I am creating a data format, which will be stored in a DS2431 1-wire EEPROM. One page will be using EPROM emulation mode (where data once written can only be modified by clearing bits). In this page I want to store a byte with an ID, which cannot be changed to another valid value (due to only allowing clearing bits).
I am considering using the set of values that have a popcount of 4 (there are 70 different values). Clearing any bits means popcount is no longer 4, so this satisfies the desired property.
But could a set of byte values be found with more than 70 different values, that satisfy the property?
No. For an 8-bit value, using four bits is optimal.
If you have your 70 4-bit values and decide to add a 5-bit value as valid, you have to give up five 4-bit values that can be created by clearing a bit. Similarly, if you want a valid 3-bit value, you also have to give up five 4-bit values.
If you could increase the number of bits, then you can increase the ratio of possible values to bits used.
Since there are only 256 possible values and 8 possible populations it is a trivial task to test all possible population counts:
#include <stdio.h>
#include <stdint.h>
int popcount( uint8_t byte )
{
int count = 0 ;
for( uint8_t b = 0x01; b != 0; b <<= 1 )
{
count = count + (((byte & b) != 0) ? 1 : 0) ;
}
return count ;
}
int main()
{
int valuecount[8] = {0} ;
for( int i = 0; i < 256; i++ )
{
valuecount[popcount(i)]++ ;
}
printf( "popcount\tvalues\n") ;
for( int p = 0; p < 9; p++ )
{
printf( " %d\t\t %d\n", p, valuecount[p] ) ;
}
return 0;
}
Result:
popcount values
0 1
1 8
2 28
3 56
4 70
5 56
6 28
7 8
8 1
The optimum population count for any word length n is always n / 2. For 16-bits the number of values with 8 1-bits is 12870.
I have an integer which is the length of a text field, lets say the length is 6. I need to find the nearest multiple of 16 to this number and then get the difference between the two numbers. So in this case it would be 8 (It could also be 4 but I'm only interested in going up).
I have an implementation of this in C#:
int padding = 16 - (txtUserPwd.TextLength % 16);
However I can't work out how to do this in Objective-C (especially without RoundUp).
It's probably quite simple to do but I can't work it out, any help is appreciated!
Try this:
-(int)differenceToNextPowerOfTwo:(int)n
{
unsigned int v = n;
v--;
v |= v >> 1;
v |= v >> 2;
v |= v >> 4;
v |= v >> 8;
v |= v >> 16;
v++;
return v - n;
}
Source: http://graphics.stanford.edu/~seander/bithacks.html#RoundUpPowerOf2
Is there any way to convert a signed integer into an array of bytes in NXC? I can't use explicit type casting or pointers either, due to language limitations.
I've tried:
for(unsigned long i = 1; i <= 2; i++)
{
MM_mem[id.idx] = ((val & (0xFF << ((2 - i) * 8)))) >> ((2 - i) * 8));
id.idx++;
}
But it fails.
EDIT: This works... It just wasn't downloading. I've wasted about an hour trying to figure it out. >_>
EDIT: In NXC, >> is a arithmetic shift. int is a signed 16-bit integer type. A byte is the same thing as unsigned char.
NXC is 'Not eXactly C', a relative of C, but distinctly different from C.
How about
unsigned char b[4];
b[0] = (x & 0xFF000000) >> 24;
b[1] = (x & 0x00FF0000) >> 16;
b[2] = (x & 0x0000FF00) >> 8;
b[3] = x & 0xFF;
The best way to do this in NXC with the opcodes available in the underlying VM is to use FlattenVar to convert any type into a string (aka byte array with a null added at the end). It results in a single VM opcode operation where any of the above options which use shifts and logical ANDs and array operations will require dozens of lines of assembly language.
task main()
{
int x = Random(); // 16 bit random number - could be negative
string data;
data = FlattenVar(x); // convert type to byte array with trailing null
NumOut(0, LCD_LINE1, x);
for (int i=0; i < ArrayLen(data)-1; i++)
{
#ifdef __ENHANCED_FIRMWARE
TextOut(0, LCD_LINE2-8*i, FormatNum("0x%2.2x", data[i]));
#else
NumOut(0, LCD_LINE2-8*i, data[i]);
#endif
}
Wait(SEC_4);
}
The best way to get help with LEGO MINDSTORMS and the NXT and Not eXactly C is via the mindboards forums at http://forums.mindboards.net/
Question originally tagged c; this answer may not be applicable to Not eXactly C.
