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I try two function for modular exponentiation for big base return wrong results,
One of the function is:
uint64_t modular_exponentiation(uint64_t x, uint64_t y, uint64_t p)
{
uint64_t res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
For input x = 1103362698 ,y = 137911680 , p=1217409241131113809;
It return the value (x^y mod p):749298230523009574(Incorrect).
The correct value is:152166603192600961
The other function i try, gave same result, What is wrong with these functions?
The other one is :
long int exponentMod(long int A, long int B, long int C)
{
// Base cases
if (A == 0)
return 0;
if (B == 0)
return 1;
// If B is even
long int y;
if (B % 2 == 0) {
y = exponentMod(A, B / 2, C);
y = (y * y) % C;
}
// If B is odd
else {
y = A % C;
y = (y * exponentMod(A, B - 1, C) % C) % C;
}
return (long int)((y + C) % C);
}
With p = 1217409241131113809, this value as well as any intermediate values for res and x will be larger than 32 bits. This means that multiplying two of these numbers could result in a value larger than 64 bits which overflows the datatype you're using.
If you restrict the parameters to 32 bit datatypes and use 64 bit datatypes for intermediate values then the function will work. Otherwise you'll need to use a big number library to get correct output.
This program works in handling positive integers, but not on negative integers. How can I fix this one? Thanks!
By the way, is my code good or not? Is there a better way on getting the quotient and remainder without using the '/', '%', and '*' operators?
#include <stdio.h>
int divide(int x, int y, int quotient);
int getRem(int x, int y, int quotient, int product, int count,
int remainder);
int main()
{
int dividend, divisor, quotient = 0, product = 0;
int remainder, count = 0;
scanf("%d %d", ÷nd, &divisor);
printf("\nQuotient: %d", divide(dividend, divisor, quotient));
quotient = divide(dividend, divisor, quotient);
printf("\nRemainder: %d", getRem(dividend, divisor, quotient, product, count, remainder));
}
int divide(int x, int y, int quotient)
{
while (x > 0)
{
x -= y;
quotient++;
}
if (x != 0)
return quotient - 1;
else
return quotient;
}
int getRem(int x, int y, int quotient, int product, int count, int remainder)
{
while (count != y)
{
product += quotient;
count++;
remainder = x - product;
}
return remainder;
}
By the way, is my code good or not?
Well, there's room for improvements...
First of all - don't pass unnecessary variables to your function!
A function that shall divide x by y shall only take x and y as arguments. Whatever variables you need inside the function shall be defined inside the function.
So the first step is to change your divide function to be:
int divide(int x, int y)
{
int quotient = 0; // use a local variable
while (x > 0)
{
x -= y;
quotient++;
}
if (x != 0)
return quotient - 1;
else
return quotient;
}
Another (minor) issue is the two return statements. With a simple change of the while statement that can be avoided.
int divide(int x, int y)
{
int quotient = 0; // use a local variable
while (x >= y) // notice this change
{
x -= y;
quotient++;
}
return quotient;
}
Also notice that a call like divide(42, 0); will cause an infinite loop. So perhaps you should check for y being zero.
The algorithm can be improved - especially for large numbers - but I guess you want a simple approach so I stick to your basic algorithm.
... but not on negative integers. How can I fix this one?
A simple approach is to convert any negative input before entering the loop and maintain a counter to remember the number of negative numbers. Something like:
int divide(int x, int y)
{
int quotient = 0;
int negative = 0;
if (x < 0)
{
x = -x; // Make x positive
++negative;
}
if (y < 0)
{
y = -y; // Make y positive
++negative;
}
while (x >= y) // Both x and y are positive here
{
x -= y;
quotient++;
}
return (negative == 1) ? -quotient : quotient;
}
int main(void)
{
printf("%d\n", divide( 5, 2));
printf("%d\n", divide( 5,-2));
printf("%d\n", divide(-5, 2));
printf("%d\n", divide(-5,-2));
printf("%d\n", divide( 6, 2));
printf("%d\n", divide( 6,-2));
printf("%d\n", divide(-6, 2));
printf("%d\n", divide(-6,-2));
return 0;
}
Output:
2
-2
-2
2
3
-3
-3
3
You can apply the same kind of changes to the function getRem and I'll leave that part for you as an exercise...
However, notice that your current function uses quotient without any benefit. The function (only handling positive numbers) could simply be:
int getRem(int x, int y) // requires x >= 0 and y > 0
{
while (x >= y)
{
x -= y;
}
return x;
}
Is there a better way of getting the quotient and remainder without using the '/', '%', and '*' operators?
The majority of the time, I think, by far the best way of computing quotient and remainder is with the / and % operators. (It's their job, after all!)
If you need both quotient and remainder at the same time, there's also the div function, which does exactly that. (Its return value is a struct containing quot and rem members.) It's a bit cumbersome to use, but it might be more efficient if it means executing a single divide instruction (which tends to give you both quotient and remainder at the machine level anyway) instead of two. (Or, these days, with a modern optimizing compiler, I suspect it wouldn't make any difference anyway.)
But no matter how you do it, there's always a question when it comes to negative numbers. If the dividend or the divisor is negative, what sign(s) should the quotient and remainder be? There are basically two answers. (Actually, there are more than two, but these are the two I think about.)
In Euclidean division, the remainder is always positive, and is therefore always in the range [0,divisor) (that is, between 0 and divisor-1).
In many programming languages (including C), the remainder always has the sign of the dividend.
See the Wikipedia article on modulo operation for much, much more information on these and other alternatives.
If your programming language gives you the second definition (as C does), but you want a remainder that's never negative, one way to fix it is just to do a regular division, test whether the remainder is negative, and if it is, increment the remainder by the divisor and decrement the quotient by 1:
quotient = x / y;
remainder = x % y;
if(remainder < 0) {
reminder += y;
quotient--;
}
This works because of the formal definition of integer division with remainder:
div(x, y) → q, r such that y × q + r = x
If you subtract 1 from q and add y to r, you get the same result, and it's still x.
