This is my first post here so please be patient with me.
I am stuck on a problem and don't know what to do. I am given a
non-overlapping substring of length 'm' that appears twice in a string of length 'n'. What is the probability of finding the substring of length 'm' in both that string and another string of length '2n'?
For argument's sake let's say that m = 4 and n = 33. I have tried to use Independent Event probability as well as Markov Chain Models, but my answer never seems to be correct.
What would be the chance that the same 2 non-overlapping substrings of length 4 that are found in the string of length 33 will be found in a string of length 66?
My question is similar to How to efficiently find identical substrings of a specified length in a collection of strings
Let's assume that I have t strings, each one is at length n
and I need to find a substring at length k that has at most one index that is not identical and it needs to be the same index in every string, for example, consider the following 4 strings:
ACTAGGGGT
TAGGAAACC
CCCGGTTGG
GTGGGACTG
The output, in this case, should be: T*GG and S = (3,1,6,2)
where S is the starting index of the substring in each string Si.
Palindrome is a language in automata. But i am unable to understand the following paragraph. I have calculated many things, and tried my best to estimate, but i couldnot.
Length of palindroma:
As we know that string is of length n and numbfer of symbols in the alphabet is 2, which shows that there are as many palindromes of length 2n as there are the strings of lenth n i.e. the required number of palindromes are 2^n.
A palindrome is a word that is the same read from the left as read from the right. So the first half determines completely the letters in the second half. This is why the number of palindromes of length $2n$ is equal to the number of words of length $n$ - the latter are all the possible first halfs of words of length $2n$.
Over an alphabet of two letters you have two choices for each of the $n$ positions, so there are $2^n$ distinct words of length $n$.
Starting with a list of integers the task is to convert each integer into a string such that the resulting list of strings will be in numeric order when sorted lexicographically.
This is needed so that a particular system that is only capable of sorting strings will produce an output that is in numeric order.
Example:
Given the integers
1, 23, 3
we could convert the to strings like this:
"01", "23", "03"
so that when sorted they become:
"01", "03", "23"
which is correct. A wrong result would be:
"1", "23", "3"
because that list is sorted in "string order", not in numeric order.
I'm looking for something more efficient than the simple zero-padding scheme. In order to cover all possible 32 bit integers we'd need to pad to 10 digits which is inefficient.
For integers, prefix each number with the length. To make it more readable, use 'a' for length 1, and 'b' for length 2. Example:
non-encoded encoded
1 "a1"
3 "a3"
23 "b23"
This scheme is a bit simpler than prefixing each digit, but only works with numbers, not numbers mixed with text. It can be made to work for negative numbers as well, and even BigDecimal numbers, using some tricks. I wrote an implementation in Apache Jackrabbit 2.x, to make BigDecimal indexable (sortable) as text. For that, I used a format that only uses the characters '0' to '9' and consists of:
one character for: signum(value) + 2
one character for: signum(exponent) + 2
one character for: length(exponent) - 1
multiple characters for: exponent
multiple characters for: value (-1 if inverted)
Only the signum is encoded if the value is zero. The exponent is not encoded if zero. Negative values are "inverted" character by character (0 => 9, 1 => 8, and so on). The same applies to the exponent.
Examples:
non-encoded encoded
0 "2"
2 "322" (signum 1; exponent 0; value 2)
120 "330212" (signum 1; exponent signum 1, length 1, value 2; value 12)
-1 "179" (signum -1, rest inverted; exponent 0; value 1 (-1, inverted))
Values between BigDecimal(BigInteger.ONE, Integer.MIN_VALUE) and BigDecimal(BigInteger.ONE, Integer.MAX_VALUE) are supported.
TL;DR
Encode digits according to their order of magnitude (OM) and other characters so they sort as desired, relative to numbers: jj-a123 would be encoded zjzjz-zaC1B2A3
Longer explanation
This would depend somewhat upon the sorting algorithm that will finally be used to sort and how one would want any given punctuation characters to be sorted in relation to letters and numbers, but if it's "ascii-betical" or similar, you could encode each digit of a number to represent its order of magnitude (OM) in the number, while encoding other characters such that they would sort according to your desired sort order.
For simplicity, I would suggest beginning with encoding every non-numeric character with a "high" value (e.g. lower case z or even ~ if final value is ASCII), so that it sorts after encoded digits. Then cache each digit encountered until another non-numeric is encountered, then encode each cached digit with a value representing its OM. If the number 12945 was encountered in between non-numerics, you would output an E to encode an OM of 5, then the digit that is that order of magnitude, 1, followed by the next OM of 4 (D) and its associated digit, 2. Continue until all numeric digits have been flushed, then continue with non-numerics.
Non-numerics would be treated individually and ranked relative to the OM of digits. If it is desired for them to sort "above" numbers (perhaps the space character or certain others deemed special) they would be encoded by prepending a low-value character (like the space character, if final value will be treated and sorted as ASCII). When/if another numeric is encountered, begin caching and encode according to OM once all consecutive numerics are cached.
Alternately, processing the string in reverse order would preclude the need to cache numbers except for a single "is it a digit?" test and "is the last character a digit?" test. If the first is not true, then use (one of?) the "non-digit" OM character(s). If the first test is true then use the lowest-OM "digit" character (A in my examples). If both tests are true, then increment your OM character (A -> B or E -> F) before use.
