How can I check if std::pow will overflow double - c++

I have a function that deals with arbitrarily large grids. I need to compute if a grid to the power of another number will fit into a double due to using std::pow. If it cannot, I want to take a different branch and use gnu multiprecision library instead of normal.
Is there a quick way to see if:
int a = 1024;
int b = 0-10;
if(checkPowFitsDouble(a, b)) {
long c = static_cast<long>(std::pow(a, b)); //this will only work if b < 6
} else {
mpz_t c; //yada yada gmp
}
I am completely stumped on checkPowFitsDouble; perhaps there is some math trick I don't know of.

A common trick to check whether exponentiations will overflow uses logarithms. The idea is based on these relationships:
a^b <= m <=> log(a^b) <= log(m) <=> b * log(a) <= log(m) <=> b <= log(m) / log(a)
For instance,
int a = 1024;
for (int b = 0; b < 10; ++b) {
if (b * std::log(a) < std::log(std::numeric_limits<long>::max())) {
long c = std::pow(a, b);
std::cout << c << '\n';
}
else
std::cout << "overflow\n";
}
This gives the idea. I hope this helps.

Unless it's particularly performance-critical, the suggestion would be to try it and see. If it overflows a double, std::pow will return HUGE_VAL. Hence something like:
double val = std::pow(a, b);
if(val != HUGE_VAL) {
...
} else {
mpz_t c;
//...
}

You can easily use the reverse functions in the test:
if ( std::log( DBL_MAX ) / std::log( a ) < b ) {
// std::pow( a, b ) will not overflow...
} else {
}
It might be just as good to just do the pow, and see if it
succeeds:
errno = 0;
double powab = std::pow( a, b );
if ( errno == 0 ) {
// std::pow succeeded (without overflow)
} else {
// some error (probably overflow) with std::pow.
}
You won't gain much time by just calculating std::log( a ).
(std::log( DBL_MAX ) is, of course, a constant, so only needs
to be calculated once.)

With a logarithm base 10, you can deduce that std:pow(a, b) has log(a^b) = b log a digits. You can then trivially see if it fits a double, which can fit values up to DBL_MAX.
However, this method performs additional computation than just computing a^b once. Measure a version with GMP first and see if checking for overflow actually provides any measurable and reproducible benefits.
EDIT: Ignore this, std::pow already returns an appropriate value in case an overflow occurs, so use that.

Related

how could I use the power function in c/c++ without pow(), functions, or recursion

