I am trying to solve a problem, for example I have a 4 point and each two point has a cost between them. Now I want to find a sequence of nodes which total cost would be less than a bound. I have written a code but it seems not working. The main problem is I have define a python function and trying to call it with in a constraint.
Here is my code: I have a function def getVal(n1,n2): where n1, n2 are Int Sort. The line Nodes = [ Int("n_%s" % (i)) for i in range(totalNodeNumber) ] defines 4 points as Int sort and when I am adding a constraint s.add(getVal(Nodes[0], Nodes[1]) + getVal(Nodes[1], Nodes[2]) < 100) then it calls getVal function immediately. But I want that, when Z3 will decide a value for Nodes[0], Nodes[1], Nodes[2], Nodes[3] then the function should be called for getting the cost between to points.
from z3 import *
import random
totalNodeNumber = 4
Nodes = [ Int("n_%s" % (i)) for i in range(totalNodeNumber) ]
def getVal(n1,n2):
# I need n1 and n2 values those assigned by Z3
cost = random.randint(1,20)
print cost
return IntVal(cost)
s = Solver()
#constraint: Each Nodes value should be distinct
nodes_index_distinct_constraint = Distinct(Nodes)
s.add(nodes_index_distinct_constraint)
#constraint: Each Nodes value should be between 0 and totalNodeNumber
def get_node_index_value_constraint(i):
return And(Nodes[i] >= 0, Nodes[i] < totalNodeNumber)
nodes_index_constraint = [ get_node_index_value_constraint(i) for i in range(totalNodeNumber)]
s.add(nodes_index_constraint)
#constraint: Problem with this constraint
# Here is the problem it's just called python getVal function twice without assiging Nodes[0],Nodes[1],Nodes[2] values
# But I want to implement that - Z3 will call python function during his decission making of variables
s.add(getVal(Nodes[0], Nodes[1]) + getVal(Nodes[1], Nodes[2]) + getVal(Nodes[2], Nodes[3]) < 100)
if s.check() == sat:
print "SAT"
print "Model: "
m = s.model()
nodeIndex = [ m.evaluate(Nodes[i]) for i in range(totalNodeNumber) ]
print nodeIndex
else:
print "UNSAT"
print "No solution found !!"
If this is not a right way to solve the problem then could you please tell me what would be other alternative way to solve it. Can I encode this kind of problem to find optimal sequence of way points using Z3 solver?
I don't understand what problem you need to solve. Definitely, the way getVal is formulated does not make sense. It does not use the arguments n1, n2. If you want to examine values produced by a model, then you do this after Z3 returns from a call to check().
I don't think you can use a python function in your SMT logic. What you could alternatively is define getVal as a Function like this
getVal = Function('getVal',IntSort(),IntSort(),IntSort())
And constraint the edge weights as
s.add(And(getVal(0,1)==1,getVal(1,2)==2,getVal(0,2)==3))
The first two input parameters of getVal represent the node ids and the last integer represents the weight.
Related
I use Maxima for calculations. I solve a system of nonlinear equations using Newton's method (mnewton()). I get the solution in the form of a list:
[[φ2=5.921818183272879,s=5.155870949147037]]
How to get the numerical value of the first (φ2) and second (s) unknown. If I substitute:
x: roz1[1][2]$
I get that x is equal to: s=5.155870949147037
What to do to make x equal to a numerical value only: 5.155870949147037
(without s=).
My code:
Maxima code
I have two ideas. (1) You can call rhs to return the right-hand side of an equation (likewise lhs for the left-hand side). E.g. rhs(s = 123) returns 123.
(2) You can call assoc to find the value associated with s (or any variable) in the mnewton results. E.g. assoc('s, [a = 1, b = 2, s = 3, u = 5]) returns 3.
I like (2) better since it is not necessary to know where in the list is the one that you're interested in.
I am building a genetic algorithm that does a time series forecast in the symbolic regression analysis. I’m trying to get the algorithm to find an equation that will match the underlying trend of the data. (predict monthly beer sales)
The idea is to use lisp like expressions, which writes the equation in a tree. This allows for branch swapping in the crossover/mating stage.
5* (5 +5)
Written as:
X = '(mul 5 (add 5 5))'
Y = parser(X)
y = ['mul', 5, ['add', 5, 5]]
I want to know how to create an initial population set where the individuals represent different expressions automatically. Where there “fitness” is related to how well each equation matches the underlying trend.
For example, one individual could be: '(add 100 (mul x (sin (mul x 3))))'
Where x is time in months.
How do I automatically generate expressions for my population? I have no idea how to do this, any help would be very appreciated.
You can easily solve this problem with recursion and a random number generator random() which returns a (pseudo-)random float between 0 and 1. Here is some pseudocode:
randomExp() {
// Choose a function(like mul or add):
func = getRandomFunction() // Just choose one of your functions randomly.
arg1 = ""
rand1 = random()
// Choose the arguments. You may choose other percentages here depending how deep you want it to be and how many 'x' you want to have.
if(rand1 < 0.2)
arg1 = randomExp() // Here add a new expression
else if(rand1 < 0.5)
arg1 = "x"
else
arg1 = randomConstant() // Get a random constant in a predefined range.
