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For a monte carlo integration process, I need to pull a lot of random samples from
a histogram that has N buckets, and where N is arbitrary (i.e. not a power of two) but
doesn't change at all during the course of the computation.
By a lot, I mean something on the order of 10^10, 10 billions, so pretty much any
kind of lengthy precomputation is likely worth it in the face of the sheer number of
samples).
I have at my disposal a very fast uniform pseudo random number generator that
typically produces unsigned 64 bits integers (all the ints in the discussion
below are unsigned).
The naive way to pull a sample : histogram[ prng() % histogram.size() ]
The naive way is very slow: the modulo operation is using an integer division (IDIV)
which is terribly expensive and the compiler, not knowing the value of histogram.size()
at compile time, can't be up to its usual magic (i.e. http://www.azillionmonkeys.com/qed/adiv.html)
As a matter of fact, the bulk of my computation time is spent extracting that darn modulo.
The slightly less naive way: I use libdivide (http://libdivide.com/) which is capable
of pulling off a very fast "divide by a constant not known at compile time".
That gives me a very nice win (25% or so), but I have a nagging feeling that I can do
better, here's why:
First intuition: libdivide computes a division. What I need is a modulo, and to get there
I have to do an additional mult and a sub : mod = dividend - divisor*(uint64_t)(dividend/divisor). I suspect there might be a small win there, using libdivide-type
techniques that produce the modulo directly.
Second intuition: I am actually not interested in the modulo itself. What I truly want is
to efficiently produce a uniformly distributed integer value that is guaranteed to be strictly smaller than N.
The modulo is a fairly standard way of getting there, because of two of its properties:
A) mod(prng(), N) is guaranteed to be uniformly distributed if prng() is
B) mod(prgn(), N) is guaranteed to belong to [0,N[
But the modulo is/does much more that just satisfy the two constraints above, and in fact
it does probably too much work.
All need is a function, any function that obeys constraints A) and B) and is fast.
So, long intro, but here comes my two questions:
Is there something out there equivalent to libdivide that computes integer modulos directly ?
Is there some function F(X, N) of integers X and N which obeys the following two constraints:
If X is a random variable uniformly distributed then F(X,N) is also unirformly distributed
F(X, N) is guranteed to be in [0, N[
(PS : I know that if N is small, I do not need to cunsume all the 64 bits coming out of
the PRNG. As a matter of fact, I already do that. But like I said, even that optimization
is a minor win when compare to the big fat loss of having to compute a modulo).
Edit : prng() % N is indeed not exactly uniformly distributed. But for N large enough, I don't think it's much of problem (or is it ?)
Edit 2 : prng() % N is indeed potentially very badly distributed. I had never realized how bad it could get. Ouch. I found a good article on this : http://ericlippert.com/2013/12/16/how-much-bias-is-introduced-by-the-remainder-technique
Under the circumstances, the simplest approach may work the best. One extremely simple approach that might work out if your PRNG is fast enough would be to pre-compute one less than the next larger power of 2 than your N to use as a mask. I.e., given some number that looks like 0001xxxxxxxx in binary (where x means we don't care if it's a 1 or a 0) we want a mask like 000111111111.
From there, we generate numbers as follows:
Generate a number
and it with your mask
if result > n, go to 1
The exact effectiveness of this will depend on how close N is to a power of 2. Each successive power of 2 is (obviously enough) double its predecessor. So, in the best case N is exactly one less than a power of 2, and our test in step 3 always passes. We've added only a mask and a comparison to the time taken for the PRNG itself.
In the worst case, N is exactly equal to a power of 2. In this case, we expect to throw away roughly half the numbers we generated.
On average, N ends up roughly halfway between powers of 2. That means, on average, we throw away about one out of four inputs. We can nearly ignore the mask and comparison themselves, so our speed loss compared to the "raw" generator is basically equal to the number of its outputs that we discard, or 25% on average.
If you have fast access to the needed instruction, you could 64-bit multiply prng() by N and return the high 64 bits of the 128-bit result. This is sort of like multiplying a uniform real in [0, 1) by N and truncating, with bias on the order of the modulo version (i.e., practically negligible; a 32-bit version of this answer would have small but perhaps noticeable bias).
Another possibility to explore would be use word parallelism on a branchless modulo algorithm operating on single bits, to get random numbers in batches.
Libdivide, or any other complex ways to optimize that modulo are simply overkill. In a situation as yours, the only sensible approach is to
ensure that your table size is a power of two (add padding if you must!)
replace the modulo operation with a bitmask operation. Like this:
size_t tableSize = 1 << 16;
size_t tableMask = tableSize - 1;
...
histogram[prng() & tableMask]
A bitmask operation is a single cycle on any CPU that is worth its money, you can't beat its speed.
--
Note:
I don't know about the quality of your random number generator, but it may not be a good idea to use the last bits of the random number. Some RNGs produce poor randomness in the last bits and better randomness in the upper bits. If that is the case with your RNG, use a bitshift to get the most significant bits:
size_t bitCount = 16;
...
histogram[prng() >> (64 - bitCount)]
This is just as fast as the bitmask, but it uses different bits.
