beginner here trying to understand the source of the bug.
I have written this recursive function for finding the binomial coefficient between two numbers which is apparently correct in concept. However, for these two numbers, n =4 and k=2, I should be getting 6 as a result whereas I actually get 16. Any idea why is that happening?
#include<stdio.h>
int binomial(int n, int k)
{
if ((k = 0) || (k == n))
return 1;
if (k>n)
return 0;
return binomial(n - 1, k - 1) + binomial(n - 1, k);
}
int main()
{
int a, b, res;
a = 4;
b = 2;
res = binomial(a, b);
printf("The result is %d", res);
return 0;
}
This line looks wrong as it assigns 0 to k:
if ((k=0) || (k==n))
You probably mean:
if ((k==0) || (k==n))
This line
if ((k = 0) || (k == n))
should be
if ((k == 0) || (k == n))
^^
You were assigning zero to k.
As #Michael Walz points out, it's good practice to compile with -Wall to turn on all compilation warnings.
Related
Here is a code to exponentiate a number to a given power:
#include <stdio.h>
int foo(int m, int k) {
if (k == 0) {
return 1;
} else if (k % 2 != 0) {
return m * foo(m, k - 1);
} else {
int p = foo(m, k / 2);
return p * p;
}
}
int main() {
int m, k;
while (scanf("%d %d", &m, &k) == 2) {
printf("%d\n", foo(m, k));
}
return 0;
}
How do I calculate the time complexity of the function foo?
I have been able to deduce that if k is a power of 2, the time complexity is O(log k).
But I am finding it difficult to calculate for other values of k. Any help would be much appreciated.
How do I calculate the time complexity of the function foo()?
I have been able to deduce that if k is a power of 2, the time complexity is O(logk).
First, I assume that the time needed for each function call is constant (this would for example not be the case if the time needed for a multiplication depends on the numbers being multiplied - which is the case on some computers).
We also assume that k>=1 (otherwise, the function will run endlessly unless there is an overflow).
Let's think the value k as a binary number:
If the rightmost bit is 0 (k%2!=0 is false), the number is shifted right by one bit (foo(m,k/2)) and the function is called recursively.
If the rightmost bit is 1 (k%2!=0 is true), the bit is changed to a 0 (foo(m,k-1)) and the function is called recursively. (We don't look at the case k=1, yet.)
This means that the function is called once for each bit and it is called once for each 1 bit. Or, in other words: It is called once for each 0 bit in the number and twice for each 1 bit.
If N is the number of function calls, n1 is the number of 1 bits and n0 is the number of 0 bits, we get the following formula:
N = n0 + 2*n1 + C
The constant C (C=(-1), if I didn't make a mistake) represents the case k=1 that we ignored up to now.
This means:
N = (n0 + n1) + n1 + C
And - because n0 + n1 = floor(log2(k)) + 1:
floor(log2(k)) + C <= N <= 2*floor(log2(k)) + C
As you can see, the time complexity is always O(log(k))
O(log(k))
Some modification added to output a statistics for spread sheet plot.
#include <stdio.h>
#include <math.h>
#ifndef TEST_NUM
#define TEST_NUM (100)
#endif
static size_t iter_count;
int foo(int m, int k) {
iter_count++;
if (k == 0) {
return 1;
} else if(k == 1) {
return m;
} else if (k % 2 != 0) {
return m * foo(m, k - 1);
} else {
int p = foo(m, k / 2);
return p * p;
}
}
int main() {
for (int i = 1; i < TEST_NUM; ++i) {
iter_count = 0;
int dummy_result = foo(1, i);
printf("%d, %zu, %f\n", i, iter_count, log2(i));
}
return 0;
}
Build it.
gcc t1.c -DTEST_NUM=10000
./a > output.csv
Now open the output file with a spread sheet program and plot the last two output columns.