Here's a sample program demonstrating three different alternatives. I've encapsulated the little "adjust quotient for nonnegative remainder" algorithm as the function euclid(), which returns the same div_t type as div() does:
#include <stdio.h>
#include <stdlib.h>
div_t euclid(int, int);
int main()
{
int x = -7, y = 3;
printf("operators: q = %d, r = %d\n", x/y, x%y);
div_t qr = div(x, y);
printf("div: q = %d, r = %d\n", qr.quot, qr.rem);
qr = euclid(x, y);
printf("euclid: q = %d, r = %d\n", qr.quot, qr.rem);
}
div_t euclid(int x, int y)
{
div_t qr = div(x, y);
if(qr.rem < 0) {
qr.quot--;
qr.rem += y;
}
return qr;
}
To really explore this, you'll want to try things out for all four cases:
positive dividend, positive divisor (the normal case, no ambiguity)
negative dividend, positive divisor (the case I showed)
positive dividend, negative divisor
negative dividend, negative divisor
This program works in handling positive integers, but not on negative integers. How can I fix this one?
When you call :
divide(-10, 3) or divide(-10, -3) the while loop condition while(x > 0) will be false
divide(10, -3) the while loop will be infinite loop because of the statement x -= y. (x) will be increase by the value of (y)
is my code good or not?
Your code is need to be organized:
in the main function you need to pass only the necessary parameters. in the divide function why you pass the quotient. the quotient will be calculated inside the function. so that you should edit the prototype to int divide(int x, int y);.
the same thing with getRem function you should edit the prototype to int getRem(int x, int y);.
in the main function you are calling the divide function twice to print the quotient and to save the quotient in the variable quotient. instead you should call the function only one time and reuse the returned value.
After the previous points your main and functions prototypes should be as follow:
#include <stdio.h>
int divide(int x, int y);
int getRem(int x, int y);
int main()
{
int dividend, divisor, quotient = 0;
scanf("%d %d", ÷nd, &divisor);
quotient = divide(dividend, divisor);
printf("\nQuotient: %d", quotient);
printf("\nRemainder: %d", getRem(dividend, divisor));
}
Now lets analyze the function divide. The first point, In math when multiplying or dividing, you actually do the operation on the sign as doing on the number. the following is the math rules for multiply or divide the sign.
negative * positive -> negative
positive * negative -> negative
negative * negative -> positive
negative / positive -> negative
positive / negative -> negative
negative / negative -> positive
The second point is the while loop. You should divide until (x) is less than (y).
As an example: suppose x = 7, y = 3 :
after first loop x -> 4 and y -> 3
after second loop x -> 1 and y -> 3 so that the condition should be while(x >= y)
so now you should first manipulate the sign division after that the numbers division. as follows.
int divide(int x, int y)
{
int quotient = 0;
//save the resulting sign in the sign variable
int sign = 1;
int temp = 0;
/* sign division */
// negative / positive -> negative
if((x < 0) && (y >= 0)){
sign = -1;
temp = x;
x -= temp;
x -= temp;
}// positive / negative -> negative
else if((x >= 0) && (y < 0)){
sign = -1;
temp = y;
y -= temp;
y -= temp;
}// negative / negative -> positive
else if((x < 0) && (y < 0)){
temp = x;
x -= temp;
x -= temp;
temp = y;
y -= temp;
y -= temp;
}
while (x >= y)
{
x -= y;
quotient++;
}
if(sign < 0){
temp = quotient;
quotient -= temp;
quotient -= temp;
}
return quotient;
}
Lets go into the getRem function. The remainder(%) operation rules is as follows:
negative % (negative or positive) -> negative
positive % (negative or positive) -> positive
note: the result follows the sign of the first operand
As an Example:
suppose x = -10 and y = 3, x / y = -3, to retrieve (x) multiply y and -3, so x = y * -3 = -9 the remainder is equal to -1
suppose x = 10 and y = -3, x / y = -3',x = y * -3 = 9` the remainder is equal to 1
now apply the previous points on the getRem function to get the following result:
int getRem(int x, int y)
{
int temp, remSign = 1;
if(x < 0){
remSign = -1;
temp = x;
x -= temp;
x -= temp;
}
if(y < 0){
temp = y;
y -= temp;
y -= temp;
}
while (x >= y)
{
x -= y;
}
if(remSign < 0){
x = -x;
}
return x;
}
You can to combine the two operation in one function using pointers for remainder as follows:
#include <stdio.h>
int divide(int x, int y,int * rem);
int getRem(int x, int y);
int main()
{
int dividend, divisor, quotient = 0;
int rem;
scanf("%d %d", ÷nd, &divisor);
quotient = divide(dividend, divisor, &rem);
printf("\nQuotient: %d", quotient);
printf("\nRemainder: %d", rem);
}
int divide(int x, int y, int * rem)
{
int quotient = 0;
int sign = 1;
int remSign = 1;
int temp = 0;
if((x < 0) && (y >= 0)){
sign = -1;
remSign = -1;
temp = x;
x -= temp;
x -= temp;
}
else if((x >= 0) && (y < 0)){
sign = -1;
temp = y;
y -= temp;
y -= temp;
}
else if((x < 0) && (y < 0)){
temp = x;
x -= temp;
x -= temp;
temp = y;
y -= temp;
y -= temp;
}
while (x >= y)
{
x -= y;
quotient++;
}
if(remSign < 0){
*rem = -x;
}else{
*rem = x;
}
if(sign < 0){
temp = quotient;
quotient -= temp;
quotient -= temp;
}
return quotient;
}
I require symmetric modulo operator for balanced ternary number system; It can be used to calculate least significant trit of trits sum (ternary addition function); My implementation seems inefficient, it uses 2 modulo(%) operators, and several addition/substraction operations. Here is my code (it works correctly):
int symmetric_modulo( int x, int fullRadix = 3 ) {
int halfRadix = fullRadix / 2;
int divRemainder = x % fullRadix;
return ( divRemainder + halfRadix + fullRadix ) % fullRadix - halfRadix;
}
x may be in range [-3..3]; and result should be -1, 0 or 1;
Can you offer more efficient code to do this?
int divRemainder = x % fullRadix; only needed if x + halfRadix + fullRadix overflows or the result is negative.
