Is there is some mathematical "optimum" base that would speed up factorial calculation?
Background:
Just for fun, I'm implementing my own bignum library. (-: Is this my first mistake? :-).
I'm experimenting with various bases used in the internal representation and regression testing by printing out exact values (in decimal) for n factorial (n!).
The way my bignum library represents integers and does multiplication, the time is proportional to the total number of "1" bits in the internal representation n!.
Using a base 2, 4, 8, 16, 2^8, 2^30, or etc. in my internal representation all give me exactly the same total number of "1" bits for any particular number.
Unless I've made some mistake, any given factorial (n!) represented in base 18 has fewer "1" bits than the same value represented in base 10 or or base 16 or base 19.
And so (in principle), using base 18 would make my bignum library run faster than using base 10 or some binary 2^w base or base 19.
I think this has something to do with the fact that n! is either shorter or has more "trailing zeros" or both when printed out in base 18 than in base 10 or or base 16 or base 19.
Is there some other base that would work even better than base 18?
In other words,
Is there a base that represents n! with even fewer "1" bits than base 18?
This is not a dup of "What is a convenient base for a bignum library & primality testing algorithm?" because I suspect "the optimum base for working with integers that are known to be large factorials, with lots of factors of 2 and 3" is different than "the optimum base for working with integers that don't have any small factors and are possibly prime".
(-: Is speeding up factorial calculations -- perhaps at the expense of other kinds of calculations -- my second mistake? :-)
edit:
For example:
(decimal) 16! ==
(decimal ) == 20,922,789,888,000 // uses decimal 14 "1" bits
(dozenal ) == 2,41A,B88,000,000 // uses decimal 10 "1" bits
(hexadecimal) == 130,777,758,000 // uses decimal 18 "1" bits
(octadecimal) == 5F,8B5,024,000 // uses decimal 14 "1" bits
(I'm more-or-less storing the digits on the right, without the commas, plus some metadata overhead).
(While one might think that "As you increase the base you will use fewer "1" bits to represent a given number", or ""As you increase the base you will use fewer nonzero digits to represent a given number", the above example shows that is not always true.)
I'm storing each digit as a small integer ("int" or "long int" or "byte"). Is there any other reasonable way to store digits?
I'm pretty sure my computer stores those integers in binary -- each "1", "2", "4", "8", and "G" digit use one "1" bit; each "3", "5", "6", "9", and "A" digit use two "1" bits; each "7" and "B" digit use three "1" bits; each "F" digit uses four "1" bits, etc.
Both the decimal and the octadecimal representation of this value (16!) require 14 "1" bits.
So I made a mistake in my earlier calculation: For every n, representing n! in octadecimal doesn't always have fewer "1" bits than representing the same value in decimal.
But the question still stands: is there some other "optimum" base that requires the fewest number of 1 bits for storing large factorials?
Someone asks: "How do you store those numbers?"
Well, that's exactly my question -- what is the best way of storing numbers of the form n! ?
I could internally use digits in base 10, or some power-of-two base, or base 18, or some other base. Which one is best?
I could store these integers internally as a 1D array of digits, with a length however long is needed to store all the digits. Is there any reasonable way of printing out 100! in decimal without such an array?
If you're just trying to optimize running time for calculating factorial, and changing the base is the only parameter you're changing, then the optimum base will likely contain small factors. 60 might be a reasonable choice. If you want to experiment, I would try various bases of the form (2^a)(3^b)(5^c)
Improving the speed of multiplication is probably the best way performance. What algorithm are you using for multiplication? (school-book, Karatsuba, Toom-Cook, FFT, ...)
There are other factors to consider, too. If you will be converting the numbers to decimal frequently, then a base that is a power of 10 will make the conversion as fast as possible.
Many(*) years ago, I wrote a base-6 floating point library specifically to solve a problem with repeated multiplication/division by 2 and/or 3. But unless you are trying to solve a specific problem, I think you will be better served by optimizing your algorithms in general than by just trying to optimize factorial.
casevh
(*) I originally said "Several years ago" until I remembered the program ran for many days on a 12Mhz 80286.
While from purely mathematical viewpoint the optimal base is e (and after rounding to nearest integer - 3), from a practical standpoint for a bignum library on a computer, pick a machine word size as the base for your numeric system ( 2^32 or 2^64 ). Yes, it's huge, but the higher abstraction layer of your bignum system is the choke point, the underlying calculations on machine words are the fast part, so delegate as much computation to the CPU low-level instructions while keeping your own work to minimum.
And no, it's not a mistake. It's a very good learning exercise.
