Is my logic on a bitshifting question right? - javascript

im just about done with an assignment and on the last question I have attempted to do it but I dont know if I used the bitshifting correctly.
For the question, I had to extract the right byte of an integer, then get the first 3 bits, next 3 bits, and last 2 bits of that right byte and assign it to unsigned integer variables.
What I have tried so far is:
int rightmost = (y>>24)&0xFF // to get rightmost byte
int first = (rightmost <<< 1)&0xFF // to get first 3 bits of that byte
int second = (rightmost >>> 3)&0xFF // to get next 3 bits
int third = (rightmost >>> 6)&0xFF // to get last 2 bits
Id just like to know if I am going in the right direction

I'd do this:
var firstByte = y & 0xff;
That's the least-significant byte. If the value in y is 12, all the bits will be in that byte.
Then, to isolate the parts of that byte, you have to use & to chop off all the bits you don't want, and then >> to get the bits into the least-significant positions. The order in which you do that doesn't matter, though it does dictate what you put on the other side of &:
var first3 = firstByte & 0x07; // no need to shift
var second3 = (firstByte >> 3) & 0x07; // shift by 3 and then mask off the rest
var last2 = (firstByte >> 6) & 0x03; // shift by 6 and mask
In binary, 0x07 looks like 00000111. Thus using & with that isolates the least-significant 3 bits in a number.
Test below.
JavaScript is kind-of weird because in between all those operations the language maintains the numbers as 64-bit floating-point values. For integer values however that doesn't really matter, and indeed an optimized runtime may not actually keep the floating point representations around, if it's really smart about things.
var y = 2359; // binary: 100100110111
var firstByte = y & 0xff;
console.log("firstByte: " + firstByte);
var first3 = firstByte & 0x07;
console.log("should be 7: " + first3); // should be 7
var second3 = (firstByte >> 3) & 0x07;
console.log("should be 6: " + second3); // should be 6
var last2 = (firstByte >> 6) & 0x03;
console.log("should be 0: " + last2); // should be 0

Related

How does bitwise AND OR and XOR works on -negative signed integers?

