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# Longest palindromic substrings

Longest palindromic substrings is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Let given a string s. The goal is to find the longest palindromic substring in s.

## F#

### Manacher Function

` // Manacher Function. Nigel Galloway: October 1st., 2020let Manacher(s:string) = let oddP,evenP=Array.zeroCreate s.Length,Array.zeroCreate s.Length                         let rec fN i g e (l:int[])=match g>=0 && e<s.Length && s.[g]=s.[e] with true->l.[i]<-l.[i]+1; fN i (g-1) (e+1) l |_->()                         let rec fGo n g Ʃ=match Ʃ<s.Length with                                            false->oddP                                           |_->if Ʃ<=g then oddP.[Ʃ]<-min (oddP.[n+g-Ʃ]) (g-Ʃ)                                               fN Ʃ (Ʃ-oddP.[Ʃ]-1) (Ʃ+oddP.[Ʃ]+1) oddP                                               match (Ʃ+oddP.[Ʃ])>g with true->fGo (Ʃ-oddP.[Ʃ]) (Ʃ+oddP.[Ʃ]) (Ʃ+1) |_->fGo n g (Ʃ+1)                         let rec fGe n g Ʃ=match Ʃ<s.Length with                                            false->evenP                                           |_->if Ʃ<=g then evenP.[Ʃ]<-min (evenP.[n+g-Ʃ]) (g-Ʃ)                                               fN Ʃ (Ʃ-evenP.[Ʃ]) (Ʃ+evenP.[Ʃ]+1) evenP                                               match (Ʃ+evenP.[Ʃ])>g with true->fGe (Ʃ-evenP.[Ʃ]+1) (Ʃ+evenP.[Ʃ]) (Ʃ+1) |_->fGe n g (Ʃ+1)                         (fGo 0 -1 0,fGe 0 -1 0) `

` let fN g=if g=[||] then (0,0) else g|>Array.mapi(fun n g->(n,g))|>Array.maxBy sndlet lpss s=let n,g=Manacher s in let n,g=fN n,fN g in if (snd n)*2+1>(snd g)*2 then s.[(fst n)-(snd n)..(fst n)+(snd n)] else s.[(fst g)-(snd g)+1..(fst g)+(snd g)]let test = ["three old rotators"; "never reverse"; "stable was I ere I saw elbatrosses"; "abracadabra"; "drome"; "the abbatial palace"; ""]test|>List.iter(fun n->printfn "A longest palindromic substring of \"%s\" is \"%s\"" n (lpss n)) `
Output:
```A longest palindromic substring of "three old rotators" is "rotator"
A longest palindromic substring of "never reverse" is "ever reve"
A longest palindromic substring of "stable was I ere I saw elbatrosses" is "table was I ere I saw elbat"
A longest palindromic substring of "abracadabra" is "aca"
A longest palindromic substring of "drome" is "d"
A longest palindromic substring of "the abbatial palace" is "abba"
A longest palindromic substring of "" is ""
```

## Go

Translation of: Wren
`package main import (    "fmt"    "sort") func reverse(s string) string {    var r = []rune(s)    for i, j := 0, len(r)-1; i < j; i, j = i+1, j-1 {        r[i], r[j] = r[j], r[i]    }    return string(r)} func longestPalSubstring(s string) []string {    var le = len(s)    if le <= 1 {        return []string{s}    }    targetLen := le    var longest []string    i := 0    for {        j := i + targetLen - 1        if j < le {            ss := s[i : j+1]            if reverse(ss) == ss {                longest = append(longest, ss)            }            i++        } else {            if len(longest) > 0 {                return longest            }            i = 0            targetLen--        }    }    return longest} func distinct(sa []string) []string {    sort.Strings(sa)    duplicated := make([]bool, len(sa))    for i := 1; i < len(sa); i++ {        if sa[i] == sa[i-1] {            duplicated[i] = true        }    }    var res []string    for i := 0; i < len(sa); i++ {        if !duplicated[i] {            res = append(res, sa[i])        }    }    return res} func main() {    strings := []string{"babaccd", "rotator", "reverse", "forever", "several", "palindrome", "abaracadaraba"}    fmt.Println("The palindromic substrings having the longest length are:")    for _, s := range strings {        longest := distinct(longestPalSubstring(s))        fmt.Printf("  %-13s Length %d -> %v\n", s, len(longest), longest)    }}`
Output:
```The palindromic substrings having the longest length are:
babaccd       Length 3 -> [aba bab]
rotator       Length 7 -> [rotator]
reverse       Length 5 -> [rever]
forever       Length 5 -> [rever]
several       Length 3 -> [eve]
palindrome    Length 1 -> [a d e i l m n o p r]
```

