* 1) **North Korean aggression will spark a Second Korean War. The war will quickly turn nuclear and end badly for North Korea.*

Well, unless there’s been some secret war that’s been going on, I think this one didn’t turn out to be true. I suppose that’s a good thing.

Score: 0/1

* 2) **A star will go supernova and will be brighter than Sirius in the night sky.*

Nope. I’m sure a star went supernova somewhere (in fact, given the sheer number of stars in the universe, about *2,500* went supernova while reading this sentence), but no visible supernovae are to be seen.

Score: 0/2

* 3) **There will be a massive earthquake in Europe.*

A cursory search reveals that the most massive earthquake that occurred in Europe was a 6.4-magnitude quake in Greece on October 12^{th}. Since that is relatively massive, I’m going to take that as a hit.

Score: 1/3

* 4) **The iPhone 6 (or 5S, whatever they call it) will be released and be a flop.*

The iPhone 5S (as well as the lower-market 5C) was indeed released in 2013, but it seemed to sell as well as its predecessors, and got generally positive reviews.

Score: 1/4

* 5) **Astronomers will measure the existence of molecular oxygen in the atmosphere of an Earth-sized extrasolar planet, thus showing that it contains life.*

This would have been massive news if it had occurred, so it’s obvious that this is a miss. Several extrasolar planets *did* have their atmospheric compositions measure, however, detecting such things as water vapor, but these were massive Jupiter-sized planets or larger.

On a slightly more serious note, I think this method is the most likely way that we’ll initially discover extraterrestrial life. All it takes is pointing it at the right star and doing the difficult work of picking out the planet’s light from the star’s. I honestly think that by 2020, we’ll probably know for relative certainty that there is extraterrestrial life, and that it exists on a given extrasolar planet.

Score: 1/5

* 6) **There will be a near-miss from an asteroid that will swing by Earth and come less than 10,000 km to the surface. It will not be discovered until after it has passed.*

I couldn’t find an asteroid that *specifically* came within 10,000 km, of course that doesn’t mean one didn’t happen. There was, obviously, a relatively large one that did *hit* us on February 15th, in Chelyabinsk Oblast, Russia, causing some 1,000 injuries to people on the ground. It wouldn’t really count as a “near-miss”, but when you think about it, doesn’t a “near-miss” imply it *didn’t *miss…:) I’m going to count it, literally in this case, as a hit.

Score: 2/6

And now on to the celebrity deaths:

* 1) **Sylvia Browne*

BOOM! Sylvia Browne died November 20^{th}. This was the one I wanted the most. It may seem awful to wish that someone would die, then be happy when they actually do, but Sylvia Browne was probably one of the worst human beings to have ever lived. She would emotionally manipulate grieving people to line her pockets with cash, and didn’t give a crap about it. She would unhesitatingly tell parents of missing children that they were dead, subjected to horrible fates like being sold into sex-slavery in Asia, or even that they were still alive when they turned out to be dead the whole time. She totally knew what she was doing, knew she was ruining people’s lives, and all she could say was, “**Screw ’em. Anybody who believes this stuff oughtta be taken.**”

Funnily enough, she predicted she would die at age 88. She was only 77 when she died.

Score: 3/7

* 2) **Joseph Ratzinger, Pope Benedict XVI*

Well, Joseph Ratzinger is still with us on this Earth, but he did resign as pope, and is now styled as Pope Emeritus. Therefore, the pope did “die” in some sort of way…

Score: 4/8

* 3) One of the Kardashians*

I’m sure if you trace the family line back far enough, you could find a relative of the family that passed away, but nobody really notable died. I was mainly thinking of someone who would otherwise make the cover of People magazine.

Final Score: 4/9

So a 44% hit rate. I suppose if you were to relax the criteria, for example counting some Kardashian relative, or saying supernova did happen out there somewhere, I could claim a higher score. To really prove that any of the predictions made were false would basically involve proving a negative, which is pretty much impossible.

That’s the crucial thing in all this. To claim something is true, you need to have evidence to back it up. If you don’t have any, then there’s no reason to believe the claim in the first place.

Coming soon, predictions for 2014!

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It’s basically a 4x, RTS space game, with game play elements taken from other similar games that I like with other features that I’ve always wanted in a space RTS that I’ve never seen. It’s basically the RTS space game I’ve always wanted to play but nobody’s ever made, as far as I know.

