Regex for binary multiple of 3 - regex

I would like to know how can I construct a regex to know if a number in base 2 (binary) is multiple of 3. I had read in this thread Check if a number is divisible by 3 but they dont do it with a regex, and the graph someone drew is wrong(because it doesn't accept even numbers). I have tried with: ((1+)(0*)(1+))(0) but it doesn't works for some values. Hope you can help me.
UPDATE:
Ok, thanks all for your help, now I know how to draw the NFA, here I left the graph and the regular expresion:
In the graph, the states are the number in base 10 mod 3.
For example: to go to state 1 you have to have 1, then you can add 1 or 0, if you add 1, you would have 11(3 in base 10), and this number mod 3 is 0 then you draw the arc to the state 0.
((0*)((11)*)((1((00) *)1) *)(101 *(0|((00) *1 *) *0)1) *(1(000)+1*01)*) *
And the other regex works, but this is shorter.
Thanks a lot :)

I know this is an old question, but an efficient answer is yet to be given and this question pops up first for "binary divisible by 3 regex" on Google.
Based on the DFA proposed by the author, a ridiculously short regex can be generated by simplifying the routes a binary string can take through the DFA.
The simplest one, using only state A, is:
0*
Including state B:
0*(11)*0*
Including state C:
0*(1(01*0)*1)*0*
And include the fact that after going back to state A, the whole process can be started again.
0*((1(01*0)*1)*0*)*
Using some basic regex rules, this simplifies to
(1(01*0)*1|0)*
Have a nice day.

If I may plug my solution for this code golf question! It's a piece of JavaScript that generates regexes (probably inefficiently, but does the job) for divisibility for each base.
This is what it generates for divisibility by 3 in base 2:
/^((((0+)?1)(10*1)*0)(0(10*1)*0|1)*(0(10*1)*(1(0+)?))|(((0+)?1)(10*1)*(1(0+)?)|(0(0+)?)))$/
Edit: comparing to Asmor's, probably very inefficient :)
Edit 2: Also, this is a duplicate of this question.

For some who is learning and searching how to do this:
see this video:
https://www.youtube.com/watch?v=SmT1DXLl3f4&t=138s
write state quations and solve them with Axden's Theorem
The way I did is visible in the image-result is the same as pointed out by user #Kert Ojasoo. I hope i did it corretly because i spent 2 days to solve it...

n+2n = 3n. Thus, 2 adjacent bits set to 1 denote a multiple of 3. If there are an odd number of adjacent 1s, that would not be 3.
So I'd propose this regex:
(0*(11)?)+

Related

Excluding % from a Regex number search

I'm attempting to create a Regex that finds only 2-digit integers or numbers with a precision of 2 decimal points.
In the example string at the bottom, I want to find only the following:
21 and 10.50
Using this expression, 100% is getting captured, in addition to the strings I desire to capture:
(\d){1,2}(\.?)([0-9]?[0-9]?){1,2}
I know I need to use ^% somewhere, but I can't figure out where it goes. Any suggestions are greatly appreciated.
Here's my sample string:
Earn Up to $21 Per Hour - Deliver Food with !!
Delivery Drivers work when they want and make great money when they do.
All orders are prepaid, just pick them up and deliver them to hungry diners. No waiting in line or fumbling with receipts and prepaid cards.
It's fast and easy to start working. Get started today.
Apply Now
Why choose ?
More orders than any other takeout platform
100% of our restaurants are official partners
Competitive pay: Per order fee + mileage + tips
We guarantee an hourly minimum of $10.50/hour*
Create your own schedule & work the hours you want
Word boundaries in your regular expression will grant you a bit more control.
Since word boundaries are a bit strict, we need to introduce an OR condition to address both cases which will satisfy your regex.
(\b[\d]{2}\.[\d]{2}\b)|(\b[\d]{2}\b)
Edit: Try this one,
\b[\d]{2}\b(\.[\d]{2})?
The first example has a chance to fail as it is order dependent due to the way it short-circuits. This I believe should address multiple cases properly.
I think this should work:
(?<!\d)((\d+\.\d\d)|(\d\d))(?!%|\d)
Demo (and explanation)
EDIT:
Improved version:
(?<!\d)(\d{1,2}(?:\.\d{1,2})?)(?!%|\d)
Demo (and explanation)
You can try this variant: (\d{1,}|[\d.])\b(?!%)
It uses negative lookahead (?!%) to exclude digits following by % sign.
Details at regex101

