Ways to speed up user implemented RK4 The Next CEO of Stack OverflowSpeed up Numerical IntegrationHow to choose MaxStepFraction for optimal speed of NDSolveNIntegrate: how to speed up code?Compiling FoldList implementation for RK4How to speed up this code?Solving an unstable BVP numerically, accurately and efficientlyHow to speed up integral of results of PDE modelSolve BVP involving user defined functionWays to speed up PickSpeed up ParametricNDSolve

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Ways to speed up user implemented RK4



The Next CEO of Stack OverflowSpeed up Numerical IntegrationHow to choose MaxStepFraction for optimal speed of NDSolveNIntegrate: how to speed up code?Compiling FoldList implementation for RK4How to speed up this code?Solving an unstable BVP numerically, accurately and efficientlyHow to speed up integral of results of PDE modelSolve BVP involving user defined functionWays to speed up PickSpeed up ParametricNDSolve










8












$begingroup$


So, I've implemented RK4, and I'm wondering what I can do to make it more efficient? What I've got so far is below. I wish to still record all steps. I think AppendTo is doing the most damage to the time, is there a faster alternative?



rk4[f_, variables_, valtinit_, tinit_, tfinal_, nsteps_] := 
Module[table, xlist, ylist, step, k1, k2, k3, k4,
xlist = tinit;
step = N[(tfinal - tinit)/(nsteps)];
ylist = valtinit;
table = xlist, ylist;
Table[
k1 = step* f /. MapThread[Rule, variables, ylist]; (*
Equivalent to step* f/.Thread[Rule[variables,ylist]]*)
k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist];
k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist];
k4 = step*f /. MapThread[Rule, variables, k3 + ylist];
ylist += 1/6 (k1 + 2 (k2 + k3) + k4);
xlist += step;
AppendTo[table, xlist, ylist];
xlist, ylist, nsteps];
table
];


Example Input:



funclist = -x + y, x - y;
initials = 1, 2;
variables = x, y;
init = 0;
final = 200;
nstep = 20000;
approx = rk4[funclist, variables, initials, init, final, nstep]//AbsoluteTiming;



3.59932,...




I'd love some suggestions!










share|improve this question











$endgroup$







  • 3




    $begingroup$
    AppendTo is quadratic time complexity. Might be better to preallocate and set by index. Also it'll be much faster to not use Rule and instead code stuff up a little bit more explicitly. As a general rule, too, use vectorized operators. Those can be very fast. And if everything can be totally functional over "packed arrays" (look them up here) it'll be very quick too.
    $endgroup$
    – b3m2a1
    2 days ago










  • $begingroup$
    I'll work on implementing it more explicity, this is what came to find first though. It'll require some changes to the inputs, I'll have to ponder this. And preallocating the list is a quick change that won't be an issue to do, I can't believe I forgot that's faster :(. Thanks though!
    $endgroup$
    – Shinaolord
    2 days ago










  • $begingroup$
    Shinaoloard, using Join[ xlist, ylist, Table[ k1 = step*f /. MapThread[Rule, variables, ylist]; k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist]; k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist]; k4 = step*f /. MapThread[Rule, variables, k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps ] ] as return value is already a first step. There is no point in appending if you use a Table anyways.
    $endgroup$
    – Henrik Schumacher
    2 days ago






  • 3




    $begingroup$
    Why not just get NDSolve[] to use fourth-order Runge-Kutta to begin with?
    $endgroup$
    – J. M. is slightly pensive
    yesterday






  • 1




    $begingroup$
    @J.M.isslightlypensive I know it can, I just wanted to make sure I could actually code it myself, instead of just using options to get Mathematica to do it for me (:. Thanks for trying to help though!!
    $endgroup$
    – Shinaolord
    yesterday















8












$begingroup$


So, I've implemented RK4, and I'm wondering what I can do to make it more efficient? What I've got so far is below. I wish to still record all steps. I think AppendTo is doing the most damage to the time, is there a faster alternative?