I will not pretend I know any math, so do not take my answer as the holy "optimum" you are probably looking for. If I would have to do factorial as fast as possible, I would either try some approximation (something like Stirling approximation) or reduce the number of multiplications, because multiplication is expensive operation. If you represent the number in k-base, you can simulate multiplication by k with help of shifting. If you choose the 2-base, half of all multiplications will be shifts. The other multiplications are shifts and one bit switch. If you aim for minimizing the number of "1"s in your representation of the number, this depends on what numbers you represent. As you increase the base you will use fewer "1"s to represent a given number, but you will need to have more bits for every order, which means more potential "1"s. I hope it helps at least a bit, if not, just ask, I will try to answer.
If by '"1" bits' you mean digits, then I suggest a base of 256 or 65536. In other words, make each byte / word a "digit" for the purposes of your math. The computer handles these numbers routinely and is optimized for doing so. Your factorials will be swift, and so will other operations.
Not to mention the computer handles a great deal of the conversions from a similar base to these with ease. (rhyming unintentional)
Related
An application I'm working on needs to store weights of the format X pounds, y.y ounces. The database is MySQL, but I imagine this is DB agnostic.
I can think of three ways to do this:
Convert the weight to decimal pounds and store in a single field. (5 lbs 6.2 oz = 5.33671875 lbs)
Convert the weight to decimal ounces and store in a single field. (5 lbs 6.2 oz = 86.2 oz)
Store the pounds portion as an integer and the ounces portion as a decimal, in two fields.
I'm thinking that #1 is not such a good idea, since decimal pounds will produce numbers of arbitrary precision, which would need to be stored as a float, which could lead to inaccuracies which are inherent in floating point numbers.
Is there a compelling reason to choose #2 over #3 or vise-versa?
TL;DR
Choose either option #1 or option #2—there's no difference between them. Don't use option #3, because it's awkward to work with.
You claim that there are inherent inaccuracies in floating point numbers. I think that this deserves to be explored a little first.
When deciding upon a numeral system for representing a number (whether on a piece of paper, in a computer circuit, or elsewhere), there are two separate issues to consider:
its basis; and
its format.
Pick a base, any base…
Limited by finite space, one cannot represent an arbitrary member of an infinite set. For example: no matter how much paper you buy or how small your handwriting, it'd always be possible to find an integer that won't fit in the given space (you could just keep appending extra digits until the paper runs out). So, with integers, we usually restrict our finite space to representing only those that fall within some particular interval—e.g. if we have space for the positive/negative sign and three digits, we might restrict ourselves to the interval [-999,+999].
Every non-empty interval contains an infinite set of real numbers. In other words, no matter what interval one takes over the real numbers—be it [-999,+999], [0,1], [0.000001,0.000002] or anything else—there is still an infinite set of reals within that interval (one need only keep appending (non-zero) fractional digits)! Therefore arbitrary real numbers must always be "rounded" to something that can be represented in finite space.
The set of real numbers that can be represented in finite space depends upon the numeral system that is used. In our (familiar) positional base-10 system, finite space will suffice for one-half (0.510) but not for one-third (0.33333…10); by contrast, in the (less familiar) positional base-9 system, it is the other way around (those same numbers are respectively 0.44444…9 and 0.39). The consequence of all this is that some numbers that can be represented using only a small amount of space in positional base-10 (and therefore appear to be very "round" to us humans), e.g. one-tenth, would actually require infinite binary circuits to be stored precisely (and therefore don't appear to be very "round" to our digital friends)! Notably, since 2 is a factor of 10, the same is not true in reverse: any number that can be represented with finite binary can also be represented with finite decimal.
We can't do any better for continuous quantities. Ultimately such quantities must use a finite representation in some numeral system: it's arbitrary whether that system happens to be easy on computer circuits, on human fingers, on something else or on nothing at all—whichever system is used, the value must be rounded and therefore it always results in "representation error".
In other words, even if one has a perfectly accurate measuring instrument (which is physically impossible), then any measurement it reports will already have been rounded to a number that happens to fit on its display (in whatever base it uses—typically decimal, for obvious reasons). So, "86.2 oz" is never actually "86.2 oz" but rather a representation of "something between 86.1500000... oz and 86.2499999... oz". (Actually, because in reality the instrument is imperfect, all we can ever really say is that we have some degree of confidence that the actual value falls within that interval—but that is definitely departing some way from the point here).
But we can do better for discrete quantities. Such values are not "arbitrary real numbers" and therefore none of the above applies to them: they can be represented exactly in the numeral system in which they were defined—and indeed, should be (as converting to another numeral system and truncating to a finite length would result in rounding to an inexact number). Computers can (inefficiently) handle such situations by representing the number as a string: e.g. consider ASCII or BCD encoding.