Anyway, see you later,

Arik

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So I decided why not come out with a fun list of predictions myself? I’m just as “psychic” as these other people claim to be (i.e. not at all) so if they can do it, I can do it too. Now, I’m not making these anywhere as strong as other predictions I’ve made. These are just for fun and to show how anyone can *claim* to be psychic with only very flimsy evidence behind it. Anyway, here they are:

1) North Korean aggression will spark a Second Korean War. The war will quickly turn nuclear and end badly for North Korea.

2) A star will go supernova and will be brighter than Sirius in the night sky.

3) There will be a massive earthquake in Europe.

4) The iPhone 6 (or 5S, whatever they call it) will be released and be a flop.

5) Astronomers will measure the existence of molecular oxygen in the atmosphere of an Earth-sized extrasolar planet, thus showing that it contains life.

6) There will be a near-miss from an asteroid that will swing by Earth and come less than 10,000 km to the surface. It will not be discovered until after it has passed.

Celebrity Deaths:

1) Sylvia Brown (please, please, please)

2) Joey Ratz, aka, Pope Benedict XVI

3) One of the Kardashians

Ok, so there it all is. Furthermore I will do something that no alleged psychic has ever done: after January 1, 2014 I will systematically go through and determine which of my predictions, if any, I got right.

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So really, I think my prediction was a bit too conservative. Here’s what I think now. If you’ll notice, I’m also moving the timeline up a few years just because I think it can likely happen by then.

My prediction is this: new cell phones will, by and large, disappear by 2025. Yes, I think that a technology like Google’s Project Glass will completely supplant cell phones. There will probably still be cell phones around, of course, but they’ll be older models, maybe also cheap pre-paid ones. But the new, high-end cell phones won’t actually be hand-held devices like today (and they very likely won’t be called “cell phones”). The common status symbol of the past 25 years or so will be relegated to history.

Here’s how they’ll work. You’ll have your headset, or whatever it’ll be called, and you’ll be able to interact with it via voice command. But there’s another way in which you’ll also control it: by pressing buttons projected onto a surface. How? Well, take a look at this video:

The technology already exists. In fact, you probably won’t even have to actually project it, enabling everyone around you to see it. The projection itself will probably be completely within the headset and only you will be able to see it. The headset will be able to see what you’re pressing and where and will be able to match it to the appropriate virtual control. To the bystander, it’ll seem as if you’re pressing a blank wall. Right now this is technology that, while possible, really doesn’t have a practical application. Headsets give it an application. No other control system (keyboards, touchscreens, mice, etc) will be as useful as this for headsets. And really, it’s not very different from using a touchscreen, so I think adoption won’t be much of a problem.

Another interesting improvement will be the elimination of “talking-to-yourself-syndrome”. We’ve all seen people on Bluetooth who are using their phone. It’s jarring but we’ve sort of gotten used to it. The advantage of being on you head, though, is that you can run a ine down the neck and detect movement. Thus, simply mouthing words (with your mouth closed, too, so others can’t even tell you’re talking), the headset will be able to understand you. Seem strange? Well, again, it’s already existing technology:

It’s a bit limited now, but given a decade of R&D and about 1000-fold increase in processing power between now and then, it will likely be as reliable as talking to someone in real-time. And again, while this may be a somewhat pointless and obtrusive technology now (“you mean I’ve got to put that collar on to use it?”) if you’ve already got something on your head it’s not as big of a step (and it will almost certainly be a lot smaller, less obtrusive, and more discreet).

The cell phones of 2025 will not be anything like the cell phones of today. They will be light-weight, unobtrusive headsets that will be able to do all the things cell phones do today: make calls, send texts (dictated), surf the internet, download new apps, etc. And it will bring us one step closer to fully immersive virtual reality. Personal headsets can be the enabling technology where true, interactive, public virtual reality will begin, once they’re adopted.

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From what I can tell, *Mathematics Teacher* is basically like an industry magazine for math teachers, going over various teaching topics and ideas and the like, which is cool. I never heard about it before but, a publication is a publication.

