Why does a 16bit LFSR pass all Diehard tests? - random

I made a implementation of an LFSR in Hardware. It was based on the LFSR from the wikipedia page. It has the same output.
It pass all the tests for dieharder, however, if I plot the pairs I get this not very random lines in 2D
2D plot of the LFSR
So, how can I have a strong statistical test that could prove this PRNG is not ideal?
SOLVED:
I have to use comand
dieharder -a -f exemple_LSRF_BS_1_DH.txt -g 202
And add a header to the output file I have my numbers.

By the looks of your plot, I'd guess that your random bitstream emits the entire register after each cycle, rather than emitting just one bit per cycle. This means that when viewed as 16-bit words, x_(n+1) is either X_n / 2 or X_n / 2 + 32768. This manifests as two diagonal lines with a gradient of 0.5 (or 2.0, depending on the order and/or shift direction).
Normal usage of an LFSR would emit either one bit per cycle, or all n bits every n cycles. This does produce some negative properties, but they're not so obvious as what you've shown.
As for why your test fails dieharder; I think there must be a flaw in your test set-up. I modified code from Wikipedia to emit the 16-bit state on every cycle on stdout, and piped that into dieharder -a -g200, and it failed immediately on the first five tests. That's what one would expect; even hexdump -C shows obvious visible patterns.
Modifying the code to emit the 16-bit state every 16 cycles, hexdump -C looks much more random, but dieharder still fails just the same.
Perhaps you didn't specify the generator source to dieharder, and so it used its default internal generator. You can confirm this on line five of the output:
rng_name |rands/second| Seed |
mt19937| 1.42e+08 |1473327481|
Even so, if you want more thorough tests for your generators, have a look at TestU01.

Related

Genetic programming: Can we get multiple bit output from each individual(tree) if we provide 5 bit input?

I want to create population of trees using genetic programming. I am using deap python framework. My trees are based on combination of different logic gates and some terminals.
I want to provide each tree/individual 5 bit input and want to get 32 bit output from each tree. I will calculate fitness from these 32 bits.
Issue: When I provide 5 bit input then Each tree evaluates to single bit output as it is based on logic gates. Is there any way I design the individuals in such a way that they provide me 32 bit output whatever the size of input is ?
Possible approach 1 : Run 5 bit input 32 times . But that will always produce the same same output bits. And I don't want to use random in terminal because I want to produce same output if the input is same.
2: get the output from each node in the tree : But don't know how to do it in deap. Also the number of nodes vary in each individual.
Solution: I think that if I provide 128 bit binary value to one tree (made of and/or/not/xor gates etc) and do bitwise operation on each gate then I might get 128 bit output.

Is it acceptable to use each byte of a PRNG-generated number separately?

Say you have a non-cryptographically secure PRNG that generates 64-bit output.
Assuming that bytes are 8 bits, is it acceptable to use each byte of the 64-bit output as separate 8-bit random numbers or would that possibly break the randomness guarantees of a good PRNG? Or does it depend on the PRNG?
Because the PRNG is not cryptographically secure, the "randomness guarantee" I am worried about is not security, but whether the byte stream has the same guarantee of randomness, using the same definition of "randomness" that PRNG authors use, that the PRNG has with respect to its 64-bit output.
This should be quite safe with a CSPRNG. For comparison it's like reading /dev/random byte by byte. With a good CSPRNG it is also perfectly acceptable to simply generate a 64bit sample 8 times and pick 8 bits per sample as well (throwing away the 56 other bits).
With PRNGs that are not CSPRNG you will have 'security' concerns in terms of the raw output of the PRNG that outweigh whether or not you chop up output into byte sized chunks.
In all cases it is vital to make sure the PRNG is seeded and periodically re-seeded correctly (so as to flush any possibly compromised internal state regularly). Security depends on the unpredictability of your internal state, which is ultimately driven by the quality of your seed input. One thing good CSPRNG implementations will do for you is to pessimistically estimate the amount of captured 'entropy' to safeguard the output from predictable internal state.
Note however that with 8 bits you only have 256 possible outputs in any case, so it becomes more of a question of how you use this. For instance, if you do something like XOR based encryption against the output of a PRNG (i.e. treating it as a one time pad based on some pre shared secret seed), then using a known plain text attack may relatively easily reveal the contents of the internal state of the PRNG. That is another type of attack which good CSPRNG implementations are supposed to guard against by their design (using e.g. a computationally secure hash function).
EDIT to add: if you don't care about 'security' but only need the output to look random, then this should be quite safe -- in theory a good PRNG is just as likely to yield a 0 as 1, and that should not vary between any octet. So you expect a linear distribution of possible output values. One thing you can do to verify whether this skews the distribution is to run a Monte Carlo simulation of some reasonably large size (e.g. 1M) and compare the histograms with 256 bins for both the raw 64 bit and the 8 * 8 bit output. You expect a roughly flat diagram for both cases if the linear distribution is preserved intact.
It depends on the generator and its parameterization. Quoting from the Wikipedia page for Linear Congruential Generators: "The low-order bits of LCGs when m is a power of 2 should never be relied on for any degree of randomness whatsoever. [...]any full-cycle LCG when m is a power of 2 will produce alternately odd and even results."