I was just solving random problems on bitwise operators and trying various other combination for making personal notes. And somehow I just cannot figure out the solution.
Say I wanted to check bitwise AND between two integers or on a ~number and -negative number(~num1 & -num2) and various other combo's. Then I can see the answer but I haven't been able to establish how this happened?
Console:
console.log(25 & 3); outputs 1 (I can solve this easily).
console.log(-25 & -3); outputs-27.
Similarly
console.log(~25 & ~3); outputs -28.
console.log(25 & ~3); outputs -24.
console.log(~25 & 3); outputs -2.
console.log(~25 & -3); outputs --28.
console.log(-25 & ~3); outputs --28.
I know the logic behind "console.log(25 & -3)".
25 is 11001
-3 is 11101(3=00011 The minus sign is like 2s compliment+1)
AND-11001 = 25.
But I cannot make it work the same way when both the numbers are negative or with the other cases mentioned above. I have tried various combinations of numbers too, not just these two. But I cannot solve the problem. Can somebody explain the binary logic used in the problems I cannot solve.
(I've spend about 2 hrs here on SO to find the answer and another 1 hr+ on google, but I still haven't found the answer).
Thanks and Regards.
JavaScript specifies that bitwise operations on integers are performed as though they were stored in two's-complement notation. Fortunately, most computer hardware nowadays uses this notation natively anyway.
For brevity's sake I'm going to show the following numbers as 8-bit binary. They're actually 32-bit in JavaScript, but for the numbers in the original question, this doesn't change the outcome. It does, however, let us drop a whole lot of leading bits.
console.log(-25 & -3); //outputs -27. How?
If we write the integers in binary, we get (11100111 & 11111101) respectively. AND those together and you get 11100101, which is -27.
In your later examples, you seem to be using the NOT operator (~) and negation (-) interchangeably. You can't do that in two's complement: ~ and - are not the same thing. ~25 is 11100110, which is -26, not -25. Similarly, ~3 is 11111100, which is -4, not -3.
But when we put these together, we can work out the examples you gave.
console.log(~25 & ~3); //outputs-28. How?
11100110 & 11111100 = 11100100, which is -28 (not 28, as you wrote)
console.log(25 & ~3);//outputs-24. How?
00011001 & 11111100 = 00011000, which is 24
console.log(~25 & 3);//outputs-2. How?
11100110 & 00000011 = 00000001, which is 2
console.log(~25 & -3);//outputs--28. How?
11100110 & 11111101 = 11100100, which is -28
console.log(-25 & ~3);//outputs--28. How?
11100111 & 11111100 = 11100100, which is -28
The real key to understanding this is that you don't really use bitwise operations on integers. You use them on bags of bits of a certain size, and these bags of bits happen to be conveniently representable as integers. This is key to understanding what's going on here, because you've stumbled across a case where the difference matters.
There are specific circumstances in computer science where you can manipulate bags of bits in ways that, by coincidence, give the same results as if you'd done particular mathematical operations on numbers. But this only works in specific circumstances, and they require you to assume certain things about the numbers you're working on, and if your numbers don't fit those assumptions, things break down.
This is one of the reasons Donald Knuth said "premature optimization is the root of all evil". If you want to use bitwise operations in place of actual integer math, you have to be absolutely certain that your inputs will actually follow the assumptions required for that trick to work. Otherwise, the results will start looking strange when you start using inputs outside of those assumptions.
25 = 16+8+1 = 0b011001, I've added another 0 digit as the sign digit. Practically you'll have at least 8 binary digits
but the two's complement math is the same. To get -25 in 6-bits two's complement, you'd do -25 = ~25 + 1=0b100111
3=2+1=0b000011; -3 = ~3+1 = 0b111101
When you & the two, you get:
-25 = ~25 + 1=0b100111
&
-3 = ~3 + 1 = 0b111101
0b100101
The leftmost bit (sign bit) is set so it's a negative number. To find what it's a negative of, you reverse the process and first subtract 1 and then do ~.
~(0b100101-1) = 0b011011
thats 1+2+0*4+8+16 = 27 so -25&-3=-27.
For 25 & ~3, it's:
25 = 16+8+1 = 0b011001
& ~3 = 0b111100
______________________
0b011000 = 24
For ~25 & 3, it's:
~25 = 0b100110
& ~3 = 0b000011
______________________
0b000010 = 2
For ~25 & -3, it's:
~25 = 0b100110
& ~3+1 = 0b111101
______________________
0b100100 #negative
#find what it's a negative of:
~(0b100100-1) =~0b100011 = 0b011100 = 4+8+16 = 28
0b100100 = -28
-27 has 6 binary digits in it so you should be using numbers with at least that many digits. With 8-bit numbers then we have:
00011001 = 25
00000011 = 3
00011011 = 27
and:
11100111 = -25
11111101 = -3
11100101 = -27
Now -25 & -3 = -27 because 11100111 & 11111101 = 11100101
The binary string representation of a 32 bit integer can be found with:
(i >>> 0).toString(2).padStart(32, '0')
The bitwise anding of two binary strings is straightforward
The integer value of a signed, 32 bit binary string is either
parseInt(bitwiseAndString, 2)
if the string starts with a '0', or
-~parseInt(bitwiseAndString, 2) - 1
if it starts with a '1'
Putting all that together:
const tests = [
['-25', '-3'],
['~25', '-3'],
['25', '~3'],
['~25', '3'],
['~25', '~3'],
['-25', '~3']
]
const output = (s,t) => { console.log(`${`${s}:`.padEnd(20, ' ')}${t}`); }
const bitwiseAnd = (i, j) => {
console.log(`Calculating ${i} & ${j}`);
const bitStringI = (eval(i) >>> 0).toString(2).padStart(32, '0');
const bitStringJ = (eval(j) >>> 0).toString(2).padStart(32, '0');
output(`bit string for ${i}`, bitStringI);
output(`bit string for ${j}`, bitStringJ);
const bitArrayI = bitStringI.split('');
const bitArrayJ = bitStringJ.split('');
const bitwiseAndString = bitArrayI.map((s, idx) => s === '1' && bitArrayJ[idx] === '1' ? '1' : '0').join('');
output('bitwise and string', bitwiseAndString);
const intValue = bitwiseAndString[0] === '1' ? -~parseInt(bitwiseAndString, 2) - 1 : parseInt(bitwiseAndString, 2);
if (intValue === (eval(i) & eval(j))) {
console.log(`integer value: ${intValue} ✓`);
} else {
console.error(`calculation failed: ${intValue} !== ${i & j}`);
}
}
tests.forEach(([i, j]) => { bitwiseAnd(i, j); })

