I want to know how I can log the values other than states during integration by odeint. I have a simulation of the satellite dynamics, which is described as differential equations of total angular momentum, L, and momentum of an internal wheel, h. My simulation is running correctly. But I need to log not only the state variables but also some other values such as external torque, N, and angular velocity, omega, that is Jinv*L, where Jinv is a 3x3 constant, satellite-inertia matrix. In a sense, the purpose of my simulator is not to calculate L and h, but to generate time-histories of "other" varialbes.
To show what I'm doing, below is a slightly simplified version of my current code.
class satellite
{
public:
Eigen::Matrix3d Jinv;
void operator()( state_type &x , state_type &dxdt , double t )
{
L << x[0], x[1], x[2];
h << x[3], x[4], x[5], x[6];
N = external_torque(t);
omega = Jinv * (L-h);
dLdt = N - omega.cross(L);
OMEGA = func1(omega(0), omega(1), omega(2));
dqdt = OMEGA * q * 0.5;
dxdt[0] = dLdt(0); dxdt[1] = dLdt(1); dxdt[2] = dLdt(2);
dxdt[3] = dqdt(0); dxdt[4] = dqdt(1); dxdt[5] = dqdt(2); dxdt[6] = dqdt(3);
}
};
class streaming_observer
{
public:
std::ostream& os;
satellite& sat;
streaming_observer( std::ostream& _os, satellite& _sat ) : os(_os), sat(_sat) { }
template<class State>
void operator() (const State& x, double t) const
{
L << x[0], x[1], x[2];
os << t << ' ' << (sat.Jinv*(L)).transpose() << std::endl;
}
};
You must do the calculation of your intermediate and the logging in the observer. To avoid redundancy it might be favourable to out the calculations in a separate function of class method and call this method from the system function (hence the operator() in your example) and from the observer. You can also record values in there and do some later analysis with these values.
Related
I'm trying to use fminunc in Octave for a logistic problem, but it doesn't work. It says that I didn't define variables, but actually I did. If I define variables directly in the costFunction,and not in the main, it doesn't give any problem, but the function doesn't work really. In fact the exitFlag is equal to -3 and it doesn't converge at all.
Here's my function:
function [jVal, gradient] = cost(theta, X, y)
X = [1,0.14,0.09,0.58,0.39,0,0.55,0.23,0.64;1,-0.57,-0.54,-0.16,0.21,0,-0.11,-0.61,-0.35;1,0.42,0.45,-0.41,-0.6,0,-0.44,0.38,-0.29];
y = [1;0;1];
theta = [0.8;0.2;0.6;0.3;0.4;0.5;0.6;0.2;0.4];
jVal = 0;
jVal = costFunction2(X, y, theta); %this is another function that gives me jVal. I'm quite sure it is
%correct because I use it also with other algorithms and it
%works perfectly
m = length(y);
xSize = size(X, 2);
gradient = zeros(xSize, 1);
sig = X * theta;
h = 1 ./(1 + exp(-sig));
for i = 1:m
for j = 1:xSize
gradient(j) = (1/m) * sum(h(i) - y(i)) .* X(i, j);
end
end
Here's my main:
theta = [0.8;0.2;0.6;0.3;0.4;0.5;0.6;0.2;0.4];
options = optimset('GradObj', 'on', 'MaxIter', 100);
[optTheta, functionVal, exitFlag] = fminunc(#cost, theta, options)
if I compile it:
optTheta =
0.80000
0.20000
0.60000
0.30000
0.40000
0.50000
0.60000
0.20000
0.40000
functionVal = 0.15967
exitFlag = -3
How can I resolve this problem?
You are not in fact using fminunc correctly. From the documentation:
-- fminunc (FCN, X0)
-- fminunc (FCN, X0, OPTIONS)
FCN should accept a vector (array) defining the unknown variables,
and return the objective function value, optionally with gradient.
'fminunc' attempts to determine a vector X such that 'FCN (X)' is a
local minimum.
What you are passing is not a handle to a function that accepts a single vector argument. Instead, what you are passing (i.e. #cost) is a handle to a function that takes three arguments.
