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Messages - phenomist
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106
« on: February 28, 2011, 10:02:35 pm »
137426454337521899137502452525605466748829534697526357946510356073439963309278587906732336287048350193565270671049611558935534759741894429581138110991019864818898792610752419053602832785246554297841481080233597667469592742563152706207467933515159107380059434427511231769817853007350965081930222294271540538551
The key in decimal. If I converted what I think is the key.
107
« on: February 28, 2011, 09:01:15 pm »
Hmm, would somebody be interested in converting the key from hexdecimal to decimal? I can't get NET 4.0 Framework running on this computer...
108
« on: February 27, 2011, 05:52:02 pm »
Some spellcheckers identify repeated words as possibly misspelled.
109
« on: February 27, 2011, 05:37:49 pm »
BTW, for a 95x63 matrix, store it as 10 pictures, considering that you can restrict yourself where each entry has a integer maximum of 1000.
If I recall this costs somewhere around 7600 bytes.
Of course, if you don't expect the user to paint the same pixel hundreds of times, you can reduce the number of pictures.
Previous example rehashed:
Onscreen
OOOOO
*O*O*
**O**
*O*O*
OOOOO
In layer 1
*OOO*
*O*O*
*****
*O*O*
*OOO*
In layer 2
O***O
*****
**O**
*****
O***O
if a pixel is written over 3 times, then it will be in both layer 1 and layer 2; if 4 times, layer 4, etc.
110
« on: February 27, 2011, 02:40:12 am »
I don't know about the feasibility about this, but what if you store the total number of times every pixel was drawn on, and turn the pixels off for only those that are undo-ed?
For example, consider
A***B
*****
*****
*****
C***D
Drawn lines: A-B, B-C, C-D, D-A. Now the total times drawn counter looks like this:
21112
01010
00200
01010
21112
Undoing it once would reduce the counters on the D-A line, namely change it to
11112
00010
00100
01000
21111
so two pixels would turn off.
How would you store a 64x96 matrix? Since hopefully you're using less than 1000 actions due to list size limits, 10 pictures would be enough to store as a binary counter.
Hopefully this works?
111
« on: February 25, 2011, 06:31:35 pm »
Using your definition: http://oeis.org/A111766EDIT: Notice that there's a closed form!
112
« on: February 25, 2011, 06:17:41 pm »
Currently, you're using the following:
Even values of 'a'
b=a^2/4-1
c=a^2/4+1
Odd values of 'a'
b=(a^2-1)/2
c=(a^2+1)/2
which means you're assuming that for odd a, b and c are 1 apart; for even a, b and c are 2 apart. This works usually, but sometimes fails. (e.g. a=20, c-b=8, not 2)
As mentioned a bit earlier in this thread, the generalized Pythagorean triple can be written in the form (m^2-n^2,2mn,m^2+n^2). If you're looking at only primitive triples, then m,n should be coprime.
Note that the first leg is (m+n)(m-n), so factoring should make this task easy.
113
« on: February 25, 2011, 02:29:48 am »
Hm... are you counting certain triples like [20,21,29]? Not sure if they're fitting in the current description.
114
« on: February 25, 2011, 01:21:17 am »
Oh, you're limiting yourself to primitive pythagorean triples only? (so [6,8,10] is invalid because it is simply 2x[3,4,5] ?)
Also, is it three exactly, or at least three?
115
« on: February 24, 2011, 09:54:54 pm »
24: (24,32,40), (18,24,30), (24,70,74), (7, 24, 25)...
840: see 24 example but multiply by 35
So it's not exactly 3.
116
« on: February 24, 2011, 07:31:07 pm »
I'm still wondering though... even given the new conditions, we should still expect to see a LOT more than just 2 numbers in the first 10,000.
For example, 60 should work with (36, 48, 60); (45, 60, 75); (60, 80, 100).
117
« on: February 24, 2011, 07:01:25 pm »
Oh... misinterpreted, sorry.
Well, I suppose you could do individual tests on m^2-n^2=N, 2mn=N, m^2+n^2=N...
In particular, the first two are factorings of N.
118
« on: February 24, 2011, 06:57:18 pm »
(m^2-n^2, 2mn, m^2+n^2) will generate all the pythagorean triples. For instance, m=2 n=1 gives (3,4,5). Also, the set of all numbers that are part of a pythagorean triple isn't terribly interesting; all numbers larger than 2 are part of a pythagorean triple  (3,4,5); (5,12,13); (7,24,25); etc. for odd numbers, even numbers just double these triples.
119
« on: February 24, 2011, 06:52:08 pm »
O_o how the hell adding an infinite number of positive numbers equals a negative one?
Math sure is weird.
Let's see if this rings a bell to programmers: It's a bit like signed numbers. 65535 = -1. 65535 = 1111111111111111 in binary. (in Axe. Sort of.) Only in math, it's a perfect computer that has an infinite number of bits. But if you turn all of the bits on, then it becomes negative  . 1+2+4+8+16+... = -1. Well, maybe not, but that's the way I think of it. Still taking calculus, so I don't know the intricacies of this. Also, strange followup: You've heard that you can't go colder below absolute zero, but it turns out that negative temperature is WARMER than an infinite amount of degrees. http://en.wikipedia.org/wiki/Negative_temperature
120
« on: February 24, 2011, 02:22:19 am »
Already heard of the trick before, so this would be an invalid result, but:
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