Just FYI there is a small flaw, because you missed a half of all possible collisions by not taking into account, that if you do a monochromatic search, a honey farm and a honey farm transposed by 1 cell to any direction are different targets. Also you could probably do somewhat better search by limiting the upper limit of the final population and increasing the number of generations or steps in return.chris_c wrote:2. Use the following shell script to iteratively generate collisions between a honey farm and more and more gliders.
I get the patterns of opposite color by reflection. E.g when looking for reflectors SE->SW I also searched for reflectors NW->SW and reflected them. That changes the color of the honey farm relative to the gliders.codeholic wrote:...if you do a monochromatic search, a honey farm and a honey farm transposed by 1 cell to any direction are different targets.
Your search is definitely more thorough, it's written to find the best patterns, whereas mine only finds patterns that are in cheap in terms of resets by chance.
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x = 444, y = 264, rule = B3/S23 407b2o$407b2o20$178b2o$178b2o8$410b2o$410b2o$441b2o$441bobo$442b2o4$ 373b3o47b2o$375bo47b2o$374bo2$431b2o$176b3o252b2o9$361b3o$363bo$362bo 3$146b3o$148bo$147bo2$414b2o$413bo2bo$413bobo$411b2obo$410bo2bo$410bob o$411bo3$160b2o$134b3o22bo2bo$136bo22bobo$135bo21b2obo$156bo2bo$156bob o$157bo2$363b3o$365bo$364bo3$134b3o$136bo$135bo$113b3o$115bo204b3o$ 114bo207bo$321bo7$111b3o$113bo204b3o$112bo207bo$319bo5$361b3o$363bo$ 362bo13$100b3o$102bo204b3o37b3o$101bo207bo39bo$308bo39bo17$68bo$68b2o$ 67bobo2$272bo$272b2o$271bobo$309b3o$311bo$310bo8$55bo249b3o$55b2o250bo $54bobo249bo3$258bo$258b2o$257bobo$35bo$35b2o265b3o$34bobo267bo$303bo 10$24bo$24b2o$23bobo7$241bo$241b2o$240bobo2$15bo$15b2o$14bobo13$8bo$8b 2o$7bobo4$198bo$198b2o$197bobo8$3o$2bo$bo13$181bo$181b2o$180bobo23$ 168bo$168b2o$167bobo!
You're right. Good point.chris_c wrote:I get the patterns of opposite color by reflection. E.g when looking for reflectors SE->SW I also searched for reflectors NW->SW and reflected them. That changes the color of the honey farm relative to the gliders..
Unless it's got bugs. Hehechris_c wrote:Your search is definitely more thorough, it's written to find the best patterns
Wow, these look pretty good! Have you checked that they're really three resets, though? Most of the glider phases look like they will work, but I'm thinking that the final glider in the left-hand recipe (for example) will need a fourth reset.chris_c wrote:Two 180 degree edge shooting turners of opposite color with full clean up and reconstruction in three resets...
Adjacent gliders within a single scan will have different phases; those last two gliders are 4fd apart, so they would have to have the same phase if they're on the same scan. They both interact with parts of the same P2 pattern, so you can't change the phase, the way you can with your nice P1 slow (-3,-6) block-splitting recipe.
Oh, no! Looks like you are right. The back end of the patterns are copy/pasted from codeholic's posts and I misunderstood the number of resets involved there. I've made an effort to count the damage and it looks like the left side pattern costs 4 resets. The right hand side still costs 3 I believe. Hopefully someone can confirm.dvgrn wrote:Adjacent gliders within a single scan will have different phases; those last two gliders are 4fd apart, so they would have to have the same phase if they're on the same scan. They both interact with parts of the same P2 pattern, so you can't change the phase, the way you can with your nice P1 slow (-3,-6) block-splitting recipe.
Anyway, I am still pleasantly surprised by how well the (3,6) block move pattern and these turners fit together.
