Four years in this house and I have yet to find a Hamiltonian circuit of the lawn. Curse you NP-completeness!
Explanation: In graph theory, a Hamiltonian circuit is a path that starts at a point in a graph, travels through all other points in the graph exactly once, and returns to the starting point. Determining whether such a path exists in a graph is an example of what's called an NP-complete problem, which means (among other things) that there is no known algorithm of polynomial time complexity for solving it. In other words, it's really hard.
Consider my lawn to be a collection of points, with each point connected to the points closest to it by an edge. Now it's no longer a lawn, it's a planar graph! Thus, we can apply graph theory, and the most efficient way to mow my lawn is the Hamiltonian circuit that starts (and ends) closest to my garage.
You know, for $30 a mow, you can just pay the local lawn guy to do it while you play computer games.
I really should be thinking of it as a Travelling Salesman Problem instead. In that case, I attach a weight to each of the edges in the graph (the length of the edge) and then solve for the shortest tour of the vertices. It's still an NP-complete problem, but now I not only mow each part of the lawn only once, I also travel a minimal distance in doing so. I can't solve it, but if I could it would give me a better answer.
You think about this stuff while vacuuming, too, don't you?
Yes. And mopping the floor. Mopping the floor is easier to visualize because the floor is neatly broken down in to tiles. It's more difficult, however, because you have to take into consideration that you can never stand in a place where you've already mopped.
This is why mathematicians never work as landscapers or housecleaners.
Interestingly, those jobs are often taken by immigrants who speak little or no English, and therefore have better communication skills than most math majors.
It kind of sucks the fun out of things when you make fun of yourself.
It sure does!
Can we talk about toasters now?
No.
Showing posts with label Mathematics. Show all posts
Showing posts with label Mathematics. Show all posts
Thursday, May 28, 2009
Wednesday, May 13, 2009
The Best
It's the return of the new and improved and I'm smooth, ain't nobody does it better, here's the proof.
Source: This is a line from The Best by 2 Skinnee J's. The chorus of the song then continues to lay out the details of the proof, which are quite elegant and simple, but unfortunately this space is not dope enough to contain it.
Special Blog Bonus: To balance this small post, I offer a link to a very large LEGO castle:
Source: This is a line from The Best by 2 Skinnee J's. The chorus of the song then continues to lay out the details of the proof, which are quite elegant and simple, but unfortunately this space is not dope enough to contain it.
Special Blog Bonus: To balance this small post, I offer a link to a very large LEGO castle:
Wednesday, April 8, 2009
Finally, A Post
Remember, past performance is not an indication of future results.
Explanation: This may be the golden rule of being the parent of a newborn child. Not only do we not have a schedule in our house, but every time I think we have a schedule, it changes on me. Let's hope the same holds true for the stock market, huh?
In Case You Care: The answer to last week's math problem is (x-1)(y-1) - 1. So, for x=5 and y=11, the biggest number that cannot be made out of 5's and 11's is 4*10-1 = 39. Sure enough, if you consider the numbers that can be made from 5's and 11's, the list looks like this:
0 = 5*0 + 11*0
5 = 5*1 + 11*0
10 = 5*2 + 11*0
11 = 5*0 + 11*1
15 = 5*3 + 11*0
16 = 5*1 + 11*1
20 = 5*4 + 11*0
21 = 5*2 + 11*1
22 = 5*0 + 11*2
25 = 5*5 + 11*0
26 = 5*3 + 11*1
27 = 5*1 + 11*2
30 = 5*6 + 11*0
31 = 5*4 + 11*1
32 = 5*2 + 11*2
33 = 5*0 + 11*3
35 = 5*7 + 11*0
36 = 5*5 + 11*1
37 = 5*3 + 11*2
38 = 5*1 + 11*3
40 = 5*8 + 11*0
41 = 5*6 + 11*1
42 = 5*4 + 11*2
43 = 5*2 + 11*3
44 = 5*0 + 11*4
At this point, we have reached a string of five consecutive numbers (our x), which means we can make all numbers greater than that by simply adding multiples of 5 to 40, 41, 42, 43, or 44.
As I can't remember enough number theory to write a proof here, I'll leave that as an exercise for the reader.
Sorry for the missing post yesterday and the late one today - it has been a busy week for me.
Explanation: This may be the golden rule of being the parent of a newborn child. Not only do we not have a schedule in our house, but every time I think we have a schedule, it changes on me. Let's hope the same holds true for the stock market, huh?
