I slept 3.5 hours last night

"If I really got mixed up, and was wrong about the size, then just multiply all the listed times by 2^32, and wake me in 8 trillion AD."
-from a poster on Slashdot, on breaking SHA-1

PBS is up a creek

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Let A(N) be the set of increasing finite sequences of natural numbers, N. Call a subset T of A(N) thin if, for all s and t in T, s is not an initial sequence of t.

Theorem: (Nash-Williams) Let m be in N, I be an infinite subset of N, and {T1, T2, ... , Tm} be a partition of a thin subset T of A(N). Then there exists an infinite subset K of I such that T intersect A(K) is contained in a single Ti.

This is freaking sweet. Here's why: look at the special case where T is the subset of A(N) with all elements the same length. It's Ramsey's theorem! Word is, there's an ultrafilter proof of this (Erdös). I'm looking for it right now.
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