using floating point arithmetic with Z3 C++ APIs - c++

I am trying to solve a problem of nonlinear real numbers using Z3. I need the Z3 to generate multiple solutions.
In the problem domain, precision is not a critical issue; I need just one or two decimal digits after the decimal point. so, I need to set Z3 not to explore all the search space of real numbers to minimize the time to find multiple solutions.
I am trying to replace the real numbers with floating point numbers. I read the fpa example in the c_api.c file but I found it a little bit confusing for me.
for example, let me assume that I want to convert the reals in the following code:
config cfg;
cfg.set("auto_config", true);
context con(cfg);
expr x = con.real_const("x");
expr y = con.real_const("y");
solver sol(con);
sol.add(x*y > 10);
std::cout << sol.check() << "\n";
std::cout << sol.get_model() << "\n";
}
I tried the following code but it didn't work
config cfg;
cfg.set("auto_config", true);
context con(cfg);
expr sign = con.bv_const("sig", 1);
expr exp = con.bv_const("exp", 10);
expr sig = con.bv_const("sig", 10);
expr x = to_expr(con, Z3_mk_fpa_fp(con, sign, exp, sig));
expr y = to_expr(con, Z3_mk_fpa_fp(con, sign, exp, sig));
solver sol(con);
sol.add(x*y > 10);
std::cout << sol.check() << "\n";
and the output is:
Assertion failed: false, file c:\users\rehab\downloads\z3-master\z3-master\src\a
pi\c++\z3++.h, line 1199
My questions are:
Are there any detailed examples or code snippets about using fpa in C++ APIs? it is not clear to me how to convert the fpa example in the C API to C++ API.
What's wrong in the above code conversion?

I'm not sure if using floats is the best way to go for your problem. But sounds like you tried all other options and non-linearity is getting in your way. Note that even if you model your problem with floats, floating-point arithmetic is quite tricky and solver may have hard time finding satisfying models. Furthermore, solutions maybe way far off from actual results due to numerical instability.
Using C
Leaving all those aside, the correct way to code your query using the C api would be (assuming we use 32-bit single-precision floats):
#include <z3.h>
int main(void) {
Z3_config cfg = Z3_mk_config();
Z3_context ctx = Z3_mk_context(cfg);
Z3_solver s = Z3_mk_solver(ctx);
Z3_solver_inc_ref(ctx, s);
Z3_del_config(cfg);
Z3_sort float_sort = Z3_mk_fpa_sort(ctx, 8, 24);
Z3_symbol s_x = Z3_mk_string_symbol(ctx, "x");
Z3_symbol s_y = Z3_mk_string_symbol(ctx, "y");
Z3_ast x = Z3_mk_const(ctx, s_x, float_sort);
Z3_ast y = Z3_mk_const(ctx, s_y, float_sort);
Z3_symbol s_x_times_y = Z3_mk_string_symbol(ctx, "x_times_y");
Z3_ast x_times_y = Z3_mk_const(ctx, s_x_times_y, float_sort);
Z3_ast c1 = Z3_mk_eq(ctx, x_times_y, Z3_mk_fpa_mul(ctx, Z3_mk_fpa_rne(ctx), x, y));
Z3_ast c2 = Z3_mk_fpa_gt(ctx, x_times_y, Z3_mk_fpa_numeral_float(ctx, 10, float_sort));
Z3_solver_assert(ctx, s, c1);
Z3_solver_assert(ctx, s, c2);
Z3_lbool result = Z3_solver_check(ctx, s);
switch(result) {
case Z3_L_FALSE: printf("unsat\n");
break;
case Z3_L_UNDEF: printf("undef\n");
break;
case Z3_L_TRUE: { Z3_model m = Z3_solver_get_model(ctx, s);
if(m) Z3_model_inc_ref(ctx, m);
printf("sat\n%s\n", Z3_model_to_string(ctx, m));
break;
}
}
return 0;
}
When run, this prints:
sat
x_times_y -> (fp #b0 #xbe #b10110110110101010000010)
y -> (fp #b0 #xb5 #b00000000000000000000000)
x -> (fp #b0 #x88 #b10110110110101010000010)
These are single-precision floating point numbers; you can read about them in wikipedia for instance. In more conventional notation, they are:
x_times_y -> 1.5810592e19
y -> 1.8014399e16
x -> 877.6642
This is quite tricky to use, but what you have asked.
Using Python
I'd heartily recommend using the Python API to at least see what the solver is capable of before investing into such complicated C code. Here's how it would look in Python:
from z3 import *
x = FP('x', FPSort(8, 24))
y = FP('y', FPSort(8, 24))
s = Solver()
s.add(x*y > 10);
s.check()
print s.model()
When run, this prints:
[y = 1.32167303562164306640625,
x = 1.513233661651611328125*(2**121)]
Perhaps not what you expected, but it is a valid model indeed.
Using Haskell
Just to give you a taste of simplicity, here's how the same problem can be expressed using the Haskell bindings (It's just a mere one liner!)
Prelude Data.SBV> sat $ \x y -> fpIsPoint x &&& fpIsPoint y &&& x * y .> (10::SFloat)
Satisfiable. Model:
s0 = 5.1129496e28 :: Float
s1 = 6.6554557e9 :: Float
Summary
Note that Floating-point also has issues regarding NaN/Infinity values, so you might have to avoid those explicitly. (This is what the Haskell expression did by using the isFPPoint predicate. Coding it in Python or C would require more code, but is surely doable.)
It should be emphasized that literally any other binding to Z3 (Python, Haskell, Scala, what have you) will give you a better experience than what you'll get with C/C++/Java. (Even direct coding in SMTLib would be nicer.)
So, I heartily recommend using some higher-level interface (Python is a good one: It is easy to learn), and once you are confident with the model and how it works, you can then start coding the same in C if necessary.

