Simple balance scale means you can't figure ratios. You just know which side is heavier, that's it.
Riddler: find the fake coins
- Thread starter TheOriginalHappyGoat
- Start date
sorry dude, but it's all there.
as for how i know if the ringer is heavier or lighter, i just went through that.
if we know the ringer is in stack 3, then we know all of stacks 1 and 2 are legit, and become "control" coins.
so when we weigh 3 from stack 3 vs 3 from the control group, they are either equal, or one of the 3 from stack 3 is heavier or lighter.
if the 3 from stack 3 is heavier than the 3 from the control group, then the ringer is heavier than the real coins. (and vice versa).
as for "alt solution", why is my solution an alt solution?
An alternative to Marvin's, which also works. Doesn't mean your solution is lesser, just that it's different.
I like your solution. The key was when you added the specification that you were taking two from the heavy and one from the light (or the other way around, as long as it's specific), rather than A or B. That's what got over that last hurdle of possibly ending up with a bad coin, but not knowing if it's lighter or heavier.
Suppose 1234=5678 so you know the fake is 90ab. It follows that if 159a=627b then we know that 0 is the fake and we can weigh it against any other coin to find out if it is heavier or lighter. But what if 159a>627b? Then we know that the fake is in 9ab but we don't know if it lighter or heavier. We only have 1 weighing left so we can't figure it out. So...I don't think your procedure works in all cases.
Marvin's layout is incorrect, because he was trying to recreate it from memory, but his process works. Switch the 8 for the first b in his layout, and I think that clears it up.
Ok Ivegotwinners does it.
3 stacks A,B, C of 4 coins each.
Let me do an example to follow along in the case of inequality.
if stacks A & B come out unequal,
Let’s say stack A=1234 > 5678=Stack B and thus we know fake is either heavy in A or light in B
remove 2 coins from stack A, and 1 coin from stack B, and set those 3 coins aside.
Ok, remove 34 from stack A=12 and remove 8 from B=567. Call OA=34 original stack A and OB=8 original stack B
stack A now has 2 coins, stack B now has 3 coins. (5 coins total).
yes
move 2 coins from stack B to stack A,
I move 56 from B to A so A=1256 and B=7
move 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.
Okay I move 2 to B so A=156 and B=27…3 coins in A 2 coins in B as you say.
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.
Stack C=09ab. I take 0 from C and let B=270. So A=156 and B=270
you now have 2 stacks of 3 coins each.
Correct. Now I do my second weighing of A and B.
if the 2 stacks are now balanced, then the ringer is one of the 3 you just removed and set aside.
Let’s consider the case that A=B…that leave me knowing that the fake is in 348 and either heavy if in 34 or light if in 8. But I only have one weighing to go. So I can weigh 38 and 56. If 38 is heavier then I know fake is 3 and heavy. If it is lighter then I know 8 is the fake and lighter. If equal then I know 4 is the fake and heavier. Check.
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.
Ok A=156<270=B. Fake is not 1 because it 1 is fake only if heavier, similarly not 7 because 7 is lighter. So fake is either 5 or 6 and lighter or 2 and heavier. So, as above, weigh 25 against fair coins 78. If lighter then fake is 5 , if heavier then fake is 2, and if equal fake is 6 and lighter. Check
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.
Ok A=156>270=B. Fake is either 1 and heavy or 7 and light. Weigh 1 against 2. If 1 is heavier then 1 is the heavy fake. If it is equal then 7 is the fake and light. Check
3 stacks A,B, C of 4 coins each.
Let me do an example to follow along in the case of inequality.
if stacks A & B come out unequal,
Let’s say stack A=1234 > 5678=Stack B and thus we know fake is either heavy in A or light in B
remove 2 coins from stack A, and 1 coin from stack B, and set those 3 coins aside.
Ok, remove 34 from stack A=12 and remove 8 from B=567. Call OA=34 original stack A and OB=8 original stack B
stack A now has 2 coins, stack B now has 3 coins. (5 coins total).
yes
move 2 coins from stack B to stack A,
I move 56 from B to A so A=1256 and B=7
move 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.
Okay I move 2 to B so A=156 and B=27…3 coins in A 2 coins in B as you say.
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.
Stack C=09ab. I take 0 from C and let B=270. So A=156 and B=270
you now have 2 stacks of 3 coins each.
Correct. Now I do my second weighing of A and B.
if the 2 stacks are now balanced, then the ringer is one of the 3 you just removed and set aside.
