I am trying to improve the speed of my code, written in C++. Based on profilers, the function cbrt()/cbrtf32x is the function I spend the most time in/on (or more specifically):
double test_func(const double &test_val){
double cbrt_test_val = cbrt(test_val);
return (1 - 1e-10*cbrt_test_val);
}
According to data, I spend more then three times the time for cbrt()/cbrtf32x() than for the closest cost-expensive function. Thus I was wondering how to improve this function, and how to speed it up? The input values range from 1e18 to 1e30.
There is little that can be done if you are doing the cubic roots one at a time, and you want the exact result.
As is, I would be surprised if you can improve the cubic root calculation more than 10-20% - if that - while getting the same result numerically. (Note: I got that 10%-20% number out of thin air; it's an opinion, not a scientific number at all.)
If you can batch up the calculations, you might be able to SIMD the operation, or multi-thread them, or if you know more about the distribution of the data (or can find out more,) you might be able to sort them and - I don't know - maybe calculate an incremental cubic root or something.
If you can get away with an approximation, then there are more things that you can do. For example, you are calculating the function f(x) = 1 - cbrt(x) / 1e10, which is the same as 1 - cbrt(x / 1e30) which is a strictly decreasing function that maps the domain [1e18..1e30] to the range [0..0.9999]. With y = x / 1e30 it becomes f(y) = 1 - cbrt(y) and now y is in the range [1e-12..1] and it can be pre-calculated and approximated using a look-up table.
Depending on the number of times you need a cubic root, how much accuracy loss you can get away with (which determines the size of the table,) and whether you can sort or bucket your input (to improve the CPU cache utilization for your LUT look-ups) you might get a nice speed boost out of this.
Short version of the question: overflow or timeout in current settings when calculating large int64_t and double, anyway to avoid these?
Test case:
If only demand is 80,000,000,000, solved with correct result. But if it's 800,000,000,000, returned incorrect 0.
If input has two or more demands (means more inequalities need to be calculated), smaller value will also cause incorrectness. e.g., three equal demands of 20,000,000,000 will cause the problem.
I'm using COIN-OR CLP linear programming solver to solve some network flow problems. I use int64_t when representing the link bandwidth. But CLP uses double most of time and cannot transfer to other types easily.
When the values of the variables are not that large (typically smaller than 10,000,000,000) and the constraints (inequalities) are relatively few, it will give the solution I want it to. But if either of the above factors increases, the tool will stop and return a 0 value solution. I think the reason is the calculation complexity is over its maximum, so program breaks at some trivial point (it uses LP simplex method).
The inequality is some kind of:
totalFlowSum <= usePercentage * demand
I changed it to
totalFlowSum - usePercentage * demand <= 0
Since totalFLowSum and demand are very large int64_t, usePercentage is double, if the constraints like this are too many (several or even more), or if the demand is larger than 100,000,000,000, the returned solution will be wrong.
Is there any way to correct this, like increase the break threshold or avoid this level of calculation magnitude?
Decrease some accuracy is acceptable. I have a possible solution is that 1,000 times smaller on inputs and 1,000 time larger on outputs. But this is kind of naïve and may cause too much code modification in the program.
Update:
I have changed the formulation to
totalFlowSum / demand - usePercentage <= 0
but the problem still exists.
Update 2:
I divided usePercentage by 1000, making its coefficient from 1 to 0.001, it worked. But if I also divide totalFlowSum/demand by 1000 simultaneously, still no result. I don't know why...
I changed the rhs of equalities from 0 to 0.1, the problem is then solved! Since the inputs are very large, 0.1 offset won't impact the solution at all.
I think the reason is that previous coeffs are badly scaled, so the complier failed to find an exact answer.
Here's the problem:
I am currently trying to create a control system which is required to find a solution to a series of complex linear equations without a unique solution.
My problem arises because there will ever only be six equations, while there may be upwards of 20 unknowns (usually way more than six unknowns). Of course, this will not yield an exact solution through the standard Gaussian elimination or by changing them in a matrix to reduced row echelon form.
However, I think that I may be able to optimize things further and get a more accurate solution because I know that each of the unknowns cannot have a value smaller than zero or greater than one, but it is free to take on any value in between them.
