i m new to yosys and arachne-pnr. Here is the snippet from .asc and .icebox_explain. I could not understand how the bits of LC_5 are derived from .logic_tile 1 11.
example.v
module top (input a, b, output y);
assign y = a & b;
endmodule
example.asc
.logic_tile 1 11
000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000
000000000000001000000000000000000000000000000000000000
000000000000000101000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000
000000000000000000000000000000011000110000000000000000
000000000000000000000000000000001110110000000000000000
000000000000000000000000000000000000000000000000000000
000000000000001111000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000
icebox_example.asc
.logic_tile 1 11
LC_5 0000000000001111 0000
buffer local_g0_4 lutff_5/in_3
buffer local_g3_0 lutff_5/in_2
buffer neigh_op_lft_4 local_g0_4
buffer sp4_h_r_24 local_g3_0
How, LC_5 0000000000001111 0000 is generated from this stream?
Given in documentation:
LC_5 B10[36] B10[37] B10[38] B10[39] B10[40] B10[41] B10[42] B10[43] B10[44] B10[45] B11[36] B11[37] B11[38] B11[39] B11[40] B11[41] B11[42] B11[43] B11[44] B11[45]
I guess it should be:
B10-36:45=1100000000
B11-36:45=1100000000
which is not correct. Can you please help and guide?
LC_5 isn't the LC_ bits directly, but the 16 LUT init bits followed by 4 flipflop config bits.
The mapping from these to LC_ bit index is another step, also described in http://www.clifford.at/icestorm/logic_tile.html right at the bottom.
Related
In 8 bits we can store 8 numbers from 0 to 1 each. We can also say that we can store 8 different piece of data in a range from 0 to 255.
0/1 0/1 0/1 0/1 0/1 0/1 0/1 0/1 → 8 different piece of data
↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓
[bit] [bit] [bit] [bit] [bit] [bit] [bit] [bit] → 8 bits total
If instead of storing 0 or 1 we need to store 0, 1 or 2, then we will end up using more bits, of course. In this condition, by the way, we can store 0, 1, 2 or 3. However, this is beyond what I need but technically it will use the same amount of data space.
0/1/2 0/1/2 0/1/2 0/1/2 → 4 different piece of data (expectative)
0/1/2/3 0/1/2/3 0/1/2/3 0/1/2/3 → 4 different piece of data (too much for me)
↓ ↓ ↓ ↓
[2bits] [2bits] [2bits] [2bits] → 8 bits total
We can agree that we are "losing" storage capacity. The solution is to do a radix conversion (I don't know if that's the right term) using the numerical base of 3, this way we will achieve the objective of storing exclusively 0, 1 or 2, and in the same 8 bits we can store more information (and, to be honest, I don't know how the bits organize themselves for this to work).
Examples:
00000₃ → int 0
01212₃ → int 50
11111₃ → int 121
12121₃ → int 151
22222₃ → int 242 (max)
In this conversion, we were able to store 5 pieces of information composed of 0, 1 or 2, instead of just 4. And there's still a "space" left, and that's what I'd like to talk about.
I was wondering if there is any way to store mixed base data instead of fixed base (eg. 3, as exemplified above). To be clearer, supposing that I wanted to store two pieces of data composed of 0, 1 or 2 each, and one piece of data composed of a number between 0 and 8. In binary, and still limited to 256, we can do it as follows:
0/1/2 0/1/2 0/1/2/3/4/5/6/7/8 → 3 piece of different data
↓ ↓ ↓
[2bits] [2bits] [ 4 bits ] → 8 bits total
But, it seems to me that we have empty "spaces" left again, since it doesn't seem to me to be using the full capacity possible in 8 bits, and maybe it's still possible to add more data if we use number base conversion instead.
The problem is: I don't know how to handle the data in this way, in a way to reliably merge and split it again.
In the real world: I need to transfer a lot of data that uses very little information (eg numbers 0 to 2, 0 to 5, 0 to 10, etc.). And I'm currently doing bitwise snapping of this data. And in my case, any optimized byte is a very nice gain (if this is really possible, maybe there is a 20~40% data saving).
I understand that it might consume more processing on rebasing, merge and split conversions, but that won't be an issue, because this processing will be done on the client side (merge when sending and split when receiving), and the server manages the same data already optimized (no additional processing).
--
Possible solutions:
Let's imagine that I have a set of three numbers that can be 0, 1 or 2, and another set of three numbers that can go from 0 to 8 (so it is 9 possibilities each). The goal is to merge all these numbers into a single integer (which should use about 2 bytes at best current method).