rk4[f_, variables_, valtinit_, tinit_, tfinal_, nsteps_] := 
Module[table, xlist, ylist, step, k1, k2, k3, k4,
xlist = tinit;
step = N[(tfinal - tinit)/(nsteps)];
ylist = valtinit;
table = xlist, ylist;
Table[
k1 = step* f /. MapThread[Rule, variables, ylist]; (*
Equivalent to step* f/.Thread[Rule[variables,ylist]]*)
k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist];
k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist];
k4 = step*f /. MapThread[Rule, variables, k3 + ylist];
ylist += 1/6 (k1 + 2 (k2 + k3) + k4);
xlist += step;
AppendTo[table, xlist, ylist];
xlist, ylist, nsteps];
table
];


Example Input:



funclist = -x + y, x - y;
initials = 1, 2;
variables = x, y;
init = 0;
final = 200;
nstep = 20000;
approx = rk4[funclist, variables, initials, init, final, nstep]//AbsoluteTiming;



3.59932,...




I'd love some suggestions!










share|improve this question











$endgroup$







  • 3




    $begingroup$
    AppendTo is quadratic time complexity. Might be better to preallocate and set by index. Also it'll be much faster to not use Rule and instead code stuff up a little bit more explicitly. As a general rule, too, use vectorized operators. Those can be very fast. And if everything can be totally functional over "packed arrays" (look them up here) it'll be very quick too.
    $endgroup$
    – b3m2a1
    2 days ago










  • $begingroup$
    I'll work on implementing it more explicity, this is what came to find first though. It'll require some changes to the inputs, I'll have to ponder this. And preallocating the list is a quick change that won't be an issue to do, I can't believe I forgot that's faster :(. Thanks though!
    $endgroup$
    – Shinaolord
    2 days ago










  • $begingroup$
    Shinaoloard, using Join[ xlist, ylist, Table[ k1 = step*f /. MapThread[Rule, variables, ylist]; k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist]; k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist]; k4 = step*f /. MapThread[Rule, variables, k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps ] ] as return value is already a first step. There is no point in appending if you use a Table anyways.
    $endgroup$
    – Henrik Schumacher
    2 days ago






  • 3




    $begingroup$
    Why not just get NDSolve[] to use fourth-order Runge-Kutta to begin with?
    $endgroup$
    – J. M. is slightly pensive
    yesterday






  • 1




    $begingroup$
    @J.M.isslightlypensive I know it can, I just wanted to make sure I could actually code it myself, instead of just using options to get Mathematica to do it for me (:. Thanks for trying to help though!!
    $endgroup$
    – Shinaolord
    yesterday













8












8








8


1



$begingroup$


So, I've implemented RK4, and I'm wondering what I can do to make it more efficient? What I've got so far is below. I wish to still record all steps. I think AppendTo is doing the most damage to the time, is there a faster alternative?



rk4[f_, variables_, valtinit_, tinit_, tfinal_, nsteps_] := 
Module[table, xlist, ylist, step, k1, k2, k3, k4,
xlist = tinit;
step = N[(tfinal - tinit)/(nsteps)];
ylist = valtinit;
table = xlist, ylist;
Table[
k1 = step* f /. MapThread[Rule, variables, ylist]; (*
Equivalent to step* f/.Thread[Rule[variables,ylist]]*)
k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist];
k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist];
k4 = step*f /. MapThread[Rule, variables, k3 + ylist];
ylist += 1/6 (k1 + 2 (k2 + k3) + k4);
xlist += step;
AppendTo[table, xlist, ylist];
xlist, ylist, nsteps];
table
];


Example Input:



funclist = -x + y, x - y;
initials = 1, 2;
variables = x, y;
init = 0;
final = 200;
nstep = 20000;
approx = rk4[funclist, variables, initials, init, final, nstep]//AbsoluteTiming;



3.59932,...