How exactly does PC/Mac generates random numbers for either 0 or 1?

This question is NOT about how to use any language to generate a random number between any interval. It is about generating either 0 or 1.
I understand that many random generator algorithm manipulate the very basic random(0 or 1) function and take seed from users and use an algorithm to generate various random numbers as needed.
The question is that how the CPU generate either 0 or 1? If I throw a coin, I can generate head or tailer. That's because I physically throw a coin and let the nature decide. But how does CPU do it? There must be an action that the CPU does (like throwing a coin) to get either 0 or 1 randomly, right?
Could anyone tell me about it?
Thanks
(This has several facets and thus several algorithms. Keep in mind that there are many different forms of randomness used for different purposes, but I understand your question in the way that you are interested in actual randomness used for cryptography.)
The fundamental problem here is that computers are (mostly) deterministic machines. Given the same input in the same state they always yield the same result. However, there are a few ways of actually gathering entropy:
User input. Since users bring outside input into the system you can take that to derive some bits from that. Similar to how you could use radioactive decay or line noise.
Network activity. Again, an outside source of stuff.
Generally interrupts (which kinda include the first two).
As alluded to in the first item, noise from peripherals, such as audio input or a webcam can be used.
There is dedicated hardware that can generate a few hundred MiB of randomness per second. Usually they give you random numbers directly instead of their internal entropy, though.
How exactly you derive bits from that is up to you but you could use time between events, or actual content from the events, etc. – generally eliminating bias from entropy sources isn't easy or trivial and a lot of thought and algorithmic work goes into that (in the case of the aforementioned special hardware this is all done in hardware and the code using it doesn't need to care about it).
Once you have a pool of actually random bits you can just use them as random numbers (/dev/random on Linux does that). But this has downsides, since there is usually little actual entropy and possibly a higher demand for random numbers. So you can invent algorithms to “stretch” that initial randomness in a manner that makes it still impossible or at least very difficult to predict anything about following numbers (/dev/urandom on Linux or both /dev/random and /dev/urandom on FreeBSD do that). Fortuna and Yarrow are so-called cryptographically secure pseudo-random number generators and designed with that in mind. You still have a very good guarantee about the quality of random numbers you generate, but have many more before your entropy pool runs out.
In any case, the CPU itself cannot give you a random 0 or 1. There's a lot more involved and this usually includes the complete computer system or special hardware built for that purpose.
There is also a second class of computational randomness: Plain vanilla pseudo-random number generators (PRNGs). What I said earlier about determinism – this is the embodiment of it. Given the same so-called seed a PRNG will yield the exact same sequence of numbers every time¹. While this sounds idiotic it has practical benefits.
Suppose you run a simulation involving lots of random numbers, maybe to simulate interaction between molecules or atoms that involve certain probabilities and unpredictable behaviour. In science you want results anyone can independently verify, given the same setup and procedure (or, with computing, the same algorithms). If you used actual randomness the only option you have would be to save every single random number used to make sure others can replicate the results independently.
But with a PRNG all you need to save is the seed and remember what algorithm you used. Others can then get the exact same sequence of pseudo-random numbers independently. Very nice property to have :-)
Footnotes
¹ This even includes the CSPRNGs mentioned above, but they are designed to be used in a special way that includes regular re-seeding with entropy to overcome that problem.
A CPU can only generate a uniform random number, U(0,1), which happens to range from 0 to 1. So mathematically, it would be defined as a random variable U in the range [0,1]. Examples of random draws of a U(0,1) random number in the range 0 to 1 would be 0.28100002, 0.34522, 0.7921, etc. The probability of any value between 0 and 1 is equal, i.e., they are equiprobable.
You can generate binary random variates that are either 0 or 1 by setting a random draw of U(0,1) to a 0 if U(0,1)<=0.5 and 1 if U(0,1)>0.5, since in theory there will be an equal number of random draws of U(0,1) below 0.5 and above 0.5.

