garbage day

[Albert Chan]

I made mistakes for a 'forever' formula, thus post removed, and start over.
I was playing with error = lo .. high, not realized Δy, Δm sign can be switched.
Maximum error is bigger, due to symmetry.

There are 97 leap years in 400 years, 97/400 = 0.2425
It so happen 365.2425 * 400 = 146097 = 7*20871 days, exactly.
If we can prove formula correct for 400 years, it is 'forever' correct.

Brute force check for days error, if key year end's in 00
Note: keydate day number doesn't matter, it can be anything!
Δd does not change. This is why I use d0=0 for brute force check.

Code:

function key_year_error(y0)
    local e1, e2, y, m, d, d1, d2 = 0, 0
    for m0 = 1,12 do
        local kd = day(y0,m0,0) -- key date
        for w=7,146097,7 do     -- weeks for 400 years
            y,m,d = date(kd+w)  -- exact date
            d1 = 365.2425*((y-y0) + (m-m0)/12) + d
            d2 = 365.2425*(y-y0) + 30.5*(m-m0) + d
            if abs(e1) < abs(w-d1) then e1=w-d1 end
            if abs(e2) < abs(w-d2) then e2=w-d2 end
        end
    end
    return e1, e2
end

lua> for y=0,300,100 do printf("%03d %8.4f %8.4f\n", y, key_year_error(y)) end
000  -2.9825   -2.9775
100  -3.6644   -3.7275
200  -3.4144   -3.4775
300   3.6644    3.7275

Both estimated weeks formulas work if key year is divisible by 200, with errors within 3.5 days.

---

It would be nice if we can just look up "good" key date from calendar.
Lets shift only months instead, and check if it would still work for 400 years.

Code:

function key_month_error(m0)
    local e1, e2, y, m, d, d1, d2 = 0, 0
    for y0 = 0,399 do
        local kd = day(y0,m0,0) -- key date
        for w=7,146097,7 do     -- weeks for 400 years
            y,m,d = date(kd+w)  -- exact date
            d1 = 365.2425*((y-y0) + (m-m0)/12) + d
            d2 = 365.2425*(y-y0) + 30.5*(m-m0) + d
            if abs(e1) < abs(w-d1) then e1=w-d1 end
            if abs(e2) < abs(w-d2) then e2=w-d2 end
        end
    end
    return e1, e2
end

lua> for m=1,12 do printf("%-3d %8.4f %8.4f\n", m, key_month_error(m)) end
1    -3.8288    -3.9550
2    -4.3919    -4.4550
3     4.3919     4.4550
4     3.8288     3.9550
5     4.2656     4.4550
6     3.7025     3.9550
7     4.1394     4.4550
8     3.5762     3.9550
9    -3.5762     3.4550
10    3.4500     3.9550
11   -3.7025     3.4550
12    3.3237     3.9550

We have a few options here, all brute force confirmed valid.
(month close to February error would be bad, as expected)

Δw = round((365.2425*(Δy + Δm/12) + Δd)/7), work if key month = 10, 12
Δw = round((365.2425*Δy + 30.5*Δm + Δd)/7), work if key month = 9, 11

It is arbitrary, I pick 'forever' formula with least days error.

Δw = round((365.2425*(Δy + Δm/12) + Δd)/7), key month = December

For previous keydate = 2024/11/5, move down calendar by 4 rows --> 2024/12/3 (no math)

Or, we keep the same keydate, but only "make" the month as December.

2024/11/5 ≡ 2024/12/(5-30) ≡ 2024/12/(-25)