- **C Programming**
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- - **a bit of a puzzle**
(*http://www.velocityreviews.com/forums/t597236-a-bit-of-a-puzzle.html*)

a bit of a puzzleHere is a little algorithm I came across whose implementation is
amusingly obscure: what simple function does the following C code compute, and why? #include <stdint.h> unsigned foo(uint32_t n) { const uint32_t a = 0x05f66a47; static const unsigned bar[32] = {0,1,2,26,23,3,15,27,24,21,19,4,12,16,28,6,31,25,2 2,14,20,18,11,5,30,13,17,10,29,9,8,7}; n = ~n; return bar[(a * (n & (-n))) >> 27]; } To save you the trouble of compiling and running it yourself, here is what it produces for n = 0,1,2,...,31: 0 -> 0, 1 -> 1, 2 -> 0, 3 -> 2, 4 -> 0, 5 -> 1, 6 -> 0, 7 -> 3, 8 -> 0, 9 -> 1, 10 -> 0, 11 -> 2, 12 -> 0, 13 -> 1, 14 -> 0, 15 -> 4, 16 -> 0, 17 -> 1, 18 -> 0, 19 -> 2, 20 -> 0, 21 -> 1, 22 -> 0, 23 -> 3, 24 - > 0, 25 -> 1, 26 -> 0, 27 -> 2, 28 -> 0, 29 -> 1, 30 -> 0, 31 -> 5 |

Re: a bit of a puzzleSteven G. Johnson said:
<snip> > To save you the trouble of compiling and running it yourself, here is > what it produces for n = 0,1,2,...,31: > > 0 -> 0, 1 -> 1, 2 -> 0, 3 -> 2, 4 -> 0, 5 -> 1, 6 -> 0, 7 -> 3, 8 -> > 0, 9 -> 1, 10 -> 0, 11 -> 2, 12 -> 0, 13 -> 1, 14 -> 0, 15 -> 4, 16 -> > 0, 17 -> 1, 18 -> 0, 19 -> 2, 20 -> 0, 21 -> 1, 22 -> 0, 23 -> 3, 24 - >> 0, 25 -> 1, 26 -> 0, 27 -> 2, 28 -> 0, 29 -> 1, 30 -> 0, 31 -> 5 Obviously I can't tell what the algorithm is actually used for, but it is isomorphic to the following usage: the Nth number represents the number of levels beneath the Nth item discovered in a perfectly balanced binary tree during a postorder traversal (left, centre, right). -- Richard Heathfield <http://www.cpax.org.uk> Email: -http://www. +rjh@ Google users: <http://www.cpax.org.uk/prg/writings/googly.php> "Usenet is a strange place" - dmr 29 July 1999 |

Re: a bit of a puzzleOn Mar 9, 1:58 pm, Richard Heathfield <r...@see.sig.invalid> wrote:
> Obviously I can't tell what the algorithm is actually used for, but it is > isomorphic to the following usage: the Nth number represents the number of > levels beneath the Nth item discovered in a perfectly balanced binary tree > during a postorder traversal (left, centre, right). There is a simpler, purely arithmetic definition of what it computes. By "why?" I don't mean "why was this particular code written" (obviously unanswerable) but "why does it compute what it does for all n (once you figure out what it does)?" i.e., how does it work? |

