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Challenges

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Re: Challenges

Postby AforAmpere » October 3rd, 2018, 9:17 pm

wwei23 wrote:Find one of those that uses a replicator not based on this one.

Do we have to use 2-sate isotropic, or can we do a non-symmetric rule with this behavior?
Things to work on:
- Find a (7,1)c/8 ship in a Non-totalistic rule (someone please search the rules)
- Find a C/10 in JustFriends
- Find a C/10 in Day and Night
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Re: Challenges

Postby PHPBB12345 » October 3rd, 2018, 9:19 pm

AforAmpere wrote:
wwei23 wrote:Find one of those that uses a replicator not based on this one.

Do we have to use 2-sate isotropic, or can we do a non-symmetric rule with this behavior?

Hint: It's use 2-state isotropic rule.
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Re: Challenges

Postby wwei23 » October 3rd, 2018, 9:20 pm

AforAmpere wrote:
wwei23 wrote:Find one of those that uses a replicator not based on this one.

Do we have to use 2-sate isotropic, or can we do a non-symmetric rule with this behavior?

2-state isotropic. After all, if one exists, there's probably another, like the huge variety of Sierpinski builders. Surely there must be a variety of these.
Also, I don't know if any exist in totalistic rules, the answer is probably no...
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Re: Challenges

Postby 77topaz » October 4th, 2018, 1:41 am

A non-isotropic example was known at least since May this year: viewtopic.php?p=60361#p60361
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Re: Challenges

Postby dvgrn » October 23rd, 2018, 3:28 pm

New small challenge:

I'm in need of a one-time attachment to a Scorbie Splitter that gets an extra glider out, but only the first time the circuit is used.

I can manage this with the three white still lifes below, which are fairly cheap for slsparse to construct. Can anyone find just one or two Spartan-ish still lifes that can accomplish the same trick?

Code: Select all
x = 143, y = 86, rule = LifeHistory
54.4B50.B$55.4B48.2B$56.4B46.3B$57.4B44.4B$58.4B42.4B$59.4B40.4B$60.
4B38.4B$61.4B36.4B$62.4B34.4B$63.4B32.4B$64.4B30.4B$65.4B28.4B$36.2A
11.A16.4B4.B21.4B$35.B2AB9.A.A16.4B2.3B19.4B$36.3B9.A.A17.9B5.2A10.4B
$35.B.B9.2A.3A2.2A12.9B4.A10.4B$35.5B8.B4.A2.A13.8B.BA.A9.4B$35.6B6.
2AB3A3.A.AB2.4B3.9B.B2A9.4B$35.8B4.2A.A6.2AB2.5B2.11B10.4B$36.13B10.
8B2.11B9.4B$34.13B12.21B8.4B$33.15B12.19B8.4B$33.15B10.B2.19B6.4B$32.
17B.B5.26B3.4B$32.51B.4B$31.13B2A13B2A21B.4B$30.14B2A13B2A25B$29.2AB
3.50B$28.A2.A4.48B$27.A.2A5.6B3.B2.2B2.33B$27.A7.6B13.4B2.7B2.16B$26.
2A6.9B17.6B3.18B$33.4B4.2A19.3B5.18B$32.4B5.A21.B4.3B.9B3.B2CB$31.4B
7.3A23.2A3.8B4.BC3B6.B$30.4B10.A24.A6.4B5.BCBC2B4.3B$29.4B33.3A5.4B6.
3B2C5B.3B$28.4B34.A7.2A8.18B$27.4B44.A8.19B$26.4B42.3A10.18B$25.4B43.
A10.19B$24.4B55.7B2C12B$23.4B56.7B2C6B.3B2C$22.3CB58.8B2.4B2.2B2C$22.
2BC61.6B9.4B$23.C62.5B11.4B$86.5B12.4B$87.3B14.4B$88.B16.4B$106.3B$
107.2B$108.B11$116.4B19.B$117.4B6.E10.2B$.2A115.4B5.3E7.4B$A.A116.4B
7.E5.4B$2.A117.4B5.2E4.4B$121.4B4.9B$122.4B5.6B$123.4B2.8B$124.15B$
125.14B$126.13B$127.10B.B2E$129.3B2EB3.BE.E$129.3B2EB6.E$131.4B6.2E$
131.3B$128.EB.2B$127.E.EB2EB$127.E.EBEBEB$124.2E.E.E.E.E2.E$124.E2.E
2.2E.4E$126.2E4.E$132.E.E$133.2E!
#C [[ THUMBNAIL THUMBSIZE 2 Z 4 ]]

