I want to check whether the code is running correctly or not. the function takes in a base number and exponent and returns the result.
suppose the user input 2 as their base number and 4 as the exponent
2^4
function return 16
I want to output 16 on the screen.
I tried outputting the result inside the function but the program quit after entering the number.
You can use the "MsgBox" function to output stuff or you could use "Debug.Print" to write in the "Immediate" window.
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."
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
I have a problem that requires searching and saving some values that prevent it from doing an infinite loop. Every possible state of this problem is expressed as an unique 8 digit code with base 6 (all digits are 0-5). When the program evaluates this position, I want a boolean to be set as true so as not to evaluate this position again. However an array 1..55555555 is too large in memory and if i convert the 8 digit code to decimal it takes too much time. Also not all combinations are possible in the problem; 11 11 11 11, 11 11 55 12 and others are not valid and i need not use extra memory. So, is there a way to store as value "true" a block of memory with adress e.g 23451211 and when i call the evaluating process check if 23451211 is true or unassigned;
6 to the power 8 = 1679616.
To mark used or not you need one bit, thus you can do with about 209952 bytes.
In recent Free Pascal's, bitpacked structures are done as follows
var
arr : bitpacked array [0..6*6*6*6*6*6*6*6-1] of boolean;
and arr[x] will give true or false.
The conversion time from base 6 to binary (not decimal!) will probably be shorter than trying to use large swaths of memory. (((digit8)*6+digit7)*6+digit6)*6 etc
p.s. FPC does have an exponent operator, but not for constants, so that's why 6^8 is written like that.
I want to generate unique code numbers (composed of 7 digits exactly). The code number is generated randomly and saved in MySQL table.
I have another requirement. All generated codes should differ in at least two digits. This is useful to prevent errors while typing the user code. Hopefully, it will prevent referring to another user code while doing some operations as it is more unlikely to miss two digits and match another existing user code.
The generate algorithm works simply like:
Retrieve all previous codes if any from MySQL table.
Generate one code at a time.
Subtract the generated code with all previous codes.
Check the number of non-zero digits in the subtraction result.
If it is > 1, accept the generated code and add it to previous codes.
Otherwise, jump to 2.
Repeat steps from 2 to 6 for the number of requested codes.
Save the generated codes in the DB table.
The algorithm works fine, but the problem is related to performance. It takes a very long to finish generating the codes when requesting to generate a large number of codes like: 10,000.
The question: Is there any way to improve the performance of this algorithm?
I am using perl + MySQL on Ubuntu server if that matters.
Have you considered a variant of the Luhn algorithm? Luhn is used to generate a check digit for strings of numbers in lots of applications, including credit card account numbers. It's part of the ISO-7812-1 standard for generating identifiers. It will catch any number that is entered with one incorrect digit, which implies any two valid numbers differ in a least two digits.
Check out Algorithm::LUHN in CPAN for a perl implementation.
Don't retrieve the existing codes, just generate a potential new code and see if there are any conflicting ones in the database:
SELECT code FROM table WHERE abs(code-?) regexp '^[1-9]?0*$';
(where the placeholder is the newly generated code).
Ah, I missed the generating lots of codes at once part. Do it like this (completely untested):
my #codes = existing_codes();
my $frontwards_index = {};
my $backwards_index = {};
for my $code (#codes) {
index_code($code, $frontwards_index);
index_code(reverse($code), $backwards_index);
}
my #new_codes = map generate_code($frontwards_index, $backwards_index), 1..10000;
sub index_code {
my ($code, $index) = #_;
push #{ $index{ substr($code, 0, length($code)/2) } }, $code;
return;
}
sub check_index {
my ($code, $index) = #_;
my $found = grep { ($_ ^ $code) =~ y/\0//c <= 1 } #{ $index{ substr($code, 0, length($code)/2 } };
return $found;
}
sub generate_code {
my ($frontwards_index, $backwards_index) = #_;
my $new_code;
do {
$new_code = sprintf("%07d", rand(10000000));
} while check_index($new_code, $frontwards_index)
|| check_index(reverse($new_code), $backwards_index);
index_code($new_code, $frontwards_index);
index_code(reverse($new_code), $backwards_index);
return $new_code;
}
Put the numbers 0 through 9,999,999 in an augmented binary search tree. The augmentation is to keep track of the number of sub-nodes to the left and to the right. So for example when your algorithm begins, the top node should have value 5,000,000, and it should know that it has 5,000,000 nodes to the left, and 4,999,999 nodes to the right. Now create a hashtable. For each value you've used already, remove its node from the augmented binary search tree and add the value to the hashtable. Make sure to maintain the augmentation.
To get a single value, follow these steps.
Use the top node to determine how many nodes are left in the tree. Let's say you have n nodes left. Pick a random number between 0 and n. Using the augmentation, you can find the nth node in your tree in log(n) time.
Once you've found that node, compute all the values that would make the value at that node invalid. Let's say your node has value 1,111,111. If you already have 2,111,111 or 3,111,111 or... then you can't use 1,111,111. Since there are 8 other options per digit and 7 digits, you only need to check 56 possible values. Check to see if any of those values are in your hashtable. If you haven't used any of those values yet, you can use your random node. If you have used any of them, then you can't.
Remove your node from the augmented tree. Make sure that you maintain the augmented information.
If you can't use that value, return to step 1.
If you can use that value, you have a new random code. Add it to the hashtable.
Now, checking to see if a value is available takes O(1) time instead of O(n) time. Also, finding another available random value to check takes O(log n) time instead of... ah... I'm not sure how to analyze your algorithm.
Long story short, if you start from scratch and use this algorithm, you will generate a complete list of valid codes in O(n log n). Since n is 10,000,000, it will take a few seconds or something.
Did I do the math right there everybody? Let me know if that doesn't check out or if I need to clarify anything.
Use a hash.
After generating a successful code (not conflicting with any existing code), but that code in the hash table, and also put the 63 other codes that differ by exactly one digit into the hash.
To see if a randomly generated code will conflict with an existing code, just check if that code exists in the hash.
Howabout:
Generate a 6 digit code by autoincrementing the previous one.
Generate a 1 digit code by incrementing the previous one mod 10.
Concatenate the two.
Presto, guaranteed to differ in two digits. :D
(Yes, being slightly facetious. I'm assuming that 'random' or at least quasi-random is necessary. In which case, generate a 6 digit random key, repeat until its not a duplicate (i.e. make the column unique, repeat until the insert doesn't fail the constraint), then generate a check digit, as someone already said.)