Apply a format…
Since it's a property of the numeral system's (somewhat arbitrary) basis, whether or not a value appears to be "round" has no bearing on its precision. That's a really important observation, which runs counter to many people's intuition (and it's the reason I spent so much time explaining numerical basis above).
Precision is instead determined by how many significant figures a representation has. We need a storage format that is capable of recording our values to at least as many significant figures as we consider them to be correct. Taking by way of example values that we consider to be correct when stated as 86.2 and 0.0000862, the two most common options are:
Fixed point, where the number of significant figures depends on magnitude: e.g. in fixed 5-decimal-point representation, our values would be stored as 86.20000 and 0.00009 (and therefore have 7 and 1 significant figures of precision respectively). In this example, precision has been lost in the latter value (and indeed, it wouldn't take much more for us to have been totally unable to represent anything of significance); and the former value stored false precision, which is a waste of our finite space (and indeed, it wouldn't take much more for the value to become so large that it overflows the storage capacity).
A common example of when this format might be appropriate is for an accounting system: monetary sums must usually be tracked to the penny irrespective of their magnitude (therefore less precision is required for small values, and more precision is required for large values). As it happens, currency is usually also considered to be discrete (pennies are indivisible), so this is also a good example of a situation where a particular basis (decimal for most modern currencies) is desirable to avoid the representation errors discussed above.
One usually implements fixed point storage by treating one's values as quotients over a common denominator and storing the numerator as an integer. In our example, the common denominator could be 105, so instead of 86.20000 and 0.00009 one would store the integers 8620000 and 9 and remember that they must be divided by 100000.
Floating point, where the number of significant figures is constant irrespective of magnitude: e.g. in 5-significant-figure decimal representation, our values would be stored as 86.200 and 0.000086200 (and, by definition, have 5 significant figures of precision both times). In this example, both values have been stored without any loss of precision; and they both also have the same amount of false precision, which is less wasteful (and we can therefore use our finite space to represent a far greater range of values—both large and small).
A common example of when this format might be appropriate is for recording any real world measurements: the precision of measuring instruments (which all suffer from both systematic and random errors) is fairly constant irrespective of scale so, given sufficient significant figures (typically around 3 or 4 digits), absolutely no precision is lost even if a change of base resulted in rounding to a different number.
One usually implements floating point storage by treating one's values as integer significands with integer exponents. In our example, the significand could be 86200 for both values whereupon the (base-10) exponents would be -4 and -9 respectively.
But how precise are the floating point storage formats used by our computers?
An IEEE754 single precision (binary32) floating point number has 24 bits, or log10(224) (over 7) digits, of significance—i.e. it has a tolerance of less than ±0.000006%. In other words, it is more precise than saying "86.20000".
An IEEE754 double precision (binary64) floating point number has 53 bits, or log10(253) (almost 16) digits, of significance—i.e. it has a tolerance of just over ±0.00000000000001%. In other words, it is more precise than saying "86.2000000000000".
The most important thing to realise is that these formats are, respectively, over ten thousand and over one trillion times more precise than saying "86.2"—even though exact conversions of the binary back into decimal happens to include erroneous false precision (which we must ignore: more on this shortly)!
Notice also that both fixed and floating point formats will result in loss of precision when a value is known more precisely than the format supports. Such rounding errors can propagate in arithmetic operations to yield apparently erroneous results (which no doubt explains your reference to the "inherent inaccuracies" of floating point numbers): for example, 1⁄3 × 3000 in 5-place fixed point would yield 999.99000 rather than 1000.00000; and 1⁄7 − 7⁄50 in 5-significant figure floating point would yield 0.0028600 rather than 0.0028571.
The field of numerical analysis is dedicated to understanding these effects, but it is important to realise that any usable system (even performing calculations in your head) is vulnerable to such problems because no method of calculation that is guaranteed to terminate can ever offer infinite precision: consider, for example, how to calculate the area of a circle—there will necessarily be loss of precision in the value used for π, which will propagate into the result.
Conclusion
Real world measurements should use binary floating point: it's fast, compact, extremely precise and no worse than anything else (including the decimal version from which you started). Since MySQL's floating-point datatypes are IEEE754, this is exactly what they offer.
Currency applications should use denary fixed point: whilst it's slow and wastes memory, it ensures both that values are not rounded to inexact quantities and that pennies are not lost on large monetary sums. Since MySQL's fixed-point datatypes are BCD-encoded strings, this is exactly what they offer.