If no overflow/negative result possible as in x may be in range [-3..3]; Simplification:
int symmetric_modulo(int x, int fullRadix) {
int halfRadix = fullRadix / 2;
return (x + halfRadix + fullRadix) % fullRadix - halfRadix;
}
How would you divide a number by 3 without using *, /, +, -, %, operators?
The number may be signed or unsigned.
This is a simple function which performs the desired operation. But it requires the + operator, so all you have left to do is to add the values with bit-operators:
// replaces the + operator
int add(int x, int y)
{
while (x) {
int t = (x & y) << 1;
y ^= x;
x = t;
}
return y;
}
int divideby3(int num)
{
int sum = 0;
while (num > 3) {
sum = add(num >> 2, sum);
num = add(num >> 2, num & 3);
}
if (num == 3)
sum = add(sum, 1);
return sum;
}
As Jim commented this works, because:
n = 4 * a + b
n / 3 = a + (a + b) / 3
So sum += a, n = a + b, and iterate
When a == 0 (n < 4), sum += floor(n / 3); i.e. 1, if n == 3, else 0
Idiotic conditions call for an idiotic solution:
#include <stdio.h>
#include <stdlib.h>
int main()
{
FILE * fp=fopen("temp.dat","w+b");
int number=12346;
int divisor=3;
char * buf = calloc(number,1);
fwrite(buf,number,1,fp);
rewind(fp);
int result=fread(buf,divisor,number,fp);
printf("%d / %d = %d", number, divisor, result);
free(buf);
fclose(fp);
return 0;
}
If also the decimal part is needed, just declare result as double and add to it the result of fmod(number,divisor).
Explanation of how it works
The fwrite writes number bytes (number being 123456 in the example above).
rewind resets the file pointer to the front of the file.
fread reads a maximum of number "records" that are divisor in length from the file, and returns the number of elements it read.
If you write 30 bytes then read back the file in units of 3, you get 10 "units". 30 / 3 = 10
log(pow(exp(number),0.33333333333333333333)) /* :-) */
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
int num = 1234567;
int den = 3;
div_t r = div(num,den); // div() is a standard C function.
printf("%d\n", r.quot);
return 0;
}
You can use (platform dependent) inline assembly, e.g., for x86: (also works for negative numbers)
#include <stdio.h>
int main() {
int dividend = -42, divisor = 5, quotient, remainder;
__asm__ ( "cdq; idivl %%ebx;"
: "=a" (quotient), "=d" (remainder)
: "a" (dividend), "b" (divisor)
: );
printf("%i / %i = %i, remainder: %i\n", dividend, divisor, quotient, remainder);
return 0;
}
Use itoa to convert to a base 3 string. Drop the last trit and convert back to base 10.
// Note: itoa is non-standard but actual implementations
// don't seem to handle negative when base != 10.
int div3(int i) {
char str[42];
sprintf(str, "%d", INT_MIN); // Put minus sign at str[0]
if (i>0) // Remove sign if positive
str[0] = ' ';
itoa(abs(i), &str[1], 3); // Put ternary absolute value starting at str[1]
str[strlen(&str[1])] = '\0'; // Drop last digit
return strtol(str, NULL, 3); // Read back result
}
(note: see Edit 2 below for a better version!)
This is not as tricky as it sounds, because you said "without using the [..] + [..] operators". See below, if you want to forbid using the + character all together.
unsigned div_by(unsigned const x, unsigned const by) {
unsigned floor = 0;
for (unsigned cmp = 0, r = 0; cmp <= x;) {
for (unsigned i = 0; i < by; i++)
cmp++; // that's not the + operator!
floor = r;
r++; // neither is this.
}
return floor;
}
then just say div_by(100,3) to divide 100 by 3.
Edit: You can go on and replace the ++ operator as well:
unsigned inc(unsigned x) {
for (unsigned mask = 1; mask; mask <<= 1) {
if (mask & x)
x &= ~mask;
else
return x & mask;
}
return 0; // overflow (note that both x and mask are 0 here)
}
Edit 2: Slightly faster version without using any operator that contains the +,-,*,/,% characters.
unsigned add(char const zero[], unsigned const x, unsigned const y) {
// this exploits that &foo[bar] == foo+bar if foo is of type char*
return (int)(uintptr_t)(&((&zero[x])[y]));
}
unsigned div_by(unsigned const x, unsigned const by) {
unsigned floor = 0;
for (unsigned cmp = 0, r = 0; cmp <= x;) {
cmp = add(0,cmp,by);
floor = r;
r = add(0,r,1);
}
return floor;
}
We use the first argument of the add function because we cannot denote the type of pointers without using the * character, except in function parameter lists, where the syntax type[] is identical to type* const.
FWIW, you can easily implement a multiplication function using a similar trick to use the 0x55555556 trick proposed by AndreyT:
int mul(int const x, int const y) {
return sizeof(struct {
char const ignore[y];
}[x]);
}
It is easily possible on the Setun computer.
To divide an integer by 3, shift right by 1 place.
I'm not sure whether it's strictly possible to implement a conforming C compiler on such a platform though. We might have to stretch the rules a bit, like interpreting "at least 8 bits" as "capable of holding at least integers from -128 to +127".
Here's my solution:
public static int div_by_3(long a) {
a <<= 30;
for(int i = 2; i <= 32 ; i <<= 1) {
a = add(a, a >> i);
}
return (int) (a >> 32);
}
public static long add(long a, long b) {
long carry = (a & b) << 1;
long sum = (a ^ b);
return carry == 0 ? sum : add(carry, sum);
}
First, note that
1/3 = 1/4 + 1/16 + 1/64 + ...
Now, the rest is simple!
a/3 = a * 1/3
a/3 = a * (1/4 + 1/16 + 1/64 + ...)
a/3 = a/4 + a/16 + 1/64 + ...
a/3 = a >> 2 + a >> 4 + a >> 6 + ...