I'd love some suggestions!










share|improve this question











$endgroup$




So, I've implemented RK4, and I'm wondering what I can do to make it more efficient? What I've got so far is below. I wish to still record all steps. I think AppendTo is doing the most damage to the time, is there a faster alternative?



rk4[f_, variables_, valtinit_, tinit_, tfinal_, nsteps_] := 
Module[table, xlist, ylist, step, k1, k2, k3, k4,
xlist = tinit;
step = N[(tfinal - tinit)/(nsteps)];
ylist = valtinit;
table = xlist, ylist;
Table[
k1 = step* f /. MapThread[Rule, variables, ylist]; (*
Equivalent to step* f/.Thread[Rule[variables,ylist]]*)
k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist];
k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist];
k4 = step*f /. MapThread[Rule, variables, k3 + ylist];
ylist += 1/6 (k1 + 2 (k2 + k3) + k4);
xlist += step;
AppendTo[table, xlist, ylist];
xlist, ylist, nsteps];
table
];


Example Input:



funclist = -x + y, x - y;
initials = 1, 2;
variables = x, y;
init = 0;
final = 200;
nstep = 20000;
approx = rk4[funclist, variables, initials, init, final, nstep]//AbsoluteTiming;



3.59932,...




I'd love some suggestions!







differential-equations numerical-integration performance-tuning






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday









xzczd

27.4k573255




27.4k573255










asked 2 days ago









ShinaolordShinaolord

1088




1088







  • 3




    $begingroup$
    AppendTo is quadratic time complexity. Might be better to preallocate and set by index. Also it'll be much faster to not use Rule and instead code stuff up a little bit more explicitly. As a general rule, too, use vectorized operators. Those can be very fast. And if everything can be totally functional over "packed arrays" (look them up here) it'll be very quick too.
    $endgroup$
    – b3m2a1
    2 days ago










  • $begingroup$
    I'll work on implementing it more explicity, this is what came to find first though. It'll require some changes to the inputs, I'll have to ponder this. And preallocating the list is a quick change that won't be an issue to do, I can't believe I forgot that's faster :(. Thanks though!
    $endgroup$
    – Shinaolord
    2 days ago










  • $begingroup$
    Shinaoloard, using Join[ xlist, ylist, Table[ k1 = step*f /. MapThread[Rule, variables, ylist]; k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist]; k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist]; k4 = step*f /. MapThread[Rule, variables, k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps ] ] as return value is already a first step. There is no point in appending if you use a Table anyways.
    $endgroup$
    – Henrik Schumacher
    2 days ago






  • 3




    $begingroup$
    Why not just get NDSolve[] to use fourth-order Runge-Kutta to begin with?
    $endgroup$
    – J. M. is slightly pensive
    yesterday






  • 1




    $begingroup$
    @J.M.isslightlypensive I know it can, I just wanted to make sure I could actually code it myself, instead of just using options to get Mathematica to do it for me (:. Thanks for trying to help though!!
    $endgroup$
    – Shinaolord
    yesterday












  • 3




    $begingroup$
    AppendTo is quadratic time complexity. Might be better to preallocate and set by index. Also it'll be much faster to not use Rule and instead code stuff up a little bit more explicitly. As a general rule, too, use vectorized operators. Those can be very fast. And if everything can be totally functional over "packed arrays" (look them up here) it'll be very quick too.
    $endgroup$
    – b3m2a1
    2 days ago










  • $begingroup$
    I'll work on implementing it more explicity, this is what came to find first though. It'll require some changes to the inputs, I'll have to ponder this. And preallocating the list is a quick change that won't be an issue to do, I can't believe I forgot that's faster :(. Thanks though!
    $endgroup$
    – Shinaolord
    2 days ago










  • $begingroup$
    Shinaoloard, using Join[ xlist, ylist, Table[ k1 = step*f /. MapThread[Rule, variables, ylist]; k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist]; k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist]; k4 = step*f /. MapThread[Rule, variables, k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps ] ] as return value is already a first step. There is no point in appending if you use a Table anyways.
    $endgroup$
    – Henrik Schumacher
    2 days ago






  • 3




    $begingroup$
    Why not just get NDSolve[] to use fourth-order Runge-Kutta to begin with?
    $endgroup$
    – J. M. is slightly pensive
    yesterday






  • 1




    $begingroup$
    @J.M.isslightlypensive I know it can, I just wanted to make sure I could actually code it myself, instead of just using options to get Mathematica to do it for me (:. Thanks for trying to help though!!
    $endgroup$
    – Shinaolord
    yesterday







3




3




$begingroup$
AppendTo is quadratic time complexity. Might be better to preallocate and set by index. Also it'll be much faster to not use Rule and instead code stuff up a little bit more explicitly. As a general rule, too, use vectorized operators. Those can be very fast. And if everything can be totally functional over "packed arrays" (look them up here) it'll be very quick too.
$endgroup$
– b3m2a1
2 days ago




