Having a (lst_v) list of BitVectors and a (lst_b) list of boolean expressions values. How do you perform the following operations using z3py:
use the lst_b to mask elements in lst_v. The masking needed to use And function since the boolean expression needs to be solved in the final step.
compute the the xor of all remaining elements
test the all the bits are set in the result by using the Solve class of z3py.
A variation of the problem is to exchange xor with addition
It's not entirely clear what you are trying to achieve; but perhaps the following will get you going:
from z3 import *
# Assume we have a list of 3 32-bit values
x, y, z = BitVecs('x y z', 32)
lst_v = [x, y, z]
# Corresponding booleans:
mx, my, mz = Bools('mx my mz')
lst_b = [mx, my, mz]
# 32-bit zero
zero = BitVecVal(0, 32)
# Mask
masked = [If(b, v, zero) for (b, v) in zip(lst_b, lst_v)]
# Xor reduce
final = reduce(lambda x, y: x^y, masked, zero)
# 32-bit all 1's
allOnes = BitVecVal(-1, 32)
s = Solver()
s.add(final == allOnes)
# make it interesting, assert some known values and constraints
s.add(x == BitVecVal(123212, 32))
s.add(UGT(x + y, z+12))
s.add(ULT(y, allOnes))
if s.check() == sat:
print s.model()
else:
print "No solution"
When I run this, I get:
[mz = True,
mx = False,
my = True,
z = 2147479427,
y = 2147487868,
x = 123212]
which suggests, I should XOR y and z as 32 bit values; which gives 4294967295, which has all its bits set as a 32-bit quantity.
Related
I have the following problem: I have a list of items
[O_0,..., O_n]
, where each item is represented by a binary power (o_0 represented by 2^0, ..., o_n by 2^n). I have constructed a list of combinations of these elements (each combination is represented by the sum of the binary representations of the items). For example I have
combs = [3, 9, 15, ......].
Given a new combination of these items say C_1, I would like to test if any of the elements of combs is included in C_1. An efficient and fast way that I thought of was to compute for each element c_i from combs, test if c_i & C_1 == c_i which means that it is true for this element. It is fast since I am doing a bitiwise and.
My problem is that instead of having 1 element C_1, I have a very big number of them C_1, ..., C_k, and I have to test for each one of them the above condition. SO I was wondering if there are any faster ways than the one I mentioned to test the condition of all of the elements (this is actually the same problem as testing if a set is a subset of another, which is why i chose binary representation from the beginning to transform the problem into a binary one).
My understanding of the problem: given a collection of k sets Y and a collection of m sets X, we would like to find a subset S of Y such that for all y in S, there exists x in X s.t. x is a subset of y.
I will assume that sets are represented by n-vectors of zeros and ones denoting inclusion. Here is the setup:
import pandas as pd # for drop_duplicates & benchmarks
import numpy as np
np.random.seed(0)
n = 100 # 100 "atomic" elements
m = 1000 # small sets
k = 1000 # large sets
X = pd.DataFrame(np.random.randint(0, 2, size=(m, n))).drop_duplicates().values
Y = pd.DataFrame(np.random.randint(0, 2, size=(k, n))).drop_duplicates().values
# For each row y in Y, we would like to check if there exists a row x in X
# s.t. x represents a subset of Y
def naive(Y, X):
# O(k^2 + m^2)
for i, y in enumerate(Y):
for x in X:
found_subset = False
if (x <= y).all():
yield i
found_subset = True
if found_subset:
break
def naive_as_array(Y, X):
return np.array(list(naive(Y, X)))
The naive function iterates over all pairs of sets that may satisfy the inclusion relation and short-circuits whenever appropriate. The runtime is O(m + k) where m = len(combs).
As an alternative, we can consider the following recursive algorithm processing each element (from 1 to n) at a time:
def contains(Y, X):
"""
Y : k x n indicator array specifying sets
X : m x n indicator array specifying sets
output: subset Z of [0..k-1] s.t. i in Z iff there exists x in X s.t.
