User blog:GamesFan2000/Redefined HNEAN (Part 1)

Redefined Hypernova Exploding Array Notation (Part 1)

I made a series of posts on my blog about an array notation. Well, I think I need to redefine a few things about it. This is the first part of the redux of Hypernova Exploding Array Notation, or HNEAN.

Part 1: What is Array Notation?

Array notation is a form of mathematical expressions where the numbers are entries in a set, and each entry has some form of effect on the outcome of the expression. The most well-known version of this was invented by Jonathan Bowers and would be expanded upon by Bowers and fellow googologist Chris Bird. Array notations are often represented by {a, b, c, …}.

Part 2: How does regular array notation work?

The smaller arrays are very easy to understand in normal array notation. {a} = a, {a, b} = a^b, and {a, b, c} = a^^^…c arrows…^^^b. The point where it always loses me is at four entries. {a, b, c, d} is usually either equal to or stronger than chained arrow notation. I don’t even bother with what happens beyond it.

Part 3: How my notation works

So, how does my array notation work? Here’s how we define it:

Define b as our first entry, p as our second entry, n as our first non-one entry after p, and o as our last entry before n. Define e as the number of entries in the array.

Rule 1: If b = 1, the array is default, which means that the answer is 1.

Rule 2: If p = 1, the array defaults to a one-entry array.

Rule 3: If n is the third entry, all ones after it are removed from the array.

Rule 4: If n isn’t the third entry, or this rule isn’t applied, then start the process. It goes as follows:

Step 1: Find n in the array. If n is the third entry, skip straight to step 4.

Step 2: If n isn’t the third entry, make every entry before it equal to b, unless p = 1. n will now become the original array. (Note: Rule 3 still applies, so remove all 1’s after n from the array before this.)

Step 3: In the array where n used to be, remove the original o from the array, and repeat steps 2 and 3 until o doesn’t exist in the arrays.

Step 4: Start with the last two entries. You will be doing an operation where the second-to-last entry is affected by the last entry. The operation level is e+1, where 1 refers to addition, 2 refers to multiplication, and so on and so forth. Call the second-to-last entry s and the last entry l. Create an l-length set of s’s where you place the operator needed in-between each s. Solve this, reduce l by 1, and replace s with the new answer. Repeat until l = 1, remove it from the array, and repeat this process until one entry remains.

Step 5: Now that we have only one entry, you will create a b-number of layers. The operation level of the last layer is equal to b+2, and each preceding layer has a lower level than the succeeding layer by 1. As such, the first layer is exponentiation. Create a b-high power tower of b’s. The answer of that is used in the second layer. We’ll call said answer a. Create an a-long set of tetrations of a. Move onto the next layer with the answer of the second layer. Repeat this process until you solve the last layer.

Part 4: The Rules in Action: Single-Entry and Two-Entry

Now that we’ve defined the rules of this notation, let’s see them in action.

The first case we’ll look at is {2}. We skip straight to step 7 and create two layers. In the first layer, we have 2^2, which equals four. And then… you create 4^^4^^4^^4 in the second layer. Yep, that’s extremely big. And remember that we solve from right to left in Knuth up-arrow notation. Since the answer to 4^^4^^4^^4 is the answer to the array, {2} = 4^^4^^4^^4. Next, we’ll look at {3}. This is where Step 5 really shows its strength. The first layer is 3^3^3, which is something in the realms of 7.6 trillion. Yeah, you see where this is going. A 3^27 long set of tetrations of 3^27. And the answer of that goes into the final layer, which uses pentation.

Now we’ll look at {3, 3}. We go to step 6. We create a three-high power tower of threes, or 3^3^3…uh oh. It’s that expression that creates that 7.6 trillion result. OH WAIT, WE’RE NOT DONE YET! We now have to exponentiate that 7.6 trillion by itself because p now equals two. Then we can begin step 7. Having to go through a (3^27)^(3^27)-high power tower of (3^27)^(3^27)’s. Just to get to layer two. And there’s (3^27)^(3^27) layers.

An extreme example of two-entry arrays is {10, 10}. Yeah, good luck. We start of by solving a ten-high power tower of tens. Just 10^10 is ten billion. After you figure out the answer to that, you move on to a nine-high tower of the answer to the first tower, an eight-high tower of the answer to the second tower… The number of layers you build from the answer of the last tower is absolutely monumental.

Part 5: Three-Entry Arrays

Now we’ve hit the point where all hope is lost for the universe… three entries in an array.

The first case is {2, 2, 2}. If you’re smart, you know that 2(any normal operation)2 is 4. So, we do 2^^2 = 4 and move on. A four-high tower of 2’s is 16, a three-high tower of 16’s can’t be expressed, and the answer of the second tower to the power of itself should be banned. I’m not even going to attempt to go into the layers from this point onwards.

{3, 3, 3} is our next case. We start with 3^^3^^3, or 3^^3^27. Then we do (3^^3^27)^^(3^^3^27). The answer of (3^^3^27)^^(3^^3^27) is the new total for p. Yep, this number would take forever for even a computer to reduce.

Of course, I’ll mention {10, 10, 10} for a split second, but I’ll leave the solving to you.

Part 6: Four-Entry Arrays

Now we get to the complicated stuff. Four-entry arrays can be very painful.

Here’s a simple one to start: {3, 3, 3, 3}. We start by solving 3^^^3^^^3, then pentating the answer of that to itself. O_O is all I can use to express my reaction.

The interesting stuff happens when arrays like {3, 3, 1, 3} occur. Steps 2 and 3 apply here. The array becomes {3, 3, 3, {3, 3, 3}}. These arrays are much larger than those that have n as the third entry. You solve the ‘contained’ array before doing anything else, then you solve the main array. By virtue of the rule set, {10, 10, 1, 10} > {10, 10, 10, 10}.

Part 5: Five Entries and Beyond

Now for the five-entry arrays, and larger.

{3, 3, 3, 3, 3} has you start with 3^^^^3^^^^3, then the answer of that hexated to itself, then so on and so forth.

{3, 3, 1, 3, 3} becomes {3, 3, 3, {3, 3, 3, 3}, 3}. Again, contained arrays are solved before anything else.

{3, 3, 1, 1, 3} becomes {3, 3, 3, 3, {3, 3, 3, {3, 3, 3}}}.

{3, 3, 3, 1, 3} defaults to {3, 3, 3, 3}.

{3, 3, 1, 1, 1, 3} becomes {3, 3, 3, 3, 3, {3, 3, 3, 3, {3, 3, 3, {3, 3, 3}}}}.

{3, 3, 1, 1, 1, 1, 3} becomes {3, 3, 3, 3, 3, 3, {3, 3, 3, 3, 3, {3, 3, 3, 3, {3, 3, 3, {3, 3, 3}}}}}.