I need to write a program that takes 2 digits(X and n) and then prints X with last n digits of X reversed.
For example
Input: 12345 3
Output: 12543
Input: 523 2
Output: 532
I already wrote a control mechanism for checking n is greater or equal than the number of digits of X
For example if inputs are 6343 and 7, program prints that inputs should be changed and takes input again.
My main problem is I couldn't find an algorithm for reversing last n digits. I can reverse any int with this code
int X, r = 0;
printf("Enter a number to reverse\n");
scanf("%d", &n);
while (X != 0)
{
r = r * 10;
r = r + n%10;
X = X/10;
}
printf("Reverse of the number = %d", r);
But I couldn't figure how two reverse just last digits. Can you give me any idea for that?
I couldn't figure how to reverse just last digits
Separate the number using pow(10,n) - see later code.
unsigned reverse_last_digits(unsigned x, unsigned n) {
unsigned pow10 = powu(10, n);
unsigned lower = x%pow10;
unsigned upper = x - lower;
return upper + reverseu(lower, n);
}
Create a loop that extracts the least-significant-digit (%10) and builds up another integer by applying that digit. (y = y*10 + new_digit)
unsigned reverseu(unsigned x, unsigned n) {
unsigned y = 0;
while (n-- > 0) {
y = y*10 + x%10;
x /= 10;
}
return y;
}
For integer type problems, consider integer helper functions and avoid floating point functions like pow() as they may provide only an approximate results. Easy enough to code an integer pow().
unsigned powu(unsigned x, unsigned expo) {
unsigned y = 1;
while (expo > 0) {
if (expo & 1) {
y = x * y;
}
expo >>= 1;
x *= x;
}
return y;
}
Test
int main() {
printf("%u\n", reverse_last_digits(12345, 3));
printf("%u\n", reverse_last_digits(523, 2));
printf("%u\n", reverse_last_digits(42001, 3));
printf("%u\n", reverse_last_digits(1, 2));
}
Output
12543
532
42100
10
Code uses unsigned rather than int to avoid undefined behavior (UB) on int overflow.
It is an easy one.
1. let say the number you want to reverse is curr_number;
2. Now, the places you want to reverse is x;
(remember to verify that x must be less than the number of digit of curr_number);
3. now, just take a temp integer and store curr_number / pow(10,x) ('/' = divide and pow(10,x) is 10 to the power x)
4. now, take a second number temp2, which will store curr_number-(temp * pow(10,x) )
5. reverse this temp2 (using your function)
6. now, answer = (temp * pow(10,x) ) + (temp2) //(note temp2 is reversed)
example with steps:
curr_number = 1234567
places you want to reverse is 3
temp = 1234567 / (10^3) i.e (1234567/1000) = 1234 (because it is int type)
temp2 = 1234567 - (1234*10^3) i.e 1234567 - 1234000 = 567
reverse(567) = 765
answer = (1234 * 10^3) + 765 = 1234765
Create two variables
lastn which stores the last n digits (345)
r which stores the reversed last n digits (543)
Subtract lastn from the original number (12345 - 345 = 12000)
Add r to the above number (12000 + 543 = 12543)
int c = 0; // count number of digits
int original = x;
int lastn = 0;
while (x != 0 && c < n) {
r = r * 10;
r = r + x % 10;
lastn += (x % 10) * pow(10, c);
x = x / 10;
c++;
}
printf("reversed: %d\n", original - lastn + r);
In case you don't have problems using char, you can do this
#include <stdio.h>
#include <string.h>
#define SIZE 10
int main() {
char n[SIZE]; // the Number;
int x; // number of last digits of n to reverse
int len; // number of digits of n
scanf("%s%d", n, &x);
len = strlen(n);
for(int i = 0; i < len; i++) {
i < len - x ? printf("%c", n[i]) : printf("%c", n[2*len -1 - i - x]);
}
return 0;
}
If you want you can make the program more readable by splitting the for in two
for(int i = 0; i < len - x; i++) {
printf("%c", n[i]);
}
for(int i = len-1; i >= len - x; i--) {
printf("%c", n[i]);
}
Note: the program won't work if n > x (i.e. if you want to swap more digits than you got)
Problem statement :
Given a 32-bit signed integer, reverse digits of an integer.