# x is a subset of Y[i]. Z is represented by a 1D numpy array.
"""
k, n = Y.shape
assert Y.shape[1] == X.shape[1]
detected = np.zeros(k, dtype=np.bool)
inds = np.arange(k)
# utility to account for sets that already have a subset detected
def account_for_detected(Y, inds):
mask = ~detected[inds]
Y = Y[mask]
inds = inds[mask]
return Y, inds
# inductively reduce Y.shape[1] (==X.shape[1])
def f(Y, X, inds):
if Y.shape[0] == 0 or X.shape[0] == 0:
# collection Y is empty; inculsions are impossible
return
# avoid redundant comparisons by dropping sets y in Y
# if it is already known that y contains some element of X
Y, inds = account_for_detected(Y, inds)
if Y.shape[1] == 1:
# Y and X are collections of singletons
Y = np.ravel(Y)
X = np.ravel(X)
X_vals = np.zeros(2, dtype=np.int)
X_vals[X] = 1
if X_vals[0] > 0:
detected[inds] = True
elif X_vals[1] > 0:
detected[inds[Y==1]] = True
return
else:
# make a recursive call
Ymask = Y[:,0] == 0
Xmask = X[:,0] == 0
# if x in X is a subset of y in Y, x[0] <= y[0]
f(Y[Ymask,1:], X[Xmask,1:], inds[Ymask])
# by now, detected is updated in the outer scope
# process the remaining Y's
f(Y[~Ymask,1:], X[:,1:], inds[~Ymask])
# done
# make call at root:
f(Y, X, inds)
# return indices
return np.where(detected)[0]
At step d between 1 and n, we split sets Y into Y0 and Y1 where Y0 contains sets in Y that do not contain element d, and Y1 contains sets in Y that do contain element d. Similarly, we define X0 and X1. A key observation is that sets in X1 cannot occur as subsets of sets in Y0. Therefore, we can reduce the number of comparisons in the recursive call.
Timings:
%timeit contains(Y, X)
%timeit naive_as_array(Y, X)
10 loops, best of 3: 185 ms per loop
1 loop, best of 3: 2.39 s per loop
I am using an RK4 algorithm:
function R=RK4_h(f,a,b,ya,h)
% Input
% - f field of the edo y'=f(t,y). A string of characters 'f'
% - a and b initial and final time
% - ya initial value y0
% - h lenght of the step
% Output
% - R=[T' Y'] where T independent variable and Y dependent variable
N = fix((b-a) / h);
T = zeros(1,N+1);
Y = zeros(1,N+1);
% Vector of the time values
T = a:h:b;
% Solving ordinary differential equation
Y(1) = ya;
for j = 1:N
k1 = h*feval(f,T(j),Y(j));
k2 = h*feval(f,T(j)+h/2,Y(j)+k1/2);
k3 = h*feval(f,T(j)+h/2,Y(j)+k2/2);
k4 = h*feval(f,T(j)+h,Y(j)+k3);
Y(j+1) = Y(j) + (k1+2*k2+2*k3+k4)/6;
end
R=[T' Y'];
In my main script I call it for every value as:
xlabel('x')
ylabel('y')
h=0.05;
fprintf ('\n First block \n');
xx = [0:h:1];
Nodes = length(xx);
yy = zeros(1,Nodes);
for i=1:Nodes
fp(i)=feval('edo',-1,xx(i));
end
E=RK4_h('edo',0,1,-1,h);
plot(E);
fprintf ('\n%f',E);
The problem is when I try to use RK4 algorithm with edo formula:
function edo = edo(y,t)
edo = 6*((exp(1))^(6*t))*(y-(2*t))^2+2;
The results are not logical, for example, the real value for are: y(0)=8, y(1)=11,53. But the estimate is not close. Any of both coordinates in E vector represents a feasible approach for the problem, so I do not know if this is the correct implementation.
There is a basic error of implementation?
The function edo takes t as the first parameter, and y as the second parameter. You have the parameters reversed.
Your function should be:
function edo = edo(t,y) % NOT edo(y,t)
edo = 6*((exp(1))^(6*t))*(y-(2*t))^2+2;
I'm trying to convert a base-10 integer k into a base-q integer, but not in the standard way. Firstly, I'd like my result to be a vectors (or a string 'a,b,c,...' so that it can be converted to a vector, but not 'abc...'). Most importantly, I'd like each 'digit' to be in base-10. As an example, suppose I have the number 23 (in base-10) and I want to convert it to base-12. This would be 1B in the standard 1,...,9,A,B notation; however, I want it to come out as [1, 11]. I'm only interested in numbers k with 0 \le k \le n^q - 1, where n is fixed in advance.
Put another way, I wish to find coefficients a(r) such that
k = \sum_{r=0}^{n-1} a(r) q^r
where each a(r) is in base-10. (Note that 0 \le a(r) \le q-1.)
I know I could do this with a for-loop -- struggling to get the exact formula at the moment! -- but I want to do it vectorised, or with a fast internal function.