For k positive, the function foo calls itself recursively p times if k is the p-th power of 2. If k is not a power of 2, the number of recursive calls is strictly inferior to 2 * p where p is the exponent of the largest power of 2 inferior to k.
Here is a demonstration:
let's expand the recursive call in the case k % 2 != 0:
int foo(int m, int k) {
if (k == 1) {
return m;
} else
if (k % 2 != 0) { /* 2 recursive calls */
// return m * foo(m, k - 1);
int p = foo(m, k / 2);
return m * p * p;
} else { /* 1 recursive call */
int p = foo(m, k / 2);
return p * p;
}
}
The total number of calls is floor(log2(k)) + bitcount(k), and bitcount(k) is by construction <= ceil(log2(k)).
There are no loops in the code and the time of each individual call is bounded by a constant, hence the overall time complexity of O(log k).
The number of times the function is called (recursively or not) per power call is proportional to the minimum number of bits in the exponent required to represent it in binary form.
Each time you enter in the function, it solves by reducing the number by one if the exponent is odd, OR reducing it to half if the exponent is even. This means that we will do n squares per significant bit in the number, and m more multiplications by the base for all the bits that are 1 in the exponent (which are, at most, n, so m < n) for a 32bit significant exponent (this is an exponent between 2^31 and 2^32 the routine will do between 32 and 64 products to get the result, and will reenter to itself a maximum of 64 times)
as in both cases the routine is tail-recursive, the code you post can be substituted with an iterative code in which a while loop is used to solve the problem.
int foo(int m, int k)
{
int prod = 1; /* last recursion foo(m, 0); */
int sq = m; /* squares */
while (k) {
if (k & 1) {
prod *= sq; /* foo(m, k); k odd */
}
k >>= 1;
sq *= sq;
}
return prod; /* return final product */
}
That's huge savings!!! (between 32 multiplications and 64 multiplications, to elevate something to 1,000,000,000 power)
I was reading about Euclid's algorithm to calculate GCD and found the following code:
#include <stdio.h>
int main()
{
int m, n;
scanf("%d%d", &n, &m);
if (n < 0) n = -n;
if (m < 0) m = -m;
while(n != 0)
{
int temp = n;
n = m % n;
m = temp;
}
if(m != 0)
printf("The gcd is %d\n", m);
return 0;
}
But I have some questions:
why if n<0 or m<0 we let n=-n or m=-m
What if m==0 what should I return? The smallest possible GCD is 1 but this function doesn't return anything in this case... (If we ignore return 0 which is necessary for main)
The second part of the algorithm only works for non-negative values of m and n. Since gcd(m,n) = gcd(|m|,|n|) they are made positive.
If m = 0 and n != 0 the gcd is n and vica versa.
If both n and m are 0 the gcd is undefined because every number is a divisor of 0. Therefore, nothing is printed in this case.
Answer 1:
This algorithm works properly for the non-negative value m and positive value n.
Example:
GCD(-6, 2) = GCD(6, -2) = GCD(-6, -2) = GCD(6, 2) = 2
Answer 2:
If m == 0 then this function will return GCD equal n
I'd like to ask the following misunderstandings of C language, which I see I'm having.
I'm sorry if the code is not properly indented, I tried as much as I could but there are not so many guides on the internet.
The program asked given a starting number 'val' and a Even-Odd or Odd-Even alternating sequence (which stops whenever this rules is violated) to print the greater prime number with 'val'.
I tried with two functions and the main: one to control the GCD between two given numbers and the other to keep tracks of the greatest one, but I think I miss something in the code or in the conception of C function,
Because when compiled it returns me 0 or great number which I'm not entering.
One example to understand what I should do:
If my sequence was 10, 7, 8, 23 and my val was 3, I had to print 23, because it is the greatest integer prime with 3.
