I've implemented the code here in C# to get the MSB of an int. I'm not certain what I need to do with the log reference table and the main code to extend the code to 64 bit.
The only thing the text says is it will take 2 more CPU ops, so I deduce the change is minor.
The table does not need to be changed. One more level of if() is needed:
if (ttt = v >> 32)
{
if (tt = ttt >> 16)
r = (t = tt >> 8) ? 56 + LogTable256[t] : 48 + LogTable256[tt]
else
r = (t = ttt >> 8) ? 40 + LogTable256[t] : 32 + LogTable256[ttt]
}
else
{
if (tt = v >> 16)
r = (t = tt >> 8) ? 24 + LogTable256[t] : 16 + LogTable256[tt];
else
r = (t = v >> 8) ? 8 + LogTable256[t] : LogTable256[v];
}
Related
I have an img in binary format format.
Each pixel is 10bit.
They are putted consecutively, like [9-2] in first byte and [1:0] in second byte.
Where 9 IS MSB bit, 0 - LSB.
How to convert them to 16bit array?
E.g:
Store like 8 bit:
0b10000011,
0b10101100,
0b10011001,
0b11000000,
0b10101000,
Actual data which contains real pixels and which I want convert to 16bit:
0b1000001110,
0b1011001001,
0b1001110000,
0b0010101000,
It seems you want to convert data like this
img[0] = AAAAAAAA
img[1] = AABBBBBB
img[2] = BBBBCCCC
img[3] = CCCCCCDD
img[4] = DDDDDDDD
to data like this:
array[0] = 000000AAAAAAAAAA
array[1] = 000000BBBBBBBBBB
array[2] = 000000CCCCCCCCCC
array[3] = 000000DDDDDDDDDD
It can be done like this:
array[0] = ((img[0] << 2) | (img[1] >> 6)) & 0x3ff;
array[1] = ((img[1] << 4) | (img[2] >> 4)) & 0x3ff;
array[2] = ((img[2] << 6) | (img[3] >> 2)) & 0x3ff;
array[3] = ((img[3] << 8) | img[4] ) & 0x3ff;
To convert multiple blocks, for loop is useful like this:
for (int i = 0; (i + 1) * 5 <= num_of_bytes_in_img; i++) {
array[i * 4 + 0] = ((img[i * 5 + 0] << 2) | (img[i * 5 + 1] >> 6)) & 0x3ff;
array[i * 4 + 1] = ((img[i * 5 + 1] << 4) | (img[i * 5 + 2] >> 4)) & 0x3ff;
array[i * 4 + 2] = ((img[i * 5 + 2] << 6) | (img[i * 5 + 3] >> 2)) & 0x3ff;
array[i * 4 + 3] = ((img[i * 5 + 3] << 8) | img[i * 5 + 4] ) & 0x3ff;
}
(this loop won't convert last a few bytes, so you should treat them separately or add padding to img not to read out-of-range while using this loop to end of the image)
I've seen the answer here: http://clc-wiki.net/wiki/K%26R2_solutions:Chapter_2:Exercise_6
and i've tested the first, but in this part:
x = 29638;
y = 999;
p = 10;
n = 8;
return (x & ((~0 << (p + 1)) | (~(~0 << (p + 1 - n)))))
in a paper it give to me a 6, but in the program it return 28678...
in this part:
111001111000110
&000100000000111
in the result, the left-most three bits has to be 1's like in x but the bitwise operator & says:
The output of bitwise AND is 1 if the corresponding bits of all operands is 1. If either bit of an operand is 0, the result of corresponding bit is evaluated to 0.
so why it returns the number with thats 3 bits in 1?
Here we go, one step at a time (using 16-bit numbers). We start with:
(x & ((~0 << (p + 1)) | (~(~0 << (p + 1 - n)))))
Substituting in numbers (in decimal):
(29638 & ((~0 << (10 + 1)) | (~(~0 << (10 + 1 - 8)))))
Totalling up the bit shift amounts gives:
(29638 & ((~0 << 11) | (~(~0 << 3))))
Rewriting numbers as binary and applying the ~0s...
(0111001111000110 & ((1111111111111111 << 1011) | (~(1111111111111111 << 0011))))
After performing the shifts we get:
(0111001111000110 & (1111100000000000 | (~ 1111111111111000)))
Applying the other bitwise-NOT (~):
(0111001111000110 & (1111100000000000 | 0000000000000111))
And the bitwise-OR (|):
0111001111000110 & 1111100000000111
And finally the bitwise-AND (&):
0111000000000110
So we then have binary 0111000000000110, which is 2 + 4 + 4096 + 8192 + 16384, which is 28678.
