I have a fixed input and output for my simulink embeded function.
However I would like to compute a variable size element inside the function, (only used for calculation).
Therefore I would prefer not to declare the block as receiving or sending dynamic size signal. (or using coder.varsize)
ex:
K = find( p_var(:) == 0 ); % p_var is a vector containing some zeros
p_var(K) = []; % p_var is a therefore a varsize vector
% the variability is something I need for reason
n = length(p_var) % n is dynamic
M = ones(10,n) % M & L are also dynamic
L = ones(n,50)
G = M*L; % G is fixed of size [10*50]
Here the variable G is always fixed... but I have this type of error :
Dimension 2 is fixed on the left-hand side but varies on the right ([1 x 9] ~= [1 x :?])
Thank you for your time.
You have to define an upper bound for the size of p_var. This can be done in a couple of ways, such as using coder.varsize as shown below.
A couple of other things to note:
If p_var is an input to your function you cannot change its size,but would need a temporary variable as shown below.
You most likely do not want to use find as you have done, but should use logical indexing instead.
function G = fcn(u)
p_var = u;
coder.varsize('p_var', [10,1]);
p_var(p_var(:) == 0) = []; % p_var is therefore a varsize vector
the variability is something I need for reason
n = length(p_var); % n is dynamic
M = ones(10,n); % M & L are also dynamic
L = ones(n,50);
G = M*L; % G is fixed of size [10*50]
Related
I'm trying to set up a ampl model which clusters given points in a 2-dimensional space according to the model of Saglam et al(2005). For testing purposes I want to generate randomly some datapoints and then calculate the euclidian distance matrix for them (since I need this one). I'm aware that I could only make the distance matrix without the data points but in a later step the data points will be given and then I need to calculate the distances between each the points.
Below you'll find the code I've written so far. While loading the model I keep getting the error message "i is not defined". Since i is a subscript that should run over x1 and x1 is a parameter which is defined over the set D and have one subscript, I cannot figure out why this code should be invalid. As far as I understand, I don't have to define variables if I use them only as subscripts?
reset;
# parameters to define clustered
param m; # numbers of data points
param n; # numbers of clusters
# sets
set D := 1..m; #points to be clustered
set L := 1..n; #clusters
# randomly generate datapoints
param x1 {D} = Uniform(1,m);
param x2 {D} = Uniform(1,m);
param d {D,D} = sqrt((x1[i]-x1[j])^2 + (x2[i]-x2[j])^2);
# variables
var x {D, L} binary;
var D_l {L} >=0;
var D_max >= 0;
#minimization funcion
minimize max_clus_dis: D_max;
# constraints
subject to C1 {i in D, j in D, l in L}: D_l[l] >= d[i,j] * (x[i,l] + x[j,l] - 1);
subject to C2 {i in D}: sum{l in L} x[i,l] = 1;
subject to C3 {l in L}: D_max >= D_l[l];
So far I tried to change the line form param x1 to
param x1 {i in D, j in D} = ...
as well as
param d {x1, x2} = ...
Alas, nothing of this helped. So, any help someone can offer is deeply appreciated. I searched the web but I found nothing useful for my task.
I found eventually what was missing. The line in which I calculated the parameter d should be
param d {i in D, j in D} = sqrt((x1[i]-x1[j])^2 + (x2[i]-x2[j])^2);
Retrospectively it's clear that the subscripts i and j should have been mentioned on the line, I don't know how I could miss that.
I am new to GNUPLOT. I am trying to plot 3d vector fields. However I am having trouble defining a function of three variables f(x,y,z). Can anyone show me how to do this correctly?
