I have the following state:
a : ℕ,
a_1 : ℕ,
nltm : of_nat a_1 + 1 < of_nat a + 1
⊢ of_nat a_1 < of_nat a
Do note that of_nat constructs Z.
Is there a painless way of discharging the goal?
This is lt_of_lt_add_right:
open int
example (a a_1 : ℕ) (nltm : of_nat a_1 + 1 < of_nat a + 1) : of_nat a_1 < of_nat a :=
lt_of_add_lt_add_right nltm
Learning the Lean naming conventions is very helpful for finding such lemmas.
Related
I want to prove the following about fraction addition. Namely j/2+k/2 = (j+k)/2. To do so, I have the following below:
lemma adding_fraction(j: nat, k: nat)
ensures j / 2 + k / 2 == (j + k) / 2
{
assert j / 2 + k / 2
== j * (1 / 2) + k * (1 / 2)
== (1/2) * (j + k)
== (j + k) / 2
assert j/2 + k/2 == (j + k) / 2;
}
I get an invalid AssertStmt error on the last like of this lemma. I'm not entirely sure why. Moreover I'm not sure why dafny is unable to prove this about addition already. Could someone provide some assistance on how I can prove this lemma?
I see one error actually now. Namely in dafny 1/2 == 0 is true. So all division seems to be done with a floor operation. That said in the code above is j%2=0 and k%2=0 then dafny's able to prove it no problem. How can I force dafny not to use floor? This is an issue since in real life: 1/2 + 5/2 = 6/2 = 3 but in dafny with floor: 1/2 + 5/2 = 0 + 2 = 2.
First, the lemma is not true. If j and k are both odd, then j / 2 + k / 2 will be one less than (j + k) / 2 because division is truncating integer division. For example, we can prove the following counterexample in Dafny
lemma counterexample()
ensures var j := 1; var k := 3;
!(j / 2 + k / 2 == (j + k) / 2)
{}
As to your other question, you are missing a semicolon on the second to last line. Dafny's error message could be improved here, but it's good to keep in mind in the future that any time Dafny reports an "invalid ", it's a syntax error and you should be on the lookout for typos. The full Dafny message for this error is (roughly)
tmp.dfy(8,4): Error: invalid AssertStmt
1 parse errors detected in tmp.dfy
You can see that it at least contains a line number (for the assertion after the missing semicolon) and the phrase "parse errors" on the second line of the error message.
After fixing this syntax error, there are several semantic errors having to do with not being able to prove the assertions. The underlying cause for these is that the lemma is false.
I am attempting a maximisation problem subject to various constraints.
i.e. max y = x1 + x2 + x3 + .... + xn
where each xi is a vector of values over time: x1 = (x11, x12, x13,...)
Some of the constraints state that specific values of xit cannot be positive in the same time period.
i.e. if(x1t > 0), x2t = 0; if(x2t > 0), x1t = 0
For context, the constraint is equivalent to "maximise the revenue of a shop, but you cant sell product A and B on the same day"
How do I go about formulating an LP model in Excel (using solver) to solve this.
This is called a complementarity constraint. One way of modeling this is:
x(1,t) * x(2,t) = 0
x(i,t) ≥ 0
However, this is nonlinear (and in a somewhat nasty way). A linear approach, using an extra binary variable δ can look like:
x(1,t) ≤ UP(1,t) * δ(t)
x(2,t) ≤ UP(2,t) * (1-δ(t))
x(i,t) ∈ [0,UP(i,t)] 'UP is an upper bound on x'
δ(t) ∈ {0,1} 'δ is a binary variable'
I constructed a finite automata for the language L of all strings made of the symbols 0, 1 and 2 (Σ = {0, 1, 2}) where the last symbol is not smaller than the first symbol. E.g., the strings 0, 2012, 01231 and 102 are in the language, but 10, 2021 and 201 are not in the language.
Then from that an GNFA so I can convert to RE.
My RE looks like this:
(0(0+1+2)* )(1(0(1+2)+1+2)* )(2((0+1)2+2))*)
I have no idea if this is correct, as I think I understand RE but not entirely sure.
Could someone please tell me if it’s correct and if not why?
There is a general method to convert any DFA into a regular expression, and is probably what you should be using to solve this homework problem.
For your attempt specifically, you can tell whether an RE is incorrect by finding a word that should be in the language, but that your RE doesn't accept, or a word that shouldn't be in the language that the RE does accept. In this case, the string 1002 should be in the language, but the RE doesn't match it.
