flat assembler
Message board for the users of flat assembler.
Index
> Main > Compute 10^n/10 using only 1 register with 2 instructions Goto page Previous 1, 2, 3 Next 
Author 

revolution
Oh, we need to do division in the ring.
So for 7 we do (6/4) mod 7 = 5 Becomes 2^32 * (6/(2^(32 mod period)) mod 7) / 7 = 0xB6DB6DB7 

28 Jan 2019, 09:46 

Tomasz Grysztar
In other words:
You could compute their truncated forms this way directly on a 32bit CPU: Code: ; compute 1/11: mov eax,1 xor cl,cl periodic: mov edx,930 shl edx,cl add eax,edx add cl,10 cmp cl,32 jb periodic The only hard part, if you wanted to compute 2adic remainders at runtime, would be to find a period (preferably the minimum one). This means finding such a multiply of denominator that it is for some k. You could apply methods discussed with Chinese remainder theorem. However, this might easily overflow on 32bit architecture. You may try to design a different algorithm computing 2adic representation, but please note that 2adic division is really something different from division of integers or standard base 2 floatingpoint numbers, even if the patterns of digits appear to be very similar. Numbers in base 2 may have infinite expansions to the right (negative powers of 2), while 2adic numbers can expand infinitely to the left and the number of digits to the right of the point is always finite. With padic arithmetic you have an exact precision on the right side. For example, consider a base 2 representation of , which is 0.01010101010101... If you try to multiply this by 3, the approximations are going to look like 0.1111...  their initial digit is wrong and there is a loss of precision. On the other hand, the 2adic representation of is ...0101011. Now if you multiply it by 3, you get ...0001. You can stop at any point, dropping the higher digits, but the ones you have on the right end are already precise and final. This is because the low n digits of the result depend only on low n digits of the factors, so as long as you know the low digits of the factors precisely, you also know the low digits of the product precisely. 

28 Jan 2019, 10:26 

revolution
I wanted to use the native division instruction.
Code: mov ebx,7 ;the divisor lea edx,[ebx1] ;numerator high xor eax,eax ;numerator low div ebx ;6/7 bad result inc eax ;+1 Code: mov ebx,7 ;the divisor lea edx,[ebx1] ;numerator high xor eax,eax ;numerator low ;code to compute (6 shr 2) mod 7 giving edx = 5 div ebx ;5/7 the correct result inc eax ;+1 Edit: Something like this can do the modulo part Code: PERIOD = 3 mov ecx,32 mod PERIOD .modulo_division: bt edx,0 sbb eax,eax and eax,ebx add edx,eax shr edx,1 loop .modulo_division 

28 Jan 2019, 11:00 

Tomasz Grysztar
OK, I think I understand now. Basically, to find the low digits of you want to compute divided by 7 (with a regular integer division with remainder), where N is at least 32 and a multiple of the period (3 in this case).
Then, because N is a multiple of period: and your division gives: . Dividing both sides by 7, we get: And thus: This number only differs from actual on digits starting from position N. Since N is at least 32, it means that low 32 digits of this number are identical to . Now to get low digits of you need to multiply this by 6 and then add 1, or alternatively just negate the number (subtract it from 0). Yes, this seems like a valid method of obtaining the low digits of 2adic fraction, but when N becomes too large it becomes problematic. And, of course, you need to know the period here. I believe there should exist an actual algorithm for padic division digit after digit, starting from the lowest one  with such one we would not have to know the period in advance. 

28 Jan 2019, 11:25 

revolution
The period for 1/37 is 36. I think that can't be done in 32bit code.


28 Jan 2019, 11:28 

Tomasz Grysztar
revolution wrote: The period for 1/37 is 36. I think that can't be done in 32bit code. Code: ; compute low 32 digits of 1/37: mov edx,1 shl (3632) xor eax,eax sub eax,1 sbb edx,0 mov ecx,37 div ecx test edx,edx jnz assertion_failed ; negate to get low 32 digits of 1/37: neg eax 

28 Jan 2019, 11:40 

Tomasz Grysztar
I patched up a code that uses chained DIVs and should work for any odd divisor:
Code: mov ecx,2 ; number of dwords, except the highest one all are 0FFFFFFFFh mov ebx,1 ; start with 2^331 search: push ecx xor edx,edx mov eax,ebx jmp first_division divide: or eax,0FFFFFFFFh first_division: div [divisor] loop divide pop ecx test edx,edx jz found stc rcl ebx,1 jnc search mov ebx,1 inc ecx jmp search found: ; eax is now low 32 digits of 1/[divisor] neg eax ; negate to get low 32 digits of 1/[divisor] Last edited by Tomasz Grysztar on 28 Jan 2019, 12:43; edited 1 time in total 

28 Jan 2019, 11:59 

revolution
Yes, for very large periods I think using a normal native divider is the way. So the code needs to split in two paths, one for short periods where we can use a single imul, and another for other values where we use the div.


