Maxima often asks positive, negative, zero ? while solving ODE. Is there a way to see all of them at once ?
Load the noninteractive package, which comes bundled with recent Maxima versions, before solving the ODE.
You can load it with:
load(noninteractive);
Related
I tried to run ENMTools on Windows. And I set ENMTools option and Maxent option as manual suggested. The “RAM to assign to Maxent” option was set as 512. But when I calculate the niche overlap from two ASCII files, the software kept saying "out of memory". I wonder what could I do now? Any help would be appreciated.
The “RAM to assign to Maxent” option was set as 512 and 1000, but they all failed.
This question is related to my previous question
Is it possible to get a legit range info when using a SMT constraint with Z3
So it seems that "efficiently" finding the maximum range info is not proper, given typical 32-bit vectors and so on. But on the other hand, I am thinking whether it is feasible to find certain "sub-maximum" range info, which hopefully becomes more efficient. Another thing is that we may want to have certain "safe" guarantee, say for all elements in the sub-maximum range, they must satisfy the constraint, but there could exist some other solutions that would satisfy the constraint as well.
I am currently exploring whether model counting technique could make sense in this setting. Any thoughts would be appreciated very much. Thanks.
General case
This is not just a question of efficiency. Consider a problem where you have two variables a and b, and a single constraint:
a != b
What's the range of b? (maximum or otherwise?)
You can say all values are legitimate. But that would be wrong, as obviously the choice of a impacts the choice of b. The more variables you have around, the more complicated the problem will become. I don't think the problem is even well defined in this case, so searching for a solution (efficient or otherwise) doesn't make much sense.
Single variable assumption
Having said that, I think you can come up with a solution if you assume there's precisely one variable in the system. (Or, alternatively, if you fix all the other variables to some predefined constants.) If you're willing to go down this path, then you can implement a binary search algorithm to find a reasonably sized range by simply proving the quantified formula
Exists([b], And(b >= minBound, b <= maxBound, Not(constraints)))
Once you get unsat for this, you have your range. So long as you get sat, you can adjust your minBound/maxBound to search within smaller ranges. In the worst case, this can turn into a linear walk, but you can "cut-down" this search by making sure you go down a significant size in each step. That could be a parameter to the whole search, depending on how large you want your intervals to be. It'll have to be a choice between trying to find a maximal range, and how long you want to spend in this search. Of course, if you cut-down too much, you can miss a big interval, but that's the cost of efficiency.
Example1 (Good case) There's a single constraint that says b != 5. Then your search will be quick and depending on which branch you'll go, you'll either find [0, 4] or [6, 255] assuming 8-bit words.
Example2 (Bad case) There's a single constraint that says b is even. Then your search will exhibit worst-case behavior, and if your "cut-down" size is 1, you'll possibly iterate 255 times before you settle down on [0, 0]; assuming z3 gives you the maximum odd number in each call.
I hope that illustrates the point. In general, though, I'd assume you'd be closer to the "good case" for practical applications and even if your cut-down size is minimal you can most likely converge in a few iterations. Of course, this entirely depends on your problem domain, but I'd expect it to hold for software analysis in general.
I saw in a previous post from last August that Z3 did not support optimizations.
However it also stated that the developers are planning to add such support.
I could not find anything in the source to suggest this has happened.
Can anyone tell me if my assumption that there is no support is correct or was it added but I somehow missed it?
Thanks,
Omer
If your optimization has an integer valued objective function, one approach that works reasonably well is to run a binary search for the optimal value. Suppose you're solving the set of constraints C(x,y,z), maximizing the objective function f(x,y,z).
Find an arbitrary solution (x0, y0, z0) to C(x,y,z).
Compute f0 = f(x0, y0, z0). This will be your first lower bound.
As long as you don't know any upper-bound on the objective value, try to solve the constraints C(x,y,z) ∧ f(x,y,z) > 2 * L, where L is your best lower bound (initially, f0, then whatever you found that was better).
Once you have both an upper and a lower bound, apply binary search: solve C(x,y,z) ∧ 2 * f(x,y,z) > (U - L). If the formula is satisfiable, you can compute a new lower bound using the model. If it is unsatisfiable, (U - L) / 2 is a new upper-bound.
Step 3. will not terminate if your problem does not admit a maximum, so you may want to bound it if you are not sure it does.
You should of course use push and pop to solve the succession of problems incrementally. You'll additionally need the ability to extract models for intermediate steps and to evaluate f on them.
We have used this approach in our work on Kaplan with reasonable success.
Z3 currently does not support optimization. This is on the TODO list, but it has not been implemented yet. The following slide decks describe the approach that will be used in Z3:
Exact nonlinear optimization on demand
Computation in Real Closed Infinitesimal and Transcendental Extensions of the Rationals
The library for computing with infinitesimals has already been implemented, and is available in the unstable (work-in-progress) branch, and online at rise4fun.
I am trying to run this code but it keeps crashing:
log10(x):=log(x)/log(10);
char(x):=floor(log10(x))+1;
mantissa(x):=x/10**char(x);
chop(x,d):=(10**char(x))*(floor(mantissa(x)*(10**d))/(10**d));
rnd(x,d):=chop(x+5*10**(char(x)-d-1),d);
d:5;
a:10;
Ibwd:[[30,rnd(integrate((x**60)/(1+10*x^2),x,0,1),d)]];
for n from 30 thru 1 step -1 do Ibwd:append([[n-1,rnd(1/(2*n-1)-a*last(first(Ibwd)),d)]],Ibwd);
Maxima crashes when it evaluates the last line. Any ideas why it may happen?
