Suffix tree: Difference between revisions

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s += this.toString_f(children[children.length - 1], pre + ' ');
s += this.toString_f(children[children.length - 1], pre + ' ');
return s;
return s;
}

getLongestRepeatedSubString() {
return this.getLongestRepeatedSubString_f(0);
}
getLongestRepeatedSubString_f(n) {
let str = '';
for (const n2 of this.nodes[n].children.filter(n2 => this.nodes[n2].children.length > 0)) { // ignore leaves
const temp = this.getLongestRepeatedSubString_f(n2);
if (temp.length > str.length) {
str = temp;
}
}
return this.nodes[n].sub + str;
}
}
}
}


const st = new SuffixTree('banana');
const st = new SuffixTree('banana');
console.log(st.toString());
console.log(st.toString());</lang>
console.log('>', st.getLongestRepeatedSubString());</lang>


{{out}}
{{out}}
Line 762: Line 747:
└─┐ na
└─┐ na
└─- na
└─- na

> ana
</pre>
</pre>


Revision as of 20:03, 8 June 2020

Suffix tree 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.

A suffix tree is a data structure commonly used in string algorithms.

Given a string S of length n, its suffix tree is a tree T such that:

  • T has exactly n leaves numbered from 1 to n.
  • Except for the root, every internal node has at least two children.
  • Each edge of T is labelled with a non-empty substring of S.
  • No two edges starting out of a node can have string labels beginning with the same character.
  • The string obtained by concatenating all the string labels found on the path from the root to leaf i spells out suffix S[i..n], for i from 1 to n.


Such a tree does not exist for all strings. To ensure existence, a character that is not found in S must be appended at its end. The character '$' is traditionally used for this purpose.

For this task, build and display the suffix tree of the string "banana$". Displaying the tree can be done using the code from the visualize a tree task, but any other convenient method is accepted.

There are several ways to implement the tree data structure, for instance how edges should be labelled. Latitude is given in this matter, but notice that a simple way to do it is to label each node with the label of the edge leading to it.

The computation time for an efficient algorithm should be , but such an algorithm might be difficult to implement. An easier, algorithm is accepted.

C#

Translation of: C++

<lang csharp>using System; using System.Collections.Generic;

namespace SuffixTree { 

   class Node {
       public string sub;                     // a substring of the input string
       public List<int> ch = new List<int>(); // vector of child nodes

       public Node() {
           sub = "";
       }

       public Node(string sub, params int[] children) {
           this.sub = sub;
           ch.AddRange(children);
       }
   }

   class SuffixTree {
       readonly List<Node> nodes = new List<Node>();

       public SuffixTree(string str) {
           nodes.Add(new Node());
           for (int i = 0; i < str.Length; i++) {
               AddSuffix(str.Substring(i));
           }
       }

       public void Visualize() {
           if (nodes.Count == 0) {
               Console.WriteLine("<empty>");
               return;
           }

           void f(int n, string pre) {
               var children = nodes[n].ch;
               if (children.Count == 0) {
                   Console.WriteLine("- {0}", nodes[n].sub);
                   return;
               }
               Console.WriteLine("+ {0}", nodes[n].sub);

               var it = children.GetEnumerator();
               if (it.MoveNext()) {
                   do {
                       var cit = it;
                       if (!cit.MoveNext()) break;

                       Console.Write("{0}+-", pre);
                       f(it.Current, pre + "| ");
                   } while (it.MoveNext());
               }

               Console.Write("{0}+-", pre);
               f(children[children.Count-1], pre+"  ");
           }

           f(0, "");
       }

       private void AddSuffix(string suf) {
           int n = 0;
           int i = 0;
           while (i < suf.Length) {
               char b = suf[i];
               int x2 = 0;
               int n2;
               while (true) {
                   var children = nodes[n].ch;
                   if (x2 == children.Count) {
                       // no matching child, remainder of suf becomes new node
                       n2 = nodes.Count;
                       nodes.Add(new Node(suf.Substring(i)));
                       nodes[n].ch.Add(n2);
                       return;
                   }
                   n2 = children[x2];
                   if (nodes[n2].sub[0] == b) {
                       break;
                   }
                   x2++;
               }
               // find prefix of remaining suffix in common with child
               var sub2 = nodes[n2].sub;
               int j = 0;
               while (j < sub2.Length) {
                   if (suf[i + j] != sub2[j]) {
                       // split n2
                       var n3 = n2;
                       // new node for the part in common
                       n2 = nodes.Count;
                       nodes.Add(new Node(sub2.Substring(0, j), n3));
                       nodes[n3].sub = sub2.Substring(j); // old node loses the part in common
                       nodes[n].ch[x2] = n2;
                       break; // continue down the tree
                   }
                   j++;
               }
               i += j; // advance past part in common
               n = n2; // continue down the tree
           }
       }
   }

   class Program {
       static void Main() {
           new SuffixTree("banana$").Visualize();
       }
   }

