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Ikosarakt1

aka Hypercube 20.

Admin Chat moderator
  • I live in the Observable Universe
  • My occupation is Googologist
  • I am Male
  • Ikosarakt1

    Small Turing machines.

    August 28, 2015 by Ikosarakt1

    Let's say that TM computes f(n) = m if for input = n consecutive 1's it outputs m 1's (may not be consecutive). Head is placed on the leftmost 1. For f(0), let's say that input is empty.

    Next I'll post some TMs found by the program. If anyone interested, you can check them at simulator.


    f(n) = n*2 for (n>=0):

    0 _ _ r halt 0 1 1 l 1 1 _ 1 r 2 1 1 _ l 1 2 _ 1 r 2 2 1 1 r 0

    f(0) = 1

    f(n) = n*2 for (n>0)

    0 _ 1 r halt 0 1 1 r 1 1 _ 1 r 0 1 1 _ l 2 2 _ 1 r 2 2 1 1 l 1

    f(n) = n*2 for (n>0, doesn't halt at n=0)

    0 _ 1 r 0 0 1 1 r 2 1 _ 1 r 0 1 1 _ l 1 2 _ 1 r halt 2 1 1 l 1

    f(n) = n*3 for (n>=0)

    0 _ _ l halt 0 1 1 l 1 1 _ _ l 2 1 1 _ l 1 2 _ 1 r 2 2 1 1 r 0 Read more >
  • Ikosarakt1

    My site

    February 13, 2015 by Ikosarakt1

    Today I've created my googological site here. It has almost no content for now, more detailed pages will be added later.

    Read more >
  • Ikosarakt1

    Algorithm for n(k)

    January 30, 2015 by Ikosarakt1

    I've decided to explore Friedman's stuff better. I should start with n(k) because it's slowest widely known function defined by him.

    Let's define the function L(n,m) as the number of Friedman strings which have n letters from alphabet {1,2,...,m}.

    For example, let's compute L(4,2). First, we write all strings with n letters from alphabet {1,2,...,m}:

    1111

    1112

    1121

    1122

    1211

    1212

    1221

    1222

    2111

    2112

    2121

    2122

    2211

    2212

    2221

    2222

    Then we remove strings written in bold text from this list because they're not Friedman. It contains 8 strings and so L(4,2) = 8.

    Now we can redefine n(k) as the largest p so that L(p,k) > 0. The problem drops to explore the behavior of L(p,k) function.

    For k = 2, we have the following sequence:

    L(1,2) = 2

    L(2,2) = 4

    L(3,2) = 8

    L(4,2) = 8

    L…

    Read more >
  • Ikosarakt1

    Something for BEAF

    January 19, 2015 by Ikosarakt1

    It's known that b^n & a is defined by Bowers, but he wrote nothing about b*n & a or b+n & a. I'm inspired by the idea to make & operator universal, so that it can take any function f(n) and return its corresponding array, f(n) & n.


    First of all, we can't say that the expression to the left from & is a number, because there's a math axiom which says:

    and do step 2.

    For example:

    (2+2)*(2+2)*(2+2) & 3 has level 1.

    ((2+2)*(2+2)*(2+2)+2)*((2+2)*(2+2)*(2+2)+2) has level 2

    (((2+1)*2+1)*2+1)*2+1 has level 3.

    ((((2+1)*2+1)*2+1)*2+1)*2+1 has level 4.

    I think we can use levels to determine which separator should be placed instead of [?].

    Read more >
  • Ikosarakt1

    Let me publish some comparisons which show that actual limit of sub-legion BEAF is

    Read more >

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