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Diophantine Fruit

54 点作者 isaac21259将近 4 年前

6 条评论

opheliate将近 4 年前
One of my favourite parodies of this &quot;brain teaser&quot; format invokes Fermat&#x27;s last theorem: [0]<p>0: <a href="https:&#x2F;&#x2F;i.imgur.com&#x2F;5QGR1Lt.jpeg" rel="nofollow">https:&#x2F;&#x2F;i.imgur.com&#x2F;5QGR1Lt.jpeg</a>
kevinventullo将近 4 年前
To get a sense of how far one could push these fruit problems, see Matiyasevich’s Theorem (<a href="http:&#x2F;&#x2F;www.scholarpedia.org&#x2F;article&#x2F;Matiyasevich_theorem" rel="nofollow">http:&#x2F;&#x2F;www.scholarpedia.org&#x2F;article&#x2F;Matiyasevich_theorem</a>).<p>My understanding is a bit fuzzy, but it basically says you can encode any recursively enumerable set in terms of solutions to Diophantine equations (i.e. integer polynomials).<p>In particular you can encode, say, the set of all Turing Machines which halt in terms of the solutions to some integer polynomial.
mjreacher将近 4 年前
See further discussion here: <a href="https:&#x2F;&#x2F;old.reddit.com&#x2F;r&#x2F;math&#x2F;comments&#x2F;osfc0x&#x2F;you_know_those_annoying_fruit_equation_memes&#x2F;" rel="nofollow">https:&#x2F;&#x2F;old.reddit.com&#x2F;r&#x2F;math&#x2F;comments&#x2F;osfc0x&#x2F;you_know_those...</a>
londons_explore将近 4 年前
You need to add brackets...<p>The audience for these fruit problems don&#x27;t &#x27;do&#x27; typical multiplication-first arithmetic.
technocratius将近 4 年前
Haha, great fun! The picture might need a few cycles of JPEG (re)compression to make it even more convincing.
joshu将近 4 年前
delightfully cruel!