An old idea, but ever since then I couldn't get around finding some reasonably good way to solve the problem it raised. So I "invented" (see below) a very compact, and in my opinion, reasonably well performing PRNG, but I can't get to figure out algorithms to build suitable seed values for it at large bit depths. My current solution is simply brute-forcing, it's running time is O(n^3).
The generator
My idea came from XOR taps (essentially LFSRs) some old 8bit machines used for sound generation. I fiddled with XOR as a base on a C64, tried to put together opcodes, and experienced with the result. The final working solution looked like this:
asl
adc #num1
eor #num2
This is 5 bytes on the 6502. With a well chosen num1 and num2, in the accumulator it iterates over all 256 values in a seemingly random order, that is, it looks reasonably random when used to fill the screen (I wrote a little 256b demo back then on this). There are 40 suitable num1 & num2 pairs for this, all giving decent looking sequences.
The concept can be well generalized, if expressed in pure C, it may look like this (BITS being the bit depth of the sequence):
r = (((r >> (BITS-1)) & 1U) + (r << 1) + num1) ^ num2;
r = r & ((1U<<BITS)-1U);
This C code is longer since it is generalized, and even if one would use the full depth of an unsigned integer, C wouldn't have the necessary carry logic to transfer the high bit of the shift to the add operation.
For some performance analysis and comparisons, see below, after the question(s).
The problem / question(s)
The core problem with the generator is finding suitable num1 and num2 which would make it iterate over the whole possible sequence of a given bit depth. At the end of this section I attach my code which just brute-forces it. It will finish in reasonable time for up to 12 bits, you may wait for all 16 bits (there are 5736 possible pairs for that by the way, acquired with an overnight full search a while ago), and you may get a few 20 bits if you are patient. But O(n^3) is really nasty...
(Who will get to find the first full 32bit sequence?)
Other interesting questions which arise:
For both num1 and num2 only odd values are able to produce full sequences. Why? This may not be hard (simple logic, I guess), but I never reasonably proved it.
There is a mirroring property along num1 (the add value), that is, if 'a' with a given 'b' num2 gives a full sequence, then the 2 complement of 'a' (in the given bit depth) with the same num2 is also a full sequence. I only observed this happening reliably with all the full generations I calculated.
A third interesting property is that for all the num1 & num2 pairs the resulting sequences seem to form proper circles, that is, at least the number zero seems to be always part of a circle. Without this property my brute force search would die in an infinite loop.
Bonus: Was this PRNG already known before? (and I just re-invented it)?
And here is the brute force search's code (C):
#define BITS 16
#include "stdio.h"
#include "stdlib.h"
int main(void)
{
unsigned int r;
unsigned int c;
unsigned int num1;
unsigned int num2;
unsigned int mc=0U;
num1=1U; /* Only odd add values produce useful results */
do{
num2=1U; /* Only odd eor values produce useful results */
do{
r= 0U;
c=~0U;
do{
r=(((r>>(BITS-1)) & 1U)+r+r+num1)^num2;
r&=(1U<<(BITS-1)) | ((1U<<(BITS-1))-1U); /* 32bit safe */
c++;
}while (r);
if (c>=mc){
mc=c;
printf("Count-1: %08X, Num1(adc): %08X, Num2(eor): %08X\n", c, num1, num2);
}
num2+=2U;
num2&=(1U<<(BITS-1)) | ((1U<<(BITS-1))-1U);
}while(num2!=1U);
num1+=2U;
num1&=((1U<<(BITS-1))-1U); /* Do not check complements */
}while(num1!=1U);
return 0;
}
This, to show it is working, after each iteration will output the pair found if it's sequence length is equal or longer than the previous. Modify the BITS constant for sequences of other depths.
Seed hunting
I did some graphing relating to the seeds. Here is a nice image showing all the 9bit sequence lengths:
The white dots are the full length sequences, X axis is for num1 (add), Y axis is for num2 (xor), the brighter the dot, the longer the sequence. Other bit depth look very similar in pattern: they all seem to be broken up to sixteen major tiles with two patterns repeating with mirroring. The similarity of the tiles is not complete, for example above a diagonal from the up-left corner to the bottom-right is clearly visible while it's opposite is absent, but for the full-length sequences this property seems to be reliable.
Relying on this it is possible to reduce the work even more than by the previous assumptions, but that's still O(n^3)...
Performance analysis
As of current the longest sequences possible to be generated are 24bits: on my computer it takes at about 5 hours to brute-force a full 24bit sequence for this. This is still just so-so for real PRNG tests such as Diehard, so as of now I rather gone by an own approach.
First it's important to understand the role of the generator. This by no means would be a very good generator for it's simplicity, it's goal is rather to produce decent numbers blazing fast. On this region not needing multiply / divide operations, a Galois LFSR can produce similar performance. So my generator is of any use if it is capable to outperform this one.