Re: a bit of a puzzleOn Sun, 09 Mar 2008 10:40:50 -0700, Steven G. Johnson wrote:
> Here is a little algorithm I came across whose implementation is > amusingly obscure: what simple function does the following C code > compute, and why? > > #include <stdint.h> > unsigned foo(uint32_t n) > { > const uint32_t a = 0x05f66a47; > static const unsigned bar[32] = > {0,1,2,26,23,3,15,27,24,21,19,4,12,16,28,6,31,25,2 2,14,20,18,11,5,30,13,17,10,29,9,8,7}; > n = ~n; > return bar[(a * (n & (-n))) >> 27]; n = ~n; n = a * (n & -n); return bar[n >> 27]; You're assuming that the multiplication (a * (n & (-n))) will take place in 32 bits. This is not the case when int or unsigned int has more than 32 bits. Store the result in n to be sure. > } > > To save you the trouble of compiling and running it yourself, here is > what it produces for n = 0,1,2,...,31: > > 0 -> 0, 1 -> 1, 2 -> 0, 3 -> 2, 4 -> 0, 5 -> 1, 6 -> 0, 7 -> 3, 8 -> 0, > 9 -> 1, 10 -> 0, 11 -> 2, 12 -> 0, 13 -> 1, 14 -> 0, 15 -> 4, 16 -> 0, > 17 -> 1, 18 -> 0, 19 -> 2, 20 -> 0, 21 -> 1, 22 -> 0, 23 -> 3, 24 - >> 0, 25 -> 1, 26 -> 0, 27 -> 2, 28 -> 0, 29 -> 1, 30 -> 0, 31 -> 5 This looks like the number of consecutive bits set, starting from the least significant. (~n) & (-~n) can be written as (n + 1) & ~n. It is the value of the lowest bit that is not set. The multiplication by 0x05f66a47 happens to produce unique values in bits 27-31 for the first 32 powers of two (I'm not seeing how this value was obtained), so after extracting those bits, you can use a lookup table to find the result. |

Re: a bit of a puzzleJust to be clear, I know what it does and how it works; I'm not asking
for programming help. I'm posing it as a puzzle for amusement and edification. Background: It's a C implementation of an algorithm I found in a rather famous book, for a rather simple and important arithmetic problem (which shows up as a subproblem in Gray codes, low-discrepancy sequences, and many other problems), and is one of the fastest and most compact solutions to this problem that I could find (without using assembly) (although there is one slightly faster method on my machine that is not so compact). What I found amusing is that, if you don't comment the code, it seems quite nonobvious what this algorithm computes, and even if you deduce that from the function outputs it seems very nonobvious why/how it works. Regards, Steven G. Johnson |

Re: a bit of a puzzle"Steven G. Johnson" wrote:
> > Here is a little algorithm I came across whose implementation is > amusingly obscure: what simple function does the following C code > compute, and why? > > #include <stdint.h> > unsigned foo(uint32_t n) > { > const uint32_t a = 0x05f66a47; > static const unsigned bar[32] = > {0,1,2,26,23,3,15,27,24,21,19,4,12,16,28,6,31,25,2 2,14,20,18,11,5,30,13,17,10,29,9,8,7}; > n = ~n; > return bar[(a * (n & (-n))) >> 27]; > } Doesn't seem to work very well on a machine with 16 bit integers. -- [mail]: Chuck F (cbfalconer at maineline dot net) [page]: <http://cbfalconer.home.att.net> Try the download section. -- Posted via a free Usenet account from http://www.teranews.com |

Re: a bit of a puzzleOn Mar 9, 2:31 pm, Harald van D©¦k <true...@gmail.com> wrote:
> (~n) & (-~n) can be written as (n + 1) & ~n. It is the value of the > lowest bit that is not set. The multiplication by 0x05f66a47 happens to > produce unique values in bits 27-31 for the first 32 powers of two (I'm > not seeing how this value was obtained), so after extracting those bits, > you can use a lookup table to find the result. Yes, the algorithm is described in Knuth volume 4A (draft fascicle) section 7.1.3 ("Bitwise tricks and techniques"), who attributes it Lauter and others in 1997. The main trick is the magic constant "0x05f66a47"; Knuth describes the properties it requires (related to de Bruijn) cycles, and the fact that there are many such constants that will work; this particular one was found by a brute-force search. It's a cute and compact way of finding the number of rightmost-1 bits (or rightmost-zero bits, without the ~n). A slightly (~15%) faster way (but less compact), at least on my machine, is to search the bytes of ~n, starting from the least-significant byte, until a nonzero byte is found, and then use a 256-element lookup table...this has slower worst-case performance (for cases where the rightmost zero bit is in the most significant byte), but slightly better average-case performance (YMMV) (since a random number has a 255/256 chance of having its rightmost zero in the least-significant byte). Of course, on processors like x86 you can do faster in assembly, or with gcc's __builtin_ctz function, but it's a fun little puzzle to do this as fast and/or as compactly as possible in plain C. Steven |

