Searching with mismatches

[Peter Simmonds Member Find all posts View Profile Close Peter Simmonds]

sorry if this has been covered before.

If I had a long text string (eg. 5000 characters), and I wanted to find out whether there was a partially matching sequence of characters in a second string of 5000 characters, are there any ways of speeding up the matching process. The basic search method, I imagine, would involve scanning every combination of consecutive letters in the two strings until a matching run was found, for example, a required match of 20 letters from 30. This is horribly time consuming, and the time taken goes up proportional to the square of the string lengths.

It occurred to me that if INSTR has an additional parameter that specified the number of mismatches that would be tolerated, it would be very easy to cut one string into 30 character partly overlapping fragments (search_fragment$), and simply test each one against the other string (search_string$):

eg.

match_pos = INSTR(search_string$, search_fragment$, 10)

where 10 is the number of mismatches allowed, and match_pos would be assigned the first matching position.

INSTR obviously doesn't work in this way, and none of the matching functions tolerate mismatches as would be required here.

I realise this is a simple problem that must have been solved elegantly before (probably in the 1950s or 1960s!), so any thoughts on quicker strategies would be useful.

Thanks

Peter

[George Bleck Member Find all posts View Profile Close George Bleck]

Might this thread help?

http://www.powerbasic.com/support/pb...on+subsequence

[Stanley Durham Member Find all posts View Profile Close Stanley Durham]

other considerations;

REGEXPR might be best
VERIFY([start&,] MainString, MatchString)

(using ANY option)
TALLY(MainString, [ANY] MatchString)
INSTR([Pos&,] Main$, [ANY] Match$)

Suffix tree; http://en.wikipedia.org/wiki/Suffix_tree
Suffix array; http://en.wikipedia.org/wiki/Suffix_array
Burrows–Wheeler transform; value

[Chris Holbrook Member Find all posts View Profile Close Chris Holbrook]

How many matching sequences are you looking for, all of them or just the first one?

If you want 'em all, then there is a good chance that quite a few single characters will have lots of matches, so you may want to specify a mimimum length.

How can 30 equally sized overlapping fragments in a string of 5000 characters contain all the possible candidates for a substring match?

[Michael Mattias Member Find all posts View Profile Close Michael Mattias]

Add one more option... "or pretty close"... and you understand why Google(r) pays its lexer coders well.

[Peter Simmonds Member Find all posts View Profile Close Peter Simmonds]

Ideally it should find all of them and record the one with the highest number of matches

The idea is to find matches of, for example, 20 matches from 30 aligned letters in the two strings. The problem is that the matched letters need not be consecutive, so its not really a case of finding minimum length matches as you would in REGEXPR.

Ill try to look at the examples of tree searching methods, but this look a bit beyond me!

Thanks for your help

Peter

[Gösta H. Lovgren-2 Member Find all posts View Profile Close Gösta H. Lovgren-2]

The problem is that the matched letters need not be consecutive...
Peter

I'm not understanding that. If the letters are not consecutive, how can they match?

Example code would be appreciated. Preferably compileable. Which would probably spawn a dozen different answers from half as many sources. (Everybody having more than one way to skin a cat.)

=========================================
A vacant mind invites dangerous inmates.
(Nicholas Hilliard)
=========================================

[Chris Holbrook Member Find all posts View Profile Close Chris Holbrook]

Peter, do you mean that, for example, ABC would match with xAxBxCx, where x represents zero or more charcters which are not A, B or C?

[Peter Simmonds Member Find all posts View Profile Close Peter Simmonds]

Hi

sorry if this was unclear. For the hypothetical example of the 20 matches from 30 numbers or letters, the following would be a match:

1234567890123456789012345678901234567890123456780
      ***   * * ****  * * *********  ***      *
3456217893563357789033315778901234523890455287744

since there is a region where enough numbers are the same within a 30 letter window.

Principle is that the matches don't have to be consecutive so the minimum length methods cannot be used.

I could post code for the slow way of doing this, but it really is rubbish when scaled up!

Peter

[Rodney Hicks Find all posts View Profile Close Rodney Hicks]

Pseudo code:

IF LEN(searchme$)<=LEN(matchme$) THEN
  FOR c=1 TO LEN(searchme$)-29
    mismatch=0
    match=0
    FOR cc=c TO c+29
      IF MID$(searchme$,cc,1)<>MID$(matchme$,cc,1) THEN
        INCR mismatch
        IF mismatch=11 THEN EXIT FOR
      ELSE
        INCR match
      END IF
    NEXT cc
    IF mismatch>10 THEN ITERATE FOR
    'at this point save your 'c' value or do whatever you want to do with a
    'matched string
  NEXT c
END IF

[Gösta H. Lovgren-2 Member Find all posts View Profile Close Gösta H. Lovgren-2]

Hi

sorry if this was unclear. For the hypothetical example of the 20 matches from 30 numbers or letters, the following would be a match:

1234567890123456789012345678901234567890123456780
      ***   * * ****  * * *********  ***      *
3456217893563357789033315778901234523890455287744

since there is a region where enough numbers are the same within a 30 letter window.