$begingroup$
AppendTo is quadratic time complexity. Might be better to preallocate and set by index. Also it'll be much faster to not use Rule and instead code stuff up a little bit more explicitly. As a general rule, too, use vectorized operators. Those can be very fast. And if everything can be totally functional over "packed arrays" (look them up here) it'll be very quick too.
$endgroup$
– b3m2a1
2 days ago












$begingroup$
I'll work on implementing it more explicity, this is what came to find first though. It'll require some changes to the inputs, I'll have to ponder this. And preallocating the list is a quick change that won't be an issue to do, I can't believe I forgot that's faster :(. Thanks though!
$endgroup$
– Shinaolord
2 days ago




$begingroup$
I'll work on implementing it more explicity, this is what came to find first though. It'll require some changes to the inputs, I'll have to ponder this. And preallocating the list is a quick change that won't be an issue to do, I can't believe I forgot that's faster :(. Thanks though!
$endgroup$
– Shinaolord
2 days ago












$begingroup$
Shinaoloard, using Join[ xlist, ylist, Table[ k1 = step*f /. MapThread[Rule, variables, ylist]; k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist]; k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist]; k4 = step*f /. MapThread[Rule, variables, k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps ] ] as return value is already a first step. There is no point in appending if you use a Table anyways.
$endgroup$
– Henrik Schumacher
2 days ago




$begingroup$
Shinaoloard, using Join[ xlist, ylist, Table[ k1 = step*f /. MapThread[Rule, variables, ylist]; k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist]; k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist]; k4 = step*f /. MapThread[Rule, variables, k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps ] ] as return value is already a first step. There is no point in appending if you use a Table anyways.
$endgroup$
– Henrik Schumacher
2 days ago




3




3




$begingroup$
Why not just get NDSolve[] to use fourth-order Runge-Kutta to begin with?
$endgroup$
– J. M. is slightly pensive
yesterday




$begingroup$
Why not just get NDSolve[] to use fourth-order Runge-Kutta to begin with?
$endgroup$
– J. M. is slightly pensive
yesterday




1




1




$begingroup$
@J.M.isslightlypensive I know it can, I just wanted to make sure I could actually code it myself, instead of just using options to get Mathematica to do it for me (:. Thanks for trying to help though!!
$endgroup$
– Shinaolord
yesterday




$begingroup$
@J.M.isslightlypensive I know it can, I just wanted to make sure I could actually code it myself, instead of just using options to get Mathematica to do it for me (:. Thanks for trying to help though!!
$endgroup$
– Shinaolord
yesterday










1 Answer
1






active

oldest

votes


















15












$begingroup$

Just to give you an impression how fast things may get when you use the right tools.



For given stepsize τ and given vector field F, this creates a CompiledFunction cStep that computes a single Runge-Kutta step



F = X [Function] -Indexed[X, 2], Indexed[X, 1];

τ = 0.01;
Block[YY, Y, k1, k2, k3, k4,

YY = Table[Compile`GetElement[Y, i], i, 1, 2];
k1 = τ F[YY];
k2 = τ F[0.5 k1 + YY];
k3 = τ F[0.5 k2 + YY];
k4 = τ F[k3 + YY];

cStep = With[code = YY + (k1 + 2. (k2 + k3) + k4)/6. ,
Compile[Y, _Real, 1,
code,
CompilationTarget -> "C",
RuntimeOptions -> "Speed"
]
]
];


Now we can apply it 20 million times with NestList and it stills takes only about 2 seconds.



nsteps = 20000000;
xlist = Range[0., τ nsteps, τ];
Ylist = NestList[cStep, 1., 0., nsteps]; // AbsoluteTiming // First



2.08678




Edit



This can be sped up even more my avoiding NestList (the loop behind it can also be compiled which saves several calls to libraries) and by utilizing that the dimension of the ODE is known at compile time. For low dimensional systems, it may be also beneficial to avoid tensor operations altogether and to perform computations in scalar registers as done below.



