I want to generate random numbers from -n to n excluding 0. Can someone provide me the code in C? How to exclude 0?
One idea might be to generate a random number x in the range [1,2n], inclusive. Then return -(x - n) for x larger than n, else just return x.
This should work:
int my_random(int n)
{
const int x = 1 + rand() / (RAND_MAX / (2 * n) + 1);
return x > n ? -(x - n) : x;
}
See the comp.lang.c FAQ for more information about how to use rand() safely; it explains the above usage.
The simplest thing I can suggest to do is to generate a Random number between 0 and 2n and then to do the math trick:
result= n - randomNumber
Although 0 might be very unlikely you can check for that using an If and redo the random number generation.
int random(int N)
{
int x;
do{
x=rand()%(N*2+1)-N;
}while(x==0);
return x;
}
It chooses a number from -N to N, but keeps on doing it if it is 0.
An alternative, as suggested in the comments, generates a number between -N and N-1 and increments it if its positive or 0:
int random(int N)
{
int x;
x=rand()%(N*2)-N;
if(x>=0) x++;
return x;
}
Related
My try
double sum_squares_from(double x, double n){
return n<=0 ? 0 : x*x + sum_squares_from((x+n-1)*(x+n-1),n-1);
}
Instead of using loops my professor wants us to write functions like this...
What the exercise asks for is a function sum_squares_from() with double x being the starting number and n is the number of number. For example if you do x = 2 and n = 4 you get 2*2+3*3+4*4+5*5. It returns zero if n == 0.
My thinking was that in my example what I have is basically x*x+(x+1)(x+1)+(x+1+1)(x+1+1)+(x+1+1+1)(x+1+1+1) = (x+0)(x+0)+(x+1)(x+1)+(x+2)(x+2)+(x+3)(x+3) = (x+n-1)^2 repeated n times where n gets decremented every time by one until it becomes zero and then you sum everything.
Did I do it right?
(if my professor seems a bit demanding... he somehow does this sort of thing all in his head without auxiliary calculations. Scary guy)
It's not recursive, but it's one line:
int
sum_squares(int x, int n) {
return ((x + n - 1) * (x + n) * (2 * (x + n - 1) + 1) / 6) - ((x - 1) * x * (2 * (x - 1) + 1) / 6);
}
Sum of squares (of integers) has a closed-form solution for 1 .. n. This code calculates the sum of squares from 1 .. (x+n) and then subtracts the sum of squares from 1 .. (x-1).
The original version of this answer used ASCII art.
So,
∑i:0..n i = n(n+1)(½)
∑i:0..n i2 = n(n+1)(2n+1)(⅙)
We note that,
∑i:0..n (x+i)2
= ∑i:0...n x2 + 2xi + i2
= (n+1)x2 + (2x)∑i:0..n i + ∑i:0..n i2
= (n+1)x2 + n(n+1)x + n(n+1)(2n+1)(⅙)
Thus, your sum has the closed form:
double sum_squares_from(double x, int n) {
return ((n-- > 0)
? (n + 1) * x * x
+ x * n * (n + 1)
+ n * (n + 1) * (2 * n + 1) / 6.
: 0);
}
If I apply some obfuscation, the one-line version becomes:
double sum_squares_from(double x, int n) {
return (n-->0)?(n+1)*(x*x+x*n+n*(2*n+1)/6.):0;
}
If the task is to implement the summation in a loop, use tail recursion. Tail recursion can be mechanically replaced with a loop, and many compilers implement this optimization.
static double sum_squares_from_loop(double x, int n, double s) {
return (n <= 0) ? s : sum_squares_from_loop(x+1, n-1, s+x*x);
}
double sum_squares_from(double x, int n) {
return sum_squares_from_loop(x, n, 0);
}
As an illustration, if you observe the generated assembly in GCC at a sufficient optimization level (-Os, -O2, or -O3), you will notice that the recursive call is eliminated (and sum_squares_from_loop is inlined to boot).
Try it online!
As mentioned in my original comment, n should not be type double, but instead be type int to avoid floating point comparison problems with n <= 0. Making the change and simplifying the multiplication and recursive call, you do:
double sum_squares_from(double x, int n)
{
return n <= 0 ? 0 : x * x + sum_squares_from (x + 1, n - 1);
}
If you think about starting with x * x and increasing x by 1, n times, then the simple x * x + sum_squares_from (x + 1, n - 1) is quite easy to understand.