The test I performed were all of 16bit generators. I chose this depth since it gives an useful sequence length while the numbers may still be broken up in two 8bit parts making it possible to present various bit-exact graphs for visual analysis.
The core of the tests were looking for correlations along previous and currently generated numbers. For this I used X:Y plots where the previous generation was the Y, the current the X, both broken up to low / high parts as above mentioned for two graphs. I created a program capable of plotting these stepped in real time so to also make it possible to roughly examine how the numbers follow each other, how the graphs fill up. Here obviously only the end results are shown as the generators ran through their full 2^16 or 2^16-1 (Galois) cycle.
The explanation of the fields:
The images consist 8x2 256x256 graphs making the total image size 2048x512 (check them at original size).
The top left graph just confirms that indeed a full sequence was plotted, it is simply an X = r % 256; Y = r / 256; plot.
The bottom left graph shows every second number only plotted the same way as the top, just confirming that the numbers occur reasonably randomly.
From the second graph the top row are the high byte correlation graphs. The first of them uses the previous generation, the next skips one number (so uses 2nd previous generation), and so on until the 7th previous generation.
From the second the bottom row are the low byte correlation graphs, organized the same way as above.
Galois generator, 0xB400 tap set
This is the generator found in the Wikipedia Galois example. It's performance is not the worst, but it is still definitely not really good.
Galois generator, 0xA55A tap set
One of the decent Galois "seeds" I found. Note that the low part of the 16bit numbers seem to be a lot better than the above, however I couldn't find any Galois "seed" which would fuzz up the high byte.
My generator, 0x7F25 (adc), 0x00DB (eor) seed
This is the best of my generators where the high byte of the EOR value is zero. Limiting the high byte is useful on 8bit machines since then this calculation can be omitted for smaller code and faster execution if the loss of randomness performance is affordable.
My generator, 0x778B (adc), 0x4A8B (eor) seed
This is one of the very good quality seeds by my measurements.
To find seeds with good correlation, I built a small program which would analyse them to some degree, the same way for Galois and mine. The "good quality" examples were pinpointed by that program, and then I tested several of them and selected one from those.
Some conclusions:
The Galois generator seems to be more rigid than mine. On all the correlation graphs definite geometrical patterns are observable (some seeds produce "checkerboard" patterns, not shown here) even if it is not composed of lines. My generator also shows patterns, but with more generations they grow less defined.
A portion of the Galois generator's result which include the bits in the high byte seems to be inherently rigid which property seems to be absent from my generator. This is a weak assumption yet probably needing some more research (to see if this is always so with the Galois generator and not with mine on other bit combinations).
The Galois generator lacks zero (maximal period being 2^16-1).
As of now it is impossible to generate a good set of seeds for my generator above 20 bits.
Later I might get in this subject deeper seeking to test the generator with Diehard, but as of now the lack of the ability of generating large enough seeds for it makes it impossible.
This is some form of a non-linear shift feedback register. I don't know if it has been used as such, but it resembles linear shift feedback registers somewhat. Read this Wikipedia page as an introduction to LSFRs. They are used frequently in pseudo random number generation.
However, your pseudo random number generator is inherently bad in that there is a linear correlation between the highest order bit of a previously generated number and the lowest order bit of a number generated next. You shift the highest bit B out, and then the lowest order bit of the new number will be the XOR or B, the lowest order bit of the additive constant num1 and the lowest order bit of the XORed constant num2, because binary addition is equivalent to exclusive or at the lowest order bit. Most likely your PRNG has other similar deficiencies. Creating good PRNGs is hard.
However, I must admit that the C64 code is pleasingly compact!
Related
I've been using Random (java.util.Random) to shuffle a deck of 52 cards. There are 52! (8.0658175e+67) possibilities. Yet, I've found out that the seed for java.util.Random is a long, which is much smaller at 2^64 (1.8446744e+19).
From here, I'm suspicious whether java.util.Random is really that random; is it actually capable of generating all 52! possibilities?
If not, how can I reliably generate a better random sequence that can produce all 52! possibilities?
Selecting a random permutation requires simultaneously more and less randomness than what your question implies. Let me explain.
The bad news: need more randomness.
The fundamental flaw in your approach is that it's trying to choose between ~2226 possibilities using 64 bits of entropy (the random seed). To fairly choose between ~2226 possibilities you're going to have to find a way to generate 226 bits of entropy instead of 64.
There are several ways to generate random bits: dedicated hardware, CPU instructions, OS interfaces, online services. There is already an implicit assumption in your question that you can somehow generate 64 bits, so just do whatever you were going to do, only four times, and donate the excess bits to charity. :)
The good news: need less randomness.
