So, here is a question when user input 3 than it prints the 7 as output and when user input 7 than it prints 3 as output, but without using any condition and loop.
here is what I done.But the problem is i'm using 2 numbers as input but i have to done it with one input.
void main() {
print(mod(7, 3));
}
int mod(int num1, int num2) {
//3*2 mod 7
//7*2 mod 3
int answer = (num1 * 2) % num2 + 1;
return answer;
}
please help how to do this.You can use any language to solve this.
Let's look at the binary representation of 3 and 7 : 3 is 0...0011 and 7 is 0...0111, which means that the only bit that differs between 3 and 7 is the 3rd bit. Therefore, we can transform 3 into 7 and vice-versa simply by "flipping" that 3rd bit.
"Flipping a bit" can be done with an "exclusive-or" operation. Xor is a very useful operation (one of my favorite): one way to look at it is to consider that you have a data input and a control input, and if the control input is 0, then the data input is left unchanged, but if the control input is 1, then the data input is flipped (you can draw the truth table of Xor to convince yourself of that interpretation). Of course, it's only a way to consider the xor operation, but it's particularly useful in our case : we want to flip the 3rd bit, and leave all of the other bit unchanged.
C offers a "bitwise xor" operation ( A ^ B), which do a xor of each bit of A and B. Therefore, what we want is to have a number, that we will use as the control number, which 3rd bit is 1 and all other bits are 0: this number is 4.
Finally, we can have this function that converts 3 into 7 and vice-versa by simply applying a Xor with 4 to the input.
#include <stdio.h>
int flip(int num1) {
return num1 ^ 4;
}
void main() {
printf("%d", flip(3));
}
Related
Is there a way to "truncate" an integer using bit twiddling, as if it floor-divided and then multiplied back, as in:
z = floor(x / y) * y
I know it is possible to do so if y is of power of two, for example:
z = floor(x / 4) * 4 == x & ~3
But what trick does one use when y is some general positive integer?
For each individual y, there is a sequence of operations (addition, subtraction, and binary shift) which divides x by y faster than the (x86) division instruction.
Finding that sequence however is not straightforward, and must be done in advance (feasible when you divide by the same y a lot).
A simple example: to divide an arbitrary uint32 x by 3, we can instead calculate x * M in uint64 type and shift it to the right by 33 bits, where M is a magic constant equal to 233 / 3 rounded up.
The following code (C) tries 20 random uint32 values with the above algorithm and checks that the result is equal to just dividing by 3:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main ()
{
int step;
unsigned x, y1, y2;
unsigned const M = (1ULL << 33) / 3 + 1;
srand (time (NULL));
for (step = 0; step < 20; step++)
{
x = (rand () << 30) | (rand () << 15) | rand ();
y1 = x / 3;
y2 = (x * 1ULL * M) >> 33;
printf ("%10u %10u %10u %s\n", x, y1, y2, y1 == y2 ? "true" : "false");
}
return 0;
}
For further information, see Hacker's Delight book in general, and the freely available addition - chapter 10 here: hackersdelight.org/divcMore.pdf.
The reason this works for powers of 2 is the way binary representations works. Dividing by 2 (or powers of 2) is identical to bit shifting. Shifting right and then back left the same amount is identical to floor-division as you put it.
Consider an arbitrary binary number: 110101010111. If you'd bit shift it 3 times to the right (division by 8), and then back again it would turn to 110101010000 which is identical to ANDing it with 111111111000. Now lets consider division by 3 of the (decimal) number 16: start with 10000. Division (not shifting!) by 3 would be 5 (101) and multiply by 3 again is 15 (1111). No bit shifting can do that.
The obvious thing to do is to convert to whatever base you are trying to work with, and then basically make the last digit 0. (Or if you are working with a kth power, then make the last k digits 0). However you asked about bit (base-2) operations. It turns out that for any desired base B (at least, that is odd), you can come up with a number in binary so that the first M digits in base B are anything you want, for any M. Thus, how could you possibly have a general method for what you want (with an odd base), that just works on bits (binary)? At the very least it would probably be a lot more complicated than simply converting your number to your desired base and setting however many last digits to 0 and then converting back to natural base-2 integer representation.
