unbalancing_factor calculations #79
|
I'm spinning this from your reply (thanks for pointing me to the calculations!) In the let unbalancing_factor = 1.0
+ (self.unbalancing_factor
* ((k1.key.unbalancing.0 - k2.key.unbalancing.0).abs()
+ (k1.key.unbalancing.1 - k2.key.unbalancing.1).abs()));If I understand correctly, the unbalancing_positions:
- [[0, 0.5], [0, 0], [0, 0], [0, 0.5], [1, 1], [-1, 1], [0, 0.5], [0, 0], [0, 0], [0, 0.5]]
- [[0, 0], [0, 0], [0, 0], [0, 0], [1, 0], [-1, 0], [0, 0], [0, 0], [0, 0], [0, 0]]
- [[0, 0], [0, 0], [0, 0], [0, 0], [1, -1], [-1, -1], [0, 0], [0, 0], [0, 0], [0, 0]]
- [ [0, 0], [0, 0], [0, 0], [0, 0], [0, 0], [0, 0], ]I understand, that if you have: k1: [0, 0.5] and k2: [1, 1], the cost would be: and the right? Then a related question: Why the same finger bigram does not have cost? Is is somehow handled differently? I mean, there is if f1 == Finger::Thumb || f2 == Finger::Thumb || h1 != h2 || f1 == f2 {
return Some(0.0);
}but how about the left index finger bigrams here: If cost is 0.0 when In addition, how important do you consider the unbalancing_factor as there's also key_costs in the keyboard configuration? |
Replies: 1 comment 1 reply
Yes, exactly. This part is supposed to evaluate how much hand movement is involved in hitting the bigram keys. In this exampe, the hand needs to move one unit to the right and half a unit up, totaling to 1.5.
The
The To which extent you consider this effect important can be encoded in the |

Yes, exactly. This part is supposed to evaluate how much hand movement is involved in hitting the bigram keys. In this exampe, the hand needs to move one unit to the right and half a unit up, totaling to 1.5.
The
movement_patternmetric is only supposed to evaluate non-SFB bigrams. All SFB-bigrams are e…