The code below models a simple SIR model (used in disease control) in Mathematica. (I copied it directly from my notebook).
The equations can be solved using NDSolve and the solutions are inserted into three different functions for further use.
As can be seen the Beta term on the first line varies depending on the value of Inf[t], which is one of the three solutions of the NDSolve function.
This code works fine and I have included this in order to better explain my quesion below.
Beta = Piecewise[{{0.01, Inf[t] > 20}, {.06, Inf[t] <= 20}}];
Mu = 0.1;
Pop = 100;
ans = NDSolve[{S'[t] == -Beta S[t] Inf[t],
Inf'[t] == Beta S[t] Inf[t] - Mu Inf[t],
R'[t] == Mu Inf[t],
S[0] == Pop - 1, Inf[0] == 1,
R[0] == 0}, {S[t], Inf[t], R[t]}, {t, 0, 10}];
Sus[t_] = S[t] /. ans[[1, 1]];
Infected[t_] = Inf[t] /. ans[[1, 2]];
Rec[t_] = R[t] /. ans[[1, 3]];
I now wanted to update the code so that instead of having an either/or value for the Beta parameter based on the Inf[t] value, I would have the Beta value being equal to the output of a function where Inf[t] is the input. This can be seen below where UpdateTransmission[] is the function.
When I try and run the code below though the Beta value remains at 0 and does not increase. The problem is not with the UpdateTransmission function as I have tested this independently.
Beta = UpdateTransmission[SpinMatrix, ThresholdMatrix, Inf[t]];
Mu = 0.1;
Pop = 100;
ans = NDSolve[{S'[t] == -Beta S[t] Inf[t],
Inf'[t] == Beta S[t] Inf[t] - Mu Inf[t],
R'[t] == Mu Inf[t], S[0] == Pop - 1, Inf[0] == 1,
R[0] == 0},
{S[t], Inf[t], R[t]}, {t, 0, 10}];
Sus[t_] = S[t] /. ans[[1, 1]];
Infected[t_] = Inf[t] /. ans[[1, 2]];
Rec[t_] = R[t] /. ans[[1, 3]];
Plot[{Sus[t], Infected[t], Rec[t]}, {t, 0, 5}]
Can anyone shed some light on why this may not be running correctly?
Edit: here is the updated function
UpdateTransmission[S_, Th_, Infect_] := Module[{BetaOverall},
P = S;
For[i = 1, i <= Pop, i++,
P[[i]] = Sign[Infect - Th[[i]]];];
BetaOverall = ((Count[P, 1]*.02) + (Count[P, -1]*.5))/Pop
]
Here are the two lists that are referred to in the code above:
SpinMatrix = Table[-1, {Pop}]
val := RandomReal[NormalDistribution[.5, .1]]
ThresholdMatrix = Table[Pop*val, {Pop}]
Edit Edit
Ok I've put everything together and tried to plot my three curves. Now as can be seen here they are all flat-lining. The Sus[t] line is staying at 100 whilst the other two seem to be staying below 1. What should be happening here is that the Sus[t] line should drop considerably and the other two lines should ramp up.