A list version, written out of curiosity. A faster approach could be made with an indexed datatype.

`-------------- LONGEST PALINDROMIC SUBSTRINGS ------------ longestPalindromes :: String -> ([String], Int)longestPalindromes [] = ([], 0)longestPalindromes s = go \$ palindromes s  where    go xs      | null xs = (return <\$> s, 1)      | otherwise = (filter ((w ==) . length) xs, w)      where        w = maximum \$ length <\$> xs palindromes :: String -> [String]palindromes = fmap go . palindromicNuclei  where    go (pivot, (xs, ys)) =      let suffix = fmap fst (takeWhile (uncurry (==)) (zip xs ys))      in reverse suffix <> pivot <> suffix palindromicNuclei :: String -> [(String, (String, String))]palindromicNuclei =  concatMap go .  init . tail . ((zip . scanl (flip ((<>) . return)) []) <*> scanr (:) [])  where    go (a@(x:_), b@(h:y:ys))      | x == h = [("", (a, b))]      | otherwise =        [ ([h], (a, y : ys))        | x == y ]    go _ = []  --------------------------- TEST -------------------------main :: IO ()main =  putStrLn \$  fTable    "Longest palindromic substrings:\n"    show    show    longestPalindromes    [ "three old rotators"    , "never reverse"    , "stable was I ere I saw elbatrosses"    , "abracadabra"    , "drome"    , "the abbatial palace"    , ""    ] ------------------------ FORMATTING ----------------------fTable :: String -> (a -> String) -> (b -> String) -> (a -> b) -> [a] -> StringfTable s xShow fxShow f xs =  unlines \$  s : fmap (((++) . rjust w ' ' . xShow) <*> ((" -> " ++) . fxShow . f)) xs  where    rjust n c = drop . length <*> (replicate n c ++)    w = maximum (length . xShow <\$> xs)`
Output:
```Longest palindromic substrings:

"three old rotators" -> (["rotator"],7)
"never reverse" -> (["ever reve"],9)
"stable was I ere I saw elbatrosses" -> (["table was I ere I saw elbat"],27)
"drome" -> (["d","r","o","m","e"],1)
"the abbatial palace" -> (["abba"],4)
"" -> ([],0)```

## Julia

`function allpalindromics(s)    list, len = String[], length(s)    for i in 1:len-1, j in i+1:len        substr = s[i:j]        if substr == reverse(substr)            push!(list, substr)        end    end    return listend for teststring in ["babaccd", "rotator", "reverse", "forever", "several", "palindrome"]    list = sort!(allpalindromics(teststring), lt = (x, y) -> length(x) < length(y))    println(isempty(list) ? "No palindromes of 2 or more letters found in \"\$teststring." :        "The longest palindromic substring of \$teststring is: \"",        join(list[findall(x -> length(x) == length(list[end]), list)], "\" or \""), "\"")end `
Output:
```The longest palindromic substring of babaccd is: "bab" or "aba"
The longest palindromic substring of rotator is: "rotator"
The longest palindromic substring of reverse is: "rever"
The longest palindromic substring of forever is: "rever"
The longest palindromic substring of several is: "eve"
No palindromes of 2 or more letters found in "palindrome."
```

### Manacher algorithm

` function manacher(str)    s =  "^" * join(split(str, ""), "#") * "\\$"    len = length(s)    pals = fill(0, len)    center, right = 1, 1    for i in 2:len-1        pals[i] = right > i && right - i > 0 && pals[2 * center - i] > 0        while s[i + pals[i] + 1] == s[i - pals[i] - 1]            pals[i] += 1        end        if i + pals[i] > right            center, right = i, i + pals[i]        end    end    maxlen, centerindex = findmax(pals)    start = isodd(maxlen) ? (centerindex-maxlen) ÷ 2 + 1 : (centerindex-maxlen) ÷ 2    return str[start:(centerindex+maxlen)÷2]end for teststring in ["babaccd", "rotator", "reverse", "forever", "several", "palindrome", "abaracadabra"]    pal = manacher(teststring)    println(length(pal) < 2 ? "No palindromes of 2 or more letters found in \"\$teststring.\"" :        "The longest palindromic substring of \$teststring is: \"\$pal\"")end `
Output:
```The longest palindromic substring of babaccd is: "aba"
The longest palindromic substring of rotator is: "rotator"
The longest palindromic substring of reverse is: "rever"
The longest palindromic substring of forever is: "rever"
The longest palindromic substring of several is: "eve"
No palindromes of 2 or more letters found in "palindrome."
The longest palindromic substring of abaracadabra is: "ara"
```

