Ways to speed up user implemented RK4
$begingroup$
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
$endgroup$
2
$begingroup$
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
13 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
13 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
13 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
13 hours ago
3
$begingroup$
Why not just getNDSolve
to use fourth-order Runge-Kutta to begin with?
$endgroup$
– J. M. is slightly pensive♦
8 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 9 hours ago
xzczd
27.4k573255
27.4k573255
asked 14 hours ago
ShinaolordShinaolord
1008
1008
2
$begingroup$
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
13 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
13 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
13 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
13 hours ago
3
$begingroup$
Why not just getNDSolve
to use fourth-order Runge-Kutta to begin with?
$endgroup$
– J. M. is slightly pensive♦
8 hours ago
|
show 2 more comments
2
$begingroup$
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
13 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
13 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
13 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
13 hours ago
3
$begingroup$
Why not just getNDSolve
to use fourth-order Runge-Kutta to begin with?
$endgroup$
– J. M. is slightly pensive♦
8 hours ago
2
2
$begingroup$
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
13 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
13 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
13 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
13 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
13 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
13 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
13 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
13 hours ago
3
3
$begingroup$
Why not just get
NDSolve
to use fourth-order Runge-Kutta to begin with?$endgroup$
– J. M. is slightly pensive♦
8 hours ago
$begingroup$
Why not just get
NDSolve
to use fourth-order Runge-Kutta to begin with?$endgroup$
– J. M. is slightly pensive♦
8 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 20 million times with NestList
and still need only 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 behing 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.
$endgroup$
1
$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
13 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
13 hours ago
$begingroup$
I'll have to play around withCompile
, it definitely seems like a massive speed up if used correctly.
$endgroup$
– Shinaolord
13 hours 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 operationsCompile
could probably be avoided altogether if one so desired.
$endgroup$
– b3m2a1
3 hours ago
$begingroup$
@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 fieldF
may be very nonlinear.
$endgroup$
– Henrik Schumacher
3 hours ago
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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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 20 million times with NestList
and still need only 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 behing 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.
$endgroup$
1
$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
13 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
13 hours ago
$begingroup$
I'll have to play around withCompile
, it definitely seems like a massive speed up if used correctly.
$endgroup$
– Shinaolord
13 hours 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 operationsCompile
could probably be avoided altogether if one so desired.
$endgroup$
– b3m2a1
3 hours ago
$begingroup$
@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 fieldF
may be very nonlinear.
$endgroup$
– Henrik Schumacher
3 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 20 million times with NestList
and still need only 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 behing 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.
$endgroup$
1
$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
13 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
13 hours ago
$begingroup$
I'll have to play around withCompile
, it definitely seems like a massive speed up if used correctly.
$endgroup$
– Shinaolord
13 hours 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 operationsCompile
could probably be avoided altogether if one so desired.
$endgroup$
– b3m2a1
3 hours ago
$begingroup$
@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 fieldF
may be very nonlinear.
$endgroup$
– Henrik Schumacher
3 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 20 million times with NestList
and still need only 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 behing 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.
$endgroup$
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 20 million times with NestList
and still need only 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 behing 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.
edited 3 hours ago
answered 13 hours ago
Henrik SchumacherHenrik Schumacher
58.1k579160
58.1k579160
1
$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
13 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
13 hours ago
$begingroup$
I'll have to play around withCompile
, it definitely seems like a massive speed up if used correctly.
$endgroup$
– Shinaolord
13 hours 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 operationsCompile
could probably be avoided altogether if one so desired.
$endgroup$
– b3m2a1
3 hours ago
$begingroup$
@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 fieldF
may be very nonlinear.
$endgroup$
– Henrik Schumacher
3 hours ago
add a comment |
1
$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
13 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
13 hours ago
$begingroup$
I'll have to play around withCompile
, it definitely seems like a massive speed up if used correctly.
$endgroup$
– Shinaolord
13 hours 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 operationsCompile
could probably be avoided altogether if one so desired.
$endgroup$
– b3m2a1
3 hours ago
$begingroup$
@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 fieldF
may be very nonlinear.
$endgroup$
– Henrik Schumacher
3 hours ago
1
1
$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
13 hours ago
$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
13 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
13 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
13 hours ago
$begingroup$
I'll have to play around with
Compile
, it definitely seems like a massive speed up if used correctly.$endgroup$
– Shinaolord
13 hours ago
$begingroup$
I'll have to play around with
Compile
, it definitely seems like a massive speed up if used correctly.$endgroup$
– Shinaolord
13 hours 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
3 hours ago
$begingroup$
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
3 hours ago
$begingroup$
@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
3 hours ago
$begingroup$
@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
3 hours ago
add a comment |
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2
$begingroup$
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
13 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
13 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
13 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
13 hours ago
3
$begingroup$
Why not just get
NDSolve
to use fourth-order Runge-Kutta to begin with?$endgroup$
– J. M. is slightly pensive♦
8 hours ago