I've got a question to which I need someone to help me verify to see if my answer is correct or not.
Showing all steps involved, convert the unsigned decimal number 28.2 to:
Hexadecimal, using 1 digit for the integer part and 2 for the fractional part.
I come up with the answer C1.33, but since I am only allowed 1 digit for the integer part I would assume I would remove the 1?
Hex, or hexadecimal, is a number system of base 16.hex digits are
0,1,2,3,4,5,6,7,8 and 9 A, B, C, D, E, F
decimal hex
C1 28
C 12
1 1
For 28 is C1 in hex, you cannot represent it with either c or 1
Please refer: link
I was looking at the Shannon's definitions if intrinsic information and entropy (of a "message").
Honestly, I fail to intuitively grasp why Shannon defined those two in terms of the logarithm (apart from the desirable "split multiplication into sum" property of logarithms, which is indeed desirable).
Can anyone help me to shed some light on this?
Thanks.
I believe that Shannon was working at Bell Labs when he developed the idea of Shannon entropy : the goal of his research was to best encode information, with bits (so 0 and 1).
This is the reason of the log2: it has to do with binary encoding of a message. If numbers that can take 8 different values are transmitted on a telecommunication line, signals of length 3 bits (log2(8) = 3) will be needed to transmit these numbers.
Shannon entropy is the minimum number of bits you will need to encode each character of a message (for any message written in any alphabet).
Let us take an example. We have the following message to encode with bits:
"0112003333".
The characters of the message are in {0,1,2,3}, so we would need at most log2(4) = 2 bits to encode the characters of this message. For example, we could use the following way to encode the characters:
0 would be coded by 00
1 would be coded by 01
2 would be coded by 10
3 would be coded by 11
The message would then be encoded like that: "00010110000011111111"
However we could do better if we chose to code the most frequent characters on only one bit and the other on two bits:
0 would be coded by 0
1 would be coded by 01
2 would be coded by 10
3 would be coded by 1
The message would then be encoded like that: "0010110001111"
So the entropy of "0112003333" is between 1 and 2 (it is 1.85, to be more precise).
11111 111111 11111 > 16 bit
R G B
how can I get the R/G/B values?
R = x/2048 (2048 are the 6+5 G+B 11 bit)
G = x-R*2048/32 (32 are the B )
is there a simple explanation of this? why should the decimal value be divided by others bit to get the single group value?
If you want to get R you have to move 11 bits on the right so that G and B disappear. So
R=x>>11
but every shift of 1 bit is equivalent to a division by 2 so the expression above is equivalent to
R=X/(2^11)=x/2048
And so you get the R.
To get the G, according to the formula, you can first remove the R, which is the R you obtained previously but shift back of 11 bits (or multiplied by 2048). Now done this you have a number like:
111111 11111
You shift left of 5 bits (or divide for 32) and you get the result.
Note that in practice you can do this easier doing this:
B=(x and 0x1F)
G=((x>>5) and 0x3F)
R=((x>>11) and 0x1F)
Additional details:
X is any positive integer 6 digits or less.
X is left-padded with zeros to maintain a width of 6.
Please explain your answer :)
(This might be better in the Math site, but figured it involves programming functions)
The picture from the german Wikipedia article is very helpful:
You see that 6 consecutive bits from the original bytes generate a Base64 value. To generate + or / (codes 62 and 63), you'd need the bitstrings 111110 and 111111, so at least 5 consecutive bits set.
However, look at the ASCII codes for 0...9:
00110000
00110001
00110010
00110011
00110100
00110101
00110110
00110111
00111000
00111001
No matter how you concatenate six of those, there won't be more than 3 consecutive bits set. So it's not possible to generate a Base64 string that contains + or / this way, Y will always be alphanumeric.
EDIT: In fact, you can even rule other Base64 values out like 000010 (C), so this leads to nice follow-up questions/puzzles like "How many of the 64 values are possible at all?".
I'm trying to learn C and have come across the inability to work with REALLY big numbers (i.e., 100 digits, 1000 digits, etc.). I am aware that there exist libraries to do this, but I want to attempt to implement it myself.
I just want to know if anyone has or can provide a very detailed, dumbed down explanation of arbitrary-precision arithmetic.
It's all a matter of adequate storage and algorithms to treat numbers as smaller parts. Let's assume you have a compiler in which an int can only be 0 through 99 and you want to handle numbers up to 999999 (we'll only worry about positive numbers here to keep it simple).
You do that by giving each number three ints and using the same rules you (should have) learned back in primary school for addition, subtraction and the other basic operations.
In an arbitrary precision library, there's no fixed limit on the number of base types used to represent our numbers, just whatever memory can hold.
Addition for example: 123456 + 78:
12 34 56
78
-- -- --
12 35 34
Working from the least significant end:
initial carry = 0.
56 + 78 + 0 carry = 134 = 34 with 1 carry
34 + 00 + 1 carry = 35 = 35 with 0 carry
12 + 00 + 0 carry = 12 = 12 with 0 carry
This is, in fact, how addition generally works at the bit level inside your CPU.