What is the problem with this:
int value;
char bytes[sizeof(int)];
bytes[0] = (value >> 0) & 0xFF;
bytes[1] = (value >> 8) & 0xFF;
bytes[2] = (value >> 16) & 0xFF;
bytes[3] = (value >> 24) & 0xFF;
You can regard it as an unrolled loop. The shift by zero could be omitted; the optimizer would certainly do so. Even though the result of right-shifting a negative value is not defined, there is no problem because this code only accesses the bits where the behaviour is defined.
This code gives the bytes in a little-endian order - the least-significant byte is in bytes[0]. Clearly, big-endian order is achieved by:
int value;
char bytes[sizeof(int)];
bytes[3] = (value >> 0) & 0xFF;
bytes[2] = (value >> 8) & 0xFF;
bytes[1] = (value >> 16) & 0xFF;
bytes[0] = (value >> 24) & 0xFF;
I need to find whether a number is divisible by 3 without using %, / or *. The hint given was to use atoi() function. Any idea how to do it?
The current answers all focus on decimal digits, when applying the "add all digits and see if that divides by 3". That trick actually works in hex as well; e.g. 0x12 can be divided by 3 because 0x1 + 0x2 = 0x3. And "converting" to hex is a lot easier than converting to decimal.
Pseudo-code:
int reduce(int i) {
if (i > 0x10)
return reduce((i >> 4) + (i & 0x0F)); // Reduces 0x102 to 0x12 to 0x3.
else
return i; // Done.
}
bool isDiv3(int i) {
i = reduce(i);
return i==0 || i==3 || i==6 || i==9 || i==0xC || i == 0xF;
}
[edit]
Inspired by R, a faster version (O log log N):
int reduce(unsigned i) {
if (i >= 6)
return reduce((i >> 2) + (i & 0x03));
else
return i; // Done.
}
bool isDiv3(unsigned i) {
// Do a few big shifts first before recursing.
i = (i >> 16) + (i & 0xFFFF);
i = (i >> 8) + (i & 0xFF);
i = (i >> 4) + (i & 0xF);
// Because of additive overflow, it's possible that i > 0x10 here. No big deal.
i = reduce(i);
return i==0 || i==3;
}
Subtract 3 until you either
a) hit 0 - number was divisible by 3
b) get a number less than 0 - number wasn't divisible
-- edited version to fix noted problems
while n > 0:
n -= 3
while n < 0:
n += 3
return n == 0
Split the number into digits. Add the digits together. Repeat until you have only one digit left. If that digit is 3, 6, or 9, the number is divisible by 3. (And don't forget to handle 0 as a special case).
While the technique of converting to a string and then adding the decimal digits together is elegant, it either requires division or is inefficient in the conversion-to-a-string step. Is there a way to apply the idea directly to a binary number, without first converting to a string of decimal digits?
It turns out, there is:
Given a binary number, the sum of its odd bits minus the sum of its even bits is divisible by 3 iff the original number was divisible by 3.
As an example: take the number 3726, which is divisible by 3. In binary, this is 111010001110. So we take the odd digits, starting from the right and moving left, which are [1, 1, 0, 1, 1, 1]; the sum of these is 5. The even bits are [0, 1, 0, 0, 0, 1]; the sum of these is 2. 5 - 2 = 3, from which we can conclude that the original number is divisible by 3.
A number divisible by 3, iirc has a characteristic that the sum of its digit is divisible by 3. For example,
12 -> 1 + 2 = 3
144 -> 1 + 4 + 4 = 9
The interview question essentially asks you to come up with (or have already known) the divisibility rule shorthand with 3 as the divisor.