Defining your own functions in gnuplot is pretty intuitive. According to the gnuplot documentation the syntax is as follows
<func-name>( <dummy1> {,<dummy2>} ... {,<dummy5>} ) = <expression>
Examples:
w = 2
q = floor(tan(pi/2 - 0.1))
f(x) = sin(w*x)
sinc(x) = sin(pi*x)/(pi*x)
delta(t) = (t == 0)
ramp(t) = (t > 0) ? t : 0
min(a,b) = (a < b) ? a : b
comb(n,k) = n!/(k!*(n-k)!)
len3d(x,y,z) = sqrt(x*x+y*y+z*z)
plot f(x) = sin(x*a), a = 0.2, f(x), a = 0.4, f(x)
There is also a large set of built-in mathematical functions which you can use (in the definition of your own function).
For piecewise defined functions you can use the fact that undefined values are ignored. Therefore, the function
y(x) = x < 0 ? 1/0 : x
is only defined for positive arguments.
Powers are defined by **. Hence f(x)=x*x is identical to f(x)=x**2
If you have still problems in defining your own function, please feel free to ask. (Shouldn't a 3d-function only depend on x and y, i.e., f(x,y)=...?)
For examples of 3d plots, also see the gnuplot demo site.
I am trying to follow the tutorial of using the optimization tool box in MATLAB. Specifically, I have a function
f = exp(x(1))*(4*x(1)^2+2*x(2)^2+4*x(1)*x(2)+2*x(2)+1)+b
subject to the constraint:
(x(1))^2+x(2)-1=0,
-x(1)*x(2)-10<=0.
and I want to minimize this function for a range of b=[0,20]. (That is, I want to minimize this function for b=0, b=1,b=2 ... and so on).
Below is the steps taken from the MATLAB's tutorial webpage(http://www.mathworks.com/help/optim/ug/nonlinear-equality-and-inequality-constraints.html), how should I change the code so that, the optimization will run for 20 times, and save the optimal values for each b?
Step 1: Write a file objfun.m.
function f = objfun(x)
f = exp(x(1))*(4*x(1)^2+2*x(2)^2+4*x(1)*x(2)+2*x(2)+1)+b;
Step 2: Write a file confuneq.m for the nonlinear constraints.
function [c, ceq] = confuneq(x)
% Nonlinear inequality constraints
c = -x(1)*x(2) - 10;
% Nonlinear equality constraints
ceq = x(1)^2 + x(2) - 1;
Step 3: Invoke constrained optimization routine.
x0 = [-1,1]; % Make a starting guess at the solution
options = optimoptions(#fmincon,'Algorithm','sqp');
[x,fval] = fmincon(#objfun,x0,[],[],[],[],[],[],...
#confuneq,options);
After 21 function evaluations, the solution produced is
x, fval
x =
-0.7529 0.4332
fval =
1.5093
Update:
I tried your answer, but I am encountering problem with your step 2. Bascially, I just fill the my step 2 to your step 2 (below the comment "optimization just like before").
%initialize list of targets
b = 0:1:20;
%preallocate/initialize result vectors using zeros (increases speed)
opt_x = zeros(length(b));
opt_fval = zeros(length(b));
>> for idx = 1, length(b)
objfun = #(x)objfun_builder(x,b)
%optimization just like before
x0 = [-1,1]; % Make a starting guess at the solution
options = optimoptions(#fmincon,'Algorithm','sqp');
[x,fval] = fmincon(#objfun,x0,[],[],[],[],[],[],...
#confuneq,options);
%end the stuff I fill in
opt_x(idx) = x
opt_fval(idx) = fval
end
However, it gave me the output is:
Error: "objfun" was previously used as a variable, conflicting
with its use here as the name of a function or command.
See "How MATLAB Recognizes Command Syntax" in the MATLAB
documentation for details.
There are two things you need to change about your code:
Creation of the objective function.
Multiple optimizations using a loop.
1st Step
For more flexibility with regard to b, you need to set up another function that returns a handle to the desired objective function, e.g.
function h = objfun_builder(x, b)
h = #(x)(objfun(x));
function f = objfun(x)
f = exp(x(1))*(4*x(1)^2+2*x(2)^2+4*x(1)*x(2)+2*x(2)+1) + b;
end
end
A more elegant and shorter approach are anonymous functions, e.g.
objfun_builder = #(x,b)(exp(x(1))*(4*x(1)^2+2*x(2)^2+4*x(1)*x(2)+2*x(2)+1) + b);
After all, this works out to be the same as above. It might be less intuitive for a Matlab-beginner, though.
