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Messages - Xeda112358

Pages: [1] 2 3 ... 295
1
General Discussion / Re: Watcha Been Listening To?
« on: August 02, 2017, 07:25:03 am »
For a moment I thought you meant  to type Otep XD I haven't heard anything about that one for a while.

2
Sorry, I'm on my phone so I'll probably not go too in-depth on this :( Bug me for details if I don't get around to it and you need them :P

So:
Given ##x\in[-.5ln2,.5ln2]##
Let ##y=x^{2}##
Let ##a=\frac{x}{2}\frac{1+\frac{5y}{156}\left(1+\frac{3y}{550}\left(1+\frac{y}{1512}\right)\right)}{1+\frac{3y}{26}\left(1+\frac{5y}{396}\left(1+\frac{y}{450}\right)\right)}##
Then ##e^{x}\approx\frac{1+a}{1-a}##

Accuracy is ~75.9666 bits.
7 increments, 1 decrement
6 constant multiplications
6 general multiplications
2 general divisions
1 div by 2 (right shift or decrement an exponent)

For comparison, that's comparable to 16 terms of the Taylor's series, or 8 terms of the standard Padé expansion (exponential is special in that it comes out to f(x)/f(-x) so it can be done even easier than most).

I basically carried out a Padé expansion for e^x to infinity, noticed that after the constant term all the even coefficients were zero, so I used a Padé expansion on that function to quickly find our approximation for a.

In my usage, I actually implemented a 2x function since I'm using binary floats with 64-bit precision. I take int(x) and save that for a final exponent for the float. Remove that value from x. By definition of int(), x is now non-negative. If x≥.5, increment that saved exponent, subtract 1 from x. Now x is on [-.5,.5]. Now we need to perform 2x, but that is equivalent to ex*ln(2). We've effectively applied range reduction to our original input and now we can rely on the algorithm at the top. That integer part that we saved earlier now gets added to the exponent byte, et voilà !

I think when I calculated max error using my floats it was accurate up to the last two bits or something. Don't quote me on that as I don't have those notes on me at the moment and it was done months ago.

3
Miscellaneous / Re: Hello again everyone
« on: May 02, 2017, 09:08:38 am »
Hey Art! It's been the same way for me, too. Adulthood and social media... ugh :P I love programming and do program on my calc most days, but it's usually just a couple of minutes here and there throughout the day :( I just don't have time for projects. I couldn't tell if I wasn't seeing you around or if it was because I don't check in regularly XD

4
TI-BASIC / Re: TI-Concours 2017 has started!
« on: March 25, 2017, 08:22:47 pm »
Oh, this looks fun! I make no promises, but I might enter!

I have a new job and my free time is much more consistent. Aaaand the competition ends on my birthday :)

5
ASM / Re: Shifting data block down a nibble?
« on: March 25, 2017, 08:18:15 pm »
harold's method is essentially what I did to shift the screen left or right by multiples of 4. It's slow, but faster than shifting by 1 four times.

6
ASM / Re: Should I learn ASM?
« on: March 15, 2017, 10:32:24 am »
I'm also guessing division would be the subtraction instead of addition, trunkating any remainder.
Division is typically performed exactly like 'schoolbook' long division (until you get to higher precision, then you can use some state-of-the-art algorithms and math magic).

Start with an accumulator and quotient set to 0.
Rotate the one bit of the numerator into the accumulator.
If accumulator>=denominator, then subtract the denominator from the accumulator and shift in a 1 to the quotient, else shift in a zero
Repeat at the "rotate" step.

In code, since we are getting rid of one bit at a time in the numerator and adding 1 bit at a time to the quotient, we can actually recycle the freed up bits in the numerator.
Here is an example of HL/C where C<128, A is the accumulator and HL doubles as the quotient and numerator:
Code: [Select]
HL_div_C:
;Input:
;  HL is the numerator
;  C is the denominator. C<128
;Output:
;  A is the remainder
;  HL is the quotient.
    xor a
    ld b,16
loop:
    add hl,hl   ;this works like shifting HL left by 1. Overflow (thus, the top bit) is left in the carry flag
    rla         ;shift a left, rotating in the c flag as the low bit.
    cp c        ;compare a to c. Basically does A-C and returns the flags. If C>A, then there will be underflow setting the C flag.
    jr c,skip   ;skip the next two bytes if c flag is set (so A<C)
    inc l       ;we know the low bit of HL is 0, so incrementing HL will set that bit.
    sub c
skip:
    djnz loop
    ret
ld b,8 - This must be telling the loop to repeat 8 times to check each bit, so, how does the loop relate to b? I'm thinking about how nested loops would work...
djnz loop - kindof like "goto" loop, with the automatic decrementing of register b?
That is correct and that's why we initially do ld b,8. Keep in mind that djnz * and jr * are intended for small (ish) loops or redirecting to code relatively close by. It can jump backwards up to 126 bytes and forward up to 128 (back 1 byte is simply a slower way of performing rst 38h and I do not suggest it, back 2 bytes creates an infinite loop, back 0 bytes does nothing but waste 7 or 9 clock cycles). Most assemblers will warn you of out-of-bounds jumps.
For longer loops or jumping to code far away, use jp *. djnz * only works with register b and always decrements.