You could extend your histogram to a "large" power of two by cycling it, filling in the trailing spaces with some dummy value (guaranteed to never occur in the real data). E.g. given a histogram
[10, 5, 6]
extend it to length 16 like so (assuming -1 is an appropriate sentinel):
[10, 5, 6, 10, 5, 6, 10, 5, 6, 10, 5, 6, 10, 5, 6, -1]
Then sampling can be done via a binary mask histogram[prng() & mask] where mask = (1 << new_length) - 1, with a check for the sentinel value to retry, that is,
int value;
do {
value = histogram[prng() & mask];
} while (value == SENTINEL);
// use `value` here
The extension is longer than necessary to make retries unlikely by ensuring that the vast majority of the elements are valid (e.g. in the example above only 1/16 lookups will "fail", and this rate can be reduced further by extending it to e.g. 64). You could even use a "branch prediction" hint (e.g. __builtin_expect in GCC) on the check so that the compiler orders code to be optimal for the case when value != SENTINEL, which is hopefully the common case.
This is very much a memory vs. speed trade-off.
Just a few ideas to complement the other good answers:
What percent of time is spent in the modulo operation, and how do you know what that percent is? I only ask because sometimes people say something is terribly slow when in fact it is less than 10% of the time and they only think it's big because they're using a silly self-time-only profiler. (I have a hard time envisioning a modulo operation taking a lot of time compared to a random number generator.)
When does the number of buckets become known? If it doesn't change too frequently, you can write a program-generator. When the number of buckets changes, automatically print out a new program, compile, link, and use it for your massive execution.
That way, the compiler will know the number of buckets.
Have you considered using a quasi-random number generator, as opposed to a pseudo-random generator? It can give you higher precision of integration in much fewer samples.
Could the number of buckets be reduced without hurting the accuracy of the integration too much?
The non-uniformity dbaupp cautions about can be side-stepped by rejecting&redrawing values no less than M*(2^64/M) (before taking the modulus).
If M can be represented in no more than 32 bits, you can get more than one value less than M by repeated multiplication (see David Eisenstat's answer) or divmod; alternatively, you can use bit operations to single out bit patterns long enough for M, again rejecting values no less than M.
(I'd be surprised at modulus not being dwarfed in time/cycle/energy consumption by random number generation.)
To feed the bucket, you may use std::binomial_distribution to directly feed each bucket instead of feeding the bucket one sample by one sample:
Following may help:
int nrolls = 60; // number of experiments
const std::size_t N = 6;
unsigned int bucket[N] = {};
std::mt19937 generator(time(nullptr));
for (int i = 0; i != N; ++i) {
double proba = 1. / static_cast<double>(N - i);
std::binomial_distribution<int> distribution (nrolls, proba);
bucket[i] = distribution(generator);
nrolls -= bucket[i];
}
Live example
Instead of integer division you can use fixed point math, i.e integer multiplication & bitshift. Say if your prng() returns values in range 0-65535 and you want this quantized to range 0-99, then you do (prng()*100)>>16. Just make sure that the multiplication doesn't overflow your integer type, so you may have to shift the result of prng() right. Note that this mapping is better than modulo since it's retains the uniform distribution.
Thanks everyone for you suggestions.
First, I am now thoroughly convinced that modulo is really evil.
It is both very slow and yields incorrect results in most cases.
After implementing and testing quite a few of the suggestions, what
seems to be the best speed/quality compromise is the solution proposed
by #Gene:
pre-compute normalizer as:
auto normalizer = histogram.size() / (1.0+urng.max());
draw samples with:
return histogram[ (uint32_t)floor(urng() * normalizer);
It is the fastest of all methods I've tried so far, and as far as I can tell,
it yields a distribution that's much better, even if it may not be as perfect
as the rejection method.
Edit: I implemented David Eisenstat's method, which is more or less the same as Jarkkol's suggestion : index = (rng() * N) >> 32. It works as well as the floating point normalization and it is a little faster (9% faster in fact). So it is my preferred way now.
I was thinking about an algorithm in division of large numbers: dividing with remainder bigint C by bigint D, where we know the representation of C in base b, and D is of form b^k-1. It's probably the easiest to show it on an example. Let's try dividing C=21979182173 by D=999.
We write the number as sets of three digits: 21 979 182 173
We take sums (modulo 999) of consecutive sets, starting from the left: 21 001 183 356
We add 1 to those sets preceding the ones where we "went over 999": 22 001 183 356
Indeed, 21979182173/999=22001183 and remainder 356.
I've calculated the complexity and, if I'm not mistaken, the algorithm should work in O(n), n being the number of digits of C in base b representation. I've also done a very crude and unoptimized version of the algorithm (only for b=10) in C++, tested it against GMP's general integer division algorithm and it really does seem to fare better than GMP. I couldn't find anything like this implemented anywhere I looked, so I had to resort to testing it against general division.