2nd Step
Instead of placing an .m-file objfun.m in your path, you will need to call
objfun = #(x)(objfun_builder(x,myB));
to create an objective function in your workspace. In order to loop over the interval b=[0,20], use the following loop
%initialize list of targets
b = 0:1:20;
%preallocate/initialize result vectors using zeros (increases speed)
opt_x = zeros(length(b))
opt_fval = zeros(length(b))
%start optimization of list of targets (`b`s)
for idx = 1, length(b)
objfun = #(x)objfun_builder(x,b)
%optimization just like before
opt_x(idx) = x
opt_fval(idx) = fval
end
Let's say we have a set : {1, 2, ..., n}.
How many subsets of order R exist S = {a_i1, a_i2, ...a_iR} that sum up to a certain number S?. What is the recursion for this problem?
Just define method to solve original problem. Parameters it receives are:
max number to use (n),
subset size (R),
subset sum (S),
and returns number of combinations.
To implement this method, first we have to check is it possible to make this request. It is not possible to fulfill task if:
subset size is larger than number of possible elements (R > n)
maximal possible sum is smaller than S. n + (n-1) + ... + (n-R+1) < S => R*((n-R) + (R+1)/2) < S
After that it is enough to try all possibilities for larger element that will go in subset. In python style it should be implemented like:
def combinations(n, R, S):
if R > n or R*((n-R) + (R+1)/2) < S:
return 0
c = 0
for i in xrange(R, n+1): # try i as maximal element in subset. It can go from R to n
# recursion n is i-1, since i is already used
# recursion R is R-1, since we put i in a set
# recursion S is S-i, since i is added to a set and we are looking for sum without it
c += combinations(i-1, R-1, S-i)
return c
I am trying to write an algorithm in python for knapsack problem. I did quite a few iterations and came to the following solution. It seems perfect for me. When I ran it on test sets,
it does not output optimal value
and sometimes gives maximum recursion depth error.
after changing the maxValues function it is outputting the optimal value, But it takes very very long time for datasets having more points. how to refine it
For the second problem, I have inspected the data for which it gives the error. The data is like huge and only just couple of them exceeds and the knapsack capacity. So it unnecessarily goes through the entire list.
So what I planned to do is at the start of running my recursive function, I tried to see the entire weights list where each weight is less than the current capacity and prune the rest. The following is the code I am planning to implement.
#~ weights_jump = find_indices(temp_w, lambda e: e < capacity)
#~ if(len(weights_jump)>0):
#~ temp_w[0:weights_jump[0]-1] = []
#~ temp_v[0:weights_jump[0]-1] = []
My main problem remains that why it is not outputting the optimal value. Please help me in this regards and also to integrate the above code into the current algorithm
The following is the main function. the input for this function is as follows,
A knapsack input contains n + 1 lines. The first line contains two integers, the first is the number
of items in the problem, n. The second number is the capacity of the knapsack, K. The remaining
lines present the data for each of the items. Each line, i ∈ 0 . . . n − 1 contains two integers, the
item’s value vi followed by its weight wi
eg input:
n K
v_0 w_0
v_1 w_1
...
v_n-1 w_n-1
def solveIt(inputData):
# parse the input
lines = inputData.split('\n')
firstLine = lines[0].split()
items = int(firstLine[0])
capacity = int(firstLine[1])
K = capacity
values = []
weights = []
for i in range(1, items+1):
line = lines[i]
parts = line.split()
values.append(int(parts[0]))
weights.append(int(parts[1]))
items = len(values)
#end of parsing
value = 0
weight = 0
print(weights)
print(values)
v = node(value,weights,values,K,0,taken);
# prepare the solution in the specified output format
outputData = str(v[0]) + ' ' + str(0) + '\n'
outputData += ' '.join(map(str, v[1]))
return outputData
The following is the recursive function
I'will try to explain this recursive function. Let's say I 'm starting off with root node and now I ill have two decisions to make either to take the first element or not.