## Phix

`function longest_palindromes(string s)--  s = lower/strip_spaces_and_punctuation/utf8_to_utf32, if rqd    integer longest = 2 -- (do not treat length 1 as palindromic)--  integer longest = 1 -- (do not treat length 0 as palindromic) [works just fine too]    sequence res = {}    for i=1 to length(s) do        for e=length(s) to i+longest-1 by -1 do            if s[e]=s[i] then                string p = s[i..e]                integer lp = length(p)                if lp>=longest and p=reverse(p) then                    if lp>longest then                        longest = lp                        res = {p}                    elsif not find(p,res) then -- (or just "else")                        res = append(res,p)                    end if                end if            end if        end for    end for    return res -- (or "sort(res)" or "unique(res)", as needed)end function constant tests = {"babaccd","rotator","reverse","forever","several","palindrome","abaracadaraba"}for i=1 to length(tests) do    printf(1,"%s: %v\n",{tests[i],longest_palindromes(tests[i])})end for`
Output:
```babaccd: {"bab","aba"}
rotator: {"rotator"}
reverse: {"rever"}
forever: {"rever"}
several: {"eve"}
palindrome: {}
```

with longest initialised to 1, you get the same except for `palindrome: {"p","a","l","i","n","d","r","o","m","e"}`

### faster

`function Manacher(string text)     -- Manacher's algorithm (linear time)    -- based on https://www.geeksforgeeks.org/manachers-algorithm-linear-time-longest-palindromic-substring-part-4    -- but with a few tweaks, renames, and bugfixes (in particular the < (positions-1), which I later found LIJIE already said)    sequence res = {}    integer positions = length(text)*2+1    if positions>1 then        sequence LPS = repeat(0,positions)                 LPS = 1        integer centerPosition = 1,                centerRightPosition = 2,                maxLPSLength = 0         for currentRightPosition=2 to positions-1 do            integer lcp = LPS[currentRightPosition+1],                    diff = centerRightPosition - currentRightPosition            -- If currentRightPosition is within centerRightPosition            if diff >= 0 then                -- get currentLeftPosition iMirror for currentRightPosition                integer iMirror = 2*centerPosition-currentRightPosition + 1                lcp = min(LPS[iMirror], diff)            end if             -- Attempt to expand palindrome centered at currentRightPosition            -- Here for odd positions, we compare characters and              -- if match then increment LPS Length by ONE             -- If even position, we just increment LPS by ONE without              -- any character comparison            while ((currentRightPosition + lcp) < (positions-1) and (currentRightPosition - lcp) > 0) and                  ((remainder(currentRightPosition+lcp+1, 2) == 0) or                   (text[floor((currentRightPosition+lcp+1)/2)+1] == text[floor((currentRightPosition-lcp-1)/2)+1] )) do                lcp += 1            end while            LPS[currentRightPosition+1] = lcp            maxLPSLength = max(lcp,maxLPSLength)             // If palindrome centered at currentRightPosition            // expand beyond centerRightPosition,             // adjust centerPosition based on expanded palindrome.             if (currentRightPosition + lcp) > centerRightPosition then                centerPosition = currentRightPosition                centerRightPosition = currentRightPosition + lcp            end if        end for        for p=1 to positions do            if LPS[p] = maxLPSLength then                integer start = floor((p-1 - maxLPSLength)/2) + 1,                        finish = start + maxLPSLength - 1                string r = text[start..finish]                if not find(r,res) then                    res = append(res,r)                end if            end if        end for    end if    return resend function include mpfr.empfr pi = mpfr_init(0,-10001) -- (set precision to 10,000 dp, plus the "3.")mpfr_const_pi(pi)string piStr =  mpfr_sprintf("%.10000Rf", pi),       s = shorten(piStr)printf(1,"%s: %v\n",{s,Manacher(piStr)})`
Output:

(Same as above if given the same inputs.)
However, while Manacher finishes 10,000 digits in 0s, longest_palindromes takes 1s for 2,000 digits, 15s for 5,000 digits, and 2 mins for 10,000 digits,
which goes to prove that longest_palindromes() above is O(n2), whereas Manacher() is O(n).

