User:Wyriel
Sailing thoughts.
Puzzle scoring is about averages. Not summation. Do a bunch of low value moves to clear up space -- tank your average.
That having been said, one should still find suitably fast ways to clear up enough room to get high scores -- just be aware that such moves should be considered 'painful'.
The puzzles are about thinking on ones feet. Doing something repeatable, mechanical, scientific, etc., goes less rewarded. Being creative and imaginative, adapting to the board at end, without having to sit for a minute thinking about it == mad bonus.
There are 1) raw points == combos 2) raw points averaged in some fashion == indicator 3) raw points averaged in some fashion and subsequently summed == DR 4) standing: some calculation on 3). Perhaps something as simple as continuing the summation for the entire puzzling session (but this allows one to sidestep the consistency thing (maybe)).
Theories about idle: 1) Spending longer than 10 seconds counts as making a move even though you didn't. [and then one has to play the move one is on] 2) Constant idle penalty, perhaps even as nasty as constant penalty straight to the efficiency, or even nastier and straight penalty to the performance level. Probably more like a -20 raw points thing.
Probably both 1) and 2), but pausing eliminates 2).
Theories about averaging: 1) Can't be over time -- playing faster would be a huge win, and it isn't. 2) Can't be over just moves -- Nemo: "All puzzles are played better faster" 3) Perhaps moves made plus a constant? So if you go faster, the constant has less effect?
Actually, it could be 2, but nominal analysis assumes perfect play. An imperfect player by a constant, not percentage, playing twice as fast, does better. That is, some player who makes a combo that doesn't use every ball given -- has 5-10 balls lying around afterwards. But suppose the next combo doesn't increase that, it just leaves it at 5-10 balls lying around (presumably different balls). One can think of it as the same 5-10 balls lying around as dead weight -- analogous to the idea of 3). Playing twice as fast (with all else remaining the same) halves the effect of those 5-10 balls.
Theories about factors: 1) Extra balls in a break (bilge style combos) -- I think these are beneficial, not just in a tactical sense, but in a raw scoring sense. Being able to get to 40s and 20s on viridian with my old style of sailing (only normal clears!) doesn't make any sense unless simultaneous clears are giving a boost. I think it is small and hard to notice without a multiplier like x5,6,7,8, being applied. Even then unclear ;). 2) Obstacle blocks. These are worth a bunch. Clearing these out with a bunch of lame doubles/triples (or even singles) does not tank ones score nearly as fast as normal clears do. Singly clearing them takes 2 moves at least (one of a double color), i.e., just as many as a normal clear does -- the extra 1 block is not helpful for making the break happen. Generally what happens is that one of the balls that was dropped just becomes a new 'free' block on the board, replacing the old one. It follows that if one is dividing by the same denominator but not getting the same effect, then the numerator is larger. 3) Booching boards. Maybe hurts score (above and beyond the moves which can no longer be cashed in)? I think its a constant score penalty so I don't notice it, as I play high score high moves, for which a constant has less effect. 4) Abandoning boards. Doesn't appear to hurt, at least as long as one is not suffering from first league woes. So if you sail for a long time, holding station, get grappled, and then after combat start with sparkly, its enough to build big combo and abandon -- that big combo is not helping the current contribution to standing, it is negating first league woes when you retake station. The combos before the grapple are what is contributing to the standing update at the time of the abandon. In fact, it appears to help one's standing (but no difference to ship performance) 5) No partial credit (like in carp).
Mathy thing. The maximum of x*y subject to the constraint x + y <= c is (c/2)^2 (setting x and y to be equal). In sailing one can only indirectly affect the products going into the average, and it is a series of products (each one with x_i increasing by one). y is the base value of the balls in that combo step, which is nominally 1 for a totally vanilla clear, and more for everything else. A V^2 on a normal ball is just 7*1. A bingo on a better ball (say a platform) is 4*k, where k is how much better a platform ball is than a normal ball. k >= 2 (most people say k=2), which gives the bingo at least 8 on that particular combo step. The shorter combo has greater contribution from the low terms, so the V^2 would be better than a vanilla bingo'd platform, but this sort of tradeoff is very important to keep in mind.