Of course, I am trying to create code that would find a correct solution, but in the case that there are multiple combinations that yield satisfactory results, I would want to minimize Sum of (value of unknown * efficiency constant) over all unknowns, i.e. Sigma[xI*eI] from I=0 to n, but finding an accurate solution is of a greater priority.
Performance is also important, due to the fact that this algorithm may need to be run several times per second.
So, does anyone have any ideas to help me on implementing this?
Edit: You might just want to stick to linear programming with equality and inequality constraints, but here's an interesting exact solution that does not incorporate the constraint that your unknowns are between 0 and 1.
Here's a powerpoint discussing your problem: http://see.stanford.edu/materials/lsoeldsee263/08-min-norm.pdf
I'll translate your problem into math to make things a bit easier to figure out:
you have a 6x20 matrix A and a vector x with 20 elements. You want to minimize (x^T)e subject to Ax=y. According to the slides, if you were just minimizing the sum of x, then the answer is A^T(AA^T)^(-1)y. I'll take another look at this as soon as I get the chance and see what the solution is to minimizing (x^T)e (ie your specific problem).
Edit: I looked in the powerpoint some more and near the end there's a slide entitled "General norm minimization with equality constraints". I am going to switch the notation to match the slide's:
Your problem is that you want to minimize ||Ax-b||, where b = 0 and A is your e vector and x is the 20 unknowns. This is subject to Cx=d. Apparently the answer is:
x=(A^T A)^-1 (A^T b -C^T(C(A^T A)^-1 C^T)^-1 (C(A^T A)^-1 A^Tb - d))
it's not pretty, but it's not as bad as you might think. There's really aren't that many calculations. For example (A^TA)^-1 only needs to be calculated once and then you can reuse the answer. And your matrices aren't that big.
Note that I didn't incorporate the constraint that the elements of x are within [0,1].
It looks like the solution for what I am doing is with Linear Programming. It is starting to come back to me, but if I have other problems I will post them in their own dedicated questions instead of turning this into an encyclopedia.
I hear this statement quite often, that multiplication on modern hardware is so optimized that it actually is at the same speed as addition. Is that true?
I never can get any authoritative confirmation. My own research only adds questions. The speed tests usually show data that confuses me. Here is an example:
#include <stdio.h>
#include <sys/time.h>
unsigned int time1000() {
timeval val;
gettimeofday(&val, 0);
val.tv_sec &= 0xffff;
return val.tv_sec * 1000 + val.tv_usec / 1000;
}
int main() {
unsigned int sum = 1, T = time1000();
for (int i = 1; i < 100000000; i++) {
sum += i + (i+1); sum++;
}
printf("%u %u\n", time1000() - T, sum);
sum = 1;
T = time1000();
for (int i = 1; i < 100000000; i++) {
sum += i * (i+1); sum++;
}
printf("%u %u\n", time1000() - T, sum);
}
The code above can show that multiplication is faster:
clang++ benchmark.cpp -o benchmark
./benchmark
746 1974919423
708 3830355456
But with other compilers, other compiler arguments, differently written inner loops, the results can vary and I cannot even get an approximation.
Multiplication of two n-bit numbers can in fact be done in O(log n) circuit depth, just like addition.
Addition in O(log n) is done by splitting the number in half and (recursively) adding the two parts in parallel, where the upper half is solved for both the "0-carry" and "1-carry" case. Once the lower half is added, the carry is examined, and its value is used to choose between the 0-carry and 1-carry case.
Multiplication in O(log n) depth is also done through parallelization, where every sum of 3 numbers is reduced to a sum of just 2 numbers in parallel, and the sums are done in some manner like the above.
I won't explain it here, but you can find reading material on fast addition and multiplication by looking up "carry-lookahead" and "carry-save" addition.
So from a theoretical standpoint, since circuits are obviously inherently parallel (unlike software), the only reason multiplication would be asymptotically slower is the constant factor in the front, not the asymptotic complexity.
Integer multiplication will be slower.
Agner Fog's instruction tables show that when using 32-bit integer registers, Haswell's ADD/SUB take 0.25–1 cycles (depending on how well pipelined your instructions are) while MUL takes 2–4 cycles. Floating-point is the other way around: ADDSS/SUBSS take 1–3 cycles while MULSS takes 0.5–5 cycles.