Maybe this?
double sum_squares_from(double x, double n) {
return n <= 0 ? 0 : (x + n - 1) * (x + n - 1) + sum_squares_from(x, n - 1);
}
One of my C assignments was it to write an approximation of arctan(x) in the language C. The equation which I should base it on is
arctan(x)=\sum {k=0}^{\infty }(-1)^{k} \tfrac{x^{2k+1}}{2k+1}
In addition x is only defined as -1<=x<=1.
Here is my code.
#include <stdio.h>
#include <math.h>
double main(void) {
double x=1;
double k;
double sum;
double sum_old;
int count;
double pw(double y, double n) {
double i;
double number = 1;
for (i = 0; i < n; i++) {
number *= y;
}
return(number);
}
double fc (double y) {
double i;
double number = 1;
for (i = 1; i <= y; i++){
number *= i;
}
return(number);
}
if(x >= (-1) && x <= 1) {
for(k=0; sum!=sum_old; k++) {
sum_old = sum;
sum += pw((-1), k) * pw(x, (2*k) + 1)/((2*k) + 1);
count++;
printf("%d || %.17lf\n", count, sum);
}
printf("My result is: %.17lf\n",sum);
printf("atan(%f) is: %.17f\n", x, atan(x));
printf("My result minus atan(x) = %.17lf\n", sum - atan(x));
} else {
printf("x is not defined. Please choose an x in the intervall [-1, 1]\n");
}
return 0;
}
It seemingly works fine with every value, except value 1 and -1. If x=1, then the output ends with:
...
7207 || 0.78543285189457468
7208 || 0.78536
Whereas the output should look more like this. In this case x=0.5.
25 || 0.46364760900080587
26 || 0.46364760900080587
My result is: 0.46364760900080587
atan(0.500000) is: 0.46364760900080609
My result minus atan(x) atan(x) = -0.00000000000000022
How can I improve my code so that it can run with x=1 and x=-1.
Thanks in advance.
PS: I use my own created pw() function instead of pow(), because I wanted to bybass the restriction of not using pow() as we didn't had that in our lectures yet.
PPS: I'd appreciate any advice as to how to improve my code.
In each iteration, you add (-1)k • x2k+1 / (2k+1), and you stop when there is no change to the sum.
If this were calculated with ideal arithmetic (exact, infinitely precise arithmetic), it would never stop for non-zero x, since you are always changing the sum. When calculating with fixed-precision arithmetic, it stops when the term is so small it does not change the sum because of the limited precision.
When |x| is less than one by any significant amount, this comes quickly because x2k+1 gets smaller. When |x| is one, the term becomes just 1 / (2k+1), which gets smaller very slowly. Not until k is around 253 would the sum stop changing.
You might consider changing your stopping condition to be when sum has not changed from sum_old very much rather than when it has not changed at all.
if(x >= (-1) && x <= 1) {
for(k=0; sum!=sum_old; k++) {
sum_old = sum;
sum += pw((-1), k) * pw(x, (2*k) + 1)/((2*k) + 1);
count++;
printf("%d || %.17lf\n", count, sum);
}
Comparing doubles can be tricky. The conventional way to compare doubles is to test within epsilon. There should be an epsilon value defined somewhere, but for your purposes how many digits are enough to approximate? If you only need like 3 or 4 digits you can instead have
#define EPSILON 0.0001 //make this however precise you need to approximate.
if(x >= (-1) && x <= 1) {
for(k=0; fabs(sum - sum_old) > EPSILON; k++) {
sum_old = sum;
sum += pw((-1), k) * pw(x, (2*k) + 1)/((2*k) + 1);
count++;
printf("%d || %.17lf\n", count, sum);
}
If the issue is that -1,1 iterate too many times either reduce the precision or increase the step per iteration. I am not sure that is what you're asking though, please clarify.
I think the cause of this is for a mathematical reason rather than a programming one.
Away from the little mistakes and adjustments that you should do to your code, putting x = 1 in the infinite series of arctan, is a boundary condition:
In this series, we add a negative value to a positive value then a negative value. This means the sum will be increasing, decreasing, increasing, ... and this will make some difference each iteration. This difference will be smaller until the preciseness of double won't catch it, so the program will stop and give us the value.
But in the sum equation. When we set z = 1 and n goes from 0 to ∞, this will make this term (-1^n) equal to 1 in one time and -1 in the next iteration. Also,
the value of the z-term will be one and the denominator value when n approaches infinity will = ∞ .