I found several articles which discuss what seem to be quite similar matters, but none of them concentrate on actual implementations, especially in bases different than 2. I suppose that's because of the way numbers are internally stored, although the mentioned algorithm seems useful for, say, b=10, even taking that into account. I also tried contacting some other people, but, again, to no avail.
Thus, my question would be: is there an article or a book or something where the aforementioned algorithm is described, possibly discussing the implementations? If not, would it make sense for me to try and implement and test such an algorithm in, say, C/C++ or is this algorithm somehow inherently bad?
Also, I'm not a programmer and while I'm reasonably OK at programming, I admittedly don't have much knowledge of computer "internals". Thus, pardon my ignorance - it's highly possible there are one or more very stupid things in this post. Sorry once again.
Thanks a lot!
Further clarification of points raised in the comments/answers:
Thanks, everyone - as I didn't want to comment on all the great answers and advice with the same thing, I'd just like to address one point a lot of you touched on.
I am fully aware that working in bases 2^n is, generally speaking, clearly the most efficient way of doing things. Pretty much all bigint libraries use 2^32 or whatever. However, what if (and, I emphasize, it would be useful only for this particular algorithm!) we implement bigints as an array of digits in base b? Of course, we require b here to be "reasonable": b=10, the most natural case, seems reasonable enough. I know it's more or less inefficient both considering memory and time, taking into account how numbers are internally stored, but I have been able to, if my (basic and possibly somehow flawed) tests are correct, produce results faster than GMP's general division, which would give sense to implementing such an algorithm.
Ninefingers notices I'd have to use in that case an expensive modulo operation. I hope not: I can see if old+new crossed, say, 999, just by looking at the number of digits of old+new+1. If it has 4 digits, we're done. Even more, since old<999 and new<=999, we know that if old+new+1 has 4 digits (it can't have more), then, (old+new)%999 equals deleting the leftmost digit of (old+new+1), which I presume we can do cheaply.
Of course, I'm not disputing obvious limitations of this algorithm nor I claim it can't be improved - it can only divide with a certain class of numbers and we have to a priori know the representation of dividend in base b. However, for b=10, for instance, the latter seems natural.
Now, say we have implemented bignums as I outlined above. Say C=(a_1a_2...a_n) in base b and D=b^k-1. The algorithm (which could be probably much more optimized) would go like this. I hope there aren't many typos.
if k>n, we're obviously done
add a zero (i.e. a_0=0) at the beginning of C (just in case we try to divide, say, 9999 with 99)
l=n%k (mod for "regular" integers - shouldn't be too expensive)
old=(a_0...a_l) (the first set of digits, possibly with less than k digits)
for (i=l+1; i < n; i=i+k) (We will have floor(n/k) or so iterations)
new=(a_i...a_(i+k-1))
new=new+old (this is bigint addition, thus O(k))
aux=new+1 (again, bigint addition - O(k) - which I'm not happy about)
if aux has more than k digits
delete first digit of aux
old=old+1 (bigint addition once again)
fill old with zeroes at the beginning so it has as much digits as it should
(a_(i-k)...a_(i-1))=old (if i=l+1, (a _ 0...a _ l)=old)
new=aux
fill new with zeroes at the beginning so it has as much digits as it should
(a_i...a_(i+k-1)=new
quot=(a_0...a_(n-k+1))
rem=new
There, thanks for discussing this with me - as I said, this does seem to me to be an interesting "special case" algorithm to try to implement, test and discuss, if nobody sees any fatal flaws in it. If it's something not widely discussed so far, even better. Please, let me know what you think. Sorry about the long post.
Also, just a few more personal comments:
#Ninefingers: I actually have some (very basic!) knowledge of how GMP works, what it does and of general bigint division algorithms, so I was able to understand much of your argument. I'm also aware GMP is highly optimized and in a way customizes itself for different platforms, so I'm certainly not trying to "beat it" in general - that seems as much fruitful as attacking a tank with a pointed stick. However, that's not the idea of this algorithm - it works in very special cases (which GMP does not appear to cover). On an unrelated note, are you sure general divisions are done in O(n)? The most I've seen done is M(n). (And that can, if I understand correctly, in practice (Schönhage–Strassen etc.) not reach O(n). Fürer's algorithm, which still doesn't reach O(n), is, if I'm correct, almost purely theoretical.)
#Avi Berger: This doesn't actually seem to be exactly the same as "casting out nines", although the idea is similar. However, the aforementioned algorithm should work all the time, if I'm not mistaken.
Your algorithm is a variation of a base 10 algorithm known as "casting out nines". Your example is using base 1000 and "casting out" 999's (one less than the base). This used to be taught in elementary school as way to do a quick check on hand calculations. I had a high school math teacher who was horrified to learn that it wasn't being taught anymore and filled us in on it.