You need to 'convert' this into a handle to a function that takes only one input, and does what you want under the hood. The easiest way to do this is by 'wrapping' your cost function into an anonymous function that only takes one argument, and calls the cost function in the appropriate way, e.g.
fminunc( #(t) cost(t, X, y), theta, options )
Note: This assumes X and y are defined in the scope where you do this 'wrapping' business
I am looking for efficient AVX (AVX512) implementation of
// Given
float u[8];
float v[8];
// Compute
float a[8];
float b[8];
// Such that
for ( int i = 0; i < 8; ++i )
{
a[i] = fabs(u[i]) >= fabs(v[i]) ? u[i] : v[i];
b[i] = fabs(u[i]) < fabs(v[i]) ? u[i] : v[i];
}
I.e., I need to select element-wise into a from u and v based on mask, and into b based on !mask, where mask = (fabs(u) >= fabs(v)) element-wise.
I had this exact same problem just the other day. The solution I came up with (using AVX only) was:
// take the absolute value of u and v
__m256 sign_bit = _mm256_set1_ps(-0.0f);
__m256 u_abs = _mm256_andnot_ps(sign_bit, u);
__m256 v_abs = _mm256_andnot_ps(sign_bit, v);
// get a mask indicating the indices for which abs(u[i]) >= abs(v[i])
__m256 u_ge_v = _mm256_cmp_ps(u_abs, v_abs, _CMP_GE_OS);
// use the mask to select the appropriate elements into a and b, flipping the argument
// order for b to invert the sense of the mask
__m256 a = _mm256_blendv_ps(u, v, u_ge_v);
__m256 b = _mm256_blendv_ps(v, u, u_ge_v);
The AVX512 equivalent would be:
// take the absolute value of u and v
__m512 sign_bit = _mm512_set1_ps(-0.0f);
__m512 u_abs = _mm512_andnot_ps(sign_bit, u);
__m512 v_abs = _mm512_andnot_ps(sign_bit, v);
// get a mask indicating the indices for which abs(u[i]) >= abs(v[i])
__mmask16 u_ge_v = _mm512_cmp_ps_mask(u_abs, v_abs, _CMP_GE_OS);
// use the mask to select the appropriate elements into a and b, flipping the argument
// order for b to invert the sense of the mask
__m512 a = _mm512_mask_blend_ps(u_ge_v, u, v);
__m512 b = _mm512_mask_blend_ps(u_ge_v, v, u);
As Peter Cordes suggested in the comments above, there are other approaches as well like taking the absolute value followed by a min/max and then reinserting the sign bit, but I couldn't find anything that was shorter/lower latency than this sequence of instructions.
Actually, there is another approach using AVX512DQ's VRANGEPS via the _mm512_range_ps() intrinsic. Intel's intrinsic guide describes it as follows:
Calculate the max, min, absolute max, or absolute min (depending on control in imm8) for packed single-precision (32-bit) floating-point elements in a and b, and store the results in dst. imm8[1:0] specifies the operation control: 00 = min, 01 = max, 10 = absolute max, 11 = absolute min. imm8[3:2] specifies the sign control: 00 = sign from a, 01 = sign from compare result, 10 = clear sign bit, 11 = set sign bit.
Note that there appears to be a typo in the above; actually imm8[3:2] == 10 is "absolute min" and imm8[3:2] == 11 is "absolute max" if you look at the details of the per-element operation:
CASE opCtl[1:0] OF
0: tmp[31:0] := (src1[31:0] <= src2[31:0]) ? src1[31:0] : src2[31:0]
1: tmp[31:0] := (src1[31:0] <= src2[31:0]) ? src2[31:0] : src1[31:0]
2: tmp[31:0] := (ABS(src1[31:0]) <= ABS(src2[31:0])) ? src1[31:0] : src2[31:0]
3: tmp[31:0] := (ABS(src1[31:0]) <= ABS(src2[31:0])) ? src2[31:0] : src1[31:0]
ESAC
CASE signSelCtl[1:0] OF
0: dst[31:0] := (src1[31] << 31) OR (tmp[30:0])
1: dst[31:0] := tmp[63:0]
2: dst[31:0] := (0 << 31) OR (tmp[30:0])
3: dst[31:0] := (1 << 31) OR (tmp[30:0])
ESAC
RETURN dst
So you can get the same result with just two instructions:
auto a = _mm512_range_ps(v, u, 0x7); // 0b0111 = sign from compare result, absolute max
auto b = _mm512_range_ps(v, u, 0x6); // 0b0110 = sign from compare result, absolute min
The argument order (v, u) is a bit unintuitive, but it's needed in order to get the same behavior that you described in the OP in the event that the elements have equal absolute value (namely, that the value from u is passed through to a, and v goes to b).