Related

Avoiding NANs for expressions that are partly imaginary in c++

I have an expression whose outcome is a real number, but is composed of imaginary terms (that cancel one another). A significantly more simple example than the one I am considering would be something like,
z = a + 1/[sqrt(a-b) - a] - f[sqrt(a-b)] = a
where a and b are real numbers and f is some function that statisfies the above expression. It would not surprise you that in some cases, say for b > a (which does not always occur, but could occur in some cases), the above expression returns nan, since some of its terms are imaginary.
Sometimes, it is possible to work out the algebra and write out this not-really-complex expression using real numbers only. However, in my case, the algebra is very messy (so messy that even Matlab's symbolic package and Mathematica are unable to trivially simplify).
I am wondering if there is some other way to work out expressions you know to be real, but are partly imaginary.
PS: not important for this question, but for more info about the expression I am dealing with, please see another question I previously asked.
tl;dr for the comment thread:
If you know you're doing something that will involve imaginary numbers, just use std::complex.
You can't avoid getting NaN if you insist on asking for a real result to something (sqrt, say) that you know will have an imaginary component. There is no real answer it can give you.
At the end of your computation, if imag(result) is zero (or within a suitable epsilon), then your imaginary parts cancelled out and you have a real(result).
As a concrete example:
#include <complex>
#include <iostream>
int main()
{
std::complex<double> a{-5, 0}; // -5 + 0i
std::complex<double> b{ 5, 0}; // +5 + 0i
auto z = b + sqrt(a) - sqrt(a);
std::cout << "z = " << real(z) << " + " << imag(z) << "i\n";
}
prints
z = 5 + 0i
With your new example
z = a + 1/(sqrt(a-b) - a) - f(sqrt(a-b)) = a
it'll be useful to make a of type std::complex in the first place, and to use a complex 1+0i for the numerator as well. This is because of the way overloaded operators are resolved:
using cx = std::complex<double>;
cx f(cx); // whatever this does, it also needs to handle complex inputs
cx foo(cx a, cx b)
{
return a + cx{1}/(sqrt(a-b) - a) - f(sqrt(a-b));
}
auto

How to implement a custom simplify tactic in Z3?