// Do the same for the second argument:
arg2 = ""
…
…
// Put everything together and return it:
return "("+func+" "+arg1+" "+arg2+")"
}
You might want to also limit the recursion depth, as this might return you a theoretically infinitely long expression.
I have a problem where I need to do a linear interpolation on some data as it is acquired from a sensor (it's technically position data, but the nature of the data doesn't really matter). I'm doing this now in matlab, but since I will eventually migrate this code to other languages, I want to keep the code as simple as possible and not use any complicated matlab-specific/built-in functions.
My implementation initially seems OK, but when checking my work against matlab's built-in interp1 function, it seems my implementation isn't perfect, and I have no idea why. Below is the code I'm using on a dataset already fully collected, but as I loop through the data, I act as if I only have the current sample and the previous sample, which mirrors the problem I will eventually face.
%make some dummy data
np = 109; %number of data points for x and y
x_data = linspace(3,98,np) + (normrnd(0.4,0.2,[1,np]));
y_data = normrnd(2.5, 1.5, [1,np]);
%define the query points the data will be interpolated over
qp = [1:100];
kk=2; %indexes through the data
cc = 1; %indexes through the query points
qpi = qp(cc); %qpi is the current query point in the loop
y_interp = qp*nan; %this will hold our solution
while kk<=length(x_data)
kk = kk+1; %update the data counter
%perform online interpolation
if cc<length(qp)-1
if qpi>=y_data(kk-1) %the query point, of course, has to be in-between the current value and the next value of x_data
y_interp(cc) = myInterp(x_data(kk-1), x_data(kk), y_data(kk-1), y_data(kk), qpi);
end
if qpi>x_data(kk), %if the current query point is already larger than the current sample, update the sample
kk = kk+1;
else %otherwise, update the query point to ensure its in between the samples for the next iteration
cc = cc + 1;
qpi = qp(cc);
%It is possible that if the change in x_data is greater than the resolution of the query
%points, an update like the above wont work. In this case, we must lag the data
if qpi<x_data(kk),
kk=kk-1;
end
end
end
end
%get the correct interpolation
y_interp_correct = interp1(x_data, y_data, qp);
%plot both solutions to show the difference
figure;
plot(y_interp,'displayname','manual-solution'); hold on;
plot(y_interp_correct,'k--','displayname','matlab solution');
leg1 = legend('show');
set(leg1,'Location','Best');
ylabel('interpolated points');
xlabel('query points');
Note that the "myInterp" function is as follows:
function yi = myInterp(x1, x2, y1, y2, qp)
%linearly interpolate the function value y(x) over the query point qp
yi = y1 + (qp-x1) * ( (y2-y1)/(x2-x1) );
end
And here is the plot showing that my implementation isn't correct :-(
Can anyone help me find where the mistake is? And why? I suspect it has something to do with ensuring that the query point is in-between the previous and current x-samples, but I'm not sure.
The problem in your code is that you at times call myInterp with a value of qpi that is outside of the bounds x_data(kk-1) and x_data(kk). This leads to invalid extrapolation results.
Your logic of looping over kk rather than cc is very confusing to me. I would write a simple for loop over cc, which are the points at which you want to interpolate. For each of these points, advance kk, if necessary, such that qp(cc) is in between x_data(kk) and x_data(kk+1) (you can use kk-1 and kk instead if you prefer, just initialize kk=2 to ensure that kk-1 exists, I just find starting at kk=1 more intuitive).
To simplify the logic here, I'm limiting the values in qp to be inside the limits of x_data, so that we don't need to test to ensure that x_data(kk+1) exists, nor that x_data(1)<pq(cc). You can add those tests in if you wish.
Here's my code:
qp = [ceil(x_data(1)+0.1):floor(x_data(end)-0.1)];
y_interp = qp*nan; % this will hold our solution
kk=1; % indexes through the data
for cc=1:numel(qp)
% advance kk to where we can interpolate
% (this loop is guaranteed to not index out of bounds because x_data(end)>qp(end),
% but needs to be adjusted if this is not ensured prior to the loop)
while x_data(kk+1) < qp(cc)
kk = kk + 1;
end
% perform online interpolation
y_interp(cc) = myInterp(x_data(kk), x_data(kk+1), y_data(kk), y_data(kk+1), qp(cc));
end
As you can see, the logic is a lot simpler this way. The result is identical to y_interp_correct. The inner while x_data... loop serves the same purpose as your outer while loop, and would be the place where you read your data from wherever it's coming from.
