If I want to compute a^b mod c then there is an efficient way of doing this, without computing a^b in full.
However, when programming, if I write f g x then g(x) is computed regardless of f.
J provides for composing f and g in special cases, and the modulo power function is one of them. For instance, the following runs extremely quickly.
1000&| # (2&^) 10000000x
This is because the 'atop' conjunction # tells the language to compose the functions if possible. If I remove it, it goes unbearably slowly.
If however I want to work with x^x , then ^~ no longer works and I get limit errors for large values. Binding those large values however does work.
So
999&| # (100333454&^) 100333454x
executes nice and fast but
999&| # ^~ 100333454x
gives me a limit error - the RHS is too big.
Am I right in thinking that in this case, J is not using an efficient power-modulo algorithm?
Per the special code page:
m&|#^ dyad avoids exponentiation for integer arguments
m&|#(n&^) monad avoids exponentiation for integer arguments
The case for ^~ is not supported by special code.
https://github.com/Pascal-J/BN-openssl-bindings-for-J
includes bindings for openssl (distributed with J on windows, and pre-installed on all other compatible systems)'s BN library including modexp. The x ^ x case will be especially efficient due to fewer parameters converted from J to "C" (BN)
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 try to eliminate "not" in the expression using the script http://rise4fun.com/Z3/XwGt however, it doesn't work, there is still not in the result, would someone help me?
The output goals in your example are in simplified formula. There is no redundant assertion in any goal. The Z3 simplified form always use non-strict inequalities such as t1 <= t2 and t1 >= t2. Strict inequalities such as t1 < 0 are encoded using negation. That is, t1 < 0 is encoded as not t1 >= 0. The idea is to reduce the number of atoms used to encode a big formula.
It would be useful to understand why you need to eliminate the nots in the output. If there is an useful application for that, we will include a new tactic (in the next release) that performs the transformation.
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.
By definition the integer division returns the quotient.
Why 4613.9145 div 100. gives an error ("bad argument") ?
For div the arguments need to be integers. / accepts arbitrary numbers as arguments, especially floats. So for your example, the following would work:
1> 4613.9145 / 100.
46.139145
To contrast the difference, try:
2> 10 / 10.
1.0
3> 10 div 10.
1
Documentation: http://www.erlang.org/doc/reference_manual/expressions.html
Update: Integer division, sometimes denoted \, can be defined as:
a \ b = floor(a / b)
So you'll need a floor function, which isn't in the standard lib.
% intdiv.erl
-module(intdiv).
-export([floor/1, idiv/2]).
floor(X) when X < 0 ->
T = trunc(X),
case X - T == 0 of
true -> T;
false -> T - 1
end;
floor(X) ->
trunc(X) .
idiv(A, B) ->
floor(A / B) .
Usage:
$ erl
...
Eshell V5.7.5 (abort with ^G)
> c(intdiv).
{ok,intdiv}
> intdiv:idiv(4613.9145, 100).
46
Integer division in Erlang, div, is defined to take two integers as input and return an integer. The link you give in an earlier comment, http://mathworld.wolfram.com/IntegerDivision.html, only uses integers in its examples so is not really useful in this discussion. Using trunc and round will allow you use any arguments you wish.
I don't know quite what you mean by "definition." Language designers are free to define operators however they wish. In Erlang, they have defined div to accept only integer arguments.
If it is the design decisions of Erlang's creators that you are interested in knowing, you could email them. Also, if you are curious enough to sift through the (remarkably short) grammar, you can find it here. Best luck!
Not sure what you're looking for, #Bertaud. Regardless of how it's defined elsewhere, Erlang's div only works on integers. You can convert the arguments to integers before calling div:
trunc(4613.9145) div 100.
or you can use / instead of div and convert the quotient to an integer afterward:
trunc(4613.9145 / 100).
And trunc may or may not be what you want- you may want round, or floor or ceiling (which are not defined in Erlang's standard library, but aren't hard to define yourself, as miku did with floor above). That's part of the reason Erlang doesn't assume something and do the conversion for you. But in any case, if you want an integer quotient from two non-integers in Erlang, you have to have some sort of explicit conversion step somewhere.
I just asked a question about how the Erlang compiler implements pattern matching, and I got some great responses, one of which is the compiled bytecode (obtained with a parameter passed to the c() directive):
{function, match, 1, 2}.
{label,1}.
{func_info,{atom,match},{atom,match},1}.
{label,2}.
{test,is_tuple,{f,3},[{x,0}]}.
{test,test_arity,{f,3},[{x,0},2]}.
{get_tuple_element,{x,0},0,{x,1}}.
{test,is_eq_exact,{f,3},[{x,1},{atom,a}]}.
return.
{label,3}.
{badmatch,{x,0}}
Its all just plain Erlang tuples. I was expecting some cryptic binary thingy, guess not. I am asking this on impulse here (I could look at the compiler source but asking questions always ends up better with extra insight), how is this output translated in the binary level?
Say {test,is_tuple,{f,3},[{x,0}]} for example. I am assuming this is one instruction, called 'test'... anyway, so this output would essentially be the AST of the bytecode level language, from which the binary encoding is just a 1-1 translation?
This is all so exciting, I had no idea that I can this easily see what the Erlang compiler break things into.
ok so I dug into the compiler source code to find the answer, and to my surprise the asm file produced with the 'S' parameter to the compile:file() function is actually consulted in as is (file:consult()) and then the tuples are checked one by one for further action(line 661 - beam_consult_asm(St) -> - compile.erl). further on then there's a generated mapping table in there (compile folder of the erlang source) that shows what the serial number of each bytecode label is, and Im guessing this is used to generate the actual binary signature of the bytecode.
great stuff. but you just gotta love the consult() function, you can almost have a lispy type syntax for a random language and avoid the need for a parser/lexer fully and just consult source code into the compiler and do stuff with it... code as data data as code...
The compiler has a so-called pattern match compiler which will take a pattern and compile it down to what is essentially a series of branches, switches and such. The code for Erlang is in v3_kernel.erl in the compiler. It uses Simon Peyton Jones, "The Implementation of Functional
Programming Languages", available online at
http://research.microsoft.com/en-us/um/people/simonpj/papers/slpj-book-1987/
Another worthy paper is the one by Peter Sestoft,
http://www.itu.dk/~sestoft/papers/match.ps.gz
which derives a pattern match compiler by inspecting partial evaluation of a simpler system. It may be an easier read, especially if you know ML.
The basic idea is that if you have, say:
% 1
f(a, b) ->
% 2
f(a, c) ->
% 3
f(b, b) ->
% 4
f(b, c) ->
Suppose now we have a call f(X, Y). Say X = a. Then only 1 and 2 are applicable. So we check Y = b and then Y = c. If on the other hand X /= a then we know that we can skip 1 and 2 and begin testing 3 and 4. The key is that if something does not match it tells us something about where the match can continue as well as when we do match. It is a set of constraints which we can solve by testing.
Pattern match compilers seek to optimize the number of tests so there are as few as possible before we have conclusion. Statically typed language have some advantages here since they may know that:
-type foo() :: a | b | c.
and then if we have
-spec f(foo() -> any().
f(a) ->
f(b) ->
f(c) ->
and we did not match f(a), f(b) then f(c) must match. Erlang has to check and then fail if it doesn't match.