compressing a string of 0's and 1's in js

Itroduction
I'm currently working on John Conway's Game of Life in js. I have the game working (view here) and i'm working on extra functionalities such as sharing your "grid / game" to your friends. To do this i'm extracting the value's of the grid (if the cell is alive or dead) into a long string of 0's and 1's.
This string has a variable length since the grid is not always the same size. for example:
grid 1 has a length and width of 30 => so the string's length is 900
grid 2 has a length and width of 50 => so the string's length is 2500
The problem
As you can see these string's of 0's and 1's are way too long to copy around and share.
However hard i try I don't seem to be able to come up with a code that would compress a string this long to a easy to handle one.
Any ideas on how to compress (and decompress) this?
I have considered simply writing down every possible grid option for the gird sizes 1x1 to 100x100 and giving them a key/reference to use as sharable code. Doing that by hand would be madness but maybe any of you has an idea on how to create an algorithm that can do this?
GitHub repository
In case it wasn't already obvious, the string you're trying to store looks like a binary string.
Counting systems
Binary is a number in base-2. This essentially means that there are two characters being used to keep count. Normally we are used to count with base-10 (decimal characters). In computer science the hexadecimal system (base-16) is also widely being used.
Since you're not storing the bits as bits but as bytes (use var a = 0b1100001; if you ever wish to store them like bits) the 'binary' you wish to store just takes as much space as any other random string with the same length.
Since you're using the binary system each position just has 2 possible values. When using the hexadecimal value a single position can hold up to 16 possible values. This is already a big improvement when it comes to storing the data compactly. As an example 0b11111111 and 0xff both represents the decimal number 255.
In your situation that'd shave 6 bytes of every 8 bytes you have to store. In the end you'd be stuck with a string just 1/4th of the length of the original string.
Javascript implementation
Essentially what we want to do is to interpret the string you store as binary and retrieve the hexadecimal value. Luckily JavaScript has built in functionality to achieve stuff like this:
var bin =
'1110101110100011' +
'0000101111100001' +
'1010010101011010' +
'0000110111011111' +
'1111111001010101' +
'0111000011100001' +
'1011010100110001' +
'0111111110010100' +
'0111110110100101' +
'0000111101100111' +
'1100001111011100' +
'0101011100001111' +
'0110011011001101' +
'1000110010001001' +
'1010100010000011' +
'0011110000000000';
var returnValue = '';
for (var i = 0; i < parseInt(bin.length / 8); i++) {
returnValue += parseInt(bin.substr(i*8, 8), 2).toString(16);
}
console.log(bin.length); // Will return 265
console.log(returnValue.length); // Will return 64
We're saying "parse this string and interpret it like a base-2 number and store it as a hexadecimal string".
Decoding is practically the same. Replace all occurrences of the number 8 in the example above with 2 and vice versa.
Please note
A prerequisite for this code to work correctly is that the binary length is dividable by 8. See the following example:
parseInt('00011110', 2).toString(16); // returns '1e'
parseInt('1e', 16).toString(2); // returns '11110'
// Technically both representations still have the same decimal value
When decoding you should add leading zeros until you have a full byte (8 bits).
In case the positions you have to store are not dividable by 8 you can, for example, add padding and add a number to the front of the output string to identify how much positions to strip.
Wait, there's more
To get even shorter strings you can build a lookup table with 265 characters in which you search for the character associated with the specific position. (This works because you're still storing the hexadecimal value as a string.) Sadly neither the ASCII nor the UTF-8 encodings are suited for this as there are blocks with values which have no characters defined.
It may look like:
// Go fill this array until you have 265 values within it.
var lookup = ['A', 'B', 'C', 'D'];
var smallerValue = lookup[0x00];