There are two primary reasons why this string isn't matched. The first is that there should be a union rather than a concatenation between the three major parts of the language (words starting with 0, 1 and 2, respectively:
(0(0+1+2)*) (1(0(1+2)+1+2)*) (2((0+1)2+2))*) // wrong
(0(0+1+2)*) + (1(0(1+2)+1+2)*) + (2((0+1)2+2))*) // better
The second problem is that in the 1 and 2 cases, the digits smaller than the starting digit need to be repeatable:
(1(0 (1+2)+1+2)*) // wrong
(1(0*(1+2)+1+2)*) // better
If you do both of those things, the RE will be correct. I'll leave it as an exercise for you to follow that step for the 2 case.
The next thing you can try is find a way to make the RE more compact:
(1(0*(1+2)+1+2)*) // verbose
(1(0*(1+2))*) // equivalent, but more compact
This last step is just a matter of preference. You don't need the trailing +1+2 because 0* can be of zero length, so 0*(1+2) covers the +1+2 case.
You can use an algorithm but this DFA might be easy enough to convert as a one-off.
First, note that if the first symbol seen in the initial state is 0, you transition to state A and remain there. A is accepting. This means any string beginning with 0 is accepted. Thus, our regular expression might as well have a term like 0(0+1+2)* in it.
Second, note that if the first symbol seen in the initial state is 1, you transition to state B and remain in states B and D from that point on. You only leave B if you see 0 and you stay out of B as long as you keep seeing 0. The only way to end on D is if the last symbol you saw was 0. Therefore, strings beginning with 1 are accepted if and only if the strings don't end in 0. We can have a term like 1(0+1+2)*(1+2) in our regular expression as well to cover these cases.
Third, note that if the first symbol seen in the initial state is 2, you transition to state C and remain in states C and E from that point on. You leave state C if you see anything but 2 and stay out of B until you see a 2 again. The only way to end up on C is if the last symbol you saw was 2. Therefore, strings beginning with 2 are accepted if and only if the strings end in 2. We can have a term like 2(0+1+2)*(2) in our regular expression as well to cover these cases.
Finally, we see that there are no other cases to consider; our three terms cover all cases and the union of them fully describes our language:
0(0+1+2)* + 1(0+1+2)*(1+2) + 2(0+1+2)*2
It was easy to just write out the answer here because this DFA is sort of like three simple DFAs put together with a start state. More complicated DFAs might be easier to convert to REs using algorithms that don't require you understand or follow what the DFA is doing.
Note that if the start state is accepting (mentioned in a comment on another answer) the RE changes as follows:
e + 0(0+1+2)* + 1(0+1+2)*(1+2) + 2(0+1+2)*2
Basically, we just tack the empty string onto it since it is not already generated by any of the other parts of the aggregate expression.
You have the equivalent of what is known as a right-linear system. It's right-linear because the variables occur on the right hand sides only to the first degree and only on the right-hand sides of each term. The system that you have may be written - with a change in labels from 0,1,2 to u,v,w - as
S ≥ u A + v B + w C
A ≥ 1 + (u + v + w) A
B ≥ 1 + u D + (v + w) B
C ≥ 1 + (u + v) E + w C
D ≥ u D + (v + w) B
E ≥ (u + v) E + w C
The underlying algebra is known as a Kleene algebra. It is defined by the following identities that serve as its fundamental properties
(xy)z = x(yz), x1 = x = 1x,
(x + y) + z = x + (y + z), x + 0 = x = 0 + x,
y0z = 0, w(x + y)z = wxz + wyz,
x + y = y + x, x + x = x,
with a partial ordering relation defined by
x ≤ y ⇔ y ≥ x ⇔ ∃z(x + z = y) ⇔ x + y = y
With respect to this ordering relation, all finite subsets have least upper bounds, including the following
0 = ⋁ ∅, x + y = ⋁ {x, y}
The sum operator "+" is the least upper bound operator.
The system you have is a right-linear fixed point system, since it expresses the variables on the left as a (right-linear) function, as given on the right, of the variables. The object being specified by the system is the least solution with respect to the ordering; i.e. the least fixed point solution; and the regular expression sought out is the value that the main variable has in the least fixed point solution.
The last axiom(s) for Kleene algebras can be stated in any of a number of equivalent ways, including the following:
0* = 1
the least fixed point solution to x ≥ a + bx + xc is x = b* a c*.