Finally, bear in mind that programming languages usually represent fractional values using binary floating-point types: so if your database stores values in another format, you need to be careful how they are brought into your application or else they may get converted (with all the ensuing issues that entails) at the interface.
Which option is best in this case?
Hopefully I've convinced you that your values can safely (and should) be stored in floating point types without worrying too much about any "inaccuracies"? Remember, they're more precise than your flimsy 3-significant-digit decimal representation ever was: you just have to ignore false precision (but one must always do that anyway, even if using a fixed-point decimal format).
As for your question: choose either option 1 or 2 over option 3—it makes comparisons easier (for example, to find the maximal mass, one could just use MAX(mass), whereas to do it efficiently across two columns would require some nesting).
Between those two, it doesn’t matter which one chooses—floating point numbers are stored with a constant number of significant bits irrespective of their scale.
Furthermore, whilst in the general case it could happen that some values are rounded to binary numbers that are closer to their original decimal representation using option 1 whilst simultaneously others are rounded to binary numbers that are closer to their original decimal representation using option 2, as we shall shortly see such representation errors only manifest within the false precision that should always be ignored.
However, in this case, because it happens that there are 16 ounces to 1 pound (and 16 is a power of 2), the relative differences between original decimal values and stored binary numbers using the two approaches is identical:
5.387510 (not 5.3367187510 as stated in your question) would be stored in a binary32 float as 101.0110001100110011001102 (which is 5.3874998092651367187510): this is 0.0000036% from the original value (but, as discussed above, the "original value" was already a pretty lousy representation of the physical quantity it represents).
Knowing that a binary32 float stores only 7 decimal digits of precision, our compiler knows for certain that everything from the 8th digit onwards is definitely false precision and therefore must be ignored in every case—thus, provided that our input value didn't require more precision than that (and if it did, binary32 was obviously the wrong choice of format), this guarantees a return to a decimal value that looks just as round as that from which we started: 5.38750010. However, we should really apply domain knowledge at this point (as we should with any storage format) to discard any further false precision that might exist, such as those two trailing zeroes.
86.210 would be stored in a binary32 float as 1010110.001100110011001102 (which is 86.199996948242187510): this is also 0.0000036% from the original value. As before, we then ignore false precision to return to our original input.
Notice how the binary representations of the numbers are identical, except for the placement of the radix point (which is four bits apart):
101.0110 00110011001100110
101 0110.00110011001100110
This is because 5.3875 × 24 = 86.2.
As an aside: being European (albeit British), I also have a strong aversion to imperial units of measurement—handling values of different scales is just so messy. I'd almost certainly store masses in SI units (e.g. kilograms or grams) and then perform conversions to imperial units as required within the presentation layer of my application. Plus rigidly adhering to SI units might one day save you from losing $125m.
I’d be tempted to store it in a metric unit, as they tend to be simple decimals and not complex values like pounds and ounces. That way, you can just store the one value (i.e. 103.25 kg) rather than the pounds–ounces equivalent, and it’s easier to perform conversions.
This is something I’ve dealt with in the past. I do a lot of work on pro wrestling and mixed martial arts (MMA) websites where fighters’ heights and weights need to be recorded. They tend to be displayed as feet and inches and pounds and ounces, but I still store the values in their centimetres and kilogram equivalents, and then do the conversion when displaying on the site.
First, I had not known about how floating point numbers were inaccurate - thankfully a search latter helps me understand: Floating Point Inaccuracy Examples
I would fully agree with #eggyal - keep the data in a single format in a single column. This allows you to expose it to the application and let the application deal with the presentation of it - be it in lbs/oz, rounded up lbs, whatever.
The database should keep the raw data while the presentation layer dictates the layout.
You can use decimal data type for weight column.
decimal('weight', 8, 2); // precision = 8, scale = 2
Storage size:
Precision 1-9 5 Bytes
Precision 10-19 9 Bytes
Precision 20-28 13 Bytes
Precision 29-38 17 Bytes
I know the "<<" is a bit operation. but I do not understand what it exactly functions in TCL, and when should we use it?
can anyone help me on this?
The << operator in Tcl's expressions is an arithmetic bit shift left. It's exceptionally similar to the equivalent in C and many other languages, and would be used in all the same places (it's logically equivalent to a multiply by a suitable power of 2, but it's usually advisable to use a shift when thinking about bits and a multiply when thinking about numbers).
Note that one key difference with many other languages (from Tcl 8.5 onwards) is that it does not “drop bits off the front”; the language implementation automatically uses wider number representations as necessary so that information is never lost. Bits are dropped by using a separate binary mask operation (e.g., & ((1 << $numBits) - 1)).