This way you can have 265 possible values at a single position, AND you have used your byte to the fullest.
Please note that no real compression is happening here. We're rather utilising data types to be used more efficiently for your current use case.
If we make the assumption than the grid contains much more 0's than 1's, you may want to try this simple compression scheme:
convert the binary string to an hexadecimal string
convert '00' sub-strings to 'z' symbol
convert 'zz' sub-strings to 'Z' symbol
we could go further, but let's stop here for the demo
Below is an example with a 16x16 grid:
var bin =
'0000000000000000' +
'0000001000000000' +
'0000011100000000' +
'0000001000000000' +
'0000000000000000' +
'0000000000111000' +
'0000100000111000' +
'0000000000111000' +
'0000000000000000' +
'0000000000000000' +
'0000000010000000' +
'0000000101000000' +
'0000000010000000' +
'0000000000000000' +
'0000100000000000' +
'0000000000000000';
var packed = bin
.match(/(.{4})/g)
.map(function(x) {
return parseInt(x, 2).toString(16);
})
.join('')
.replace(/00/g, 'z')
.replace(/zz/g, 'Z');
This will produce the string "Z02z07z02ZZ380838z38ZZz8z14z08Zz8Zz".
The unpacking process is doing the exact opposite:
var bin = packed
.replace(/Z/g, 'zz')
.replace(/z/g, '00')
.split('')
.map(function(x) {
return ('000' + parseInt(x, 16).toString(2)).substr(-4, 4);
})
.join('');
Note that this code will only work correctly if the length of the input string is a multiple of 4. If it's not the case, you'll have to pad the input and crop the output.
EDIT : 2nd method
If the input is completely random -- with roughly as many 0's as 1's and no specific repeating patterns -- the best you can do is probably to convert the binary string to a BASE64 string. It will be significantly shorter (this time with a fixed compression ratio of about 17%) and can still be copied/pasted by the user.
Packing:
var bin =
'1110101110100011' +
'0000101111100001' +
'1010010101011010' +
'0000110111011111' +
'1111111001010101' +
'0111000011100001' +
'1011010100110001' +
'0111111110010100' +
'0111110110100101' +
'0000111101100111' +
'1100001111011100' +
'0101011100001111' +
'0110011011001101' +
'1000110010001001' +
'1010100010000011' +
'0011110000000000';
var packed =
btoa(
bin
.match(/(.{8})/g)
.map(function(x) {
return String.fromCharCode(parseInt(x, 2));
})
.join('')
);
Will produce the string "66ML4aVaDd/+VXDhtTF/lH2lD2fD3FcPZs2MiaiDPAA=".
Unpacking:
var bin =
atob(packed)
.split('')
.map(function(x) {
return ('0000000' + x.charCodeAt(0).toString(2)).substr(-8, 8);
})
.join('');
Or if you want to go a step further, you can consider using something like base91 instead, for a reduced encoding overhead.
LZ-string
Using LZ-string I was able to compress the "code" quite a bit.
By simply compressing it to base64 like this:
var compressed = LZString.compressToBase64(string)
Decompressing is also just as simple as this:
var decompressed = LZString.decompressFromBase64(compressed)
However the length of this compressed string is still pretty long given that you have about as many 0s as 1s (not given in the example)
example
But the compression does work.
ANSWER
For any of you who are wondering how exactly I ended up doing it, here's how:
First I made sure every string passed in would be padded with leading 0s untill it was devidable by 8. (saving the amount of 0s used to pad, since they're needed while decompressing)
I used Corstian's answer and functions to compress my string (interpreted as binary) into a hexadecimal string. Although i had to make one slight alteration.
Not every binary substring with a lenght of 8 will return exactly 2 hex characters. so for those cases i ended up just adding a 0 in front of the substring. The hex substring will have the same value but it's length will now be 2.
Next up i used a functionality from Arnaulds answer. Taking every double character and replacing it with a single character (one not used in the hexadecimal alphabet to avoid conflict). I did this twice for every hexadecimal character.
For example:
the hex string 11 will become h and hh will become H
01101111 will become 0h0H
Since most grids are gonna have more dead cells then alive ones, I made sure the 0s would be able to compress even further, using Arnaulds method again but going a step further.
00 -> g | gg -> G | GG -> w | ww -> W | WW -> x | xx -> X | XX-> y | yy -> Y | YY -> z | zz -> Z
This resulted in Z representing 4096 (binary) 0s