before this I will call maxValue function to see the maximum value that can be obtained following this branch. If it is less than existing_max no need to search, so prune.
i will follow left branch if the weight of the first element is less than capacity. so append(1).
updating values,weights list etc and again call node function.
so it first transverses entire left branch and then transverses right branch.
in the right im just updating the values,weights lists and calling node function.
For this function inputs are
value -- The current value of the problem, It is initially set to zero and is it goes it gets increased
weights list
values list
current capacity
current max value found by algo. If this existing_max is greater than the maximum value that can be obtained by following a branch,
there is no need to search that branch. so entire branch is pruned
existing_nodes is the list which tells whether a particular item is taken (1) or not (0)
def node(value,weights,values,capacity,existing_max,existing_nodes):
v1=[];e1=[]; #values we get from left branch
v2=[];e2=[]; #values we get from right branch
e=[];
e = existing_nodes[:];
temp_w = weights[:]
temp_v = values[:];
#first check if the list is empty
if(len(values)==0):
r = [value,existing_nodes[:]]
return r;
#centre check if this entire branch could be pruned. it checks for max value that can be obtained is more than the max value inputted to this
max_value = value+maxValue(weights,values,capacity);
print('existing _max is '+str(existing_max))
print('weight in concern '+str(weights[0])+' value is '+str(value))
if(max_value<=existing_max):
return [0,[]];
#Transversing the left branch
#Transverse only if the weight does not exceed the capacity
print colored('leftbranch','red');
#search for indices of weights where weight < capacity
#~ weights_jump = find_indices(temp_w, lambda e: e < capacity)
#~ if(len(weights_jump)>0):
#~ temp_w[0:weights_jump[0]-1] = []
#~ temp_v[0:weights_jump[0]-1] = []
if(temp_w[0]<=capacity):
updated_value = temp_v[0]+value;
k = capacity-temp_w[0];
temp_w.pop(0);
temp_v.pop(0);
e1 =e[:]
e1.append(1);
print(str(updated_value)+' '+str(k)+' ')
raw_input('press ')
v1= node(updated_value,temp_w,temp_v,k,existing_max,e1);
#after transversing left node update existing_max
if(v1[0]>existing_max):
existing_max = v1[0];
else:
v1 = [0,[]]
#Transverse the right branch
#it implies we are not including the current value so remove that from weights and values.
print('rightbranch')
#~ print(str(value)+' '+str(capacity)+' ')
raw_input("Press Enter to continue...")
weights.pop(0);
values.pop(0);
e2 =e[:];
e2.append(0);
v2 = node(value,weights,values,capacity,existing_max,e2);
if(v1[0]>v2[0]):
return v1;
else:
return v2;
The following is the helper function maxValue which is called from recursive function
def maxValues(weights,values,K):
weights = weights;
values = values;
a=[];
l = 0;
#~ print('hello');
items = len(weights);
#~ print(items);
max = 0;k = K;
for i in range(0,items):
t = (i,float(values[i])/weights[i]);
a.append(t);
#~ print(i);
a = sorted(a,key=operator.itemgetter(1),reverse=True);
#~ print(a);
length = len(a);
for (i,v) in a:
#~ print('this is loop'+str(l)+'and k is '+str(k)+'weight is '+str(weights[i]) );
if weights[i]<=k:
max = max+values[i];
w = weights[i];
#~ print('this'+str(w));
k = k-w;
if(k==0):
break;
else:
max = max+ (float(k)/weights[i])*values[i];
w = k
k = k-w;
#~ print('this is w '+str(max));
if(k==0):
break;
l= l+1;
return max;
I spent days on this and could not do anything.