# Difference between revisions of "TADM2E 3.5"

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(Undo revision 457 by Letientai299 (talk), the hypothesis is that all node is identical, so the child pointers in the leaf node count as overhead even if not connected) |
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1: Each node is identical, so the ratio of data over total should be: <br> | 1: Each node is identical, so the ratio of data over total should be: <br> | ||

− | 4 | + | <math> |

+ | \frac{Data}{Data + 3*Pointers} = \frac{4}{4 + 3*4} = \frac{1}{4} | ||

+ | </math> | ||

2: In a full tree, given n leaf nodes, there are n-1 internal nodes. Both leaf and internal nodes are worth 4 bytes: <br> | 2: In a full tree, given n leaf nodes, there are n-1 internal nodes. Both leaf and internal nodes are worth 4 bytes: <br> | ||

− | 4*n | + | <math> |

+ | \frac{\text{Size of Leafs}}{\text{Size of Leaf} + \text{Size of Internal Nodes}} = \frac{4 * n}{4*n + 4*(n-1)} = \frac{n}{2n-1} = \frac{1}{2} (n\to\infty) | ||

+ | </math> |

## Latest revision as of 05:06, 29 December 2016

1: Each node is identical, so the ratio of data over total should be:

$ \frac{Data}{Data + 3*Pointers} = \frac{4}{4 + 3*4} = \frac{1}{4} $

2: In a full tree, given n leaf nodes, there are n-1 internal nodes. Both leaf and internal nodes are worth 4 bytes:

$ \frac{\text{Size of Leafs}}{\text{Size of Leaf} + \text{Size of Internal Nodes}} = \frac{4 * n}{4*n + 4*(n-1)} = \frac{n}{2n-1} = \frac{1}{2} (n\to\infty) $