}</lang>

Output:
+
+-- banana$
+-+ a
| +-+ na
| | +-- na$
| | +-- $
| +-- $
+-+ na
| +-- na$
| +-- $
+-- $

C++

Translation of: D

<lang cpp>#include <functional>

  1. include <iostream>
  2. include <vector>

struct Node { 

   std::string sub = "";   // a substring of the input string
   std::vector<int> ch;    // vector of child nodes

   Node() {
       // empty
   }

   Node(const std::string& sub, std::initializer_list<int> children) : sub(sub) {
       ch.insert(ch.end(), children);
   }

};

struct SuffixTree { 

   std::vector<Node> nodes;

   SuffixTree(const std::string& str) {
       nodes.push_back(Node{});
       for (size_t i = 0; i < str.length(); i++) {
           addSuffix(str.substr(i));
       }
   }

   void visualize() {
       if (nodes.size() == 0) {
           std::cout << "<empty>\n";
           return;
       }

       std::function<void(int, const std::string&)> f;
       f = [&](int n, const std::string & pre) {
           auto children = nodes[n].ch;
           if (children.size() == 0) {
               std::cout << "- " << nodes[n].sub << '\n';
               return;
           }
           std::cout << "+ " << nodes[n].sub << '\n';

           auto it = std::begin(children);
           if (it != std::end(children)) do {
               if (std::next(it) == std::end(children)) break;
               std::cout << pre << "+-";
               f(*it, pre + "| ");
               it = std::next(it);
           } while (true);

           std::cout << pre << "+-";
           f(children[children.size() - 1], pre + "  ");
       };

       f(0, "");
   }

private: 

   void addSuffix(const std::string & suf) {
       int n = 0;
       size_t i = 0;
       while (i < suf.length()) {
           char b = suf[i];
           int x2 = 0;
           int n2;
           while (true) {
               auto children = nodes[n].ch;
               if (x2 == children.size()) {
                   // no matching child, remainder of suf becomes new node
                   n2 = nodes.size();
                   nodes.push_back(Node(suf.substr(i), {}));
                   nodes[n].ch.push_back(n2);
                   return;
               }
               n2 = children[x2];
               if (nodes[n2].sub[0] == b) {
                   break;
               }
               x2++;
           }
           // find prefix of remaining suffix in common with child
           auto sub2 = nodes[n2].sub;
           size_t j = 0;
           while (j < sub2.size()) {
               if (suf[i + j] != sub2[j]) {
                   // split n2
                   auto n3 = n2;
                   // new node for the part in common
                   n2 = nodes.size();
                   nodes.push_back(Node(sub2.substr(0, j), { n3 }));
                   nodes[n3].sub = sub2.substr(j); // old node loses the part in common
                   nodes[n].ch[x2] = n2;
                   break; // continue down the tree
               }
               j++;
           }
           i += j; // advance past part in common
           n = n2; // continue down the tree
       }
   }

};

int main() { 

   SuffixTree("banana$").visualize();

}</lang>

Output:
+
+-- banana$
+-+ a
| +-+ na
| | +-- na$
| | +-- $
| +-- $
+-+ na
| +-- na$
| +-- $
+-- $

D

Translation of: Kotlin

<lang D>import std.stdio;

struct Node { 

   string sub = ""; // a substring of the input string
   int[] ch;        // array of child nodes

   this(string sub, int[] children ...) {
       this.sub = sub;
       ch = children;
   }

