C code to Haskell - c

So, i would like to convert a part of C code to Haskell. I wrote this part (it's a simplified example of what I want to do) in C, but being the newbie I am in Haskell, I can't really make it work.
float g(int n, float a, float p, float s)
{
int c;
while (n>0)
{
c = n % 2;
if (!c) s += p;
else s -= p;
p *= a;
n--;
}
return s;
}
Anyone got any ideas/solutions?

Lee's translation is already pretty good (well, he confused the odd and even cases(1)), but he fell into a couple of performance traps.
g n a p s =
if n > 0
then
let c = n `mod` 2
s' = (if c == 0 then (-) else (+)) s p
p' = p * a
in g (n-1) a p' s'
else s
He used mod instead of rem. The latter maps to machine division, the former performs additional checks to ensure a non-negative result. Thus mod is a bit slower than rem, and if either satisfies the needs - because they yield identical results in the case where both arguments are non-negative; or because the result is only compared to 0 (both conditions are satisfied here) - rem is preferable. Even better, and a bit more idiomatic is to use even (which uses rem for the reasons mentioned above). The difference is not huge, though.
No type signature. That means that the code is (type-class) polymorphic, and thus no strictness analysis is possible, nor any specialisations. If the code is used in the same module at a specific type, GHC can (and usually will, if optimisations are enabled) create a specialised version for that specific type that allows strictness analysis and some other optimisations (inlining of class methods like (+) etc.), in that case, one does not pay the polymorhism penalty. But if the use site is in a different module, that cannot happen. If (type-class) polymorphic code is desired, one should mark it INLINABLE or INLINE (for GHC < 7), so that its unfolding is exposed in the .hi file and the function can be specialised and optimised at the use site.
Since g is recursive, it cannot be inlined [meaning, GHC cannot inline it; in principle it is possible] at use sites, which often would enable more optimisations than a mere specialisation.
One technique that often allows better optimisation for recursive functions is the worker/wrapper transformation. One creates a wrapper that calls a recursive (local) worker, then the non-recursive wrapper can be inlined, and when the worker is called with known arguments, that can enable further optimisations like constant folding or, in the case of function arguments, inlining. In particular the latter often has an enormous impact, when combined with a static-argument-transformation (arguments that never change in the recursive calls are not passed as arguments to the recursive worker).
In this case, we only have one static argument of type Float, so a worker/wrapper transformation with a SAT typically makes no difference (as a rule of thumb, a SAT pays off when
the static argument is a function
several non-function arguments are static
so by this rule, we shouldn't expect any benefit from w/w + SAT, and in general, there is none). Here we have one special case where w/w + SAT can make a big difference, and that is when the factor a is 1. GHC has {-# RULES #-} that eliminate multiplication by 1 for various types, and with such a short loop body, a multiplication more or less per iteration makes a difference, the running time is reduced by about 40% after points 3 and 4 have been applied. (There are no RULES for multiplication by 0 or by -1 for floating point types because 0*x = 0 resp. (-1)*x = -x don't hold for NaNs.) For all other a, the w/w + SATed
{-# INLINABLE g #-}
g n a p s = worker n p s
where
worker n p s
| n <= 0 = s
| otherwise = let s' = if even n then s + p else s - p
in worker (n-1) a (p*a) s'
does not perform measurably different from the top-level recursive version with the same optimisations done.
Strictness. GHC's strictness analyser is good, but not perfect. It cannot see far enough through the algorithm to determine that the function is
strict in p if n >= 1 (assuming addition - (+) - is strict in both arguments)
also strict in a if n >= 2 (assuming strictness of (*) in both arguments)
and then produce a worker that is strict in both. Instead you get a worker that uses an unboxed Int# for n and an unboxed Float# for s (I'm using the type Int -> Float -> Float -> Float -> Float here, corresponding to the C), and boxed Floats for a and p. Thus in each iteration you get two unboxings and a re-boxing. That costs (relatively) a lot of time, since besides that it's just a bit of simple arithmetic and tests.
Help GHC along a bit, and make the worker (or g itself, if you don't do the worker/wrapper transform) strict in p (bang pattern for example). That is enough to allow GHC producing a worker using unboxed values throughout.
Using division to test parity (not applicable if the type is Int and the LLVM backend is used).
GHC's optimiser hasn't got down to the low-level bits very much yet, so the native code generator emits a division instruction for
x `rem` 2 == 0
and, when the rest of the loop body is as cheap as it is here, that costs a lot of time. LLVM's optimiser has already been taught to replace that with a bitmasking at type Int, so with ghc -O2 -fllvm you don't need to do that manually. With the native code generator, substituting that with
x .&. 1 == 0
(needs import Data.Bits of course) produces a significant speedup (on normal platforms where a bitwise and is much faster than a division).
The final result
{-# INLINABLE g #-}
g n a p s = worker n p s
where
worker k !ap acc
| k > 0 = worker (k-1) (ap*a) (if k .&. (1 :: Int) == 0 then acc + ap else acc - ap)
| otherwise = acc
performs not measurably different (for the tested values) from the result of gcc -O3 -msse2 loop.c, except for a = -1, where gcc replaces the multiplication with a negation (assuming all NaNs equivalent).
(1) He's not alone in that,
c = n % 2;
if (!c) s += p;
else s -= p;
seems to be really tricky, as far as I can see everybody(2) got that wrong.
(2) With one exception ;)

As a first step, let's simplify your code:
float g(int n, float a, float p, float s) {
if (n <= 0) return s;
float s2 = n % 2 == 0 ? s + p : s - p;
return g(n - 1, a, a*p, s2)
}
We have turned your original function into a recursive one that exhibits a certain structure. It's a sequence! We can turn this into Haskell conveniently:
gs :: Bool -> Float -> Float -> Float -> [Float]
gs nb a p s = s : gs (not nb) a (a*p) (if nb then s - p else s + p)
Finally we just need to index this list:
g :: Integer -> Float -> Float -> Float -> Float
g n a p s = gs (even n) a p s !! (n - 1)
The code is not tested, but it should work. If not, it's probably just an off-by-one error.

Here is how I would tackle this problem in Haskell. First, I observe that there are several loops merged into one here: we are
forming a geometric sequence (whose factor is a suitably negative version of p)
taking a prefix of the sequence
summing the result
So my solution follows this structure as well, with a tiny bit of s and p thrown in for good measure because that's what your code does. In a from-scratch version, I'd probably drop those two parameters entirely.
g n a p s = sum (s : take n (iterate (*(-a)) start)) where
start | odd n = -p
| otherwise = p

A fairly direct translation would be:
g n a p s =
if n > 0
then
let c = n `mod` 2
s' = (if c == 0 then (-) else (+)) s p
p' = p * a
in g (n-1) a p' s'
else s

Look at the signature of the g function (i.e., float g (int n, float a, float p, float s)) you know that your Haskell function will receive 4 elements and return a float, thus:
g :: Integer -> Float -> Float -> Float -> Float
let us now look into the loop, we see that n > 0 is the stop case, and n--; will be the decreasing step used on the recursive call. Therefore:
g :: Integer -> Float -> Float -> Float -> Float
g n a p s | n <= 0 = s
to n > 0, you have another conditional if (!(n % 2)) s += p; else s -= p; inside the loop. If n is odd than you will do s += p, p *= a and n--. In Haskell it will be:
g :: Integer -> Float -> Float -> Float -> Float
g n a p s | n <= 0 = s
| odd n = g (n-1) a (p*a) (s+p)
If n is even than you will do s-=p, p*=a; and n--. Thus:
g :: Integer -> Float -> Float -> Float -> Float
g n a p s | n <= 0 = s
| odd n = g (n-1) a (p*a) (s+p)
| otherwise = g (n-1) a (p*a) (s-p)

To expand on #Landei and #MathematicalOrchid 's comments below the question: The algorithm proposed to solve the problem at hand is always O(n). However, if you realize that what you're actually doing is computing a partial sum of the geometric series, you can use the well-known summation formula:
g n a p s = s + (-1)**n * p * ((-a)**n-1) / (-a-1)
This will be faster as the exponentiation can be done faster than O(n) by repeated squaring or other clever methods, which are likely automatically employed for integer powers by modern compilers.

