What is a Proof?

1. Great Theoretical Ideas In Computer Science What is a Proof? ? Recitation 3 CS 15-251

2. A Game with Quarters Steven Rudich: www.discretemath.com www.rudich.net

3. A Game with Quarters • A circular table • Players take turns placing quarters on the table • Quarters cannot overlap • Last player to place a quarter wins Question: Is there a winning strategy for the first player? Steven Rudich: www.discretemath.com www.rudich.net

4. An Attempt • Yes! Here’s a strategy: • Place the first quarter at the center of the table • Each turn after that, mirror the moves of your opponent • Thus, you can always make a move Steven Rudich: www.discretemath.com www.rudich.net

5. An Objection • Each turn after that, mirror the moves of your opponent • What do you mean by “mirror”? • The proof is ambiguous. It can be interpreted in multiple ways. It fails to be clear. Steven Rudich: www.discretemath.com www.rudich.net

6. Clarification We will choose an axis of symmetry and reflect around it. Steven Rudich: www.discretemath.com www.rudich.net

7. A Problem Steven Rudich: www.discretemath.com www.rudich.net

8. A Problem • The proof assumes that we can always place a quarter in the mirror position. • The proof is takes a leap that is too big. It fails to be rigorous. Steven Rudich: www.discretemath.com www.rudich.net

9. Solution: Radial Symmetry Steven Rudich: www.discretemath.com www.rudich.net

10. A Little Skepticism • A failure of clarity and rigor made our previous attempts bogus. • Let’s take a skeptical eye to this latest attempt. How do we justify: “For each moves our opponent makes, we can make the symmetric move”? Steven Rudich: www.discretemath.com www.rudich.net

11. More Rigor • Let’s be much more careful about our statements: • For all n: After the opponent makes move n, we can still make another move. • Proof by induction: • Base Case: n = 0. On our first turn, we can certainly make a move. • Induction Step: n > 0. Suppose our opponent has just put down a quarter. We simply put a quarter in the spot radially opposite. Steven Rudich: www.discretemath.com www.rudich.net

12. More Rigor • Induction Step: n > 0. Suppose our opponent has just put down a quarter. We simply put a quarter in the spot radially opposite. How do we know we can do this? Steven Rudich: www.discretemath.com www.rudich.net

13. Strengthening the Induction Hypothesis • For all n: After opponent makes move n, we can still make another move, and retain the symmetry of the table, and cover the center point. Steven Rudich: www.discretemath.com www.rudich.net

14. Strengthening the Induction Hypothesis • Proof by induction: • Base Case: n = 0. Place quarter at center. • IS: n > 0. Opponent places quarter in some position. By IH, the table was symmetric before, and the center point was covered, so the radially opposite position is open. Place the quarter there. Steven Rudich: www.discretemath.com www.rudich.net

15. Lessons Learned • Clarity • Having one exact meaning, being unambiguous • “What does that mean?” • Ask yourself: “Is this ambiguous? Could it be interpreted in two different ways?” • Rigor • Proving each step explicitly, avoiding big leaps • “Does that really follow?” • Ask yourself: “Am I taking too big a leap? Is that really obvious? Have I considered all the possible cases?” Steven Rudich: www.discretemath.com www.rudich.net

16. A Curiosity Steven Rudich: www.discretemath.com www.rudich.net

17. The Power of Proofs • The primary purpose of proofs is to validate our ideas. But a powerful secondary purpose is to show us which aspects of our conjectures are essential and which are not. When we begin to refine and analyze proofs, we understand our original conjectures much more deeply and can generalize them. Steven Rudich: www.discretemath.com www.rudich.net

18. A Bogus Example • Theorem: All horses are the same color (!) • Proof by induction! Steven Rudich: www.discretemath.com www.rudich.net

19. Equestrian Chromatics • Base Case: One horse is the same as itself. • IS: Look at a group of n horses. By induction, horses 1..n-1 are the same color, and horses 2..n are the same color. Therefore horse 1 is the same as 2..n-1, which is the same as n. Thus the entire group is the same color. Steven Rudich: www.discretemath.com www.rudich.net

20. Equestrian Chromatics • Base Case: One horse is the same as itself. • IS: Look at a group of n horses. By induction, the color of horse C(1) = C(2) = … = C(n-1). Also, C(n-1) = C(n). Thus C(1) = C(2) = … = C(n), and the entire group is the same color. Steven Rudich: www.discretemath.com www.rudich.net

21. Tournaments • Consider a chess tournament: • There are n players • Each player plays a game against every other one • Show: The players can be lined up 1, 2, …, n such that for all i, player i beat player i+1. Steven Rudich: www.discretemath.com www.rudich.net

22. First Attempt • “Simply line up the players in order of skill. The best guy beat the next best, and he beat the next best, so on to the worst.” • Objection: “in order of skill” is unclear. “Skill” is not a well-defined notion. Steven Rudich: www.discretemath.com www.rudich.net

23. Another Attempt • Let’s clarify: Give each player a score equal to the number of games he won. Sort the players by score. Clearly each player is better than the next, so player i beat player i+1 all down the line. • Objection: It is not at all obvious that sorting the players results in the desired ordering. And what do we do if there are ties? Steven Rudich: www.discretemath.com www.rudich.net

24. Try, Try Again • How about this: Start with the guy who won the most games. Find a guy who he beat and put him next. Find a guy who he beat and put him next. Continue until you've lined up everyone. • Objection: It is not at all obvious that you can always find the next player to put in line. What if you get stuck? Steven Rudich: www.discretemath.com www.rudich.net

25. One More Time • Start with any player. Now take another player. If the second guy beat the first, then put them in the order 2, 1; otherwise make it 1, 2. Now grab a third guy. If he beat the first guy on the list, put him at the beginning; otherwise, if he lost to the last guy, put him on the end; otherwise, put him in the middle. Now take the fourth guy, and likewise find a spot where he belongs. For each player up to n, put him in the spot where he belongs. Thus you create the list. Steven Rudich: www.discretemath.com www.rudich.net

26. One More Time • Objection: In the first place, this is very verbose. Strive to make your proofs concise and succinct. • More importantly, the “proof” doesn’t explicitly identify the pattern. What is the “spot where he belongs”? What if no such spot exists? Steven Rudich: www.discretemath.com www.rudich.net

27. A Happy Ending • Proof by induction on n: • BC: n = 1. The list is of length one. • IS: Take the first n-1 players; by induction, we can order them appropriately. Now take player n. If he lost to everyone, put him at the end. Otherwise, put him in front of the first player he beat. He must have lost to whoever is before him (if anyone). Steven Rudich: www.discretemath.com www.rudich.net

28. A Happy Ending • This proof works. It is clear: Each statement is unambiguous. It is rigorous: There are no questionable steps or big leaps. Steven Rudich: www.discretemath.com www.rudich.net

29. The Moral • Devising correct proofs is a matter of vigilance. You must constantly ask yourself: • “Is this statement ambiguous? Could it be interepreted multiple ways?” • “Am I making too big a leap? Have I considered all possible cases?” Steven Rudich: www.discretemath.com www.rudich.net