## Wolves And Rabbits

### December 1, 2009

A nearby meadow supports a large population of rabbits; the rabbits eat the grass that grows abundantly in the sunny meadow, and reproduce quickly, as rabbits do. The meadow also supports a population of wolves that eat the rabbits. As the population of rabbits grows, so does the population of wolves, until there are so many wolves that they overeat the rabbits, whereupon the wolf population begins to diminish. But once the wolf population diminishes, the rabbit population is able to begin growing again, and of course as it does so does the wolf population, in a cycle that never ends. The graph at right shows the cycles, with the predator’s cycle offset and trailing the prey’s cycles.

Ecologists who study such prey-predator populations have determined that there is never a stable equilibrium, but that the populations are cyclical. In the 1920s, an American demographer named Alfred Lotka and an Italian physicist name Vito Volterra, working independently, discovered the differential equations that govern the size of the prey and predator populations:

dR/dt = R_{g} · R – R_{d} · R · W

dW/dt = W_{g} · R · W – W_{d} · W

Where:

R = number of rabbits at beginning of time period

W = number of wolves at beginning of time period

R_{g} = growth rate of rabbits

R_{d} = death rate of rabbits

W_{g} = growth rate of wolves

W_{d} = death rate of wolves

dR/dt = change in rabbit population during current time period

dW/dt = change in wolf population during current time period

Your task is to write a function that models the wolf and rabbit populations; your function should take the initial populations, growth rates, and death rates of wolves and rabbits and calculate the populations at each time step. When you are finished, you are welcome to read or run a suggested solution, or to post your solution or discuss the exercise in the comments below.

Pages: 1 2

My Haskell solution (see http://bonsaicode.wordpress.com/2009/12/01/programming-praxis-wolves-and-rabbits/ for a version with comments):

The reason, if you are curious, that the Runge-Kutta example appears in the Scheme reports is that it appeared in the Algol 60 report (which is also responsible for the title template “Revised^N Report on the Algorithmic Language X”, though neither Algol 60 nor Algol 68 never got beyond Revised^1).

The Algol 60 version is no beauty by modern standards, but here it is for comparison:

procedure RK (x,y,n,FKT,eps,eta,xE,yE,fi); value x,y; integer n;

Boolean fi; real x,eps,eta,xE; array y,yE; procedure FKT;

comment RK integrates the system y’k=fk(x,y1,y2,…,yn)(k=1,2,…n)

of differential equations with the method of Runge-Kutta with automatic

search for appropriate length of integration step. Parameters are: The

initial values x and y[k] for x and the unknown functions yk(x). The

order n of the system. The procedure FKT(x,y,n,z) which represents the

system to be integrated, i.e. the set of functions fk. The tolerance values eps

and eta which govern the accuracy of the numerical integration. The end

of the integration interval xE; The output parameter yE which represents

the solution x=xE. The Boolean variable fi, which must always be given

the value true for an isolated or first entry into RK. If however the functions

y must be available at several meshpoints x0,x1,…,xn, then the procedure

must be called repeatedly (with x=xk, xE=x(k+1), for k=0,1,…,n-1)

and then the later calls may occur with fi=false which saves computing

time. The input parameters of FKT must be x,y,z,n, the output parameter z

represents the set of derivatives z[k]=fk(x,y[1],y[2],…,y[n]) for x and

the actual y’s. A procedure comp enters as a non-local identifier;

begin

array z,y1,y2,y3[1:n]; real x1,x2,x3,H; Boolean out;

integer k,j; own real s,Hs;

procedure RK1ST (x,y,h,xe,ye); real x,h,xe; array y,ye;

comment RK1ST integrates one single Runge-Kutta step with

initial values x, y[k] which yields the output parameters xe=x+h

and ye[k], the latter being the solution at xe. Important: the

parameters n, FKT, z enter RK1ST as nonlocal entities;

begin

array w[1:n], a[1:5]; integer k,j;

a[1]:=a[2]:=a[5]:=h/2; a[3]:=a[4]:=h;

xe:=x;

for k:=1 step 1 until n do ye[k]:=w[k]:=y[k];

for j:=1 step 1 until 4 do

begin

FKT(xe,w,n,z);

xe:=x+a[j];

for k:=1 step 1 until n do

begin

w[k]:=y[k]+a[j]TIMESz[k];

ye[k]:=ye[k]+a[j+1]TIMESz[k]/3

end k

end j

end RK1ST;

