User blog:GamesFan2000/HNEAN (Part 5: The "array of" Operator)

Hello! Welcome back to the wonderful and weird world of HNEAN! We are now ready to take the next step in expanding the definition. Now I will introduce you to a symbol that should be very familiar to those who are well aquainted with BEAF.

The & Operator
The & symbol in BEAF is the "array of" operator. Well, we can use that in HNEAN as well.

3&3 = {3, 3, 3}, where the first number determines the entry value and the second number determines the length.

Simple enough, you may say. Oh, but don't be foolish.

(3^3)&(3^3)

That is not solved the way you think it is solved. In the "array of" rule set, 3^3 actually means {3, 3, 3}. So a {3, 3, 3} length array of {3, 3, 3}'s...wait a minute...I think we've seen that exact same description before. Oh, I remember, the array that is built from 3^3&3^3 is the exact same as the array built from {3{1}3}! Now, the up-arrows are VERY different in this operation than in Knuth notation. 3^^3 actually means {3, 3, 3}^{3, 3, 3}^...{3, 3, 3}, or a {3, 3, 3} length single up-arrow set of {3, 3, 3}'s. You can use this to create absolutely massive determinant numbers. Of course, 3^^^3 creates a same size set with double arrows, quadruple arrows does it for triple arrows, and so on.

3&3&3

These are always solved right to left, and arrows before &. So, what does adding a second & do? The second & is solved like usual, so {3, 3, 3}. What it then does to the second determinant number of the first &, which, I'll remind you is the length determiner, is create a set that has {3, 3, 3} arrays of {3, 3, 3}, and {3, 3, 3} arrows inbetween each array. Lovely! So big! You can have numerous &'s in a chain, of course.

3&&3

"Oh, so a {3, 3, 3} length set of &'s with {3, 3, 3} length sets of {3, 3, 3} length arrays with entry values of {3, 3, 3} surrounding the &'s and {3, 3, 3} arrows between each array?" No. It's even crazier. The length and entry value of the determinant array is actually a^b for &&. So replace those arrays you just mentioned with 27-length arrays of 27's. In chained &&'s, whatever you get for the && you solved is the amount of single &'s in the second determinant of the && immediately after the one you just solved, in addition to the number of arrays surrounding each &, the length and entry values of those arrays, and how many arrows are inbetween each array.

3&&&3

As you can imagine, the determinate numbers are now tetrated, a^^b. I'll remind you that that is 3^27 for the example expression above. A 3^27 length array of 3^27's that determines how many &&'s there are, how many &'s are inbetween each &&, how many arrays are surrounding each separator, the length and entry value of each array, and the number of arrows between each array. I'll let you take that in. If that was part of a chain, then the answer would determine how many &&'s are in the second determinant of the &&& after it, as well as the number of &'s between each &&, the number of arrays surrounding each separator, the length and entry values of those arrays, and the number of arrows between each array.

That's where I'll leave it for now.