I'm learning C programming and made the algorithm below to solve this problem:
The code actually works, but initially the loop was with only 10 repetitions (rep <= 10), and the anwer for p = 3 was almost correct, so I changed rep <= 20. And It gave me just the exact answer from my calculator. And then I tried with a higher number, 12, and the output again was inaccurate. So I ended raising rep <= 35. If I get the loop for higher repetitions I get "-nan", and if the input for p is too high it will be the same. So just have to see the pattern to know that the problem of inaccuracy will get back as I input higher numbers which is not the case because the output will be NaN if I input a high value.
Is it possible to solve it without higher level functions? just want to know if my program is ok for the level in which I am now...
#include <stdio.h>
int main()
{
float p; //the power for e
float power; //the copy of p for the loop
float e = 1; //the e number I wanna raise to the power of p
int x = 1; //the starting number for each factorial generation
float factorial = 1;
int rep = 1; //the repeater for the loop
printf( "Enter the power you want to raise: " );
scanf( "%f", &p );
power = p;
while ( rep <= 35) {
while ( x > 1) {
factorial *= x;
x--;
}
e += p / factorial;
//printf("\nthe value of p: %f", p); (TESTER)
//printf("\nthe value of factorial: %f", factorial); (TESTER)
p *= power; //the new value for p
rep++;
factorial = 1;
x = rep; //the new value for the next factorial to be generated
//printf("\n%f", e); (TESTER)
}
printf("%.3f", e);
return 0;
}
Sorry if I had syntax/orthography errors, I'm still learning the language.
Before we begin, let's write your original code as a function, with some clean-ups:
float exp_original(float x, int rep = 35)
{
float sum = 1.0f;
float power = 1.0f;
for (int i = 1; i <= rep; i++)
{
float factorial = 1.0f;
for (int j = 2; j <= i; j++)
factorial *= j;
power *= x;
sum += power / factorial;
}
return sum;
}
There were some unnecessary variables you used which were removed, but otherwise the procedure is the same: compute the factorial from scratch.
Let's look at the ratio between successive terms in the series:
We can thus simply multiply the current term by this expression to get the next term:
float exp_iterative(float x, int rep = 35)
{
float sum = 1.0f;
float term = 1.0f;
for (int i = 1; i <= rep; i++)
{
term *= x / i;
sum += term;
}
return sum;
}
Seems much simpler, but is it better? Comparison against the C-library exp function (which we assume to be maximally precise):
x exp (C) exp_orig exp_iter
-------------------------------------------
1 2.7182817 2.718282 2.718282
2 7.3890562 7.3890567 7.3890567
3 20.085537 20.085539 20.085539
4 54.598148 54.598152 54.598152
5 148.41316 148.41318 148.41316
6 403.4288 403.42871 403.42877
7 1096.6332 1096.6334 1096.6334
8 2980.958 2980.9583 2980.9587
9 8103.084 8103.083 8103.083
10 22026.465 22026.467 22026.465
11 59874.141 59874.148 59874.152
12 162754.8 162754.77 162754.78
13 442413.41 -nan(ind) 442413.38
14 1202604.3 -nan(ind) 1202603.5
15 3269017.3 -nan(ind) 3269007.3
16 8886111 -nan(ind) 8886009
17 24154952 -nan(ind) 24153986
18 65659968 -nan(ind) 65652048
19 1.784823e+08 -nan(ind) 1.7842389e+08
20 4.8516518e+08 -nan(ind) 4.8477536e+08
The two custom implementations are neck-and-neck in-terms of precision, until x = 13 where the original gives NaN. This is because the highest power term 13^35 = 9.7278604e+38 exceeds the maximum value FLT_MAX = 3.40282e+38. The accumulated term in the iterative version never reaches anywhere near the limit.
I'm trying to solve a code to run this series pi = 4 - 4/3 + 4/5 - 4/7 + 4/9 ... and so on. The thing is, as i run with higher values, the function tends to 4, not 3,1415.... The program seem to be running only the greatest value that n assumes. Also i cant make %%.lflf work to set decimals according to a variable, (p). The algorithm seems correct but after days i'm desperate for any help, greatly appreciated.
if ( 2 == route ) {
printf("piseries calculator\n");
double pi,n,p;
printf("define precision");
scanf("%lf",&p);
n=0;
while (n++ <= p ) {
pi = (4) - (n * ( 4 / ( 1 + 2 *(n)))) +( n * ( 4 / ( 3 + 2 * (n))));
}
printf("%%.lflf",p,pi);
return 0;
}
First of all, make n and p integers.