I'm using a C++ compiler but writing code in C (if that helps)
There's a series of numbers
(-1^(a-1)/2a-1)B^(2a-1)
A and X are user defined... A must be positive, but X can be anything (+,-)...
to decode this sequence... I need use exponents/powers, but was given some restrictions... I can't make another function, use recursion, or pow() (among other advanced math functions that come with cmath or math.h).
There were plenty of similar questions, but many answers have used functions and recursion which aren't directly relevant to this question.
This is the code that works perfectly with pow(), I spent a lot of time trying to modify it to replace pow() with my own code, but nothing seems to be working... mainly getting wrong results. X and J are user inputted variables
for (int i = 1; i < j; i++)
sum += (pow(-1, i - 1)) / (5 * i - 1) * (pow(x, 5 * i - 1));
}
You can use macros to get away with no function calls restriction as macros will generate inline code which is technically not a function call
however in case of more complex operations macro can not have return value so you need to use some local variable for the result (in case of more than single expression) like:
int ret;
#define my_pow_notemp(a,b) (b==0)?1:(b==1)?a:(b==2)?a*a:(b==3)?a*a*a:0
#define my_pow(a,b)\
{\
ret=1;\
if (int(b& 1)) ret*=a;\
if (int(b& 2)) ret*=a*a;\
if (int(b& 4)) ret*=a*a*a*a;\
if (int(b& 8)) ret*=a*a*a*a*a*a*a*a;\
if (int(b&16)) ret*=a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a;\
if (int(b&32)) ret*=a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a;\
}
void main()
{
int a=2,b=3,c;
c=my_pow_notemp(a,b); // c = a^b
my_pow(a,b); c = ret; // c = a^b
}
as you can see you can use my_pow_notemp directly but the code is hardcoded so only up to a^3 if you want more you have to add it to code. The my_pow is accepting exponents up to a^63 and its also an example on how to return value in case of more complex code inside macro. Here are some (normal) ways on how to compute powers in case you need non integer or negative exponents (but to convert it to unrolled code will be insanely hard without loops/recursion):
Power by squaring for negative exponents
In case you want to get away with recursion and function calls you can use templates instead of macros but that is limited to C++.
template<class T> T my_pow(T a,T b)
{
if (b==0) return 1;
if (b==1) return a;
return a*my_pow(a,b-1);
}
void main()
{
int a=2,b=3,c;
c=my_pow(a,b);
}
As you can see templates have return value so no problem even with more complex code (more than single expression).
To avoid loops you can use LUT tables
int my_pow[4][4]=
{
{1,0,0,0}, // 0^
{1,1,1,1}, // 1^
{1,2,4,8}, // 2^
{1,3,9,27}, // 3^
};
void main()
{
int a=2,b=3,c;
c=my_pow[a][b];
}
If you have access to FPU or advanced math assembly you can use that as asm instruction is not a function call. FPU usually have log,exp,pow functions natively. This however limits the code to specific instruction set !!!
Here some examples:
How to: pow(real, real) in x86
So when I consider your limitation I think the best way is:
#define my_pow(a,b) (b==0)?1:(b==1)?a:(b==2)?a*a:(b==3)?a*a*a:0
void main()
{
int a=2,b=3,c;
c=my_pow(a,b); // c = a^b
}
Which will work on int exponents b up to 3 (if you want more just add (b==4)?a*a*a*a: ... :0) and both int and float bases a. If you need much bigger exponent use the complicated version with local temp variable for returning result.
[Edit1] ultimative single expression macro with power by squaring up to a^15
#define my_pow(a,b) (1* (int(b&1))?a:1* (int(b&2))?a*a:1* (int(b&4))?a*a*a*a:1* (int(b&8))?a*a*a*a*a*a*a*a:1)
void main()
{
int a=2,b=3,c;
c=my_pow(a,b); // c = a^b
}
In case you want more than a^15 just add sub term (int(b&16))?a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a:1 and so on for each bit of exponent.
It is a series. Replace pow() based on the previous iteration. #Bathsheba
Code does not need to call pow(). It can form pow(x, 5 * i - 1) and pow(-1, i - 1), since both have an int exponent based on the iterator i, from the prior loop iteration.
Example:
Let f(x, i) = pow(x, 5 * i - 1)
Then f(x, 1) = x*x*x*x
and f(x, i > 1) = f(x, i-1) * x*x*x*x*x
double power_n1 = 1.0;
double power_x5 = x*x*x*x;
for (int i = 1; i < j + 1; i++)
// sum += (pow(-1, i - 1)) / (5 * i - 1) * (pow(x, 5 * i - 1));
sum += power_n1 / (5 * i - 1) * power_x5;
power_n1 = -power_n1;
power_x5 *= x*x*x*x*x;
}