(I tried to insert and image but I can't as I don't have the reputation points required so I'll just past in the code and you can see the plot yourself on your own machine)
Pop = 100;
SpinMatrix = Table[-1, {Pop}];
val := RandomReal[NormalDistribution[.5, .1]];
ThresholdMatrix = Table[Pop*val, {Pop}];
updateTransmission[S_, Th_, Infect_] := Module[{}, P = S;
For[i = 1, i <= Pop, i++, P[[i]] = Sign[Infect - Th[[i]]];];
Return[((Count[P, 1]*.02) + (Count[P, -1]*.5))/Pop]];
beta[t_] := updateTransmission[SpinMatrix, ThresholdMatrix, Inf[t]];
mu = 0.1;
ans = NDSolve[{S'[t] == -beta[t] S[t] Inf[t],
Inf'[t] == beta[t] S[t] Inf[t] -
mu Inf[t], R'[t] == mu Inf[t], S[0] == Pop - 1, Inf[0] == 1,
R[0] == 0}, {S[t], Inf[t], R[t]}, {t, 0, 10}];
Sus[t_] = S[t] /. First#ans;
Infected[t_] = Inf[t] /. First#ans;
Rec[t_] = R[t] /. First#ans;
Plot[{Sus[t], Infected[t], Rec[t]}, {t, 0, 10}]
The output that I am expecting should look similar to that of the code given below:
Beta = Piecewise[{{0.5, Inf[t] > 20}, {.02, Inf[t] <= 20}}];
Mu = 0.1;
Pop = 100;
ans = NDSolve[{S'[t] == -Beta S[t] Inf[t],
Inf'[t] == Beta S[t] Inf[t] - Mu Inf[t],
R'[t] == Mu Inf[t], S[0] == Pop - 1, Inf[0] == 1,
R[0] == 0}, {S[t], Inf[t], R[t]}, {t, 0, 10}];
Sus[t_] = S[t] /. ans[[1, 1]];
Infected[t_] = Inf[t] /. ans[[1, 2]];
Rec[t_] = R[t] /. ans[[1, 3]];
Plot[{Sus[t], Infected[t], Rec[t]}, {t, 0, 10}]
The culprit is Sign[ ]
I don't know why, but I traced the problem to the Sign[ ] function that is not working properly inside NDSolve!
Removing it:
Pop = 100;
SpinMatrix = Table[-1, {Pop}];
val := RandomReal[NormalDistribution[.5, .1]];
ThresholdMatrix = Table[Pop*val, {Pop}];
updateTransmission[Th_, Inf_] :=
Total[Table[If[Inf >= Th[[i]], 2/100, 1/2]/Pop , {i, Pop}]];
beta[t_] := updateTransmission[ThresholdMatrix, Inf[t]];
mu = 0.1;
ans = NDSolve[{
S'[t] == -beta[t] S[t] Inf[t],
Inf'[t] == beta[t] S[t] Inf[t] - mu Inf[t],
R'[t] == mu Inf[t],
S[0] == Pop - 1,
R[0] == 0,
Inf[0] == 1}, {S[t], Inf[t], R[t]}, {t, 0, 10}];
Sus[t_] = S[t] /. First#ans;
Infected[t_] = Inf[t] /. First#ans;
Rec[t_] := R[t] /. First#ans;
Plot[{Sus[t], Infected[t], Rec[t]}, {t, 0, 10}]
Gives:
Probably someone with better knowledge of Mma could explain what is happening in your code.
HTH!
In some ways, you are encountering the difference between Set (=) and SetDelayed (:=). For instance, if you wrote f = 7, f becomes 7 in all occurrences of f after it was initialized. But, if you wrote f = 7 t instead, and tried to use it as you would a function, i.e. f[8], you'd get (7 t)[8] because Set says that the value of f is unchanging. SetDelayed, however, implies that the value of f will change and must be reevaluated every time it occurs. Your initial case, though, is special.
When you wrote
Beta = Piecewise[{{0.01, Inf[t] > 20}, {.06, Inf[t] <= 20}}]
Inf[t] was undefined, so that it remained unevaluated. But, every occurence of Beta in your differential equations was replaced by the above formula, courtesy of Set, so NDSolve only saw the Piecewise functions. In your second case, you wrote
Beta = UpdateTransmission[Inf[t]]
Here the problem is that UpdateTransmission is executed only when Beta is initially defined, and while Piecewise remains unevaluated, UpdateTransmission most likely still gives a result for a purely symbolic input. I'd try one of three things,
replace every occurrence of Beta in you equations with UpdateTransmission[Inf[t]],
redefine Beta using SetDelayed, e.g.
Beta := UpdateTransmission[Inf[t]]
so that it will be reevaluated every time it is encountered, or
redefine UpdateTransmission to not accept symbols via either
UpdateTransmission[x_?(Head[#]=!=Symbol&)] := ...
or
UpdateTransmission[x_] /; Head[x]=!= Symbol := ...
Option 3 works by forcing UpdateTransmission[Inf[t]] to remain unevaluated, and effectively does the same thing as option 1. But, it requires a minimum of change. Personally, I'm in favor of options 1 or 3, as I don't know how many times Beta will need to be reevaluated as NDSolve operates.
