I'm trying to write the master equations for genetic networks, as they are many equations I'm trying to make a table for writing all of them at time. However, I don't know how to adjust the boundaries, I mean:
I wrote a matrix with all the variables that I need:
p={{p11,p12},{p21,p22}}
Then I wrote a table for creating the differential equations:
Table[p[[i,j]]'[t]== p[[i-1,j]][t]+p[[i,j-1]][t]+p[[i+1,j]][t]+p[[i,j+1]][t],{i,1,2},{j,1,2}]
However the part p[[i-1,j]] when i=1 is p[[0,1]]but it doesn't exist and I need to put 0 instead of this but I dont not how. I tried with If but it doesn't work well. What can i do?
Will this work for you?
pf[i_,j_]:=If[i<1||i>2||j<1||j>2,0,p[[i,j]][t]];
Table[p[i,j]'[t]== pf[i-1,j]+pf[i,j-1]+pf[i+1,j]+pf[i,j+1],{i,1,2},{j,1,2}]
which returns
{{p[1, 1]]'[t] == p[[1,2]][t] + p[[2,1]][t], p[1, 2]]'[t] == p[[1,1]][t] + p[[2,2]][t]},
{p[2, 1]]'[t] == p[[1,1]][t] + p[[2,2]][t], p[2, 2]]'[t] == p[[1,2]][t] + p[[2,1]][t]}}
Related
I'm relatively new to Z3 and experimenting with it in python. I've coded a program which returns the order in which different actions is performed, represented with a number. Z3 returns an integer representing the second the action starts.
Now I want to look at the model and see if there is an instance of time where nothing happens. To do this I made a list with only 0's and I want to change the index at the times where each action is being executed, to 1. For instance, if an action start at the 5th second and takes 8 seconds to be executed, the index 5 to 12 would be set to 1. Doing this with all the actions and then look for 0's in the list would hopefully give me the instances where nothing happens.
The problem is: I would like to write something like this for coding the problem
list_for_check = [0]*total_time
m = s.model()
for action in actions:
for index in range(m.evaluate(action.number) , m.evaluate(action.number) + action.time_it_takes):
list_for_check[index] = 1
But I get the error:
'IntNumRef' object cannot be interpreted as an integer
I've understood that Z3 isn't returning normal ints or bools in their models, but writing
if m.evaluate(action.boolean):
works, so I'm assuming the if is overwritten in a way, but this doesn't seem to be the case with range. So my question is: Is there a way to use range with Z3 ints? Or is there another way to do this?
The problem might also be that action.time_it_takes is an integer and adding a Z3int with a "normal" int doesn't work. (Done in the second part of the range).
I've also tried using int(m.evaluate(action.number)), but it doesn't work.
Thanks in advance :)
When you call evaluate it returns an IntNumRef, which is an internal z3 representation of an integer number inside z3. You need to call as_long() method of it to convert it to a Python number. Here's an example:
from z3 import *
s = Solver()
a = Int('a')
s.add(a > 4);
s.add(a < 7);
if s.check() == sat:
m = s.model()
print("a is %s" % m.evaluate(a))
print("Iterating from a to a+5:")
av = m.evaluate(a).as_long()
for index in range(av, av + 5):
print(index)
When I run this, I get:
a is 5
Iterating from a to a+5:
5
6
7
8
9
which is exactly what you're trying to achieve.
The method as_long() is defined here. Note that there are similar conversion functions from bit-vectors and rationals as well. You can search the z3py api using the interface at: https://z3prover.github.io/api/html/namespacez3py.html
I´m having serious performance issues with a job that is running everyday and I think i cannot improve the algorithm; so I´m gonnga explain you what is the problem to solve and the algorithm we have, and maybe you have some other ideas to solve the problem better.
So the problem we have to solve is:
There is a set of Rules, ~ 120.000 Rules.
Every rule has a set of combinations of Codes. Codes are basically strings. So we have ~8 combinations per rule. Example of a combination: TTAAT;ZZUHH;GGZZU;WWOOF;SSJJW;FFFOLL
There is a set of Objects, ~800 objects.
