Binary slow salvos

For discussion of specific patterns or specific families of patterns, both newly-discovered and well-known.
gameoflifemaniac
Posts: 1224
Joined: January 22nd, 2017, 11:17 am
Location: There too

Re: Binary slow salvos

Will there be an algorithm that will be optimized for the RCT and run billions or trillions of orders of magnitude faster than HashLife does?
I was so socially awkward in the past and it will haunt me for the rest of my life.

Code: Select all

``````b4o25bo\$o29bo\$b3o3b3o2bob2o2bob2o2bo3bobo\$4bobo3bob2o2bob2o2bobo3bobo\$
4bobo3bobo5bo5bo3bobo\$o3bobo3bobo5bo6b4o\$b3o3b3o2bo5bo9bobo\$24b4o!``````

calcyman
Posts: 2352
Joined: June 1st, 2009, 4:32 pm

Re: Binary slow salvos

I've found a universal set for MathAndCode's original 17-glider RCT. Specifically, we have:
• a 3fd pull and 4fd push, which together allow arbitrary-integer fd moves;
• a nondestructive glider emission reaction;
• a destructive block creation reaction which throws the block beyond any of the lanes used for either of the above;
• a 14fd pull which can follow an elbow-duplication operation.

Code: Select all

``````x = 139813, y = 139605, rule = LifeHistory
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99.A199.A199.A510\$92472.3A197.3A\$92474.A199.A\$92473.A199.A212\$92164.
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80440.3A\$80442.A\$80441.A254\$80184.3A197.3A\$80186.A199.A\$80185.A199.A
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199.A\$48953.A199.A724\$48232.2A198.2A\$48231.A.A197.A.A\$48233.A199.A40\$
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46336.3A\$46338.A\$46337.A254\$46080.3A\$46082.A\$46081.A724\$45360.2A\$
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43776.3A\$43778.A\$43777.A254\$43520.3A\$43522.A\$43521.A510\$43008.3A\$
43010.A\$43009.A724\$42288.2A\$42287.A.A\$42289.A40\$42240.3A\$42242.A\$
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40752.2A\$40751.A.A\$40753.A40\$40704.3A\$40706.A\$40705.A254\$40448.3A\$
40450.A\$40449.A510\$39936.3A\$39938.A\$39937.A510\$39424.3A\$39426.A\$
39425.A510\$38912.3A\$38914.A\$38913.A510\$38400.3A\$38402.A\$38401.A724\$
37680.2A\$37679.A.A\$37681.A40\$37632.3A\$37634.A\$37633.A254\$37376.3A\$
37378.A\$37377.A510\$36864.3A\$36866.A\$36865.A724\$36144.2A\$36143.A.A\$
36145.A40\$36096.3A\$36098.A\$36097.A254\$35840.3A\$35842.A\$35841.A510\$
35328.3A\$35330.A\$35329.A510\$34816.3A\$34818.A\$34817.A510\$34304.3A\$
34306.A\$34305.A724\$33584.2A\$33583.A.A\$33585.A40\$33536.3A\$33538.A\$
33537.A254\$33280.3A\$33282.A\$33281.A510\$32768.3A\$32770.A\$32769.A510\$
32256.3A\$32258.A\$32257.A724\$31536.2A\$31535.A.A\$31537.A40\$31488.3A\$
31490.A\$31489.A468\$31024.2A\$31023.A.A\$31025.A40\$30976.3A\$30978.A\$
30977.A254\$30720.3A\$30722.A\$30721.A510\$30208.3A\$30210.A\$30209.A724\$
29488.2A\$29487.A.A\$29489.A40\$29440.3A\$29442.A\$29441.A254\$29184.3A\$
29186.A\$29185.A510\$28672.3A\$28674.A\$28673.A724\$27952.2A\$27951.A.A\$
27953.A40\$27904.3A\$27906.A\$27905.A254\$27648.3A\$27650.A\$27649.A510\$
27136.3A\$27138.A\$27137.A724\$26416.2A\$26415.A.A\$26417.A40\$26368.3A\$
26370.A\$26369.A254\$26112.3A\$26114.A\$26113.A510\$25600.3A\$25602.A\$
25601.A510\$25088.3A\$25090.A\$25089.A510\$24576.3A\$24578.A\$24577.A510\$
24064.3A\$24066.A\$24065.A724\$23344.2A\$23343.A.A\$23345.A40\$23296.3A\$
23298.A\$23297.A254\$23040.3A\$23042.A\$23041.A510\$22528.3A\$22530.A\$
22529.A510\$22016.3A\$22018.A\$22017.A724\$21296.2A\$21295.A.A\$21297.A40\$
21248.3A\$21250.A\$21249.A254\$20992.3A\$20994.A\$20993.A510\$20480.3A\$
20482.A\$20481.A510\$19968.3A\$19970.A\$19969.A724\$19248.2A\$19247.A.A\$
19249.A40\$19200.3A\$19202.A\$19201.A254\$18944.3A\$18946.A\$18945.A510\$
18432.3A\$18434.A\$18433.A724\$17712.2A\$17711.A.A\$17713.A40\$17664.3A\$
17666.A\$17665.A468\$17200.2A\$17199.A.A\$17201.A40\$17152.3A\$17154.A\$
17153.A468\$16688.2A\$16687.A.A\$16689.A40\$16640.3A\$16642.A\$16641.A254\$
16384.3A\$16386.A\$16385.A724\$15664.2A\$15663.A.A\$15665.A40\$15616.3A\$
15618.A\$15617.A254\$15360.3A\$15362.A\$15361.A510\$14848.3A\$14850.A\$
14849.A724\$14128.2A\$14127.A.A\$14129.A40\$14080.3A\$14082.A\$14081.A254\$
13824.3A\$13826.A\$13825.A510\$13312.3A\$13314.A\$13313.A724\$12592.2A\$
12591.A.A\$12593.A40\$12544.3A\$12546.A\$12545.A254\$12288.3A\$12290.A\$
12289.A510\$11776.3A\$11778.A\$11777.A724\$11056.2A\$11055.A.A\$11057.A40\$
11008.3A\$11010.A\$11009.A254\$10752.3A\$10754.A\$10753.A510\$10240.3A\$
10242.A\$10241.A510\$9728.3A\$9730.A\$9729.A510\$9216.3A\$9218.A\$9217.A724\$
8496.2A\$8495.A.A\$8497.A40\$8448.3A\$8450.A\$8449.A254\$8192.3A\$8194.A\$
8193.A510\$7680.3A\$7682.A\$7681.A724\$6960.2A\$6959.A.A\$6961.A40\$6912.3A\$
6914.A\$6913.A254\$6656.3A\$6658.A\$6657.A510\$6144.3A\$6146.A\$6145.A724\$
5424.2A\$5423.A.A\$5425.A40\$5376.3A\$5378.A\$5377.A254\$5120.3A\$5122.A\$
5121.A510\$4608.3A\$4610.A\$4609.A510\$4096.3A\$4098.A\$4097.A724\$3376.2A\$
3375.A.A\$3377.A40\$3328.3A\$3330.A\$3329.A254\$3072.3A\$3074.A\$3073.A510\$
2560.3A\$2562.A\$2561.A724\$1840.2A\$1839.A.A\$1841.A40\$1792.3A\$1794.A\$
1793.A468\$1328.2A\$1327.A.A\$1329.A40\$1280.3A\$1282.A\$1281.A254\$1024.3A\$
1026.A\$1025.A510\$512.3A\$514.A\$513.A510\$3A\$2.A\$.A!``````
This double-blinker elbow can itself be reached from the standard honeyfarm in MathAndCode's original RCT. Here's a demonstration which leaves some junk behind but converts to a 'three-quarter traffic light' in the same location as used in all of the above recipes:

Code: Select all

``````x = 14173, y = 14173, rule = LifeHistory
14166.C\$14165.C.C\$14165.C.C\$14166.C2\$14161.2C7.2C\$14160.C2.C5.C2.C\$
14161.2C7.2C2\$14166.C\$14165.C.C\$14165.C.C\$14166.C334\$13824.3A\$13826.A
\$13825.A510\$13312.3A\$13314.A\$13313.A510\$12800.3A\$12802.A\$12801.A596\$
12208.2A\$12207.A.A\$12209.A40\$12160.3A\$12162.A\$12161.A382\$11776.3A\$
11778.A\$11777.A596\$11184.2A\$11183.A.A\$11185.A40\$11136.3A\$11138.A\$
11137.A468\$10672.2A\$10671.A.A\$10673.A40\$10624.3A\$10626.A\$10625.A382\$
10240.3A\$10242.A\$10241.A596\$9648.2A\$9647.A.A\$9649.A40\$9600.3A\$9602.A\$
9601.A468\$9136.2A\$9135.A.A\$9137.A40\$9088.3A\$9090.A\$9089.A468\$8624.2A\$
8623.A.A\$8625.A40\$8576.3A\$8578.A\$8577.A382\$8192.3A\$8194.A\$8193.A510\$
7680.3A\$7682.A\$7681.A510\$7168.3A\$7170.A\$7169.A510\$6656.3A\$6658.A\$
6657.A596\$6064.2A\$6063.A.A\$6065.A40\$6016.3A\$6018.A\$6017.A468\$5552.2A\$
5551.A.A\$5553.A40\$5504.3A\$5506.A\$5505.A382\$5120.3A\$5122.A\$5121.A510\$
4608.3A\$4610.A\$4609.A596\$4016.2A\$4015.A.A\$4017.A40\$3968.3A\$3970.A\$
3969.A382\$3584.3A\$3586.A\$3585.A596\$2992.2A\$2991.A.A\$2993.A40\$2944.3A\$
2946.A\$2945.A468\$2480.2A\$2479.A.A\$2481.A40\$2432.3A\$2434.A\$2433.A382\$
2048.3A\$2050.A\$2049.A510\$1536.3A\$1538.A\$1537.A596\$944.2A\$943.A.A\$945.
A40\$896.3A\$898.A\$897.A382\$512.3A\$514.A\$513.A510\$3A\$2.A\$.A!
``````
We can turn this into a double-elbow using the rest of the elbow duplication reaction, then pull that elbow using the 14hd pull, then use the power of universal construction to clean up the junk that was left behind when converting from the initial honeyfarm.

I think this proves the universality of MathAndCode's 17-glider RCT, which (as Macbi pointed out) has the corollary that all glider-constructible N-bit still-lifes can be constructed in at most N-1 gliders. (For N <= 17, we have explicit constructions; for N >= 18, we can appeal to the RCT.)

In particular, all 18- and 19-bit still-lifes have been synthesised, so they all have 17-glider constructions.
What do you do with ill crystallographers? Take them to the mono-clinic!

Moosey
Posts: 4027
Joined: January 27th, 2019, 5:54 pm
Location: A house, or perhaps the OCA board. Or [click to not expand]
Contact:

Re: Binary slow salvos

dang, congratulations!
My CA rules can be found here

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Nanho walåt derwo esaato?

MathAndCode
Posts: 2745
Joined: August 31st, 2020, 5:58 pm

Re: Binary slow salvos

calcyman wrote:
September 19th, 2020, 4:29 am
Very impressive work!

Your two construction lanes are the same phase and 6hd apart. Is it possible to make them 3hd apart instead? I don't think we have a universal 6hd same-phase set of Fibonacci slow salvos.

