-- Сигнал: день > 3 средних за предыдущие 7 календарных дней.
-- Требуется 7 дней истории и ненулевая базовая сумма; пропущенные дни = 0.
-- Целочисленное сравнение через quotient/remainder исключает overflow
-- при умножении большой SQLite INTEGER суммы на 3 или 7.
WITH RECURSIVE calendar(day) AS (
    SELECT MIN(order_date) FROM facts HAVING COUNT(*) > 0
    UNION ALL
    SELECT date(day, '+1 day') FROM calendar
    WHERE day < (SELECT MAX(order_date) FROM facts)
), totals AS (
    SELECT order_date, SUM(revenue_kopecks) AS revenue_kopecks
    FROM daily_category GROUP BY order_date
), filled AS (
    SELECT day AS order_date, COALESCE(revenue_kopecks, 0) AS revenue_kopecks
    FROM calendar LEFT JOIN totals ON totals.order_date = calendar.day
), baseline AS (
    SELECT *, SUM(revenue_kopecks) OVER w AS previous_7_sum_kopecks,
              COUNT(*) OVER w AS history_days
    FROM filled WINDOW w AS (ORDER BY order_date ROWS BETWEEN 7 PRECEDING AND 1 PRECEDING)
), thresholds AS (
    SELECT *, (previous_7_sum_kopecks / 7) * 3
               + ((previous_7_sum_kopecks % 7) * 3) / 7 AS threshold_floor_kopecks
    FROM baseline
)
SELECT order_date, revenue_kopecks, previous_7_sum_kopecks,
       threshold_floor_kopecks
FROM thresholds
WHERE history_days = 7 AND previous_7_sum_kopecks > 0
  AND revenue_kopecks > threshold_floor_kopecks
ORDER BY order_date;