Solution 1: each number being stored in a single complete byte:
byte A from 0 to 2
byte B from 0 to 2
byte C from 0 to 2
byte D from 0 to 8
byte E from 0 to 8
byte F from 0 to 8
Problem: it will be necessary to consume 6 bytes to store these 6 numbers.
Solution 2: storage through bits:
2 bits to store A
2 bits to store B
2 bits to store C
4 bits to store F
4 bits to store G
4 bits to store H
6 unused bits to complete the last byte
Problems: in addition to 2 bits being "more than necessary" to store numbers from 0 to 2 (A, B and C), and same for 4 bits from 0 to 8, we will use a total of 3 bytes and still have 6 "dead" (unused) bits at end to complete the last byte.
Solution 3: Separate into two sets and convert their respective bases individually:
A, B and C to base 3 consumes 1 byte
D, E and F to base 9 consumes 2 bytes
Problem: despite being a great solution at the moment (and despite the example above, in more complex situations it can be more optimized than solution 2), and that's what I'm using, I believe there's a lot of "left over" space in this union, and maybe it's still possible to squeeze everything into 2 bytes only.
For example, the conversion of 222₃ consumes up to 5 bits, which means that in this byte we still have 3 unused bits. The 888₉ conversion consumes 10 bits. This makes me see the possibility of using only 15 bits with only 1 unused bit (2 bytes).
Then we come to the next solution.
Solution 4: move the bits to further optimize space:
higher bits stores A, B and C
lower bits stores D, E and F
Example:
[5 bits of numbers of base 3] +
[10 bits of numbers of base 9] +
[1 unused bit]
Problem: currently this solution is even better than the one I am currently using. However, I still see a possibility to improve in situations where more information may be available through number base conversion and union.
You pack your values in the following way:
Example:
possible values for
a = [0,2] -> 3 states
b = [0,2] -> 3 states
c = [0,8] -> 9 states
lets assume you have following values
a := 1
b := 2
c := 8
than you can calculate the final number the following way
int number = 0;
int multi = 1;
number = number.add(multi.multiply(a));
multi = multi.multiply(3));
number = number.add(multi.multiply(b));
multi = multi.multiply(3));
number = number.add(multi.multiply(c));
number holds now your packed numbers
to unpack you just need todo
a = number.mod(3)
number = number.divide(3)
b = number.mod(3)
number = number.divide(3)
c = number.mod(9)
for example
> SETBIT bitmapsarestrings 2 1
> SETBIT bitmapsarestrings 3 1
> SETBIT bitmapsarestrings 5 1
> SETBIT bitmapsarestrings 10 1
> SETBIT bitmapsarestrings 11 1
> SETBIT bitmapsarestrings 14 1
> GET bitmapsarestrings
"42"
Binary storage should not like this: 0010 0110 0001 1100 ?
stored in this way, why value is 42?
These SETBIT operations will make the value as a binary string, whose length is 2 bytes or 16 bits. After the settings, the value will be 0b 00110100 00110010 in binary format.
The first byte (0b 00110100) is 52, which is the ascii code of '4', and the second byte (0b 00110010) is 50, which is the ascii code of '2'. So when you get the value of the string, it returns "42".
What #for_stack said, or just refer to the lines immediately above that example (https://redis.io/commands/setbit):
Bitmaps are not an actual data type, but a set of bit-oriented operations defined on the String type (for more information refer to the Bitmaps section of the Data Types Introduction page). This means that bitmaps can be used with string commands, and most importantly with SET and GET.
Because Redis' strings are binary-safe, a bitmap is trivially encoded as a bytes stream. The first byte of the string corresponds to offsets 0..7 of the bitmap, the second byte to the 8..15 range, and so forth.
I am currently putting an engine into another car and I want to keep the fuel economy calulation inside the boardcomputer working. I managed to recode this part sucessfully, but I have been trying to figure out the (simple?) two byte dataformat they used without success. I assume it is fixed point notation, but no matter how I shift it around, it does not line up. How do the two bytes represent the right number?