```3.141592653589793238...05600101655256375679 (10,002 digits): {"398989893","020141020"}
```

Then again, this is also pretty fast (same output):

Translation of: Raku
`function longest_palindromes_raku(string s)--  s = lower/strip_spaces_and_punctuation/utf8_to_utf32, if rqd    integer longest = 2 -- (do not treat length 1 as palindromic)--  integer longest = 1 -- (do not treat length 0 as palindromic) [works just fine too]    sequence res = {}    for i=1 to length(s) do        for j=0 to iff(i>1 and s[i-1]=s[i]?2:1) do            integer rev = j,                    fwd = 1            while rev<i and i+fwd<=length(s) and s[i-rev]=s[i+fwd] do                rev += 1                fwd += 1            end while            string p = s[i-rev+1..i+fwd-1]            integer lp = length(p)            if lp>=longest then                if lp>longest then                    longest = lp                    res = {p}                elsif not find(p,res) then -- (or just "else")                    res = append(res,p)                end if            end if        end for    end for    return res -- (or "sort(res)" or "unique(res)", as needed)end function printf(1,"%s: %v\n",{s,longest_palindromes_raku(piStr)})s = "abbbc"printf(1,"%s: %v\n",{s,longest_palindromes_raku(s)})`
Output:

(first line matches the above, the second was a initially a bug)

