Generate all combinations from list of characters - perl

I am busy implementing a lab for pen testers to create MD5 hashes from 4 letter words. I need the words to have a combination of lower and uppercase letters as well as numeric and special characters, but I just do not seem to find out how to combine any given characters in all orders. So currently I have this:
my $str = 'aaaa';
print $str++, $/ while $str le 'dddd';
Which will do:
aaaa
aaab
aaac
aaad
...
...
dddd
There is no way however how I can make it do:
Aaaa
AAaa
aAaa
...
dddD
Not even to mention adding numbers and special characters. What I really wanted to do was to make the characters to create words based on a given list. So if I feel I want to use abeDod## it should create all combinations from those characters.
Edit to clarify.
Let's say I give the characters aBc# I need it to give it a a count to say it must have maximum of 4 letters per word and with combination of all the given characters, like:
aBc#
Bac#
caB#
#Bca
...
I hope that clarifies the question.

Use a list of integers that are ASCII codes for the characters you accept, to sample from it using your favorite (pseudo-)random number generator. Then convert each to its character using chr and concatenate them.
Like
perl -wE'$rw .= chr( 32+(int rand 126-32) ) for 1..4; say $rw'
Notes
I use a one-liner merely for easy copy-paste testing. Write this nicely in a script, please
I use the sketchy rand, good for shuffling things a bit. Replace with a better one if needed
Glueing four (pseudo-)random numbers does not build a good distribution; even as each letter on its own does, the whole thing does not. But the four should satisfy most needs.
If not, I think that you'd need to produce a far longer list (range of allowed chars repeated four times perhaps) and randomize it, then draw four-letter subsequences. A lot more work
I need to tap dance a little to produce (random-ish) integers from 32 to 126 using rand, since it takes only the end of range. Also, this takes all of them from that range, likely not what you want; so specify subranges, or specific lists that you want to draw from

Related

Does Perl's Glob have a limitation?

I am running the following expecting return strings of 5 characters:
while (glob '{a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z}'x5) {
print "$_\n";
}
but it returns only 4 characters:
anbc
anbd
anbe
anbf
anbg
...
However, when I reduce the number of characters in the list:
while (glob '{a,b,c,d,e,f,g,h,i,j,k,l,m}'x5) {
print "$_\n";
}
it returns correctly:
aamid
aamie
aamif
aamig
aamih
...
Can someone please tell me what I am missing here, is there a limit of some sort? or is there a way around this?
If it makes any difference, It returns the same result in both perl 5.26 and perl 5.28
The glob first creates all possible file name expansions, so it will first generate the complete list from the shell-style glob/pattern it is given. Only then will it iterate over it, if used in scalar context. That's why it's so hard (impossible?) to escape the iterator without exhausting it; see this post.
In your first example that's 265 strings (11_881_376), each five chars long. So a list of ~12 million strings, with (naive) total in excess of 56Mb ... plus the overhead for a scalar, which I think at minimum is 12 bytes or such. So at the order of a 100Mb's, at the very least, right there in one list.†
I am not aware of any formal limits on lengths of things in Perl (other than in regex) but glob does all that internally and there must be undocumented limits -- perhaps some buffers are overrun somewhere, internally? It is a bit excessive.
As for a way around this -- generate that list of 5-char strings iteratively, instead of letting glob roll its magic behind the scenes. Then it absolutely should not have a problem.
However, I find the whole thing a bit big for comfort, even in that case. I'd really recommend to write an algorithm that generates and provides one list element at a time (an "iterator"), and work with that.
There are good libraries that can do that (and a lot more), some of which are Algorithm::Loops recommended in a previous post on this matter (and in a comment), Algorithm::Combinatorics (same comment), Set::CrossProduct from another answer here ...
Also note that, while this is a clever use of glob, the library is meant to work with files. Apart from misusing it in principle, I think that it will check each of (the ~ 12 million) names for a valid entry! (See this page.) That's a lot of unneeded disk work. (And if you were to use "globs" like * or ? on some systems it returns a list with only strings that actually have files, so you'd quietly get different results.)
† I'm getting 56 bytes for a size of a 5-char scalar. While that is for a declared variable, which may take a little more than an anonymous scalar, in the test program with length-4 strings the actual total size is indeed a good order of magnitude larger than the naively computed one. So the real thing may well be on the order of 1Gb, in one operation.
Update A simple test program that generates that list of 5-char long strings (using the same glob approach) ran for 15-ish minutes on a server-class machine and took 725 Mb of memory.
It did produce the right number of actual 5-char long strings, seemingly correct, on this server.
Everything has some limitation.
Here's a pure Perl module that can do it for you iteratively. It doesn't generate the entire list at once and you start to get results immediately:
use v5.10;
use Set::CrossProduct;
my $set = Set::CrossProduct->new( [ ([ 'a'..'z' ]) x 5 ] );
while( my $item = $set->get ) {
say join '', #$item
}

Why does this line return sum of integers 1-10?

