1.1: To determine if a pattern is order ≥n, run a single cell n generations. Now copy that, and xor-paste it so that the top left corner matches up with the leftmost cell on the highest row. Repeat this until either the pattern is empty or a cell appears to the right of all the live cells in a row or below the lowest row. For 2n, this splits the pattern into 4 subsets and finds a k-generation predecessor for each of them. See the diagram (for n=1):
Code: Select all
#C [[ VIEWONLY ]]
x = 148, y = 97, rule = LifeHistory
13.6A$6.2A2.3A.2A2.A33.3D3A$4.5A.A4.A.A2.5A20.2A2.3AEAD2.A$4.A2.A3.A.
A4.2A.2A20.5A.A2.2ED.A2.5A31.6D$3.2A2.4A.3A2.2A3.3A18.A2.A3.A.A4.2A.
2A26.2A2.3AEADE.D$3.4A2.A.2A2.5A2.A19.2A2.4A.3A2.2A3.3A22.5A.A2.2EDED
E.5A$3.A.A2.4A3.A3.3A.2A8.A8.4A2.A.2A2.5A2.A24.A2.A3.A.A4.2A.2A$3.A.
4A2.5A.4A.A11.A7.A.A2.4A3.A3.3A.2A21.2A2.4A.3A2.2A3.3A$3.3A4.5A3.3A.A
6.7A6.A.4A2.5A.4A.A11.A11.4A2.A.2A2.5A2.A$3.2A.4A3.2A.2A.3A12.A7.3A4.
5A3.3A.A12.A10.A.A2.4A3.A3.3A.2A$4.A2.2A.2A.3A3.A13.A8.2A.4A3.2A.2A.
3A8.7A9.A.4A2.5A.4A.A$4.4A.A.3A4.A24.A2.2A.2A.3A3.A15.A10.3A4.5A3.3A.
A$9.2A.A4.A25.4A.A.3A4.A15.A11.2A.4A3.2A.2A.3A$6.2A.A.2A3.A31.2A.A4.A
29.A2.2A.2A.3A3.A$7.A.A2.A2.A29.2A.A.2A3.A30.4A.A.3A4.A$6.A.A3.A.A31.
A.A2.A2.A36.2A.A4.A$6.2A4.2A31.A.A3.A.A34.2A.A.2A3.A$45.2A4.2A36.A.A
2.A2.A$88.A.A3.A.A$88.2A4.2A22$12.6A$5.2A2.3A.2A2.A42.CECECE$3.5A.A4.
A.A2.5A29.DA2.DAD.DA2.D30.A3.A3.A32.C.C.C$3.A2.A3.A.A4.2A.2A29.ECECE.
E4.C.C2.ECECE$2.2A2.4A.3A2.2A3.3A27.D2.A3.A.A4.DA.AD26.A3.A.A5.A.A18.
C.C7.C.C3.C.C$2.4A2.A.2A2.5A2.A28.CE2.CECE.ECE2.CE3.ECE$2.A.A2.4A3.A
3.3A.2A26.ADAD2.A.AD2.ADADA2.D22.A.A3.A.A3.A.A.A18.C3.C.C3.C3.C5.C$2.
A.4A2.5A.4A.A28.C.C2.ECEC3.C3.CEC.CE48.A$2.3A4.5A3.3A.A14.5A9.A.ADAD
2.ADADA.ADAD.D22.A.A.A3.A.A.A.A.A10.3A5.C.C3.C.C3.C3.C.C.C$2.2A.4A3.
2A.2A.3A29.CEC4.ECECE3.ECE.E14.A35.A$3.A2.2A.2A.3A3.A17.5A9.AD.DADA3.
AD.DA.ADA16.A6.A3.A.A3.A3.A.A.A16.C.C5.C.C5.C$3.4A.A.3A4.A33.E2.CE.EC
.CEC3.C13.7A$8.2A.A4.A34.DADA.A.ADA4.D19.A8.A.A.A.A.A28.C3.C.C.C3.C$
5.2A.A.2A3.A40.CE.E4.C19.A$6.A.A2.A2.A38.DA.A.AD3.D32.A.A.A32.C7.C$5.
A.A3.A.A40.C.C2.E2.C$5.2A4.2A40.D.D3.D.D69.C.C5.C$53.EC4.EC$131.C5.C
8$137.E.E.E$11.A19.A9.A.A.A$12.A6.3A10.A61.D3.D.D.D3.D20.E.E.E.E9.E.E
.E$5.A3.5A5.A.A7.5A7.A3.A$12.A6.3A10.A59.D13.D3.D16.E3.E.E.E.E3.E3.E.