In my tool, I use conditions that compare constants to integer variables (for example y < 100). Often, there are multiple conditions for one variable and I want to simplify those cases. For example: y < 100 && y != 99 should become y < 99. The simplify tactic does not do this and none of the arguments for simplify sound like they can help.
In Code:
context c;
goal g(c);
expr x = c.int_const("x");
expr y = c.int_const("y");
solver s(c);
expr F = y < 100 && y != 99;
g.add(F);
tactic t = tactic(c, "simplify");
apply_result r = t(g);
for (unsigned i = 0; i < r.size(); i++) {
std::cout << "subgoal " << i << "\n" << r[i] << "\n";
}
The output in the end returns: subgoal 0
(goal
(not (<= 100 y))
(not (= y 99)))
and not subgoal 0(goal(not(<= 99 y)) or something similar as I want it to be.
Therefore I want to implement my own simplify tactic. Unfortunately I cannot find how to do this. I am aware, that the tactic needs to be implemented in C++, but how can I introduce my tactic to Z3?
The Z3 tactics are stored in the directory: src/tactic. The arithmetic related tactics are in the subdirectory arith. You should use an existing tactic as a "template" for implementing your tactic.
A good example is
https://z3.codeplex.com/SourceControl/latest#src/tactic/arith/normalize_bounds_tactic.cpp
To make the new tactic available in the API and SMT 2.0 front-end, we have to include a comment containing a ADD_TACTIC command. This command instructs the Z3 mk_make script to glue everything together. The arguments are: name of the tactic, description, and C++ code for creating the tactic.
/*
ADD_TACTIC("normalize-bounds",
"replace a variable x with lower bound k <= x with x' = x - k.",
"mk_normalize_bounds_tactic(m, p)")
*/
BTW, you may also try to implement the new feature by extending an existing tactic such as:
https://z3.codeplex.com/SourceControl/latest#src/tactic/arith/propagate_ineqs_tactic.cpp

Solving a linear equation in one variable

What would be the most efficient algorithm to solve a linear equation in one variable given as a string input to a function? For example, for input string:
"x + 9 – 2 - 4 + x = – x + 5 – 1 + 3 – x"
The output should be 1.
I am considering using a stack and pushing each string token onto it as I encounter spaces in the string. If the input was in polish notation then it would have been easier to pop numbers off the stack to get to a result, but I am not sure what approach to take here.
It is an interview question.
Solving the linear equation is (I hope) extremely easy for you once you've worked out the coefficients a and b in the equation a * x + b = 0.
So, the difficult part of the problem is parsing the expression and "evaluating" it to find the coefficients. Your example expression is extremely simple, it uses only the operators unary -, binary -, binary +. And =, which you could handle specially.
It is not clear from the question whether the solution should also handle expressions involving binary * and /, or parentheses. I'm wondering whether the interview question is intended:
to make you write some simple code, or
to make you ask what the real scope of the problem is before you write anything.
Both are important skills :-)
It could even be that the question is intended:
to separate those with lots of experience writing parsers (who will solve it as fast as they can write/type) from those with none (who might struggle to solve it at all within a few minutes, at least without some hints).