Secondly n starts at 3 doesn't it?
Third the Pi series is 4.0 (- fraction + fraction)...
Finally you can printf using %.*lf to increase/limit the precision of the output.
if ( 2 == route )
{
printf("piseries calculator\n");
double pi=4.0;
int n,p;
printf("define precision");
scanf("%d",&p);
for(n=3;n<p;n+=2)
{
pi -= 4.0/n;
n+=2;
pi += 4.0/n;
}
printf("%.*lf",p,pi);
return 0;
}
I would scale it as well for those calculations
#include <stdio.h>
#define Niterations 1000
#define SCALE 1000
int main(void) {
double pi = 4.0 * SCALE;
for(unsigned i = 0; i < Niterations; i ++)
{
double f = 4.0 * SCALE / (3 + i * 2);
pi += (i & 1) ? f : -f;
}
printf("N iterations = %u, pi = %lf", Niterations, pi / SCALE);
// your code goes here
return 0;
}
Test it https://ideone.com/mSZNyg
Result: N iterations = 10000, pi = 3.141692643590519029572760700830258429050445556640625000000000 ...
I'm looking for solution to find the sum of numbers. Input will be given has n in integer and problem is to find Sum of the values of sum(1)+ sum(1+2) + sum(1+2+3) + ... + sum(1+2+..+n). I need a very optimised solution using dynamic programming or any math calculation.
int main()
{
int sum = 0;
int i = 0, n = 6;
for( i = 1; i < n; i++ )
sum = sum + findSumN( i );
printf( "%d",sum );
}
You can often find a formula for series like this by calculating the first few terms and using the results to search the On-Line Encyclopedia of Integer Sequences.
1 = 1
1 + (1+2) = 4
4 + (1+2+3) = 10
10 + (1+2+3+4) = 20
20 + (1+2+3+4+5) = 35
35 + (1+2+3+4+5+6) = 56
The sequence you're trying to calculate (1, 4, 10, 20, 35, 56, ...) is A000292, which has the following formula:
a(n) = n × (n + 1) × (n + 2) / 6
If you play with the number you can find some patterns. Starts with
sum(1 + 2 + 3 ... + N) = ((1 + N) * N) /2
Then there is a relationship between the max number and the value above, that is from 1 the difference step 1/3 everytime the max number increase by 1. So get:
(1 + ((1.0 / 3.0) * (max - 1)))
I am not good enough at math to explain why this pattern occurs. Perhaps someone can explain it in a math way.
The following is my solution, no iteration needed.
int main()
{
int min = 1;
int max = 11254;
double sum = ((min + max) * max / 2) * (1 + ((1.0 / 3.0) * (max - 1)));
printf("%.f", sum);
}
Look at the closed form of sum(n)=1+2+…+n and look up the Pascal's triangle identities. This gives immediately a very fast computation method.
As
binom(k,2) + binom(k,3) = binom(k+1,3)
binom(k,2) = binom(k+1,3) - binom(k,3)
the summation of binom(k+1,2) from k=M to N results in the sum value
binom(N+2,3)-binom(M+1,3)=(N+2)*(N+1)*N/6-(M+1)*M*(M-1)/6
= (N+1-M) * ((N+1)²+(N+1)M+M²-1)/6
Days ago I had a job interview were they ask me how I would calculate the sum of all the numbers multiples of 2 or 5 from 1 to 10000 using a the c language. I did this:
int mult2_5()
{
int i, sum=0;
for(i = 1; i <= 10000; i++)
if(i % 2 == 0 || i % 5 == 0)
sum += i;
return sum;
}
I as wonder if it was any faster implementation that this one?
