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Ways to speed up user implemented RK4

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3

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!

differential-equations numerical-integration

share|improve this question

asked 2 hours ago

ShinaolordShinaolord

808

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

 | 
show 1 more comment

3

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!

differential-equations numerical-integration

share|improve this question

asked 2 hours ago

ShinaolordShinaolord

808

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

 | 
show 1 more comment

$begingroup$

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

share|improve this question

asked 2 hours ago

ShinaolordShinaolord

808

$endgroup$

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

   ];

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;

share|improve this question

asked 2 hours ago

ShinaolordShinaolord

808

share|improve this question

asked 2 hours ago

ShinaolordShinaolord

808

asked 2 hours ago

ShinaolordShinaolord

808

asked 2 hours ago

ShinaolordShinaolord

808

1 Answer
1

active

oldest

votes

4

$begingroup$

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

share|improve this answer

answered 2 hours ago

Henrik SchumacherHenrik Schumacher

57.9k579159

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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4

$begingroup$

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

share|improve this answer

answered 2 hours ago

Henrik SchumacherHenrik Schumacher

57.9k579159

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

4

$begingroup$

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

share|improve this answer

answered 2 hours ago

Henrik SchumacherHenrik Schumacher

57.9k579159

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

share|improve this answer

answered 2 hours ago

Henrik SchumacherHenrik Schumacher

57.9k579159

$endgroup$

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"

     ]

    ]

  ];

nsteps = 20000000;

xlist = Range[0., step nsteps, step];

Ylist = NestList[cStep, initials, nsteps]; // AbsoluteTiming // First

share|improve this answer

answered 2 hours ago

Henrik SchumacherHenrik Schumacher

57.9k579159

answered 2 hours ago

Henrik SchumacherHenrik Schumacher

57.9k579159

answered 2 hours ago

Henrik SchumacherHenrik Schumacher

57.9k579159