In Case You Care: The answer to last week's math problem is (x-1)(y-1) - 1. So, for x=5 and y=11, the biggest number that cannot be made out of 5's and 11's is 4*10-1 = 39. Sure enough, if you consider the numbers that can be made from 5's and 11's, the list looks like this:
0 = 5*0 + 11*0
5 = 5*1 + 11*0
10 = 5*2 + 11*0
11 = 5*0 + 11*1
15 = 5*3 + 11*0
16 = 5*1 + 11*1
20 = 5*4 + 11*0
21 = 5*2 + 11*1
22 = 5*0 + 11*2
25 = 5*5 + 11*0
26 = 5*3 + 11*1
27 = 5*1 + 11*2
30 = 5*6 + 11*0
31 = 5*4 + 11*1
32 = 5*2 + 11*2
33 = 5*0 + 11*3
35 = 5*7 + 11*0
36 = 5*5 + 11*1
37 = 5*3 + 11*2
38 = 5*1 + 11*3
40 = 5*8 + 11*0
41 = 5*6 + 11*1
42 = 5*4 + 11*2
43 = 5*2 + 11*3
44 = 5*0 + 11*4
At this point, we have reached a string of five consecutive numbers (our x), which means we can make all numbers greater than that by simply adding multiples of 5 to 40, 41, 42, 43, or 44.
As I can't remember enough number theory to write a proof here, I'll leave that as an exercise for the reader.
Sorry for the missing post yesterday and the late one today - it has been a busy week for me.
Thursday, April 2, 2009
Math Makes Me Happy!
If x and y are relatively prime positive integers, what is the greatest value z such that there do not exist integers A and B (>=0) to make the equation Ax + By = z true?
Explanation: I think the explanation of this status will perhaps be more disturbing than the actual message. Then again, if you're a returning visitor to this blog, you probably already know how disturbed I am.
There is a game I play on Facebook in which (among other things) you have an allotment of energy with which to do jobs. The jobs each have an energy cost associated with them. So, let's say you have 20 energy and there are jobs that require 1 energy, 5 energy, and 6 energy. You can do 20 of the first job, or 4 of the second job, or 3 of the third job (with 2 energy left over, perhaps to do the first job twice as well). That part is pretty simple.
There is also a feature called an "energy boost" which will automatically refill your energy AND gives you 25% more on top of that. Because the energy boost refills your energy for free, it is beneficial to use all of your energy up before you apply the energy boost. For example, if you have 4 energy points left (with a max of 20) and you apply the energy boost, you end up with 25 energy points, with a net gain of 21 points. If you use the 4 energy points before applying the energy boost, you go from 0 to 25, netting you 25 points. The less energy you have, the more you benefit from the energy boost.
From this logic arose some interesting situations. If I have an energy boost available and I still have 137 out of 150 possible energy points, how do I break down the jobs I can do such that I end up with 0 energy. It's not as easy to compute when your jobs take 5, 8, 13, 22, and 27 energy. So I started to wonder...
Can I know without a doubt that, if my energy total is high enough, it is possible to get down to 0 before I start to do my jobs? I figured I'd start with the case where only two jobs are available. Clearly, the job costs have to be relatively prime. If they weren't, 4 and 6 for example, then number that isn't a multiple of their shared factor (odd numbers in this case) would leave a remainder.
I started to do the math and found a neat little formula that tells me without a doubt that if an energy total is above a certain level, I am guaranteed to be able to zero out my energy using relatively prime job costs. I'm not sure why it works yet, but I know I found the solution.
Can you find an answer? Can you tell me why it works? I'm almost certain there's a fundamental theorem in number theory lying around which covers this, but I haven't the time to look it up right now. Do you think Euler had to interrupt his work to change diapers? I think not.
Explanation: I think the explanation of this status will perhaps be more disturbing than the actual message. Then again, if you're a returning visitor to this blog, you probably already know how disturbed I am.
There is a game I play on Facebook in which (among other things) you have an allotment of energy with which to do jobs. The jobs each have an energy cost associated with them. So, let's say you have 20 energy and there are jobs that require 1 energy, 5 energy, and 6 energy. You can do 20 of the first job, or 4 of the second job, or 3 of the third job (with 2 energy left over, perhaps to do the first job twice as well). That part is pretty simple.
There is also a feature called an "energy boost" which will automatically refill your energy AND gives you 25% more on top of that. Because the energy boost refills your energy for free, it is beneficial to use all of your energy up before you apply the energy boost. For example, if you have 4 energy points left (with a max of 20) and you apply the energy boost, you end up with 25 energy points, with a net gain of 21 points. If you use the 4 energy points before applying the energy boost, you go from 0 to 25, netting you 25 points. The less energy you have, the more you benefit from the energy boost.
From this logic arose some interesting situations. If I have an energy boost available and I still have 137 out of 150 possible energy points, how do I break down the jobs I can do such that I end up with 0 energy. It's not as easy to compute when your jobs take 5, 8, 13, 22, and 27 energy. So I started to wonder...
