Count Relationships, Complements, and Extreme Cases
Lesson 11.1 — 关系计数、补集与极端情况
Use mutual relationships, forbidden pairs, and carefully chosen maximum or minimum cases to turn wordy logic into countable mathematics.
Count who is eligible. Replace “every pair” with what is forbidden. Push a condition to its edge.
Count only the people who are eligible to connect
Throughout this lesson, count different people met, not the number of meeting events. Meeting the same person twice still counts as one person. “Isolated” means meeting nobody in the group during the time being considered. A meeting is a mutual relationship: if A met B, then B met A. A person cannot meet themselves, and an isolated person cannot be counted as anyone else's meeting.
Mutual means one undirected link
An edge between two people represents one meeting relationship. It has no arrow because the relationship works both ways.
The symbol ⇔ means “if and only if”: both statements describe the same relationship.
Solve the 30-worker office puzzle
Use the complete office scenario below.
Worked example — one isolated office worker
An office has 30 workers. On one day, one worker met none of the other office workers. What is the greatest number of colleagues that any one worker could have met that day?
Start with all workers
Do not count the chosen worker
Do not count the isolated worker
Turn an “every pair” clue into a forbidden pair
In every hat puzzle on this page, each person wears exactly one hat, either red or blue. A forbidden pair tells us what cannot happen. “Every pair contains at least one red hat” is easier to use as “two blue hats can never be chosen together.”
If there are 0 or 1 blue hats
You cannot select two blue-hat people, so every pair contains a red hat.
If there are 2 or more blue hats
Select any two blue-hat people. That pair contains no red hat, so the rule fails.
Apply the forbidden-pair idea to 68 guests
This guided practice makes one intended condition explicit so the answer is unique.
original-based practice — red and blue hats
At a gathering there are 68 people. Everyone wears either a red hat or a blue hat. Among any two people, at least one wears a red hat. Both hat colors are present. By how many does the number of red hats exceed the number of blue hats?
Maximum blue hats
Two blue hats would create a forbidden pair.
Red hats
Difference
Push a group condition to its extreme
The pair rule is part of a wider pattern. Here a group means any choice of g different people, and g is no larger than the total number of people. If every group of g people must contain at least one red hat, then an all-blue group of size g is forbidden.
Use consecutive relationship counts
The practice problem combines a fixed total, a consecutive list, and an extreme description.
original exercise — a celebration with 69 guests
There are 69 male and female guests. Every female guest knows a different number of male guests. When these counts are listed from smallest to largest, they increase by 1 each time, with no gaps or repeats. The female guest who knows the fewest men knows 16. The female guest who knows the most men knows all but 2 of the male guests. How many female guests are there?
Let M be the number of male guests
The largest acquaintance count is:
The consecutive list is:
Count the terms
The number of female guests is:
Then use the total:
Build a candidate ledger instead of guessing
A candidate must satisfy the total number of people, the first and last acquaintance counts, and the whole-number condition at the same time.
| Male guests M | Maximum known M − k | Sequence length | Female guests T − M | Works? |
|---|
Transfer the structure and repair common errors
The story can change while the counting structure stays the same.
New transfer puzzle — camp acquaintances
At a camp there are 50 senior and junior campers. The numbers of juniors known by the seniors, listed in increasing order, rise by 1 each time, with no gaps or repeats. The least-connected senior knows 8 juniors. The most-connected senior knows all but 3 juniors. How many seniors and how many juniors are there?
Translate the endpoint
If there are J juniors, the maximum known is:
The number of seniors is the number of terms from 8 to J − 3:
Use the total
Independent workshop
Solve at least 6 of 8 fresh problems. Each has a local hint.
Hint
Exclude the focus person and both isolated people.
Hint
An all-blue group of five is forbidden.
Hint
Use as many blue hats as the rule permits.
Hint
The rule allows at most one blue; the two-color condition ensures one.
Hint
Include both endpoints: last minus first plus one.
Hint
For J juniors, seniors = (J − 2) − 5 + 1.
Hint
The resulting equation is 2J − 6 = 41.
Hint
Test the all-red arrangement too.
Independent exit ticket
Solve all five fresh problems. The certificate also requires Missions 1–9.
Hint
Count only eligible others.
Hint
How many blue hats would create a forbidden group?
Hint
Number of seniors is (J − 4) − 9 + 1 = J − 12.
Hint
Solve 2J − 12 = 55, then check the unit.
Hint
The plan must reach the bound without creating six blue hats.
This writing is saved for the student or teacher, but the lesson never grades its wording or length.
Certificate of Completion
Relationship & Extreme-Case Navigator
This certifies that the student can count mutual relationships, use complements, identify forbidden groups, and test whole-number extreme cases.
Teaching notes
The office-worker example and red/blue-hat practice are based on Chapter 11, page 70.
The worked example's 68-person hat practice states that every pair contains a red hat, which implies at most one blue hat. Its intended numerical answer assumes that a blue hat is present. To make the student task logically complete, this lesson states explicitly that both colors are present.
The live generalizations, transfer puzzle, candidate ledger, workshop, and exit ticket are new instructional scaffolds.