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Improvement for Yau Half Centers for 4x4

Lykos

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Feb 27, 2016
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I recently found an improvement over the way I do Yau half centers for 4x4. I am not sure whether this is a novelty or whether everyone already does it this way, but I haven't seen it in any tutorials yet.

The first 3 half centers are unchanged. You just do 3 half centers and put them next to your cross edges. (Of course they don't necessarily have to match the cross edges, they just should have the correct order according to the color scheme) As an example, let's assume we have the green, red and blue half centers.

The second part is where my improvement comes in. Jsust to put it into perspective, I'll describe how I understood how half centers are done and how I used to do it:
Since I already have a green, red and blue half center, I complete the other half of the green, red and blue half centers in any order. After I finished one half center, I put into it's place. E.g. I create a blue half center, then I put it next to the first blue half center to finish the blue center. At the end, the orange center will be solved automatically.

My idea is the following. You waste moves by putting the half inside. It's like a beginner who always inserts cross pieces in the right place. Intermediate solvers don't do that any more. You can insert the cross piece at any point as long as the order of the pieces is correct. In the end, the cross is adjusted.

I now do the same for the second half of the half centers. I solve them in the correct order, but I don't put them into their place. I just adjust them at the end if all of them are finished. Also, I don't follow the restriction that I should do the green, red and blue half centers and not the orange one. Instead, I do any 3 half centers. I could do orange, blue red for example. Typically, I just reuse whatever bars already exist. If the orange bar exists, I use that one. In the end, I adjust and thereby finish the centers. This is about 3 moves shorter for me on average.

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scottishcuber

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I see your point. But it seems difficult to implement as smoothly as the 'easy' way. Also, you would probably have to solve 4 sets of centers, whereas if you solve centers the normal way then you only have to solve 3 sets of centers and by default the last one will be finished.

(I feel I explained that poorly, let me know)
 

Lykos

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Feb 27, 2016
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For the first half centers, I solve 3 as well, nothing changes about that. For the second half, if you don't do the one that you left out in the first half (i.e. the orange one in my example), then you also solve 3. So it's the same in that case, the only difference is that you adjust the centers at the end. I am pretty convinced at this point that this optimization is an improvement.

I am less sure about my other proposal, i.e. you don't necessarily solve the same 3 half centers, but just 3 arbitrary ones. There is a chance that the last one is not solved but a checker pattern and then you occasionally have to solve 4. So I am not sure whether this optimization is worth it. However, since there are 4 orange centers around and only 10 places where they could be, it is very likely that you already have a bar somewhere in the beginning. I wouldn't start building orange if there is no bar yet. But if there is already a bar and I don't have to invest any moves or maybe just one move to use it, I would use it. You solve 2 more centers and then you have a 50% chance that the last center is skipped. If you get a checker pattern, well, then you have to solve that one instead. So you build one more center in 3 moves. But since you didn't invest moves for the orange bar (or maybe just one) I think it should still be an advantage on average.
 

TDM

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This is what I've always done... I never knew I was doing half centres differently up until now.

E: I always solve the same colours for my second three pairs e.g. if I solve blue-red-green for my first three, I'll solve blue-red-green for my second three. Or sometimes I'll solve orange-blue-red-green for my second three since there's usually an orange (or whatever the fourth colour is) pair solved and it makes lookahead easier, at the cost of maybe 1-2 moves.
 
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