Landscape of Round Holes

Save vs. Blog

Well, I haven't posted in a while, and for that i apologize. I've been busy with other things, mostly helping my father restore an old Victorian home. But that is another post. For now i would like to introduce the topic of RPGs, that's Role Playing Games. It's most well known example is D&D, yes who would have thought that a guy that does math for fun would have played D&D, right? But actually they two are rather co-incidental, the math satisfies a certain logical, structural inclination, and the role-playing a more creative, spontaneous side - it's a yin and yang thang.

There are, in the broadest sense, two types of role-players: the player and the person who moderates the game, known, generally, as the GM. Now, most GMs are creative enough that they want to add to, subtract from, and/or modify the existing rules, which is entirely acceptable, almost expected. Many times this leads to a desire to create an entirely new setting. I known i have had this desire myself and even my most recent opportunity to play was in such a "home-made" setting. And in my opinion is the pinnacle and ultimate expression of role-playing. The dream, of course, would be to see your creation published, which i have not known anyone to do - until recently.

Pembrose, a old role-playing buddy, has been asked to create a supplement for an established, but still growing, role-playing setting called Lejendary Adventures, by none other that the father of role-playing Gary Gygax. Just to be clear, as grammar is funny sometimes, not only is the game by Gary Gygax, but he was asked by Gary Gygax... personally. Pardon me while i nerd out for a second, but... that is just fucking cool! He is new to, but intrigued by, the blogging thing so, I would encourage all who, like me, would enjoy the vicarious thrill of what it's like to create official role-playing material to visit Pembrose's site and welcome him to the blog-o-sphere.

Now that i have marked myself as a D&D geek, i will field the next (rolling a d20... "crap") 18 questions about my role-playing past. Damn, this really makes me want to play again - tom, are you reading this?

OK, if i have a number expressed in this form:

a₁kⁿ + a₂k^n-1 + a₃k^n-2 ... + a_nk¹

then the coeffcients and the exponents can replace the x and y from the previous posts table/program. So, a₁ for x and n for y will be answer one, then a₂ for x and n-1 for y will be answer two, and so on up to a_n for x and 1 for y. Answer one is z₁ and answer two is z₂ then the sum of all the answers will be called z' [pronounced zee prime, no fantasy island jokes please]

Now, n is only one variable, but each of the a's is it's own variable so z' can be practically anything you want. So, question one: How many z' 's are there if an a can't be more than n? and question two: if given a number for z' how many ways, expressed in the above form, will produce that z' if an a cannot be more than n?

strange table

y
:
5	||	0	5	6	7	8	9
4	||	0	4	5	6	7	8
3	||	0	3	4	5	6	7
2	||	0	2	3	4	5	6
1	||	0	1	2	3	4	5
0	||	0	0	0	0	0	0
	||=	=	=	=	=	=	=
		0	1	2	3	4	5	..	x

Open call for help. What formula creates the above table. It has two variables, x and y, which will always be greater than or equal to 0. If x and y are not zero then the formula is x+y-1, but if either x or y is zero then the results is 0.

Anime for Beginners

Best Science Fiction Anime

1. Cowboy Bebob (espesially the series)
2. Ghost in the Shell
3. Battle Angle Alita
4. Patlabor
5. Macross Plus

Best Fantasy Anime

1. Ninja Scroll
2. Record of Lodoss War (series)
3. Read Or Die (espesially the movie)
4. Samurai X (aka Ruroni Kenshin) - warning loooong series
5. Vampire Hunter D

Best Anime Movies

1. Spirited Away
2. Princess Mononoke
3. Grave of the Fireflies
4. Nuasica of the Valley of the Wind
5. Metropolis

Anime that is recommended but i haven't seen yet

1. Perfect Blue
2. Neon Genisis: Eveangelion
3. Escaflowne
4. Samurai Champloo
5. FLCL (pronounced 'fooly cooly') - supposed to be weird

Best Overall

1. Cowboy Bebob, the series (watch it, love it, best animation ever created since DaVinci dropped the Mona Lisa)
2. Ghost in the Shell
3. Princess Mononoke (i prefer Nausica, but i think this is better for the beginner)

10 most random songs list

In due compliance with very metal's meme I have listed the first ten songs from my entire music library when placed on shuffle, I took the liberty of excluding the duplicates of artists already listed.