EDIT: There would be an advantage if the output glider went southeast but didn't hit the Snark I've added in yellow in the southeast. So my solution really counts as 5sL, not 3sL, because I have to add a couple of boats or eaters or something to bounce the glider out of the way of the Snark. Southeast is the ideal direction for the glider, though, so if it goes some other direction then add 1sL to get it back on track.

Really "Spartan" is pretty much an out-of-date concept now, unless we modernize it to mean just "anything that slsparse has no problem constructing". The slsparse construction database has a large number of relatively rare still lifes and constellations that I would never have dared to try to include in old pre-slmake self-constructing circuitry.

For example, when I needed to change a glider's color in the most recent Orthogonoids, I didn't have to think twice about it, just threw in an aircraft carrier and slsparse happily compiled it for me along with everything else.
Last edited by dvgrn on October 23rd, 2018, 4:02 pm, edited 2 times in total.
Reason: added a Snark on the Scorbie Splitter's transparent lane
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Re: Challenges

Postby BlinkerSpawn » October 23rd, 2018, 11:01 pm

4-sL-equivalent solution which can be flipped to go either way:
x = 88, y = 47, rule = LifeHistory
32.4B50.B$33.4B48.2B$34.4B46.3B$35.4B44.4B$36.4B42.4B$37.4B40.4B$38.
4B38.4B$39.4B36.4B$40.4B34.4B$41.4B32.4B$42.4B30.4B$43.4B28.4B$14.2A
11.A16.4B4.B21.4B$13.B2AB9.A.A16.4B2.3B19.4B$14.3B9.A.A17.9B5.2A10.4B
$13.B.B9.2A.3A2.2A12.9B4.A10.4B$13.5B8.B4.A2.A13.8B.BA.A9.4B$13.6B6.
2AB3A3.A.AB2.4B3.9B.B2A9.4B$13.8B4.2A.A6.2AB2.5B2.11B10.4B$14.13B10.
8B2.11B9.4B$12.13B12.21B8.4B$11.15B12.19B8.4B$11.15B10.B2.19B6.4B$10.
17B.B5.26B3.4B$10.51B.4B$9.13B2A13B2A21B.4B$8.14B2A13B2A25B$7.2AB3.
50B$6.A2.A4.48B$5.A.2A5.6B3.B2.2B2.33B$5.A7.6B13.4B2.7B2.16B$4.2A6.9B
17.6B3.16B$11.4B4.2A19.3B5.14B$10.4B5.A21.B4.3B.9B4.2C$9.4B7.3A23.2A
3.8B3.C2.C$8.4B10.A24.A6.4B4.C2.C$7.4B33.3A5.4B7.2C$6.4B34.A7.2A$5.4B
44.A$4.4B42.3A$3.4B43.A14.2C9.2C$2.4B59.C.C8.2C$.4B61.C$3CB$2BC$.C!

EDIT: Got Seeds of Destruction so I could search for a better solution but it's adding extra linebreaks to the pattern, argh!
EDIT 2: One manually-added not-Syringe later...
x = 87, y = 46, rule = LifeHistory
32.4B50.B$33.4B48.2B$34.4B46.3B$35.4B44.4B$36.4B42.4B$37.4B40.4B$38.
4B38.4B$39.4B36.4B$40.4B34.4B$41.4B32.4B$42.4B30.4B$43.4B28.4B$14.2A
11.A16.4B4.B21.4B$13.B2AB9.A.A16.4B2.3B19.4B$14.3B9.A.A17.9B5.2A10.4B
$13.B.B9.2A.3A2.2A12.9B4.A10.4B$13.5B8.B4.A2.A13.8B.BA.A9.4B$13.6B6.
2AB3A3.A.AB2.4B3.9B.B2A9.4B$13.8B4.2A.A6.2AB2.5B2.11B10.4B$14.13B10.
8B2.11B9.4B$12.13B12.21B8.4B$11.15B12.19B8.4B$11.15B10.B2.19B6.4B$10.
17B.B5.26B3.4B$10.51B.4B$9.13B2A13B2A21B.4B$8.14B2A13B2A25B$7.2AB3.
50B$6.A2.A4.48B$5.A.2A5.6B3.B2.2B2.33B$5.A7.6B13.4B2.7B2.16B$4.2A6.9B
17.6B3.16B$11.4B4.2A19.3B5.14B$10.4B5.A21.B4.3B.9B4.C$9.4B7.3A23.2A3.
8B3.C.C$8.4B10.A24.A6.4B4.C2.C$7.4B33.3A5.4B7.2C$6.4B34.A7.2A$5.4B44.
A$4.4B42.3A$3.4B43.A11.2C$2.4B55.C.C$.4B56.2C$3CB$2BC$.C!
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Re: Challenges

