I'm trying to invert a string in go but I'm having trouble handling the characters. Unlike C, GO treats strings as vectors of bytes, rather than characters, which are called runes here. I tried to do some type conversions to do the assignments, but so far I could not.
The idea here is to generate 5 strings with random characters of sizes 100, 200, 300, 400 and 500 and then invert their characters. I was able to make C work with ease, but in GO, the language returns an error saying that it is not possible to perform the assignment.
func inverte() {
var c = "A"
var strs, aux string
rand.Seed(time.Now().UnixNano())
// Gera 5 vetores de 100, 200, 300, 400, e 500 caracteres
for i := 1; i < 6; i++ {
strs = randomString(i * 100)
fmt.Print(strs)
for i2, j := 0, len(strs); i2 < j; i2, j = i+1, j-1 {
aux = strs[i2]
strs[i2] = strs[j]
strs[j] = aux
}
}
}
If you want to take into account unicode combining characters (characters that are intended to modify other characters, like an acute accent ´ + e = é), Andrew Sellers has an interesting take in this gist.
It starts by listing the Unicode block range for all combining diacritical marks (CDM) (the Unicode block containing the most common combining characters)
regulars (inherited), so the usual ◌̀ ◌́ ◌̂ ◌̃ ◌̄ ◌̅ ◌̆ ◌̇ ◌̈, ...;
extended (containing diacritical marks used in German dialectology -- Teuthonista)
supplement (or the Uralic Phonetic Alphabet, Medievalist notations, and German dialectology -- again, Teuthonista)
for symbols (arrows, dots, enclosures, and overlays for modifying symbol characters)
Half Marks (diacritic mark parts for spanning multiple characters, as seen here)
var combining = &unicode.RangeTable{
R16: []unicode.Range16{
{0x0300, 0x036f, 1}, // combining diacritical marks
{0x1ab0, 0x1aff, 1}, // combining diacritical marks extended
{0x1dc0, 0x1dff, 1}, // combining diacritical marks supplement
{0x20d0, 0x20ff, 1}, // combining diacritical marks for symbols
{0xfe20, 0xfe2f, 1}, // combining half marks
},
}
You can then read, rune after rune, your initial string:
sv := []rune(s)
But if you do so in reverse order, you will encounter combining diacritical marks (CDMs) first, and those need to preserve their order, to not be reversed
for ix := len(sv) - 1; ix >= 0; ix-- {
r := sv[ix]
if unicode.In(r, combining) {
cv = append(cv, r)
fmt.Printf("Detect combining diacritical mark ' %c'\n", r)
}
(note the space around the %c combining rune: '%c' without space would means combining the mark with the first 'ͤ': instead of ' ͤ '. I tried to use the CGJ Combining Grapheme Joiner \u034F, but that does not work)
If you encounter finally a regular rune, you need to combine with those CDMs, before adding it to your reverse final rune array.
} else {
rrv := make([]rune, 0, len(cv)+1)
rrv = append(rrv, r)
rrv = append(rrv, cv...)
fmt.Printf("regular mark '%c' (with '%d' combining diacritical marks '%s') => '%s'\n", r, len(cv), string(cv), string(rrv))
rv = append(rv, rrv...)
cv = make([]rune, 0)
}
Where it gets even more complex is with emojis, and, for instance more recently, modifiers like the Medium-Dark Skin Tone, the type 5 on the Fitzpatrick Scale of skin tones.
If ignored, Reverse '👩🏾🦰👱🏾🧑🏾⚖️' will give '️⚖🏾🧑🏾👱🦰🏾👩', loosing the skin tone on the last two emojis.
And don't get me started on the ZERO WIDTH JOINER (200D), which, from Wisdom/Awesome-Unicode, forces adjacent characters to be joined together (e.g., Arabic characters or supported emoji). It Can be used this to compose sequentially combined emoji.
