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Tomasz Grysztar
If you look at my post about division by multiplication you may find a link to a very old discussion on the other message board where I wrote about such trick:
Tomasz Grysztar wrote: If you want to divide by the odd number, and you know the number you want to divide is divisible without remainder, you can do this by multiplying by "magic" number and take the LOW half of the result, without need to do any shifts etc. In the field of 2adic numbers the representation of number 1/5 is an infinite sequence of digits: ...00110011001100110011001101. if we group them in fours and convert to hexadecimal, this is: ...CCCCCCCCCD. Now, in general when you multiply any padic numbers, the low n digits of product only depend on the low n digits of factors. So if you only take low n digits of a rational number like 1/5 and multiply another number by it, the low n digits of the result are going to be exactly the same as in actual product of 1/5 and the other number. Therefore this method should always give a perfect result when the other number is divisible by 5, because then the result is an integer and everything above the low n digits is (in 2adic case) an infinite sequence of 0s or 1s (depending whether the integer was positive or negative). However, when the number is not divisible by 5, the result is a rational number with a more complex infinite expansion, for example: 9*...CCCCCCCD. = ...33333335. The low digits of 2adic representation of 9/5 do not resemble a small integer at all. 

27 Jan 2019, 14:18 

revolution
I'm still trying to understand the whole 2adic thing. The highorder bits I can understand how that works, but the low order bits ... they are still a mystery to me about what happens there.


27 Jan 2019, 15:15 

Tomasz Grysztar
Let me try to explain it as simply as I can.
For certain you are already familiar with positional numeral systems like decimal, binary, hexadecimal, etc. They can represent any positive number in form: where We usually write down such representation as a sequence with the highest power to the left: Now, padic numbers have analogous representations, but they are potentially infinite: Normally this kind of infinite sum would be divergent. But when the base is a prime number, it is possible to define such function of distance between any two numbers (based on padic norm) that these sums become convergent (with padic norm the powers of p are closer and closer to zero the higher the exponent is). Since this only works for prime numbers, we use "p" to denote such base. 2 is prime, so we can have 2adic numbers. They are like binary numbers, but including infinite ones. As we usually write higher powers on the left, we have to append an infinite number of digits on the left side. Any regular positive integer has a representation that just has infinitely many zero digits appended on the left: What about the other sequences? Let's consider a number that has all binary digits set to 1: So this is a representation of a negative number. In fact, all negative numbers are going to have representations that have all digits set to 1 starting at some point. (Note that the calculation we did would normally be invalid, but it is correct under a 2adic metric, all these sums are convergent then). It is possible to add and subtract these numbers just like you would do it with finite ones (with carry), except that here you have to repeat carrying till infinity. If you add one to , the carry is going to repeat endlessly, leaving just zero digits behind. It confirms that this number is the same as minus one. This nicely relates to the two's complement interpretation of binary numbers in computer architectures. Any such number can be seen as a lowest n digits of a 2adic number with assumption that all higher digits are the same as the highest digit (called "sign bit") of these n. For example, with 8bit representations: The two's complement arithmetic is a kind of projection of a more general 2adic arithmetic, and this is what makes it work so well. There are still many other infinite sequences possible. Let's take a look at one that has 0s and 1s interleaved: This time we got a fractional number. We can verify this through multiplication. Just like addition and subtraction, multiplication of padic numbers (which under the hood is a multiplication of infinite series) can be done in the same manner as classic long multiplication, except that it needs to be prolonged infinitely to the left: This one turned out quite simple, because number 3 has just two bits set, so multiplication by 3 ends up being just a sum of a number and its copy multiplied by 2 (that is: shifted one digit to the left). And thus we have verified that the representation of multiplied by 3 indeed gives 1 as a result. I encourage to play with this a little, for example subtract from 0 to see what looks like and then multiply that by 3, which should result in plain 1. Looking at this long multiplication algorithm, we see that low n digits of result only depend on the low n bits of the factors. So, for example, when we know that the result of multiplication is going to be small positive integer, we only need to compute the rightmost digits, as we know that the remaining ones to the left are going to be an infinite chain of zeros. Every rational number with an odd denominator has a periodic 2adic representation. What about even denominators? Dividing a number by 2 is equivalent to shifting its representation right by on digit and we may end up shifting some digits to the negative powers. This is analogous to formal Laurent series. For example: ________________________________________________________________________ Now, how can we find 2adic representation of any fraction? Or padic in general? For a number in form we already know the drill: This gives us, for example: The value of k gives us the length of a period, and the numbers of this form have a representation of mostly zeros, with 1 every k digits. Multiplying by p is the same as shifting left, so numbers like also have a very similar simple form: This kind of number has a period k, with just a single 1 digit repeated over this period. The ones that have numerator smaller than denominator do not overlap with each other. We can therefore split a numerator into parts according to its positional representation, and the terms we get can be easily merged with each other. For example, let's take 5 = 101b = 4 + 1: This demonstrates that negative fractions of such form (with denominator equal to for some k) have a periodic structure and the repeated pattern has the same digits as a representation of numerator (as long as it fits within the number of digits provided by the length of the period, which is true when numerator is smaller than denominator). To make positive fractions out of these negative ones, it is enough to add 1: If we have a denominator n that is not in such nice form, but at least is not divisible by p, we just need to find k such that is a multiple of n. As p and n are relatively prime, such k must exist (it follows from Chinese remainder theorem). In particular, when n is also a prime, such exponent may be quickly obtained with aid of Fermat's little theorem, which states that is then a multiple of n. For example, with n=5 (and p=2) we get , which indeed is a multiple of 5. Then to convert a fraction we just need to multiply numerator and denominator with appropriate value (in this case 3) to obtain denominator of form : Finally, if a denominator is divisible by p, we should factorize it into a power of p and a number not divisible by p: . Then, after finding a representation of , we shift it right by s digits: This is a Laurentserieslike representation, it has a digit corresponding to a negative power of 2. Last edited by Tomasz Grysztar on 14 Feb 2019, 09:21; edited 7 times in total 