Postby chris_c » October 24th, 2018, 9:51 am

dvgrn wrote:I'm in need of a one-time attachment to a Scorbie Splitter that gets an extra glider out, but only the first time the circuit is used.


You don't state how useful a glider heading NW would be. Maybe because it is useless or maybe it didn't occur to you as an option (like me). Anyway, I found this:

x = 87, y = 46, rule = LifeHistory
32.4B50.B$33.4B48.2B$34.4B46.3B$35.4B44.4B$36.4B42.4B$37.4B40.4B$38.
4B38.4B$39.4B36.4B$40.4B34.4B$41.4B32.4B$42.4B30.4B$43.4B28.4B$14.2A
11.A16.4B4.B21.4B$13.B2AB9.A.A16.4B2.3B19.4B$14.3B9.A.A17.9B5.2A10.4B
$13.B.B9.2A.3A2.2A12.9B4.A10.4B$13.5B8.B4.A2.A13.8B.BA.A9.4B$13.6B6.
2AB3A3.A.AB2.4B3.9B.B2A9.4B$13.8B4.2A.A6.2AB2.5B2.11B10.4B$14.13B10.
8B2.11B9.4B$12.13B12.21B8.4B$11.15B12.19B8.4B$11.15B10.B2.19B6.4B$10.
17B.B5.26B3.4B$10.51B.4B$9.13B2A13B2A21B.4B$8.14B2A13B2A25B$7.2AB3.
50B$6.A2.A4.48B$5.A.2A5.6B3.B2.2B2.33B$5.A7.6B13.4B2.7B2.14B$4.2A6.9B
17.6B3.14B7.2A$11.4B4.2A19.3B5.13B7.A.A$10.4B5.A21.B4.3B.9B10.A$9.4B
7.3A23.2A3.8B3.2A$8.4B10.A24.A6.4B3.A2.A$7.4B33.3A5.4B6.2A$6.4B34.A7.
2A$5.4B44.A$4.4B42.3A$3.4B43.A$2.4B$.4B$3CB$2BC$.C!

as well as this:

x = 87, y = 46, rule = LifeHistory
32.4B50.B$33.4B48.2B$34.4B46.3B$35.4B44.4B$36.4B42.4B$37.4B40.4B$38.
4B38.4B$39.4B36.4B$40.4B34.4B$41.4B32.4B$42.4B30.4B$43.4B28.4B$14.2A
11.A16.4B4.B21.4B$13.B2AB9.A.A16.4B2.3B19.4B$14.3B9.A.A17.9B5.2A10.4B
$13.B.B9.2A.3A2.2A12.9B4.A10.4B$13.5B8.B4.A2.A13.8B.BA.A9.4B$13.6B6.
2AB3A3.A.AB2.4B3.9B.B2A9.4B$13.8B4.2A.A6.2AB2.5B2.11B10.4B$14.13B10.
8B2.11B9.4B$12.13B12.21B8.4B$11.15B12.19B8.4B$11.15B10.B2.19B6.4B$10.
17B.B5.26B3.4B$10.51B.4B$9.13B2A13B2A21B.4B$8.14B2A13B2A25B$7.2AB3.
50B$6.A2.A4.48B$5.A.2A5.6B3.B2.2B2.33B$5.A7.6B13.4B2.7B2.14B$4.2A6.9B
17.6B3.14B$11.4B4.2A19.3B5.13B3.A$10.4B5.A21.B4.3B.9B4.A.A$9.4B7.3A
23.2A3.8B4.A.A$8.4B10.A24.A6.4B6.A$7.4B33.3A5.4B$6.4B34.A7.2A$5.4B44.
A$4.4B42.3A$3.4B43.A$2.4B57.A$.4B57.A.A$3CB59.2A$2BC$.C!
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Re: Challenges