All of the bit shuffling algorithms I've found deal with 16-bit or 32-bit, which means that even if I use only the first 25-bits of an int, the shuffle will leave bits outside. This function is in an inner loop of a CPU-intensive process so I'd prefer it to be as fast as possible. I've tried modifying the code of the Hacker's Delight 32-bit shuffle algorithm
x = (x & 0x0000FF00) << 8 | (x >> 8) & 0x0000FF00 | x & 0xFF0000FF;
x = (x & 0x00F000F0) << 4 | (x >> 4) & 0x00F000F0 | x & 0xF00FF00F;
x = (x & 0x0C0C0C0C) << 2 | (x >> 2) & 0x0C0C0C0C | x & 0xC3C3C3C3;
x = (x & 0x22222222) << 1 | (x >> 1) & 0x22222222 | x & 0x99999999;
but am having difficulty in doing some partly because I'm not sure where the masks come from. I tried shifting the number and re-shuffling but so far the results are all for naught. Any help would be GREATLY appreciated!
(I am using C but I can convert an algorithm from another language)
First, for the sake of evenness, we can extend the problem to a 26-bit shuffle by remembering that bit 25 will appear at the end of the interleaved list, so we can trim it off after the interleaving operation without affecting the positions of the other bits.
Now we want to interleave the first and second sets of 13 bits; but we only have an algorithm to interleave the first and second sets of 16 bits.
A straightfoward approach might be to just move the high and low parts of x into more workable positions before applying the standard algorithm:
x = (x & 0x1ffe000) << 3 | x & 0x00001fff;
x = (x & 0x0000FF00) << 8 | (x >> 8) & 0x0000FF00 | x & 0xFF0000FF;
x = (x & 0x00F000F0) << 4 | (x >> 4) & 0x00F000F0 | x & 0xF00FF00F;
x = (x & 0x0C0C0C0C) << 2 | (x >> 2) & 0x0C0C0C0C | x & 0xC3C3C3C3;
x = (x & 0x22222222) << 1 | (x >> 1) & 0x22222222 | x & 0x99999999;
The zeroes at the top of each half will be interleaved and appear at the top of the result.
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Closed 10 years ago.
Possible Duplicate:
Finding consecutive bit string of 1 or 0
Is it possible to count, from left, consecutive 1's in an integer?
So: total number of consecutive set bits starting from the top bit.
Using only:
! ~ & ^ | + << >>
-1= 0xFFFFFFFF would return 32
0xFFF0F0F0 would return 12 (FFF = 111111111111)
No loops, unfortunately.
Can assume the machine:
Uses 2s complement, 32-bit representations of integers.
Performs right shifts arithmetically.
Has unpredictable behavior when shifting an integer by more
than the word size.
I'm forbidden to:
Use any control constructs such as if, do, while, for, switch, etc.
Define or use any macros.
Define any additional functions in this file.
Call any functions.
Use any other operations, such as &&, ||, -, or ?:
Use any form of casting.
Use any data type other than int. This implies that you
cannot use arrays, structs, or unions.
I've looked at
Finding consecutive bit string of 1 or 0
It's using loops, which I can't use. I don't even know where to start.
(Yes, this is an assignment, but I'm simply asking those of you skilled enough for help. I've done pretty much all of those I need to do, but this one just won't work.)
(For those downvoting simply because it's for school:
FAQ:
1 a specific programming problem, check
2 However, if your motivation is “I would like others to explain ______ to me”, then you are probably OK.)
You can do it like this:
int result = clz(~x);
i.e. invert all the bits and then count leading zeroes.
clz returns the number of leading zero bits (also commonly known as ffs or nlz) - see here for implementation details: http://en.wikipedia.org/wiki/Find_first_set#Algorithms
Here you are. The function argument may be signed or unsigned. The alg is independent on signedness.
