Ways to speed up user implemented RK4Speed 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
Student evaluations of teaching assistants
Displaying the order of the columns of a table
Is a roofing delivery truck likely to crack my driveway slab?
Where in the Bible does the greeting ("Dominus Vobiscum") used at Mass come from?
Was the picture area of a CRT a parallelogram (instead of a true rectangle)?
The plural of 'stomach"
What are the ramifications of creating a homebrew world without an Astral Plane?
apt-get update is failing in debian
How do I rename a LINUX host without needing to reboot for the rename to take effect?
Curses work by shouting - How to avoid collateral damage?
Valid Badminton Score?
What will be the benefits of Brexit?
How does a character multiclassing into warlock get a focus?
Why "be dealt cards" rather than "be dealing cards"?
Should my PhD thesis be submitted under my legal name?
Irreducibility of a simple polynomial
Tiptoe or tiphoof? Adjusting words to better fit fantasy races
Hide Select Output from T-SQL
Finding all intervals that match predicate in vector
What is the term when two people sing in harmony, but they aren't singing the same notes?
Implement the Thanos sorting algorithm
Everything Bob says is false. How does he get people to trust him?
Greatest common substring
Was Spock the First Vulcan in Starfleet?
Ways to speed up user implemented RK4
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
$begingroup$
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!
differential-equations numerical-integration performance-tuning
$endgroup$
|
show 2 more comments
$begingroup$
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!
differential-equations numerical-integration performance-tuning
$endgroup$
1
$begingroup$
AppendTo
is quadratic time complexity. Might be better to preallocate and set by index. Also it'll be much faster to not useRule
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
5 hours 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
5 hours ago
$begingroup$
Shinaoloard, usingJoin[ 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 aTable
anyways.
$endgroup$
– Henrik Schumacher
5 hours ago
$begingroup$
@HenrikSchumacher do you think it would be faster to Pre-allocate a list of length nsteps, and append the values, or to join the values using table? I can obviously changeTable
toDo
to remove the time it takes to make the table list, going by b3m2a1's method, or I could useJoin
as you have suggested. I'm thinking your method may be faster, though. I've already removed theMapThread
part, I am testing the speed increase granted by that at the moment. Just curious which path you think will be faster.
$endgroup$
– Shinaolord
5 hours ago
$begingroup$
I am currently testing the speed difference between the one in the post andrk4t2[f_, valtinit_, tinit_, tfinal_, nsteps_] := Module[test, table, xlist, ylist, step, k1, k2, k3, k4, xlist = tinit; step = N[(tfinal - tinit)/(nsteps)]; ylist = valtinit; table = xlist, ylist; test = Table[ k1 = step* f[ylist] ; k2 = step*f[k1/2 + ylist]; k3 = step*f[k2/2 + ylist]; k4 = step*f[k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps]; Join[table, test] ];
$endgroup$
– Shinaolord
4 hours ago
|
show 2 more comments
$begingroup$
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!
differential-equations numerical-integration performance-tuning
$endgroup$
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
differential-equations numerical-integration performance-tuning
edited 30 mins ago
xzczd
27.4k573255
27.4k573255
asked 5 hours ago
ShinaolordShinaolord
908
908
1
$begingroup$
AppendTo
is quadratic time complexity. Might be better to preallocate and set by index. Also it'll be much faster to not useRule
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
5 hours 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
5 hours ago
$begingroup$
Shinaoloard, usingJoin[ 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 aTable
anyways.
$endgroup$
– Henrik Schumacher
5 hours ago
$begingroup$
@HenrikSchumacher do you think it would be faster to Pre-allocate a list of length nsteps, and append the values, or to join the values using table? I can obviously changeTable
toDo
to remove the time it takes to make the table list, going by b3m2a1's method, or I could useJoin
as you have suggested. I'm thinking your method may be faster, though. I've already removed theMapThread
part, I am testing the speed increase granted by that at the moment. Just curious which path you think will be faster.