}

struct SuffixTree { 

   Node[] nodes;

   this(string str) {
       nodes ~= Node();
       for (int i=0; i<str.length; ++i) {
           addSuffix(str[i..$]);
       }
   }

   private void addSuffix(string suf) {
       int n = 0;
       int i = 0;
       while (i < suf.length) {
           char b  = suf[i];
           int x2 = 0;
           int n2;
           while (true) {
               auto children = nodes[n].ch;
               if (x2 == children.length) {
                   // no matching child, remainder of suf becomes new node.
                   n2 = nodes.length;
                   nodes ~= Node(suf[i..$]);
                   nodes[n].ch ~= n2;
                   return;
               }
               n2 = children[x2];
               if (nodes[n2].sub[0] == b) {
                   break;
               }
               x2++;
           }
           // find prefix of remaining suffix in common with child
           auto sub2 = nodes[n2].sub;
           int j = 0;
           while (j < sub2.length) {
               if (suf[i + j] != sub2[j]) {
                   // split n2
                   auto n3 = n2;
                   // new node for the part in common
                   n2 = nodes.length;
                   nodes ~= Node(sub2[0..j], n3);
                   nodes[n3].sub = sub2[j..$];  // old node loses the part in common
                   nodes[n].ch[x2] = n2;
                   break;  // continue down the tree
               }
               j++;
           }
           i += j;  // advance past part in common
           n = n2;  // continue down the tree
       }
   }

   void visualize() {
       if (nodes.length == 0) {
           writeln("<empty>");
           return;
       }

       void f(int n, string pre) {
           auto children = nodes[n].ch;
           if (children.length == 0) {
               writefln("╴ %s", nodes[n].sub);
               return;
           }
           writefln("┐ %s", nodes[n].sub);
           foreach (c; children[0..$-1]) {
               write(pre, "├─");
               f(c, pre ~ "│ ");
           }
           write(pre, "└─");
           f(children[$-1], pre ~ "  ");
       }

       f(0, "");
   }

}

void main() { 

   SuffixTree("banana$").visualize();

}</lang>

Output:
┐ 
├─╴ banana$
├─┐ a
│ ├─┐ na
│ │ ├─╴ na$
│ │ └─╴ $
│ └─╴ $
├─┐ na
│ ├─╴ na$
│ └─╴ $
└─╴ $

Go

Vis function from Visualize_a_tree#Unicode. <lang go>package main

import "fmt"

func main() { 

   vis(buildTree("banana$"))

}

type tree []node

type node struct { 

   sub string // a substring of the input string
   ch  []int  // list of child nodes

}

func buildTree(s string) tree { 

   t := tree{node{}} // root node
   for i := range s {
       t = t.addSuffix(s[i:])
   }
   return t

}

func (t tree) addSuffix(suf string) tree { 

   n := 0
   for i := 0; i < len(suf); {
       b := suf[i]
       ch := t[n].ch
       var x2, n2 int
       for ; ; x2++ {
           if x2 == len(ch) {
               // no matching child, remainder of suf becomes new node.
               n2 = len(t)
               t = append(t, node{sub: suf[i:]})
               t[n].ch = append(t[n].ch, n2)
               return t
           }
           n2 = ch[x2]
           if t[n2].sub[0] == b {
               break
           }
       }
       // find prefix of remaining suffix in common with child
       sub2 := t[n2].sub
       j := 0
       for ; j < len(sub2); j++ {
           if suf[i+j] != sub2[j] {
               // split n2
               n3 := n2
               // new node for the part in common
               n2 = len(t)
               t = append(t, node{sub2[:j], []int{n3}})
               t[n3].sub = sub2[j:] // old node loses the part in common
               t[n].ch[x2] = n2
               break // continue down the tree
           }
       }
       i += j // advance past part in common
       n = n2 // continue down the tree
   }
   return t

}

func vis(t tree) { 

   if len(t) == 0 {
       fmt.Println("<empty>")
       return
   }
   var f func(int, string)
   f = func(n int, pre string) {
       children := t[n].ch
       if len(children) == 0 {
           fmt.Println("╴", t[n].sub)
           return
       }
       fmt.Println("┐", t[n].sub)
       last := len(children) - 1
       for _, ch := range children[:last] {
           fmt.Print(pre, "├─")
           f(ch, pre+"│ ")
       }
       fmt.Print(pre, "└─")
       f(children[last], pre+"  ")
   }
   f(0, "")

}</lang>

Output:
┐ 
├─╴ banana$
├─┐ a
│ ├─┐ na
│ │ ├─╴ na$
│ │ └─╴ $
│ └─╴ $
├─┐ na
│ ├─╴ na$
│ └─╴ $
└─╴ $

J

Implementation:

<lang J>classify=: {.@> </. ]

build_tree=:3 :0 

 tree=. ,:_;_;
 if. 0=#y do. tree return.end.
 if. 1=#y do. tree,(#;y);0;y return.end.
 for_box.classify y do.
   char=. {.>{.>box
   subtree=. }.build_tree }.each>box
   ndx=.I.0=1&{::"1 subtree
   n=.#tree
   if. 1=#ndx do.
     counts=. 1 + 0&{::"1 subtree
     parents=. (n-1) (+*]&*) 1&{::"1 subtree
     edges=. (ndx}~ <@(char,ndx&{::)) 2&{"1 subtree
     tree=. tree, counts;"0 1 parents;"0 edges
   else.
     tree=. tree,(__;0;,char),(1;n;0) + ::]&.>"1 subtree
   end.
 end.

)

suffix_tree=:3 :0 

 assert. -.({:e.}:)y
 tree=. B=:|:build_tree <\. y
 ((1+#y)-each {.tree),}.tree

)</lang>

Task example:

<lang J> suffix_tree 'banana$' ┌──┬───────┬─┬──┬───┬─┬─┬──┬───┬─┬─┐ │__│1 │_│_ │2 │4│6│_ │3 │5│7│ ├──┼───────┼─┼──┼───┼─┼─┼──┼───┼─┼─┤ │_ │0 │0│2 │3 │3│2│0 │7 │7│0│ ├──┼───────┼─┼──┼───┼─┼─┼──┼───┼─┼─┤ │ │banana$│a│na│na$│$│$│na│na$│$│$│ └──┴───────┴─┴──┴───┴─┴─┴──┴───┴─┴─┘</lang>

The first row is the leaf number (_ for internal nodes).

The second row is parent index (_ for root node).

The third row is the edge's substring (empty for root node).

Visualizing, using showtree and prefixing the substring leading to each leaf with the leaf number (in brackets):

<lang J>fmttree=: ;@(1&{) showtree~ {: (,~ }.`('[','] ',~":)@.(_>|))each {. 

  fmttree suffix_tree 'banana$'
   ┌─ [1] banana$                    
   │                       ┌─ [2] na$
   │             ┌─ na ────┴─ [4] $

────┼─ a ─────────┴─ [6] $ 

   │             ┌─ [3] na$          
   ├─ na ────────┴─ [5] $            
   └─ [7] $

</lang>

Java

Translation of: Kotlin

<lang Java>import java.util.ArrayList; import java.util.List;

public class SuffixTreeProblem { 

   private static class Node {
       String sub = "";                       // a substring of the input string
       List<Integer> ch = new ArrayList<>();  // list of child nodes
   }

   private static class SuffixTree {
       private List<Node> nodes = new ArrayList<>();

       public SuffixTree(String str) {
           nodes.add(new Node());
           for (int i = 0; i < str.length(); ++i) {
               addSuffix(str.substring(i));
           }
       }

       private void addSuffix(String suf) {
           int n = 0;
           int i = 0;
           while (i < suf.length()) {
               char b = suf.charAt(i);
               List<Integer> children = nodes.get(n).ch;
               int x2 = 0;
               int n2;
               while (true) {
                   if (x2 == children.size()) {
                       // no matching child, remainder of suf becomes new node.
                       n2 = nodes.size();
                       Node temp = new Node();
                       temp.sub = suf.substring(i);
                       nodes.add(temp);
                       children.add(n2);
                       return;
                   }
                   n2 = children.get(x2);
                   if (nodes.get(n2).sub.charAt(0) == b) break;
                   x2++;
               }
               // find prefix of remaining suffix in common with child
               String sub2 = nodes.get(n2).sub;
               int j = 0;
               while (j < sub2.length()) {
                   if (suf.charAt(i + j) != sub2.charAt(j)) {
                       // split n2
                       int n3 = n2;
                       // new node for the part in common
                       n2 = nodes.size();
                       Node temp = new Node();
                       temp.sub = sub2.substring(0, j);
                       temp.ch.add(n3);
                       nodes.add(temp);
                       nodes.get(n3).sub = sub2.substring(j);  // old node loses the part in common
                       nodes.get(n).ch.set(x2, n2);
                       break;  // continue down the tree
                   }
                   j++;
               }
               i += j;  // advance past part in common
               n = n2;  // continue down the tree
           }
       }

       public void visualize() {
           if (nodes.isEmpty()) {
               System.out.println("<empty>");
               return;
           }
           visualize_f(0, "");
       }

       private void visualize_f(int n, String pre) {
           List<Integer> children = nodes.get(n).ch;
           if (children.isEmpty()) {
               System.out.println("- " + nodes.get(n).sub);
               return;
           }
           System.out.println("┐ " + nodes.get(n).sub);
           for (int i = 0; i < children.size() - 1; i++) {
               Integer c = children.get(i);
               System.out.print(pre + "├─");
               visualize_f(c, pre + "│ ");
           }
           System.out.print(pre + "└─");
           visualize_f(children.get(children.size() - 1), pre + "  ");
       }
   }

   public static void main(String[] args) {
       new SuffixTree("banana$").visualize();
   }

