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revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 17269
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revolution
Borsuc wrote:
revolution wrote:
Okay, but you have to pay for shipping. Deal?
If I buy 3, will shipping be free?
windwakr wrote:
You can't buy revolution prize stars, you have to earn them, duh.
Actually I am open to bribes. Didn't you know?

Borsuc I can understand that you are really desperate and excited at the prospect of having a star for yourself. I know how much they are coveted and cherished by the current winners (i.e. not at all). However, three stars??? Isn't that just a tad bit too suspicious? People will start to wonder. How about you just start at one star and work your way up slowly.

Now remember that stars (the real type) are heavy and shipping costs are equal to mass(kg)*distance(km)/time(days). If you can just let me know which star(s) you prefer and the time you are prepared to wait for delivery then I will arrange it all and will deliver for free if the shipping comes to less than $50 (in any currency you choose).

eg. Let's say you fancied Proxima Centauri (2.446e+29 kg, 4.0142e+13 km). Then if you are prepared to wait 1.964e+41 days I will deliver for free. If you want it faster then I am sorry but I will have to charge you.
Post 21 Aug 2009, 01:30
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windwakr



Joined: 30 Jun 2004
Posts: 827
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windwakr
http://board.flatassembler.net/topic.php?p=82994#82994
Been trying to look into that one....UGH, my brain hurts.... Crying or Very sad
It's so hard for me to grasp an irrational number base, I'm so used to integer bases.

EDIT: I need to stop using posts as scratch pads...

At least I've learned a few interesting things about a number I'll never care about in my life after this. Laughing
Stupid stuff like Phi/(Phi-1)=Phi², and Phi+1=Phi²


Right now I can do addition and subtraction fine in base PHI, next onto multiplication and then division....UGH


EDIT: LOL, nevermind the question I had here about multiplication in base PHI, it's stupidly simple!...Now on to divison Sad

I can count in base PHI, can you? It's actually pretty simple when you know what to do.
Base Phi is represented as additions of powers of Phi.
Before the decimal point are rising powers. For example, 1001.0(approx. 5.2360679...) would be Phi^3+Phi^0 . After the decimal it's negative powers, starting at -1.
Code:
 1=1
 2=10.01
 3=100.01
 4=101.01
 5=1000.1001
 6=1010.0001
 7=10000.0001
 8=10001.0001
 9=10010.0101
10=10100.0101
    

Easy stuff.....


It doesn't look like that method of repeating numbers would work in an irrational number base.

Even if I don't come up with an answer for your question I'm at least giving my brain a good workout.

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Post 21 Aug 2009, 03:40
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windwakr



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Location: Michigan, USA
windwakr
revolution wrote:

Hmm, does it work in phinary (base φ)?

I don't have any mathematical proof or anything, but I have done some testing(unless I'm dividing wrong...). When you take the number to convert to repeating as phi, and divide it by (Phi-1), you get Phi², which is NOT a repeating decimal as Tomasz's post says it should be. I think the reason why is because Phi is a special IRRATIONAL number.

Ok, just found out that any number divided by (Phi-1) equals xPhi, where x is that number. Very Interesting.
EDIT: My first edit left in stuff I meant to take out, lol...fixed

Example:
Phi-1=0.1 in base Phi, 1 in base phi is 1.0..now divide 1.0 by 0.1....answer is 10, thats 1phi.

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Last edited by windwakr on 21 Aug 2009, 19:25; edited 2 times in total
Post 21 Aug 2009, 17:39
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windwakr



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windwakr
A better example. Lets looks at base 10 for a second. Lets take x/((b^5)-1), which is equal to x/9999. Let x equal 5. Doing the division you get 0.000500050005 repeating. So, using the x/((b^5)-1), lets look at base Phi. Again, x is equal to 5, which is 1000.1001 in base Phi. Now lets take Phi^5, which is 100000.0 in base Phi, and subtract 1 off of that. It comes out to 010100.1. Ok, now lets divide 1000.1001(5) by 10100.1(Phi^5-1) in base Phi. Ok, it DOES stabilize repetition, but not until a little ways in. 0.0100100100001000010000100001 I think this does show that it DOES NOT work in base Phi, or probably any irrational base.