Here are two examples of composed emojis, whose inner elements order should remain in the same order when "reversed":
👩🏾🦰 alone is (from Unicode to code points converter):
👩: women (1f469)
dark skin (1f3fe)
ZERO WIDTH JOINER (200d)
🦰red hair (1f9b0)
Those should remain in the exact same order.
The "character" "judge" (meaning an abstract idea of the semantic value for "judge") can be represented with several glyphs or one glyph.
🧑🏾⚖️ is actually one composed glyph (composed here of two emojis), representing a judge. That sequence should not be inverted.
The program below correctly detect the "zero width joiner" and do not invert the emojis it combines.
It you inspect that emoji, you will find it composed of:
🧑: Adult (1F9D1)
🏾: dark skin (1f3fe)
ZERO WIDTH JOINER (200d) discussed above
⚖: scale (2696)
VARIATION SELECTOR (FE0F), part of the unicode combining characters (characters that are intended to modify other characters), here requesting that character 'scale' to be displayed emoji-style (with color) ⚖️, using VS16 (U+FE0F), instead of text style (monochrome) '⚖', using VS15 (U+FE0E).
Again, that sequence order needs to be preserved.
Note: the actual judge emoji 👨🏾⚖️ uses a MAN 🧑 (1F468), instead of an Adult 🧑 (1F9D1) (plus the other characters listed above: dark skin, ZWJ, scale), and is therefore represented as one glyph, instead of a cluster of graphemes.
Meaning: the single glyph, official emoji for "judge", needs to combine "man" with "scale" (resulting in one glyph 👨🏾⚖️) instead of "adult" + "scale".
The latter, "adult" + "scale", is still considered as "one character": you cannot select just the scale, because of the ZWJ (Zero Width Joiner).
But that "character" is represented as a composed glyph 🧑🏾⚖️, two glyphs, each one a concrete written representations a corresponding grapheme through codepoint+font)
Obviously, using the first combination ("man"+"scale") results in a more expressive character 👨🏾⚖️.
See "The relationship between graphemes and abstract characters for textual representation"
Graphemes and orthographic characters are fairly concrete objects, in the sense that they are familiar to common users—non-experts, who are typically taught to work in terms of them from the time they first learn their “ABCs” (or equivalent from their writing system, of course).
In the domain of information systems, however, we have a different sense of character: abstract characters which are minimal units of textual representation within a given system.
These are, indeed, abstract in two important senses:
first, some of these abstract characters may not correspond to anything concrete in an orthography, as we saw above in the case of HORIZONTAL TAB.
Secondly, the concrete objects of writing (graphemes and orthographic characters) can be represented by abstract characters in more than one way, and not necessarily in a one-to-one manner, as we saw above in the case of “ô” being represented by a sequence <O, CIRCUMFLEX>.
Then: "From grapheme to codepoint to glyph":
Graphemes are the units in terms of which users are usually accustomed to thinking.
Within the computer, however, processes are done in terms of characters.
We don’t make any direct connection between graphemes and glyphs.
As we have defined these two notions here, there is no direct connection between them. They can only be related indirectly through the abstract characters.
This is a key point to grasp: the abstract characters are the element in common through which the others relate.
Full example in Go playground.
Reverse 'Hello, World' => 'dlroW ,olleH'
Reverse '👽👶⃠🎃' => '🎃👶⃠👽'
Reverse '👩🏾🦰👱🏾🧑🏾⚖️' => '🧑🏾⚖️👱🏾👩🏾🦰'
Reverse 'aͤoͧiͤ š́ž́ʟ́' => 'ʟ́ž́š́ iͤoͧaͤ'
Reverse 'H̙̖ell͔o̙̟͚͎̗̹̬ ̯W̖͝ǫ̬̞̜rḷ̦̣̪d̰̲̗͈' => 'd̰̲̗͈ḷ̦̣̪rǫ̬̞̜W̖͝ ̯o̙̟͚͎̗̹̬l͔leH̙̖'
As you correctly identified, go strings are immutable, so you cannot assign to rune/character values at given indices.