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x = 145, y = 152, rule = B3/S23 138bo$137bobo$137bobo$138bo2$133b2o7b2o$132bo2bo5bo2bo$133b2o7b2o2$ 138bo$137bobo$137bobo$138bo12$126bo$126b2o$125bobo24$94b3o$96bo$95bo 22$78bo$78b2o$77bobo23$41bo$41b2o$40bobo23$22bo$22b2o$21bobo23$bo$b2o$ obo!
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x = 295, y = 26, rule = B3/S23 184bo$o133bo49bo$o45bo87bo49bo$o45bo49bo37bo149bo$46bo49bo187bo$96bo 134bo52bo$190b2o39bo$6b2o132b2o48bobo38bo$6bobo43b2o86bobo47bo$6bo45bo bo47b2o36bo149b2o$52bo49bobo185bobo$102bo134b2o51bo$237bobo$237bo$5b2o 51b2o$4b2o51b2o40b2o$6bo52bo38b2o$100bo2$195b3o$144b2o49bo$143b2o51bo$ 145bo$242b3o48bo$242bo49b2o$243bo48bobo!
Nice one! But this one comes out to a delay of 5 relative to the reference 180-turn... just like two previous 180-turners. In fact, the seven-blinker 180-turner comes out to exactly the same spacing and horizontal offset as this new one:codeholic wrote:Another good even-lane one...
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#C reference pattern for 180-turner delays -- #C red gliders show reference position after 2400 ticks. #C The southeastward offset is arbitrarily chosen to be big enough #C that all known 180-turners can be tested against each other. x = 5569, y = 1531, rule = LifeHistory 1716.2A$1716.2A5$1719.2D$1719.2D13$1718.2A$1718.2A5$1721.2D$1721.2D4$ 1723.2A$1723.2A7$1725.A$1725.A$1725.A4$1699.2D$1698.D2.D$1699.2D$ 1730.2A$1730.2A$1705.D$1704.D.D$1704.D.D$1705.D$1733.2D$1733.2D4$ 1705.2D$1705.2D8$1732.2A$1732.2A$1757.2D$1757.D.D$1758.D2$1735.2D$ 1735.2D2$1749.2A$1748.A2.A$1748.A2.AD.2A$1749.2A.D.2A$1752.D4$1758.CD $1757.ADC$1757.A.A$1751.2A5.A$1750.A2.A$1751.A.A$1752.A2$1755.2D$ 1755.2D5$2029.2A$2029.2A11$2029.A$2028.A.A$2028.A2.A$2029.2A2$2027.A$ 2026.A.A$2027.A10$2037.A$2036.A.A$2036.A.A$2037.A2$2033.3A51$2384.2A$ 2383.A2.A$2384.2A$2375.2A$2375.A.A$721.A.A6.A.A4.A7.A2.A.A.A.A1621.2A 2.2A$719.A5.A1654.2A$729.A3.A3.A.A3.A.A2.A$719.A$728.A5.A2.A2.A.A2.A 2.A$720.A.A.A25.A.A$728.A.A.A.A2.A3.A3.A2.A$725.A$728.A5.A2.A7.A2.A$ 719.A5.A410.2A$721.A.A4.A5.A2.A7.A2.A.A.A.A381.2A3$1137.2A17.2A$1137. 2A17.2A364.A.A.A$1921.A5.A393.A5.A393.A5.A$1528.A$1157.2A762.A5.A393. A5.A393.A5.A8.2A$1157.2A369.A1206.A2.A$758.A.A1160.A5.A393.A5.A393.A 5.A8.2A$719.A.A.A5.A.A.A.A3.A.A.A.A10.A5.A760.A.A.A$725.A1195.A.A.A.A 393.A.A.A.A393.A.A.A.A$719.A9.A9.A16.A5.A765.A$725.A1201.A399.A399.A$ 719.A.A.A5.A9.A16.A5.A765.A$731.A.A7.A.A380.A.A395.A.A.A400.A399.A 399.A$719.A3.A5.A9.A16.A5.A359.A5.A2$719.A4.A4.A9.A16.A5.A359.A5.A$ 758.A.A$719.A5.A3.A.A.A.A3.A382.A5.A2$1122.A5.A2$1122.A5.A1605.2A$ 1124.A.A1606.A.A$2734.A2$2725.2A$2724.A2.A$2725.2A3$328.2A398.A.A$ 328.2A399.2A$729.A$324.3C397.3C397.3C397.3C397.3C397.3C397.3C$326.C 399.C399.C399.C399.C399.C399.C$325.C399.C399.C399.C399.C399.C399.C12$ 712.2A$711.A2.A$711.A.A$709.2A.A$708.A2.A$708.A.A$709.A5$2303.3A$702. 2A1601.A$701.A2.A1599.A$701.A.A$699.2A.A$698.A2.A$698.A.A$699.A6$692. 