Related
Below code is used to solve a stochastic equation numerically in Mathematica for one particle. I wonder if there is a way to generalize it to the case of more than one particle and average over them. Is there anyone who knows how to do that?
Clear["Global`*"]
{ a = Pi, , b = 2 Pi, l = 5, k = 1};
ic = x#tbegin == 1;
tbegin = 1;
tend = 400;
interval = {1, 25};
lst := NestWhileList[(# + RandomVariate#TruncatedDistribution[interval,
StableDistribution[1, 0.3, 0, 0, 1]]) &, tbegin, # < tend &];
F[t_] := Piecewise[{{k, Or ## #}}, -k] &[# <= t < #2 & ###
Partition[lst, 2]];
eqn := x'[t] == (F#(t)) ;
sol = NDSolveValue[{eqn, ic}, x, {t, tbegin, tend},
MaxSteps -> Infinity];
Plot[sol#t, {t, tbegin, tend}]
First[First[sol]]
Plot[sol'[t], {t, tbegin, tend}]
Plot[F[t], {t, tbegin, tend}]
I am solving a project in Mathematica 10 and I think that the best way to do it is using a loop like For or Do. After build it I obtain the results I looking for but with a to much big multiplicity. Here is the isolated part of the code:
(*Initializing variables*)
epot[0] = 1; p[0] = 1; \[Psi][0] = HermiteH[0, x] E^(-(x^2/2));
e[n_] := e[n] = epot[n];
(*Defining function*)
\[Psi][n_] := \[Psi][n] = (Sum[p[k]*x^k,{k,0,4*n}]) [Psi][0];
(*Differential equation*)
S = - D[D[\[Psi][n], x], x] + x^2 \[Psi][n] + x^4 \[Psi][n - 1] - Sum[e[n-k]*\[Psi][k],{k,0,n}];
(*Construction of the loop*)
S1 = Collect[E^(x^2/2) S, x, Simplify];
c = Coefficient[S1, x, 0];
sol = Solve[c == 0, epot[n]]; e[n] = epot[n] /. sol;
For[j = 1, j <= 4 n, j++,
c = Coefficient[S1, x, j];
sol = Solve[c == 0, p[j]];
p[j] = p[j] /. sol;];
(*Results*)
Print[Subscript[e, n], "= ", e[n] // InputForm];
Subscript[e, 1]= {{{3/4}}}
Print[ArrayDepth[e[n]]];
3 (*Multiplicity, it should be 1*)
Print[Subscript[\[Psi], n], "= ", \[Psi][n]];
Subscript[\[Psi], 1]= {{E^(-(x^2/2)) (1-(3 x^2)/8-x^4/8)}}
Print[ArrayDepth[\[Psi][n]]];
2 (*Multiplicity, it should be 1*)
After this calculation, the question remaining is how do i substitute this results in the original functions. Thank you very much.
I am trying to solve an D-equation and do not know y[0], but I know y[x1]=y1.
I want to solve the DSolve only in the relevant xrange x=[x1, infinitny].
How could it work?
Attached the example that does not work
dsolv2 = DSolve[{y'[x] == c*0.5*t12[x, l2]^2 - alpha*y[x], y[twhenrcomesin] == zwhenrcomesin, x >= twhenrcomesin}, y[x], x]
dsolv2 = Flatten[dsolv2]
zsecondphase[x_] = y[x] /. dsolv2[[1]]
I am aware that DSolve does not allow the inequality condition but I put it in to explain you what I am looking for (t12[x,l2] will give me a value only depending on x since l2 is known).