Postby dvgrn » October 24th, 2018, 1:57 pm

chris_c wrote:You don't state how useful a glider heading NW would be. Maybe because it is useless or maybe it didn't occur to you as an option (like me)...

Well, it will have to be turned with one or two extra still lifes, and if it's turned 180 with a long boat it fails to miss the Snark-obstacle. So I think one of the 2sL SW or NE options, plus a boat, will be the winner, unless somebody turns up a 2sL SE output that avoids the Snark. Preferably the glider would pass southwest of the Snark, while I'm wishing.

Thanks!

(This will be used for a simple first-draft memory loop spaceship. Looks like slmake might be able to compile most of it in one step, once calcyman makes a few updates, and I get a few one-time turners set up to catch that glider.)
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Re: Challenges

Postby dvgrn » October 25th, 2018, 6:52 pm

Okay, here's a trickier question: starting from a block, what's the smallest total number of slow gliders needed to produce from a block (each):

1) a splitter with one 0-degree output on the same lane as the input, and one at 90 degrees -- where the input comes from the same direction as the slow salvo;
2) two clean color-changing turners with different output parities, where the input is 90 degrees from the slow salvo;
3) two clean color-preserving one-time turners with different output parities, where the input is 90 degrees from the slow salvo

?

We have 1sL and 2sL one-time turners that do all these things, but the odds are high that other constellations will be much cheaper to construct with slow salvos. Quite possibly they'll be much larger and much uglier constellations, but if the goal is to minimize the construction cost of a freeze-dried slow salvo, that might not matter.

I guess it's probably better to add the requirement that the splitters and turners should be still-life constellations, so that there won't be any possible timing problems when it's time to trigger the seed constellation.
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Re: Challenges

Postby danny » June 5th, 2019, 7:33 am

dvgrn wrote:A glider-to-c/3-spaceship converter whose RLE fits in a forum posting and so can be run in LifeViewer.


Is this possible with the new LifeViewer format?
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Re: Challenges

Postby Moosey » June 5th, 2019, 8:00 am

I think it’s worth mentioning the stuff going on in the thread for academic questions— testitemqlstudop and I are attempting to be (possibly) the youngest people to devise a computable function which exceeds TREE(3) at 3, without using a preconstructed function (that is, any idiot could just say f(n) = TREE(n) + 1). Why?
aforampere sort of wrote:Either way, it’s rather nontrivial to devise a computable system which exceeds TREE(3)

(In response to a few functions I had devised, the last of which, σM(n), far exceeded Graham’s number at n=2, as it was equivalent to a huge stack of only somewhat fast growing functions, but the stack was far far far larger than graham’s number’s stack.)

I think this is a challenge (defining a function f(n) where f(n) grows absurdly fast) that may interest others.
Regardless, I think it’s appropriate to post it here.

So here are the rules of “the function-inventing tournament”:
You should attempt to devise a function f(n) such that:
f grows faster than TREE, or at the least, exceeds TREE(3) at n=3
f must not be based upon another predefined fast-growing function (I.e. no functions based on TREE(n) or any other functions of similar caliber), however, one may define a function or multiple functions which are necessary for whatever reason. You may also use more basic operations, like addition or multiplication or factorials, etc. I think you get the point of this rule, but I’m bad with words.
f can be defined in any way, be it in words or operators or both (however, if you are a beginner, please note that functions defined verbally are generally easier to make faster-growing than those defined other ways, however those defined simply numerically are far more easy to compose— but far more difficult to make huge! At least, that was my experience)
My rules:
They can be found here

Also, the tree game
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Re: Challenges