However, I want to be able to take n to be large, so would prefer a faster way than this. (Of course, I could change this to a parfor-loop or do it on the GPU; these aren't practical for my current situation, so I'd prefer a more direct version.)
I've looked at stuff like dec2base, num2str, str2num, base2dec and so on, but with no luck. Any suggestion would be most appreciated.
Regarding speed and space, any preallocation for integers in the range [0, q-1] or similar would also be good.
To be clear, I am looking for an algorithm that works for any q and n, converting any number in the range [0,q^n - 1].
You can use dec2base and replace the characters by numbers:
x = 23;
b = 12;
[~, result] = ismember(dec2base(x,b), ['0':'9' 'A':'Z']);
result = result -1;
gives
>> result
result =
1 11
This works for base up to 36 only, due to dec2base limitations.
For any base (possibly above 36) you need to do the conversion manually. I once wrote a base2base function to do that (it's essentially long division). The number should be input as a vector of digits in the origin base, so you need dec2base(...,10) first. For example:
x = 125;
b = 6;
result = base2base(dec2base(x,10), '0':'9', b); % origin nunber, origin base, target base
gives
result =
3 2 5
Or if you need to specify the number of digits:
x = 125;
b = 6;
d = 5;
result = base2base(dec2base(x,10), '0':'9', b, d)
result =
0 0 3 2 5
EDIT (August 15, 2017): Corrected two bugs: handling of input consisting of all "zeros" (thanks to #Sanchises for noticing), and properly left-padding the output with "zeros" if needed.
function Z = base2base(varargin)
% Three inputs: origin array, origin base, target base
% If a base is specified by a number, say b, the digits are [0,1,...,d-1].
% The base can also be directly an array with the digits
% Fourth input, optional: how many digits the output should have as a
% minimum (padding with leading zeros, i.e with the first digit)
% Non-valid digits in origin array are discarded.
% It works with cell arrays. In this case it gives a matrix in which each
% row is padded with leading zeros if needed
% If the base is specified as a number, digits are numbers, not
% characters as in `dec2base` and `base2dec`
if ~iscell(varargin{1}), varargin{1} = varargin(1); end
if numel(varargin{2})>1, ax = varargin{2}; bx=numel(ax); else bx = varargin{2}; ax = 0:bx-1; end
if numel(varargin{3})>1, az = varargin{3}; bz=numel(az); else bz = varargin{3}; az = 0:bz-1; end
Z = cell(size(varargin{1}));
for c = 1:numel(varargin{1})
x = varargin{1}{c}; [valid, x] = ismember(x,ax); x = x(valid)-1;
if ~isempty(x) && ~any(x) % Non-empty input, all zeros
z = 0;
elseif ~isempty(x) % Non-empty input, at least a nonzero
z = NaN(1,ceil(numel(x)*log2(bx)/log2(bz))); done_outer = false;
n = 0;
while ~done_outer
n = n + 1;
x = [0 x(find(x,1):end)];
y = NaN(size(x)); done_inner = false;
m = 0;
while ~done_inner
m = m + 1;
t = x(1)*bx+x(2);
r = mod(t, bz); q = (t-r)/bz;
y(m) = q; x = [r x(3:end)];
done_inner = numel(x) < 2;
end
y = y(1:m);
z(n) = r; x = y; done_outer = ~any(x);
end
z = z(n:-1:1);
else % Empty input
z = []; % output will be empty (unless user has required left-padding) with the
% appropriate class
end
if numel(varargin)>=4 && numel(z)<varargin{4}, z = [zeros(1,varargin{4}-numel(z)) z]; end
% left-pad if required by user
Z{c} = z;
end
L = max(cellfun(#numel, Z));
Z = cellfun(#(x) [zeros(1, L-numel(x)) x], Z, 'uniformoutput', false); % left-pad so that
% result will be a matrix
Z = vertcat(Z{:});
Z = az(Z+1);
Matlab's internal dec2base command contains essentially what you are asking for.
It actually creates an array of base-10 digits before they are converted to a character array of '0'-'9' and 'A'-'Z' which is the reason for its limitation to bases <= 36.
So after removing the last step of character conversion from dec2base and modifying the error checking accordingly gives the function dec2basevect you were asking for.
The result will be a base-10 vector and you are no longer limited to bases <= 36. The most significant digit will be in index one of this vector. If you need it the other way round, i.e. least significant digit in index one, just do a fliplr to the result.
Due to copyrights by MathWorks, you have to make the necessary modifications to dec2baseon your own.
I am just a beginner of programming, and sorry in advance for bothering you by a (presumably) basic question.