Casting out 999's in base 1000 won't work as a general division algorithm. It will generate values that are congruent modulo 999 to the actual quotient and remainder - not the actual values. Your algorithm is a bit different and I haven't checked if it works, but it is based on effectively using base 1000 and the divisor being 1 less than the base. If you wanted to try it for dividing by 47, you would have to convert to a base 48 number system first.
Google "casting out nines" for more information.
Edit: I originally read your post a bit too quickly, and you do know of this as a working algorithm. As #Ninefingers and #Karl Bielefeldt have stated more clearly than me in their comments, what you aren't including in your performance estimate is the conversion into a base appropriate for the particular divisor at hand.
I feel the need to add to this based on my comment. This isn't an answer, but an explanation as to the background.
A bignum library uses what are called limbs - search for mp_limb_t in the gmp source, which are usually a fixed-size integer field.
When you do something like addition, one way (albeit inefficient) to approach it is to do this:
doublelimb r = limb_a + limb_b + carryfrompreviousiteration
This double-sized limb catches the overflow of limb_a + limb_b in the case that the sum is bigger than the limb size. So if the total is bigger than 2^32 if we're using uint32_t as our limb size, the overflow can be caught.
Why do we need this? Well, what you typically do is loop through all the limbs - you've done this yourself in dividing your integer up and going through each one - but we do it LSL first (so the smallest limb first) just as you'd do arithmetic by hand.
This might seem inefficient, but this is just the C way of doing things. To really break out the big guns, x86 has adc as an instruction - add with carry. What this does is an arithmetic and on your fields and sets the carry bit if the arithmetic overflows the size of the register. The next time you do add or adc, the processor factors in the carry bit too. In subtraction it's called the borrow flag.
This also applies to shift operations. As such, this feature of the processor is crucial to what makes bignums fast. So the fact is, there's electronic circuitry in the chip for doing this stuff - doing it in software is always going to be slower.
Without going into too much detail, operations are built up from this ability to add, shift, subtract etc. They're crucial. Oh and you use the full width of your processor's register per limb if you're doing it right.
Second point - conversion between bases. You cannot take a value in the middle of a number and change it's base, because you can't account for the overflow from the digit beneath it in your original base, and that number can't account for the overflow from the digit beneath... and so on. In short, every time you want to change base, you need to convert the entire bignum from the original base to your new base back again. So you have to walk the bignum (all the limbs) three times at least. Or, alternatively, detect overflows expensively in all other operations... remember, now you need to do modulo operations to work out if you overflowed, whereas before the processor was doing it for us.
I should also like to add that whilst what you've got is probably quick for this case, bear in mind that as a bignum library gmp does a fair bit of work for you, like memory management. If you're using mpz_ you're using an abstraction above what I've described here, for starters. Finally, gmp uses hand optimised assembly with unrolled loops for just about every platform you've ever heard of, plus more. There's a very good reason it ships with Mathematica, Maple et al.
Now, just for reference, some reading material.
Modern Computer Arithmetic is a Knuth-like work for arbitrary precision libraries.
Donald Knuth, Seminumerical Algorithms (The Art of Computer Programming Volume II).
William Hart's blog on implementing algorithm's for bsdnt in which he discusses various division algorithms. If you're interested in bignum libraries, this is an excellent resource. I considered myself a good programmer until I started following this sort of stuff...
To sum it up for you: division assembly instructions suck, so people generally compute inverses and multiply instead, as you do when defining division in modular arithmetic. The various techniques that exist (see MCA) are mostly O(n).
Edit: Ok, not all of the techniques are O(n). Most of the techniques called div1 (dividing by something not bigger than a limb are O(n). When you go bigger you end up with O(n^2) complexity; this is hard to avoid.
Now, could you implement bigints as an array of digits? Well yes, of course you could. However, consider the idea just under addition
/* you wouldn't do this just before add, it's just to
show you the declaration.
*/
uint32_t* x = malloc(num_limbs*sizeof(uint32_t));
uint32_t* y = malloc(num_limbs*sizeof(uint32_t));
uint32_t* a = malloc(num_limbs*sizeof(uint32_t));
uint32_t m;
for ( i = 0; i < num_limbs; i++ )
{
m = 0;
uint64_t t = x[i] + y[i] + m;
/* now we need to work out if that overflowed at all */
if ( (t/somebase) >= 1 ) /* expensive division */
{
m = t % somebase; /* get the overflow */
}
}
/* frees somewhere */
That's a rough sketch of what you're looking at for addition via your scheme. So you have to run the conversion between bases. So you're going to need a conversion to your representation for the base, then back when you're done, because this form is just really slow everywhere else. We're not talking about the difference between O(n) and O(n^2) here, but we are talking about an expensive division instruction per limb or an expensive conversion every time you want to divide. See this.
Next up, how do you expand your division for general case division? By that, I mean when you want to divide those two numbers x and y from the above code. You can't, is the answer, without resorting to bignum-based facilities, which are expensive. See Knuth. Taking modulo a number greater than your size doesn't work.