The last step of the compression was adding the amount of leading 0s in front of the compressed string, so we can shave those off at the end of decompressing.
This is how the returned string looks in the end.
amount of leading 0s-compressed string so a 64*64 empty grid, will result in 0-Z
Decompressing is practically doing everything the other way around.
Firstly splitting the number that represents how many leading 0s we've used as padding from the compressed string.
Then using Arnaulds functionality, turning the further "compressed" characters back into hexadecimal code.
Taking this hex string and turning it back into binary code. Making sure, as Corstian pointed out, that every binary substring will have a length of 8. (ifnot we pad the substrings with leading 0s untill the do, exactly, have a length of 8)
And then the last step is to shave off the leading 0s we've used as padding to make the begin string devidable by 8.
The functions
Function I use to compress:
/**
* Compresses the a binary string into a compressed string.
* Returns the compressed string.
*/
Codes.compress = function(bin) {
bin = bin.toString(); // To make sure the binary is a string;
var returnValue = ''; // Empty string to add our data to later on.
// If the lenght of the binary string is not devidable by 8 the compression
// won't work correctly. So we add leading 0s to the string and store the amount
// of leading 0s in a variable.
// Determining the amount of 'padding' needed.
var padding = ((Math.ceil(bin.length/8))*8)-bin.length;
// Adding the leading 0s to the binary string.
for (var i = 0; i < padding; i++) {
bin = '0'+bin;
}
for (var i = 0; i < parseInt(bin.length / 8); i++) {
// Determining the substring.
var substring = bin.substr(i*8, 8)
// Determining the hexValue of this binary substring.
var hexValue = parseInt(substring, 2).toString(16);
// Not all binary values produce two hex numbers. For example:
// '00000011' gives just a '3' while what we wand would be '03'. So we add a 0 in front.
if(hexValue.length == 1) hexValue = '0'+hexValue;
// Adding this hexValue to the end string which we will return.
returnValue += hexValue;
}
// Compressing the hex string even further.
// If there's any double hex chars in the string it will take those and compress those into 1 char.
// Then if we have multiple of those chars these are compressed into 1 char again.
// For example: the hex string "ff will result in a "v" and "ffff" will result in a "V".
// Also: "11" will result in a "h" and "1111" will result in a "H"
// For the 0s this process is repeated a few times.
// (string with 4096 0s) (this would represent a 64*64 EMPTY grid)
// will result in a "Z".
var returnValue = returnValue.replace(/00/g, 'g')
.replace(/gg/g, 'G')
// Since 0s are probably more likely to exist in our binary and hex, we go a step further compressing them like this:
.replace(/GG/g, 'w')
.replace(/ww/g, 'W')
.replace(/WW/g, 'x')
.replace(/xx/g, 'X')
.replace(/XX/g, 'y')
.replace(/yy/g, 'Y')
.replace(/YY/g, 'z')
.replace(/zz/g, 'Z')
//Rest of the chars...
.replace(/11/g, 'h')
.replace(/hh/g, 'H')
.replace(/22/g, 'i')
.replace(/ii/g, 'I')
.replace(/33/g, 'j')
.replace(/jj/g, 'J')
.replace(/44/g, 'k')
.replace(/kk/g, 'K')
.replace(/55/g, 'l')
.replace(/ll/g, 'L')
.replace(/66/g, 'm')
.replace(/mm/g, 'M')
.replace(/77/g, 'n')
.replace(/nn/g, 'N')
.replace(/88/g, 'o')
.replace(/oo/g, 'O')
.replace(/99/g, 'p')
.replace(/pp/g, 'P')
.replace(/aa/g, 'q')
.replace(/qq/g, 'Q')
.replace(/bb/g, 'r')
.replace(/rr/g, 'R')
.replace(/cc/g, 's')
.replace(/ss/g, 'S')
.replace(/dd/g, 't')
.replace(/tt/g, 'T')
.replace(/ee/g, 'u')
.replace(/uu/g, 'U')
.replace(/ff/g, 'v')
.replace(/vv/g, 'V');
// Adding the number of leading 0s that need to be ignored when decompressing to the string.
returnValue = padding+'-'+returnValue;
// Returning the compressed string.
return returnValue;
}
The function I use to decompress:
/**
* Decompresses the compressed string back into a binary string.
* Returns the decompressed string.
*/
Codes.decompress = function(compressed) {
var returnValue = ''; // Empty string to add our data to later on.
// Splitting the input on '-' to seperate the number of paddin 0s and the actual hex code.
var compressedArr = compressed.split('-');
var paddingAmount = compressedArr[0]; // Setting a variable equal to the amount of leading 0s used while compressing.
