I'm getting involved with ZPL (a little bit) since a few days, so I'm sorry if the questions will look stupid.
I've got to build a bar code 128 and I finally realized: I got to make it as shorter as possible.
My main question is: is it possible to switch to subset C and then back to B for just 2 digits? I read the documentation and subset C will ready digits from 00 to 99, so in theory it should work, practically, will it be worth it?
Basically when I translate a bar code with Zebra designer, and print it to a file, it doesn't bother to switch to subset C for just a couple of digits.
This is the text I need to see in the bar code:
AB1C234D567890123456
By the documentation I read, I would build something like this:
FD>:AB1C>523>64D>5567890123456
Instead Zebra Designer does:
FD>:AB1C234D>5567890123456
So the other question is, will the bar code be the same length? Actually, will mine be shorter? [I don't have a printer with me at the moment]
Last question:
Let's say I don't want to spend much time scripting this up, will the following work ok, or will it make the bar code larger?
AB1C>523>64D>556>578>590>512>534>556
So I can just build a very simple script which checks two chars per time, if they're both numbers, then add >5 to the string.
Thank you :)
Ah, some nice loose terminology. Do you mean couple="exactly 2" or couple="a few"?
Changing from one subset to another takes one code element, so for exactly 2 digits, you'd need one element to change and one to represent the 2 digits in subset C. On the other hand, staying with your original subset would take 2 elements - so no, it's not worth the change.
Further, if you were to change to C for 2 digits and then back to your original, the change would actually be costly - C(12)B = 3 elements whereas 12 would only be 2.
If you repeat the exercise for 4 digits, then switching to C would generate C(12)(34) = 3 elements against 4 to stay with what you have; or C(12)(34)B = 4 elements if you switch and change back, or 4 elements if you stick - so no gain.
With 6 or more successive numerics, then you gain regardless of whether or not you switch back.
So overall,
2-digit terminal : No difference
2-digit other : code is longer
4-digit terminal : code is shorter
4-digit other : no difference
more than 4 digits : code is shorter.
And an ODD number of digits would need to be output in code A or B for the first digit and then the above table applies to the remainder.
This may not be the answer you're looking for, but specifying A (Automatic Mode) as the final parameter to the ^BC command will make the printer do this for you.
Example:
^XA
^FO100,100
^BY3
^BCN,100,N,N,A
^FD0123456789^FS
^XZ
Related
I am currently studying pseudo code, and despite my 3 year history in programming, this 1 particular practice-exam question has me perplexed with its unconventional code (shown below):
Highlighted in Pink, are my 2 main problems with the code. I have experience across 3 languages, yet I have never encountered the control flow method <>, and cannot imagine exactly what it would be used for. In addition to this, the variable average appears in the code in the form of "average:6:2", for which I am equally clueless.
To Summarise:
What is the function of the control flow method "<>"
As is stated in question (a) in the image above, what is the purpose of 'average:6:2'?
<> is common for not equal
While number is not equal to 999
number:filed_width:precision is pascal formatter for real number with filed_width being the space for field and precision is numer of digits after dot. so 3.141519:4:1 will print <space>3.1
What is the function of the control flow method "<>"
- It is "less than or greater than". If the input is equal to 999 then the loop ends. The number 999 is used as a Sentinel Value.
What is the purpose of 'average:6:2'?
- I'm thinking this is 6 digits with 2 decimal places.
In my app I will have several lines of numbers from 4 to 10
characters in length. How would I go around separating the generated
numbers every 3 numbers (at thousand, million, billion).
I will not know the length of the numbers beforehand so the same .text
label can sometimes be longer or shorter.
e.g. from 1234567 12345 1234567890
to 1,234,567 12,345 1,234,567,890
The separators can be a space, a dot or a comma (any is fine). I am
quite familiar with AI2 but can't figure this one out.
EDIT: Seems like I managed to segment the text (yay!)
What would I have to do now if I have let's say 10 labels with all different lengths? It would be extremely excessive to check length of each label using an if and then segmenting it using corresponding length option. Is there a way to shorten the process?
Thanks a lot :)
You can write your own custom procedure, for example see the following App Inventor Classic example, which uses a for range loop:
The algorithm works like this (example=2134256):
result after first loop: ,256
result after second loop: ,134,256
then add the remaining digits of the number, in this case 2
result in the end: 2,134,256
Edit: For App Inventor 2, you also have to use a procedure with result. Also let me recommend to use a local variable result instead of a global variable.
[Background Story]
I am working with a 5 year old user identification system, and I am trying to add IDs to the database. The problem I have is that the system that reads the ID numbers requires some sort of checksum, and no-one working here now has ever worked with it, so no-one knows how it works.
I have access to the list of existing IDs, which already have correct checksums. Also, as the checksum only has 16 possible values, I can create any ID I want and run it through the authentication system up to 16 times until I get the correct checksum (but this is quite time consuming)
[Question]
What methods can I use to help guess the checksum algorithm of used for some data?
