I am trying to implement the idea found in this paper:
http://crypto.stanford.edu/craig/easy-fhe.pdf.
However, I do not know how the compute the circuit representation of an algorithm.
Suppose I have a function that takes a list of exactly 32, 32 bit signed integers, and returns a 64 bit signed integer representing the sum of the integers. How can I convert this function to a Boolean function? That is, I need to design a circuit where each output wire is a boolean function of the ands/ ors/ and nots of the 1024 input wires.
Notice that the function will take a fixed width input and produce a fixed width output.
Are there any techniques from electrical engineering or math that I can use?
Consider the logic in an FPGA. I think this will help you get an idea of the kind of circuit needed to sum exactly 32 32-bit inputs into a 64-bit output.
Related
I have a 32-bit number that is created by encoding some data, I want to be more confident that the data (a max 32-bit number) is not changed when decoding it, so I am going to add some error detection bits.
I need to keep the data as short as possible, so I can only add a few bits for error detection, in some cases just 1 bit.
I'm looking for an algorithm that detects more bit changes and needs fewer extra bits.
I was thinking of calculating a checksum or CRC and just dropping extra bits or maybe xor the result to make it shorter but I'm not sure if the error detection remains good enough.
Thanks in advance for any help.
A 1-bit CRC, with polynomial x+1 would simply be the parity of your 32 message bits. That will detect any one-bit error in the resulting 32 bits. For a 2-bit CRC, you can use x2+1. You can define a CRC of any length. See Koopman's list for good CRC polynomials for CRCs of degree 3 and higher.
I am currently working on a school project and one of my tasks is to implement a 16-bit by 16-bit 2's complement integer divider as a digital logic circuit (in other words 16-bit input divided by another 16-bit input). The output is straight forward where it shows quotient Q and remainder R. Also special cases like dividing by zero are taken care of with preset conditions.
My primary issue here is that the only way that I am able to implement this is by using long division or a very long recurring subtraction. Even then, I'm not sure how to implement long division without creating a messy circuit. Open to suggestions in case there is no other way.
Because of this, I have looked into other division algorithms like the Newton-Raphson division, but I don't think those algorithms are possible to implement as a logic circuit (and I just don't know and understand how too..). So I was wondering if there were any speed-friendly division algorithms to do this.
My application requires a fractional quantity multiplied by a monetary value.
For example, $65.50 × 0.55 hours = $36.025 (rounded to $36.03).
I know that floats should not be used to represent money, so I'm storing all of my monetary values as cents. $65.50 in the above equation is stored as 6550 (integer).
For the fractional coefficient, my issue is that 0.55 does not have a 32-bit float representation. In the use case above, 0.55 hours == 33 minutes, so 0.55 is an example of a specific value that my application will need to account for exactly. The floating point representation of 0.550000012 is insufficient, because the user will not understand where the additional 0.000000012 came from. I cannot simply call a rounding function on 0.550000012 because it will round to the whole number.
Multiplication solution
To solve this, my first idea was to store all quantities as integers and multiply × 1000. So 0.55 entered by the user would become 550 (integer) when stored. All calculations would happen without floats, and then simply divide by 1000 (integer division, not float) when presenting the result to the user.
I realize that this would permanently limit me to 3 decimal places of
precision. If I decide that 3 is adequate for the lifetime of my
application, does this approach make sense?
Are there potential rounding issues if I were to use integer division?
Is there a name for this process? EDIT: As indicated by #SergGr, this is fixed-point arithmetic.
Is there a better approach?
EDIT:
I should have clarified, this is not time-specific. It is for generic quantities like 1.256 pounds of flour, 1 sofa, or 0.25 hours (think invoices).
What I'm trying to replicate here is a more exact version of Postgres's extra_float_digits = 0 functionality, where if the user enters 0.55 (float32), the database stores 0.550000012 but when queried for the result returns 0.55 which appears to be exactly what the user typed.
I am willing to limit this application's precision to 3 decimal places (it's business, not scientific), so that's what made me consider the × 1000 approach.
I'm using the Go programming language, but I'm interested in generic cross-language solutions.
Another solution to store the result is using the rational form of the value. You can explain the number by two integer value which the number is equal p/q, such that both p and q are integers. Hence, you can have more precision for your numbers and do some math with the rational numbers in the format of two integers.
Note: This is an attempt to merge different comments into one coherent answer as was requested by Matt.
