The more specific the better i really need help!!!
2006-10-30 08:16:08 · 5 answers · asked by Greg 1 in Science & Mathematics ➔ Mathematics
Standart Answers:
- Maximal Chain by Counting Method -
By opening n links we get n+1 parts (plus the n open links).
There are 2^(n+1) possible subsets of this parts times
n+1 ways to add open links which gives the total of
(n+1) 2^(n+1) possebilities (including zero). This can
be reached by a chain of length (n+1) 2^(n+1) - 1
which is dissected into parts of length (n+1) 2^k with 0<=k<=n.
Non Standart Answers:
Assume that a chain of length k for every 1<=k<=length(chain)
must be doable with the open links. Then you can reach
links length dissection
0 1 1
1 5 1-(1)-3
2 13 1-(1)-3-(1)-7
3 29 1-(1)-3-(1)-7-(1)-15
4 61 1-(1)-3-(1)-7-(1)-15-(1)-31
n 2^(n+2)-3
2006-10-30 08:29:48 · answer #1 · answered by Clayton A 2 · 0⤊ 0⤋
Make the first cut at link 32. And then make the 2nd and 3rd cuts at the link 16 of the two remaining changes.
2006-10-30 16:38:20 · answer #2 · answered by Chris J 6 · 0⤊ 0⤋
cut anywhere you feel like. whats your question?
2006-10-30 16:19:24 · answer #3 · answered by Anonymous · 0⤊ 0⤋
What is your question?
2006-10-30 16:19:00 · answer #4 · answered by vanman8u 5 · 0⤊ 0⤋
Your question please
2006-10-30 16:26:33 · answer #5 · answered by prodigy 1 · 0⤊ 0⤋