28 Jan 2019, 12:01 

Tomasz Grysztar
revolution wrote: Yes, for very large periods I think using a normal native divider is the way. So the code needs to split in two paths, one for short periods where we can use a single imul, and another for other values where we use the div. Code: ; assuming EAX is divisible by 137, compute the quotient: imul eax,077975B9h ; multiply by low 32 digits of 2adic 1/137 

28 Jan 2019, 12:05 

revolution
Tomasz Grysztar wrote: When divisor is an odd number, once you have low 32 bits of its reciprocal, you always use just a single IMUL. 

28 Jan 2019, 12:13 

Tomasz Grysztar
revolution wrote:
The only hard part is to find the reciprocal. The algorithms we have here are not efficient. I may think about finding something better. On the other hand, when you know the divisor in advance (which is usual case for replacing division with multiplication), you can take some time to find its 2adic reciprocal and then just use it. 

28 Jan 2019, 12:22 

Tomasz Grysztar
All right, I designed much more efficient algorithm for computing 2adic division. Just put A=1 to compute a reciprocal:
Code: divide_2adic: ; in: ; edx = low 32 bits of A ; ebx = low 32 bits of B (must be odd) ; on success: ; cf = 0 ; eax = low 32 bits of 2adic A/B test ebx,1 jz .invalid xor eax,eax .iteration: mov ecx,eax imul ecx,ebx xor ecx,edx bsf ecx,ecx jz .done bts eax,ecx jmp .iteration .done: clc ret .invalid: stc ret Code: ; a few tests: mov edx,1 mov ebx,3 call divide_2adic ; result: 0xAAAAAAAB mov edx,1 mov ebx,3 call divide_2adic ; result: 0x55555555 mov edx,1 mov ebx,5 call divide_2adic ; result: 0xCCCCCCCD mov edx,1 mov ebx,7 call divide_2adic ; result: 0xB6DB6DB7 mov edx,1 mov ebx,37 call divide_2adic ; result: 0x914C1BAD mov edx,1 mov ebx,137 call divide_2adic ; result: 0x077975B9 

28 Jan 2019, 13:20 

revolution
I did some more tests. In fact I brute forced the entire 32bit space. Didn't take too long either, just a few minutes. Using a normal div instead of imul would have taken much longer.
Code: format ELF executable 0 entry main SYS_EXIT = 1 SYS_WRITE = 4 STD_OUTPUT = 1 segment executable main: mov esi,1 .loop_divisor: ;limit the print output to avoid spamming the console mov edx,esi and edx,not 1 bsf eax,edx bsr edx,edx sub edx,eax cmp edx,4  1 ;if we have no more than the top four bits set then print something ja .done_print mov edx,esi mov ecx,trial.at call write_hex mov ecx,trial mov edx,trial.len call out_message .done_print: ;divide_2adic: ;see post from Tomasz Grysztar at https://board.flatassembler.net/topic.php?p=208280#208280 xor edi,edi .divide_2adic_test: mov ecx,edi imul ecx,esi xor ecx,1 ;fixed numerator of 1 bsf ecx,ecx jz .divide_2adic_done bts edi,ecx jmp .divide_2adic_test .divide_2adic_done: ;edi = the 2adic divisor xor eax,eax .loop_trials: mov ecx,eax imul ecx,edi imul ecx,esi cmp eax,ecx jne .failed add eax,esi ;only trial proper multiples jnc .loop_trials add esi,2 ;only trial odd numbers jnc .loop_divisor mov ecx,success mov edx,success.len call out_message .close: mov eax,SYS_EXIT xor ebx,ebx int 0x80 .failed: mov edx,esi mov ecx,fail.at call write_hex mov ecx,fail mov edx,fail.len call out_message jmp .close out_message: mov eax,SYS_WRITE mov ebx,STD_OUTPUT int 0x80 retn write_hex: ;ecx = address ;edx = value .next_nibble: mov eax,edx shr eax,28 cmp al,10 sbb al,0x69 das mov [ecx],al inc ecx shl edx,4 jnz .next_nibble retn segment readable writeable trial: db 'Trialing 0x' trial.at db '00000000',10 trial.len = $  trial fail: db 'Failed at 0x' fail.at db '00000000',10 fail.len = $  fail success: db 'All values passed okay',10 success.len = $  success Quote: All values passed okay 

