Most frequent k chars distance

Most frequent k chars distance 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.

The string distance function (SDF) that is the subject of this entry was proposed in a paper published in 2014 as a method for quickly estimating how similar two ordered sets or strings are. This SDF has also been proposed as being suitable in bioinformatics, e.g. for comparing sequences in the wp:fasta format.

Unfortunately, the original paper by Sadi Evren Seker et al. (arXiv:1401.6596 [cs.DS]) has a number of internal inconsistencies and for this and other reasons is not entirely clear in several key respects, but the paper does give some worked examples and several of these (notably the two given under the "Worked Examples" heading below) agree with the interpretation used by the authors of the Python entry herein, so that interpretation is used in the present task description.

The Task

(a) Show a time-efficient implementation of the SDF as if by a function mostFreqKSDF(s1, s2, k, n), where s1 and s2 are arbitrary strings, and k and n are non-negative integers as explained below;

(b) Show the values produced by the SDF for the following values of s1 and s2, with k = 2 and n = 10:

String Inputs	"Top Two"	SDF Output (k=2, n=10)
'night' 'nacht'	n:1 i:1 n:1 a:1	8
'my' 'a'	m:1 y:1 a:1	10
‘research’ ‘research’	r:2 e:2 r:2 e:2	2
‘significant’ ‘capabilities’	i:3 n:2 i:3 a:2	4

(c) Show the value produced by the SDF for k = 2, n = 100, and the two strings:

"LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG"

Reference Algorithm

The original paper employs an unusual scheme for encoding the frequencies of single letters as a string of letters and numbers. This scheme has several disadvantages, and so in the following we take a more abstract view and simply assume, solely for the purposes of this description, that two functions are available:

1) MostFreqK(s, k) returns (in decreasing order of their frequency of occurrence) the k most frequently occurring letters in s, with any ties broken by the position of the first occurrence in s;

2) For any letter c in MostFreqK(s, k), Freq(s,c) returns the frequency of c in s.

MFKsimilarity

Next define MFKsimilarlity(s1, s2, k) as if by the following pseudo-code:

int function MFKsimilarity(string s1, string s2, int k):
    let similarity = 0
    let mfk1 = MostFreqK(s1, k)
    let mfk2 = MostFreqK(s2, k)
    for each c in mfk1
        if (c occurs in mfk2) 
        then similarity += Freq(s1, c) + Freq(s2, c)
	end
    return similarity

String Distance Wrapper Function

Now define MostFreqKSDF as if by:

int function MostFreqKSDF(string s1, string s2, int k, int n)
    return n - MFKsimilarity(s1, s2, k)

Worked Examples

The original paper gives several worked examples, including these two:

(i) MostFreqKSDF("night", "nacht", 2, 10)

For "night", the "top two" letters with their frequencies are: n:1 and i:1.

For "nacht", the "top two" letters with their frequencies are: n:1 and a:1.

Since "n" is the only shared letter amongst these candidates, the SDF is 10 - 1 - 1 = 8, as per the original paper.

(2) MostFreqKSDF( "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV", "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG", 2, 100)

For the first string, the "top two" letters with their frequencies are: L:9 and T:8; and for the second, the "top two" are F:9 and L:8.

Since "L" is the only shared letter amongst these top-k characters, and since the sum of the corresponding frequencies is 9+8 = 17, the SDF is 83.

11l

Translation of: Python

<lang 11l>F most_freq_khashing(inputString, K)

  DefaultDict[Char, Int] occuDict
  L(c) inputString
     occuDict[c]++
  V occuList = sorted(occuDict.items(), key' x -> x[1], reverse' 1B)
  V outputDict = Dict(occuList[0 .< K])
  R outputDict

F most_freq_ksimilarity(inputStr1, inputStr2)

  V similarity = 0
  L(c, cnt1) inputStr1
     I c C inputStr2
        V cnt2 = inputStr2[c]
        similarity += cnt1 + cnt2
  R similarity

F most_freq_ksdf(inputStr1, inputStr2, K, maxDistance)

  R maxDistance - most_freq_ksimilarity(most_freq_khashing(inputStr1, K), most_freq_khashing(inputStr2, K))

V str1 = ‘LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV’ V str2 = ‘EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG’ V K = 2 V maxDistance = 100 V dict1 = most_freq_khashing(str1, 2) print(dict1, end' ":\n") print(dict1.map((c, cnt) -> c‘’String(cnt)).join(‘’)) V dict2 = most_freq_khashing(str2, 2) print(dict2, end' ":\n") print(dict2.map((c, cnt) -> c‘’String(cnt)).join(‘’)) print(most_freq_ksdf(str1, str2, K, maxDistance))</lang>

Output:

[L = 9, T = 8]:
L9T8
[F = 9, L = 8]:
F9L8
83

C++

<lang cpp>#include <string>

1. include <vector>
2. include <map>
3. include <iostream>
4. include <algorithm>
5. include <utility>
6. include <sstream>

std::string mostFreqKHashing ( const std::string & input , int k ) {

  std::ostringstream oss ;
  std::map<char, int> frequencies ;
  for ( char c : input ) {
     frequencies[ c ] = std::count ( input.begin( ) , input.end( ) , c ) ;
  }
  std::vector<std::pair<char , int>> letters ( frequencies.begin( ) , frequencies.end( ) ) ;
  std::sort ( letters.begin( ) , letters.end( ) , [input] ( std::pair<char, int> a ,

std::pair<char, int> b ) { char fc = std::get<0>( a ) ; char fs = std::get<0>( b ) ; int o = std::get<1>( a ) ; int p = std::get<1>( b ) ; if ( o != p ) { return o > p ; } else { return input.find_first_of( fc ) < input.find_first_of ( fs ) ; } } ) ;

  for ( int i = 0 ; i < letters.size( ) ; i++ ) {
     oss << std::get<0>( letters[ i ] ) ;
     oss << std::get<1>( letters[ i ] ) ;
  }
  std::string output ( oss.str( ).substr( 0 , 2 * k ) ) ;
  if ( letters.size( ) >= k ) {
     return output ;
  }
  else {
     return output.append( "NULL0" ) ;
  }

}

int mostFreqKSimilarity ( const std::string & first , const std::string & second ) {

  int i = 0 ;
  while ( i < first.length( ) - 1  ) {
     auto found = second.find_first_of( first.substr( i , 2 ) ) ;
     if ( found != std::string::npos ) 

return std::stoi ( first.substr( i , 2 )) ;

     else 

i += 2 ;

  }
  return 0 ;

}

int mostFreqKSDF ( const std::string & firstSeq , const std::string & secondSeq , int num ) {

  return mostFreqKSimilarity ( mostFreqKHashing( firstSeq , num ) , mostFreqKHashing( secondSeq , num ) ) ;

