Integer approximation of a segment

Let 𝑂𝑋𝑌 be a Cartesian coordinate system with an integer lattice whose unit squares are staggered. The integer approximation of the segment 𝐴𝐵 is given using the cellular domain S𝐴𝐵 of (colored) cells, the interior of each of which.has a non-empty intersection with 𝐴𝐵.
If 𝑃± 𝐴𝐵 — right and left closed half-planes defined by the line 𝑙𝐴𝐵 by the point 𝐴 and 𝐵, then S± 𝐴𝐵 = S𝐴𝐵 ∩ 𝑃± 𝐴𝐵 — its right and left areas. (There are no integer points inside S𝐴𝐵.)
Polyline L±(𝐴±,𝐵±) from S± 𝐴𝐵 with ends 𝐴± and 𝐵± and whole vertices — right and left by (integer) approximations of the segment 𝐴𝐵 — the ends are selected from the vertices of the extreme cells. If 𝑙𝐴𝐵 is parallel to one of the coordinate axes, then we assume S𝐴𝐵 = ? and then approximation of the segment 𝐴𝐵 is minimum segment with integer ends containing 𝐴𝐵.
Such approximations are constructed using the algorithm ‘pulling noses", which is a geometric interpretation of the chain fraction of the angular coefficient of the straight line 𝑙𝐴𝐵. Based on this construction method, an exact formula for calculating the number of integer points inside an arbitrary triangle is obtained, and the problem of S.V. Konyagin is partially solved about chess coloring: If U(𝑡) is the set of all colored cells from a triangle cut off by a straight line 𝑓𝑡 : 𝑦 = −𝛼𝑥 + 𝑡, 𝛼, 𝑡 > 0, then the difference 𝑢(𝑡) between white and black cells from U(𝑡)for every positive irrational 𝛼 is bounded neither from below nor from above when 𝑡 → ∞. The solution is obtained for numbers of the form: 𝑒±1, tg±1, [𝑎−0 ; 𝑎−1 , 𝑎−2 , . . .]±1, [𝑎+0 ; 𝑎+1 , 𝑎+2 , . . .]±1,
[𝑎+0 ; 𝑎−1 , 𝑎+2 , 𝑙𝑑𝑜𝑡𝑠]±1, where the superscript plus (minus) indicates on the parity (odd) of the element of the continued fraction defined by 𝛼.
The method of constructing an approximation of the segment was used to solve the problem of chess coloring for the numbers
(√5+1)/2 , [𝑎+0 ; 𝑎+1 , 𝑎+2 , . . .], 𝑎−2𝑛+1 and [𝑎−0 ; 𝑎−1 , 𝑎−
2 , 𝑙𝑑𝑜𝑡𝑠], if limited 2𝑘−1𝑏3𝑏9 · · · 𝑏6(𝑘−1)+3 + · · · + 22Σ︀𝑘
𝑖1>𝑖2>𝑖3=1 𝑏6(𝑘−𝑖1)+3𝑏6(𝑘−𝑖2)+3 𝑏6(𝑘−𝑖3)+3 + 2
Σ︀𝑘 𝑖1>𝑖2=1 𝑏6(𝑘−𝑖1)+3𝑏6(𝑘−𝑖2)+3 + Σ︀𝑘 𝑖=1 𝑏6(𝑘−𝑖1)+3 + 1,
for some 𝑏𝑛 = ⌊︁𝑎−(𝑛−1)/2⌋︁ representing the whole part of 𝑎−
(𝑛−1)/2 . So for 𝑏𝑛 = 0 the chain fraction is [𝑎− 0 ; 𝑎−1 , 𝑎−2 , . . .] =
(√5+1)/2 .

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