```3.141592653589793238...05600101655256375679 (10,002 digits): {"398989893","020141020"}
abbbc: {"bbb"}
```

## Python

Defines maximal expansions of any two or three character palindromic nuclei in the string.

(This version ignores case but allows non-alphanumerics).

`'''Longest palindromic substrings'''  # longestPalindrome :: String -> ([String], Int)def longestPalindromes(s):    '''All palindromes of the maximal length       drawn from a case-flattened copy of       the given string, tupled with the       maximal length.       Non-alphanumerics are included here.    '''    k = s.lower()    palindromes = [        palExpansion(k)(ab) for ab        in palindromicNuclei(k)    ]    maxLength = max([        len(x) for x in palindromes    ]) if palindromes else 1    return (        [            x for x in palindromes if maxLength == len(x)        ] if palindromes else list(s),        maxLength    ) if s else ([], 0)  # palindromicNuclei :: String -> [(Int, Int)]def palindromicNuclei(s):    '''Ranges of all the 2 or 3 character       palindromic nuclei in s.    '''    cs = list(s)    return [        # Two-character nuclei.        (i, 1 + i) for (i, (a, b))        in enumerate(zip(cs, cs[1:]))        if a == b    ] + [        # Three-character nuclei.        (i, 2 + i) for (i, (a, b, c))        in enumerate(zip(cs, cs[1:], cs[2:]))        if a == c    ]  # palExpansion :: String -> (Int, Int) -> Stringdef palExpansion(s):    '''Full expansion of the palindromic       nucleus with the given range in s.    '''    iEnd = len(s) - 1     def limit(ij):        i, j = ij        return 0 == i or iEnd == j or s[i-1] != s[j+1]     def expansion(ij):        i, j = ij        return (i - 1, 1 + j)     def go(ij):        ab = until(limit)(expansion)(ij)        return s[ab:ab + 1]    return go  # ------------------------- TEST -------------------------# main :: IO ()def main():    '''Longest palindromic substrings'''    print(        fTable(main.__doc__ + ':\n')(repr)(repr)(            longestPalindromes        )([            'three old rotators',            'never reverse',            'stable was I ere I saw elbatrosses',            'abracadabra',            'drome',            'the abbatial palace',            ''        ])    )  # ----------------------- GENERIC ------------------------ # until :: (a -> Bool) -> (a -> a) -> a -> adef until(p):    '''The result of repeatedly applying f until p holds.       The initial seed value is x.    '''    def go(f):        def g(x):            v = x            while not p(v):                v = f(v)            return v        return g    return go  # ---------------------- FORMATTING ---------------------- # fTable :: String -> (a -> String) -># (b -> String) -> (a -> b) -> [a] -> Stringdef fTable(s):    '''Heading -> x display function -> fx display function ->       f -> xs -> tabular string.    '''    def gox(xShow):        def gofx(fxShow):            def gof(f):                def goxs(xs):                    ys = [xShow(x) for x in xs]                    w = max(map(len, ys))                     def arrowed(x, y):                        return y.rjust(w, ' ') + ' -> ' + (                            fxShow(f(x))                        )                    return s + '\n' + '\n'.join(                        map(arrowed, xs, ys)                    )                return goxs            return gof        return gofx    return gox  # MAIN ---if __name__ == '__main__':    main()`
Output:
```Longest palindromic substrings:

'three old rotators' -> (['rotator'], 7)
'never reverse' -> (['ever reve'], 9)
'stable was I ere I saw elbatrosses' -> (['table was i ere i saw elbat'], 27)
'drome' -> (['d', 'r', 'o', 'm', 'e'], 1)
'the abbatial palace' -> (['abba'], 4)
'' -> ([], 0)```

## Raku

Works with: Rakudo version 2020.09

This version regularizes (ignores) case and ignores non alphanumeric characters. It is only concerned with finding the longest palindromic substrings so does not exhaustively find all possible palindromes. If a palindromic substring is found to be part of a longer palindrome, it is not captured separately. Showing the longest 5 palindromic substring groups. Run it with no parameters to operate on the default; pass in a file name to run it against that instead.