Here's the code :
#include <stdio.h>
int mcd(int a, int b)
{ // Gcd function
if (a == 0)
return b;
else
return mcd(b % a, b);
}
int valuta(int val, int h) // Valuing Max function
{
int temp = 0;
if (mcd(val, h) == 1 && h > temp)
temp = h;
return temp;
}
int main()
{
int val, d, x, y, z, t, contatore = 1;
scanf("%d", &val);
scanf("%d%d", &x, &y);
if (x > y && mcd(val, x) == 1)
{ // Two options
t = x;
}
else if (y > x && mcd(val, y) == 1)
{
t = y;
}
if ((x % 2 == 0 && y % 2 == 0) || (x % 2 == 1 && y % 2 == 1))
{ // Bad case
if (x > y && mcd(val, x) == 1)
{
t = x;
contatore = 0;
}
else if (y > x && mcd(val, y) == 1)
{
t = y;
contatore = 0;
}
}
else
{
while (contatore == 1)
{
scanf("%d", &z);
t = valuta(val, z);
if (x % 2 == 0 && z % 2 == 0)
{ // Even- Odd - Even
scanf("%d", &d);
t = valuta(val, d);
if (d % 2 == 0)
{
contatore = 0;
}
else
{
contatore = 0;
}
}
if (x % 2 == 1 && z % 2 == 1)
{ //Odd- Even- Odd
scanf("%d", &d);
t = valuta(val, d);
if (d % 2 == 1)
{
contatore = 0;
}
else
{
contatore = 0;
}
}
}
}
printf("%d\n", t);
return 0;
}
PS. Is there any way to reduce the number of lines of code or to reduce the effort in coding? I mean, a straightforward solution will be helpful.
Your valuta() function is flawed in that it needs to return the maximum qualifying value so far but has no knowledge of the previous maximum - temp is always zero. The following takes the previous maximum as an argument:
int valuta(int val, int h, int previous )
{
return ( mcd(val, h) == 1 && h > previous ) ? h : previous ;
}
And is called from main() thus:
t = valuta( val, x, t ) ;
The test mcd(val, h) == 1 is flawed, because mcd() only ever returns the value of parameter b which is not modified in the recursion, so will never return 1, unless the argument b is 1. Since I have no real idea what mcd() is intended to do, I cannot tell you how to fix it. It appear to be a broken implementation of Euclid's greatest common divisor algorithm, which correctly implemented would be:
int mcd(int a, int b)
{
if(b == 0)
return a;
else
return mcd(b, a % b);
}
But I cannot see how that relates to:
"[...] he greatest integer prime with 3 [...]
The odd/even even/odd sequence handling can be drastically simplified to the extent that it is shorter and simpler than your method (as requested) - and so that it works!
The following is a clearer starting point, but may not be a solution since it is unclear what it is it is supposed to do.
#include <stdio.h>
#include <stdbool.h>
int mcd(int a, int b)
{
if(b == 0)
return a;
else
return mcd(b, a % b);
}
int valuta(int val, int h, int previous )
{
return ( mcd(val, h) && h > previous ) ? h : previous ;
}
int main()
{
int val, x, t ;
printf( "Enter value:") ;
scanf("%d", &val);
typedef enum
{
EVEN = 0,
ODD = 1,
UNDEFINED
} eOddEven ;
eOddEven expect = UNDEFINED ;
bool sequence_valid = true ;
printf( "Enter sequence in odd/even or even/odd order (break sequence to exit):\n") ;
while( sequence_valid )
{
scanf("%d", &x);
if( expect == UNDEFINED )
{
// Sequence order determined by first value
expect = (x & 1) == 0 ? EVEN : ODD ;
}
else
{
// Switch expected odd/even
expect = (expect == ODD) ? EVEN : ODD ;
// Is new value in the expected sequence?
sequence_valid = (expect == ((x & 1) == 0 ? EVEN : ODD)) ;
}
// If the sequence is valid...
if( sequence_valid )
{
// Test if input is largest qualifying value
t = valuta( val, x, t ) ;
}
}
// Result
printf("Result: %d\n", t);
return 0;
}
Is there any way to find nth root of the number without any external library in C? I'm working on a bare metal code so there is no OS. Also, no complete C is there.