int leftmost_ones(int x)
{
x = ~x;
x = x | x >> 1 | x >> 2 | x >> 3 | x >> 4 | x >> 5 | x >> 6 | x >> 7 |
x >> 8 | x >> 9 | x >> 10 | x >> 11 | x >> 12 | x >> 13 | x >> 14 |
x >> 15 | x >> 16 | x >> 17 | x >> 18 | x >> 19 | x >> 20 | x >> 21 |
x >> 22 | x >> 23 | x >> 24 | x >> 25 | x >> 26 | x >> 27 | x >> 28 |
x >> 29 | x >> 30 | x >> 31;
x = ~x;
return (x & 1) + (x >> 1 & 1) + (x >> 2 & 1) + (x >> 3 & 1) + (x >> 4 & 1) +
(x >> 5 & 1) + (x >> 6 & 1) + (x >> 7 & 1) + (x >> 8 & 1) + (x >> 9 & 1) +
(x >> 10 & 1) + (x >> 11 & 1) + (x >> 12 & 1) + (x >> 13 & 1) + (x >> 14 & 1) +
(x >> 15 & 1) + (x >> 16 & 1) + (x >> 17 & 1) + (x >> 18 & 1) + (x >> 19 & 1) +
(x >> 20 & 1) + (x >> 21 & 1) + (x >> 22 & 1) + (x >> 23 & 1) + (x >> 24 & 1) +
(x >> 25 & 1) + (x >> 26 & 1) + (x >> 27 & 1) + (x >> 28 & 1) + (x >> 29 & 1) +
(x >> 30 & 1) + (x >> 31 & 1);
}
A version with some optimization:
int leftmost_ones(int x)
{
x = ~x;
x |= x >> 16;
x |= x >> 8;
x |= x >> 4;
x |= x >> 2;
x |= x >> 1;
x = ~x;
return (x & 1) + (x >> 1 & 1) + (x >> 2 & 1) + (x >> 3 & 1) + (x >> 4 & 1) +
(x >> 5 & 1) + (x >> 6 & 1) + (x >> 7 & 1) + (x >> 8 & 1) + (x >> 9 & 1) +
(x >> 10 & 1) + (x >> 11 & 1) + (x >> 12 & 1) + (x >> 13 & 1) + (x >> 14 & 1) +
(x >> 15 & 1) + (x >> 16 & 1) + (x >> 17 & 1) + (x >> 18 & 1) + (x >> 19 & 1) +
(x >> 20 & 1) + (x >> 21 & 1) + (x >> 22 & 1) + (x >> 23 & 1) + (x >> 24 & 1) +
(x >> 25 & 1) + (x >> 26 & 1) + (x >> 27 & 1) + (x >> 28 & 1) + (x >> 29 & 1) +
(x >> 30 & 1) + (x >> 31 & 1);
}
Can you use a loop?
int mask = 0x80000000;
int count = 0;
while (number & mask) {
count += 1;
mask >>= 1;
}
I think it's doable, by basically unrolling the typical loop and being generally annoying.
How about this: an expression that is 1 if and only if the answer is 1? I offer:
const int ok1 = !((number & 0xc0000000) - 0x800000000);
The ! and subtraction are to work around that someone broke the == key on our keyboard, of course.
And then, an expression that is 1 if and only if the anwer is 2:
const int ok2 = !((number & 0xe0000000) - 0xc0000000);
If you continue to form these, the final answer is their sum:
const int answer = ok1 + ok2 + ... + ok32;
By the way, I can't seem to remember being given these weirdly restricted assignments when I was in school, I guess times have changed. :)
int count_consecutive_bits(unsigned int x) {
int res = 0;
while (x & 0x80000000) { ++res; x <<= 1; }
return res;
}
~ & ^ | + << >> are the only operations I can use
Before I continue, this is a homework question, I've been stuck on this for a really long time.
My original approach: I thought that !x could be done with two's complement and doing something with it's additive inverse. I know that an xor is probably in here but I'm really at a loss how to approach this.
For the record: I also cannot use conditionals, loops, ==, etc, only the functions (bitwise) I mentioned above.
For example:
!0 = 1
!1 = 0
!anything besides 0 = 0
Assuming a 32 bit unsigned int:
(((x>>1) | (x&1)) + ~0U) >> 31
should do the trick
Assuming x is signed, need to return 0 for any number not zero, and 1 for zero.
A right shift on a signed integer usually is an arithmetical shift in most implementations (e.g. the sign bit is copied over). Therefore right shift x by 31 and its negation by 31. One of those two will be a negative number and so right shifted by 31 will be 0xFFFFFFFF (of course if x = 0 then the right shift will produce 0x0 which is what you want). You don't know if x or its negation is the negative number so just 'or' them together and you will get what you want. Next add 1 and your good.
implementation:
int bang(int x) {
return ((x >> 31) | ((~x + 1) >> 31)) + 1;
}
The following code copies any 1 bit to all positions. This maps all non-zeroes to 0xFFFFFFFF == -1, while leaving 0 at 0. Then it adds 1, mapping -1 to 0 and 0 to 1.
x = x | x << 1 | x >> 1
x = x | x << 2 | x >> 2
x = x | x << 4 | x >> 4
x = x | x << 8 | x >> 8
x = x | x << 16 | x >> 16
x = x + 1
For 32 bit signed integer x
// Set the bottom bit if any bit set.
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
x ^= 1; // Toggle the bottom bit - now 0 if any bit set.
x &= 1; // Clear the unwanted bits to leave 0 or 1.
Assuming e.g. an 8-bit unsigned type:
~(((x >> 0) & 1)
| ((x >> 1) & 1)
| ((x >> 2) & 1)
...
| ((x >> 7) & 1)) & 1
You can just do ~x & 1 because it yields 1 for 0 and 0 for everything else