This is an even more complex answer than simply multiplication versus addition. In reality the answer will most likely NEVER be yes. Multiplication, electronically, is a much more complicated circuit. Most of the reasons why, is that multiplication is the act of a multiplication step followed by an addition step, remember what it was like to multiply decimal numbers prior to using a calculator.
The other thing to remember is that multiplication will take longer or shorter depending on the architecture of the processor you are running it on. This may or may not be simply company specific. While an AMD will most likely be different than an Intel, even an Intel i7 may be different from a core 2 (within the same generation), and certainly different between generations (especially the farther back you go).
In all TECHNICALITY, if multiplies were the only thing you were doing (without looping, counting etc...), multiplies would be 2 to (as ive seen on PPC architectures) 35 times slower. This is more an exercise in understanding your architecture, and electronics.
In Addition:
It should be noted that a processor COULD be built for which ALL operations including a multiply take a single clock. What this processor would have to do is, get rid of all pipelining, and slow the clock so that the HW latency of any OPs circuit is less than or equal to the latency PROVIDED by the clock timing.
To do this would get rid of the inherent performance gains we are able to get when adding pipelining into a processor. Pipelining is the idea of taking a task and breaking it down into smaller sub-tasks that can be performed much quicker. By storing and forwarding the results of each sub-task between sub-tasks, we can now run a faster clock rate that only needs to allow for the longest latency of the sub-tasks, and not from the overarching task as a whole.
Picture of time through a multiply:
|--------------------------------------------------| Non-Pipelined
|--Step 1--|--Step 2--|--Step 3--|--Step 4--|--Step 5--| Pipelined
In the above diagram, the non-pipelined circuit takes 50 units of time. In the pipelined version, we have split the 50 units into 5 steps each taking 10 units of time, with a store step in between. It is EXTREMELY important to note that in the pipelined example, each of the steps can be working completely on their own and in parallel. For an operation to be completed, it must move through all 5 steps in order but another of the same operation with operands can be in step 2 as one is in step 1, 3, 4, and 5.
With all of this being said, this pipelined approach allows us to continuously fill the operator each clock cycle, and get a result out on each clock cycle IF we are able to order our operations such that we can perform all of one operation before we switch to another operation, and all we take as a timing hit is the original amount of clocks necessary to get the FIRST operation out of the pipeline.
Mystical brings up another good point. It is also important to look at the architecture from a more systems perspective. It is true that the newer Haswell architectures was built to better the Floating Point multiply performance within the processor. For this reason as the System level, it was architected to allow multiple multiplies to occur in simultaneity versus an add which can only happen once per system clock.
All of this can be summed up as follows:
Each architecture is different from a lower level HW perspective as well as from a system perspective
FUNCTIONALLY, a multiply will always take more time than an add because it combines a true multiply along with a true addition step.
Understand the architecture you are trying to run your code on, and find the right balance between readability and getting truly the best performance from that architecture.
Intel since Haswell has
add performance of 4/clock throughput, 1 cycle latency. (Any operand-size)
imul performance of 1/clock throughput, 3 cycle latency. (Any operand-size)
Ryzen is similar. Bulldozer-family has much lower integer throughput and not-fully-pipelined multiply, including extra slow for 64-bit operand-size multiply. See https://agner.org/optimize/ and other links in https://stackoverflow.com/tags/x86/info
But a good compiler could auto-vectorize your loops. (SIMD-integer multiply throughput and latency are both worse than SIMD-integer add). Or simply constant-propagate through them to just print out the answer! Clang really does know the closed-form Gauss's formula for sum(i=0..n) and can recognize some loops that do that.
You forgot to enable optimization so both loops bottleneck on the ALU + store/reload latency of keeping sum in memory between each of sum += independent stuff and sum++. See Why does clang produce inefficient asm with -O0 (for this simple floating point sum)? for more about just how bad the resulting asm is, and why that's the case. clang++ defaults to -O0 (debug mode: keep variables in memory where a debugger can modify them between any C++ statements).
Store-forwarding latency on a modern x86 like Sandybridge-family (including Haswell and Skylake) is about 3 to 5 cycles, depending on timing of the reload. So with a 1-cycle latency ALU add in there, too, you're looking at about two 6-cycle latency steps in the critical path for this loop. (Plenty to hide all the store / reload and calculation based on i, and the loop-counter update).