On Skylake and Ice Lake Xeon platforms (probably any of the Xeons that have dual FMA units, probably?), VRANGEPS has throughput 2, so the two checks can issue and execute simultaneously, with latency of 4 cycles. This is only a modest latency improvement on the original approach, but the throughput is better and it requires fewer instructions/uops/instruction cache space.
clang does a pretty reasonable job of auto-vectorizing it with -ffast-math and the necessary __restrict qualifiers: https://godbolt.org/z/NMvN1u. and both inputs to ABS them, compare once, vblendvps twice on the original inputs with the same mask but the other sources in the opposite order to get min and max.
That's pretty much what I was thinking before checking what compilers did, and looking at their output to firm up the details I hadn't thought through yet. I don't see anything more clever than that. I don't think we can avoid abs()ing both a and b separately; there's no cmpps compare predicate that compares magnitudes and ignores the sign bit.
// untested: I *might* have reversed min/max, but I think this is right.
#include <immintrin.h>
// returns min_abs
__m256 minmax_abs(__m256 u, __m256 v, __m256 *max_result) {
const __m256 signbits = _mm256_set1_ps(-0.0f);
__m256 abs_u = _mm256_andnot_ps(signbits, u);
__m256 abs_v = _mm256_andnot_ps(signbits, v); // strip the sign bit
__m256 maxabs_is_v = _mm256_cmp_ps(abs_u, abs_v, _CMP_LT_OS); // u < v
*max_result = _mm256_blendv_ps(v, u, maxabs_is_v);
return _mm256_blendv_ps(u, v, maxabs_is_v);
}
You'd do the same thing with AVX512 except you compare into a mask instead of another vector.
// returns min_abs
__m512 minmax_abs512(__m512 u, __m512 v, __m512 *max_result) {
const __m512 absmask = _mm512_castsi512_ps(_mm512_set1_epi32(0x7fffffff));
__m512 abs_u = _mm512_and_ps(absmask, u);
__m512 abs_v = _mm512_and_ps(absmask, v); // strip the sign bit
__mmask16 maxabs_is_v = _mm512_cmp_ps_mask(abs_u, abs_v, _CMP_LT_OS); // u < v
*max_result = _mm512_mask_blend_ps(maxabs_is_v, v, u);
return _mm512_mask_blend_ps(maxabs_is_v, u, v);
}
Clang compiles the return statement in an interesting way (Godbolt):
.LCPI2_0:
.long 2147483647 # 0x7fffffff
minmax_abs512(float __vector(16), float __vector(16), float __vector(16)*): # #minmax_abs512(float __vector(16), float __vector(16), float __vector(16)*)
vbroadcastss zmm2, dword ptr [rip + .LCPI2_0]
vandps zmm3, zmm0, zmm2
vandps zmm2, zmm1, zmm2
vcmpltps k1, zmm3, zmm2
vblendmps zmm2 {k1}, zmm1, zmm0
vmovaps zmmword ptr [rdi], zmm2 ## store the blend result
vmovaps zmm0 {k1}, zmm1 ## interesting choice: blend merge-masking
ret
Instead of using another vblendmps, clang notices that zmm0 already has one of the blend inputs, and uses merge-masking with a regular vector vmovaps. This has zero advantage of Skylake-AVX512 for 512-bit vblendmps (both single-uop instructions for port 0 or 5), but if Agner Fog's instruction tables are right, vblendmps x/y/zmm only ever runs on port 0 or 5, but a masked 256-bit or 128-bit vmovaps x/ymm{k}, x/ymm can run on any of p0/p1/p5.
Both are single-uop / single-cycle latency, unlike AVX2 vblendvps based on a mask vector which is 2 uops. (So AVX512 is an advantage even for 256-bit vectors). Unfortunately, none of gcc, clang, or ICC turn the _mm256_cmp_ps into _mm256_cmp_ps_mask and optimize the AVX2 intrinsics to AVX512 instructions when compiling with -march=skylake-avx512.)
s/512/256/ to make a version of minmax_abs512 that uses AVX512 for 256-bit vectors.
Gcc goes even further, and does the questionable "optimization" of
vmovaps zmm2, zmm1 # tmp118, v
vmovaps zmm2{k1}, zmm0 # tmp118, tmp114, tmp118, u
instead of using one blend instruction. (I keep thinking I'm seeing a store followed by a masked store, but no, neither compiler is blending that way).
So the Fibonacci number for log (N) — without matrices.