((b^n)-1) Where n is the number of digits is the same as Tomasz's way of just repeating (b-1) n times.


EDIT: This below has nothing to do with the question, so I'll put in in quotes
windwakr wrote:

base 10, (10^4)-1=9999
base 16, (10^4)-1=FFFF
base 8, (10^4)-1=7777
base 2, (10^4)-1=1111
base phi, (10^4)-1=1010

Notice something interesting? The base when represented in its own base is "10", that seems to be true for ANY base.
EDIT: Ya, I see why now, gotta remember that numbers are added powers.
10 would be equal to b^1...the base you're in.


Heres some other ((b^n)-1) for base Phi, weird results.

(phi^0)-1=0
(phi^1)-1=0.1
(phi^2)-1=10
(phi^3)-1=100.1
(phi^4)-1=1010
(phi^5)-1=10100.1
(phi^6)-1=101010
(phi^7)-1=1010100.1

I DO see a pattern, but it's not as easy as a normal number base, like 10(9,99,999,9999,99999, etc.)


EDIT: Another example. 10.01(2) / ((Phi^7)-1)
answer is 0.0000001100001001000100100010010001 with 0010001 repeating

In base ten, the same example would be 2 / 999999999((10^7)-1), answer 0.000000200000020000002 with 0000002 repeating.

See, it just doesn't repeat like it should in base Phi, I don't think it will work.

This base Phi is actually pretty neat. Maybe I should do everything in base Phi. : )

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Last edited by windwakr on 21 Aug 2009, 23:02; edited 6 times in total
Post 21 Aug 2009, 18:27
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revolution
When all else fails, read the source


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revolution
windwakr wrote:
Even if I don't come up with an answer for your question I'm at least giving my brain a good workout.
And there is not even a prize for this so it is purely an intellectual pursuit. Wink
Post 21 Aug 2009, 22:37
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windwakr



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Location: Michigan, USA
windwakr
No prize? You called it an "open puzzle not solved", so I assumed there would be a prize involved. I guess I shouldn't ASSuME.
revolution wrote:

Ah, such impatience. There are currently three open puzzles not solved in this thread. To which are you referring?

http://board.flatassembler.net/topic.php?p=82994#82994 (no answers yet)
http://board.flatassembler.net/topic.php?p=83742#83742 (half solved)
http://board.flatassembler.net/topic.php?p=88366#88366 (half solved)

Maybe you should say which ones give prizes and which ones don't before you get peoples hopes up.


But, I guess I don't mind. I learned a neat base that no one cares about. lol
I also learned A LOT of things about a number I didn't know much about before last night.

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Post 21 Aug 2009, 22:39
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revolution
When all else fails, read the source


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revolution
Nope, no prize. This one was just flippantly thrown out into the wind with nothing attached.
Post 21 Aug 2009, 22:53
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revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 17269
Location: In your JS exploiting you and your system
revolution
windwakr wrote:
Maybe you should say which ones give prizes and which ones don't before you get peoples hopes up.
Okay.

http://board.flatassembler.net/topic.php?p=82994#82994 (no prize, but interesting)
http://board.flatassembler.net/topic.php?p=83742#83742 (half solved, prize pending)
http://board.flatassembler.net/topic.php?p=88366#88366 (solved, prize issued)
Post 21 Aug 2009, 22:56
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windwakr



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Posts: 827
Location: Michigan, USA
windwakr
What I can't believe is that some 12-year old boy thought up this crazy base back in 1957. Geeze, 12-year old boys now(just 50 years later) don't even understand base 10 very good, they're too reliant on calculators. They just punch in what they're told, not understanding it, and "magically" see the answer.