TL;DR
Yes, this approach makes sense but most probably is not the best choice
Yes, there are rounding issues but there inevitably will be some no matter what representation you use
What you suggest using is called Decimal fixed point numbers
I'd argue yes, there is a better approach and it is to use some standard or popular decimal floating point numbers library for your language (Go is not my native language so I can't recommend one)
In PostgreSQL it is better to use Numeric (something like Numeric(15,3) for example) rather than a combination of float4/float8 and extra_float_digits. Actually this is what the first item in the PostgreSQL doc on Floating-Point Types suggests:
If you require exact storage and calculations (such as for monetary amounts), use the numeric type instead.
Some more details on how non-integer numbers can be stored
First of all there is a fundamental fact that there are infinitely many numbers in the range [0;1] so you obviously can't store every number there in any finite data structure. It means you have to make some compromises: no matter what way you choose, there will be some numbers you can't store exactly so you'll have to round.
Another important point is that people are used to 10-based system and in that system only results of division by numbers in a form of 2^a*5^b can be represented using a finite number of digits. For every other rational number even if you somehow store it in the exact form, you will have to do some truncation and rounding at the formatting for human usage stage.
Potentially there are infinitely many ways to store numbers. In practice only a few are widely used:
floating point numbers with two major branches of binary (this is what most today's hardware natively implements and what is support by most of the languages as float or double) and decimal. This is the format that store mantissa and exponent (can be negative), so the number is mantissa * base^exponent (I omit sign and just say it is logically a part of the mantissa although in practice it is usually stored separately). Binary vs. decimal is specified by the base. For example 0.5 will be stored in binary as a pair (1,-1) i.e. 1*2^-1 and in decimal as a pair (5,-1) i.e. 5*10^-1. Theoretically you can use any other base as well but in practice only 2 and 10 make sense as the bases.
fixed point numbers with the same division in binary and decimal. The idea is the same as in floating point numbers but some fixed exponent is used for all the numbers. What you suggests is actually a decimal fixed point number with the exponent fixed at -3. I've seen a usage of binary fixed-point numbers on some embedded hardware where there is no built-in support of floating point numbers, because binary fixed-point numbers can be implemented with reasonable efficiency using integer arithmetic. As for decimal fixed-point numbers, in practice they are not much easier to implement that decimal floating-point numbers but provide much less flexibility.
rational numbers format i.e. the value is stored as a pair of (p, q) which represents p/q (and usually q>0 so sign stored in p and either p=0, q=1 for 0 or gcd(p,q) = 1 for every other number). Usually this requires some big integer arithmetic to be useful in the first place (here is a Go example of math.big.Rat). Actually this might be an useful format for some problems and people often forget about this possibility, probably because it is often not a part of a standard library. Another obvious drawback is that as I said people are not used to think in rational numbers (can you easily compare which is greater 123/456 or 213/789?) so you'll have to convert the final results to some other form. Another drawback is that if you have a long chain of computations, internal numbers (p and q) might easily become very big values so computations will be slow. Still it may be useful to store intermediate results of calculations.
In practical terms there is also a division into arbitrary length and fixed length representations. For example:
IEEE 754 float or double are fixed length floating-point binary representations,
Go math.big.Float is an arbitrary length floating-point binary representations
.Net decimal is a fixed length floating-point decimal representations
Java BigDecimal is an arbitrary length floating-point decimal representations
In practical terms I'd says that the best solution for your problem is some big enough fixed length floating point decimal representations (like .Net decimal). An arbitrary length implementation would also work. If you have to make an implementation from scratch, than your idea of a fixed length fixed point decimal representation might be OK because it is the easiest thing to implement yourself (a bit easier than the previous alternatives) but it may become a burden at some point.
As mentioned in the comments, it would be best to use some builtin Decimal module in your language to handle exact arithmetic. However, since you haven't specified a language, we cannot be certain that your language may even have such a module. If it does not, here is how to go about doing so.
