Robin Houston

@robinhouston

Cofounder of . Also maths. Blogging sometimes at . My name is an anagram of “No enthusiasm or job”.

London
Vrijeme pridruživanja: travanj 2009.

Medijski sadržaj

  1. 2. velj

    One of those apt juxtapositions that Twitter occasionally throws up.

  2. 26. sij
    Odgovor korisnicima

    Aha! Found it. I just love Cassels’ style.

  3. 26. sij
    Odgovor korisniku/ci

    Here’s the one sentence version.

  4. 26. sij

    This was posted to Math Overflow in 2018, but I just found it via a comment on ’s video , which also explains this proof.

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  5. 26. sij

    You know Zagier’s brilliant-but-baffling “one sentence proof” that every prime of the form 4k+1 is the sum of two squares? It turns out there’s a lovely intuitive explanation of it!

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  6. 14. sij
    Odgovor korisnicima i sljedećem broju korisnika:

    Diagonal paths vs crenellated orthogonal paths:

  7. 14. sij
    Odgovor korisnicima i sljedećem broju korisnika:

    (It’s even simpler if we use crenellations rather than pyramids.)

  8. 14. sij
    Odgovor korisnicima

    It occurs to me that your example has a simpler solution. I wonder how generalisable this is.

  9. 14. sij
    Odgovor korisnicima i sljedećem broju korisnika:

    A version with fewer black cells:

  10. 14. sij
    Odgovor korisnicima i sljedećem broju korisnika:

    I was quite wrong! That grid *is* a subset of an infinite grid with finitely many black cells. I wonder if all finite rectangular grids are.

  11. 14. sij
    Odgovor korisnicima i sljedećem broju korisnika:

    But this grid can nevertheless be reduced to a white grid (assuming no special boundary conditions).

  12. 14. sij
    Odgovor korisnicima

    I suspect there are some finite grids, like this one, that are not a subset of any infinite grid with finitely many black cells.

  13. 13. sij

    It is connected in the 2×2 case, at any rate.

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  14. 13. sij
    Odgovor korisniku/ci

    I don’t think all the proofs were published, no. Zeilberger later reproved the Cosmological Theorem.

  15. 13. sij

    Don’t miss the portrait of Georges Perec as a prime number. (Would Jacques Roubaud have been a more apt choice?)

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  16. 9. sij

    If we have a node of the quotient graph, and an edge leading out of it, that determines a unique node of the original graph. Suppose not. Then we would have equivalent nodes (red and blue) connected by different edges.

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  17. 9. sij

    and every N-shaped subgraph actually has the dotted edge too:

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  18. 9. sij

    As long as the graph has no subgraph like this, which I’ll call a diamond:

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  19. 9. sij

    Here (from Wikipedia) is a graph – showing the weight-1 edges only – for de Bruijn sequences with Σ = {0, 1} and s=3.

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  20. 9. sij

    Here’s the graph for superpermutations with n=3, for example, where each edge is labelled with its weight. Weight-3 edges – such as 231→321 – are not shown.

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