Prev: Symmetry and Abstraction
Next: how to approach proofs?
From: Risto Lankinen on 3 May 2010 04:31 Hi! Take an arbitrarily large two dimensional table and place pegs onto a finite number of its cells while observing the following rules: 1. Cell (0,0) contains a peg. 2. Cell (0,n) contains a peg if and only if Cell (n,0) contains a peg. 3. Cell (p,q) contains a peg if and only if both Cell (p,0) and Cell (0,q) contain a peg. The resultant peg configuration necessarily consists of a square (n^2) number of pegs having a symmetric pattern wrt/ the main diagonal (e.g. cells (0,0), (1,1), (2,2), , (n,n)) . Next, assign values to the cells so that Cell (p,q) == Re((i-1)^(p+q)) where Re means real part, and i is the imaginary unit. Examples of some cell values follow: Cell (0,0) == 1 Cell (0,1) == Cell (1,0) == -1 Cell (0,2) == Cell (1,1) == Cell (2,0) == 0 Cell (0,3) == Cell (1,2) == Cell (2,1) == Cell (3,0) == 2 etc the sequence continues 1,-1,0,2,-4,4,0,-8,16,-16,0,32,-64,64,0,-128,256, See also A009116 in The On-Line Encyclopedia of Integer Sequences: http://www.research.att.com/~njas/sequences/A009116 The sum of the values of all pegged cells is some integer, and the sum of the values of pegged cells on the first row is the arithmetic average of (one pair of) its integer factors (including negatives, e.g. 45 = 5*9 = -5*-9 ). Knowing N (the grand sum) and A (the first row sum) it is easy to factorize N . - - - Explain the underlying mechanism. - Risto -
From: adacrypt on 3 May 2010 06:20 On May 3, 9:31 am, Risto Lankinen <rlank...(a)gmail.com> wrote: > Hi! > > Take an arbitrarily large two dimensional table and place pegs onto a > finite number of its cells while observing the following rules: > > 1. Cell (0,0) contains a peg. > 2. Cell (0,n) contains a peg if and only if Cell (n,0) contains a peg. > 3. Cell (p,q) contains a peg if and only if both Cell (p,0) and Cell > (0,q) contain a peg. > > The resultant peg configuration necessarily consists of a square (n^2) > number of pegs having a symmetric pattern wrt/ the main diagonal (e.g. > cells (0,0), (1,1), (2,2), , (n,n)) . > > Next, assign values to the cells so that Cell (p,q) == Re((i-1)^(p+q)) > where Re means real part, and i is the imaginary unit. Examples of > some cell values follow: > > Cell (0,0) == 1 > Cell (0,1) == Cell (1,0) == -1 > Cell (0,2) == Cell (1,1) == Cell (2,0) == 0 > Cell (0,3) == Cell (1,2) == Cell (2,1) == Cell (3,0) == 2 > etc the sequence continues > 1,-1,0,2,-4,4,0,-8,16,-16,0,32,-64,64,0,-128,256, > > See also A009116 in The On-Line Encyclopedia of Integer Sequences:http://www.research.att.com/~njas/sequences/A009116 > > The sum of the values of all pegged cells is some integer, and the sum > of the values of pegged cells on the first row is the arithmetic > average of (one pair of) its integer factors (including negatives, > e.g. 45 = 5*9 = -5*-9 ). Knowing N (the grand sum) and A (the first > row sum) it is easy to factorize N . > > - - - > > Explain the underlying mechanism. > > - Risto - Hi , On first inspection this seems a bit like a Vigenere Cipher that uses a square in which the plaintext is valid for a given key if and only if the plaintext in hand for the key in hand is also the key elsewhere for the plaintext elsewhere. This means that key/plaintext combinations are pegged inseperably to each other. This convolution of the Vigenere square sounds a bit tortuous and unnecessary as an embellishment of a well known existing cipher. It could cause much difficulty of encryption without any apparent increase in security but don't let me discourage from trying something new - Good Luck with everything. - adacrypt
|
Pages: 1 Prev: Symmetry and Abstraction Next: how to approach proofs? |