Solution for Section Worksheet 4, #7b & #7c Section Worksheet 4 #3a & #3b (same problem)
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Solution for • Section Worksheet 4, #7b & #7c • Section Worksheet 4 #3a & #3b (same problem)
description
Solution for Section Worksheet 4, #7b & #7c Section Worksheet 4 #3a & #3b (same problem). M=3, L=3. Insert 12. 12. M=3, L=3. Insert 24. 12. 24. M=3, L=3. Insert 36. 12. 24. 36. M=3, L=3. Insert 17. 24. 12. 12. 24. 12. 24. Create root. split. 17. 17. 36. 17. 36. 24. - PowerPoint PPT Presentation
Transcript of Solution for Section Worksheet 4, #7b & #7c Section Worksheet 4 #3a & #3b (same problem)
Solution for • Section Worksheet 4, #7b &
#7c• Section Worksheet 4 #3a &
#3b (same problem)
M=3, L=3Insert 12
12
M=3, L=3Insert 24
12
24
M=3, L=3Insert 36
12
24
36
M=3, L=3Insert 17
12
17
24
36
36 overflows!
split
12
17
24
36
Old node needs to have ceiling( (L+1)/2 ) = 2 items, new node needs to have floor( (L+1)/2 ) = 2 items.
Create root
12
17
24
36
24
First key in new root is the smallest item of right child
M=3, L=3Insert 18
12
17
18
24
36
24
M=3, L=3Insert 5
18
18 overflows!
split
Old node needs to have ceiling( (L+1)/2 ) = 2 items, new node needs to have floor( (L+1)/2 ) = 2 items. Then fix pointer and internal node keys.
5
12
17
24
36
24
175
12
24
36
17 24
18
M=3, L=3Insert 22
175
12
24
36
17 24
18
22
M=3, L=3Insert 20 (step 1)
22
22 overflows!
Split internal node175
12
24
36
17 24
18
20
After splitting (17,18,20,22) into (17,18) & (20,22) we have overflow in the internal (parent) node
175
12
24
36
17 20
18
20
22
Overflow!
M=3, L=3Insert 20 (step 2)
Create root
Old internal node needs to have ceiling( (M+1)/2 ) = 2 pointers, new node needs to have floor( (M+1)/2 ) = 2 pointers. This means they each have 1 key, since each key has a left and a right pointer.
175
12
24
36
17
18
20
22
24
175
12
24
36
17
18
20
22
24
20
Root’s 1st key should be the smallest item in its right child, so smallest item of (20,22,24,36) is 20.
M=3, L=3Delete 17 (step 1)
Delete 17
175
12
24
36
17
18
20
22
24
20
Original Tree
175
12
24
36
17
20
22
24
20
Underflow, because leaf now has ceiling(L/2) - 1 items in leaf node.
M=3, L=3Delete 17 (step 2)
merge
If (17)’s neighbor, (5,12), has enough items such that it won’t underflow, we could adopt 12, but that would cause underflow, so need to merge
175
12
24
36
17
20
22
24
20
Merging leaf nodes caused underflow in parent internal node, which now only ceiling(M/2)-1 pointers (i.e. only 1 pointer). So need to fix!
5
12
17
24
36
20
22
24
20
M=3, L=3Delete 17 (step 3)
merge
Internal node’s neighbor, (24), does not have enough pointers such that it will not underflow if we adopt one, so we must merge these two internal nodes.
But now, the root is underflowed (only 1 pointer), so…
5
12
17
24
36
20
22
24
20
5
12
17
24
36
20 24
20
22
M=3, L=3Delete 17 (step 4)
….we can just delete the root.
5
12
17
24
36
20 24
20
22
M=3, L=3Delete 12
No underflow, so we’re done.
5
17
24
36
20 24
20
22
M=3, L=3Delete 22
Deleting 22 causes underflow in middle child. We could adopt one of neighbor’s children to fix underflow, except then, neighbor would underflow, so we need to merge.
5
17
24
36
20 24
20merge
After merging, fix the keys in the internal node.
5
17
20
24
36
24
M=3, L=3Delete 5
Deleting 5 doesn’t cause underflow, so we’re fine.
17
20
24
36
24
M=3, L=3Delete 36 (step 1)
Deleting 36 causes an underflow, i.e. leaf now has ceiling(L/2)-1 items. Adopting from neighbor (17,20) would cause underflow in the neighbor, so need to merge.
17
20
24
24
merge 17
20
24
Now we have an underflow of only 1 pointer in the parent node, so we need to fix it…
M=3, L=3Delete 36 (step 2)
17
20
24
…so we delete it, and we’re done.
