So I would like to print polynomials in one variable (s) with one parameter (a), say
a·s^3 − s^2 - a^2·s − a + 1.
Sage always displays it with decreasing degree, and I would like to get something like
1 - a - a^2·s - s^2 + a·s^3
to export it to LaTeX. I can't figure out how to do this... Thanks in advance.
As an alternative to string manipulation, one can use the series expansion.
F = a*s^3 - s^2 - a^2*s - a + 1
F.series(s, F.degree(s)+1)
returns
(-a + 1) + (-a^2)*s + (-1)*s^2 + (a)*s^3
which appears to be what you wanted, save for some redundant parentheses.
This works because (a) a power series is ordered from lowest to highest coefficients; (b) making the order of remainder greater than the degree of the polynomial ensures that the series is just the polynomial itself.
This is not easy, because the sort order is defined in Pynac, a fork of Ginac, which Sage uses for its basic symbolic manipulation. However, depending on what you need, it is possible programmatically:
sage: F = 1 + x + x^2
sage: "+".join(map(str,sorted([f for f in F.operands()],key=lambda exp:exp.degree(x))))
'1+x+x^2'
I don't know whether this sort of thing is powerful enough for your needs, though. You may have to traverse the "expression tree" quite a bit but at least your sort of example seems to work.
sage: F = a + a^2*x + x^2 - a*x^2
sage: "+".join(map(str,sorted([f for f in F.operands()],key=lambda exp:exp.degree(x))))
'a+a^2*x+-a*x^2+x^2'
Doing this in a short statement requires a number of Python tricks like this, which are very well worth learning if you are going to use Sage (or Numpy, or pandas, or ...) a fair amount.
Related
I have a fairly large expression that involves a lot of subexpressions of the form (100*A^3 + 200*A^2 + 100*A)*x or (-A^2 - A)*y or (100*A^2 + 100*A)*z
I know, but I don't know how to tell Maxima this, that it in this case is valid to make the approximation A+1 ~ A, thereby effectively removing anything but the highest power of A in each coefficient.
I'm now looking for functions, tools, or methods that I can use to guide Maxima in dropping various terms that aren't important.
I have attempted with subst, but that requires me to specify each and every factor separately, because:
subst([A+1=B], (A+2)*(A+1)*2);
subst([A+1=B], (A+2)*(A*2+2));
(%o1) 2*(A+2)*B
(%o2) (A+2)*(2*A+2)
(that is, I need to add one expression for each slightly different variant)
I tried with ratsimp, but that's too eager to change every occurrence:
ratsubst(B, A+1, A*(A+1)*2);
ratsubst(B, A+1, A*(A*2+2));
(%o3) 2*B^2-2*B
(%o4) 2*B^2-2*B
which isn't actually simpler, as I would have preferred the answer to have been given as 2*B^2.
In another answer, (https://stackoverflow.com/a/22695050/5999883) the functions let and letsimp were suggested for the task of substituting values, but I fail to get them to really do anything:
x:(A+1)*A;
let ( A+1, B );
letsimp(x);
(x)A*(A+1)
(%o6) A+1 --\> B
(%o7) A^2+A
Again, I'd like to approximate this expression to A^2 (B^2, whatever it's called).
I understand that this is, in general, a hard problem (is e.g. A^2 + 10^8*A still okay to approximate as A^2?) but I think that what I'm looking for is a function or method of calculation that would be a little bit smarter than subst and can recognize that the same substitution could be done in the expression A^2+A as in the expression 100*A^2+100*A or -A^2-A instead of making me create a list of three (or twenty) individual substitutions when calling subst. The "nice" part of the full expression that I'm working on is that each of these A factors are of the form k*A^n*(A+1)^m for various small integers n, m, so I never actually end up with the degenerate case mentioned above.
(I was briefly thinking of re-expressing my expression as a polynomial in A, but this will not work as the only valid approximation of the expression (A^3+A^2+A)*x + y is A^3*x + y -- I know nothing about the relative sizes of x and y.
I have a panel data set for which I would like to calculate moving averages across years.
Each year is a variable for which there is an observation for each state, and I would like to create a new variable for the average of every three year period.
For example:
P1947=rmean(v1943 v1944 v1945), P1947=rmean(v1944 v1945 v1946)
I figured I should use a foreach loop with the egen command, but I'm not sure about how I should refer to the different variables within the loop.
I'd appreciate any guidance!
This data structure is quite unfit for purpose. Assuming an identifier id you need to reshape, e.g.
reshape long v, i(id) j(year)
tsset id year
Then a moving average is easy. Use tssmooth or just generate, e.g.
gen mave = (L.v + v + F.v)/3
or (better)
gen mave = 0.25 * L.v + 0.5 * v + 0.25 * F.v
More on why your data structure is quite unfit: Not only would calculation of a moving average need a loop (not necessarily involving egen), but you would be creating several new extra variables. Using those in any subsequent analysis would be somewhere between awkward and impossible.