Once you have those 226 random bits, the rest can be done deterministically and so the properties of java.util.Random can be made irrelevant. Here is how.
Let's say we generate all 52! permutations (bear with me) and sort them lexicographically.
To choose one of the permutations all we need is a single random integer between 0 and 52!-1. That integer is our 226 bits of entropy. We'll use it as an index into our sorted list of permutations. If the random index is uniformly distributed, not only are you guaranteed that all permutations can be chosen, they will be chosen equiprobably (which is a stronger guarantee than what the question is asking).
Now, you don't actually need to generate all those permutations. You can produce one directly, given its randomly chosen position in our hypothetical sorted list. This can be done in O(n2) time using the Lehmer[1] code (also see numbering permutations and factoriadic number system). The n here is the size of your deck, i.e. 52.
There is a C implementation in this StackOverflow answer. There are several integer variables there that would overflow for n=52, but luckily in Java you can use java.math.BigInteger. The rest of the computations can be transcribed almost as-is:
public static int[] shuffle(int n, BigInteger random_index) {
int[] perm = new int[n];
BigInteger[] fact = new BigInteger[n];
fact[0] = BigInteger.ONE;
for (int k = 1; k < n; ++k) {
fact[k] = fact[k - 1].multiply(BigInteger.valueOf(k));
}
// compute factorial code
for (int k = 0; k < n; ++k) {
BigInteger[] divmod = random_index.divideAndRemainder(fact[n - 1 - k]);
perm[k] = divmod[0].intValue();
random_index = divmod[1];
}
// readjust values to obtain the permutation
// start from the end and check if preceding values are lower
for (int k = n - 1; k > 0; --k) {
for (int j = k - 1; j >= 0; --j) {
if (perm[j] <= perm[k]) {
perm[k]++;
}
}
}
return perm;
}
public static void main (String[] args) {
System.out.printf("%s\n", Arrays.toString(
shuffle(52, new BigInteger(
"7890123456789012345678901234567890123456789012345678901234567890"))));
}
[1] Not to be confused with Lehrer. :)
Your analysis is correct: seeding a pseudo-random number generator with any specific seed must yield the same sequence after a shuffle, limiting the number of permutations that you could obtain to 264. This assertion is easy to verify experimentally by calling Collection.shuffle twice, passing a Random object initialized with the same seed, and observing that the two random shuffles are identical.
A solution to this, then, is to use a random number generator that allows for a larger seed. Java provides SecureRandom class that could be initialized with byte[] array of virtually unlimited size. You could then pass an instance of SecureRandom to Collections.shuffle to complete the task:
byte seed[] = new byte[...];
Random rnd = new SecureRandom(seed);
Collections.shuffle(deck, rnd);
In general, a pseudorandom number generator (PRNG) can't choose from among all permutations of a 52-item list if its maximum cycle length is less than 226 bits.
java.util.Random implements an algorithm with a modulus of 248 and a maximum cycle length of not more than that, so much less than 2226 (corresponding to the 226 bits I referred to). You will need to use another PRNG with a bigger cycle length, specifically one with a maximum cycle length of 52 factorial or greater.
See also "Shuffling" in my article on random number generators.
This consideration is independent of the nature of the PRNG; it applies equally to cryptographic and noncryptographic PRNGs (of course, noncryptographic PRNGs are inappropriate whenever information security is involved).
Although java.security.SecureRandom allows seeds of unlimited length to be passed in, the SecureRandom implementation could use an underlying PRNG (e.g., "SHA1PRNG" or "DRBG"). And it depends on that PRNG's maximum cycle length whether it's capable of choosing from among 52 factorial permutations.
Let me apologize in advance, because this is a little tough to understand...
First of all, you already know that java.util.Random is not completely random at all. It generates sequences in a perfectly predictable way from the seed. You are completely correct that, since the seed is only 64 bits long, it can only generate 2^64 different sequences. If you were to somehow generate 64 real random bits and use them to select a seed, you could not use that seed to randomly choose between all of the 52! possible sequences with equal probability.
However, this fact is of no consequence as long as you're not actually going to generate more than 2^64 sequences, as long as there is nothing 'special' or 'noticeably special' about the 2^64 sequences that it can generate.
Lets say you had a much better PRNG that used 1000-bit seeds. Imagine you had two ways to initialize it -- one way would initialize it using the whole seed, and one way would hash the seed down to 64 bits before initializing it.
If you didn't know which initializer was which, could you write any kind of test to distinguish them? Unless you were (un)lucky enough to end up initializing the bad one with the same 64 bits twice, then the answer is no. You could not distinguish between the two initializers without some detailed knowledge of some weakness in the specific PRNG implementation.