Subtraction is similar (using subtraction of the base type and borrow instead of carry), multiplication can be done with repeated additions (very slow) or cross-products (faster) and division is trickier but can be done by shifting and subtraction of the numbers involved (the long division you would have learned as a kid).
I've actually written libraries to do this sort of stuff using the maximum powers of ten that can be fit into an integer when squared (to prevent overflow when multiplying two ints together, such as a 16-bit int being limited to 0 through 99 to generate 9,801 (<32,768) when squared, or 32-bit int using 0 through 9,999 to generate 99,980,001 (<2,147,483,648)) which greatly eased the algorithms.
Some tricks to watch out for.
1/ When adding or multiplying numbers, pre-allocate the maximum space needed then reduce later if you find it's too much. For example, adding two 100-"digit" (where digit is an int) numbers will never give you more than 101 digits. Multiply a 12-digit number by a 3 digit number will never generate more than 15 digits (add the digit counts).
2/ For added speed, normalise (reduce the storage required for) the numbers only if absolutely necessary - my library had this as a separate call so the user can decide between speed and storage concerns.
3/ Addition of a positive and negative number is subtraction, and subtracting a negative number is the same as adding the equivalent positive. You can save quite a bit of code by having the add and subtract methods call each other after adjusting signs.
4/ Avoid subtracting big numbers from small ones since you invariably end up with numbers like:
10
11-
-- -- -- --
99 99 99 99 (and you still have a borrow).
Instead, subtract 10 from 11, then negate it:
11
10-
--
1 (then negate to get -1).
Here are the comments (turned into text) from one of the libraries I had to do this for. The code itself is, unfortunately, copyrighted, but you may be able to pick out enough information to handle the four basic operations. Assume in the following that -a and -b represent negative numbers and a and b are zero or positive numbers.
For addition, if signs are different, use subtraction of the negation:
-a + b becomes b - a
a + -b becomes a - b
For subtraction, if signs are different, use addition of the negation:
a - -b becomes a + b
-a - b becomes -(a + b)
Also special handling to ensure we're subtracting small numbers from large:
small - big becomes -(big - small)
Multiplication uses entry-level math as follows:
475(a) x 32(b) = 475 x (30 + 2)
= 475 x 30 + 475 x 2
= 4750 x 3 + 475 x 2
= 4750 + 4750 + 4750 + 475 + 475
The way in which this is achieved involves extracting each of the digits of 32 one at a time (backwards) then using add to calculate a value to be added to the result (initially zero).
ShiftLeft and ShiftRight operations are used to quickly multiply or divide a LongInt by the wrap value (10 for "real" math). In the example above, we add 475 to zero 2 times (the last digit of 32) to get 950 (result = 0 + 950 = 950).
Then we left shift 475 to get 4750 and right shift 32 to get 3. Add 4750 to zero 3 times to get 14250 then add to result of 950 to get 15200.
Left shift 4750 to get 47500, right shift 3 to get 0. Since the right shifted 32 is now zero, we're finished and, in fact 475 x 32 does equal 15200.
Division is also tricky but based on early arithmetic (the "gazinta" method for "goes into"). Consider the following long division for 12345 / 27:
457
+-------
27 | 12345 27 is larger than 1 or 12 so we first use 123.
108 27 goes into 123 4 times, 4 x 27 = 108, 123 - 108 = 15.
---
154 Bring down 4.
135 27 goes into 154 5 times, 5 x 27 = 135, 154 - 135 = 19.
---
195 Bring down 5.
189 27 goes into 195 7 times, 7 x 27 = 189, 195 - 189 = 6.
---
6 Nothing more to bring down, so stop.
Therefore 12345 / 27 is 457 with remainder 6. Verify:
457 x 27 + 6
= 12339 + 6
= 12345
This is implemented by using a draw-down variable (initially zero) to bring down the segments of 12345 one at a time until it's greater or equal to 27.
Then we simply subtract 27 from that until we get below 27 - the number of subtractions is the segment added to the top line.
When there are no more segments to bring down, we have our result.
Keep in mind these are pretty basic algorithms. There are far better ways to do complex arithmetic if your numbers are going to be particularly large. You can look into something like GNU Multiple Precision Arithmetic Library - it's substantially better and faster than my own libraries.
It does have the rather unfortunate misfeature in that it will simply exit if it runs out of memory (a rather fatal flaw for a general purpose library in my opinion) but, if you can look past that, it's pretty good at what it does.
If you cannot use it for licensing reasons (or because you don't want your application just exiting for no apparent reason), you could at least get the algorithms from there for integrating into your own code.
I've also found that the bods over at MPIR (a fork of GMP) are more amenable to discussions on potential changes - they seem a more developer-friendly bunch.
While re-inventing the wheel is extremely good for your personal edification and learning, its also an extremely large task. I don't want to dissuade you as its an important exercise and one that I've done myself, but you should be aware that there are subtle and complex issues at work that larger packages address.
For example, multiplication. Naively, you might think of the 'schoolboy' method, i.e. write one number above the other, then do long multiplication as you learned in school. example:
123
x 34
-----
492
+ 3690
---------
4182
but this method is extremely slow (O(n^2), n being the number of digits). Instead, modern bignum packages use either a discrete Fourier transform or a Numeric transform to turn this into an essentially O(n ln(n)) operation.