Now all we have to do is add together these bit shifted values of a! Oops! We can't add though, so instead, we'll have to write an add function using bit-wise operators! If you're familiar with bit-wise operators, my solution should look fairly simple... but just in-case you aren't, I'll walk through an example at the end.
Another thing to note is that first I shift left by 30! This is to make sure that the fractions don't get rounded off.
11 + 6
1011 + 0110
sum = 1011 ^ 0110 = 1101
carry = (1011 & 0110) << 1 = 0010 << 1 = 0100
Now you recurse!
1101 + 0100
sum = 1101 ^ 0100 = 1001
carry = (1101 & 0100) << 1 = 0100 << 1 = 1000
Again!
1001 + 1000
sum = 1001 ^ 1000 = 0001
carry = (1001 & 1000) << 1 = 1000 << 1 = 10000
One last time!
0001 + 10000
sum = 0001 ^ 10000 = 10001 = 17
carry = (0001 & 10000) << 1 = 0
Done!
It's simply carry addition that you learned as a child!
111
1011
+0110
-----
10001
This implementation failed because we can not add all terms of the equation:
a / 3 = a/4 + a/4^2 + a/4^3 + ... + a/4^i + ... = f(a, i) + a * 1/3 * 1/4^i
f(a, i) = a/4 + a/4^2 + ... + a/4^i
Suppose the reslut of div_by_3(a) = x, then x <= floor(f(a, i)) < a / 3. When a = 3k, we get wrong answer.
To divide a 32-bit number by 3 one can multiply it by 0x55555556 and then take the upper 32 bits of the 64 bit result.
Now all that's left to do is to implement multiplication using bit operations and shifts...
Yet another solution. This should handle all ints (including negative ints) except the min value of an int, which would need to be handled as a hard coded exception. This basically does division by subtraction but only using bit operators (shifts, xor, & and complement). For faster speed, it subtracts 3 * (decreasing powers of 2). In c#, it executes around 444 of these DivideBy3 calls per millisecond (2.2 seconds for 1,000,000 divides), so not horrendously slow, but no where near as fast as a simple x/3. By comparison, Coodey's nice solution is about 5 times faster than this one.
public static int DivideBy3(int a) {
bool negative = a < 0;
if (negative) a = Negate(a);
int result;
int sub = 3 << 29;
int threes = 1 << 29;
result = 0;
while (threes > 0) {
if (a >= sub) {
a = Add(a, Negate(sub));
result = Add(result, threes);
}
sub >>= 1;
threes >>= 1;
}
if (negative) result = Negate(result);
return result;
}
public static int Negate(int a) {
return Add(~a, 1);
}
public static int Add(int a, int b) {
int x = 0;
x = a ^ b;
while ((a & b) != 0) {
b = (a & b) << 1;
a = x;
x = a ^ b;
}
return x;
}
This is c# because that's what I had handy, but differences from c should be minor.
It's really quite easy.
if (number == 0) return 0;
if (number == 1) return 0;
if (number == 2) return 0;
if (number == 3) return 1;
if (number == 4) return 1;
if (number == 5) return 1;
if (number == 6) return 2;
(I have of course omitted some of the program for the sake of brevity.) If the programmer gets tired of typing this all out, I'm sure that he or she could write a separate program to generate it for him. I happen to be aware of a certain operator, /, that would simplify his job immensely.
Using counters is a basic solution:
int DivBy3(int num) {
int result = 0;
int counter = 0;
while (1) {
if (num == counter) //Modulus 0
return result;
counter = abs(~counter); //++counter
if (num == counter) //Modulus 1
return result;
counter = abs(~counter); //++counter
if (num == counter) //Modulus 2
return result;
counter = abs(~counter); //++counter
result = abs(~result); //++result
}
}
It is also easy to perform a modulus function, check the comments.
This one is the classical division algorithm in base 2:
#include <stdio.h>
#include <stdint.h>
int main()
{
uint32_t mod3[6] = { 0,1,2,0,1,2 };
uint32_t x = 1234567; // number to divide, and remainder at the end
uint32_t y = 0; // result
int bit = 31; // current bit
printf("X=%u X/3=%u\n",x,x/3); // the '/3' is for testing
while (bit>0)
{
printf("BIT=%d X=%u Y=%u\n",bit,x,y);
// decrement bit
int h = 1; while (1) { bit ^= h; if ( bit&h ) h <<= 1; else break; }
uint32_t r = x>>bit; // current remainder in 0..5
x ^= r<<bit; // remove R bits from X
if (r >= 3) y |= 1<<bit; // new output bit
x |= mod3[r]<<bit; // new remainder inserted in X
}
printf("Y=%u\n",y);
}
Write the program in Pascal and use the DIV operator.
Since the question is tagged c, you can probably write a function in Pascal and call it from your C program; the method for doing so is system-specific.
But here's an example that works on my Ubuntu system with the Free Pascal fp-compiler package installed. (I'm doing this out of sheer misplaced stubbornness; I make no claim that this is useful.)
divide_by_3.pas :
unit Divide_By_3;
interface
function div_by_3(n: integer): integer; cdecl; export;
implementation
function div_by_3(n: integer): integer; cdecl;
begin
div_by_3 := n div 3;
end;
end.
main.c :
#include <stdio.h>
#include <stdlib.h>
extern int div_by_3(int n);
int main(void) {
int n;
fputs("Enter a number: ", stdout);
fflush(stdout);
scanf("%d", &n);
printf("%d / 3 = %d\n", n, div_by_3(n));
return 0;
}
To build:
fpc divide_by_3.pas && gcc divide_by_3.o main.c -o main
Sample execution:
$ ./main
Enter a number: 100
100 / 3 = 33
int div3(int x)
{
int reminder = abs(x);
int result = 0;
while(reminder >= 3)
{
result++;
reminder--;
reminder--;
reminder--;
}
return result;
}
Didn't cross-check if this answer is already published. If the program need to be extended to floating numbers, the numbers can be multiplied by 10*number of precision needed and then the following code can be again applied.