There are a number of uses for the << shift left operator. Some that come to my mind are :
Bit by bit processing. Shift a number and observe highest order bit etc. It comes in more handy than you might think.
If you add a zero to a number in the decimal number system you effectively multiply it by 10. shifting bits effectively means multiplying by 2. This actually translated into a low level assembly command of bit shifting which has lower compute cycles than multiplication by 2. This is used for efficiency in the gaming industry. Shift if twice (<< 2) to multiply it by 4 and so on.
I am sure there are many others.
The << operation is not much different from C's, for instance. And it's used when you need to shift bits of an integer value to the left. This can be occasionally useful when doing subtle number crunching like implemening a hash function or deserialising something from an input bytestream (but note that [binary scan] covers almost all of what's needed for this sort of thing). For a more general info refer to this Wikipedia article or something like this, this is not really Tcl-related.
The '<<' is a left bit shift. You must apply it to an integer. This arithmetic operator will shift the bits to left.
For example, if you want to shifted the number 1 twice to the left in the Tcl interpreter tclsh, type:
expr { 1 << 2 }
The command will return 4.
Pay special attention to the maximum integer the interpreter hold on your platform.
We were asked to find a way to compress a square binary matrix as much as possible, and if possible, to add redundancy bits to check and maybe correct errors.
The redundancy thing is easy to implement in my opinion. The complicated part is compressing the matrix. I thought about using run-length after reshaping the matrix to a vector because there will be more zeros than ones, but I only achieved a 40bits compression (we are working on small sizes) although I thought it'd be better.
Also, after run-length an idea was Huffman coding the matrix, but a dictionary must be sent in order to recover the original information.
I'd like to know what would be the best way to compress a binary matrix?
After reading some comments, yes #Adam you're right, the 14x14 matrix should be compressed in 128bits, so if I only use the coordinates (rows&cols) for each non-zero element, still it would be 160bits (since there are twenty ones). I'm not looking for an exact solution but for a useful idea.
You can only talk about compressing something if you have a distribution and a representation. That's the issue of the dictionary you have to send along: you always need some sort of dictionary of protocol to uncompress something. It just so happens that things like .zip and .mpeg already have those dictionaries/codecs. Even something as simple as Huffman-encoding is an algorithm; on the other side of the communication channel (you can think of compression as communication), the other person already has a bit of code (the dictionary) to perform the Huffman decompression scheme.
Thus you cannot even begin to talk about compressing something without first thinking "what kinds of matrices do I expect to see?", "is the data truly random, or is there order?", and if so "how can I represent the matrices to take advantage of order in the data?".
You cannot compress some matrices without increasing the size of other objects (by at least 1 bit). This is bad news if all matrices are equally probable, and you care equally about them all.
Addenda:
The answer to use sparse matrix machinery is not necessarily the right answer. The matrix could for example be represented in python as [[(r+c)%2 for c in range (cols)] for r in range(rows)] (a checkerboard pattern), and a sparse matrix wouldn't compress it at all, but the Kolmogorov complexity of the matrix is the above program's length.
Well, I know every matrix will have the same number of ones, so this is kind of deterministic. The only think I don't know is where the 1's will be. Also, if I transmit the matrix with a dictionary and there are burst errors, maybe the dictionary gets affected so... wouldnt be the resulting information corrupted? That's why I was trying to use lossless data compression such as run-length, the decoder just doesnt need a dictionary. --original poster
How many 1s does the matrix have as a fraction of its size, and what is its size (NxN -- what is N)?
Furthermore, this is an incorrect assertion and should not be used as a reason to desire run-length encoding (which still requires a program); when you transmit data over a channel, you can always add error-correction to this data. "Data" is just a blob of bits. You can transmit both the data and any required dictionaries over the channel. The error-correcting machinery does not care at all what the bits you transmit are for.
Addendum 2:
There are (14*14) choose 20 possible arrangements, which I assume are randomly chosen. If this number was larger than 128^2 what you're trying to do would be impossible. Fortunately log_2((14*14) choose 20) ~= 90bits < 128bits so it's possible.
The simple solution of writing down 20 numbers like 32,2,67,175,52,...,168 won't work because log_2(14*14)*20 ~= 153bits > 128bits. This would be equivalent to run-length encoding. We want to do something like this but we are on a very strict budget and cannot afford to be "wasteful" with bits.
Because you care about each possibility equally, your "dictionary"/"program" will simulate a giant lookup table. Matlab's sparse matrix implementation may work but is not guaranteed to work and is thus not a correct solution.