2A$691.A2.A$691.A.A876.2C$689.2A.A876.C2.C$688.A2.A877.C.C$688.A.A 876.2C.C$689.A876.C2.C$1566.C.C$1567.C3$2289.A$682.2A1605.2A$681.A2.A 1603.A.A$681.A.A$679.2A.A$678.A2.A$678.A.A$679.A6$672.2A1998.2A$671.A 2.A1996.A2.A$671.A.A1997.A.A$669.2A.A1996.2A.A$668.A2.A1996.A2.A$668. 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As of 6 July 2014, we have delay-0, delay-3, and multiple delay-4 same-color 180-turners. For color-changing turners we have every possible option except for delay-3. Good enough!
(I believe codeholic has been calling color-changing turners "even", so presumably same-color turners would be "odd".)
That's right. It's still easier to me to think in terms of half-diagonals rather than colors (though 90-degree turners would certainly raise a problem.)dvgrn wrote:(I believe codeholic has been calling color-changing turners "even", so presumably same-color turners would be "odd".)
Do we actually need all possible equivalence classes? For the "classic" setup with 1 NW and 2 SE rakes we'd need just one of each type, if I get it right. If we are to build 1 NW reflector + 3 SE rakes then we'll need at most 2 same-color turners and 1 anti-color turner.
I'm judging just by distances between glider lanes in the synchronized salvo, so I may be easily wrong if I'm missing something.
That seems mostly right. If I have all the math figured out right (and please don't assume that!) then we can perfectly well get away with just one same-color and one anti-color 180-turner no matter what the rake geometry is.codeholic wrote:Do we actually need all possible equivalence classes? For the "classic" setup with 1 NW and 2 SE rakes we'd need just one of each type, if I get it right. If we are to build 1 NW reflector + 3 SE rakes then we'll need at most 2 same-color turners and 1 anti-color turner.
We can freely substitute any same-color rake for any other same-color rake, and the same with the anti-color ones. The only caveat is that we have to adjust the number of resets, to produce a glider with the right timing mod 8 so that the 180-turned glider goes to the correct spacetime location.
Our anti-color rake collection includes a 0-delay and a 4-delay rake. I believe this means that we can get away with no more than three resets before one of those two rakes will produce its output in the correct phase mod 8. If we had 2-delay and 6-delay anti-color rakes as well, then we'd need just one reset at most.
Depending on the timing, we might not need any resets at all -- I haven't tried building any of the actual HB-trail geometries yet to see how the current collection works out. A knightship might work out perfectly for some carefully chosen period P+8N, but you'd need seven resets if you wanted to build period P+8N+1 (or whatever).
So it seems to make sense to keep collecting and classifying 180-turners just until it gets too annoying to do any more -- and then we'll just pick a knightship period that works best for the equivalence classes we have available, to find a front end that needs only a reasonable number of resets.