You can encode loops almost-naturally with the Haskell Prelude function until :: (a -> Bool) -> (a -> a) -> a -> a:
g :: Int -> Float -> Float -> Float -> Float
g n a p s =
fst.snd $
until ((<= 0).fst)
(\(n,(!s,!p)) -> (n-1, (if even n then s+p else s-p, p*a)))
(n,(s,p))
The bang-patterns !s and !p mark strictly-calculated intermediate variables, to prevent excessive laziness which would otherwise harm efficiency.
until pred step start repeatedly applies the step function until pred called with the last generated value will hold, starting with initial value start. It can be represented by the pseudocode:
def until (pred, step, start): // well, actually,
while( true ): def until (pred, step, start):
if pred(start): return(start) if pred(start): return(start)
start := step(start) call until(pred, step, step(start))
The first pseudocode is equivalent to the second (which is how until is actually implemented) in the presence of tail call optimization, which is why in many functional languages where TCO is present loops are encoded via recursion.
So in Haskell, until is coded as
until p f x | p x = x
| otherwise = until p f (f x)
But it could have been coded differently, making explicit the interim results:
until p f x = last $ go x -- or, last (go x)
where go x | p x = [x]
| otherwise = x : go (f x)
Using the Haskell standard higher-order functions break and iterate this could be written as a stream-processing code,
until p f x = let (_,(r:_)) = break p (iterate f x) in r
-- or: span (not.p) ....
or just
until p f x = head $ dropWhile (not.p) $ iterate f x -- or, equivalently,
-- head . dropWhile (not.p) . iterate f $ x
If TCO weren't present in a given Haskell implementation, the last version would be the one to use.
Hopefully this makes clearer how the stream-processing code from Daniel Wagner's answer comes about,
g n a p s = s + (sum . take n . iterate (*(-a)) $ if odd n then (-p) else p)
because the predicate involved is about counting down from n, and
fst . snd . head . dropWhile ((> 0).fst) $
iterate (\(n,(!s,!p)) -> (n-1, (if even n then s+p else s-p, p*a)))
(n,(s,p))
===
fst . snd . head . dropWhile ((> 0).fst) $
iterate (\(n,(!s,!p)) -> (n-1, (s+p, p*(-a))))
(n,(s, if odd n then (-p) else p)) -- 0 is even
===
fst . (!! n) $
iterate (\(!s,!p) -> (s+p, p*(-a)))
(s, if odd n then (-p) else p)
===
foldl' (+) s . take n . iterate (*(-a)) $ if odd n then (-p) else p
In pure FP, the stream-processing paradigm makes all history of a computation available, as a stream (list) of values.

