Re: simplifying ulam spiral code

*To*: mathgroup at smc.vnet.net*Subject*: [mg56435] Re: simplifying ulam spiral code*From*: "Carl K. Woll" <carlw at u.washington.edu>*Date*: Mon, 25 Apr 2005 01:30:51 -0400 (EDT)*Organization*: University of Washington*References*: <d4cmlr$39e$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

"zak" <chocolatez at gmail.com> wrote in message news:d4cmlr$39e$1 at smc.vnet.net... > hi > in the site: > http://teachers.crescentschool.org/weissworld/m3/spiral/spiral.html > there is a mathematica code for drawing ULAM'S SPIRAL > the code is: > [snip] I have to confess that I don't understand how zak's code relates to the above link. The text at the link says that 1 is placed at the origin, 2 is placed to the right of 1, and succeeding integers are placed in a counterclockwise spiral. Hence, 3 ought to have the coordinates {1,1}, 4 ought to have the coordinates {0,1}, etc. At any rate, it is not too difficult to program a function to determine the coordinates of an integer in the above spiral. If one notices that the bottom right and top left corners have the integer values n^2+1, then one eventually gets coords[k_Integer] := Module[{n, a, b}, n = Floor[Sqrt[k - .9]]; a = k - n^2 - n - 1; b = Quotient[2n + 1 - (-1)^n, 4]; (-1)^n {Quotient[Abs[a] + a, 2] - b, Quotient[Abs[a] - a, 2] - b}] coords[k_List] := Module[{n, a, b, c}, n = Floor[Sqrt[k - .9]]; a = k - n^2 - n - 1; c = (-1)^n; b = Quotient[2n + 1 - c, 4]; c Transpose[{Quotient[Abs[a] + a, 2] - b, Quotient[Abs[a] - a, 2] - b}]] A few comments may be in order. Concentrating on the second function definition, I used a number of ideas to speed up it's execution. I used Sqrt[k-.9] so that Mathematica is taking square roots of real numbers instead of integers, which perhaps surprisingly is much faster. I used Sqrt[k-.9] instead of Sqrt[k-1.] to avoid spurious cancellation errors when the Floor of the result is evaluated. I wanted to make sure that all my arrays were packed, so I used Quotient instead of dividing two integers. Even though 2n+1-(-1)^n is always divisible by 4, (2n+1-(-1)^n)/4 is not packed even when 2n+1-(-1)^n is packed. Finally, I used (Abs[a]+a)/2 (with Quotient instead of using division) to change all negative values in the list a to 0 and (Abs[a]-a)/2 to change all positive values in the list a to 0. At any rate, using coords on a list of the first million integers takes a bit less than 2 seconds on my machine. Now, we are ready to used coords to find the coordinates of the primes. For example, if we are interested in the first million primes: data = Prime[Range[10^6]]; // Timing {9.547 Second, Null} Now, we use coords to get the coordinates of the primes. pts = coords[data]; // Timing {2.375 Second, Null} Applying Developer`ToPackedArray to the data would speed up the pts computation by a bit. Looking at the first few points of pts reveals Take[pts, 10] {{1, 0}, {1, 1}, {-1, 1}, {-1, -1}, {2, 0}, {2, 2}, {-2, 2}, {-2, 0}, {0, -2}, {3, 1}} which looks correct to me. Carl Woll

**Follow-Ups**:**Re: Re: simplifying ulam spiral code***From:*DrBob <drbob@bigfoot.com>