Can I know without a doubt that, if my energy total is high enough, it is possible to get down to 0 before I start to do my jobs? I figured I'd start with the case where only two jobs are available. Clearly, the job costs have to be relatively prime. If they weren't, 4 and 6 for example, then number that isn't a multiple of their shared factor (odd numbers in this case) would leave a remainder.
I started to do the math and found a neat little formula that tells me without a doubt that if an energy total is above a certain level, I am guaranteed to be able to zero out my energy using relatively prime job costs. I'm not sure why it works yet, but I know I found the solution.
Can you find an answer? Can you tell me why it works? I'm almost certain there's a fundamental theorem in number theory lying around which covers this, but I haven't the time to look it up right now. Do you think Euler had to interrupt his work to change diapers? I think not.
Thursday, March 5, 2009
Starbursts + Math = Fun
Most people just eat the candy. Me? I do the math.
Explanation: Not so long ago, I was eating from a box of Starbursts. While not a huge fan of candy, I seem to have a bit of a weakness for their chewy goodness. Anyway, the Starbursts were individually wrapped and inside a box, so I could reach in and just grab a bunch. My normal Starburst eating style is to grab 4 or 5, vow that I'll stop after those, and repeat until the box is empty.
At one point, I took 4 Starbursts out and they were all red. As the red ones are my favorites, I didn't have much of a complaint, but I did notice how unusual it seemed that all four were red. Most people would stop there, but me? I did the math.
Embarrassingly, I had to look to see how many varieties of Starburst were in the box. The answer was five: red, pink, yellow, orange, and green. Furthermore, I assumed that I would have been impressed to pick any four of the same color, not just reds. That means the probability of this occurrence was really just the probability that I picked four Starbursts, where the color of one didn't matter and the color of the other three were the same as the first.*
From there, the math was straightforward. Five to the third power is 125, so the probability was 1/125, or 0.8%. Not inconceivable odds, but not very likely either. I then ate my little math project.
*People tend not to consider this assumption, but it is quite important. For instance, you may look at a family with four daughters and say that it's amazing because the probability that all four children were girls is 1/16, or 6.25%. In reality, though, you would have thought the same thing about four boys. The probability you would be computing is NOT the probability that there would be four girls, it's the probability that you'd be amazed by the genders of the four children. The probability that four kids are all the same sex is just 1/8 or 12.5%. I once demonstrated this to somebody by flipping a coin three times. The coin came up heads all three times. Admittedly, my demonstration could have gone worse, but luckily, it was a friendly coin.**
**Yes, I just said "luckily" in a post about probability. Deal with it.
Explanation: Not so long ago, I was eating from a box of Starbursts. While not a huge fan of candy, I seem to have a bit of a weakness for their chewy goodness. Anyway, the Starbursts were individually wrapped and inside a box, so I could reach in and just grab a bunch. My normal Starburst eating style is to grab 4 or 5, vow that I'll stop after those, and repeat until the box is empty.
At one point, I took 4 Starbursts out and they were all red. As the red ones are my favorites, I didn't have much of a complaint, but I did notice how unusual it seemed that all four were red. Most people would stop there, but me? I did the math.
Embarrassingly, I had to look to see how many varieties of Starburst were in the box. The answer was five: red, pink, yellow, orange, and green. Furthermore, I assumed that I would have been impressed to pick any four of the same color, not just reds. That means the probability of this occurrence was really just the probability that I picked four Starbursts, where the color of one didn't matter and the color of the other three were the same as the first.*
From there, the math was straightforward. Five to the third power is 125, so the probability was 1/125, or 0.8%. Not inconceivable odds, but not very likely either. I then ate my little math project.
*People tend not to consider this assumption, but it is quite important. For instance, you may look at a family with four daughters and say that it's amazing because the probability that all four children were girls is 1/16, or 6.25%. In reality, though, you would have thought the same thing about four boys. The probability you would be computing is NOT the probability that there would be four girls, it's the probability that you'd be amazed by the genders of the four children. The probability that four kids are all the same sex is just 1/8 or 12.5%. I once demonstrated this to somebody by flipping a coin three times. The coin came up heads all three times. Admittedly, my demonstration could have gone worse, but luckily, it was a friendly coin.**
**Yes, I just said "luckily" in a post about probability. Deal with it.
Friday, February 6, 2009
Time is Money
Were those five minutes of my time really worth the 20 cents?
Explanation: Last night I got a text message. It wasn't from someone I knew - it was some MP3 service sending me spam of some sort. I've never seen a text like that before. The details don't really matter. The important fact is that I have text messaging blocked on my phone... or at least I thought I did.
This is the kind of thing that really drives me nuts. I'm supposed to have text messages blocked. I clearly don't anymore. There's no way I'm going to let Verizon charge me 20 cents for a text message that I shouldn't have received in the first place. No way. Plus, this message might be the tip of the iceberg. I had to nip it in the bud, so I gave Verizon a call.
