I like this. Here is a one-liner I wrote recently that calculates the 10th Fibonacci number:

    (lambda n: n(lambda x: x+1)(0))(((lambda f: (lambda x: x(x))(lambda y: f(lambda z: y(y)(z))))(lambda f: (lambda f: lambda n: (lambda f: lambda x: x) if (lambda b: b(True)(False))((lambda n: n(lambda x: (lambda x: lambda y: y))(lambda x: lambda y: x))(n)) else f(n))(lambda n:(lambda m: lambda n: (lambda m: lambda n: (lambda n: n(lambda x: (lambda x: lambda y: y))(lambda x: lambda y: x))((lambda m: lambda n: n((lambda n: lambda f: lambda x: n(lambda g: lambda h: h(g(f)))(lambda u: x)(lambda u: u)))(m))(m)(n)))((lambda n: lambda f: lambda x: f(n(f)(x)))(m))(n))(n)((lambda n: lambda f: lambda x: f(n(f)(x)))(lambda f: lambda x: f(x)))(n)((lambda m: lambda n: lambda f: lambda x: m(f)(n(f)(x)))(f((lambda m: lambda n: n((lambda n: lambda f: lambda x: n(lambda g: lambda h: h(g(f)))(lambda u: x)(lambda u: u)))(m))(n)(lambda f: lambda x: f(x))))(f((lambda m: lambda n: n((lambda n: lambda f: lambda x: n(lambda g: lambda h: h(g(f)))(lambda u: x)(lambda u: u)))(m))(n)((lambda n: lambda f: lambda x: f(n(f)(x)))(lambda f: lambda x: f(x)))))))))(lambda f: lambda x: f(f(f(f(f(f(f(f(f(f(x))))))))))))

Paste that into a Python 3 interpreter, you should see "55". Somewhere in there is the Y-contaminator and several other tricks similar to what this article describes.