C++ calling single helper function with *this attributes

Edit: I’m a beginner to C++, and I’d like to understand more about how to optimize my code.
I have created a Fraction object in C++ as well as overloaded +, - operations etc. When I came to the unary operators, however, I realized I didn't know how to reduce the fraction in the most efficient manner. So I have a function gcd that finds the greatest divisor:
int gcd (int n, int m) {
int newN = n < 0 ? -n : n;
int newM = m < 0 ? -m : m;
if (newM <= newN && newN % newM == 0) { return newM; }
else if (newN < newM) { return gcd(newM, newN); }
else { return gcd(newM, newN%newM); }
}
and then I have an overloaded operator, for example, incrementation:
Fraction& Fraction::operator++() {
num = num + denom;
//reduce fraction
int divisor = gcd(denom,num);
num = num/divisor;
denom = denom/divisor;
if (num < 0 && denom < 0) {num *= (-1);}
if (denom < 0) {denom *= (-1);}
return *this;
}
For efficiency, I would like to put the reduce fraction part of the code in a separate single helper function so the final function would look like this:
Fraction& Fraction::operator++() {
num = num + denom;
//reduce fraction
reduce(num, denom);
return *this;
}
That way I don't have to copy and paste whatever is in //reduce fraction everytime I overload a unary operator for example. However, I'm not sure how the reduce(Fraction num, Fraction& denom) function should look like. At most I can implement it like this:
void reduce(int& num, int& denom) {
int divisor = gcd(denom,num);
num = num/divisor;
denom = denom/divisor;
if (num < 0 && denom < 0) {num *= (-1);}
if (denom < 0) {denom *= (-1);}
}
I'm sure the code above will run into issues during compilation, so I was wondering if I could be suggested any pointers as to efficiently create the reduce fraction function. This is maybe being a bit nitpicky since my original code runs fine, but since I am new to C++, I'd like to learn more about how I can make my code more efficient. Thanks a lot! Let me know if more information is needed.
Edit: The above code does not work. Compiles correctly, but does not reduce fraction properly. So 1/2 + 1/4 results in 6/8, not 3/4.
Well on a high level your gcd function is too complicated and the last part of reduce is a bit wrong. If only denom is negative you invert it.
Nicely shows why it's always a good idea to put code into proper functions because they can also be separately tested. So I'd suggest writing some unit tests for your reduce and gcd functions.
Start with a simple solution like
static int gcd(int a, int b)
{
return b == 0 ? a : gcd(b, a % b);
}
and adapt if needed for negative numbers considering % semantics. Thinking about it the function should already be fine like that and you just need to call std::abs(gcd(n,d)) in reduce.
In general you should ask yourself if you really want to pay the renormalization cost at every single operation or if you let the user decide when to call reduce.
For lower-level optimizations here are some hints:
Always test/measure, e.g. by looking at what the compiler actually produces with godbolt.org.
The recursion in gcd is not a problem from a performance point of view in this case as it's tail recursive and the compiler will turn it into a loop for you.
The out parameters in reduce are bad for optimizations cause the compiler has to prove they don't point to the same object. Returning a std::pair and using C++11 std::tie or C++17 structured bindings at the callsite if possible is way more elegant.

Comparing floating-point numbers (floats or doubles) for min/max

How to compare two different floats, to a certain degree of accuracy. I know that there are very slight imprecisions while using floats or doubles in any programming language, but which may however, be enough to cause this comparison of floats a < b to return a different value than it actually should.
I was solving a problem from the UVa Online Judge which gave me a Wrong Answer many times. It took few float values as input, albeit, to 2 decimal places. I figured out the makeshift solution, which was by splitting the input and converting it to ints, but I wouldn't prefer to use that always.
So my question is, what's the best way to compare whether a float a is lesser (or greater) than a float b, provided that the inputs of a and b are given correct to n decimal places, in this case, 2?
The language I prefer is C++.
Use std::numeric_limits<T>::epsilon() to check whether two numbers are almost equal. If you want to know whether one is greater/less you should also take into account the relative tolerance.
#include <cmath>
#include <limits>
template < typename T >
bool fuzzy_compare(T a, T b)
{
return std::abs(a - b) < std::numeric_limits<T>::epsilon();
};
Just use math:
#define PREC 0.01 //(1/pow(10,n)), n = 2
float c = a-b;
if (abs(c) < PREC) {
printf("a equals b");
} else if(c < 0){
printf("b is grater than a");
} else
printf("a is grater than b");
}
Use the setprecison() operator. The number you put in between the parentheses will determine how many numbers pass the decimal will be included in the output. Be sure to include the iomanip library.
Comparing floats is alway a tricky Here is a more complicated example, showing why you should use std::numeric_limits<T>::epsilon().
The first line returns true, but the second returns false (on my machine).
float64_t CalculateEpsilon ()
{
float64_t l_AllowedInaccuray = 1; // 1.1, 0.9
int32_t significantDecimalPlaces = 2;
return ( l_AllowedInaccuray * pow ( 0.1, significantDecimalPlaces ) );
}
bool IsEqual ( float64_t lhs, float64_t rhs )
{
float64_t l_Epsilon = CalculateEpsilon ();
float64_t l_Delta = std::abs ( lhs - rhs );
return l_Delta <= l_Epsilon;
}
int32_t main ()
{
std::cout << IsEqual ( 107.35999999999999, 107.350 ); //returns true
std::cout << IsEqual ( 107.359999999999999, 107.350 ); //returns false
return 0;
}