}</lang>

Output:
┐ 
├─- banana$
├─┐ a
│ ├─┐ na
│ │ ├─- na$
│ │ └─- $
│ └─- $
├─┐ na
│ ├─- na$
│ └─- $
└─- $

JavaScript

Translation of: Java

<lang JavaScript>class Node { 

 sub = ; // a substring of the input string
 children = []; // list of child nodes

}

class SuffixTree { 

 nodes = [];

 constructor(str) {
   this.nodes.push(new Node());
   for (let i = 0; i < str.length; ++i) {
     this.addSuffix(str.slice(i));
   }
 }

 addSuffix(suf) {
   let n = 0;
   let i = 0;
   while (i < suf.length) {
     const b = suf.charAt(i);
     const children = this.nodes[n].children;
     let x2 = 0;
     let n2;
     while (true) {
       if (x2 === children.length) {
         // no matching child, remainder of suf becomes new node.
         n2 = this.nodes.length;
         const temp = new Node();
         temp.sub = suf.slice(i);
         this.nodes.push(temp);
         children.push(n2);
         return;
       }
       n2 = children[x2];
       if (this.nodes[n2].sub.charAt(0) === b) break;
       x2++;
     }
     // find prefix of remaining suffix in common with child
     const sub2 = this.nodes[n2].sub;
     let j = 0;
     while (j < sub2.length) {
       if (suf.charAt(i + j) !== sub2.charAt(j)) {
         // split n2
         const n3 = n2;
         // new node for the part in common
         n2 = this.nodes.length;
         const temp = new Node();
         temp.sub = sub2.slice(0, j);
         temp.children.push(n3);
         this.nodes.push(temp);
         this.nodes[n3].sub = sub2.slice(j);  // old node loses the part in common
         this.nodes[n].children[x2] = n2;
         break;  // continue down the tree
       }
       j++;
     }
     i += j;  // advance past part in common
     n = n2;  // continue down the tree
   }
 }

 toString() {
   if (this.nodes.length === 0) {
     return '<empty>';
   }
   return this.toString_f(0, );
 }

 toString_f(n, pre) {
   const children = this.nodes[n].children;
   if (children.length === 0) {
     return '- ' + this.nodes[n].sub + '\n';
   }
   let s = '┐ ' + this.nodes[n].sub + '\n';
   for (let i = 0; i < children.length - 1; i++) {
     const c = children[i];
     s += pre + '├─';
     s += this.toString_f(c, pre + '│ ');
   }
   s += pre + '└─';
   s += this.toString_f(children[children.length - 1], pre + '  ');
   return s;
 }

}

const st = new SuffixTree('banana'); console.log(st.toString());</lang>

Output:
┐ 
├─- banana
├─┐ a
│ └─┐ na
│   └─- na
└─┐ na
  └─- na

Julia

Translation of: Go

<lang julia>import Base.print

mutable struct Node 

   sub::String
   ch::Vector{Int}
   Node(str, v=Int[]) = new(str, v)

end

struct SuffixTree 

   nodes::Vector{Node}
   function SuffixTree(s::String)
       nod = [Node("", Int[])]
       for i in 1:length(s)
           addSuffix!(nod, s[i:end])
       end
       return new(nod)
   end

end

function addSuffix!(tree::Vector{Node}, suf::String) 

   n, i = 1, 1
   while i <= length(suf)
       x2, n2, b = 1, 1, suf[i]
       while true
           children = tree[n].ch
           if x2 > length(children)
               push!(tree, Node(suf[i:end]))
               push!(tree[n].ch, length(tree))
               return
           end
           n2 = children[x2]
           (tree[n2].sub[1] == b) && break
           x2 += 1
       end
       sub2, j = tree[n2].sub, 0
       while j < length(sub2)
           if suf[i + j] != sub2[j + 1]
               push!(tree, Node(sub2[1:j], [n2]))
               tree[n2].sub = sub2[j+1:end]
               n2 = length(tree)
               tree[n].ch[x2] = n2
               break
           end
           j += 1
       end
       i += j
       n = n2
   end