Alternatively, imagine that the Random class had an array of 2^64 sequences that were selected completely and random at some time in the distant past, and that the seed was just an index into this array.
So the fact that Random uses only 64 bits for its seed is actually not necessarily a problem statistically, as long as there is no significant chance that you will use the same seed twice.
Of course, for cryptographic purposes, a 64 bit seed is just not enough, because getting a system to use the same seed twice is computationally feasible.
EDIT:
I should add that, even though all of the above is correct, that the actual implementation of java.util.Random is not awesome. If you are writing a card game, maybe use the MessageDigest API to generate the SHA-256 hash of "MyGameName"+System.currentTimeMillis(), and use those bits to shuffle the deck. By the above argument, as long as your users are not really gambling, you don't have to worry that currentTimeMillis returns a long. If your users are really gambling, then use SecureRandom with no seed.
I'm going to take a bit of a different tack on this. You're right on your assumptions - your PRNG isn't going to be able to hit all 52! possibilities.
The question is: what's the scale of your card game?
If you're making a simple klondike-style game? Then you definitely don't need all 52! possibilities. Instead, look at it like this: a player will have 18 quintillion distinct games. Even accounting for the 'Birthday Problem', they'd have to play billions of hands before they'd run into the first duplicate game.
If you're making a monte-carlo simulation? Then you're probably okay. You might have to deal with artifacts due to the 'P' in PRNG, but you're probably not going to run into problems simply due to a low seed space (again, you're looking at quintillions of unique possibilities.) On the flip side, if you're working with large iteration count, then, yeah, your low seed space might be a deal-breaker.
If you're making a multiplayer card game, particularly if there's money on the line? Then you're going to need to do some googling on how the online poker sites handled the same problem you're asking about. Because while the low seed space issue isn't noticeable to the average player, it is exploitable if it's worth the time investment. (The poker sites all went through a phase where their PRNGs were 'hacked', letting someone see the hole cards of all the other players, simply by deducing the seed from exposed cards.) If this is the situation you're in, don't simply find a better PRNG - you'll need to treat it as seriously as a Crypto problem.
Short solution which is essentially the same of dasblinkenlight:
// Java 7
SecureRandom random = new SecureRandom();
// Java 8
SecureRandom random = SecureRandom.getInstanceStrong();
Collections.shuffle(deck, random);
You don't need to worry about the internal state. Long explanation why:
When you create a SecureRandom instance this way, it accesses an OS specific
true random number generator. This is either an entropy pool where values are
accessed which contain random bits (e.g. for a nanosecond timer the nanosecond
precision is essentially random) or an internal hardware number generator.
This input (!) which may still contain spurious traces are fed into a
cryptographically strong hash which removes those traces. That is the reason those CSPRNGs are used, not for creating those numbers themselves! The SecureRandom has a counter which traces how many bits were used (getBytes(), getLong() etc.) and refills the SecureRandom with entropy bits when necessary.
In short: Simply forget objections and use SecureRandom as true random number generator.
If you consider the number as just an array of bits (or bytes) then maybe you could use the (Secure)Random.nextBytes solutions suggested in this Stack Overflow question, and then map the array into a new BigInteger(byte[]).
A very simple algorithm is to apply SHA-256 to a sequence of integers incrementing from 0 upwards. (A salt can be appended if desired to "get a different sequence".) If we assume that the output of SHA-256 is "as good as" uniformly distributed integers between 0 and 2256 - 1 then we have enough entropy for the task.
To get a permutation from the output of SHA256 (when expressed as an integer) one simply needs to reduce it modulo 52, 51, 50... as in this pseudocode:
deck = [0..52]
shuffled = []
r = SHA256(i)
while deck.size > 0:
pick = r % deck.size
r = floor(r / deck.size)
shuffled.append(deck[pick])
delete deck[pick]
My Empirical research results are Java.Random is not totally truly random. If you try yourself by using Random class "nextGaussian()"-method and generate enough big sample population for numbers between -1 and 1, the graph is normal distbruted field know as Gaussian Model.
Finnish goverment owned gambling-bookmarker have a once per day whole year around every day drawn lottery-game where winning table shows that the Bookmarker gives winnings in normal distrbuted way. My Java Simulation with 5 million draws shows me that with nextInt() -methdod used number draw, winnings are normally distributed same kind of like the my Bookmarker deals the winnings in each draw.
My best picks are avoiding numbers 3 and 7 in each of ending ones and that's true that they are rarely in winning results. Couple of times won five out of five picks by avoiding 3 and 7 numbers in ones column in Integer between 1-70 (Keno).
Finnish Lottery drawn once per week Saturday evenings If you play System with 12 numbers out of 39, perhaps you get 5 or 6 right picks in your coupon by avoiding 3 and 7 values.