Let me explain. Try 21979182173 mod 1099. Let's assume here for simplicity's sake that the biggest size field we can have is three digits. This is a contrived example, but the biggest field size I know if uses 128 bits using gcc extensions. Anyway, the point is, you:
21 979 182 173
Split your number into limbs. Then you take modulo and sum:
21 1000 1182 1355
It doesn't work. This is where Avi is correct, because this is a form of casting out nines, or an adaption thereof, but it doesn't work here because our fields have overflowed for a start - you're using the modulo to ensure each field stays within its limb/field size.
So what's the solution? Split your number up into a series of appropriately sized bignums? And start using bignum functions to calculate everything you need to? This is going to be much slower than any existing way of manipulating the fields directly.
Now perhaps you're only proposing this case for dividing by a limb, not a bignum, in which case it can work, but hensel division and precomputed inverses etc do to without the conversion requirement. I have no idea if this algorithm would be faster than say hensel division; it would be an interesting comparison; the problem comes with a common representation across the bignum library. The representation chosen in existing bignum libraries is for the reasons I've expanded on - it makes sense at the assembly level, where it was first done.
As a side note; you don't have to use uint32_t to represent your limbs. You use a size ideally the size of the registers of the system (say uint64_t) so that you can take advantage of assembly-optimised versions. So on a 64-bit system adc rax, rbx only sets the overflow (CF) if the result overspills 2^64 bits.
tl;dr version: the problem isn't your algorithm or idea; it's the problem of converting between bases, since the representation you need for your algorithm isn't the most efficient way to do it in add/sub/mul etc. To paraphrase knuth: This shows you the difference between mathematical elegance and computational efficiency.
If you need to frequently divide by the same divisor, using it (or a power of it) as your base makes division as cheap as bit-shifting is for base 2 binary integers.
You could use base 999 if you want; there's nothing special about using a power-of-10 base except that it makes conversion to decimal integer very cheap. (You can work one limb at a time instead of having to do a full division over the whole integer. It's like the difference between converting a binary integer to decimal vs. turning every 4 bits into a hex digit. Binary -> hex can start with the most significant bits, but converting to non-power-of-2 bases has to be LSB-first using division.)
For example, to compute the first 1000 decimal digits of Fibonacci(109) for a code-golf question with a performance requirement, my 105 bytes of x86 machine code answer used the same algorithm as this Python answer: the usual a+=b; b+=a Fibonacci iteration, but divide by (a power of) 10 every time a gets too large.
Fibonacci grows faster than carry propagates, so discarding the low decimal digits occasionally doesn't change the high digits long-term. (You keep a few extra beyond the precision you want).
Dividing by a power of 2 doesn't work, unless you keep track of how many powers of 2 you've discarded, because the eventual binary -> decimal conversion at the end would depend on that.
So for this algorithm, you have to do extended-precision addition, and division by 10 (or whatever power of 10 you want).
I stored base-109 limbs in 32-bit integer elements. Dividing by 109 is trivially cheap: just a pointer increment to skip the low limb. Instead of actually doing a memmove, I just offset the pointer used by the next add iteration.
I think division by a power of 10 other than 10^9 would be somewhat cheap, but would require an actual division on each limb, and propagating the remainder to the next limb.
Extended-precision addition is somewhat more expensive this way than with binary limbs, because I have to generate the carry-out manually with a compare: sum[i] = a[i] + b[i]; carry = sum < a; (unsigned comparison). And also manually wrap to 10^9 based on that compare, with a conditional-move instruction. But I was able to use that carry-out as an input to adc (x86 add-with-carry instruction).
You don't need a full modulo to handle the wrapping on addition, because you know you've wrapped at most once.
This wastes a just over 2 bits of each 32-bit limb: 10^9 instead of 2^32 = 4.29... * 10^9. Storing base-10 digits one per byte would be significantly less space efficient, and very much worse for performance, because an 8-bit binary addition costs the same as a 64-bit binary addition on a modern 64-bit CPU.
I was aiming for code-size: for pure performance I would have used 64-bit limbs holding base-10^19 "digits". (2^64 = 1.84... * 10^19, so this wastes less than 1 bit per 64.) This lets you get twice as much work done with each hardware add instruction. Hmm, actually this might be a problem: the sum of two limbs might wrap the 64-bit integer, so just checking for > 10^19 isn't sufficient anymore. You could work in base 5*10^18, or in base 10^18, or do more complicated carry-out detection that checks for binary carry as well as manual carry.
Storing packed BCD with one digit per 4 bit nibble would be even worse for performance, because there isn't hardware support for blocking carry from one nibble to the next within a byte.
Overall, my version ran about 10x faster than the Python extended-precision version on the same hardware (but it had room for significant optimization for speed, by dividing less often). (70 seconds or 80 seconds vs. 12 minutes)
Still, I think for this particular implementation of that algorithm (where I only needed addition and division, and division happened after every few additions), the choice of base-10^9 limbs was very good. There are much more efficient algorithms for the Nth Fibonacci number that don't need to do 1 billion extended-precision additions.