EDIT: it looks as though they can be made 9hd apart, which might be sufficient (Chris mentioned that (9, 0) looked promising):

Code: Select all

``````x = 625, y = 613, rule = B3/S23
23bo\$21bobo\$22b2o62\$87bo\$85bobo\$86b2o62\$151bo\$149bobo\$150b2o62\$215bo\$
213bobo\$214b2o62\$279bo\$277bobo\$278b2o36\$310bo\$310bo\$310bo19\$286b2o\$
285bobo\$287bo34\$366b2o\$366bobo\$366bo\$256b3o\$258bo\$257bo23\$222b2o\$221bo
bo\$223bo34\$430b2o\$430bobo\$430bo\$192b3o\$194bo\$193bo23\$158b2o\$157bobo\$
159bo34\$494b2o\$494bobo\$494bo\$128b3o\$130bo\$129bo23\$94b2o\$93bobo\$95bo34\$
558b2o\$558bobo\$558bo\$64b3o\$66bo\$65bo23\$30b2o\$29bobo\$31bo34\$622b2o\$622b
obo\$622bo\$3o\$2bo\$bo!``````
The reason that I used gliders with the same parity six lanes apart was because I thought that that was what was used in the 35-glider universal constructor. However, looking back at it, I see that the gliders were actually seven lanes apart. I'm sure that the general design could work with other offset-parity pairs, so there's no need to prove this one universal, and indeed, you seem to have already found another pair that works.
gameoflifemaniac wrote:
September 19th, 2020, 7:59 am
Will there be an algorithm that will be optimized for the RCT and run billions or trillions of orders of magnitude faster than HashLife does?
HashLife should be able to handle an RCT-based universal constructor well. Most of the cells will be from the GPSEs' trails of ash, which repeat every 32 full diagonals or every two generations. Both are powers of two, which means that they are HashLife-friendly. Most of the cells that are not part of the GPSEs' ash are from the GPSEs' glider streams, which repeat every 64 full diagonals or every 256 generations. Both of these are powers of two, so these will also be HashLife-friendly. The circuitry repeats every 256 generations, so it will be HashLife-friendly. The target will likely only consist of still-lifes and p2 oscillators for most of the time, so because changes in the target (caused by construction gliders) will be exponentially sparse for most of the time, the target will be HashLife-friendly. The only part that does not seem HashLife-friendly to me is the GPSEs themselves: They repeat every 384 generations displaced by 32 full diagonals. Because 384 is not a power of two, but rather three times a power of two, it is not a factor of any power of two, so it is not HashLife-friendly under my current understanding of HashLife. Compared to the number of calculations required without HashLife, the GPSEs themselves represent too small of a fraction of the total number of cells (except at the very beginning) to result in significant inefficiency, but because they will be the only parts that are not HashLife-friendly (assuming that my current understanding of HashLife is correct), they will account for a significant proportion of total calculations when HashLife is enabled.
However, it's possible that HashLife has some feature to account for this. After all, historically, many large patterns in ConwayLife have relied on patterns repeating every 15, 30, 60, or 120 generations, such as the pentadecathlon and the queen bee shuttle. If HashLife can be told to search for patterns repeating after a number of generations that is either a power of two or three times a power of two, then it will be able to handle GPSEs efficiently. If not, that feature could theoretically be added.
I have reduced the cost of universal construction to seventeen gliders and probably to sixteen. All that remains is for the universal operations to be found.

calcyman
Posts: 2352
Joined: June 1st, 2009, 4:32 pm

Re: Potential universal construction in 16±2 gliders

MathAndCode wrote:
September 10th, 2020, 1:05 pm
Twenty-first edit: I found a method that results in a minimal change in ash and no backward gliders.

Code: Select all

``````x = 285, y = 226, rule = B3/S23
43b2o\$43b2o2\$6b2o\$5bobo\$6bo\$17bo\$17bo\$17bo4\$24b3o\$35b2o33b2o\$35b2o32bobo\$69b2o\$38bo\$38bo\$38bo2\$40b3o2\$37b2o\$36bo2bo\$37b2o40b2o\$79b2o4\$7b2o\$6bo2bo\$7bobo\$8bo66b2o\$75b2o2\$38b2o\$37bobo\$38bo4\$b2o47b3o\$o2bo48b3o\$b2o42b2o2b2o4bo\$45b2o2bo2bo3bo\$50bo4bo11b2o\$53bo13b2o\$51bobo\$52bo17bo\$70bo\$70bo2\$72b3o2\$69b2o\$68bo2bo\$69b2o5\$39b2o\$38bo2bo\$39bobo\$40bo3\$70b2o\$69bobo\$70bo4\$33b2o44b2o\$32bo2bo42bobo\$33b2o30b2o12bo\$64bobo\$65bo3\$53b2o\$53b2o5\$100b2o49bobo\$91b2o7b2o50b2o\$90bo2bo5bo2bo49bo\$79b6o6bobo5bo2bo\$78bo5bo7bo5bo2bo\$78bo20bo2bo\$70bo8bo4bo14b4o\$69bobo10b3o15bo\$70b2o31bo\$101bo\$101b2o4\$105b2o\$105b2o3\$77b2o\$77b2o3\$107bo\$107b2o\$109bo2\$85b2o7b3o\$85b2o\$92bo5bo12bo\$92bo5bo\$92bo5bo2b3o\$100bo2bo\$94b3o3bo3bo\$99b2obobo\$99b2ob2o12b4o\$100b3o12b2o4bo\$115b3o4bo\$117bo5bo\$115b2o\$115b2o2b2o4bo\$101b3o11bo9bo\$116bo2bo3bo\$117bo\$118b2obo\$120bo20\$215bobo\$216b2o\$216bo62\$279bobo\$280b2o\$280bo7\$283bo\$282b2o\$282bobo!``````
I plan to use this for a universal constructor that can be synthesized with seventeen gliders soon.
Unfortunately, that doesn't seem to work: delaying the glider by 256 generations causes an explosion which generates a retrograde glider heading back towards the RCT.

Now, we can flip over the GPSE and cap the glider stream with a distant crystal; unfortunately, this costs an extra glider (so N = 18 instead of 17).

Anyway, this is still a huge improvement over 32 -- many congratulations!

EDIT: actually, we can just make this north-westerly GPSE have a much much much longer glider trail than the south-easterly GPSE (by synthesising it earlier), so that by the time that the retrograde gliders reach the synthesis site, everything else has completed and the recipe has built a couple of convenient eaters.
What do you do with ill crystallographers? Take them to the mono-clinic!

MathAndCode
Posts: 2745
Joined: August 31st, 2020, 5:58 pm

Re: Potential universal construction in 16±2 gliders

calcyman wrote:
September 19th, 2020, 2:03 pm
We can just make this north-westerly GPSE have a much much much longer glider trail than the south-easterly GPSE (by synthesising it earlier), so that by the time that the retrograde gliders reach the synthesis site, everything else has completed and the recipe has built a couple of convenient eaters.
I suggested that except that only the eater needs to be completed by the time that the first C₁ glider returns.
MathAndCode wrote:
September 10th, 2020, 1:05 pm
However, this can be solved by constructing a fishhook in the appropriate place before any C₁ gliders arrive.
Now that the 4-GPSE solution is done besides attempting to synthesize the southwest pair of GPSEs with less than eight gliders, what do you think of my idea for a 3-GPSE solution?
MathAndCode wrote:
September 10th, 2020, 1:05 pm
Tenth edit: I have been intermittently pondering the possibility of a 3-GPSE solution, and I may have just conceived one: Glider stream A₀ will feed into GPSE C in addition to glider stream C₀ feeding into GPSE A. The timing will be such so that GPSE A always sends back a glider on lane A₁ when it receives a signal, and GPSE C either sends back a glider on lane C₁ or doesn't send back any signal when it releases a signal. Ordinarily, glider B will destroy gliders A₀ and C₀. However, when glider A₁ arrives, it will shift glider B's path via a logic gate, preventing it from crashing into gliders A₀ and C₀, letting them go past each other. If there is no glider C₁, then glider B will proceed to construction on its path, but if there is a glider C₁, then it will shift glider B via another logic gate, making it take a different path to construction.
I have reduced the cost of universal construction to seventeen gliders and probably to sixteen. All that remains is for the universal operations to be found.

gameoflifemaniac
Posts: 1224
Joined: January 22nd, 2017, 11:17 am
Location: There too

Re: Binary slow salvos

MathAndCode wrote:
September 19th, 2020, 12:28 pm
calcyman wrote:
September 19th, 2020, 4:29 am
Very impressive work!

Your two construction lanes are the same phase and 6hd apart. Is it possible to make them 3hd apart instead? I don't think we have a universal 6hd same-phase set of Fibonacci slow salvos.

EDIT: it looks as though they can be made 9hd apart, which might be sufficient (Chris mentioned that (9, 0) looked promising):

Code: Select all

``````x = 625, y = 613, rule = B3/S23
23bo\$21bobo\$22b2o62\$87bo\$85bobo\$86b2o62\$151bo\$149bobo\$150b2o62\$215bo\$
213bobo\$214b2o62\$279bo\$277bobo\$278b2o36\$310bo\$310bo\$310bo19\$286b2o\$
285bobo\$287bo34\$366b2o\$366bobo\$366bo\$256b3o\$258bo\$257bo23\$222b2o\$221bo
bo\$223bo34\$430b2o\$430bobo\$430bo\$192b3o\$194bo\$193bo23\$158b2o\$157bobo\$
159bo34\$494b2o\$494bobo\$494bo\$128b3o\$130bo\$129bo23\$94b2o\$93bobo\$95bo34\$
558b2o\$558bobo\$558bo\$64b3o\$66bo\$65bo23\$30b2o\$29bobo\$31bo34\$622b2o\$622b
obo\$622bo\$3o\$2bo\$bo!``````
The reason that I used gliders with the same parity six lanes apart was because I thought that that was what was used in the 35-glider universal constructor. However, looking back at it, I see that the gliders were actually seven lanes apart. I'm sure that the general design could work with other offset-parity pairs, so there's no need to prove this one universal, and indeed, you seem to have already found another pair that works.
gameoflifemaniac wrote:
September 19th, 2020, 7:59 am
Will there be an algorithm that will be optimized for the RCT and run billions or trillions of orders of magnitude faster than HashLife does?
HashLife should be able to handle an RCT-based universal constructor well. Most of the cells will be from the GPSEs' trails of ash, which repeat every 32 full diagonals or every two generations. Both are powers of two, which means that they are HashLife-friendly. Most of the cells that are not part of the GPSEs' ash are from the GPSEs' glider streams, which repeat every 64 full diagonals or every 256 generations. Both of these are powers of two, so these will also be HashLife-friendly. The circuitry repeats every 256 generations, so it will be HashLife-friendly. The target will likely only consist of still-lifes and p2 oscillators for most of the time, so because changes in the target (caused by construction gliders) will be exponentially sparse for most of the time, the target will be HashLife-friendly. The only part that does not seem HashLife-friendly to me is the GPSEs themselves: They repeat every 384 generations displaced by 32 full diagonals. Because 384 is not a power of two, but rather three times a power of two, it is not a factor of any power of two, so it is not HashLife-friendly under my current understanding of HashLife. Compared to the number of calculations required without HashLife, the GPSEs themselves represent too small of a fraction of the total number of cells (except at the very beginning) to result in significant inefficiency, but because they will be the only parts that are not HashLife-friendly (assuming that my current understanding of HashLife is correct), they will account for a significant proportion of total calculations when HashLife is enabled.
However, it's possible that HashLife has some feature to account for this. After all, historically, many large patterns in ConwayLife have relied on patterns repeating every 15, 30, 60, or 120 generations, such as the pentadecathlon and the queen bee shuttle. If HashLife can be told to search for patterns repeating after a number of generations that is either a power of two or three times a power of two, then it will be able to handle GPSEs efficiently. If not, that feature could theoretically be added.
Ok but if you take a pattern (any pattern, actually) and run it at a step of 2^1000000, Golly can't handle it and crashes (I've tested it). To see anything happening in a RCT that is synthesizing some more complicated pattern, like an expensive spaceship, we would need a step of 2^billions where the exponent would be most likely exceed the 32-bit integer limit. We would need an algorithm to even handle patterns with sizes of billions orders of magnitude, too.
I was so socially awkward in the past and it will haunt me for the rest of my life.

Code: Select all

``````b4o25bo\$o29bo\$b3o3b3o2bob2o2bob2o2bo3bobo\$4bobo3bob2o2bob2o2bobo3bobo\$
4bobo3bobo5bo5bo3bobo\$o3bobo3bobo5bo6b4o\$b3o3b3o2bo5bo9bobo\$24b4o!``````

MathAndCode
Posts: 2745
Joined: August 31st, 2020, 5:58 pm

Re: Binary slow salvos

gameoflifemaniac wrote:
September 19th, 2020, 2:38 pm
Ok but if you take a pattern (any pattern, actually) and run it at a step of 2^1000000, Golly can't handle it and crashes (I've tested it). To see anything happening in a RCT that is synthesizing some more complicated pattern, like an expensive spaceship, we would need a step of 2^billions where the exponent would be most likely exceed the 32-bit integer limit. We would need an algorithm to even handle patterns with sizes of billions orders of magnitude, too.
That's a fair point. I would ask people who have already ran universal constructors on Golly.
I have reduced the cost of universal construction to seventeen gliders and probably to sixteen. All that remains is for the universal operations to be found.

calcyman
Posts: 2352
Joined: June 1st, 2009, 4:32 pm

Re: Binary slow salvos

A 3-GPSE solution would be very exciting!