end

function Base.print(io::IO, suffixtree::SuffixTree) 

   function treeprint(n::Int, pre::String)
       children = suffixtree.nodes[n].ch
       if isempty(children)
           println("╴ ", suffixtree.nodes[n].sub)
       else
           println("┐ ", suffixtree.nodes[n].sub)
           for c in children[1:end-1]
               print(pre, "├─")
               treeprint(c, pre * "│ ")
           end
           print(pre, "└─")
           treeprint(children[end], pre * "  ")
       end
   end
   if isempty(suffixtree.nodes)
       println("<empty>")
   else
       treeprint(1, "")
   end

end

println(SuffixTree("banana\$"))

</lang>

Output:
┐
├─╴ banana$
├─┐ a
│ ├─┐ na
│ │ ├─╴ na$
│ │ └─╴ $
│ └─╴ $
├─┐ na
│ ├─╴ na$
│ └─╴ $
└─╴ $

Kotlin

Translation of: Go

<lang scala>// version 1.1.3

class Node { 

   var sub = ""                    // a substring of the input string
   var ch  = mutableListOf<Int>()  // list of child nodes

}

class SuffixTree(val str: String) { 

   val nodes = mutableListOf<Node>(Node())

   init {
       for (i in 0 until str.length) addSuffix(str.substring(i))
   }

   private fun addSuffix(suf: String) {
       var n = 0
       var i = 0
       while (i < suf.length) {
           val b  = suf[i]
           val children = nodes[n].ch
           var x2 = 0
           var n2: Int
           while (true) {
               if (x2 == children.size) {
                   // no matching child, remainder of suf becomes new node.
                   n2 = nodes.size
                   nodes.add(Node().apply { sub = suf.substring(i) } )
                   children.add(n2)
                   return
               }
               n2 = children[x2]
               if (nodes[n2].sub[0] == b) break
               x2++
           }
           // find prefix of remaining suffix in common with child
           val sub2 = nodes[n2].sub
           var j = 0
           while (j < sub2.length) {
               if (suf[i + j] != sub2[j]) {
                   // split n2
                   val n3 = n2
                   // new node for the part in common
                   n2 = nodes.size
                   nodes.add(Node().apply {
                       sub = sub2.substring(0, j)
                       ch.add(n3)
                   })
                   nodes[n3].sub = sub2.substring(j)  // old node loses the part in common
                   nodes[n].ch[x2] = n2
                   break  // continue down the tree
               }
               j++
           }
           i += j  // advance past part in common
           n = n2  // continue down the tree
       }
   }

   fun visualize() {
       if (nodes.isEmpty()) {
           println("<empty>")
           return
       }

       fun f(n: Int, pre: String) {
           val children = nodes[n].ch
           if (children.isEmpty()) {
               println("╴ ${nodes[n].sub}")
               return
           }
           println("┐ ${nodes[n].sub}")
           for (c in children.dropLast(1)) {
               print(pre + "├─")
               f(c, pre + "│ ")
           }
           print(pre + "└─")
           f(children.last(), pre + "  ")
       }

       f(0, "")
   }

}

fun main(args: Array<String>) { 

   SuffixTree("banana$").visualize()

}</lang>

Output:
┐ 
├─╴ banana$
├─┐ a
│ ├─┐ na
│ │ ├─╴ na$
│ │ └─╴ $
│ └─╴ $
├─┐ na
│ ├─╴ na$
│ └─╴ $
└─╴ $

Perl

Translation of: Raku

<lang Perl>use strict; use warnings; use Data::Dumper;

sub classify { 

   my $h = {};
   for (@_) { push @{$h->{substr($_,0,1)}}, $_ }
   return $h;

} sub suffixes { 

   my $str = shift;
   map { substr $str, $_ } 0 .. length($str) - 1;

} sub suffix_tree { 

   return +{} if @_ == 0;
   return +{ $_[0] => +{} } if @_ == 1;
   my $h = {};
   my $classif = classify @_;
   for my $key (keys %$classif) {
       my $subtree = suffix_tree(
           map { substr $_, 1 } @{$classif->{$key}}
       );
       my @subkeys = keys %$subtree;
       if (@subkeys == 1) {
           my ($subkey) = @subkeys;
           $h->{"$key$subkey"} = $subtree->{$subkey};
       } else { $h->{$key} = $subtree }
   }
   return $h;