Finnish Lottery have numbers 1-40 to choose and it takes 4 coupon to cover all the nnumbers with 12 number system. The total cost is 240 euros and in long term it's too expensive for the regural gambler to play without going broke. Even if you share coupons to other customers available to buy still you have to be quite a lucky if you want to make profit.
I have a very specific problem:
I have uniformly random values spread on a 15x50 grid and the sample I want to hash corresponds to a square of 5x5 cells centered around any possible grid position.
The number of samples can thus vary from 25 (away from borders, most cases) to 20, 15 (near a border) down to a minimum of 9 (in a corner).
So even though the cell values are random, the location introduces a deterministic variation in the sequence length.
The hash table size is a small number, typically between 50 and 20.
The function will operate on a large set of randomly generated grids (a few hundreds/thousands), and might be called a few thousands times per grid. The positions on the grid can be considered random.
I would like a function that could spread the 15x50 possible samples as evenly as possible.
I have tried the following pseudo-code:
int32 hash = 0;
int i = 0; // I guess i could take any initial value and even be left uninitialized, but fixing one makes the function deterministic
foreach (value in block)
{
hash ^= (value << (i%28))
i++
}
hash %= table_size
but the results, though not grossly imbalanced, do not seem very smooth to me. Maybe it's because the sample is too small, but the circumstances make it difficult to run the code on a bigger sample, and I would rather not have to write a complete test harness if some computer savvy has an answer ready for me :).
I am not sure pairing the values two by two and using a general purpose byte hashing strategy would be the best solution, especially since the number of values might be odd.
I have tought of using a 17th value to represent off-grid cells, but that seems to introduce a bias (the sequences from cells near a border will have a lot of "off grid" values).
I am not sure either what would be the best way to test the efficiency of various solutions (how many grids shall I generate to have an idea of the performances, for instance).
http://www.partow.net/programming/hashfunctions/
Here are few different hash function from experts on various fields. Functions are designed for 8bit values, but I am sure you can extend for your case. I dont know what to suggest, but I think that any of them should work better than your current idea.
Problem with current approach you propose is that values are cyclic in field 2^n and if you make mod 64 at the end for example you lost most values out and only last 3 values remains in final result.
Despite your scepticism I would just shove them through a standard hash function.
If they are well randomised (and relatively independent - you don't say) to begin with you probably don't need to do too much work. Fowler-Noll-Vo (FNV) is a good candidate in these circumstances.
FNV operates on a series of 8-bit input and your input is (logically) 4-bit.
I would start without even bothering to pack 'two by two' as you describe.
If you feel like trying that, just logically pad odd length series with the message length (reduced to a 4 bit value obviously).
I wouldn't expect that packing to improve the hash. It may save you a tiny number of cycles because it swaps a relatively expensive * with a << and a |.
Try both and report back!
Here are implementations of packed and 'normal' versions of FNV1a in C:
#include <inttypes.h>
static const uint32_t sFNVOffsetBasis=2166136261;
static const uint32_t sFNVPrime= 16777619;
const uint32_t FNV1aPacked4Bit(const uint8_t*const pBytes,const size_t pSize) {
uint32_t rHash=sFNVOffsetBasis;
for(size_t i=0;i<pSize;i+=2){
rHash=rHash^(pBytes[i]|(pBytes[i+1]<<4));
rHash=rHash*sFNVPrime;
}
if(pSize%2){//Length is odd. The loop missed the last element.
rHash=rHash^(pBytes[pSize-1]|((pSize&0x1E)<<3));
rHash=rHash*sFNVPrime;
}
return rHash;
}
const uint32_t FNV1a(const uint8_t*const pBytes,const size_t pSize) {
uint32_t rHash=sFNVOffsetBasis;
for(size_t i=0;i<pSize;++i){
rHash=(rHash^pBytes[i])*sFNVPrime;
}
return rHash;
}
NB: I've edited it to skip the first bit when adding in the length. Obviously the bottom bit of an odd length is 100% biased to 1. I don't know how length is distributed. It may be wiser to put it in at the start than the end.
I'm programming a PLC with some legacy software (RSLogix 500, don't ask) and it does not natively support a modulus operation, but I need one. I do not have access to: modulus, integer division, local variables, a truncate operation (though I can hack it with rounding). Furthermore, all variables available to me are laid out in tables sorted by data type. Finally, it should work for floating point decimals, for example 12345.678 MOD 10000 = 2345.678.
If we make our equation:
dividend / divisor = integer quotient, remainder
There are two obvious implementations.
Implementation 1:
Perform floating point division: dividend / divisor = decimal quotient. Then hack together a truncation operation so you find the integer quotient. Multiply it by the divisor and find the difference between the dividend and that, which results in the remainder.
