In an ILP, given two variables x and y, is it possible to define a variable z where
z = (x==y)?
Meaning, if x equals y then z = 1.
Else, z = 0.
x and y are integer variables bounded between (1, M).
If you know that x <= y, then you can use z in {0,1}, x+Mz >= y, x+z <= y.
If you don't know which of x and y is minimum, you can do it with more work, by adding a variable minxy which takes the value of the minimum. You need to introduce another new variable (which I call a) to do this.
Introduce a variable a that's 0 if x<=y, 1 if y<=x (it could be either 0 or 1 if x==y):
a in {0,1}, x-y <= Ma, y-x <= M(1-a)
Introduce a variable minxy that's the minimum of x and y:
minxy <= x
minxy <= y
minxy >= x - Ma
minxy >= y - M(1-a)
Then you can define your z:
minxy + Mz >= x + y - minxy
minxy + z <= x + y - minxy
(Noting that max(x,y) is x + y - minxy).
Related
In an ILP, How can I write a constraint "X is not in the interval [Y, Y+10]" ?
Is it even possible?
You didn't say, but I'm assuming x and y are non-negative integers and that the range you indicate to avoid is inclusive of endpoints...
You have an "or" condition there, so you will need to introduce a helper or "indicator" variable to handle the or condition, call it z, and a (constant) parameter for a reasonable upper bound on y, call it M:
z ∈ {0, 1} # 0 if below y, 1 if above y+10
M > reasonable_upper_bound(y)
Then 2 constraints based on that info to either constrain x to the lower set of values or the upper, excluding the interval (of disinterest):
x <= (y - 1) + z * M
x >= (y + 10 + 1) - (1 - z) * M
Truth Table:
z=0 z=1
x <= y-1 ~M
x >= ~0 y + 11
I've been given the following function written in pseudocode:
P:
{
int x, y, z;
read (x, y, z);
while (x != y) {
x = x - y;
z = z + y
};
write z;
}
Given that f(x,y,z) is the function calculated by P, I would like to know if the function "g(x,y,z)=1 if f(x,y,z) is not a total function or g(x,y,z)=0 otherwise", is computable.
My first guess is: yes, it is computable (for example for x=y).
Is there a more rigorous general approach to prove that?
P does not change the value of y, and the only way it changes the value of x is to subtract y from x until x = y. If subtracting y from x does not eventually result in x = y, then the loop continues forever. We know that subtracting y from x repeatedly can only cause x = y if initially x = cy for natural numbers c >= 1. So, g(x,y,z) = 1 because f(x,y,z) is not a total function; it is undefined when x != cy for any natural number c >= 1. Even if what you meant is that g(x,y,z) = 1 whenever f(x,y,z) is defined, it is still computable, since g(x,y,z) is the function:
g(x,y,z) = { 1, if x = cy for some natural number c >= 1 }
{ 0, otherwise }
The condition x = cy for some natural number c >= 1 is itself computable since this is equivalent to "x >= y" and "GCD(x, y) = y".
I am currently using Z3py to to deduce some invariants which are encoded as a conjunction of horn-clauses whilst also providing a template for the invariant. I'm starting with a simple example first if you see the code snippet below.
x = 0;
while(x < 5){
x += 1
}
assert(x == 5)
This translates into the horn clauses
x = 0 => Inv(x)
x < 5 /\ Inv(x) => Inv(x +1)
Not( x < 5) /\ Inv(x) => x = 5
The invariant here is x <= 5.
I have provided a template for the invariant of the form a*x + b <= c
so that all the solver has to do is guess a set of values for a,b and c that can reduce to x <= 5.
However when I encode it up I keep getting unsat. If try to assert Not (x==5) I get a=2 , b = 1/8 and c = 2 which makes little sense to me as a counterexample.