There's actually another potential optimisation which might be able to save a glider from the 4-GPSE RCT. Specifically, if the first northwest-bound glider collides like so:

Code: Select all

``````x = 685, y = 638, rule = B3/S23
8b3o2\$5b2o\$4bo2bo\$5b2o40b2o\$47b2o7\$43b2o\$43b2o2\$6b2o\$5bobo\$6bo\$17bo\$
17bo\$17bo4\$24b3o\$35b2o33b2o\$35b2o32bobo\$69b2o\$38bo\$38bo\$38bo2\$40b3o2\$
37b2o\$36bo2bo\$37b2o40b2o\$79b2o4\$7b2o\$6bo2bo\$7bobo\$8bo66b2o\$75b2o2\$38b
2o\$37bobo\$38bo\$49bo\$49bo\$49bo\$b2o\$o2bo\$b2o\$56b3o\$67b2o33b2o\$67b2o32bob
o\$101b2o\$70bo\$70bo\$70bo2\$72b3o2\$69b2o\$68bo2bo\$69b2o40b2o\$111b2o4\$39b2o
\$38bo2bo\$39bobo\$40bo66b2o\$107b2o2\$70b2o\$69bobo\$70bo\$81bo\$81bo\$81bo\$33b
2o\$32bo2bo\$33b2o\$88b3o\$99b2o33b2o\$99b2o32bobo\$133b2o\$102bo\$102bo\$102bo
2\$104b3o2\$101b2o\$100bo2bo\$101b2o40b2o\$143b2o4\$71b2o\$70bo2bo\$71bobo\$72b
o66b2o\$139b2o2\$102b2o\$101bobo\$102bo\$113bo\$113bo\$113bo\$65b2o\$64bo2bo\$
65b2o\$120b3o\$131b2o33b2o\$131b2o32bobo\$165b2o\$134bo\$134bo\$134bo2\$136b3o
2\$133b2o\$132bo2bo\$133b2o40b2o\$175b2o4\$103b2o\$102bo2bo\$103bobo\$104bo66b
2o\$171b2o2\$134b2o\$133bobo\$134bo\$145bo\$145bo\$145bo\$97b2o\$96bo2bo\$97b2o\$
152b3o\$163b2o33b2o\$163b2o32bobo\$197b2o\$166bo\$166bo\$166bo2\$168b3o2\$165b
2o\$164bo2bo\$165b2o40b2o\$207b2o4\$135b2o\$134bo2bo\$135bobo\$136bo66b2o\$
203b2o2\$166b2o\$165bobo\$166bo\$177bo\$177bo\$177bo\$129b2o\$128bo2bo\$129b2o\$
184b3o\$195b2o33b2o\$195b2o32bobo\$229b2o\$198bo\$198bo\$198bo2\$200b3o2\$197b
2o\$196bo2bo\$197b2o40b2o\$239b2o4\$167b2o\$166bo2bo\$167bobo\$168bo66b2o\$
235b2o2\$198b2o\$197bobo\$198bo\$209bo\$209bo\$209bo\$161b2o\$160bo2bo\$161b2o\$
216b3o\$227b2o33b2o\$227b2o32bobo\$261b2o\$230bo\$230bo\$230bo2\$232b3o2\$229b
2o\$228bo2bo\$229b2o40b2o\$271b2o4\$199b2o\$198bo2bo\$199bobo\$200bo66b2o\$
267b2o2\$230b2o\$229bobo\$230bo\$241bo\$241bo\$241bo\$193b2o\$192bo2bo\$193b2o\$
248b3o\$259b2o\$259b2o2\$262bo\$262bo\$262bo2\$264b3o2\$261b2o\$260bo2bo\$261b
2o5\$231b2o59bo\$230bo2bo56b6o\$231bobo56b2o2b2o\$232bo\$294bo\$290b2o2bo\$
262b2o26b2o3bo\$261bobo30bo\$262bo29bobo\$293bo3\$225b2o44b2o\$224bo2bo42bo
bo\$225b2o30b2o12bo\$256bobo\$257bo33b2o\$291b2o2\$245b2o47bo\$245b2o47bo\$
294bo2\$296b3o\$275b2o\$275b3o15b2o\$292bo2bo\$272b3o18b2o\$279bobo\$271bo2bo
4bobo2b2o\$274bo3bo2bo2b2o\$262bo7bo2bo5b2o\$261bobo6b3o6b3o\$262b2o6\$297b
2o\$297b2o\$359bobo\$360b2o\$269b2o89bo\$269b2o2\$289b2o\$288bobo\$289bo3\$277b
2o\$277b2o29bobo\$307bo3bo\$306bo5bo\$306b3o4bo\$297bo9bob2obo\$296bobo\$294b
2o2bo\$297bo2\$294b3o4\$298b3o2\$307bo\$298b3o6bo\$312b3o\$313bo\$305bo6b3o\$
305b4o4b5o\$304bo3b4ob2o\$305bo4b2o\$313b2o30\$423bobo\$424b2o\$424bo62\$487b
obo\$488b2o\$488bo62\$551bobo\$552b2o\$552bo62\$615bobo\$616b2o\$616bo71\$683bo
\$682b2o\$682bobo!``````
Then the retrograde glider is in the correct phase to hit the first glider from the south-west:

Code: Select all

``````x = 667, y = 655, rule = B3/S23
21bobo\$22b2o\$22bo62\$85bobo\$86b2o\$86bo62\$149bobo\$150b2o\$150bo62\$213bobo
\$214b2o\$214bo25\$276bobo\$277b2o\$277bo35\$277bobo\$278b2o\$278bo115\$273bo\$
273b2o\$272bobo18\$409bo\$408b2o\$408bobo\$256b2o\$257b2o\$256bo39\$209bo\$209b
2o\$208bobo18\$473bo\$472b2o\$472bobo\$192b2o\$193b2o\$192bo39\$145bo\$145b2o\$
144bobo18\$537bo\$536b2o\$536bobo\$128b2o\$129b2o\$128bo39\$81bo\$81b2o\$80bobo
18\$601bo\$600b2o\$600bobo\$64b2o\$65b2o\$64bo39\$17bo\$17b2o\$16bobo18\$665bo\$
664b2o\$664bobo\$2o\$b2o\$o!``````
This traffic light is then going to be slowly bombarded with RCT-encoded gliders from two different directions. It's a long shot, but it might be possible to encode a recipe that creates a target for the construction arm and capping off the retrograde gliders (most probably with a crystal).
gameoflifemaniac wrote:
September 19th, 2020, 2:38 pm
Ok but if you take a pattern (any pattern, actually) and run it at a step of 2^1000000, Golly can't handle it and crashes (I've tested it). To see anything happening in a RCT that is synthesizing some more complicated pattern, like an expensive spaceship, we would need a step of 2^billions where the exponent would be most likely exceed the 32-bit integer limit. We would need an algorithm to even handle patterns with sizes of billions orders of magnitude, too.
Yes, my best estimate for the cost of an RCT cleaning up after itself is 10^9 bits, most of which just construct a Chapman-Greene construction arm into which the remainder of the recipe bits are fed verbatim.
MathAndCode wrote:
September 19th, 2020, 2:47 pm
That's a fair point. I would ask people who have already ran universal constructors on Golly.
Ordinary universal constructors take polynomial time as a function of the recipe length. The RCT is in a class of its own, taking exponential time as a function of the recipe length.
What do you do with ill crystallographers? Take them to the mono-clinic!

Macbi
Posts: 782
Joined: March 29th, 2009, 4:58 am

Re: Binary slow salvos

calcyman wrote:
September 19th, 2020, 3:02 pm
Ordinary universal constructors take polynomial time as a function of the recipe length. The RCT is in a class of its own, taking exponential time as a function of the recipe length.
But with HashLife we should be back down to linear time as a function of recipe length. So maybe Golly could run it in a human amount of time (say, less than a year).

MathAndCode
Posts: 2745
Joined: August 31st, 2020, 5:58 pm

Re: Binary slow salvos

calcyman wrote:
September 19th, 2020, 3:02 pm
A 3-GPSE solution would be very exciting!
I suggested a way to look for possibilities.
MathAndCode wrote:
September 10th, 2020, 1:05 pm
I'll wait for someone to do some automated searches in order to find out which relative positions and timings of GPSEs A and C work and then which of these will let the logic gates function for some relative position and timing of GPSE B.
The reason that I'm not willing to do another manual search for this is because the number of cases that I'd have to search is much higher. Let's call the number of timings with which a glider can hit a GPSE on the particular lane that I've found to be the most useful X. (X is equal to 256, but the important part is that it's fairly large and can therefore be used to indicate the growth rate.) For the case that I've already searched, I needed to search X/2 different possibilities (because I was looking for pairs that worked, not single timings that worked). For each pair, I needed to make sure that both timings resulted in a return signal and no sideways gliders, so that would result in a theoretical maximum of two simulations per pair, for X simulations total. However, if one glider in a pair resulted in no return signal or at least one sideways glider, then I didn't need to check the other member of the pair, and there was a region of consecutive timings where I didn't need to check each timing individually because they all resulted in the same ∏-∏-LOM-forwards glider sequence (The LOM is not directly necessary for the forwards glider, but its formation indicates that the second ∏ heptomino did not interact with the next Herschel before it finished settling into ash, which is necessary for the formation of a forwards glider on a particular lane.), so the number of manual simulations that I did was closer to X/2 than to X. If I want to check all of the possibilities where two GPSEs are feeding gliders into each other, then I'll need to manually check about X²/8 possibilities (closer to X²/16 if one accounts for redundancies due to the symmetry of the design) because I had no reason to record timing pairs where both gliders returned the same signal or at least one glider resulted in no return signal when I did my manual search. This is going to be much higher than X/2. Each possibility will require a maximum of four simulations, and making sure that my search is comprehensive will be more difficult. Therefore, I'm going to let a computer do this. I'm willing to use my own computer, but I'd prefer to not have to write the code because someone else has probably already figured out how to do things like check whether or not there are any sideways gliders. Once the program goes through all of those possibilities and reduces them to a more manageable number, then I will work on developing the 3-GPSE solution.
Also, I only searched for all of the possibilities for one lane. There are at least a dozen other potential lanes where I searched for only a minority of the possibilities or none at all. It would be nice to search all of those as well, but I will not do that manually because that would be impractical.