Instead of reversing the string in-place one must create a copy of the runes in the string and reverse those instead, and then return the resulting string.
For example (Go Playground):
func reverse(s string) string {
rs := []rune(s)
for i, j := 0, len(rs)-1; i < j; i, j = i+1, j-1 {
rs[i], rs[j] = rs[j], rs[i]
}
return string(rs)
}
func main() {
fmt.Println(reverse("Hello, World!"))
// !dlroW ,olleH
fmt.Println(reverse("Hello, 世界!"))
// !界世 ,olleH
}
There are problems with this approach due to the intricacies of Unicode (e.g. combining diacritical marks) but this will get you started.
Suppose we are given two strings s1 and s2(both lowercase). We have two find the minimal lexographic string that can be formed by merging two strings.
At the beginning , it looks prettty simple as merge of the mergesort algorithm. But let us see what can go wrong.
s1: zyy
s2: zy
Now if we perform merge on these two we must decide which z to pick as they are equal, clearly if we pick z of s2 first then the string formed will be:
zyzyy
If we pick z of s1 first, the string formed will be:
zyyzy which is correct.
As we can see the merge of mergesort can lead to wrong answer.
Here's another example:
s1:zyy
s2:zyb
Now the correct answer will be zybzyy which will be got only if pick z of s2 first.
There are plenty of other cases in which the simple merge will fail. My question is Is there any standard algorithm out there used to perform merge for such output.
You could use dynamic programming. In f[x][y] store the minimal lexicographical string such that you've taken x charecters from the first string s1 and y characters from the second s2. You can calculate f in bottom-top manner using the update:
f[x][y] = min(f[x-1][y] + s1[x], f[x][y-1] + s2[y]) \\ the '+' here represents
\\ the concatenation of a
\\ string and a character
You start with f[0][0] = "" (empty string).
For efficiency you can store the strings in f as references. That is, you can store in f the objects
class StringRef {
StringRef prev;
char c;
}
To extract what string you have at certain f[x][y] you just follow the references. To udapate you point back to either f[x-1][y] or f[x][y-1] depending on what your update step says.
It seems that the solution can be almost the same as you described (the "mergesort"-like approach), except that with special handling of equality. So long as the first characters of both strings are equal, you look ahead at the second character, 3rd, etc. If the end is reached for some string, consider the first character of the other string as the next character in the string for which the end is reached, etc. for the 2nd character, etc. If the ends for both strings are reached, then it doesn't matter from which string to take the first character. Note that this algorithm is O(N) because after a look-ahead on equal prefixes you know the whole look-ahead sequence (i.e. string prefix) to include, not just one first character.
EDIT: you look ahead so long as the current i-th characters from both strings are equal and alphabetically not larger than the first character in the current prefix.
I simply want to check if all the letters that occur in a string are upper-case (if they have lower- and upper-case variants). Tcl's built-in procs don't behave quite as desired, e.g.,
string is upper "123A"
returns false, but I would want it to return true. I would also want it to return true if the A were replaced with, say, an upper-case Cyrillic letter, or a letter from another popular alphabet that doesn't have a case. I could simply filter out all non-letters from the string, but that's not so simple I think when you're trying to handle letters from languages other than just English.
In this case, you don't want string is upper as that checks if the string is just upper case letters. (Numbers aren't letters.)
Instead, you want to do:
set str "123A"
if {$str eq [string toupper $str]} {
# It's upper-case by your definition...
}
I would like to implement a function with R that removes repeated characters in a string. For instance, say my function is named removeRS, so it is supposed to work this way:
removeRS('Buenaaaaaaaaa Suerrrrte')
Buena Suerte
removeRS('Hoy estoy tristeeeeeee')
Hoy estoy triste
My function is going to be used with strings written in spanish, so it is not that common (or at least correct) to find words that have more than three successive vowels. No bother about the possible sentiment behind them. Nonetheless, there are words that can have two successive consonants (especially ll and rr), but we could skip this from our function.