I want to store billions (10^9) of double precision floating point numbers in memory and save space. These values are grouped in thousands of ordered sets (they are time series), and within a set, I know that the difference between values is usually not large (compared to their absolute value). Also, the closer to each other, the higher the probability of the difference being relatively small.
A perfect fit would be a delta encoding that stores only the difference of each value to its predecessor. However, I want random access to subsets of the data, so I can't depend on going through a complete set in sequence. I'm therefore using deltas to a set-wide baseline that yields deltas which I expect to be within 10 to 50 percent of the absolute value (most of the time).
I have considered the following approaches:
divide the smaller value by the larger one, yielding a value between 0 and 1 that could be stored as an integer of some fixed precision plus one bit for remembering which number was divided by which. This is fairly straightforward and yields satisfactory compression, but is not a lossless method and thus only a secondary choice.
XOR the IEEE 754 binary64 encoded representations of both values and store the length of the long stretches of zeroes at the beginning of the exponent and mantissa plus the remaining bits which were different. Here I'm quite unsure how to judge the compression, although I think it should be good in most cases.
Are there standard ways to do this? What might be problems about my approaches above? What other solutions have you seen or used yourself?
Rarely are all the bits of a double-precision number meaningful.
If you have billions of values that are the result of some measurement, find the calibration and error of your measurement device. Quantize the values so that you only work with meaningful bits.
Often, you'll find that you only need 16 bits of actual dynamic range. You can probably compress all of this into arrays of "short" that retain all of the original input.
Use a simple "Z-score technique" where every value is really a signed fraction of the standard deviation.
So a sequence of samples with a mean of m and a standard deviation of s gets transformed into a bunch of Z score. Normal Z-score transformations use a double, but you should use a fixed-point version of that double. s/1000 or s/16384 or something that retains only the actual precision of your data, not the noise bits on the end.
for u in samples:
z = int( 16384*(u-m)/s )
for z in scaled_samples:
u = s*(z/16384.0)+m
Your Z-scores retain a pleasant easy-to-work with statistical relationship with the original samples.
Let's say you use a signed 16-bit Z-score. You have +/- 32,768. Scale this by 16,384 and your Z-scores have an effective resolution of 0.000061 decimal.
If you use a signed 24-but Z-score, you have +/- 8 million. Scale this by 4,194,304 and you have a resolution of 0.00000024.
I seriously doubt you have measuring devices this accurate. Further, any arithmetic done as part of filter, calibration or noise reduction may reduce the effective range because of noise bits introduced during the arithmetic. A badly thought-out division operator could make a great many of your decimal places nothing more than noise.
Whatever compression scheme you pick, you can decouple that from the problem of needing to be able to perform arbitrary seeks by compressing into fixed-size blocks and prepending to each block a header containing all the data required to decompress it (e.g. for a delta encoding scheme, the block would contain deltas enconded in some fashion that takes advantage of their small magnitude to make them take less space, e.g. fewer bits for exponent/mantissa, conversion to fixed-point value, Huffman encoding etc; and the header a single uncompressed sample); seeking then becomes a matter of cheaply selecting the appropriate block, then decompressing it.
If the compression ratio is so variable that much space is being wasted padding the compressed data to produce fixed size blocks, a directory of offsets into the compressed data could be built instead and the state required to decompress recorded in that.
If you know a group of doubles has the same exponent, you could store the exponent once, and only store the mantissa for each value.
I need to evaluate the sum of the row: 1/1+1/2+1/3+...+1/n. Considering that in C++ evaluations are not complete accurate, the order of summation plays important role. 1/n+1/(n-1)+...+1/2+1/1 expression gives the more accurate result.
So I need to find out the order of summation, which provides the maximum accuracy.
I don't even know where to begin.
Preferred language of realization is C++.
Sorry for my English, if there are any mistakes.
For large n you'd better use asymptotic formulas, like the ones on http://en.wikipedia.org/wiki/Harmonic_number;
Another way is to use exp-log transformation. Basically:
H_n = 1 + 1/2 + 1/3 + ... + 1/n = log(exp(1 + 1/2 + 1/3 + ... + 1/n)) = log(exp(1) * exp(1/2) * exp(1/3) * ... * exp(1/n)).
Exponents and logarithms can be calculated pretty quickly and accuratelly by your standard library. Using multiplication you should get much more accurate results.
If this is your homework and you are required to use simple addition, you'll better add from the smallest one to the largest one, as others suggested.
The reason for the lack of accuracy is the precision of the float, double, and long double types. They only store so many "decimal" places. So adding a very small value to a large value has no effect, the small term is "lost" in the larger one.
The series you're summing has a "long tail", in the sense that the small terms should add up to a large contribution. But if you sum in descending order, then after a while each new small term will have no effect (even before that, most of its decimal places will be discarded). Once you get to that point you can add a billion more terms, and if you do them one at a time it still has no effect.
