Make the largest infinity that you can!

Question

The ordinal number system is a system with infinite numbers. A lot of infinite numbers. So many infinite numbers that it literally does not have an infinity to represent its own infiniteness. The image above gives a little idea of how they work. An ordinal number (Von Neumann construction) is a set of previous ordinals. For example, 0 is the empty set, 1 is the set {0}, 2 is the set {0, 1} and etc. Then we get to ω, which is {0, 1, 2, 3 ...}. ω+1 is {0, 1, 2, 3 ... ω}, ω times two is {0, 1, 2 ... ω, ω + 1, ω + 2...} and you just keep going like that.

Your program will output a set of ordinals, such {0, 1, 4}. Your score will then be the least ordinal more than all the ordinal in your set. For {0, 1, 4} the score would be 5. For {0, 1, 2 ...}, the score would be ω.

How do you output your ordinals you ask. Code of course. Namely, your program will output a potentially infinite list of other programs, in quotes, one on each line (use the literal string "\n" to represent new lines). A program corresponds to its score as indicated above. For example, if you output

"A"
"B"
"C"

where A, B, and C are themselves valid answers and have scores {0, 1, 4}, the score of your program would be 5. Note that A, B, and C, have to be full programs, not fragments.

Based on the above rules, a program that outputs nothing has a score of 0 (the least ordinal greater than all {} is 0). Also, remember a set cannot contain itself, via the axiom of foundation. Namely, every set (and therefore ordinal) has a path down to zero. This means the a full-quine would be invalid as its not a set.

Also, no program is allowed to access outside resources (its own file, the internet etc...). Also, when you list your score, put the cantor normal form of the score alongside it if it isn't in cantor normal form already, if you can (if not, someone else can).

After taking all the above into account, the actual answer you post must be under 1,000,000 bytes (not counting comments). (This upper bound will likely only come into play for automatically generated code). Also, you get to increment your score for each byte you don't use (since we are dealing with infinities, this will probably only come into account when ordinals are very close or the same). Again, this paragraph only applies to the posted answer, not the generated ones, or the ones the generated ones generate, and so on.

This has the quine tag, because it may be helpful to generate at least part of the sources own code, for use in making large ordinals. It is by no means required though (for example, a submission with score 5 would probably not need its own source code).

For a worked out and annotated example, see here.

Answer 1

Haskell: ψ(Ω^Ωω) + 999672 points

main=print$f++shows f"$iterate(Z:@)Z";f="data N=Z|N:@N deriving Show;r=repeat;s Z=[];s(a:@Z)=r a;s(Z:@b)=($u b)<$>h b;s(a:@b)=do(f,c)<-h b`zip`s a;f<$>[c:@b,a]\nh=scanl(flip(.))id.t;t(Z:@Z)=r(:@Z);t(Z:@b)=((Z:@).)<$>t b;t(a:@b)=flip(:@).(:@b)<$>s a;u Z=Z;u(Z:@b)=u b;u(a:@_)=a;g f=mapM_$print.((f++shows f\"$s$\")++).show;main=g"

329 bytes of code with ordinal ψ(Ω^Ωω) + 1. Uses a tree-based representation of ordinals discovered by Jervell (2005). The tree with children α, β, …, γ is denoted α :@ (β :@ (… :@ (γ :@ Z)…)). This left-to-right order agrees with Jervell, although note that Madore flips it right-to-left.

Haskell: Γ₀ + 999777 points

main=print$f++shows f"$iterate((:[]).T)[]";f="data T=T[T]deriving(Eq,Ord,Show);r=reverse;s[]=[];s(T b:c)=(++c)<$>scanl(++)[](b>>cycle[fst(span(>=T b)c)++[T$r d]|d<-s$r b]);g f=mapM_$print.((f++shows f\"$s$\")++).show;main=g"

224 bytes of code with ordinal Γ₀ + 1. This is based on a generalization of Beklemishev’s Worm to ordinal-valued elements, which are themselves recursively represented by worms.

