Comments for João F. Ferreira

Comment on A square grid path problem by João Ferreira

João Ferreira — Sun, 19 Feb 2012 22:57:13 +0000

Hi Paul, thanks for reading. No, I don’t have any Mathematica code. Sorry.

Comment on A square grid path problem by João Ferreira

João Ferreira — Sun, 19 Feb 2012 22:55:22 +0000

Hi Barry, replace m and n by 4 and 3 in the last formula.

Comment on A square grid path problem by Barry

Barry — Fri, 03 Feb 2012 02:56:35 +0000

Sorry 4 (horizontal) x 3 (vertical) grid.

Comment on A square grid path problem by Barry

Barry — Fri, 03 Feb 2012 02:55:52 +0000

We have college professors and students trying to solve this problem and I’m trying to help my 6th grader solve it for a 2 (horizontal) x 3 (vertical) grid. Any help?

Comment on On Peirce’s law and the law of the excluded middle by João Ferreira

João Ferreira — Tue, 31 Jan 2012 09:17:26 +0000

Hi Brent, thanks for your comment. Perhaps I should have made it clear that my proof uses only classical logic. For me, this was an exercise on formulae manipulation: how to get from one formula to another by syntactic manipulation. Of course that the result by itself is not very interesting; I could have just proved that Peirce’s law is a theorem in classical logic and then conclude that it must be the same as any other theorem. (By the way, I wouldn’t use the proof above to show that it is a theorem; I would rather use the connecting lemma and the fact that distributivity distributes over equality.)

If I have time, I’ll think about calculational approaches to the intuitionistic proof.

Comment on A square grid path problem by Saurabh

Saurabh — Mon, 30 Jan 2012 16:53:48 +0000

Could you please help me in solving the following problem. We have a 4 X 4 square grid. The allowed moves are Up or Right. How many distinct paths are possible from (0,0) to (4,4) always excluding the points
(1,2) and (3,3)?

Comment on On Peirce’s law and the law of the excluded middle by Brent Yorgey

Brent Yorgey — Mon, 30 Jan 2012 13:35:06 +0000

Peirce’s law and LEM are two characterizations of classical logic: the point of the Coq proof is to show that they are equivalent in intuitionistic logic. That is, if either one was added as an axiom on top of intuitionistic logic, it would be possible to derive the other. Your proof uses classical logic; both the equivalence of p -> q to ~p \/ q and also the version of De Morgan’s law you use (which involves double negation elimination) are only valid in a classical setting. Proving that Peirce’s law and LEM are equivalent using classical logic is not very interesting, because both are independently derivable from the axioms — it’s like proving “true iff true”.

I would, however, be interested in seeing a calculational presentation of the intuitionistic proof, which is notoriously tricky to grok. The video doesn’t really give much insight into how the proof actually goes since it uses some high-powered Coq automation.

Comment on A square grid path problem by AbdulFattah Popoola

AbdulFattah Popoola — Sun, 15 Jan 2012 19:31:17 +0000

Great tutorial! Very clear and straight to the point. Thanks man!!

Comment on A square grid path problem by Fred

Fred — Wed, 07 Dec 2011 12:23:16 +0000

I saw the same problem in statistic class, though I’m interested in the total number of paths available including going “up” and “left”. For a 2×2 square with paths going from top left to lower right would have another 6 paths going like this (D means down, U means up, R means right, L means left) DRURDD, DDRURD, DDRUURDD, and RDLDRR, RRDLDR, RRDLLDRR. Of these paths there are four 6 units paths and two 8 units paths. I’m not sure if there’s a way to compute this, thinking of writing a minimum distance algorithm for this problem.

Comment on A square grid path problem by paul

paul — Wed, 23 Nov 2011 02:32:46 +0000

…continued from previous e-mail
http://mathforum.org/advanced/robertd/manhattan.html