Edit: I just realized that while the search program cannot use the results from my manual search because I didn't record the results that my 3-GPSE idea requires, it can use the results from an earlier part of its search. This will reduce the number of simulations necessary from a quadratic function of X to a linear function of X, but if done that way, a manual search would probably result in confusion about which relative timings for glider C₀ and GPSE A are compatible with which relative timings for glider A₀ and GPSE C. Also, I don't want to do the same manual search again. The first search took up a significant amount of time. Frankly, if I had not felt that I had to prove the merit of my system myself, I would have asked someone else to code an automated search the first time (although, looking back at it, it may have been more efficient to write an automated search myself the first time, depending on how long it would have taken for me to get the code to work).
I have reduced the cost of universal construction to seventeen gliders and probably to sixteen. All that remains is for the universal operations to be found.

calcyman
Posts: 2352
Joined: June 1st, 2009, 4:32 pm

Re: Binary slow salvos

I'm going to give a general non-technical overview of the timeline of the RCT cleanly synthesising a spider.

Firstly, I'll introduce some nomenclature:
• epicentre: the area of interest where all of the gliders collide;
• parsec: the distance between the construction site of the south-easterly GPSE and the epicentre;
• aeon: the time it takes for a lightspeed diagonal signal to traverse a parsec;
The number of ticks in an aeon (or equivalently the number of fd in a parsec) is some small multiple of the recipe (represented as an integer), which is roughly 2^(number of bits in the recipe).

The timeline begins as follows:
• 24 aeons BS: the north-westerly GPSE and south-westerly GPSE pair are simultaneously constructed 4 parsecs away from the epicentre;
• 12 aeons BS: the south-easterly GPSE is constructed one parsec south-east of the epicentre, heading north-west towards the epicentre;
• 8 aeons BS: the four streams of GPSE-generated gliders simultaneously hit the epicentre, together with the 17th glider to construct the honeyfarm. A sole signal-carrying glider heads south-east from the epicentre;
• 6 aeons BS: the signal is reflected from the south-easterly GPSE, either as a hole in the glider stream or as a glider on a different lane (encoding the 1st bit of the recipe);
• 4 aeons BS: the 1st bit reaches the epicentre, releasing either a single glider or a glider pair towards the honeyfarm;
• 2 aeons BS: the 2nd bit reaches the epicentre;
• 1 aeon BS: the 3rd bit reaches the epicentre;
• 0.5 aeons BS: the 4th bit reaches the epicentre;
• 0.25 aeons BS: the 5th bit reaches the epicentre;
• ...
This rapid acceleration of the bits heading towards the construction arm is reminiscent of the technological singularity -- https://en.wikipedia.org/wiki/Technological_singularity -- so we'll follow suit and call it 'the singularity' and measure time as either BS or AS. At the singularity, the construction arm has built an assembly of various still-lifes in time for the south-easterly GPSE to be cleanly absorbed. The other three glider streams are cleanly eatered.
• 0 - 24 aeons AS: in this time period, we have some occasional retrograde gliders from the north-west collide into an eater, but otherwise nothing particularly interesting happens.
• 24 aeons AS: this is the exciting point where the remaining three GPSEs are cleanly absorbed. A sole glider is emitted, which is the only remaining activity in the universe. It courses through a maze of one-time turners and splitters, sending slow-salvos of gliders towards the origins of the four GPSEs followed by a wave of Corder-rakes.
• 28 aeons AS: the slow-salvos reach the source of the the south-easterly GPSE.
• 36 aeons AS: the Corder-rakes have cleaned up the south-easterly GPSE and are absorbed;
• 40 aeons AS: the slow-salvos reach the sources of the remaining GPSEs;
• 72 aeons AS: the Corder-rakes have cleaned up the remaining GPSEs and are absorbed, sending a single glider returning to the epicentre;
• 84 aeons AS: the returning glider hits the epicentre, triggering the seed of one-time reflectors and splitters to emplace the few hundred gliders necessary to cleanly synthesise a spider. At this point, the spider is the only thing remaining in the universe.
For something more sophisticated, such as an RCT-based knightship, we carefully ensure that we leave 'target' blocks behind at 36 and 72 aeons AS, and then choreograph some arrangement of gliders and standard spaceships which propagate around to ensure that all 17 gliders are constructed in the right place at the right time, translated (2, 1) away from their original position. This will take a few dozen more aeons.
What do you do with ill crystallographers? Take them to the mono-clinic!

wwei23
Posts: 2449
Joined: May 22nd, 2017, 6:14 pm
Location: The (Life?) Universe

Re: Binary slow salvos

From Simkin:

Code: Select all

``````x = 98, y = 77, rule = B3/S23
97bo\$95b2o\$96b2o69\$4bo\$3bo\$3b3o\$bo\$2o\$obo!
``````
13 gliders.
@Hunting so we're making a referential signature chain?
Stop being mean to MathAndCode!
LeapLife has fun drifters!

MathAndCode
Posts: 2745
Joined: August 31st, 2020, 5:58 pm

Re: Binary slow salvos

wwei23 wrote:
September 19th, 2020, 7:01 pm
From Simkin:

Code: Select all

``````x = 98, y = 77, rule = B3/S23
97bo\$95b2o\$96b2o69\$4bo\$3bo\$3b3o\$bo\$2o\$obo!
``````
13 gliders.
I already knew about that, but it can't be used in this case because it has escaping gliders. However, given Michael Simkin's history of finding efficient ways to construct GPSEs, I think that he could probably find a way to construct the southwest pair in six or seven gliders instead of eight. Does anyone know how to ping him?
I have reduced the cost of universal construction to seventeen gliders and probably to sixteen. All that remains is for the universal operations to be found.

wwei23
Posts: 2449
Joined: May 22nd, 2017, 6:14 pm
Location: The (Life?) Universe

Re: Binary slow salvos

Also, forget corderrakes. Just use 2-engine corderships and 3-engine corderships to clean up the glider-producing switch engine debris.

