So, in the other emulator, I just do the same thing with the proper functions. Another good side of this is that I don't have to memorize a big name like "mainmemory.read_s16_le".Language: lua-- Compatibility local u8 = mainmemory.read_u8 local s8 = mainmemory.read_s8 local u16 = mainmemory.read_u16_le local s16 = mainmemory.read_s16_le -- Example local RNG = u8(WRAM.RNG)
EDIT: The sample you posted yields this error:LuaInterface.LuaScriptException: E:\Games\Emulation\Lua\sample.lua:7: attempt to call field 'register' (a nil value)
The PAL version has the locking up issue on ALL emulators. Similar to Crash Bash PAL locking up issue when trying to start "Adventure Mode" (which also occurs on all emu).
By the way, Amaraticando, you said that this was a question on a math olympiad that you did. Did you do and/or solve this question during the olympiad?
I have tried in BizHawk SVN r8610 (and other previous versions) and in mednafen-0.9.37.1-win64: the results are similar.
I have used correct/pure dumps of the game: EDC and no EDC.
I have also used a few different PAL bios.
By the way, the NTSC-U and NTSC-J versions work really fine, AFAIK.
Is there a table which shows what powerup you get at each PI number?
-Prove that a circle can be surrounded by exactly 6 others of its own radius. IE The above picture might actually be incomplete, there may be indistinguishable gaps in between each one; so prove it to make sure. -If surrounding the central circle with smaller ones, what radius would they need to be in order to have exactly 8 surrounding it instead?
So, if n = 6:
csc(pi/6) = 1/sin(pi/6) = 2
Then, r = R.
If n = 8:
csc(pi/8) = sqrt(4 + 2*sqrt(2))
Then, r = R/(sqrt(4 + 2*sqrt(2)) - 1) = R*0.6199..
This problem gets really interesting in higher dimensions: http://en.wikipedia.org/wiki/Kissing_number_problem
PS: sorry if my r looks like the symbol for pi.
EDIT: whoa, FractalFusion and me posted pretty much the same thing at the same time.
I don't know if it can be something on my end, but here's the bonus fanfare;
Keep in mind that these functions are relatively slow, so if you're doing a lot of drawing, it's probably faster to calculate your x/y offsets/scaling at the start of a frame, then use those for all the rest. Though it's perfectly possible your computer has no trouble just calling everything all the time.
even if you have all possible 256 characters in C++ program, you can reach maximum 256^1000 different possible outputs.
Consider any conjecture that could be disproven via a counterexample among a countable number of cases (e.g. Goldbach's conjecture). Write a computer program that sequentially tests this conjecture for increasing values. In the case of Goldbach's conjecture, we would consider every even number ≥ 4 sequentially and test whether or not it is the sum of two prime numbers. Suppose this program is simulated on an n-state Turing machine. If it finds a counterexample (an even number ≥ 4 that is not the sum of 2 primes in our example), it halts and notifies us. However, if the conjecture is true, then our program will never halt. (This program halts only if it finds a counterexample.) Now, this program is simulated by an n-state Turing machine, so if we know S(n) we can decide (in a finite amount of time) whether or not it will ever halt by simply running the machine that many steps. And if, after S(n) steps, the machine does not halt, we know that it never will and thus that there are no counterexamples to the given conjecture (i.e., no even numbers that are not the sum of two primes). This would prove the conjecture to be true.
Add client.borderheight(), client.borderwidth(), client.bufferwidth(), client.bufferheight(), client.transformPointX(), client.transformPointY()
"What is the largest integer that can be computed by a terminating C++ program of at most 1000 characters?" The condition that the program must terminate actually puts an upper limit to how big the computer number can be. Since the program has a limited size, the computed number also must have a limited size. Thus the core question is whether this problem is unsolvable or just very hard.
Since no one knows whether some odd perfect number exists, we don't know if this program will stop. Of course there's a difference between unknowledge and undecidability. So, is this problem decidable? Possibly, only if there's no odd perfect number. I have to investigate that...Language: pseudocodefunction is_perfect_number(int n) blablabla return bool_perfect_number // true if n is a perfect number, false otherwise end int n = 1; while true do print(n); if is_perfect_number(n) == true then break end n = n + 2; end
I've stopped using VLC, my audio cuts out every 10 seconds. MPC-HC works perfectly.