Haskell: ε₀ + 999853 points

main=print$f++shows f"$map(:[])[0..]";f="s[]=[];s(0:a)=[a];s(a:b)=b:s(a:fst(span(>=a)b)++a-1:b);g f=mapM_$print.((f++shows f\"$s$\")++).show;main=g"

148 bytes of code with ordinal ε₀ + 1. This is based on Beklemishev’s Worm. The list

map(+1)α ++ [0] ++ map(+1)β ++ [0] ++ ⋯ ++ [0] ++ map(+1)γ

represents the ordinal (ω^γ + ⋯ + ω^β + ω^α) − 1. So the second-level outputs [0], [1], [2], [3], … represent 1, ω, ω^ω, ω^ωω, …, the first-level output represents ε₀, and the initial program represents ε₀ + 1.

Haskell: ε₀ + 999807 points

main=print$f++shows f"$iterate(:@Z)Z";f="data N=Z|N:@N deriving Show;s Z=[];s(Z:@b)=repeat b;s(a:@Z)=scanl(flip(:@))Z$s a;s(a:@b)=(a:@)<$>s b;g f=mapM_$print.((f++shows f\"$s$\")++).show;main=g"

194 bytes of code with ordinal ε₀ + 1.

Z represents 0, and we can prove by transfinite induction on α, then on β, that α :@ β ≥ ω^α + β. So there are second-level outputs at least as big as any tower ω^ω⋰ω, which means the first-level output is at least ε₀ and the initial program is at least ε₀ + 1.

Answer 2

Ruby, ψ₀(ψ_{X(ψM+1(ΩM+1^ΩM+1))}(0)) + 999663 points

Assuming I understand my program properly, my score is ψ₀(ψ_{X(ψM+1(ΩM+1^ΩM+1))}(0)) + 999663 points where ψ is an ordinal collapsing function, X is the chi function (Mahlo collapsing function), and M is the first Mahlo 'ordinal'.

This program is an extension of the one I wrote on Golf a number bigger than TREE(3) and utterly trumps all of the other solutions here for now.

z=->x,k{f=->a,n,b=a,q=n{c,d,e=a;!c ?q:a==c ?a-1:e==0||e&&d==0?c:e ?[[c,d,f[e,n,b,q]],f[d,n,b,q],c]:n<1?9:!d ?[f[b,n-1],c]:c==0?n:[t=[f[c,n],d],n,d==0?n:[f[d,n,b,t]]]};x==0??p:(w="\\"*k+"\"";y="";(0..1.0/0).map{|g|y+="x=#{f[x,g]};puts x==0??p:"+w+"#{z[f[x,g],k*2+1]}"+w+";"};y)};h=-1;(0..1.0/0).map{|i|puts "puts \"#{z[[i,h=[h,h,h]],1]}\""}

Try it online!

Ruby, ψ₀(ψ_I(I^I)) + 999674 points

z=->x,k{f=->a,n,b=a{c,d,e,q=a;a==c ?a-1:e==0||d==0?(q ?[c]:c):e ?[[c,d,f[e,n,b],q],f[d,n,b],c,q]:[n<1?9:d&&c==0?[d]:d ?[f[c,n],d]:[f[b,n-1],f[c,n,b]],n,n]};x==0??p:(w="\\"*k+"\"";y="";(0..1.0/0).map{|g|y+="x=#{f[x,g]};puts(x==0??p:"+w+"#{z[f[x,g],k*2+1]}"+w+");"};y)};h=0;(0..1.0/0).map{|i|puts("puts(\"#{z[h=[h,h,h,0],1]}\")")}

Try it online! (warning: it won't do much, since clearly (0..1.0/0).map{...} cannot terminate. It's how I print infinitely many programs as well.)

Ruby, ψ₀(ψ_I(0)) + 999697 points

z=->x,k{f=->a,n,b=a{c,d,e=a;a==c ?a-1:e==0||d==0?c:e ?[[c,d,f[e,n,b]],d-1,c]:[n<1?9:d&&c==0?[d]:d ?[f[c,n-1],d]:[f[b,n-=1],f[c,n,b]],n,n]};x==0??p:(w="\\"*k+"\"";y="";(0..1.0/0).map{|g|y+="x=#{f[x,g]};puts(x==0??p:"+w+"#{z[f[x,g],k*2+1]}"+w+");"};y)};h=0;(0..1.0/0).map{|i|h=[h];puts|i|puts("puts(\"#{z[hz[h=[h],1]}\")")}

Try it online!