Every object has a set of ~200 codes.
We have to check for every Rule, if there is at least one Combination of Codes that is fully contained in the Objects. It means =>
loop in Rules
Loop in Combinations of the rule
Loop in Objects
every code of the combination found in the Object? => create relationship rule/object and continue with the next object
end of loop
end of loop
end of loop
For example, if we have the Rule with this combination of two codes: HHGGT; ZZUUF
And let´s say we have an object with this codes: HHGGT; DHZZU; OIJUH; ZHGTF; HHGGT; JUHZT; ZZUUF; TGRFE; UHZGT; FCDXS
Then we create a relationship between the Object and the Rule because every code of the combination of the rule is contained in the codes of the object => this is what the algorithm has to do.
As you can see this is quite expensive, because we need 120.000 x 8 x 800 = 750 millions of times in the worst-case scenario.
This is a simplified scenario of the real problem; actually what we do in the loops is a little bit more complicated, that´s why we have to reduce this somehow.
I tried to think in a solution but I don´t have any ideas!
Do you see something wrong here?
Best regards and thank you for the time :)
Something like this might work better if I'm understanding correctly (this is in python):
RULES = [
['abc', 'def',],
['aaa', 'sfd',],
['xyy', 'eff',]]
OBJECTS = [
('rrr', 'abc', 'www', 'def'),
('pqs', 'llq', 'aaa', 'sdr'),
('xyy', 'hjk', 'fed', 'eff'),
('pnn', 'rrr', 'mmm', 'qsq')
]
MapOfCodesToObjects = {}
for obj in OBJECTS:
for code in obj:
if (code in MapOfCodesToObjects):
MapOfCodesToObjects[code].add(obj)
else:
MapOfCodesToObjects[code] = set({obj})
RELATIONS = []
for rule in RULES:
if (len(rule) == 0):
continue
if (rule[0] in MapOfCodesToObjects):
ValidObjects = MapOfCodesToObjects[rule[0]]
else:
continue
for i in range(1, len(rule)):
if (rule[i] in MapOfCodesToObjects):
codeObjects = MapOfCodesToObjects[rule[i]]
else:
ValidObjects = set()
break
ValidObjects = ValidObjects.intersection(codeObjects)
if (len(ValidObjects) == 0):
break
for vo in ValidObjects:
RELATIONS.append((rule, vo))
for R in RELATIONS:
print(R)
First you build a map of codes to objects. If there are nObj objects and nCodePerObj codes on average per object, this takes O(nObj*nCodePerObj * log(nObj*nCodePerObj).
Next you iterate through the rules and look up each code in each rule in the map you built. There is a relation if a certain object occurs for every code in the rule, i.e. if it is in the set intersection of the objects for every code in the rule. Since hash lookups have O(1) time complexity on average, and set intersection has time complexity O(min of the lengths of the 2 sets), this will take O(nRule * nCodePerRule * nObjectsPerCode), (note that is nObjectsPerCode, not nCodePerObj, the performance gets worse when one code is included in many objects).
I am relatively new to Modelica (Dymola-environment) and I am getting very desperate/upset that I cannot solve such a simple problem as a random number generation in Modelica and I hope that you can help me out.
The simple function random produces a random number between 0 and 1 with an input seed seedIn[3] and produces the output seed seedOut[3] for the next time step or event. The call
(z,seedOut) = random(seedIn);
works perfectly fine.
The problem is that I cannot find a way in Modelica to compute this assignment over time by using the seedOut[3] as the next seedIn[3], which is very frustrating.
My simple program looks like this:
*model Randomgenerator
Real z;
Integer seedIn[3]( start={1,23,131},fixed=true), seedOut[3];
equation
(z,seedOut) = random(seedIn);
algorithm
seedIn := seedOut;
end Randomgenerator;*
I have tried nearly all possibilities with algorithm assignments, initial conditions and equations but none of them works. I just simply want to use seedOut in the next time step. One problem seems to be that when entering into the algorithm section, neither the initial conditions nor the values from the equation section are used.