So the sum several iterations will be like +1/∞ -1/∞ +1/∞ -1/∞ ... (where ∞ here represents a big number). That way the series will not reach a specific number. This is because z = 1 is a boundary in this equation. And that is causing infinite iterations in your solution without reaching a number.
If you need to calculate arctan(1) I think you should use this formula:
All formulas are from this Wikipedia article.
Here is some modifications that make your code more compact and has less errors:
#include <stdio.h>
#include <math.h>
#define x 0.5 //here x is much easier to change
double pw(double, double); //declaration of the function should be done
int main() { //the default return type of main is int.
double k;
double sum = 0 ; //you should initiate your variables.
double sum_old = 1 ; //=1 only to pass the for condition first time.
//you don't need to define counter here
if(x < -1 || x > 1){
printf("x is not defined. Please choose an x in the interval [-1, 1]\n");
return 0;
}
for(k=0; sum!=sum_old; k++) {
sum_old = sum;
sum += pw((-1), k) * pw(x, (2*k) + 1)/((2*k) + 1);
printf("%.0f || %.17lf\n", k, sum);
}
printf("My result is: %.17lf\n",sum);
printf("atan(%f) is: %.17f\n", x, atan(x));
printf("My result minus atan(x) = %.17lf\n", sum - atan(x));
return 0;
}
double pw(double y, double n) { //functions should be declared out of the main function
double i;
double number = 1;
for (i = 0; i < n; i++) {
number *= y;
}
return(number);
}
double fc (double y) {
double i;
double number = 1;
for (i = 1; i <= y; i++){
number *= i;
}
return(number);
}
I wanna make a funcion that will take a natural number and make a new number so every digit in the old number will be incremented and if the digit is 9 it will become zero, but not to check specificly if the digit is 9.
example:
930 will return 41
9999 will return 0
879021 will return 980132.
This is what i got so far:
int newNumber(int n)
{
int dig;
if (n < 9)
return n + 1;
dig = n % 10;
dig++;
n = n / 10;
n = n * 10 + dig;
return newNumber(n/10);
}
There are a couple of issues with your code:
It doesn't handle a single digit of 9 (which cause a stack overflow).
Adding 1 to 9 makes 10 not 0.
I've run it through the sample data you supplied and it seems to work (in C#) and it has a hard core recursive line at the end.
int newNumber(int n)
{
if (n == 9)
return 0;
if (n < 9)
return n + 1;
return (newNumber(n / 10) * 10) + newNumber(n % 10);
}
Here's to avoid the check for n == 9:
int newNumber(int n)
{
static int table[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 };
return (n <= 9) ? table[n] : (newNumber(n / 10) * 10) + newNumber(n % 10);
}
A lookup table seems the most appropriate and does exactly what the requirements describe. Trying to use the non-compatible arithmetic operators results in side effects (as we see in Bathsheba's answer for example), that then need to be corrected.
unsigned newNumber(unsigned n, unsigned c = 0)
{
return n ? (n + 1) % 10 + 10 * newNumber(n / 10, 1 + c) : !c;
}
is one way, and it will treat 0 as 1, via the !c branch where c counts the number of recursions. Note the tail recursion in the ternary conditional branch - some compilers will optimise a tail recursion out to a simple loop, see What is tail recursion?
Bathsheba's solution posted above is very elegant by using the ternary operator, but it will give you a wrong result if the input is zero. To avoid that you may use a stub function:
#include <stdio.h>
int incDigits(int n)
{
return n ? (n + 1) % 10 + incDigits(n / 10) * 10 : 0;
}
int newNumber(int n)
{
return n ? incDigits(n) : 1;
}
int main()
{
for(int i = 0; i <= 100; ++i)
{
int n = newNumber(i);
printf("%d -> %d\n", i, n);
}
}
EDIT: user meaning-matters also posted a way to fix the input value problem using a lookup table, but he still has to check if n equals 9, which is something you don't want. So I believe using a stub function still is the best way.
Two ternary operator has been used to take care of the two cases:
i) number equal to 9
ii) number not equal to 9 => Another ternary operator is used to take care of further two possible cases:
a) number greater than 9( return sum of num(n/10)*10 and num(n%10) ); this can be further elaborated based on the argument fed to the num function.
b)number smaller than 9(return number plus one(n+1))
Once this function is called from the main function with argument equal to the number to be transformed in the manner asked in the question, each call in line4 will undergo recursion until they pass the argument to the subsequent iteration less than or equal to 9(which leads to termination of the recursion). With basic understanding of recursion, the above para can easily be understood in context to the subroutine below.
