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{{Infobox
 
{{Infobox
 
|Box title = Extensible Illion System
 
|Box title = Extensible Illion System
|Row 1 title = state
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|Row 1 title = Growth
|Row 1 info = ill-defined
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|Row 1 info = <math>f_{\omega^2}(x)</math><ref name="placeholder"/>
 
|Row 2 title = Based On
 
|Row 2 title = Based On
 
|Row 2 info = [[-illion]]s
 
|Row 2 info = [[-illion]]s
 
}}
 
}}
   
'''Extensible Illion System''' is a notation by [[User:Nirvana Supermind|Nirvana Supermind]] (for generating large -illions)<ref>https://integralview.wordpress.com/2020/10/12/extensible-illion-system-index/</ref>. It is divided into extensions, and there is currently just one:
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'''Extensible Illion System''' is a notation by [[User:Nirvana Supermind|Nirvana Supermind]] (for generating large -illions)<ref name="placeholder">https://integralview.wordpress.com/2020/10/12/extensible-illion-system-index/</ref>. It is divided into extensions, and there is currently just one:
   
  +
# Primitive Illion System
# Primitive Illion System <ref name="orginal">[https://integralview.wordpress.com/2020/10/13/primitive-illion-system/ The original definition (Retrieved at UTC 7:00 on 15/10/2020)]</ref><ref name="current">[https://integralview.wordpress.com/2020/10/13/primitive-illion-system/ The current definition (Retrieved at UTC 23:00 on 15/10/2020)]</ref>
 
 
This is ill-defined, as is explained later.
 
   
 
== Primitive Illion System ==
 
== Primitive Illion System ==
The primitive notation takes a base and any amount of arguments, which are non-negative integers. The original rules for it<ref name="original"> are:
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The primitive notation takes a base and any amount of arguments, which are non-negative integers. The rules for it are:
   
 
# 0[b] = 10<sup>3b+3</sup>
 
# 0[b] = 10<sup>3b+3</sup>
 
#a[] = 1000
 
# a[b] = a-1[b,b,b,b…] with b “b”s for a>0
 
# a[b] = a-1[b,b,b,b…] with b “b”s for a>0
 
# a[b,c] = a[a[b,c-1]/1000]
 
# a[b,c] = a[a[b,c-1]/1000]
 
# a[#,0] = a[#]
 
# a[#,0] = a[#]
 
# a[#,b,c] = a[#,a[#,b,c-1]] for c>0
 
# a[#,b,c] = a[#,a[#,b,c-1]] for c>0
Here, # is a portion of the array, which can be empty.
 
   
  +
Here, # is a portion of the array, which can be empty. If there are two or more distinct rules to apply to a single expression, the upper rule will be applied if it is well-defined, or else the lower rule. Note that an old version of the notation<ref name="current">[https://integralview.wordpress.com/2020/10/13/primitive-illion-system/ The original definition (Retrieved at UTC 7:00 on 15/10/2020)]</ref> excluded the second rule and did not include the precedence, making some expressions such as ill-defined:
This notation qas intended to reach w^2 in the fast-growing hierarchy, but was ill-defined because of many errors:
 
# There is no rule applicable to 0[] and 1[].
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*There is no rule applicable to 0[] and 1[].
# There are two distinct ways to solve 1[0].
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*There are two distinct ways to solve 1[0].
## If you apply the rule 2, then the result will be 0[], which is ill-defined by the reason above.
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**If you apply the rule 2, then the result will be 0[], which is ill-defined by the reason above.
## If you apply the rule 4, then the result will be 1[], which is ill-defined by the reason above.
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**If you apply the rule 4, then the result will be 1[], which is ill-defined by the reason above.
# There are two distinct ways to solve 1[0,0].
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*There are two distinct ways to solve 1[0,0].
## If you apply the rule 3, then the result will be 1[1[0,-1]/1000], which is ill-defined because -1 is negative.
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**If you apply the rule 3, then the result will be 1[1[0,-1]/1000], which is ill-defined because -1 is negative.
## If you apply the rule 4, then the result will be 1[0], which is ill-defined by the reason above.
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**If you apply the rule 4, then the result will be 1[0], which is ill-defined by the reason above.
  +
# The computation of 0[4,1] = 0[0[4,0]/1000] is intended to be 0[1.015], which is ill-defined because 1.015 is not an integer.
 