$endgroup$
– Shinaolord
5 hours ago
$begingroup$
I am currently testing the speed difference between the one in the post andrk4t2[f_, valtinit_, tinit_, tfinal_, nsteps_] := Module[test, table, xlist, ylist, step, k1, k2, k3, k4, xlist = tinit; step = N[(tfinal - tinit)/(nsteps)]; ylist = valtinit; table = xlist, ylist; test = Table[ k1 = step* f[ylist] ; k2 = step*f[k1/2 + ylist]; k3 = step*f[k2/2 + ylist]; k4 = step*f[k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps]; Join[table, test] ];
$endgroup$
– Shinaolord
4 hours ago
|
show 2 more comments
1
$begingroup$
AppendTo
is quadratic time complexity. Might be better to preallocate and set by index. Also it'll be much faster to not useRule
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
5 hours 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
5 hours ago
$begingroup$
Shinaoloard, usingJoin[ 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 aTable
anyways.
$endgroup$
– Henrik Schumacher
5 hours ago
$begingroup$
@HenrikSchumacher do you think it would be faster to Pre-allocate a list of length nsteps, and append the values, or to join the values using table? I can obviously changeTable
toDo
to remove the time it takes to make the table list, going by b3m2a1's method, or I could useJoin
as you have suggested. I'm thinking your method may be faster, though. I've already removed theMapThread
part, I am testing the speed increase granted by that at the moment. Just curious which path you think will be faster.
$endgroup$
– Shinaolord
5 hours ago
$begingroup$
I am currently testing the speed difference between the one in the post andrk4t2[f_, valtinit_, tinit_, tfinal_, nsteps_] := Module[test, table, xlist, ylist, step, k1, k2, k3, k4, xlist = tinit; step = N[(tfinal - tinit)/(nsteps)]; ylist = valtinit; table = xlist, ylist; test = Table[ k1 = step* f[ylist] ; k2 = step*f[k1/2 + ylist]; k3 = step*f[k2/2 + ylist]; k4 = step*f[k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps]; Join[table, test] ];
$endgroup$
– Shinaolord
4 hours ago
1
1
$begingroup$
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
5 hours 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
5 hours 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
5 hours 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
5 hours 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
5 hours 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
5 hours ago
$begingroup$
@HenrikSchumacher do you think it would be faster to Pre-allocate a list of length nsteps, and append the values, or to join the values using table? I can obviously change
Table
to Do
to remove the time it takes to make the table list, going by b3m2a1's method, or I could use Join
as you have suggested. I'm thinking your method may be faster, though. I've already removed the MapThread
part, I am testing the speed increase granted by that at the moment. Just curious which path you think will be faster.$endgroup$
– Shinaolord
5 hours ago
$begingroup$
@HenrikSchumacher do you think it would be faster to Pre-allocate a list of length nsteps, and append the values, or to join the values using table? I can obviously change
Table
to Do
to remove the time it takes to make the table list, going by b3m2a1's method, or I could use Join
as you have suggested. I'm thinking your method may be faster, though. I've already removed the MapThread
part, I am testing the speed increase granted by that at the moment. Just curious which path you think will be faster.$endgroup$
– Shinaolord
5 hours ago
$begingroup$
I am currently testing the speed difference between the one in the post and
rk4t2[f_, valtinit_, tinit_, tfinal_, nsteps_] := Module[test, table, xlist, ylist, step, k1, k2, k3, k4, xlist = tinit; step = N[(tfinal - tinit)/(nsteps)]; ylist = valtinit; table = xlist, ylist; test = Table[ k1 = step* f[ylist] ; k2 = step*f[k1/2 + ylist]; k3 = step*f[k2/2 + ylist]; k4 = step*f[k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps]; Join[table, test] ];
$endgroup$
– Shinaolord
4 hours ago
$begingroup$
I am currently testing the speed difference between the one in the post and
rk4t2[f_, valtinit_, tinit_, tfinal_, nsteps_] := Module[test, table, xlist, ylist, step, k1, k2, k3, k4, xlist = tinit; step = N[(tfinal - tinit)/(nsteps)]; ylist = valtinit; table = xlist, ylist; test = Table[ k1 = step* f[ylist] ; k2 = step*f[k1/2 + ylist]; k3 = step*f[k2/2 + ylist]; k4 = step*f[k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps]; Join[table, test] ];
$endgroup$
– Shinaolord
4 hours ago
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Just to give you an impression how fast things may get when you use the right tools.
For given stepsize τ
and given vectorfield 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 2 million times with NestList
and still need only 2 seconds.
nsteps = 20000000;
xlist = Range[0., step nsteps, step];
Ylist = NestList[cStep, initials, nsteps]; // AbsoluteTiming // First
2.08678
$endgroup$
$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
4 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
4 hours ago
$begingroup$
I'll have to play around withCompile
, it definitely seems like a massive speed up if used correctly.