} print +Dumper suffix_tree suffixes 'banana$';</lang>

Output:
$VAR1 = {
          '$' => {},
          'a' => {
                   '$' => {},
                   'na' => {
                             'na$' => {},
                             '$' => {}
                           }
                 },
          'banana$' => {},
          'na' => {
                    'na$' => {},
                    '$' => {}
                  }
        };

Phix

Translation of: D

<lang Phix>-- tree nodes are simply {string substr, sequence children_idx} enum SUB=1, CHILDREN=2

function addSuffix(sequence t, string suffix) 

   int n = 1, i = 1
   while i<=length(suffix) do
       integer ch = suffix[i], x2 = 1, n2
       while (true) do
           sequence children = t[n][CHILDREN]
           if x2>length(children) then
               -- no matching child, remainder of suffix becomes new node.
               t = append(t,{suffix[i..$],{}})
               t[n][CHILDREN] &= length(t)
               return t
           end if
           n2 = children[x2]
           if t[n2][SUB][1]==ch then exit end if
           x2 += 1
       end while
       -- find prefix of remaining suffix in common with child
       string prefix = t[n2][SUB]
       int j = 0
       while j<length(prefix) do
           if suffix[i+j]!=prefix[j+1] then
               -- split n2: new node for the part in common
               t = append(t,{prefix[1..j],{n2}})
               -- old node loses the part in common
               t[n2][SUB] = prefix[j+1..$]
               -- and child idx moves to newly created node
               n2 = length(t)
               t[n][CHILDREN][x2] = n2
               exit    -- continue down the tree
           end if
           j += 1
       end while
       i += j  -- advance past part in common
       n = n2  -- continue down the tree
   end while
   return t

end function

function SuffixTree(string s) 

   sequence t = {{"",{}}}
   for i=1 to length(s) do
       t = addSuffix(t,s[i..$])
   end for
   return t

end function

procedure visualize(sequence t, integer n=1, string pre="") 

   if length(t)=0 then
       printf(1,"<empty>\n");
       return;
   end if
   sequence children = t[n][CHILDREN]
   if length(children)=0 then
       printf(1,"- %s\n", {t[n][SUB]})
       return
   end if
   printf(1,"+ %s\n", {t[n][SUB]})
   integer l = length(children)
   for i=1 to l do
       puts(1,pre&"+-")
       visualize(t,children[i],pre&iff(i=l?"  ":"| "))
   end for

end procedure

sequence t = SuffixTree("banana$") visualize(t)</lang>

Output:
+
+-- banana$
+-+ a
| +-+ na
| | +-- na$
| | +-- $
| +-- $
+-+ na
| +-- na$
| +-- $
+-- $

Python

Translation of: D

<lang python>class Node: 

   def __init__(self, sub="", children=[]):
       self.sub = sub
       self.ch = children

class SuffixTree: 

   def __init__(self, str):
       self.nodes = [Node()]
       for i in range(len(str)):
           self.addSuffix(str[i:])

   def addSuffix(self, suf):
       n = 0
       i = 0
       while i < len(suf):
           b = suf[i]
           x2 = 0
           while True:
               children = self.nodes[n].ch
               if x2 == len(children):
                   # no matching child, remainder of suf becomes new node
                   n2 = len(self.nodes)
                   self.nodes.append(Node(suf[i:], []))
                   self.nodes[n].ch.append(n2)
                   return
               n2 = children[x2]
               if self.nodes[n2].sub[0] == b:
                   break
               x2 = x2 + 1

           # find prefix of remaining suffix in common with child
           sub2 = self.nodes[n2].sub
           j = 0
           while j < len(sub2):
               if suf[i + j] != sub2[j]:
                   # split n2
                   n3 = n2
                   # new node for the part in common
                   n2 = len(self.nodes)
                   self.nodes.append(Node(sub2[:j], [n3]))
                   self.nodes[n3].sub = sub2[j:] # old node loses the part in common
                   self.nodes[n].ch[x2] = n2
                   break # continue down the tree
               j = j + 1
           i = i + j   # advance past part in common
           n = n2      # continue down the tree

   def visualize(self):
       if len(self.nodes) == 0:
           print "<empty>"
           return

       def f(n, pre):
           children = self.nodes[n].ch
           if len(children) == 0:
               print "--", self.nodes[n].sub
               return
           print "+-", self.nodes[n].sub
           for c in children[:-1]:
               print pre, "+-",
               f(c, pre + " | ")
           print pre, "+-",
           f(children[-1], pre + "  ")

       f(0, "")

