Split into Cases and Follow If–Then Clues
Lesson 10.4 · 分类讨论与条件推理
Learn to divide all possibilities into complete, non-overlapping cases, then follow an “if … then …” clue in the direction it actually points.
Split all possibilities into complete cases
A useful case split must cover every possibility and must not count any possibility twice.
Rule 1 — Exhaustive
Every possible value must belong to at least one case. If a value is left out, the proof has a gap.
Rule 2 — Non-overlapping
No possible value should belong to two cases. Otherwise the same branch is counted twice.
Eight children: possible numbers of red hats
Let r be the number of red hats. Since there are eight children, r can be any whole number from 0 through 8.
Turn the hat rule into a live model
Read the complete hat-and-balloon rules below before testing cases.
The eight-child hat and balloon puzzle
Eight children each wear either a red hat or a blue hat. Each child can see the other seven hats but cannot see their own hat.
- If a child sees at least three red hats, they take a red balloon.
- Otherwise, they take a blue balloon.
- At the end, the group contains both red and blue balloons.
Question: How many children wear red hats?
A blue-hat child sees all r red hats. A red-hat child sees only r − 1 red hats because they cannot see their own hat.
Reject the low case: r ≤ 2
The values 0, 1, and 2 all lead to the same balloon result, so one argument handles all three.
Reject the high case: r ≥ 4
The values 4 through 8 also behave alike.
Keep the boundary case: r = 3
After rejecting the low and high groups, the only untested case is the boundary value.
• A blue-hat child sees all 3 red hats and takes a red balloon.
• A red-hat child sees only the other 2 red hats and takes a blue balloon.
Read “if … then …” precisely
An if–then clue is a one-way promise. It is broken only when the first part happens but the promised result does not.
The arrow
P → Q
Read: “If P is true, then Q is true.”
- If P happens, Q is forced.
- If P does not happen, the clue makes no promise about Q.
- Q being true does not prove P.
P
Q
| P | Q | P → Q | Reason |
|---|---|---|---|
| True | True | True | Promise kept |
| True | False | False | Promise broken |
| False | True | True | P is false, so the rule makes no demand on Q |
| False | False | True | P is false, so the rule makes no demand on Q |
Follow the worked example’s excellent-grade chain
All three if–then statements are true, but exactly two of the four students earned an excellent grade.
The complete puzzle
- A says: “If I am excellent, then B is excellent.”
- B says: “If I am excellent, then C is excellent.”
- C says: “If I am excellent, then D is excellent.”
All three statements are correct. Exactly two students are excellent. Who are they?
Test every possible pair
Fast deduction
If A were excellent, the arrows would force B, C, and D too—four excellent students. So A is not excellent.
Continue
If B were excellent, the arrows would force C and D—three excellent students. So B is not excellent. The required two are C and D.
Avoid reverse-arrow mistakes
The arrow can be followed forward. If Q is false, the true rule P → Q tells us P must be false too. But it cannot simply be reversed.
Added transfer challenge
Four students, named P, Q, R, and S, join a project under these true rules. Here P → Q means “if student P joins, student Q joins”:
Exactly three students join. Which three?
Fresh workshop
Try all six fresh checks. Use a hint after your first attempt.
Six children wear red or blue hats. Each sees all five other hats but cannot see their own hat. Each takes a red balloon if they see at least two red hats; otherwise blue. Both balloon colors appear.
One hint
Group counts as at most 1, exactly 2, and at least 3.
Worked reasoning — open after trying
At most 1 red hat gives all blue balloons; at least 3 gives all red. At 2 red hats, those two children see one red hat and take blue; the other four see two and take red.
Fresh exit challenge
Complete Missions 2 and 4, this workshop, and all five exit checks to earn your certificate. Reflection and extra exploration are optional.
Five students follow true joining rules A → B → C → D → E. Exactly two join. Separately, consider one true rule P → Q.
One hint
For each possible joiner, count all students forced to join farther along the chain.
Worked reasoning — open after trying
A, B, or C joining would force too many. D and E are the two. P true with Q false breaks the rule, so Q false forces P false. Q true alone does not determine P.
Lesson checkpoints completed
Case-and-Implication Strategist
You completed the checkpoints on case splits, thresholds and if–then reasoning. Revisit any steps for which you needed solution help.
Teaching notes
The eight-child hat-and-balloon puzzle and the A→B→C→D excellent-grade puzzle are fully restated in English.The live case models, implication truth table, transfer challenge, feedback, and assessments are added instructional scaffolds.