The modulus operator is inefficient. A more faster implementation would be something like this:
int multiply2_5(int max)
{
int i, x2 = 0,x5 = 0,x10 = 0;
for(i = 2; i < max; i+=2) x2 += i; // Store all multiples of 2 O(max/2)
for(i = 5; i < max; i+=5) x5 += i; // Store all multiples of 3 O(max/5)
for(i = 10; i < max; i+=10) x10 += i; // Store all multiples 10; O(max/10)
return x2+x5-x10;
}
In this solution I had to take out multiples of 10 because, 2 and 5 have 10 as multiple so on the second loop it will add multiples of 10 that already been added in the first loop; The three loops combine have O(8/10 max).
Another even better solution is if you take a mathematical approach.
You are trying to sum all numbers like this 2 + 4 + 6 + 8 ... 10000 and 5 + 10 + 15 +20 + ... 10000 this is the same of having 2 * (1 + 2 + 3 + 4 + … + 5000) and 5 * ( 1 + 2 + 3 + 4 + ... + 2000), the sum of 'n' natural number is (n * (n + 1)) (source) so you can calculate in a constant time, as it follows:
int multiply2_5(int max)
{
// x = 2 + 4 + 6 + ... = 2 * (1 + 2 + 3 +...)
// y = 5 + 10 + 15 + ... = 5 * (1 + 2 + 3 +...)
// The sun of n natural numbers is sn = (n (n + 1)) / 2
int x2 = max/ 2; // 2 * ( 1 +2 + 3 … max/2)
int x5 = max /5; // 5 * ( 1 +2 + 3 … max/5)
int x10 = max/ 10;
int sn2 = 0.5 * (x2 * (x2+1)); // (n * (n + 1)) / 2
int sn5 = 0.5 * (x5 * (x5+1));
int sn10 = 0.5 * (x10 * (x10+1));
return (2*sn2) + (5 *sn5) - (10*sn10);
}
As mentioned in an earlier answer, explicitly looping through the relevant multiples is better than testing the remainder each loop. But it is not necessary to calculate the multiples of 10 and subtract. Just start at 5 and step by 10 to skip them all together.
int multiply2_5b(int max)
{
int i, x2 = 0,x5 = 0;
for (i = 2; i < max; i += 2) x2 += i; // Sum all multiples of 2
for (i = 5; i < max; i += 10) x5 += i; // Sum all odd multiples of 5
return x2 + x5;
}
Just work it out on paper first, if that's what you mean by "faster".
$2\sum_{1<=2k<=10000}k + 5\sum_{1<=5k<=10000} - 10\sum_{1<=10k<=10000}k$
Sorry, my SO equation-fu is weak...Anyway this route will give you something you can almost handle on paper: 5000*6001 after reducing a few steps
int
mult2_5(void)
{
return 5000*6001;
}
Project Euler problem 1 is very similar. There's lots of folks who've posted their solution to this one.
This can just be done using math. Something like 2 * sum(1 to 5000) + 5 * sum(1 to 2000) - 10 * sum(1 to 1000). off-by-one errors left as exercise.
I almost got to a pure and simple multiplication by doing a simple loop that starts with 35 (sum of 2 + 4 + 5 + 6 + 8 + 10) with a step of 60, as that's how much your result will increase by when you take the next lot e.g. 12 + 14 + 15 + 16 + 18 + 20 etc. for(int i=35;i<5976;i=i+60) { sum=sum+i }
The 5976 comes from 5975 being the last row of numbers that end in 1000 i.e. 992 + 994 + 995 + 996 + 998 + 1000.
So it turns out that this loop runs 100 times, increasing the sum by 35 the first turn and increasing by 60 the remaining 99 times. Which is now reducable to a simple multiplication or such.
I'm looking for some nice C code that will accomplish effectively:
while (deltaPhase >= M_PI) deltaPhase -= M_TWOPI;
while (deltaPhase < -M_PI) deltaPhase += M_TWOPI;
What are my options?
Edit Apr 19, 2013:
Modulo function updated to handle boundary cases as noted by aka.nice and arr_sea:
static const double _PI= 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348;
static const double _TWO_PI= 6.2831853071795864769252867665590057683943387987502116419498891846156328125724179972560696;
// Floating-point modulo
// The result (the remainder) has same sign as the divisor.