I'm coding an LED display (7x48) and the language I'm working in is BASIC (no former experience in that language, but in C/C++) and I have a small issue.
I have an array (red[20] of byte) and an example of a current state is:
to make it easier here lets say its red[3]
10011010 01011100 01011101
and now i need to shift the array by 1 so in next cycle its supposed to be
00110100 10111000 10111011
so what happened is that the whole array shifted for 1 bit to left
the BASIC I'm working with doesn't have any .NET APIs so I need the total low level code (doesn't have to be BASIC, I can translate it, I just need an idea how to do it as I'm limited to 8KB code memory so I have to fully optmize it)
If most significant bit is 1:
subtract value of most significant bit
multiply by 2
add 1
otherwise:
multiply by 2
You should be able to use bit shift operations:
http://msdn.microsoft.com/en-us/library/2d9yb87a.aspx
Let x be the element you want to shift:
x = (x<<1) | (x>>23)
or in general, if you want to shift left by y bits and there are a total of n bits:
x = (x<<y) | (x>>(n-y))
I don't know basic well, but here's what I would do in a C++/Java/C# language:
Assuming you have red[] of length n:
int b = 32; //Number of bits per byte (your example showed 24, but usually there are 32)
int y = 1; //Number of bytes to shift to the left
int carry = 0; //The bytes to carry over (I'm assuming that they move up the array from red[0] to red[1], etc.
for (int i=0;i<n;i++)
{
int newCarry = (red[i]>>(n-y));
red[i] = (red[i]<<y) | carry;
carry = newCarry;
}
//Complete the loop
red[0]|=carry;
We have two N-bit numbers (0< N< 100000). We have to perform q queries (0< q<500000) over these numbers. The query can be of following three types:
set_a idx x: Set A[idx] to x, where 0 <= idx < N, where A[idx] is idx'th least significant bit of A.
set_b idx x: Set B[idx] to x, where 0 <= idx < N.
get_c idx: Print C[idx], where C=A+B, and 0<=idx
Now, I have optimized the code to the best extent I can.
First, I tried with an int array for a, b and c. For every update, I calculate c and return the ith bit when queried. It was damn slow. Cleared 4/11 test cases only.
I moved over to using boolean array. It was around 2 times faster than int array approach. Cleared 7/11 testcases.
Next, I figured out that I need not calculate c for calculating idx th bit of A+B. I will just scan A and B towards right from idx until I find either a[i]=b[i]=0 or a[i]=b[i]=1. If a[i]=b[i]=0, then I just add up towards left to idx th bit starting with initial carry=0. And if a[i]=b[i]=1, then I just add up towards left to idx th bit starting with initial carry=1.
This was faster but cleared only 8/11 testcases.
Then, I figured out once, I get to the position i, a[i]=b[i]=0 or a[i]=b[i]=1, then I need not add up towards idx th position. If a[i]=b[i]=0, then answer is (a[idx]+b[idx])%2 and if a[i]=b[i]=1, then the answer is (a[idx]+b[idx]+1)%2. It was around 40% faster but still cleared only 8/11 testcases.
Now my question is how do get down those 3 'hard' testcases? I dont know what they are but the program is taking >3 sec to solve the problem.
Here is the code: http://ideone.com/LopZf
One possible optimization is to replace
(a[pos]+b[pos]+carry)%2
with
a[pos]^b[pos]^carry
The XOR operator (^) performs addition modulo 2, making the potentially expensive mod operation (%) unnecessary. Depending on the language and compiler, the compiler may make optimizations for you when doing a mod with a power of 2. But since you are micro-optimizing it is a simple change to make that removes dependence on that optimization being made for you behind the scenes.
http://en.wikipedia.org/wiki/Exclusive_or
This is just one suggestion that is simple to make. As others have suggested, using packed ints to represent your bit array will likely also improve what is probably the worst case test for your code. That would be the get_c function of the most significant bit, with either A or B (but not both) being 1 for all the other positions, requiring a scan of every bit position to the least significant bit to determine carry. If you were using packed ints for your bits, there would only be approximately 1/32 as many operations neccessary (assuming 32 bit ints). Using packed ints however would be a somewhat more complicated than your use of a simple boolean array (which really is likely just an array of bytes).