  +
Later, the creator updated the definition by adding the case for the empty array:
 
  +
However, these issues got fixed.
  +
   
# 0[b] = 10<sup>3b+3</sup>
 
# a[] = 1000
 
# a[b] = a-1[b,b,b,b…] with b “b”s for a>0
 
# a[b,c] = a[a[b,c-1]/1000]
 
# a[#,0] = a[#]
 
# a[#,b,c] = a[#,a[#,b,c-1]] for c>0
 
Here, # is a portion of the array, which can be empty. If there are two or more distinct rules to apply to a single expression, the upper rule will be applied.
 
   
  +
<br />
It is still ill-defined because of the same issues for 1[0,0] (the invalidity of 1[0,-1]) and 0[4,1] (the invalidity of 0[1.015]).
 
   
 
=== Example ===
 
=== Example ===
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0[4,1] = 0[0[4,0]/1000]
 
0[4,1] = 0[0[4,0]/1000]
   
0[4,0] = 0[4] = 1015
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0[4,0] = 0[4] = 10<sup>15</sup>
   
 
0[4,1] = 0[10<sup>15</sup>]
0[4,1] = 0[1.015], which is ill-defined because 1.015 is not an integer.
 
   
0[4,1,1] is intended to be 1010^(3*10^12+3)3*10^12+3, but is ill-defined because 0[4,0[4,1]] includes the ill-defined value 0[4,1].
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0[4,1,1] = 10<sup><sup>10^(3*10^12+3)3</sup>3*10^12+3</sup>.
   
 
== Sources ==
 
== Sources ==

Revision as of 01:55, 16 October 2020

Not to be confused with Extensible-E System.
Extensible Illion System
Growth
[1]
Based On


Extensible Illion System is a notation by Nirvana Supermind (for generating large -illions)[1]. It is divided into extensions, and there is currently just one:

  1. Primitive Illion System

Primitive Illion System

The primitive notation takes a base and any amount of arguments, which are non-negative integers. The rules for it are:

  1. 0[b] = 103b+3
  2. a[] = 1000
  3. a[b] = a-1[b,b,b,b…] with b “b”s for a>0
  4. a[b,c] = a[a[b,c-1]/1000]
  5. a[#,0] = a[#]
  6. a[#,b,c] = a[#,a[#,b,c-1]] for c>0

Here, # is a portion of the array, which can be empty. If there are two or more distinct rules to apply to a single expression, the upper rule will be applied if it is well-defined, or else the lower rule. Note that an old version of the notation[2] excluded the second rule and did not include the precedence, making some expressions such as ill-defined:

  • There is no rule applicable to 0[] and 1[].
  • There are two distinct ways to solve 1[0].
    • If you apply the rule 2, then the result will be 0[], which is ill-defined by the reason above.
    • If you apply the rule 4, then the result will be 1[], which is ill-defined by the reason above.
  • There are two distinct ways to solve 1[0,0].
    • If you apply the rule 3, then the result will be 1[1[0,-1]/1000], which is ill-defined because -1 is negative.
    • If you apply the rule 4, then the result will be 1[0], which is ill-defined by the reason above.


However, these issues got fixed.



Example

0[4,1,1] = 0[4,0[4,1]]

0[4,1] = 0[0[4,0]/1000]

0[4,0] = 0[4] = 1015

0[4,1] = 0[1015]

0[4,1,1] = 1010^(3*10^12+3)33*10^12+3.

Sources