E$11.A19.A9.A.A.A$92.D.D5.D3.D.D3.D20.E.E9.E3.E$86.A32.A$85.3A6.D.D3.
D.D3.D.D.D7.3A6.E5.E.E.E3.E.E.E$86.A32.A$92.D.D.D5.D.D3.D18.E3.E.E3.E
2$92.D.D5.D5.D26.E.E2$94.D5.D3.D30.E2$94.D.D3.D.D26.E5.E$4.D.E.D3.A2$
4.D3.D.A5.A.A2$A.A.E.D.D3.A.A.A10.A5.A$26.A7.A8.DED.A$A.A.A3.A.A.A.A.
A8.11A7.D.DA2.2A$26.A7.A6.2AE2D.3A$A3.A.A3.A3.A.A.A8.A5.A7.3A.5A$41.A
.2A.A.3A$2.A.A.A.A.A31.5A$43.3A$4.A.A.A!
This can also be used to find the n-generation predecessor of a pattern.
For both Fredkin and Replicator, this even works on any infinite pattern where every live cell is below a given line of positive slope.
1.2: For bounding box, this is definitely the case. If you run some pattern n generations, it increases each side of the bounding box by 2n. Obviously the smallest this can be is (2n+1)*(2n+1), corresponding to the single cell. For population, I am not so certain, but for a pair of cells to have a smaller generation-n pattern, over half of their cells would need to overlap, which seems impossible.
1.3: Yes. The method in 1.1 generates a unique predecessor, which we can prove easily. Consider the leftmost cell of the top row (the first cell to be removed). There is a unique cell which can create it in n generations without creating cells to the left of or above it, namely the cell n cells to the right and n cells down from it, which must be on. After xoring the n-generation image of a single cell, consider the cell directly to the right of the original cell. If it is off, then the cell n cells down and to the right must also be off, and if it is on, then the cell n cells diagonally must also be on. By induction, every cell on the top row is uniquely determined. We can repeat the same process on the next row, and this shows that every cell must be uniquely determined as well.
2.1: No. In Fredkin you can xor lines of dominoes in arbitrary positions over the pattern and it will still work:
Code: Select all
x = 96, y = 40, rule = B1357/S02468
7b2o$o4b2ob2o$o2b2o3bo48b2o$ob2o3bo42bo4b2ob2o$obo3bo32b2ob2ob2ob6obob
o2b2ob2ob2ob2o$o4bo7bo36bob2o3bo$o3bo7b3o3b2ob2ob2ob2ob2ob2ob2o12bobo
3bo$o3bo8bo36bo4bo$ob2o46bo3bo$2o48bo3bo$o49bob2o$50b2o$50bo2$86bo$86b
o2$86bo$86bo2$86bo$39b2ob2ob2ob2ob2ob2o4b2ob2ob2ob2o15bo6b2o$50bo4b2ob
2o26bo4b2ob2o$50bo2b2o3bo30b2o3bo$50bob2o3bo30b2o3bo$50bobo3bo29bobo3b
o$50bo4bo35bo$50bo3bo35bo$39b2ob2ob2ob5o2bob2ob2ob2ob2ob2o15bo3bo$50bo
b2o34b2o$50b2o35bo$50bo35bo$86bo$86bo2$86bo$86bo2$86bo$86bo!