I don't like this because it involves a bunch of variables of different types. I can't 'pass' variables to a subroutine, so I just have to allocate some of the global variables located in multiple different variable tables, and it's difficult to follow. Unfortunately, 'difficult to follow' counts, because it needs to be simple enough for a maintenance worker to mess with.
Implementation 2:
Create a loop such that while dividend > divisor divisor = dividend - divisor. This is very clean, but it violates one of the big rules of PLC programming, which is to never use loops, since if someone inadvertently modifies an index counter you could get stuck in an infinite loop and machinery would go crazy or irrecoverably fault. Plus loops are hard for maintenance to troubleshoot. Plus, I don't even have looping instructions, I have to use labels and jumps. Eww.
So I'm wondering if anyone has any clever math hacks or smarter implementations of modulus than either of these. I have access to + - * /, exponents, sqrt, trig functions, log, abs value, and AND/OR/NOT/XOR.
How many bits are you dealing with? You could do something like:
if dividend > 32 * divisor dividend -= 32 * divisor
if dividend > 16 * divisor dividend -= 16 * divisor
if dividend > 8 * divisor dividend -= 8 * divisor
if dividend > 4 * divisor dividend -= 4 * divisor
if dividend > 2 * divisor dividend -= 2 * divisor
if dividend > 1 * divisor dividend -= 1 * divisor
quotient = dividend
Just unroll as many times as there are bits in dividend. Make sure to be careful about those multiplies overflowing. This is just like your #2 except it takes log(n) instead of n iterations, so it is feasible to unroll completely.
If you don't mind overly complicating things and wasting computer time you can calculate modulus with periodic trig functions:
atan(tan(( 12345.678 -5000)*pi/10000))*10000/pi+5000 = 2345.678
Seriously though, subtracting 10000 once or twice (your "implementation 2") is better. The usual algorithms for general floating point modulus require a number of bit-level manipulations that are probably unfeasible for you. See for example http://www.netlib.org/fdlibm/e_fmod.c (The algorithm is simple but the code is complex because of special cases and because it is written for IEEE 754 double precision numbers assuming there is no 64-bit integer type)
This all seems completely overcomplicated. You have an encoder index that rolls over at 10000 and objects rolling along the line whose positions you are tracking at any given point. If you need to forward project stop points or action points along the line, just add however many inches you need and immediately subtract 10000 if your target result is greater than 10000.
Alternatively, or in addition, you always get a new encoder value every PLC scan. In the case where the difference between the current value and last value is negative you can energize a working contact to flag the wrap event and make appropriate corrections for any calculations on that scan. (**or increment a secondary counter as below)
Without knowing more about the actual problem it is hard to suggest a more specific solution but there are certainly better solutions. I don't see a need for MOD here at all. Furthermore, the guys on the floor will thank you for not filling up the machine with obfuscated wizard stuff.
I quote :
Finally, it has to work for floating point decimals, for example
12345.678 MOD 10000 = 2345.678
There is a brilliant function that exists to do this - it's a subtraction. Why does it need to be more complicated than that? If your conveyor line is actually longer than 833 feet then roll a second counter that increments on a primary index roll-over until you've got enough distance to cover the ground you need.
For example, if you need 100000 inches of conveyor memory you can have a secondary counter that rolls over at 10. Primary encoder rollovers can be easily detected as above and you increment the secondary counter each time. Your working encoder position, then, is 10000 times the counter value plus the current encoder value. Work in the extended units only and make the secondary counter roll over at whatever value you require to not lose any parts. The problem, again, then reduces to a simple subtraction (as above).
I use this technique with a planetary geared rotational part holder, for example. I have an encoder that rolls over once per primary rotation while the planetary geared satellite parts (which themselves rotate around a stator gear) require 43 primary rotations to return to an identical starting orientation. With a simple counter that increments (or decrements, depending on direction) at the primary encoder rollover point it gives you a fully absolute measure of where the parts are at. In this case, the secondary counter rolls over at 43.
This would work identically for a linear conveyor with the only difference being that a linear conveyor can go on for an infinite distance. The problem then only needs to be limited by the longest linear path taken by the worst-case part on the line.
With the caveat that I've never used RSLogix, here is the general idea (I've used generic symbols here and my syntax is probably a bit wrong but you should get the idea)
With the above, you end up with a value ENC_EXT which has essentially transformed your encoder from a 10k inch one to a 100k inch one. I don't know if your conveyor can run in reverse, if it can you would need to handle the down count also. If the entire rest of your program only works with the ENC_EXT value then you don't even have to worry about the fact that your encoder only goes to 10k. It now goes to 100k (or whatever you want) and the wraparound can be handled with a subtraction instead of a modulus.
