B tree B+ tree

Question

Comments

51

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Could anyone explain the meaning of 500k records of 200 bytes each,here why he is mentioning such things which has no use in solving the problem.

Answer 1

let order of tree be p.

case 1 : for internal node : 

size of key * #keys + tree ptr * #child pointer <=1024

15(p-1) +5p<=1024

20p<=1039

p<=51.98

p=51

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case 2 :for leaf node

(key pointer +data pointer)(p-1)+ tree pointer <=1024

20(p-1)+5<=1024

20p<=1029

p<=51.98

p=51

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so in both cases we get 51 so order of tree is 51.

Comments

case 1 : for internal node : let p be the order i.e no. of child pointers.

size of key * #keys + tree ptr * #child pointer <=1024

15(p-1) +5p<=1024

20p<=1039

p<=51.98

p=51

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case 2 :for leaf node let p be the order i.e. no. of keys in the leaf node

(key +data pointer)p+ tree pointer <=1024

20p+5<=1024

20p<=1019

p<=50.95

p=50

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so for internal node order is 51 and for leaf node order is 50.

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for leaf node let p be the order i.e. no. of keys in the leaf node

but order = #keys+1...

why u have considered different orders... for internal node and leaf node .

one as #pointer other as #keys. 

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I calculated order for internal node as 50 and leaf node as 51.  what will be the answer to this question 50 or 51?