Ni // i-th Fibonacci number
= Ni-1 + Ni-2 // by definition
= (Ni-2 + Ni-3) + Ni-2 // unwrap Ni-1
= 2*Ni-2 + Ni-3 // reduce the equation
= 2*(Ni-3 + Ni-4) + Ni-3 //unwrap Ni-2
// And so on
= 3*Ni-3 + 2*Ni-4
= 5*Ni-4 + 3*Ni-5
= 8*Ni-5 + 5*Ni-6
= Nk*Ni-k + Nk-1*Ni-k-1
Now we write a recursive function, where at each step we take k~=I/2.
static long N(long i)
{
if (i < 2) return 1;
long k=i/2;
return N(k) * N(i - k) + N(k - 1) * N(i - k - 1);
}
Where is the fault?
You get a recursion formula for the effort: T(n) = 4T(n/2) + O(1). (disregarding the fact that the numbers get bigger, so the O(1) does not even hold). It's clear from this that T(n) is not in O(log(n)). Instead one gets by the master theorem T(n) is in O(n^2).
Btw, this is even slower than the trivial algorithm to calculate all Fibonacci numbers up to n.
The four N calls inside the function each have an argument of around i/2. So the length of the stack of N calls in total is roughly equal to log2N, but because each call generates four more, the bottom 'layer' of calls has 4^log2N = O(n2) Thus, the fault is that N calls itself four times. With only two calls, as in the conventional iterative method, it would be O(n). I don't know of any way to do this with only one call, which could be O(log n).
An O(n) version based on this formula would be:
static long N(long i) {
if (i<2) {
return 1;
}
long k = i/2;
long val1;
long val2;
val1 = N(k-1);
val2 = N(k);
if (i%2==0) {
return val2*val2+val1*val1;
}
return val2*(val2+val1)+val1*val2;
}
which makes 2 N calls per function, making it O(n).
public class fibonacci {
public static int count=0;
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int i = scan.nextInt();
System.out.println("value of i ="+ i);
int result = fun(i);
System.out.println("final result is " +result);
}
public static int fun(int i) {
count++;
System.out.println("fun is called and count is "+count);
if(i < 2) {
System.out.println("function returned");
return 1;
}
int k = i/2;
int part1 = fun(k);
int part2 = fun(i-k);
int part3 = fun(k-1);
int part4 = fun(i-k-1);
return ((part1*part2) + (part3*part4)); /*RESULT WILL BE SAME FOR BOTH METHODS*/
//return ((fun(k)*fun(i-k))+(fun(k-1)*fun(i-k-1)));
}
}
I tried to code to problem defined by you in java. What i observed is that complexity of above code is not completely O(N^2) but less than that.But as per conventions and standards the worst case complexity is O(N^2) including some other factors like computation(division,multiplication) and comparison time analysis.
The output of above code gives me information about how many times the function
fun(int i) computes and is being called.
OUTPUT
So including the time taken for comparison and division, multiplication operations, the worst case time complexity is O(N^2) not O(LogN).
Ok if we use Analysis of the recursive Fibonacci program technique.Then we end up getting a simple equation
T(N) = 4* T(N/2) + O(1)
where O(1) is some constant time.
So let's apply Master's method on this equation.
According to Master's method
T(n) = aT(n/b) + f(n) where a >= 1 and b > 1
There are following three cases:
If f(n) = Θ(nc) where c < Logba then T(n) = Θ(nLogba)
If f(n) = Θ(nc) where c = Logba then T(n) = Θ(ncLog n)
If f(n) = Θ(nc) where c > Logba then T(n) = Θ(f(n))
And in our equation a=4 , b=2 & c=0.
As case 1 c < logba => 0 < 2 (which is log base 2 and equals to 2) is satisfied
hence T(n) = O(n^2).
For more information about how master's algorithm works please visit: Analysis of Algorithms
Your idea is correct, and it will perform in O(log n) provided you don't compute the same formula
over and over again. The whole point of having N(k) * N(i-k) is to have (k = i - k) so you only have to compute one instead of two. But if you only call recursively, you are performing the computation twice.
What you need is called memoization. That is, store every value that you already have computed, and
if it comes up again, then you get it in O(1).