How to make two double variables with different length become equal

I'm new to Visual C++, i have two double variables, let say AA = 10.650406 while b = 10.65040606439, how can i make them equal?
here is an example
AA = 10.650406;
if (a == tempfunc(zz))
execute TRUEFUNCTION
else
exceute FALSEFUNCTION
the variable AA is a double while the function tempfunc return double, if the value AA is 10.650406, while the return value of tempfunc is 10.65040606439. the question is how can make these value equal so i can execute the function TRUEFUNCTION
The typical solution is to compare the difference, using an "epsilon value". Something like
const double eps = 0.000001; // Adjust this to match the "perecision" you need.
if (abs(a-b) < eps)
{
// Values are almost equal
}
I guess your question is: how to have a check that would be true for these two numbers, but not for all..:
10.123 == 10.1234 (TRUE)
10.123 == 11.123 (FALSE)
if you have a fixed number of digits after the decimal separator (6 in your example), you could do this:
int a2 = a * 10e6;
int b2 = b * 10e6; // (conversion to integers; so the digits smaller than the 6th digit will be removed)
Now you can check:
if (a2 == b2)
{
//do your thing
}
in short:
if ( ((int) a*10e6) == ((int) b*10e6))
{
//do your thing
}

How can I obtain the cube root in C++?

I know how to obtain the square root of a number using the sqrt function.
How can I obtain the cube root of a number?
sqrt stands for "square root", and "square root" means raising to the power of 1/2. There is no such thing as "square root with root 2", or "square root with root 3". For other roots, you change the first word; in your case, you are seeking how to perform cube rooting.
Before C++11, there is no specific function for this, but you can go back to first principles:
Square root: std::pow(n, 1/2.) (or std::sqrt(n))
Cube root: std::pow(n, 1/3.) (or std::cbrt(n) since C++11)
Fourth root: std::pow(n, 1/4.)
etc.
If you're expecting to pass negative values for n, avoid the std::pow solution — it doesn't support negative inputs with fractional exponents, and this is why std::cbrt was added:
std::cout << std::pow(-8, 1/3.) << '\n'; // Output: -nan
std::cout << std::cbrt(-8) << '\n'; // Output: -2
N.B. That . is really important, because otherwise 1/3 uses integer division and results in 0.
in C++11 std::cbrt was introduced as part of math library, you may refer
include <cmath>
std::pow(n, 1./3.)
Also, in C++11 there is cbrt in the same header.
Math for Dummies.
The nth root of x is equal to x^(1/n), so use std::pow. But I don't see what this has to with operator overloading.
Just to point this out, though we can use both ways but
long long res = pow(1e9, 1.0/3);
long long res2 = cbrt(1e9);
cout<<res<<endl;
cout<<res2<<endl;
returns
999
1000
So, in order to get the correct results with pow function we need to add an offset of 0.5 with the actual number or use a double data type i.e.
long long res = pow(1e9+0.5, 1.0/3)
double res = pow(1e9, 1.0/3)
more detailed explanation here C++ pow unusual type conversion
Actually the round must go for the above solutions to work.
The Correct solution would be
ans = round(pow(n, 1./3.));
The solution for this problem is
cube_root = pow(n,(float)1/3);
and you should #include <math.h> library file
Older standards of C/C++ don't support cbrt() function.
When we write code like cube_root = pow(n,1/3); the compiler thinks 1/3 = 0 (division problem in C/C++), so you need to do typecasting using (float)1/3 in order to get the correct answer
#include<iostream.h>
#include<conio.h>
#include<math.h>
using namespace std;
int main(){
float n = 64 , cube_root ;
clrscr();
cube_root = pow(n , (float)1/3);
cout<<"cube root = "<<cube_root<<endl;
getch();
return 0;
}
cube root = 4
You can try this C algorithm :
// return a number that, when multiplied by itself twice, makes N.
unsigned cube_root(unsigned n){
unsigned a = 0, b;
for (int c = sizeof(unsigned) * CHAR_BIT / 3 * 3 ; c >= 0; c -= 3) {
a <<= 1;
b = 3 * a * (a + 1) + 1;
if (n >> c >= b)
n -= b << c, ++a;
}
return a;
}
I would discourage any of the above methods as they didn't work for me. I did pow(64, 1/3.) along with pow(64, 1./3.) but the answer I got was 3
Here's my logic.
ans = pow(n, 1/3.);
if (pow(ans, 3) != n){
ans++;
}