// Similar to matlab's mod(); Not similar to fmod() - Mod(-3,4)= 1 fmod(-3,4)= -3
template<typename T>
T Mod(T x, T y)
{
static_assert(!std::numeric_limits<T>::is_exact , "Mod: floating-point type expected");
if (0. == y)
return x;
double m= x - y * floor(x/y);
// handle boundary cases resulted from floating-point cut off:
if (y > 0) // modulo range: [0..y)
{
if (m>=y) // Mod(-1e-16 , 360. ): m= 360.
return 0;
if (m<0 )
{
if (y+m == y)
return 0 ; // just in case...
else
return y+m; // Mod(106.81415022205296 , _TWO_PI ): m= -1.421e-14
}
}
else // modulo range: (y..0]
{
if (m<=y) // Mod(1e-16 , -360. ): m= -360.
return 0;
if (m>0 )
{
if (y+m == y)
return 0 ; // just in case...
else
return y+m; // Mod(-106.81415022205296, -_TWO_PI): m= 1.421e-14
}
}
return m;
}
// wrap [rad] angle to [-PI..PI)
inline double WrapPosNegPI(double fAng)
{
return Mod(fAng + _PI, _TWO_PI) - _PI;
}
// wrap [rad] angle to [0..TWO_PI)
inline double WrapTwoPI(double fAng)
{
return Mod(fAng, _TWO_PI);
}
// wrap [deg] angle to [-180..180)
inline double WrapPosNeg180(double fAng)
{
return Mod(fAng + 180., 360.) - 180.;
}
// wrap [deg] angle to [0..360)
inline double Wrap360(double fAng)
{
return Mod(fAng ,360.);
}
One-liner constant-time solution:
Okay, it's a two-liner if you count the second function for [min,max) form, but close enough — you could merge them together anyways.
/* change to `float/fmodf` or `long double/fmodl` or `int/%` as appropriate */
/* wrap x -> [0,max) */
double wrapMax(double x, double max)
{
/* integer math: `(max + x % max) % max` */
return fmod(max + fmod(x, max), max);
}
/* wrap x -> [min,max) */
double wrapMinMax(double x, double min, double max)
{
return min + wrapMax(x - min, max - min);
}
Then you can simply use deltaPhase = wrapMinMax(deltaPhase, -M_PI, +M_PI).
The solutions is constant-time, meaning that the time it takes does not depend on how far your value is from [-PI,+PI) — for better or for worse.
Verification:
Now, I don't expect you to take my word for it, so here are some examples, including boundary conditions. I'm using integers for clarity, but it works much the same with fmod() and floats:
Positive x:
wrapMax(3, 5) == 3: (5 + 3 % 5) % 5 == (5 + 3) % 5 == 8 % 5 == 3
wrapMax(6, 5) == 1: (5 + 6 % 5) % 5 == (5 + 1) % 5 == 6 % 5 == 1
Negative x:
Note: These assume that integer modulo copies left-hand sign; if not, you get the above ("Positive") case.
wrapMax(-3, 5) == 2: (5 + (-3) % 5) % 5 == (5 - 3) % 5 == 2 % 5 == 2
wrapMax(-6, 5) == 4: (5 + (-6) % 5) % 5 == (5 - 1) % 5 == 4 % 5 == 4
Boundaries:
wrapMax(0, 5) == 0: (5 + 0 % 5) % 5 == (5 + 0) % 5 == 5 % 5 == 0
wrapMax(5, 5) == 0: (5 + 5 % 5) % 5 == (5 + 0) % 5== 5 % 5 == 0
wrapMax(-5, 5) == 0: (5 + (-5) % 5) % 5 == (5 + 0) % 5 == 5 % 5 == 0
Note: Possibly -0 instead of +0 for floating-point.
The wrapMinMax function works much the same: wrapping x to [min,max) is the same as wrapping x - min to [0,max-min), and then (re-)adding min to the result.
I don't know what would happen with a negative max, but feel free to check that yourself!
If ever your input angle can reach arbitrarily high values, and if continuity matters, you can also try
atan2(sin(x),cos(x))
This will preserve continuity of sin(x) and cos(x) better than modulo for high values of x, especially in single precision (float).