Postby fluffykitty » June 5th, 2019, 5:29 pm

Here's a somewhat boring entry:
f(0,x)=x
f(a,x)=f(g(a,a,x),x+1)
g(4b+2,x,n)=b
g(a,x,0)=0
g(4b+1,x,n+1)=2^g(x,x,n)*(4b+2)
g(2^a*(4b+2),x,n+1)=2^g(a,a,n+1)*(4*g(2^a*(4b+2),x,n)+2)
g(2^a*(4b+2)-1,x,n+1)=2^g(a,x,n+1)*(4*g(2^a*(4b+2)-1,x,n)+2)-1
Submit f(2^2^16,n)
If you want to test this function, use something like f(8,n) since even f(2^2^16,1) is massive.
Approximations of other functions:
Successor: f(2,n)
Weak Goodstein sequences/Ackermann function: f(512,n)
Strong Goodstein sequences/Simple hydras/tree: f(4,n)
TREE: f(2^(2^2046-1),n)
Something stronger than TREE: f(2^2^16,n)
Odd first arguments of f are not guaranteed to work correctly.
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Re: Challenges

Postby Moosey » June 5th, 2019, 5:37 pm

fluffykitty wrote:Here's a somewhat boring entry:
f(0,x)=x
f(a,x)=f(g(a,a,x),x+1)
g(4b+2,x,n)=b
g(a,x,0)=0
g(4b+1,x,n+1)=2^g(x,x,n)*(4b+2)
g(2^a*(4b+2),x,n+1)=2^g(a,a,n+1)*(4*g(2^a*(4b+2),x,n)+2)
g(2^a*(4b+2)-1,x,n+1)=2^g(a,x,n+1)*(4*g(2^a*(4b+2)-1,x,n)+2)-1
Submit f(2^2^16,n)
If you want to test this function, use something like f(8,n) since even f(2^2^16,1) is massive.
Approximations of other functions:
Successor: f(2,n)
Weak Goodstein sequences/Ackermann function: f(512,n)
Strong Goodstein sequences/Simple hydras/tree: f(4,n)
TREE: f(2^(2^2046-1),n)
Something stronger than TREE: f(2^2^16,n)
Odd first arguments of f are not guaranteed to work correctly.


That’s pretty good.

What about the revised MATRIXPARTY function I submitted on that thread? How do the two options work?

I wish I could find out how good my function is relative to TREE, but I need someone else to find it out.
My rules:
They can be found here

Also, the tree game
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Re: Challenges

Postby fluffykitty » June 5th, 2019, 5:50 pm

Done, but you won't like it
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Re: Challenges

Postby Moosey » June 5th, 2019, 6:08 pm

fluffykitty wrote:Done, but you won't like it

Dang, I sure don’t.

What if I make a function δM(m,n) (deltmoose)
Based on sigmoose(n) (http://www.conwaylife.com/forums/viewto ... 150#p76892)

Such that δM(m,n)=
σM(δM(m-1,σM(n))), m>0
σM(n), m=0


How big is
δM(δM(δM(σM(σM(10^100)),σM(σM(10^100))),σM(σM(10^100))),δM(δM(σM(σM(10^100)),σM(σM(10^100))),σM(σM(10^100))))
?

We have already established that sigmoose(2) > g_64 by far, since it’s Mltrs(f(3,1,4))
My rules:
They can be found here

Also, the tree game
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Re: Challenges

Postby fluffykitty » June 5th, 2019, 6:50 pm

deltmoose only adds 1 more FGH level (w+3), so your number is easily beaten by f_(w+4)(4) or f(2^2219,4) in my function. To gain more power, you need to do something like Mltrs again.
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Re: Challenges

Postby Moosey » June 5th, 2019, 7:35 pm

fluffykitty wrote:deltmoose only adds 1 more FGH level (w+3), so your number is easily beaten by f_(w+4)(4) or f(2^2219,4) in my function. To gain more power, you need to do something like Mltrs again.

Alright, so deltmooseextended (δME_n(p,q)) =
δM(δME_(n-1)(p,q),δME_(n-1)(p,q)), n > 0
δM(p,q), n = 0


And then Mgltrs(n,p,q) (Moosey Greek letter function stacked)
=
δME_(Mgltrs(n-1,p,q))(p,q), n > 0
δΜ(p,q), n = 0


And then, there is (I have the Cyrillic keyboard for the sole purpose of defining functions and making emoticons— (°Д°))
дM(n) (demoose) =
Mgltrs(дM(n-1),дM(n-1),дM(n-1)), n>0
Mgltrs(googol,googol,googol), n=0

The googol thing was rather arbitrarily chosen but whatever.