After five minutes on the phone with Verizon, we established that my block had been removed for some reason. The block has now been re-applied to my account. They were unsure if the 20 cents would be refunded to me automatically, so I have to check my next bill and let them know if the charge is still there (more time and effort) to make sure I'm not paying for a text message I didn't want in the first place.
From a sheer economic standpoint, this drives me insane. My time is money. Now, how much money one's time is worth is a different debate, but I like to base decisions on an amount that I think I could easily earn in an hour's time. The New Jersey minimum wage is currently $7.15, so that's a safe place to start. I like to think I bring a bit more to the table, so I generally assume $10 an hour. Certainly I could make more than that if I was working harder, but I feel like that's a wage I could pull in without really exerting myself too much.
Once you have a dollar value, decisions are easier. Do I pay the neighborhood guy $30 each time he mows my lawn? Well, I can mow my entire lawn in about an hour, so I can consider that as $10 worth of time. Is it worth paying $30 to preserve $10 worth of time? No way. I'll mow my own lawn.
Now, obviously some experiences have some intrinsic value (or lack thereof) that you have to include as well. Would I spend $30 for the neighborhood guy to spend an hour removing the poison ivy from my backyard? Sure I would. My time is worth $10, and not spending an hour handling poison ivy is worth much more than $20 to me.*
On the other side of the coin, I wouldn't pay the guy $5 to wash my car for an hour on a hot summer afternoon. Sure, my time is worth $10, but spending an hour in the cold mist of the hose is worth way more than the $5 I'd save by letting him do it.
Now, let's consider my phone call. I spent 5 minutes of my time in an effort to save 20 cents. That's an hourly savings of $2.40. My time is worth about $10 an hour. You do the math. Even worse is the fact that A) you just know they're going to screw up my bill anyway, so I have to keep an eye on it and B) when they do screw it up I'm going to have to spend ANOTHER 5 minutes on the phone with them. That takes me up to about 15 minutes of total effort, taking the hourly savings down to 80 cents an hour.
So, in short, I hate Verizon. And now I have the math to prove it!
*Plus, I have to add value for the pleasure of knowing that guy is going to be itching for the next two weeks. That'll teach him to up his rates!
Explanation: Last night I got a text message. It wasn't from someone I knew - it was some MP3 service sending me spam of some sort. I've never seen a text like that before. The details don't really matter. The important fact is that I have text messaging blocked on my phone... or at least I thought I did.
This is the kind of thing that really drives me nuts. I'm supposed to have text messages blocked. I clearly don't anymore. There's no way I'm going to let Verizon charge me 20 cents for a text message that I shouldn't have received in the first place. No way. Plus, this message might be the tip of the iceberg. I had to nip it in the bud, so I gave Verizon a call.
After five minutes on the phone with Verizon, we established that my block had been removed for some reason. The block has now been re-applied to my account. They were unsure if the 20 cents would be refunded to me automatically, so I have to check my next bill and let them know if the charge is still there (more time and effort) to make sure I'm not paying for a text message I didn't want in the first place.
From a sheer economic standpoint, this drives me insane. My time is money. Now, how much money one's time is worth is a different debate, but I like to base decisions on an amount that I think I could easily earn in an hour's time. The New Jersey minimum wage is currently $7.15, so that's a safe place to start. I like to think I bring a bit more to the table, so I generally assume $10 an hour. Certainly I could make more than that if I was working harder, but I feel like that's a wage I could pull in without really exerting myself too much.
Once you have a dollar value, decisions are easier. Do I pay the neighborhood guy $30 each time he mows my lawn? Well, I can mow my entire lawn in about an hour, so I can consider that as $10 worth of time. Is it worth paying $30 to preserve $10 worth of time? No way. I'll mow my own lawn.
Now, obviously some experiences have some intrinsic value (or lack thereof) that you have to include as well. Would I spend $30 for the neighborhood guy to spend an hour removing the poison ivy from my backyard? Sure I would. My time is worth $10, and not spending an hour handling poison ivy is worth much more than $20 to me.*
On the other side of the coin, I wouldn't pay the guy $5 to wash my car for an hour on a hot summer afternoon. Sure, my time is worth $10, but spending an hour in the cold mist of the hose is worth way more than the $5 I'd save by letting him do it.
Now, let's consider my phone call. I spent 5 minutes of my time in an effort to save 20 cents. That's an hourly savings of $2.40. My time is worth about $10 an hour. You do the math. Even worse is the fact that A) you just know they're going to screw up my bill anyway, so I have to keep an eye on it and B) when they do screw it up I'm going to have to spend ANOTHER 5 minutes on the phone with them. That takes me up to about 15 minutes of total effort, taking the hourly savings down to 80 cents an hour.
So, in short, I hate Verizon. And now I have the math to prove it!