One of the divisibility rule for 3 is as follows:
Take any number and add together each digit in the number. Then take that sum and determine if it is divisible by 3 (repeating the same procedure as necessary). If the final number is divisible by 3, then the original number is divisible by 3.
Example:
16,499,205,854,376
=> 1+6+4+9+9+2+0+5+8+5+4+3+7+6 sums to 69
=> 6 + 9 = 15 => 1 + 5 = 6, which is clearly divisible by 3.
See also
Wikipedia/Divisibility rule - has many rules for many divisors
Given a number x.
Convert x to a string. Parse the string character by character. Convert each parsed character to a number (using atoi()) and add up all these numbers into a new number y.
Repeat the process until your final resultant number is one digit long. If that one digit is either 3,6 or 9, the origional number x is divisible by 3.
My solution in Java only works for 32-bit unsigned ints.
static boolean isDivisibleBy3(int n) {
int x = n;
x = (x >>> 16) + (x & 0xffff); // max 0x0001fffe
x = (x >>> 8) + (x & 0x00ff); // max 0x02fd
x = (x >>> 4) + (x & 0x000f); // max 0x003d (for 0x02ef)
x = (x >>> 4) + (x & 0x000f); // max 0x0011 (for 0x002f)
return ((011111111111 >> x) & 1) != 0;
}
It first reduces the number down to a number less than 32. The last step checks for divisibility by shifting the mask the appropriate number of times to the right.
You didn't tag this C, but since you mentioned atoi, I'm going to give a C solution:
int isdiv3(int x)
{
div_t d = div(x, 3);
return !d.rem;
}
bool isDiv3(unsigned int n)
{
unsigned int n_div_3 =
n * (unsigned int) 0xaaaaaaab;
return (n_div_3 < 0x55555556);//<=>n_div_3 <= 0x55555555
/*
because 3 * 0xaaaaaaab == 0x200000001 and
(uint32_t) 0x200000001 == 1
*/
}
bool isDiv5(unsigned int n)
{
unsigned int n_div_5 =
i * (unsigned int) 0xcccccccd;
return (n_div_5 < 0x33333334);//<=>n_div_5 <= 0x33333333
/*
because 5 * 0xcccccccd == 0x4 0000 0001 and
(uint32_t) 0x400000001 == 1
*/
}
Following the same rule, to obtain the result of divisibility test by 'n', we can :
multiply the number by 0x1 0000 0000 - (1/n)*0xFFFFFFFF
compare to (1/n) * 0xFFFFFFFF
The counterpart is that for some values, the test won't be able to return a correct result for all the 32bit numbers you want to test, for example, with divisibility by 7 :
we got 0x100000000- (1/n)*0xFFFFFFFF = 0xDB6DB6DC
and 7 * 0xDB6DB6DC = 0x6 0000 0004,
We will only test one quarter of the values, but we can certainly avoid that with substractions.
Other examples :
11 * 0xE8BA2E8C = A0000 0004, one quarter of the values
17 * 0xF0F0F0F1 = 10 0000 0000 1
comparing to 0xF0F0F0F
Every values !
Etc., we can even test every numbers by combining natural numbers between them.
A number is divisible by 3 if all the digits in the number when added gives a result 3, 6 or 9. For example 3693 is divisible by 3 as 3+6+9+3 = 21 and 2+1=3 and 3 is divisible by 3.
inline bool divisible3(uint32_t x) //inline is not a must, because latest compilers always optimize it as inline.
{
//1431655765 = (2^32 - 1) / 3
//2863311531 = (2^32) - 1431655765
return x * 2863311531u <= 1431655765u;
}
On some compilers this is even faster then regular way: x % 3. Read more here.
well a number is divisible by 3 if all the sum of digits of the number are divisible by 3. so you could get each digit as a substring of the input number and then add them up. you then would repeat this process until there is only a single digit result.
if this is 3, 6 or 9 the number is divisable by 3.
Here is a pseudo-algol i came up with .