Afterword :
PLCs are first and foremost state machines. The best solutions for PLC programs are usually those that are in harmony with this idea. If your hardware is not sufficient to fully represent the state of the machine then the PLC program should do its best to fill in the gaps for that missing state information with the information it has. The above solution does this - it takes the insufficient 10000 inches of state information and extends it to suit the requirements of the process.
The benefit of this approach is that you now have preserved absolute state information, not just for the conveyor, but also for any parts on the line. You can track them forward and backward for troubleshooting and debugging and you have a much simpler and clearer coordinate system to work with for future extensions. With a modulus calculation you are throwing away state information and trying to solve individual problems in a functional way - this is often not the best way to work with PLCs. You kind of have to forget what you know from other programming languages and work in a different way. PLCs are a different beast and they work best when treated as such.
You can use a subroutine to do exactly what you are talking about. You can tuck the tricky code away so the maintenance techs will never encounter it. It's almost certainly the easiest for you and your maintenance crew to understand.
It's been a while since I used RSLogix500, so I might get a couple of terms wrong, but you'll get the point.
Define a Data File each for your floating points and integers, and give them symbols something along the lines of MOD_F and MOD_N. If you make these intimidating enough, maintenance techs leave them alone, and all you need them for is passing parameters and workspace during your math.
If you really worried about them messing up the data tables, there are ways to protect them, but I have forgotten what they are on a SLC/500.
Next, defined a subroutine, far away numerically from the ones in use now, if possible. Name it something like MODULUS. Again, maintenance guys almost always stay out of SBRs if they sound like programming names.
In the rungs immediately before your JSR instruction, load the variables you want to process into the MOD_N and MOD_F Data Files. Comment these rungs with instructions that they load data for MODULUS SBR. Make the comments clear to anyone with a programming background.
Call your JSR conditionally, only when you need to. Maintenance techs do not bother troubleshooting non-executing logic, so if your JSR is not active, they will rarely look at it.
Now you have your own little walled garden where you can write your loop without maintenance getting involved with it. Only use those Data Files, and don't assume the state of anything but those files is what you expect. In other words, you cannot trust indirect addressing. Indexed addressing is OK, as long as you define the index within your MODULUS JSR. Do not trust any incoming index. It's pretty easy to write a FOR loop with one word from your MOD_N file, a jump and a label. Your whole Implementation #2 should be less than ten rungs or so. I would consider using an expression instruction or something...the one that lets you just type in an expression. Might need a 504 or 505 for that instruction. Works well for combined float/integer math. Check the results though to make sure the rounding doesn't kill you.
After you are done, validate your code, perfectly if possible. If this code ever causes a math overflow and faults the processor, you will never hear the end of it. Run it on a simulator if you have one, with weird values (in case they somehow mess up the loading of the function inputs), and make sure the PLC does not fault.
If you do all that, no one will ever even realize you used regular programming techniques in the PLC, and you will be fine. AS LONG AS IT WORKS.
This is a loop based on the answer by #Keith Randall, but it also maintains the result of the division by substraction. I kept the printf's for clarity.
#include <stdio.h>
#include <limits.h>
#define NBIT (CHAR_BIT * sizeof (unsigned int))
unsigned modulo(unsigned dividend, unsigned divisor)
{
unsigned quotient, bit;
printf("%u / %u:", dividend, divisor);
for (bit = NBIT, quotient=0; bit-- && dividend >= divisor; ) {
if (dividend < (1ul << bit) * divisor) continue;
dividend -= (1ul << bit) * divisor;
quotient += (1ul << bit);
}
printf("%u, %u\n", quotient, dividend);
return dividend; // the remainder *is* the modulo
}
int main(void)
{
modulo( 13,5);
modulo( 33,11);
return 0;
}
For several valid reasons I have to use BSD's random() to generate awfully large amounts of random numbers, and since its cycle is quite short (~2^69, if I'm not mistaken) the quality of such numbers degrades pretty quickly for my use case. I could use the rng board I have access to but it's painfully slow so I thought I could do this trick: take one number from the board, use it to seed random(), use random() to draw numbers and reseed it when the board says a new number is available. The board generates about 100 numbers per second so my guess is that random() hardly gets to cycle over and the generation rate easily keeps up with my requirements of several millions numbers per second.
Anyway, the problem is that random() claims to uniformly draw numbers between 0 and (2^31)-1, but I've been drawing an uncountable amount of numbers and I've never ever seen a 0 nor a (2^31)-1 so far. Maybe some 1 and (2^31)-2, but I've never seen the extremes. Now, I know the problem with random numbers is that you can never be sure (see Dilbert, Debian), but this seem extremely odd nonetheless. Moreover I tried analysing the generated datasets with Octave using the histc() function, and the lowest and the highest bins contain between half and three quarter the amount of numbers of the middle bins (which in turn are uniformly filled, so I guess in some sense the distribution is "uniform").