Note: Assume we are dealing with an environment that could only store
integers within the 32-bit signed integer range: [ −2^31, 2^31 − 1]. For
the purpose of this problem, assume that your function returns 0 when
the reversed integer overflows.
I'm trying to implement the recursive function reverseRec(), It's working for smaller values but it's a mess for the edge cases.
int reverseRec(int x)
{
if(abs(x)<=9)
{
return x;
}
else
{
return reverseRec(x/10) + ((x%10)*(pow(10, (floor(log10(abs(x)))))));
}
}
I've implemented non recursive function which is working just fine :
int reverse(int x)
{
long long val = 0;
do{
val = val*10 + (x%10);
x /= 10;
}while(x);
return (val < INT_MIN || val > INT_MAX) ? 0 : val;
}
Here I use variable val of long long type to check the result with MAX and MIN of signed int type but the description of the problem specifically mentioned that we need to deal within the range of 32-bit integer, although somehow it got accepted but I'm just curious If there is a way to implement a recursive function using only int datatype ?
One more thing even if I consider using long long I'm failing to implement it in the recursive function reverseRec().
If there is a way to implement a recursive function using only int datatype ?
(and) returns 0 when the reversed integer overflows
Yes.
For such +/- problems, I like to fold the int values to one side and negate as needed. The folding to one side (- or +) simplifies overflow detection as only a single side needs testing
I prefer folding to the negative side as there are more negatives, than positives. (With 32-bit int, really didn't make any difference for this problem.)
As code forms the reversed value, test if the following r * 10 + least_digit may overflow before doing it.
An int only recursive solution to reverse an int. Overflow returns 0.
#include <limits.h>
#include <stdio.h>
static int reverse_recurse(int i, int r) {
if (i) {
int least_digit = i % 10;
if (r <= INT_MIN / 10 && (r < INT_MIN / 10 || least_digit < INT_MIN % 10)) {
return 1; /// Overflow indication
}
r = reverse_recurse(i / 10, r * 10 + least_digit);
}
return r;
}
// Reverse an int, overflow returns 0
int reverse_int(int i) {
// Proceed with negative values, they have more range than + side
int r = reverse_recurse(i > 0 ? -i : i, 0);
if (r > 0) {
return 0;
}
if (i > 0) {
if (r < -INT_MAX) {
return 0;
}
r = -r;
}
return r;
}
Test
int main(void) {
int t[] = {0, 1, 42, 1234567890, 1234567892, INT_MAX, INT_MIN};
for (unsigned i = 0; i < sizeof t / sizeof t[0]; i++) {
printf("%11d %11d\n", t[i], reverse_int(t[i]));
if (t[i] != INT_MIN) {
printf("%11d %11d\n", -t[i], reverse_int(-t[i]));
}
}
}
Output
0 0
0 0
1 1
-1 -1
42 24
-42 -24
1234567890 987654321
-1234567890 -987654321
1234567892 0
-1234567892 0
2147483647 0
-2147483647 0
-2147483648 0
You could add a second parameter:
int reverseRec(int x, int reversed)
{
if(x == 0)
{
return reversed;
}
else
{
return reverseRec(x/10, reversed * 10 + x%10);
}
}
And call the function passing the 0 for the second parameter. If you want negative numbers you can check the sign before and pass the absolute value to this function.