Let us follow binary progress of multiples of 3
000 011
000 110
001 001
001 100
001 111
010 010
010 101
011 000
011 011
011 110
100 001
100 100
100 111
101 010
101 101
just have a remark that, for a binary multiple of 3 x=abcdef in following couples abc=(000,011),(001,100),(010,101) cde doest change , hence, my proposed algorithm:
divisible(x):
y = x&7
z = x>>3
if number_of_bits(z)<4
if z=000 or 011 or 110 , return (y==000 or 011 or 110) end
if z=001 or 100 or 111 , return (y==001 or 100 or 111) end
if z=010 or 101 , return (y==010 or 101) end
end
if divisible(z) , return (y==000 or 011 or 110) end
if divisible(z-1) , return (y==001 or 100 or 111) end
if divisible(z-2) , return (y==010 or 101) end
end
C# Solution for checking if a number is divisible by 3
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
int num = 33;
bool flag = false;
while (true)
{
num = num - 7;
if (num == 0)
{
flag = true;
break;
}
else if (num < 0)
{
break;
}
else
{
flag = false;
}
}
if (flag)
Console.WriteLine("Divisible by 3");
else
Console.WriteLine("Not Divisible by 3");
Console.ReadLine();
}
}
}
Here is your optimized solution that every one should know.................
Source: http://www.geeksforgeeks.org/archives/511
#include<stdio.h>
int isMultiple(int n)
{
int o_count = 0;
int e_count = 0;
if(n < 0)
n = -n;
if(n == 0)
return 1;
if(n == 1)
return 0;
while(n)
{
if(n & 1)
o_count++;
n = n>>1;
if(n & 1)
e_count++;
n = n>>1;
}
return isMultiple(abs(o_count - e_count));
}
int main()
{
int num = 23;
if (isMultiple(num))
printf("multiple of 3");
else
printf(" not multiple of 3");
return 0;
}
Since I am new to cuda .. I need your kind help
I have this long vector, for each group of 24 elements, I need to do the following:
for the first 12 elements, the even numbered elements are multiplied by -1,
for the second 12 elements, the odd numbered elements are multiplied by -1 then the following swap takes place:
Graph: because I don't yet have enough points, I couldn't post the image so here it is:
http://www.freeimagehosting.net/image.php?e4b88fb666.png
I have written this piece of code, and wonder if you could help me further optimize it to solve for divergence or bank conflicts ..
//subvector is a multiple of 24, Mds and Nds are shared memory
____shared____ double Mds[subVector];
____shared____ double Nds[subVector];
int tx = threadIdx.x;
int tx_mod = tx ^ 0x0001;
int basex = __umul24(blockDim.x, blockIdx.x);
Mds[tx] = M.elements[basex + tx];
__syncthreads();
// flip the signs
if (tx < (tx/24)*24 + 12)
{
//if < 12 and even
if ((tx & 0x0001)==0)
Mds[tx] = -Mds[tx];
}
else
if (tx < (tx/24)*24 + 24)
{
//if >12 and < 24 and odd
if ((tx & 0x0001)==1)
Mds[tx] = -Mds[tx];
}
__syncthreads();
if (tx < (tx/24)*24 + 6)
{
//for the first 6 elements .. swap with last six in the 24elements group (see graph)
Nds[tx] = Mds[tx_mod + 18];
Mds [tx_mod + 18] = Mds [tx];
Mds[tx] = Nds[tx];
}
else
if (tx < (tx/24)*24 + 12)
{
// for the second 6 elements .. swp with next adjacent group (see graph)
Nds[tx] = Mds[tx_mod + 6];
Mds [tx_mod + 6] = Mds [tx];
Mds[tx] = Nds[tx];
}
__syncthreads();
Thanks in advance ..
paul gave you pretty good starting points you previous questions.
couple things to watch out for: you are doing non-base 2 division which is expensive.
Instead try to utilize multidimensional nature of the thread block. For example, make the x-dimension of size 24, which will eliminate need for division.
in general, try to fit thread block dimensions to reflect your data dimensions.
simplify sign flipping: for example, if you do not want to flip sign, you can still multiplied by identity 1. Figure out how to map even/odd numbers to 1 and -1 using just arithmetic: for example sign = (even*2+1) - 2 where even is either 1 or 0.