Another correction to the talent show: When I say that "Sam" was using a 4x4x4 emulating a 3x3x3, obviously by "Sam" I really mean "Michael Bilow". :P I don't even know who "Sam" is. I think there's a first-year named Sam.
On the subject of Rubik's Cubes, you should probably all see this: Rubik's Hypercube. Got it from Xiao. Brain-breaking, no?
I'd add in something about how Abandon Dan had his detergent "stolen" but I really don't see how I can make it funny, so skip right ahead to...
Evan's topology minicourse! Evan teaches us point-set topology - which he doesn't really know. Or rather which he learned just now so he could teach it to us, apparently so he could do other topology more rigorously. I think.
The lecture was unfortunately slowed down a lot by - well, he started with metric spaces, and then defined topological spaces generally later, and one of the first things he did was to ask people for examples of metric spaces. Enough answers to that to slow everything down quite a bit. Then as an example of a weird sort of metric he talks about the p-adic metric. Enough people not quite understanding that to slow everything down a bit. Then again I suppose all you really need is Avi and Abandon Dan, but anyway...
The problem set. Some of it is very easy (Every metric space is Hausdorff[0], is regular, is normal) (some of this looks not-so-easy, and some of it... well, here's problem 3 for you.
3a. Show that if X is a topological space and S⊆X, then it is possible to obtain at most 14 distinct sets (counting S itself) from S by the operations of taking the complement and taking the closure.
3b. Find a subset of R which actually attains the maximum of 14 distinct subsets.
Now that is just a "WTF?" question. Especially considering we've just been introduced to the concept of a topology that day. And why, of all possible places to put it, did he make it problem 3?!
...and then of course he didn't get as far as actually defining what a continuous function is, so some of them are, for now, impossible.
As for the geometry problem set.
Unlike the other geometry problem sets, which have pretty much followed what we're doing in class, this one is just all about CP1 (or PC1, if you prefer; I'll use the former as it's easier in HTML). It's really neat.
Although, the exploration question is totally unrelated.
Now as for lab!
As you probably don't know because I've never mentioned the labs before, I'm in the Arithmetree lab. The basic idea of this is as follows: This guy, Loday, has defined addition and multiplication on planar binary trees. Actually on sets of pb trees of all the same degree (number of internal vertices), called "groves". Investigate the properties.
Now before I go any further it should be noted that your knowledge of algebra may not be of much use here because no, it's not a ring, it's not even a rig. Addition does not commute and distributivity works only on one side. Kind of like the ordinals in that respect, I suppose. Chris, who's not in our lab, was wondering if they could be related to the ordinals somehow. "Um... Chris... there are countably many pb trees. The ordinals are a *proper class*. And, it seems, quite a bit nicer."
One of the first problems in doing this is just writing up programs to add and multiply and do other operations defined on the things. Not on the actual tree structure, but on this neat little way Loday came up with of representing them which is just a list of integers. Because that makes it that much more readable.
Oh, how are addition and multiplication actually defined?
You don't want to know.
...Trust me. You don't.
...Well, you might, in which case I'll explain it, but not here.
So anyway. One question we came up with is, when we add[4] trees, we of course get a grove. What's the cardinality of this grove? A few days ago we came up with the conjecture:
|x + y|=(mr(x)+ml(y) choose mr(x)) (=(mr(x)+ml(y) choose ml(y)))[3]
and also... oh, did I mention that in addition to addition (no pun intended) you also have "left addition" and "right addition"? The symbols for them are not in HTML, unless I want to use Unicode, which I don't, so I'll just denote them by "-|" for left addition and "|-" for right addition , as that's what the symbols look like, because pictorially the "+" symbol splits into those left and right halves. No, I'm not going to define those either. So anyway, continuing,
|x -| y|=(mr(x) multichoose ml(y))
|x |- y|=(ml(y) multichoose mr(x))
So today we actually went and proved those. Hm... nice combinatorial expressions for the cardinality of these grove... how do you think we proved it? Beautiful combinatorial proof, or messy strong induction?
That's right! Messy strong induction! Why? Because the beautiful combinatorial proof, if it exists, would require some actual understanding of the underlying combinatorics of these things.
...yes, it would be better if we did understand it, but do you realize how utterly messy it must be?
See now, Loday gives an actual definition of addition, and then later proves a recursion for it. *Always* use the recursion. Always. The actual definition is ridiculous. Hence the use of messy strong induction rather than "beautiful" combinatorics.
Now I tell Tom about this and he says about how you should always give the combinatorial proof - he says how (he's in the Polyominoes lab) he was proving how many stacks there are of a given perimeter, apparently it's the (p-3)th Fibonacci. Now, he could have proved it using the recursion, but instead he decided to do it using the the combinatorial definition of the Fibonaccis, just to make the proof nicer.
...I show him the definition of addition. He quickly drops that.
So today Dina asks if I know any of the first-years' LJs. I know that at least one of them has one, but that's all.
So yesterday ODan, Rich (another Honors kid) actually started a game of Monopoly with Mung. PROMYS kids playing Monopoly on a weekday?! Well eventually Eric took over for Mung... and then eventually Brian Lee took over for Eric. Only way it could have worked, I suppose. :P ODan won by just "letting" people give him one of their properties rather than pay his exorbitant rent which would have required them to sell houses.
Hm. Looks like I'm missing Dustin's lecture on straightedge and compass constructions. Ah well.
-Sniffnoy
[0]Isn't "Hausdorffness" such a great word?
[3]No, I'm not going to tell you what the mr() and ml() functions are.