Code: Select all

``````x = 618, y = 603, rule = B3/S23
556bo\$555bobo\$555bobo\$556bo3\$568bo\$567bobo\$566bo2bo\$567b2o7\$591bo\$591b
o\$591bo7\$585b3o7\$524bo\$523bobo\$523bobo\$524bo59b2o\$575bo8b2o\$562bo11bob
o4bo33b2o\$536bo24bobo10bobo4bo33bobo\$535bobo24b2o11bo5bo34b2o\$534bo2bo
\$535b2o40b3o2\$597b2o\$597b2o3\$605b2o\$559bo45b2o\$559bo\$559bo7\$553b3o7\$
492bo\$491bobo\$491bobo\$492bo59b2o\$543bo8b2o\$530bo11bobo4bo33b2o\$504bo
24bobo10bobo4bo33bobo\$503bobo24b2o11bo5bo34b2o\$502bo2bo\$503b2o40b3o2\$
565b2o\$565b2o3\$573b2o\$527bo45b2o\$527bo\$527bo7\$521b3o7\$460bo\$459bobo\$
459bobo\$460bo59b2o\$511bo8b2o\$498bo11bobo4bo33b2o\$472bo24bobo10bobo4bo
33bobo\$471bobo24b2o11bo5bo34b2o\$470bo2bo\$471b2o40b3o2\$533b2o\$533b2o3\$
541b2o\$495bo45b2o\$495bo\$495bo7\$489b3o7\$428bo\$427bobo\$427bobo\$428bo59b
2o\$479bo8b2o\$466bo11bobo4bo33b2o\$440bo24bobo10bobo4bo33bobo\$439bobo24b
2o11bo5bo34b2o\$438bo2bo\$439b2o40b3o2\$501b2o\$501b2o3\$509b2o\$463bo45b2o\$
463bo\$463bo7\$457b3o7\$396bo\$395bobo\$395bobo\$396bo59b2o\$447bo8b2o\$434bo
11bobo4bo33b2o\$408bo24bobo10bobo4bo33bobo\$407bobo24b2o11bo5bo34b2o\$
406bo2bo\$407b2o40b3o2\$469b2o\$469b2o3\$477b2o\$431bo45b2o\$431bo\$431bo7\$
425b3o7\$364bo\$363bobo\$363bobo\$364bo59b2o\$415bo8b2o\$308bo93bo11bobo4bo
33b2o\$300bobob2obo68bo24bobo10bobo4bo33bobo\$300bo3b4o3b3o61bobo24b2o
11bo5bo34b2o\$300b2ob2o2bo66bo2bo\$308bo66b2o40b3o\$308b3o\$289b2o17b3o
126b2o\$289b2o146b2o3\$445b2o\$327b3o69bo45b2o\$330bo68bo\$325b2o4bo67bo\$
281b2o27b3o15bo3bo\$281b2o35b2o4bo4bo2bo\$315bob3o4bo3bo3bo\$324b3obob2o\$
288bo39bo\$283bobob3o34b4o\$281b2obo5b2o17b2obo11b2o67b3o\$282bo26bo2bo\$
283b3o5bo18b3o\$287bo\$288b2o\$289bo\$290bo4bo\$290bo4bo36bo\$331bobo\$331bob
o\$332bo59b2o\$322b2o59bo8b2o\$322b2o46bo11bobo4bo33b2o\$344bo24bobo10bobo
4bo33bobo\$343bobo24b2o11bo5bo34b2o\$342bo2bo\$343b2o40b3o2\$405b2o\$314b2o
89b2o\$314b2o2\$413b2o\$367bo45b2o\$367bo\$367bo5\$427bo\$427bo\$361b3o63bo\$
423b2o\$409b2o12b2o\$271bo138b2o11b3o3bo\$263bobob2obo139bo15b3o\$263bo3b
4o3b3o132b3o15bo\$263b2ob2o2bo138b3o15b2o\$271bo137bo16b3o\$271b3o134bo2b
o14bo\$252b2o17b3o128b2o5b3o16bo\$252b2o106b2o40b2o6bo15bo\$351bo8b2o64b
3o\$338bo11bobo4bo\$312bo24bobo10bobo4bo\$290b3o18bobo24b2o11bo5bo\$293bo
16bo2bo127bo\$288b2o4bo16b2o40b3o84bobo\$244b2o27b3o15bo3bo115b2o26bo3bo
\$244b2o35b2o4bo4bo2bo77b2o18b3o13b2o2bo25bo3bo\$278bob3o4bo3bo3bo77b2o
17bo16b5o23b3obobo\$287b3obob2o97bo3bo2b2o8bob3o22bo2bob2o\$251bo39bo
100bo2b2obo8b2o2b2o23bobo\$246bobob3o34b4o90b2o24bo3bo25b2o\$244b2obo5b
2o17b2obo11b2o46bo45b2o11b2o10bob2o\$245bo26bo2bo59bo71bobo\$246b3o5bo
18b3o59bo72b2o\$250bo158b2o2bo\$251b2o160b3o\$252bo154b3o3b4o\$253bo4bo
148b3o4bobo\$253bo4bo149b4ob3o\$411b2o3bo\$329b3o80bo3bo\$413bo2bo\$285b2o
128bo4b2o13b2o\$285b2o133b2o13b2o3\$421b3o\$421b2o3\$277b2o49b2o97b2o\$277b
2o40bo8b2o97b2o\$306bo11bobo4bo\$280bo24bobo10bobo4bo\$279bobo24b2o11bo5b
o\$278bo2bo\$279b2o40b3o2\$341b2o\$341b2o3\$349b2o\$303bo45b2o\$303bo\$303bo
47b3o\$350bo3bo\$350bo4bo\$352bo3bo\$349b3obo2bo\$348bo7bo\$348bobo3bo\$297b
3o47b2obo3bo\$217bo129b2ob3o\$209bobob2obo118b2o\$209bo3b4o3b3o112b2o\$
209b2ob2o2bo\$217bo\$217b3o131b2o\$198b2o17b3o131b2o\$198b2o151bo2\$296b2o
29b2o22bo\$287bo8b2o29b2o\$236b3o35bo11bobo4bo71bo\$239bo33bobo10bobo4bo
51b3o5bo11bo\$234b2o4bo33b2o11bo5bo51bo7bo11bo\$190b2o27b3o15bo3bo103bob
o5bo7b2o\$190b2o35b2o4bo4bo2bo47b3o54b2o13b2o\$224bob3o4bo3bo3bo119b3o3b
o\$233b3obob2o2b2o64b2o53b3o\$197bo39bo5b2o64b2o54bo\$192bobob3o34b4o128b
2o\$190b2obo5b2o17b2obo11b2o76b3o50b3o\$191bo26bo2bo88bo3bo49bo\$192b3o5b
o18b3o49bo38bo4bo50bo\$196bo74bo40bo3bo47bo\$197b2o72bo37b3obo2bo47b3o\$
198bo109bo7bo\$199bo4bo103bobo3bo\$199bo4bo102b2obo3bo\$307b2ob3o\$295b2o
43b2o\$295b2o\$231b2o32b3o\$231b2o\$311b2o32bobo\$311b2o27b2o5bo12b2o\$311bo
30b2o4bo11b2o\$343bo4b2o\$287b2o22bo32bo3bo\$287b2o\$223b2o100bo19bo2bo\$
223b2o80b3o5bo11bo19bobo\$264b2o39bo7bo11bo19bo2bo\$255bo8b2o39bobo5bo7b
2o23b2o4b2o\$242bo11bobo4bo44b2o13b2o24bo4b2o\$241bobo10bobo4bo59b3o3bo\$
242b2o11bo5bo62b3o\$325bo\$257b3o65b2o\$324b3o\$324bo\$326bo\$324bo\$324b3o\$
152b3o\$155bo83bo\$150b2o4bo82bo\$135b3o15bo3bo81bo\$143b2o4bo4bo2bo142b2o
\$140bob3o4bo3bo3bo102b3o\$149b3obob2o102bo3bo\$153bo105bo4bo\$127b2o20b4o
108bo3bo39bobo\$127b2o5b2obo11b2o107b3obo2bo34b2o5bo12b2o\$134bo2bo95b3o
21bo7bo36b2o4bo11b2o\$135b3o119bobo3bo39bo4b2o\$128bo127b2obo3bo40bo3bo\$
123bo4bo127b2ob3o\$122bobo2bobo114b2o59bo2bo\$121bo3bobobo114b2o59bobo\$
119b2o2b3o4bo174bo2bo\$119bo3bobo180b2o4b2o\$125b2o133b2o45bo4b2o\$120bob
2o4bo131b2o\$133bo13b2o83b2o26bo\$134bo6b3o3b2o74bo8b2o\$132bo2bo5bobo16b
o49bo11bobo4bo6b2o22bo\$132bo2bo4bo2bo16b2o47bobo10bobo4bo6b2o\$133b2o6b
3o14b5o47b2o11bo5bo44bo\$141bo21bo90b3o5bo11bo\$142bo14bo5b2o60b3o26bo7b
o11bo\$141bobo16b2ob3o88bobo5bo7b2o\$142b2o11b2o2b2obobo90b2o13b2o\$142bo
13b2o2bo2bo106b3o3bo\$160b2o20b3o88b3o\$156bob2o25bo88bo\$156bo23b2o4bo
87b2o\$165b3o15bo3bo19bo65b3o\$173b2o4bo4bo2bo19bo65bo\$170bob3o4bo3bo3bo
19bo67bo\$179b3obob2o86bo\$183bo89b3o\$179b4o\$164b2obo11b2o\$164bo2bo\$165b
3o\$201b3o45b2o4\$254bobo\$249b2o5bo12b2o\$251b2o4bo11b2o\$252bo4b2o\$177b2o
74bo3bo\$177b2o\$254bo2bo\$191bo62bobo\$190bobo4bo56bo2bo\$190bobo4bo57b2o
4b2o\$191bo5bo58bo4b2o2\$169b2o22b3o\$169b2o6\$175bo\$175bo\$175bo2\$195b3o\$
194bo3bo\$194bo4bo\$196bo3bo\$193b3obo2bo\$169b3o20bo7bo\$192bobo3bo\$191b2o
bo3bo\$191b2ob3o\$179b2o\$179b2o3\$195b2o\$195b2o\$195bo\$159bo\$158bobo4bo5b
2o22bo\$158bobo4bo5b2o\$159bo5bo43bo\$189b3o5bo11bo\$161b3o25bo7bo11bo\$
189bobo5bo7b2o\$190b2o13b2o\$205b3o3bo\$208b3o\$209bo\$209b2o\$143bo64b3o\$
143bo64bo\$143bo66bo\$208bo\$208b3o5\$137b3o44b2o4\$189bobo\$184b2o5bo12b2o\$
186b2o4bo11b2o\$187bo4b2o\$188bo3bo2\$189bo2bo\$127bo61bobo\$126bobo60bo2bo
\$126bobo61b2o4b2o\$127bo63bo4b2o2\$129b3o7\$111bo\$111bo\$111bo131b3o\$242bo
3bo\$242bo4bo\$244bo3bo\$241b3obo2bo\$240bo7bo\$240bobo3bo\$105b3o131b2obo3b
o\$239b2ob3o\$227b2o\$227b2o3\$243b2o\$118bo124b2o\$110bobob2obo125bo\$110bo
3b4o3b3o\$110b2ob2o2bo101b2o22bo\$118bo100b2o\$118b3o136bo\$99b2o17b3o116b
3o5bo11bo\$99b2o136bo7bo11bo\$237bobo5bo7b2o\$238b2o13b2o\$253b3o3bo\$137b
3o116b3o\$140bo116bo\$135b2o4bo115b2o\$91b2o27b3o15bo3bo113b3o\$91b2o35b2o
4bo4bo2bo113bo\$79bo45bob3o4bo3bo3bo115bo\$79bo54b3obob2o114bo\$79bo18bo
39bo117b3o\$93bobob3o34b4o\$91b2obo5b2o17b2obo11b2o\$92bo26bo2bo\$93b3o5bo
18b3o\$97bo134b2o\$98b2o\$73b3o23bo\$100bo4bo\$25bo74bo4bo131bobo\$24bobo
205b2o5bo12b2o\$23b2ob2o206b2o4bo11b2o\$8b2o16b2o207bo4b2o\$7bo2bo9b2o
110b2o102bo3bo\$6bo2bo12bo2b3o104b2o\$20b2obo213bo2bo\$22bo2bo211bobo\$10b
o11bo214bo2bo\$6bo3b2o11b2o213b2o4b2o\$5b2o3bo228bo4b2o\$2o3b2obobo13bo\$
2o3b2o3bo11b2ob2o44b3o50b2o\$7b3o13bo46bo3bo49b2o\$7b3o16bo43bo4bo\$24bo
14b2o31bo3bo\$9bo31bo27b3obo2bo\$8b2o28bo2bo26bo7bo\$7bob2o26b4ob2o24bobo
3bo\$6b2o36bo22b2obo3bo\$5bo3bobo23bo2b3o3bo22b2ob3o\$5b7o22b3obo2bo2bo2b
o7b2o\$3b2o3bobo23b3o4b2o4bo7b2o\$3b3o29bobob2o6bo\$3bo32bob2o\$4bo4bo26bo
34b2o\$4bo4bo61b2o\$6b3o62bo2\$9bo3bobo31b2o22bo\$7b2obob2o2bo30b2o\$8bo2bo
bobo69bo\$8bo2bo53b3o5bo11bo\$8bo2b2o52bo7bo11bo\$8bo2b4o50bobo5bo7b2o\$9b
2o55b2o13b2o\$33b2o46b3o3bo\$19bobo11b2o49b3o\$18bo2bo63bo\$18bo2bo63b2o\$
19b2o63b3o\$84bo\$86bo\$84bo\$25b2o57b3o\$25b2o4\$60b2o4\$65bobo\$60b2o5bo12b
2o\$62b2o4bo11b2o\$63bo4b2o\$64bo3bo2\$65bo2bo\$65bobo\$65bo2bo\$66b2o4b2o\$
67bo4b2o!
``````
@Hunting so we're making a referential signature chain?
Stop being mean to MathAndCode!
LeapLife has fun drifters!

MathAndCode
Posts: 2745
Joined: August 31st, 2020, 5:58 pm

Re: Binary slow salvos

wwei23 wrote:
September 19th, 2020, 8:01 pm
Also, forget corderrakes. Just use 2-engine corderships and 3-engine corderships to clean up the glider-producing switch engine debris.
Do they also work with the debris from signal reflection?
I have reduced the cost of universal construction to seventeen gliders and probably to sixteen. All that remains is for the universal operations to be found.

wwei23
Posts: 2449
Joined: May 22nd, 2017, 6:14 pm
Location: The (Life?) Universe

Re: Binary slow salvos

MathAndCode wrote:
September 19th, 2020, 9:07 pm
Do they also work with the debris from signal reflection?
No. I don't think that can be predicted in advance, even.
@Hunting so we're making a referential signature chain?
Stop being mean to MathAndCode!
LeapLife has fun drifters!

MathAndCode
Posts: 2745
Joined: August 31st, 2020, 5:58 pm

Re: Binary slow salvos

wwei23 wrote:
September 19th, 2020, 9:25 pm
MathAndCode wrote:
September 19th, 2020, 9:07 pm
Do they also work with the debris from signal reflection?
No. I don't think that can be predicted in advance, even.
There are only two reflection mechanisms, so there are only three possibilities, including the standard ash.

Also, here's a partial idea for a 3-GPSE RCT-based universal constructor dependent on an applicable return signal from a signal-reflecting search for all lanes: Under normal circumstances, glider streams A₀ and B are fed into a logic gate, but there is no output due to glider stream C₀. A gap in glider stream A₀ lets a B glider through as a construction glider and lets a C₀ through to send another signal to GPSE A. A gap in glider stream B (due to an A₁ glider) lets lets a C₀ through to send another signal to GPSE A and lets an A₀ through to GPSE C. This might result in a gap in glider stream C₀ or a C₁ glider. The former will allow a B′ glider through as a construction glider, and I'm not sure what to do with the latter yet.
I have reduced the cost of universal construction to seventeen gliders and probably to sixteen. All that remains is for the universal operations to be found.

dvgrn
Moderator
Posts: 7578
Joined: May 17th, 2009, 11:00 pm
Contact:

Re: Binary slow salvos

MathAndCode wrote:
September 19th, 2020, 9:36 pm
wwei23 wrote:
September 19th, 2020, 9:25 pm
MathAndCode wrote:
September 19th, 2020, 9:07 pm
Do they also work with the debris from signal reflection?
No. I don't think that can be predicted in advance, even.
There are only two reflection mechanisms, so there are only three possibilities, including the standard ash.
Quite possibly we can drive a fleet of various kinds of Corderships past the three types of debris, and clean each kind up if it's there. Corderrakes would be kind of a problem for the signal-reflection debris, since gliders intended for a specific type of debris would probably wreak havoc when that debris wasn't present. A Cordership-spark cleanup is much safer.

All of the optional debris will always appear at the same offset mod 8 along the GPSE path, right? Somebody could design that Corderfleet without too much trouble, and it's not all that difficult to build a stable seed constellation for a Corderfleet like that, since the distance between them is arbitrary.

After they've done their work, all the Corderships will have to crash into something, without releasing any gliders. Maybe the simplest idea would be some set of bait still lifes that participates in complete mutual destruction with each Cordership. Anyone interested in hunting for a workable set of baits for 2- and 3-engine Corderships?

