Mar 02, 2013
RAY: Imagine that you have in front of you fifty coins. They all look exactly alike except one of them is a fake. Because it's a fake, it weighs a couple of grams more than a real coin. So, if you had a balance scale, and you knew which was the bogus coin, you would put it on one side of the scale, a good coin on the other side...
TOM: ...and it would be immediately obvious from this imbalance which was the phony coin, because it's heavier than a real coin.
RAY: Right. Knowing that, you have in front of you fifty coins -- one of which is bogus. The question is, what is the fewest number of weighings on a balance scale,that you need to perform, to determine which coin is bogus?
TOM: And, Part B of the puzzler: How can you do it in four?
TOM: ...and it would be immediately obvious from this imbalance which was the phony coin, because it's heavier than a real coin.
RAY: Right. Knowing that, you have in front of you fifty coins -- one of which is bogus. The question is, what is the fewest number of weighings on a balance scale,that you need to perform, to determine which coin is bogus?
TOM: And, Part B of the puzzler: How can you do it in four?
Answer:
RAY: Here’s the answer. At first blush you would think that you would divide the 50 coins in half.
TOM: Yeah. So, you'd do 25 and 25. That's weighing number one.
RAY: Right. You find out that the bogus coin is on the left side. Then you weigh the coins on the left, 12 and 12 with one left over.
Now, Assume the worst case scenario. One of them's heavier. So now that’s two weighings. Now you divide it six and six, making it three weighings. Then three and three, for four weighings.
TOM: And you're done for. It takes five.
RAY: Right. No matter how you do it using that system, it won't work. So you had to come up with something a little more clever. And what you do is divide the coins into three piles. Two piles of 17 and one of 16.
You take the two piles of 17 and you put those on the scale, and you keep the 16 pile aside,. Right away, you can see that you're going to eliminate not half the coins, but two thirds of the coins.
TOM: Oh, man. It's so beautiful, isn't it? It's beautiful.
RAY: So, let's assume that one of the 17 is the heavier one. You throw everything else away.
You've only made one weighing and you've narrowed it down to 17 coins. Now, you could divide the 17 in half, but better still, divide it thirds so you've got six and six and five.
OK? And you can see very clearly that you're going to be able to do this because the next group would be three and three, and then one and one. And bingo! And the key is, once you figure out the idea that you're going to divide it into three piles and not two, it jumps right out at you.
TOM: I always thought that the binary search was the only way to go.
RAY: Well, there you go. Who's our winner this week?
TOM: The winner is Laura Leonard from Charleston, SC. Congratulations, Laura!
TOM: Yeah. So, you'd do 25 and 25. That's weighing number one.
RAY: Right. You find out that the bogus coin is on the left side. Then you weigh the coins on the left, 12 and 12 with one left over.
Now, Assume the worst case scenario. One of them's heavier. So now that’s two weighings. Now you divide it six and six, making it three weighings. Then three and three, for four weighings.
TOM: And you're done for. It takes five.
RAY: Right. No matter how you do it using that system, it won't work. So you had to come up with something a little more clever. And what you do is divide the coins into three piles. Two piles of 17 and one of 16.
You take the two piles of 17 and you put those on the scale, and you keep the 16 pile aside,. Right away, you can see that you're going to eliminate not half the coins, but two thirds of the coins.
TOM: Oh, man. It's so beautiful, isn't it? It's beautiful.
RAY: So, let's assume that one of the 17 is the heavier one. You throw everything else away.
You've only made one weighing and you've narrowed it down to 17 coins. Now, you could divide the 17 in half, but better still, divide it thirds so you've got six and six and five.
OK? And you can see very clearly that you're going to be able to do this because the next group would be three and three, and then one and one. And bingo! And the key is, once you figure out the idea that you're going to divide it into three piles and not two, it jumps right out at you.
TOM: I always thought that the binary search was the only way to go.
RAY: Well, there you go. Who's our winner this week?
TOM: The winner is Laura Leonard from Charleston, SC. Congratulations, Laura!