τ = 0.01;
cFlow = Block[YY, Y, k1, k2, k3, k4, τ, Ylist, j,
YY = Table[Compile`GetElement[Ylist, j, i], i, 1, 2];
k1 = τ F[YY];
k2 = τ F[0.5 k1 + YY];
k3 = τ F[0.5 k2 + YY];
k4 = τ F[k3 + YY];
With[
code1 = (YY + (k1 + 2. (k2 + k3) + k4)/6)[[1]],
code2 = (YY + (k1 + 2. (k2 + k3) + k4)/6)[[2]]
,
Compile[Y0, _Real, 1, τ, _Real, n, _Integer,
Block[Ylist,
Ylist = Table[0., n + 1, Length[Y0]];
Ylist[[1]] = Y0;
Do[
Ylist[[j + 1, 1]] = code1;
Ylist[[j + 1, 2]] = code2;
,
j, 1, n];
Ylist
],
CompilationTarget -> "C", RuntimeOptions -> "Speed"
]
]
];
Ylist2 = cFlow[1., 0., τ, nsteps]; // AbsoluteTiming // First



1.06549




Don't be too upset by parts of the code being highlighted in red; this is on purpose.






share|improve this answer











$endgroup$








  • 1




    $begingroup$
    Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
    $endgroup$
    – Shinaolord
    2 days ago










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    2 days ago










  • $begingroup$
    I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
    $endgroup$
    – Shinaolord
    2 days ago






  • 1




    $begingroup$
    This is exactly the kind of thing I like to show people when they complain about the slowness of Mathematica. Of course with some cleverness in vectorized operations Compile could probably be avoided altogether if one so desired.
    $endgroup$
    – b3m2a1
    yesterday











  • $begingroup$
    @b3m2a1 Yeah, right. However, with ODE systems of this small dimension (dim = 2) I am not sure how to utilize vectorization since the ODE has to be solved in serial and because the vector field F may be very nonlinear.
    $endgroup$
    – Henrik Schumacher
    yesterday











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









15












$begingroup$

Just to give you an impression how fast things may get when you use the right tools.



For given stepsize τ and given vector field F, this creates a CompiledFunction cStep that computes a single Runge-Kutta step



F = X [Function] -Indexed[X, 2], Indexed[X, 1];

τ = 0.01;
Block[YY, Y, k1, k2, k3, k4,

YY = Table[Compile`GetElement[Y, i], i, 1, 2];
k1 = τ F[YY];
k2 = τ F[0.5 k1 + YY];
k3 = τ F[0.5 k2 + YY];
k4 = τ F[k3 + YY];

cStep = With[code = YY + (k1 + 2. (k2 + k3) + k4)/6. ,
Compile[Y, _Real, 1,
code,
CompilationTarget -> "C",
RuntimeOptions -> "Speed"
]
]
];


Now we can apply it 20 million times with NestList and it stills takes only about 2 seconds.



nsteps = 20000000;
xlist = Range[0., τ nsteps, τ];
Ylist = NestList[cStep, 1., 0., nsteps]; // AbsoluteTiming // First



2.08678




Edit



This can be sped up even more my avoiding NestList (the loop behind it can also be compiled which saves several calls to libraries) and by utilizing that the dimension of the ODE is known at compile time. For low dimensional systems, it may be also beneficial to avoid tensor operations altogether and to perform computations in scalar registers as done below.



τ = 0.01;
cFlow = Block[YY, Y, k1, k2, k3, k4, τ, Ylist, j,
YY = Table[Compile`GetElement[Ylist, j, i], i, 1, 2];
k1 = τ F[YY];
k2 = τ F[0.5 k1 + YY];
k3 = τ F[0.5 k2 + YY];
k4 = τ F[k3 + YY];
With[
code1 = (YY + (k1 + 2. (k2 + k3) + k4)/6)[[1]],
code2 = (YY + (k1 + 2. (k2 + k3) + k4)/6)[[2]]
,
Compile[Y0, _Real, 1, τ, _Real, n, _Integer,
Block[Ylist,
Ylist = Table[0., n + 1, Length[Y0]];
Ylist[[1]] = Y0;
Do[
Ylist[[j + 1, 1]] = code1;
Ylist[[j + 1, 2]] = code2;
,
j, 1, n];
Ylist
],
CompilationTarget -> "C", RuntimeOptions -> "Speed"
]
]
];
Ylist2 = cFlow[1., 0., τ, nsteps]; // AbsoluteTiming // First