I provide my code below and would be grateful for any help on correcting my encoding.
x = Real('x')
x_2 = Real('x_2')
a = Real('a')
b = Real('b')
c = Real('c')
s = Solver()
s.add(ForAll([x],And(
Implies(x == 0 , a*x + b <= c),
Implies(And(x_2 == x + 1, x < 5, a*x + b <= c), a*x_2 + b <= c),
Implies(And(a*x + b <= c, Not(x < 5)), x==5)
)))
if (s.check() == sat):
print(s.model())
Edit: it gets stranger for me. If I remove the x_2 definition and just replace x_2 with (x + 1) in the second horn clause as well as delete the x_2 = x_2 + 1, I get unsat whether I write Not( x==5) or x==5 in the final horn clause.
There were two things preventing your original encoding from working:
1) It's not possible to satisfy x_2 == x + 1 for all x for a single value of x_2. Thus, if you're going to write x_2 == x + 1, both x and x_2 need to be universally quantified.
2) Somewhat surprisingly, this problem is satisfiable in the integers but not in the reals. You can see the problem with the clause x < 5 /\ Inv(x) => Inv(x + 1). If x is an integer, then this is satisfied by x <= 5. However, if x is allowed to be any real value, then you could have x == 4.5, which satisfies both x < 5 and x <= 5, but not x + 1 <= 5, so Inv(x) = (x <= 5) does not satisfy this problem in the reals.
Also, you might find it helpful to define Inv(x), it cleans up the code quite a bit. Here is the encoding of your problem with those changes:
from z3 import *
# Changing these from 'Int' to 'Real' changes the problem from sat to unsat.
x = Int('x')
x_2 = Int('x_2')
a = Int('a')
b = Int('b')
c = Int('c')
def Inv(x):
return a*x + b <= c
s = Solver()
# I think this is the simplest encoding for your problem.
clause1 = Implies(x == 0 , Inv(x))
clause2 = Implies(And(x < 5, Inv(x)), Inv(x + 1))
clause3 = Implies(And(Inv(x), Not(x < 5)), x == 5)
s.add(ForAll([x], And(clause1, clause2, clause3)))
# Alternatively, if clause2 is specified with x_2, then x_2 needs to be
# universally quantified. Note the ForAll([x, x_2]...
#clause2 = Implies(And(x_2 == x + 1, x < 5, Inv(x)), Inv(x_2))
#s.add(ForAll([x, x_2], And(clause1, clause2, clause3)))
# Print result all the time, to avoid confusing unknown with unsat.
result = s.check()
print result
if (result == sat):
print(s.model())
One more thing: it's a bit strange to me to write a*x + b <= c as a template, because this is the same as a*x <= d for some integer d.
I need to draw circles using pixels with these constraints:
the total of pixels across the diameter is an even number,
there is no empty pixels between two circles of radius R and R+1 (R is an integer).
The midpoint algorithm can’t be used but I found out that Eric Andres wrote the exact thing I want. The algorithm can be found in this article under the name of “half integer centered circle”. For those who don’t have access to it, I put the interesting part is at the end of the question.
I encounter difficulties to implement the algorithm. I copied the algorithm in Processing using the Python syntax (for the ease of visualisation):
def half_integer_centered_circle(xc, yc, R):
x = 1
y = R
d = R
while y >= x:
point(xc + x, yc + y)
point(xc + x, yc - y + 1)
point(xc - x + 1, yc + y)
point(xc - x + 1, yc - y + 1)
point(xc + y, yc + x)
point(xc + y, yc - x + 1)
point(xc - y + 1, yc + x)
point(xc - y + 1, yc - x + 1)
if d > x:
d = d - x
x = x + 1
elif d < R + 1 - y:
d = d + y - 1
y = y - 1
else:
d = d + y - x - 1
x = x + 1
y = y - 1
The point() function just plot a pixel at the given coordinates. Please also note that in the article, x is initialised as S, which is strange because there is no S elsewhere (it’s not explained at all), however it is said that the circle begins at (x, y) = (1, R), so I wrote x = 1.