A more reasonable test program that implements (0..5).map{...} instead:

z=->x,k{f=->a,n,b=a{c,d,e=a;a==c ?a-1:e==0||d==0?c:e ?[[c,d,f[e,n,b]],d-1,c]:[n<1?9:d&&c==0?[d]:d ?[f[c,n-1],d]:[f[b,n-=1],f[c,n,b]],n,n]};x==0??p:(w="\\"*k+"\"";y="";(0..5).map{|g|y+="x=#{f[x,g]};puts(x==0??p:"+w+"#{z[f[x,g],k*2+1]}"+w+");"};y)};h=0;(0..5).map{|i|h=[h];puts("puts(\"#{z[h,1]}\")")}

Try it online!

Unfortunately, even with (1..1).map{...}, you will find this program to be extremely memory intensive. I mean, the length of the output exceeds things like SCG(13).

By using the simpler program, we may consider a few values:

a = value you want to enter in.
z=->x,k{f=->a,n,b=a{c,d,e=a;a==c ?a-1:e==0||d==0?c:e ?[[c,d,f[e,n,b]],d-1,c]:[n<1?9:d&&c==0?[d]:d ?[f[c,n-1],d]:[f[b,n-=1],f[c,n,b]],n,n]};x==0??p:(w="\\"*k+"\"";y="x=#{f[x,k]};puts(x==0??p:"+w+"#{z[f[x,k],k*2+1]}"+w+");";y)};puts("puts(\"#{z[a,1]}\")")

Try it online!

It basically prints the same program repeatedly, in the format of:

x=...;puts(x==0??p:"...");

where the initialized x records the ordinal the program is related to, and the "..." holds the programs after x has been reduced. If x==0, then it prints

p

which is a program that prints nothing with a score of zero, hence

x=0;puts(x==0??p:"p");

has a score of 1, and

x=1;puts(x==0??p:"x=0;puts(x==0??p:\"p\");");

has a score of 2, and

x=2;puts(x==0??p:"x=1;puts(x==0??p:\"x=0;puts(x==0??p:\\\"p\\\");\");");

has a score of 3, etc., and my original program prints these programs in the format

puts("...")

where ... are the programs listed above.

My actual program actually prints these programs in the format

x=0;puts(x==0??p:"p;p;p;p;p;...");

Infinitely many times, and for values such as ω, it does something akin to

x=ω;puts(x==0??p:"(x=0 program); (x=1 program); (x=2 program); ...")

And thus, the score of the program

x=(some_ordinal);puts(x==0??p:"...")

is 1+some_ordinal, and the score of

puts("x=(some_ordinal);puts(x==0??p:\"...\")")

is 1+some_ordinal+1, and the score of

z=->x,k{f=->a,n,b=a{c,d,e=a;a==c ?a-1:e==0||d==0?c:e ?[[c,d,f[e,n,b]],d-1,c]:[n<1?9:d&&c==0?[d]:d ?[f[c,n-1],d]:[f[b,n-=1],f[c,n,b]],n,n]};x==0??p:(w="\\"*k+"\"";y="";(0..1.0/0).map{|g|y+="x=#{f[x,g]};puts(x==0??p:"+w+"#{z[f[x,g],k*2+1]}"+w+");"};y)};puts("puts(\"#{z[some_ordinal,1]}\")")

is 1+some_ordinal+2.

---

Explanation of ordinals:

f[a,n,b] reduces an ordinal a.

