Binary Calculator using Picaxe microchip? - binary

I want to be a complete nerd and make a very simple binary calculator.
It will be two rows of 8 switches, each switch representing a bit, so a row is a byte (number), the two rows are added together, and a row of 9 LED's will display the result in binary.
Is this possible to do with a picaxe microchip?
If not, what could I do it with?
Cheers,
Nick

Your problem would be data input/output lines. The basic idea is trivial in any microcontroller, but it's the number of input/output pins available.
You might want to look into several shift registers (one per row and one per output) so you can marshal the bits in on a single pin or two and out on a single pin.
Specifically:
74hc165n parallel-in/serial-out for the inputs
74hc595 for the output.

Related

Nand2Tetris-Obtaining Register from RAM chips

I've recently completed Chapter 3 of the associated textbook for this course: The Elements of Computing, Second Edition.
While I was able to implement all of the chips described in this chapter, I am still trying to wrap my head around how exactly the RAM chips work. I think I understand them in theory (e.g. a Ram4K chip stores a set of 8 RAM512 chips, which itself is a set of 8 RAM64 chips).
What I am unsure about is actually using the chips. For example, suppose I try to output a single register from RAM16K using this code, given an address:
CHIP RAM16K {
IN in[16], load, address[14];
OUT out[16];
PARTS:
Mux4Way16(a=firstRam, b=secondRam, c=thirdRam, d=fourthRam, sel=address[12..13], out=out);
And(a=load, b=load, out=shouldLoad);
DMux4Way(in=shouldLoad, sel=address[12..13], a=setRamOne, b=setRamTwo, c=setRamThree, d=setRamFour);
RAM4K(in=in, load=setRamOne, address=address[0..11], out=firstRam);
RAM4K(in=in, load=setRamTwo, address=address[0..11], out=secondRam);
RAM4K(in=in, load=setRamThree, address=address[0..11], out=thirdRam);
RAM4K(in=in, load=setRamFour, address=address[0..11], out=fourthRam);
}
How does the above code get the underlying register? If I understand the description of the chip correctly, it is supposed to return a single register. I can see that it outputs a RAM4K based on a series of address bits -- does it also get the base register itself recursively through the chips at the bottom? Why doesn't this code have an error if it's outputting a RAM4K when we expect a register?
It's been a while since I did the course so please excuse any minor errors below.
Each RAM chip (whatever the size) consists of an array of smaller chips. If you are implementing a 16K chip with 4K subchips, then there will be 4 of them.
So you would use 2 bits of the incoming address to select what sub-chip you need to work with, and the remaining 12 bits are sent on to all the sub-chip. It doesn't matter how you divide up the bits, as long as you have a set of 2 and a set of 12.
Specifically, the 2 select bits are used to route the load signal to just one sub-chip (ie: using a DMux4Way), so loads only affect that one sub-chip, and they are also used to pick which of the sub-chips outputs are used (ie: a Mux4Way16).
When I was doing it, I found that the simplest way to do things was always use the least-significant bits as the select bits. So for example, my RAM64 chip used address[0..2] as the select bits, and passed address[3..5] to the RAM8 sub-chips.
The thing that may be confusing you is that in these kinds of circuits, all of the sub-chips are activated. It's just that you use the select bits to decide which sub-chip's output to pass on to the outputs, and also as a filter to decide which sub-chip might perform a load.
As the saying goes, "It's turtles (or ram chips) all the way down."

Convert EM4x02 ID to Hitag2 Value

I've been working on an RFID project to produce our own RFID cards to work on our existing timeclocks and readers.
I've got most of the work done, and have been able to successfully write a Hitag2 card using the value of page 4 & 5 from another card (so basically copying the card) then changing the config bit which makes it act like an EM4x02 which allows our readers to read it.
What I'm struggling with is trying to relate the hex code on page4/5 to the output you get when scanning as an EM4x..
The values of the hitag page 4/5 are FF800000/003EDF10. This translates to 0000001EBC when read as an EM4x.
Does anybody have an idea on how this translation is done? I've tried using the methods in RFIDIOT but that doesn't seem to work for this.
I've managed to find how this is done after finding a hitag2 datasheet from 1999 (the only one I could find that explains the bits when hitag is in public mode A)
Firstly, convert the number you want on the EM4 card to hex.
Convert that hex into binary.
Split the binary into 4 bit chunks, then work out the even parity for each section and add it to the end of each chunk. (So you'll end up with 5 bits per chunk)
Then, work out the even parity of each column in the data (i.e first character of all chunks, then second etc. But ignoring the parity bit you added) and add these 4 bytes to the binary string.
Then add the correct amount of zeros at the start to ensure the data section has 50 bits.
Once you have the data section sorted, add 9 bits of 1 to the beginning (header) and a final 0 to the very end of the binary.
Your whole binary string should be 64 bits long.
Convert this to hex and split it in half. You can then write these onto pages 4/5 of a Hitag2 card.
You then need to change the configuration bit to 0x02 for the tag to work in public mode a.
Just thought I would send you the diagram of how this works.Em4X tag data