Related

Efficiently computing (a - K) / (a + K) with improved accuracy

In various contexts, for example for the argument reduction for mathematical functions, one needs to compute (a - K) / (a + K), where a is a positive variable argument and K is a constant. In many cases, K is a power of two, which is the use case relevant to my work. I am looking for efficient ways to compute this quotient more accurately than can be accomplished with the straightforward division. Hardware support for fused multiply-add (FMA) can be assumed, as this operation is provided by all major CPU and GPU architectures at this time, and is available in C/C++ via the functionsfma() and fmaf().
For ease of exploration, I am experimenting with float arithmetic. Since I plan to port the approach to double arithmetic as well, no operations using higher than the native precision of both argument and result may be used. My best solution so far is:
/* Compute q = (a - K) / (a + K) with improved accuracy. Variant 1 */
m = a - K;
p = a + K;
r = 1.0f / p;
q = m * r;
t = fmaf (q, -2.0f*K, m);
e = fmaf (q, -m, t);
q = fmaf (r, e, q);
For arguments a in the interval [K/2, 4.23*K], code above computes the quotient almost correctly rounded for all inputs (maximum error is exceedingly close to 0.5 ulps), provided that K is a power of 2, and there is no overflow or underflow in intermediate results. For K not a power of two, this code is still more accurate than the naive algorithm based on division. In terms of performance, this code can be faster than the naive approach on platforms where the floating-point reciprocal can be computed faster than the floating-point division.
I make the following observation when K = 2n: When the upper bound of the work interval increases to 8*K, 16*K, ... maximum error increases gradually and starts to slowly approximate the maximum error of the naive computation from below. Unfortunately, the same does not appear to be true for the lower bound of the interval. If the lower bound drops to 0.25*K, the maximum error of the improved method above equals the maximum error of the naive method.
Is there a method to compute q = (a - K) / (a + K) that can achieve smaller maximum error (measured in ulp vs the mathematical result) compared to both the naive method and the above code sequence, over a wider interval, in particular for intervals whose lower bound is less than 0.5*K? Efficiency is important, but a few more operations than are used in the above code can likely be tolerated.
In one answer below, it was pointed out that I could enhance accuracy by returning the quotient as an unevaluated sum of two operands, that is, as a head-tail pair q:qlo, i.e. similar to the well-known double-float and double-double formats. In my code above, this would mean changing the last line to qlo = r * e.
This approach is certainly useful, and I had already contemplated its use for an extended-precision logarithm for use in pow(). But it doesn't fundamentally help with the desired widening of the interval on which the enhanced computation provides more accurate quotients. In a particular case I am looking at, I would like to use K=2 (for single precision) or K=4 (for double precision) to keep the primary approximation interval narrow, and the interval for a is roughly [0,28]. The practical problem I am facing is that for arguments < 0.25*K the accuracy of the improved division is not substantially better than with the naive method.
If a is large compared to K, then (a-K)/(a+K) = 1 - 2K / (a + K) will give a good approximation. If a is small compared to K, then 2a / (a + K) - 1 will give a good approximation. If K/2 ≤ a ≤ 2K, then a-K is an exact operation, so doing the division will give a decent result.
One possibility is to track error of m and p into m1 and p1 with classical Dekker/Schewchuk:
m=a-k;
k0=a-m;
a0=k0+m;
k1=k0-k;
a1=a-a0;
m1=a1+k1;
p=a+k;
k0=p-a;
a0=p-k0;
k1=k-k0;
a1=a-a0;
p1=a1+k1;
Then, correct the naive division:
q=m/p;
r0=fmaf(p,-q,m);
r1=fmaf(p1,-q,m1);
r=r0+r1;
q1=r/p;
q=q+q1;
That'll cost you 2 divisions, but should be near half ulp if I didn't screw up.
But these divisions can be replaced by multiplications with inverse of p without any problem, since the first incorrectly rounded division will be compensated by remainder r, and second incorrectly rounded division does not really matter (the last bits of correction q1 won't change anything).
I don't really have an answer (proper floating point error analyses are very tedious) but a few observations:
Fast reciprocal instructions (such as RCPSS) are not as accurate as division, so you may see a reduction in accuracy if using these.
m is computed exactly if a &in; [0.5×Kb, 21+n×Kb), where Kb is the power of 2 below K (or K itself if K is a power of 2), and n is the number of trailing zeros in the significand of K (i.e. if K is a power of 2, then n=23).
This is similar to a simplified form of the div2 algorithm from Dekker (1971): to expand the range (particularly the lower bound), you'll probably have to incorporate more correction terms from this (i.e. store m as the sum of 2 floats, or use a double).
Since my goal is to merely widen the interval on which accurate results are achieved, rather than to find a solution that works for all possible values of a, making use of double-float arithmetic for all intermediate computation seems too costly.
Thinking some more about the problem, it is clear that the computation of the remainder of the division, e in the code from my question, is the crucial part of achieving more accurate result. Mathematically, the remainder is (a-K) - q * (a+K). In my code, I simply used m to represent (a-K) and represented (a+k) as m + 2*K, as this delivers numerically superior results to the straightforward representation.
With relatively small additional computational cost, (a+K) can be represented as a double-float, that is, a head-tail pair p:plo, which leads to the following modified version of my original code:
/* Compute q = (a - K) / (a + K) with improved accuracy. Variant 2 */
m = a - K;
p = a + K;
r = 1.0f / p;
q = m * r;
mx = fmaxf (a, K);
mn = fminf (a, K);
plo = (mx - p) + mn;
t = fmaf (q, -p, m);
e = fmaf (q, -plo, t);
q = fmaf (r, e, q);
Testing shows that this delivers nearly correctly rounded results for a in [K/2, 224*K), allowing for a substantial increase to the upper bound of the interval on which accurate results are achieved.
Widening the interval at the lower end requires the more accurate representation of (a-K). We can compute this as a double-float head-tail pair m:mlo, which leads to the following code variant:
/* Compute q = (a - K) / (a + K) with improved accuracy. Variant 3 */
m = a - K;
p = a + K;
r = 1.0f / p;
q = m * r;
plo = (a < K) ? ((K - p) + a) : ((a - p) + K);
mlo = (a < K) ? (a - (K + m)) : ((a - m) - K);
t = fmaf (q, -p, m);
e = fmaf (q, -plo, t);
e = e + mlo;
q = fmaf (r, e, q);
Exhaustive testing hows that this delivers nearly correctly rounded results for a in the interval [K/224, K*224). Unfortunately, this comes at a cost of ten additional operations compared to the code in my question, which is a steep price to pay to get the maximum error from around 1.625 ulps with the naive computation down to near 0.5 ulp.
As in my original code from the question, one can express (a+K) in terms of (a-K), thus eliminating the computation of the tail of p, plo. This approach results in the following code:
/* Compute q = (a - K) / (a + K) with improved accuracy. Variant 4 */
m = a - K;
p = a + K;
r = 1.0f / p;
q = m * r;
mlo = (a < K) ? (a - (K + m)) : ((a - m) - K);
t = fmaf (q, -2.0f*K, m);
t = fmaf (q, -m, t);
e = fmaf (q - 1.0f, -mlo, t);
q = fmaf (r, e, q);
This turns out to be advantageous if the main focus is decreasing the lower limit of the interval, which is my particular focus as explained in the question. Exhaustive testing of the single-precision case shows that when K=2n nearly correctly rounded results are produced for values of a in the interval [K/224, 4.23*K]. With a total of 14 or 15 operations (depending on whether an architecture supports full predication or just conditional moves), this requires seven to eight more operations than my original code.
Lastly, one might base the residual computation directly on the original variable a to avoid the error inherent in the computation of m and p. This leads to the following code that, for K = 2n, computes nearly correctly rounded results for a in the interval [K/224, K/3):
/* Compute q = (a - K) / (a + K) with improved accuracy. Variant 5 */
m = a - K;
p = a + K;
r = 1.0f / p;
q = m * r;
t = fmaf (q + 1.0f, -K, a);
e = fmaf (q, -a, t);
q = fmaf (r, e, q);
If you can relax the API to return another variable that models the error, then the solution becomes much simpler:
float foo(float a, float k, float *res)
{
float ret=(a-k)/(a+k);
*res = fmaf(-ret,a+k,a-k)/(a+k);
return ret;
}
This solution only handles truncation error of division, but does not handle the loss of precision of a+k and a-k.
To handle those errors, I think I need to use double precision, or bithack to use fixed point.
Test code is updated to artificially generate non zero least significant bits
in the input
test code
https://ideone.com/bHxAg8
The problem is the addition in (a + K). Any loss of precision in (a + K) is magnified by the division. The problem isn't the division itself.
If the exponents of a and K are the same (almost) no precision is lost, and if the absolute difference between the exponents is greater than the significand size then either (a + K) == a (if a has larger magnitude) or (a + K) == K (if K has larger magnitude).
There is no way to prevent this. Increasing the significand size (e.g. using 80-bit "extended double" on 80x86) only helps widen the "accurate result range" slightly. To understand why, consider smallest + largest (where smallest is the smallest positive denormal a 32-bit floating point number can be). In this case (for 32-bit floats) you'd need a significand size of about 260 bits for the result to avoid precision loss completely. Doing (e.g.) temp = 1/(a + K); result = a * temp - K / temp; won't help much either because you've still got exactly the same (a + K) problem (but it would avoid a similar problem in (a - K)). Also you can't do result = anything / p + anything_error/p_error because division doesn't work like that.
There are only 3 alternatives I can think of to get close to 0.5 ulps for all possible positive values of a that can fit in 32-bit floating point. None are likely to be acceptable.
The first alternative involves pre-computing a lookup table (using "big real number" maths) for every value of a, which (with some tricks) ends up being about 2 GiB for 32-bit floating point (and completely insane for 64-bit floating point). Of course if the range of possible values of a is smaller than "any positive value that can fit in a 32-bit float" the size of the lookup table would be reduced.
The second alternative is to use something else ("big real number") for the calculation at run-time (and convert to/from 32-bit floating point).
The third alternative involves, "something" (I don't know what it's called, but it's expensive). Set the rounding mode to "round to positive infinity" and calculate temp1 = (a + K); if(a < K) temp2 = (a - K); then switch to "round to negative infinity" and calculate if(a >= K) temp2 = (a - K); lower_bound = temp2 / temp1;. Next do a_lower = a and decrease a_lower by the smallest amount possible and repeat the "lower_bound" calculation, and keep doing that until you get a different value for lower_bound, then revert back to the previous value of a_lower. After that you do essentially the same (but opposite rounding modes, and incrementing not decrementing) to determine upper_bound and a_upper (starting with the original value of a). Finally, interpolate, like a_range = a_upper - a_lower; result = upper_bound * (a_upper - a) / a_range + lower_bound * (a - a_lower) / a_range;. Note that you will want to calculate an initial upper and lower bound and skip all of this if they're equal. Also be warned that this is all "in theory, completely untested" and I probably borked it somewhere.
Mainly what I'm saying is that (in my opinion) you should give up and accept that there's nothing that you can do to get close to 0.5 ulp. Sorry.. :)