}

int main( ) {

  std::string s1("LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" ) ;
  std::string s2( "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" ) ;
  std::cout << "MostFreqKHashing( s1 , 2 ) = " << mostFreqKHashing( s1 , 2 ) << '\n' ;
  std::cout << "MostFreqKHashing( s2 , 2 ) = " << mostFreqKHashing( s2 , 2 ) << '\n' ;
  return 0 ;

} </lang>

Output:

MostFreqKHashing( s1 , 2 ) = L9T8
MostFreqKHashing( s2 , 2 ) = F9L8

Go

Translation of: Kotlin

<lang go>package main

import (

   "fmt"
   "sort"

)

type cf struct {

   c rune
   f int

}

func reverseStr(s string) string {

   runes := []rune(s)
   for i, j := 0, len(runes)-1; i < j; i, j = i+1, j-1 {
       runes[i], runes[j] = runes[j], runes[i]
   }
   return string(runes)

}

func indexOfCf(cfs []cf, r rune) int {

   for i, cf := range cfs {
       if cf.c == r {
           return i
       }
   }
   return -1

}

func minOf(i, j int) int {

   if i < j {
       return i
   }
   return j

}

func mostFreqKHashing(input string, k int) string {

   var cfs []cf
   for _, r := range input {
       ix := indexOfCf(cfs, r)
       if ix >= 0 {
           cfs[ix].f++
       } else {
           cfs = append(cfs, cf{r, 1})
       }
   }
   sort.SliceStable(cfs, func(i, j int) bool {
       return cfs[i].f > cfs[j].f // descending, preserves order when equal
   })
   acc := ""
   min := minOf(len(cfs), k)
   for _, cf := range cfs[:min] {
       acc += fmt.Sprintf("%c%c", cf.c, cf.f)
   }
   return acc

}

func mostFreqKSimilarity(input1, input2 string) int {

   similarity := 0
   runes1, runes2 := []rune(input1), []rune(input2)
   for i := 0; i < len(runes1); i += 2 {
       for j := 0; j < len(runes2); j += 2 {
           if runes1[i] == runes2[j] {
               freq1, freq2 := runes1[i+1], runes2[j+1]
               if freq1 != freq2 {
                   continue // assuming here that frequencies need to match
               }
               similarity += int(freq1)
           }
       }
   }
   return similarity

}

func mostFreqKSDF(input1, input2 string, k, maxDistance int) {

   fmt.Println("input1 :", input1)
   fmt.Println("input2 :", input2)
   s1 := mostFreqKHashing(input1, k)
   s2 := mostFreqKHashing(input2, k)
   fmt.Printf("mfkh(input1, %d) = ", k)
   for i, c := range s1 {
       if i%2 == 0 {
           fmt.Printf("%c", c)
       } else {
           fmt.Printf("%d", c)
       }
   }
   fmt.Printf("\nmfkh(input2, %d) = ", k)
   for i, c := range s2 {
       if i%2 == 0 {
           fmt.Printf("%c", c)
       } else {
           fmt.Printf("%d", c)
       }
   }
   result := maxDistance - mostFreqKSimilarity(s1, s2)
   fmt.Printf("\nSDF(input1, input2, %d, %d) = %d\n\n", k, maxDistance, result)

}

func main() {

   pairs := [][2]string{
       {"research", "seeking"},
       {"night", "nacht"},
       {"my", "a"},
       {"research", "research"},
       {"aaaaabbbb", "ababababa"},
       {"significant", "capabilities"},
   }
   for _, pair := range pairs {
       mostFreqKSDF(pair[0], pair[1], 2, 10)
   }

   s1 := "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV"
   s2 := "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG"
   mostFreqKSDF(s1, s2, 2, 100)
   s1 = "abracadabra12121212121abracadabra12121212121"
   s2 = reverseStr(s1)
   mostFreqKSDF(s1, s2, 2, 100)

}</lang>

Output:

input1 : research
input2 : seeking
mfkh(input1, 2) = r2e2
mfkh(input2, 2) = e2s1
SDF(input1, input2, 2, 10) = 8

input1 : night
input2 : nacht
mfkh(input1, 2) = n1i1
mfkh(input2, 2) = n1a1
SDF(input1, input2, 2, 10) = 9

input1 : my
input2 : a
mfkh(input1, 2) = m1y1
mfkh(input2, 2) = a1
SDF(input1, input2, 2, 10) = 10

input1 : research
input2 : research
mfkh(input1, 2) = r2e2
mfkh(input2, 2) = r2e2
SDF(input1, input2, 2, 10) = 6

input1 : aaaaabbbb
input2 : ababababa
mfkh(input1, 2) = a5b4
mfkh(input2, 2) = a5b4
SDF(input1, input2, 2, 10) = 1

input1 : significant
input2 : capabilities
mfkh(input1, 2) = i3n2
mfkh(input2, 2) = i3a2
SDF(input1, input2, 2, 10) = 7

input1 : LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV
input2 : EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG
mfkh(input1, 2) = L9T8
mfkh(input2, 2) = F9L8
SDF(input1, input2, 2, 100) = 100

input1 : abracadabra12121212121abracadabra12121212121
input2 : 12121212121arbadacarba12121212121arbadacarba
mfkh(input1, 2) = 112a10
mfkh(input2, 2) = 112210
SDF(input1, input2, 2, 100) = 88

Haskell

<lang Haskell>module MostFrequentK

  where

import Data.List ( nub , sortBy ) import qualified Data.Set as S

count :: Eq a => [a] -> a -> Int count [] x = 0 count ( x:xs ) k

  |x == k = 1 + count xs k
  |otherwise = count xs k

orderedStatistics :: String -> [(Char , Int)] orderedStatistics s = sortBy myCriterion $ nub $ zip s ( map (\c -> count s c ) s )

  where
     myCriterion :: (Char , Int) -> (Char , Int) -> Ordering
     myCriterion (c1 , n1) (c2, n2) 

|n1 > n2 = LT |n1 < n2 = GT |n1 == n2 = compare ( found c1 s ) ( found c2 s )

     found :: Char -> String -> Int
     found e s = length $ takeWhile (/= e ) s

mostFreqKHashing :: String -> Int -> String mostFreqKHashing s n = foldl ((++)) [] $ map toString $ take n $ orderedStatistics s

  where
     toString :: (Char , Int) -> String
     toString ( c , i ) = c : show i

mostFreqKSimilarity :: String -> String -> Int mostFreqKSimilarity s t = snd $ head $ S.toList $ S.fromList ( doublets s ) `S.intersection`