So, to sum up, this function should replace the letters that appear at least three times in a row with just that letter. In one of the examples above, aaaaaaaaa is replaced with a.
Could you give me any hints to carry out this task with R?
I did not think very carefully on this, but this is my quick solution using references in regular expressions:
gsub('([[:alpha:]])\\1+', '\\1', 'Buenaaaaaaaaa Suerrrrte')
# [1] "Buena Suerte"
() captures a letter first, \\1 refers to that letter, + means to match it once or more; put all these pieces together, we can match a letter two or more times.
To include other characters besides alphanumerics, replace [[:alpha:]] with a regex matching whatever you wish to include.
I think you should pay attention to the ambiguities in your problem description. This is a first stab, but it clearly does not work with "Good Luck" in the manner you desire:
removeRS <- function(str) paste(rle(strsplit(str, "")[[1]])$values, collapse="")
removeRS('Buenaaaaaaaaa Suerrrrte')
#[1] "Buena Suerte"
Since you want to replace letters that appear AT LEAST 3 times, here is my solution:
gsub("([[:alpha:]])\\1{2,}", "\\1", "Buennaaaa Suerrrtee")
#[1] "Buenna Suertee"
As you can see the 4 "a" have been reduced to only 1 a, the 3 r have been reduced to 1 r but the 2 n and the 2 e have not been changed.
As suggested above you can replace the [[:alpha:]] by any combination of [a-zA-KM-Z] or similar, and even use the "or" operator | inside the squre brackets [y|Q] if you want your code to affect only repetitions of y and Q.
gsub("([a|e])\\1{2,}", "\\1", "Buennaaaa Suerrrtee")
# [1] "Buenna Suerrrtee"
# triple r are not affected and there are no triple e.
Now I'm a programmer who's recently discovered how bad he is when it comes to mathematics and decided to focus a bit on it from that point forward, so I apologize if my question insults your intelligence.
In mathematics, is there the concept of strings that is used in programming? i.e. a permutation of characters.
As an example, say I wanted to translate the following into mathematical notation:
let s be a string of n number of characters.
Reason being I would want to use that representation in find other things about string s, such as its length: len(s).
How do you formally represent such a thing in mathematics?
Talking more practically, so to speak, let's say I wanted to mathematically explain such a function:
fitness(s,n) = 1 / |n - len(s)|
Or written in more "programming-friendly" sort of way:
fitness(s,n) = 1 / abs(n - len(s))
I used this function to explain how a fitness function for a given GA works; the question was about finding strings with 5 characters, and I needed the solutions to be sorted in ascending order according to their fitness score, given by the above function.
So my question is, how do you represent the above pseudo-code in mathematical notation?
You can use the notation of language theory, which is used to discuss things like regular languages, context free grammars, compiler theory, etc. A quick overview:
A set of characters is known as an alphabet. You could write: "Let A be the ASCII alphabet, a set containing the 128 ASCII characters."
A string is a sequence of characters. ε is the empty string.
A set of strings is formally known as a language. A common statement is, "Let s ∈ L be a string in language L."
Concatenating alphabets produces sets of strings (languages). A represents all 1-character strings, AA, also written A2, is the set of all two character strings. A0 is the set of all zero-length strings and is precisely A0 = {ε}. (It contains exactly one string, the empty string.)
A* is special notation and represents the set of all strings over the alphabet A, of any length. That is, A* = A0 ∪ A1 ∪ A2 ∪ A3 ... . You may recognize this notation from regular expressions.
For length use absolute value bars. The length of a string s is |s|.
So for your statement:
let s be a string of n number of characters.
You could write:
Let A be a set of characters and s ∈ An be a string of n characters. The length of s is |s| = n.
Mathematically, you have explained fitness(s, n) just fine as long as len(s) is well-defined.
In CS texts, a string s over a set S is defined as a finite ordered list of elements of S and its length is often written as |s| - but this is only notation, and doesn't change the (mathematical) meaning behind your definition of fitness, which is pretty clear just how you've written it.