I'm finding the new A1:BCDA2 design to be a bit of a headache, in terms of figuring out how far away to put A2 at the northwest end so that it will end up at exactly the right place in the southeast when it's time for the suppression reactions to happen.
It looks to me as if we'll have to send the C trigger gliders, and maybe the A2 ones also, to their 180-turners right away, and then wait for quite a few resets before we can send the B trigger gliders. The C gliders will get slowed down a lot by the very long string of half-bakeries they'll need to reach their initial diagonal -- but we need them to arrive a good while before the B gliders, so we'll have to wait around before triggering the Bs.
(We can't just move the B 180-turners farther northeast in this case, because the trigger and return glider paths would cross the C gliders' half-bakery chains, and of course those aren't permeable.)
We can be building a superslow elbow on the A1 track in the NW while all these resets are going on, so we're not exactly wasting HBK length. But if lots of resets have to happen anyway, then we definitely don't need representatives of all 16 180-turner equivalence classes. The two that chris_c has already made into (-3,-6)-compatible recipes should be all we need...!
See any holes in my wild theories so far?
Well I can't conceive of how the the clean up pattern for the A1:BA2C version was engineered so I can't comment on that part, but the rest seems pretty reasonable.dvgrn wrote: See any holes in my wild theories so far?
I've been working on the turner patterns and I reckon I have made a full toolkit where all patterns take at most three resets.
First of all the 180 degrees turners (I managed to improve the first one down to three resets with a bit of re-jigging).
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x = 549, y = 273, rule = B3/S23 512b2o$512b2o6$221b2o$221b2o17$207b3o$209bo$208bo14b2o$223b2o2$515b2o$ 515b2o$546b2o$546bobo$547b2o4$478b3o47b2o$195b3o282bo47b2o$197bo281bo$ 196bo$536b2o$536b2o$224b3o8$466b3o$468bo$467bo7$519b2o$518bo2bo$191b3o 324bobo$193bo322b2obo$192bo322bo2bo$515bobo$516bo2$159b3o48b2o$161bo 47bo2bo$160bo48bobo$207b2obo$206bo2bo$206bobo$207bo$190b3o$192bo$191bo 276b3o$470bo$469bo$155b3o$157bo$156bo4$425b3o$427bo$426bo8$423b3o$425b o$424bo2$143bo$143b2o$142bobo$466b3o$468bo$467bo14$412b3o37b3o$414bo 39bo$126b3o284bo39bo$128bo$127bo17$117bo$117b2o$116bobo258bo$377b2o$ 376bobo$414b3o$416bo$415bo8$410b3o$412bo$411bo3$99b3o261bo$101bo261b2o $100bo261bobo2$407b3o$409bo$408bo17$64b3o$66bo$65bo280bo$346b2o$345bob o9$53b3o$55bo$54bo11$44b3o$46bo256bo$45bo257b2o$302bobo12$37b3o$39bo$ 38bo9$286bo$286b2o$285bobo9$b3o$3bo$2bo12$273bo$273b2o$272bobo7$2o$b2o $o!
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x = 807, y = 730, rule = B3/S23 obo$b2o$bo270$653bo$651bobo$652b2o70$348b2o$348b2o15$372b2o$372bobo$ 373bo2$350b2o$350b2o4$367bo$367bo$367bo4$373b2o$373b2o7$370b2o$370b2o 9$803b2o$803b2o10$346b2o$291b3o51bo2bo$293bo51bobo$292bo50b2obo$342bo 2bo$342bobo$343bo3$805b2o$805b2o5$279b3o$281bo$280bo19$278b3o$280bo$ 279bo8$789b2o$788bo2bo$788bobo$786b2obo$746b3o36bo2bo$748bo36bobo$747b o38bo2$278b3o$280bo$279bo5$277b3o$279bo$278bo$734b3o$736bo$735bo14$ 226b3o$228bo$227bo2$708b3o23b2o$710bo22bobo$709bo25bo8$706b3o$708bo$ 707bo3$215bo$215b2o$214bobo6$703b2o$702bobo$704bo10$707b2o$706bobo$ 708bo25$674b2o$673bobo$675bo6$172bo$172b2o$171bobo7$665b2o$664bobo$ 666bo10$157b3o$159bo$158bo17$140bo$140b2o$139bobo13$129bo$129b2o$128bo bo8$69b3o$71bo$70bo8$67b3o$69bo$68bo21$56b3o$58bo$57bo18$540b3o$542bo$ 541bo2$50b2o$49bobo$51bo6$530b3o$532bo$44b2o485bo$43bobo$45bo8$520b3o$ 522bo$521bo14$16b2o$15bobo$17bo9$13b2o$12bobo$14bo!