For the grammar given, provide 3 valid example strings

today I got a homework on my 'Programming Languages' class and I'm having trouble. Here is the full question;
For the grammars given below, draw transition diagram and transition
table. Provide 3 valid example strings.
And this is the one I'm having trouble with;
1?1.(0|1)+
I don't really know what the question mark (?) stands for on this example, and I couldn't find an online paper. I don't want any help on diagram or the table, I could make them if I knew what '?' means. Please help me with this, thanks in advance.
Usually the question mark ? stands for 0 or 1 time
So 1? will match "" (empty) and "1"

Complex regex to check for two words and a quanity

Ok what I'm trying to do is to check for the presence of
"TestItem-1"
a number greater then 1
one of the possible words in the list of "KG. Kg, kg, Kilo(s) or Kilogram(s)"
Where any of the items could be in any order and within a 6 word limit of each other.
Has to be done in regex as there is no access to the underlying scripting engine
This is what I've got as there a way of checking greater then I decided to use a range of 1-999 for the number check.
\b(?:[T|t]estItem-1\W+(?:\w+\W+){1,6}(^[0-9]|[1-9][0-9]|[1-9][0-9][0-9])$)\W+(?:\w+\W+){1,6}[K|k]il[o|os]|[K|k][[G|GS]|[g|gs]]|[|K|k]ilogra[m|ms]\b
Examples of what I need to find would be like -
"TestItem-1 is unstable in quanties above 12 Kilograms"
"1 Kilogram of TestItem-1"
While I wouldn't want to find -
"15 units of TestItem-1"
I know that what I got isn't working each section appears to work independently of each other but not together.
I pass this over to far greater minds then mine :)
You can try something like this:
\b(?:[2-9]|\d\d+)\b\s\b(?:KG.|Kg,|kg,|Kilos?|Kilograms?)\b(?:\S+\s){0,6}\bTestItem-1\b|\bTestItem-1\b(?:\S+\s){0,6}\b(?:[2-9]|\d\d+)\b\s\b(?:KG.|Kg,|kg,|Kilos?|Kilograms?)\b
Not ideal with the duplication but without lookarounds that's the best I could think of. I'll try and improve it in a bit.

find a string with at least n matching elements

I have a list of numbers that I want to find at least 3 of...
here is an example
I have a large list of numbers in a sql database in the format of (for example)
01-02-03-04-05-06
06-08-19-24-25-36
etc etc
basically 6 random numbers between 0 and 99.
Now I want to find the strings where at least 3 of a set of given numbers occurs.
For example:
given: 01-02-03-10-11-12
return the strings that have at least 3 of those numbers in them.
eg
01-05-06-09-10-12 would match
03-08-10-12-18-22 would match
03-09-12-18-22-38 would not
I am thinking that there might be some algorithm or even regular expression that could match this... but my lack of computer science textbook experience is tripping me up I think.
No - this is not a homework question! This is for an actual application!
I am developing in ruby, but any language answer would be appreciated
You can use a string replacement to replace - with | to turn 01-02-03-10-11-12 into 01|02|03|10|11|12. Then wrap it like this:
((01|02|03|10|11|12).*){3}
This will find any of the digit pairs, then ignore any number of characters... 3 times. If it matches, then success.

