Ways to speed up user implemented RK4 Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Numerical solution using RK4 to solve nonlinear ODE?Speed 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

Can a Wizard take the Magic Initiate feat and select spells from the Wizard list?

Does GDPR cover the collection of data by websites that crawl the web and resell user data

“Since the train was delayed for more than an hour, passengers were given a full refund.” – Why is there no article before “passengers”?

"Destructive force" carried by a B-52?

Can the van der Waals coefficients be negative in the van der Waals equation for real gases?

Can 'non' with gerundive mean both lack of obligation and negative obligation?

How to create a command for the "strange m" symbol in latex?

Why do some non-religious people reject artificial consciousness?

Marquee sign letters

A German immigrant ancestor has a "Registration Affidavit of Alien Enemy" on file. What does that mean exactly?

What is the difference between 准时 and 按时?

Recursive calls to a function - why is the address of the parameter passed to it lowering with each call?

Is "ein Herz wie das meine" an antiquated or colloquial use of the possesive pronoun?

Who's this lady in the war room?

Why these surprising proportionalities of integrals involving odd zeta values?

What could prevent concentrated local exploration?

Coin Game with infinite paradox

Does the Pact of the Blade warlock feature allow me to customize the properties of the pact weapon I create?

Are Flameskulls resistant to magical piercing damage?

Will I be more secure with my own router behind my ISP's router?

What is the evidence that custom checks in Northern Ireland are going to result in violence?

Short story about an alien named Ushtu(?) coming from a future Earth, when ours was destroyed by a nuclear explosion

Are there any AGPL-style licences that require source code modifications to be public?

Compiling and throwing simple dynamic exceptions at runtime for JVM



Ways to speed up user implemented RK4



Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Numerical solution using RK4 to solve nonlinear ODE?Speed 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
    Mar 26 at 21:31










  • $begingroup$
    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
    Mar 26 at 21:33










  • $begingroup$
    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
    Mar 26 at 21:40






  • 3




    $begingroup$
    Why not just get NDSolve[] to use fourth-order Runge-Kutta to begin with?
    $endgroup$
    – J. M. is away
    Mar 27 at 2:46






  • 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
    Mar 27 at 13:27















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
    Mar 26 at 21:31










  • $begingroup$
    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
    Mar 26 at 21:33










  • $begingroup$
    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
    Mar 26 at 21:40






  • 3




    $begingroup$
    Why not just get NDSolve[] to use fourth-order Runge-Kutta to begin with?
    $endgroup$
    – J. M. is away
    Mar 27 at 2:46






  • 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
    Mar 27 at 13:27













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 Mar 27 at 2:17









xzczd

27.8k576258




27.8k576258










asked Mar 26 at 21:21









ShinaolordShinaolord

30519




30519







  • 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
    Mar 26 at 21:31










  • $begingroup$
    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
    Mar 26 at 21:33










  • $begingroup$
    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
    Mar 26 at 21:40






  • 3




    $begingroup$
    Why not just get NDSolve[] to use fourth-order Runge-Kutta to begin with?
    $endgroup$
    – J. M. is away
    Mar 27 at 2:46






  • 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
    Mar 27 at 13:27












  • 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
    Mar 26 at 21:31










  • $begingroup$
    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
    Mar 26 at 21:33










  • $begingroup$
    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
    Mar 26 at 21:40






  • 3




    $begingroup$
    Why not just get NDSolve[] to use fourth-order Runge-Kutta to begin with?
    $endgroup$
    – J. M. is away
    Mar 27 at 2:46






  • 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
    Mar 27 at 13:27







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
Mar 26 at 21:31




$begingroup$
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
Mar 26 at 21:31












$begingroup$
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
Mar 26 at 21:33




$begingroup$
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
Mar 26 at 21:33












$begingroup$
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
Mar 26 at 21:40




$begingroup$
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
Mar 26 at 21:40




3




3




$begingroup$
Why not just get NDSolve[] to use fourth-order Runge-Kutta to begin with?
$endgroup$
– J. M. is away
Mar 27 at 2:46




$begingroup$
Why not just get NDSolve[] to use fourth-order Runge-Kutta to begin with?
$endgroup$
– J. M. is away
Mar 27 at 2:46




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
Mar 27 at 13:27




$begingroup$
@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
Mar 27 at 13:27