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x = 559, y = 338, rule = B3/S23 555b2o$555b2o19$557b2o$557b2o4$206b2o$206b2o13$196b3o336b2o$198bo335bo 2bo$197bo337b2o2$516b2o$515bobo23bo$208b2o307bo22bobo$208b2o330bobo$ 541bo6$184b3o354b2o$186bo354b2o$185bo2$504b2o$503bobo$505bo17$501b2o$ 180b3o317bobo$182bo319bo$181bo17$163bo$163b2o$162bobo15$118b3o$120bo 344b2o$119bo344bobo$466bo13$452b2o$451bobo$453bo$115bo$115b2o$114bobo 15$440b2o$439bobo$441bo2$102b3o$104bo$103bo14$422b2o$421bobo$423bo$85b o$85b2o$84bobo11$370b2o$369bobo$371bo2$410b2o$409bobo$411bo2$70b3o$72b o$71bo296b2o$367bobo$369bo13$27b3o$29bo$28bo8$361b2o$360bobo$17b3o342b o$19bo$18bo8$351b2o$350bobo$352bo17$3o$2bo$bo18$300b2o$299bobo$301bo 14$286b2o$285bobo$287bo14$272b2o$271bobo$273bo30$254b2o$253bobo$255bo!
1. Realising that starting from the final pattern of the (3,6) block push usually allows extra parallelism because both pieces can be worked on independently.
2. Used codeholic's search program to find cheap turners that come directly from a block.
3. Used codeholic's search program to find some cheap ways of converting a block into a HF with large displacement. This helps to avoid any dependency between the two parts of the pattern.
Crossing my fingers that there are no huge mistakes.
Impressive! I'll double-check the glider parities sometime tomorrow, and see if I can combine the 90-degree turners into four 180-degree turners to add to the reference pattern I posted above.chris_c wrote:I've been working on the turner patterns and I reckon I have made a full toolkit where all patterns take at most three resets...
Crossing my fingers that there are no huge mistakes.
Then it's probably time to start picking out 180-turners that will work well together in the front end of a 7-synchronized-glider HBK. Has anyone done any work on a recipe for the superslow elbow on the NW side, or shall I try stringing something together for that?
So I did. Unfortunately, wouldn't you know it, that one is also a delay-4, so it doesn't improve the toolkit any. I've updated the reference pattern for completeness' sake.codeholic wrote:It seems you forgot this one.dvgrn wrote:As of 4 July 2014, we have just one same-color 180-turner, which is a delay-4.
With any luck at all, stringing together 90-turners should give us a few new same-color options. Further bulletins as events warrant.
I've checked the period-2 phases of the critical (those that either collide with or produce period-2 oscillators) gliders in your 180° turners, and they're all consistent. Specifically, in pass n, the gliders on paths 2n (mod 4) must be in one phase, and the gliders on paths 2n + 2 (mod 4) must be in the other phase. Has this already been taken into consideration, or is it purely serendipitous that your 180° turners actually work without introducing extra 'NOP' resets?chris_c wrote:First of all the 180 degrees turners (I managed to improve the first one down to three resets with a bit of re-jigging).