C/C++ Bit Array or Bit Vector
Convert bit array to uint or similar packed value
http://en.wikipedia.org/wiki/Bit_array
There are lots of other examples on Stackoverflow and the net for using ints as if they were bit arrays.
Here is a solution that looks a bit like your algorithm. I demonstrate it with bytes, but of course you can easily optimize the algorithm using 32 bit words (I suppose your machine has 64 bits arithmetic nowadays).
void setbit( unsigned char*x,unsigned int idx,unsigned int bit)
{
unsigned int digitIndex = idx>>3;
unsigned int bitIndex = idx & 7;
if( ((x[digitIndex]>>bitIndex)&1) ^ bit) x[digitIndex]^=(1u<<bitIndex);
}
unsigned int getbit(unsigned char *a,unsigned char *b,unsigned int idx)
{
unsigned int digitIndex = idx>>3;
unsigned int bitIndex = idx & 7;
unsigned int c = a[digitIndex]+b[digitIndex];
unsigned int bit = (c>>bitIndex) & 1;
/* a zero bit on the right will absorb a carry, let's check if any */
if( (c^(c+1))>>bitIndex )
{
/* none, we must check if there's a carry propagating from the right digits */
for(;digitIndex-- > 0;)
{
c=a[digitIndex]+b[digitIndex];
if( c > 255 ) return bit^1; /* yes, a carry */
if( c < 255 ) return bit; /* no carry possible, a zero bit will absorb it */
}
}
return bit;
}
If you find anything cryptic, just ask.
Edit: oops, I inverted the zero bit condition...
int x = n / 3; // <-- make this faster
// for instance
int a = n * 3; // <-- normal integer multiplication
int b = (n << 1) + n; // <-- potentially faster multiplication
The guy who said "leave it to the compiler" was right, but I don't have the "reputation" to mod him up or comment. I asked gcc to compile int test(int a) { return a / 3; } for an ix86 and then disassembled the output. Just for academic interest, what it's doing is roughly multiplying by 0x55555556 and then taking the top 32 bits of the 64 bit result of that. You can demonstrate this to yourself with eg:
$ ruby -e 'puts(60000 * 0x55555556 >> 32)'
20000
$ ruby -e 'puts(72 * 0x55555556 >> 32)'
24
$
The wikipedia page on Montgomery division is hard to read but fortunately the compiler guys have done it so you don't have to.
This is the fastest as the compiler will optimize it if it can depending on the output processor.
int a;
int b;
a = some value;
b = a / 3;
There is a faster way to do it if you know the ranges of the values, for example, if you are dividing a signed integer by 3 and you know the range of the value to be divided is 0 to 768, then you can multiply it by a factor and shift it to the left by a power of 2 to that factor divided by 3.
eg.
Range 0 -> 768
you could use shifting of 10 bits, which multiplying by 1024, you want to divide by 3 so your multiplier should be 1024 / 3 = 341,
so you can now use (x * 341) >> 10
(Make sure the shift is a signed shift if using signed integers), also make sure the shift is an actually shift and not a bit ROLL
This will effectively divide the value 3, and will run at about 1.6 times the speed as a natural divide by 3 on a standard x86 / x64 CPU.
Of course the only reason you can make this optimization when the compiler cant is because the compiler does not know the maximum range of X and therefore cannot make this determination, but you as the programmer can.
Sometime it may even be more beneficial to move the value into a larger value and then do the same thing, ie. if you have an int of full range you could make it an 64-bit value and then do the multiply and shift instead of dividing by 3.
I had to do this recently to speed up image processing, i needed to find the average of 3 color channels, each color channel with a byte range (0 - 255). red green and blue.
At first i just simply used:
avg = (r + g + b) / 3;
(So r + g + b has a maximum of 768 and a minimum of 0, because each channel is a byte 0 - 255)
After millions of iterations the entire operation took 36 milliseconds.
I changed the line to:
avg = (r + g + b) * 341 >> 10;
And that took it down to 22 milliseconds, its amazing what can be done with a little ingenuity.
