I have an example of a code and not sure what way is the best to use.
For example I have
if (x = 1) and (y = 2) and (if abc = false then check if z = 3) then
begin
...
check only
if x = 1
if y = 2
if abc = false check z = 3. if abc = true then dont check z = 3
i am not sure if i am explaining the best but hopefuly people will understand.
I want to know if this is possible or the best way to do it. Keeping in mind that rather than in example where its x, y, z and abc. there can be more in my use.
I currently have structure as...which i dont think is practical, and think theres a better way but i am not sure
if (abc = false) then
begin
if (x = 1) and (y = 2) and (z = 3) then
begin
...
end
else
begin
if (x = 1) and (y = 2) then
begin
...
Thanks in advance
I think you're looking for or. Now you will check that x must be 1, y must be 2, and if abc is false, z must be 3.
If abc = true, z can still be three, but it won't be checked.
Note that I just wrote abc instead of abc = true. Since it's a Boolean (true/false) already, that's allowed.
Also note how the operations are grouped using parentheses. The total sub-expression abc or (z=3) must return true for the total expression to return true.
Furthermore the sequence of the terms is significant - they are evaluated left-to-right. If the term (abc or (z=3)) is replaced by the logically-equivalent term ((z=3) or abc) then z=3 will be evaluated.
if (x = 1) and (y = 2) and (abc or (z = 3)) then
// Your magic goes here
Test program body to prove sequence is important
function z : Integer;
begin
writeln('Z being evaluated');
result := x + y;
end;
begin
x := 1;y := 2;
abc := true;
if (x=1) and (y=2) and (abc or (z=3)) then
writeln ('evaluated true')
else
writeln ('evaluated false');
writeln('done');
readln;
end.
Neither of your code samples compile, because neither is using the proper syntax.
This should get you started:
if (x = 1) and (y = 2) then
begin
if (abc) then
// Handle abc = True
else
begin
if (z = 3) then
// Handle abc = false and z = 3
else
// Handle abc = false and z <> 3
end;
end;
Related
I am trying to prove a property in Dafny, which makes use of powers.
Concretely, this one: forall x,y in Reals : 2xy <= x^2+y^2. I implemented this idea in the following lemma:
lemma product2_lessEqual_powProduct (x:real, y:real)
requires 0.0<x<=1.0 && 0.0<y<=1.0
ensures 2.0*x*y <= (x*x)+(y*y)
{}
Which is verified with no problem (I guess some automatic induction is performed below).
However, I would like to use an own power function in order to make power(x,2) instead of x*x. Thus, I took a power function from https://github.com/bor0/dafny-tutorial/blob/master/pow.dfy, which is as follows:
function method power(A:int, N:nat):int
{
if (N==0) then 1 else A * power(A,N-1)
}
method pow(A:int, N:int) returns (x:int)
requires N >= 0
ensures x == power(A, N)
{
x := 1;
var i := N;
while i != 0
invariant x == power(A, (N-i))
{
x := x * A;
i := i - 1;
}
}
However, since I am using real values for the basis of the exponential, I modified it a bit so that it works for exponentials:
function method power(A:real, N:nat):real
{
if (N==0) then 1.0 else A * power(A,N-1)
}
method pow(A:real, N:int) returns (x:real)
requires N >= 0
ensures x == power(A, N)
{
x := 1.0;
var i := N;
while i != 0
invariant x == power(A, (N-i))
{
x := x * A;
i := i - 1;
}
}
Thus, I wanted to test it with the previous lemma:
lemma product2_lessEqual_powProduct (x:real, y:real)
requires 0.0<x<=1.0 && 0.0<y<=1.0
ensures 2.0*x*y <= power(x,2)+power(y,2)
{}
Surprisingly, it tells me the typical A postcondition might not hold on this return path.Verifier.
Can anyone explain why this happens? Why is it verifying with primitive operations of Dafny, but not when I define them functions? And how could I prove this lemma now?
Even though second parameter of power is concrete and small, Dafny is not doing enough unrolling to prove desired fact. Adding {:fuel 2} to power makes proof go through. You can read more about fuel here https://dafny.org/dafny/DafnyRef/DafnyRef.html#sec-fuel
function method {:fuel 2} power(A:real, N:nat):real
{
if (N==0) then 1.0 else A * power(A,N-1)
}
method pow(A:real, N:int) returns (x:real)
requires N >= 0
ensures x == power(A, N)
{
x := 1.0;
var i := N;
while i != 0
invariant x == power(A, (N-i))
{
x := x * A;
i := i - 1;
}
}
lemma product2_lessEqual_powProduct (x:real, y:real)
requires 0.0<x<=1.0 && 0.0<y<=1.0
ensures 2.0*x*y <= power(x,2)+power(y,2)
{}
It's surprising until you realize that there is a mathematical theory for A*A, but power(A, 2) requires two unfolding of power to have a meaning.
If you want your function to work seamlessly with the theory and prove your last lemma, you can give it precise postconditions:
function method power(A:real, N:nat): (result: real)
ensures N == 1 ==> result == A
ensures N == 2 ==> result == A*A
{
if (N==0) then 1.0 else A * power(A,N-1)
}
I tested it, your second lemma verifies.
I have the following functions:
P[t_] := P[t] = P[t-1] +a*ED[t-1];
ED[t_] := ED[t] = DF[t] + DC[t];
DF[t_] := DF[t] = b (F - P[t]);
DC[t_] := DC[t] = c (P[t] - F);
And the following parameters:
a=1;
c=0.2;
b = 0.75;
F=100;
In Mathematica I use the function "ListLinePlot" in order to plot P[t] and F:
ListLinePlot[{Table[P[t], {t, 0, 25}], Table[F, {t, 0, 25}]}, PlotStyle → {Black, Red},Frame → True, FrameLabel → {"time", "price"}, AspectRatio → 0.4, PlotRange → All]
How can I do this in wxMaxima? Is there a similar function or an alternative to ListLinePlot?