You can write a program like this for nth root. This program is for square root.
int floorSqrt(int x)
{
// Base cases
if (x == 0 || x == 1)
return x;
// Staring from 1, try all numbers until
// i*i is greater than or equal to x.
int i = 1, result = 1;
while (result < x)
{
if (result == x)
return result;
i++;
result = i*i;
}
return i-1;
}
You can use the same approach for nth root.
Here there is a C implementation of the the nth root algorithm you can find in wikipedia. It needs an exponentiation algorithm, so I also include an implementation of a basic method for exponentiation by squaring that you can find also find in wikipedia.
double npower(double const base, int const n)
{
if (n < 0) return npower(1/base, -n)
else if (n == 0) return 1.0;
else if (n == 1) return base;
else if (n % 2) return base*npower(base*base, n/2);
else return npower(base*base, n/2);
}
double nroot(double const base, int const n)
{
if (n == 1) return base;
else if (n <= 0 || base < 0) return NAN;
else {
double delta, x = base/n;
do {
delta = (base/npower(x,n-1)-x)/n;
x += delta;
} while (fabs(delta) >= 1e-8);
return x;
}
}
Some comments on this:
The nth root algorithm in wikipedia leaves freedom for the initial guess. In this example I set it up to be base/n, but this was just a guess.
The macro NAN is usually defined in <math.h>, so you would need to define it to be suitable for your needs.
Both functions are implemented in a very rough and simple way, and their performance can be greatly improved with careful thought.
The tolerance in this example is set to 1e-8 and should be changed to something different. It should probably be proportional to the value of the base.
You can try the nth_root C function :
// return a number that, when multiplied by itself nth times, makes N.
unsigned nth_root(const unsigned n, const unsigned nth) {
unsigned a = n, b, c, r = nth ? n + (n > 1) : n == 1 ;
for (; a < r; b = a + (nth - 1) * r, a = b / nth)
for (r = a, a = n, c = nth - 1; c && (a /= r); --c);
return r;
}
Source
I am attempting to write a simple recursive function to evaluate Associated Legendre Polynomials in C. As of now, I only have it written up to L=4 and M=3. I am using the seventh recurrence formula from the following Wikipedia article:
https://en.wikipedia.org/wiki/Associated_Legendre_polynomials
Here is my code:
#include <stdio.h>
#include <math.h>
long double Legendre(int L, int M, double x){
if((M == 0) && (L == 0)){
return 1;
}
if((M == 0) && (L == 1)){
return x;
}
if((M == 0) && (L == 2)){
return (1/2) * ((3*x*x) - 1);
}
if((M == 0) && (L == 3)){
return (1/2) * ((5*pow(x,3)) - (3*x));
}
if((M == 0) && (L == 4)){
return (1/8) * (3 - (30*x*x) + (35*pow(x,4)));
}
else if(M > 0){
return (((L - M + 1)*x*Legendre(L,M-1,x))
- ((L + M - 1)*Legendre(L-1,M-1,x)))
/ sqrt(1-(x*x));
}
return 0;
}
int main(void){
int L = 4;
int M = 2;
double x = 0.6;
printf("%Lf\n",Legendre(L,M,x));
}
As per Mathematica, the Associated Legendre Polynomial with L=4, M=2, and x=0.6 should be 7.3. However, when I run the above program, I get the following result:
$ gcc Rodrigues.c -o Rodrigues
$ ./Rodrigues
0.000000
$
(My source file name is Rodrigues.c). The output should be around 7.3, not 0. The code also produces other non-zero (though still incorrect) values for different Ls and Ms. I am suspecting it is because I have somehow messed up the recursion (as you can tell, I am relatively new to programming). What am I missing? Thanks.