1 Answer
1






active

oldest

votes


















17












$begingroup$

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
    Mar 26 at 22:07










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    Mar 26 at 22:08










  • $begingroup$
    I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
    $endgroup$
    – Shinaolord
    Mar 26 at 22:08






  • 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
    Mar 27 at 7:45







  • 1




    $begingroup$
    Might be of interest: JM's blog article tpfto.wordpress.com/2019/04/03/the-runge-kutta-gill-method
    $endgroup$
    – anderstood
    Apr 3 at 14:58











Your Answer








StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













draft saved

draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194002%2fways-to-speed-up-user-implemented-rk4%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









17












$begingroup$

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
    Mar 26 at 22:07










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    Mar 26 at 22:08










  • $begingroup$
    I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
    $endgroup$
    – Shinaolord
    Mar 26 at 22:08






  • 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
    Mar 27 at 7:45







  • 1




    $begingroup$
    Might be of interest: JM's blog article tpfto.wordpress.com/2019/04/03/the-runge-kutta-gill-method
    $endgroup$
    – anderstood
    Apr 3 at 14:58















17












$begingroup$

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
    Mar 26 at 22:07










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    Mar 26 at 22:08










  • $begingroup$
    I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
    $endgroup$
    – Shinaolord
    Mar 26 at 22:08






  • 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
    Mar 27 at 7:45







  • 1




    $begingroup$
    Might be of interest: JM's blog article tpfto.wordpress.com/2019/04/03/the-runge-kutta-gill-method
    $endgroup$
    – anderstood
    Apr 3 at 14:58













17












17








17





$begingroup$

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 Mar 27 at 14:19

























answered Mar 26 at 22:05









Henrik SchumacherHenrik Schumacher

60.8k585171




60.8k585171







  • 1




    $begingroup$
    Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
    $endgroup$
    – Shinaolord
    Mar 26 at 22:07










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    Mar 26 at 22:08










  • $begingroup$
    I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
    $endgroup$
    – Shinaolord
    Mar 26 at 22:08






  • 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
    Mar 27 at 7:45







  • 1




    $begingroup$
    Might be of interest: JM's blog article tpfto.wordpress.com/2019/04/03/the-runge-kutta-gill-method
    $endgroup$
    – anderstood
    Apr 3 at 14:58












  • 1




    $begingroup$
    Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
    $endgroup$
    – Shinaolord
    Mar 26 at 22:07










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    Mar 26 at 22:08










  • $begingroup$
    I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
    $endgroup$
    – Shinaolord
    Mar 26 at 22:08






  • 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
    Mar 27 at 7:45







  • 1




    $begingroup$
    Might be of interest: JM's blog article tpfto.wordpress.com/2019/04/03/the-runge-kutta-gill-method
    $endgroup$
    – anderstood
    Apr 3 at 14:58







1




1




$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
Mar 26 at 22:07




$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
Mar 26 at 22:07












$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Mar 26 at 22:08




$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Mar 26 at 22:08












$begingroup$
I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
$endgroup$
– Shinaolord
Mar 26 at 22:08




$begingroup$
I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
$endgroup$
– Shinaolord
Mar 26 at 22:08




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
Mar 27 at 7:45





$begingroup$
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
Mar 27 at 7:45





1




1




$begingroup$
Might be of interest: JM's blog article tpfto.wordpress.com/2019/04/03/the-runge-kutta-gill-method
$endgroup$
– anderstood
Apr 3 at 14:58




$begingroup$
Might be of interest: JM's blog article tpfto.wordpress.com/2019/04/03/the-runge-kutta-gill-method
$endgroup$
– anderstood
Apr 3 at 14:58

















draft saved

draft discarded
















































Thanks for contributing an answer to Mathematica Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid


  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194002%2fways-to-speed-up-user-implemented-rk4%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Færeyskur hestur Heimild | Tengill | Tilvísanir | LeiðsagnarvalRossið - síða um færeyska hrossið á færeyskuGott ár hjá færeyska hestinum

He _____ here since 1970 . Answer needed [closed]What does “since he was so high” mean?Meaning of “catch birds for”?How do I ensure “since” takes the meaning I want?“Who cares here” meaningWhat does “right round toward” mean?the time tense (had now been detected)What does the phrase “ring around the roses” mean here?Correct usage of “visited upon”Meaning of “foiled rail sabotage bid”It was the third time I had gone to Rome or It is the third time I had been to Rome

Bunad