Integer to Binary Conversion in Simulink

This might look a repetition to my earlier question. But I think its not.
I am looking for a technique to convert the signal in the Decimal format to binary format.
I intend to use the Simulink blocks in the Xilinx Library to convert decimal to binary format.
So if the input is 3, the expected output should in 11( 2 Clock Cycles). I am looking for the output to be obtained serially.
Please suggest me how to do it or any pointers in the internet would be helpful.
Thanks
You are correct, what you need is the parallel to serial block from system generator.
It is described in this document:
http://www.xilinx.com/support/documentation/sw_manuals/xilinx13_1/sysgen_ref.pdf
This block is a rate changing block. Check the mentions of the parallel to serial block in these documents for further descriptions:
http://www.xilinx.com/support/documentation/sw_manuals/xilinx13_1/sysgen_gs.pdf
http://www.xilinx.com/support/documentation/sw_manuals/xilinx13_1/sysgen_user.pdf
Use a normal constant block with a Matlab variable in it, this already gives the output in "normal" binary (assuming you set the properties on it to be unsigned and the binary point at 0.
Then you need to write a small serialiser block, which takes that input, latches it into a shift register and then shifts the register once per clock cycle with the bit that "falls off the end" becoming your output bit. Depending on which way your shift, you can make it come MSB first of LSB first.
You'll have to build the shift register out of ordinary registers and a mux before each one to select whether you are doing a parallel load or shifting. (This is the sort of thing which is a couple of lines of code in VHDL, but a right faff in graphics).
If you have to increase the serial rate, you need to clock it from a faster clock - you could use a DCM to generate this.
Matlab has a dec2bin function that will convert from a decimal number to a binary string. So, for example dec2bin(3) would return 11.
There's also a corresponding bin2dec which takes a binary string and converts to a decimal number, so that bin2dec('11') would return 3.
If you're wanting to convert a non-integer decimal number to a binary string, you'll first want to determine what's the smallest binary place you want to represent, and then do a little bit of pre- and post-processing, combined with dec2bin to get the results you're looking for. So, if the smallest binary place you want is the 1/512th place (or 2^-9), then you could do the following (where binPrecision equals 1/512):
function result = myDec2Bin(decNum, binPrecision)
isNegative=(decNum < 0);
intPart=floor(abs(decNum));
binaryIntPart=dec2bin(intPart);
fracPart=abs(decNum)-intPart;
scaledFracPart=round(fracPart / binPrecision);
scaledBinRep=dec2bin(scaledFracPart);
temp=num2str(10^log2(1/binPrecision)+str2num(scaledBinRep),'%d');
result=[binaryIntPart,'.',temp(2:end)];
if isNegative
result=['-',result];
end
end
The result of myDec2Bin(0.256, 1/512) would then be 0.010000011, and the result of myDec2Bin(-0.984, 1/512) would be -0.111111000. (Note that the output is a string.)

What is redundant versus non-redundant number format?

I'm having trouble understanding the algorithm being used in this FPGA circuit. It deals with redundant versus non-redundant number format. I have seen some mathematical (formal) definitions of non-redundant format but I just can't really grasp it.
Excerpt from this paper describing the algorithm:
Figure 3 shows a block diagram of the scalable Montgomery multiplier. The kernel contains p w-bit PEs for a total of wp bit cells. Z is stored in carry-save redundant form. If PE p completes Z^0 before PE1 has finished Z^(e-1), the result must be queued until PE1 becomes available again. The design in [5] queues the results in redundant form, requiring 2w bits per entry. For large n the queue consumes significant area, so we propose converting Z to nonredundant form to save half the queue space, as shown in Figure 4. On the first cycle, Z is initialized to 0. When no queuing is needed, the carry-save redundant Z' is bypassed directly to avoid the latency of the carry-propagate adder. The nonredundant Z result is also an output of the system.
And the diagrams:
And here is the "improved" PE block diagram. This shows the 'improved' PE block diagram - 'improved' has to do with some unrelated aspects.
I don't have a picture of the 'not improved' FIFO but I think it is just a straight normal FIFO. What I don't understand is, does the FIFO's CPA and 3 input MUX somehow convert between formats?
Understanding redundant versus non-redundant formats (in concrete examples) is the first step, understanding how this circuit achieves it would be step 2..
A bit of background and a look at users.ece.utexas.edu/~adnan/vlsi-05-backup/lec12Datapath.ppt suggests the following:
Doing a proper binary add is relatively slow and/or area-consuming, because of the time that it takes to propagate the carries properly.
If you work bit-wise in parallel you can take three binary numbers, sum the bits at the same location in each number, and produce two binary numbers.
Slide 27 points out that 0001 + 0111 + 1101 = 1011 + 0101(0).
Since a multiplier needs to do a LOT of additions, you build the adder tree as a collection of reductions of 3 numbers to 2 numbers, eventually ending up with two numbers as output, abcde....z
and ABCDE...Z0. This is your output in redundant form, and the true answer is in fact abcde...z + ABCDE...Z0

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