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Computing the Parity
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Closed 3 years ago.
So I am supposed to find the parity of a number where parity is odd if there are odd number of set bits in the binary representation of the given number and even otherwise. I have a solution but the code is not very understandable and thus am looking for an explanation to it.
int y = x ^ (x >> 1); //given number is x.
y = y ^ (y >> 2);
y = y ^ (y >> 4);
y = y ^ (y >> 8);
y = y ^ (y >> 16);
// Rightmost bit of y holds the parity value
// if (y&1) is 1 then parity is odd else even
if (y & 1)
return odd;
return even;
This is quite simple. This is how bits are merged to get parity bit:
Bit numbers:
0 1 2 3 4 5 6 7
\ / \ / \ / \ /
01 23 45 56 x ^ (x >> 1)
\ / \ /
\ / \ /
0123 4567 y ^ (y >> 2);
\ /
\ /
\ /
\ /
01234567 y ^ (y >> 4);
Simply this tree shows how the youngest bit is effectively evaluated.
Note that there are extra pairs on each step, but they are simply do not have impact on bit which is a final result.
Related
Given a long int x, count the number of values of a that satisfy the following conditions:
a XOR x > x
0 < a < x
where a and x are long integers and XOR is the bitwise XOR operator
How would you go about completing this problem?
I should also mentioned that the input x can be as large as 10^10
I have managed to get a brute force solution by iterating over 0 to x checking the conditions and incrementing a count value.. however this is not an optimal solution...
This is the brute force that I tried. It works but is extremely slow for large values of x.
for(int i =0; i < x; i++)
{
if((0 < i && i < x) && (i ^ x) > x)
count++;
}
long long NumberOfA(long long x)
{
long long t = x <<1;
while(t^(t&-t)) t ^= (t&-t);
return t-++x;
}
long long x = 10000000000;
printf("%lld ==> %lld\n", 10LL, NumberOfA(10LL) );
printf("%lld ==> %lld\n", x, NumberOfA(x) );
Output
10 ==> 5
10000000000 ==> 7179869183
Link to IDEOne Code
Trying to explain the logic (using example 10, or 1010b)
Shift x to the left 1. (Value 20 or 10100b)
Turn off all low bits, leaving just the high bit (Value 16 or 10000b)
Subtract x+1 (16 - 11 == 5)
Attempting to explain
(although its not easy)
Your rule is that a ^ x must be bigger than x, but that you cannot add extra bits to a or x.
(If you start with a 4-bit value, you can only use 4-bits)
The biggest possible value for a number in N-bits is 2^n -1.
(eg. 4-bit number, 2^4-1 == 15)
Lets call this number B.
Between your value x and B (inclusive), there are B-x possible values.
(back to my example, 10. Between 15 and 10, there are 5 possible values: 11, 12, 13, 14, 15)
In my code, t is x << 1, then with all the low bits turned off.
(10 << 1 is 20; turn off all the low bits to get 16)
Then 16 - 1 is B, and B - x is your answer:
(t - 1 - x, is the same as t - ++x, is the answer)
One way to look at this is to consider each bit in x.
If it's 1, then flipping it will yield a smaller number.
If it's 0, then flipping it will yield a larger number, and we should count it - and also all the combinations of bits to the right. That conveniently adds up to the mask value.
long f(long const x)
{
// only positive x can have non-zero result
if (x <= 0) return 0;
long count = 0;
// Iterate from LSB to MSB
for (long mask = 1; mask < x; mask <<= 1)
count += x & mask
? 0
: mask;
return count;
}
We might suspect a pattern here - it looks like we're just copying x and flipping its bits.