Probability mass of summing two discrete random variables, in linearithmic time

Given two discrete random variables, their (arbitrary) probability mass functions a and b and a natural-number N such that both of the variables have the domain [0..N] (therefore the functions can be represented as arrays), the probability that the functions' corresponding random variables have a given sum (i.e. P(A+B==target)) can be computed, in O(N) time, by treating the arrays as vectors and using their dot product, albeit with one of the inputs reversed and both inputs re-sliced in order to align them and eliminate bounds errors; thus each position i of a is matched with a position j of b such that i+j==target. Such an algorithm looks something like this:
-- same runtime as dotProduct and sum; other components are O(1)
P :: Vector Int -> Vector Int -> Int -> Ratio Int
P a b target | length a /= length b = undefined
| 0 <= target && target <= 2 * length a
= (dotProduct (shift target a) (reverse (shift target b)))
%
(sum a * sum b) -- O(length a)
-- == sum $ map (\x -> (a!x)*(b!(target-x))) [0..length a]
| otherwise = 0
where
-- O(1)
shift t v = slice start' len' v
where start = t - length v - 1
len = length v - abs start
-- unlike `drop ... $ take ... v`,
-- slice does not simply `id` when given out-of-bounds indices
start' = min (V.length v) (max 0 start)
len' = min (V.length v) (max 0 len)
-- usual linear-algebra definition
-- O(length a); length inequality already guarded-away by caller
dotProduct a b = sum $ zipWith (*) a b
Given the same information, one might treat the variables' sum as its own discrete random variable, albeit one whose probability mass function is unknown. Evaluating the entirety of this probability mass function (and thereby producing the array that corresponds thereto) can be done in O(N²) time by performing N dot-products, with each product having its operands differently-shifted; i.e.:
pm :: Vector Int -> Vector Int -> Vector (Ratio Int)
pm a b = map (P a b) $ generate (2 * length a + 1) id
I am told, however, that producing such a table of values of this probability mass function can actually be done in O(N*log(N)) time. As far as I can tell, no two of the multiplications across all of the involved dot-products share the same ordered pair of indices, and I do not think that I can, e.g., combine two dot-subproducts in any useful way to form a T(n)=2T(n/2)+O(n)-type recursion; therefore I am curious as to how and why, exactly, such a runtime is possible.
In a nutshell, you have a transformation F (called discrete Fourier transform) that maps the set of vectors of size N onto itself and such that
F(a*b) = F(a).F(b)
where * is the convolution operator you just described and . is the standard dot product.
Moreover Fis invertible and you can therefore recover a*b as
a*b = F^{-1}(F(a).F(b))
Now this is all very nice but the key point is that F (and F^{-1}) can be computed in O(N log(N)) time using something called Fast Fourier Transform (FFT). Thereby, because the usual dot product . can be computed in O(N), you obtain a O(N log(N)) algorithm for computing the convolution of two distributions.
I therefore suggest you look up this and that.

Big integer addition code

I am trying to immplement big integer addition in CUDA using the following code
__global__ void add(unsigned *A, unsigned *B, unsigned *C/*output*/, int radix){
int id = blockIdx.x * blockDim.x + threadIdx.x;
A[id ] = A[id] + B[id];
C[id ] = A[id]/radix;
__syncthreads();
A[id] = A[id]%radix + ((id>0)?C[id -1]:0);
__syncthreads();
C[id] = A[id];
}
but it does not work properly and also i don't now how to handle the extra carry bit. Thanks
TL;DR build a carry-lookahead adder where each individual additionner adds modulo radix, instead of modulo 2
Additions need incoming carries
The problem in your model is that you have a rippling carry. See Rippling carry adders.
If you were in an FPGA that wouldn't be a problem because they have dedicated logic to do that fast (carry chains, they're cool). But alas, you're on a GPU !
That is, for a given id, you only know the input carry (thus whether you are going to sum A[id]+B[id] or A[id]+B[id]+1) when all the sums with smaller id values have been computed. As a matter of fact, initially, you only know the first carry.
A[3]+B[3] + ? A[2]+B[2] + ? A[1]+B[1] + ? A[0]+B[0] + 0
| | | |
v v v v
C[3] C[2] C[1] C[0]
Characterize the carry output
And each sum also has a carry output, which isn't on the drawing. So you have to think of the addition in this larger scheme as a function with 3 inputs and 2 outputs : (C, c_out) = add(A, B, c_in)
In order to not wait O(n) for the sum to complete (where n is the number of items your sum is cut into), you can precompute all the possible results at each id. That isn't such a huge load of work, since A and B don't change, only the carries. So you have 2 possible outputs : (c_out0, C) = add(A, B, 0) and (c_out1, C') = add(A, B, 1).
Now with all these results, we need to basically implement a carry lookahead unit.
For that, we need to figure out to functions of each sum's carry output P and G :
P a.k.a. all of the following definitions
Propagate
"if a carry comes in, then a carry will go out of this sum"
c_out1 && !c_out0
A + B == radix-1
G a.k.a. all of the following definitions
Generate
"whatever carry comes in, a carry will go out of this sum"
c_out1 && c_out0
c_out0
A + B >= radix
So in other terms, c_out = G or (P and c_in). So now we have a start of an algorithm that can tell us easily for each id the carry output as a function of its carry input directly :
At each id, compute C[id] = A[id]+B[id]+0
Get G[id] = C[id] > radix -1
Get P[id] = C[id] == radix-1
Logarithmic tree
Now we can finish in O(log(n)), even though treeish things are nasty on GPUs, but still shorter than waiting. Indeed, from 2 additions next to each other, we can get a group G and a group P :
For id and id+1 :
step = 2
if id % step == 0, do steps 6 through 10, otherwise, do nothing
group_P = P[id] and P[id+step/2]
group_G = (P[id+step/2] and G[id]) or G[id+step/2]
c_in[id+step/2] = G[id] or (P[id] and c_in[id])
step = step * 2
if step < n, go to 5
At the end (after repeating steps 5-10 for every level of your tree with less ids every time), everything will be expressed in terms of Ps and Gs which you computed, and c_in[0] which is 0. On the wikipedia page there are formulas for the grouping by 4 instead of 2, which will get you an answer in O(log_4(n)) instead of O(log_2(n)).
Hence the end of the algorithm :
At each id, get c_in[id]
return (C[id]+c_in[id]) % radix
Take advantage of hardware
What we really did in this last part, was mimic the circuitry of a carry-lookahead adder with logic. However, we already have additionners in the hardware that do similar things (by definition).
Let us replace our definitions of P and G based on radix by those based on 2 like the logic inside our hardware, mimicking a sum of 2 bits a and b at each stage : if P = a ^ b (xor), and G = a & b (logical and). In other words, a = P or G and b = G. So if we create a intP integer and a intG integer, where each bit is respectively the P and G we computed from each ids sum (limiting us to 64 sums), then the addition (intP | intG) + intG has the exact same carry propagation as our elaborate logical scheme.
The reduction to form these integers will still be a logarithmic operation I guess, but that was to be expected.
The interesting part, is that each bit of the sum is function of its carry input. Indeed, every bit of the sum is eventually function of 3 bits a+b+c_in % 2.
If at that bit P == 1, then a + b == 1, thus a+b+c_in % 2 == !c_in
Otherwise, a+b is either 0 or 2, and a+b+c_in % 2 == c_in
Thus we can trivially form the integer (or rather bit-array) int_cin = ((P|G)+G) ^ P with ^ being xor.
Thus we have an alternate ending to our algorithm, replacing steps 4 and later :
at each id, shift P and G by id : P = P << id and G = G << id
do an OR-reduction to get intG and intP which are the OR of all the P and G for id 0..63
Compute (once) int_cin = ((P|G)+G) ^ P
at each id, get `c_in = int_cin & (1 << id) ? 1 : 0;
return (C[id]+c_in) % radix
PS : Also, watch out for integer overflow in your arrays, if radix is big. If it isn't then the whole thing doesn't really make sense I guess...
PPS : in the alternate ending, if you have more than 64 items, characterize them by their P and G as if radix was 2^64, and re-run the same steps at a higher level (reduction, get c_in) and then get back to the lower level apply 7 with P+G+carry in from higher level

How to improve the performance of this Haskell program?

