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Messages - Xeda112358
Pages: 1 ... 35 36 [37] 38 39 ... 317
541
« on: February 28, 2014, 07:31:45 am »
I've noticed with the latest update (the one that WabbitEmu prompted for an update) that having the skin on really eats up resources and causes lots of lag with keypresses. Turning the skin off, it works just fine.
542
« on: February 27, 2014, 10:36:17 am »
Haha, as I expected, I would be too busy for coding That's alright, though, I like the experience Also, the new TI-BASIC project seems fun!
543
« on: February 24, 2014, 09:57:23 am »
Yep, that is correct. 5k+1 mod 100 = 26, so (5k+1)/2 mod 10 = 3. Then (5k+1)/2=5m is never true, so the only way to take care of that is if (5k+1)/2=(4n+1)2r, but since it is odd, we know it must be that r=0, so (5k+1)/2=(4n+1)
Also, sorry if I am wrong about anything, I haven't slept much.
544
« on: February 24, 2014, 07:47:53 am »
Oops, yeah.
545
« on: February 24, 2014, 07:05:30 am »
You only need to worry about when n(2n-1) is not of the form 2n5m.
2n-1 is always odd, so that must always be a power of 5.
2n-1=5k ==> n=(5k-1)/2
So then you need (5k-1)/2 to be factored as powers of 2 and 5. 5 will never be a factor, so we need (5k-1)/2=2m ==> 5k-1=2m+1
This is true for k=1, which is n=2.
All other cases are when (4N+1) happens to cancel any non-2 and non-5 factors from the denominator, which happens at N=8. (2n-1 = 15, 4n+1=33, so the 3 cancels leaving 2^3*5 in the denominator).
546
« on: February 16, 2014, 03:36:32 pm »
Well, it isn't actually asking to pay fort a game anymore than a game requiring you to reach a certain level before unlocking a mini game.
547
« on: February 15, 2014, 10:30:31 am »
Well, the Nspire is a whole different system than the Z80 calcs, so I have only done BASIC for the most part on it. LZ compression followed by statistical compression would probably get the best results. LZ compression would compress any repeated words or phrases to 4 bytes (maybe 5 depending on how large the file actually is). This might not seem good, but the bytes are just a pointer to the existing phrase. For example, the following: gird yourselves, and ye shall be broken in pieces; gird yourselves, and ye shall be broken in pieces. Since it is less than 256 bytes would be compressed to something like the following: [0+49]gird yourselves, and ye shall be broken in pieces[0+2]; [128+0][0+1].
Everything in [] is a single byte. So the original phrase would have been 101 bytes, but the encoded string would have been 1+49+1+2+1+1+1=56 bytes. However, in the much larger file of a book, words like "ye", "be", and "yourselves" might be encoded in a single byte (like BASIC tokenization as codebender described) so the compression would be significantly better. If you have a compressed copy on your computer, chances are software can be easily written to decompress on your calc.
548
« on: February 08, 2014, 04:54:32 pm »
Heh, I started mine a few hours ago I won't have a working program, so I'm just going to make it draw hearts everywhere and get 1 point.
549
« on: February 08, 2014, 04:18:24 pm »
Do you know if Axe will ever be able to work in WabbitEmu and jsTIfied again? It always crashes for me.
550
« on: February 06, 2014, 10:58:18 am »
I haven't started mine, yet
551
« on: February 03, 2014, 07:23:53 am »
Yup, I am in each category. I don't know Lua, but the worst that can happen is that I won't win
552
« on: February 01, 2014, 03:00:02 pm »
I'm not getting the right dates according to Wikipedia Starting in 2012, it is off by 1, before that, it's just off. I copied it to Source Coder and edited the E2 things to be the correct one. EDIT: Also, for suggestions, sub() can be used to divide by 100, using the fPart() method to do modulo has issues with rounding error, causing it to pass unexpected results, so you should use the int() method: X-Nint(X/N
And you will save some more bytes by computing some of the constants and using the inverse token. For example, see these two lines: 50-30fPart((11(1+19fPart(Y/19))-Ans+int(Ans/4)+int(8(Ans+11)/25))/30 50-30fPart(30{^-1}(11+209fPart(Y/19)-Ans+int(Ans/4)+int(.32(Ans+11
553
« on: February 01, 2014, 08:42:11 am »
Question: The english documentation says to name the file MINDXX, but the French says TOMEKXX. Which one should we use?
554
« on: January 31, 2014, 07:07:03 am »
I've had this problem, too. I think it is partially the OS that isn't accepting the apps, but I'm not sure. My fix has been to make a dummy program (or real program) archive it, then garbage collect, then send the app.
555
« on: January 28, 2014, 12:29:06 pm »
I managed to optimize the first iteration. I thought to analyze it while in class, so I saved two bytes, 8 cycles.
EDIT: Here is some pseudocode for getting the full square (not just the lower bits):
A_sqrd: ;Input: A is an x-bit number (example, 8-bit, 16-bit, et-cetera) ;Output: A*A A<<1→B ;Store A shifted once into B. 0→ACC ;Initialise the accumulator with 0 1<<(x-1)→MASK ;Set the last bit of MASK For (x-1) Iterations: ACC<<1→ACC ;shift ACC (same as *2 or ACC+ACC) B<<1→B ;shift left once, keep carry ((SBC(0,0)^A)&MASK)+ACC→ACC ;SBC(0,0) is "subtract with carry" and will yield either -1 (carry flag set) or 0 (carry flag reset). (MASK>>1)|MASK→MASK ;arithmetic right shift Return (A<<x-A)-ACC
Implemented on the Z80, it kind of sucks taking twice as long as a traditional routine. However, it is novel, I think, taking 7 iterations:
L_sqrd: ;Input: L ;Output: L*L->HL ;It's really slow compared to a generic 8-bit multiplication ;It's also pretty big ;But it is a novel approach! ld e,a ;this just stays a constant for the algorithm rlca ld d,a ;this will be rotated to get the bits of the input ld l,0 ;the accumulator ld bc,0780h ;c is a mask loop: add hl,hl rlc d sbc a,a xor e and c add a,l ld l,a jr nc,$+3 inc h sra c djnz loop ;HL is 255*input-input*input ;add input ld c,e add hl,bc ;HL is 256*input-input*input ;negate HL xor a sub l ld l,a sbc a,a sub h ld h,a ;HL is input*input-256*input ;add input*256 ld e,b add hl,de ;HL is input*input ret
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