Let's confirm, using a minimal test program:
#include <cstdlib>
#include <iostream>
int main(int, char **argv)
{
while (*++argv)
std::cout << *argv << " -> " << f(std::atol(*argv)) << std::endl;
}
0 -> 0
1 -> 0
2 -> 1
3 -> 0
4 -> 3
5 -> 2
6 -> 1
7 -> 0
8 -> 7
9 -> 6
10 -> 5
11 -> 4
12 -> 3
13 -> 2
14 -> 1
15 -> 0
So all we have to do is 'smear' the value so that all the zero bits after the most-significant 1 are set, then xor with that:
long f(long const x)
{
if (x <= 0) return 0;
long mask = x;
while (mask & (mask+1))
mask |= mask+1;
return mask ^ x;
}
This is much faster, and still O(log n).
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F is a function that number x has been repeated in an ascending order f(x).
x : 1 2 3 4 5 6 7 8 9 10
f(x): 1 2 2 3 3 3 4 4 4 4
my function gets 'x' and gives 'f(x)' and it has to do it without array but it goes wrong in high numbers.
int main()
{
int n;
cin>>n;
int i=1,a=1;
if(n==1)
cout<<'1';
else{
while(true){
a++;
i=i+a;
if(i>=n)
break;
}
}
cout<<a;
return 0;
}
TL;DR
f(x) = floor(0.5 + sqrt(1 + 8 * (x - 1)) / 2)
Explanation
Well, since this is a mathematical problem, just solve it with math ;)
One thing to notice is the correlation between the table and the triangular numbers:
h(x) = sum(range(1, x)) = x*(x + 1)/2 //triangular number
x 1 2 3 4 5 6 7 8 9 10
f(x) 1 2 2 3 3 3 4 4 4 4
h(f(x)) 1 3 3 6 6 6 10 10 10 10
So how does that help us? Well, we can write a new equation:
h(f(x)) = x | x = max({n | f(n) = f(x)})
And logically for the inverse the following should apply:
h^-1(x) = f(x)
No we've got two options:
Call it a day and just solve the rest via brute-force:
i = 1
sum = 0
while sum < x:
sum += i
i++
return i - 1
Or build our function h^-1(x):
h(x) = y = (x+1)x/2
h^-1(y) = x with h(x) = y
x ^ 2 + x - 2y = 0
solve for x using the quadratic formula:
x = 0.5 +/- sqrt(1 + 8y) / 2
Now this formula still lacks a few things:
we get two results, one of which is negative. We can just throw the negative result away, so +/- turns into +
this formula is 0-based. To be honest, I'm still trying to figure out why. Solution: simply decrement y by 1 to get the proper result
while this formula returns the correct result for the matching numbers, i.e. y = 3 -> x = 3, it returns floating-point numbers for other input, so we'll have to round down appropriately
Putting it together:
f(x) = floor(0.5 + sqrt(1 + 8 * (x - 1)) / 2)
int f(int x) {
return (x * (x + 1)) / 2;
}
int main() {
int n;
cin >> n;
int left = 1, right = n;
while(left < right) {
int mid = left + (right - left) / 2;
int val = f(mid);
if(val >= n) {
right = mid;
}
else {
left = mid + 1;
}
}
cout << left;
return 0;
}
Use binary search. Right now I am in mobile. I will add the explanation later if needed. Let me know if you don't understand anything.
Suppose X and Y are two positive integers and Y is a power of two. Then what does this expression calculate?
(X+Y-1) & ~(Y-1)
I found this expression appearing in certain c/c++ implementation of Memory Pool (X represents the object size in bytes and Y represents the alignment in bytes, the expression returns the block size in bytes fit for use in the Memory Pool).
&~(Y-1) where Y is a power of 2, zeroes the last n bits, where Y = 2n: Y-1 produces n 1-bits, inverting that via ~ gives you a mask with n zeroes at the end, anding via bit-level & zeroes the bits where the mask is zero.
Effectively that produces a number that is some multiple of Y's power of 2.
It can maximally have the effect of subtracting Y-1 from the number, so add that first, giving (X+Y-1) & ~(Y-1). This is a number that's not less than X, and is a multiple of Y.