[4]Or multiply, but multiplication is even more ridiculous than addition.
--
Artificial Intelligence is no match for Natural Stupidity.
On the subject of Rubik's Cubes, you should probably all see this: Rubik's Hypercube. Got it from Xiao. Brain-breaking, no?
I'd add in something about how Abandon Dan had his detergent "stolen" but I really don't see how I can make it funny, so skip right ahead to...
Evan's topology minicourse! Evan teaches us point-set topology - which he doesn't really know. Or rather which he learned just now so he could teach it to us, apparently so he could do other topology more rigorously. I think.
The lecture was unfortunately slowed down a lot by - well, he started with metric spaces, and then defined topological spaces generally later, and one of the first things he did was to ask people for examples of metric spaces. Enough answers to that to slow everything down quite a bit. Then as an example of a weird sort of metric he talks about the p-adic metric. Enough people not quite understanding that to slow everything down a bit. Then again I suppose all you really need is Avi and Abandon Dan, but anyway...
The problem set. Some of it is very easy (Every metric space is Hausdorff[0], is regular, is normal) (some of this looks not-so-easy, and some of it... well, here's problem 3 for you.
3a. Show that if X is a topological space and S⊆X, then it is possible to obtain at most 14 distinct sets (counting S itself) from S by the operations of taking the complement and taking the closure.
3b. Find a subset of R which actually attains the maximum of 14 distinct subsets.
Now that is just a "WTF?" question. Especially considering we've just been introduced to the concept of a topology that day. And why, of all possible places to put it, did he make it problem 3?!
...and then of course he didn't get as far as actually defining what a continuous function is, so some of them are, for now, impossible.
As for the geometry problem set.
Unlike the other geometry problem sets, which have pretty much followed what we're doing in class, this one is just all about CP1 (or PC1, if you prefer; I'll use the former as it's easier in HTML). It's really neat.
Although, the exploration question is totally unrelated.
Now as for lab!
As you probably don't know because I've never mentioned the labs before, I'm in the Arithmetree lab. The basic idea of this is as follows: This guy, Loday, has defined addition and multiplication on planar binary trees. Actually on sets of pb trees of all the same degree (number of internal vertices), called "groves". Investigate the properties.
Now before I go any further it should be noted that your knowledge of algebra may not be of much use here because no, it's not a ring, it's not even a rig. Addition does not commute and distributivity works only on one side. Kind of like the ordinals in that respect, I suppose. Chris, who's not in our lab, was wondering if they could be related to the ordinals somehow. "Um... Chris... there are countably many pb trees. The ordinals are a *proper class*. And, it seems, quite a bit nicer."
One of the first problems in doing this is just writing up programs to add and multiply and do other operations defined on the things. Not on the actual tree structure, but on this neat little way Loday came up with of representing them which is just a list of integers. Because that makes it that much more readable.
Oh, how are addition and multiplication actually defined?
You don't want to know.
...Trust me. You don't.
...Well, you might, in which case I'll explain it, but not here.
So anyway. One question we came up with is, when we add[4] trees, we of course get a grove. What's the cardinality of this grove? A few days ago we came up with the conjecture:
|x + y|=(mr(x)+ml(y) choose mr(x)) (=(mr(x)+ml(y) choose ml(y)))[3]
and also... oh, did I mention that in addition to addition (no pun intended) you also have "left addition" and "right addition"? The symbols for them are not in HTML, unless I want to use Unicode, which I don't, so I'll just denote them by "-|" for left addition and "|-" for right addition , as that's what the symbols look like, because pictorially the "+" symbol splits into those left and right halves. No, I'm not going to define those either. So anyway, continuing,
|x -| y|=(mr(x) multichoose ml(y))
|x |- y|=(ml(y) multichoose mr(x))
So today we actually went and proved those. Hm... nice combinatorial expressions for the cardinality of these grove... how do you think we proved it? Beautiful combinatorial proof, or messy strong induction?
That's right! Messy strong induction! Why? Because the beautiful combinatorial proof, if it exists, would require some actual understanding of the underlying combinatorics of these things.
...yes, it would be better if we did understand it, but do you realize how utterly messy it must be?
See now, Loday gives an actual definition of addition, and then later proves a recursion for it. *Always* use the recursion. Always. The actual definition is ridiculous. Hence the use of messy strong induction rather than "beautiful" combinatorics.
Now I tell Tom about this and he says about how you should always give the combinatorial proof - he says how (he's in the Polyominoes lab) he was proving how many stacks there are of a given perimeter, apparently it's the (p-3)th Fibonacci. Now, he could have proved it using the recursion, but instead he decided to do it using the the combinatorial definition of the Fibonaccis, just to make the proof nicer.
...I show him the definition of addition. He quickly drops that.
So today Dina asks if I know any of the first-years' LJs. I know that at least one of them has one, but that's all.
So yesterday ODan, Rich (another Honors kid) actually started a game of Monopoly with Mung. PROMYS kids playing Monopoly on a weekday?! Well eventually Eric took over for Mung... and then eventually Brian Lee took over for Eric. Only way it could have worked, I suppose. :P ODan won by just "letting" people give him one of their properties rather than pay his exorbitant rent which would have required them to sell houses.
Hm. Looks like I'm missing Dustin's lecture on straightedge and compass constructions. Ah well.
-Sniffnoy
[0]Isn't "Hausdorffness" such a great word?
[3]No, I'm not going to tell you what the mr() and ml() functions are.
[4]Or multiply, but multiplication is even more ridiculous than addition.
--
Artificial Intelligence is no match for Natural Stupidity.