compressed = compressedArr[1]; // Setting the compressed variable to the actual hex code.
// Decompressing further compressed characters.
compressed = compressed// Decompressing the further compressed 0s. (even further then the rest of the chars.)
.replace(/Z/g, 'zz')
.replace(/z/g, 'YY')
.replace(/Y/g, 'yy')
.replace(/y/g, 'XX')
.replace(/X/g, 'xx')
.replace(/x/g, 'WW')
.replace(/W/g, 'ww')
.replace(/w/g, 'GG')
.replace(/G/g, 'gg')
.replace(/g/g, '00')
// Rest of chars...
.replace(/H/g, 'hh')
.replace(/h/g, '11')
.replace(/I/g, 'ii')
.replace(/i/g, '22')
.replace(/J/g, 'jj')
.replace(/j/g, '33')
.replace(/K/g, 'kk')
.replace(/k/g, '44')
.replace(/L/g, 'll')
.replace(/l/g, '55')
.replace(/M/g, 'mm')
.replace(/m/g, '66')
.replace(/N/g, 'nn')
.replace(/n/g, '77')
.replace(/O/g, 'oo')
.replace(/o/g, '88')
.replace(/P/g, 'pp')
.replace(/p/g, '99')
.replace(/Q/g, 'qq')
.replace(/q/g, 'aa')
.replace(/R/g, 'rr')
.replace(/r/g, 'bb')
.replace(/S/g, 'ss')
.replace(/s/g, 'cc')
.replace(/T/g, 'tt')
.replace(/t/g, 'dd')
.replace(/U/g, 'uu')
.replace(/u/g, 'ee')
.replace(/V/g, 'vv')
.replace(/v/g, 'ff');
for (var i = 0; i < parseInt(compressed.length / 2); i++) {
// Determining the substring.
var substring = compressed.substr(i*2, 2);
// Determining the binValue of this hex substring.
var binValue = parseInt(substring, 16).toString(2);
// If the length of the binary value is not equal to 8 we add leading 0s (js deletes the leading 0s)
// For instance the binary number 00011110 is equal to the hex number 1e,
// but simply running the code above will return 11110. So we have to add the leading 0s back.
if (binValue.length != 8) {
// Determining how many 0s to add:
var diffrence = 8 - binValue.length;
// Adding the 0s:
for (var j = 0; j < diffrence; j++) {
binValue = '0'+binValue;
}
}
// Adding the binValue to the end string which we will return.
returnValue += binValue
}
var decompressedArr = returnValue.split('');
returnValue = ''; // Emptying the return variable.
// Deleting the not needed leading 0s used as padding.
for (var i = paddingAmount; i < decompressedArr.length; i++) {
returnValue += decompressedArr[i];
}
// Returning the decompressed string.
return returnValue;
}
URL shortener
I still found the "compressed" strings a little long for sharing / pasting around. So i used a simple URL shortener (view here) to make this process a little easier for the user.
Now you might ask, then why did you need to compress this string anyway?
Here's why:
First of all, my project is hosted on github pages (gh-pages). The info page of gh-pages tells us that the url can't be any longer than 2000 chars. This would mean that the max grid size would be the square root of 2000 - length of the base url, which isn't that big. By using this "compression" we are able to share much larger grids.
Now the second reason why is that, it's a challange. I find dealing with problems like these fun and also helpfull since you learn a lot.
Live
You can view the live version of my project here. and/or find the github repository here.
Thankyou
I want to thank everyone who helped me with this problem. Especially Corstian and Arnauld, since i ended up using their answers to reach my final functions.
Sooooo.... thanks guys! apriciate it!
In the Game of Life there is a board of ones and zeros. I want to back up to previous generation - size 4800 - save each 16 cells as hexadecimal = 1/4 the size. http://innerbeing.epizy.com/cwebgl/gameoflife.html [g = Go] [b = Backup]
function drawGen(n) {
stop(); var i = clamp(n,0,brw*brh-1), hex = gensave[i].toString();
echo(":",i, n,nGEN); nGEN = i; var str = '';
for (var i = 0; i < parseInt(hex.length / 4); i++)
str = str + pad(parseInt(hex.substr(i*4,4), 16).toString(2),16,'0');
for (var j=0;j<Board.length;j++) Board[j] = intr(str.substr(j,1));
drawBoard();
}
function Bin2Hex(n) {
var i = n.indexOf("1"); /// leading Zeros = NAN
if (i == -1) return "0000";
i = right(n,i*-1);
return pad(parseInt(i,2).toString(16),4,'0');
}
function saveGen(n) {
var b = Board.join(''), str = ''; /// concat array to string 10101
for (var i = 0; i < parseInt(b.length / 16); i++)
str = str + Bin2Hex(b.substr(i*16,16));
gensave[n] = str;
}
function right(st,n) {
var s = st.toString();
if (!n) return s;
if (n < 0) return s.substr(n * -1,s.length + n);
return s.substr(s.length - n,n);
}
function pad(str, l, padwith) {
var s = str;
while (s.length < l) s = padwith + s;
return s;
}