There are other ways to express it. A consequence is that one has identities such as the following:
1 + a a* = a* = 1 + a* a
(a + b)* = a* (b a*)*
(a b)* a = a (b a)*
In general, right linear systems, such as the one corresponding to your problem may be written in vector-matrix form as 𝐪 ≥ 𝐚 + A 𝐪, with the least fixed point solution given in matrix form as 𝐪 = A* 𝐚. The central theorem of Kleene algebras is that all finite right-linear systems have least fixed point solutions; so that one can actually define matrix algebras over Kleene algebras with product and sum given respectively as matrix product and matrix sum, and that this algebra can be made into a Kleene algebra with a suitably-defined matrix star operation through which the least fixed point solution is expressed. If the matrix A decomposes into block form as
B C
D E
then the star A* of the matrix has the block form
(B + C E* D)* (B + C E* D)* C E*
(E + D B* C)* D B* (E + D B* C)*
So, what this is actually saying is that for a vector-matrix system of the form
x ≥ a + B x + C y
y ≥ b + D x + E y
the least fixed point solution is given by
x = (B + C E* D)* (a + C E* b)
y = (E + D B* C)* (D B* a + b)
The star of a matrix, if expressed directly in terms of its components, will generally be huge and highly redundant. For an n×n matrix, it has size O(n³) - cubic in n - if you allow for redundant sub-expressions to be defined by macros. Otherwise, if you in-line insert all the redundancy then I think it blows up to a highly-redundant mess that is exponential in n in size.
So, there's intelligence required and involved (literally meaning: AI) in finding or pruning optimal forms that avoid the blow-up as much as possible. That's a non-trivial job for any purported matrix solver and regular expression synthesis compiler.
An heuristic, for your system, is to solve for the variables that don't have a "1" on the right-hand side and in-line substitute the solutions - and to work from bottom-up in terms of the dependency chain of the variables. That would mean starting with D and E first
D ≥ u* (v + w) B
E ≥ (u + v)* w C
In-line substitute into the other inequations
S ≥ u A + v B + w C
A ≥ 1 + (u + v + w) A
B ≥ 1 + u u* (v + w) B + (v + w) B
C ≥ 1 + (u + v) (u + v)* w C + w C
Apply Kleene algebra identities (e.g. x x* y + y = x* y)
S ≥ u A + v B + w C
A ≥ 1 + (u + v + w) A
B ≥ 1 + u* (v + w) B
C ≥ 1 + (u + v)* w C
Solve for the next layer of dependency up: A, B and C:
A ≥ (u + v + w)*
B ≥ (u* (v + w))*
C ≥ ((u + v)* w)*
Apply some more Kleene algebra (e.g. (x* y)* = 1 + (x + y)* y) to get
B ≥ 1 + N (v + w)
C ≥ 1 + N w
where, for convenience we set N = (u + v + w)*. In-line substitute at the top-level:
S ≥ u N + v (1 + N (v + w)) + w (1 + N w).
The least fixed point solution, in the main variable S, is thus:
S = u N + v + v N (v + w) + w + w N w.
where
N = (u + v + w)*.
As you can already see, even with this simple example, there's a lot of chess-playing to navigate through the system to find an optimally-pruned solution. So, it's certainly not a trivial problem. What you're essentially doing is synthesizing a control-flow structure for a program in a structured programming language from a set of goto's ... essentially the core process of reverse-compiling from assembly language to a high level language.
One measure of optimization is that of minimizing the loop-depth - which here means minimizing the depth of the stars or the star height. For example, the expression x* (y x*)* has star-height 2 but reduces to (x + y)*, which has star height 1. Methods for reducing star-height come out of the research by Hashiguchi and his resolution of the minimal star-height problem. His proof and solution (dating, I believe, from the 1980's or 1990's) is complex and to this day the process still goes on of making something more practical of it and rendering it in more accessible form.
Hashiguchi's formulation was cast in the older 1950's and 1960's formulation, predating the axiomatization of Kleene algebras (which was in the 1990's), so to date, nobody has rewritten his solution in entirely algebraic form within the framework of Kleene algebras anywhere in the literature ... as far as I'm aware. Whoever accomplishes this will have, as a result, a core element of an intelligent regular expression synthesis compiler, but also of a reverse-compiler and programming language synthesis de-compiler. Essentially, with something like that on hand, you'd be able to read code straight from binary and the lid will be blown off the world of proprietary systems. [Bite tongue, bite tongue, mustn't reveal secret yet, must keep the ring hidden.]
