Sturmian word: Difference between revisions
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imported>CosmiaNebula No edit summary |
imported>CosmiaNebula (Continued fraction convergents) |
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* Given a positive rational number <math>\frac mn</math>, specified by two positive integers <math>m, n</math>, output its entire Sturmian word. |
* Given a positive rational number <math>\frac mn</math>, specified by two positive integers <math>m, n</math>, output its entire Sturmian word. |
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* Given a quadratic real number <math>\frac{\sqrt{a} + m}{n}</math>, specified by |
* Given a quadratic real number <math>\frac{b\sqrt{a} + m}{n} > 0</math>, specified by integers <math>a, b, m, n </math>, where <math>a</math> is not a perfect square, output the first <math>k</math> letters of Sturmian words when given a positive number <math>k</math>. |
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(If the programming language can represent infinite data structures, then that works too.) |
(If the programming language can represent infinite data structures, then that works too.) |
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The key difficulty is accurately calculating <math>floor(k\sqrt a) </math> for large <math>k</math>. Floating point arithmetic would lose precision. One can either do this simply by directly searching for some integer <math>a'</math> such that <math>a'^2 \leq k^2a < (a'+1)^2</math>, or by more trickly methods, such as the continued fraction approach. |
The key difficulty is accurately calculating <math>floor(k\sqrt a) </math> for large <math>k</math>. Floating point arithmetic would lose precision. One can either do this simply by directly searching for some integer <math>a'</math> such that <math>a'^2 \leq k^2a < (a'+1)^2</math>, or by more trickly methods, such as the continued fraction approach. |
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First calculate the continued fraction |
First calculate the [[continued fraction convergents]] to <math>\sqrt a</math>. Let <math>\frac mn </math> be a convergent to <math>\sqrt a</math>, such that <math>n \geq k</math>, then since the convergent sequence is the '''best rational approximant''' for denominators up to that point, we know for sure that, if we write out <math>\frac{0}{k}, \frac{1}{k}, \dots</math>, the sequence would stride right across the gap <math>(m/n, 2x - m/n)</math>. Thus, we can take the largest <math>l</math> such that <math>l/k \leq m/n</math>, and we would know for sure that <math>l = floor(k\sqrt a)</math>. |
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In summary,<math display="block">floor(k\sqrt a) = floor(mk/n)</math>where <math>m/n</math> is the first continued fraction approximant to <math>\sqrt a</math> with a denominator <math>n \geq k</math> |
In summary,<math display="block">floor(k\sqrt a) = floor(mk/n)</math>where <math>m/n</math> is the first continued fraction approximant to <math>\sqrt a</math> with a denominator <math>n \geq k</math> |
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where <math>m/n</math> is the first continued fraction approximant to <math>\sqrt a</math> with a denominator <math>n \geq k</math> |