#include <stdio.h>
int main()
{
int aNumber = 500;
int gResult = 0;
int aLoop = 0;
int i = 0;
for(i = 0; i < aNumber; i++)
{
if(aLoop == 3)
{
gResult++;
aLoop = 0;
}
aLoop++;
}
printf("Reulst of %d / 3 = %d", aNumber, gResult);
return 0;
}
This should work for any divisor, not only three. Currently only for unsigned, but extending it to signed should not be that difficult.
#include <stdio.h>
unsigned sub(unsigned two, unsigned one);
unsigned bitdiv(unsigned top, unsigned bot);
unsigned sub(unsigned two, unsigned one)
{
unsigned bor;
bor = one;
do {
one = ~two & bor;
two ^= bor;
bor = one<<1;
} while (one);
return two;
}
unsigned bitdiv(unsigned top, unsigned bot)
{
unsigned result, shift;
if (!bot || top < bot) return 0;
for(shift=1;top >= (bot<<=1); shift++) {;}
bot >>= 1;
for (result=0; shift--; bot >>= 1 ) {
result <<=1;
if (top >= bot) {
top = sub(top,bot);
result |= 1;
}
}
return result;
}
int main(void)
{
unsigned arg,val;
for (arg=2; arg < 40; arg++) {
val = bitdiv(arg,3);
printf("Arg=%u Val=%u\n", arg, val);
}
return 0;
}
Would it be cheating to use the / operator "behind the scenes" by using eval and string concatenation?
For example, in Javacript, you can do
function div3 (n) {
var div = String.fromCharCode(47);
return eval([n, div, 3].join(""));
}
First that I've come up with.
irb(main):101:0> div3 = -> n { s = '%0' + n.to_s + 's'; (s % '').gsub(' ', ' ').size }
=> #<Proc:0x0000000205ae90#(irb):101 (lambda)>
irb(main):102:0> div3[12]
=> 4
irb(main):103:0> div3[666]
=> 222
EDIT: Sorry, I didn't notice the tag C. But you can use the idea about string formatting, I guess...
Using BC Math in PHP:
<?php
$a = 12345;
$b = bcdiv($a, 3);
?>
MySQL (it's an interview from Oracle)
> SELECT 12345 DIV 3;
Pascal:
a:= 12345;
b:= a div 3;
x86-64 assembly language:
mov r8, 3
xor rdx, rdx
mov rax, 12345
idiv r8
The following script generates a C program that solves the problem without using the operators * / + - %:
#!/usr/bin/env python3
print('''#include <stdint.h>
#include <stdio.h>
const int32_t div_by_3(const int32_t input)
{
''')
for i in range(-2**31, 2**31):
print(' if(input == %d) return %d;' % (i, i / 3))
print(r'''
return 42; // impossible
}
int main()
{
const int32_t number = 8;
printf("%d / 3 = %d\n", number, div_by_3(number));
}
''')
Using Hacker's Delight Magic number calculator
int divideByThree(int num)
{
return (fma(num, 1431655766, 0) >> 32);
}
Where fma is a standard library function defined in math.h header.
How about this approach (c#)?
private int dividedBy3(int n) {
List<Object> a = new Object[n].ToList();
List<Object> b = new List<object>();
while (a.Count > 2) {
a.RemoveRange(0, 3);
b.Add(new Object());
}
return b.Count;
}
I think the right answer is:
Why would I not use a basic operator to do a basic operation?
Solution using fma() library function, works for any positive number:
#include <stdio.h>
#include <math.h>
int main()
{
int number = 8;//Any +ve no.
int temp = 3, result = 0;
while(temp <= number){
temp = fma(temp, 1, 3); //fma(a, b, c) is a library function and returns (a*b) + c.
result = fma(result, 1, 1);
}
printf("\n\n%d divided by 3 = %d\n", number, result);
}
See my another answer.
First:
x/3 = (x/4) / (1-1/4)
Then figure out how to solve x/(1 - y):
x/(1-1/y)
= x * (1+y) / (1-y^2)
= x * (1+y) * (1+y^2) / (1-y^4)
= ...
= x * (1+y) * (1+y^2) * (1+y^4) * ... * (1+y^(2^i)) / (1-y^(2^(i+i))
= x * (1+y) * (1+y^2) * (1+y^4) * ... * (1+y^(2^i))
with y = 1/4:
int div3(int x) {
x <<= 6; // need more precise
x += x>>2; // x = x * (1+(1/2)^2)
x += x>>4; // x = x * (1+(1/2)^4)
x += x>>8; // x = x * (1+(1/2)^8)
x += x>>16; // x = x * (1+(1/2)^16)
return (x+1)>>8; // as (1-(1/2)^32) very near 1,
// we plus 1 instead of div (1-(1/2)^32)
}
Although it uses +, but somebody already implements add by bitwise op.
Use cblas, included as part of OS X's Accelerate framework.
[02:31:59] [william#relativity ~]$ cat div3.c
#import <stdio.h>
#import <Accelerate/Accelerate.h>
int main() {
float multiplicand = 123456.0;
float multiplier = 0.333333;
printf("%f * %f == ", multiplicand, multiplier);
cblas_sscal(1, multiplier, &multiplicand, 1);
printf("%f\n", multiplicand);
}
[02:32:07] [william#relativity ~]$ clang div3.c -framework Accelerate -o div3 && ./div3
123456.000000 * 0.333333 == 41151.957031
Generally, a solution to this would be:
log(pow(exp(numerator),pow(denominator,-1)))
Okay I think we all agree that this isn't a real world problem. So just for fun, here's how to do it with Ada and multithreading:
with Ada.Text_IO;
procedure Divide_By_3 is
protected type Divisor_Type is
entry Poke;
entry Finish;
private
entry Release;
entry Stop_Emptying;
Emptying : Boolean := False;
end Divisor_Type;
protected type Collector_Type is
entry Poke;
entry Finish;
private
Emptying : Boolean := False;
end Collector_Type;
task type Input is
end Input;
task type Output is
end Output;
protected body Divisor_Type is
entry Poke when not Emptying and Stop_Emptying'Count = 0 is
begin
requeue Release;
end Poke;
entry Release when Release'Count >= 3 or Emptying is
New_Output : access Output;
begin
if not Emptying then
New_Output := new Output;
Emptying := True;
requeue Stop_Emptying;
end if;
end Release;
entry Stop_Emptying when Release'Count = 0 is
begin
Emptying := False;
end Stop_Emptying;
entry Finish when Poke'Count = 0 and Release'Count < 3 is
begin
Emptying := True;
requeue Stop_Emptying;
end Finish;
end Divisor_Type;
protected body Collector_Type is
entry Poke when Emptying is
begin
null;
end Poke;
entry Finish when True is
begin
Ada.Text_IO.Put_Line (Poke'Count'Img);
Emptying := True;
end Finish;
end Collector_Type;
Collector : Collector_Type;
Divisor : Divisor_Type;
task body Input is
begin
Divisor.Poke;
end Input;
task body Output is
begin
Collector.Poke;
end Output;
Cur_Input : access Input;
-- Input value:
Number : Integer := 18;
begin
for I in 1 .. Number loop
Cur_Input := new Input;
end loop;
Divisor.Finish;
Collector.Finish;
end Divide_By_3;
What is the most efficient way given to raise an integer to the power of another integer in C?
