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Re: simplifying ulam spiral code
*To*: mathgroup at smc.vnet.net
*Subject*: [mg56456] Re: simplifying ulam spiral code
*From*: "Carl K. Woll" <carlw at u.washington.edu>
*Date*: Tue, 26 Apr 2005 01:33:03 -0400 (EDT)
*Organization*: University of Washington
*References*: <d4cmlr$39e$1@smc.vnet.net> <d4hvsv$1jp$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
"Carl K. Woll" <carlw at u.washington.edu> wrote in message
news:d4hvsv$1jp$1 at smc.vnet.net...
> "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
>
>
I see now that I started with 1 at the origin, and others started with 0 at
the origin. Hence, a small modification of my code will return results that
agree with e.g., Daniel Lichtblau.
coords[k_List] := Module[{n, a, b, c},
n = Floor[Sqrt[k + .1]];
a = k - n(n + 1) ;
c = 2Mod[n, 2, -1] + 1;
b = Quotient[2n + 1 - c, 4];
c Transpose[{Quotient[Abs[a] + a, 2] - b, Quotient[Abs[a] - a, 2] - b}]]
Basically, I substituted k+1 for k. I also got rid of Power by using n(n+1)
to evaluate n^2+n and I by using Mod to find (-1)^n. These latter changes
increased the speed of coords by about 33%.
Carl Woll
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