27 Jan 2019, 17:15 

revolution
So the 2adic representation of 1/5 (hex ...cccccccc) can be multiplied by any other number with a factor of 5 and we always get a whole number result.
Let's try it out 555555555₁₀ × cccccccd₁₆ = {...<nonzerobits>...}069F6BC7₁₆ which truncates to 111111111₁₀ It works. But I have to make the value cccccccd₁₆ 

27 Jan 2019, 17:39 

revolution
Can we also use that to check for divisibility by 5 by multiplying the result by 5 and see if it equals the original number?
Edit: Of course not, we will get back the same result since 5 * 1/5. Doh. Edit2: Brute force to the rescue Code: format ELF executable 0 entry main SYS_EXIT = 1 SYS_WRITE = 4 STD_OUTPUT = 1 segment executable main: xor eax,eax .loop: mov ecx,eax imul ecx,0xcccccccd imul ecx,5 cmp eax,ecx jne .failed inc eax jnz .loop mov ecx,success mov edx,success.len jmp .close .failed: mov edx,eax mov ecx,fail_at call write_hex mov ecx,failure mov edx,failure.len .close: mov eax,SYS_WRITE mov ebx,STD_OUTPUT int 0x80 mov eax,SYS_EXIT xor ebx,ebx int 0x80 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 failure: db 'Failed at 0x' fail_at db '00000000',10 failure.len = $  failure success: db 'All values passed okay',10 success.len = $  success Quote: All values passed okay 

27 Jan 2019, 17:42 

Tomasz Grysztar
revolution wrote: It works. But I have to make the value cccccccd₁₆ Tomasz Grysztar wrote: 1/5 is an infinite sequence of digits: ...00110011001100110011001101. 

27 Jan 2019, 18:18 

Tomasz Grysztar
I have edited my long post above and added a section explaining how to obtain a 2adic representation of any rational number.
Because numbers that have denominator divisible by 2 have representations that include negative powers of 2, only fractions with an odd denominator can be used in the form of their low 32/64 digits for computations like yours. This is why to divide by 10 you have to use 1/5 and shift. As long as you know that your number is exactly divisible by an odd denominator, you can use the 2adic reciprocal safely. It is not hard to prove that the low 32/64 bits of the result are then going to be correct (even for negative numbers!). I believe that even a simplified explanation that I provided here should be enough of a foundation for a proof. 