Certain levels of additional filtering - or even translation - could be applied. If one wanted to allow accurate sorting based upon Roman numerals, one could encode them as decimal (or even hexadecimal) numbers with an appropriate OM.
Treating decimal points (either periods or commas, depending) as actual decimal separators, and distinct from other punctuation would probably be beyond the true utility of this encoding scheme, as alphanumeric fields seldom use a period or comma as a decimal separator. If it is desired to use them that way, the algorithm would simply detect a decimal separator (either period or comma as appropriate, in between digits) and not encode the numeric portion after that separator as anything but normal text. Fractional portions are actually sorted correctly during a normal ASCII based sort, because more digits represents greater precision - not greater magnitude.
Examples
non-encoded encoded
----------- -------
12345 E1D2C3B4A5
a100 zaC1B0A0
a20 zaB2A0
a2000 zaD2C0B0A0
x100.5 zxC1B0A0z.A5
x100.23 zxC1B0A0z.B2A3
1, 23, 3 A1z,z B2A1z,z A3
1, 2, 3 A1z,z A2z,z A3
1,2,3 A1z,A2z,A3
Potential advantages
Going somewhat beyond simple numeric sorting, some advantages to this encoding method would be several aspects of flexibility with final effective sort order - you are essentially encoding a category for each character - digits get a category based upon their position within the greater string of digits known as a number, while other characters are simply told to sort in their normal way (e.g. ASCII), but after numbers. Any exceptions that should sort before numbers or in other orders would be in one or more additional categories. ASCII can effectively be re-encoded to sort in a non-ASCII way:
You could encode lower case letters to sort before or along with upper case letters. To switch the lower and upper cases, you encode lower case letters with a y and upper case letters with a z. For a pseudo-case-insensitive sort, categorizing both A and a with the same encoding character would sort both of them before B and b, though A would nonetheless always sort before a
If you want Extended ASCII characters (e.g. with diacritics) to sort along with their ASCII cousins, you encode À, Á, Â, Ã, Ä, Å, and Æ along with A by using an a as the OM character, encode B, C, and Ç with a b, and E, È, É, Ê, and Ë with a c, etc. The same intra-category sort order caveat still applies, and some decisions need to be made on characters like capital Eth, and to a certain extent others like Thorn, and Sharp S (Ð, Þ, and ß respectively) as to whether they will sort based on similarities in appearance or pronunciation, or instead more properly perhaps, alphabetical order.
Small advantage of being basically human-readable, with effort
Caveats
Though this allows many 'categories' of characters to be defined, be sure to remember that each order of magnitude for digits is its own category - you need to know that the data will not contain numbers that are greater in OM than approximately 250, depending upon how many other categories you wish to define (ASCII 0 is reserved for storing strings, and there needs to be at least one other character to indicate "not a digit" - at least for alphanumeric data - making the maximum perhaps 254 orders of magnitude), but that should be plenty for any situation I can imagine. I'm not sure what other issues quantum computing will bring about, but there's probably a quantum solution to it, whatever it is.
Finally, if the hyphen is encoded as a non-numeric character, and all non-numerics are encoded with a higher OM than digits, negative numbers would be encoded as greater than any positive number. The hyphen should be encoded as a lower-than-digit-OM (perhaps only when preceding a digit) if negative numbers need to be sorted correctly according to magnitude.
Since the ASCII code of A is greater than 9, you could encode them as hexadecimal strings.
The integers
1, 23, 3
can be encoded as
00000001, 00000017, 00000003
and 32-bit integers can always be encoded as 8-character strings. (assume unsigned)
I'm playing with some variant of Hadamard matrices. I want to generate all n-bit binary strings which satisfy these requirements:
You can assume that n is a multiple of 4.
The first string is 0n.→ a string of all 0s.
The remaining strings are sorted in alphabetic order.→ 0 comes before 1.
Every two distinct n-bit strings have Hamming distance n/2.→ Two distinct n-bit strings agree in exactly n/2 positions and disagree in exactly n/2 positions.
Due to the above condition, every string except for the first string must have the same number of 0s and 1s. → Every string other than the first string must have n/2 ones and n/2 zeros.
(Updated) All the n-bit strings begin with 0.
For example, this is the list that I want for when n=4.
0000
0011
0101
0110
You can easily see that every two distinct rows have hamming distance n/2 = 4/2 = 2 and the list satisfies all the other requirements as well.
Note that I want to generate all such strings. My algorithm may just output three strings 0000, 0011, and 0101 before terminating. This list satisfies all the requirements above but it misses 0110.
What would be a good way to generate such sets? A python pseudo-code is preferred but any high-level description will do.
What is the maximum number of such strings for a given n?For example, when n=4, the max number of such strings happen to be 4. I'm wondering whether there can be any closed form solution for this upper bound.
Thanks.
To answer question 1,
Starting with a string of n zeros (let's call it s0) and a string of n/2 zeros followed by n/2 1's (call it s1), generate the next permutation (call it p):
scan string from right to left
replace first occurrence of "01" with "10"
(unless the first occurrence is at the string start)
move all "1"'s that are on the right of the "01" to the string end
return replaced string
Use the permutation generation order to keep a record of permutations added to sets. If the number of bits set in xoring p with each number currently in the set is n/2, add p to the list; otherwise, if the number of bits set in xoring p with s1 is n/2 and p has not been recorded, start a new set search with s0, s1; and p only as an additional condition for the xor test (since the primary search will review all permutations, this set need not generate additional sets). Use p to generate the next permutation.