SuffixTree("banana$").visualize()</lang>

Output:
+-
 +- -- banana$
 +- +- a
 |  +- +- na
 |  |  +- -- na$
 |  |  +- -- $
 |  +- -- $
 +- +- na
 |  +- -- na$
 |  +- -- $
 +- -- $

Racket

See Suffix trees with Ukkonen’s algorithm by Danny Yoo for more information on how to use suffix trees in Racket.

<lang racket>#lang racket (require (planet dyoo/suffixtree)) (define tree (make-tree)) (tree-add! tree (string->label "banana$"))

(define (show-node nd dpth) 

 (define children (node-children nd))
 (printf "~a~a ~a~%" (match dpth
                       [(regexp #px"(.*) $" (list _ d)) (string-append d "`")]
                       [else else]) (if (null? children) "--" "-+") (label->string (node-up-label nd)))
 (let l ((children children))
   (match children
     ((list) (void))
     ((list c) (show-node c (string-append dpth "  ")))
     ((list c ct ...) (show-node c (string-append dpth " |")) (l ct)))))

(show-node (tree-root tree) "")</lang>

Output:
-+ 
 |-- $
 |-+ a
 | |-- $
 | `-+ na
 |   |-- $
 |   `-- na$
 |-+ na
 | |-- $
 | `-- na$
 `-- banana$

Raku

(formerly Perl 6)

Works with: Rakudo version 2018.04

Here is quite a naive algorithm, probably .

The display code is a variant of the visualize a tree task code.

<lang perl6>multi suffix-tree(Str $str) { suffix-tree flat map &flip, [\~] $str.flip.comb } multi suffix-tree(@a) { 

   hash
   @a == 0 ?? () !!
   @a == 1 ?? ( @a[0] => [] ) !!
   gather for @a.classify(*.substr(0, 1)) {
       my $subtree = suffix-tree(grep *.chars, map *.substr(1), .value[]);
       if $subtree == 1 {
           my $pair = $subtree.pick;
           take .key ~ $pair.key => $pair.value;
       } else {
           take .key => $subtree;
       }
   }

}

my $tree = root => suffix-tree 'banana$';

.say for visualize-tree $tree, *.key, *.value.List;

sub visualize-tree($tree, &label, &children, 

                  :$indent = ,
                  :@mid = ('├─', '│ '),
                  :@end = ('└─', '  '),

) { 

   sub visit($node, *@pre) {
       gather {
           take @pre[0] ~ $node.&label;
           my @children = sort $node.&children;
           my $end = @children.end;
           for @children.kv -> $_, $child {
               when $end { take visit($child, (@pre[1] X~ @end)) }
               default   { take visit($child, (@pre[1] X~ @mid)) }
           }
       }
   }
   flat visit($tree, $indent xx 2);

}</lang>

Output:
root
├─$
├─a
│ ├─$
│ └─na
│   ├─$
│   └─na$
├─banana$
└─na
  ├─$
  └─na$

Sidef

Translation of: Raku

<lang ruby>func suffix_tree(Str t) { 

   suffix_tree(^t.len -> map { t.substr(_) })

}

func suffix_tree(a {.len == 1}) { 

   Hash(a[0] => nil)

}

func suffix_tree(Arr a) { 

   var h = Hash()
   for k,v in (a.group_by { .char(0) }) {
       var subtree = suffix_tree(v.map { .substr(1) })
       var subkeys = subtree.keys
       if (subkeys.len == 1) {
           var subk = subkeys[0]
           h{k + subk} = subtree{subk}
       }
       else {
           h{k} = subtree
       }
   }
   return h

}

say suffix_tree('banana$')</lang>

Output:
Hash(
    "$" => nil,
    "a" => Hash(
        "$" => nil,
        "na" => Hash(
            "$" => nil,
            "na$" => nil
        )
    ),
    "banana$" => nil,
    "na" => Hash(
        "$" => nil,
        "na$" => nil
    )
)
Retrieved from "https://rosettacode.org/wiki/Suffix_tree?oldid=231290"