For example, P^4 would be 10*k/8, whereas N^7 would have value 28/14. Since k is at least 2, the P^4 is bigger. P^4,N is no better than P^4, unless k is exactly 2 and the resultant value/move is higher than one's normal value/move -- the last term is a measly 5*1 raw, so it would be (10k+5)/10, i.e., k+1/2, which for k=2, is the same as 5/4*k. Any greater k and 5/4*k will exceed k+1/2. Now, if k+1/2 is better than one's normal average, then this is a contribution of 10 moves above average instead of 8 moves at even more above average, so even if k > 2 the combo P^4,N might be good for you personally, but it depends upon where one's average is, and how much weight is given to one's past performance. The difference of contributing 10 moves worth of value above average would not, however, often beat contributing 8 moves of even higher value.
NP^4 is better than P^4 under any reasonable assumptions. This would be the series: 1*1,2*k,3*k,4*k,5*k=14k+1 and 10 moves, so, (14k+1)/10 = 7/5*k + 0.1 7/5 > 5/4, and the .1 only helps the already winning combo (but obviously not by much).
The value of that initial clear is to bump up the multipliers on the platforms. Any number of initial clears, for k >= 2, are superior to less initial clears. For N^mP^4, we have sum(1..m) + k*(m+1,m+2,m+3,m+4) = m*(m+1)/2 + (4m+10)k and 2m+8 moves, so, (m*(m+1)/2 + (4m+10)k) / (2m+8) = f which we want to compare to l=m-1: (l*(l+1)/2 + (4l+10)k) / (2l+8) = (m*(m-1)/2 + (4m+6)k) / (2m+6) = e and we want to show f > e so (m*(m+1)/2 + (4m+10)k) / (2m+8) > (m*(m-1)/2 + (4m+6)k) / (2m+6) <=> (2m+6)(m(m+1)/2 + (4m+10)k) > (2m+8)(m(m-1)/2 + (4m+6)k <=> (2m+6)(m^2/2 + m/2 + 4mk+10k) > (2m+8)(m^2/2 - m/2 + 4mk+6k <=> m^3 + m^2 + 8m^2k + 20mk + 3m^2 + 3m + 24mk + 60k > m^3 - m^2 + 8m^2k + 12mk + 4m^2 - 4m + 32mk + 48k <=> 0 + 2m^2 + 0 + 8mk - m^2 + 7m - 8mk + 12k > 0 <=> m^2 + 7m + 12k > 0 which is true since m,k > 0.
This is easy to see for large m, where the difference between 2m+8 and 2m+6 is neglible, but the difference between m(m+1)/2 and m(m-1)/2 as well as 10k vs 6k, is large.
Eventually P^4N^m is superior to P^4 (consider m=100). Breaking the two pieces separately would be almost as good as joining them; and doing P^4 and then N^mP^1 is probably better than P^4N^m, and almost certainly better than P^4N^m followed by P^1 (but who would do that?). If one's board has only 4 platforms, then adding clears to the end to get rid of any excess balls (as much as possible) is good, as those balls will be 0 otherwise -- any value is better than 0. If the board will have just one platform left, and the waste balls won't be clearable before that platform, then P^4N^m followed by N^lP^1 is better than launching the P^4 immediately and doing N^jP^1N^m, where j is presumably fairly small (because one has to fit the followup m clears for cleanup). If j is 4 or more, then the second way is better; for exampe, if there are very few waste balls, so that minimum m is something like 2, then P^4;N^4P^1N^2 is better than P^4N^2;N^4P^1, since the waste balls have greater value. If clearing up the waste allows one to pull off an N^5P^1, however, then P^4N^2;N^5P^1 is going to be better than P^4;N^4P^1N^2 -- increasing the value of the waste balls is not going to compensate for the decrease (by 1) of the platform in the second clear.
I'm reminded of pool -- really expert pool players are not just contemplating the geometry of the current shot, but how to setup the ball for the next shot, and the shot after that...