Some examples:
Bytes (Dec) --> Result
174,10 -> 2,67
92,11 -> 2,84
128,22 -> 3,75
25,29 -> 4,85
225,23 -> 3,98
00,40 -> 5,00
128,34 -> 5,75
Here's a partial solution:
First, swap the bytes. Then join them:
The result (in hex) is:
0AAE
0B5C
1680
1D19
17E1
2800
2280
Then split the into the first digit (4 bits), the remaining three digits (12 bits) and keep the entire number (16 bits) as well. The result (in decimal) is:
0 2734 2734
0 2908 2908
1 1664 5760
1 3353 7449
1 2017 6113
2 2048 10240
2 640 8832
The first digits seems to be a multiplication factor. 0 stands for 1024, 1 for 1536, 2 for 2048. The formula possibly is f = (1024 + n * 512).
Now divide the entire number by the multiplication factor. The result, rounded to two decimal places, is:
2734 / 1024 = 2.67
2908 / 1024 = 2.84
5760 / 1536 = 3.75
7449 / 1536 = 4.85
6113 / 1536 = 3.98
10240 / 2048 = 5.00
8832 / 2048 = 4.31
It works for all except the last number, which might contain a mistake.
So it seems to be some sort of floating-point number but I don't recoginze the specific format. Possibly, there is a simpler formula the explains the number.
I changed the last 4 bits of all pixels in a png image to zero. Why was the size of image reduced to less than half?. I believe png is lossless compression. Will I be able to retrieve the embedded data.(here I added 0000 to all pixels, but in actual case i need to add different data) I'm using vb 2015. The code I used is attached below.
Dim image1 As Bitmap = New Bitmap("C:\STEG\originalimage.png")
For i As Integer = 0 To image1.Height - 1
For j As Integer = 0 To image1.Width - 1
Dim value As Color = image1.GetPixel(j, i)
Dim t1 As Byte = value.R
Dim t2 As Byte = value.G
Dim t3 As Byte = value.B
t1 = t1 And 240
t2 = t2 And 240
t3 = t3 And 240
image1.SetPixel(j, i, Color.FromArgb(255, t1, t2, t3))
Next
Next
image1.Save("C:\STEG\originalimageedited.png")
PNG is a lossless image compression format, yes. However if it's lossless or not doesn't anything have to do with the great reduction in size. What affects it is how the compression algorithm works.
I don't know how PNG's algorithm work, but basically compression is about repeating data. For instance if the last 4 bits of every byte is 0000, like so:
1000 0000
0000 0000
1101 0000
0110 0000
...then a compression algorithm could for example express it as:
repeat 0000 every 4 bits
or:
1000 4-0
4-0 4-0
1101 4-0
0110 4-0
(4-0 indicates that it should write four '0' bits)
which might only take up a few bytes in the compressed file, while in the uncompressed file these four bits make up half of every byte.
Thus if you have 1 KB = 1024 bytes of color data and compress it like the first example above, the file size will (roughly speaking) shrink to 512 bytes + the few bytes that indicate the repeat.
For Objective-C:
Hi everyone, I'm trying to convert a hex input into binary. For example, someone enters in :
A55
I want that to convert to
101001010101
I've tried looking through past posts and none seem to be working for me. Any help would be greatly appreciated.
Use a lookup table: there are only 16 possible characters in a HEX representation, each corresponding to a four-character binary code group. Go through the HEX character-by-character, obtain a lookup, and put it in the resultant NSString.
Here is a copy of the lookup table for you.
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
A 1010
B 1011
C 1100
D 1101
E 1110
F 1111
There are multiple options as to how to do lookups. The simplest way would be making a 128-element array, and placing NSStrings at the elements corresponding to codes of the characters (i.e. at positions '0', '1', ..., 'E', 'F', with single quotes; these are very important).
I believe there is a built-in function for this. If not, you should at least be able to go hex->dec then dec->bin
You can write the conversion from scratch if you know the number of characters, bin to hex is common enough algorithmically.
A mathematical look at the algorithms
SO Answers in C/C++ Another
Base 10 to base n in Objective C
C Hex->Bin
Build a lookup table (an array where you can supply a value between 0 and 15 to get the binary for that hex digit):
char *hex_to_bin[] = {
"0000", "0001", "0010", "0011",
/* ... */
"1100", "1101", "1110", "1111"
};
There should be 16 elements in that table. The conversion process for multiple digits is to handle one digit at a time, appending the results onto the end of your result storage.
Use getchar() to read a char:
int c = getchar();
if (c < 0) { puts("Error: Invalid input or premature closure."); }
Use strchr() to determine which array index to retrieve:
char *digits = "00112233445566778899AaBbCcDdEeFf";
size_t digit = (strchr(digits, c) - digits) / 2;
Look up the corresponding binary values for digit:
printf("%s", hex_to_bin[digit]); // You'll want to use strcat here.