                          S.fromList ( doublets t )
  where
     toPair :: String -> (Char , Int)
     toPair s = ( head s , fromEnum ( head $ tail s ) - 48 )
     doublets :: String -> [(Char , Int)]
     doublets str = map toPair [take 2 $ drop start str | start <- [0 , 2 ..length str - 2]]
                          

mostFreqKSDF :: String -> String -> Int ->Int mostFreqKSDF s t n = mostFreqKSimilarity ( mostFreqKHashing s n ) (mostFreqKHashing t n ) </lang>

Output:

mostFrequentKHashing "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" 2
"L9T8"
*MostFrequentK> mostFrequentKHashing "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" 2
"F9L8"

J

Solution:<lang j>NB. String Distance Wrapper Function mfksDF =: {:@:[ - (mfks@:(mfkh&.>)~ {.)~

NB. Most Frequent K Distance mfks =: score@:(charMap@:[ {"1 charVals@:])/@:kHashes

 score    =.  ([ +/ .* =)/                  NB. (+ +/ .* *.&:*)/  for sum += freq_in_left + freq_in_right
 charMap  =.  (,&< i.&> <@:~.@:,)&;/
 charVals =.  (; , 0:)"1
 kHashes  =.  0 1 |: ({.&>~ [: <./ #&>)
 

NB. Most Frequent K Hashing mfkh =: _&$: : (takeK freqHash) NB. Default LHA of _ means "return complete frequency table"

 takeK    =.  (<.#) {. ]
 freqHash =.  ~. (] \:~ ,.&:(<"0)) #/.~ 

NB. No need to fix mfksDF mfkh =: mfkh f. mfks =: mfks f.</lang>

Examples:<lang j>verb define fkh =. ;@:,@:(":&.>) NB. format k hash

assert. 'r2e2 e2s1' (-: [: fkh 2&mfkh)&>&;: 'research seeking' assert. 2 = mfks 2 mfkh&.> 'research';'seeking'

assert. 'n1i1 n1a1' (-: [: fkh 2&mfkh)&>&;: 'night nacht' assert. 9 = 2 10 mfksDF 'night';'nacht'

assert. 'm1y1 a1' (-: [: fkh 2&mfkh)&>&;: 'my a' assert. 10 = 2 10 mfksDF 'my';'a'

assert. 'r2e2' -: fkh 2 mfkh 'research' assert. 6 = 2 10 mfksDF 'research';'research' NB. task says 8; right answer is 6

assert. 'a5b4 a5b4' (-: [: fkh 2&mfkh)&>&;: 'aaaaabbbb ababababa' assert. 1 = 2 10 mfksDF 'aaaaabbbb';'ababababa'

assert. 'i3n2 i3a2' (-: [: fkh 2&mfkh)&>&;: 'significant capabilities' assert. 7 = 2 10 mfksDF 'significant';'capabilities' NB. task says 5; right answer is 7

assert. 'L9T8 F9L8' (-: [: fkh 2&mfkh)&>&;: 'LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG' assert. 100 = 2 100 mfksDF 'LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV';'EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG' NB. task says 83; right answer is 100

'pass' )

  pass</lang>

Notes: As of press time, there are significant discrepancies between the task description, its pseudocode, the test cases provided, and the two other existing implementations. See the talk page for the assumptions made in this implementation to reconcile these discrepancies (in particular, in the scoring function).

Java

This example is incomplete. Works now when there more than 10 occurences of a character in string. Please check, there can be more elegant way to deal with it. Comment was : This will fail catastrophically on “ABACADAEAFAEADACABA” as that has 10 ‘A’ characters in it. Please ensure that it meets all task requirements and remove this message.

Translation of the pseudo-code of the Wikipedia article wp:Most frequent k characters to wp:java implementation of three functions given in the definition section are given below with wp:JavaDoc comments:

<lang java>import java.util.Collections; import java.util.Comparator; import java.util.HashMap; import java.util.LinkedHashMap; import java.util.ArrayList; import java.util.List; import java.util.Map;

public class SDF {

   /** Counting the number of occurrences of each character
    * @param character array
    * @return hashmap : Key = char, Value = num of occurrence
    */
   public static HashMap<Character, Integer> countElementOcurrences(char[] array) {

       HashMap<Character, Integer> countMap = new HashMap<Character, Integer>();

       for (char element : array) {
           Integer count = countMap.get(element);
           count = (count == null) ? 1 : count + 1;
           countMap.put(element, count);
       }

       return countMap;
   }
   
   /**
    * Sorts the counted numbers of characters (keys, values) by java Collection List
    * @param HashMap (with key as character, value as number of occurrences)
    * @return sorted HashMap
    */
   private static <K, V extends Comparable<? super V>>
           HashMap<K, V> descendingSortByValues(HashMap<K, V> map) { 

List<Map.Entry<K, V>> list = new ArrayList<Map.Entry<K, V>>(map.entrySet()); // Defined Custom Comparator here Collections.sort(list, new Comparator<Map.Entry<K, V>>() { public int compare(Map.Entry<K, V> o1, Map.Entry<K, V> o2) { return o2.getValue().compareTo(o1.getValue()); } });

// Here I am copying the sorted list in HashMap // using LinkedHashMap to preserve the insertion order HashMap<K, V> sortedHashMap = new LinkedHashMap<K, V>(); for (Map.Entry<K, V> entry : list) { sortedHashMap.put(entry.getKey(), entry.getValue()); } return sortedHashMap;

   }
   /**
    * get most frequent k characters
    * @param array of characters
    * @param limit of k
    * @return hashed String 
    */
   public static String mostOcurrencesElement(char[] array, int k) {
       HashMap<Character, Integer> countMap = countElementOcurrences(array);
       System.out.println(countMap);
       Map<Character, Integer> map = descendingSortByValues(countMap); 
       System.out.println(map);
       int i = 0;
       String output = "";
       for (Map.Entry<Character, Integer> pairs : map.entrySet()) {

if (i++ >= k) break;

           output += "" + pairs.getKey() + pairs.getValue();
       }
       return output;
   }
   /**
    * Calculates the similarity between two input strings
    * @param input string 1
    * @param input string 2
    * @param maximum possible limit value 
    * @return distance as integer
    */
   public static int getDiff(String str1, String str2, int limit) {
       int similarity = 0;

int k = 0; for (int i = 0; i < str1.length() ; i = k) { k ++; if (Character.isLetter(str1.charAt(i))) { int pos = str2.indexOf(str1.charAt(i));

if (pos >= 0) { String digitStr1 = ""; while ( k < str1.length() && !Character.isLetter(str1.charAt(k))) { digitStr1 += str1.charAt(k); k++; }

int k2 = pos+1; String digitStr2 = ""; while (k2 < str2.length() && !Character.isLetter(str2.charAt(k2)) ) { digitStr2 += str2.charAt(k2); k2++; }

similarity += Integer.parseInt(digitStr2) + Integer.parseInt(digitStr1);

} } } return Math.abs(limit - similarity);

   }
   /**
    * Wrapper function 
    * @param input string 1
    * @param input string 2
    * @param maximum possible limit value 
    * @return distance as integer
    */
   public static int SDFfunc(String str1, String str2, int limit) {
       return getDiff(mostOcurrencesElement(str1.toCharArray(), 2), mostOcurrencesElement(str2.toCharArray(), 2), limit);
   }

   public static void main(String[] args) {
       String input1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV";
       String input2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG";
       System.out.println(SDF.SDFfunc(input1,input2,100));

   }

}</lang>

JavaScript

Defines the functions for counting and frequency outside of the hashing function, as it is easier to calculate the comparison from a structured result rather than parsing a hash.