1. "Quincy Punk Episode" - Spoon - Series of Sneaks
2. "Men in Black" - Will Smith - Greatest Hits
3. "Lovefool" - The Cardigans - single
4. "Gasoline" - Audioslave - Audioslave
5. "Lay It Down" - Aerosmith
6. "Battle Flag" - Lo Fidelity Allstars - How to Operate on a Blown Mind
7. "Cowgirl" - Underworld - ???
8. "Westbound Sign" - Greenday - insomniac
9. "Shake Your Rump" - Beastie Boys - The Sounds of Science
10. "Question" - Old 97's - Satellite Rides
11. "Belleville Rendevouz" - ??? - Soundtrack (this was just so random i had to include it)

If you are reading this and you have not already tagged by very metal - you are now tagged. Open your music player, list all your songs, shuffle them, take the first ten - no matter what they are. If you have a blog, send me a trackback ping for the, uh scientific tracking of memes.

P.S. Man i am so glad that my Aqua's "Barbie Girl" didn't show up. :)

I drink therefore I am

Last week i had a brief, if somewhat inebriated, philosophical discussion with a stranger in a bar, it was about the classic question of free will. I took the pro side and he took the con position, that was not necessarily his own, so that we could have a dialog. I would like to introduce the topic here, now.

I will offer this for the argument against free will, until someone else takes up the mantle, it's a simple and powerful argument, especially if repeated - the brain is the brain. Meaning, that the brain and it's products, namely the mind, are set in place. It can't change itself. If a person commits a crime, he can't be held accountable - the brain is the brain. Anything that brought him to that point was an outside factor beyond his control, e.g. an abusive father, alcohol addiction, or even a physiological proclivity towards crime. All either set from the beginning or from cause and effect.

Now, my argument for free will, that i have only recently begun to flesh out is this. The brain is a dynamic system. A non-deterministic 'machine' if you will, that even knowing all the inputs, the output cannot be guessed - with 100% accuracy. So, if the brain, and subsequently the mind, cannot be predicted, then it cannot be said, in hind site, that a particular outcome was inevitable.

By analogy with the weather, if we can't tell if it is going to rain or not one month from now (even if given the present conditions in perfect accuracy), how can we say, a month later when it does rain, that it couldn't have happened any other way?

I don't know, what do you think?

Update: There is already a school of thought with a similar line of reasoning, and an awarded book on the subject, The Significance of Free Will, by Robert Kane, a professor at the University of Texas, Austin. That just made no. 2 on my reading list.

Did you get that thlog I sent you?

Since construction takes too long to do with each number lets construct them using the variable k:

m=1

k = k

m=2

k*k = k^2

k+k = 2k

k/k = 1

k-k = 0

m=3

k^2 *k = k^3                  k^2 +k = k^2+k              [k^2 /k] = repeat               |k^2 -k| = k^2-k

2k *k = 2k^2                  2k +k = 3k                     [2k /k] = 2                      |2k -k| = repeat

1 *k = repeat                 1 +k = k+1                     [1 /k] = repeat               |1 -k| = k-1

0 *k = repeat                 0 +k = repeat                [0 /k] = undefined            |0 -k| = repeat

m=4 (discarding all the non-useful results)

n^4

2n^3

n^3+n^2

n^3-n^2

n^3+n

n^3-n

4n^2

3n^2

2n^2+n

2n^2-n

n^2+2n

n^2-2n

n^2+1

n^2-1

4n

2n+1

2n-1

3

n/2

Ordering all of the results (ignoring n/2 for now) give this sequence, also shown are the number, m, of k's used:

n                      m

0                      2

1                      2

2                      3

3                      4

:

k-1                   3

k                       1

k+1                   3

:

2k-1                  4

2k                     2

2k+1                 4

:

3k                     3

:

4k                     4

:

k^2-2k              4

k^2-k                3

k^2-1                4

k^2                   2

k^2+1               4

k^2+k               3

k^2+2k             4

:

2k^2-k              4

2k^2                 3

2k^2+k             4

:

3k^2                 4

:

4k^2                 4

:

k^3-k^2             4

k^3-k                4

k^3                   3

k^3-k                4

k^3+k^2            4

:

2k^3                 4

:

k^4                   4

The patterns a little hard to see, but it's there. Concentrating on just the numbers between 1 and k (assuming k is relatively large) gives this table, where m is the number of k's used in that number, t is the total number of k's used up to that number and a is the average:

n                      m         t           a

1                      2          2          2

2                      3          5          2.5

3                      4          9          3

4                      5          14        3.5

:

k-4                   6          ?          ?

k-3                   5          ?          ?

k-2                   4          ?          ?

k-1                   3          ?          ?

k                      1          ?          ?

Now as n increase from one, the number of k's increase by one each time; likewise as n decreases from (k-1), the number of k's also increases by one each time. So somewhere around the middle will be a number that uses the maximum number of k's. If k is odd then the number will be (k+1)/2 and m will be (k+3)/2; and if k is even there will be two numbers, k/2 and (k+2)/2, m will be (k/2)+1 for both. Introducing those into the table gives:

(k is odd)

n                      m                     t                       a

1                      2                      2                      2

2                      3                      5                      2.5

3                      4                      9                      3

4                      5                      14                    3.25

:                       :                       :                       :

(k+1)/2             (k+3)/2             sum(k+3)/2 -1   (sum(k+3)/2 -1) / n

:                       :                       :                       :

k-3                   5                      ?                      ?

k-2                   4                      ?                      ?

k-1                   3                      ?                      ?

k                      1                      ?                      ?