rrc e - you say this rotates register e, is this the same as changing the register input from left to right?
For example, if e=01111011, then rrc e would change it to e=10111101. The bottom bit gets returned in the c flag as well as being returned to the top bit of e. I used rrc * since rotating 8 times would leave it exactly how it was input. I could have used rr * or even sra * or srl *, but I prefer to avoid destroying registers if I can.

7
ASM / Re: Should I learn ASM?
« on: March 13, 2017, 11:25:06 pm »
But the multiplication isn't convoluted! It's exactly how most of us are taught in grade school, except instead of multiplying digits 0~9, it's just multiplying by 0 or 1 which is super trivial. Like:

         01110101
        x10101101
        ---------
         01110101
        000000000
       0111010100
      01110101000
     000000000000
    0111010100000
   00000000000000
  011101010000000

Or removing the multiplies by zero:

         01110101
        x10101101
        ---------
         01110101
       0111010100
      01110101000
    0111010100000
  011101010000000




So suppose bit 3 is set. Then you basically add your top number, shifted left three times. As an example, suppose you wanted to multiply C*E (ignoring the top 8 bits):
Code: [Select]
;C is our "top" number.
;E is our "bottom" number.
;A will be our "accumulator"

    ld a,0

    rrc e    ;this rotates register 'e' right, putting the bottom bit as "carry" [out of the register].
    jr nc,checkbit1   ;nc == not carry. If "carry" out was zero, skip this step.
    add a,c    ;if carry out was 1, then add to our accumulator.
checkbit1:
    sla c    ;finally, shift our "top number" to the left in case we need to add this to the accumulator, too. Then [REPEAT] 7 more times.
    rrc e
    jr nc,checkbit2
    add a,c
checkbit2:
    sla c
    rrc e
    jr nc,checkbit3
    add a,c
checkbit3:
    sla c
    rrc e
    jr nc,checkbit4
    add a,c
checkbit4:
    sla c
    rrc e
    jr nc,checkbit5
    add a,c
checkbit5:
    sla c
    rrc e
    jr nc,checkbit6
    add a,c
checkbit6:
    sla c
    rrc e
    jr nc,checkbit7
    add a,c
checkbit7:
    sla c
    rrc e
    jr nc,all_done
    add a,c
all_done:
    ret
If you can see how that relates to the school book algorithm, then just know that the following does practically the same thing:
Code: [Select]
;C is our "top" number.
;E is our "bottom" number.
;A will be our "accumulator"

    xor a    ;mad hax to set A to zero. Faster, smaller.
    ld b,8
loop:
    rrc e    ;this rotates register 'e' right, putting the bottom bit as "carry" [out of the register].
    jr nc,no_add   ;nc == not carry. If "carry" out was zero, skip this step.
    add a,c    ;if carry out was 1, then add to our accumulator.
no_add:
    sla c    ;finally, shift our "top number" to the left in case we need to add this to the accumulator, too.
    djnz loop  ;aaand repeat, decrementing register B until zero (this is a specialized instruction on the Z80)
    ret

But please don't use that in a real program :P It's terribly inefficient. If you understand how the register pairing works, you can come up with a much better AND more convoluted algorithm:
Spoiler For "Step-by-Step How to derive the 'best' 8-bit Multiplication Algorithm":
Let's start by rearranging the above code. The way we do 'schoolbook' multiplication starts at the least significant digit, but we can just as easily start from the most significant digit. So let's do an example in base 10:

    377
   x613
   ----
=1*3*377+10*1*377+100*6*377
=1(3*377+10(1*377+10(6*377)))