I think that summing in ascending order should give best accuracy for this kind of series, although it's possible there are some odd corner cases where errors due to rounding to powers of (1/2) might just so happen to give a closer answer for some addition orders than others. You probably can't really predict this, though.
I don't even know where to begin.
Here: What Every Computer Scientist Should Know About Floating-Point Arithmetic
Actually, if you're doing the summation for large N, adding in order from smallest to largest is not the best way -- you can still get into a situation where the numbers you're adding are too small relative to the sum to produce an accurate result.
Look at the problem this way: You have N summations, regardless of ordering, and you wish to have the least total error. Thus, you should be able to get the least total error by minimizing the error of each summation -- and you minimize the error in a summation by adding values as nearly close to each other as possible. I believe that following that chain of logic gives you a binary tree of partial sums:
Sum[0,i] = value[i]
Sum[1,i/2] = Sum[0,i] + Sum[0,i+1]
Sum[j+1,i/2] = Sum[j,i] + Sum[j,i+1]
and so on until you get to a single answer.
Of course, when N is not a power of two, you'll end up with leftovers at each stage, which you need to carry over into the summations at the next stage.
(The margins of StackOverflow are of course too small to include a proof that this is optimal. In part because I haven't taken the time to prove it. But it does work for any N, however large, as all of the additions are adding values of nearly identical magnitude. Well, all but log(N) of them in the worst not-power-of-2 case, and that's vanishingly small compared to N.)
http://en.wikipedia.org/wiki/Arbitrary-precision_arithmetic
You can find libraries with ready for use implementation for C/C++.
For example http://www.apfloat.org/apfloat/
Unless you use some accurate closed-form representation, a small-to-large ordered summation is likely to be most accurate simple solution (it's not clear to me why a log-exp would help - that's a neat trick, but you're not winning anything with it here, as far as I can tell).
You can further gain precision by realizing that after a while, the sum will become "quantized": Effectively, when you have 2 digits of precision, adding 1.3 to 41 results in 42, not 42.3 - but you achieve almost a precision doubling by maintaining an "error" term. This is called Kahan Summation. You'd compute the error term (42-41-1.3 == -0.3) and correct that in the next addition by adding 0.3 to the next term before you add it in again.
Kahan Summation in addition to a small-to-large ordering is liable to be as accurate as you'll ever need to get. I seriously doubt you'll ever need anything better for the harmonic series - after all, even after 2^45 iterations (crazy many) you'd still only be dealing with a numbers that are at least 1/2^45 large, and a sum that's on the order of 45 (<2^6), for an order of magnitude difference of 51 powers-of-two - i.e. even still representable in a double precision variable if you add in the "wrong" order.
If you go small-to-large, and use Kahan Summation, the sun's probably going to extinguish before today's processors reach a percent of error - and you'll run into other tricky accuracy issues just due to the individual term error on that scale first anyhow (being that a number of the order of 2^53 or larger cannot be represented accurately as a double at all anyhow.)
I'm not sure about the order of summation playing an important role, I havent heard that before. I guess you want to do this in floating point arithmetic so the first thing is to think more inline of (1.0/1.0 + 1.0/2.0+1.0/3.0) - otherwise the compiler will do integer division
to determine order of evaluation, maybe a for loop or brackets?
e.g.
float f = 0.0;
for (int i=n; i>0; --i)
{
f += 1.0/static_cast<float>(i);
}
oh forgot to say, compilers will normally have switches to determine floating point evaluation mode. this is maybe related to what you say on order of summation - in visual C+ these are found in code-generation compile settings, in g++ there're options -float that handle this
actually, the other guy is right - you should do summation in order of smallest component first; so
1/n + 1/(n-1) .. 1/1
this is because the precision of a floating point number is linked to the scale, if you start at 1 you'll have 23 bits of precision relative to 1.0. if you start at a smaller number the precision is relative to the smaller number, so you'll get 23 bits of precision relative to 1xe-200 or whatever. then as the number gets bigger rounding error will occur, but the overall error will be less than the other direction
As all your numbers are rationals, the easiest (and also maybe the fastest, as it will have to do less floating point operations) would be to do the computations with rationals (tuples of 2 integers p,q), and then do just one floating point division at the end.
update to use this technique effectively you will need to use bigints for p & q, as they grow quite fast...