Unfortunately, my contribution seems to be almost entire cut out of the article entirely. It’s not too surprising given that I only solved on specific problem in a textbook and the article is about teaching Mayan numerals in general, but is still a bit disappointing. At least my name is still on the byline:

So anyway, what was the problem I solved? Well to understand that, let’s delve a little bit into the Mayan number system. Mayan numbers are actually not too different from ours. They are positional, so each digit is a multiple of some base unit like ours. For example, in our system, each number is multiplied by a factor of ten. So 50 is 5 × 10, 3,000 is 3 × 10^{3}_{}, and so on. Each time you move a digit to the left, you’re multiplying it by 10. Same thing in the Mayan system, except that for them, you’re multiplying by twenty. Also, the Mayans wrote their numbers from top to bottom, instead of left to right, but that’s just a small detail.

The Mayans had unique symbols for each digit from 0 to 19 to stick in for each digit (well really, they only had three symbols, the dot (for 1), the line (for 5), and the shell (for 0). Kind of similar to tally marks). I’m not going to use the symbols for this post since it would be too confusing and too much work for me (you can see them here), so I’m going to use the typical notation of A.B.C.D… and so on, where each letter is a digit and the larger digit places are towards the left. So, for a quick example, 1024 in our system would be 2.15.4.

The only really weird thing about the Mayan numeral system is that there is a little “jog” in it. While almost every place is twenty times the value of the previous place, the third place (which would be the 20^{2}, or 400s place) is actually only 18 times the previous one. Why? Who knows? Although 18 × 20 is 360 which is very close to the number of days per year, so it’s possible that’s why. In the Mayan Long Count Calendar (responsible for the 12/21/2012 end of the world nonsense), 1.0.0 days is called a *tun*, about equal to a year, and is obviously a nice round number, rather than 18.0 (or really, 18.5 to be more accurate) if it were truly a pure base-20 system.

So what was the exact problem I solved? Well, given what I said about the numbering system with its strange twist, is there an easy and quick way to multiply any given Mayan number by 20? If it were purely base-20 the answer would be trivial: just stick a zero on the end (like multiplying any number by 10 in our system). But if you actually do that, you haven’t actually multiplied the entire number by 20. The second digit moves into the third digit which means it’s only been multiplied by 18, so the resulting number is a bit smaller than 20 times the number.

Nobody in any class the professor has taught had ever figured this out, nor had she, nor even the book publishers apparently had an answer (the professor had called and asked them). So here’s the answer:

1) Stick a zero on the end.

2) Take what is now the third digit and multiply it by two (don’t change it in the actual number, just see what it is and multiply it by two in your head)

3) Add that number to the second digit.

4) Regroup to account for possible roll-over in the second digit (i.e. carry the one…or two as the case may be)

5) There is no step 5 because you’ve already finished!

It really is that easy. Basically, what you’re doing is this: When you add a zero on the end every digit is indeed multiplied by 20, *except for the second digit*. Therefore, you need to add that difference back into the number to get the correct answer. This can result in the second digit being to large for its place (i.e. being 18 or over), so if you need to, you can subtract 18 one or two times (you won’t need to do it more than twice, I promise), and add the number of times you had to subtract 18 to the third digit. It’s like if you were to add 9 + 7. Well, you can’t cram “16” into a single digit, so you subtract 10 once, and add the one to next place.

Let’s do two examples:

A) Multiply 8.2.13 by 20

1) Add a zero: 8.2.13.0

2) Take the third digit and multiply by two (so that would be 4)

3) Add that to the second digit: 8.2.17.0

4) Regroup if necessary (since the second digit is less than 18, it doesn’t roll over)

Therefore 8.2.13 × 20 = 8.2.17.0

B) Multiply 7.5.12.16 by 20

1) Add a zero: 7.5.12.16.0

2) Take the third digit and multiply by two (so, 24)

3) Add that to the second digit: 7.5.12.40.0

4) Regroup if necessary (40 goes into 18 twice, with a remainder of 4, so we make the second digit 4 and add 2 to the third digit): 7.5.14.4.0

So anyway, that’s it. The only problem left is to see if the rolling-over ever stops. I mean, what if the number you’re dealing with keeps making higher digits continue to roll over all the way up? For example, if you have the number (in our system) 9999999999999 and add 1, the ones’ place becomes 0 and a 1 rolls over into the tens’ place, which also becomes zero and rolls over into the hundreds’ place…which *also* becomes zero and rolls over into the thousands’ place, and so on. It would be unfortunate if this were to happen in Mayan numerals, since all that writing would take time.