See also Adding a redundant assignment speeds up code when compiled without optimization for another no-optimization benchmark. In that one, store-forwarding latency is actually reduced by having more independent work in the loop, delaying the reload attempt.
Modern x86 CPUs have 1/clock multiply throughput so even with optimization you wouldn't see a throughput bottleneck from it. Or on Bulldozer-family, not fully pipelined with 1 per 2-clock throughput.
More likely you'd bottleneck on the front-end work of getting all the work issued every cycle.
Although lea does allow very efficient copy-and-add, and doing i + i + 1 with a single instruction. Although really a good compiler would see that the loop only uses 2*i and optimize to increment by 2. i.e. a strength-reduction to do repeated addition by 2 instead of having to shift inside the loop.
And of course with optimization the extra sum++ can just fold into the sum += stuff where stuff already includes a constant. Not so with the multiply.
I came to this thread to get an idea of what the modern processors are doing in regard to integer math and the number of cycles required to do them. I worked on this problem of speeding up 32-bit integer multiplies and divides on the 65c816 processor in the 1990's. Using the method below, I was able to triple the speed of the standard math libraries available in the ORCA/M compilers at the time.
So the idea that multiplies are faster than adds is simply not the case (except rarely) but like people said it depends upon how the architecture is implemented. If there are enough steps being performed available between clock cycles, yes a multiply could effectively be the same speed as an add based on the clock, but there would be a lot of wasted time. In that case it would be nice to have an instruction that performs multiple (dependent) adds / subtracts given one instruction and multiple values. One can dream.
On the 65c816 processor, there were no multiply or divide instructions. Mult and Div were done with shifts and adds.
To perform a 16 bit add, you would do the following:
LDA $0000 - loaded a value into the Accumulator (5 cycles)
ADC $0002 - add with carry (5 cycles)
STA $0004 - store the value in the Accumulator back to memory (5 cycles)
15 cycles total for an add
If dealing with a call like from C, you would have additional overhead of dealing with pushing and pulling values off the stack. Creating routines that would do two multiples at once would save overhead for example.
The traditional way of doing the multiply is shifts and adds through the entire value of the one number. Each time the carry became a one as it is shifted left would mean you needed to add the value again. This required a test of each bit and a shift of the result.
I replaced that with a lookup table of 256 items so as the carry bits would not need to be checked. It was also possible to determine overflow before doing the multiply to not waste time. (On a modern processor this could be done in parallel but I don't know if they do this in the hardware). Given two 32 bit numbers and prescreened overflow, one of the multipliers is always 16 bits or less, thus one would only need to run through 8 bit multiplies once or twice to perform the entire 32 bit multiply. The result of this was multiplies that were 3 times as fast.
the speed of the 16 bit multiplies ranged from 12 cycles to about 37 cycles
multiply by 2 (0000 0010)
LDA $0000 - loaded a value into the Accumulator (5 cycles).
ASL - shift left (2 cycles).
STA $0004 - store the value in the Accumulator back to memory (5 cycles).
12 cycles plus call overhead.
multiply by (0101 1010)
LDA $0000 - loaded a value into the Accumulator (5 cycles)
ASL - shift left (2 cycles)
ASL - shift left (2 cycles)
ADC $0000 - add with carry for next bit (5 cycles)
ASL - shift left (2 cycles)
ADC $0000 - add with carry for next bit (5 cycles)
ASL - shift left (2 cycles)
ASL - shift left (2 cycles)
ADC $0000 - add with carry for next bit (5 cycles)
ASL - shift left (2 cycles)
STA $0004 - store the value in the Accumulator back to memory (5 cycles)
37 cycles plus call overhead
Since the databus of the AppleIIgs for which this was written was only 8 bits wide, to load 16 bit values required 5 cycles to load from memory, one extra for the pointer, and one extra cycle for the second byte.
LDA instruction (1 cycle because it is an 8 bit value)
$0000 (16 bit value requires two cycles to load)
memory location (requires two cycles to load because of an 8 bit data bus)
Modern processors would be able to do this faster because they have a 32 bit data bus at worst. In the processor logic itself the system of gates would have no additional delay at all compared to the data bus delay since the whole value would get loaded at once.
To do the complete 32 bit multiply, you would need to do the above twice and add the results together to get the final answer. The modern processors should be able to do the two in parallel and add the results for the answer. Combined with the overflow precheck done in parallel, it would minimize the time required to do the multiply.