Re: a bit of a puzzleCBFalconer <cbfalconer@yahoo.com> writes:
> "Steven G. Johnson" wrote: >> Here is a little algorithm I came across whose implementation is >> amusingly obscure: what simple function does the following C code >> compute, and why? >> >> #include <stdint.h> >> unsigned foo(uint32_t n) >> { >> const uint32_t a = 0x05f66a47; >> static const unsigned bar[32] = >> {0,1,2,26,23,3,15,27,24,21,19,4,12,16,28,6,31,25,2 2,14,20,18,11,5,30,13,17,10,29,9,8,7}; >> n = ~n; >> return bar[(a * (n & (-n))) >> 27]; >> } > > Doesn't seem to work very well on a machine with 16 bit integers. Most machines have 16-bit integers; often the 16-bit integer type is called "short". If you mean 16-bit ints, I don't see how that would cause a problem, since the fucntion's argument is of type uint32_t. (The unsigned result shouldn't be a problem unless you're worried about integers with more than 32767 bits.) Or am I missing something? -- Keith Thompson (The_Other_Keith) <kst-u@mib.org> Nokia "We must do something. This is something. Therefore, we must do this." -- Antony Jay and Jonathan Lynn, "Yes Minister" |

Re: a bit of a puzzleKeith Thompson wrote:
> CBFalconer <cbfalconer@yahoo.com> writes: >> "Steven G. Johnson" wrote: > >>> Here is a little algorithm I came across whose implementation is >>> amusingly obscure: what simple function does the following C code >>> compute, and why? >>> >>> #include <stdint.h> >>> unsigned foo(uint32_t n) { >>> const uint32_t a = 0x05f66a47; >>> static const unsigned bar[32] = >>> {0,1,2,26,23,3,15,27,24,21,19,4,12,16,28,6,31,25,2 2,14,20,18,11,5,30,13,17,10,29,9,8,7}; >>> n = ~n; >>> return bar[(a * (n & (-n))) >> 27]; >>> } >> >> Doesn't seem to work very well on a machine with 16 bit integers. > > Most machines have 16-bit integers; often the 16-bit integer type is > called "short". > > If you mean 16-bit ints, I don't see how that would cause a problem, > since the fucntion's argument is of type uint32_t. (The unsigned > result shouldn't be a problem unless you're worried about integers > with more than 32767 bits.) > > Or am I missing something? Yeah, I meant int, and was sloppy. To me, uint32_t doesn't exist, since it is not guaranteed. Lets make the complaint about a machine with an 18 bit int. -- [mail]: Chuck F (cbfalconer at maineline dot net) [page]: <http://cbfalconer.home.att.net> Try the download section. -- Posted via a free Usenet account from http://www.teranews.com |

Re: a bit of a puzzleCBFalconer wrote:
> "Steven G. Johnson" wrote: >> >> Here is a little algorithm I came across whose implementation is >> amusingly obscure: what simple function does the following C code >> compute, and why? >> >> #include <stdint.h> >> unsigned foo(uint32_t n) >> { >> const uint32_t a = 0x05f66a47; >> static const unsigned bar[32] = >> {0,1,2,26,23,3,15,27,24,21,19,4,12,16,28,6,31,25,2 2,14,20,18,11,5,30,13,17,10,29,9,8,7}; >> n = ~n; >> return bar[(a * (n & (-n))) >> 27]; >> } > > Doesn't seem to work very well on a machine with 16 bit integers. Mind to enlighten me why? Bye, Jojo |

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