And this is just for integers. When you get into more complicated functions on some type of real representation of number (log, sqrt, exp, etc.) things get even more complicated.
If you'd like some theoretical background, I highly recommend reading the first chapter of Yap's book, "Fundamental Problems of Algorithmic Algebra". As already mentioned, the gmp bignum library is an excellent library. For real numbers, I've used MPFR and liked it.
Don't reinvent the wheel: it might turn out to be square!
Use a third party library, such as GNU MP, that is tried and tested.
You do it in basically the same way you do with pencil and paper...
The number is to be represented in a buffer (array) able to take on an arbitrary size (which means using malloc and realloc) as needed
you implement basic arithmetic as much as possible using language supported structures, and deal with carries and moving the radix-point manually
you scour numeric analysis texts to find efficient arguments for dealing by more complex function
you only implement as much as you need.
Typically you will use as you basic unit of computation
bytes containing with 0-99 or 0-255
16 bit words contaning wither 0-9999 or 0--65536
32 bit words containing...
...
as dictated by your architecture.
The choice of binary or decimal base depends on you desires for maximum space efficiency, human readability, and the presence of absence of Binary Coded Decimal (BCD) math support on your chip.
You can do it with high school level of mathematics. Though more advanced algorithms are used in reality. So for example to add two 1024-byte numbers :
unsigned char first[1024], second[1024], result[1025];
unsigned char carry = 0;
unsigned int sum = 0;
for(size_t i = 0; i < 1024; i++)
{
sum = first[i] + second[i] + carry;
carry = sum - 255;
}
result will have to be bigger by one place in case of addition to take care of maximum values. Look at this :
9
+
9
----
18
TTMath is a great library if you want to learn. It is built using C++. The above example was silly one, but this is how addition and subtraction is done in general!
A good reference about the subject is Computational complexity of mathematical operations. It tells you how much space is required for each operation you want to implement. For example, If you have two N-digit numbers, then you need 2N digits to store the result of multiplication.
As Mitch said, it is by far not an easy task to implement! I recommend you take a look at TTMath if you know C++.
One of the ultimate references (IMHO) is Knuth's TAOCP Volume II. It explains lots of algorithms for representing numbers and arithmetic operations on these representations.
#Book{Knuth:taocp:2,
author = {Knuth, Donald E.},
title = {The Art of Computer Programming},
volume = {2: Seminumerical Algorithms, second edition},
year = {1981},
publisher = {\Range{Addison}{Wesley}},
isbn = {0-201-03822-6},
}
Assuming that you wish to write a big integer code yourself, this can be surprisingly simple to do, spoken as someone who did it recently (though in MATLAB.) Here are a few of the tricks I used:
I stored each individual decimal digit as a double number. This makes many operations simple, especially output. While it does take up more storage than you might wish, memory is cheap here, and it makes multiplication very efficient if you can convolve a pair of vectors efficiently. Alternatively, you can store several decimal digits in a double, but beware then that convolution to do the multiplication can cause numerical problems on very large numbers.
Store a sign bit separately.
Addition of two numbers is mainly a matter of adding the digits, then check for a carry at each step.
Multiplication of a pair of numbers is best done as convolution followed by a carry step, at least if you have a fast convolution code on tap.
Even when you store the numbers as a string of individual decimal digits, division (also mod/rem ops) can be done to gain roughly 13 decimal digits at a time in the result. This is much more efficient than a divide that works on only 1 decimal digit at a time.
To compute an integer power of an integer, compute the binary representation of the exponent. Then use repeated squaring operations to compute the powers as needed.
Many operations (factoring, primality tests, etc.) will benefit from a powermod operation. That is, when you compute mod(a^p,N), reduce the result mod N at each step of the exponentiation where p has been expressed in a binary form. Do not compute a^p first, and then try to reduce it mod N.
Here's a simple ( naive ) example I did in PHP.
I implemented "Add" and "Multiply" and used that for an exponent example.
http://adevsoft.com/simple-php-arbitrary-precision-integer-big-num-example/
Code snip
// Add two big integers
function ba($a, $b)
{
if( $a === "0" ) return $b;
else if( $b === "0") return $a;
$aa = str_split(strrev(strlen($a)>1?ltrim($a,"0"):$a), 9);
$bb = str_split(strrev(strlen($b)>1?ltrim($b,"0"):$b), 9);
$rr = Array();
$maxC = max(Array(count($aa), count($bb)));
$aa = array_pad(array_map("strrev", $aa),$maxC+1,"0");
$bb = array_pad(array_map("strrev", $bb),$maxC+1,"0");
for( $i=0; $i<=$maxC; $i++ )
{
$t = str_pad((string) ($aa[$i] + $bb[$i]), 9, "0", STR_PAD_LEFT);
if( strlen($t) > 9 )
{
$aa[$i+1] = ba($aa[$i+1], substr($t,0,1));
$t = substr($t, 1);
}
array_unshift($rr, $t);
}
return implode($rr);
}