EDIT
t12[j24_, lambda242_] := (cinv1 - cinv2)/(cop2 - cop1 + (h2*lambda242)*E^(p*j24));
cinv1 = 30; cinv2 = 4; cinv3 = 3; h2 = 1.4; h3 = 1.2; alpha = 0.04; z = 50; p = 0.06; cop1 = 0; cop2 = 1; cop3 = 1.3; teta2 = 0.19; teta3 =0.1; co2 = -0.6; z0 = 10;l2 = 0.1;
Your equation is first order and linear, so you can get a very general solution :
generic = DSolve[{y'[x] == f[x] - alpha*y[x], y[x0] == y0}, y[x], x]
Then you can substitute your specific term :
c = 1;
x0 = 1;
y0 = 1;
solution[x_] = generic[[1, 1, 2]] /. {f[x_] -> c*0.5*t12[x, l2]^2}
Plot[solution[x], {x, x0, 100}]
What is wrong with this example?
t12[x_] := Exp[-x .01] Sin[x];
dsolv2 = Chop#DSolve[{y'[x] == c*0.5*t12[x]^2 - alpha*y[x], y[1] == 1}, y[x], x];
Plot[y[x] /. dsolv2[[1]] /. {alpha -> 1, c -> 1}, {x, 1, 100}, PlotRange -> Full]
Edit
Regarding your commentary:
Try using a piecewise function to restrict the domain:
t12[x_] := Piecewise[{{ Exp[-x .01] Sin[x], x >= 1}, {Indeterminate, True}}] ;
dsolv2 = Chop#DSolve[{y'[x] == c*0.5*t12[x]^2 - alpha*y[x], y[1] == 1}, y[x], x];
Plot[y[x] /. dsolv2[[1]] /. {alpha -> 1, c -> 1}, {x, 1, 100}, PlotRange -> Full]
I'm learning about dynamic programming via the 0-1 knapsack problem.
I'm getting some weird Nulls out from the function part1. Like 3Null, 5Null etc. Why is this?
The code is an implementation of:
http://www.youtube.com/watch?v=EH6h7WA7sDw
I use a matrix to store all the values and keeps, dont know how efficient this is since it is a list of lists(indexing O(1)?).
This is my code:
(* 0-1 Knapsack problem
item = {value, weight}
Constraint is maxweight. Objective is to max value.
Input on the form:
Matrix[{value,weight},
{value,weight},
...
]
*)
lookup[x_, y_, m_] := m[[x, y]];
part1[items_, maxweight_] := {
nbrofitems = Dimensions[items][[1]];
keep = values = Table[0, {j, 0, nbrofitems}, {i, 1, maxweight}];
For[j = 2, j <= nbrofitems + 1, j++,
itemweight = items[[j - 1, 2]];
itemvalue = items[[j - 1, 1]];
For[i = 1, i <= maxweight, i++,
{
x = lookup[j - 1, i, values];
diff = i - itemweight;
If[diff > 0, y = lookup[j - 1, diff, values], y = 0];
If[itemweight <= i ,
{If[x < itemvalue + y,
{values[[j, i]] = itemvalue + y; keep[[j, i]] = 1;},
{values[[j, i]] = x; keep[[j, i]] = 0;}]
},
y(*y eller x?*)]
}
]
]
{values, keep}
}
solvek[keep_, items_, maxweight_] :=
{
(*w=remaining weight in knapsack*)
(*i=current item*)
w = maxweight;
knapsack = {};
nbrofitems = Dimensions[items][[1]];
For[i = nbrofitems, i > 0, i--,
If[keep[[i, w]] == 1, {Append[knapsack, i]; w -= items[[i, 2]];
i -= 1;}, i - 1]];
knapsack
}
Clear[keep, v, a, b, c]
maxweight = 5;
nbrofitems = 3;
a = {5, 3};
b = {3, 2};
c = {4, 1};
items = {a, b, c};
MatrixForm[items]
Print["Results:"]
results = part1[items, 5];
keep = results[[1]];
Print["keep:"];
Print[keep];
Print["------"];
results2 = solvek[keep, items, 5];
MatrixForm[results2]
(*MatrixForm[results[[1]]]
MatrixForm[results[[2]]]*)
{{{0,0,0,0,0},{0,0,5 Null,5 Null,5 Null},{0,3 Null,5 Null,5 Null,8 Null},{4 Null,4 Null,7 Null,9 Null,9 Null}},{{0,0,0,0,0},{0,0,Null,Null,Null},{0,Null,0,0,Null},{Null,Null,Null,Null,Null}}}
While your code gives errors here, the Null problem occurs because For[] returns Null. So add a ; at the end of the outermost For statement in part1 (ie, just before {values,keep}.