<lang javascript>//returns an object of counts keyed by character const kCounts = str => {

 const counts = {};
 for (let char of str) {
   counts[char] = counts[char] ? counts[char] + 1 : 1;
 }
 return counts;

};

//returns an array of length k containing the characters with the highest counts const frequentK = (counts, k) => {

 //note that this is written for clarity rather than speed,
 //as it sorts all of the counts when only the top k are needed
 return Object.keys(counts)
   .sort((a, b) => counts[b] - counts[a])
   .slice(0, k);

};

//returns a hashed string of the most frequent k characters and their frequencies const mostFreqKHashing = (str, k) => {

 const counts = kCounts(str);
 return frequentK(counts, k)
   .map(char => char + counts[char])
   .join("");

};

//numeric score of similarity based on the sum of counts of characters appearing in the top k of both strings const mostFreqKSimilarity = (str1, str2, k) => {

 const counts1 = kCounts(str1);
 const counts2 = kCounts(str2);
 const freq1 = frequentK(counts1, k);
 const freq2 = frequentK(counts2, k);
 let similarity = 0;
 for (let char of freq1) {
   //only considers a character if it is in the top k of both strings
   if (freq2.includes(char)) {
     //the examples on this page are inconsistent in whether the similarity
     //should be the sum of the two counts or only the shared count (the minimum of the two)
     //this code uses the sum
     similarity += counts1[char] + counts2[char];
   }
 }
 return similarity;

};

//subtracts the similarity score from the maxDifference const mostFreqKSDF = (str1, str2, k, maxDistance) => {

 return maxDistance - mostFreqKSimilarity(str1, str2, k);

};</lang>

Testing with Jest <lang javascript>test('hash of "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" is "L9T8"', () => {

 expect(mostFreqKHashing(str1, 2)).toBe("L9T8");

});

test('hash of "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" is "F9L8"', () => {

 expect(mostFreqKHashing(str2, 2)).toBe("F9L8");

});

test("SDF of strings 1 and 2 with k=2 and max=100 is 83", () => {

 expect(mostFreqKSDF(str1, str2, 2, 100)).toBe(83);

});</lang>

Testing in Console <lang javascript>const str1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV"; const str2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG"; const K = 2; const maxDistance = 100; console.log(mostFreqKHashing(str1, K)); console.log(mostFreqKHashing(str2, K)); console.log(mostFreqKSDF(str1, str2, K, maxDistance));</lang>

Output:

L9T8 
F9L8 
83

Kotlin

The code for the MostFreqKSimilarity function differs in this task from that in the associated Wikipedia article. Also the description is inconsistent with the code in both cases.

What I've decided to assume is that the frequency for commonly occurring characters must be the same in both strings for it to be added to the 'similarity' variable which seems to be the implication of both descriptions. This gives the same results as the Wikipedia article for all except the last example where it gives 100 rather than 83.

It's also evident that you can't just add the frequency of each character to the output string of the MostFreqKHashing function and then expect to be able to parse it afterwards. This is because, in the general case, any printing characters (including digits) could occur in the input string and the frequency could be more than 9. I've therefore encoded the frequency as the character with the same unicode value rather than the frequency itself.

I have no idea what NULL0 is supposed to mean so I've ignored that. <lang scala>// version 1.1.51

fun mostFreqKHashing(input: String, k: Int): String =

   input.groupBy { it }.map { Pair(it.key, it.value.size) }
                       .sortedByDescending { it.second } // preserves original order when equal
                       .take(k)
                       .fold("") { acc, v -> acc + "${v.first}${v.second.toChar()}" }

fun mostFreqKSimilarity(input1: String, input2: String): Int {

   var similarity = 0
   for (i in 0 until input1.length step 2) {
       for (j in 0 until input2.length step 2) {
           if (input1[i] == input2[j]) {
               val freq1 = input1[i + 1].toInt()
               val freq2 = input2[j + 1].toInt()
               if (freq1 != freq2) continue  // assuming here that frequencies need to match
               similarity += freq1
           }
       }
   }
   return similarity

}

fun mostFreqKSDF(input1: String, input2: String, k: Int, maxDistance: Int) {

   println("input1 : $input1")
   println("input2 : $input2")
   val s1 = mostFreqKHashing(input1, k)
   val s2 = mostFreqKHashing(input2, k)
   print("mfkh(input1, $k) = ")
   for ((i, c) in s1.withIndex()) print(if (i % 2 == 0) c.toString() else c.toInt().toString())
   print("\nmfkh(input2, $k) = ")
   for ((i, c) in s2.withIndex()) print(if (i % 2 == 0) c.toString() else c.toInt().toString())
   val result = maxDistance - mostFreqKSimilarity(s1, s2)
   println("\nSDF(input1, input2, $k, $maxDistance) = $result\n")

}

fun main(args: Array<String>) {

   val pairs = listOf(
       Pair("research", "seeking"),
       Pair("night", "nacht"),
       Pair("my", "a"),
       Pair("research", "research"),
       Pair("aaaaabbbb", "ababababa"),
       Pair("significant", "capabilities")
   )
   for (pair in pairs) mostFreqKSDF(pair.first, pair.second, 2, 10)

   var s1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV"
   var s2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG"
   mostFreqKSDF(s1, s2, 2, 100)
   s1 = "abracadabra12121212121abracadabra12121212121"
   s2 = s1.reversed()
   mostFreqKSDF(s1, s2, 2, 100)

}</lang>

Output:

input1 : research
input2 : seeking
mfkh(input1, 2) = r2e2
mfkh(input2, 2) = e2s1
SDF(input1, input2, 2, 10) = 8

input1 : night
input2 : nacht
mfkh(input1, 2) = n1i1
mfkh(input2, 2) = n1a1
SDF(input1, input2, 2, 10) = 9

input1 : my
input2 : a
mfkh(input1, 2) = m1y1
mfkh(input2, 2) = a1
SDF(input1, input2, 2, 10) = 10

input1 : research
input2 : research
mfkh(input1, 2) = r2e2
mfkh(input2, 2) = r2e2
SDF(input1, input2, 2, 10) = 6

input1 : aaaaabbbb
input2 : ababababa
mfkh(input1, 2) = a5b4
mfkh(input2, 2) = a5b4
SDF(input1, input2, 2, 10) = 1

input1 : significant
input2 : capabilities
mfkh(input1, 2) = i3n2
mfkh(input2, 2) = i3a2
SDF(input1, input2, 2, 10) = 7

input1 : LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV
input2 : EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG
mfkh(input1, 2) = L9T8
mfkh(input2, 2) = F9L8
SDF(input1, input2, 2, 100) = 100

input1 : abracadabra12121212121abracadabra12121212121
input2 : 12121212121arbadacarba12121212121arbadacarba
mfkh(input1, 2) = 112a10
mfkh(input2, 2) = 112210
SDF(input1, input2, 2, 100) = 88

Nim

It is impossible to get consistent results. We chose to implement Wikipedia algorithm (note that the page is no longer available, except in Internet Archive). With this algorithm, we get the same results as those of Wikipedia except for the last example where we get 91 instead of 83.