// 2^3
pow(2,3) == 8
// 5^5
pow(5,5) == 3125
Exponentiation by squaring.
int ipow(int base, int exp)
{
int result = 1;
for (;;)
{
if (exp & 1)
result *= base;
exp >>= 1;
if (!exp)
break;
base *= base;
}
return result;
}
This is the standard method for doing modular exponentiation for huge numbers in asymmetric cryptography.
Note that exponentiation by squaring is not the most optimal method. It is probably the best you can do as a general method that works for all exponent values, but for a specific exponent value there might be a better sequence that needs fewer multiplications.
For instance, if you want to compute x^15, the method of exponentiation by squaring will give you:
x^15 = (x^7)*(x^7)*x
x^7 = (x^3)*(x^3)*x
x^3 = x*x*x
This is a total of 6 multiplications.
It turns out this can be done using "just" 5 multiplications via addition-chain exponentiation.
n*n = n^2
n^2*n = n^3
n^3*n^3 = n^6
n^6*n^6 = n^12
n^12*n^3 = n^15
There are no efficient algorithms to find this optimal sequence of multiplications. From Wikipedia:
The problem of finding the shortest addition chain cannot be solved by dynamic programming, because it does not satisfy the assumption of optimal substructure. That is, it is not sufficient to decompose the power into smaller powers, each of which is computed minimally, since the addition chains for the smaller powers may be related (to share computations). For example, in the shortest addition chain for a¹⁵ above, the subproblem for a⁶ must be computed as (a³)² since a³ is re-used (as opposed to, say, a⁶ = a²(a²)², which also requires three multiplies).
If you need to raise 2 to a power. The fastest way to do so is to bit shift by the power.
2 ** 3 == 1 << 3 == 8
2 ** 30 == 1 << 30 == 1073741824 (A Gigabyte)
Here is the method in Java
private int ipow(int base, int exp)
{
int result = 1;
while (exp != 0)
{
if ((exp & 1) == 1)
result *= base;
exp >>= 1;
base *= base;
}
return result;
}
An extremely specialized case is, when you need say 2^(-x to the y), where x, is of course is negative and y is too large to do shifting on an int. You can still do 2^x in constant time by screwing with a float.
struct IeeeFloat
{
unsigned int base : 23;
unsigned int exponent : 8;
unsigned int signBit : 1;
};
union IeeeFloatUnion
{
IeeeFloat brokenOut;
float f;
};
inline float twoToThe(char exponent)
{
// notice how the range checking is already done on the exponent var
static IeeeFloatUnion u;
u.f = 2.0;
// Change the exponent part of the float
u.brokenOut.exponent += (exponent - 1);
return (u.f);
}
You can get more powers of 2 by using a double as the base type.
(Thanks a lot to commenters for helping to square this post away).
There's also the possibility that learning more about IEEE floats, other special cases of exponentiation might present themselves.
power() function to work for Integers Only
int power(int base, unsigned int exp){
if (exp == 0)
return 1;
int temp = power(base, exp/2);
if (exp%2 == 0)
return temp*temp;
else
return base*temp*temp;
}
Complexity = O(log(exp))
power() function to work for negative exp and float base.
float power(float base, int exp) {
if( exp == 0)
return 1;
float temp = power(base, exp/2);
if (exp%2 == 0)
return temp*temp;
else {
if(exp > 0)
return base*temp*temp;
else
return (temp*temp)/base; //negative exponent computation
}
}
Complexity = O(log(exp))
If you want to get the value of an integer for 2 raised to the power of something it is always better to use the shift option:
pow(2,5) can be replaced by 1<<5
This is much more efficient.
int pow( int base, int exponent)
{ // Does not work for negative exponents. (But that would be leaving the range of int)
if (exponent == 0) return 1; // base case;
int temp = pow(base, exponent/2);
if (exponent % 2 == 0)
return temp * temp;
else
return (base * temp * temp);
}
Just as a follow up to comments on the efficiency of exponentiation by squaring.
The advantage of that approach is that it runs in log(n) time. For example, if you were going to calculate something huge, such as x^1048575 (2^20 - 1), you only have to go thru the loop 20 times, not 1 million+ using the naive approach.
Also, in terms of code complexity, it is simpler than trying to find the most optimal sequence of multiplications, a la Pramod's suggestion.
Edit:
I guess I should clarify before someone tags me for the potential for overflow. This approach assumes that you have some sort of hugeint library.
Late to the party:
Below is a solution that also deals with y < 0 as best as it can.
