I am doing a project in C++ and I'm having hard time with substraction of binary numbers. It's easy doing it on paper but in code it's a bit hard for me.
Can someone please give me the algorithm for the subtraction of two binary numbers? It's supposed to be done WITHOUT conversion to the decimal system.
Thanks!
int subtractBinaries(int x, int y) {
while (y != 0) {
int borrow = (~x) & y;
x = x ^ y;
y = borrow << 1;
}
return x;
}
Related
I'm having trouble trying to find a function to look at a certain bit. If, for example, I had a binary number of 1111 1111 1111 1011, and I wanted to just look at the most significant bit ( the bit all the way to the left, in this case 1) what function could I use to just look at that bit?
The program is to test if a binary number is positive or negative. I started off by using hex number 0x0005, and then using a two's compliment function to make it negative. But now, I need a way to check if the first bit is 1 or 0 and to return a value out of that. The integer n would be equal to 1 or 0 depending on if it is negative or positive. My code is as follows:
#include <msp430.h>
signed long x=0x0005;
int y,i,n;
void main(void)
{
y=~x;
i=y+1;
}
There are two main ways I have done something like this in the past. The first is a bit mask which you would use if you always are checking the exact same bit(s). For example:
#define MASK 0x80000000
// Return value of "0" means the bit wasn't set, "1" means the bit was.
// You can check as many bits as you want with this call.
int ApplyMask(int number) {
return number & MASK;
}
Second is a bit shift, then a mask (for getting an arbitrary bit):
int CheckBit(int number, int bitIndex) {
return number & (1 << bitIndex);
}
One or the other of these should do what you are looking for. Best of luck!
bool isSetBit (signed long number, int bit)
{
assert ((bit >= 0) && (bit < (sizeof (signed long) * 8)));
return (number & (((signed long) 1) << bit)) != 0;
}
To check the sign bit:
if (isSetBit (y, sizeof (y) * 8 - 1))
...
I have a vector, and I would like to do the following, using CUDA and Thrust transformations:
// thrust::device_vector v;
// for k times:
// calculate constants a and b as functions of k;
// for (i=0; i < v.size(); i++)
// v[i] = a*v[i] + b*v[i+1];
How should I correctly implement this? One way I can do it is to have vector w, and apply thrust::transform onto v and save the results to w. But k is unknown ahead of time, and I don't want to create w1, w2, ... and waste a lot of GPU memory space. Preferably I want to minimize the amount of data copying. But I'm not sure how to implement this using one vector without the values stepping on each other. Is there something Thrust provides that can do this?
If the v.size() is large enough to fully utilize the GPU, you could launch k kernels to do this, with a extra buffer mem and no extra data transfer.
thrust::device_vector u(v.size());
for(k=0;;)
{
// calculate a & b
thrust::transform(v.begin(), v.end()-1, v.begin()+1, u.begin(), a*_1 + b*_2);
k++;
if(k>=K)
break;
// calculate a & b
thrust::transform(u.begin(), u.end()-1, u.begin()+1, v.begin(), a*_1 + b*_2);
k++;
if(k>=K)
break;
}
I don't actually understand the "k times", but the following code may help you.
struct OP {
const int a, b;
OP(const int p, const int q): a(p), b(q){};
int operator()(const int v1, const int v2) {
return a*v1+b*v2;
}
}
thrust::device_vector<int> w(v.size());
thrust::transform(v.begin(), v.end()-1, //input_1
v.begin()+1, //input_2
w.begin(), //output
OP(a, b)); //functor
v = w;
I think learning about "functor", and several examples of thrust will give you a good guide.
Hope this will help you to solve your problem. :)
Which is the best way to store a symmetric matrix in memory?
It would be good to save half of the space without compromising speed and complexity of the structure too much. This is a language-agnostic question but if you need to make some assumptions just assume it's a good old plain programming language like C or C++..
It seems a thing that has a sense just if there is a way to keep things simple or just when the matrix itself is really big, am I right?
Just for the sake of formality I mean that this assertion is always true for the data I want to store
matrix[x][y] == matrix[y][x]
Here is a good method to store a symmetric matrix, it requires only N(N+1)/2 memory:
int fromMatrixToVector(int i, int j, int N)
{
if (i <= j)
return i * N - (i - 1) * i / 2 + j - i;
else
return j * N - (j - 1) * j / 2 + i - j;
}
For some triangular matrix
0 1 2 3
4 5 6
7 8
9
1D representation (stored in std::vector, for example) looks like as follows:
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
And call fromMatrixToVector(1, 2, 4) returns 5, so the matrix data is vector[5] -> 5.
For more information see http://www.codeguru.com/cpp/cpp/algorithms/general/article.php/c11211/TIP-Half-Size-Triangular-Matrix.htm
I find that many high performance packages just store the whole matrix, but then only read the upper triangle or lower triangle. They might then use the additional space for storing temporary data during the computation.
However if storage is really an issue then just store the n(n+1)/2 elements making the upper triangle in a one-dimensional array. If that makes access complicated for you, just define a set of helper functions.
In C to access a matrix matA you could define a macro:
#define A(i,j, dim) ((i <= j)?matA[i*dim + j]:matA[j*dim + i])
then you can access your array nearly normally.
Well I would try a triangular matrix, like this:
int[][] sym = new int[rows][];
for( int i = 0; i < cols; ++i ) {
sym=new int[i+1];
}
But then you wil have to face the problem when someone wants to access the "other side". Eg he wants to access [0][10] but in your case this val is stored in[10][0] (assuming 10x10).
The probably "best" way is the lazy one - dont do anything until the user requests. So you could load the specific row if the user types somethin like print(matrix[4]).
