Sets
i / 1, 2 /;
Parameters
j(i) / 2, 1 /;
Variables
x(i);
So, here I have an index i, a parameter that depends on i which gives the same values as i, and a variable that depends on x.
If I want to get x(2), I could of course write x(2), but what if I wanted to write x(j(1)). Since j(1) = 2, this ought to be the same, but GAMS doesn't like it, and says it expected a set.
How can I do this?
You could use a mapping:
see example below which maps x(1) to P(2) and x(2) to P(1), using the mapping j which maps 1 to 2 and 2 to 1.
Set i / 1, 2 /;
alias(k,i);
set j(i,k) /
1.2
2.1
/;
parameter P(i);
P("1") = 10;
P("2") = 20;
Variables
x(i);
loop(j(i,k), x.l(k) =P(i));
execute_unload "test.gdx";
Not sure if I understand the question correctly, but maybe you just mean x(i) = j(i) ? Then for all set elements of i x will take the same value of j. If you only want the first set element: x("1") = j("1"). j("1") = 2, so x("1") will equal 2 too.
Related
In this transformer tutorial:
https://www.tensorflow.org/text/tutorials/transformer
def get_angles(pos, i, d_model):
angle_rates = 1 / np.power(10000, (2 * (i//2)) / np.float32(d_model))
return pos * angle_rates
I don't understand why 'i//2' is used, since in the original formula there is no specification of the integer division.
So what's the purpose of i//2?
according to the formula
PE(pos, 2i) = sin(1/100000^(2i /D) , PE(pos, 2i+1) = cos(1/100000^(2i
/D)
As you can see, for odd row, 2i -> 2i, for even row 2i+1 -> 2i, for example, a word embedding with [0,1,2,3,4,5,6,7] should transfer to [0,0,1,1,2,2,3,3], there is where i//2 comes from.
The last line here results in a incorrect signature to the map call:
my #array=[0,1,2];
say "String Repetition";
say #array.map({($_ x 2)});
say #array.map: * x 2;
say "\nCross product ";
say #array.map({($_ X 2)});
say #array.map: * X 2;
say "\nList Repetition";
say #array.map({$_ xx 2});
say #array.map: * xx 2;
The output being:
String Repetition
(00 11 22)
(00 11 22)
Cross product
(((0 2)) ((1 2)) ((2 2)))
(((0 2)) ((1 2)) ((2 2)))
List Repetition
((0 0) (1 1) (2 2))
Cannot resolve caller map(Array:D: Seq:D); none of these signatures match:
($: Hash \h, *%_)
(\SELF: █; :$label, :$item, *%_)
The x operator returns a Str, the X returns a List of Lists and the xx return a List.
Is this changed somehow using the Whatever. Why is this error happening? Thanks in advance
Let me see if I can get this through clearly. If I don't, please ask.
Short answer: xx has a special meaning together with Whatever, so it's not creating a WhateverCode as in the rest of the examples.
Let's see if I can get this straight with the long answer.
First, definitions. * is called Whatever. It's generally used in situations in which it's curried
I'm not too happy with this name, which points at functional-language-currying, but does not seem to be used in that sense, but in the sense of stewing or baking. Anyway.
Currying it turns it into WhateverCode. So an * by itself is Whatever, * with some stuff is WhateverCode, creating a block out of thin air.
However, that does not happen automatically, because some times we need Whatever just be Whatever. You have a few exceptions listed on Whatever documentation. One of them is using xx, because xx together with Whatever should create infinite lists.
But that's not what I'm doing, you can say. * is in front of the number to multiply. Well, yes. But this code in Actions.nqp (which generates code from the source) refers to infix xx. So it does not really matter.
So, back to the short answer: you can't always use * together with other elements to create code. Some operators, such as that one, .. or ... will have special meaning in the proximity of *, so you'll need to use something else, like placeholder arguments.
The xx operator is “thunky”.
say( rand xx 2 );
# (0.7080396712923503 0.3938678220039854)
Notice that rand got executed twice. x and X don't do that.
say( rand x 2 );
0.133525574759261740.13352557475926174
say( rand X 1,2 );
((0.2969453468495996 1) (0.2969453468495996 2))
That is xx sees each side as something sort of like a lambda on their own.
(A “thunk”)
say (* + 1 xx 2);
# ({ ... } { ... })
say (* + 1 xx 2)».(5);
# (6 6)
So you get a sequence of * repeated twice.
say (* xx 2).map: {.^name}
# (Whatever Whatever)
(The term *, is an instance of Whatever)
This also means that you can't create a WhateverCode closure with && / and, || / or, ^^ / xor, or //.
say (* && 1);
# 1
Note that * also does something different on the right side of xx.
It creates an infinite sequence.
say ( 2 xx * ).head(20);
# (2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2)
If xx wasn't “thunky”, then this would also have created a WhateverCode lambda.
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.
First of all I see the number of strings as the following:
1 (epsilon 0 length string) + 3 (pick one letter) + 9 (3 options for first letter, 3 options for second)
For a total of 13 strings. Now as far as I know a language can pick any combination of this for example l1 = {ab,a,ac} l2 = {c}
I'm not sure how to calculate the total number of languages there could be here. Any Help?
So you have a set with 13 elements. A particular language could be any subset of this set. How many subsets does this set have?
This is called the power set of that set, and it has 213 elements.
Cardinality of character set, say d = 3.
Total words possible of length (<= k), say w = (d^(k+1) - 1)/(d-1) = 13.
Total languages possible = Power Set {Each word can be included or not} = 2^w = 8192.
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