Begin of program:

if fi then begin H:=xE-x; s:=0 end else H:=Hs;

out:=false;

AA: if (x+2.01TIMESH-xE)>0) EQUIVALENCE (H>0) then

begin Hs:=H; out:=true; H:=(xE-x)/2 end if;

RK1ST (x,y,2TIMESH,x1,y1);

BB: RK1ST (x,y,H,x2,y2); RK1ST (x2,y2,H,x3,y3);

for k:=1 step 1 until n do

if comp (y1[k],y3[k],eta)>eps then goto CC;

comment comp(a,b,c) is a function designator, the value of

which is the absolute value of the difference of the mantissae of a

and b, after the exponents of these quantities have been made equal

to the largest of the exponents of the originally given parameters

a, b, c;

x:=x3; if out then goto DD;

for k:=1 step 1 until n do y[k]:=y3[k];

if s=5 then begin s:=0; H:=2TIMESH end if;

s:=s+1; goto AA;

CC: H:=0.5TIMESX; out:=false; x1:=x2;

for k:=1 step 1 until n do y1[k]:=y2[k];

goto BB;

DD: for k:=1 step 1 until n do yE[k]:=y3[k]

end RK

Arrgh, indentation lost. Well, you can read it at http://www.masswerk.at/algol60/report.htm .

I must admit that I cheated a tiny bit. I took a stare at your Haskell solution, and basically did a straightforward port to Python (my “just muck around” language of choice for tiny problems like that). I used generators in lieu of laziness, but other than that, you should recognize it. It’s been quite some time since I had to implement Runge-Kutta integrators (the most recent application of this kind of stuff that I’ve done would have been simple particle systems where just Euler steps are okay for most of my purposes) so it was a nice review.

Anyway, here’s the code:

[...] [...]

here’s some Matlab code to solve the problem using the 2nd order Runge-Kutta

you can use a variable time step if you like (deltaT)

function [t, wolves, rabbits]=rkWR(R0, W0, Rg, Wg, Rd, Wd, deltaT, maxT )

h=deltaT;

tmax= maxT;

t=0:h:tmax;

nsteps = tmax/h;

wolves = zeros(nsteps+1,1);

rabbits = zeros(nsteps+1,1);

wolves(1)=W0;

rabbits(1)=R0;

for i=1:nsteps;

rk1 = h*rabbitsFunc(Rg,Rd, rabbits(i), wolves(i));

wk1 = h*wolvesFunc(Wg,Wd, rabbits(i), wolves(i));

rk2 = h*rabbitsFunc(Rg,Rd, rabbits(i)+rk1/2, wolves(i)+wk1/2);

wk2 = h*wolvesFunc(Wg,Wd, rabbits(i)+rk1/2, wolves(i)+wk1/2);

rabbits(i+1)=rabbits(i)+rk2;

wolves(i+1)=wolves(i)+wk2;

end

end

function [dWdt] = rabbitsFunc(Rg, Rd, R, W )

dWdt =Rg*R-Rd*R*W;

end

function [dRdt] = wolvesFunc(Wg, Wd, R, W)

dRdt = Wg*R*W-Wd*W;

end

to run for this example

[t,w,r] = rkWR(40, 15, 0.1, 0.005, 0.01, 0.1,1, 200);

figure, plot(t, w,’-+r’);

hold on, plot(t, r, ‘-ob’);