Fortunately, it doesn’t. Let’s take a Mayan number that would be like 999999etc: 19.19.17.19. If you were to add 1 to this number, the places would keep rolling over until you were left with 1.0.0.0.0. So, let’s see if that happens when you multiply by 20:

1) Add a zero: 19.19.17.19.0

2) Take the third place and multiply by two (17 × 2 = 34)

3) Add that number into the second place digit (19 + 34 = 53): 19.19.17.53.0

4) Regroup: 53 ÷ 18 = 2 remainder 17. So the third digit becomes 19 and the second becomes 17: 19.19.19.17.0

Phew! The rollover *stops* at the third place in all cases (I can say this since this is the case where the number are at their largest).

So that’s my solution. As my professor might say “it’s good clean fun!” So anyway…wait…hold on, what’s that at the end of the article?…

I have a program that can convert our number system into Mayan numbers? I don’t have a problem that can convert our number system into Mayan numbers. That’s means I need to make one, quick! My e-mail’s on the thing! And the deadline where everyone loses interest is in three days! ARRRRRGGGG!!!!!

See you on the 22^{nd}!

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This past quarter, I took a class called “History of Mathematics”. It seemed interesting, there was nothing else to fill my schedule, and, as it turns out, it will could towards my elective credit for my degree. During the class we had a homework problem which asked to find a simple way of multiplying a Mayan number by 20. Nobody could solve it, and the teacher (Dr. Shirley Gray) stated that in all her years of teaching the class, no student of hers had ever solved the problem. She had even written the textbook’s authors and *they* didn’t have a solution either. Well, I solved it.

It seems like a simple problem. But to understand it, we must understand how Mayan numbers work. Mayan numbers are, essentially, a base-20 positional system. It is like ours, except that each successive digit it twenty times greater than the previous one (in ours, of course, each digit is only ten times greater).

Based on this, it should be trivial to multiply by 20: just add a zero on the end (like multiplying by 10 in our system). However, there is a slight problem. For some reason, the third digit in a Mayan number is only **18** times the previous digit (meaning that 1-0-0 in their system is equal to 360, not 400). Adding a zero on the end won’t result in a number twenty times greater.

So, how do you do it? Well, I can’t write the answer here because I don’t want to inadvertently mess up getting the paper published. Dr. Gray did grill me on whether I found the solution online (which I didn’t, of course. Plagiarism is a very serious academic offense and if I had found it online, I would have stated so in the first place. I understand her caution completely). What I will say, though, is that the solution is simple, quick, and you don’t even have to convert the Mayan number into our own system to do it (they didn’t use Hindu-Arabic numbers, naturally. It was system of lines and dots written vertically, with a strange shell shape for the number zero. See here).

Where the paper comes in is that Dr. Gray was already writing a paper on Mayan numbers and wants to include my finding in it. So it’s not really* **my* paper, but it’s still a mention in some official publication. Which is more than I’ve ever done so far.

So, when the paper is published, I will announce it here, as well as the actual solution with the proof I developed. It will interesting to put on my resumé, right below ‘recording “Hollaback Girl” with Gwen Stefani’ (true story! Also, sorry…)

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Google has announced that they are working on, essentially, augmented reality glasses. Of course, technology like this has been worked on for decades with little to show for it. But this seems, possibly, different. This this time, it is a large high-tech company with a track record of producing practical and highly profitable technological innovations. If anyone has a shot to making these glasses widely available it’s them.

Second, take a look at this video:

Absolutely everything that is shown is what smart phones do today: listening to music, taking pictures, making phone/video calls, getting weather, looking up info online, *texting…*it’s like wearing an iPhone on your head.

Furthermore, and something I stressed in my former post, it is very lightweight, and not cumbersome at all to use, like a real pair of glasses. *This is essential*. If it was a giant bulky headset, no one would want to buy it, no matter what it did. Plus, when not using it, you can still see where you’re going. You don’t even have to take it off.

Now, what needs to happen is for someone to make a virtual reality app. Something like Second Life that you can stream through the glasses. True it won’t be fully immersive, and you might need to have some sort of controller hooked into the glasses (note to Google: be sure to include USB or microSD slots), but it is a first step. From there you can build onto it to eventually create fully immersive virtual reality.

Anyway, based on all this, I predict two possible outcomes: 1) Google will introduce these within a few years (2014-15 frame) and make billions, or 2) This will flop by Apple will come out with their own version a little later (2014-17 frame) and make billions (maybe they can call it the iEye ;)) Either way, the next wave of personal technology is rapidly approaching.

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