This speed up occurred in C# even though I had optimisations turned on and was running the program natively without debugging info and not through the IDE.
See How To Divide By 3 for an extended discussion of more efficiently dividing by 3, focused on doing FPGA arithmetic operations.
Also relevant:
Optimizing integer divisions with Multiply Shift in C#
Depending on your platform and depending on your C compiler, a native solution like just using
y = x / 3
Can be fast or it can be awfully slow (even if division is done entirely in hardware, if it is done using a DIV instruction, this instruction is about 3 to 4 times slower than a multiplication on modern CPUs). Very good C compilers with optimization flags turned on may optimize this operation, but if you want to be sure, you are better off optimizing it yourself.
For optimization it is important to have integer numbers of a known size. In C int has no known size (it can vary by platform and compiler!), so you are better using C99 fixed-size integers. The code below assumes that you want to divide an unsigned 32-bit integer by three and that you C compiler knows about 64 bit integer numbers (NOTE: Even on a 32 bit CPU architecture most C compilers can handle 64 bit integers just fine):
static inline uint32_t divby3 (
uint32_t divideMe
) {
return (uint32_t)(((uint64_t)0xAAAAAAABULL * divideMe) >> 33);
}
As crazy as this might sound, but the method above indeed does divide by 3. All it needs for doing so is a single 64 bit multiplication and a shift (like I said, multiplications might be 3 to 4 times faster than divisions on your CPU). In a 64 bit application this code will be a lot faster than in a 32 bit application (in a 32 bit application multiplying two 64 bit numbers take 3 multiplications and 3 additions on 32 bit values) - however, it might be still faster than a division on a 32 bit machine.
On the other hand, if your compiler is a very good one and knows the trick how to optimize integer division by a constant (latest GCC does, I just checked), it will generate the code above anyway (GCC will create exactly this code for "/3" if you enable at least optimization level 1). For other compilers... you cannot rely or expect that it will use tricks like that, even though this method is very well documented and mentioned everywhere on the Internet.
Problem is that it only works for constant numbers, not for variable ones. You always need to know the magic number (here 0xAAAAAAAB) and the correct operations after the multiplication (shifts and/or additions in most cases) and both is different depending on the number you want to divide by and both take too much CPU time to calculate them on the fly (that would be slower than hardware division). However, it's easy for a compiler to calculate these during compile time (where one second more or less compile time plays hardly a role).
For 64 bit numbers:
uint64_t divBy3(uint64_t x)
{
return x*12297829382473034411ULL;
}
However this isn't the truncating integer division you might expect.
It works correctly if the number is already divisible by 3, but it returns a huge number if it isn't.
For example if you run it on for example 11, it returns 6148914691236517209. This looks like a garbage but it's in fact the correct answer: multiply it by 3 and you get back the 11!
If you are looking for the truncating division, then just use the / operator. I highly doubt you can get much faster than that.
Theory:
64 bit unsigned arithmetic is a modulo 2^64 arithmetic.
This means for each integer which is coprime with the 2^64 modulus (essentially all odd numbers) there exists a multiplicative inverse which you can use to multiply with instead of division. This magic number can be obtained by solving the 3*x + 2^64*y = 1 equation using the Extended Euclidean Algorithm.
What if you really don't want to multiply or divide? Here is is an approximation I just invented. It works because (x/3) = (x/4) + (x/12). But since (x/12) = (x/4) / 3 we just have to repeat the process until its good enough.
#include <stdio.h>
void main()
{
int n = 1000;
int a,b;
a = n >> 2;
b = (a >> 2);
a += b;
b = (b >> 2);
a += b;
b = (b >> 2);
a += b;
b = (b >> 2);
a += b;
printf("a=%d\n", a);
}
The result is 330. It could be made more accurate using b = ((b+2)>>2); to account for rounding.
If you are allowed to multiply, just pick a suitable approximation for (1/3), with a power-of-2 divisor. For example, n * (1/3) ~= n * 43 / 128 = (n * 43) >> 7.
This technique is most useful in Indiana.