Now, if I define дME_m(n) like so:
дM(дME_(m-1)(n)) n > 0
дM(n), n = 0


And then Mcltrs(n) in the very same way—
Mcltrs(n,m) =
дME_(Mcltrs(n-1))(m), n>0
дM(googol,googol,googol), n = 0


I’m assuming it’s along the lines of ω^3.

While I’m at it, I might as well point out that the [new keyboard letter]M(n) is just the iterated version of the previous M[keyboard first letter]ltrs(n), and then the next [same keyboard letter]ME_m(n) is the iterated version of that, and then the stacked one is that sub that sub that sub that sub that...

How would I define a function which is to Mltrs, Mgltrs, and Mcltrs what Mltrs is to the Moosey letter functions? I’m assuming that would be ω^ω ish.

How big is Mcltrs(2)?
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Re: Challenges

Postby fluffykitty » June 5th, 2019, 7:44 pm

δME is FGH level w+4, Mgltrs is FGH level w+5, дM is FGH level w+6, дME is infinite but if defined correctly would be FGH level w+7, Mcltrs would be w+8, and the diagonalizer of M*ltrs would be w2. Try comparing the definitions of Mlrts and Mgltrs to see why one is a much larger step up than the other.
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Re: Challenges

Postby Moosey » June 5th, 2019, 8:05 pm

fluffykitty wrote:δME is FGH level w+4, Mgltrs is FGH level w+5, дM is FGH level w+6, дME is infinite but if defined correctly would be FGH level w+7, Mcltrs would be w+8, and the diagonalizer of M*ltrs would be w2. Try comparing the definitions of Mlrts and Mgltrs to see why one is a much larger step up than the other.

I don’t get it.
Is it not actually δΜ sub (δΜ sub δΜ....?
That’s how I meant to define it.
Or did I somehow define Mltrs far better than I intended to?
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Re: Challenges

Postby fluffykitty » June 5th, 2019, 8:45 pm

The problem is each function is nesting the previous function in a simple fashion, so each step adds 1 to the FGH level. However, the ltr_n function adds 2 to the FGH level for each increment of n, so something like ltr_n(4,4,4) has FGH level w.
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Re: Challenges

Postby Moosey » June 5th, 2019, 8:47 pm

fluffykitty wrote:The problem is each function is nesting the previous function in a simple fashion, so each step adds 1 to the FGH level. However, the ltr_n function adds 2 to the FGH level for each increment of n, so something like ltr_n(4,4,4) has FGH level w.

so how do I redefine mgltrs so that it creates an equal leap to that of mltrs ?
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Re: Challenges

Postby fluffykitty » June 5th, 2019, 8:51 pm

Well first you need an analog of ltr_n at the f_(w+n) level. Then you can nest that to get a much stronger Mgltrs function, make another ltr_n analog... and then diagonalize over all of these levels to reach w^2.
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Re: Challenges

Postby Moosey » June 5th, 2019, 8:58 pm

fluffykitty wrote:Well first you need an analog of ltr_n at the f_(w+n) level. Then you can nest that to get a much stronger Mgltrs function, make another ltr_n analog... and then diagonalize over all of these levels to reach w^2.

Okay, walk me through it. But I'm about to go to sleep!
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Re: Challenges

Postby fluffykitty » June 5th, 2019, 9:20 pm

You can do exactly the same thing as ltr_n but with ltr_n(n,n,n) (or similar) as the base function instead of f. From there extension to w^2 should be easy.
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Re: Challenges

Postby PkmnQ » June 6th, 2019, 4:10 am

み(x,y,z) = (x+y)! + み(xz,y,z-1)
み(x,y,0) = x! + み(x,y-1,xy)
み(x,0,0) = x!

ミ(x) = み(x+1,x+1,x^2)

み(1,1,0) = 3

み(2,2,1) = 24 + 2 + 6 + 9! + 25! + 49! + 48! + み(48,0,48)

Yeah...even with 4 statements it's powerful.
Don't know how it compares to tree though.
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