Anyway, to allow for future more complicated requirements, there are two common approaches to parsing arithmetic expressions: recursive descent or Dijkstra's shunting-yard algorithm. You can look these up, and if you only need the simple expressions in version 1.0 then you can use a simplified form of Dijkstra's algorithm. Then once you've parsed the expression, you need to evaluate it: use values that are linear expressions in x and interpret = as an operator with lowest possible precedence that means "subtract". The result is a linear expression in x that is equal to 0.
If you don't need complicated expressions then you can evaluate that simple example pretty much directly from left-to-right once you've tokenised it[*]:
x
x + 9
// set the "we've found minus sign" bit to negate the first thing that follows
x + 7 // and clear the negative bit
x + 3
2 * x + 3
// set the "we've found the equals sign" bit to negate everything that follows
3 * x + 3
3 * x - 2
3 * x - 1
3 * x - 4
4 * x - 4
Finally, solve a * x + b = 0 as x = - b/a.
[*] example tokenisation code, in Python:
acc = None
for idx, ch in enumerate(input):
if ch in '1234567890':
if acc is None: acc = 0
acc = 10 * acc + int(ch)
continue
if acc != None:
yield acc
acc = None
if ch in '+-=x':
yield ch
elif ch == ' ':
pass
else:
raise ValueError('illegal character "%s" at %d' % (ch, idx))
Alternative example tokenisation code, also in Python, assuming there will always be spaces between tokens as in the example. This leaves token validation to the parser:
return input.split()
ok some simple psuedo code that you could use to solve this problem
function(stinrgToParse){
arrayoftokens = stringToParse.match(RegexMatching);
foreach(arrayoftokens as token)
{
//now step through the tokens and determine what they are
//and store the neccesary information.
}
//Use the above information to do the arithmetic.
//count the number of times a variable appears positive and negative
//do the arithmetic.
//add up the numbers both positive and negative.
//return the result.
}
The first thing is to parse the string, to identify the various tokens (numbers, variables and operators), so that an expression tree can be formed by giving operator proper precedences.
Regular expressions can help, but that's not the only method (grammar parsers like boost::spirit are good too, and you can even run your own: its all a "find and recourse").
The tree can then be manipulated reducing the nodes executing those operation that deals with constants and by grouping variables related operations, executing them accordingly.
This goes on recursively until you remain with a variable related node and a constant node.
At the point the solution is calculated trivially.
They are basically the same principles that leads to the production of an interpreter or a compiler.
Consider:
from operator import add, sub
def ab(expr):
a, b, op = 0, 0, add
for t in expr.split():
if t == '+': op = add
elif t == '-': op = sub
elif t == 'x': a = op(a, 1)
else : b = op(b, int(t))
return a, b
Given an expression like 1 + x - 2 - x... this converts it to a canonical form ax+b and returns a pair of coefficients (a,b).
Now, let's obtain the coefficients from both parts of the equation:
le, ri = equation.split('=')
a1, b1 = ab(le)
a2, b2 = ab(ri)
and finally solve the trivial equation a1*x + b1 = a2*x + b2:
x = (b2 - b1) / (a1 - a2)
Of course, this only solves this particular example, without operator precedence or parentheses. To support the latter you'll need a parser, presumable a recursive descent one, which would be simper to code by hand.