In trying to learn C programming I programed this question and get some correct results and some incorrect. I don't see the reason for the difference.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h> // requires adding link to math -lm as in: gcc b.c -lm -o q11
int ReverseInt(int startValue, int decimalPlace)
{
if(decimalPlace == 0) // if done returns value
{
return startValue;
}
int temp = startValue % 10; // gets units digit
int newStart = (startValue -temp)/10; // computes new starting value after removing one digit
int newDecimal = decimalPlace -1;
int value = temp*pow(10,decimalPlace);
return value + ReverseInt(newStart,newDecimal); // calls itself recursively until done
}
int main()
{
int x, decimalP, startValue;
printf("Input number to be reversed \n Please note number must be less than 214748364 :");
scanf("%d", &x);
if (x > 214748364)
{
printf("Input number to be reversed \n Please note number must be less than 214748364 :");
scanf("%d", &x);
}
decimalP = round(log10(x)); // computes the number of powers of 10 - 0 being units etc.
startValue = ReverseInt(x, decimalP); // calls function with number to be reversed and powers of 10
printf("\n reverse of %d is %d \n", x, startValue);
}
Output is: reverse of 1234 is 4321 but then reverse of 4321 is 12340
It's late and nothing better does not come into my mind. No float calculations. Of course, integer has to be big enough to accommodate the result. Otherwise it is an UB.
int rev(int x, int partial, int *max)
{
int result;
if(x / partial < 10 && (int)(x / partial) > -10)
{
*max = partial;
return abs(x % 10) * partial;
}
result = rev(x, partial * 10, max) + abs(((x / (int)(*max / partial)) % 10) * partial);
return result;
}
int reverse(int x)
{
int max;
return rev(x, 1, &max) * ((x < 0) ? -1 : 1);
}
int main(void){
printf("%d", reverse(-456789));
}
https://godbolt.org/z/M1eezf
unsigned rev(unsigned x, unsigned partial, unsigned *max)
{
unsigned result;
if(x / partial < 10)
{
*max = partial;
return (x % 10) * partial;
}
result = rev(x, partial * 10, max) + (x / (*max / partial) % 10) * partial;
return result;
}
unsigned reverse(unsigned x)
{
unsigned max;
return rev(x, 1, &max);
}
int main(void){
printf("%u", reverse(123456));
}
when using long long to store the result all possible integers can be reversed
long long rev(int x, long long partial, long long *max)
{
long long result;
if(x / partial < 10 && (int)(x / partial) > -10)
{
*max = partial;
return abs(x % 10) * partial;
}
result = rev(x, partial * 10, max) + abs(((x / (int)(*max / partial)) % 10) * partial);
return result;
}
long long reverse(int x)
{
long long max;
return rev(x, 1, &max) * ((x < 0) ? -1 : 1);
}
int main(void){
printf("%d reversed %lld\n", INT_MIN, reverse(INT_MIN));
printf("%d reversed %lld\n", INT_MAX, reverse(INT_MAX));
}
https://godbolt.org/z/KMfbxz
I am assuming by reversing an integer you mean turning 129 to 921 or 120 to 21.
You need an initial method to initialize your recursive function.
Your recursive function must figure out how many decimal places your integer uses. This can be found by using log base 10 with the value and then converting the result to a integer.
log10 (103) approx. 2.04 => 2
Modulus the initial value by 10 to get the ones place and store it in a variable called temp
Subtract the ones place from the initial value and store that in a variable called newStart.
divide this value by 10
Subtract one from the decimal place and store in another variable called newDecimal.
Return the ones place times 10 to the power of the decimal place and add it to the function where the initial value is newStart and the decimalPlace is newDecimal.