In order to avoid the limitation to 10 occurrences, we chose to use an ordered table (equivalent to Python OrderedDict). We could have used Kotlin solution or used a sequence of tuples (char, count) but the ordered table gives better performance.

<lang Nim>import algorithm, sugar, tables

type CharCounts = OrderedTable[char, int]

func `$`(counts: CharCounts): string =

 for (c, count) in counts.pairs:
   result.add $c & $count

func mostFreqKHashing(str: string; k: Positive): CharCounts =

 var counts: CharCounts                 # To get the counts in apparition order.
 for c in str: inc counts.mgetOrPut(c, 0)
 counts.sort((x, y) => cmp(x[1], y[1]), Descending)  # Note that sort is stable.
 var count = 0
 for c, val in counts.pairs:
   inc count
   result[c] = val
   if count == k: break

func mostFreqKSimilarity(input1, input2: CharCounts): int =

 for c, count in input1.pairs:
   if c in input2:
     result += count

func mostFreqKSDF(str1, str2: string; k, maxDistance: Positive): int =

 maxDistance - mostFreqKSimilarity(mostFreqKHashing(str1, k), mostFreqKHashing(str2, k))

const

 Pairs = [("night", "nacht"),
          ("my", "a"),
          ("research", "research"),
          ("aaaaabbbb", "ababababa"),
          ("significant", "capabilities")]

for (str1, str2) in Pairs:

 echo "str1: ", str1
 echo "str2: ", str2
 echo "mostFreqKHashing(str1, 2) = ", mostFreqKHashing(str1, 2)
 echo "mostFreqKHashing(str2, 2) = ", mostFreqKHashing(str2, 2)
 echo "mostFreqKSDF(str1, str2, 2, 10) = ", mostFreqKSDF(str1, str2, 2, 10)
 echo()

const

 S1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV"
 S2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG"

echo "str1: ", S1 echo "str2: ", S2 echo "mostFreqKHashing(str1, 2) = ", mostFreqKHashing(S1, 2) echo "mostFreqKHashing(str2, 2) = ", mostFreqKHashing(S2, 2) echo "mostFreqKSDF(str1, str2, 2, 100) = ", mostFreqKSDF(S1, S2, 2, 100)</lang>

Output:

str1: night
str2: nacht
mostFreqKHashing(str1, 2) = n1i1
mostFreqKHashing(str2, 2) = n1a1
mostFreqKSDF(str1, str2, 2, 10) = 9

str1: my
str2: a
mostFreqKHashing(str1, 2) = m1y1
mostFreqKHashing(str2, 2) = a1
mostFreqKSDF(str1, str2, 2, 10) = 10

str1: research
str2: research
mostFreqKHashing(str1, 2) = r2e2
mostFreqKHashing(str2, 2) = r2e2
mostFreqKSDF(str1, str2, 2, 10) = 6

str1: aaaaabbbb
str2: ababababa
mostFreqKHashing(str1, 2) = a5b4
mostFreqKHashing(str2, 2) = a5b4
mostFreqKSDF(str1, str2, 2, 10) = 1

str1: significant
str2: capabilities
mostFreqKHashing(str1, 2) = i3n2
mostFreqKHashing(str2, 2) = i3a2
mostFreqKSDF(str1, str2, 2, 10) = 7

str1: LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV
str2: EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG
mostFreqKHashing(str1, 2) = L9T8
mostFreqKHashing(str2, 2) = F9L8
mostFreqKSDF(str1, str2, 2, 100) = 91

Perl

<lang Perl>#!/usr/bin/perl use strict ; use warnings ;

sub mostFreqHashing {

  my $inputstring = shift ;
  my $howmany = shift ;
  my $outputstring ;
  my %letterfrequencies = findFrequencies ( $inputstring ) ;
  my @orderedChars = sort { $letterfrequencies{$b} <=> $letterfrequencies{$a} ||
     index( $inputstring , $a ) <=> index ( $inputstring , $b ) } keys %letterfrequencies ;
  for my $i ( 0..$howmany - 1 ) {
     $outputstring .= ( $orderedChars[ $i ] . $letterfrequencies{$orderedChars[ $i ]} ) ;
  }
  return $outputstring ;

}

sub findFrequencies {

  my $input = shift ;
  my %letterfrequencies ;
  for my $i ( 0..length( $input ) - 1 ) {
     $letterfrequencies{substr( $input , $i , 1 ) }++ ;
  }
  return %letterfrequencies ;

}

sub mostFreqKSimilarity {

  my $first = shift ;
  my $second = shift ;
  my $similarity = 0 ;
  my %frequencies_first = findFrequencies( $first ) ;
  my %frequencies_second = findFrequencies( $second ) ;
  foreach my $letter ( keys %frequencies_first ) {
     if ( exists ( $frequencies_second{$letter} ) ) {

$similarity += ( $frequencies_second{$letter} + $frequencies_first{$letter} ) ;

     }
  }
  return $similarity ;

}

sub mostFreqKSDF {

  (my $input1 , my $input2 , my $k , my $maxdistance ) = @_ ;
  return $maxdistance - mostFreqKSimilarity( mostFreqHashing( $input1 , $k) ,

mostFreqHashing( $input2 , $k) ) ; }

my $firststring = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" ; my $secondstring = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" ; print "MostFreqKHashing ( " . '$firststring , 2)' . " is " . mostFreqHashing( $firststring , 2 ) . "\n" ; print "MostFreqKHashing ( " . '$secondstring , 2)' . " is " . mostFreqHashing( $secondstring , 2 ) . "\n" ; </lang>

Output:

MostFreqKHashing ( $firststring , 2) is L9T8
MostFreqKHashing ( $secondstring , 2) is F9L8