*Plus, I have to add value for the pleasure of knowing that guy is going to be itching for the next two weeks. That'll teach him to up his rates!
Tuesday, January 13, 2009
Commuting?
Oddly, I'm working from home and yet my commute was much longer than usual this morning.
Explanation: To make our afternoon scheduling easier, I drove my wife to work this morning. I understand that most people do this sort of thing every day, but I am not most people. Furthermore, if I do have to leave the house first thing in the morning, I would prefer that it wasn't during the coldest spell we've had all winter. Anyway, I am now back at home, but a bit more cold and confused than normal.
Special Blog Bonus: Ever wonder how to make a Penrose Triangle out of LEGOs? Here are the simple instructions:
If you still have difficulty, you might want to consider this model for help:
Explanation: To make our afternoon scheduling easier, I drove my wife to work this morning. I understand that most people do this sort of thing every day, but I am not most people. Furthermore, if I do have to leave the house first thing in the morning, I would prefer that it wasn't during the coldest spell we've had all winter. Anyway, I am now back at home, but a bit more cold and confused than normal.
Special Blog Bonus: Ever wonder how to make a Penrose Triangle out of LEGOs? Here are the simple instructions:
If you still have difficulty, you might want to consider this model for help:
Monday, September 8, 2008
The Fraternal Order of Delta Hand
"So, you're dealt a hand, right?" "No, I'm Alpha Omicron Pi."
Explanation: Recently, I've been pondering a probability issue in poker. I attempted to explain it to my wife last night, resulting in the above conversation. Now, to her credit, she was joking, and it was so funny I left the room to go write it down. That pretty much ended our conversation, though. Things like this happen when you marry someone who was in a sorority.
Now Down To The Nitty-Gritty: My problem is this: In Texas Hold'Em, what is the probability that my starting hand will win against n other starting hands. For example, if you are dealt pocket aces and are playing heads up, you have about an 85% chance of beating a random hand.
It's easy to design an algorithm for this. For a given starting hand, iterate through all other possible starting hands (or combinations thereof) and generate all possible combinations of community cards for those starting hands, keeping track of the total wins for your starting hand and the total number of permutations considered. On my home machine, given the best algorithm I've come up with so far, I think I can perform this computation for n=1 for all possible starting hands in under a day. Last time I did this was a while ago, so I don't remember exactly how long it took, but I've done it before in a reasonable amount of time.
Using the same algorithm for n=2, however, increases the complexity to the point where it can take days to consider a single starting hand. There are 169 possible starting hands, so the computation for all possible starting hands cannot be completed in a reasonable amount of time. I need a better algorithm, and that's what I've been working on.
Special Blog Bonus: This problem mocks me, as the snake mocks the snake charmer* in this Far Side:
*Yeah, my segue is a stretch. So what? You try connecting a snake in Groucho glasses to a complex combinatorics problem.
Explanation: Recently, I've been pondering a probability issue in poker. I attempted to explain it to my wife last night, resulting in the above conversation. Now, to her credit, she was joking, and it was so funny I left the room to go write it down. That pretty much ended our conversation, though. Things like this happen when you marry someone who was in a sorority.
Now Down To The Nitty-Gritty: My problem is this: In Texas Hold'Em, what is the probability that my starting hand will win against n other starting hands. For example, if you are dealt pocket aces and are playing heads up, you have about an 85% chance of beating a random hand.
It's easy to design an algorithm for this. For a given starting hand, iterate through all other possible starting hands (or combinations thereof) and generate all possible combinations of community cards for those starting hands, keeping track of the total wins for your starting hand and the total number of permutations considered. On my home machine, given the best algorithm I've come up with so far, I think I can perform this computation for n=1 for all possible starting hands in under a day. Last time I did this was a while ago, so I don't remember exactly how long it took, but I've done it before in a reasonable amount of time.
Using the same algorithm for n=2, however, increases the complexity to the point where it can take days to consider a single starting hand. There are 169 possible starting hands, so the computation for all possible starting hands cannot be completed in a reasonable amount of time. I need a better algorithm, and that's what I've been working on.
Special Blog Bonus: This problem mocks me, as the snake mocks the snake charmer* in this Far Side:
*Yeah, my segue is a stretch. So what? You try connecting a snake in Groucho glasses to a complex combinatorics problem.
Friday, May 16, 2008
Fermat's Last Status Message
I have discovered a truly marvelous status message today, which this area is too narrow to contain.
Explanation: Those of you who are familiar with the history of number theory (or, more likely, with my blog) recognize this as a knockoff of Pierre de Fermat's words describing his Last Theorem. And yes, I've mentioned this theorem before. Some topics are too good to let go of.
Special Blog Announcement: Weekend posts are back! Through the power of scheduled posts (it's about time) Jeremy's Status Message will come to you with posts every day! We apologize for our recent weekend laziness.