1.06549




Don't be too upset by parts of the code being highlighted in red; this is on purpose.






share|improve this answer











$endgroup$








  • 1




    $begingroup$
    Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
    $endgroup$
    – Shinaolord
    2 days ago










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    2 days ago










  • $begingroup$
    I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
    $endgroup$
    – Shinaolord
    2 days ago






  • 1




    $begingroup$
    This is exactly the kind of thing I like to show people when they complain about the slowness of Mathematica. Of course with some cleverness in vectorized operations Compile could probably be avoided altogether if one so desired.
    $endgroup$
    – b3m2a1
    yesterday











  • $begingroup$
    @b3m2a1 Yeah, right. However, with ODE systems of this small dimension (dim = 2) I am not sure how to utilize vectorization since the ODE has to be solved in serial and because the vector field F may be very nonlinear.
    $endgroup$
    – Henrik Schumacher
    yesterday















15












$begingroup$

Just to give you an impression how fast things may get when you use the right tools.



For given stepsize τ and given vector field F, this creates a CompiledFunction cStep that computes a single Runge-Kutta step



F = X [Function] -Indexed[X, 2], Indexed[X, 1];

τ = 0.01;
Block[YY, Y, k1, k2, k3, k4,

YY = Table[Compile`GetElement[Y, i], i, 1, 2];
k1 = τ F[YY];
k2 = τ F[0.5 k1 + YY];
k3 = τ F[0.5 k2 + YY];
k4 = τ F[k3 + YY];

cStep = With[code = YY + (k1 + 2. (k2 + k3) + k4)/6. ,
Compile[Y, _Real, 1,
code,
CompilationTarget -> "C",
RuntimeOptions -> "Speed"
]
]
];


Now we can apply it 20 million times with NestList and it stills takes only about 2 seconds.



nsteps = 20000000;
xlist = Range[0., τ nsteps, τ];
Ylist = NestList[cStep, 1., 0., nsteps]; // AbsoluteTiming // First



2.08678




Edit



This can be sped up even more my avoiding NestList (the loop behind it can also be compiled which saves several calls to libraries) and by utilizing that the dimension of the ODE is known at compile time. For low dimensional systems, it may be also beneficial to avoid tensor operations altogether and to perform computations in scalar registers as done below.



τ = 0.01;
cFlow = Block[YY, Y, k1, k2, k3, k4, τ, Ylist, j,
YY = Table[Compile`GetElement[Ylist, j, i], i, 1, 2];
k1 = τ F[YY];
k2 = τ F[0.5 k1 + YY];
k3 = τ F[0.5 k2 + YY];
k4 = τ F[k3 + YY];
With[
code1 = (YY + (k1 + 2. (k2 + k3) + k4)/6)[[1]],
code2 = (YY + (k1 + 2. (k2 + k3) + k4)/6)[[2]]
,
Compile[Y0, _Real, 1, τ, _Real, n, _Integer,
Block[Ylist,
Ylist = Table[0., n + 1, Length[Y0]];
Ylist[[1]] = Y0;
Do[
Ylist[[j + 1, 1]] = code1;
Ylist[[j + 1, 2]] = code2;
,
j, 1, n];
Ylist
],
CompilationTarget -> "C", RuntimeOptions -> "Speed"
]
]
];
Ylist2 = cFlow[1., 0., τ, nsteps]; // AbsoluteTiming // First



1.06549




Don't be too upset by parts of the code being highlighted in red; this is on purpose.






share|improve this answer











$endgroup$








  • 1




    $begingroup$
    Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
    $endgroup$
    – Shinaolord
    2 days ago










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    2 days ago










  • $begingroup$
    I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
    $endgroup$
    – Shinaolord
    2 days ago






  • 1




    $begingroup$
    This is exactly the kind of thing I like to show people when they complain about the slowness of Mathematica. Of course with some cleverness in vectorized operations Compile could probably be avoided altogether if one so desired.
    $endgroup$
    – b3m2a1
    yesterday











  • $begingroup$
    @b3m2a1 Yeah, right. However, with ODE systems of this small dimension (dim = 2) I am not sure how to utilize vectorization since the ODE has to be solved in serial and because the vector field F may be very nonlinear.
    $endgroup$
    – Henrik Schumacher
    yesterday













15












15








15





$begingroup$

Just to give you an impression how fast things may get when you use the right tools.