I would like to perform the following task:
(I apologize for inconvenience; I don't know how to input a TeX-y formula in Stack Overflow ). I am primarily considering an implementation on MATLAB or Scilab, but language does not matter so much.
The most naive approach to perform this, I think, is to form an n-nested for loop, that is (the case n=2 on MATLAB is shown for example),
n=2;
x=[x1,x2];
for u=0:1
y(1)=u;
if x(1)>0 then
y(1)=1;
end
for v=0:1
y(2)=v;
if x(2)>0 then
y(2)=1;
end
z=Function(y);
end
end
However, this implementation is too laborious for large n, and more importantly, it causes 2^n-2^k abundant evaluations of the function, where k is a number of negative elements in x. Also, naively forming a k-nested for loop with knowledge of which element in x is negative, e.g.
n=2;
x=[-1,2];
y=[1,1];
for u=0:1
y(1)=u;
z=Function(y);
end
doesn't seem to be a good way; if we want to perform the task for different x, we have to rewrite a code.
I would be grateful if you provide an idea to implement a code such that (a) evaluates the function only 2^k times (possible minimum number of evaluations) and (b) we don't have to rewrite a code even if we change x.
You can evaluate Function on y in Ax easily using recursion
function eval(Function, x, y, i, n) {
if(i == n) {
// break condition, evaluate Function
Function(y);
} else {
// always evaluate y(i) == 1
y(i) = 1;
eval(Function, x, y, i + 1, n);
// eval y(i) == 0 only if x(i) <= 0
if(x(i) <= 0) {
y(i) = 0;
eval(Function, x, y, i + 1, n);
}
}
}
Turning that into efficient Matlab code is another problem.
As you've stated the number of evaluations is 2^k. Let's sort x so that only the last k elements are non-positive. To evaluate Function index y using the reverse of the permutation of the sort of x: Function(y(perm)). Even better the same method allows us to build Ax directly using dec2bin:
// every column of the resulting matrix is a member of Ax: y_i = Ax(:,i)
function Ax = getAx(x)
n = length(x);
// find the k indices of non-positives in x
is = find(x <= 0);
k = length(is);
// construct Y (last k rows are all possible combinations of [0 1])
Y = [ones(n - k, 2 ^ k); (dec2bin(0:2^k-1)' - '0')];
// re-order the rows in Y to get Ax according to the permutation is (inverse is)
perm([setdiff(1:n, is) is]) = 1:n;
Ax = Y(perm, :);
end
Now rewrite Function to accept a matrix or iterate over the columns in Ax = getAx(x); to evaluate all Function(y).
I want to write some functions as follows
y = f(x) and another function,
x = g(y) that acts as a reversible, where
y = f(g(y)) and where x and y are permutated integers.
For very simple example in the range of integers in 0 to 10 it would look like this:
0->1
1->2
2->3
...
9->10
10->0
but this is the simplest method by adding 1 and reversing by subtracting 1.
I want to have a more sofisticated algorithm that can do the following,
234927773->4299
34->33928830
850033->23234243423
but the reverse can be obtained by conversion
The solution could be obtained with a huge table storing pairs of unique integers but this will not be correct. This must be a function.
You could just XOR.
y = x XOR p
x = y XOR p
Though not my area of expertise, I think that cryptography should provide some valuable answers to your question.
If the domain of your permutation is a power of 2, you can use any block cipher: 'f' is encryption with a specific key, and 'g' is decryption with the same key. If your domain is not a power of 2, you can probably still use a block cipher: see this article.
You could use polynomial interpolation methods to interpolate a function one way, then do reverse interpolation to find the inverse function.
Here is some example code in MATLAB:
function [a] = Coef(x, y)
n = length(x);
a = y;
for j = 2:n
for i = n:-1:j
a(i) = (a(i) - a(i-1)) / (x(i) - x(i-j+1));
end
end
end
function [val] = Eval(x, a, t)
n = length(x);
val = a(n);
for i = n-1:-1:1
val = a(i) + val*(t-x(i));
end
end
It builds a Divided Difference table and evaluates a function based on Newtons Interpolation.
Then if your sets of points are x, and y (as vectors of the same length, where x(i) matches to y(i), your forward interpolation function at value n would be Eval(x, Coef(x, y), n) and the reverse interpolation function would be Eval(y, Coef(y, x), n).
Depending on your language, there are probably much cleaner ways to do this, but this gets down and dirty with the maths.
Here is an excerpt from the Text Book which is used in my Numerical Methods class: Google Book Link