Reading variable length bits from a binary string

Im new to javascript and node.js, I have a base64 encoded string of data that I need to parse several values from which are of various bit lengths.
I figured I would start by using the Buffer object to read the b64 string but from there I am completely lost.
The data are a series of unsigned integers, The format is something akin to this:
Header:
8 bits - uint
3 bits - uint
2 bits - uint
3 bits - unused padding
6 bits - uint
After that there are recurring sections of either 23 bit or 13 bit length of data each with a couple of fields I need to extract.
An example of a 23 bit section:
3 bit - uint
10 bit - uint
10 bit - uint
My question is this, What is the best way to take an arbitrary number of bits and put the resulting value in a separate uint? Note that some of the values are multi-byte (> 8 bits) so I cant step byte for byte.
I apologize if my explanation is kind of vague but hopefully it will suffice.
One simple way to read any amount of bits is e.g.
function bufferBitReader(buffer) {
var bitPos = 0;
function readOneBit() {
var offset = Math.floor(bitPos / 8),
shift = 7 - bitPos % 8;
bitPos += 1;
return (buffer[offset] >> shift) & 1;
}
function readBits(n) {
var i, value = 0;
for (i = 0; i < n; i += 1) {
value = value << 1 | readOneBit();
}
return value;
}
function isEnd() {
return Math.floor(bitPos / 8) >= buffer.length;
}
return {
readOneBit: readOneBit,
readBits: readBits,
isEnd: isEnd
};
}
You just take your but buffer and initialize the reader by
var bitReader = bufferBitReader(buffer);
Then you can read any number of bits by calling
bitReader.readBits(8);
bitReader.readBits(3);
bitReader.readBits(2);
...
You can test whether you already read all bits by
bitReader.isEnd()
One thing to make sure is the actual order of bit that is expected... some 'bit streams' are expected to get bits from the least significant to the most significant.. this code expects the opposite that the first bit you read is the most significant of the first byte...

adding string numbers (math) with local storage

I am trying to add numbers as strings using basic math. I first set the local storage to "0" then add "1" to it each time. I feel I am on the right path, but when I run this my result is not 0 + 1 = 1 rather I get "01" in my local storage. I want to be able to add 1 to the existing local storage each time so 0 + 1 I get 1. Next time around 1 + 1 I get 2, and 2 + 1 I get 3 and so on.
// sets "points" to 0 when user first loads page.
if (localStorage.getItem("points") === null){
localStorage.setItem("points", "0");
}
// get points
var totalPoints = localStorage.getItem("points");
// add 1 points to exisiting total
var addPoint = totalPoints +"1";
// set new total
localStorage.setItem("points", addPoint);
You can convert a string to a number in several ways (not an exhaustive list):
var n = s * 1; // s is the string
var n = s - 0;
var n = parseFloat(s);
var n = Number(s);
var n = ~~s; // force to 32-bit integer
var n = parseInt(s, 10); // also integer, precise up to 53 bits
Convert your strings to numbers when you fetch them from local storage, do the math, and then put the results back.
edit — the thing to keep in mind is that + is more "interesting" than the other arithmetic operators because it has meaning for string-valued operands. In fact, JavaScript tends to prefer string interpretation of the + operator, so if there's a string on one side and a number on the other, the operation is string concatenation and not arithmetic addition.