Similar String algorithm

I'm looking for an algorithm, or at least theory of operation on how you would find similar text in two or more different strings...
Much like the question posed here: Algorithm to find articles with similar text, the difference being that my text strings will only ever be a handful of words.
Like say I have a string:
"Into the clear blue sky"
and I'm doing a compare with the following two strings:
"The color is sky blue" and
"In the blue clear sky"
I'm looking for an algorithm that can be used to match the text in the two, and decide on how close they match. In my case, spelling, and punctuation are going to be important. I don't want them to affect the ability to discover the real text. In the above example, if the color reference is stored as "'sky-blue'", I want it to still be able to match. However, the 3rd string listed should be a BETTER match over the second, etc.
I'm sure places like Google probably use something similar with the "Did you mean:" feature...
* EDIT *
In talking with a friend, he worked with a guy who wrote a paper on this topic. I thought I might share it with everyone reading this, as there are some really good methods and processes described in it...
Here's the link to his paper, I hope it is helpful to those reading this question, and on the topic of similar string algorithms.
Levenshtein distance will not completely work, because you want to allow rearrangements. I think your best bet is going to be to find best rearrangement with levenstein distance as cost for each word.
To find the cost of rearrangement, kinda like the pancake sorting problem. So, you can permute every combination of words (filtering out exact matches), with every combination of other string, trying to minimize a combination of permute distance and Levenshtein distance on each word pair.
edit:
Now that I have a second I can post a quick example (all 'best' guesses are on inspection and not actually running the algorithms):
original strings | best rearrangement w/ lev distance per word
Into the clear blue sky | Into the c_lear blue sky
The color is sky blue | is__ the colo_r blue sky
R_dist = dist( 3 1 2 5 4 ) --> 3 1 2 *4 5* --> *2 1 3* 4 5 --> *1 2* 3 4 5 = 3
L_dist = (2D+S) + (I+D+S) (Total Subsitutions: 2, deletions: 3, insertion: 1)
(notice all the flips include all elements in the range, and I use ranges where Xi - Xj = +/- 1)
Other example
original strings | best rearrangement w/ lev distance per word
Into the clear blue sky | Into the clear blue sky
In the blue clear sky | In__ the clear blue sky
R_dist = dist( 1 2 4 3 5 ) --> 1 2 *3 4* 5 = 1
L_dist = (2D) (Total Subsitutions: 0, deletions: 2, insertion: 0)
And to show all possible combinations of the three...
The color is sky blue | The colo_r is sky blue
In the blue clear sky | the c_lear in sky blue
R_dist = dist( 2 4 1 3 5 ) --> *2 3 1 4* 5 --> *1 3 2* 4 5 --> 1 *2 3* 4 5 = 3
L_dist = (D+I+S) + (S) (Total Subsitutions: 2, deletions: 1, insertion: 1)
Anyway you make the cost function the second choice will be lowest cost, which is what you expected!
One way to determine a measure of "overall similarity without respect to order" is to use some kind of compression-based distance. Basically, the way most compression algorithms (e.g. gzip) work is to scan along a string looking for string segments that have appeared earlier -- any time such a segment is found, it is replaced with an (offset, length) pair identifying the earlier segment to use. You can use measures of how well two strings compress to detect similarities between them.
Suppose you have a function string comp(string s) that returns a compressed version of s. You can then use the following expression as a "similarity score" between two strings s and t:
len(comp(s)) + len(comp(t)) - len(comp(s . t))
where . is taken to be concatenation. The idea is that you are measuring how much further you can compress t by looking at s first. If s == t, then len(comp(s . t)) will be barely any larger than len(comp(s)) and you'll get a high score, while if they are completely different, len(comp(s . t)) will be very near len(comp(s) + comp(t)) and you'll get a score near zero. Intermediate levels of similarity produce intermediate scores.
Actually the following formula is even better as it is symmetric (i.e. the score doesn't change depending on which string is s and which is t):
2 * (len(comp(s)) + len(comp(t))) - len(comp(s . t)) - len(comp(t . s))
This technique has its roots in information theory.
Advantages: good compression algorithms are already available, so you don't need to do much coding, and they run in linear time (or nearly so) so they're fast. By contrast, solutions involving all permutations of words grow super-exponentially in the number of words (although admittedly that may not be a problem in your case as you say you know there will only be a handful of words).
One way (although this is perhaps better suited a spellcheck-type algorithm) is the "edit distance", ie., calculate how many edits it takes to transform one string to another. A common technique is found here:
http://en.wikipedia.org/wiki/Levenshtein_distance
You might want to look into the algorithms used by biologists to compare DNA sequences, since they have to cope with many of the same things (chunks may be missing, or have been inserted, or just moved to a different position in the string.
The Smith-Waterman algorithm would be one example that'd probably work fairly well, although it might be too slow for your uses. Might give you a starting point, though.
i had a similar problem, i needed to get the percentage of characters in a string that were similar. it needed exact sequences, so for example "hello sir" and "sir hello" when compared needed to give me five characters that are the same, in this case they would be the two "hello"'s. it would then take the length of the longest of the two strings and give me a percentage of how similar they were. this is the code that i came up with
int compare(string a, string b){
return(a.size() > b.size() ? bigger(a,b) : bigger(b,a));
}
int bigger(string a, string b){
int maxcount = 0, currentcount = 0;//used to see which set of concurrent characters were biggest
for(int i = 0; i < a.size(); ++i){
for(int j = 0; j < b.size(); ++j){
if(a[i+j] == b[j]){
++currentcount;
}
else{
if(currentcount > maxcount){
maxcount = currentcount;
}//end if
currentcount = 0;
}//end else
}//end inner for loop
}//end outer for loop
return ((int)(((float)maxcount/((float)a.size()))*100));
}
I can't mark two answers here, so I'm going to answer and mark my own. The Levenshtein distance appears to be the correct method in most cases for this. But, it is worth mentioning j_random_hackers answer as well. I have used an implementation of LZMA to test his theory, and it proves to be a sound solution. In my original question I was looking for a method for short strings (2 to 200 chars), where the Levenshtein Distance algorithm will work. But, not mentioned in the question was the need to compare two (larger) strings (in this case, text files of moderate size) and to perform a quick check to see how similar the two are. I believe that this compression technique will work well but I have yet to study it to find at which point one becomes better than the other, in terms of the size of the sample data and the speed/cost of the operation in question. I think a lot of the answers given to this question are valuable, and worth mentioning, for anyone looking to solve a similar string ordeal like I'm doing here. Thank you all for your great answers, and I hope they can be used to serve others well too.
There's another way. Pattern recognition using convolution. Image A is run thru a Fourier transform. Image B also. Now superimposing F(A) over F(B) then transforming this back gives you a black image with a few white spots. Those spots indicate where A matches B strongly. Total sum of spots would indicate an overall similarity. Not sure how you'd run an FFT on strings but I'm pretty sure it would work.
The difficulty would be to match the strings semantically.
You could generate some kind of value based on the lexical properties of the string. e.g. They bot have blue, and sky, and they're in the same sentence, etc etc... But it won't handle cases where "Sky's jean is blue", or some other odd ball English construction that uses same words, but you'd need to parse the English grammar...
To do anything beyond lexical similarity, you'd need to look at natural language processing, and there isn't going to be one single algorith that would solve your problem.
Possible approach:
Construct a Dictionary with a string key of "word1|word2" for all combinations of words in the reference string. A single combination may happen multiple times, so the value of the Dictionary should be a list of numbers, each representing the distance between the words in the reference string.
When you do this, there will be duplication here: for every "word1|word2" dictionary entry, there will be a "word2|word1" entry with the same list of distance values, but negated.
For each combination of words in the comparison string (words 1 and 2, words 1 and 3, words 2 and 3, etc.), check the two keys (word1|word2 and word2|word1) in the reference string and find the closest value to the distance in the current string. Add the absolute value of the difference between the current distance and the closest distance to a counter.
If the closest reference distance between the words is in the opposite direction (word2|word1) as the comparison string, you may want to weight it smaller than if the closest value was in the same direction in both strings.
When you are finished, divide the sum by the square of the number of words in the comparison string.
This should provide some decimal value representing how closely each word/phrase matches some word/phrase in the original string.
Of course, if the original string is longer, it won't account for that, so it may be necessary to compute this both directions (using one as the reference, then the other) and average them.
I have absolutely no code for this, and I probably just re-invented a very crude wheel. YMMV.