Let’s consider the case that A=B…that leave me knowing that the fake is in 348 and either heavy if in 34 or light if in 8. But I only have one weighing to go. So I can weigh 38 and 56. If 38 is heavier then I know fake is 3 and heavy. If it is lighter then I know 8 is the fake and lighter. If equal then I know 4 is the fake and heavier. Check.
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.
Ok A=156<270=B. Fake is not 1 because it 1 is fake only if heavier, similarly not 7 because 7 is lighter. So fake is either 5 or 6 and lighter or 2 and heavier. So, as above, weigh 25 against fair coins 78. If lighter then fake is 5 , if heavier then fake is 2, and if equal fake is 6 and lighter. Check
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.
Ok A=156>270=B. Fake is either 1 and heavy or 7 and light. Weigh 1 against 2. If 1 is heavier then 1 is the heavy fake. If it is equal then 7 is the fake and light. Check
The official answer is very similar to IGW's (congrats, @i'vegotwinners ), although it's a little more elegant:
You now have 12 gold coins, but again, one has been doctored. It’s known to have a different weight than the others, but it could be heavier or lighter. How can you determine the doctored coin, and whether it’s heavier or lighter, with only three weighings?
The goal is to tease out as much information as you can from a single weighing, so you can get down to three candidates that you can figure out in a single weighing, like we did in the Express.
Start by putting four coins on each side of the scale. Assume that your first weighing balances. You now have four impure candidates (the four coins you did not weigh) and eight control coins (the eight you did weigh). Weigh three of the candidates against three of the controls, setting aside one candidate. If your second weighing balances, you’re almost done: The culprit is the candidate you left aside and you can use your third weighing to test whether it’s heavier or lighter than the pure coins. If your second weighing tips, you know whether the culprit is heavier or lighter based on how it tips, and you know it is one of three candidate coins. Your third weighing is the same as in the Riddler Express: Weigh one candidate against one other. If they balance, the one you left aside is the culprit. If it tips, the culprit is the one causing the tipping, and you already know if it’s heavier or lighter.
But what if your first weighing tips? Things get a little trickier here, as the puzzle’s submitter, Josh Kaplan, explains. You now have four heavy candidate coins, four light candidate coins, and four control coins. Keep good track of these! For your second weighing, place the coins like so: HHLL vs. HLCC. If this second weighing balances, you are left with one heavy candidate coin and one light candidate coin. Use your third weighing to compare a heavy candidate with a control coin — if it tips, you found the culprit heavy coin, and if it balances, the culprit is a light coin you set aside. If this second weighing tips, it’ll tip either left-side up or right-side up. If it tips left-side up, you are left with only the two light candidates on the left and the one heavy candidate on the right. For the third weighing, pit the two light candidates against each other, and you’re done. If it tips right-side up, you are left with only the two heavy candidates on the left and the one light candidate on the right. For the third weighing, pit the two heavy candidates each other, and you’re done.
The goal is to tease out as much information as you can from a single weighing, so you can get down to three candidates that you can figure out in a single weighing, like we did in the Express.
Start by putting four coins on each side of the scale. Assume that your first weighing balances. You now have four impure candidates (the four coins you did not weigh) and eight control coins (the eight you did weigh). Weigh three of the candidates against three of the controls, setting aside one candidate. If your second weighing balances, you’re almost done: The culprit is the candidate you left aside and you can use your third weighing to test whether it’s heavier or lighter than the pure coins. If your second weighing tips, you know whether the culprit is heavier or lighter based on how it tips, and you know it is one of three candidate coins. Your third weighing is the same as in the Riddler Express: Weigh one candidate against one other. If they balance, the one you left aside is the culprit. If it tips, the culprit is the one causing the tipping, and you already know if it’s heavier or lighter.
But what if your first weighing tips? Things get a little trickier here, as the puzzle’s submitter, Josh Kaplan, explains. You now have four heavy candidate coins, four light candidate coins, and four control coins. Keep good track of these! For your second weighing, place the coins like so: HHLL vs. HLCC. If this second weighing balances, you are left with one heavy candidate coin and one light candidate coin. Use your third weighing to compare a heavy candidate with a control coin — if it tips, you found the culprit heavy coin, and if it balances, the culprit is a light coin you set aside. If this second weighing tips, it’ll tip either left-side up or right-side up. If it tips left-side up, you are left with only the two light candidates on the left and the one heavy candidate on the right. For the third weighing, pit the two light candidates against each other, and you’re done. If it tips right-side up, you are left with only the two heavy candidates on the left and the one light candidate on the right. For the third weighing, pit the two heavy candidates each other, and you’re done.
Good job I've got winners. Good job Goat for a fun thread!
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