Anyway it is readily apparent that multiplies require significantly more effort than an add. How many steps to process the operation between cpu clock cycles would determine how many cycles of the clock would be required. If the clock is slow enough, then the adds would appear to be the same speed as a multiply.
Regards,
Ken
A multiplication requires a final step of an addition of, at minimum, the same size of the number; so it will take longer than an addition. In decimal:
123
112
----
+246 ----
123 | matrix generation
123 ----
-----
13776 <---------------- Addition
Same applies in binary, with a more elaborate reduction of the matrix.
That said, reasons why they may take the same amount of time:
To simplify the pipelined architecture, all regular instructions can be designed to take the same amount of cycles (exceptions are memory moves for instance, that depend on how long it takes to talk to external memory).
Since the adder for the final step of the multiplier is just like the adder for an add instruction... why not use the same adder by skipping the matrix generation and reduction? If they use the same adder, then obviously they will take the same amount of time.
Of course, there are more complex architectures where this is not the case, and you might obtain completely different values. You also have architectures that take several instructions in parallel when they don't depend on each other, and then you are a bit at the mercy of your compiler... and of the operating system.
The only way to run this test rigorously you would have to run in assembly and without an operating system - otherwise there are too many variables.
Even if it were, that mostly tells us what restriction the clock puts on our hardware. We can't clock higher because heat(?), but the number of ADD instruction gates a signal could pass during a clock could be very many but a single ADD instruction would only utilize one of them. So while it may at some point take equally many clock cycles, not all of the propagation time for the signals is utilized.
If we could clock higher we could def. make ADD faster probably by several orders of magnitude.
This really depends on your machine. Of course, integer multiplication is quite complex compared to addition, but quite a few AMD CPU can execute a multiplication in a single cycle. That is just as fast as addition.
Other CPUs take three or four cycles to do a multiplication, which is a bit slower than addition. But it's nowhere near the performance penalty you had to suffer ten years ago (back then a 32-Bit multiplication could take thirty-something cycles on some CPUs).
So, yes, multiplication is in the same speed class nowadays, but no, it's still not exactly as fast as addition on all CPUs.
Even on ARM (known for its high efficiency and small, clean design), integer multiplications take 3-7 cycles and than integer additions take 1 cycle.
However, an add/shift trick is often used to multiply integers by constants faster than the multiply instruction can calculate the answer.
The reason this works well on ARM is that ARM has a "barrel shifter", which allows many instructions to shift or rotate one of their arguments by 1-31 bits at zero cost, i.e. x = a + b and x = a + (b << s) take exactly the same amount of time.
Utilizing this processor feature, let's say you want to calculate a * 15. Then since 15 = 1111 (base 2), the following pseudocode (translated into ARM assembly) would implement the multiplication:
a_times_3 = a + (a << 1) // a * (0011 (base 2))
a_times_15 = a_times_3 + (a_times_3 << 2) // a * (0011 (base 2) + 1100 (base 2))
Similarly you could multiply by 13 = 1101 (base 2) using either of the following:
a_times_5 = a + (a << 2)
a_times_13 = a_times_5 + (a << 3)
a_times_3 = a + (a << 1)
a_times_15 = a_times_3 + (a_times_3 << 2)
a_times_13 = a_times_15 - (a << 1)
The first snippet is obviously faster in this case, but sometimes subtraction helps when translating a constant multiplication into add/shift combinations.
This multiplication trick was used heavily in the ARM assembly coding community in the late 80s, on the Acorn Archimedes and Acorn RISC PC (the origin of the ARM processor). Back then, a lot of ARM assembly was written by hand, since squeezing every last cycle out of the processor was important. Coders in the ARM demoscene developed many techniques like this for speeding up code, most of which are probably lost to history now that almost no assembly code is written by hand anymore. Compilers probably incorporate many tricks like this, but I'm sure there are many more that never made the transition from "black art optimization" to compiler implementation.
You can of course write explicit add/shift multiplication code like this in any compiled language, and the code may or may not run faster than a straight multiplication once compiled.
x86_64 may also benefit from this multiplication trick for small constants, although I don't believe shifting is zero-cost on the x86_64 ISA, in either the Intel or AMD implementations (x86_64 probably takes one extra cycle for each integer shift or rotate).