Consider using Binary Coded Decimal to store your values. The way it works is by restricting the values that can be stored per byte to 0 through 9 (inclusive), "wasting" the rest. You can encode a decimal representation of a number byte by byte that way. For example, 613 would become
6 -> 0000 0110
1 -> 0000 0001
3 -> 0000 0011
613 -> 0000 0110 0000 0001 0000 0011
Where each grouping of 4 digits above is a "nibble" of a byte. In practice, a packed variant is used, where two decimal digits are packed into a byte (one per nibble) to be less "wasteful". You can then implement a few methods to do your basic addition, subtract, multiplication, etc. Just iterate over an array of bytes, and perform your classic grade school addition / multiplication algorithms (keep in mind for the packed variant that you may need to pad a zero to get an even number of nibbles). You just need to keep a variable to store where the decimal point is, and remember to carry where necessary to preserve the encoding.
This might look a repetition to my earlier question. But I think its not.
I am looking for a technique to convert the signal in the Decimal format to binary format.
I intend to use the Simulink blocks in the Xilinx Library to convert decimal to binary format.
So if the input is 3, the expected output should in 11( 2 Clock Cycles). I am looking for the output to be obtained serially.
Please suggest me how to do it or any pointers in the internet would be helpful.
Thanks
You are correct, what you need is the parallel to serial block from system generator.
It is described in this document:
http://www.xilinx.com/support/documentation/sw_manuals/xilinx13_1/sysgen_ref.pdf
This block is a rate changing block. Check the mentions of the parallel to serial block in these documents for further descriptions:
http://www.xilinx.com/support/documentation/sw_manuals/xilinx13_1/sysgen_gs.pdf
http://www.xilinx.com/support/documentation/sw_manuals/xilinx13_1/sysgen_user.pdf
Use a normal constant block with a Matlab variable in it, this already gives the output in "normal" binary (assuming you set the properties on it to be unsigned and the binary point at 0.
Then you need to write a small serialiser block, which takes that input, latches it into a shift register and then shifts the register once per clock cycle with the bit that "falls off the end" becoming your output bit. Depending on which way your shift, you can make it come MSB first of LSB first.
You'll have to build the shift register out of ordinary registers and a mux before each one to select whether you are doing a parallel load or shifting. (This is the sort of thing which is a couple of lines of code in VHDL, but a right faff in graphics).
If you have to increase the serial rate, you need to clock it from a faster clock - you could use a DCM to generate this.
Matlab has a dec2bin function that will convert from a decimal number to a binary string. So, for example dec2bin(3) would return 11.
There's also a corresponding bin2dec which takes a binary string and converts to a decimal number, so that bin2dec('11') would return 3.
If you're wanting to convert a non-integer decimal number to a binary string, you'll first want to determine what's the smallest binary place you want to represent, and then do a little bit of pre- and post-processing, combined with dec2bin to get the results you're looking for. So, if the smallest binary place you want is the 1/512th place (or 2^-9), then you could do the following (where binPrecision equals 1/512):
function result = myDec2Bin(decNum, binPrecision)
isNegative=(decNum < 0);
intPart=floor(abs(decNum));
binaryIntPart=dec2bin(intPart);
fracPart=abs(decNum)-intPart;
scaledFracPart=round(fracPart / binPrecision);
scaledBinRep=dec2bin(scaledFracPart);
temp=num2str(10^log2(1/binPrecision)+str2num(scaledBinRep),'%d');
result=[binaryIntPart,'.',temp(2:end)];
if isNegative
result=['-',result];
end
end
The result of myDec2Bin(0.256, 1/512) would then be 0.010000011, and the result of myDec2Bin(-0.984, 1/512) would be -0.111111000. (Note that the output is a string.)
How the heck does Ruby do this? Does Jörg or anyone else know what's happening behind the scenes?
Unfortunately I don't know C very well so bignum.c is of little help to me. I was just kind of curious it someone could explain (in plain English) the theory behind whatever miracle algorithm its using.