Efficient Exponentiation For HUGE Numbers (I'm Talking Googols)

I am in the midst of solving a simple combination problem whose solution is 2^(n-1).
The only problem is 1 <= n <= 2^31 -1 (max value for signed 32 bit integer)
I tried using Java's BigInteger class but It times out for numbers 2^31/10^4 and greater, so that clearly doesn't work out.
Furthermore, I am limited to using only built-in classes for Java or C++.
Knowing I require speed, I chose to build a class in C++ which does arithmetic on strings.
Now, when I do multiplication, my program multiplies similarly to how we multiply on paper for efficiency (as opposed to repeatedly adding the strings).
But even with that in place, I can't multiply 2 by itself 2^31 - 1 times, it is just not efficient enough.
So I started reading texts on the problem and I came to the solution of...
2^n = 2^(n/2) * 2^(n/2) * 2^(n%2) (where / denotes integer division and % denotes modulus)
This means I can solve exponentiation in a logarithmic number of multiplications. But to me, I can't get around how to apply this method to my code? How do I choose a lower bound and what is the most efficient way to keep track of the various numbers that I need for my final multiplication?
If anyone has any knowledge on how to solve this problem, please elaborate (example code is appreciated).
UPDATE
Thanks to everyone for all your help! Clearly this problem is meant to be solved in a realistic way, but I did manage to outperform java.math.BigInteger with a power function that only performs ceil(log2(n)) iterations.
If anyone is interested in the code I've produced, here it is...
using namespace std;
bool m_greater_or_equal (string & a, string & b){ //is a greater than or equal to b?
if (a.length()!=b.length()){
return a.length()>b.length();
}
for (int i = 0;i<a.length();i++){
if (a[i]!=b[i]){
return a[i]>b[i];
}
}
return true;
}
string add (string& a, string& b){
if (!m_greater_or_equal(a,b)) return add(b,a);
string x = string(a.rbegin(),a.rend());
string y = string(b.rbegin(),b.rend());
string result = "";
for (int i = 0;i<x.length()-y.length()+1;i++){
y.push_back('0');
}
int carry = 0;
for (int i =0;i<x.length();i++){
char c = x[i]+y[i]+carry-'0'-'0';
carry = c/10;
c%=10;
result.push_back(c+'0');
}
if (carry==1) result.push_back('1');
return string(result.rbegin(),result.rend());
}
string multiply (string&a, string&b){
string row = b, tmp;
string result = "0";
for (int i = a.length()-1;i>=0;i--){
for (int j= 0;j<(a[i]-'0');j++){
tmp = add(result,row);
result = tmp;
}
row.push_back('0');
}
return result;
}
int counter = 0;
string m_pow (string&a, int exp){
counter++;
if(exp==1){
return a;
}
if (exp==0){
return "1";
}
string p = m_pow(a,exp/2);
string res;
if (exp%2==0){
res = "1"; //a^exp%2 is a^0 = 1
} else {
res = a; //a^exp%2 is a^1 = a
}
string x = multiply(p,p);
return multiply(x,res);
//return multiply(multiply(p,p),res); Doesn't work because multiply(p,p) is not const
}
int main(){
string x ="2";
cout<<m_pow(x,5000)<<endl<<endl;
cout<<counter<<endl;
return 0;
}
As mentioned by #Oli's answer, this is not a question of computing 2^n as that's trivially just a 1 followed by 0s in binary.
But since you want to print them out in decimal, this becomes a question of how to convert from binary to decimal for very large numbers.
My answer to that is that it's not realistic. (I hope this question just stems from curiosity.)
You mention trying to compute 2^(2^31 - 1) and printing that out in decimal. That number is 646,456,993 digits long.
Java BigInteger can't do it. It's meant for small numbers and uses O(n^2) algorithms.
As mentioned in the comments, there are no built-in BigNum libraries in C++.
Even Mathematica can't handle it: General::ovfl : Overflow occurred in computation.
Your best bet is to use the GMP library.
If you're just interested in seeing part of the answer:
2^(2^31 - 1) = 2^2147483647 =
880806525841981676603746574895920 ... 7925005662562914027527972323328
(total: 646,456,993 digits)
This was done using a close-sourced library and took roughly 37 seconds and 3.2 GB of memory on a Core i7 2600K # 4.4 GHz including the time needed to write all 646 million digits to a massive text file.
(It took notepad longer to open the file than needed to compute it.)
Now to answer your question of how to actually compute such a power in the general case, #dasblinkenlight has the answer to that which is a variant of Exponentiation by Squaring.
Converting from binary to decimal for large numbers is a much harder task. The standard algorithm here is Divide-and-Conquer conversion.
I do not recommend you try to implement the latter - as it's far beyond the scope of starting programmers. (and is also somewhat math-intensive)
You don't need to do any multiplication at all. 2^(n-1) is just 1 << (n-1), i.e. 1 followed by (n-1) zeros (in binary).
The easiest way to apply this method in your code is to apply it the most direct way - recursively. It works for any number a, not only for 2, so I wrote code that takes a as a parameter to make it more interesting:
MyBigInt pow(MyBigInt a, int p) {
if (!p) return MyBigInt.One;
MyBigInt halfPower = pow(a, p/2);
MyBigInt res = (p%2 == 0) ? MyBigInt.One : a;
return res * halfPower * halfPower;
}