Phix

Translation of: Go

<lang Phix>function mostFreqKHashing(string input, integer k)

   string cfs = ""
   sequence cfsnx = {}
   for i=1 to length(input) do
       integer r = input[i],
               ix = find(r,cfs)
       if ix>0 then
           cfsnx[ix][1] -= 1
       else
           cfs &= r
           cfsnx = append(cfsnx,{-1,length(cfs)})
       end if
   end for
   cfsnx = sort(cfsnx) -- (aside: the idx forces stable sort)
   sequence acc := {}
   for i=1 to min(length(cfs),k) do
       integer {n,ix} = cfsnx[i]
       acc &= {cfs[ix], -n}
   end for
   return acc

end function

function mostFreqKSimilarity(sequence input1, input2)

   integer similarity := 0
   for i=1 to length(input1) by 2 do
       for j=1 to length(input2) by 2 do
           if input1[i] == input2[j] then
               integer freq1 = input1[i+1],
                       freq2 = input2[j+1]
               if freq1=freq2 then
                   similarity += freq1
               end if
           end if
       end for
   end for
   return similarity

end function

function flat(sequence s)

   string res = ""
   for i=1 to length(s) by 2 do
       res &= sprintf("%c%d",s[i..i+1])
   end for 
   return res

end function

procedure mostFreqKSDF(string input1, input2, integer k, maxDistance)

   printf(1,"input1 : %s\n", {input1})
   printf(1,"input2 : %s\n", {input2})
   sequence s1 := mostFreqKHashing(input1, k),
            s2 := mostFreqKHashing(input2, k)
   printf(1,"mfkh(input1, %d) = %s\n", {k,flat(s1)})
   printf(1,"mfkh(input2, %d) = %s\n", {k,flat(s2)})
   integer result := maxDistance - mostFreqKSimilarity(s1, s2)
   printf(1,"SDF(input1, input2, %d, %d) = %d\n\n", {k, maxDistance, result})

end procedure

constant tests = {{"research", "seeking"},

                 {"night", "nacht"},
                 {"my", "a"},
                 {"research", "research"},
                 {"aaaaabbbb", "ababababa"},
                 {"significant", "capabilities"}}

for i=1 to length(tests) do

   string {t1,t2} = tests[i]
   mostFreqKSDF(t1, t2, 2, 10)

end for

string s1 := "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV",

      s2 := "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG"

mostFreqKSDF(s1, s2, 2, 100) s1 = "abracadabra12121212121abracadabra12121212121" s2 = reverse(s1) mostFreqKSDF(s1, s2, 2, 100)</lang> Output matches Go and Kotlin.

Python

Works with: Python version 2.7+

unoptimized and limited <lang python>import collections def MostFreqKHashing(inputString, K):

   occuDict = collections.defaultdict(int)
   for c in inputString:
       occuDict[c] += 1
   occuList = sorted(occuDict.items(), key = lambda x: x[1], reverse = True)
   outputStr = .join(c + str(cnt) for c, cnt in occuList[:K])
   return outputStr 

1. If number of occurrence of the character is not more than 9

def MostFreqKSimilarity(inputStr1, inputStr2):

   similarity = 0
   for i in range(0, len(inputStr1), 2):
       c = inputStr1[i]
       cnt1 = int(inputStr1[i + 1])
       for j in range(0, len(inputStr2), 2):
           if inputStr2[j] == c:
               cnt2 = int(inputStr2[j + 1])
               similarity += cnt1 + cnt2
               break
   return similarity

def MostFreqKSDF(inputStr1, inputStr2, K, maxDistance):

   return maxDistance - MostFreqKSimilarity(MostFreqKHashing(inputStr1,K), MostFreqKHashing(inputStr2,K))</lang>

optimized

A version that replaces the intermediate string with OrderedDict to reduce the time complexity of lookup operation: <lang python>import collections def MostFreqKHashing(inputString, K):

   occuDict = collections.defaultdict(int)
   for c in inputString:
       occuDict[c] += 1
   occuList = sorted(occuDict.items(), key = lambda x: x[1], reverse = True)
   outputDict = collections.OrderedDict(occuList[:K])
   #Return OrdredDict instead of string for faster lookup.
   return outputDict 

def MostFreqKSimilarity(inputStr1, inputStr2):

   similarity = 0
   for c, cnt1 in inputStr1.items():
       #Reduce the time complexity of lookup operation to about O(1).
       if c in inputStr2: 
           cnt2 = inputStr2[c]
           similarity += cnt1 + cnt2
   return similarity

def MostFreqKSDF(inputStr1, inputStr2, K, maxDistance):

   return maxDistance - MostFreqKSimilarity(MostFreqKHashing(inputStr1,K), MostFreqKHashing(inputStr2,K))</lang>

Test: <lang python>str1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" str2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" K = 2 maxDistance = 100 dict1 = MostFreqKHashing(str1, 2) print("%s:"%dict1) print(.join(c + str(cnt) for c, cnt in dict1.items())) dict2 = MostFreqKHashing(str2, 2) print("%s:"%dict2) print(.join(c + str(cnt) for c, cnt in dict2.items())) print(MostFreqKSDF(str1, str2, K, maxDistance))</lang>

Output:

OrderedDict([('L', 9), ('T', 8)]):
L9T8
OrderedDict([('F', 9), ('L', 8)]):
F9L8
83

Racket

<lang Racket>#lang racket

(define (MostFreqKHashing inputString K)

 (define t (make-hash))
 (for ([c (in-string inputString)] [i (in-naturals)])
   (define b (cdr (hash-ref! t c (λ() (cons i (box 0))))))
   (set-box! b (add1 (unbox b))))
 (define l (for/list ([(k v) (in-hash t)]) (list (car v) k (unbox (cdr v)))))
 (map cdr (take (sort (sort l < #:key car) > #:key caddr) K)))

(define (MostFreqKSimilarity inputStr1 inputStr2) ; not strings in this impl.

 (for*/sum ([c1 (in-list inputStr1)] [c2 (in-value (assq (car c1) inputStr2))]
            #:when c2)
   (+ (cadr c1) (cadr c2))))

(define (MostFreqKSDF inputStr1 inputStr2 K maxDistance)

 (- maxDistance (MostFreqKSimilarity (MostFreqKHashing inputStr1 K)
                                     (MostFreqKHashing inputStr2 K))))

(MostFreqKSDF

"LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV"
"EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG"
2 100)

=> 83

(Should add more tests, but it looks like there's a bunch of mistakes

in the given tests...)

</lang>

Raku

(formerly Perl 6)

Works with: Rakudo version 2017.09

My initial impressions of this task are registered on the discussion page under "Prank Page?".

The "most frequent k characters" hashing function is straightforward enough to implement. The distance function is incomprehensible though. The description doesn't match the pseudo-code and neither of them match the examples. I'm not going to bother trying to figure it out unless there is some possible value.