Every natural number reduces to the natural number below it.

f[k,n,b] = k-1

[c,0,e] is the successor of c, and it always reduces down to c.

f[[c,0,e],n,b] = c

[c,d,e] is a backwards associative hyperoperation, are reduces as follows:

f[[c,d,0],n,b] = c
f[[c,d,e],n,b] = [[c,d,f[e,n,b]],f[d,n,b],c]

[0] is the first infinite ordinal (equivalent to ω) and reduces as follows:

f[[0],0,b] = [9,0,0]
f[[0],n,b] = [n,n,n]

[c] is the cth omega ordinal and reduces as follows:

f[[c],0,b] = [9,0,0]
f[[c],n,b] = [[f[b,n-1,b],f[c,n,b]],n,n]
Note the two-argument array here.

[c,d] is ψ_d(c) and reduces as follows:

f[[c,d],0,b] = [9,0,0]
f[[0,d],n,b] = [[d],n,n]
f[[c,d],n,b] = [[f[c,n,c],d],n,n]

[c,d,e,0] is basically the same as [c,d,e], except it enumerates over the operation [c] instead of the successor operation.

f[[c,0,e,0],n,b] = [c]
f[[c,d,0,0],n,b] = [c]
f[[c,d,e,0],n,b] = [[c,d,f[e,n,b],0],f[d,n,b],c,0]

Answer 3

Java + Brainf***, ω+999180 points

A java program that produces infinitely many Brainf*** programs, of which each produces the last as output.

Why? Because I can.

Any improvements to the Brainf*** generation part are definitely welcome.

import java.io.*;import java.util.*;import static java.lang.System.*;
class O{
    static String F(String s){
        String t="++++++++++++++++++++";
        List<Character> c=Arrays.asList('+','-','.','<','>','[',']');
        t+="[->++>++>++>+++>+++>++++>++++<<<<<<<]>+++>+++++>++++++>>++>+++++++++++>+++++++++++++";
        for(int x=0,i=6;x<s.length();x++){
            int d=c.indexOf(s.charAt(x)),k=d;
            while(d>i){d--;t+=">";}
            while(d<i){d++;t+="<";}
            t+=".";i=k;
        }return t;
    };
    static void p(String a){
        int i=48;while(i<a.length()){out.println(a.substring(i-48,i));i+=48;}
        out.println(a.substring(i-48));out.println();
    }
    public static void main(String[]a) throws Exception{setOut(new PrintStream("o.txt"));String b="";while(true)p(b=F(b));}
}

Answer 4

Literate Haskell (GHC 7.10): ω² + 999686 points.

This will serve as an example answer. Since its an example, it only makes sense to use literate programming. It won't score well though. The birdies will decrease my score, but oh well. First, let's make a helper function s. If x is an ordinal, s x = x+1, but we find unexpected uses of s.

> s x="main=putStrLn "++show x

The show function luckily does all the string sanitization for us. Its also worth making 0. Zero is not the successor of anything, but s"" will equal "main=putStrLn """, which equals 0.

> z=s""

Now we will make a function that takes n, and returns ω*n.

> o 0=z
> o n=q++"\nmain=putStrLn$unlines$map show$iterate s$o "++show(n-1)

It works by making ω*n by {ω*(n-1), ω*(n-1)+1, ω*(n-1)+2, ...}. What is this q? Why, its the reason we have a quine tag. q is the above helper functions so far.

> h x = x ++ show x
> q = h "s x=\"main=putStrLn \"++show x\nz=s\"\"\no 0=z\no n=q++\"\\nmain=putStrLn$unlines$map show$iterate s$o \"++show(n-1)\nh x = x ++ show x\nq = h "

Note that only o n needs q, not any of its successors (s(o(n)), s(s(o(n)))), since they don't need the helper functions (except indirectly from o n.) Now we only have to output all of ω*n for finite n.

> main=mapM(print.o)[0..]

There we go. ω^2 Having used only 314 code character, I claim a final bonus of 999686, giving me a final score of ω^2 + 999686, which is already in cantor's normal form.