A fast prototype in Lisp, that has built in rationals shows:
(defun sum_harmonic (n acc)
(if (= n 0) acc (sum_harmonic (- n 1) (+ acc (/ 1 n)))))
(sum_harmonic 10 0)
7381/2520
[2.9289682]
(sum_harmonic 100 0)
14466636279520351160221518043104131447711/278881500918849908658135235741249214272
[5.1873775]
(sum_harmonic 1000 0)
53362913282294785045591045624042980409652472280384260097101349248456268889497101
75750609790198503569140908873155046809837844217211788500946430234432656602250210
02784256328520814055449412104425101426727702947747127089179639677796104532246924
26866468888281582071984897105110796873249319155529397017508931564519976085734473
01418328401172441228064907430770373668317005580029365923508858936023528585280816
0759574737836655413175508131522517/712886527466509305316638415571427292066835886
18858930404520019911543240875811114994764441519138715869117178170195752565129802
64067621009251465871004305131072686268143200196609974862745937188343705015434452
52373974529896314567498212823695623282379401106880926231770886197954079124775455
80493264757378299233527517967352480424636380511370343312147817468508784534856780
21888075373249921995672056932029099390891687487672697950931603520000
[7.485471]
So, the next better option could be to mantain the list of floating points and to reduce it summing the two smallest numbers in each step...
my understanding of the entropy formula is that it's used to compute the minimum number of bits required to represent some data. It's usually worded differently when defined, but the previous understanding is what I relied on until now.
Here's my problem. Suppose I have a sequence of 100 '1' followed by 100 '0' = 200 bits. The alphabet is {0,1}, base of entropy is 2. Probability of symbol "0" is 0.5 and "1" is 0.5. So the entropy is 1 or 1 bit to represent 1 bit.
However you can run-length encode it with something like 100 / 1 / 100 / 0 where it's number of bits to output followed by the bit. It seems like I have a representation smaller than the data. Especially if you increase the 100 to much larger number.
I'm using: http://en.wikipedia.org/wiki/Information_entropy as reference at the moment.
Where did I go wrong? Is it the probability assigned to symbols? I don't think it's wrong. Or did I get the connection between compression and entropy wrong? Anything else?
Thanks.
Edit
Following some of the answers my followup are: would you apply the entropy formula to a particular instance of a message to try to find out its information content? Would it be valid to take the message "aaab" and say the entropy is ~0.811. If yes then what's the entropy of 1...10....0 where 1s and 0s are repeated n times using the entropy formula. Is the answer 1?
Yes I understand that you are creating a random variable of your input symbols and guessing at the probability mass function based on your message. What I'm trying to confirm is the entropy formula does not take into account the position of the symbols in the message.
Or did I get the connection between compression and entropy wrong?
You're pretty close, but this last question is where the mistake was. If you're able to compress something into a form that was smaller than its original representation, it means that the original representation had at least some redundancy. Each bit in the message really wasn't conveying 1 bit of information.
Because redundant data does not contribute to the information content of a message, it also does not increase its entropy. Imagine, for example, a "random bit generator" that only returns the value "0". This conveys no information at all! (Actually, it conveys an undefined amount of information, because any binary message consisting of only one kind of symbol requires a division by zero in the entropy formula.)
By contrast, had you simulated a large number of random coin flips, it would be very hard to reduce the size of this message by much. Each bit would be contributing close to 1 bit of entropy.
When you compress data, you extract that redundancy. In exchange, you pay a one-time entropy price by having to devise a scheme that knows how to compress and decompress this data; that itself takes some information.
However you can run-length encode it with something like 100 / 1 / 100 / 0 where it's number of bits to output followed by the bit. It seems like I have a representation smaller than the data. Especially if you increase the 100 to much larger number.
To summarize, the fact that you could devise a scheme to make the encoding of the data smaller than the original data tells you something important. Namely, it says that your original data contained very little information.
Further reading
For a more thorough treatment of this, including exactly how you'd calculate the entropy for any arbitrary sequence of digits with a few examples, check out this short whitepaper.
Have a look at Kolmogorov complexity
The minimum number of bits into which a string can be compressed without losing information. This is defined with respect to a fixed, but universal decompression scheme, given by a universal Turing machine.
And in your particular case, don't restrict yourself to alphabet {0,1}. For your example use {0...0, 1...1} (hundred of 0's and hundred of 1's)
Your encoding works in this example, but it is possible to conceive an equally valid case: 010101010101... which would be encoded as 1 / 0 / 1 / 1 / ...
Entropy is measured across all possible messages that can be constructed in the given alphabet, and not just pathological examples!
John Feminella got it right, but I think there is more to say.
Shannon entropy is based on probability, and probability is always in the eye of the beholder.
You said that 1 and 0 were equally likely (0.5). If that is so, then the string of 100 1s followed by 100 0s has a probability of 0.5^200, of which -log(base 2) is 200 bits, as you expect. However, the entropy of that string (in Shannon terms) is its information content times its probability, or 200 * 0.5^200, still a really small number.
This is important because if you do run-length coding to compress the string, in the case of this string it will get a small length, but averaged over all 2^200 strings, it will not do well. With luck, it will average out to about 200, but not less.
On the other hand, if you look at your original string and say it is so striking that whoever generated it is likely to generate more like it, then you are really saying its probability is larger than 0.5^200, so you are making a different assumptions about the original probability structure of the generator of the string, namely that it has lower entropy than 200 bits.
Personally, I find this subject really interesting, especially when you look into Kolmogorov (Algorithmic) information. In that case, you define the information content of a string as the length of the smallest program that could generate it. This leads to all sorts of insights into software engineering and language design.
I hope that helps, and thanks for your question.