I'm working through the problems in Project Euler as a way of learning Haskell, and I find that my programs are a lot slower than a comparable C version, even when compiled. What can I do to speed up my Haskell programs?
For example, my brute-force solution to Problem 14 is:
import Data.Int
import Data.Ord
import Data.List
searchTo = 1000000
nextNumber :: Int64 -> Int64
nextNumber n
| even n = n `div` 2
| otherwise = 3 * n + 1
sequenceLength :: Int64 -> Int
sequenceLength 1 = 1
sequenceLength n = 1 + (sequenceLength next)
where next = nextNumber n
longestSequence = maximumBy (comparing sequenceLength) [1..searchTo]
main = putStrLn $ show $ longestSequence
Which takes around 220 seconds, while an "equivalent" brute-force C version only takes 1.2 seconds.
#include <stdio.h>
int main(int argc, char **argv)
{
int longest = 0;
int terms = 0;
int i;
unsigned long j;
for (i = 1; i <= 1000000; i++)
{
j = i;
int this_terms = 1;
while (j != 1)
{
this_terms++;
if (this_terms > terms)
{
terms = this_terms;
longest = i;
}
if (j % 2 == 0)
j = j / 2;
else
j = 3 * j + 1;
}
}
printf("%d\n", longest);
return 0;
}
What am I doing wrong? Or am I naive to think that Haskell could even approach C's speed?
(I'm compiling the C version with gcc -O2, and the Haskell version with ghc --make -O).
For testing purpose I have just set searchTo = 100000. The time taken is 7.34s. A few modification leads to some big improvement:
Use an Integer instead of Int64. This improves the time to 1.75s.
Use an accumulator (you don't need sequenceLength to be lazy right?) 1.54s.
seqLen2 :: Int -> Integer -> Int
seqLen2 a 1 = a
seqLen2 a n = seqLen2 (a+1) (nextNumber n)
sequenceLength :: Integer -> Int
sequenceLength = seqLen2 1
Rewrite the nextNumber using quotRem, thus avoiding computing the division twice (once in even and once in div). 1.27s.
nextNumber :: Integer -> Integer
nextNumber n
| r == 0 = q
| otherwise = 6*q + 4
where (q,r) = quotRem n 2
Use Schwartzian transform instead of maximumBy. The problem of maximumBy . comparing is that the sequenceLength function is called more than once for each value. 0.32s.
longestSequence = snd $ maximum [(sequenceLength a, a) | a <- [1..searchTo]]
Note:
I check the time by compiling with ghc -O and run with +RTS -s)
My machine is running on Mac OS X 10.6. The GHC version is 6.12.2. The compiled file is in i386 architecture.)
The C problem runs at 0.078s with the corresponding parameter. It is compiled with gcc -O3 -m32.
Although this is already rather old, let me chime in, there's one crucial point that hasn't been addressed before.
First, the timings of the different programmes on my box. Since I'm on a 64-bit linux system, they show somewhat different characteristics: using Integer instead of Int64 does not improve performance as it would with a 32-bit GHC, where each Int64 operation would incur the cost of a C-call while the computations with Integers fitting in signed 32-bit integers don't need a foreign call (since only few operations exceed that range here, Integer is the better choice on a 32-bit GHC).
C: 0.3 seconds
Original Haskell: 14.24 seconds, using Integer instead of Int64: 33.96 seconds
KennyTM's improved version: 5.55 seconds, using Int: 1.85 seconds
Chris Kuklewicz's version: 5.73 seconds, using Int: 1.90 seconds
FUZxxl's version: 3.56 seconds, using quotRem instead of divMod: 1.79 seconds
So what have we?
Calculate the length with an accumulator so the compiler can transform it (basically) into a loop
Don't recalculate the sequence lengths for the comparisons
Don't use div resp. divMod when it's not necessary, quot resp. quotRem are much faster
What is still missing?
if (j % 2 == 0)
j = j / 2;
else
j = 3 * j + 1;
Any C compiler I have used transforms the test j % 2 == 0 into a bit-masking and doesn't use a division instruction. GHC does not (yet) do that. So testing even n or computing n `quotRem` 2 is quite an expensive operation. Replacing nextNumber in KennyTM's Integer version with
nextNumber :: Integer -> Integer
nextNumber n
| fromInteger n .&. 1 == (0 :: Int) = n `quot` 2
| otherwise = 3*n+1
reduces its running time to 3.25 seconds (Note: for Integer, n `quot` 2 is faster than n `shiftR` 1, that takes 12.69 seconds!).
Doing the same in the Int version reduces its running time to 0.41 seconds. For Ints, the bit-shift for division by 2 is a bit faster than the quot operation, reducing its running time to 0.39 seconds.
Eliminating the construction of the list (that doesn't appear in the C version either),
module Main (main) where
import Data.Bits
result :: Int
result = findMax 0 0 1
findMax :: Int -> Int -> Int -> Int
findMax start len can
| can > 1000000 = start
| canlen > len = findMax can canlen (can+1)
| otherwise = findMax start len (can+1)
where
canlen = findLen 1 can
findLen :: Int -> Int -> Int
findLen l 1 = l
findLen l n
| n .&. 1 == 0 = findLen (l+1) (n `shiftR` 1)
| otherwise = findLen (l+1) (3*n+1)
main :: IO ()
main = print result
yields a further small speedup, resulting in a running time of 0.37 seconds.
So the Haskell version that's in close correspondence to the C version doesn't take that much longer, it's a factor of ~1.3.
Well, let's be fair, there's an inefficiency in the C version that's not present in the Haskell versions,
if (this_terms > terms)
{
terms = this_terms;
longest = i;
}
appearing in the inner loop. Lifting that out of the inner loop in the C version reduces its running time to 0.27 seconds, making the factor ~1.4.
The comparing may be recomputing sequenceLength too much. This is my best version:
type I = Integer
data P = P {-# UNPACK #-} !Int {-# UNPACK #-} !I deriving (Eq,Ord,Show)
searchTo = 1000000
nextNumber :: I -> I
nextNumber n = case quotRem n 2 of
(n2,0) -> n2
_ -> 3*n+1
sequenceLength :: I -> Int
sequenceLength x = count x 1 where
count 1 acc = acc
count n acc = count (nextNumber n) (succ acc)
longestSequence = maximum . map (\i -> P (sequenceLength i) i) $ [1..searchTo]
main = putStrLn $ show $ longestSequence
The answer and timing are slower than C, but it does use arbitrary precision integers (through the Integer type):
ghc -O2 --make euler14-fgij.hs
time ./euler14-fgij
P 525 837799
real 0m3.235s
user 0m3.184s
sys 0m0.015s
Haskell's lists are heap-based, whereas your C code is exceedingly tight and makes no heap use at all. You need to refactor to remove the dependency on lists.
Even if I'm a bit late, here is mine, I removed the dependency on lists and this solution uses no heap at all too.
{-# LANGUAGE BangPatterns #-}
-- Compiled with ghc -O2 -fvia-C -optc-O3 -Wall euler.hs
module Main (main) where
searchTo :: Int
searchTo = 1000000
nextNumber :: Int -> Int
nextNumber n = case n `divMod` 2 of
(k,0) -> k
_ -> 3*n + 1
sequenceLength :: Int -> Int
sequenceLength n = sl 1 n where
sl k 1 = k
sl k x = sl (k + 1) (nextNumber x)
longestSequence :: Int
longestSequence = testValues 1 0 0 where
testValues number !longest !longestNum
| number > searchTo = longestNum
| otherwise = testValues (number + 1) longest' longestNum' where
nlength = sequenceLength number
(longest',longestNum') = if nlength > longest
then (nlength,number)
else (longest,longestNum)
main :: IO ()
main = print longestSequence
I compiled this piece with ghc -O2 -fvia-C -optc-O3 -Wall euler.hs and it runs in 5 secs, compared to 80 of the beginning implementation. It doesn't uses Integer, but because I'm on a 64-bit machine, the results may be cheated.
The compiler can unbox all Ints in this case, resulting in really fast code. It runs faster than all other solutions I've seen so far, but C is still faster.

