Lesson 20. Stop on conflicts and expired records

Why this matters

If a notice board has two different notices for one activity, choosing the first is unsafe. Once a notice has expired, its old schedule cannot be presented as current. Combine question parsing, lookup, and these two checks.

Before running

valid_until is the last calendar day on which the card is valid, inclusive. It uses the same year-month-day form. from datetime import date imports Python’s calendar date type. date.today() reads the computer’s date; date.fromisoformat(...) converts text into a date we can compare. The computer clock must be correct. != means “not equal”; or means either condition is enough.

[
  {"club": "шахмат", "kind": "уақыт", "value": "бейсенбі, 15:00", "source": "club-sheet-01", "checked_on": "2026-09-01", "valid_until": "2026-12-31"},
  {"club": "шахмат", "kind": "орын", "value": "203-бөлме", "source": "club-sheet-01", "checked_on": "2026-09-01", "valid_until": "2026-12-31"}
]

From the project root, enter the step folder and run the program:

cd course/kazakh-ai/step-20
python3 model.py

On Windows, use py model.py instead of the last command.

import json
from datetime import date

with open("facts.json", encoding="utf-8") as file:
    facts = json.load(file)
question = input("Сұрақ: ").lower().replace("?", "").replace(",", "")
words = question.split()
entities = []
intents = []
for word in words:
    if word in ["шахмат", "сурет"]:
        entities.append(word)
    if word == "қашан":
        intents.append("уақыт")
    if word == "қайда":
        intents.append("орын")
if len(entities) != 1 or len(intents) != 1:
    print("нақтылаңыз: бір үйірме және бір сұрақ түрі керек")
else:
    club = entities[0]
    kind = intents[0]
    matches = []
    for fact in facts:
        if fact["club"] == club and fact["kind"] == kind:
            matches.append(fact)
    if len(matches) == 0:
        print("білмеймін: дерек жоқ")
    elif len(matches) > 1:
        print("тоқта: бірнеше дерек табылды")
    else:
        fact = matches[0]
        today = date.today()
        until = date.fromisoformat(fact["valid_until"])
        if today > until:
            print("білмеймін: дерек ескірген")
        elif kind == "уақыт":
            print(fact["club"], "үйірмесінің уақыты:", fact["value"])
            print("Дереккөз:", fact["source"], "| тексерілген күні:", fact["checked_on"])
        else:
            print(fact["club"], "үйірмесінің орны:", fact["value"])
            print("Дереккөз:", fact["source"], "| тексерілген күні:", fact["checked_on"])

Step files.

How the program works

First, the lesson 15 logic extracts exactly one activity and one information type. Then we scan the whole catalog. Zero records means missing information; two or more mean a possible conflict or duplicate that a person should review. Only for one card do we compare dates: today > until becomes true the day after valid_until. Only then do we print the answer and source. checked_on records a past check; valid_until sets the expiry. They serve different purposes. A bad source sheet, a wrong computer clock, and limited vocabulary can still cause errors; the model does not guarantee truth.

Support map

Question → exactly two keys → 0/1/many records → expiry of one record → sourced answer or clear stop.

Recall and check

Try Шахмат қашан?, then Сурет қайда?. Through 31 December 2026, the first answer is шахмат үйірмесінің уақыты: бейсенбі, 15:00 plus a source line; afterwards it is білмеймін: дерек ескірген. The second is always білмеймін: дерек жоқ. To test a conflict, add a second шахмат / уақыт card with another value: expect тоқта: бірнеше дерек табылды. To test expiry, set valid_until to a past date. Common mistake: deleting an old card before finding the cause of the conflict.