(k is even)

n                      m                     t                       a

1                      2                      2                      2

2                      3                      5                      2.5

3                      4                      9                      3

4                      5                      14                    3.25

:                       :                       :                       :

k/2                    (k+2)/2             sum(k+2)/2 -1   (sum(k+2)/2 -1) / n

(k+2)/2             (k+2)/2             -                       -

:                       :                       :                       :

k-3                   5                      ?                      ?

k-2                   4                      ?                      ?

k-1                   3                      ?                      ?

k                      1                      ?                      ?

I encourage you to try it out, use 9 and 10 as k for odd and even and fill in the tables - see if it works.

So, if t is a running total up to 'the halfway point' and there is a similar running total from (k-1) down to 'the halfway point', then the grand total is both of those together. If k is odd the total, t when n=k is {(sum(k+3)/2) + (sum(k+1)/2) -3} which after simplifying is (k^2 + 6k -3)/4; and if k is even, the total, t when n=k is, after simplifying, (k^2 + 6k -4)/4. The respective averages would then be those totals divided by n, which in this case is equal to k.

Test question: What is the average number of fours needed to make all the numbers between one and four (using only +,-,*, and /)?

Test answer: if k=4 then

1) a = [ (k^2 + 6k - 3)/4 ] / k

2) a = [ (16 + 24 - 3)/4 ] / 4

3) a = [ (37)/4 ] / 4

4) a = [ 9.25 ] / 4

5) a = 9 / 4

6) a = 2.25

Conclusion for part 2: the average number of k's needed to make up all the numbers between 1 and k is equal to k squared, plus six k, minus three; divided by 4, rounded down, divided by k. Now do you remember the n/2 i left out of consideration a while back, well that guy is going to complicate things a bit, hopefully nothing i can't finish, but that will have to wait until the next post. As always comments are open, any observations or ideas will be considered.

Mr. Roboto Strikes Again.

Your Brain's Pattern

Structured and organized, you have a knack for thinking clearly. You are very logical - and you don't let your thoughts get polluted with emotions. And while your thoughts are pretty serious, they're anything from boring. It's minds like yours that have built the great cities of the world!

Yes, who could have guessed, right? Well apparently my structured, clear, logical, unemotional brain had pretty much guessed that this was going to be the results i would get by picking this picture. I don't mind so much being "logical" as much as just being so predictably logical. But, dammit, my logical brain says they go hand in hand.

Here is my second choice,

Your Brain's Pattern

Your brain is always looking for the connections in life. You always amaze your friends by figuring out things first. You're also good at connecting people - and often play match maker. You see the world in fluid, flexible terms. Nothing is black or white.

What Pattern Is Your Brain?

Oh yea, i almost forgot, i guess is should start ending my posts with, "Live long, and prosper." It is the only logical thing to do. Instead I'll give you a quote from a different TV series from the same period, of which i was reminded of recently.

"I will not be pushed, filed, stamped, indexed, briefed, debriefed or numbered."  -- The Prisoner.

Gentleman, BEHOLD! ...a Thlog.

I have decided to blog one of my math distractions, a kind of peek inside my head as I go through the steps of trying to understand something just over my intellectual horizon. The problem I have chosen to inflict upon you is a generalization of the four 4 problem introduced to me a while back. Now this is a live production, you will be getting the info pretty much as I come up with it, so it will probably take several posts and may not have a conclusion - many times I either find a more interesting problem to work on or I the problem becomes too difficult for me to finish.

OK, here is the general idea: if given a number k what is the least number of k's needed to make any Natural number n, and what is the average number of k's used for all the numbers up to n? In other words if given any number, say 7, how many 7's does it take to make each of the numbers from 1 to whatever. Now in order to make the problem workable within my amateur time frame, I also decided to restrict the operations, allowing only arithmetic operators i.e. addition, subtraction, multiplication, division, exponents, and roots (higher level operators like tetration are also allowed however they probably won't be needed). To be clear factorials are not used, neither is concatenation (44 is not used as two fours) or decimals (it's a form of concatenation), so repeating decimals is right out. Just digits and simple math.