`my @chars = ( @*ARGS ?? @*ARGS.IO.slurp !! q:to/BOB/ ) .lc.comb: /\w/;    Lyrics to "Bob" copyright Weird Al Yankovic    https://www.youtube.com/watch?v=JUQDzj6R3p4     I, man, am regal - a German am I    Never odd or even    If I had a hi-fi    Madam, I'm Adam    Too hot to hoot    No lemons, no melon    Too bad I hid a boot    Lisa Bonet ate no basil    Warsaw was raw    Was it a car or a cat I saw?     Rise to vote, sir    Do geese see God?    "Do nine men interpret?" "Nine men," I nod    Rats live on no evil star    Won't lovers revolt now?    Race fast, safe car    Pa's a sap    Ma is as selfless as I am    May a moody baby doom a yam?     Ah, Satan sees Natasha    No devil lived on    Lonely Tylenol    Not a banana baton    No "x" in "Nixon"    O, stone, be not so    O Geronimo, no minor ego    "Naomi," I moan    "A Toyota's a Toyota"    A dog, a panic in a pagoda     Oh no! Don Ho!    Nurse, I spy gypsies - run!    Senile felines    Now I see bees I won    UFO tofu    We panic in a pew    Oozy rat in a sanitary zoo    God! A red nugget! A fat egg under a dog!    Go hang a salami, I'm a lasagna hog!    BOB#" my @cpfoa = flat(1 ..^ @chars).race(:1000batch).map: -> \idx {    my @s;    for 1, 2 {       my int (\$rev, \$fwd) = \$_, 1;       loop {            quietly last if (\$rev > idx) || (@chars[idx - \$rev] ne @chars[idx + \$fwd]);            \$rev = \$rev + 1;            \$fwd = \$fwd + 1;        }        @s.push: @chars[idx - \$rev ^..^ idx + \$fwd].join if \$rev + \$fwd > 2;        last if @chars[idx - 1] ne @chars[idx];    }    next unless +@s;    @s} "{.key} ({+.value})\t{.value.unique.sort}".put for @cpfoa.classify( *.chars ).sort( -*.key ).head(5);`
Output:

Returns the length, (the count) and the list:

```29 (2)	doninemeninterpretninemeninod godarednuggetafateggunderadog
26 (1)	gohangasalamiimalasagnahog
23 (1)	arwontloversrevoltnowra
21 (4)	imanamregalagermanami mayamoodybabydoomayam ootnolemonsnomelontoo oozyratinasanitaryzoo
20 (1)	ratsliveonnoevilstar
```

This isn't intensively optimised but isn't too shabby either. When run against the first million digits of pi: 1000000 digits of pi text file (Pass in the file path/name at the command line) we get:

```13 (1)	9475082805749
12 (1)	450197791054
11 (8)	04778787740 09577577590 21348884312 28112721182 41428782414 49612121694 53850405835 84995859948
10 (9)	0045445400 0136776310 1112552111 3517997153 5783993875 6282662826 7046006407 7264994627 8890770988
9 (98)	019161910 020141020 023181320 036646630 037101730 037585730 065363560 068363860 087191780 091747190 100353001 104848401 111262111 131838131 132161231 156393651 160929061 166717661 182232281 193131391 193505391 207060702 211878112 222737222 223404322 242424242 250171052 258232852 267919762 272636272 302474203 313989313 314151413 314424413 318272813 323212323 330626033 332525233 336474633 355575553 357979753 365949563 398989893 407959704 408616804 448767844 450909054 463202364 469797964 479797974 480363084 489696984 490797094 532121235 546000645 549161945 557040755 559555955 563040365 563828365 598292895 621969126 623707326 636414636 636888636 641949146 650272056 662292266 667252766 681565186 684777486 712383217 720565027 726868627 762727267 769646967 777474777 807161708 819686918 833303338 834363438 858838858 866292668 886181688 895505598 896848698 909565909 918888819 926676629 927202729 929373929 944525449 944848449 953252359 972464279 975595579 979202979 992868299```

in right around 7 seconds on my system.

## REXX

`/*REXX program finds and displays the  longest palindromic string(s) in a given string. */parse arg s                                      /*obtain optional argument from the CL.*/if s=='' | s==","  then s= 'babaccd rotator reverse forever several palindrome abaracadaraba'                                                 /* [↑] the case of strings is respected*/       do i=1  for words(s);    say;        say  /*search each of the  (S)  strings.    */       x= word(s, i);           L= length(x)     /*get a string to be examined & length.*/       m= 0                   do LL=2  for L-1              /*start with palindromes of length two.*/                   if find(1)  then m= max(m,LL) /*Found a palindrome?  Set M=new length*/                   end   /*LL*/       LL= max(1,m)       call find .       say ' longest palindromic substrings for string: '        x       say '────────────────────────────────────────────'copies('─', 2 + L)             do n=1  for words(@)                /*show longest palindromic substrings. */             say '    (length='LL")  "       word(@, n)             end   /*n*/       end         /*i*/exit 0                                           /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/find: parse arg short                            /*if SHORT==1,  only find 1 palindrome.*/      @=                                         /*initialize palindrome list to a null.*/            do j=1  for L-LL+1                   /*obtain length of possible palindromes*/            \$= substr(x, j, LL)                  /*obtain a possible palindromic substr.*/            if \$\==reverse(\$)  then iterate      /*Not a palindrome?       Then skip it.*/            @= @ \$                               /*add a palindromic substring to a list*/            if short==1  then return 1           /*have found  one   palindrome.        */            end   /*j*/;      return 0           /*  "    "    some  palindrome(s).     */`
output   when using the default input:
``` longest palindromic substrings for string:  babaccd
─────────────────────────────────────────────────────
(length=3)   bab
(length=3)   aba