There is the result I get for a radii between 1 pixel and 20 pixels:
As you can see, there are holes between circles and the circle with R = 3 is different from the given example (see below). Also, the circles are not really round compared to what you get with the midpoint algorithm.
How can I get the correct result?
Original Eric Andres’ algorithm:
I don't understand the way in which the algorithm has been presented in that paper. As I read it the else if clause associated with case (b) doesn't have a preceding if. I get the same results as you when transcribing it as written
Looking at the text, rather than the pseudocode, the article seems to be suggesting an algorithm of the following form:
x = 1
y = R
while x is less than or equal to y:
draw(x, y)
# ...
if the pixel to the right has radius between R - 1/2 and R + 1/2:
move one pixel to the right
if the pixel below has radius between R - 1/2 and R + 1/2:
move one pixel down
else:
move one pixel diagonally down and right
Which seems plausible. In python:
#!/usr/bin/python3
import numpy as np
import matplotlib.pyplot as pp
fg = pp.figure()
ax = fg.add_subplot(111)
def point(x, y, c):
xx = [x - 1/2, x + 1/2, x + 1/2, x - 1/2, x - 1/2 ]
yy = [y - 1/2, y - 1/2, y + 1/2, y + 1/2, y - 1/2 ]
ax.plot(xx, yy, 'k-')
ax.fill_between(xx, yy, color=c, linewidth=0)
def half_integer_centered_circle(R, c):
x = 1
y = R
while y >= x:
point(x, y, c)
point(x, - y + 1, c)
point(- x + 1, y, c)
point(- x + 1, - y + 1, c)
point(y, x, c)
point(y, - x + 1, c)
point(- y + 1, x, c)
point(- y + 1, - x + 1, c)
def test(x, y):
rSqr = x**2 + y**2
return (R - 1/2)**2 < rSqr and rSqr < (R + 1/2)**2
if test(x + 1, y):
x += 1
elif test(x, y - 1):
y -= 1
else:
x += 1
y -= 1
for i in range(1, 5):
half_integer_centered_circle(2*i - 1, 'r')
half_integer_centered_circle(2*i, 'b')
pp.axis('equal')
pp.show()
This seems to work as intended. Note that I removed the circle centre for simplicity. It should be easy enough to add in again.
Edit Realised I could match the radius 3 image if I tweaked the logic a bit.
I have been looking into this matter and observed three issues in the original paper:
The arithmetic circle copied here (Figure 10.a in the paper) is not consistent with the formal definition of the "half integer centered circle". In one case the distance to the center must be between R-1/2 and R+1/2 and in the other between integer values. The consequence is that this specific algorithm, if properly implemented, can never generate the circle of Figure 10.a.
There is a mistake in one of the inequalities of the algorithm pseudo code: the test for case (b) should be d <= (R + 1 - y) instead of d < (R + 1 - y).
All those pixels that satisfy x==y have only 4-fold symmetry (not 8-fold) and are generated twice by the algorithm. Although producing duplicated pixels may not be a problem for a drawing routine, it is not acceptable for the application that I am interested in. However this can be easily fixed by adding a simple check of the x==y condition and skipping the four duplicated pixels.
The python code of the original question includes the inequality error mentioned above and an additional mistake due to missing parenthesis in one of the expressions that should read d = d + (y - x - 1).
