sbi

Heapsort has poor cache performance though. Its asymptotic complexity depends on the assumption that memory access is O(1), which is not borne out in practice.

Webster's Dictionary does not discuss the pejorative use of the word "ghetto" and in any case is not an arbiter of racism.

What charming casual racism. "It's pretty ghetto" ... "The ghetto way is just to run this on a machine" ... "The ghetto reversing is to run strings."

Thanks for explaining that to me. To add to the confusion, tabasco is also a pepper cultivar named after the Mexican state.

Tabasco is a state in Mexico, though; how were its trademark owners able to obtain the name?

There are many different versions of awk: gawk, (BSD) nawk, mawk, etc. I think OS X uses nawk, but mawk is reputedly faster. Gawk is definitely slower than both mawk. I'm not surprised that Perl is faster though.

Probably not, but in some cases sed and AWK might be faster. I am a big fan of AWK. It is limited in some ways that make it impractical to use for really serious programs, but it is very expressive. Look at [1, 2].

[1]: http://c2.com/doc/expense/

[2]: http://www.pement.org/awk/awk1line.txt

The data from the study are available on the PLoS website. The authors coded incomplete documents as mistakes---some participants submitted empty LaTeX documents for the table exercise with hundreds of "mistakes." Since the LaTeX users mean percentage completion was lower than that of the Word users, this may explain part of the authors' finding that the LaTeX users were more error prone as well.

Looking over their data, it seems that they coded incomplete text as an error. So, for example, participants 34 and 37 completed none of the table exercise and were coded as having produced 513 errors (both were LaTeX users). This accounts for the vast majority of the variation in the observed errors. I think they've convincingly shown that Word is more productive than LaTeX for some of the study tasks, but not that the resulting output is more correct.

That being said, I don't doubt that LaTeX is harder to use and more error prone than Word. Since I find myself frequently writing mathematics-heavy text, I personally prefer LaTeX ...

Doesn't std::list<T> generate different code for each type T, at least naively? The generated code could end up taking up more space even if the data structures on which it operates look the same in memory.

This article correctly observes that "many resources are not plain memory" and then goes on to advise users to close() a file in a destructor. However, one of the ways that files are not like memory is that while free() can never fail, fclose() can fail if writing is buffered and the final write fails. This is especially problematic in C++ since throwing from a destructor is dangerous. Whether you rely on RAII or GC to free files, you will not be able to catch this sort of error. It is precisely for this reason that GC is a mechanism for memory management, not resource management in general.

The Kelly criterion doesn't quite apply in this problem, since you are asked to optimize a more conservative strategy, namely to guarantee (with high probability) a certain amount of winnings rather than maximize log(winnings). In fact the Kelly value for f (in the problem's notation) is 0.25.

Typically you wouldn't use lgamma to compute exact values of factorials at all. For a real world example of this function in action, look at John D. Cook's blog:

[1]: http://www.johndcook.com/blog/2012/07/14/log-gamma-differenc...

[2]: http://www.johndcook.com/blog/2010/08/16/how-to-compute-log-...

[3]: http://www.johndcook.com/blog/2008/04/24/how-to-calculate-bi...

I explained this in a reply below, but this is a linear-algebraic technique. The matrix A = [0 1; 1 1] satisfies the relation A^2 - A - 1 = 0. (In fact, if you think of A as a lag operator, this is the definition of Fibonacci numbers!) You can use this relation to compute products of matrices of the form aA + bI where I is the identity matrix---the comment below has the derivation. This is a bit faster than using matrix multiplication because there are only two coefficients to keep track of.

This might seem like a pointless optimization, but it is actually useful. One advantage is that you don't need to factor the polynomial x^2 - x - 1 or work with square roots. The closed-form for Fibonacci numbers is a bit of a red herring because to work with it exactly, you will end up doing the same kinds of computations anyway. For recurrences of higher degree, factorization might be impossible and this is the only route.

When you're working over a finite field, even if you don't start with a recurrence, there are fast techniques for finding factors of the characteristic polynomial of a sparse matrix (the Berlekamp-Massey algorithm comes to mind).

You are correct that this code uses "Russian Peasant Multiplication." The question is: what is being exponentiated? One choice is the matrix A = [0 1; 1 1], whose powers are A^n = [F_{n-1} F_n; F_n F_{n+1] if we compute those powers treating A as a matrix. This involves some redundant computations since F_n is being computed twice.