MathAndCode
Posts: 2745
Joined: August 31st, 2020, 5:58 pm

Re: Binary slow salvos

dvgrn wrote:
September 19th, 2020, 10:01 pm
Quite possibly we can drive a fleet of various kinds of Corderships past the three types of debris, and clean each kind up if it's there. Corderrakes would be kind of a problem for the signal-reflection debris, since gliders intended for a specific type of debris would probably wreak havoc when that debris wasn't present. A Cordership-spark cleanup is much safer.
Actually, there is a method that could be used for detecting reflection ash.
MathAndCode wrote:
September 1st, 2020, 12:48 pm
This seems like an efficient way of burning through some of the glider-producing switch engine's Herschel ash.

Code: Select all

``````x = 581, y = 546, rule = B3/S23
obo\$b2o\$bo19\$23bo\$24bo\$22b3o233\$258bo\$259bo\$257b3o100\$421bo\$420b3o15bo\$419b2ob2o14bo\$418b2o3bo7b4o3bo\$419bobobobo5bo3bo\$420b2obo2bo4b2o3bo\$423b2obo7b3o\$424b2o10b2obo\$437b3o\$438bo10\$426b2o\$426b2o3\$462b2o\$462b2o2\$373b2o9bobo\$373bobo9b2o64bo\$373bo11bo64bobo\$450b2o3\$469bo\$468b3o\$469b2o\$442b2o\$442b2o22b2o\$464b2o5bo\$464bo5bo\$464b2o\$466bo3bo\$467bo2bo\$468bo8b2o\$477bobo\$478bo3\$407bo\$408bo37b2o\$406b3o36bo2bo\$446b2o\$443bo\$443bo\$443bo2\$445b3o46b2o\$494b2o2\$448b2o\$448b2o33bo\$482bobo\$469bo12b2o30b2o\$468bobo42bo2bo\$468b2o44b2o2\$450bo\$450b2o\$449bobo26bo\$477bobo\$477b2o2\$440b2o\$441bo66bo\$507bobo\$507bo2bo\$508b2o\$442b2obo\$443b2o\$444bo2\$446bo31b2o\$442b2ob2o30bo2bo\$443b3o32b2o\$444bo30bo\$475bo\$475bo2\$477b3o3\$480b2o\$480b2o9bo\$491bo\$491bo54b2o\$545bo2bo\$546b2o2\$498b3o2\$510bo\$509bobo\$509b2o2\$472b2o\$472b2o66bo\$539bobo\$539bo2bo\$540b2o4\$468b2o\$468b2o40b2o\$509bo2bo\$510b2o\$507bo\$507bo\$507bo2\$509b3o2\$478b2o\$477bobo32b2o\$477b2o33b2o9bo\$523bo\$523bo54b2o\$577bo2bo\$578b2o2\$530b3o2\$542bo\$541bobo\$541b2o2\$504b2o\$504b2o66bo\$571bobo\$571bo2bo\$572b2o4\$500b2o\$500b2o40b2o\$541bo2bo\$542b2o\$539bo\$539bo\$539bo2\$541b3o2\$510b2o\$509bobo32b2o\$509b2o33b2o9bo\$555bo\$555bo4\$562b3o2\$574bo\$573bobo\$573b2o2\$536b2o\$536b2o7\$532b2o\$532b2o40b2o\$573bo2bo\$574b2o\$571bo\$571bo\$571bo2\$573b3o2\$542b2o\$541bobo32b2o\$541b2o33b2o!``````
One could send pairs of gliders. However, this has several problems, such as not being able to tell which reflection ash is present and requiring a mechanism to store how long the previous glider took to return in order to know how long to wait for the current glider.
Another possibility is to enter the string of bits twice, the first time for construction and the second time for telling some mechanism what sequence of ash-destruction salvos to send. (The problem of having to clean up the second round of bits is avoided by programming the destruction salvo sender to send two sequences of destruction salvos instead of only one, but doing that would also be complicated.)
I have reduced the cost of universal construction to seventeen gliders and probably to sixteen. All that remains is for the universal operations to be found.

dvgrn
Moderator
Posts: 7578
Joined: May 17th, 2009, 11:00 pm
Contact:

Re: Binary slow salvos

MathAndCode wrote:
September 19th, 2020, 10:10 pm
One could send pairs of gliders. However, this has several problems, such as not being able to tell which reflection ash is present and requiring a mechanism to store how long the previous glider took to return in order to know how long to wait for the current glider.
Another possibility is to enter the string of bits twice, the first time for construction and the second time for telling some mechanism what sequence of ash-destruction salvos to send. (The problem of having to clean up the second round of bits is avoided by programming the destruction salvo sender to send two sequences of destruction salvos instead of only one, but doing that would also be complicated.)
For years this RCT project has been cursed by the specter of unnecessary complexity. The idea starts out relatively simple, but then it turns out let's say that the number of bits on the tape can be hugely reduced -- maybe by using the first N bits to build a block-keeper, or build a universal computer that can store the second half of the bits, or build some other decision-making cleanup device that can help make the cleanup "easier".

All of these additional structures add design complexity, because you have to figure out exactly how to build them -- what's the circuit layout, where will it be built where it won't get in the way, how will it be started up, how will it be shut down, where will the self-destruct signal come from to clean it all up completely when its work is done? Based on experience so far, there isn't really anyone in the world with enough time, inclination and expertise to solve all those problems and make a working demonstration pattern.

By contrast, if cleanup can be done by a fleet of Corderships, then it's fairly trivial to build a seed constellation that can produce that fleet of Corderships. No matter how many Corderships it takes, it will still require orders of magnitude less problem-solving than any design involving unbuildable vaporware decision-making mechanisms. It's also much easier to prove to everyone's satisfaction that a Cordership-fleet seed will work; you can just trigger it and see.

Even if the geometry turns out to require building a seed constellation for a slow salvo, that then fires its gliders back into the "epicentre" to build the Cordership-fleet seed along with the actual intended target object, and also shoots gliders and *WSSes all around the epicentre to clean up all the non-repetitive debris from the crashed GPSEs and so on ... that's still relatively simple, conceptually, and it uses technology that has already been demonstrated in the volatility-1 oscillator. We could easily build a proof-of-concept, which isn't really something you can say for more complex designs.

-- Maybe it's worth waiting a bit on the Cordership cleanup research, though, in case the 3 GPSE RCT design turns out to be workable and a different cleanup is needed?

MathAndCode
Posts: 2745
Joined: August 31st, 2020, 5:58 pm

Re: Binary slow salvos

dvgrn wrote:
September 20th, 2020, 8:18 am
By contrast, if cleanup can be done by a fleet of Corderships, then it's fairly trivial to build a seed constellation that can produce that fleet of Corderships. No matter how many Corderships it takes, it will still require orders of magnitude less problem-solving than any design involving unbuildable vaporware decision-making mechanisms. It's also much easier to prove to everyone's satisfaction that a Cordership-fleet seed will work; you can just trigger it and see.

Even if the geometry turns out to require building a seed constellation for a slow salvo, that then fires its gliders back into the "epicentre" to build the Cordership-fleet seed along with the actual intended target object, and also shoots gliders and *WSSes all around the epicentre to clean up all the non-repetitive debris from the crashed GPSEs and so on ... that's still relatively simple, conceptually, and it uses technology that has already been demonstrated in the volatility-1 oscillator. We could easily build a proof-of-concept, which isn't really something you can say for more complex designs.

-- Maybe it's worth waiting a bit on the Cordership cleanup research, though, in case the 3 GPSE RCT design turns out to be workable and a different cleanup is needed?
Yes; using Cordership sparks would be best. I only suggested the other things in case they didn't work on the variable ash or there was no way to stop them.
I think that it's worth investigating Cordership cleanup for the 4-GPSE solution for two reasons: First, we'll probably have to clean up the ash from the ∏-∏-LoM-forwards glider sequence in both cases. Secondly, any 3-GPSE solution will probably use at least one logic gate, and there will be no chance of constructing two GPSEs with less than eight gliders, so the 3-GPSE solution's glider cost won't be much less than the 4-GPSE solution's glider cost.
Also, if you want a 3-GPSE solution, you should do at least one of the following:
• Convince me that it would not be so bad to do the same manual search again
• Code and do this automated search (or code it then give it to me so I can run it on my computer)
• Convince me that it would not be so wasteful to reinvent things like sideways glider detection
• Show me examples of other people developing things like sideways glider detection and hope that I happen to know the programming language that they're in or am willing to learn it
Last edited by MathAndCode on September 20th, 2020, 5:56 pm, edited 1 time in total.
I have reduced the cost of universal construction to seventeen gliders and probably to sixteen. All that remains is for the universal operations to be found.

calcyman
Posts: 2352
Joined: June 1st, 2009, 4:32 pm

Re: Potential universal construction in 16±2 gliders

calcyman wrote:
September 19th, 2020, 2:03 pm
EDIT: actually, we can just make this north-westerly GPSE have a much much much longer glider trail than the south-easterly GPSE (by synthesising it earlier), so that by the time that the retrograde gliders reach the synthesis site, everything else has completed and the recipe has built a couple of convenient eaters.
Here's a demonstration that the 17 gliders can come from infinity whilst being synchronised so that (at generation 50000 in the pattern below) the minimum population temporarily drops to 64:

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``````x = 40758, y = 40976, rule = B3/S23
2.A\$A.A\$.2A414\$132.A\$133.A\$131.3A9219\$40755.A.A\$40755.2A\$40756.A
17\$9513.A\$9514.2A\$9513.2A\$40725.A\$40725.A.A\$40725.2A6168\$99.A.A\$
100.2A\$100.A17\$25131.A\$25129.2A\$25130.2A\$131.A\$129.A.A\$130.2A90\$
25185.A.A\$25185.2A\$25186.A9467\$135.2A\$134.A.A\$136.A\$25135.2A\$
25134.2A\$25136.A17\$105.A\$105.2A\$104.A.A15411\$126.3A\$128.A\$127.A18\$
40728.3A\$40728.A\$40729.A74\$25182.3A\$25182.A\$25183.A\$187.2A\$186.A.A
\$188.A25\$25205.2A\$25204.2A\$25206.A!``````
What do you do with ill crystallographers? Take them to the mono-clinic!

MathAndCode
Posts: 2745
Joined: August 31st, 2020, 5:58 pm

3-GPSE ideas

Here is a list of the 3-GPSE ideas that I have had so far:

1.
MathAndCode wrote:
September 10th, 2020, 1:05 pm
Tenth edit: … Glider stream A₀ will feed into GPSE C in addition to glider stream C₀ feeding into GPSE A. The timing will be such so that GPSE A always sends back a glider on lane A₁ when it receives a signal, and GPSE C either sends back a glider on lane C₁ or doesn't send back any signal when it releases a signal. Ordinarily, glider B will destroy gliders A₀ and C₀. However, when glider A₁ arrives, it will shift glider B's path via a logic gate, preventing it from crashing into gliders A₀ and C₀, letting them go past each other. If there is no glider C₁, then glider B will proceed to construction on its path, but if there is a glider C₁, then it will shift glider B via another logic gate, making it take a different path to construction.
• We know that the two construction lanes would be twelve lanes apart with opposite parity, which is common to some other designs, so we can start work on ensuring that that parity-displacement pair is universal (if that has not already been done) as soon as we want.
• There might not be any position-phase pairs for GPSEs A and C that will work.
• It requires multiple logic gates, which will increase the glider cost.
• No one has volunteered to code this automated search yet.
2a.
MathAndCode wrote:
September 10th, 2020, 1:05 pm
Eighteenth edit: … Under normal circumstances, gliders A₀ and B collide and, with the help of a logic gate, make another glider (which shall be called glider B′) that then collides with and annihilates glider C. If there is a gap in glider stream A₀, then glider B is let through as a construction glider, and glider C is let through in order to send another signal to GPSE A. If a glider comes on lane A₁, then it will annihilate glider B, letting glider C be sent to GPSE A and letting glider A₀ be sent to GPSE C. The signal to GPSE C will return first as a gap in glider stream C that will let glider B′ through as a construction glider. Then the signal sent to GPSE A will return either as a gap in glider stream A₀ or as a glider on lane A₁, and the cycle will proceed again.
• We know that the two construction lanes would be twelve lanes apart with opposite parity, which is common to some other designs, so we can start work on ensuring that that parity-displacement pair is universal (if that has not already been done) as soon as we want.
• I currently do not know a possible displacement of glider A₁ relative to glider A₀ so that gliders A₀ and B create a traffic light and glider and gliders A₁ and B create nothing.
• There might not be a way for glider C to come so that it cleanly prevents the logic gate from having any output.
• Ensuring that no C₁ gliders occur might not be possible.
2b.
MathAndCode wrote:
September 19th, 2020, 9:36 pm
Under normal circumstances, glider streams A₀ and B are fed into a logic gate, but there is no output due to glider stream C₀. A gap in glider stream A₀ lets a B glider through as a construction glider and lets a C₀ through to send another signal to GPSE A. A gap in glider stream B (due to an A₁ glider) lets a C₀ through to send another signal to GPSE A and lets an A₀ through to GPSE C. This might result in a gap in glider stream C₀ or a C₁ glider. The former will allow a B′ glider through as a construction glider, and I'm not sure what to do with the latter yet.
This is the same as 2a except that it gives up on preventing any C₁ gliders and figures out a way to deal with them instead, which is why it doesn't get a separate number.
• We know that the two construction lanes from the same direction would be twelve lanes apart with opposite parity, which is common to some other designs, so we can start work on ensuring that that parity-displacement pair is universal (if that has not already been done) as soon as we want.
• Construction with three lanes from two different directions will likely be easier.
• We might be able to create the target by colliding a B or B′ glider with a C₁ glider.
• I currently do not know a possible displacement of glider A₁ relative to glider A₀ so that gliders A₀ and B create a traffic light and glider and gliders A₁ and B create nothing.
• There might not be a way for glider C₀ to come so that it cleanly prevents the logic gate from having any output.
• Controlling how the signal sent to GPSE C returns will be at best difficult and at worst impossible. Even if it's impossible, universal construction will probably still be possible (by catching the C₁ gliders with a crystal and sending extra signals to GPSE C), but it will definitely be more inconvenient.
3. Under normal circumstances, glider streams B and C are fed into a logic gate, but there is no output due to glider stream A₀. A gap in glider stream A₀ lets a B′ glider through as a construction glider. A gap in glider stream B due to an (due to an A₁ glider) C₀ through to send another signal to GPSE A. A gap in glider stream B (due to an A₁ glider) lets lets a C₀ through to send another signal to GPSE A and lets an A₀ through to GPSE C. This might result in a gap in glider stream C₀ or a C₁ glider. The former will allow a B glider through as a construction glider, and I'm not sure what to do with the latter yet, but we might be able to avoid it. This is similar to 2.
• We know that the two construction lanes from the same direction would be twelve lanes apart with opposite parity, which is common to some other designs, so we can start work on ensuring that that parity-displacement pair is universal (if that has not already been done) as soon as we want.
• Construction with three lanes from two different directions will likely be easier.
• We might be able to create the target by colliding a B or B′ glider with a C₁ glider.
• This could provide more flexibility in terms of making sure that there is a way for an A₁ glider to annihilate a B glider
• There might not be a way for an A₀ glider to come so that it cleanly prevents the logic gate from having any output.
• Controlling how the signal sent to GPSE C returns will be at best difficult and at worst impossible. Even if it's impossible, universal construction will probably still be possible (by catching the C₁ gliders with a crystal and sending extra signals to GPSE C), but it will definitely be more inconvenient.
4.
MathAndCode wrote:
September 1st, 2020, 12:48 pm
These two signals create extra gliders on different lanes without removing any second natural gliders.

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``````x = 966, y = 908, rule = B3/S23
o\$b2o\$2o657\$651b3o\$651bo\$652bo62\$715b3o\$715bo\$716bo39\$839bo\$838b3o2\$838b3o\$839b2ob2o\$841bo\$855b2o\$855b2o3\$839bo\$838bobo\$839bo2\$838b2o\$837bobo\$856bo\$837b3obobo11bo\$841b3o13b2o\$841b2o14b2o\$858bo\$855bo2bo\$855bobo\$779b3o71b2o\$779bo72bo\$780bo73bo\$843b2o5b4o\$843b2o4bo2bo\$849bo2bo\$850b3o\$879b2o\$879b2o3\$868bo\$867bobo\$867b2o6\$859b2o4bobo20bo\$859b2o3bo3bo18b2obobo\$863bo4bo17bo4bo\$863bo4bo8b2o6bo3bobo\$863bo3bo9b2o5bo3bo2b2o\$864bo2bo15bo3bo\$883bo5b4obo\$883bo2bobo4bob2o\$884b3o6bo\$887bo3b2o\$888bo3\$863b2o\$862bo2bo\$863b2o\$860bo\$860bo\$860bo2\$862b3o12b2o32b2o\$878b2o31b2o\$877bo\$865b2o\$865b2o33bo\$899bobo\$832bo53bo12b2o30b2o\$833b2o50bobo42bo2bo\$832b2o51b2o44b2o4\$895bo\$894bobo\$894b2o2\$857b2o\$857b2o66bo\$924bobo\$924bo2bo\$925b2o\$872b3o\$872bob2o\$868bo3bo2b2o\$867b2o4b4o\$843b3o5bo7b2o5bobo3bobo20b2o\$843bo5bobo6b3o4b2o5b2o20bo2bo\$844bo6b2obo6bo4b2o5b2o20b2o\$848bo2bo3bobobobo5bo4b3o17bo\$848bo7bo3bo31bo\$850bo4b2obo33bo\$850b2o6bo2bo\$859b2o33b3o\$860bo2\$897b2o\$897b2o9bo\$908bo\$908bo54b2o\$962bo2bo\$963b2o2\$915b3o2\$927bo\$926bobo\$926b2o2\$889b2o\$889b2o66bo\$956bobo\$956bo2bo\$957b2o4\$885b2o\$885b2o40b2o\$926bo2bo\$927b2o\$924bo\$924bo\$924bo2\$926b3o2\$895b2o\$894bobo32b2o\$894b2o33b2o9bo\$940bo\$940bo4\$947b3o2\$959bo\$958bobo\$958b2o2\$921b2o\$921b2o!``````

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``````x = 966, y = 927, rule = B3/S23
o\$b2o\$2o651\$646b2o\$646bobo\$646bo62\$710b2o\$710bobo\$710bo45\$836bo\$836bo5b3o\$835bobo3b2o2bo\$836bo3bo5bo\$836bo3bo\$841bo2b2o\$842b2o11b2o\$855b2o8\$849bo\$839b2o9b2o\$774b2o62bob3o8b2o\$774bobo61bo3bo10bo\$774bo64b2obo5b5o6b2o\$840bo10bo6bo2bo\$857b2ob2o\$856bobo\$849b2o4b2o\$849b2o4bo3bo\$856bo2bo\$843b2o11b3o\$843b2o3\$879b2o\$845b2o32b2o\$847bo\$845b3o5b2o\$868bo\$867bobo\$867b2o3\$889bo\$866b2o20bobo\$861b2o3b2o22bo\$859b2o2bo22bob2o5b2o\$858bob3obo20b2obo6bob2o\$857bo2bo3bo33bo\$858bo2b2ob2o11b2o10bobo4bobo\$859b2o3bo12b2o9bo2bo4b2o\$860b3obobo22bobo\$865b2o26b2o\$893b3o\$892bo2bo\$892bobo\$893bo3\$863b2o\$862bo2bo17bo\$863b2o18b2o\$860bo21bobo\$860bo\$860bo2\$862b3o46b2o\$911b2o2\$865b2o\$865b2o33bo\$899bobo\$832bo53bo12b2o30b2o\$833b2o50bobo42bo2bo\$832b2o51b2o44b2o4\$895bo\$894bobo\$894b2o2\$857b2o\$857b2o66bo\$924bobo\$838b2o84bo2bo\$838bobo84b2o\$838bo\$850b4o9b3o\$854bo8bobo\$848bo6bo6b2o2b2o\$847b3o4bo7b3o30b2o\$845bo5bobo6b2o32bo2bo\$845bobo3b2o7b4o2bobo26b2o\$845bo3b2o13b2obobo5bo16bo\$847b3o11bo8bo4bo16bo\$861b3o3b4o4bo16bo\$865b2ob2o\$855b3o2b2obobo28b3o\$849b2ob4o2b4o2bo\$852b5obo3b2o\$857bo39b2o\$897b2o9bo\$908bo\$908bo54b2o\$962bo2bo\$963b2o2\$915b3o2\$927bo\$926bobo\$926b2o2\$889b2o\$889b2o66bo\$956bobo\$956bo2bo\$957b2o4\$885b2o\$885b2o40b2o\$926bo2bo\$927b2o\$924bo\$924bo\$924bo2\$926b3o2\$895b2o\$894bobo32b2o\$894b2o33b2o9bo\$940bo\$940bo4\$947b3o2\$959bo\$958bobo\$958b2o2\$921b2o\$921b2o7\$917b2o\$917b2o40b2o\$958bo2bo\$959b2o\$956bo\$956bo\$956bo2\$958b3o2\$927b2o\$926bobo32b2o\$926b2o33b2o!``````
Here is a mechanism where the return signals are gliders on different outer lanes, but a backwards glider from one of the signals leads to the same restriction as with the signal pairs listed at the beginning of this post.

Code: Select all