If you want to use a one dimensional array the code would look something like this:
int[] new matrix[(rows * (rows + 1 )) >> 1];
int z;
matrix[ ( ( z = ( x < y ? y : x ) ) * ( z + 1 ) >> 1 ) + ( y < x ? y : x ) ] = yourValue;
You can get rid of the multiplications if you create an additional look-up table:
int[] new matrix[(rows * (rows + 1 )) >> 1];
int[] lookup[rows];
for ( int i= 0; i < rows; i++)
{
lookup[i] = (i * (i+1)) >> 1;
}
matrix[ lookup[ x < y ? y : x ] + ( x < y ? x : y ) ] = yourValue;
If you're using something that supports operator overloading (e.g. C++), it's pretty easy to handle this transparently. Just create a matrix class that checks the two subscripts, and if the second is greater than the first, swap them:
template <class T>
class sym_matrix {
std::vector<std::vector<T> > data;
public:
T operator()(int x, int y) {
if (y>x)
return data[y][x];
else
return data[x][y];
}
};
For the moment I've skipped over everything else, and just covered the subscripting. In reality, to handle use as both an lvalue and an rvalue correctly, you'll typically want to return a proxy instead of a T directly. You'll want a ctor that creates data as a triangle (i.e., for an NxN matrix, the first row will have N elements, the second N-1, and so on -- or, equivalantly 1, 2, ...N). You might also consider creating data as a single vector -- you have to compute the correct offset into it, but that's not terribly difficult, and it will use a bit less memory, run a bit faster, etc. I'd use the simple code for the first version, and optimize later if necessary.
You could use a staggered array (or whatever they're called) if your language supports it, and when x < y, switch the position of x and y. So...
Pseudocode (somewhat Python style, but not really) for an n x n matrix:
matrix[n][]
for i from 0 to n-1:
matrix[i] = some_value_type[i + 1]
[next, assign values to the elements of the half-matrix]
And then when referring to values....
if x < y:
return matrix[y][x]
else:
return matrix[x][y]
Multiplying two numbers in log space means adding them:
log_multiply(x, y) = log( exp(x) * exp(y) )
= x + y
Adding two numbers in log space means you do a special log-add operation:
log_add(x, y) = log( exp(x) + exp(y) )
which is implemented in the following code, in a way that doesn't require us to take the two exponentials (and lose runtime speed and precision):
double log_add(double x, double y) {
if(x == neginf)
return y;
if(y == neginf)
return x;
return max(x, y) + log1p(exp( -fabs(x - y) ));
}
(Here is another one.)
But here is the question:
Is there a trick to do it for subtraction as well?
log_subtract(x, y) = log( exp(x) - exp(y) )
without having to take the exponents and lose precision?
double log_subtract(double x, double y) {
// ?
}
How about
double log_subtract(double x, double y) {
if(x <= y)
// error!! computing the log of a negative number
if(y == neginf)
return x;
return x + log1p(-exp(y-x));
}
That's just based on some quick math I did...
The library functions for exp and log lose precision for extreme values.
log1p gets you half way there, but what you need is a function that treats the error for both the log and the exp parts.
See this article: http://cran.r-project.org/web/packages/Rmpfr/vignettes/log1mexp-note.pdf
The title is "Accurately Computing log(1 - exp(-|a|))".
The article discusses how to seemlessly merge different algorithms to create good error bounds for a larger range of inputs.
How can the XOR operation (on two 32 bit ints) be implemented using only basic arithmetic operations? Do you have to do it bitwise after dividing by each power of 2 in turn, or is there a shortcut? I don't care about execution speed so much as about the simplest, shortest code.
Edit:
This is not homework, but a riddle posed on a hacker.org. The point is to implement XOR on a stack-based virtual machine with very limited operations (similar to the brainfuck language and yes - no shift or mod). Using that VM is the difficult part, though of course made easier by an algorithm that is short and simple.
While FryGuy's solution is clever, I'll have to go with my original ideal (similar to litb's solution) because comparisons are difficult to use as well in that environment.
I would do it the simple way:
uint xor(uint a, uint b):
uint ret = 0;
uint fact = 0x80000000;
while (fact > 0)
{
if ((a >= fact || b >= fact) && (a < fact || b < fact))
ret += fact;
if (a >= fact)
a -= fact;
if (b >= fact)
b -= fact;
fact /= 2;
}
return ret;
There might be an easier way, but I don't know of one.
I don't know whether this defeats the point of your question, but you can implement XOR with AND, OR, and NOT, like this:
uint xor(uint a, uint b) {
return (a | b) & ~(a & b);
}
In english, that's "a or b, but not a and b", which maps precisely to the definition of XOR.
Of course, I'm not sticking strictly to your stipulation of using only the arithmetic operators, but at least this a simple, easy-to-understand reimplementation.
I'm sorry i only know the straight forward one in head:
uint32_t mod_op(uint32_t a, uint32_t b) {
uint32_t int_div = a / b;
return a - (b * int_div);
}
uint32_t xor_op(uint32_t a, uint32_t b) {
uint32_t n = 1u;
uint32_t result = 0u;
while(a != 0 || b != 0) {
// or just: result += n * mod_op(a - b, 2);
if(mod_op(a, 2) != mod_op(b, 2)) {
result += n;
}
a /= 2;
b /= 2;
n *= 2;
}
return result;
}
The alternative in comments can be used instead of the if to avoid branching. But then again, the solution isn't exactly fast either and it makes it look stranger :)
It's easier if you have the AND because
A OR B = A + B - (A AND B)
A XOR B = A + B - 2(A AND B)
int customxor(int a, int b)
{
return a + b - 2*(a & b);
}