If we want to convert that last line to pseudo-code:
Code: [Select]
0->acc
10*acc+6*377->acc
10*acc+1*377->acc
10*acc+3*377->acc
So if we wanted something like that in assembly:
Code: [Select]
H*E -> A
;H is the "bottom" number that we will no be checking the top digit down to the bottom digit.
;E is the 'Top" number.
;A is the accumulator
;basic algo, after initializing A to zero.
;    multiply A by 2.
;    shift H left by 1
;    if this results in a bit carried out (so a 1 carried out), then add E ('top' number) to A (the accumulator)
;    repeat 7 more times for all bits in H.
    xor a
    ld b,8
loop:
    add a,a
    sla h        ;shifts H left by 1, bringing in a 0 for the low bit. mathematically the same as H*2 -> H
    jr nc,skip_add
    add a,e
skip_add:
    djnz loop
    ret
That's faster, but not optimal! To get the optimal way, lets stray from optimality a little to make 'L' our accumulator. Since we can't directly add another register to L, we'll have to juggle with register A making it slower:
Code: [Select]
    ld l,0
    ld b,8
loop:
    sla l
    sla h
    jr nc,skip_add
    ld a,l \ add a,e \  ld l,a
skip_add:
    djnz loop
    ret
But since we know that 'sla l' will spit out a zero-bit for the first 8 iterations (all of them), we can do 'sla l \ rl h' which is the same size and speed. However, this is the same as just doing doing "add hl,hl" ! This is where you'll have to learn how register pairs work :P
Code: [Select]
    ld l,0
    ld b,8
loop:
    add hl,hl
    jr nc,skip_add
    ld a,l \ add a,e \  ld l,a
skip_add:
    djnz loop
    ret
But wait, there is more! We don't have an "add l,e" instruction, but we do have an "add hl,de" instruction. If we make D==0, then we can change 'ld a,l \ add a,e \  ld l,a' to 'add hl,de'. The problem is, if the lower byte of HL, (so L) overflows, then the upper byte H, our 'bottom' number. We can't have it changing our input value halfway through the algorithm! Thankfully, the changes never propagate into those upper bits. This requires some tedious work to prove, but if you are cool with taking that at face value, then our last piece of the puzzle gives us:
Code: [Select]
    ld d,0
    ld l,d    ;since D=0 already, this sets L to 0, as we want. It's smaller and faster than ld l,0.
    ld b,8
loop:
    add hl,hl
    jr nc,skip_add
    add hl,de
skip_add:
    djnz loop
    ret
Even better, this actually givers us the full 16-bit result instead of just the lower 8 bits :)

8
General Calculator Help / Re: Ti84+SE misbehaving
« on: January 25, 2017, 10:08:12 am »
Which OS version? 2.53MP is known to be fairly buggy, 2.55MP is pretty stable in my experience, but I have gotten it to randomly crash before.

9
The Axe Parser Project / Re: Need Help With Collision
« on: January 18, 2017, 02:53:22 pm »
Have you tried swapping X and Y ? I haven't really dissected the code so this is a shot in the dark, but it's an issue I've frequently encountered.

10
TI-BASIC / Re: Fastest way to render random points and lines?
« on: January 15, 2017, 09:27:27 am »
Of course assembly would be way faster, but have you tried the standard built in Pxl-On(, Pxl-Off(, Pxl-Change(, Pt-On(, Pt-Off(, Pt-Change(, Line(, and Circle( commands? Also, if you include an extra argument for Circle( of {i it will perform faster (i being the imaginary i).

If you haven't tried Axe yet, I urge you to try that if you want much faster graphics. Just be warned that Axe works a lot closer to assembly, so you can't just press [ON] to break out of infinite loops. For example, if you do While 1:End, you will need to pull a battery and get a RAM clear.

Edit: Also, if you want a pixel based circle (instead of TIs point-based), I think KermM wrote an entirely TI-BASIC routine that performs faster than TIs.

11
News / Re: Reuben Quest: Lost Between Times
« on: January 09, 2017, 09:03:48 pm »
This is.....odd.
Which calculator?
Did you have an item selected? Which one? Was it a bottle? What was inside the bottle?
TI-84+SE OS 2.55MP (shush, I use it for the extra built in functions :P )
I tried it at various points in the game including with water selected. It also crashes when I go to use magic, then press down. And occasionally if I die and hit [2nd] too quickly, it glitches to an odd map area and when I move it causes it to crash.

12
News / Re: Reuben Quest: Lost Between Times
« on: January 01, 2017, 09:22:20 am »
I haven't played an awful lot, but it's fun so far!
I have one major bug to report, though: Whenever I press the down arrow when I'm charged for an attack, I get an error (Err:Link I believe) and it causes my calc to freeze and I have to pull a battery.

13
News / Re: Reuben Quest: Lost Between Times
« on: December 19, 2016, 09:39:32 am »
Holy canoli, +1 for that, for sure! It looks sooo good! Gonna start playing when I boot up my pi to send to my calc!

I wonder if this will make it into next year's POTY then? I sure hope so XD

15
TI-BASIC / Re: Help overcoming Symbolic bug
« on: November 12, 2016, 07:07:13 am »
That's really clever! Is this bug only an issue when Batlib is installed? I'm worried that it might be a bug with Batlib or Batlib+Symbolic.

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