Optimize me! (C, performance) -- followup to bit-twiddling question

Thanks to some very helpful stackOverflow users at Bit twiddling: which bit is set?, I have constructed my function (posted at the end of the question).
Any suggestions -- even small suggestions -- would be appreciated. Hopefully it will make my code better, but at the least it should teach me something. :)
Overview
This function will be called at least 1013 times, and possibly as often as 1015. That is, this code will run for months in all likelihood, so any performance tips would be helpful.
This function accounts for 72-77% of the program's time, based on profiling and about a dozen runs in different configurations (optimizing certain parameters not relevant here).
At the moment the function runs in an average of 50 clocks. I'm not sure how much this can be improved, but I'd be thrilled to see it run in 30.
Key Observation
If at some point in the calculation you can tell that the value that will be returned will be small (exact value negotiable -- say, below a million) you can abort early. I'm only interested in large values.
This is how I hope to save the most time, rather than by further micro-optimizations (though these are of course welcome as well!).
Performance Information
smallprimes is a bit array (64 bits); on average about 8 bits will be set, but it could be as few as 0 or as many as 12.
q will usually be nonzero. (Notice that the function exits early if q and smallprimes are zero.)
r and s will often be 0. If q is zero, r and s will be too; if r is zero, s will be too.
As the comment at the end says, nu is usually 1 by the end, so I have an efficient special case for it.
The calculations below the special case may appear to risk overflow, but through appropriate modeling I have proved that, for my input, this will not occur -- so don't worry about that case.
Functions not defined here (ugcd, minuu, star, etc.) have already been optimized; none take long to run. pr is a small array (all in L1). Also, all functions called here are pure functions.
But if you really care... ugcd is the gcd, minuu is the minimum, vals is the number of trailing binary 0s, __builtin_ffs is the location of the leftmost binary 1, star is (n-1) >> vals(n-1), pr is an array of the primes from 2 to 313.
The calculations are currently being done on a Phenom II 920 x4, though optimizations for i7 or Woodcrest are still of interest (if I get compute time on other nodes).
I would be happy to answer any questions you have about the function or its constituents.
What it actually does
Added in response to a request. You don't need to read this part.
The input is an odd number n with 1 < n < 4282250400097. The other inputs provide the factorization of the number in this particular sense:
smallprimes&1 is set if the number is divisible by 3, smallprimes&2 is set if the number is divisible by 5, smallprimes&4 is set if the number is divisible by 7, smallprimes&8 is set if the number is divisible by 11, etc. up to the most significant bit which represents 313. A number divisible by the square of a prime is not represented differently from a number divisible by just that number. (In fact, multiples of squares can be discarded; in the preprocessing stage in another function multiples of squares of primes <= lim have smallprimes and q set to 0 so they will be dropped, where the optimal value of lim is determined by experimentation.)
q, r, and s represent larger factors of the number. Any remaining factor (which may be greater than the square root of the number, or if s is nonzero may even be less) can be found by dividing factors out from n.
Once all the factors are recovered in this way, the number of bases, 1 <= b < n, to which n is a strong pseudoprime are counted using a mathematical formula best explained by the code.
Improvements so far
Pushed the early exit test up. This clearly saves work so I made the change.
The appropriate functions are already inline, so __attribute__ ((inline)) does nothing. Oddly, marking the main function bases and some of the helpers with __attribute ((hot)) hurt performance by almost 2% and I can't figure out why (but it's reproducible with over 20 tests). So I didn't make that change. Likewise, __attribute__ ((const)), at best, did not help. I was more than slightly surprised by this.
Code
ulong bases(ulong smallprimes, ulong n, ulong q, ulong r, ulong s)
{
if (!smallprimes & !q)
return 0;
ulong f = __builtin_popcountll(smallprimes) + (q > 1) + (r > 1) + (s > 1);
ulong nu = 0xFFFF; // "Infinity" for the purpose of minimum
ulong nn = star(n);
ulong prod = 1;
while (smallprimes) {
ulong bit = smallprimes & (-smallprimes);
ulong p = pr[__builtin_ffsll(bit)];
nu = minuu(nu, vals(p - 1));
prod *= ugcd(nn, star(p));
n /= p;
while (n % p == 0)
n /= p;
smallprimes ^= bit;
}
if (q) {
nu = minuu(nu, vals(q - 1));
prod *= ugcd(nn, star(q));
n /= q;
while (n % q == 0)
n /= q;
} else {
goto BASES_END;
}
if (r) {
nu = minuu(nu, vals(r - 1));
prod *= ugcd(nn, star(r));
n /= r;
while (n % r == 0)
n /= r;
} else {
goto BASES_END;
}
if (s) {
nu = minuu(nu, vals(s - 1));
prod *= ugcd(nn, star(s));
n /= s;
while (n % s == 0)
n /= s;
}
BASES_END:
if (n > 1) {
nu = minuu(nu, vals(n - 1));
prod *= ugcd(nn, star(n));
f++;
}
// This happens ~88% of the time in my tests, so special-case it.
if (nu == 1)
return prod << 1;
ulong tmp = f * nu;
long fac = 1 << tmp;
fac = (fac - 1) / ((1 << f) - 1) + 1;
return fac * prod;
}
You seem to be wasting much time doing divisions by the factors. It is much faster to replace a division with a multiplication by the reciprocal of divisor (division: ~15-80(!) cycles, depending on the divisor, multiplication: ~4 cycles), IF of course you can precompute the reciprocals.
While this seems unlikely to be possible with q, r, s - due to the range of those vars, it is very easy to do with p, which always comes from the small, static pr[] array. Precompute the reciprocals of those primes and store them in another array. Then, instead of dividing by p, multiply by the reciprocal taken from the second array. (Or make a single array of structs.)
Now, obtaining exact division result by this method requires some trickery to compensate for rounding errors. You will find the gory details of this technique in this document, on page 138.
EDIT:
After consulting Hacker's Delight (an excellent book, BTW) on the subject, it seems that you can make it even faster by exploiting the fact that all divisions in your code are exact (i.e. remainder is zero).
It seems that for every divisor d which is odd and base B = 2word_size, there exists a unique multiplicative inverse d⃰ which satisfies the conditions: d⃰ < B and d·d⃰ ≡ 1 (mod B). For every x which is an exact multiple of d, this implies x/d ≡ x·d⃰ (mod B). Which means you can simply replace a division with a multiplication, no added corrections, checks, rounding problems, whatever. (The proofs of these theorems can be found in the book.) Note that this multiplicative inverse need not be equal to the reciprocal as defined by the previous method!
How to check whether a given x is an exact multiple of d - i.e. x mod d = 0 ? Easy! x mod d = 0 iff x·d⃰ mod B ≤ ⌊(B-1)/d⌋. Note that this upper limit can be precomputed.
So, in code:
unsigned x, d;
unsigned inv_d = mulinv(d); //precompute this!
unsigned limit = (unsigned)-1 / d; //precompute this!