Here's an example
const int MAX = 10000;
// memoization array
int f[MAX] = {0};
// Return nth fibonacci number using memoization
int fib(int n) {
// Base case
if (n == 0)
return 0;
if (n == 1 || n == 2)
return (f[n] = 1);
// If fib(n) is already computed
if (f[n]) return f[n];
// (n & 1) is 1 iff n is odd
int k = n/2;
// Applying your formula
f[n] = fib(k) * fib(n - k) + fib(k - 1) * fib(n - k - 1);
return f[n];
}
The statements in procedural blocks execute seqeuntially so why aren't any of the block1, block2 or block3 inferring a latch?
module testing(
input logic a, b, c,
output logic x, y, z, v
);
logic tmp_ref, tmp1, tmp2, tmp3;
//reference
always_comb begin: ref_block
tmp_ref = a & b;
x = tmp_ref ^ c;
end
always_comb begin: block1
y = tmp1 ^ c;
tmp1 = a & b;
end
always #(*) begin: block2
tmp2 <= a & b;
z = tmp2 ^ c;
end
always #(c) begin: block3
tmp3 = a & b;
v = tmp3 ^ c;
end
endmodule: testing
In block1 y is calculated using the blocking assignment before the new value of tmp1 is available.
In block2 tmp2 is calculated using a non-blocking assignment, which should postpone the assignment for when the always block finishes. Meanwhile, z is calculated using the blocking assignment and the new value of tmp2 is not yet available.
In block3 there is an incomplete sensitivity list and still no latch.
Here is the synthesis result from Quartus II 14.1:
Only when I add this block a latch is inferred:
//infers a latch
always #(*) begin: block4
if (c == 1'b1) begin
tmp4 = a & b;
w = tmp4 ^ c;
end
end
Can someone please explain why incomplete sensitivity list or using a variable before the value is updated does not infer a latch in a combinatorial block?
The type of assignment used in a combinatorial block will not effect synthesis. The use of non-blocking (<=) may result in RTL (pre-synthesis) to gates (post-synthesis) simulator mismatches.
The same is true for sensitivity lists, synthesis will give the behaviour of auto generated or complete list.
In a clocked process (#(posedge clk)) use non-blocking (<=) to get the simulation behaviour of a flip-flop. It is possible to use blocking (=) as well to have combinatorial code inside the clocked process but mixing styles is considered a bad coding practice. The combinatorial part code just be moved to a separate combinatorial block (always #*).
A latch is a basic memory element, if the circuit does not need memory then it will not be inferred.
For example:
always #* begin:
v = (a & b) ^ c;
end
v is completely defined by inputs, there is no memory involved. In comparison to :
always #* begin
if (c == 1'b1) begin
w = (a & b) ^ c;
end
end
When c is 0 w must hold its value, therefore a latch is inferred.
It is worth noting that while latches are not bad, care must be taken with the timing of when the open and close to ensure they capture the correct data. Therefore inferred latch are typically seen as bad and are from poor coding.
SystemVerilog has the following syntax for semantically implying design intent:
always_latch begin
if (c == 1'b1) begin
w = (a & b) ^ c;
end
end
I've been struggling with the following code. It's an F# implementation of the Forward-Euler algorithm used for modelling stars moving in a gravitational field.
let force (b1:Body) (b2:Body) =
let r = (b2.Position - b1.Position)
let rm = (float32)r.MagnitudeSquared + softeningLengthSquared
if (b1 = b2) then
VectorFloat.Zero
else
r * (b1.Mass * b2.Mass) / (Math.Sqrt((float)rm) * (float)rm)
member this.Integrate(dT, (bodies:Body[])) =
for i = 0 to bodies.Length - 1 do
for j = (i + 1) to bodies.Length - 1 do
let f = force bodies.[i] bodies.[j]
bodies.[i].Acceleration <- bodies.[i].Acceleration + (f / bodies.[i].Mass)
bodies.[j].Acceleration <- bodies.[j].Acceleration - (f / bodies.[j].Mass)
bodies.[i].Position <- bodies.[i].Position + bodies.[i].Velocity * dT
bodies.[i].Velocity <- bodies.[i].Velocity + bodies.[i].Acceleration * dT
While this works it isn't exactly "functional". It also suffers from horrible performance, it's 2.5 times slower than the equivalent c# code. bodies is an array of structs of type Body.
The thing I'm struggling with is that force() is an expensive function so usually you calculate it once for each pair and rely on the fact that Fij = -Fji. But this really messes up any loop unfolding etc.
Suggestions gratefully received! No this isn't homework...
Thanks,
Ade
UPDATED: To clarify Body and VectorFloat are defined as C# structs. This is because the program interops between F#/C# and C++/CLI. Eventually I'm going to get the code up on BitBucket but it's a work in progress I have some issues to sort out before I can put it up.
