83bismuth38 wrote:Yeah, it just plows right through.

Okay, let's see how long it takes. We can start with ThreeDeadlyEnemies, but we might need some new variables, or we might not need some of the variables we have, depending on which rule variant we set up. There are still a couple of reasonable possibilities:

1) State 1 kills State 2 when it touches it, and thinks it's an OFF state, but State 2 thinks State 1 is an ON state (so three-neighbor births will still happen when an OFF cell has a combination of State 1 and State 2 neighbors)

or

2) State 1 kills State 2, but State 2 thinks State 1 is an OFF state

I'll try #2, because there are just a couple of complicated parts. Here's a test pattern:

`x = 94, y = 40, rule = RockScissorsPaperLife`

38.B.B$38.2B$39.B5$82.A.A$82.A.A$82.3A6$23.A.A$23.2A$24.A$73.B.B15.C.

C$73.B.B15.C.C$73.3B15.3C$17.B$17.2B$16.B.B14$.C$.2C$C.C!

With the three pi explosions all near each other, they all affect one color and are affected by another, so you don't get any standard pi-debris shapes. But if you delete any one out of the three, one of the other two should dominate and not even notice that the remaining color is there, so you'll get one standard pi debris shape.

Here's what I think is the rule --

EDIT: patched 4/19/2017 in response to

bug report below.

`@RULE RockScissorsPaperLife`

@TABLE

n_states:4

neighborhood:Moore

symmetries:permute

var a={0,1,2,3}

var b={0,1,2,3}

var c={0,1,2,3}

var d={0,1,2,3}

var e={0,1,2,3}

var f={0,1,2,3}

var g={0,1,2,3}

var h={0,1,2,3}

# Dead cells for State 1

var i={0,2,3}

var j={0,2,3}

var k={0,2,3}

var l={0,2,3}

var m={0,2,3}

var n={0,2,3}

# Dead cells for State 2

var o={0,1,3}

var p={0,1,3}

var q={0,1,3}

var r={0,1,3}

var s={0,1,3}

var t={0,1,3}

# Dead cells for State 3

var u={0,1,2}

var v={0,1,2}

var w={0,1,2}

var x={0,1,2}

var y={0,1,2}

var z={0,1,2}

var any={0,1,2,3}

# Birth in competitive circumstances

0,1,1,1,3,3,3,u,v,3

1,1,1,1,3,3,3,u,v,3

0,2,2,2,1,1,1,i,j,1

2,2,2,2,1,1,1,i,j,1

0,3,3,3,2,2,2,o,p,2

3,3,3,3,2,2,2,o,p,2

# Birth

0,1,1,1,i,j,k,l,m,1

2,1,1,1,i,j,k,l,m,1

0,2,2,2,o,p,q,r,s,2

3,2,2,2,o,p,q,r,s,2

0,3,3,3,u,v,w,x,y,3

1,3,3,3,u,v,w,x,y,3

# the RockScissorsPaper rules

3,2,a,b,c,d,e,f,g,0

2,1,a,b,c,d,e,f,g,0

1,3,a,b,c,d,e,f,g,0

# Three-neighbor survival

1,1,1,1,i,j,k,l,m,1

2,2,2,2,o,p,q,r,s,2

3,3,3,3,u,v,w,x,y,3

# Two-neighbor survival

1,1,1,i,j,k,l,m,n,1

2,2,2,o,p,q,r,s,t,2

3,3,3,u,v,w,x,y,z,3

# Death

any,a,b,c,d,e,f,g,h,0

We just had to add rules for

-- State 3 with any State 2 neighbor dies (overrules any other thing that might happen);

-- State 2 with any State 1 neighbor dies;

-- State 1 with any State 3 neighbor dies;

`# the RockScissorsPaper rules`

3,2,a,b,c,d,e,f,g,0

2,1,a,b,c,d,e,f,g,0

1,3,a,b,c,d,e,f,g,0

Then we had to take away the rules that say that multiple colors can collaborate on new births: replace the 'i's with 1, 2, and 3 respectively, in "0,1,1,i,0,0,0,0,0,1", "0,2,2,i,0,0,0,0,0,2", and "0,3,3,i,0,0,0,0,0,3".

But then we also had to add variables to allow births to still happen when there are more than three neighbors, as long as they're different colors:

`# Dead cells for State 1`

var i={0,2,3}

var j={0,2,3}

var k={0,2,3}

var l={0,2,3}

var m={0,2,3}

var n={0,2,3}

# Dead cells for State 2

var o={0,1,3}

var p={0,1,3}

var q={0,1,3}

var r={0,1,3}

var s={0,1,3}

var t={0,1,3}

# Dead cells for State 3

var u={0,1,2}

var v={0,1,2}

var w={0,1,2}

var x={0,1,2}

var y={0,1,2}

var z={0,1,2}

var any={0,1,2,3}

# Birth

0,1,1,1,h,i,j,k,l,1

0,2,2,2,m,n,o,p,q,2

0,3,3,3,r,s,t,u,v,3

There are just three more annoyances:

1) This last "Birth" addition will be asymmetric, because "0,1,1,1,h,i,j,k,l,1" will produce a State 1 cell even when the surroundings are something like "1,1,1,3,3,3,0,0". Paper covers rock, so presumably when both paper and rock are trying for a new birth in the same cell, paper ought to win...! These rules are more specific cases, so they have to be listed above the regular birth rules, above, or they'll never get triggered:

`# Birth in competitive circumstances`

0,1,1,1,3,3,3,h,i,3

0,2,2,2,1,1,1,m,n,1

0,3,3,3,2,2,2,r,s,2

2) Births still happen even if the cell is a losing color instead of really OFF:

[

EDIT: I picked the wrong variables in the "Birth in competitive circumstances" section, so the text below is wrong -- compare the current rule text, or see

here]

`# Birth in competitive circumstances`

0,1,1,1,3,3,3,i,j,3

1,1,1,1,3,3,3,i,j,3

0,2,2,2,1,1,1,o,p,1

2,2,2,2,1,1,1,o,p,1

0,3,3,3,2,2,2,u,v,2

3,3,3,3,2,2,2,u,v,2

# Birth

0,1,1,1,i,j,k,l,m,1

2,1,1,1,i,j,k,l,m,1

0,2,2,2,o,p,q,r,s,2

3,2,2,2,o,p,q,r,s,2

0,3,3,3,u,v,w,x,y,3

1,3,3,3,u,v,w,x,y,3

3) The rock/scissors/paper death rules have to come after the above birth rules, because a cell shouldn't actually die if it's due to be born as another color on the next tick. If the death rules come first, you get non-Life-standard behavior in in two-color situations, in the winning color as well as the losing color.