This is my attempt in wxMaxima:
P[t] := P[t-1] + a * ED[t-1];
ED[t] := DF[t] + DC[t];
DF[t] := b*[F-P[t]];
DC[t] := c*[P[t]-F];
a=1;
c=0.2;
b=0.75;
F=100;
And then I tried:
draw2d(points(P[t], [t,0,25]))
The plotted function should look like this:
OK, I've adapted the code you showed above. This works for me. I'm working with Maxima 5.44 on macOS.
P[t] := P[t-1] + a * ED[t-1];
ED[t] := DF[t] + DC[t];
DF[t] := b*(F-P[t]);
DC[t] := c*(P[t]-F);
a:1;
c:0.2;
b:0.75;
F:100;
P[0]: F + 1;
Pt_list: makelist (P[t], t, 0, 25);
load (draw);
set_draw_defaults (terminal = qt);
draw2d (points_joined = true, points(Pt_list));
Notes. (1) There needs to be a base case for the recursion on P. I put P[0]: F + 1. (2) Assignments are : instead of =. Note that x = y is a symbolic equation instead of an assignment. (3) Square brackets [ ] are only for subscripts and lists. Use parentheses ( ) for grouping expressions. (4) Syntax for draw2d is a little different, I fixed it up. (I put a default for terminal since the built-in value is incorrect for Maxima on macOS; if you are working on Linux or Windows, you can omit that.)
EDIT: Try this to draw a horizontal line as well.
draw2d (points_joined = true, points(Pt_list),
color = red, points([[0, F], [25, F]]),
yrange = [F - 1, P[0] + 1]);
Probably a basic question related to Z3: i am trying to get all solutions of a boolean expression, e.g. for a OR b, i want to get {(true, true),(false,true),(true,false)}
Based on other responses found, e.g. Z3: finding all satisfying models, i have the following code:
a = Bool('a')
b = Bool('b')
f1=Or(a,b)
s=Solver()
s.add(f1)
while s.check() == sat:
print s
s.add(Not(And(a == s.model()[a], b == s.model()[b])))
The issue is that it enters an infinite loop as at the second iteration: the constraint a == s.model()[a] is evaluated to false b/c s.model()[a] does not exist anymore.
Can someone tell what i am doing wrong?
I would advice you to try writing your loop like this instead:
from z3 import *
a = Bool('a')
b = Bool('b')
f1 = Or(a,b)
s = Solver()
s.add(f1)
while s.check() == sat:
m = s.model()
v_a = m.eval(a, model_completion=True)
v_b = m.eval(b, model_completion=True)
print("Model:")
print("a := " + str(v_a))
print("b := " + str(v_b))
bc = Or(a != v_a, b != v_b)
s.add(bc)
The output is:
Model:
a := True
b := False
Model:
a := False
b := True
Model:
a := True
b := True
The argument model_completion=True is necessary because otherwise m.eval(x) behaves like the identity relation for any x Boolean variable with a don't care value in the current model m and it returns x as a result instead of True/False. (See related Q/A)
NOTE: since z3 kindly marks don't care Boolean variables, an alternative option would be to write your own model generator that auto-completes any partial model. This would reduce the number of calls to s.check(). The performance impact of this implementation is hard to gauge, but it might be slightly faster.
I've progressed steadily on this issue, however I'm coming acrossed a problem in validating if a spot is clear or not... Perhaps my explanation isn't good so I'll just leave it in code:
func.CheckNear = function(field, pos)
local x, y = pos[1], pos[2];
local coordinates = {{x + 1, y}, {x - 1, y}, {x, y + 1}, {x, y - 1}}
for key, array in next, coordinates do
local field = field[array[2]];
if field then
if (field[coordinates[1]]) then
if field[coordinates[1]] == "*" then
coordinates[key] = nil;
end;
else
coordinates[key] = nil;
end;
else
coordinates[key] = nil
end;
end;
return coordinates;
end;
I found that sometimes I must give types explicitly for pattern variables, otherwise Rascal would not work as expected. The following session in the Console says it all:
rascal>data foo = bar(int);
ok
rascal>int x = 1;
int: 1
rascal>[x | bar(x) <- [bar(2), bar(3)]];
list[void]: []
rascal>[x | bar(int x) <- [bar(2), bar(3)]];
list[int]: [2,3]
Why did this happen?
In the current version of Rascal it is such that variables in patterns that exist in the surrounding scope are not matched and shadowed, but rather checked for equality.
So:
<int x, x> := <2,2> => true // x is first introduced and then checked for equality
<int x, x> := <2,3> => false // x is first introduced and then checked for equality
{ int x = 1; if (x := 2) println("true"); else println("false"); // false!
And this holds for all places where we use pattern matching.
We have had several complaints about this particular design of "non-linear matching", and we intend to add an operator soon ($) to identify the intention of taking something from the surround scope. If the operator is not used, then shadowing will occur:
<int x, $x> := <2,2> => true // x is first introduced and then checked for equality
<int x, $x> := <2,3> => false // x is first introduced and then checked for equality
<int x, x> := <2,3> // static error due to illegal shadowing
<int x, y> := <2,3> => true // x and y are both introduced
{ int x = 1; if ($x := 2) println("true"); else println("false"); // false!
{ int x = 1; if (x := 2) println("true <x>"); else println("false"); // true, prints 2! or perhaps a static error.
Might also add some additional power to get expressions into patterns as in:
<1, ${1 + 2 + 3}> := <1,6> // true