Indeed, exact_value_of_pi - double_precision_approximation ~= 1.22e-16
On the other hand, most library/hardware use a high precision approximation of PI for applying the modulo when evaluating trigonometric functions (though x86 family is known to use a rather poor one).
Result might be in [-pi,pi], you'll have to check the exact bounds.
Personaly, I would prevent any angle to reach several revolutions by wrapping systematically and stick to a fmod solution like the one of boost.
There is also fmod function in math.h but the sign causes trouble so that a subsequent operation is needed to make the result fir in the proper range (like you already do with the while's). For big values of deltaPhase this is probably faster than substracting/adding `M_TWOPI' hundreds of times.
deltaPhase = fmod(deltaPhase, M_TWOPI);
EDIT:
I didn't try it intensively but I think you can use fmod this way by handling positive and negative values differently:
if (deltaPhase>0)
deltaPhase = fmod(deltaPhase+M_PI, 2.0*M_PI)-M_PI;
else
deltaPhase = fmod(deltaPhase-M_PI, 2.0*M_PI)+M_PI;
The computational time is constant (unlike the while solution which gets slower as the absolute value of deltaPhase increases)
I would do this:
double wrap(double x) {
return x-2*M_PI*floor(x/(2*M_PI)+0.5);
}
There will be significant numerical errors. The best solution to the numerical errors is to store your phase scaled by 1/PI or by 1/(2*PI) and depending on what you are doing store them as fixed point.
Instead of working in radians, use angles scaled by 1/(2π) and use modf, floor etc. Convert back to radians to use library functions.
This also has the effect that rotating ten thousand and a half revolutions is the same as rotating half then ten thousand revolutions, which is not guaranteed if your angles are in radians, as you have an exact representation in the floating point value rather than summing approximate representations:
#include <iostream>
#include <cmath>
float wrap_rads ( float r )
{
while ( r > M_PI ) {
r -= 2 * M_PI;
}
while ( r <= -M_PI ) {
r += 2 * M_PI;
}
return r;
}
float wrap_grads ( float r )
{
float i;
r = modff ( r, &i );
if ( r > 0.5 ) r -= 1;
if ( r <= -0.5 ) r += 1;
return r;
}
int main ()
{
for (int rotations = 1; rotations < 100000; rotations *= 10 ) {
{
float pi = ( float ) M_PI;
float two_pi = 2 * pi;
float a = pi;
a += rotations * two_pi;
std::cout << rotations << " and a half rotations in radians " << a << " => " << wrap_rads ( a ) / two_pi << '\n' ;
}
{
float pi = ( float ) 0.5;
float two_pi = 2 * pi;
float a = pi;
a += rotations * two_pi;
std::cout << rotations << " and a half rotations in grads " << a << " => " << wrap_grads ( a ) / two_pi << '\n' ;
}
std::cout << '\n';
}}
Here is a version for other people finding this question that can use C++ with Boost:
#include <boost/math/constants/constants.hpp>
#include <boost/math/special_functions/sign.hpp>
template<typename T>
inline T normalizeRadiansPiToMinusPi(T rad)
{
// copy the sign of the value in radians to the value of pi
T signedPI = boost::math::copysign(boost::math::constants::pi<T>(),rad);
// set the value of rad to the appropriate signed value between pi and -pi
rad = fmod(rad+signedPI,(2*boost::math::constants::pi<T>())) - signedPI;
return rad;
}
C++11 version, no Boost dependency:
#include <cmath>
// Bring the 'difference' between two angles into [-pi; pi].
template <typename T>
T normalizeRadiansPiToMinusPi(T rad) {
// Copy the sign of the value in radians to the value of pi.