EDIT I'll give a sample loop, while not moving from my stance that it is poor technique. I don't see a reason behind your naming convention whereby P1947 is a mean for 1943-1945; I assume that's just a typo. Let's suppose that we have data for 1913-2012. For means of 3 years, we lose one year at each end.
forval j = 1914/2011 {
local i = `j' - 1
local k = `j' + 1
gen P`j' = (v`i' + v`j' + v`k') / 3
}
That could be written more concisely, at the expense of a flurry of macros within macros. Using unequal weights is easy, as above. The only reason to use egen is that it doesn't give up if there are missings, which the above will do.
FURTHER EDIT
As a matter of completeness, note that it is easy to handle missings without resorting to egen.
The numerator
(v`i' + v`j' + v`k')
generalises to
(cond(missing(v`i'), 0, v`i') + cond(missing(v`j'), 0, v`j') + cond(missing(v`k'), 0, v`k')
and the denominator
3
generalises to
!missing(v`i') + !missing(v`j') + !missing(v`k')
If all values are missing, this reduces to 0/0, or missing. Otherwise, if any value is missing, we add 0 to the numerator and 0 to the denominator, which is the same as ignoring it. Naturally the code is tolerable as above for averages of 3 years, but either for that case or for averaging over more years, we would replace the lines above by a loop, which is what egen does.
There is a user written program that can do that very easily for you. It is called mvsumm and can be found through findit mvsumm
xtset id time
mvsumm observations, stat(mean) win(t) gen(new_variable) end
I have to partition a multiset into two sets who sums are equal. For example, given the multiset:
1 3 5 1 3 -1 2 0
I would output the two sets:
1) 1 3 3
2) 5 -1 2 1 0
both of which sum to 7.
I need to do this using Z3 (smt2 input format) and "Linear Arithmetic Logic", which is defined as:
formula : formula /\ formula | (formula) | atom
atom : sum op sum
op : = | <= | <
sum : term | sum + term
term : identifier | constant | constant identifier
I honestly don't know where to begin with this and any advice at all would be appreciated.
Regards.
Here is an idea:
1- Create a 0-1 integer variable c_i for each element. The idea is c_i is zero if element is in the first set, and 1 if it is in the second set. You can accomplish that by saying that 0 <= c_i and c_i <= 1.
2- The sum of the elements in the first set can be written as 1*(1 - c_1) + 3*(1 - c_2) + ... +
3- The sum of the elements in the second set can be written as 1*c1 + 3*c2 + ...
While SMT-Lib2 is quite expressive, it's not the easiest language to program in. Unless you have a hard requirement that you have to code directly in SMTLib2, I'd recommend looking into other languages that have higher-level bindings to SMT solvers. For instance, both Haskell and Scala have libraries that allow you to script SMT solvers at a much higher level. Here's how to solve your problem using the Haskell, for instance: https://gist.github.com/1701881.
The idea is that these libraries allow you to code at a much higher level, and then perform the necessary translation and querying of the SMT solver for you behind the scenes. (If you really need to get your hands onto the SMTLib encoding of your problem, you can use these libraries as well, as they typically come with the necessary API to dump the SMTLib they generate before querying the solver.)
While these libraries may not offer everything that Z3 gives you access to via SMTLib, they are much easier to use for most practical problems of interest.
I'm writing a simple parser/interpreter for a language. The instructions keep mentioning 'deferred substitution', as in
Extend the fun language feature described so that functions
can accept a list of zero or more arguments instead of just one. All
arguments to the function must evaluate with the same deferred
substitutions.
I don't need any help with implementing this, I'm just confused about what 'deferred substitution' means. Any thoughts?
Deferred substitution refers to the practice of substituting the values of variables at the latest step possible. By doing so, you are deferring the substitution of it!
Here's an example that might help you understand what it means:
Suppose that you have the following function:
f(x) = 500 + 300 + 2x + 45x
Let's say that x = 1
If you want to defer the substitution of x, you would probably do:
f(x) = 800 + 2x + 45x
f(x) = 800 + 47x
f(1) = 800 + 47(1)
Notice that we have substituted the values of x at the latest step possible, after simplifying everything that is not a variable in this function.
Suppose you've got a single linked list of size N, and you want to perform an operation on every element, beginning at the end.
I've come up with the following pseudocode:
while N > 0
Current = LinkedList
for 0 to N
Current = Current.tail
end
Operation(Current.head)
N := N-1
end
Now I've got to determine which Big-O this algorithm is.
Supposing that Operation() is O(1), I think it's something like this:
N + (N-1) + (N-2) + ... + (N-(N-1)) + 1
But I'm not sure what Big-O that actually is. I think it is definitely smaller than O(N^2), but I don't think you can say its O(N) either ...
Your equation is basically that of the triangular numbers, and sums to N(N+1)/2. I'll leave you to determine the O() from that!
A quicker way to do this is to construct a new list that is the reverse of the original list, and then perform the operations on that.
Your algorithm is O(n^2) as you suggest in your post. You can do it in O(n), though.
It's important to remember that Big-O notation is an upper bound on the algorithm's time complexity.
1+2+3+...+n = n*(n+1)/2 = 0.5*n^2+O(n)
This is O(n^2), and O(n^2) is tight, i.e. there is no lower runtime order that'd contain your runtime.
A faster algorithm that works from front-to-back could have O(n) instead of O(n^2)
Your runtime analysis is correct, the runtime is 1 + 2 + ... + N which is a sum of the arithmetic progression and therefore = (N²-N) / 2.