Explanation: Those of you who are familiar with the history of number theory (or, more likely, with my blog) recognize this as a knockoff of Pierre de Fermat's words describing his Last Theorem. And yes, I've mentioned this theorem before. Some topics are too good to let go of.
Special Blog Announcement: Weekend posts are back! Through the power of scheduled posts (it's about time) Jeremy's Status Message will come to you with posts every day! We apologize for our recent weekend laziness.
Monday, May 12, 2008
Call Me When He Turns 720
In honor of Tony Hawk's 40th birthday, I did a switch-footed 720 hyper-reverse aerial ollie kickflip faceplant off of my roof this morning.
Explanation: Yeah, that's right. Tony Hawk is 40 years old today. Oh, and in case you're wondering, the key to this complicated skating maneuver is landing on the soft part of your face. If I can do it, I'm sure you can learn it with only a little practice.
In researching the gibberish used in this post, I found online instructions (by the birthday boy himself) on how to do Ollie Fingerflips and Impossibles, Acid Drops and Bomb Drops, Ollie Norths, No Complys, Half Cabs, Bonelesses, and of course, Bam Bologna.
Special Blog Bonus: From skateboarding to LEGO dinosaurs (I'm sure there's a segue in there somewhere) please take a moment to appreciate the mating rituals of the panacotos:
Explanation: Yeah, that's right. Tony Hawk is 40 years old today. Oh, and in case you're wondering, the key to this complicated skating maneuver is landing on the soft part of your face. If I can do it, I'm sure you can learn it with only a little practice.
In researching the gibberish used in this post, I found online instructions (by the birthday boy himself) on how to do Ollie Fingerflips and Impossibles, Acid Drops and Bomb Drops, Ollie Norths, No Complys, Half Cabs, Bonelesses, and of course, Bam Bologna.
Special Blog Bonus: From skateboarding to LEGO dinosaurs (I'm sure there's a segue in there somewhere) please take a moment to appreciate the mating rituals of the panacotos:
Wednesday, January 9, 2008
Da Vinci's Last Deathtrap
"I have devised a cunning and inescapable deathtrap. Unfortunately, this room is too small to contain it."
Source: This is the punchline from yesterday's Irregular Webcomic. Say what you want about internet comics, but there's no way you're finding a good number theory joke in Family Circus or Shoe.
Explanation: For those of you who do not remember (and shame on you if you don't) I first mentioned Fermat's last theorem nine days ago. You can read about it under the explanation for item "2".
Source: This is the punchline from yesterday's Irregular Webcomic. Say what you want about internet comics, but there's no way you're finding a good number theory joke in Family Circus or Shoe.
Explanation: For those of you who do not remember (and shame on you if you don't) I first mentioned Fermat's last theorem nine days ago. You can read about it under the explanation for item "2".
Monday, December 31, 2007
The Twelve Days of Christmas: Day 7
Jeremy's Status Message Proudly Presents The Twelve Days of Christmas:
On the seventh day of Christmas,
My new iPod gave to me,
Seven Mary Three
Sixpence None The Richer
Five for Fighting,
The Four Tops,
Three Doors Down
Two Skinnee J's,
And nothing by the Partridge Family, so don't even joke about it.
Explanation: Also on my new iPod are:
Seven One Eight, by the 2 Skinnee J's
Six Underground, by the Sneaker Pimps
Five Piece Chicken Dinner, by the Beastie Boys
Four Sticks, by Led Zeppelin
Three MCs and one DJ, by the Beastie Boys
Two Step, by the Dave Matthews Band
And The Humpty Dance, by Digital Underground.
Yeah, that's right - The Humpty Dance. It has nothing to do with partridges or pear trees. It's just a cool song.
Special Blog Bonus: Feeling numerical? Don't forget 5150 by Van Halen, #41 by the Dave Matthews Band, 50 Ways to Leave Your Lover by Paul Simon, 8 Mile by Eminem, 316 by Van Halen, 99 Problems by Jay-Z, 1979 by the Smashing Pumpkins, 1984 by Van Halen, Nineteen Naughty Three by Naughty By Nature, 100 Years by Five for Fighting, One Thing by Finger Eleven, 3 A.M. Eternal by the KLF, Ten by Pearl Jam, and 6th Avenue Heartache by the Wallflowers. Speaking of numerical...
Solutions to Yesterday's Evil Number Theory Questions:
6 is the smallest perfect number, because it is equal to the sum of its proper positive divisors: 1, 2, and 3. Can you find the next even perfect number? Can you find an odd perfect number?
The next even perfect number is 28, which is the sum of 1, 2, 4, 7, and 14. The perfect numbers after that are 496 and 8128. All known perfect numbers fit the formula 2^(n-1) * (2^n - 1) where 2^(n-1) is a Mersenne prime. It is unknown whether there are any odd perfect numbers. If you managed to find one, you should tell me what it is and never speak of it again.