Hey, it says my post is 99924, does that mean we're just 76 posts from 100,000?? I hope people don't start spamming posts just to try and be the 100,000th poster. Ya, if they don't, that gives me a better chance! Laughing Even if it's not really the 100,000th post, it would still be cool to show that your post was "http://board.flatassembler.net/topic.php?p=100000#100000"

Ok, just did some testing in the Test forum, and yes, it would technically be the 100,000th post when the number reaches 100,000. The board keeps track of EVERY post, even deleted ones. That's why my post right now is 99,924 even though the post count on the front page of the forums add up to around 91,000

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Post 21 Aug 2009, 23:15
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windwakr



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windwakr
Hmmm, getting annoying now.
When I use 1 as x in x/((b^7)-1) in base Phi(The power I use, 7, is just some random number I chose, no meaning to it in case you're wondering), I get 0.00000010000001 repeating just as i would in base 10....But any number other than 1 has junk before the repeating starts....Maybe it only works on 1.

EDIT:
Works also with 1/((Phi^3)-1)....0.001001001001
1/((Phi^4)-1).....0.000100010001
1/((Phi^5)-1).....0.000010000100001
1/((Phi^6)-1).....0.000001000001000001

Ok, it seems that it works fine with 1, probably because in Phinary the only numbers you Physically represent are 0 and 1, just like Binary...

So, is that the answer? Does it work, but only on 1? Or do you not know, and just asked that question thinking no one would attempt to answer it?

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Post 22 Aug 2009, 04:15
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revolution
When all else fails, read the source


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revolution
Clue: What is (b-1) in Phinary? b is as shown here.
Post 22 Aug 2009, 05:04
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windwakr



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windwakr
(b-1) in Phinary is 0.1, but since you cant really do 0.1 next to itself repeatedly as in (b-1)_(b-1)_(b-1)_(b-1)_etc., I do the equivalent (B^n)-1

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Last edited by windwakr on 22 Aug 2009, 05:12; edited 1 time in total
Post 22 Aug 2009, 05:09
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revolution
When all else fails, read the source


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Location: In your JS exploiting you and your system
revolution
windwakr wrote:
(b-1) in Phinary is 0.1
Yes, and how would you represent (b-1)_(b-1)_(b-1), like 999 is in decimal?
Post 22 Aug 2009, 05:12
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windwakr



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windwakr
You do the equivelent to (b-1)_(b-1)_(b-1), (b^3)-1, which in Phinary is 100.1

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Post 22 Aug 2009, 05:13
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revolution
When all else fails, read the source


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revolution
So what would be {base phi of course} 0.(101000010) as a fraction?
Post 22 Aug 2009, 05:19
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windwakr



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windwakr
I don't know.
Post 22 Aug 2009, 05:34
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revolution
When all else fails, read the source


Joined: 24 Aug 2004
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revolution
Is it (φ^8+φ^6+φ^1)/(φ^9-1)?
Post 22 Aug 2009, 05:36
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windwakr



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windwakr
I tried that, I can't get anywhere with it.

Code:
101000010 / 0101010100.1



                          .
              ____________
1010101001.  |  1010000100.
    


I guess my Phinary division skills just aren't what I thought them to be. I probably am not expanding it right when trying to do it.

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Post 22 Aug 2009, 05:39
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revolution
When all else fails, read the source


Joined: 24 Aug 2004
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Location: In your JS exploiting you and your system
revolution
Okay, try with a smaller number, 0.(10010)
Post 22 Aug 2009, 05:41
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windwakr



Joined: 30 Jun 2004
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windwakr
Nope, I must be doing it wrong. I need to read up on expansion in that article I linked to, Tomorrow though, I'm too tired to think right now.

Code:
My horrible, screwed up Phinary division attempt. 
BTW: I work on just 1 line, modifying it as I go, lol

                      .000010010000001
              ____________
1010101001.  |  000000.000011010101011
    

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Post 22 Aug 2009, 05:46
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