Friday, May 19, 2006
number theory
its a short double bill... this is ty the mathematician in action. do i feel regretful not majoring in maths? actually to be honest, no. maths is not really scientific, and its the only part of science which i do well in. most of the time, i am more social science. but here's a mathematical formula which i thought of on my way home from work. ironic, if you ask me. i work in a civil service, in the social work lines. minimal math involved, yet the thing i came up with is math, a subject which i haven't touch for a long time.
here's the formula:
(2^0, 2^1, 2^2...., 2^(n-2), 2^(n-1)) is able to permuate all integers (2^n) - 1 without overlapping. the prove of it is by induction.
consider the case for n=0.
2^0 is able to permuate 1 without overlap
assume that k is true, for some k which is an integer number:
(2^0, 2^1, 2^2...., 2^(k-2), 2^(k-1)) permuates all natural number up to (2^k) -1
so now consider k+1,
(2^0, 2^1, 2^2...., 2^(k-2), 2^(k-1)) permuates all natural number up to (2^k) -1
all combination is k is accounted for. the only way to permutate for k+1 is to add 2^k to all the terms, starting from
0. hence we have the following:
(2^0, 2^1, 2^2...., 2^(k-2), 2^(k-1), 2^k) permuates all natural number up to (2^k) -1+ 2^k= 2^(k+1) -1
therefore if K is true, K+1 is also true. and 0 is true, therefore the proof is true by induction.
ok, to put it across in a less technical way.
imagine the following number sequence:
1=1 2=2 3= 2+1 4= 4 5=4+1 6= 4+2 7= 4+2+1 8=8
9=8+1 10=8+2 11=8+2+1 12=8+4 13=8+4+1 14= 8+4+2 15= 8+4+2+1
each time a power of 2 is reach, start afresh and work backworks. the reason why i thought of this is because i need to come up with a system to allow people of choose in a "binary" manner and yet produced unique combination at the end of it. what better way to use than to revert back to number. so i was wondering how i can use numbers to permutate different outcome such that at one look at each number, i know what is the answer for each question. for example, if i give question 1 1 for a yes, 0 for a no. question 2, 2 for a yes, 0 for no. question 3, 4 for yes and 0 for no. question 4 8 for a yes and 0 for a no. if my final sum is 13, i will know that the answer for question 1,3,4 are yes and 2 is a no.
somehow, the whole explanation seem to lack articulation. haha, after so many years, i realise the academic blood still runs in me. the strong passion to do research, to learn still flows in me. i have always maintain the fact that ri guys are pretty much geeks and i realise i am one of them, who is unable to break away. like it or not, i like to think of something new to write about, i mean just how many people have a blog like mine?
posted @ 9:41 PM ||