It uses a result of intmax_t for maximum range. There is no provision for answers that do not fit in intmax_t.
powjii(0, 0) --> 1 which is a common result for this case.
pow(0,negative), another undefined result, returns INTMAX_MAX
intmax_t powjii(int x, int y) {
if (y < 0) {
switch (x) {
case 0:
return INTMAX_MAX;
case 1:
return 1;
case -1:
return y % 2 ? -1 : 1;
}
return 0;
}
intmax_t z = 1;
intmax_t base = x;
for (;;) {
if (y % 2) {
z *= base;
}
y /= 2;
if (y == 0) {
break;
}
base *= base;
}
return z;
}
This code uses a forever loop for(;;) to avoid the final base *= base common in other looped solutions. That multiplication is 1) not needed and 2) could be int*int overflow which is UB.
more generic solution considering negative exponenet
private static int pow(int base, int exponent) {
int result = 1;
if (exponent == 0)
return result; // base case;
if (exponent < 0)
return 1 / pow(base, -exponent);
int temp = pow(base, exponent / 2);
if (exponent % 2 == 0)
return temp * temp;
else
return (base * temp * temp);
}
The O(log N) solution in Swift...
// Time complexity is O(log N)
func power(_ base: Int, _ exp: Int) -> Int {
// 1. If the exponent is 1 then return the number (e.g a^1 == a)
//Time complexity O(1)
if exp == 1 {
return base
}
// 2. Calculate the value of the number raised to half of the exponent. This will be used to calculate the final answer by squaring the result (e.g a^2n == (a^n)^2 == a^n * a^n). The idea is that we can do half the amount of work by obtaining a^n and multiplying the result by itself to get a^2n
//Time complexity O(log N)
let tempVal = power(base, exp/2)
// 3. If the exponent was odd then decompose the result in such a way that it allows you to divide the exponent in two (e.g. a^(2n+1) == a^1 * a^2n == a^1 * a^n * a^n). If the eponent is even then the result must be the base raised to half the exponent squared (e.g. a^2n == a^n * a^n = (a^n)^2).
//Time complexity O(1)
return (exp % 2 == 1 ? base : 1) * tempVal * tempVal
}
int pow(int const x, unsigned const e) noexcept
{
return !e ? 1 : 1 == e ? x : (e % 2 ? x : 1) * pow(x * x, e / 2);
//return !e ? 1 : 1 == e ? x : (((x ^ 1) & -(e % 2)) ^ 1) * pow(x * x, e / 2);
}
Yes, it's recursive, but a good optimizing compiler will optimize recursion away.
One more implementation (in Java). May not be most efficient solution but # of iterations is same as that of Exponential solution.
public static long pow(long base, long exp){
if(exp ==0){
return 1;
}
if(exp ==1){
return base;
}
if(exp % 2 == 0){
long half = pow(base, exp/2);
return half * half;
}else{
long half = pow(base, (exp -1)/2);
return base * half * half;
}
}
I use recursive, if the exp is even,5^10 =25^5.
int pow(float base,float exp){
if (exp==0)return 1;
else if(exp>0&&exp%2==0){
return pow(base*base,exp/2);
}else if (exp>0&&exp%2!=0){
return base*pow(base,exp-1);
}
}
In addition to the answer by Elias, which causes Undefined Behaviour when implemented with signed integers, and incorrect values for high input when implemented with unsigned integers,
here is a modified version of the Exponentiation by Squaring that also works with signed integer types, and doesn't give incorrect values:
#include <stdint.h>
#define SQRT_INT64_MAX (INT64_C(0xB504F333))
int64_t alx_pow_s64 (int64_t base, uint8_t exp)
{
int_fast64_t base_;
int_fast64_t result;
base_ = base;
if (base_ == 1)
return 1;
if (!exp)
return 1;
if (!base_)
return 0;
result = 1;
if (exp & 1)
result *= base_;
exp >>= 1;
while (exp) {
if (base_ > SQRT_INT64_MAX)
return 0;
base_ *= base_;
if (exp & 1)
result *= base_;
exp >>= 1;
}
return result;
}
Considerations for this function:
(1 ** N) == 1
(N ** 0) == 1
(0 ** 0) == 1
(0 ** N) == 0
If any overflow or wrapping is going to take place, return 0;
I used int64_t, but any width (signed or unsigned) can be used with little modification. However, if you need to use a non-fixed-width integer type, you will need to change SQRT_INT64_MAX by (int)sqrt(INT_MAX) (in the case of using int) or something similar, which should be optimized, but it is uglier, and not a C constant expression. Also casting the result of sqrt() to an int is not very good because of floating point precission in case of a perfect square, but as I don't know of any implementation where INT_MAX -or the maximum of any type- is a perfect square, you can live with that.
I have implemented algorithm that memorizes all computed powers and then uses them when need. So for example x^13 is equal to (x^2)^2^2 * x^2^2 * x where x^2^2 it taken from the table instead of computing it once again. This is basically implementation of #Pramod answer (but in C#).
The number of multiplication needed is Ceil(Log n)
public static int Power(int base, int exp)
{
int tab[] = new int[exp + 1];
tab[0] = 1;
tab[1] = base;
return Power(base, exp, tab);
}
public static int Power(int base, int exp, int tab[])
{
if(exp == 0) return 1;
if(exp == 1) return base;
int i = 1;
while(i < exp/2)
{
if(tab[2 * i] <= 0)
tab[2 * i] = tab[i] * tab[i];
i = i << 1;
}
if(exp <= i)
return tab[i];
else return tab[i] * Power(base, exp - i, tab);
}
Here is a O(1) algorithm for calculating x ** y, inspired by this comment. It works for 32-bit signed int.
For small values of y, it uses exponentiation by squaring. For large values of y, there are only a few values of x where the result doesn't overflow. This implementation uses a lookup table to read the result without calculating.
On overflow, the C standard permits any behavior, including crash. However, I decided to do bound-checking on LUT indices to prevent memory access violation, which could be surprising and undesirable.
Pseudo-code:
If `x` is between -2 and 2, use special-case formulas.
Otherwise, if `y` is between 0 and 8, use special-case formulas.