I have tried a few simple methods such as XORing and summing, but these have not worked.
So my question is: if I have data (in hexadecimal) like this:
data checksum
00029921 1
00013481 B
00026001 3
00004541 8
What methods can I use work out what sort of checksum is used?
i.e. should I try sequential numbers such as 00029921,00029922,00029923,... or 00029911,00029921,00029931,... If I do this what patterns should I look for in the changing checksum?
Similarly, would comparing swapped digits tell me anything useful about the checksum?
i.e. 00013481 and 00031481
Is there anything else that could tell me something useful? What about inverting one bit, or maybe one hex digit?
I am assuming that this will be a common checksum algorithm, but I don't know where to start in testing it.
I have read the following links, but I am not sure if I can apply any of this to my case, as I don't think mine is a CRC.
stackoverflow.com/questions/149617/how-could-i-guess-a-checksum-algorithm
stackoverflow.com/questions/2896753/find-the-algorithm-that-generates-the-checksum
cosc.canterbury.ac.nz/greg.ewing/essays/CRC-Reverse-Engineering.html
[ANSWER]
I have now downloaded a much larger list of data, and it turned out to be simpler than I was expecting, but for completeness, here is what I did.
data:
00024901 A
00024911 B
00024921 C
00024931 D
00042811 A
00042871 0
00042881 1
00042891 2
00042901 A
00042921 C
00042961 0
00042971 1
00042981 2
00043021 4
00043031 5
00043041 6
00043051 7
00043061 8
00043071 9
00043081 A
00043101 3
00043111 4
00043121 5
00043141 7
00043151 8
00043161 9
00043171 A
00044291 E
From these, I could see that when just one value was increased by a value, the checksum was also increased by the same value as in:
00024901 A
00024911 B
Also, two digits swapped did not change the checksum:
00024901 A
00042901 A
This means that the polynomial value (for these two positions at least) must be the same
Finally, the checksum for 00000000 was A, so I calculated the sum of digits plus A mod 16:
( (Σxi) +0xA )mod16
And this matched for all the values I had. Just to check that there was nothing sneaky going on with the first 3 digits that never changed in my data, I made up and tested some numbers as Eric suggested, and those all worked with this too!
Many checksums I've seen use simple weighted values based on the position of the digits. For example, if the weights are 3,5,7 the checksum might be 3*c[0] + 5*c[1] + 7*c[2], then mod 10 for the result. (In your case, mod 16, since you have 4 bit checksum)
To check if this might be the case, I suggest that you feed some simple values into your system to get an answer:
1000000 = ?
0100000 = ?
0010000 = ?
... etc. If there are simple weights based on position, this may reveal it. Even if the algorithm is something different, feeding in nice, simple values and looking for patterns may be enlightening. As Matti suggested, you/we will likely need to see more samples before decoding the pattern.
I receive encoded PDF files regularly. The encoding works like this:
the PDFs can be displayed correctly in Acrobat Reader
select all and copy the test via Acrobat Reader
and paste in a text editor
will show that the content are encoded
so, examples are:
13579 -> 3579;
hello -> jgnnq
it's basically an offset (maybe swap) of ASCII characters.
The question is how can I find the offset automatically when I have access to only a few samples. I cannot be sure whether the encoding offset is changed. All I know is some text will usually (if not always) show up, e.g. "Name:", "Summary:", "Total:", inside the PDF.
Thank you!
edit: thanks for the feedback. I'd try to break the question into smaller questions:
Part 1: How to detect identical part(s) inside string?
You need to brute-force it.
If those patterns are simple like +2 character code like in your examples (which is +2 char codes)
h i j
e f g
l m n
l m n
o p q
1 2 3
3 4 5
5 6 7
7 8 9
9 : ;
You could easily implement like this to check against knowns words
>>> text='jgnnq'
>>> knowns=['hello', '13579']
>>>
>>> for i in range(-5,+5): #check -5 to +5 char code range
... rot=''.join(chr(ord(j)+i) for j in text)
... for x in knowns:
... if x in rot:
... print rot
...
hello
Is the PDF going to contain symbolic (like math or proofs) or natural language text (English, French, etc)?
If the latter, you can use a frequency chart for letters (digraphs, trigraphs and a small dictionary of words if you want to go the distance). I think there are probably a few of these online. Here's a start. And more specifically letter frequencies.
Then, if you're sure it's a Caesar shift, you can grab the first 1000 characters or so and shift them forward by increasing amounts up to (I would guess) 127 or so. Take the resulting texts and calculate how close the frequencies match the average ones you found above. Here is information on that.
The linked letter frequencies page on Wikipedia shows only letters, so you may want to exclude them in your calculation, or better find a chart with them in it. You may also want to transform the entire resulting text into lowercase or uppercase (your preference) to treat letters the same regardless of case.