If you can create a bijection between the number range [0,2^128) and subsets of size 20, you're good to go. This corresponds to enumerating ways to descend the pyramid in http://en.wikipedia.org/wiki/Binomial_coefficient to the 20th element of row 196. This is the same as enumerating all "k-combinations". See http://en.wikipedia.org/wiki/Combination#Enumerating_k-combinations
Fortunately I know that Mathematica and Sage and other CAS software can apparently generate the "5th" or "12th" or arbitrarily numbered k-subset. Looking through their documentation, we come upon a function called "rank", e.g. http://www.sagemath.org/doc/reference/sage/combinat/subset.html
So then we do some more searching, and come across some arcane Fortran code like http://people.sc.fsu.edu/~jburkardt/m_src/subset/ksub_rank.m and http://people.sc.fsu.edu/~jburkardt/m_src/subset/ksub_unrank.m
We could reverse-engineer it, but it's kind of dense. But now we have enough information to search for k-subset rank unrank, which leads us to http://www.site.uottawa.ca/~lucia/courses/5165-09/GenCombObj.pdf -- see the section
"Generating k-subsets (of an n-set): Lexicographical
Ordering" and the rank and unrank algorithms on the next few pages.
In order to achieve the exact theoretically optimal compression, in the case of a uniformly random distribution of 1s, we must thus use this technique to biject our matrices to our output number of range <2^128. It just so happens that combinations have a natural ordering, known as ranking and unranking of combinations. You assign a number to each combination (ranking), and if you know the number you automatically know the combination (unranking). Googling k-subset rank unrank will probably yield other algorithms.
Thus your solution would look like this:
serialize the matrix into a list
e.g. [[0,0,1][0,1,1][1,0,0]] -> [0,0,1,0,1,1,1,0,0]
take the indices of the 1s:
e.g. [0,0,1,0,1,1,1,0,0] -> [3,5,6,7]
1 2 3 4 5 6 7 8 9 a k=4-subset of an n=9 set
take the rank
e.g. compressed = rank([3,5,6,7], n=9)
compressed==412 (or something, I made that up)
you're done!
e.g. 412 -binary-> 110011100 (at most n=9bits, less than 2^n=2^9=512)
to uncompress, unrank it
I'll get to 128 bits in a sec, first here's how you fit a 14x14 boolean matrix with exactly 20 nonzeros into 136 bits. It's based on the CSC sparse matrix format.
You have an array c with 14 4-bit counters that tell you how many nonzeros are in each column.
You have another array r with 20 4-bit row indices.
56 bits (c) + 80 bits (r) = 136 bits.
Let's squeeze 8 bits out of c:
Instead of 4-bit counters, use 2-bit. c is now 2*14 = 28 bits, but can't support more than 3 nonzeros per column. This leaves us with 128-80-28 = 20 bits. Use that space for array a4c with 5 4-bit elements that "add 4 to an element of c" specified by the 4-bit element. So, if a4c={2,2,10,15, 15} that means c[2] += 4; c[2] += 4 (again); c[10] += 4;.
The "most wasteful" distribution of nonzeros is one where the column count will require an add-4 to support 1 extra nonzero: so 5 columns with 4 nonzeros each. Luckily we have exactly 5 add-4s available.
Total space = 28 bits (c) + 20 bits
(a4c) + 80 bits (r) = 128 bits.
Your input is a perfect candidate for a sparse matrix. You said you're using Matlab, so you already have a good sparse matrix built for you.
spm = sparse(dense_matrix)
Matlab's sparse matrix implementation uses Compressed Sparse Columns, which has memory usage on the order of 2*(# of nonzeros) + (# of columns), which should be pretty good in your case of 20 nonzeros and 14 columns. Storing 20 values sure is better than storing 196...
Also remember that all matrices in Matlab are going to be composed of doubles. Just because your matrix can be stored as a 1-bit boolean doesn't mean Matlab won't stick it into a 64-bit floating point value... If you do need it as a boolean you're going to have to make your own type in C and use .mex files to interface with Matlab.
After thinking about this again, if all your matrices are going to be this small and they're all binary, then just store them as a binary vector (bitmask). Going off your 14x14 example, that requires 196 bits or 25 bytes (plus n, m if your dimensions are not constant). That same vector in Matlab would use 64 bits per element, or 1568 bytes. So storing the matrix as a bitmask takes as much space as 4 elements of the original matrix in Matlab, for a compression ratio of 62x.
Unfortunately I don't know if Matlab supports bitmasks natively or if you have to resort to .mex files. If you do get into C++ you can use STL's vector<bool> which implements a bitmask for you.
