm too busy to sit down and pencil-and-paper all this right now.)

 

[i'vegotwinners]

Sorry, you're skipping a lot of steps with your "BINGO." I got confused. Either way, you never explained how you find the bad coin if you know it's in stack 3.

I'll have to think about your heavy-light flipping some more. That might work, but you need to show your work for how to resolve the problem in cases where the bad coin ends up in stack 3. (EDIT: Actually, I think I have that part worked out, but I'm too busy to sit down and pencil-and-paper all this right now.)

already done, see my edit above.

BINGO

 

[iu_a_att]

The 538 Riddler this week is a good one. The easy version is this:

You have nine gold coins, but one isn’t pure. One has been minted with a cheap alloy, and is known to be heavier than the others. You have a simple balance scale. How do you determine the impure coin with only two weighings?​

If you think about it, you should get that one without too much trouble. But the hard version is a lot trickier:

You have 12 gold coins — or so you think! One is fake and is known to have a different weight than the others. It could be heavier or lighter; you only know it’s wrong. Using the same simple balance scale, how do you determine the incorrect coin, and whether it is heavier or lighter, in only three weighings?​

I thought I had it, but as I was typing the answer, I realized that I needed four weighings to be 100% sure. I still haven't figured it out, and I've been bouncing it around the ol' noggin for a while.

Any takers?

The first one is easy. Split into three sets of three. Set one of the three aside and weigh the remaining two sets. If one is heavier than the other then put the other coins aside and focus on those three. If both set of three you weigh are the same then focus on the set you didn't weigh. For the three coins you are focusing on take one coin out of that set and weigh the other two. If the two weigh the same then the third coin is the fake. if one is heavier then it is the fake.

12 coins. (Assumes I can get ratios from the balance (which should be doable but the procedure might be more involved) Each time I take a weighing I put the number in brackets [] e.g., [2] means it is the second weighing.

Put 6 on (L)eft and 3 on (R)ight set 3 (A)side.
[!.] Weigh L against R
Three cases.
I. L weighs exactly 2xR
II. L is heavier than 2xR
III. L is lighter than 2xR

I. Fake is known to be in A.
[2] weigh A against R. if A heavier (lighter) than R then fake is heavier (lighter)
[3] weigh two of the three coins in A. if one is heavier (lighter) it is the fake. If they are the same then the fake is the unweighed coin is the heavier (lighter) fake.

II. Fake is known to be in the 9. Set aside R and
[2]. Split L into two sets of three and weigh
i. if the sets are not equal the heavier set must contain the fake which is heavier
ii. the sets are equal in which case the set R contains the fake and it is lighter.
In either case we can determine the fake using the information from i or ii and using the same procedure as in [3] with one weighing as above.
Case III. Is analogous to case II.

 

[TheOriginalHappyGoat]

already done, see my edit above.

BINGO

Jesus Christ dude, you need to learn how to show your work.

FWIW, I've just been through your permutations, and I think you might have a legit alternate solution here, but you are skipping a lot of steps when trying to explain it.