The following implementation fixes all this and is compatible with python2 and python3 (no integer division issues in the point() function):
import numpy as np
import matplotlib.pyplot as pp
fg = pp.figure()
ax = fg.add_subplot(111)
def point(x, y, c):
xx = [x - 0.5, x + 0.5, x + 0.5, x - 0.5, x - 0.5 ]
yy = [y - 0.5, y - 0.5, y + 0.5, y + 0.5, y - 0.5 ]
ax.plot(xx, yy, 'k-')
ax.fill_between(xx, yy, color=c, linewidth=0)
def half_integer_centered_circle(R, c):
x = 1
y = R
d = R
while y >= x:
point(x, y, c)
point(x, - y + 1, c)
point(- x + 1, y, c)
point(- x + 1, - y + 1, c)
if y != x:
point(y, x, c)
point(y, - x + 1, c)
point(- y + 1, x, c)
point(- y + 1, - x + 1, c)
if d > x:
d = d - x
x = x + 1
elif d <= R + 1 - y:
d = d + y - 1
y = y - 1
else:
d = d + (y - x - 1)
x = x + 1
y = y - 1
for i in range(1, 5):
half_integer_centered_circle(2*i - 1, 'r')
half_integer_centered_circle(2*i, 'b')
pp.axis('equal')
pp.show()
This is a follow up question for: Subtraction operation using only increment, loop, assign, zero
We're only allowed to use the following operations:
incr(x) - Once this function is called it will assign x + 1 to x
assign(x, y) - This function will assign the value of y to x (x = y)
zero(x) - This function will assign 0 to x (x = 0)
loop X { } - operations written within brackets will be executed X times
For example, addition can be implemented as follows:
add(x, y) {
loop x
{ y = incr(y) }
return y
}
How do I implement the relational operators using these four operations? The relational operations are:
eq(x, y) - Is x equal to y?
lt(x, y) - Is x lesser than y?
gt(x, y) - Is x greater than y?
We also have their opposites:
ne(x, y) - Is x not equal to y?
gte(x, y) - Is x greater than or equal to y?
lte(x, y) - Is x lesser than or equal to y?
Any help will be appreciated.
The set of natural numbers N is closed under addition and subtraction:
N + N = N
N - N = N
This means that the addition or subtraction of two natural numbers is also a natural number (considering 0 - 1 is 0 and not -1, we can't have negative natural numbers).
However, the set of natural numbers N is not closed under relational operations:
N < N = {0, 1}
N > N = {0, 1}
This means that the result of comparing two natural numbers is either truthfulness (i.e. 1) or falsehood (i.e. 0).
So, we treat the set of booleans (i.e. {0, 1}) as a restricted set of the natural numbers (i.e. N).
false = 0
true = incr(false)
The first question we must answer is “how do we encode if statements so that we may branch based on either truthfulness or falsehood?” The answer is simple, we use the loop operation:
isZero(x) {
y = true
loop x { y = false }
return y
}
If the loop condition is true (i.e. 1) then the loop executes exactly once. If the loop condition is false (i.e. 0) then the loop doesn't execute. We can use this to write branching code.
So, how do we define the relational operations? Turns out, everything can be defined in terms of lte:
lte(x, y) {
z = sub(x, y)
z = isZero(z)
return z
}
We know that x ≥ y is the same as y ≤ x. Therefore:
gte(x, y) {
z = lte(y, x)
return z
}
We know that if x > y is true then x ≤ y is false. Therefore:
gt(x, y) {
z = lte(x, y)
z = not(z)
return z
}
We know that x < y is the same as y > x. Therefore:
lt(x, y) {
z = gt(y, x)
return z
}
We know that if x ≤ y and y ≤ x then x = y. Therefore:
eq(x, y) {
l = lte(x, y)
r = lte(y, x)
z = and(l, r)
return z
}
Finally, we know that if x = y is true then x ≠ y is false. Therefore:
ne(x, y) {
z = eq(x, y)
z = not(z)
return z
}
Now, all we need to do is define the following functions:
The sub function is defined as follows:
sub(x, y) {
loop y
{ x = decr(x) }
return x
}
decr(x) {
y = 0
z = 0
loop x {
y = z
z = incr(z)
}
return y
}
The not function is the same as the isZero function:
not(x) {
y = isZero(x)
return y
}
The and function is the same as the mul function:
and(x, y) {
z = mul(x, y)
return z
}
mul(x, y) {
z = 0
loop x { z = add(y, z) }
return z
}
add(x, y) {
loop x
{ y = incr(y) }
return y
}
That's all you need. Hope that helps.