How to let Boost::random and Matlab produce the same random numbers

To check my C++ code, I would like to be able to let Boost::Random and Matlab produce the same random numbers.
So for Boost I use the code:
boost::mt19937 var(static_cast<unsigned> (std::time(0)));
boost::uniform_int<> dist(1, 6);
boost::variate_generator<boost::mt19937&, boost::uniform_int<> > die(var, dist);
die.engine().seed(0);
for(int i = 0; i < 10; ++i) {
std::cout << die() << " ";
}
std::cout << std::endl;
Which produces (every run of the program):
4 4 5 6 4 6 4 6 3 4
And for matlab I use:
RandStream.setDefaultStream(RandStream('mt19937ar','seed',0));
randi(6,1,10)
Which produces (every run of the program):
5 6 1 6 4 1 2 4 6 6
Which is bizarre, since both use the same algorithm, and same seed.
What do I miss?
It seems that Python (using numpy) and Matlab seems comparable, in the random uniform numbers:
Matlab
RandStream.setDefaultStream(RandStream('mt19937ar','seed',203));rand(1,10)
0.8479 0.1889 0.4506 0.6253 0.9697 0.2078 0.5944 0.9115 0.2457 0.7743
Python:
random.seed(203);random.random(10)
array([ 0.84790006, 0.18893843, 0.45060688, 0.62534723, 0.96974765,
0.20780668, 0.59444858, 0.91145688, 0.24568615, 0.77430378])
C++Boost
0.8479 0.667228 0.188938 0.715892 0.450607 0.0790326 0.625347 0.972369 0.969748 0.858771
Which is identical to ever other Python and Matlab value...
I have to agree with the other answers, stating that these generators are not "absolute". They may produce different results according to the implementation. I think the simplest solution would be to implement your own generator. It might look daunting (Mersenne twister sure is by the way) but take a look at Xorshift, an extremely simple though powerful one. I copy the C implementation given in the Wikipedia link :
uint32_t xor128(void) {
static uint32_t x = 123456789;
static uint32_t y = 362436069;
static uint32_t z = 521288629;
static uint32_t w = 88675123;
uint32_t t;
t = x ^ (x << 11);
x = y; y = z; z = w;
return w = w ^ (w >> 19) ^ (t ^ (t >> 8));
}
To have the same seed, just put any values you want int x,y,z,w (except(0,0,0,0) I believe). You just need to be sure that Matlab and C++ use both 32 bit for these unsigned int.
Using the interface like
randi(6,1,10)
will apply some kind of transformation on the raw result of the random generator. This transformation is not trivial in general and Matlab will almost certainly do a different selection step than Boost.
Try comparing raw data streams from the RNGs - chances are they are the same
In case this helps anyone interested in the question:
In order to the get the same behavior for the Twister algorithm:
Download the file
http://www.math.sci.hiroshima-u.ac.jp/~m-mat/MT/MT2002/CODES/mt19937ar.c
Try the following:
#include <stdint.h>
// mt19937ar.c content..
int main(void)
{
int i;
uint32_t seed = 100;
init_genrand(seed);
for (i = 0; i < 100; ++i)
printf("%.20f\n",genrand_res53());
return 0;
}
Make sure the same values are generated within matlab:
RandStream.setGlobalStream( RandStream.create('mt19937ar','seed',100) );
rand(100,1)
randi() seems to be simply ceil( rand()*maxval )
Thanks to Fezvez's answer I've written xor128 in matlab:
function [ w, state ] = xor128( state )
%XOR128 implementation of Xorshift
% https://en.wikipedia.org/wiki/Xorshift
% A starting state might be [123456789, 362436069, 521288629, 88675123]
x = state(1);
y = state(2);
z = state(3);
w = state(4);
% t1 = (x << 11)
t1 = bitand(bitshift(x,11),hex2dec('ffffffff'));
% t = x ^ (x << 11)
t = bitxor(x,t1);
x = y;
y = z;
z = w;
% t2 = (t ^ (t >> 8))
t2 = bitxor(t, bitshift(t,-8));
% t3 = w ^ (w >> 19)
t3 = bitxor(w, bitshift(w,-19));
% w = w ^ (w >> 19) ^ (t ^ (t >> 8))
w = bitxor(t3, t2);
state = [x y z w];
end
You need to pass state in to xor128 every time you use it. I've written a "tester" function which simply returns a vector with random numbers. I tested 1000 numbers output by this function against values output by cpp with gcc and it is perfect.
function [ v ] = txor( iterations )
%TXOR test xor128, returns vector v of length iterations with random number
% output from xor128
% output
v = zeros(iterations,1);
state = [123456789, 362436069, 521288629, 88675123];
i = 1;
while i <= iterations
disp(i);
[t,state] = xor128(state);
v(i) = t;
i = i + 1;
end
I would be very careful assuming that two different implementations of pseudo random generators (even though based on the same algorithms) produce the same result. There could be that one of the implementations use some sort of tweak, hence producing different results. If you need two equal "random" distributions I suggest you either precalculate a sequence, store and access from both C++ and Matlab or create your own generator. It should be fairly easy to implement MT19937 if you use the pseudocode on Wikipedia.
Take care ensuring that both your Matlab and C++ code runs on the same architecture (that is, both runs on either 32 or 64-bit) - using a 64 bit integer in one implementation and a 32 bit integer in the other will lead to different results.