It gives you the next Y-aligned address of current address X.
Say, your current address X is 0x10000, and your alignment is 0x100, it will give you 0x10000. But if your current address X is 0x10001, you will get "next" aligned address of 0x10100.
This is useful in the scenario that you want your new object always to be aligned to blocks in memory, but not leaving any block unused. So you want to know what is the next available block-aligned address.
Why don't you just try some input and observe what happens?
#include <iostream>
unsigned compute(unsigned x, unsigned y)
{
return (x + y - 1) & ~(y - 1);
}
int main()
{
std::cout << "(x + y - 1) & ~(y - 1)" << std::endl;
for (unsigned x = 0; x < 9; ++x)
{
std::cout << "x=" << x << ", y=2 -> " << compute(x, 2) << std::endl;
}
std::cout << "----" << std::endl;
std::cout << "(x + y - 1) & ~(y - 1)" << std::endl;
for (unsigned x = 0; x < 9; ++x)
{
std::cout << "(x=" << x << ", y=2) -> " << compute(x, 2) << std::endl;
}
return 0;
}
Live Example
Output:
First set uses x in [0, 8] and y is constant 2. Second set uses x in [0, 8] and y is constant 4.
(x + y - 1) & ~(y - 1)
x=0, y=2 -> 0
x=1, y=2 -> 2
x=2, y=2 -> 2
x=3, y=2 -> 4
x=4, y=2 -> 4
x=5, y=2 -> 6
x=6, y=2 -> 6
x=7, y=2 -> 8
x=8, y=2 -> 8
----
(x + y - 1) & ~(y - 1)
(x=0, y=2) -> 0
(x=1, y=2) -> 2
(x=2, y=2) -> 2
(x=3, y=2) -> 4
(x=4, y=2) -> 4
(x=5, y=2) -> 6
(x=6, y=2) -> 6
(x=7, y=2) -> 8
(x=8, y=2) -> 8
It's easy to see the output (i.e., result right of ->) is always a multiple of y such that the output is greater than or equal to x.
First I assume that X and Y are unsigned integers.
Let's have a look at the right part:
If Y is a power of 2, it is represented in binary by one bit to 1 and all the others to 0. Example 8 will be binary 00..01000.
If you substract 1 the highest bit will be 0 and all the bits to its right will become 1. Example 8-1= 7 and in binary 00..00111
If you ~ negate this number you will make sure that all highest bit (including the original one will turn to 1 and the lovest to 0. Example: ~7 will be 11..11000
Now if you do a binary AND (&) with any number, you will set to 0 all the lower bits, in our example, the 3 lower bits. THe resulting number is hence a multiple of Y.
Let's look at the left side:
We've already analysed Y-1. In our example we had 7, that is 00..00111
If you add this to any number, you make sure that the result is greater than or equal to Y. Example with 5: 5+7=12 so 00..01100 and example with 10: 10+7=17 so 00..10001
If you then perform the AND, you'll erase the lower bits. so in our example with 5, we come to 00..01000 = 8 and in our example with 10 we get 00..10000 16.
Conclusion, it's the smallest multiple of Y wich is greater or equal to X.
Let's break it down, piece by piece.
(X+Y-1) & ~(Y-1)
Let's suppose that X = 11 and Y = 16 in accordance with your rules and that the integers are 8 bits.
(11+16-1) & ~(16-1)
Do the Addition and Subtraction
(26) & ~(15)
Translate this into binary
(0001 1010) & ~(0000 1111)
~ means not or to invert the zeros and ones
(0001 1010) & (1111 0000)
& means only to take the bits that are both ones
0001 0000
convert back to decimal
16
other examples
X = 78, Y = 32 results in 96
X = 25, Y = 64 results in 64
X = 47, Y = 16 results in 48
So, it would seem to me that the purpose of this is to find lowest multiple of Y that is equal to or greater than X. This could be used for finding the start/end address of a block of memory, or it could be used for positioning items on the screen, or any number of other possible answers as well. But without context and possibly even a full code example. There's no guarantee.