In Replicator the situation is a bit more complicated because I don't know if there are any dying wicks. However, you can xor the following infinite patterns on any pattern (tile it so there is gutter symmetry):
Code: Select all
x = 109, y = 31, rule = B1357/S1357
24b2ob2ob2ob2ob2obob2ob2ob2ob2ob2o23b2ob2ob2ob2ob2obob2ob2ob2ob2ob2o$
24bo3bobo3bobo5bobo3bobo3bo23bo3bobo3bobo5bobo3bobo3bo$26b3ob3o5b3o5b
3ob3o27b3ob3o5b3o5b3ob3o$24bobo5bobobobobobobobo5bobo23bobo5bobobobobo
bobobo5bobo$24b3ob2ob2o5b3o5b2ob2ob3o23b3ob2ob2o5b3o5b2ob2ob3o$28bo5bo
bo5bobo5bo31bo5bobo5bobo5bo$7b3o14b3o3b2ob2ob2obob2ob2ob2o3b3o23b3o3b
2ob2ob2obob2ob2ob2o3b3o$5b3o16bobobobo17bobobobo23bobobobo17bobobobo$
3b4o19b3o3b2ob2ob3ob2ob2o3b3o27b3o3b2ob2ob3ob2ob2o3b3o$2bobo19bo5bobo
3bobobobo3bobo5bo23bo5bobo3bobobobo3bobo5bo$2b2obo18b2obob2o3b3ob3ob3o
3b2obob2o23b2obob2o3b4o2bob3o3b2obob2o$2b4o9bo16bobo9bobo39bob4o6bobo$
b2o11b3o7b2obob2ob3ob2obob2ob3ob2obob2o23b2obob2obo2bobobob2ob3ob2obob
2o$2o13bo8bo5bo5bo5bo5bo5bo23bo5bobobobo5bo5bo5bo$o25b3o3b3o3b3o3b3o3b
3o27b3o5b2o2b3o3b3o3b3o$24bobobobobobobobobobobobobobobobo23bobobobo2b
ob2obobobobobobobobobo$26b3o3b3o3b3o3b3o3b3o27b3o2bob2o3b3o3b3o3b3o$
24bo5bo5bo5bo5bo5bo23bo6bo4bo5bo5bo5bo$24b2obob2ob3ob2obob2ob3ob2obob
2o23b2obobo2b3ob2obob2ob3ob2obob2o$32bobo9bobo39bobo9bobo$24b2obob2o3b
3ob3ob3o3b2obob2o23b2obob2o3b3ob3ob3o3b2obob2o$24bo5bobo3bobobobo3bobo
5bo23bo5bobo3bobobobo3bobo5bo$26b3o3b2ob2ob3ob2ob2o3b3o27b3o3b2ob2ob3o
b2ob2o3b3o$24bobobobo17bobobobo23bobobobo17bobobobo$24b3o3b2ob2ob2obob
2ob2ob2o3b3o23b3o3b2ob2ob2obob2ob2ob2o3b3o$28bo5bobo5bobo5bo31bo5bobo
5bobo5bo$24b3ob2ob2o5b3o5b2ob2ob3o23b3ob2ob2o5b3o5b2ob2ob3o$24bobo5bob
obobobobobobo5bobo23bobo5bobobobobobobobo5bobo$26b3ob3o5b3o5b3ob3o27b
3ob3o5b3o5b3ob3o$24bo3bobo3bobo5bobo3bobo3bo23bo3bobo3bobo5bobo3bobo3b
o$24b2ob2ob2ob2ob2obob2ob2ob2ob2ob2o23b2ob2ob2ob2ob2obob2ob2ob2ob2ob2o
!
This can be done on any run of a dot to (2^n)-1 generations and repeating it forever, or using the infinite limit, which can be pasted over any set of points {b}. Both methods give you an uncountable number of infinite predecessors.
2.2: See above.
3:
Code: Select all
@RULE ReplicatorReverser
@TABLE
n_states: 9
neighborhood: Moore
symmetries: none
var a={0,1}
var b=a
var c=a
var d=a
var e=a
var f=a
var g=a
var i={5,7}
var j={1,3}
var k={1,3}
var l={0,3}
# Motion of state 2 cells
0,0,0,a,b,c,d,2,0,2
2,0,0,a,b,c,d,0,0,0
# After touching a corner, the 8 cells are inverted. A state 6 cell escapes to the top.
1,0,0,a,b,c,d,2,0,4
0,a,b,c,d,e,f,g,4,5
1,a,b,c,d,e,f,g,4,7
4,0,0,a,b,c,d,0,0,0
0,0,0,0,a,4,0,0,0,6
0,0,0,0,0,6,0,0,0,6
6,0,0,0,a,0,0,0,0,0
1,0,0,a,b,i,c,0,6,0
0,0,0,a,b,i,c,0,6,1
1,0,0,a,b,c,i,d,0,0
0,0,0,a,b,c,i,d,0,1
1,a,b,c,d,e,f,i,g,0
0,a,b,c,d,e,f,i,g,1
1,a,b,c,d,e,f,g,i,0
0,a,b,c,d,e,f,g,i,1
1,i,a,b,c,d,e,f,g,0
0,i,a,b,c,d,e,f,g,1
1,a,i,b,c,d,e,f,g,0
0,a,i,b,c,d,e,f,g,1
1,0,a,i,b,c,d,e,f,0
0,0,a,i,b,c,d,e,f,1
5,a,b,c,d,e,f,g,0,0
7,a,b,c,d,e,f,g,0,1
# State 2 cell resumes its course and bounces off state 3 ends.
0,6,0,a,i,b,c,d,0,2
0,0,0,3,0,0,0,2,0,2
0,0,0,0,3,0,2,0,0,6
2,6,0,3,0,0,0,0,0,0
0,0,6,2,0,0,0,0,0,2
0,0,0,6,2,0,0,0,0,2
0,0,2,2,a,b,c,0,0,2
0,0,0,2,2,0,0,0,0,2
3,0,0,0,0,0,0,2,6,0
0,3,0,0,0,0,0,0,2,3
6,0,0,0,3,2,0,0,0,0
6,0,0,0,0,0,2,0,0,0
2,2,0,0,a,b,c,0,0,0
2,0,6,0,0,2,0,0,0,0
2,0,0,0,0,2,0,0,0,0
0,0,2,0,0,0,0,0,3,2
2,0,0,0,0,0,0,0,3,0
0,3,0,2,0,0,0,0,0,3
3,0,0,0,2,0,0,0,0,0
0,0,0,0,2,0,3,0,0,2
3,3,0,0,0,0,0,0,l,0
2,0,0,0,0,0,3,0,0,0
# When a state 6 cell hits the state 3 row at the top, it turns that cell into 1.
0,3,3,0,0,6,0,0,j,6
6,3,3,0,0,0,0,0,j,0
3,a,b,3,0,6,0,j,c,1
0,3,0,0,6,0,0,0,3,2
# The state 6 cell at the far right makes a state 2 cell, which recreates the state 3 row 1 cell downward.
3,a,b,c,l,2,0,j,d,0
2,j,a,l,0,0,0,0,k,3
0,j,k,2,0,0,0,0,1,2
0,j,k,2,0,0,0,0,3,2
0,3,0,3,0,0,0,0,0,3
3,0,0,0,3,0,0,0,0,0
@COLORS
1 150 49 104
2 14 208 240
5 114 44 186
4 125 245 65
6 215 80 172
7 199 45 27
Code: Select all
@RULE FredkinReverser
@TABLE
n_states: 9
neighborhood: Moore
symmetries: none
var a={0,1}
var b=a
var c=a
var d=a
var e=a
var f=a
var g=a
var i={5,7}
var j={1,3}
var k={1,3}
var l={0,3}
# Motion of state 2 cells
0,0,0,a,b,c,d,2,0,2
2,0,0,a,b,c,d,0,0,0
# After touching a corner, the 8 cells are inverted. A state 6 cell escapes to the top.
1,0,0,a,b,c,d,2,0,4
0,a,b,c,d,e,f,g,4,7
1,a,b,c,d,e,f,g,4,5
4,0,0,a,b,c,d,0,0,0
