Coconut Problem of the Week
Card Signaling Problem of the Week
Winter Reflection on the Year
- A clear explanation of the problem/skill/concept that anyone could understand.
I am going to talk about two problems of the week. One I got right, and one I got wrong. First, I will talk about the one I got wrong. In the card signaling problem of the week, we are asked to do some card finagling. With a partner, we were given a random set of any five cards out of fifty-two (no Jokers). We then carefully choose one of these five cards and put it in our pocket without letting the other person see. With the remaining four cards, we must tell the other person exactly which card we have in our pocket. In the Coconut weighing problem, we have been stranded on an island from a cruise vacation gone awry. We have been on this island for a couple days, and were beginning to lose hope of ever being found. But lo and behold! There be a ship on ye horizon! Be it friend or foe, you ask? We shall soon see. As the great vessel of hope draws near, our collective heart dropped as soon as we saw Ye Ole Jolly Roger waving from the top of the ship. We have no choice but to watch with dread as the ship anchors and a dinghy comes ashore. We quake in our boots as we watch a boatful of salty dogs land on the beach in front of us. The next thing we know, we are on their ship with a
scale and six coconuts in front of us. They tell us, “Yarr, we have a test for ye. We
have here in front of ye six coconuts. Two are Shamrock Green, two are Deep-Ocean
Blue, and two are Purple Pizzazz Magenta. In each color pair, one coconut is almost
imperceptibly heavier than the other. All the heavy rocks weigh the same, and all of
the light rocks weigh the same, to be sure. In two weightings or less, you must find
which are the heavy coconuts and which are the light coconuts. If ye can’t complete
this task, ye’ll walk the plank!”
For the card signaling problem, we are asked to do some card finagling. With a partner, we were given a random set of any five cards out of fifty-two (no Jokers). We then carefully choose one of these five cards and put it in our pocket without letting the other person see. With the remaining four cards, we must tell the other person exactly which card we have in our pocket.
- Evidence of the process you went through to solve it/master it/understand it.
For the card signaling problem, the first thing that we decided was that the first card must tell suit. If you draw five cards and there are four suits, there must be at least two cards that share the same suit. Once that was decided, we tried to figure out what else the first card should signify. We had the idea of denoting that the hidden card was less than the value of the first card. This rule can cut down on the possible cards significantly. The only problem is if the two cards that share a suit are the king and the two, or something like that. Having such a wide gap doesn't help us at all and the rest of the sequence is a little useless. So we decided to toss that idea. Then we though of having the first two cards signify a range. The first signifies the suit and the second signifies that the hidden card is some value between the first and second value. This helped a lot because now you don’t get trapped with having such big ranges, since with four cards the widest gap you can have between cards is about three. Now we just need to shorten the gap a little more. So we determined that the third card should signify whether the card is odd or even. This effectively cuts the amount of possible cards that were before in half. So now instead of three or four options, we have two or three options. And all we have left is the last card. We first tried having the last card being the card closest to the hidden card, but that didn’t work at all. The last card is most definitely not always the closest. Even if it was, then we would run into the same problem as before where if we have cards that are spread apart on the scale. So we scrapped that idea. But then we thought that the last card could decide on which half of the range the hidden card was. So if the last card was a face card or something high like that, then the hidden card would be on the high end of the range between the first two cards, and vice versa. Using this, we could usually get the possible cards down to one choice. Sometimes there would be two choices, and sometimes the last card didn’t align with the range at all. So with this system we still run into the same problem as before. If the cards are too spread out and the ranges are too big you can’t narrow it down into a single definite number. So then Kyle gave us the idea of allowing the ranges to wrap around. So if the first two cards were a three and a queen, the range could be three thru queen or three two ace king queen. But then there was the problem of knowing which one it was. So we decided to make it always the shorter range. Whichever range had the least space in between numbers, that's the range it was.
For the coconut weighing problem, we tried to think about it conceptually. We found that if we weighed two of different color together and it came out uneven, we could figure out the rest no problem. We would just put the heavy one in the heavy pile, the light in the light
pile, the others of the same color in the opposite piles, weigh the other color against
each other, and be done. The problem is if they come out even. Then you can sort
them into piles but you don't know which one is heavy and which one is light. So
then we tried to figure out a way where you weighed two against two, but couldn't
really come up with a solution using that method. In order to visualize it, my mom
and I used six grapes of approximately equal size and wrote a color on each,
differentiating the two of the same colour with a prime mark. Then we worked
through the different possibilities you could try. One way we were trying was to use
the first step of the first method to separate them into groups. Then we would
switch one of the colours and weigh both sets with the other colour set, so there is
one from each grouping on each side plus one from the leftover colour set. Then we
realized that this wouldn't work because if you switch one colour, then they negate
each other, because each has one light and one heavy. It would be the same as
weighing the last ones against each other. So then in class Natalie gave us the idea of
weighing one red and one blue on one side, and then one red and one green on the
other. This way, if they are even, you can weigh the ones that didn't match and then
know all of them by process of elimination. The only problem is, if the scale goes to
one side or the other during the first weighing. Then you know that the non-paired
coconuts are either either lights or both heavies, and you can’t tell this from
weighing them against each other. So we played around with which weighing to do second until we found one that works.
- Proof that you completely understand the ideas discussed (don’t be afraid to get technical here, perhaps with annotated problems that you solved).
I know my solution to the card signaling problem is not correct. I know this because I have done several sets where the sequence does not work. The issue is usually one of three things: the range is not narrowed close enough, the card is directly in the middle of the range, or the last card tells for the wrong side of halfway. I think that the way to fix this sequence would be to figure out a way to use the last card more effectively. With my sequence, it can be wrong or not substantial enough to solve the problem. I think another issue is the range system. I think that it is a good concept but could be applied in a different way to make the sequence work more consistently.
My solution to the coconut weighing problem is this: You have two reds, two blues, and two greens. First, you weigh a red and a blue against a red and a green. (P.S. the colors don’t
matter; just make sure you have two of the same color on opposite sides
If they are equal, you can weigh the green against the blue, find which is the heavy
one, put the red that was on the opposite side with it along with the green that
wasn't weighed into the heavy pile, and then put the rest into the light pile.
However if they are unequal, the above method will not work. What you must do
instead is weigh the two reds against the blue and the green.
What this allows you to do is see if the blue and green are two lights or two heavies
or one light and one heavy. If they are the lights, put them with the red that was on
the lighter side in the first weighing and the others in the heavy pile, and vice versa.
If they are one light and one heavy, put the one that was on the heavy side with the
red that was also on the heavy side with the one of the other colour that wasn't
weighed in the heavy pile and put the others in the light pile.
If the two sides are equal, you know you have one heavy and one light on each side
since there is one heavy red and one light red on each side. By weighing the blue and
green against each other, you can decipher which is the heavy and which is the light
and put them into their groups accordingly. By knowing which coconut was with
which red coconut, you can place the reds by putting them with the opposite colours they were weighed with.
- A general discussion of how this problem frames your work in math looking forward, both towards next semester and beyond. This is a good place for self reflection - what are you talented at and how will you maximize those gifts? What is a challenge and how will you account for that?
I am a visual learner and I like having props to work through problems. By working through these problems with tangible props, I was able to think about these problems visually. I could put my hands on the problem and manipulate it to find solutions. This will help me later on in life because it teaches me to think about things in a way that I can understand them better. I have to visualize things in my head and then try and work a solution from there.