27 Jan 2019, 19:45 

Tomasz Grysztar
A bonus question: do you see a relation between this topic and the fact that x86 has "IMUL dest,src" but no "MUL dest,src"?


27 Jan 2019, 20:20 

revolution
So the following bit patterns for divisors should always give correct results.
Code: 0x00000001 0x00000003 0x00000005 < the one I used above 0x0000000f 0x00000011 0x00000033 0x00000055 0x000000ff 0x00000101 0x00000303 0x00000505 0x00000f0f 0x00001111 0x00003333 0x00005555 0x0000ffff 0x00010001 0x00030003 0x00050005 0x000f000f 0x00110011 0x00330033 0x00550055 0x00ff00ff 0x01010101 0x03030303 0x05050505 0x0f0f0f0f 0x11111111 0x33333333 0x55555555 0xffffffff 

27 Jan 2019, 20:27 

revolution
Tomasz Grysztar wrote: A bonus question: do you see a relation between this topic and the fact that x86 has "IMUL dest,src" but no "MUL dest,src"? 

27 Jan 2019, 20:28 

Tomasz Grysztar
revolution wrote: So the following bit patterns for divisors should always give correct results (...) 

27 Jan 2019, 20:31 

revolution
Tomasz Grysztar wrote:


27 Jan 2019, 20:36 

Tomasz Grysztar
revolution wrote: I meant for 32bit integers. The period of the divisor needs to be a factor of 32 or we can't use it to find divisors unless we have 36bit (or some other length) CPUs. Also, I think there are other possible algorithms to obtain a 2adic reciprocal. I have chosen this one because I believe it shows nicely various interactions and allows to quickly get a better grasp of what is going on with these patterns. 

27 Jan 2019, 20:41 

revolution
Tomasz Grysztar wrote: Oh, that is you want to find divisor at runtime 

27 Jan 2019, 20:53 

Tomasz Grysztar
revolution wrote:
This in hexadecimal form becomes: ...DB6DB6DB6DB7h If you cut it down to the lowest 32 bits, you get 0B6DB6DB7h. Please run your tests and see that as long as dividend is a multiple of 7, multiplication by this reciprocal gives the right result. A 2adic reciprocal of any odd number, cut down to its lowest 32 digits, is going to work correctly for all 32bit inputs divisible by that odd number. 

27 Jan 2019, 21:01 

revolution
Tomasz Grysztar wrote:


28 Jan 2019, 04:10 

revolution
So if 1/3 = ...55555555₁₆
We can square that ...55555555₁₆^2 = ...e38e38e38e39₁₆ = +1/9 Therefore to divide by 9 we can use this: Code: imul ebx,0x38E38E39 ;divide multiples of 9 by 9 In the case of 1/7 = 0x0.249249249249... I need to multiply by 5 to get 0x0.B6DB6DB6D... 

28 Jan 2019, 07:01 

Tomasz Grysztar
I have shown the entire procedure in the second half of my edited post (I added that part later). was among the examples there.
For denominator 9, you first need to find its multiple that has form . One that works is . 2adic representation of a number with such denominator is going to have a period of 6: (the digits within a single block are the same as binary representation of 7) (because 56 = 111000b, but it is also just a simple shift left by 3 digits) (with hex digits it is ...E38,E38,E38,E39) 

28 Jan 2019, 07:57 

revolution
Tomasz Grysztar wrote: For denominator 9, you first need to find its multiple that has form . Period of 7 = 3 Period of 9 = 6 Period of 11 = 10 For 7: divisor = 1 + 2^(32  32 mod period) * 6 / 7 = 0x36DB6DB7 Oops, fails For 9: divisor = 1 + 2^(32  32 mod period) * 8 / 9 = 0x38E38E39 For 11: divisor = 1 + 2^(32  32 mod period) * 10 / 11 = 0x3A2E8BA3 Oops, fails. We lost too many bits a the top. Try again For 11: divisor = 1 + 2^(32 + period  32 mod period) * 10 / 11 = 0x(E8)BA2E8BA3 Works. But it needs more than 32 bits to compute the value. However looking at the result we can just set the top bit manually and we get a working value. 

28 Jan 2019, 09:34 

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