algorithm - Mathematical proof for a binary tree -


i not hiding part of homework i've tried enough before posting here.

so...
need prove binary tree node k have left child on 2k , right child on 2k + 1 position. i've proved induction.

now need prove binary tree node k have parent on (floor)(k/2) position. took 2 cases.
tried induction well. it's true tree of 3 nodes.
if it's true node k i'll prove node k + 1.

  1. if node k+1 shares parent node k it's true.
  2. if node k+1 has difference parent node k....

i trying make general binary tree types won't me use induction assumption. assume maybe i'll have use proved before child's position.

any help?

so you've proven kth node has children @ 2k , 2k+1. let's divide children 2 cases, , odd.

for children they're located @ i=2k k. can see that means parent @ k, or i/2, or floor(i/2).

for odd children they're located @ i=2k+1 k. can see means parent @ k. floor(i/2) in case equals floor(k+1/2), equals floor(k) = k since k integer, here parent node @ floor(i/2) also.

since set of odd , children make children, parent of ith child floor(i/2)

qed? sorry if isn't rigorous or formal enough..


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