(I'm guessing that it probably is just serendipity, since some of the non-critical gliders are in the wrong phases -- but of course this doesn't matter.)
I particularly enjoy how the first of your turners forces the knightship to have an odd period. Very nice. Indeed, that means that we can *easily* force the period of the knightship to be prime, by Dirichlet's theorem (mod 8 ).
Personally, I favour the A1:BCA2 design, since I don't think that it's worth considering the excess gliders defecated by the spaceship until after it has been built, at which point we can just bolt on a bespoke clean-up mechanism.
So, we're creating the 6 right SW-directed gliders with combinations of 180° turners and bunches of half-bakeries, and the 1 left SW-directed glider with an (expensive!) pair of 90° turners. Indeed, I have a feeling that the left glider may cost more than the 6 right gliders combined. Unless, of course, we can think of another mechanism to generate the left glider, possibly using the right gliders.
When we come to generate the 6 right SW-directed gliders, they must be done in the order (outermost, ..., innermost). This not only reduces the number of resets, but also means that we can freely and painlessly adjust the macroscopic timings of the gliders arbitrarily in either direction.
I think that even your estimate of 10^6 cells per side is a little pessimistic. I suspect we can make this thing shorter than the Caterpillar.
Well, "forces" may be a little too strong -- you could throw in an extra reset or two and end up with an even period after all, couldn't you? But it would be nice to aim for a prime period; we'll see what significant primes are in the neighborhood when we get to that stage.calcyman wrote:I particularly enjoy how the first of your turners forces the knightship to have an odd period. Very nice. Indeed, that means that we can *easily* force the period of the knightship to be prime, by Dirichlet's theorem (mod 8 ).
...Wait, is there an "A1:BCA2 design"? I didn't see how to move A2 southeast of C. Maybe you mean an A:BCDG design, with an extra guard glider or two? The plain A:BC needs three synchronized gliders on the NW side, so it doesn't seem likely that you're voting for that... though come to think of it, three superslow elbows can perfectly well be built in parallel, so they'll end up costing very little more than the single superslow elbow we have to build anyway. (Right?)calcyman wrote:Personally, I favour the A1:BCA2 design, since I don't think that it's worth considering the excess gliders defecated by the spaceship until after it has been built, at which point we can just bolt on a bespoke clean-up mechanism.
Hmm, it probably is worth thinking hard about alternate mechanisms -- things like *WSS seeds in the shooting range making *WSS+glider collisions at A, creating SW-traveling gliders and no debris that has to be moved.calcyman wrote:So, we're creating the 6 right SW-directed gliders with combinations of 180° turners and bunches of half-bakeries, and the 1 left SW-directed glider with an (expensive!) pair of 90° turners. Indeed, I have a feeling that the left glider may cost more than the 6 right gliders combined. Unless, of course, we can think of another mechanism to generate the left glider, possibly using the right gliders.
Agreed. I haven't re-done that estimate recently, but with all the new ideas flying around lately, my mental picture of it has been getting shorter and shorter...!calcyman wrote:I think that even your estimate of 10^6 cells per side is a little pessimistic. I suspect we can make this thing shorter than the Caterpillar.
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x = 165, y = 178, rule = B3/S23 158bo$157bobo$157bobo$158bo2$153b2o7b2o$152bo2bo5bo2bo$153b2o7b2o2$ 158bo$157bobo$157bobo$158bo13$130b3o$132bo$131bo22$112bo$112b2o$111bob o23$79bo$79b2o$78bobo24$57b3o$59bo$58bo48$31b3o$33bo$32bo23$3o$2bo$bo!
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x = 206, y = 228, rule = B3/S23 201bo$201bo$201bo2$197b3o3b3o2$201bo$201bo$201bo16$173bo$173b2o$172bob o23$156bo$156b2o$155bobo24$134b3o$136bo$135bo48$86b3o$88bo$87bo23$69b 3o$71bo$70bo23$30b3o$32bo$31bo48$3o$2bo$bo!