An alternative method is to use the following identity:

  (aA + b)(cA + d) = acA^2 + (ad + bc)A + bd
                   = (ad + bc + ac)A + (bd + ac)
                   = ((a + b)(c + d) - bd)A + (bd + ac)

(In fact, A^n = F_n A + F_{n-1} I_2). This step requires 3 multiplications and 4 additions and is more-or-less equivalent to the "fast doubling" algorithm in reference [11], which is more optimized. But this derivation is completely generic---it just relies on the characteristic polynomial of A, namely A^2 - A - 1 = 0.

In the case where A is not the Fibonacci matrix but something much larger, working with polynomials modulo the characteristic polynomial of A is much faster than working with A as a matrix.

Thank you for the reference. The algorithm in the parent comment is not actually matrix exponentiation but the "fast doubling algorithm" in Minase's terminology, where you exponentiate the matrix algebraically using the Cayley-Hamilton theorem rather than using matrix multiplication.

Which reference?

The "right" way to generate Fibonacci numbers is not via the closed form, but by exponentiation of the matrix A = [0 1; 1 1] that defines Fibonacci numbers as an element of its endomorphism ring Z[x]/(x^2 - x - 1). This doesn't rely on explicitly computing the eigenvalues of A and scales well to more complex recursions. Sample Haskell:

  fib :: Integer -> Integer
  fib n = fst . loop (0, 1) base . abs $ n
    where pmul (a, b) (c, d) = ((a + b)*(c + d) - b*d, b*d + a*c)
          base = if n >= 0 then (1, 0) else (1, -1)
          loop acc sq m
            | m == 0         = acc
            | m `mod` 2 == 0 = loop acc (pmul sq sq) (m `div` 2)
            | otherwise      = loop (pmul sq acc) (pmul sq sq) (m `div` 2)

There are tools for standardizing C formatting. Another poster mentioned indent. Clang-format [1] is very good, works with C++ as well as C, and has great editor integration. You can choose among predefined style (e.g. Google, Mozilla, LLVM style guides) or create your own and then and enforce it within a source tree very easily.

[1]: http://clang.llvm.org/docs/ClangFormat.html

Thanks; I had searched on pgp.mit.edu for the primary key but forgot to add the hex prefix 0x.

Curiously, the 1024 bit DSA key used for some previous releases (0x7f4d66451ebcab3a) has been signed by "Someone at Mozilla Should Sign the Release Key (so users can verify the key's owner!) <anonymous@lulz.example.com>".

I see that Mozilla is now signing hashes of Firefox downloads with a 4096-bit RSA key. But the key used to sign firefox-23.0.tar.bz2 (id 0x057CC3EB15A0A4BC) is only self-signed and was created three weeks ago. It's not on pgp.mit.edu. Is there any actual way to "validate the authenticity of these keys in an out-of-band manner" as the KEYS file recommends?

The "5,031" statistic is ludicrous. At least that many people were killed by lynching alone, not to mention Wounded Knee.

These are quotes from the Bolt drama "A Man for All Seasons," not Sir Thomas More himself.

I'm surprised nobody has mentioned Koblitz and Menezes in this thread. Unfortunately, the mathematical literature is littered with insecure, but "provably secure," cryptosystems.

[1] http://anotherlook.ca/ [2] http://www.ams.org/notices/201003/rtx100300357p.pdf

NRO has some Flash games. There's also ATF for kids, CIA for kids, US Marshals Service for kids ...

The campy NSA kids' website has nothing to do with recruitment; the Clinton administration required all Federal agencies to create websites for children in an April 18, 1997 memorandum [1]. This is clearly spelled out in the terms-of-use at nsa.gov [2].

[1]: https://www.cia.gov/kids-page/privacy-statement/whmemo.txt

[2]: http://www.nsa.gov/terms_of_use.shtml#kids

Two notes: factoring polynomials over finite fields isn't hard (look up the Berlekamp algorithm or Cantor-Zassenhaus for starters). Second, I believe the system you've described doesn't satisfy your property (b). Quantities such as M[1]C[0] - C[1]M[0] + M[0]^2 and C[1]M[2] - C[2]M[1] + M[1]^2 - M[0]M[2] are congruent to 0 mod p; compute them over the integers and then take the gcd. By computing enough of these you should recover the prime p. Once you have p, recovering z is very easy (e.g., take the gcd of C and C'-M, where C' is the derivative of C).

If you believe that Lie theory is "pratical," then yes. See John Baez's excellent survey: http://math.ucr.edu/home/baez/octonions/

This is a pretty solution, and it gives you the probability generating function for free: by some power series manipulation, the pgf of the number of draws needed is g(z) = 1 + (z-1)*exp(z). So you can show fairly easily that Var(draws needed) = 3e - e^2, for example ...

I love spicy food, but I find Sriracha almost sickly-sweet. Anyone else prefer El Yucateco or Salsa Valentina?