CS473-Algorithms

1
CS473-Algorithms

• Lecture
• RBT - DELETION

2
Deletion

• SENTINELS ( Dummy Records)
• Sentinel is a technique for handling boundary
  condition in a natural way.
• So that, you dont need special cases in the
  code.
• Example Merging lists for merge-sort.
• Put the sentinel 8 at the end of the lists to be
  merged
• Just keep merging until you write an 8 to the
  answer list.
• No need,
• -To check one list running out, and
• -Specially handle moving other list to answer

3
Deletion

• In Red-Black Trees
• Use sentinels to treat a NIL child of a node x as
  ordinary node whose parent is x.
• Could add a distinct sentinel node for each NIL
  in the tree.
• Advantage Parent of each node is well-defined.
• Disadvantage Waste of space.

4
Deletion

• In Red-Black Trees
• Instead, use the unique sentinel nilT to
  represent all NILs.
• Caution when you want to manipulate a child of a
  node x, set pnilT to x first.

5

• RB-DELETE (T,z)
• if leftz nilT or rightz nilT then
• y z
• else
• y ? SUCCESSOR(z)
• if lefty ? NIL then
• x ? lefty
• else
• x ? righty
• px ? py
• if y leftpy then
• leftpy ? x
• else
• rightpy ? x
• if y ? z then
• keyz ? keyy
• if colory BLACK then
• RB-DELETE-FIXUP(T,x)
• return y

Determine a node y to splice out
Non-nil child of y nilT if y has no child
x
Splice out node y
If successor of z was spliced out
then, move contents of y into z
Property 4 is disturbed
Non-nil child of y nilT if y has no child
x
6

• RB-DELETE-FIXUP(x)
• while x rootT and colorx BLACK
• if x ? leftpx then
• w ? rightpx
• if colorw RED then
• colorw ? BLACK
• colorpx ? RED
• LEFT-ROTATE( px )
• w ? rightpx Case 1
• if colorleftw BLACK and
  colorrightw BLACK then
• colorw ? RED
• x ? px
• else if colorrightw BLACK
  then
• colorleftw ? BLACK
• colorw ? RED
• RIGHT-ROTATE(w)
• w ? rightpx
• colorw ? colorpx
• colorpx ? BLACK

Case 1
Case 2
Case 3
Case 4
7
Deletion

• When the spliced-out node y is black, its removal
  causes
• Any path previously contained y to have one fewer
  black node.
• Thus, property 4 is violated by any ancestor of
  y.
• Imagine node x having an extra black
• This extra black assumption makes property 4 to
  hold.
• Then we splice-out black node y
• We push its blackness onto its child x
• The problem that may arise now
• Node x may be doubly-black, thus violating
  property 1

8
Deletion

• Procedure DELETE-FIXUP attemps to restore
  property 1
• Goal of while-loop is to move extra black up the
  tree until
• x points to a red node.
• - Color the node x black.
• x points to the root.
• - Extra black can be simply removed.
• Suitable rotations and recolorings can be
  performed.
• Within the while-loop
• x always points to a doubly-black, non-root node.

9
Deletion

• ???? Cases to consider in the while-loop.
• ??? 4 of them are symmetric to the other 4
  depending on
• Whether x is a left or right child of its parent
  px
• Analyze only the 4 cases in which x is a left
  child.
• Maintain a pointer w to xs sibling
• Since x is doubly-black, w ? nilT
• Otherwise, property 4 was violated before the
  delete
• of blacks on the path px NIL leaf w
• Would be smaller than the number on the path
  px-x

10
Cases of the While Loop
-switch colors of w pxpw
-LEFT(px)
-Conver to case 2, 3, 4
B
D
Case 1
-Take onw black off both x of w
-Then repeat the while-loop with new x
D
Case 2
11
Cases of the While Loop
-switch colors of w leftw
-RIGHT(w)
Case 3
D
C
- Make some color changes
-LEFT(px)
E
Case 4
12
Deletion

• ???? idea In each case,
• Number of black nodes from (and including) root
  r of the subtree
• To each of the subtrees a,ß,....., d
• Is preserved by the transformation
• Case 1 Both before and after the transformation
• nb(r?a) nb(r?ß) 3
• nb(r??) nb(r?d) nb(r?e) nb(r? ?) 2
• Case 2 Both before after the transformation
• nb(r?a) nb(r?ß) 2 c
• nb(r??) nb(r?d) nb(r?e) nb(r??) 1 c
• 1 if colorr black
• 0 if colorr red

Where c
13
Deletion

• Case 3 Both before and after the transformation
• nb(r?a) nb(r?ß) 2 c
• nb(r??) nb(r?d) 1 c
• nb(r?e) nb(r??) 2 c
• Case 4 Both before after the transformation
• nb(r?a) nb(r?ß) 2 c
• nb(r??) nb(r?d) 1 c c
• nb(r?e) nb(r??) 1 c

14
Deletion

• Case 1 w ( sibling of x) is red
• W must have black children and a black parent
• Switch colors of w and px pw
• Then, perform a left-rotation on px
• Without violating any R-B properties
• New sibling new w of x, one of old ws
  children, is now black
• Thus we have converted Case 1 into Case 2, 3 or 4
• Case 2, 3 or 4 w is black
• These cases are distibuished by colors of ws
  children
• Both of ws children are black (Case 2)
• ws left child is red, right child is black (Case
  3)
• ws right child is red.

15
Deletion

• Case 2 w is black leftw rightw are both
  black
• Take one black off both x and w leaving
• x with only one black an w red.
• Add an extra black to px
• Then repeat the while-loop with px as the new
  node x
• If we enter Case 2 thru Case 1
• Color c of new node x is red since original px
  was red.
• Thus loop terminates when it tests the loop
  condition

16
Deletion

• Case 3 w is black leftw is red rightw is
  black
• Switch colors of w and its left child leftw
• Perform a right rotation on w
• Without violating any R-B properties.
• New sibling new w of x
• Is now black node with a red right child
• Thus we have transformed Case 3 into Case 4

17
Deletion

• Case 4 w is black rightw is red
• By marking the following color changes
• Set color of ws right child to black
• Set ws color to its parents (px pw) color
• Set pxs color to black.
• and performing a left-rotation on px
• We can remove the extra black on x
• Without violating any R-B properties
• Setting x to be the root causes
• While-loop to terminate when it tests the loop
  condition.