I know that Knapsack is NP-complete while it can be solved by DP. They say that the DP solution is pseudo-polynomial, since it is exponential in the "length of input" (i.e. the numbers of bits required to encode the input). Unfortunately I did not get it. Can anybody explain that pseudo-polynomial thing to me slowly ?
The running time is O(NW) for an unbounded knapsack problem with N items and knapsack of size W. W is not polynomial in the length of the input though, which is what makes it pseudo-polynomial.
Consider W = 1,000,000,000,000. It only takes 40 bits to represent this number, so input size = 40, but the computational runtime uses the factor 1,000,000,000,000 which is O(240).
So the runtime is more accurately said to be O(N.2bits in W), which is exponential.
Also see:
How to understand the knapsack problem is NP-complete?
The NP-Completeness of Knapsack
Complexity of dynamic programming algorithm for the 0-1 knapsack problem
Pseudo-polynomial time
In most of our problems, we're dealing with large lists of numbers which fit comfortably inside standard int/float data types. Because of the way most processors are built to handle 4-8 byte numbers at a time at no additional cost (relative to numbers than fit in, say, 1 byte), we rarely encounter a change in running time from scaling our numbers up or down within ranges we encounter in real problems - so the dominant factor remains just the sheer quantity of data points, the n or m factors that we're used to.
(You can imagine that the Big-O notation is hiding a constant factor that divides-out 32 or 64 bits-per-datum, leaving only the number-of-data-points whenever each of our numbers fit in that many bits or less)
But try reworking with other algorithms to act on data sets involving big ints - numbers that require more than 8 bytes to represent - and see what that does to the runtime. The magnitude of the numbers involved always makes a difference, even in the other algorithms like binary sort, once you expand beyond the buffer of safety conventional processors give us "for-free" by handling 4-8 byte batches.
The trick with the Knapsack algorithm that we discussed is that it's unusually sensitive (relative to other algorithms ) to the magnitude of a particular parameter, W. Add one bit to W and you double the running time of the algorithm. We haven't seen that kind of dramatic response to changes in value in other algorithms before this one, which is why it might seem like we're treating Knapsack differently - but that's a genuine analysis of how it responds in a non-polynomial fashion to changes in input size.
The way I understand this is that the capacity would've been O(W) if the capacity input were an array of [1,2,...,W], which has a size of W. But the capacity input is not an array of numbers, it's instead a single integer. The time complexity is about the relationship to the size of input. The size of an integer is NOT the value of the integer, but the number of bits representing it. We do later convert this integer W into an array [1,2,...,W] in the algorithm, leading people into mistakenly thinking W is the size, but this array is not the input, the integer itself is.
Think of input as "an array of stuff", and the size as "how many stuff in the array". The item input is actually an array of n items in the array so size=n. The capacity input is NOT an array of W numbers in it, but a single integer, represented by an array of log(W) bits. Increase the size of it by 1 (adding 1 meaningful bit), W doubles so run time doubles, hence the exponential time complexity.
The Knapsack algorithm's run-time is bound not only on the size of the input (n - the number of items) but also on the magnitude of the input (W - the knapsack capacity) O(nW) which is exponential in how it is represented in computer in binary (2^n) .The computational complexity (i.e how processing is done inside a computer through bits) is only concerned with the size of the inputs, not their magnitudes/values.
Disregard the value/weight list for a moment. Let's say we have an instance with knapsack capacity 2. W would take two bits in the input data. Now we shall increase the knapsack capacity to 4, keeping the rest of the input. Our input has only grown by one bit, but the computational complexity has increased twofold. If we increase the capacity to 1024, we would have just 10 bits of the input for W instead of 2, but the complexity has increased by a factor of 512. Time complexity grows exponentially in the size of W in binary (or decimal) representation.
Another simple example that helped me understand the pseudo-polynomial concept is the naive primality testing algorithm. For a given number n we are checking if it's divided evenly by each integer number in range 2..√n, so the algorithm takes √(n−1) steps. But here, n is the magnitude of the input, not it's size.
Now The regular O(n) case
By contrast, searching an array for a given element runs in polynomial time: O(n). It takes at most n steps and here n is the size of the input (the length of the array).
[ see here ]
Calculating bits required to store decimal number
Complexity is based on input. In knapsack problem, Inputs are size, max Capacity, and profit, weight arrays. We construct dp table as size * W so we feel as its of polynomial time complexity. But, input W is an integer, not an array. So, it will be O(size*(no Of bits required to store given W)). If no of bits increase by 1, then running time doubles. Thus it is exponential, thereby pseudo-polynomial.