Thank you so much.
The problem is that the difference becomes negative and your rounding function dies horribly with a negative argument. To find this out, I changed your loop to:
for n from 30 thru 1 step -1 do
block([],
print (1/(2*n-1)-a*last(first(Ibwd))),
print (a*last(first(Ibwd))),
Ibwd: append([[n-1,rnd(1/(2*n-1)-a*last(first(Ibwd)),d)]],Ibwd),
print (Ibwd));
The last difference printed before everything fails miserably is -316539/6125000. So now try
rnd(-1,3)
and see the same problem. This all stems from the fact that you're taking the log of a negative number, which Maxima interprets as a complex number by analytic continuation. Maxima doesn't evaluate this until it absolutely has to and, somewhere in the evaluation code, something's dying horribly.
I don't know the "fix" for your specific example, since I'm not exactly sure what you're trying to do, but hopefully this gives you enough info to find it yourself.
If you want to deconstruct a floating point number, let's first make sure that it is a bigfloat.
say z: 34.1
You can access the parts of a bigfloat by using lisp, and you can also access the mantissa length in bits by ?fpprec.
Thus ?second(z)*2^(?third(z)-?fpprec) gives you :
4799148352916685/140737488355328
and bfloat(%) gives you :
3.41b1.
If you want the mantissa of z as an integer, look at ?second(z)
Now I am not sure what it is that you are trying to accomplish in base 10, but Maxima
does not do internal arithmetic in base 10.
If you want more bits or fewer, you can set fpprec,
which is linked to ?fpprec. fpprec is the "approximate base 10" precision.
Thus fpprec is initially 16
?fpprec is correspondingly 56.
You can easily change them both, e.g. fpprec:100
corresponds to ?fpprec of 335.
If you are diddling around with float representations, you might benefit from knowing
that you can look at any of the lisp by typing, for example,
?print(z)
which prints the internal form using the Lisp print function.
You can also trace any function, your own or system function, by trace.
For example you could consider doing this:
trace(append,rnd,integrate);
If you want to use machine floats, I suggest you use, for the last line,
for n from 30 thru 1 step -1 do :
Ibwd:append([[n-1,rnd(1/(2.0*n- 1.0)-a*last(first(Ibwd)),d)]],Ibwd);
Note the decimal points. But even that is not quite enough, because integration
inserts exact structures like atan(10). Trying to round these things, or compute log
of them is probably not what you want to do. I suspect that Maxima is unhappy because log is given some messy expression that turns out to be negative, even though it initially thought otherwise. It hands the number to the lisp log program which is perfectly happy to return an appropriate common-lisp complex number object. Unfortunately, most of Maxima was written BEFORE LISP HAD COMPLEX NUMBERS.
Thus the result (log -0.5)= #C(-0.6931472 3.1415927) is entirely unexpected to the rest of Maxima. Maxima has its own form for complex numbers, e.g. 3+4*%i.
In particular, the Maxima display program predates the common lisp complex number format and does not know what to do with it.
The error (stack overflow !!!) is from the display program trying to display a common lisp complex number.
How to fix all this? Well, you could try changing your program so it computes what you really want, in which case it probably won't trigger this error. Maxima's display program should be fixed, too. Also, I suspect there is something unfortunate in simplification of logs of numbers that are negative but not obviously so.
This is probably waaay too much information for the original poster, but maybe the paragraph above will help out and also possibly improve Maxima in one or more places.
It appears that your program triggers an error in Maxima's simplification (algebraic identities) code. We are investigating and I hope we have a bug fix soon.
In the meantime, here is an idea. Looks like the bug is triggered by rnd(x, d) when x < 0. I guess rnd is supposed to round x to d digits. To handle x < 0, try this:
rnd(x, d) := if x < 0 then -rnd1(-x, d) else rnd1(x, d);
rnd1(x, d) := (... put the present definition of rnd here ...);
When I do that, the loop runs to completion and Ibwd is a list of values, but I don't know what values to expect.
Has anybody got any ideas on this one?
When we run:
printf("%.0f", 40.5)
On a windows box the return is "41" but on our production ubuntu server we're getting "40"
How about using .round instead? Rails even enhances it so that you can specify the precision (see API doc).
use %.0g instead
ScottJ's answer does not explain your problem -- 40.5 is exactly representable in binary. What's probably happening is this: Windows' printf "rounds half up," and Ubuntu's printf "rounds half even."
Looks like a simple case of binary floating point imprecision. On one machine you get 40.499999999 which rounds to 40; on the other machine you get 40.500000000001 which rounds to 41.
If you need exact numbers then you should not use binary floating point. You can use fixed point decimal, or decimal floating point.
Edit: you're using BigDecimal, you say. Why not avoid any conversion to float by using #round and then #to_i? (Or #floor or #ceil instead of #round... it's not clear what your goal is.)
b = BigDecimal.new("40.5")
print b.round.to_i # => 41