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  • Alemagno12

    Here's the blog post.


    Let addR(n) = running total after the game ends if it starts with n and addL(n) = length of the string after the game ends if it starts with n.

    We start with R = 0. Then, we add n to the running total. Then, as the last number in the string decreases, the number we add to the running total decreases. So we add n-1 to the running total, then n-2, then n-3, all the way up to 0, which is when the game ends. So addR(n) = nth triangular number.

    Since only 1 character is added to the string at each step, and the starting string is 1 character long, addL(n) = n+1.

    Now, instead of addR(n) and addL(n), we have add saddR(n) and saddL(n).

    We start with n. Then, we add n-1 to the string 2 times. Then, we add n-2 to the string 2 ti…

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  • Alemagno12

    Some people have tried to define their own interpretations of BEAF beyond ε0. The problem with those interpretations is that they use X arrays instead of the array of operator, which works fine for sub-legion arrays, but makes BEAF much weaker starting from {n,n/2}.

    Using X arrays, {n,n/2} has growth rate ψ(Ωω), but using the array of operator, the growth rate of {n,m/2} depends on m:

    • {n,m/2} (m = n) has growth rate ψ(Ωω).
    • {n,m/2} (m = n+1) = n&{n,m/2} (m = n), so the m turns into an X, and {n,m/2} (m = n+1) has growth rate ψ(Ωω)+1.
    • {n,m/2} (m = n+2) = n&n&{n,m/2} (m = n), so the m turns into an X2, and {n,m/2} (m = n+1) has growth rate ψ(ΩΩ).
    • {n,m/2} (m = n+3) = n&n&n&{n,m/2} (m = n), so the m turns into an X3, and {n,m/2} (m = n+1) has growth…
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  • Alemagno12

    • Empty graph = 0
    • 1 vertex = 1
    • 2 vertices = 2
    • 3 vertices = 3
    • n vertices = n
    • 1 stick = ω
    • 1 stick + vertex = ω+1
    • 1 stick + n vertices = ω+n
    • 2 sticks = ω2
    • 2 sticks + vertex = ω2+1
    • 3 sticks = ω3
    • n sticks = ωn
    • 3-path = ω2
    • 3-path + stick = ω2
    • 2 3-paths = ω22
    • 3 3-paths = ω23
    • 4-path = ω3
    • 2 4-paths = ω32
    • 5-path = ω4
    • (n+1)-path = ωn

    === n-paths and 2

    • [WIP]
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  • Alemagno12

    Not Turing Machines

    October 10, 2017 by Alemagno12

    (Can someone change the name of the blog post? I made a typo)

    Not Turing Machines are networks of computers.

    Each computer has 3 states (the + state, the - state, and the 0 state), an infinite tape of cells (which can store any integer, which starts at 0), and a pointer. Each computer can either be active or inactive. There is also a group of computers (which can be made up of one or more computers) that are active at the start.

    When a computer is active, it can be on one of it's 3 states, depending on the integer stored by the cell that the pointer is on:

    • If the integer stored by the cell that the pointer is on is positive, the computer will be in the + state.
    • If the integer stored by the cell that the pointer is on is negative, the computer w…

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  • Alemagno12

    Ordinals in CGT

    October 7, 2017 by Alemagno12

    In Combinatorial Game Theory (CGT), we can assign values to games that satisfy the following conditions:

    1. The game is a 2-player game (the players are called Left and Right)
    2. There are many possible positions, and often a particular starting position.
    3. There are rules that specify the moves that either player can make from it's current position to it's options.
    4. Left and Right alternate turns through the whole game.
    5. Both players know what is going on. There is no missing information.
    6. There are no chance moves.
    7. If a player is unable to move in its turn, that player loses.
    8. There is always a winner.

    There are a lot of values that can be assigned to positions in those games, but for now, we will only need to know two values:

    • A position X has value 0 if whoe…
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