For given stepsize τ and given vector field F, this creates a CompiledFunction cStep that computes a single Runge-Kutta step



F = X [Function] -Indexed[X, 2], Indexed[X, 1];

τ = 0.01;
Block[YY, Y, k1, k2, k3, k4,

YY = Table[Compile`GetElement[Y, i], i, 1, 2];
k1 = τ F[YY];
k2 = τ F[0.5 k1 + YY];
k3 = τ F[0.5 k2 + YY];
k4 = τ F[k3 + YY];

cStep = With[code = YY + (k1 + 2. (k2 + k3) + k4)/6. ,
Compile[Y, _Real, 1,
code,
CompilationTarget -> "C",
RuntimeOptions -> "Speed"
]
]
];


Now we can apply it 20 million times with NestList and it stills takes only about 2 seconds.



nsteps = 20000000;
xlist = Range[0., τ nsteps, τ];
Ylist = NestList[cStep, 1., 0., nsteps]; // AbsoluteTiming // First



2.08678




Edit



This can be sped up even more my avoiding NestList (the loop behind it can also be compiled which saves several calls to libraries) and by utilizing that the dimension of the ODE is known at compile time. For low dimensional systems, it may be also beneficial to avoid tensor operations altogether and to perform computations in scalar registers as done below.



τ = 0.01;
cFlow = Block[YY, Y, k1, k2, k3, k4, τ, Ylist, j,
YY = Table[Compile`GetElement[Ylist, j, i], i, 1, 2];
k1 = τ F[YY];
k2 = τ F[0.5 k1 + YY];
k3 = τ F[0.5 k2 + YY];
k4 = τ F[k3 + YY];
With[
code1 = (YY + (k1 + 2. (k2 + k3) + k4)/6)[[1]],
code2 = (YY + (k1 + 2. (k2 + k3) + k4)/6)[[2]]
,
Compile[Y0, _Real, 1, τ, _Real, n, _Integer,
Block[Ylist,
Ylist = Table[0., n + 1, Length[Y0]];
Ylist[[1]] = Y0;
Do[
Ylist[[j + 1, 1]] = code1;
Ylist[[j + 1, 2]] = code2;
,
j, 1, n];
Ylist
],
CompilationTarget -> "C", RuntimeOptions -> "Speed"
]
]
];
Ylist2 = cFlow[1., 0., τ, nsteps]; // AbsoluteTiming // First



1.06549




Don't be too upset by parts of the code being highlighted in red; this is on purpose.






share|improve this answer











$endgroup$



Just to give you an impression how fast things may get when you use the right tools.



For given stepsize τ and given vector field F, this creates a CompiledFunction cStep that computes a single Runge-Kutta step



F = X [Function] -Indexed[X, 2], Indexed[X, 1];

τ = 0.01;
Block[YY, Y, k1, k2, k3, k4,

YY = Table[Compile`GetElement[Y, i], i, 1, 2];
k1 = τ F[YY];
k2 = τ F[0.5 k1 + YY];
k3 = τ F[0.5 k2 + YY];
k4 = τ F[k3 + YY];

cStep = With[code = YY + (k1 + 2. (k2 + k3) + k4)/6. ,
Compile[Y, _Real, 1,
code,
CompilationTarget -> "C",
RuntimeOptions -> "Speed"
]
]
];


Now we can apply it 20 million times with NestList and it stills takes only about 2 seconds.



nsteps = 20000000;
xlist = Range[0., τ nsteps, τ];
Ylist = NestList[cStep, 1., 0., nsteps]; // AbsoluteTiming // First



2.08678




Edit



This can be sped up even more my avoiding NestList (the loop behind it can also be compiled which saves several calls to libraries) and by utilizing that the dimension of the ODE is known at compile time. For low dimensional systems, it may be also beneficial to avoid tensor operations altogether and to perform computations in scalar registers as done below.