unsigned q = x*inv_d;
if(q <= limit)
{
//x % d == 0
//q == x/d
} else {
//x % d != 0
//q is garbage
}
Assuming the pr[] array becomes an array of struct prime:
struct prime {
ulong p;
ulong inv_p; //equal to mulinv(p)
ulong limit; //equal to (ulong)-1 / p
}
the while(smallprimes) loop in your code becomes:
while (smallprimes) {
ulong bit = smallprimes & (-smallprimes);
int bit_ix = __builtin_ffsll(bit);
ulong p = pr[bit_ix].p;
ulong inv_p = pr[bit_ix].inv_p;
ulong limit = pr[bit_ix].limit;
nu = minuu(nu, vals(p - 1));
prod *= ugcd(nn, star(p));
n *= inv_p;
for(;;) {
ulong q = n * inv_p;
if (q > limit)
break;
n = q;
}
smallprimes ^= bit;
}
And for the mulinv() function:
ulong mulinv(ulong d) //d needs to be odd
{
ulong x = d;
for(;;)
{
ulong tmp = d * x;
if(tmp == 1)
return x;
x *= 2 - tmp;
}
}
Note you can replace ulong with any other unsigned type - just use the same type consistently.
The proofs, whys and hows are all available in the book. A heartily recommended read :-).
If your compiler supports GCC function attributes, you can mark your pure functions with this attribute:
ulong star(ulong n) __attribute__ ((const));
This attribute indicates to the compiler that the result of the function depends only on its argument(s). This information can be used by the optimiser.
Is there a reason why you've opencoded vals() instead of using __builtin_ctz() ?
It is still somewhat unclear, what you are searching for. Quite frequently number theoretic problems allow huge speedups by deriving mathematical properties that the solutions must satisfiy.
If you are indeed searching for the integers that maximize the number of non-witnesses for the MR test (i.e. oeis.org/classic/A141768 that you mention) then it might be possible to use that the number of non-witnesses cannot be larger than phi(n)/4 and that the integers for which have this many non-witnesses are either are the product of two primes of the form
(k+1)*(2k+1)
or they are Carmichael numbers with 3 prime factors.
I'd think above some limit all integers in the sequence have this form and that it is possible to verify this by proving an upper bound for the witnesses of all other integers.
E.g. integers with 4 or more factors always have at most phi(n)/8 non-witnesses. Similar results can be derived from you formula for the number of bases for other integers.
As for micro-optimizations: Whenever you know that an integer is divisible by some quotient, then it is possible to replace the division by a multiplication with the inverse of the quotient modulo 2^64. And the tests n % q == 0 can be replaced by a test
n * inverse_q < max_q,
where inverse_q = q^(-1) mod 2^64 and max_q = 2^64 / q.
Obviously inverse_q and max_q need to be precomputed, to be efficient, but since you are using a sieve, I assume this should not be an obstacle.
Small optimization but:
ulong f;
ulong nn;
ulong nu = 0xFFFF; // "Infinity" for the purpose of minimum
ulong prod = 1;
if (!smallprimes & !q)
return 0;
// no need to do this operations before because of the previous return
f = __builtin_popcountll(smallprimes) + (q > 1) + (r > 1) + (s > 1);
nn = star(n);
BTW: you should edit your post to add star() and other functions you use definition
Try replacing this pattern (for r and q too):
n /= p;
while (n % p == 0)
n /= p;
With this:
ulong m;
...
m = n / p;
do {
n = m;
m = n / p;
} while ( m * p == n);
In my limited tests, I got a small speedup (10%) from eliminating the modulo.
Also, if p, q or r were constant, the compiler will replace the divisions by multiplications. If there are few choices for p, q or r, or if certain ones are more frequent, you might gain something by specializing the function for those values.
Have you tried using profile-guided optimisation?
Compile and link the program with the -fprofile-generate option, then run the program over a representative data set (say, a day's worth of computation).
Then re-compile and link it with the -fprofile-use option instead.
1) I would make the compiler spit out the assembly it generates and try and deduce if what it does is the best it can do... and if you spot problems, change the code so the assembly looks better. This way you can also make sure that functions you hope it'll inline (like star and vals) are really inlined. (You might need to add pragma's, or even turn them into macros)
2) It's great that you try this on a multicore machine, but this loop is singlethreaded. I'm guessing that there is an umbrella functions which splits the load across a few threads so that more cores are used?
3) It's difficult to suggest speed ups if what the actual function tries to calculate is unclear. Typically the most impressive speedups are not achieved with bit twiddling, but with a change in the algorithm. So a bit of comments might help ;^)
4) If you really want a speed up of 10* or more, check out CUDA or openCL which allows you to run C programs on your graphics hardware. It shines with functions like these!
5) You are doing loads of modulo and divides right after each other. In C this is 2 separate commands (first '/' and then '%'). However in assembly this is 1 command: 'DIV' or 'IDIV' which returns both the remainder and the quotient in one go:
B.4.75 IDIV: Signed Integer Divide
IDIV r/m8 ; F6 /7 [8086]
IDIV r/m16 ; o16 F7 /7 [8086]
IDIV r/m32 ; o32 F7 /7 [386]
IDIV performs signed integer division. The explicit operand provided is the divisor; the dividend and destination operands are implicit, in the following way:
For IDIV r/m8, AX is divided by the given operand; the quotient is stored in AL and the remainder in AH.
For IDIV r/m16, DX:AX is divided by the given operand; the quotient is stored in AX and the remainder in DX.
For IDIV r/m32, EDX:EAX is divided by the given operand; the quotient is stored in EAX and the remainder in EDX.
So it will require some inline assembly, but I'm guessing there'll be a significant speedup as there are a few places in your code which can benefit from this.
Make sure your functions get inlined. If they're out-of-line, the overhead might add up, especially in the first while loop. The best way to be sure is to examine the assembly.
Have you tried pre-computing star( pr[__builtin_ffsll(bit)] ) and vals( pr[__builtin_ffsll(bit)] - 1) ? That would trade some simple work for an array lookup, but it might be worth it if the tables are small enough.
Don't compute f until you actually need it (near the end, after your early-out). You can replace the code around BASES_END with something like
BASES_END:
ulong addToF = 0;
if (n > 1) {
nu = minuu(nu, vals(n - 1));
prod *= ugcd(nn, star(n));
addToF = 1;
}
// ... early out if nu == 1...
// ... compute f ...
f += addToF;
Hope that helps.
First some nitpicking ;-) you should be more careful about the types that you are using. In some places you seem to assume that ulong is 64 bit wide, use uint64_t there. And also for all other types, rethink carefully what you expect of them and use the appropriate type.
The optimization that I could see is integer division. Your code does that a lot, this is probably the most expensive thing you are doing. Division of small integers (uint32_t) maybe much more efficient than by big ones. In particular for uint32_t there is an assembler instruction that does division and modulo in one go, called divl.