Maybe I am too hasty though. Lets give it a try. Implement a MFKC routine and run an assortment of words through it to get a feel for how it hashes different words.

<lang perl6># Fairly straightforward implementation, actually returns a list of pairs

1. which can be joined to make a string or manipulated further.

sub mfkh ($string, \K = 2) {

   my %h;
   $string.comb.kv.map: { %h{$^v}[1] //= $^k; %h{$^v}[0]++ };
   %h.sort( { -$_.value[0], $_.value[1] } ).head(K).map( { $_.key => $_.value[0] } );

}

1. lets try running 150 or so words from unixdic.txt through it to see
2. how many unique hash values it comes up with.

my @test-words = <

   aminobenzoic arginine asinine biennial biennium brigantine brinkmanship
   britannic chinquapin clinging corinthian declination dickinson dimension
   dinnertime dionysian diophantine dominican financial financier finessing
   fingernail fingerprint finnish giovanni hopkinsian inaction inalienable
   inanimate incaution incendiary incentive inception incident incidental
   incinerate incline inclusion incommunicable incompletion inconceivable
   inconclusive incongruity inconsiderable inconsiderate inconspicuous
   incontrovertible inconvertible incurring incursion indefinable indemnify
   indemnity indeterminacy indian indiana indicant indifferent indigene
   indigenous indigent indispensable indochina indochinese indoctrinate
   indonesia inequivalent inexplainable infantile inferential inferring
   infestation inflammation inflationary influential information infringe
   infusion ingenious ingenuity ingestion ingredient inhabitant inhalation
   inharmonious inheritance inholding inhomogeneity inkling inoffensive
   inordinate inorganic inputting inseminate insensible insincere insinuate
   insistent insomnia insomniac insouciant installation instinct instinctual
   insubordinate insubstantial insulin insurrection intangible intelligent
   intensify intensive interception interruption intestinal intestine
   intoxicant introduction introversion intrusion invariant invasion inventive
   inversion involution justinian kidnapping kingpin lineprinter liniment
   livingston mainline mcginnis minion minneapolis minnie pigmentation
   pincushion pinion quinine quintessential resignation ruination seminarian
   triennial wilkinson wilmington wilsonian wineskin winnie winnipeg  

>;

say @test-words.map( { join , mfkh($_)».kv } ).Bag;</lang>

Output:

Bag(i2n2(151))

One... Nope, I was right, it is pretty much worthless.

Ring

<lang ring>

1. Project : Most frequent k chars distance

load "stdlib.ring" str1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" str2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" see "Str1 = " + str1 + nl see "Str2 = " + str2 + nl

MostFreqKHashing(str1,"str1") MostFreqKHashing(str2,"str2")

func MostFreqKHashing(str3,strp)

       chr = newlist(26,2)
       for n = 1 to 26
            str = char(n+64)
            cstr = count(str3,str)
            chr[n][1] = str
            chr[n][2] = cstr
       next
       chr = sortsecond(chr)
       chr = reverse(chr)
       see "MostFreqKHashing(" + strp + ",2) = "
       see chr[1][1] + chr[1][2] + chr[2][1] + chr[2][2] + nl

func count(cString,dString)

      sum = 0
      while substr(cString,dString) > 0
              sum++
              cString = substr(cString,substr(cString,dString)+len(string(sum)))
      end
      return sum

func sortsecond(alist)

       aList = sort(alist,2)
       for n=1 to len(alist)-1
            for m=n to len(aList)-1 
                  if alist[m+1][2] = alist[m][2] and alist[m+1][1] < alist[m][1]
                     temp = alist[m+1]
                     alist[m+1] = alist[m]
                     alist[m] = temp ok
            next
      next
      return aList

</lang> Output:

Str1 = LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV
Str2 = EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG
MostFreqKHashing(str1,2) = L9T8
MostFreqKHashing(str2,2) = F9L8

Sidef

<lang ruby>func _MostFreqKHashing(string, k) {

   var seen = Hash()
   var chars = string.chars
   var freq = chars.freq
   var schars = freq.keys.sort_by {|c| -freq{c} }

   var mfkh = []
   for i in ^k {
       chars.each { |c|
           seen{c} && next
           if (freq{c} == freq{schars[i]}) {
               seen{c} = true
               mfkh << Hash(c => c, f => freq{c})
               break
           }
       }
   }

   mfkh << (k-seen.len -> of { Hash(c => :NULL, f => 0) }...)
   mfkh

}

func MostFreqKSDF(a, b, k, d) {

   var mfkh_a = _MostFreqKHashing(a, k);
   var mfkh_b = _MostFreqKHashing(b, k);

   d - gather {
       mfkh_a.each { |s|
           s{:c} == :NULL && next
           mfkh_b.each { |t|
               s{:c} == t{:c} &&
                   take(s{:f} + (s{:f} == t{:f} ? 0 : t{:f}))
           }
       }
   }.sum

}

func MostFreqKHashing(string, k) {

   gather {
       _MostFreqKHashing(string, k).each { |h|
           take("%s%d" % (h{:c}, h{:f}))
       }
   }.join

}

var str1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" var str2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG"

say "str1 = #{str1.dump}" say "str2 = #{str2.dump}"

say

say("MostFreqKHashing(str1, 2) = ", MostFreqKHashing(str1, 2)) say("MostFreqKHashing(str2, 2) = ", MostFreqKHashing(str2, 2)) say("MostFreqKSDF(str1, str2, 2, 100) = ", MostFreqKSDF(str1, str2, 2, 100))

say

var arr = [

   %w(night nacht),
   %w(my a),
   %w(research research),
   %w(aaaaabbbb ababababa),
   %w(significant capabilities),

]

var k = 2 var limit = 10

for s,t in arr {

   "mfkh(%s, %s, #{k}) = [%s, %s] (SDF: %d)\n".printf(
       s.dump, t.dump,
       MostFreqKHashing(s, k).dump,
       MostFreqKHashing(t, k).dump,
       MostFreqKSDF(s, t, k, limit),
   )

}</lang>

Output:

str1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV"
str2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG"

MostFreqKHashing(str1, 2) = L9T8
MostFreqKHashing(str2, 2) = F9L8
MostFreqKSDF(str1, str2, 2, 100) = 83

mfkh("night", "nacht", 2) = ["n1i1", "n1a1"] (SDF: 9)
mfkh("my", "a", 2) = ["m1y1", "a1NULL0"] (SDF: 10)
mfkh("research", "research", 2) = ["r2e2", "r2e2"] (SDF: 6)
mfkh("aaaaabbbb", "ababababa", 2) = ["a5b4", "a5b4"] (SDF: 1)
mfkh("significant", "capabilities", 2) = ["i3n2", "i3a2"] (SDF: 7)