Can anybody explain this?
EDIT Some code
The board outputs this structure with the three components, and then I do some mumbo-jumbo combining them to produce the seed. I have no specs about this board, it's an ancient piece of hardware thrown together by a previous student some years ago, there's little documentation and this formula I'm using is one of those suggested in the docs. The STEP parameter tells me how may numbers I can draw using one seed so I can optimise performance and throttle down CPU usage at the same time.
float n = fabsf(fmod(sqrt(a.s1*a.s1 + a.s2*a.s2 + a.s3*a.s3), 1.0));
unsigned int seed = n * UINT32_MAX;
srandom(seed);
for(int i = 0; i < STEP; i++) {
long r = random();
n = (float)r / (UINT32_MAX >> 1);
[_numbers addObject:[NSNumber numberWithFloat:n]];
}
Are you certain that
void main() {
while (random() != 0L);
}
hangs indefinitely? On my linux machine (the Gnu C library uses the same linear feedback shift register as BSD, albeit with a different seeding procedure) it doesn't.
According to this reference the algorithm produces 'runs' of consecutive zeroes or ones up to length n-1 where n is the size of the shift register. When this has a size of 31 integers (the default case) we can even be certain that, eventually, random() will return 0 a whopping 30 (but never 31) times in a row! Of course, we may have to wait a few centuries to see it happening...
To extend the cycle length, one method is to run two RNGs, with different periods, and XOR their output. See L'Ecuyer 1988 for some examples.
I have five colors stored in the format #AARRGGBB as unsigned ints, and I need to take the average of all five. Obviously I can't simply divide each int by five and just add them, and the only way I thought of so far is to bitmask them, do each channel separately, and then OR them together again. Is there a clever or concise way of averaging all five of them?
Half way between your (OP) proposed solution and Patrick's solution looks quite neat:
Color colors[5]={ 0xAARRGGBB,...};
unsigned long sum1=0,sum2=0;
for (int i=0;i<5;i++)
{
sum1+= colors[i] &0x00FF00FF; // 0x00RR00BB
sum2+=(colors[i]>>8)&0x00FF00FF; // 0x00AA00GG
}
unsigned long output=0;
output|=(((sum1&0xFFFF)/5)&0xFF);
output|=(((sum2&0xFFFF)/5)&0xFF)<<8;
sum1>>=16;sum2>>=16; // and now the top halves
output|=(((sum1&0xFFFF)/5)&0xFF)<<16;
output|=(((sum2&0xFFFF)/5)&0xFF)<<24;
I don't think you could really divide sum1/sum2 by 5, because the bits from the top half would spill down...
If an approximation would be valid, you could try a multiplication by something like, 0.1875 (0.125+0.0625), (this means: multiply by 3 and shift down by 4 places. This you could do with bitmasking and care.)
The problem is, 0.2 has a crappy binary representation, so multiplying by it is an ass.
As ever, accuracy or speed. Your choice.
When using x86 machines with at least SSE, and if you need to approximate only, you could use the assembly instruction PAVGB (Packed Average Byte), which averages bytes. See http://www.tommesani.com/SSEPrimer.html for explanation.
Since you've got 5 values, you would need to be creative in calling PAVGB, since PAVGB will only do two values at a time.
I found smart solution of your problem, sadly it is only applicable if number of colors is power of 2. I'll show it in case of two colors:
mask = 01010101
pom = ~(a^b & mask) # ^ means xor here, ~ negation
a = a & pom
b = b & pom
avg = (a+b) >> 1
The trick of this method is — when you count average, LSB of sum (in case of two numbers) has no meaning, as it will be dropped in division (we're talking integers here, of course). In your problem, LSB of partial sums is at the same moment carry bit of sum of adjacent color. Provided, that LSB of every color sum will be 0 you can safely add those two integers — additions won't interfere with each other. Bit shift divides every color by two.
This method can be used with 4 colors as well, but you have to implement finding out the carry flag of sum of numbers made of two last bits of every color. It is also possible to omit this part and just zero last two bits of every color — biggest mistake made with this omission is 1 for every component.
EDIT I'll leave this attempt for posterity, but please note that it is incorrect and will not work.
One "clever" way you could do it would be to insert zeros between the components, parse into an unsigned long, average the numbers, convert back to a hex string, remove the zeros and finally parse into an unsigned int.
i.e. convert #AARRGGBB to #AA00RR00GG00BB
This method involves parsing and string manipulations, so will undoubtedly be slower than the method you proposed.
If you were to factor your own solution carefully, it might actually look quite clever itself.