I don't know if it's faster but if you want to use a bitwise operator to perform binary division you can use the shift and subtract method described at this page:
Set quotient to 0
Align leftmost digits in dividend and divisor
Repeat:
If that portion of the dividend above the divisor is greater than or equal to the divisor:
Then subtract divisor from that portion of the dividend and
Concatentate 1 to the right hand end of the quotient
Else concatentate 0 to the right hand end of the quotient
Shift the divisor one place right
Until dividend is less than the divisor:
quotient is correct, dividend is remainder
STOP
For really large integer division (e.g. numbers bigger than 64bit) you can represent your number as an int[] and perform division quite fast by taking two digits at a time and divide them by 3. The remainder will be part of the next two digits and so forth.
eg. 11004 / 3 you say
11/3 = 3, remaineder = 2 (from 11-3*3)
20/3 = 6, remainder = 2 (from 20-6*3)
20/3 = 6, remainder = 2 (from 20-6*3)
24/3 = 8, remainder = 0
hence the result 3668
internal static List<int> Div3(int[] a)
{
int remainder = 0;
var res = new List<int>();
for (int i = 0; i < a.Length; i++)
{
var val = remainder + a[i];
var div = val/3;
remainder = 10*(val%3);
if (div > 9)
{
res.Add(div/10);
res.Add(div%10);
}
else
res.Add(div);
}
if (res[0] == 0) res.RemoveAt(0);
return res;
}
If you really want to see this article on integer division, but it only has academic merit ... it would be an interesting application that actually needed to perform that benefited from that kind of trick.
Easy computation ... at most n iterations where n is your number of bits:
uint8_t divideby3(uint8_t x)
{
uint8_t answer =0;
do
{
x>>=1;
answer+=x;
x=-x;
}while(x);
return answer;
}
A lookup table approach would also be faster in some architectures.
uint8_t DivBy3LU(uint8_t u8Operand)
{
uint8_t ai8Div3 = [0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, ....];
return ai8Div3[u8Operand];
}
I want to be able to take, as input, a character pointer to a number in base 2 through 16 and as a second parameter, what base the number is in and then convert that to it's representation in base 2. The integer can be of arbitrary length. My solution now does what the atoi() function does, but I was curious purely out of academic interest if a lookup table solution is possible.
I have found that this is simple for binary, octal, and hexadecimal. I can simply use a lookup table for each digit to get a series of bits. For instance:
0xF1E ---> (F = 1111) (1 = 0001) (E = 1110) ---> 111100011110
0766 ---> (7 = 111) (6 = 110) (6 = 110) ---> 111110110
1000 ---> ??? ---> 1111101000
However, my problem is that I want to do this look up table method for odd bases, like base 10. I know that I could write the algorithm like atoi does and do a bunch of multiplies and adds, but for this specific problem I'm trying to see if I can do it with a look up table. It's definitely not so obvious with base 10, though. I was curious if anyone had any clever way to figure out how to generate a generic look up table for Base X -> Base 2. I know that for base 10, you can't just give it one digit at a time, so the solution would likely have to lookup a group of digits at a time.
I am aware of the multiply and add solution but since these are arbitrary length numbers, the multiply and add operations are not free so I'd like to avoid them, if at all possible.
You will have to use a look up table with an input width of m base b symbols returning n bits so that
n = log2(b) * m
for positive integers b, n and m. So if b is not a power of two, there will be no (simple) look up table solution.
I do not think that there is a solution. The following example with base 10 illustrates why.
65536 = 1 0000 0000 0000 0000
Changing the last digit from 6 to 5 will flip all bits.
65535 = 0 1111 1111 1111 1111
And almost the same will hold if you process the input starting from the end. Changing the first digit from 6 to 5 flips a significant number of bits.
55535 = 0 1101 1000 1111 0000
This is not possible in bases that aren't powers of two to convert to base-2. The reason that it is possible for base 8 (and 16) is that the way the conversion works is following:
octal ABC = 8^2*A + 8^1*B + 8^0*C (decimal)
= 0b10000000*A + 0b1000*B + C (binary)
so if you have the lookup table of A = (0b000 to 0b111), then the multiplication is always by 1 and some trailing zeros, so the multiplication is simple (just shifting left).