Blockquote
int num(int n)//line1
{//line2
int t;//line3
t=(n==9?0:(n>9?num(n/10)*10+num(n%10):n+1));//line4
return t;/line5
}//line6
Blockquote
I want to generate a random number in which every higher number has half the probability of its predecessor that is if 1 appears 4 times 2 should appear 2 times and three should appear one time and so on
the programming language must be c
int getValue(int min,int max) {
int i=0;
while(i==0)
i=rand()%max;
int j=0;
while(j==0) {
printf("%d",j);
j=rand()%i;
}
return j;
}
I'd start by normalizing the requested range (later just add min to the random generated number):
int max_rnd = max - min;
Then generate a equally distributed random number in the range 1 to pow(2, max_rnd+1) - 1
int rnd_limit = (1 << (max_rnd + 1)) - 1;
int rnd = (rand() % rnd_limit) + 1;
Round down the base-2-logarithm of the generated random number. The translation of rnd will follow this pattern:
floor(log2(1)) -> 0
floor(log2(2)) -> 1
floor(log2(3)) -> 1
floor(log2(4)) -> 2
...
floor(log2(7)) -> 2
floor(log2(8)) -> 3
...
floor(log2(15)) -> 3
...
So the distribution of results will be the reverse of the expected (double instead of half for higher numbers) but thats easy to change. Effectively, the floor(log2(rnd)) is computing the position of the highest 1 bit in rnd, so it can be implemented with a bitshift loop:
int log2_rnd = 0;
while ((rnd >> log2_rnd) != 1)
{
++log2_rnd;
}
Time for the actual result:
return (max_rnd - log2_rnd) + min;
Not covered here: Accuracy of "equally distributed" numbers generated by rand() (specially in combination with modulo) and upper limit of generated numbers.
Following function may be used:
int prioritizdRand(int min, int max)
{
int result=min;
int randNum=rand();
long int start=RAND_MAX/2;
while(randNum > start)
{
if(result> max )
{
return min;
}
result++;
start= (start + RAND_MAX)/2;
}
return result;
}
a simplistic idea: start with value one, take a random number, add one to
value if random number is odd, bitshift one right, repeat
unsigned randhalf()
{
unsigned ret = 1;
int rnd = rand();
while (rnd&1)
{
++ret;
rnd >>= 1;
}
return ret;
}
Or if upper and lower bounds are required:
int randhalf(int min, int max) // bounds are inclusive
{
int ret = min;
int rnd = rand();
while (rnd&1 && ret<max)
{
++ret;
rnd >>= 1;
}
return ret;
}
This will only provide values that are larger than min by however many bits
rand() uses.
How can I generate random numbers from 0 to 1000000?
I already tried the code below, but it still gives me numbers from 0 to 32767 (RAND_MAX):
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main(){
int i,x;
srand(time(NULL));
for(i=0; i<10000; i++){
int x = rand() % 10000000 + 1;
printf("%d\n",x);
}
return 0;
}
[Edit] The initial answer was for 0 to 1,000,000. I now see it should be 0 to 10,000,000.
As rand() will give an answer of at least 15 bits, call rand() multiple times, shift by 15 and XOR the results. Finally mod by 10,000,001.
unsigned long x;
x = rand();
x <<= 15;
x ^= rand();
x %= 10000001;
The distribution is very flat, but does introduce a very small bias. After 32768*32768 iterations, each value of x 0 to 10,000,000 to occur about 107.37 times. Instead they range from 107 to 108 times.
Combining multiple rand() call results with +, * or | will cause a significant bias in the distribution of the results.
[Edit]
RAND_MAX is 32767 (0x7FFF) for OP's platform. The C spec says "value of the RAND_MAX macro shall be at least 32767". Because RAND_MAX may be longer than 15 bits, it is important to use the ^ operator above rather than | for this code when used on other platforms.
Calculate with % 1000001 if you want numbers between 0 and 1000000.
Also RAND_MAX is conly guaranteed to be at least 32767
int main(){
int i, x;
srand(time(NULL));
for(i=0; i<10000; i++){
x = (rand() * rand()) % 1000001;
printf("%d\n",x);
}
return 0;
}
Use this function, it will give you random number between two number (min and max) :
unsigned long int my_rand (unsigned long int Min, unsigned long int Max)
{
static int first = 0;
if (first == 0)
{
srand (time (NULL)); //initialize generator of random number
first = 1;
}
return ((unsigned long int)(rand() * (Max+1 - Min) / RAND_MAX + Min));
}