I'd like to understand how unpack is returning the sum in the given perl one-liner.
I've looked at pack man page and mostly understood that it is simply formatting the given array into a scalar of ten doubles.
However, I couldn't find proper documentation for unpack with %123. Looking for help here.
print unpack "%123d*" , pack( "d*", (1..10));
This line correctly outputs 55 which is 1+2+3+...+10.
From perldoc -f unpack:
In addition to fields allowed in pack(), you may prefix a field with a % to indicate that you want a <number>-bit checksum of the items instead of the items themselves.
Thus %123d* means to add all the input integers 1..10 and then take the first 123 bit of this result in order to construct the "<number>-bit checksum". Note that %8d* or just %d* (which is equivalent to %16d*) would suffice too given that the sum is small enough.

Can I directly load text with numbers in CCC,CC format ? (K4)

I have input with floats stored like 1000,50, ie. the decimal points are replaced by commas.
Is there an option in K to load these numbers directly into floats ?
When using
data:("SFF" ;";",";") 0:. filename
I get 0ns, of course, because the numbers are not recognized as floats.
I load them as strings now, and convert them using ssr like
c:.:' .q.ssr'[data;",";"."]
but that is extremely slow.
Is there an option somewhere to have K load these numbers in CCC,CC format as floats directly ? Normal format and ccc,cc format are not mixed, any file has just one of them.
If there is not, I imagine that it must by quite easy to replace a "." somewhere in the Q-binary where the load-function sits, with a ",", to get a version which loads these numbers. Has anybody tried that ? Or any other tip to load big files with these numbers in reasonable time ?
Cheers,
Co
If ssr' is slow for your task you may find this tiny function useful:
c2p:{c:-1_sums count each x;p:ss[r:raze x;","];r[p]:".";(0,c) _ r}
Update: an alternative version:
c2p:{p:ss[r:raze x;","];r[p]:".";(0,-1_sums count'[x])_r}
It concatenates all strings into a single long string, finds positions of commas, replaces commas with periods then splits that long string:
q)N:1000000
q)s:string[N?100000],'",",'string N?1000
q)\t r1:ssr'[s;",";"."]
4284
q)\t r2:c2p s
242
q)r1~r2
1b
I was thinking something like find (?) combined with indexing/applying
q)N:1000000
q)s:string[N?100000],'",",'string N?1000
q)\ts {s[x;y]:"."}./:flip(til count s;s?\:",")
967 52972144
q)s
"93912.794"
"57144.788"
"77809.659"
"7839.47"
"6363.523"
"44761.244"
"65699.712"
It's not perfect but that's the general idea. I'm sure there is an easier way...

How does this Perl one-liner actually work?

So, I happened to notice that last.fm is hiring in my area, and since I've known a few people who worked there, I though of applying.
But I thought I'd better take a look at the current staff first.
Everyone on that page has a cute/clever/dumb strapline, like "Is life not a thousand times too short for us to bore ourselves?". In fact, it was quite amusing, until I got to this:
perl -e'print+pack+q,c*,,map$.+=$_,74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34'
Which I couldn't resist pasting into my terminal (kind of a stupid thing to do, maybe), but it printed:
Just another Last.fm hacker,
I thought it would be relatively easy to figure out how that Perl one-liner works. But I couldn't really make sense of the documentation, and I don't know Perl, so I wasn't even sure I was reading the relevant documentation.
So I tried modifying the numbers, which got me nowhere. So I decided it was genuinely interesting and worth figuring out.
So, 'how does it work' being a bit vague, my question is mainly,
What are those numbers? Why are there negative numbers and positive numbers, and does the negativity or positivity matter?
What does the combination of operators +=$_ do?
What's pack+q,c*,, doing?
This is a variant on “Just another Perl hacker”, a Perl meme. As JAPHs go, this one is relatively tame.
The first thing you need to do is figure out how to parse the perl program. It lacks parentheses around function calls and uses the + and quote-like operators in interesting ways. The original program is this:
print+pack+q,c*,,map$.+=$_,74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34
pack is a function, whereas print and map are list operators. Either way, a function or non-nullary operator name immediately followed by a plus sign can't be using + as a binary operator, so both + signs at the beginning are unary operators. This oddity is described in the manual.
If we add parentheses, use the block syntax for map, and add a bit of whitespace, we get:
print(+pack(+q,c*,,
map{$.+=$_} (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21,
18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34)))
The next tricky bit is that q here is the q quote-like operator. It's more commonly written with single quotes:
print(+pack(+'c*',
map{$.+=$_} (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21,
18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34)))
Remember that the unary plus is a no-op (apart from forcing a scalar context), so things should now be looking more familiar. This is a call to the pack function, with a format of c*, meaning “any number of characters, specified by their number in the current character set”. An alternate way to write this is
print(join("", map {chr($.+=$_)} (74, …, -34)))
The map function applies the supplied block to the elements of the argument list in order. For each element, $_ is set to the element value, and the result of the map call is the list of values returned by executing the block on the successive elements. A longer way to write this program would be
#list_accumulator = ();
for $n in (74, …, -34) {
$. += $n;
push #list_accumulator, chr($.)
}
print(join("", #list_accumulator))
The $. variable contains a running total of the numbers. The numbers are chosen so that the running total is the ASCII codes of the characters the author wants to print: 74=J, 74+43=117=u, 74+43-2=115=s, etc. They are negative or positive depending on whether each character is before or after the previous one in ASCII order.
For your next task, explain this JAPH (produced by EyesDrop).
''=~('(?{'.('-)#.)#_*([]#!#/)(#)#-#),#(##+#)'
^'][)#]`}`]()`#.#]#%[`}%[#`#!##%[').',"})')
Don't use any of this in production code.
The basic idea behind this is quite simple. You have an array containing the ASCII values of the characters. To make things a little bit more complicated you don't use absolute values, but relative ones except for the first one. So the idea is to add the specific value to the previous one, for example:
74 -> J
74 + 43 -> u
74 + 42 + (-2 ) -> s
Even though $. is a special variable in Perl it does not mean anything special in this case. It is just used to save the previous value and add the current element:
map($.+=$_, ARRAY)
Basically it means add the current list element ($_) to the variable $.. This will return a new array with the correct ASCII values for the new sentence.
The q function in Perl is used for single quoted, literal strings. E.g. you can use something like
q/Literal $1 String/
q!Another literal String!
q,Third literal string,
This means that pack+q,c*,, is basically pack 'c*', ARRAY. The c* modifier in pack interprets the value as characters. For example, it will use the value and interpret it as a character.
It basically boils down to this:
#!/usr/bin/perl
use strict;
use warnings;
my $prev_value = 0;
my #relative = (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34);
my #absolute = map($prev_value += $_, #relative);
print pack("c*", #absolute);