The script tries to find recipes that need as few resets as possible. Glider phase parity has been taken into consideration. You can see it here: https://github.com/codeholic/hbk/blob/1 ... #L229-L239calcyman wrote:Has this already been taken into consideration, or is it purely serendipitous that your 180° turners actually work without introducing extra 'NOP' resets?
The obvious reaction to use would be MWSS+G, which leaves just one out-of-the-way block that can be cleaned up later -- even a cycle and a half later if necessary:dvgrn wrote:Hmm, it probably is worth thinking hard about alternate mechanisms -- things like *WSS seeds in the shooting range making *WSS+glider collisions at A, creating SW-traveling gliders and no debris that has to be moved.
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x = 74, y = 52, rule = LifeHistory 15.2B$14.4B$13.5B.B10.2D$13.9B8.2D$13.10B$12.11B$11.13B2.5B$10.24B$9. 24B2C$8.14B3A8B2C$7.15BA11B$6.4B2.11BA7B$5.4B2.16B2.2B$4.4B4.15B$3.4B 3.3A4B.B2.6B9.2C$2.4B4.A.BA4B4.6B8.2C$.D3B5.A6B8.3B$D3B6.A.2BA3B8.B$ 3D7.A6B$11.ABA4B$11.6B22.2D$12.6B21.2D$11.6B$12.5B$12.3B$14.B.B24$71. 2A$71.A.A$71.A!
If it works, this could make the Half-Baked Knightship really short (relatively speaking) -- and I don't see why it wouldn't work perfectly well. So... what's our cheapest known monochromatic-recipe forward MWSS seed?
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x = 344, y = 352, rule = B3/S23 337bo$336bobo$336bobo$337bo2$332b2o7b2o$331bo2bo5bo2bo$332b2o7b2o2$ 337bo$336bobo$336bobo$337bo13$309b3o$311bo$310bo22$291bo$291b2o$290bob o24$271b3o$273bo$272bo22$233bo$233b2o$232bobo23$214bo$214b2o$213bobo 24$198b3o$200bo$199bo22$176bo$176b2o$175bobo49$103b3o$105bo$104bo23$ 82b3o$84bo$83bo23$65b3o$67bo$66bo72$bo$b2o$obo!
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x = 282, y = 303, rule = B3/S23 275bo$274bobo$274bobo$275bo2$270b2o7b2o$269bo2bo5bo2bo$270b2o7b2o2$ 275bo$274bobo$274bobo$275bo13$251b3o$253bo$252bo23$230b3o$232bo$231bo 23$209b3o$211bo$210bo23$170b3o$172bo$171bo23$157b3o$159bo$158bo48$111b 3o$113bo$112bo48$69b3o$71bo$70bo23$34b3o$36bo$35bo23$17b3o$19bo$18bo 23$3o$2bo$bo!
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x = 200, y = 228, rule = B3/S23 198bo$197bobo$197bobo$198bo22$168b3o$170bo$169bo22$154bo$154b2o$153bob o48$112bo$112b2o$111bobo24$90b3o$92bo$91bo23$43b3o$45bo$44bo23$30b3o$ 32bo$31bo23$21b3o$23bo$22bo23$3o$2bo$bo!
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x = 186, y = 203, rule = B3/S23 181bo$181bo$181bo2$177b3o3b3o2$181bo$181bo$181bo17$151b3o$153bo$152bo 23$136b3o$138bo$137bo48$98b3o$100bo$99bo23$51b3o$53bo$52bo23$30b3o$32b o$31bo23$9b3o$11bo$10bo23$3o$2bo$bo!
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x = 241, y = 252, rule = B3/S23 234bo$233bobo$233bobo$234bo2$229b2o7b2o$228bo2bo5bo2bo$229b2o7b2o2$ 234bo$233bobo$233bobo$234bo13$219b3o$221bo$220bo23$198b3o$200bo$199bo 23$153b3o$155bo$154bo23$134b3o$136bo$135bo22$124bo$124b2o$123bobo23$ 71bo$71b2o$70bobo23$56bo$56b2o$55bobo24$44b3o$46bo$45bo47$bo$b2o$obo!