longest palindromic substrings for string:  rotator
─────────────────────────────────────────────────────
(length=7)   rotator

longest palindromic substrings for string:  reverse
─────────────────────────────────────────────────────
(length=5)   rever

longest palindromic substrings for string:  forever
─────────────────────────────────────────────────────
(length=5)   rever

longest palindromic substrings for string:  several
─────────────────────────────────────────────────────
(length=3)   eve

longest palindromic substrings for string:  palindrome
────────────────────────────────────────────────────────
(length=1)   p
(length=1)   a
(length=1)   l
(length=1)   i
(length=1)   n
(length=1)   d
(length=1)   r
(length=1)   o
(length=1)   m
(length=1)   e

longest palindromic substrings for string:  abaracadaraba
───────────────────────────────────────────────────────────
(length=3)   aba
(length=3)   ara
(length=3)   aca
(length=3)   ara
(length=3)   aba
```

## Ring

` load "stdlib.ring" st = "babaccd"palList = [] for n = 1 to len(st)-1    for m = n+1 to len(st)        sub = substr(st,n,m-n)        if ispalindrome(sub) and len(sub) > 1           add(palList,[sub,len(sub)])        ok    nextnext palList = sort(palList,2)palList = reverse(palList)resList = []add(resList,palList) for n = 2 to len(palList)    if palList = palList[n]       add(resList,palList[n])    oknext see "Input: " + st + nlsee "Longest palindromic substrings:" + nlsee resList `
Output:
```Input: babaccd
Longest palindromic substrings:
bab
aba
```

## Wren

Library: Wren-seq
Library: Wren-fmt

I've assumed that the expression 'substring' includes the string itself and that substrings of length 1 are considered to be palindromic. Also that if there is more than one palindromic substring of the longest length, then all such distinct ones should be returned.

The Phix entry examples have been used.

`import "/seq" for Lstimport "/fmt" for Fmt var longestPalSubstring = Fn.new { |s|    var len = s.count    if (len <= 1) return [s]    var targetLen = len    var longest = []    var i = 0    while (true) {        var j = i + targetLen - 1        if (j < len) {            var ss = s[i..j]            if (ss == ss[-1..0]) longest.add(ss)            i = i + 1        } else {            if (longest.count > 0) return longest            i = 0            targetLen = targetLen - 1        }    }} var strings = ["babaccd", "rotator", "reverse", "forever", "several", "palindrome", "abaracadaraba"]System.print("The palindromic substrings having the longest length are:")for (s in strings) {    var longest = Lst.distinct(longestPalSubstring.call(s))    Fmt.print("  \$-13s Length \$d -> \$n", s, longest.count, longest)}`
Output:
```The palindromic substrings having the longest length are:
babaccd       Length 3 -> [bab, aba]
rotator       Length 7 -> [rotator]
reverse       Length 5 -> [rever]
forever       Length 5 -> [rever]
several       Length 3 -> [eve]
palindrome    Length 1 -> [p, a, l, i, n, d, r, o, m, e]