(X+Y-1) & ~(Y-1)
x = 7 = 0b0111
y = 4 = 0b0100
x+y-1 = 0b1010
y-1 = 3 = 0b0011
~(y-1) = 0b1100
(x+y-1) & ~(y-1) = 0b1000 = 8
--
x = 12 = 0b1100
y = 2 = 0b0010
x+y-1 = 13 = 0b1101
y-1 = 1 = 0b0001
~(y-1) = 0b1110
(x+y-1) & ~(y-1) = 0b1100 = 12
(x+y-1) & ~(y-1) is the smallest multiple of y greater than or equal to x
It seems provides a specified alignment of a value for example of a memory address (for example when you want to get the next aligned address).
For example if you want that a memory address would be aligned at the paragraph bound you can write
( address + 16 - 1 ) & ~( 16 - 1 )
or
( address + 15 ) & ~15
or
( address + 15 ) & ~0xf
In this case all bits before 16 will be zeroed.
This part of expression
( address + alignment - )
is used for rounding.
and this part of expression
~( alignment - 1 )
is used to build a mask thet zeroes low bits.
I need to implement a 32-bit arithmetic right shift from logical shifts, and, or, xor and normal integer arithmetic operations.
I read somewhere the following is supposed to work:
(x>>N)|(((1<<N)-1)<<(32-N))
x is the integer that will be shifted and N is the amount of bits to shift.
This works for negative (msb is 1) numbers but not for positive numbers (msb is 0).
Does anyone know an efficient algorithm that always produces the right result?
You can use this
(x >> N) | (-(x < 0) << (32 - N))
If x is negative then -(x < 0) returns -1, which have a bit pattern of all 1s, assuming 2's complement. -1 << (32 - N) will make a value which has all 1s in the top N bits and 0s in the remaining part. If x is non-negative then the latter part will always be zero, and the result will be the same as a logical shift. Alternatively it can be modified to
(x >> N) | ~(((x < 0) << (32 - N)) - 1)
Note that it won't work for N <= 0 or N >= 32 (since shifting more than the width of type invokes UB) so you should treat those cases specifically if needed
If you're not allowed to use comparison then you can change x < 0 to (unsigned)x >> 31 and get the below equivalent ways
(x >> N) | (-((unsigned)x >> 31) << (32 - N))
(x >> N) | ((~0*(unsigned)x >> 31) << (32 - N))
(x >> N) | ~((((unsigned)x >> 31) << (32 - N)) - 1)
LSR 1:
+---+---+---+---
| | | | ...
+---+---+---+---
\ \ \ \
\ \ \
\ \ \
+---+---+---+---
| 0 | | | ...
+---+---+---+---
ASR 1:
+---+---+---+---
| | | | ...
+---+---+---+---
|\ \ \ \
| \ \ \
| \ \ \
+---+---+---+---
| | | | ...
+---+---+---+---
An ASR consists of an LSR plus an OR.
So you want to replicate bit31 N times. The efficient solution is probably an efficient implementation of
( bit31 ? 0xFFFFFFFF : 0x00000000 ) << ( 32 - N ) )
I've come up with
LSL(SUB(LSR(NOT(X), 31), 1), SUB(32, N))
The whole thing
OR(LSL(SUB(LSR(NOT(X), 31), 1), SUB(32, N)), LSR(X, N))
That doesn't seem very efficient.
Please excuse the disgustingly noobish nature of this post, but I have a question for those who program in C++ and R on their personal computer.
Question: Why are these random numbers produced from the two programs below not equal, and how do I resolve this issue?
Firstly, I suspect that I have misused the local function and the <<- operator in the R program.
Secondly, I suspect it may be a floating-accuracy issue. It's not immediately obvious to me how the two programs are different, so I don't know how to go about this problem.