What is the best way to constrain the values of a PRNG to a smaller range? If you use modulus and the old max number is not evenly divisible by the new max number you bias toward the 0 through (old_max - new_max - 1). I assume the best way would be something like this (this is floating point, not integer math)
random_num = PRNG() / max_orginal_range * max_smaller_range
But something in my gut makes me question that method (maybe floating point implementation and representation differences?).
The random number generator will produce consistent results across hardware and software platforms, and the constraint needs to as well.
I was right to doubt the pseudocode above (but not for the reasons I was thinking). MichaelGG's answer got me thinking about the problem in a different way. I can model it using smaller numbers and test every outcome. So, let's assume we have a PRNG that produces a random number between 0 and 31 and you want the smaller range to be 0 to 9. If you use modulus you bias toward 0, 1, 2, and 3. If you use the pseudocode above you bias toward 0, 2, 5, and 7. I don't think there can be a good way to map one set into the other. The best that I have come up with so far is to regenerate the random numbers that are greater than old_max/new_max, but that has deep problems as well (reducing the period, time to generate new numbers until one is in the right range, etc.).
I think I may have naively approached this problem. It may be time to start some serious research into the literature (someone has to have tackled this before).
I know this might not be a particularly helpful answer, but I think the best way would be to conceive of a few different methods, then trying them out a few million times, and check the result sets.
When in doubt, try it yourself.
EDIT
It should be noted that many languages (like C#) have built in limiting in their functions
int maximumvalue = 20;
Random rand = new Random();
rand.Next(maximumvalue);
And whenever possible, you should use those rather than any code you would write yourself. Don't Reinvent The Wheel.
This problem is akin to rolling a k-sided die given only a p-sided die, without wasting randomness.
In this sense, by Lemma 3 in "Simulating a dice with a dice" by B. Kloeckner, this waste is inevitable unless "every prime number dividing k also divides p". Thus, for example, if p is a power of 2 (and any block of random bits is the same as rolling a die with a power of 2 number of faces) and k has prime factors other than 2, the best you can do is get arbitrarily close to no waste of randomness, such as by batching multiple rolls of the p-sided die until p^n is "close enough" to a power of k.
Let me also go over some of your concerns about regenerating random numbers:
"Reducing the period": Besides batching of bits, this concern can be dealt with in several ways:
Use a PRNG with a bigger "period" (maximum cycle length).
Add a Bays–Durham shuffle to the PRNG's implementation.
Use a "true" random number generator; this is not trivial.
Employ randomness extraction, which is discussed in Devroye and Gravel 2015-2020 and in my Note on Randomness Extraction. However, randomness extraction is pretty involved.
Ignore the problem, especially if it isn't a security application or serious simulation.
"Time to generate new numbers until one is in the right range": If you want unbiased random numbers, then any algorithm that does so will generally have to run forever in the worst case. Again, by Lemma 3, the algorithm will run forever in the worst case unless "every prime number dividing k also divides p", which is not the case if, say, k is 10 and p is 32.
See also the question: How to generate a random integer in the range [0,n] from a stream of random bits without wasting bits?, especially my answer there.
If PRNG() is generating uniformly distributed random numbers then the above looks good. In fact (if you want to scale the mean etc.) the above should be fine for all purposes. I guess you need to ask what the error associated with the original PRNG() is, and whether further manipulating will add to that substantially.
If in doubt, generate an appropriately sized sample set, and look at the results in Excel or similar (to check your mean / std.dev etc. for what you'd expect)
If you have access to a PRNG function (say, random()) that'll generate numbers in the range 0 <= x < 1, can you not just do:
random_num = (int) (random() * max_range);
to give you numbers in the range 0 to max_range?
Here's how the CLR's Random class works when limited (as per Reflector):
long num = maxValue - minValue;
if (num <= 0x7fffffffL) {
return (((int) (this.Sample() * num)) + minValue);
}
return (((int) ((long) (this.GetSampleForLargeRange() * num))) + minValue);
Even if you're given a positive int, it's not hard to get it to a double. Just multiply the random int by (1/maxint). Going from a 32-bit int to a double should provide adequate precision. (I haven't actually tested a PRNG like this, so I might be missing something with floats.)
Psuedo random number generators are essentially producing a random series of 1s and 0s, which when appended to each other, are an infinitely large number in base two. each time you consume a bit from you're prng, you are dividing that number by two and keeping the modulus. You can do this forever without wasting a single bit.
If you need a number in the range [0, N), then you need the same, but instead of base two, you need base N. It's basically trivial to convert the bases. Consume the number of bits you need, return the remainder of those bits back to your prng to be used next time a number is needed.