There are lots of good answers here about your main question, but I just wanted to point out that your code is not a good way to measure operation performance.
For starters, modern cpus adjust freqyuencies all the time, so you should use rdtsc to count the actual number of cycles instead of elapsed microseconds.
But more importantly, your code has artificial dependency chains, unnecessary control logic and iterators that will make your measure into an odd mix of latency and throughtput plus some constant terms added for no reason.
To really measure throughtput you should significantly unroll the loop and also add several partial sums in parallel (more sums than steps in the add/mul cpu pipelines).
No it's not, and in fact it's noticeably slower (which translated into a 15% performance hit for the particular real-world program I was running).
I realized this myself when asking this question from just a few days ago here.
Since the other answers deal with real, present-day devices -- which are bound to change and improve as time passes -- I thought we could look at the question from the theoretical side.
Proposition: When implemented in logic gates, using the usual algorithms, an integer multiplication circuit is O(log N) times slower than an addition circuit, where N is the number of bits in a word.
Proof: The time for a combinatorial circuit to stabilise is proportional to the depth of the longest sequence of logic gates from any input to any output. So we must show that a gradeschool multiply circuit is O(log N) times deeper than an addition circuit.
Addition is normally implemented as a half adder followed by N-1 full adders, with the carry bits chained from one adder to the next. This circuit clearly has depth O(N). (This circuit can be optimized in many ways, but the worst case performance will always be O(N) unless absurdly large lookup tables are used.)
To multiply A by B, we first need to multiply each bit of A with each bit of B. Each bitwise multiply is simply an AND gate. There are N^2 bitwise multiplications to perform, hence N^2 AND gates -- but all of them can execute in parallel, for a circuit depth of 1. This solves the multiplication phase of the gradeschool algorithm, leaving just the addition phase.
In the addition phase, we can combine the partial products using an inverted binary tree-shaped circuit to do many of the additions in parallel. The tree will be (log N) nodes deep, and at each node, we will be adding together two numbers with O(N) bits. This means each node can be implemented with an adder of depth O(N), giving a total circuit depth of O(N log N). QED.
Code pasted here
Hello SO. I just wrote my first semi-significant PC program, written purely for fun/to solve a problem, having not been assigned the problem in a programming class. I'm sure many of you remember the first significant program that you wrote for fun.
My issue is, I am not satisfied with the efficiency of my code. I'm not sure if it's the I/O limitation of my terminal or my code itself, but it seems to run quite slow for DAC resolutions of 8 bits or higher.
I haven't commented the code, so here is an explanation of the problem that I was attempting to solve with this program:
The output voltage of a DAC is determined by a binary number having bits Bn, Bn-1 ... B0, and a full-scale voltage.
The output voltage has an equation of the form:
Vo = G( (1/(2^(0)))*(Bn) + (1/2^(0+1))*(Bn-1) + ... + (1/2^(0+n))*(B0) )
Where G is the gain that would make an input B of all bits high the full-scale voltage.
If you run the code the idea will be quite clear.
My issue is, I think that what I'm outputting to the console can be achieved in much less than 108 lines of C++. Yes, it can easily be done by precomputing the step voltage and simply rendering the table by incrementation, but a "self-requirement" that I have for this program is that on some level it performs the series calculation described above for each binary represented input.
I'm not trying to be crazy with that requirement. I'd like this program to prove the nature of the formula, which it currently does. What I'm looking for is some suggestions on how to make my implementation generally cleaner and more efficient.
pow(2.0, x) is almost always a bad idea - especially if you're iterating over x. pow(2.0,0) == 1, and pow(2.0,x) == 2 * pow(2.0,x-1)
DtoAeqn returns an unbalanced string (one more )), that hurts my brain.
You could use Horner's method to evaluate the formula efficiently. Here's an example I use to demonstrate it for converting binary strings to decimal:
0.1101 = (1*2-1 + 1*2-2 + 0*2-3 + 1*2-4).
To evaluate this expression efficiently, rewrite the terms from right to left, then nest and evaluate as follows:
(((1 * 2-1 + 0) * 2-1 + 1) * 2-1 + 1) * 2-1 = 0.8125.
This way, you've removed the "pow" function, and you only have to do one multiplication (division) for each bit.
By the way, I see your code allows up to 128 bits of precision. You won't be able to compute that accurately in a double.