#include <stdio.h>
#include <math.h>
int ReverseInt(int startValue, int decimalPlace);
int main()
{
int i = -54;
int positive = i < 0? i*-1 : i;
double d = log10(positive);
int output = ReverseInt(positive,(int)d);
int correctedOutput = i < 0? output*-1 : output;
printf("%d \n",correctedOutput);
return 0;
}
int ReverseInt(int startValue, int decimalPlace)
{
if(decimalPlace == 0)
{
return startValue;
}
int temp = startValue % 10;
int newStart = (startValue -temp)/10;
int newDecimal = decimalPlace -1;
int value = temp*pow(10,decimalPlace);
return value + ReverseInt(newStart,newDecimal);
}
I want to make a bitwise AND computation over integers, but without converting them to binary numbers. For example, I have a integer "10111" (it is integer, not binary) and another integer "01001". I want bitwise AND of these numbers without converting them to binary and then making bitwise AND. I know it is not bitwise what I ask, but I want something similar to this. I know it can be interpreted initially as binary, converted to decimal and then do bitwise AND, but I do not want that. I want something like this:
int a;
int b;
int temp;
double result;
temp = a & b;
while (result != 0) {
if (result % 10 == 1)
count++;
result /= 10;
}
int length = floor(log10(abs(a))) + 1;
result = count / length;
return result;
I want this to check similarity of the Bag of Words(from natural language processing, string of 0s and 1s). I am importing Bag of Words in Monetdb, Column type should be Integer (Not string). If I have for example "10111" and "01001" in the Integer type cells, I want to get "00001" and fraction 1/5, because only 1 positions matches.
Thanks in advance
Might be a bit bulky, but it kind of works) You can optimize it yourself. I hope that I get you correctly.
IDEOne demo
#include <stdio.h>
unsigned int weirdAnd(unsigned int a, unsigned int b) {
unsigned int result = 0;
unsigned int coef = 1;
while (a && b) {
result += ((a % 10) && (b % 10)) * coef;
coef *= 10;
a /= 10;
b /= 10;
}
return result;
}
unsigned int weirdOr(unsigned int a, unsigned int b) {
unsigned int result = 0;
unsigned int coef = 1;
while (a || b) {
result += ((a % 10) || (b % 10)) * coef;
coef *= 10;
a /= 10;
b /= 10;
}
return result;
}
int main(void) {
// your code goes here
unsigned int a = 10110;
unsigned int b = 10011;
printf("%u and \n%u = \n%u\n\n", a, b, weirdAnd(a, b));
printf("%u or \n%u = \n%u\n\n", a, b, weirdOr(a, b));
return 0;
}
Output:
10110 and 10011 = 10010
10110 or 10011 = 10111
The problem is that and works on bits only, and it does not care if the input numbers are given in decimals, octal, hexadecimal, or any other way. To force and to work correctly, you must give it an input in 'binary', that is, only ones and zeroes. To do so you need to grab each digit of the input numbers as if they are binary digits.
The following works as expected:
#include <stdio.h>
int cheap_pow (int base, int exponent)
{
int result = base;
while (exponent-- > 0)
result *= base;
return result;
}
int main (void)
{
int a = 10111, b = 1001 ;
int result, factor;
printf ("BAD: %05d AND %05d = %05d\n", a, b, a & b);
printf ("GOOD: %05d AND %05d = ");
result = 0;
factor = 1;
while (a | b)
{
result += factor * ((a & 1) and (b & 1));
factor *= 10;
a /= 10;
b /= 10;
}
printf ("%05d\n", result);
}
but you must be careful when defining your inputs. When written directly into your source code, a value b = 01001 will be interpreted by the C compiler as octal (base 8), rather than decimal; and it will have the (decimal) value 513. It's just a Rule of C.
If your input comes from elsewhere, you don't need to take this into account – except when you use a standard function such as strtol, where you should carefully read the documentation because it does the same thing:
If base is zero or 16, the string may then include a "0x"
prefix, and the number will be read in base 16; otherwise, a zero
base is taken as 10 (decimal) unless the next character is '0', in
which case it is taken as 8 (octal).
An additional note is this program only will work in the range for signed int, that is (currently) up to 10 "binary" digits. If you need some more, you could switch to a larger type; but beyond that, you are better off with a not-numerical solution and use a character strings throughout.
int a, b;
int tmp = a & b;
int res = 0;
while ((tmp != 0 &&) (tmp / 10 != 0)){
int dig = tmp % 10;
res = (dig == 1)? ++res: res;
tmp /= 10;
}
Try this.
Suppose I have an array of bytes from a secure PRNG, and I need to generate a number between 1 and 10 using that data, how would I do that correctly?