I.e., given a 6 platform board, one could try to shoot for some variation on P^6. If the pieces are not in one's favor, then you might have to break it early at something like NP^3NP^2, and be left with one platform. It is likely better to deliberately do NP^3 leaving one with 3 platforms and more space -- perhaps enough space to pull off N^3P^3 as the second clear. NP^3;N^3P^3 is likely better than NP^3NP^2;N^3P^1, but that second clear isn't really realistic since with just one platform it wouldn't be hard to pull off at least an N^4P^1 if not an N^5P^1. If one, from the beginning, plans on doing 3 and 3 on this board, then one might be able to do N^2P^3;N^3P^3, and this is a very good set of combos on a board. They're also both smaller (than the monster NP^3NP^2), which is advantageous in pillage situations. But I haven't run any numbers; and certainly NP^3NP^2 is a combo to be proud of. Just trying to make the point that it could be advantageous to break earlier and leave oneself better placed platforms for the second combo, rather than trying to the make the first as good as possible. If one really practiced this you might be able to reliably do N^2P^3;N^4P^3, or maybe even N^3P^2;N^3P^4 -- either of which is amazing, the second combo set is probably a little better.
N^3P^2 as first combo on a board is really tough though. One would almost certainly have to do down up -- a triple from the top clears a platform on the bottom which has a stack leading up to a top platform. That stack is mostly dead weight (one could do N^3P^2N^2 if there aren't too many obstacles in the down-up fashion), but, one could structure it so that it serves as the backbone of a followup N^3P^4. After the first 3 minutes this would be good.
Of course, if you pull of the NP^6, or the more believable NP^3NP^3 (or perhaps NP^3NPNPNP, which is easier to construct (sortof) because it gives you more time to get just the right pieces for the platforms), that is huge. So ideally one wants to play so as to leave oneself the chance to backout and do an NP^3, if the P^6 variation doesn't work out. Also, one can be ultimate by just doing NP^2 over and over again. One can probably be #1 by just doing N^2P^2 or NP^3 or P^3 over and over. The thing about the P^6 way is that it takes time; time during which scores could be falling out of one's queue. This doesn't matter too much to one's standing, but it does affect the ship. So in order to have the time one has to regularly do big infrequent combos (so that only 0s are falling out of the queue). Backing out and doing two P^3 variations increases the frequency of scoring, making it more difficult to go the P^6 way on the next board. Playing faster helps a little, in that it minimizes the absolute amount of time spent at less than sparkly (if one is okay with continuing to build combos while not sparkly). Ramping up to infrequent scores from frequent scores is tough, and I think one can show that no matter what one has to pay an equivalent penalty of just building the vegas as the very first combo (but doing it as many leagues of excellent before incred is better for the ship, perhaps, then one league of booched followed by increds). Standings-wise I think it doesn't matter at all, although if one gets engaged within the first league this could be fairly awful (since it will force an early score submission...).
The ideal sloop probably has one sailor going for the first league excellent and one sailor going for the second league incred. Two increds isn't much better than one incred, so this gives max acceleration from the starting line, followed by max acceleration when the excellent starts wearing thin (and if one gets engaged, then the vegas builder will hit the incred rate of token generation first, providing max token generation as soon as possible).
It is unclear, to me, if token generation is tied to performance level or raw performance. If performance level, then a NP^2 ult can be just as good for the ship as a vegas-ult. If raw performance, there is a maximum cap that a triple-player can't just get past (the value of P^3), so that the vegas-ult will surpass the triple ult not just on the DR, but also in the absolute amount of token generation. An interesting test would be if the raw numbers from sails are adjusted by bilge levels before being converted into token generation rates, so that at no bilge, it doesn't matter if one is doing low or high incred -- i.e., there is a cap on token generation that a low incred reaches. If when there is bilge the number is adjusted before being capped, then a high incred has value. To my knowledge this test has not been done.
Two ult bilgers can keep the water out at full damage. I wonder if two, or three, ult sailors (on a sloop) can keep up max token generation at full bilge? And whether or not it matters if they are doing high vs low increds? That would be a nice symmetry. Although I'd prefer it if it just wasn't possible to be that good -- infinite move sloops, no matter the damage, are quite irritating.