Turing Machine to compare multiple binary numbers

I was wondering how to construct a Turing Machine for
A<B<C<D...<N
with all numbers (A,B,C,D,...,N) being positive binary numbers.
These are a couple examples of how the machine should work:
1001 - Accepts because there is only one number
0<1 - Accepts
0010<1000<0001 - Doesn't accept because 1000!<0001
0100<1010<1010<1000 - Doesn't accept because 1010!<1010
I've tried methods that work to compare only two numbers but I can't seem to find a way to compare multiple (should work for infinite number of inputs) numbers.
Here is a high-level block diagram for solving this problem. You can implement these blocks by using block feature of JFLAP.
Blocks Description
Done? : This block decides whether all comparisons are done, if yes, it accepts, otherwise, it goes to compare the very next two numbers.
A<B : This block is responsible to compare two binary numbers, the one that cursor is pointing at the first digit and the next one. You can use '<' as the separator between A and B and the next number.
cleanup: during the comparison, you might marked 0's and 1's to something else. This block clean up everything and prepare everything for the next comparison.
Hopefully, this gives you an idea to solve the problem.

How to do a base conversion with Little Man Computer?

I need to convert a decimal number to a base between 2 and 9 using Little Man Computer. How do I proceed?
I believe successive divisions are the best method. In my opinion, I must write a code which divides two numbers, then save the integer ratio for the next division, as well as all of the remainders in an array of indefinite size, but I've been struggling with the division code for hours now. I tried searching for a code which divides two numbers, but all the ones I tried have mistakes/don't work. I'm stuck at the easiest part of the problem, I can't imagine how I'm ever going to be able to write a self-modifying code which manages an array of ever-increasing line positions and backtracks through it at the end to extract all the remainders. I'm at a loss here, any help would be appreciated.

Of Ways to Count the Limitless Primes [closed]

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Alright, so maybe I shouldn't have shrunk this question sooo much... I have seen the post on the most efficient way to find the first 10000 primes. I'm looking for all possible ways. The goal is to have a one stop shop for primality tests. Any and all tests people know for finding prime numbers are welcome.
And so:
What are all the different ways of finding primes?
Some prime tests only work with certain numbers, for instance, the Lucas–Lehmer test only works for Mersenne numbers.
Most prime tests used for big numbers can only tell you that a certain number is "probably prime" (or, if the number fails the test, it is definitely not prime). Usually you can continue the algorithm until you have a very high probability of a number being prime.
Have a look at this page and especially its "See Also" section.
The Miller-Rabin test is, I think, one of the best tests. In its standard form it gives you probable primes - though it has been shown that if you apply the test to a number beneath 3.4*10^14, and it passes the test for each parameter 2, 3, 5, 7, 11, 13 and 17, it is definitely prime.
The AKS test was the first deterministic, proven, general, polynomial-time test. However, to the best of my knowledge, its best implementation turns out to be slower than other tests unless the input is ridiculously large.
For a given integer, the fastest primality check I know is:
Take a list of 2 to the square root of the integer.
Loop through the list, taking the remainder of the integer / current number
If the remainder is zero for any number in the list, then the integer is not prime.
If the remainder was non-zero for all numbers in the list, then the integer is prime.
It uses significantly less memory than The Sieve of Eratosthenes and is generally faster for individual numbers.
The Sieve of Eratosthenes is a decent algorithm:
Take the list of positive integers 2 to any given Ceiling.
Take the next item in the list (2 in the first iteration) and remove all multiples of it (beyond the first) from the list.
Repeat step two until you reach the given Ceiling.
Your list is now composed purely of primes.
There is a functional limit to this algorithm in that it exchanges speed for memory. When generating very large lists of primes the memory capacity needed skyrockets.
#akdom's question to me:
Looping would work fine on my previous suggestion, and you don't need to do any calculations to determine if a number is even; in your loop, simply skip every even number, as shown below:
//Assuming theInteger is the number to be tested for primality.
// Check if theInteger is divisible by 2. If not, run this loop.
// This loop skips all even numbers.
for( int i = 3; i < sqrt(theInteger); i + 2)
{
if( theInteger % i == 0)
{
//getting here denotes that theInteger is not prime
// somehow indicate that some number, i, divides it and break
break;
}
}
A Rutgers grad student recently found a recurrence relation that generates primes. The difference of its successive numbers will generate either primes or 1's.
a(1) = 7
a(n) = a(n-1) + gcd(n,a(n-1)).
It makes a lot of crap that needs to be filtered out. Benoit Cloitre also has this recurrence that does a similar task:
b(1) = 1
b(n) = b(n-1) + lcm(n,b(n-1))
then the ratio of successive numbers, minus one [b(n)/b(n-1)-1] is prime. A full account of all this can be read at Recursivity.
For the sieve, you can do better by using a wheel instead of adding one each time, check out the Improved Incremental Prime Number Sieves. Here is an example of a wheel. Let's look at the numbers, 2 and 5 to ignore. Their wheel is, [2,4,2,2].
In your algorithm using the list from 2 to the root of the integer, you can improve performance by only testing odd numbers after 2. That is, your list only needs to contain 2 and all odd numbers from 3 to the square root of the integer. This cuts the number of times you loop in half without introducing any more complexity.
#theprise
If I were wanting to use an incrementing loop instead of an instantiated list (problems with memory for massive numbers...), what would be a good way to do that without building the list?
It doesn't seem like it would be cheaper to do a divisibility check for the given integer (X % 3) than just the check for the normal number (N % X).
If you're wanting to find a way of generating prime numbers, this have been covered in a previous question.