Tcl

Works with: Tcl version 8.6

<lang tcl>package require Tcl 8.6

proc MostFreqKHashing {inputString k} {

   foreach ch [split $inputString ""] {dict incr count $ch}
   join [lrange [lsort -stride 2 -index 1 -integer -decreasing $count] 0 [expr {$k*2-1}]] ""

} proc MostFreqKSimilarity {hashStr1 hashStr2} {

   while {$hashStr2 ne ""} {

regexp {^(.)(\d+)(.*)$} $hashStr2 -> ch n hashStr2 set lookup($ch) $n

   }
   set similarity 0
   while {$hashStr1 ne ""} {

regexp {^(.)(\d+)(.*)$} $hashStr1 -> ch n hashStr1 if {[info exist lookup($ch)]} { incr similarity $n incr similarity $lookup($ch) }

   }
   return $similarity

} proc MostFreqKSDF {inputStr1 inputStr2 k limit} {

   set h1 [MostFreqKHashing $inputStr1 $k]
   set h2 [MostFreqKHashing $inputStr2 $k]
   expr {$limit - [MostFreqKSimilarity $h1 $h2]}

}</lang> Demonstrating: <lang tcl>set str1 "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" set str2 "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" puts [MostFreqKHashing $str1 2] puts [MostFreqKHashing $str2 2] puts [MostFreqKSDF $str1 $str2 2 100]</lang>

Output:

L9T8
F9L8
83

A more efficient metric calculator

This version is appreciably more efficient because it does not compute the intermediate string representation “hash”, instead working directly on the intermediate dictionaries and lists: <lang tcl>proc MostFreqKSDF {inputStr1 inputStr2 k limit} {

   set c1 [set c2 {}]
   foreach ch [split $inputStr1 ""] {dict incr c1 $ch}
   foreach ch [split $inputStr2 ""] {dict incr c2 $ch}
   set c2 [lrange [lsort -stride 2 -index 1 -integer -decreasing $c2[set c2 {}]] 0 [expr {$k*2-1}]]
   set s 0
   foreach {ch n} [lrange [lsort -stride 2 -index 1 -integer -decreasing $c1[set c1 {}]] 0 [expr {$k*2-1}]] {

if {[dict exists $c2 $ch]} { incr s [expr {$n + [dict get $c2 $ch]}] }

   }
   return [expr {$limit - $s}]

}</lang> It computes the identical value on the identical inputs.

Typescript

Translation of: Javascript

with added type annotations

<lang typescript>//returns an object of counts keyed by character const kCounts = (str: string): Record<string, number> => {

 const counts: Record<string, number> = {};
 for (let char of str) {
   counts[char] = counts[char] ? counts[char] + 1 : 1;
 }
 return counts;

};

//returns an array of length k containing the characters with the highest counts const frequentK = ( counts: Record<string, number>, k: number ): string[] => {

 //note that this is written for clarity rather than speed,
 //as it sorts all of the counts when only the top k are needed
 return Object.keys(counts)
   .sort((a, b) => counts[b] - counts[a])
   .slice(0, k);

};

//returns a hashed string of the most frequent k characters and their frequencies const mostFreqKHashing = (str: string, k: number): string => {

 const counts = kCounts(str);
 return frequentK(counts, k)
   .map(char => char + counts[char])
   .join("");

};

//numeric score of similarity based on the sum of counts of characters appearing in the top k of both strings const mostFreqKSimilarity = ( str1: string, str2: string, k: number ): number => {

 const counts1 = kCounts(str1);
 const counts2 = kCounts(str2);
 const freq1 = frequentK(counts1, k);
 const freq2 = frequentK(counts2, k);
 let similarity = 0;
 for (let char of freq1) {
   //only considers a character if it is in the top k of both strings
   if (freq2.includes(char)) {
     //the examples on this page are inconsistent in whether the similarity
     //should be the sum of the two counts or only the shared count (the minimum of the two)
     //this code uses the sum
     similarity += counts1[char] + counts2[char];
   }
 }
 return similarity;

};

//subtracts the similarity score from the maxDifference const mostFreqKSDF = ( str1: string, str2: string, k: number, maxDistance: number ): number => {

 return maxDistance - mostFreqKSimilarity(str1, str2, k);

};</lang>

Wren

Translation of: Kotlin

Library: Wren-seq

Library: Wren-sort

Library: Wren-trait

<lang ecmascript>import "/seq" for Lst import "/sort" for Sort import "/trait" for Stepped

var mostFreqKHashing = Fn.new { |input, k|

  var indivs = Lst.individuals(input.toList).map { |indiv| [indiv[0], indiv[1]] }.toList
  var cmp = Fn.new { |p1, p2| (p2[1] - p1[1]).sign }
  Sort.insertion(indivs, cmp)
  return indivs.take(k).reduce("") { |acc, p| acc + "%(p[0])%(String.fromByte(p[1]))" }

}

var mostFreqKSimilarity = Fn.new { |input1, input2|

   var similarity = 0
   for (i in Stepped.new(0...input1.count, 2)) {
       for (j in Stepped.new(0...input2.count, 2)) {
           if (input1[i] == input2[j]) {
               var freq1 = input1[i + 1].bytes[0]
               var freq2 = input2[j + 1].bytes[0]
               if (freq1 == freq2) similarity = similarity + freq1
           }
       }
   }
   return similarity

}

var mostFreqKSDF = Fn.new { |input1, input2, k, maxDistance|

   System.print("input1 : %(input1)")
   System.print("input2 : %(input2)")
   var s1 = mostFreqKHashing.call(input1, k)
   var s2 = mostFreqKHashing.call(input2, k)
   System.write("mfkh(input1, %(k)) = ")
   var i = 0
   for (c in s1) {
       System.write((i % 2 == 0) ? c : c.bytes[0])
       i = i + 1
   }
   System.write("\nmfkh(input2, %(k)) = ")
   i = 0
   for (c in s2) {
       System.write((i % 2 == 0) ? c : c.bytes[0])
       i = i + 1
   }
   var result = maxDistance - mostFreqKSimilarity.call(s1, s2)
   System.print("\nSDF(input1, input2, %(k), %(maxDistance)) = %(result)\n")

}

var pairs = [

   ["research", "seeking"],
   ["night", "nacht"],
   ["my", "a"],
   ["research", "research"],
   ["aaaaabbbb", "ababababa"],
   ["significant", "capabilities"]

] for (pair in pairs) mostFreqKSDF.call(pair[0], pair[1], 2, 10)

var s1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" var s2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" mostFreqKSDF.call(s1, s2, 2, 100) s1 = "abracadabra12121212121abracadabra12121212121" s2 = s1[-1..0] mostFreqKSDF.call(s1, s2, 2, 100)</lang>

Output:

Sane as Kotlin entry.

References