$endgroup$
– Shinaolord
4 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
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
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
$begingroup$
Just to give you an impression how fast things may get when you use the right tools.
For given stepsize τ
and given vectorfield 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 2 million times with NestList
and still need only 2 seconds.
nsteps = 20000000;
xlist = Range[0., step nsteps, step];
Ylist = NestList[cStep, initials, nsteps]; // AbsoluteTiming // First
2.08678
$endgroup$
$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
4 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
4 hours ago
$begingroup$
I'll have to play around withCompile
, it definitely seems like a massive speed up if used correctly.
$endgroup$
– Shinaolord
4 hours ago
add a comment |
$begingroup$
Just to give you an impression how fast things may get when you use the right tools.
For given stepsize τ
and given vectorfield 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 2 million times with NestList
and still need only 2 seconds.
nsteps = 20000000;
xlist = Range[0., step nsteps, step];
Ylist = NestList[cStep, initials, nsteps]; // AbsoluteTiming // First
2.08678
$endgroup$
$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
4 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
4 hours ago
$begingroup$
I'll have to play around withCompile
, it definitely seems like a massive speed up if used correctly.
$endgroup$
– Shinaolord
4 hours ago
add a comment |
$begingroup$
Just to give you an impression how fast things may get when you use the right tools.
For given stepsize τ
and given vectorfield 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 2 million times with NestList
and still need only 2 seconds.
nsteps = 20000000;
xlist = Range[0., step nsteps, step];
Ylist = NestList[cStep, initials, nsteps]; // AbsoluteTiming // First
2.08678
$endgroup$
Just to give you an impression how fast things may get when you use the right tools.
For given stepsize τ
and given vectorfield 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 2 million times with NestList
and still need only 2 seconds.
nsteps = 20000000;
xlist = Range[0., step nsteps, step];
Ylist = NestList[cStep, initials, nsteps]; // AbsoluteTiming // First
2.08678
answered 4 hours ago
Henrik SchumacherHenrik Schumacher
57.9k579159
57.9k579159
$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
4 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
4 hours ago
$begingroup$
I'll have to play around withCompile
, it definitely seems like a massive speed up if used correctly.
$endgroup$
– Shinaolord
4 hours ago
add a comment |
$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
4 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
4 hours ago
$begingroup$
I'll have to play around withCompile
, it definitely seems like a massive speed up if used correctly.
$endgroup$
– Shinaolord
4 hours ago
$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
4 hours ago
$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
4 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
4 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
4 hours ago
$begingroup$
I'll have to play around with
Compile
, it definitely seems like a massive speed up if used correctly.$endgroup$
– Shinaolord
4 hours ago
$begingroup$
I'll have to play around with
Compile
, it definitely seems like a massive speed up if used correctly.$endgroup$
– Shinaolord
4 hours ago
add a comment |
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.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
1
$begingroup$
AppendTo
is quadratic time complexity. Might be better to preallocate and set by index. Also it'll be much faster to not useRule
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
5 hours 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
5 hours 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 aTable
anyways.$endgroup$
– Henrik Schumacher
5 hours ago
$begingroup$
@HenrikSchumacher do you think it would be faster to Pre-allocate a list of length nsteps, and append the values, or to join the values using table? I can obviously change
Table
toDo
to remove the time it takes to make the table list, going by b3m2a1's method, or I could useJoin
as you have suggested. I'm thinking your method may be faster, though. I've already removed theMapThread
part, I am testing the speed increase granted by that at the moment. Just curious which path you think will be faster.$endgroup$
– Shinaolord
5 hours ago
$begingroup$
I am currently testing the speed difference between the one in the post and
rk4t2[f_, valtinit_, tinit_, tfinal_, nsteps_] := Module[test, table, xlist, ylist, step, k1, k2, k3, k4, xlist = tinit; step = N[(tfinal - tinit)/(nsteps)]; ylist = valtinit; table = xlist, ylist; test = Table[ k1 = step* f[ylist] ; k2 = step*f[k1/2 + ylist]; k3 = step*f[k2/2 + ylist]; k4 = step*f[k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps]; Join[table, test] ];
$endgroup$
– Shinaolord
4 hours ago