5 is a prime number whose binary representation, "101" is palindromic, meaning its digits read the same backward as forward. What is the largest known prime number that is palindromic in binary?
The largest known prime number which is palindromic in binary is 2^32582657, which is a 9.8 million digit number and happens to be the largest prime number known to date. It is a Mersenne prime, meaning it is one less than a power of two, hence the binary representation of the number is a string of ones, and is therefore the same both backward and forward. All of the largest known primes are Mersenne primes.
4 is an even integer that can be written as the sum of two primes, namely 2 and 2. Can you find an even integer that cannot?
This little gem is called Goldbach's conjecture. It is a conjecture because it is unproven. If you have found an integer that cannot, again, please contact me immediately with that number and then treat yourself to a nice long vacation in a place without telephones while I hold onto it all safe and sound for you. Don't worry. I won't tell anyone.
3 is a triangular number, because it belongs to the sequence (1, 3, 6, 10, 15, 21, ...). The number 3 can be represented as a sum of three triangular numbers: 3 = 1 + 1 + 1. Can you find an integer that cannot?
No you cannot. This is a subset of the Fermat polygonal number theorem, which states that every positive integer is the sum of at most n n-polygonal numbers. So, in this case, n=3.
2 is a value for n such that there exist nonzero integers x, y, and z where x^n + y^n = z^n. For example, 3^2 + 4^2 = 5^2. Can you find a value of n greater than 2 for which this is also true?
This would be Fermat's Last Theorem. Pierre de Fermat famously wrote the following words in the margin of a book: "It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second into two like powers. I have discovered a truly marvelous proof of this, which this margin is too narrow to contain." For over 350 years, mathematicians cursed this man and his "marvelous proof", until Andrew Wiles of Princeton published a proof in 1995. Anyway, in short, the answer to this question is "No, you cannot".
1 is the difference between 2 and 3. There also exist powers of 2 and 3 such that their difference is also 1, namely 2^3=8 and 3^2=9. Can you find another pair of consecutive integers who have powers that are also consecutive?
This problem was known as Catalan's conjecture for over 150 years until it was solved by Preda Mihăilescu in 2002, and it is now referred to by the much more pronounceable "Mihăilescu's theorem." The example I cited shows the only pair of consecutive integers (2 and 3) for which this is possible. So, if you found another pair of consecutive integers, I suggest you check your math.
On the seventh day of Christmas,
My new iPod gave to me,
Seven Mary Three
Sixpence None The Richer
Five for Fighting,
The Four Tops,
Three Doors Down
Two Skinnee J's,
And nothing by the Partridge Family, so don't even joke about it.
Explanation: Also on my new iPod are:
Seven One Eight, by the 2 Skinnee J's
Six Underground, by the Sneaker Pimps
Five Piece Chicken Dinner, by the Beastie Boys
Four Sticks, by Led Zeppelin
Three MCs and one DJ, by the Beastie Boys
Two Step, by the Dave Matthews Band
And The Humpty Dance, by Digital Underground.
Yeah, that's right - The Humpty Dance. It has nothing to do with partridges or pear trees. It's just a cool song.
Special Blog Bonus: Feeling numerical? Don't forget 5150 by Van Halen, #41 by the Dave Matthews Band, 50 Ways to Leave Your Lover by Paul Simon, 8 Mile by Eminem, 316 by Van Halen, 99 Problems by Jay-Z, 1979 by the Smashing Pumpkins, 1984 by Van Halen, Nineteen Naughty Three by Naughty By Nature, 100 Years by Five for Fighting, One Thing by Finger Eleven, 3 A.M. Eternal by the KLF, Ten by Pearl Jam, and 6th Avenue Heartache by the Wallflowers. Speaking of numerical...
Solutions to Yesterday's Evil Number Theory Questions:
6 is the smallest perfect number, because it is equal to the sum of its proper positive divisors: 1, 2, and 3. Can you find the next even perfect number? Can you find an odd perfect number?
The next even perfect number is 28, which is the sum of 1, 2, 4, 7, and 14. The perfect numbers after that are 496 and 8128. All known perfect numbers fit the formula 2^(n-1) * (2^n - 1) where 2^(n-1) is a Mersenne prime. It is unknown whether there are any odd perfect numbers. If you managed to find one, you should tell me what it is and never speak of it again.
5 is a prime number whose binary representation, "101" is palindromic, meaning its digits read the same backward as forward. What is the largest known prime number that is palindromic in binary?
The largest known prime number which is palindromic in binary is 2^32582657, which is a 9.8 million digit number and happens to be the largest prime number known to date. It is a Mersenne prime, meaning it is one less than a power of two, hence the binary representation of the number is a string of ones, and is therefore the same both backward and forward. All of the largest known primes are Mersenne primes.
