Question

There is a wonderful problem set called Ninety-Nine Prolog Problems. Problem P70 is the one referred to in the title. And here is a great Prolog solution of this problem which takes only 5 lines. However, my understanding of Prolog is limited.

How does this solution look like in a C-like form (no itertools available)?

Edited by request. I hope I do not violate copyright.

The problem:

Syntax in BNF:

tree ::= letter forest '^'
forest ::= | tree forest

A nice solution using difference lists:

tree(TS,T) :- atom(TS), !, atom_chars(TS,TL), tree_d(TL-[ ],T). % (+,?)
tree(TS,T) :- nonvar(T), tree_d(TL-[ ],T), atom_chars(TS,TL).   % (?,+)
tree_d([X|F1]-T, t(X,F)) :- forest_d(F1-['^'|T],F).
forest_d(F-F,[ ]).
forest_d(F1-F3,[T|F]) :- tree_d(F1-F2,T), forest_d(F2-F3,F).

Answer 1

Problem Definition

_{(taken from P-99: Ninety-Nine Prolog Problems)}

We suppose that the nodes of a multiway tree contain single characters. In the depth-first order sequence of its nodes, a special character ^ has been inserted whenever, during the tree traversal, the move is a backtrack to the previous level.

By this rule, the tree in the figure is represented as: afg^^c^bd^e^^^

_{(source: ti.bfh.ch)}

Define the syntax of the string and write a predicate tree(String,Tree) to construct the Tree when the String is given. Work with atoms (instead of strings). Make your predicate work in both directions.

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Solution Part 1: String2Tree

This is easy with a stack. Here’s the pseudocode:

FUNCTION String2Tree(String str) : Tree
   LET st BE New-Stack<Node>
   LET root BE New-Node
   st.push(root)

   FOREACH el IN str
      IF el IS '^'
         st.pop()
      ELSE
         LET child BE New-Node(el)
         LET top BE st.top()
         top.adopt(child)
         st.push(child)

   RETURN New-Tree(root)

The use of a dummy root node simplifies matters. Essentially the algorithm is as follows:

• Scan the string left to right
• Whenever we encounter a node label, we create a new node
  • That node is adopted by the top of the stack
  • That node is then pushed to the the stack and becomes the new top
• When we encounter a '^', we simply pop off the top of the stack

---

Solution Part 2: Tree2String

The opposite direction is a matter of simple recursion:

FUNCTION string(Tree t) : String
   LET sb BE New-StringBuffer

   visit(t.root, sb)

   RETURN New-String(sb)

PROCEDURE visit(Node n, StringBuffer sb)
   sb.append(n.label)

   FOREACH child IN n.children()
      visit(child, sb)

   sb.append('^')

As specified in the problem, we insert ^ whenever we backtrack to the previous level.