irb(main):001:0> 999**999
368063488259223267894700840060521865838338232037353204655959621437025609300472231530103873614505175218691345257589896391130393189447969771645832382192366076536631132001776175977932178658703660778465765811830827876982014124022948671975678131724958064427949902810498973271030787716781467419524180040734398996952930832508934116945966120176735120823151959779536852290090377452502236990839453416790640456116471139751546750048602189291028640970574762600185950226138244530187489211615864021135312077912018844630780307462205252807737757672094320692373101032517459518497524015120165166724189816766397247824175394802028228160027100623998873667435799073054618906855460488351426611310634023489044291860510352301912426608488807462312126590206830413782664554260411266378866626653755763627796569082931785645600816236891168141774993267488171702172191072731069216881668294625679492696148976999868715671440874206427212056717373099639711168901197440416590226524192782842896415414611688187391232048327738965820265934093108172054875188246591760877131657895633586576611857277011782497943522945011248430439201297015119468730712364007639373910811953430309476832453230123996750235710787086641070310288725389595138936784715274150426495416196669832679980253436807864187160054589045664027158817958549374490512399055448819148487049363674611664609890030088549591992466360050042566270348330911795487647045949301286614658650071299695652245266080672989921799342509291635330827874264789587306974472327718704306352445925996155619153783913237212716010410294999877569745287353422903443387562746452522860420416689019732913798073773281533570910205207767157128174184873357050830752777900041943256738499067821488421053870869022738698816059810579221002560882999884763252161747566893835178558961142349304466506402373556318707175710866983035313122068321102457824112014969387225476259342872866363550383840720010832906695360553556647545295849966279980830561242960013654529514995113584909050813015198928283202189194615501403435553060147713139766323195743324848047347575473228198492343231496580885057330510949058490527738662697480293583612233134502078182014347192522391449087738579081585795613547198599661273567662441490401862839817822686573112998663038868314974259766039340894024308383451039874674061160538242392803580758232755749310843694194787991556647907091849600704712003371103926967137408125713631396699343733288014254084819379380555174777020843568689927348949484201042595271932630685747613835385434424807024615161848223715989797178155169951121052285149157137697718850449708843330475301440373094611119631361702936342263219382793996895988331701890693689862459020775599439506870005130750427949747071390095256759203426671803377068109744629909769176319526837824364926844730545524646494321826241925107158040561607706364484910978348669388142016838792902926158979355432483611517588605967745393958061959024834251565197963477521095821435651996730128376734574843289089682710350244222290017891280419782767803785277960834729869249991658417000499998999
Simple: it does it the same way you do, ever since first grade. Except it doesn't compute in base 10, it computes in base 4 billion (and change).
Think about it: with our number system, we can only represent numbers from 0 to 9. So, how can we compute 6+7 without overflowing? Easy: we do actually overflow! We cannot represent the result of 6+7 as a number between 0 and 9, but we can overflow to the next place and represent it as two numbers between 0 and 9: 3×100 + 1×101. If you want to add two numbers, you add them digit-wise from the right and overflow ("carry") to the left. If you want to multiply two numbers, you have to multiply every digit of one number individually with the other number, then add up the intermediate results.
BigNum arithmetic (this is what this kind of arithmetic where the numbers are bigger than the native machine numbers is usually called) works basically the same way. Except that the base is not 10, and its not 2, either – it's the size of a native machine integer. So, on a 32 bit machine, it would be base 232 or 4 294 967 296.
Specifically, in Ruby Integer is actually an abstract class that is never instianted. Instead, it has two subclasses, Fixnum and Bignum, and numbers automagically migrate between them, depending on their size. In MRI and YARV, Fixnum can hold a 31 or 63 bit signed integer (one bit is used for tagging) depending on the native word size of the machine. In JRuby, a Fixnum can hold a full 64 bit signed integer, even on an 32 bit machine.
The simplest operation is adding two numbers. And if you look at the implementation of + or rather bigadd_core in YARV's bignum.c, it's not too bad to follow. I can't read C either, but you can cleary see how it loops over the individual digits.
You could read the source for bignum.c...
At a very high level, without going into any implementation details, bignums are calculated "by hand" like you used to do in grade school. Now, there are certainly many optimizations that can be applied, but that's the gist of it.
I don't know of the implementation details so I'll cover how a basic Big Number implementation would work.
Basically instead of relying on CPU "integers" it will create it's own using multiple CPU integers. To store arbritrary precision, well lets say you have 2 bits. So the current integer is 11. You want to add one. In normal CPU integers, this would roll over to 00
But, for big number, instead of rolling over and keeping a "fixed" integer width, it would allocate another bit and simulate an addition so that the number becomes the correct 100.
Try looking up how binary math can be done on paper. It's very simple and is trivial to convert to an algorithm.
Beaconaut APICalc 2 just released on Jan.18, 2011, which is an arbitrary-precision integer calculator for bignum arithmetic, cryptography analysis and number theory research......
http://www.beaconaut.com/forums/default.aspx?g=posts&t=13
It uses the Bignum class
irb(main):001:0> (999**999).class
=> Bignum
Rdoc is available of course