Lecture 53 · Mathematics
Why the suit-and-clock strategy always works
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Prove the same-suit and short-clock-distance guarantees for every five-card hand.
Before this lecture
In this lecture
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Turn performance into a universal claim
- For every five-card hand
- There exists a suitable pair and direction
- Assistant chooses; decoder reconstructs
Prove pigeonhole and inspect suit distributions
- Five cards, four suits
- At most one per suit gives at most four
- Same suit + distinct cards→different ranks
Define directed rank distance rigorously
- d(a, b)+d(b, a)=13
- J→Q→K→A→2: four steps
- Count moves, not the starting rank
Prove short direction bound and unique orientation
- Exactly one direction has length 1–6
- Anchor=start; hidden=end
- Ordinary rank comparison is insufficient
Work extreme hands and nonunique choices
- Many valid encodings can coexist
- Four aces +2♣ works
- Equal ranks remain distinct by suit
Generalize clock lemma without overclaiming whole trick
- Odd clock 2 r+1→short distance≤r
- Here r=6
- Pair guarantee ≠ message capacity
Specify adversarial hand and orientation game
- Try to avoid a suit pair
- Orient the short clockwise arc
- Every valid pair is accepted
Retrieve proof and bridge to permutation language
- Pigeonhole→short arc→six-valued message
- Hide endpoint of the short arc
- Next: permutation language
Companion interactive
Planned
Find and orient a valid pair in every supplied hand.
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Notes
Written notes for this lecture will appear here.
Sources
- Mathematical Card Magic, Chapter K, pp. 307–309
How these sources were used
- Pigeonhole and Poles Apart principles; examples independently specified.