T signed_pi = std::copysign(M_PI,rad);
// Set the value of difference to the appropriate signed value between pi and -pi.
rad = std::fmod(rad + signed_pi,(2 * M_PI)) - signed_pi;
return rad;
}
I encountered this question when searching for how to wrap a floating point value (or a double) between two arbitrary numbers. It didn't answer specifically for my case, so I worked out my own solution which can be seen here. This will take a given value and wrap it between lowerBound and upperBound where upperBound perfectly meets lowerBound such that they are equivalent (ie: 360 degrees == 0 degrees so 360 would wrap to 0)
Hopefully this answer is helpful to others stumbling across this question looking for a more generic bounding solution.
double boundBetween(double val, double lowerBound, double upperBound){
if(lowerBound > upperBound){std::swap(lowerBound, upperBound);}
val-=lowerBound; //adjust to 0
double rangeSize = upperBound - lowerBound;
if(rangeSize == 0){return upperBound;} //avoid dividing by 0
return val - (rangeSize * std::floor(val/rangeSize)) + lowerBound;
}
A related question for integers is available here:
Clean, efficient algorithm for wrapping integers in C++
A two-liner, non-iterative, tested solution for normalizing arbitrary angles to [-π, π):
double normalizeAngle(double angle)
{
double a = fmod(angle + M_PI, 2 * M_PI);
return a >= 0 ? (a - M_PI) : (a + M_PI);
}
Similarly, for [0, 2π):
double normalizeAngle(double angle)
{
double a = fmod(angle, 2 * M_PI);
return a >= 0 ? a : (a + 2 * M_PI);
}
In the case where fmod() is implemented through truncated division and has the same sign as the dividend, it can be taken advantage of to solve the general problem thusly:
For the case of (-PI, PI]:
if (x > 0) x = x - 2PI * ceil(x/2PI) #Shift to the negative regime
return fmod(x - PI, 2PI) + PI
And for the case of [-PI, PI):
if (x < 0) x = x - 2PI * floor(x/2PI) #Shift to the positive regime
return fmod(x + PI, 2PI) - PI
[Note that this is pseudocode; my original was written in Tcl, and I didn't want to torture everyone with that. I needed the first case, so had to figure this out.]
deltaPhase -= floor(deltaPhase/M_TWOPI)*M_TWOPI;
The way suggested you suggested is best. It is fastest for small deflections. If angles in your program are constantly being deflected into the proper range, then you should only run into big out of range values rarely. Therefore paying the cost of a complicated modular arithmetic code every round seems wasteful. Comparisons are cheap compared to modular arithmetic (http://embeddedgurus.com/stack-overflow/2011/02/efficient-c-tip-13-use-the-modulus-operator-with-caution/).
In C99:
float unwindRadians( float radians )
{
const bool radiansNeedUnwinding = radians < -M_PI || M_PI <= radians;
if ( radiansNeedUnwinding )
{
if ( signbit( radians ) )
{
radians = -fmodf( -radians + M_PI, 2.f * M_PI ) + M_PI;
}
else
{
radians = fmodf( radians + M_PI, 2.f * M_PI ) - M_PI;
}
}
return radians;
}
If linking against glibc's libm (including newlib's implementation) you can access
__ieee754_rem_pio2f() and __ieee754_rem_pio2() private functions:
extern __int32_t __ieee754_rem_pio2f (float,float*);
float wrapToPI(float xf){
const float p[4]={0,M_PI_2,M_PI,-M_PI_2};
float yf[2];
int q;
int qmod4;
q=__ieee754_rem_pio2f(xf,yf);
/* xf = q * M_PI_2 + yf[0] + yf[1] /
* yf[1] << y[0], not sure if it could be ignored */
qmod4= q % 4;
if (qmod4==2)
/* (yf[0] > 0) defines interval (-pi,pi]*/
return ( (yf[0] > 0) ? -p[2] : p[2] ) + yf[0] + yf[1];
else
return p[qmod4] + yf[0] + yf[1];
}
Edit: Just realised that you need to link to libm.a, I couldn't find the symbols declared in libm.so
I have used (in python):
def WrapAngle(Wrapped, UnWrapped ):
TWOPI = math.pi * 2
TWOPIINV = 1.0 / TWOPI
return UnWrapped + round((Wrapped - UnWrapped) * TWOPIINV) * TWOPI
c-code equivalent:
#define TWOPI 6.28318531
double WrapAngle(const double dWrapped, const double dUnWrapped )
{
const double TWOPIINV = 1.0/ TWOPI;
return dUnWrapped + round((dWrapped - dUnWrapped) * TWOPIINV) * TWOPI;
}
notice that this brings it in the wrapped domain +/- 2pi so for +/- pi domain you need to handle that afterward like:
if( angle > pi):
angle -= 2*math.pi