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x = 240, y = 253, rule = B3/S23 233bo$232bobo$232bobo$233bo2$228b2o7b2o$227bo2bo5bo2bo$228b2o7b2o2$ 233bo$232bobo$232bobo$233bo13$209b3o$211bo$210bo23$178b3o$180bo$179bo 23$157b3o$159bo$158bo23$136b3o$138bo$137bo23$115b3o$117bo$116bo23$94b 3o$96bo$95bo23$51b3o$53bo$52bo23$30b3o$32bo$31bo23$17b3o$19bo$18bo23$ 3o$2bo$bo!
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x = 286, y = 302, rule = B3/S23 284b2o$283bobo$284bo23$257b3o$259bo$258bo73$190b3o$192bo$191bo22$168bo $168b2o$167bobo23$129bo$129b2o$128bobo24$107b3o$109bo$108bo22$85bo$85b 2o$84bobo23$64bo$64b2o$63bobo23$43bo$43b2o$42bobo48$bo$b2o$obo!
These match up with each other nicely to make just two new 180-turners. You can't pair them the other way because then polychromatic salvos would be needed to build them. They're in the reference pattern now -- a "same-color 3-delay" and an "anti-color 1-delay".chris_c wrote:I've been working on the turner patterns and I reckon I have made a full toolkit where all patterns take at most three resets...
Next two 90 degree edge shooters that take three resets each...
Last of all, two 90 degree turners that take just two resets each:
-- I hope I'm doing these delay calculations right, but so far no inconsistencies have appeared. The new oblique line of red gliders shows all the different equivalent places that a 180-turner could put a glider in 2400+8N ticks, to get a 0-delay rating. Let me know if anything looks wrong.
That should be plenty -- thanks! I've updated the reference pattern now. There's a delay-0 and delay-4 for same-color 180-turners, and every possible delay except 3 for the anti-color turners. Should be fairly easy to pick out a set of compatible turners that allow an odd-period knightship. It doesn't look like we'll really need to use the 90-turners; they work fine, but they definitely take up more of the shooting range than most of the other choices.codeholic wrote:That's more or less all essentially different 180-degree turners up to 10 gliders and 3 scans.
(Go right ahead and build a good HBK front end, somebody -- I'm doing a lot of farm work this week, which is not very compatible with sitting still and sorting out 180-turner math...!)
Yes, exactly. The clever business in the patterns is lifted directly from codeholic's search output. All of the moving and junk destruction was added by me using copy/paste in golly and I was too lazy to make all of the phases correct in the cases where the phase is irrelevant.codeholic wrote:The script tries to find recipes that need as few resets as possible. Glider phase parity has been taken into consideration. You can see it here: https://github.com/codeholic/hbk/blob/1 ... #L229-L239calcyman wrote:Has this already been taken into consideration, or is it purely serendipitous that your 180° turners actually work without introducing extra 'NOP' resets?
I think any two 90-turners should be able to make any desired 180-turner (of one particular color) . Increasing the gap between the 90-turners requires an extra HB for one of the outbound gliders, which gives an odd delay. In this example input gliders of delay 100 on the same lane are converted into gliders of delay 101 on the same lane. The trick can be repeated to get any delay mod 8.dvgrn wrote: These match up with each other nicely to make just two new 180-turners.
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x = 53, y = 45, rule = B3/S23 41bo$40bobo$41b2o3$36bo$35bobo$36b2o10$25b3o23bo$27bo22bobo$26bo23b2o 2$45bo$44bobo$44b2o3$32b2o$31bo2bo$31bobo$29b2obo$28bo2bo$28bobo$29bo 10$3o$2bo$bo!