However, consider the 'odd' base of 10. When you look at the powers of 10:
10^1 = 0b1010
10^2 = 0b1100100
10^3 = 0b1111101000
10^4 = 0b10011100010000
..etc
You'll notice that the multiplication never gets simple, so you can't have any lookup tables and do bitshifts and ors, no matter how big you group them. It will always overlap. The best you can do is have a lookup table of the form: (a,b) where a is the digit position, and b is the digit (0..9). Then, you are only reduced to adding n numbers, rather than multiplying and adding n numbers (plus the cost of the memory of the lookup table)
How big are the strings? You can potentially convert the multiply-and-add to a lookup-and-add by doing something like this:
Store the numbers 0-9, 10, 20, 30, 40, ... 90, 100, 200, ... 900, 1000, 2000, ... , 9000, 10000, ... in the target base in a table.
For each character starting with the rightmost, index appropriately into the table and add it to a running result.
Of course I'm not sure how well this will actually perform, but it's a thought.
The algorithm is quite simple. Language agnostic would be:
total = 0
base <- input_base
for each character in input:
total <- total*base + number(char)
In C++:
// Helper to convert a digit to a number
unsigned int number( char ch )
{
if ( ch >= '0' && ch <= '9' ) return ch-'0';
ch = toupper(ch);
if ( ch >= 'A' && ch <= 'F' ) return 10 + (ch-'A');
}
unsigned int parse( std::string const & input, unsigned int base )
{
unsigned int total = 0;
for ( int i = 0; i < input.size(); ++i )
{
total = total*base + number(input[i]);
}
return total;
}
Of course, you should take care of possible errors (incoherent input: base 2 and input string 'af12') or any other exceptional condition.
Start with a running count of 0.
For each character in the string (reading left to right)
Multiply count by base.
Convert character to int value (0 through base)
Add character value to running count.
How accurate do you need to be?
If you're looking for perfection, then multiply-and-add is really your only recourse. And I'd be very surprised if it's the slowest part of your application.
If order-of-magnitude is good enough, use a lookup table to find the closest power of 2.
Example 1: 1234, closest power of 2 is 1024.
Example 2: 98765, closest is 65536
You could also drive this by counting the number of digits, and multiplying the appropriate power of 2 by the leftmost digit. This can be implemented as a left-shift:
Example 3: 98765 has 5 digits, closest power of 2 to 10000 is 8192 (2^13), so result is 9 << 13
I wrote this before your clarifying comment so it probably isn't quite is applicable. I'm not sure if a lookup table approach is possible or not. If you really don't need arbitrary precision, then take advantage of the runtime.
If a C/C++ solution is acceptable, I believe that the following is what you are looking for is something like the following. It probably contains bugs in edge cases, but it does compile and work as expected at least for positive numbers. Making it really work is an exercise for the reader.
/*
* NAME
* convert_num - convert a numerical string (str) of base (b) to
* a printable binary representation
* SYNOPSIS
* int convert_num(char const* s, int b, char** o)
* DESCRIPTION
* Generates a printable binary representation of an input number
* from an arbitrary base. The input number is passed as the ASCII
* character string `s'. The input string consists of characters
* from the ASCII character set {'0'..'9','A'..('A'+b-10)} where
* letter characters may be in either upper or lower case.
* RETURNS
* The number of characters from the input string `s' which were
* consumed by this operation. The output string is placed into
* newly allocated storage which is pointed to by `*o' upon successful
* completion. An error is signalled by returning `-1'.
*/
int
convert_num(char const *str, int b, char **out)
{
int rc = -1;
char *endp = NULL;
char *outp = NULL;
unsigned long num = strtoul(str, &endp, b);
if (endp != str) { /* then we have some numbers */
int numdig = -1;
rc = (endp - str); /* we have this many base `b' digits! */
frexp((double)num, &numdig); /* we need this many base 2 digits */
if ((outp=malloc(numdig+1)) == NULL) {
return -1;
}
*out = outp; /* return the buffer */
outp += numdig; /* make sure it is NUL terminated */
*outp-- = '\0';
while (numdig-- != 0) { /* fill it in from LSb to MSb */
*outp-- = ((num & 1) ? '1' : '0');
num >>= 1;
}
}
return rc;
}