Where to start, well, the Zen Buddhist in me says - at the beginning. So I’ll list all the ways to make each of the numbers from 1 to what ever using 1's:

k=1

n                       m         t           a

1= 1                  1          1          1

2= 1+1              2          3          1.5

3= 1+1+1          3          6          2

4= 1+1+1+1      4          10        2.5

:                        :           :           :

n= 1+1+1...       n          sum(n) (n+1)/2

Key: k is the number to make the n's out of, m is the number of k's used to make that number, t is the total number of k's used up to that number, and a is the average number of k's used. (editor's note: the 's is probably not grammatically correct but the spacing makes it easier for me to read and should be read as the plural not the possessive).

So for k=1 the minimum number of ones needed to make any number n is n and the average number of 1's needed to make all the numbers from 1 to n is (n+1)/2. Simple right? :) Ah, but 1 is often a trivial case, lets go on to 2. I know, I know, I want to jump to the bigger numbers like 4 too, but lets build a case by pseudo-induction.

k=2

n                                    m         t           a                       l

  1= 2/2                           2          2          2                       2

  2= 2                              1          3          1.5                    2

  3= 2 + (2/2)                   3          6          2                       3

  4= 2x2                           2          8          2                       3

  5= (2x2) + (2/2)             4          12        2.4                    4

  6= 2+2+2                      3          15        2.5                    4

  7= (2x2) + (2/2) + 2       5          20        2.857~              5

  8= 2x2x2                       3          23        2.875                5

  9= (2x2x2) + (2/2)         5          28        3.111~              5

10= (2x2x2) + 2             4          32        3.2                    5

11= (2x2x2) + (2/2) + 2  6          38        3.455~              6

12= (2+2+2) x 2             4          42        3.5                    6

13= (2^(2+2)) - (2/2) - 2 6          48        3.692~              6

14= (2^(2 + 2)) - 2          4          52        3.714~              6

15= (2^(2+2)) - (2/2)      5          57        3.8                    6

16= 2^(2+2)                   3          60        3.75                  6

17= 2^(2+2) + (2/2)        5          65        3.823~              6

18= 2^(2+2) + 2             4          69        3.833~              6

19= 2^(2+2) + (2/2) + 2  6          75        3.947~              6

20= 2^(2+2) + (2+2)       5          80        4                       6

Key: "~" denotes that the number has been approximated by rounding. l is the first appearance of a higher m, this shows how many k's are needed to make all the numbers up to that n, or put simply if I wanted to make all the numbers between 1 and 20 using five 2's I could not because the first number that has to use a minimum of 6 is 11. The sixes would continue until the first seven is encountered.

So how do I know that m is the minimum number of k’s for n? Here it was done by construction. I know that the only number with one 2 is -- 2. So I used that number twice with each of the operators and got all of the numbers with two 2's -- 2+2, 2-2, 2x2, 2/2, etc. which, disregarding duplicate and non-natural answers are 0, 1, and 4. Next we want to find all the numbers which use three 2s, so we use the numbers which use one 2 (2); the numbers which use two 2's (0,1,4); and use each of the operators. Again disregarding non-applicable answers I get: 3, 6, 8, 16, 256. So, next is, you guessed it, numbers with four 2's are made up by using each of the operators on numbers with one 2 and three 2's and on numbers with two 2's and other numbers with two 2's. Now I only see hints of patterns peeking out from behind this forest of twos so lets do k=3 and see what we can find.

k=3

n                                    m         t           a                       l

  1= 3/3                           2          2          2                       2

  2= 3 - (3/3)                    3          5          2.5                    3

  3= 3                              1          6          2                       3

  4= 3 + (3/3)                   3          9          2.25                  3

  5= (3+3) - (3/3)             4          13        2.6                    4

  6= 3+3                          2          15        2.5                    4

  7= (3+3) + (3/3)            4          19        2.714~              4

  8= (3x3) - (3/3)              4          23        2.875                4

  9= 3x3                           2          25        2.778~              4

10= (3x3) + (3/3)            4          29        2.9                    4

11= (3x3) + 3 - (3/3)       5          34        3.091~              5

12= (3x3) + 3                 3          37        3.083~              5

13= (3x3) + 3 + (3/3)      5          42        3.231~              5

14= 3x(3+3) - 3+(3/3)    6          48        3.429~              6

15= (3x3) + (3+3)           4          52        3.467~              6

16= (3x3)+(3+3)+(3/3)   6          58        3.625                6

17= (3^3) - (3x3) + (3/3) 6          64        3.765~              6

18= 3 x (3+3)                 3          67        3.722~              6

19= (3x3)+ (3x3)+(3/3)   6          73        3.842~              6

20= 3x(3+3) + 3 - (3/3)   6          79        3.95                  6

21= 3x(3+3) + 3             4          83        3.952~              6

Observation: construction takes too long (unless tom wants to write me a brute force computer program :) )

So what is the 'formula' for finding the minimum number, m, of k’s for a given n? That I will have to leave for next time, as I must go. Please post any observations on patterns or suggestions as to a direction I might take.