The long Tahua text probably is using 2 glyphs per day, and maybe in the short K text there are 2 days per glyph. Twins are lucky. There is a.m. and p.m., there is one calendar for the daylight and another (separate) for the night time. The night is divided by midnight into two halves, Matariki is is likewise a pair, i nika and i raro, etc.

The 'odd' text then becomes G, with 1 glyph per day. But in G it is quite evident that the text on side a is different from that on side b, and that way there are twins.

If K also has 1 glyph per day, it would result in a calendar which is covering only half the year. Or it should be read twice - not likely. Therefore, the main question to consider is whether it is a text with 2 days per glyph or not. If it has 2 days per glyph, then the end of the text can be seen to have a non-existent glyph where we would expect to find the last days of the year (365-366):

...
Kb5-10 Kb5-12 Kb5-13 Kb5-14 Kb5-15 Kb5-16 Kb5-17
364 368 370 372 374 376 378

8 'feathers' in haś (Kb5-14) presumably does not mean that we should count 372 / 8, because the result (46.5) is not a whole number. Instead, we could count 372 / 31 = 12 (i.e. Kb5-14 could show the end of a year with 12 months ą 31 days).

Kb5-12 could be the first glyph beyond the end of the solar year measured as 365.25 days. 5-12 is perhaps alluding to 5 * 12 = 60, and maybe 368 is to be read as 360 and 8. The double 'eyes' probably indicate a 'Janus situation'. The glyph resembles Ga1-26 in general outline:

Ga1-26 Kb5-12

In the complex Kb5-15 the structure in the center is similar to that in Eb6-1:

Eb5-35 Eb6-1 Eb6-2

Assembling the main maitaki glyphs in K it comes last:

50 90 20
Ka1-24 Ka4-8 (75) Kb4-13 (166) Kb5-15 (187)
52 = 4 * 13 22
164 = 8 * 8 + 100

The total number of glyphs in the text, 192, will then be equal to 164 + 28. However, the table illustrates but one of several possible alternatives.

It is quite possible, for instance, to maintain that the parallel with G makes it obvious that only half the year is covered by the text in K. The problem of balancing the text by incorporating twins can be solved as in G - by dividing the text into 'halves':

22 50
Ka1-1 Ka1-24 Ka4-8 (75)
23 52 = 4 * 13
90 20 4
Kb4-13 (166) Kb5-15 Kb5-20 (192)
91 = 7 * 13 26 = 2 * 13

13 * 13 = 169, a 'square' measuring out the important spring, and we can add this alternative to our table:

182 169 13 26
182 168 14 28
182 167 15 30
182 166 16 32