4 is an even integer that can be written as the sum of two primes, namely 2 and 2. Can you find an even integer that cannot?
This little gem is called Goldbach's conjecture. It is a conjecture because it is unproven. If you have found an integer that cannot, again, please contact me immediately with that number and then treat yourself to a nice long vacation in a place without telephones while I hold onto it all safe and sound for you. Don't worry. I won't tell anyone.
3 is a triangular number, because it belongs to the sequence (1, 3, 6, 10, 15, 21, ...). The number 3 can be represented as a sum of three triangular numbers: 3 = 1 + 1 + 1. Can you find an integer that cannot?
No you cannot. This is a subset of the Fermat polygonal number theorem, which states that every positive integer is the sum of at most n n-polygonal numbers. So, in this case, n=3.
2 is a value for n such that there exist nonzero integers x, y, and z where x^n + y^n = z^n. For example, 3^2 + 4^2 = 5^2. Can you find a value of n greater than 2 for which this is also true?
This would be Fermat's Last Theorem. Pierre de Fermat famously wrote the following words in the margin of a book: "It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second into two like powers. I have discovered a truly marvelous proof of this, which this margin is too narrow to contain." For over 350 years, mathematicians cursed this man and his "marvelous proof", until Andrew Wiles of Princeton published a proof in 1995. Anyway, in short, the answer to this question is "No, you cannot".
1 is the difference between 2 and 3. There also exist powers of 2 and 3 such that their difference is also 1, namely 2^3=8 and 3^2=9. Can you find another pair of consecutive integers who have powers that are also consecutive?
This problem was known as Catalan's conjecture for over 150 years until it was solved by Preda Mihăilescu in 2002, and it is now referred to by the much more pronounceable "Mihăilescu's theorem." The example I cited shows the only pair of consecutive integers (2 and 3) for which this is possible. So, if you found another pair of consecutive integers, I suggest you check your math.
Saturday, December 29, 2007
The Twelve Days of Christmas: Day 6
Jeremy's Status Message Proudly Presents The Twelve Days of Christmas:
On the sixth day of Christmas, an evil number theorist said to me,
6 is the smallest perfect number, because it is equal to the sum of its proper positive divisors: 1, 2, and 3. Can you find the next even perfect number? Can you find an odd perfect number?
5 is a prime number whose binary representation, "101" is palindromic, meaning its digits read the same backward as forward. What is the largest known prime number that is palindromic in binary?
4 is an even integer that can be written as the sum of two primes, namely 2 and 2. Can you find a greater even integer that cannot?
3 is a triangular number, because it belongs to the sequence (1, 3, 6, 10, 15, 21, ...). The number 3 can be represented as a sum of three triangular numbers: 3 = 1 + 1 + 1. Can you find a greater integer that cannot?
2 is a value for n such that there exist nonzero integers x, y, and z where x^n + y^n = z^n. For example, 3^2 + 4^2 = 5^2. Can you find a value of n greater than 2 for which this is also true?
1 is the difference between 2 and 3. There also exist powers of 2 and 3 such that their difference is also 1, namely 2^3=8 and 3^2=9. Can you find another pair of consecutive integers who have powers that are also consecutive?
Explanation: I love number theory because it is primarily concerned with integers, which seem quite simple, and yet can be extremely complex. These are just a few examples. I will post the solutions tomorrow.
On the sixth day of Christmas, an evil number theorist said to me,
6 is the smallest perfect number, because it is equal to the sum of its proper positive divisors: 1, 2, and 3. Can you find the next even perfect number? Can you find an odd perfect number?
5 is a prime number whose binary representation, "101" is palindromic, meaning its digits read the same backward as forward. What is the largest known prime number that is palindromic in binary?
4 is an even integer that can be written as the sum of two primes, namely 2 and 2. Can you find a greater even integer that cannot?
3 is a triangular number, because it belongs to the sequence (1, 3, 6, 10, 15, 21, ...). The number 3 can be represented as a sum of three triangular numbers: 3 = 1 + 1 + 1. Can you find a greater integer that cannot?
2 is a value for n such that there exist nonzero integers x, y, and z where x^n + y^n = z^n. For example, 3^2 + 4^2 = 5^2. Can you find a value of n greater than 2 for which this is also true?
1 is the difference between 2 and 3. There also exist powers of 2 and 3 such that their difference is also 1, namely 2^3=8 and 3^2=9. Can you find another pair of consecutive integers who have powers that are also consecutive?
Explanation: I love number theory because it is primarily concerned with integers, which seem quite simple, and yet can be extremely complex. These are just a few examples. I will post the solutions tomorrow.
Subscribe to:
Posts (Atom)