Lecture 07 · Mathematics
A proof in two lanes
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Prove Hummer face-count balance for 2n cards, identify the even-size hypothesis, and strengthen the statement using lane difference.
Before this lecture
In this lecture
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Generalize the ten-card picture carefully
- 2n cards; n positions per lane
- Before finish: odd up = even up
- After even-position flip: n up
Prove the base and cut step for every even size
- Cut 1: odd ↔ even
- Wrapped card: 1→2n
- Equality survives every cut
Prove the turnover step without enumerating histories
- Changes to (odd up, even up)
- DD:(+1,+1)
- UU:(−1,−1)
- DU/UD:(0, 0)
Derive the finish and extend valid starting states
- a+(n−a)=n
- Valid start: odd up = even up
- All down is sufficient, not necessary
Strengthen to lane difference and reveal sign changes
- D=a−b
- Turn 2: D→D
- Cut 1: D→−D
- Finish: n+D
Give a concrete odd-packet counterexample
- Five cards: odd lane 3, even lane 2
- 00000→11000→10001
- Equal counts fail after Cut 1
- Finish totals: 2 or 4
Extend legal turnovers to even prefixes with a pairing proof
- Even prefix: 2 m cards
- Old block difference: x−y
- New:(m−y)−(m−x)=x−y
Provide theorem-building challenge with exact modes
- Choose the claim, then test
- Balance →half up
- Arbitrary start →|D| preserved
- Odd size: inspect wraparound
Companion interactive
Planned
Prove or refute chosen lane-count claims under explicitly selected packet sizes and move sets.
The interactive for this lecture is being built and will appear here.
Notes
Written notes for this lecture will appear here.
Sources
- Magical Mathematics, Chapter 1, pp. 6–8, 12–14
- Mathematical Card Magic, Chapter 9, p. 212
How these sources were used
- Even-packet proof and discussion of odd packets.
- Two equivalent forms of Hummer CATO invariant.