Think of the array as one big unsigned integer. Then the answer is simple:
(Big_Number % 10) + 1
So all that is needed is a method to find the modulus 10 of big integers. Using modular exponentiation:
#include <limits.h>
#include <stdlib.h>
int ArrayMod10(const unsigned char *a, size_t n) {
int mod10 = 0;
int base = (UCHAR_MAX + 1) % 10;
for (size_t i = n; i-- > 0; ) {
mod10 = (base*mod10 + a[i]) % 10;
base = (base * base) % 10;
}
return mod10;
}
void test10(size_t n) {
unsigned char a[n];
// fill array with your secure PRNG
for (size_t i = 0; i<n; i++) a[i] = rand();
return ArrayMod10(a, n) + 1;
}
There will be a slight bias as 256^n is not a power of 10. With large n, this will rapidly decrease in significance.
Untested code: Detect if a biased result occurred. Calling code could repeatedly call this function with new a array values to get an unbiased result on the rare occasions when bias occurs.
int ArrayMod10BiasDetect(const unsigned char *a, size_t n, bool *biasptr) {
bool bias = true;
int mod10 = 0;
int base = (UCHAR_MAX + 1) % 10; // Note base is usually 6: 256%10, 65536%10, etc.
for (size_t i = n; i-- > 0; ) {
mod10 = (base*mod10 + a[i]) % 10;
if (n > 0) {
if (a[i] < UCHAR_MAX) bias = false;
} else {
if (a[i] < UCHAR_MAX + 1 - base) bias = false;
}
base = (base * base) % 10;
}
*biaseptr = bias;
return mod10;
}
As per the comments follow-up, it seems what you need is modulus operator [%].
You may also need to check the related wiki.
Note: Every time we use the modulo operator on a random number, there is a probability that we'll be running into modulo bias, which ends up in disbalancing the fair distribution of random numbers. You've to take care of that.
For a detailed discussion on this, please see this question and related answers.
It depends on a bunch of things. Secure PRNG sometimes makes long byte arrays instead of integers, let's say it is 16 bytes long array, then extract 32 bit integer like so: buf[0]*0x1000000+buf[1]*0x10000+buf[2]*0x100+buf[3] or use shift operator. This is random so big-endian/little-endian doesn't matter.
char randbytes[16];
//...
const char *p = randbytes;
//assumes size of int is 4
unsigned int rand1 = p[0] << 24 + p[1] << 16 + p[2] << 8 + p[3]; p += 4;
unsigned int rand2 = p[0] << 24 + p[1] << 16 + p[2] << 8 + p[3]; p += 4;
unsigned int rand3 = p[0] << 24 + p[1] << 16 + p[2] << 8 + p[3]; p += 4;
unsigned int rand4 = p[0] << 24 + p[1] << 16 + p[2] << 8 + p[3];
Then use % on the integer
ps, I think that's a long answer. If you want number between 1 and 10 then just use % on first byte.
OK, so this answer is in Java until I get to my Eclipse C/C++ IDE:
public final static int simpleBound(Random rbg, int n) {
final int BYTE_VALUES = 256;
// sanity check, only return positive numbers
if (n <= 0) {
throw new IllegalArgumentException("Oops");
}
// sanity check: choice of value 0 or 0...
if (n == 1) {
return 0;
}
// sanity check: does not fit in byte
if (n > BYTE_VALUES) {
throw new IllegalArgumentException("Oops");
}
// optimization for n = 2^y
if (Integer.bitCount(n) == 1) {
final int mask = n - 1;
return retrieveRandomByte(rbg) & mask;
}
// you can skip to this if you are sure n = 10
// z is upper bound, and contains floor(z / n) blocks of n values
final int z = (BYTE_VALUES / n) * n;
int x;
do {
x = retrieveRandomByte(rbg);
} while (x >= z);
return x % n;
}
So n is the maximum value in a range [0..n), i.e. n is exclusive. For a range [1..10] simply increase the result with 1.
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;