If you use the appropriate types your compiler might do that all for you. But you'd better check the assembler (option -S to gcc) as somebody already said. Otherwise it is easy to include some little assembler fragments here and there. I found something like that in some code of mine:
register uint32_t a asm("eax") = 0;
register uint32_t ret asm("edx") = 0;
asm("divl %4"
: "=a" (a), "=d" (ret)
: "0" (a), "1" (ret), "rm" (divisor));
As you can see this uses special registers eax and edx and stuff like that...
Did you try a table lookup version of the first while loop? You could divide smallprimes in 4 16 bit values, look up their contribution and merge them. But maybe you need the side effects.
Did you try passing in an array of primes instead of splitting them in smallprimes, q, r and s? Since I don't know what the outer code does, I am probably wrong, but there is a chance that you also have a function to convert some primes to a smallprimes bitmap, and inside this function, you convert the bitmap back to an array of primes, effecively. In addition, you seem to do identical processing for elements of smallprimes, q, r, and s. It should save you a tiny amount of processing per call.
Also, you seem to know that the passed in primes divide n. Do you have enough knowledge outside about the power of each prime that divides n? You could save a lot of time if you can eliminate the modulo operation by passing in that information to this function. In other words, if n is pow(p_0,e_0)*pow(p_1,e_1)*...*pow(p_k,e_k)*n_leftover, and if you know more about these e_is and n_leftover, passing them in would mean a lot of things you don't have to do in this function.
There may be a way to discover n_leftover (the unfactored part of n) with less number of modulo operations, but it is only a hunch, so you may need to experiment with it a bit. The idea is to use gcd to remove known factors from n repeatedly until you get rid of all known prime factors. Let me give some almost-c-code:
factors=p_0*p_1*...*p_k*q*r*s;
n_leftover=n/factors;
do {
factors=gcd(n_leftover, factors);
n_leftover = n_leftover/factors;
} while (factors != 1);
I am not at all certain this will be better than the code you have, let alone the combined mod/div suggestions you can find in other answers, but I think it is worth a try. I feel that it will be a win, especially for numbers with high numbers of small prime factors.
You're passing in the complete factorization of n, so you're factoring consecutive integers and then using the results of that factorization here. It seems to me that you might benefit from doing some of this at the time of finding the factors.
BTW, I've got some really fast code for finding the factors you're using without doing any division. It's a little like a sieve but produces factors of consecutive numbers very quickly. Can find it and post if you think it may help.
edit had to recreate the code here:
#include
#define SIZE (1024*1024) //must be 2^n
#define MASK (SIZE-1)
typedef struct {
int p;
int next;
} p_type;
p_type primes[SIZE];
int sieve[SIZE];
void init_sieve()
{
int i,n;
int count = 1;
primes[1].p = 3;
sieve[1] = 1;
for (n=5;SIZE>n;n+=2)
{
int flag = 0;
for (i=1;count>=i;i++)
{
if ((n%primes[i].p) == 0)
{
flag = 1;
break;
}
}
if (flag==0)
{
count++;
primes[count].p = n;
sieve[n>>1] = count;
}
}
}
int main()
{
int ptr,n;
init_sieve();
printf("init_done\n");
// factor odd numbers starting with 3
for (n=1;1000000000>n;n++)
{
ptr = sieve[n&MASK];
if (ptr == 0) //prime
{
// printf("%d is prime",n*2+1);
}
else //composite
{
// printf ("%d has divisors:",n*2+1);
while(ptr!=0)
{
// printf ("%d ",primes[ptr].p);
sieve[n&MASK]=primes[ptr].next;
//move the prime to the next number it divides
primes[ptr].next = sieve[(n+primes[ptr].p)&MASK];
sieve[(n+primes[ptr].p)&MASK] = ptr;
ptr = sieve[n&MASK];
}
}
// printf("\n");
}
return 0;
}
The init function creates a factor base and initializes the sieve. This takes about 13 seconds on my laptop. Then all numbers up to 1 billion are factored or determined to be prime in another 25 seconds. Numbers less than SIZE are never reported as prime because they have 1 factor in the factor base, but that could be changed.
The idea is to maintain a linked list for every entry in the sieve. Numbers are factored by simply pulling their factors out of the linked list. As they are pulled out, they are inserted into the list for the next number that will be divisible by that prime. This is very cache friendly too. The sieve size must be larger than the largest prime in the factor base. As is, this sieve could run up to 2**40 in about 7 hours which seems to be your target (except for n needing to be 64 bits).
Your algorithm could be merged into this to make use of the factors as they are identified rather than packing bits and large primes into variables to pass to your function. Or your function could be changed to take the linked list (you could create a dummy link to pass in for the prime numbers outside the factor base).
Hope it helps.
BTW, this is the first time I've posted this algorithm publicly.
just a thought but maybe using your compilers optimization options would help, if you haven't already. another thought would be that if money isn't an issue you could use the Intel C/C++ compiler, assuming your using an Intel processor. I'd also assume that other processor manufacturers (AMD, etc.) would have similar compilers
If you are going to exit immediately on (!smallprimes&!q) why not do that test before even calling the function, and save the function call overhead?
Also, it seems like you effectively have 3 different functions which are linear except for the smallprimes loop.
bases1(s,n,q), bases2(s,n,q,r), and bases3(s,n,q,r,s).
It might be a win to actually create those as 3 separate functions without the branches and gotos, and call the appropriate one:
if (!(smallprimes|q)) { r = 0;}
else if (s) { r = bases3(s,n,q,r,s);}
else if (r) { r = bases2(s,n,q,r); }
else { r = bases1(s,n,q);
This would be most effective if previous processing has already given the calling code some 'knowledge' of which function to execute and you don't have to test for it.
If the divisions you're using are with numbers that aren’t known at compile time, but are used frequently at runtime (dividing by the same number many times), then I would suggest using the libdivide library, which basically implements at runtime the optimisations that compilers do for compile time constants (using shifts masks etc.). This can provide a huge benefit. Also avoiding using x % y == 0 for something like z = x/y, z * y == x as ergosys suggested above should also have a measurable improvement.
Does the code on your top post is the optimized version? If yes, there is still too many divide operations which greatly eat CPU cycles.
This code is overexecute innecessarily a bit
if (!smallprimes & !q)
return 0;
change to logical and &&
if (!smallprimes && !q)
return 0;
will make it short circuited faster without eveluating q
And the following code
ulong bit = smallprimes & (-smallprimes);
ulong p = pr[__builtin_ffsll(bit)];
which is used to find the last set bit of smallprimes. Why don't you use the simpler way
ulong p = pr[__builtin_ctz(smallprimes)];
Another culprit for decreased performance maybe too many program branching. You may consider changing to some other less-branch or branch-less equivalents

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