0,0,0,0,a,4,0,0,0,6
0,0,0,0,0,6,0,0,0,6
6,0,0,0,a,0,0,0,0,0
1,0,0,a,b,i,c,0,6,0
0,0,0,a,b,i,c,0,6,1
1,0,0,a,b,c,i,d,0,0
0,0,0,a,b,c,i,d,0,1
1,a,b,c,d,e,f,i,g,0
0,a,b,c,d,e,f,i,g,1
1,a,b,c,d,e,f,g,i,0
0,a,b,c,d,e,f,g,i,1
1,i,a,b,c,d,e,f,g,0
0,i,a,b,c,d,e,f,g,1
1,a,i,b,c,d,e,f,g,0
0,a,i,b,c,d,e,f,g,1
1,0,a,i,b,c,d,e,f,0
0,0,a,i,b,c,d,e,f,1
5,a,b,c,d,e,f,g,0,0
7,a,b,c,d,e,f,g,0,1
# State 2 cell resumes its course and bounces off state 3 ends.
0,6,0,a,i,b,c,d,0,2
0,0,0,3,0,0,0,2,0,2
0,0,0,0,3,0,2,0,0,6
2,6,0,3,0,0,0,0,0,0
0,0,6,2,0,0,0,0,0,2
0,0,0,6,2,0,0,0,0,2
0,0,2,2,a,b,c,0,0,2
0,0,0,2,2,0,0,0,0,2
3,0,0,0,0,0,0,2,6,0
0,3,0,0,0,0,0,0,2,3
6,0,0,0,3,2,0,0,0,0
6,0,0,0,0,0,2,0,0,0
2,2,0,0,a,b,c,0,0,0
2,0,6,0,0,2,0,0,0,0
2,0,0,0,0,2,0,0,0,0
0,0,2,0,0,0,0,0,3,2
2,0,0,0,0,0,0,0,3,0
0,3,0,2,0,0,0,0,0,3
3,0,0,0,2,0,0,0,0,0
0,0,0,0,2,0,3,0,0,2
3,3,0,0,0,0,0,0,l,0
2,0,0,0,0,0,3,0,0,0
# When a state 6 cell hits the state 3 row at the top, it turns that cell into 1.
0,3,3,0,0,6,0,0,j,6
6,3,3,0,0,0,0,0,j,0
3,a,b,3,0,6,0,j,c,1
0,3,0,0,6,0,0,0,3,2
# The state 6 cell at the far right makes a state 2 cell, which recreates the state 3 row 1 cell downward.
3,a,b,c,l,2,0,j,d,0
2,j,a,l,0,0,0,0,k,3
0,j,k,2,0,0,0,0,1,2
0,j,k,2,0,0,0,0,3,2
0,3,0,3,0,0,0,0,0,3
3,0,0,0,3,0,0,0,0,0
@COLORS
1 150 49 104
2 14 208 240
5 114 44 186
4 125 245 65
6 215 80 172
7 199 45 27
The only change in the Fredkin version is that the 5 and 7 in 0,a,b,c,d,e,f,g,4,5 and 1,a,b,c,d,e,f,g,4,7 are swapped.
Example usage:
Code: Select all
x = 158, y = 155, rule = ReplicatorReverser
11.145C47$C2.B11.A3.5A2.2A2.A2.7A.3A3.3A4.3A3.4A.4A.A.3A.A.A3.2A2.A.A
.A.A.3A2.3A.A.2A.6A.A.A2.A.A2.6A2.3A2.3A2.A13.C$13.A4.A2.2A2.A.6A.A.
2A.A.A2.2A2.3A2.4A5.3A4.3A2.A.A6.5A2.2A2.2A.A.2A2.2A3.2A3.A3.A.A.A3.A
2.6A.A.3A$13.2A.A.A5.2A2.A5.A3.A3.5A.4A3.9A.2A.A3.A2.A.A4.2A.4A.2A.A.
2A3.4A.2A4.2A.A3.A2.4A.2A2.A.3A.A.A$15.A2.2A.A.3A.A.3A2.2A5.A.5A.3A3.
A4.4A2.2A.A.A2.A2.A3.2A.A3.A2.A2.A.2A.2A2.A2.A.2A.A.A3.A10.A.A.A.A3.A
.4A$13.2A3.A3.A4.3A3.2A4.2A.2A.2A2.A4.A.2A.2A2.3A.A5.A2.A.3A.2A.2A2.
3A.A3.3A.2A2.2A2.A3.A.2A.A2.2A2.4A.A.2A2.A2.2A.A$14.3A.5A.3A6.5A.4A2.
A5.8A5.A.A2.A.2A2.3A.5A3.A.A.A4.2A2.2A.2A.3A.2A.A.A6.2A.2A.3A.3A3.4A.
A$14.2A2.A2.A.A2.A.2A3.2A2.2A.3A2.3A.3A.7A3.3A.2A.4A3.A2.A2.A.2A.A4.
