- #1
andrewr
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I'm going to construct an ordered set, and I'd like to ask some questions about it; and in particular consider coding problems about this set and sets in general (Turing tape encoding).
Start with:
A={ }
And, allow a mutable temporary set, initialized with:
T={ }
and an iterator: set n=0
For reference, "x", is a character;
Now to construct the set:
1) T = T [itex]\bigcup[/itex] "x"
2) A = A [itex]\bigcup[/itex] T
3) n = n+1
4) Test if |T| ?= |A| and T ?[itex]\in[/itex] A
5) Goto 1.
The Algorithm at each iteration (step 4), will yield a pattern:
n=1, A = { {x} } , T ={ x }, Test=True
n=2, A = { {x}, {x,x} }, T= {x.x}, Test=True
n=3, A = { {x}, {x,x}, {x,x,x} }, T={x,x,x}, Test=True
n=4, A = { {x}, {x,x}, {x,x,x}, {x,x,x,x} }, T={x,x,x,x}, Test = True
... etc
The ordering of the set A is the order in which the elements were added algorithmically.
Now, using strong induction; I can say:
Base case: for n=1, [itex]\exists[/itex] y [itex]\in[/itex] A : |y| = |A|
For case n+1, exactly one element is added to T, and one element added to A, hence:
[itex]\exists[/itex] y [itex]\in[/itex] A : |y| = |A| BY 1:1 correspondence (it is not needed to count n items.)
This is inductively true, because of algorithmic step #4.
∴ For all steps, n, even an infinite number of steps (n+1...) -- [itex]\exists[/itex] y [itex]\in[/itex] A : |y| = |A|
Now, does anyone see a problem with what I have done? If so, what?
Start with:
A={ }
And, allow a mutable temporary set, initialized with:
T={ }
and an iterator: set n=0
For reference, "x", is a character;
Now to construct the set:
1) T = T [itex]\bigcup[/itex] "x"
2) A = A [itex]\bigcup[/itex] T
3) n = n+1
4) Test if |T| ?= |A| and T ?[itex]\in[/itex] A
5) Goto 1.
The Algorithm at each iteration (step 4), will yield a pattern:
n=1, A = { {x} } , T ={ x }, Test=True
n=2, A = { {x}, {x,x} }, T= {x.x}, Test=True
n=3, A = { {x}, {x,x}, {x,x,x} }, T={x,x,x}, Test=True
n=4, A = { {x}, {x,x}, {x,x,x}, {x,x,x,x} }, T={x,x,x,x}, Test = True
... etc
The ordering of the set A is the order in which the elements were added algorithmically.
Now, using strong induction; I can say:
Base case: for n=1, [itex]\exists[/itex] y [itex]\in[/itex] A : |y| = |A|
For case n+1, exactly one element is added to T, and one element added to A, hence:
[itex]\exists[/itex] y [itex]\in[/itex] A : |y| = |A| BY 1:1 correspondence (it is not needed to count n items.)
This is inductively true, because of algorithmic step #4.
∴ For all steps, n, even an infinite number of steps (n+1...) -- [itex]\exists[/itex] y [itex]\in[/itex] A : |y| = |A|
Now, does anyone see a problem with what I have done? If so, what?
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