28 Jan 2019, 14:13 

Tomasz Grysztar
This also seems like a good place to try out a BMI instruction:
Code: divide_2adic: ; in: ; edx = low 32 bits of A ; ebx = low 32 bits of B (must be odd) ; on success: ; cf = 0 ; eax = low 32 bits of 2adic A/B test ebx,1 jz .invalid xor eax,eax .iteration: mov ecx,eax imul ecx,ebx xor ecx,edx jz .done blsi ecx,ecx or eax,ecx jmp .iteration .done: clc ret .invalid: stc ret 

28 Jan 2019, 15:22 

revolution
Tomasz Grysztar wrote: This also seems like a good place to try out a BMI instruction: Quote: Illegal instruction (core dumped) 

28 Jan 2019, 15:38 

revolution
We can also get other ratios with this.
For example for division by 5: To get a ratio of 0.2 (1/5) for numbers of the form 5*n use the value 0xcccccccd To get a ratio of 0.4 (2/5) for numbers of the form 5*n use the value 0x9999999a To get a ratio of 0.6 (3/5) for numbers of the form 5*n use the value 0x66666667 To get a ratio of 0.8 (4/5) for numbers of the form 5*n use the value 0x33333334 

28 Jan 2019, 16:47 

Tomasz Grysztar
Wait, even IMUL is not really needed there. The product can be accumulated instead of being recomputed every iteration:
Code: divide_2adic: ; in: ; edx = low 32 bits of A ; ebx = low 32 bits of B (must be odd) ; on success: ; cf = 0 ; eax = low 32 bits of 2adic A/B test ebx,1 jz .invalid xor eax,eax xor esi,esi .iteration: mov ecx,esi xor ecx,edx jz .done bsf ecx,ecx bts eax,ecx mov edi,ebx shl edi,cl add esi,edi jmp .iteration .done: clc ret .invalid: stc ret 

28 Jan 2019, 22:30 

revolution
Tomasz Grysztar wrote: Wait, even IMUL is not really needed there. The product can be accumulated instead of being recomputed every iteration: Surprisingly the runtimes are about evenly split between the 2adic computation and the range trialling checker. But in the final application the 2adic computation time will be insignificant, which is why I didn't really care about using div. Just that using div was tricky with the period determination. 

29 Jan 2019, 06:06 

Tomasz Grysztar
You could also use this routine directly, to divide any two numbers. Well, except that B needs to be odd, but if it's even you can start with shifting both A and B right in parallel until B becomes odd:
Code: divide_2adic: ; in: ; edx = low 32 bits of A ; ebx = low 32 bits of B ; on success: ; cf = 0 ; eax = low 32 bits of 2adic A/B ; on failure: ; cf = 1 test ebx,1 jz .shift xor eax,eax xor esi,esi .iteration: mov ecx,esi xor ecx,edx jz .done bsf ecx,ecx bts eax,ecx mov edi,ebx shl edi,cl add esi,edi jmp .iteration .done: clc ret .shift: shr ebx,1 jz .invalid ; divide by 0 shr edx,1 jnc divide_2adic ; underflow .invalid: stc ret In fact, with this function you get access to all basic arithmetic operations on 2adic numbers (addition, subtraction, multiplication and division), as long as you can go by having just low 32 digits of the result. ADD, SUB and IMUL work just as well for truncated 2adic numbers as for integers, so the division is the only operation that needs a specialized routine. What is especially nice about truncated 2adic arithmetics is that you can use it to perform several operations on rational numbers, and as long as you know that the final result is going to be an integer in 32bit range, you know that result is going to be precise. You need such additional knowledge because just by looking at the lowest 32 bits of 2adic result you are not able to tell an integer from a fraction. Within this truncated 32bit section, arithmetic operations do not lose precision at any point, except when you divide by 2, which is a shift right and can lose a set bit by shifting it out of register (it is a failure condition in the above function). But the number of bits that are shifted into negative range is always finite, so you can get around the problem by shifting your numbers left far enough (and then shift the final result back right). Other than that, you can accumulate as many arithmetic operations on rational numbers as you wish and the result is still going to be exact, no matter how many additions/multiplications/divisions you do: Code: ; compute (1/13+8/17)*(210/11+1) = 11 mov edx,1 mov ebx,13 call divide_2adic jc failed push eax mov edx,8 mov ebx,17 call divide_2adic jc failed pop ebx add eax,ebx push eax mov edx,210 mov ebx,11 call divide_2adic jc failed inc eax pop ebx imul eax,ebx cmp eax,11 jne failed 

29 Jan 2019, 09:30 

Goto page Previous 1, 2, 3 Next < Last Thread  Next Thread > 
Forum Rules:

Copyright © 19992020, Tomasz Grysztar. Also on GitHub, YouTube, Twitter.
Website powered by rwasa.