Principle is that the matches don't have to be consecutive so the minimum length methods cannot be used.

I could post code for the slow way of doing this, but it really is rubbish when scaled up!

Peter

Post the code for the "slow" way. I, for one, still don't see what you are getting at. Probably because I am a "slow" myself. Do the stars represent where there is a match in the two strings?

Okay, I printed in a text editor and think I see what you mean.

1234567890123456789012345678901234567890123456780
3456217893563357789033315778901234523890455287744
------***-------****------*********--***------*--



Still code would be a better place to start.

======================================
"Few things are harder to put up with
than a good example."
Mark Twain (1835-1910)
======================================

[Erich Schulman Member Find all posts View Profile Close Erich Schulman]

agrep might be useful in this situation.

[Paul Dixon Member Find all posts View Profile Close Paul Dixon]

Something like this?

PBCC5 program
#COMPILE EXE


#REGISTER NONE
FUNCTION PBMAIN () AS LONG

'create a test string
FOR r& = 1 TO 5000
    a$=a$+CHR$(RND(97,122))
NEXT

'create a similar string with a mismatches
b$=a$
FOR r& = 1 TO 3000
    MID$(b$,RND(1,LEN(b$)),1)=CHR$(RND(97,122))
NEXT
k&= GoodMatch(a$,b$,30,6)
PRINT k&
PRINT MID$(a$,k&,30)
PRINT MID$(b$,k&,30)

k&= GoodMatch(a$,b$,30,7)
PRINT k&
PRINT MID$(a$,k&,30)
PRINT MID$(b$,k&,30)

k&= GoodMatch(a$,b$,30,8)
PRINT k&
PRINT MID$(a$,k&,30)
PRINT MID$(b$,k&,30)

k&= GoodMatch(a$,b$,30,9)
PRINT k&
PRINT MID$(a$,k&,30)
PRINT MID$(b$,k&,30)

k&= GoodMatch(a$,b$,30,10)
PRINT k&
PRINT MID$(a$,k&,30)
PRINT MID$(b$,k&,30)



WAITKEY$
END FUNCTION


FUNCTION GoodMatch(StringToSearch AS STRING, WhatToSearchFor AS STRING, SegmentSize AS LONG,MismatchesAllowed AS LONG) AS LONG
    
    FOR r& = 1 TO SegmentSize
        IF MID$(StringToSearch,r&,1) = MID$(WhatToSearchFor,r&,1) THEN
            MatchCount& = MatchCount& + 1
        END IF
    NEXT
    
    FOR r& = SegmentSize+ 1 TO LEN(StringToSearch)
        IF MID$(StringToSearch,r&,1) = MID$(WhatToSearchFor,r&,1) THEN
            MatchCount& = MatchCount& + 1
        END IF
        IF MID$(StringToSearch,r&-SegmentSize,1) = MID$(WhatToSearchFor,r&-SegmentSize,1) THEN
            MatchCount& = MatchCount& - 1

        END IF

        IF SegmentSize - MatchCount& < MismatchesAllowed THEN
            FUNCTION = r&-SegmentSize
            EXIT FUNCTION
        END IF
    NEXT
        
    FUNCTION = 0
    
END FUNCTION

[Peter Simmonds Member Find all posts View Profile Close Peter Simmonds]

Paul

It's more complicated than that, the program you wrote assumes that the two text strings are aligned, whereas what it really needs to do is find a 30/20 match anywhere in the StringToSearch and WhatToSearchFor strings, for example there may be match between a window starting at position 500 in StringToSearch and at 2544 in WhatToSearchFor.

Unless I am stupidly missing something in your program which is always possible!

Peter

[Peter Simmonds Member Find all posts View Profile Close Peter Simmonds]

The slow way to do it!

As requested, here is the simple way to find the match, but you'll notice the three nested loops which makes the search time proportion to:

the length of search string * length of the query string * required match length

which means that it scales very poorly for large searches.

#COMPILE EXE
DEFLNG a-z

FUNCTION PBMAIN () AS LONG

    CLS: PRINT "Sequence scan"

    REM Just some example text from the helpfile
    query$ = "If the ANY keyword is included, MatchStr  specifies a list of single characters to be searched for individually, a match on any one of which will cause the"
    search$ = "starting with its first character (or the character specified by start). EXTRACT$ returns single characters of MainStr, "

    REM specifying what match is required, ie at least 21 characters out of a text window of 30
    match_length = 30: mismatch_target = 9: REM maximum number of tolerated mismatches, ie. 30 minus 21

    REM scan individual characters within the window
    FOR a = 1 TO LEN(query$) - match_length

        query_fragment$ = MID$(query$, a, match_length)

        FOR b = 1 TO LEN(search$) - match_length

            search_fragment$ = MID$(search$, b, match_length)

            REM Initalise counters
            mismatch_total = 0: scan_total = 0: match_found = 0: REM to keep running totals

            REM Scan subfragments from search$ and query$
            FOR c = 1 TO match_length

                INCR scan_total
                IF UCASE$(MID$(query_fragment$, c, 1)) <> UCASE$(MID$(search_fragment$, c, 1)) THEN INCR mismatch_total