bitwise AND in Javascript with a 64 bit integer

I am looking for a way of performing a bitwise AND on a 64 bit integer in JavaScript.
JavaScript will cast all of its double values into signed 32-bit integers to do the bitwise operations (details here).
Javascript represents all numbers as 64-bit double precision IEEE 754 floating point numbers (see the ECMAscript spec, section 8.5.) All positive integers up to 2^53 can be encoded precisely. Larger integers get their least significant bits clipped. This leaves the question of how can you even represent a 64-bit integer in Javascript -- the native number data type clearly can't precisely represent a 64-bit int.
The following illustrates this. Although javascript appears to be able to parse hexadecimal numbers representing 64-bit numbers, the underlying numeric representation does not hold 64 bits. Try the following in your browser:
<html>
<head>
<script language="javascript">
function showPrecisionLimits() {
document.getElementById("r50").innerHTML = 0x0004000000000001 - 0x0004000000000000;
document.getElementById("r51").innerHTML = 0x0008000000000001 - 0x0008000000000000;
document.getElementById("r52").innerHTML = 0x0010000000000001 - 0x0010000000000000;
document.getElementById("r53").innerHTML = 0x0020000000000001 - 0x0020000000000000;
document.getElementById("r54").innerHTML = 0x0040000000000001 - 0x0040000000000000;
}
</script>
</head>
<body onload="showPrecisionLimits()">
<p>(2^50+1) - (2^50) = <span id="r50"></span></p>
<p>(2^51+1) - (2^51) = <span id="r51"></span></p>
<p>(2^52+1) - (2^52) = <span id="r52"></span></p>
<p>(2^53+1) - (2^53) = <span id="r53"></span></p>
<p>(2^54+1) - (2^54) = <span id="r54"></span></p>
</body>
</html>
In Firefox, Chrome and IE I'm getting the following. If numbers were stored in their full 64-bit glory, the result should have been 1 for all the substractions. Instead, you can see how the difference between 2^53+1 and 2^53 is lost.
(2^50+1) - (2^50) = 1
(2^51+1) - (2^51) = 1
(2^52+1) - (2^52) = 1
(2^53+1) - (2^53) = 0
(2^54+1) - (2^54) = 0
So what can you do?
If you choose to represent a 64-bit integer as two 32-bit numbers, then applying a bitwise AND is as simple as applying 2 bitwise AND's, to the low and high 32-bit 'words'.
For example:
var a = [ 0x0000ffff, 0xffff0000 ];
var b = [ 0x00ffff00, 0x00ffff00 ];
var c = [ a[0] & b[0], a[1] & b[1] ];
document.body.innerHTML = c[0].toString(16) + ":" + c[1].toString(16);
gets you:
ff00:ff0000
Here is code for AND int64 numbers, you can replace AND with other bitwise operation
function and(v1, v2) {
var hi = 0x80000000;
var low = 0x7fffffff;
var hi1 = ~~(v1 / hi);
var hi2 = ~~(v2 / hi);
var low1 = v1 & low;
var low2 = v2 & low;
var h = hi1 & hi2;
var l = low1 & low2;
return h*hi + l;
}
This can now be done with the new BigInt built-in numeric type. BigInt is currently (July 2019) only available in certain browsers, see the following link for details:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/BigInt
I have tested bitwise operations using BigInts in Chrome 67 and can confirm that they work as expected with up to 64 bit values.
Javascript doesn't support 64 bit integers out of the box. This is what I ended up doing:
Found long.js, a self contained Long implementation on github.
Convert the string value representing the 64 bit number to a Long.
Extract the high and low 32 bit values
Do a 32 bit bitwise and between the high and low bits, separately
Initialise a new 64 bit Long from the low and high bit
If the number is > 0 then there is correlation between the two numbers
Note: for the code example below to work you need to load
long.js.
// Handy to output leading zeros to make it easier to compare the bits when outputting to the console
function zeroPad(num, places){
var zero = places - num.length + 1;
return Array(+(zero > 0 && zero)).join('0') + num;
}
// 2^3 = 8
var val1 = Long.fromString('8', 10);
var val1High = val1.getHighBitsUnsigned();
var val1Low = val1.getLowBitsUnsigned();
// 2^61 = 2305843009213693960
var val2 = Long.fromString('2305843009213693960', 10);
var val2High = val2.getHighBitsUnsigned();
var val2Low = val2.getLowBitsUnsigned();
console.log('2^3 & (2^3 + 2^63)')
console.log(zeroPad(val1.toString(2), 64));
console.log(zeroPad(val2.toString(2), 64));
var bitwiseAndResult = Long.fromBits(val1Low & val2Low, val1High & val2High, true);
console.log(bitwiseAndResult);
console.log(zeroPad(bitwiseAndResult.toString(2), 64));
console.log('Correlation betwen val1 and val2 ?');
console.log(bitwiseAndResult > 0);
Console output:
2^3
0000000000000000000000000000000000000000000000000000000000001000
2^3 + 2^63
0010000000000000000000000000000000000000000000000000000000001000
2^3 & (2^3 + 2^63)
0000000000000000000000000000000000000000000000000000000000001000
Correlation between val1 and val2?
true
The Closure library has goog.math.Long with a bitwise add() method.
Unfortunately, the accepted answer (and others) appears not to have been adequately tested. Confronted by this problem recently, I initially tried to split my 64-bit numbers into two 32-bit numbers as suggested, but there's another little wrinkle.
Open your JavaScript console and enter:
0x80000001
When you press Enter, you'll obtain 2147483649, the decimal equivalent. Next try:
0x80000001 & 0x80000003
This gives you -2147483647, not quite what you expected. It's clear that in performing the bitwise AND, the numbers are treated as signed 32-bit integers. And the result is wrong. Even if you negate it.
My solution was to apply ~~ to the 32-bit numbers after they were split off, check for a negative sign, and then deal with this appropriately.
This is clumsy. There may be a more elegant 'fix', but I can't see it on quick examination. There's a certain irony that something that can be accomplished by a couple of lines of assembly should require so much more labour in JavaScript.

Categories

Resources