2A2.5A.A3.A6.2A.2A.3A2.3A.4A3.2A.3A$13.A6.A.A2.5A2.2A6.4A.A.2A2.A3.2A
.A.A.3A2.6A.2A.2A.A.2A5.2A.2A2.3A3.2A.A3.2A.2A.A3.A2.2A.2A2.7A2.A.4A$
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@RULE ReplicatorReverser
@TABLE
n_states: 9
neighborhood: Moore
symmetries: none
var a={0,1}
var b=a
var c=a
var d=a
var e=a
var f=a
var g=a
var i={5,7}
var j={1,3}
var k={1,3}
var l={0,3}
# Motion of state 2 cells
0,0,0,a,b,c,d,2,0,2
2,0,0,a,b,c,d,0,0,0
# After touching a corner, the 8 cells are inverted. A state 6 cell escapes to the top.
1,0,0,a,b,c,d,2,0,4
0,a,b,c,d,e,f,g,4,5
1,a,b,c,d,e,f,g,4,7
4,0,0,a,b,c,d,0,0,0
0,0,0,0,a,4,0,0,0,6
0,0,0,0,0,6,0,0,0,6
6,0,0,0,a,0,0,0,0,0
1,0,0,a,b,i,c,0,6,0
0,0,0,a,b,i,c,0,6,1
1,0,0,a,b,c,i,d,0,0
0,0,0,a,b,c,i,d,0,1
1,a,b,c,d,e,f,i,g,0
0,a,b,c,d,e,f,i,g,1
1,a,b,c,d,e,f,g,i,0
0,a,b,c,d,e,f,g,i,1
1,i,a,b,c,d,e,f,g,0
0,i,a,b,c,d,e,f,g,1
1,a,i,b,c,d,e,f,g,0
0,a,i,b,c,d,e,f,g,1
1,0,a,i,b,c,d,e,f,0
0,0,a,i,b,c,d,e,f,1
5,a,b,c,d,e,f,g,0,0
7,a,b,c,d,e,f,g,0,1
# State 2 cell resumes its course and bounces off state 3 ends.
0,6,0,a,i,b,c,d,0,2
0,0,0,3,0,0,0,2,0,2
0,0,0,0,3,0,2,0,0,6
2,6,0,3,0,0,0,0,0,0
0,0,6,2,0,0,0,0,0,2
0,0,0,6,2,0,0,0,0,2
0,0,2,2,a,b,c,0,0,2
0,0,0,2,2,0,0,0,0,2
3,0,0,0,0,0,0,2,6,0
0,3,0,0,0,0,0,0,2,3
6,0,0,0,3,2,0,0,0,0
6,0,0,0,0,0,2,0,0,0
2,2,0,0,a,b,c,0,0,0
2,0,6,0,0,2,0,0,0,0
2,0,0,0,0,2,0,0,0,0
0,0,2,0,0,0,0,0,3,2
2,0,0,0,0,0,0,0,3,0
0,3,0,2,0,0,0,0,0,3
3,0,0,0,2,0,0,0,0,0
0,0,0,0,2,0,3,0,0,2
3,3,0,0,0,0,0,0,l,0
2,0,0,0,0,0,3,0,0,0
# When a state 6 cell hits the state 3 row at the top, it turns that cell into 1.
0,3,3,0,0,6,0,0,j,6
6,3,3,0,0,0,0,0,j,0
3,a,b,3,0,6,0,j,c,1
0,3,0,0,6,0,0,0,3,2
# The state 6 cell at the far right makes a state 2 cell, which recreates the state 3 row 1 cell downward.
3,a,b,c,l,2,0,j,d,0
2,j,a,l,0,0,0,0,k,3
0,j,k,2,0,0,0,0,1,2
0,j,k,2,0,0,0,0,3,2
0,3,0,3,0,0,0,0,0,3
3,0,0,0,3,0,0,0,0,0
@COLORS
1 150 49 104
2 14 208 240
5 114 44 186
4 125 245 65
6 215 80 172
7 199 45 27
The predecessor is shifted 1 cell to the left compared to the original, which I probably could have fixed.
{a}: By definition, there is only one finite predecessor. We can find this by assuming that there is a 2-cell wide empty box around it.
{b}: This is a fractal, which can be constructed because given a pattern run 2^n-1 generations, you can construct the pattern at 2^(n+1)-1 by xoring copies over the 8 dots. You would repeat this forever. See the diagram:
Code: Select all
x = 91, y = 17, rule = B1357/S1357
74bo7bo7bo$19bo3bo3bo21b2ob2ob3ob2ob2o$2obob2o42bo3bobobobo3bo$o5bo44b
3ob3ob3o$2b3o44bobo9bobo$obobobo12bo7bo21b3ob2obob2ob3o$2b3o48bo5bo$o
5bo42b3o3b3o3b3o$2obob2o42bobobobobobobobo10bo15bo$19bo3bo3bo21b3o3b3o
3b3o$53bo5bo$49b3ob2obob2ob3o$49bobo9bobo$51b3ob3ob3o$49bo3bobobobo3bo
$49b2ob2ob3ob2ob2o$74bo7bo7bo!
There is probably a simple way to determine the infinite limit without doing that.
Any sufficiently advanced software is indistinguishable from malice.