                REM early exits if too many mismatches or match total gone beyond maximum needed
                IF mismatch_total > mismatch_target THEN EXIT FOR
                IF scan_total - mismatch_total >= match_length - mismatch_target THEN match_found = 1: EXIT FOR

            NEXT c
                    
            REM Found a match
            IF match_found THEN GOSUB Report_match

        NEXT b

    NEXT a

    PRINT: PRINT "The end"
    WAITKEY$

    EXIT FUNCTION

Report_match:

    PRINT: PRINT "Match found at"; a; "in the query sequence and"; b; "in the search sequence"
    PRINT query_fragment$

    FOR d = 1 TO match_length
        IF UCASE$(MID$(query_fragment$, d, 1)) = UCASE$(MID$(search_fragment$, d, 1)) THEN PRINT "|"; ELSE PRINT " ";
    NEXT d

    PRINT: PRINT search_fragment$
    
    WAITKEY$

RETURN

END FUNCTION

(Written in PBCC 4.0)

Peter

[John Gleason Member Find all posts View Profile Close John Gleason]

A faster "slow" way, but I can't see a better "not slow" way right now.

#COMPILE EXE
DEFLNG a-z

FUNCTION PBMAIN () AS LONG
   LOCAL searchFragPtr, queryFragPtr AS BYTE PTR, printThin AS LONG

    CLS: PRINT "Sequence scan"

    REM Just some example text from the helpfile
    'Make all capital from the beginning and save a lot of calculating time.
    query$ = REPEAT$(50, UCASE$("If the ANY keyword is included, MatchStr  specifies a list of single characters to be searched for individually, a match on any one of which will cause the"))
    search$ = REPEAT$(50, UCASE$("starting with its first character (or the character specified by start). EXTRACT$ returns single characters of MainStr, "))

    REM specifying what match is required, ie at least 21 characters out of a text window of 30
    match_length = 30: mismatch_target = 9: REM maximum number of tolerated mismatches, ie. 30 minus 21

    REM scan individual characters within the window
    FOR a = 1 TO LEN(query$) - match_length

        query_fragment$ = MID$(query$, a, match_length)
        queryFragPtr = STRPTR(query_fragment$)

        FOR b = 1 TO LEN(search$) - match_length

            search_fragment$ = MID$(search$, b, match_length)
            searchFragPtr = STRPTR(search_fragment$)

            REM Initalise counters
            mismatch_total = 0: scan_total = 0: match_found = 0: REM to keep running totals

            REM Scan subfragments from search$ and query$
            FOR c = 1 TO match_length

                INCR scan_total
                'You have a choice here of ASC or even faster, index pointers.
                IF @searchFragPtr[c-1] <> @queryFragPtr[c-1]  THEN INCR mismatch_total
'               IF ASC(query_fragment$, c) <> ASC(search_fragment$, c) THEN INCR mismatch_total

                REM early exits if too many mismatches or match total gone beyond maximum needed
                IF mismatch_total > mismatch_target THEN EXIT FOR
                IF scan_total - mismatch_total >= match_length - mismatch_target THEN match_found = 1: EXIT FOR

            NEXT c

            REM Found a match
            IF match_found THEN GOSUB Report_match

        NEXT b

    NEXT a

    PRINT: PRINT "The end"
    WAITKEY$

    EXIT FUNCTION

Report_match:

    INCR printThin
    IF (printThin AND 31) = 0 THEN    'only print 1 of every 32 matches
       PRINT: PRINT "Match found at"; a; "in the query sequence and"; b; "in the search sequence"
       PRINT query_fragment$

       FOR d = 1 TO match_length
           IF UCASE$(MID$(query_fragment$, d, 1)) = UCASE$(MID$(search_fragment$, d, 1)) THEN PRINT "|"; ELSE PRINT " ";
       NEXT d

       PRINT: PRINT search_fragment$
       WAITKEY$
    END IF


RETURN

END FUNCTION

[John Gleason Member Find all posts View Profile Close John Gleason]

btw, it is usually difficult to break out of the O(n²) time complexity prison. Ya just don't know tho, a lot of ideas arise from a lot o' gray matter here.

[Rodney Hicks Find all posts View Profile Close Rodney Hicks]

How many distinct characters are there in either string? 10 number characters, Upper case letters, lower case letters, punctuation characters, etc.?
You might try a density function to verify that there are 21 characters in the match string segment that match the search string segment before checking for matching positions.

[Peter Simmonds Member Find all posts View Profile Close Peter Simmonds]

Hi

I used a text example in the posted program but actually the real implementation is for DNA sequence matching where there are only 4 different characters (G, A, T and C and perhaps a fifth for undetermined). This might simplify the problem although a general algorithm might be better in the long run.

Peter

[Michael Mattias Member Find all posts View Profile Close Michael Mattias]

This might simplify the problem although a general algorithm might be better in the long run.

Better for whom? Optimization is always application-specific.

I'm not a pattern-match guy, but I am cerrtain limiting the target character set to this small subset of the ANSI characters can vastly simplify the solution, but it may require an entirely fresh approach to take advantage of this previously-unstated fact.

MCM