``````x = 1414, y = 1375, rule = B3/S23
o\$b2o\$2o766\$768bo\$769b2o\$768b2o325\$1089bo\$1088b2o\$1088bobo62\$1153bo\$1152b2o\$1152bobo51\$1292bo\$1278bobo6bo2b4o\$1277bo9bob4o\$1278bo2bo6b3o3b2o\$1216bo63b3o9bo2b2o\$1217b2o74bob2o\$1216b2o76bo8b2o\$1303b2o4\$1217bo\$1216b2o\$1216bobo2\$1305bo\$1304bobo\$1297bo8bo\$1283b2o11b2o5b2ob3o\$1283b2o11bo3b2o3b2ob2o\$1299b3o2b2o2bo\$1299bo2bo2b3o\$1302bo3bo\$1302bo2bo\$1298bo2bo3bo\$1305bo\$1291b2o\$1291b2o3\$1303bobo2bo18b2o\$1304b2o2bo18b2o\$1303b2o3bo\$1305bo2bo\$1308bo7bo\$1307bo7bobo\$1304bob2o7b2o\$1304b3o\$1305bo36b3o\$1342b2ob2o\$1341bobob2o\$1330b3o8bobob2o3b2o\$1311bo17bo2bo9bobo5b2o\$1310bobo16bo3b2o8bo\$1310b2o24bo\$1325b2o9bo8b2o\$1325b2o3b4o11b2o\$1334b3o\$1336bo3\$1341bo\$1334b4obo2bo\$1334bob2o3bo5bo\$1337b2o2bo6b2o\$1311b2o26bo7b2o\$1310bo2bo\$1311b2o2\$1307b3o2\$1311bo\$1311bo47b2o\$1311bo47b2o2\$1313b2o\$1313b2o33bo\$1347bobo\$1334bo12b2o30b2o\$1333bobo42bo2bo\$1333b2o44b2o4\$1343bo\$1281bo60bobo\$1280b2o60b2o\$1280bobo\$1305b2o\$1305b2o66bo\$1298b2o72bobo\$1298bo73bo2bo\$1297bo75b2o\$1297bo\$1297b4o\$1296bo3bo\$1296bo2bo2bo\$1301b2o40b2o\$1321b3o18bo2bo\$1322bo20b2o\$1322bo\$1321b3o15b3o2\$1343bo\$1343bo\$1343bo\$1311b2o\$1310bobo32b2o\$1310b2o33b2o\$1355b3o\$1411b2o\$1410bo2bo\$1411b2o\$1364bo\$1364bo\$1364bo\$1375bo\$1374bobo\$1374b2o2\$1337b2o\$1337b2o66bo\$1404bobo\$1404bo2bo\$1405b2o4\$1333b2o\$1333b2o40b2o\$1374bo2bo\$1375b2o2\$1371b3o2\$1375bo\$1375bo\$1375bo\$1343b2o\$1342bobo32b2o\$1342b2o33b2o\$1387b3o4\$1396bo\$1396bo\$1396bo\$1407bo\$1406bobo\$1406b2o2\$1369b2o\$1369b2o7\$1365b2o\$1365b2o40b2o\$1406bo2bo\$1407b2o2\$1403b3o2\$1407bo\$1407bo\$1407bo\$1375b2o\$1374bobo32b2o\$1374b2o33b2o!``````
If we use this for GPSE A, then those could be our construction gliders. However, if we use them as the construction gliders, then they likely cannot be part of the circuitry, so the construction recipe would have to be encoded in the position of another GPSE.
• This seems that most likely to be able to turned into a 2-GPSE solution (although the chance of that is still small).
• The glider lanes are rather far apart, which might make construction difficult—or even prohibit universality.
• This is only a partial idea, so I am not sure that there is a convenient way to feed signals to GPSE A with only 3 GPSEs total.