First four lines of output (0, ω, ω*2, ω*3):

"main=putStrLn \"\""
"s x=\"main=putStrLn \"++show x\nz=s\"\"\no 0=z\no n=q++\"\\nmain=putStrLn$unlines$map show$iterate s$o \"++show(n-1)\nh x = x ++ show x\nq = h \"s x=\\\"main=putStrLn \\\"++show x\\nz=s\\\"\\\"\\no 0=z\\no n=q++\\\"\\\\nmain=putStrLn$unlines$map show$iterate s$o \\\"++show(n-1)\\nh x = x ++ show x\\nq = h \"\nmain=putStrLn$unlines$map show$iterate s$o 0"
"s x=\"main=putStrLn \"++show x\nz=s\"\"\no 0=z\no n=q++\"\\nmain=putStrLn$unlines$map show$iterate s$o \"++show(n-1)\nh x = x ++ show x\nq = h \"s x=\\\"main=putStrLn \\\"++show x\\nz=s\\\"\\\"\\no 0=z\\no n=q++\\\"\\\\nmain=putStrLn$unlines$map show$iterate s$o \\\"++show(n-1)\\nh x = x ++ show x\\nq = h \"\nmain=putStrLn$unlines$map show$iterate s$o 1"
"s x=\"main=putStrLn \"++show x\nz=s\"\"\no 0=z\no n=q++\"\\nmain=putStrLn$unlines$map show$iterate s$o \"++show(n-1)\nh x = x ++ show x\nq = h \"s x=\\\"main=putStrLn \\\"++show x\\nz=s\\\"\\\"\\no 0=z\\no n=q++\\\"\\\\nmain=putStrLn$unlines$map show$iterate s$o \\\"++show(n-1)\\nh x = x ++ show x\\nq = h \"\nmain=putStrLn$unlines$map show$iterate s$o 2"

Answer 5

GolfScript, ε₀+1 + 999537 points

{{
{"{"lib`+"}:lib~~"@++"~"+p}:output;
{{}}:nothing;
{{1{).2$`+output 1}do}}:list;
{{\.(`2$`+" iter"+@`if output}}:iter;
{{1${@`@(2$`{rep~}+++\}{;;}if~}}:rep;
{\`\+}:combine;
{{\@@~}}:swap;
{{1${1$(1$`{rangemap~}+[+@@[\\]]{~}/+}{;;}if}}:rangemap;
{`{}+}:wrap;
}}:lib~~
nothing {iter combine list~} {rep combine} {~swap combine combine} rep combine swap combine combine {~list~} combine rangemap {~~} combine combine swap combine combine swap combine combine list~

It can probably be better, but I became too lazy to debug and prove larger ordinals.

Smaller ordinals

#ω^ω^ω
#nothing {iter combine list~} {rep combine swap combine combine list~} {rep combine swap combine combine swap combine combine list~} rep combine swap combine combine swap combine combine swap combine combine list~
#ω^ω^2
#nothing {iter combine list~} { {rep combine swap combine combine list~} rep combine swap combine combine swap combine combine list~} rep combine swap combine combine swap combine combine list~
#ω^nω
#nothing {iter combine list~} 2 { {rep combine swap combine combine list~} rep combine swap combine combine swap combine combine list~} rep~
#ω^ω
#nothing {iter combine list~} {rep combine swap combine combine list~} rep combine swap combine combine swap combine combine list~
#ω^n
#nothing {iter combine list~} 3 {rep combine swap combine combine list~} rep~
#ω^3
#nothing {{iter combine list~} rep combine swap combine combine list~} rep combine swap combine combine list~
#ω^2
#nothing {iter combine list~} rep combine swap combine combine list~
#nω
#nothing 3 {iter combine list~} rep~
#2ω
#nothing iter combine list combine iter combine list~
#ω
#nothing iter combine list~
#finite
#2 nothing iter~

Answer 6

Javascript (Nashorn), ω2+999807 points

Nashorn is the Javascript engine which comes built into Java. This may also work in Rhino, but I haven't tested it in that yet.

c="(b=function(a){print(a?\"(b=\"+b+\")(\"+(a-1)+\")\":\"c=\\\"(b=function(a){print(a?'(b='+b+')'+'('+(a-1)+')':'')})(\\\";a=0;while(++a)print(c+a+\\\")\\\")\")})(";a=0;while(++a)print(c+a+")")