How can I convert the tiger hash values from the official implementations into the form used by Direct Connect?

I am trying to implement a Direct Connect Client, and I am currently stuck at a point where I need to hash the files in order to be able to upload them to other clients.
As the all other clients require a TTHL (Tiger Tree Hashing Leaves) support for verification of the downloaded data. I have searched for implementations of the algorithm, and found tiger-hash-python.
I have implemented a routine that uses the hash function from before, and is able to hash large files, according to the logic specified in Tree Hash EXchange format (THEX) (basically, the tree diagram is the important part on that page).
However, the value produced by it is similar to those shown on Wikipedia, a hex digest, but is different from those shown in the DC clients I'm using for reference.
I have been unable to find out how the hex digest form is converted to this other one (39 characters, A-Z, 0-9). Could someone please explain how that is done?
Well ... I tried what Paulo Ebermann said, using the following functions:
def strdivide(list,length):
result = []
# Calculate how many blocks there are, using the condition: i*length = len(list).
# The additional maths operations are to deal with the last block which might have a smaller size
for i in range(0,int(math.ceil(float(len(list))/length))):
result.append(list[i*length:(i+1)*length])
return result
def dchash(data):
result = tiger.hash(data) # From the aformentioned tiger-hash-python script, 48-char hex digest
result = "".join([ "".join(strdivide(result[i:i+16],2)[::-1]) for i in range(0,48,16) ]) # Representation Transform
bits = "".join([chr(int(c,16)) for c in strdivide(result,2)]) # Converting every 2 hex characters into 1 normal
result = base64.b32encode(bits) # Result will be 40 characters
return result[:-1] # Leaving behind the trailing '='
The TTH for an empty file was found to be 8B630E030AD09E5D0E90FB246A3A75DBB6256C3EE7B8635A, which after the transformation specified here, becomes 5D9ED00A030E638BDB753A6A24FB900E5A63B8E73E6C25B6. Base-32 encoding this result yielded LWPNACQDBZRYXW3VHJVCJ64QBZNGHOHHHZWCLNQ, which was found to be what DC++ generates.
The only mention of the format of the hash in the Direct Connect protocol I found is on the $SR page on the NMDC Protocol wiki:
For files containing TTH, the <hub_name> parameter is replaced with TTH:<base32_encoded_tth_hash> (ref: TTH_Hash).
So, it is Base32-encoding. This is defined in RFC 4648 (and some earlier ones), section 6.
Basically, you are using the capital letters A-Z and the decimal digits 2 to 7, and one base32 digit represents 5 bits, while one base16 (hexadecimal) digit represents only 4 ones.
This means, each 5 hex digits map to 4 base32-digits, and for a Tiger hash (192 bits) you will need 40 base32-digits (in the official encoding, the last one would be a = padding, which seems to be omitted if you say that there are always 39 characters).
I'm not sure of an implementation of a conversion from hex (or bytes) to base32, but it shouldn't be too complicated with a lookup table and some bit-shifting.