I am working on programming a Markov chain in Lua, and one element of this requires me to uniformly generate random numbers. Here is a simplified example to illustrate my question:
example = function(x)
local r = math.random(1,10)
print(r)
return x[r]
end
exampleArray = {"a","b","c","d","e","f","g","h","i","j"}
print(example(exampleArray))
My issue is that when I re-run this program multiple times (mash F5) the exact same random number is generated resulting in the example function selecting the exact same array element. However, if I include many calls to the example function within the single program by repeating the print line at the end many times I get suitable random results.
This is not my intention as a proper Markov pseudo-random text generator should be able to run the same program with the same inputs multiple times and output different pseudo-random text every time. I have tried resetting the seed using math.randomseed(os.time()) and this makes it so the random number distribution is no longer uniform. My goal is to be able to re-run the above program and receive a randomly selected number every time.
You need to run math.randomseed() once before using math.random(), like this:
math.randomseed(os.time())
From your comment that you saw the first number is still the same. This is caused by the implementation of random generator in some platforms.
The solution is to pop some random numbers before using them for real:
math.randomseed(os.time())
math.random(); math.random(); math.random()
Note that the standard C library random() is usually not so uniformly random, a better solution is to use a better random generator if your platform provides one.
Reference: Lua Math Library
Standard C random numbers generator used in Lua isn't guananteed to be good for simulation. The words "Markov chain" suggest that you may need a better one. Here's a generator widely used for Monte-Carlo calculations:
local A1, A2 = 727595, 798405 -- 5^17=D20*A1+A2
local D20, D40 = 1048576, 1099511627776 -- 2^20, 2^40
local X1, X2 = 0, 1
function rand()
local U = X2*A2
local V = (X1*A2 + X2*A1) % D20
V = (V*D20 + U) % D40
X1 = math.floor(V/D20)
X2 = V - X1*D20
return V/D40
end
It generates a number between 0 and 1, so r = math.floor(rand()*10) + 1 would go into your example.
(That's multiplicative random number generator with period 2^38, multiplier 5^17 and modulo 2^40, original Pascal code by http://osmf.sscc.ru/~smp/)
math.randomseed(os.clock()*100000000000)
for i=1,3 do
math.random(10000, 65000)
end
Always results in new random numbers. Changing the seed value will ensure randomness. Don't follow os.time() because it is the epoch time and changes after one second but os.clock() won't have the same value at any close instance.
There's the Luaossl library solution: (https://github.com/wahern/luaossl)
local rand = require "openssl.rand"
local randominteger
if rand.ready() then -- rand has been properly seeded
-- Returns a cryptographically strong uniform random integer in the interval [0, n−1].
randominteger = rand.uniform(99) + 1 -- randomizes an integer from range 1 to 100
end
http://25thandclement.com/~william/projects/luaossl.pdf
I need a base converter function for Lua. I need to convert from base 10 to base 2,3,4,5,6,7,8,9,10,11...36 how can i to this?
In the string to number direction, the function tonumber() takes an optional second argument that specifies the base to use, which may range from 2 to 36 with the obvious meaning for digits in bases greater than 10.
In the number to string direction, this can be done slightly more efficiently than Nikolaus's answer by something like this:
local floor,insert = math.floor, table.insert
function basen(n,b)
n = floor(n)
if not b or b == 10 then return tostring(n) end
local digits = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
local t = {}
local sign = ""
if n < 0 then
sign = "-"
n = -n
end
repeat
local d = (n % b) + 1
n = floor(n / b)
insert(t, 1, digits:sub(d,d))
until n == 0
return sign .. table.concat(t,"")
end
This creates fewer garbage strings to collect by using table.concat() instead of repeated calls to the string concatenation operator ... Although it makes little practical difference for strings this small, this idiom should be learned because otherwise building a buffer in a loop with the concatenation operator will actually tend to O(n2) performance while table.concat() has been designed to do substantially better.
There is an unanswered question as to whether it is more efficient to push the digits on a stack in the table t with calls to table.insert(t,1,digit), or to append them to the end with t[#t+1]=digit, followed by a call to string.reverse() to put the digits in the right order. I'll leave the benchmarking to the student. Note that although the code I pasted here does run and appears to get correct answers, there may other opportunities to tune it further.
For example, the common case of base 10 is culled off and handled with the built in tostring() function. But similar culls can be done for bases 8 and 16 which have conversion specifiers for string.format() ("%o" and "%x", respectively).
Also, neither Nikolaus's solution nor mine handle non-integers particularly well. I emphasize that here by forcing the value n to an integer with math.floor() at the beginning.
Correctly converting a general floating point value to any base (even base 10) is fraught with subtleties, which I leave as an exercise to the reader.
you can use a loop to convert an integer into a string containting the required base. for bases below 10 use the following code, if you need a base larger than that you need to add a line that mapps the result of x % base to a character (usign an array for example)
x = 1234
r = ""
base = 8
while x > 0 do
r = "" .. (x % base ) .. r
x = math.floor(x / base)
end
print( r );