Re: Challenges
Posted: October 3rd, 2018, 9:17 pm
Do we have to use 2-sate isotropic, or can we do a non-symmetric rule with this behavior?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?wwei23 wrote: Find one of those that uses a replicator not based on this one.
Hint: It's use 2-state isotropic rule.AforAmpere wrote:Do we have to use 2-sate isotropic, or can we do a non-symmetric rule with this behavior?wwei23 wrote: Find one of those that uses a replicator not based on this one.
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.AforAmpere wrote:Do we have to use 2-sate isotropic, or can we do a non-symmetric rule with this behavior?wwei23 wrote: Find one of those that uses a replicator not based on this one.
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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 ]]Code: Select all
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!
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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!
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: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.
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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!
Code: Select all
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!
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.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)...
Is this possible with the new LifeViewer format?dvgrn wrote:A glider-to-c/3-spaceship converter whose RLE fits in a forum posting and so can be run in LifeViewer.
(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.)aforampere sort of wrote:Either way, it’s rather nontrivial to devise a computable system which exceeds TREE(3)
That’s pretty good.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.
Dang, I sure don’t.fluffykitty wrote:Done, but you won't like it
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σM(δM(m-1,σM(n))), m>0
σM(n), m=0Code: Select all
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))))Alright, so deltmooseextended (δME_n(p,q)) =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.
Code: Select all
δM(δME_(n-1)(p,q),δME_(n-1)(p,q)), n > 0
δM(p,q), n = 0Code: Select all
δME_(Mgltrs(n-1,p,q))(p,q), n > 0
δΜ(p,q), n = 0Code: Select all
Mgltrs(дM(n-1),дM(n-1),дM(n-1)), n>0
Mgltrs(googol,googol,googol), n=0Code: Select all
дM(дME_(m-1)(n)) n > 0
дM(n), n = 0Code: Select all
дME_(Mcltrs(n-1))(m), n>0
дM(googol,googol,googol), n = 0I don’t get it.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.
so how do I redefine mgltrs so that it creates an equal leap to that of mltrs ?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.
Okay, walk me through it. But I'm about to go to sleep!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.