Edit - saw comment about character swapping
In this case, it's a substitution cipher, which can still be broken automatically, though this time you will probably want to have a digraph chart handy to do extra analysis. This is useful because there will quite possibly be a substitution that is "closer" to average language in terms of letter analysis than the correct one, but comparing digraph frequencies will let you rule it out.
Also, I suggested shifting the characters, then seeing how close the frequencies matched the average language frequencies. You can actually just calculate the frequencies in your ciphertext first, then try to line them up with the good values. I'm not sure which is better.
Hmmm, thats a tough one.
The only thing I can suggest is using a dictionary (along with some substitution cipher algorithms) may help in decoding some of the text.
But I cannot see a solution that will decode everything for you with the scenario you describe.
Why don't you paste some sample input and we can have ago at decoding it.
It's only possible then you have a lot of examples (examples count stops then: possible to get all the combinations or just an linear values dependency or idea of the scenario).
also this question : How would I reverse engineer a cryptographic algorithm? have some advices.
Do the encoded files open correctly in PDF readers other than Acrobat Reader? If so, you could just use a PDF library (e.g. PDF Clown) and use it to programmatically extract the text you need.
I am looking for an existign path truncation algorithm (similar to what the Win32 static control does with SS_PATHELLIPSIS) for a set of paths that should focus on the distinct elements.
For example, if my paths are like this:
Unit with X/Test 3V/
Unit with X/Test 4V/
Unit with X/Test 5V/
Unit without X/Test 3V/
Unit without X/Test 6V/
Unit without X/2nd Test 6V/
When not enough display space is available, they should be truncated to something like this:
...with X/...3V/
...with X/...4V/
...with X/...5V/
...without X/...3V/
...without X/...6V/
...without X/2nd ...6V/
(Assuming that an ellipsis generally is shorter than three letters).
This is just an example of a rather simple, ideal case (e.g. they'd all end up at different lengths now, and I wouldn't know how to create a good suggestion when a path "Thingie/Long Test/" is added to the pool).
There is no given structure of the path elements, they are assigned by the user, but often items will have similar segments. It should work for proportional fonts, so the algorithm should take a measure function (and not call it to heavily) or generate a suggestion list.
Data-wise, a typical use case would contain 2..4 path segments anf 20 elements per segment.
I am looking for previous attempts into that direction, and if that's solvable wiht sensible amount of code or dependencies.
I'm assuming you're asking mainly about how to deal with the set of folder names extracted from the same level of hierarchy, since splitting by rows and path separators and aggregating by hierarchy depth is simple.
Your problem reminds me a lot of the longest common substring problem, with the differences that:
You're interested in many substrings, not just one.
You care about order.
These may appear substantial, but if you examine the dynamic-programming solution in the article you can see that it revolves around creating a table of "character collisions" and then looking for the longest diagonal in this table. I think that you could instead enumerate all diagonals in the table by the order in which they appear, and then for each path replace, by order, all appearances of these strings with ellipses.
Enforcing a minimal substring length of 2 will return a result similar to what you've outlined in your question.
It does seem like it requires some tinkering with the algorithm (for example, ensuring a certain substring is first in all strings), and then you need to invoke it over your entire set... I hope this at least gives you a possible direction.
Well, the "natural number" ordering part is actually easy, simply replace all numbers with formatted number where there is enough leading zeroes, eg. Test 9V -> Test 000009V and Test 12B -> Test 000012B. These are now sortable by standard methods.
For the actual ellipsisizing. Unless this is actually a huge system, I'd just add manual ellipsisizing "list" (of regexes, for flexibility and pain) that'd turn certain words into ellipses. This does requires continuous work, but coming up with the algorithm eats your time too; there are myriads of corner cases.
I'd probably try a "Floodfill" approach. Arrange first level of directories as you would a bitmap, every letter is a pixel. iterate over all characters that are in names of directories. with all of them, "paint" this same character, then "paint" the next character from first string such that it follows this previous character (and so on etc.) Then select the longest painted string that you find.
Example (if prefixed with *, it's painted)
Foo
BarFoo
*Foo
Bar*Foo
*F*oo
Bar*F*oo
...
note that:
*ofoo
b*oo
*o*foo
b*oo
.. painting of first 'o' stops since there are no continuing characters.
of*oo
b*oo
...
And then you get to to second "o" and it will find a substring of at least 2.
So you will have to iterate over most possible character instances (one optimization is to stop in each string at position Length-n, where n is the longest already found common substring. But then there is yet another problem (here with "Beta Beta")
| <- visibility cutout
Alfa Beta Gamma Delta 1
Alfa Beta Gamma Delta 2
Alfa Beta Beta 1
Alfa Beta Beta 2
Beta Beta 1
Beta Beta 2
Beta Beta 3
Beta Beta 4
What do you want to do? Cut Alfa Beta Gamma Delta or Alfa Beta or Beta Beta or Beta?
This is a bit rambling, but might be entertaining :).