Given an expression in sympy, how do I re-write the expression as a polynomial defined as [1]:
D11*(omega**2/k**2)**3 + D22*(omega**2/k**2)**2 + D33*(omega**2/k**2) + D44 = 0
Note that this is different from the similar question asked here (Rewrite equation as polynomial).
Let
x=(omega**2/k**2)
then
D11*x**3 + D22*x**2 + D33*x + D44 = 0
I would like to find D11, D22, D33, and D44, given that x=omega**2/k**2
Normally, the collect function (http://docs.sympy.org/latest/tutorial/simplification.html) would collect similar terms, but in this situation, it does not seem to work well.
Here is a simple example that helps to explain what I am trying to accomplish. The output should be in the form D11*(omega**2/k**2)**3 + D22*(omega**2/k**2)**2 + D33*(omega**2/k**2) + D44 = 0
from sympy import symbols, collect
from IPython.display import display
omega = symbols('omega')
k = symbols('k')
a = symbols('a')
b = symbols('b')
c = symbols('c')
D11 = a*b*c
D22 = c+b
D33 = a+c*b + b
D44 = a+b
x = omega**2/k**2
expr = (D11*x**3 + D22*x**2 + D33*x + D44)
expr0 = expr.expand()
expr1 = collect(expr0, x)
display(expr1)
The output is:
a*b*c*omega**6/k**6 + a + b + omega**2*(a + b*c + b)/k**2 + omega**4*(b + c)/k**4
Although numerically correct, I would like the polynomial in the form [1] given above, and once it is in the form, I would like to extract the D11, D22, D33, and D44 coefficients.
Using evaluate=False in the collect function gets me closer to the goal, since the output now becomes:
{omega**2/k**2: a + b*c + b, omega**4/k**4: b + c, omega**6/k**6: a*b*c, 1: a + b}
Starting with your expr1,
expr1 = a*b*c*omega**6/k**6 + a + b + omega**2*(a + b*c + b)/k**2 + omega**4*(b + c)/k**4
it seems the easiest way to get the coefficients is to turn x into a symbol, at least temporarily:
Poly(expr1.subs(x, Symbol('x')), Symbol('x')).all_coeffs()
returns [a*b*c, b + c, a + b*c + b, a + b] (the coefficients are listed starting with the highest degree).
I would probably have x = Symbol('x') there to begin with, and only use expr.subs(x, omega**2/k**2) when needed.
SymPy's internal order of terms in a sum cannot be changed. To "rearrange" a SymPy expression means to print it in a more human-friendly form. This is largely a string manipulation problem, as we are no longer producing a SymPy object.
str(Poly(expr1.subs(x, Symbol('x')), Symbol('x'))).replace('x', '(' + str(x) + ')')
returns Poly(a*b*c*(omega**2/k**2)**3 + (b + c)*(omega**2/k**2)**2 + (a + b*c + b)*(omega**2/k**2) + a + b, (omega**2/k**2), domain='ZZ[a,b,c]')
Adding .split(',')[0].replace('Poly(', '') to the above removes the meta-data of a polynomial, leaving a*b*c*(omega**2/k**2)**3 + (b + c)*(omega**2/k**2)**2 + (a + b*c + b)*(omega**2/k**2) + a + b
When defining the verb centigrade to convert its argument from a Fahrenheit to a centigrade, the code in the book 'J Primer' is:
centigrade =. 3 : 0
t1 =. y. - 32
t2 =. t1 * 5
t3 =. t2 % 9
)
However, I had the spelling error in 'y.' part.
But when I type 'y' instead of 'y.' in the definition, it works.
centigrade =. 3 : 0
t1 =. y - 32
t2 =. t1 * 5
t3 =. t2 % 9
)
Why is that? Thanks!
Originally, the J language used x. and y. for the internal arguments to an explicit verb. You can reset to allow for y. or x. if you want to use legacy code by using the foreign conjunction 9!:49
9!:49 [ 0 for not accepting y. 9!:49 [ 1 for accepting y. 9!:48 '' gives the current status of the flag. http://www.jsoftware.com/help/dictionary/dx009.htm (bottom of page)
test=: 3 : 'x. + y.'
|spelling error
| x. + y.
| ^
| test=:3 :'x. + y.'
9!:49 [ 1
test=: 3 : 'x. + y.'
9!:49 [ 0
test=: 3 : 'x. + y.'
|spelling error
| x. + y.
| ^
| test=:3 :'x. + y.'