08/09/2007, 07:05 PM
How about a terminology based on the hyper-n terminologies:
and a similar terminology for the reverse operations:
and provisions should be made for the lesser and mixed hyper-operators, like those based on left-associative or right-associative iteration or left-right and right-left associative iteration. Since these only appear after hyper-3, I would imagine a logical way to name these would be to assign hyper-R = hyper-4 hyper-RR = hyper-5, hyper-RRR = hyper-6 since they all depend on right-associative iteration, and perhaps hyper-L, hyper-LL, and hyper-LLL for the "lower" or "hypo"-operator sequence. Using this terminology, we can refer to all of the above, for example, as hyper-LR-power, hyper-LR-ational, hyper-LR-root, hyper-LR-log, and so on. This works well for the hyper-operators based on simple iteration, but not for your Binary-Tree hyper-operators. How would that work for the Binary-Tree hyper-operators?
Andrew Robbins
- hyper-n-power (1-variable) (or tetra-power, penta-power, ...)
- hyper-n-ational (1-variable) (or tetrational, pentational, ...)
- hyper-n (2-variables) (or tetration, pentation, ...)
and a similar terminology for the reverse operations:
- hyper-n-root (inverse of hyper-powers) (or tetra-root, penta-root, ...)
- hyper-n-log (inverse of hyper-ationals) (i still don't like tetra-log)
and provisions should be made for the lesser and mixed hyper-operators, like those based on left-associative or right-associative iteration or left-right and right-left associative iteration. Since these only appear after hyper-3, I would imagine a logical way to name these would be to assign hyper-R = hyper-4 hyper-RR = hyper-5, hyper-RRR = hyper-6 since they all depend on right-associative iteration, and perhaps hyper-L, hyper-LL, and hyper-LLL for the "lower" or "hypo"-operator sequence. Using this terminology, we can refer to all of the above, for example, as hyper-LR-power, hyper-LR-ational, hyper-LR-root, hyper-LR-log, and so on. This works well for the hyper-operators based on simple iteration, but not for your Binary-Tree hyper-operators. How would that work for the Binary-Tree hyper-operators?
Andrew Robbins