As I said though, the code snippet gives errors when I run it.
In case my answer isn't clear, here is how the problem occurs:
(
Do[i, {i, 1, 10}]
3
)
(*3 Null*)
while
(
Do[i, {i, 1, 10}];
3
)
(*3*)
The Null error has been reported by acl. There are more errors though.
Your keep matrix actually contains two matrices. You need to call solvek with the second one: solvek[keep[[2]], items, 5]
Various errors in solvek:
i -= 1 and i - 1 are more than superfluous (the latter one is a coding error anyway). The i-- in the beginning of the For is sufficient. As it is now you're decreasing i twice per iteration.
Append must be AppendTo
keep[[i, w]] == 1 must be keep[[i + 1, w]] == 1 as the keep matrix has one more row than there are items.
Not wrong but superfluous: nbrofitems = Dimensions[items][[1]]; nbrofitems is already globally defined
The code of your second part could look like:
solvek[keep_, items_, maxweight_] :=
Module[{w = maxweight, knapsack = {}, nbrofitems = Dimensions[items][[1]]},
For[i = nbrofitems, i > 0, i--,
If[keep[[i + 1, w]] == 1, AppendTo[knapsack, i]; w -= items[[i, 2]]]
];
knapsack
]
Can you do something like Python's yield statement in Mathematica, in order to create generators? See e.g. here for the concept.
Update
Here's an example of what I mean, to iterate over all permutations, using only O(n) space: (algorithm as in Sedgewick's Algorithms book):
gen[f_, n_] := Module[{id = -1, val = Table[Null, {n}], visit},
visit[k_] := Module[{t},
id++; If[k != 0, val[[k]] = id];
If[id == n, f[val]];
Do[If[val[[t]] == Null, visit[t]], {t, 1, n}];
id--; val[[k]] = Null;];
visit[0];
]
Then call it it like:
gen[Print,3], printing all 6 permutations of length 3.
As I have previously stated, using Compile will given faster code. Using an algorithm from fxtbook, the following code generates a next partition in lexicographic ordering:
PermutationIterator[f_, n_Integer?Positive, nextFunc_] :=
Module[{this = Range[n]},
While[this =!= {-1}, f[this]; this = nextFunc[n, this]];]
The following code assumes we run version 8:
ClearAll[cfNextPartition];
cfNextPartition[target : "MVM" | "C"] :=
cfNextPartition[target] =
Compile[{{n, _Integer}, {this, _Integer, 1}},
Module[{i = n, j = n, ni, next = this, r, s},
While[Part[next, --i] > Part[next, i + 1],
If[i == 1, i = 0; Break[]]];
If[i == 0, {-1}, ni = Part[next, i];
While[ni > Part[next, j], --j];
next[[i]] = Part[next, j]; next[[j]] = ni;
r = n; s = i + 1;
While[r > s, ni = Part[next, r]; next[[r]] = Part[next, s];
next[[s]] = ni; --r; ++s];
next
]], RuntimeOptions -> "Speed", CompilationTarget -> target
];
Then
In[75]:= Reap[PermutationIterator[Sow, 4, cfNextPartition["C"]]][[2,
1]] === Permutations[Range[4]]
Out[75]= True
This is clearly better in performance than the original gen function.
In[83]:= gen[dummy, 9] // Timing
Out[83]= {26.067, Null}
In[84]:= PermutationIterator[dummy, 9, cfNextPartition["C"]] // Timing
Out[84]= {1.03, Null}
Using Mathematica's virtual machine is not much slower:
In[85]:= PermutationIterator[dummy, 9,
cfNextPartition["MVM"]] // Timing
Out[85]= {1.154, Null}
Of course this is nowhere near C code implementation, yet provides a substantial speed-up over pure top-level code.
You probably mean the question to be more general but the example of iterating over permutations as given on the page you link to happens to be built in to Mathematica:
Scan[Print, Permutations[{1, 2, 3}]]
The Print there can be replaced with any function.