Nico Benschop wrote:
> Henry wrote:
> > While reading Paul Erdos' story in "My brain is open" I found
> > Cantor's diagonal argument which he used to 'prove' that the
> > decimals between 0 and 1 are uncountable. [...]
> [...]
> His diagonal argument to show this is very simple, and has little
> to do with reals on [0,1) which is special interpretation, but
> everything with strings over some alphabet A of at least 2 symbols.
> In fact he shows that whatever list of (binary) strings of lentgh k
> you take (for finite k there are 2^k): any square k x k table of
> such strings has a diagonal that is a string (length k) which is
> not in that listing if you 'swap bitwise' all bits in the diagonal.
> For k=6 there are 2^6= 64 binary strings, and take any selection of 6:
> 0 1 1 1 0 0 Diagonal: 0 1 0 0 0 1 (left top to right bottom)
> 1 1 0 1 0 1 swapped: 1 0 1 1 1 0 is not a row in this 6 x 6 table,
> 0 1 0 1 0 1 because it differs in each position
> 1 0 1 0 1 0 from each row, at that position!
> 0 0 0 0 0 0
> 1 1 1 0 1 1
> This holds for finite k, no matter how large, and in fact:
> -------- _independent_ of length k ------
> hence (by some axiom) also for infinite k,
> called omega w, the countable ('linear') infinite of Peano's naturals,
> or the integers, including the negative, which also are countable.
Rather than just *one* extra diagonal, I just wondered for the
finite case of order k: is there a minimum number (maybe < k ?) of
such k x k binary tables which, by permuting rows of each table,
generate as CD's *all* 2^k binary strings of length k ?
There are k! > 2^k (k>3) permutations, but not all generate different
diagonals. A table like the next (k=6) seems to be very 'prolific':
\
0 0 0 0 0 1 Note: a diagonal entry is not changed if its row
0 0 0 0 1 1 is replaced by a row with the same entry
0 0 0 1 1 1 in that column. E.g: pair swapping involes
0 0 1 1 1 1 4 entries: in rows i,j and cols i,j -
0 1 1 1 1 1 producing a new CD only if the row distance
1 1 1 1 1 1 (symm.difference) \neq 0 there.
\
Or equivalently: its symmetric image (while also columns may be used):
\
1. 1 0 0 0 0 0 If 'weight' w(i) is the number of 1's in row i,
2. 1 1 0 0 0 0 and R(w) the set of all k_strings of weight w,
3. 1 1 1 0 0 0 then symmetric analysis would require to
4. 1 1 1 1 0 0 generate all Ranks R(w) for w=0..k
5. 1 1 1 1 1 0 E.g. swap rows (2,3): CD= 0 0 1 0 0 0
6. 1 1 1 1 1 1 or if i<j swap (i,j): CD= 1 in place j, rest 0
\ So swaps yield ranks R(1), and also R(k-1) by
taking -CD (bitwise inverse CD).
Q: Does an m_cycle permution yield CD \in R(m-1) ?
\
Like (1,2,3) --> 1 1 1 0 ... CD= 0 1 1 0 0 0
1 0 0 0 ...
1 1 0 0 ...
1 1 1 1 0 .
\ ...etc