Using the 'sample' and 'reinit' functions the code below will calculate a new random number at the frequency specified in 'sample'. Note the way of defining the "start value" of seedIn.
model Randomgenerator
Real seedIn[3] = {1,23,131};
Real z;
Real[3] seedOut;
equation
(z,seedOut) = random(seedIn);
when sample(1,1) then
reinit(seedIn,pre(seedOut));
end when;
end Randomgenerator;
The 'pre' function allows the use of the previous value of the variable. If this was not used, the output 'z' would have returned a constant value. Two things regarding the 'reinint' function, it requires use of 'when' and requires 'Real' variables/expressions hence seedIn and seedOut are now defined as 'Real'.
The simple "random" generator I used was:
function random
input Real[3] seedIn;
output Real z;
output Real[3] seedOut;
algorithm
seedOut[1] :=seedIn[1] + 1;
seedOut[2] :=seedIn[2] + 5;
seedOut[3] :=seedIn[3] + 10;
z :=(0.1*seedIn[1] + 0.2*seedIn[2] + 0.3*seedIn[3])/(0.5*sum(seedIn));
end random;
Surely there are other ways depending on the application to perform this operation. At least this will give you something to start with. Hope it helps.
I have a huge matrix on which I need to do some matching operation. Here's the code I have written and it works fine, but I think there is some room to make it more optimized or write another code that does the matching in less time. Could you please help me with that?
rowsMatched = find(bigMatrix(:, 1) == matchingRow(1, 1)
& bigMatrix(:, 2) == matchingRow(1, 2)
& bigMatrix(:, 3) == matchingRow(1, 3))
The problem with this code is that I cannot use && operand. So, in case one of the columns do not match, the program still checks the next condition. How can I avoid this?
Update: Here's the solution to this problem:
rowsMatched = find(all(bsxfun(#eq, bigMatrix, matchingRow),2));
Thank you
You can use BSXFUN to do it in a vectorized manner:
rowsMatched = find(all(bsxfun(#eq, bigMatrix, matchingRow),2));
Notice that it will work for any number of columns, just matchingRow should have the same number of columns as bigMatrix.
When I run the following code, the two output matrices (diffInDiffOne & diffInDiffTwo) are the same. My guess is that coeffs is not being replaced after each loop but I have no idea why . I think that the coefficients matrix is being overwritten but I have no idea how. I tried changing the for loop order but this surprisingly didn't solve my issue either:
local treatments treat_one treat_two
matrix diffInDiffOne = J(1,9,.)
matrix diffInDiffTwo = J(1,9,.)
foreach treatment in `treatments' {
reg science inSchool#`treatment'#male
matrix coeffs=e(b)
if treat_one==`treatment'{
matrix diffInDiffOne = diffInDiffOne\coeffs
}
if treat_two==`treatment'{
matrix diffInDiffTwo = diffInDiffTwo\coeffs
}
}
matrix list diffInDiffOne
matrix list diffInDiffTwo
When I list the matrix they are both the same, depsite the fact that two regressions give different answers. Any help with this issue is much appreciated. Thanks
This code appears at first sight to reduce to
reg science inSchool#treat_one#male
matrix li e(b)
reg science inSchool#treat_two#male
matrix li e(b)
apart from the detail of adding nine missing values to the matrix.
However, that is not your code, so what is biting you? I guess at something much more subtle.
You should need to be very careful with the if command. Variables evaluated in if commands are evaluated in their first observation. So, the first time round the loop
the conditions are
if treat_one[1] == treat_one[1]
if treat_two[1] == treat_one[1]
The second time, it is
if treat_one[1] == treat_two[1]
if treat_two[1] == treat_two[1]
If it is true in your data that treat_one[1] == treat_two[1] the effect will not be as you may imagine.
If you want to test for equality of strings, do something like
if "`treatment'" == "treat_one"
You may have in mind something more like
foreach treatment in treat_one treat_two {
reg science inSchool#`treatment'#male
matrix `treatment' = e(b)
matrix list `treatment`
}
You seem to be wanting to write very complicated code for rather simple problems. A while back, I recommended thinking in terms of do-files rather than programs. That may be advice to reconsider.