Otherwise:
Set x = abs(x); remember if x was negative
If x <= 10 and y <= 19:
Load precomputed result from a lookup table
Otherwise:
Set result to 0 (overflow)
If x was negative and y is odd, negate the result
C code:
#define POW9(x) x * x * x * x * x * x * x * x * x
#define POW10(x) POW9(x) * x
#define POW11(x) POW10(x) * x
#define POW12(x) POW11(x) * x
#define POW13(x) POW12(x) * x
#define POW14(x) POW13(x) * x
#define POW15(x) POW14(x) * x
#define POW16(x) POW15(x) * x
#define POW17(x) POW16(x) * x
#define POW18(x) POW17(x) * x
#define POW19(x) POW18(x) * x
int mypow(int x, unsigned y)
{
static int table[8][11] = {
{POW9(3), POW10(3), POW11(3), POW12(3), POW13(3), POW14(3), POW15(3), POW16(3), POW17(3), POW18(3), POW19(3)},
{POW9(4), POW10(4), POW11(4), POW12(4), POW13(4), POW14(4), POW15(4), 0, 0, 0, 0},
{POW9(5), POW10(5), POW11(5), POW12(5), POW13(5), 0, 0, 0, 0, 0, 0},
{POW9(6), POW10(6), POW11(6), 0, 0, 0, 0, 0, 0, 0, 0},
{POW9(7), POW10(7), POW11(7), 0, 0, 0, 0, 0, 0, 0, 0},
{POW9(8), POW10(8), 0, 0, 0, 0, 0, 0, 0, 0, 0},
{POW9(9), 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
{POW9(10), 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
};
int is_neg;
int r;
switch (x)
{
case 0:
return y == 0 ? 1 : 0;
case 1:
return 1;
case -1:
return y % 2 == 0 ? 1 : -1;
case 2:
return 1 << y;
case -2:
return (y % 2 == 0 ? 1 : -1) << y;
default:
switch (y)
{
case 0:
return 1;
case 1:
return x;
case 2:
return x * x;
case 3:
return x * x * x;
case 4:
r = x * x;
return r * r;
case 5:
r = x * x;
return r * r * x;
case 6:
r = x * x;
return r * r * r;
case 7:
r = x * x;
return r * r * r * x;
case 8:
r = x * x;
r = r * r;
return r * r;
default:
is_neg = x < 0;
if (is_neg)
x = -x;
if (x <= 10 && y <= 19)
r = table[x - 3][y - 9];
else
r = 0;
if (is_neg && y % 2 == 1)
r = -r;
return r;
}
}
}
My case is a little different, I'm trying to create a mask from a power, but I thought I'd share the solution I found anyway.
Obviously, it only works for powers of 2.
Mask1 = 1 << (Exponent - 1);
Mask2 = Mask1 - 1;
return Mask1 + Mask2;
In case you know the exponent (and it is an integer) at compile-time, you can use templates to unroll the loop. This can be made more efficient, but I wanted to demonstrate the basic principle here:
#include <iostream>
template<unsigned long N>
unsigned long inline exp_unroll(unsigned base) {
return base * exp_unroll<N-1>(base);
}
We terminate the recursion using a template specialization:
template<>
unsigned long inline exp_unroll<1>(unsigned base) {
return base;
}
The exponent needs to be known at runtime,
int main(int argc, char * argv[]) {
std::cout << argv[1] <<"**5= " << exp_unroll<5>(atoi(argv[1])) << ;std::endl;
}
I've noticed something strange about the standard exponential squaring algorithm with gnu-GMP :
I implemented 2 nearly-identical functions - a power-modulo function using the most vanilla binary exponential squaring algorithm,
labeled ______2()
then another one basically the same concept, but re-mapped to dividing by 10 at each round instead of dividing by 2,
labeled ______10()
.
( time ( jot - 1456 9999999999 6671 | pvE0 |
gawk -Mbe '
function ______10(_, __, ___, ____, _____, _______) {
__ = +__
____ = (____+=_____=____^= \
(_ %=___=+___)<_)+____++^____—
while (__) {
if (_______= __%____) {
if (__==_______) {
return (_^__ *_____) %___
}
__-=_______
_____ = (_^_______*_____) %___
}
__/=____
_ = _^____%___
}
}
function ______2(_, __, ___, ____, _____) {
__=+__
____+=____=_____^=(_%=___=+___)<_
while (__) {
if (__ %____) {
if (__<____) {
return (_*_____) %___
}
_____ = (_____*_) %___
--__
}
__/=____
_= (_*_) %___
}
}
BEGIN {
OFMT = CONVFMT = "%.250g"
__ = (___=_^= FS=OFS= "=")(_<_)
_____ = __^(_=3)^--_ * ++_-(_+_)^_
______ = _^(_+_)-_ + _^!_
_______ = int(______*_____)
________ = 10 ^ 5 + 1
_________ = 8 ^ 4 * 2 - 1
}
GNU Awk 5.1.1, API: 3.1 (GNU MPFR 4.1.0, GNU MP 6.2.1)
.
($++NF = ______10(_=$___, NR %________ +_________,_______*(_-11))) ^!___'
out9: 48.4MiB 0:00:08 [6.02MiB/s] [6.02MiB/s] [ <=> ]
in0: 15.6MiB 0:00:08 [1.95MiB/s] [1.95MiB/s] [ <=> ]
( jot - 1456 9999999999 6671 | pvE 0.1 in0 | gawk -Mbe ; )
8.31s user 0.06s system 103% cpu 8.058 total
ffa16aa937b7beca66a173ccbf8e1e12 stdin
($++NF = ______2(_=$___, NR %________ +_________,_______*(_-11))) ^!___'
out9: 48.4MiB 0:00:12 [3.78MiB/s] [3.78MiB/s] [<=> ]
in0: 15.6MiB 0:00:12 [1.22MiB/s] [1.22MiB/s] [ <=> ]
( jot - 1456 9999999999 6671 | pvE 0.1 in0 | gawk -Mbe ; )
13.05s user 0.07s system 102% cpu 12.821 total
ffa16aa937b7beca66a173ccbf8e1e12 stdin
For reasons extremely counter-intuitive and unknown to me, for a wide variety of inputs i threw at it, the div-10 variant is nearly always faster. It's the matching of hashes between the 2 that made it truly baffling, despite computers obviously not being built in and for a base-10 paradigm.
Am I missing something critical or obvious in the code/approach that might be skewing the results in a confounding manner ? Thanks.