[StructLayout(LayoutKind.Sequential)]
public struct Body
{
public VectorFloat Position;
public float Size;
public uint Color;
public VectorFloat Velocity;
public VectorFloat Acceleration;
'''
}
[StructLayout(LayoutKind.Sequential)]
public partial struct VectorFloat
{
public System.Single X { get; set; }
public System.Single Y { get; set; }
public System.Single Z { get; set; }
}
The vector defines the sort of operators you'd expect for a standard Vector class. You could probably use the Vector3D class from the .NET framework for this case (I'm actually investigating cutting over to it).
UPDATE 2: Improved code based on the first two replies below:
for i = 0 to bodies.Length - 1 do
for j = (i + 1) to bodies.Length - 1 do
let r = ( bodies.[j].Position - bodies.[i].Position)
let rm = (float32)r.MagnitudeSquared + softeningLengthSquared
let f = r / (Math.Sqrt((float)rm) * (float)rm)
bodies.[i].Acceleration <- bodies.[i].Acceleration + (f * bodies.[j].Mass)
bodies.[j].Acceleration <- bodies.[j].Acceleration - (f * bodies.[i].Mass)
bodies.[i].Position <- bodies.[i].Position + bodies.[i].Velocity * dT
bodies.[i].Velocity <- bodies.[i].Velocity + bodies.[i].Acceleration * dT
The branch in the force function to cover the b1 == b2 case is the worst offender. You do't need this if softeningLength is always non-zero, even if it's very small (Epsilon). This optimization was in the C# code but not the F# version (doh!).
Math.Pow(x, -1.5) seems to be a lot slower than 1/ (Math.Sqrt(x) * x). Essentially this algorithm is slightly odd in that it's perfromance is dictated by the cost of this one step.
Moving the force calculation inline and getting rid of some divides also gives some improvement, but the performance was really being killed by the branching and is dominated by the cost of Sqrt.
WRT using classes over structs: There are cases (CUDA and native C++ implementations of this code and a DX9 renderer) where I need to get the array of bodies into unmanaged code or onto a GPU. In these scenarios being able to memcpy a contiguous block of memory seems like the way to go. Not something I'd get from an array of class Body.
I'm not sure if it's wise to rewrite this code in a functional style. I've seen some attempts to write pair interaction calculations in a functional manner and each one of them was harder to follow than two nested loops.
Before looking at structs vs. classes (I'm sure someone else has something smart to say about this), maybe you can try optimizing the calculation itself?
You're calculating two acceleration deltas, let's call them dAi and dAj:
dAi = r*m1*m2/(rm*sqrt(rm)) / m1
dAj = r*m1*m2/(rm*sqrt(rm)) / m2
[note: m1 = bodies.[i].mass, m2=bodies.[j].mass]]
The division by mass cancels out like this:
dAi = rm2 / (rmsqrt(rm))
dAj = rm1 / (rmsqrt(rm))
Now you only have to calculate r/(rmsqrt(rm)) for each pair (i,j).
This can be optimized further, because 1/(rmsqrt(rm)) = 1/(rm^1.5) = rm^-1.5, so if you let r' = r * (rm ** -1.5), then Edit: no it can't, that's premature optimization talking right there (see comment). Calculating r' = 1.0 / (r * sqrt r) is fastest.
dAi = m2 * r'
dAj = m1 * r'
Your code would then become something like
member this.Integrate(dT, (bodies:Body[])) =
for i = 0 to bodies.Length - 1 do
for j = (i + 1) to bodies.Length - 1 do
let r = (b2.Position - b1.Position)
let rm = (float32)r.MagnitudeSquared + softeningLengthSquared
let r' = r * (rm ** -1.5)
bodies.[i].Acceleration <- bodies.[i].Acceleration + r' * bodies.[j].Mass
bodies.[j].Acceleration <- bodies.[j].Acceleration - r' * bodies.[i].Mass
bodies.[i].Position <- bodies.[i].Position + bodies.[i].Velocity * dT
bodies.[i].Velocity <- bodies.[i].Velocity + bodies.[i].Acceleration * dT
Look, ma, no more divisions!
Warning: untested code. Try at your own risk.
I'd like to play arround with your code, but it's difficult since the definition of Body and FloatVector is missing and they also seem to be missing from the orginal blog post you point to.
I'd hazard a guess that you could improve your performance and rewrite in a more functional style using F#'s lazy computations:
http://msdn.microsoft.com/en-us/library/dd233247(VS.100).aspx
The idea is fairly simple you wrap any expensive computation that could be repeatedly calculated in a lazy ( ... ) expression then you can force the computation as many times as you like and it will only ever be calculated once.