τ = 0.01;
cFlow = Block[YY, Y, k1, k2, k3, k4, τ, Ylist, j,
YY = Table[Compile`GetElement[Ylist, j, i], i, 1, 2];
k1 = τ F[YY];
k2 = τ F[0.5 k1 + YY];
k3 = τ F[0.5 k2 + YY];
k4 = τ F[k3 + YY];
With[
code1 = (YY + (k1 + 2. (k2 + k3) + k4)/6)[[1]],
code2 = (YY + (k1 + 2. (k2 + k3) + k4)/6)[[2]]
,
Compile[Y0, _Real, 1, τ, _Real, n, _Integer,
Block[Ylist,
Ylist = Table[0., n + 1, Length[Y0]];
Ylist[[1]] = Y0;
Do[
Ylist[[j + 1, 1]] = code1;
Ylist[[j + 1, 2]] = code2;
,
j, 1, n];
Ylist
],
CompilationTarget -> "C", RuntimeOptions -> "Speed"
]
]
];
Ylist2 = cFlow[1., 0., τ, nsteps]; // AbsoluteTiming // First



1.06549




Don't be too upset by parts of the code being highlighted in red; this is on purpose.







share|improve this answer














share|improve this answer



share|improve this answer








edited yesterday

























answered 2 days ago









Henrik SchumacherHenrik Schumacher

58.3k580160




58.3k580160







  • 1




    $begingroup$
    Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
    $endgroup$
    – Shinaolord
    2 days ago










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    2 days ago










  • $begingroup$
    I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
    $endgroup$
    – Shinaolord
    2 days ago






  • 1




    $begingroup$
    This is exactly the kind of thing I like to show people when they complain about the slowness of Mathematica. Of course with some cleverness in vectorized operations Compile could probably be avoided altogether if one so desired.
    $endgroup$
    – b3m2a1
    yesterday











  • $begingroup$
    @b3m2a1 Yeah, right. However, with ODE systems of this small dimension (dim = 2) I am not sure how to utilize vectorization since the ODE has to be solved in serial and because the vector field F may be very nonlinear.
    $endgroup$
    – Henrik Schumacher
    yesterday












  • 1




    $begingroup$
    Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
    $endgroup$
    – Shinaolord
    2 days ago










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    2 days ago










  • $begingroup$
    I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
    $endgroup$
    – Shinaolord
    2 days ago






  • 1




    $begingroup$
    This is exactly the kind of thing I like to show people when they complain about the slowness of Mathematica. Of course with some cleverness in vectorized operations Compile could probably be avoided altogether if one so desired.
    $endgroup$
    – b3m2a1
    yesterday











  • $begingroup$
    @b3m2a1 Yeah, right. However, with ODE systems of this small dimension (dim = 2) I am not sure how to utilize vectorization since the ODE has to be solved in serial and because the vector field F may be very nonlinear.
    $endgroup$
    – Henrik Schumacher
    yesterday







1




1




$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
2 days ago




$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
2 days ago












$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 days ago




$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 days ago












$begingroup$
I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
$endgroup$
– Shinaolord
2 days ago




$begingroup$
I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
$endgroup$
– Shinaolord
2 days ago




1




1




$begingroup$
This is exactly the kind of thing I like to show people when they complain about the slowness of Mathematica. Of course with some cleverness in vectorized operations Compile could probably be avoided altogether if one so desired.
$endgroup$
– b3m2a1
yesterday





$begingroup$
This is exactly the kind of thing I like to show people when they complain about the slowness of Mathematica. Of course with some cleverness in vectorized operations Compile could probably be avoided altogether if one so desired.
$endgroup$
– b3m2a1
yesterday













$begingroup$
@b3m2a1 Yeah, right. However, with ODE systems of this small dimension (dim = 2) I am not sure how to utilize vectorization since the ODE has to be solved in serial and because the vector field F may be very nonlinear.
$endgroup$
– Henrik Schumacher
yesterday




$begingroup$
@b3m2a1 Yeah, right. However, with ODE systems of this small dimension (dim = 2) I am not sure how to utilize vectorization since the ODE has to be solved in serial and because the vector field F may be very nonlinear.
$endgroup$
– Henrik Schumacher
yesterday

















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