I have tried casting all my calculations in C++ to double/float (even long double), and using fmod instead of the modulus operator %: different outputs again, but still not similar to the output in R. I do not know if it of any significant importance, but I want to add that I am compiling the C++ code using the G++ compiler.
Algorithm: The following algorithm can be used in any standard personal computer. It was proposed to use in parallel three word generators,
mk = 171 mk-1 (mod 30269)
m'k = 172 m'k-1 (mod 30307)
m''k = 172 m''k-1 (mod 30323)
and to use as pseudorandom numbers the fractional parts
gk = {mk / 30269 + m'k / 30307 + m''k / 30323}
I have used the initial values m0 = 5, m'0 = 11, and m''0 = 17.
Programs: I have the following program in C++:
//: MC:Uniform.cpp
// Generate pseudo random numbers uniformly between 0 and 1
#include <iostream>
#include <math.h> // For using "fmod()"
using namespace std;
float uniform(){
// A sequence of initial values
static int x = 5;
static int y = 11;
static int z = 17;
// Some integer arithmetic required
x = 171 * (x % 177) - 2 * (x / 177);
y = 172 * (x % 176) - 35 * (y / 176);
z = 170 * (x % 178) - 63 * (z / 178);
/* If both operands are nonnegative then the
remainder is nonnegative; if not, the sign of
the remainder is implementation-defined. */
if(x < 0)
x = x + 30269;
if(y < 0)
y = y + 30307;
if(z < 0)
z = z + 30323;
return fmod(x / 30269. + y / 30307. + z / 30323., 1.);
}
int main(){
// Print 5 random numbers
for(int i = 0; i < 5; i++){
cout << uniform() << ", ";
}
}///:~
The program exites with code and outputs the following:
0.686912, 0.329174, 0.689649, 0.753722, 0.209394,
I also have a program in R, that looks like the following:
## Generate pseudo random numbers uniformly between 0 and 1
uniform <- local({
# A sequence of initial values
x = 5
y = 11
z = 17
# Use the <<- operator to make x, y and z local static
# variables in R.
f <- function(){
x <<- 171 * (x %% 177) - 2 * (x / 177)
y <<- 172 * (y %% 176) - 35 * (y / 176)
z <<- 170 * (z %% 178) - 63 * (z / 178)
return((x / 30269. + y / 30307. + z / 30323.)%%1.)
}
})
# Print 5 random numbers
for(i in 1:5){
print(uniform())
}
This program exites with code as well and produces the output
[1] 0.1857093
[1] 0.7222047
[1] 0.05103441
[1] 0.7375034
[1] 0.2065817
Any suggestions are appreciated, thanks in advance.
You need a few more %/%'s (integer division) in your R code. Remember that numeric variables in R are floating-point, not integer, by default; so / will do ordinary division with a non-integral quotient. You've also left out the part where you deal with negative x/y/z.
f <- function(){
x <<- 171 * (x %% 177) - 2 * (x %/% 177)
y <<- 172 * (y %% 176) - 35 * (y %/% 176)
z <<- 170 * (z %% 178) - 63 * (z %/% 178)
if(x < 0)
x <<- x + 30269;
if(y < 0)
y <<- y + 30307;
if(z < 0)
z <<- z + 30323;
return((x / 30269. + y / 30307. + z / 30323.)%%1)
}
After making those changes, there doesn't seem to be anything seriously wrong with the result. A quick histogram of 100000 random draws looks very uniform, and there's no autocorrelation I can find. Still doesn't match your C++ result though....
There's a simple copy/paste error in your C++ code. This
x = 171 * (x % 177) - 2 * (x / 177);
y = 172 * (x % 176) - 35 * (y / 176);
z = 170 * (x % 178) - 63 * (z / 178);
should be this.
x = 171 * (x % 177) - 2 * (x / 177);
y = 172 * (y % 176) - 35 * (y / 176);
z = 170 * (z % 178) - 63 * (z / 178);