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comp.lang.ruby

Re: [QUIZ] Editing Text (#145

James Koppel

11/5/2007 10:25:00 PM

(I thought I had sent in my solution yesterday, but it didn't send, since I didn't realize that Yahoo now sometimes demands a CAPTCHA.)


I've been busy these past few weeks, and thus didn't get around to this quiz until yesterday (Hurrah for the extension!). Even then, I only implemented the minimum (save for the inspect/to_s method to make testing easier, which doesn't count).

My solution uses a doubly-linked list of lines. Each line is guaranteed to be terminated by a newline; each newline is guaranteed to terminate a line. This means that strings that don't end in a newline will have one added.

The cursor should be thought to lie between two characters rather than on a character. Because of this, insertign before and after is the same. The cursor is represented by the node containing the line that the cursor is on, and the index of the character to the right of the cursor.

All insertions and deletions should be O(n), where n is the length of the line. Moving the cursor should be O(1).

class ListNode
attr_accessor :prev,:nxt,:val

def initialize(prev=nil,nxt=nil,val="")
@prev,@nxt,@val = prev,nxt,val
end

def to_a
first = self
first = first.prev while first.prev

arr = []
node = first
while node.nxt
arr << node
node = node.nxt
end
arr << node
arr
end
end


###Partitions string into a linked list of lines
###For all lines, the last character is guaranteed to be a newline
###That includes the last, for simplicity's sake
###
###The cursor is between characters rather than on a character. Thus, inserting after
###is the same as inserting before
class LinkedListBuffer
def initialize(str="")
str << "\n" unless str[-1,1] == "\n"
list = str.scan(/[^\n]*?\n/).map{|s| ListNode.new(nil,nil,s)}
list.each_with_index do |node, idx|
node.prev = list[idx-1] unless idx == 0
node.nxt = list[idx+1]
end
@cur_line = list.first
@line_idx = 0
end

def at_end?
!@cur_line.nxt and @line_idx == @cur_line.val.length-1
end

def at_begin?
!@cur_line.prev and @line_idx == 0
end

##The construct ""<<ch is used so that ch may be either a string or an integer

def insert(ch)
unless "\n" == "" << ch
@cur_line.val[@line_idx,1] = ""<<ch<<@cur_line.val[@line_idx]
else
@cur_line.val[@line_idx,1] = ""<<ch<<@cur_line.val[@line_idx]
line = ListNode.new(@cur_line,@cur_line.nxt,@cur_line.val.slice!((@line_idx+1)..-1))
@cur_line.nxt.prev = line if @cur_line.nxt
@cur_line.nxt = line
end
end

def del_after
unless @cur_line.val[@line_idx] == ?\n or at_end?
@cur_line.val.slice!(@line_idx,1)
else
if at_end?
nil
else
@cur_line.val[-1,1] = @cur_line.nxt.val
@cur_line.nxt = @cur_line.nxt.nxt
"\n"
end
end
end

def del_before
unless at_begin?
cursor_left
del_after
else
nil
end
end

def cursor_left
if at_begin?
nil
elsif @line_idx == 0
@cur_line = @cur_line.prev
@line_idx = @cur_line.val.length-1
"\n"
else
@line_idx -= 1
@cur_line.val[@line_idx,1]
end
end

def cursor_right
if at_end?
nil
elsif @line_idx + 1 == @cur_line.val.length
@line_idx = 0
@cur_line = @cur_line.nxt
"\n"
else
@line_idx += 1
@cur_line.val[@line_idx-1,1]
end
end

def cursor_up
if @cur_line.prev
@cur_line = @cur_line.prev
@line_idx = [@line_idx,@cur_line.val.length-1].min
else
nil
end
end

def cursor_down
if @cur_line.nxt
@cur_line = @cur_line.nxt
@line_idx = [@line_idx,@cur_line.val.length-1].min
else
nil
end
end

def to_s
@cur_line.val = @cur_line.val[0...@line_idx]+"|"+@cur_line.val[@line_idx..-1]
str = @cur_line.to_a.map{|node| node.val}.join
@cur_line.val.slice!(@line_idx)
str
end
alias inspect to_s
end


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1 Answer

Eric Mahurin

11/6/2007 8:14:00 PM

0

On 11/5/07, James Koppel <jamesbkoppel@yahoo.com> wrote:
> My solution uses a doubly-linked list of lines. Each line is guaranteed to be terminated by a newline; each newline is guaranteed to terminate a line. This means that strings that don't end in a newline will have one added.

I'm glad to see a linked-list solution. Another approach to use is to
leave the newlines out of the "lines". Newlines would be implied to
be between them (no implied newline after the last line).

> The cursor should be thought to lie between two characters rather than on a character. Because of this, inserting before and after is the same. The cursor is represented by the node containing the line that the cursor is on, and the index of the character to the right of the cursor.

Actually, they are different. Both insert the character at the same
point in the text. The difference is the final position of the
cursor/caret. insert_after is not a normal editor operation though.
It is equivalent to typing right-to-left instead of left-to-right
(what insert_before does). Since insert_after isn't a common
operation, and it can be done using insert_before followed by left, it
isn't needed. The main reason I threw it in was symmetry.

> All insertions and deletions should be O(n), where n is the length of the line. Moving the cursor should be O(1).

You could make it completely O(1) by using something like a gap buffer
for a line (or while editing one). But, it may not matter too much as
long as the line size is small enough (ruby overhead may dominate the
O(n) C code).

Also, the API given in the quiz was really only meant to be a
suggestion. Your implementation is probably better off taking a
string instead of a character to insert.

Eric