

The (lambda calculus) Y-combinator as a tattoo - jlangenauer
http://blogs.discovermagazine.com/loom/science-tattoo-emporium/?nggpage=19&pid=110

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daeken
I considered doing a Y-combinator implementation as a tattoo, but I ended up
going for a simple lambda. Why? It's more generic, and simply perfectly suited
to my work and interests. <http://img129.yfrog.com/i/9e0.jpg/>

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camccann
Just be careful, if you're considering doing something like this: Make sure
the tattoo artist is sufficiently lazy, as Y diverges under strict evaluation.
Nontermination is typically undesirable in a tattoo (except in some
subcultures).

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elrodeo
I know, it depends on the syntax, but I'd say some parenthesis are missing
(around the third lambda)... :)

~~~
eru
Not if they take lambda to extent all the way to the right, I guess. I'd
rather there are too many parens Like around the last (f (x x).

I'd prefer a graphical representation along the lines of
([http://www.cs.virginia.edu/~evans/cs655/readings/mockingbird...](http://www.cs.virginia.edu/~evans/cs655/readings/mockingbird_files/Graphical_lambda27.gif))
on my skin.

(The image is taken from the excellent article
[http://www.cs.virginia.edu/~evans/cs655/readings/mockingbird...](http://www.cs.virginia.edu/~evans/cs655/readings/mockingbird.html))

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jacquesm
Another one:

[http://blogs.discovermagazine.com/loom/science-tattoo-
empori...](http://blogs.discovermagazine.com/loom/science-tattoo-
emporium/?nggpage=22&pid=147)

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fexl
However, when building functions with combinators it's clearer to use bound
lambdas along the way:

\I = (S C C)

\U = (S I I)

\R = (S (C S) C)

\Q = (S R (C U))

\Y = (S Q Q)

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fexl
I might prefer this:

S S I (S (S (C S) C) (C (S I I)))

But then I'd be tempted to abstract the I out and apply to (S C C).