Edit: I think that the return signal from GPSE C would be correlated with the next return signal from GPSE A, so if one makes sure that the signal from GPSE A that results in an A₀ being sent to GPSE C never occurs twice in a row, then one can place GPSE C so that there will be no C₁ gliders.

Another edit: Here's another way to get rid of an A₀ glider without sending any C₁ gliders.

Code: Select all

``````x = 136, y = 155, rule = B3/S23
22bobo\$8bobo7bo2bo2bo\$7b2ob2o5bobob2obo\$8bob2o6bo\$12bo11b3o\$11bo13b2o\$11bo12b2o10\$24bo\$23bobo\$13b2o\$13b2o3\$49b2o\$49b2o3\$38bo\$37bobo\$37b2o2\$57b2o\$57b2o3\$29b2o25bo\$29b2o24b3o\$53b4obo\$53b4o2bo\$52bo\$52b2o\$57bo\$53b5o6b2o\$53b3o8bobo\$65bo4\$33b2o\$32bo2bo\$33b2o\$30bo\$30bo\$30bo2\$32b3o46b2o\$81b2o2\$35b2o\$35b2o33bo\$69bobo\$56bo12b2o30b2o\$55bobo42bo2bo\$55b2o44b2o4\$35b2o28bo\$36b2o26bobo\$35bo28b2o3\$28b2o65bo\$bo26bo2bo62bobo\$2bo24bo2b3o61bo2bo\$3o24b2o66b2o\$30b3o\$27bo5bo\$28b2o2b2o\$30bo2bo\$30b3o32b2o\$64bo2bo\$65b2o\$62bo\$62bo\$62bo2\$64b3o3\$67b2o\$67b2o9bo\$78bo\$78bo54b2o\$132bo2bo\$133b2o2\$85b3o2\$97bo\$96bobo\$96b2o2\$59b2o\$59b2o66bo\$126bobo\$126bo2bo\$127b2o4\$55b2o\$55b2o40b2o\$96bo2bo\$97b2o\$94bo\$94bo\$94bo2\$96b3o2\$65b2o\$64bobo32b2o\$64b2o33b2o9bo\$110bo\$110bo4\$117b3o6\$91b2o\$91b2o7\$87b2o\$87b2o9\$97b2o\$96bobo\$96b2o!``````
It only works for one out of two possible timings, but the ash can be used to create a crystal that can be used for the other relative timing.

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``````x = 1711, y = 1664, rule = B3/S23
3o\$o\$bo62\$64b3o\$64bo\$65bo43\$104bo\$105bo\$103b3o17\$128b3o\$128bo\$129bo62\$192b3o\$192bo\$193bo43\$232bo\$233bo\$231b3o17\$256b3o\$256bo\$257bo62\$320b3o\$320bo\$321bo43\$360bo\$361bo\$359b3o17\$384b3o\$384bo\$385bo62\$448b3o\$448bo\$449bo43\$488bo\$489bo\$487b3o17\$512b3o\$512bo\$513bo62\$576b3o\$576bo\$577bo43\$616bo\$617bo\$615b3o17\$640b3o\$640bo\$641bo62\$704b3o\$704bo\$705bo43\$744bo\$745bo\$743b3o17\$768b3o\$768bo\$769bo62\$832b3o\$832bo\$833bo43\$872bo\$873bo\$871b3o17\$896b3o\$896bo\$897bo62\$960b3o\$960bo\$961bo43\$1000bo\$1001bo\$999b3o17\$1024b3o\$1024bo\$1025bo62\$1088b3o\$1088bo\$1089bo43\$1128bo\$1129bo\$1127b3o17\$1152b3o\$1152bo\$1153bo62\$1216b3o\$1216bo\$1217bo43\$1256bo\$1257bo\$1255b3o17\$1280b3o\$1280bo\$1281bo62\$1344b3o\$1344bo\$1345bo43\$1384bo\$1385bo\$1383b3o17\$1408b3o\$1408bo\$1409bo62\$1472b3o\$1472bo\$1473bo35\$1597bobo\$1583bobo7bo2bo2bo\$1582b2ob2o5bobob2obo\$1583bob2o6bo\$1587bo11b3o\$1586bo13b2o\$1586bo12b2o2\$1512bo\$1513bo\$1511b3o6\$1599bo\$1598bobo\$1588b2o\$1588b2o3\$1624b2o\$1624b2o3\$1613bo\$1536b3o73bobo\$1536bo75b2o\$1537bo\$1632b2o\$1632b2o3\$1604b2o25bo\$1604b2o24b3o\$1628b4obo\$1628b4o2bo\$1627bo\$1627b2o\$1632bo\$1628b5o6b2o\$1628b3o8bobo\$1640bo4\$1608b2o\$1607bo2bo\$1608b2o\$1605bo\$1605bo\$1605bo2\$1607b3o46b2o\$1656b2o2\$1610b2o\$1610b2o33bo\$1644bobo\$1631bo12b2o30b2o\$1630bobo42bo2bo\$1630b2o44b2o4\$1610b2o28bo\$1611b2o26bobo\$1610bo28b2o3\$1603b2o65bo\$1576bo26bo2bo62bobo\$1577bo24bo2b3o61bo2bo\$1575b3o24b2o66b2o\$1605b3o\$1602bo5bo\$1603b2o2b2o\$1605bo2bo\$1605b3o32b2o\$1639bo2bo\$1640b2o\$1637bo\$1637bo\$1637bo2\$1639b3o3\$1642b2o\$1642b2o9bo\$1653bo\$1653bo54b2o\$1707bo2bo\$1708b2o2\$1660b3o2\$1672bo\$1671bobo\$1671b2o2\$1634b2o\$1634b2o66bo\$1701bobo\$1701bo2bo\$1702b2o4\$1630b2o\$1630b2o40b2o\$1671bo2bo\$1672b2o\$1669bo\$1669bo\$1669bo2\$1671b3o2\$1640b2o\$1639bobo32b2o\$1639b2o33b2o9bo\$1685bo\$1685bo4\$1692b3o6\$1666b2o\$1666b2o7\$1662b2o\$1662b2o9\$1672b2o\$1671bobo\$1671b2o!``````
Eventually, the crystal will interact with the Herschel debris and emit a sideways glider, but there's plenty of time to start a new crystal before that happens. There are probably a few other reactions like this one as well.

Yet another edit: Here are possible ways of canceling out the logic gate's output.

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``````x = 38, y = 49, rule = B3/S23
35bo\$35bobo\$35b2o40\$3o\$2bo22b3o\$bo23bo\$26bo2\$10b2o\$10b2o!``````

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``````x = 30, y = 45, rule = B3/S23
27bo\$27bobo\$27b2o32\$2o\$b2o15b2o\$o16b2o\$19bo6\$6b2o\$6b2o!``````

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``````x = 31, y = 45, rule = B3/S23
28bo\$28bobo\$28b2o32\$b2o\$obo16b2o\$2bo16bobo\$19bo6\$7b2o\$7b2o!``````

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``````x = 32, y = 45, rule = B3/S23
29bo\$29bobo\$29b2o33\$bo\$b2o19bo\$obo18b2o\$21bobo5\$8b2o\$8b2o!``````

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``````x = 32, y = 45, rule = B3/S23
29bo\$29bobo\$29b2o34\$3o\$2bo18b3o\$bo19bo\$22bo4\$8b2o\$8b2o!``````

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``````x = 32, y = 45, rule = B3/S23
29bo\$29bobo\$29b2o34\$2o\$b2o19b2o\$o20b2o\$23bo4\$8b2o\$8b2o!``````

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``````x = 33, y = 45, rule = B3/S23
30bo\$30bobo\$30b2o34\$b2o\$obo20b2o\$2bo20bobo\$23bo4\$9b2o\$9b2o!``````
This does not cancel the logic gate's input but might be useful.

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``````x = 20, y = 31, rule = B3/S23
13bo\$13bobo\$13b2o17\$b2o\$obo14b2o\$2bo14bobo\$17bo7\$6b2o\$6b2o!``````
I just realized that that was the wrong side for the particular 3-GPSE idea that I was trying to achieve, but I'm posting it for reference because it could be used for another 3-GPSE idea. Here are the options relevant to the 3-GPSE idea that I'm currently working on.

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``````x = 18, y = 22, rule = B3/S23
16bo\$15bo\$15b3o6\$b2o\$obo10b2o\$2bo10bobo\$13bo9\$4b2o\$4b2o!``````

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``````x = 33, y = 30, rule = B3/S23
31bo\$30bo\$30b3o21\$bo\$b2o25bo\$obo24b2o\$27bobo2\$11b2o\$11b2o!``````

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``````x = 32, y = 30, rule = B3/S23
30bo\$29bo\$29b3o20\$bo\$b2o23bo\$obo22b2o\$25bobo3\$10b2o\$10b2o!``````

Fourth edit: Actually, those last three methods don't let through a B glider unless a gap in glider stream A₀ occurs at the same time. This could theoretically be done with the right relative positioning of GPSEs A and C, but that might restrict using a C₁ glider to help create the target. The last method results in a kickback reaction that would let a couple of B′ gliders through, but that would be useless in this case.

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``````x = 161, y = 158, rule = B3/S23
159bo\$158bo\$158b3o62\$95bo\$94bo\$94b3o62\$31bo\$30bo\$30b3o20\$2bo\$2b2o23bo\$bobo22b2o\$o25bobo3\$11b2o\$11b2o!``````
Last edited by MathAndCode on October 22nd, 2020, 2:17 pm, edited 2 times in total.
I have reduced the cost of universal construction to seventeen gliders and probably to sixteen. All that remains is for the universal operations to be found.

wzkchem5
Posts: 127
Joined: April 17th, 2020, 2:19 am
Location: Valles Marineris, Mars

Re: Binary slow salvos

I wonder if either the 4-GPSE and the 3-GPSE universal constructors can be applied to the synthesis of patterns that expand sufficiently fast, for example spacefiller 1? The pattern may eventually collide with the cleanup Corderships before the Corderships die, and in this case one may no longer say that the constructor can construct anything that is glider-constructible, only that it can construct anything that is glider-constructible AND not expanding very fast in all directions. (Spacefiller 1 does not seem to be proved glider-constructible yet, but glider-constructible spacefillers whose expansion rate exceed the speed of Corderships are certainly a theoretical possibility.)
The Red Phoenix, The Yellow Phoenix, The Pink Phoenix And The Multicolored Phoenix