Padovan sequence: Difference between revisions
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end.</syntaxhighlight> |
end.</syntaxhighlight> |
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=={{header|EasyLang}}== |
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{{trans|Swift}} |
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<syntaxhighlight> |
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p[] = [ 1 1 1 ] |
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padorn = 1 |
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func padorec . |
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if padorn <= 3 |
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h = p[padorn] |
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else |
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h = p[1] + p[2] |
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. |
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p[1] = p[2] |
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p[2] = p[3] |
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p[3] = h |
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padorn += 1 |
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return h |
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. |
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pfn = 1 |
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P = 1.32471795724474602596 |
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S = 1.0453567932525329623 |
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# |
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func padofloor . |
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p = floor ((pow P (pfn - 2)) / S + 0.5) |
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pfn += 1 |
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return p |
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. |
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str$ = "A" |
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func$ padolsys . |
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r$ = str$ |
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for c$ in strchars str$ |
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if c$ = "A" |
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nxt$ &= "B" |
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elif c$ = "B" |
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nxt$ &= "C" |
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else |
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nxt$ &= "AB" |
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. |
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. |
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str$ = nxt$ |
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return r$ |
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. |
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# |
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print "First 20 terms of the Padovan sequence:" |
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for i to 64 |
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par[] &= padorec |
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. |
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for i to 20 |
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write par[i] & " " |
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. |
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print "" |
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for i to 64 |
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paf[] &= padofloor |
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. |
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if par[] = paf[] |
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print "\nRecurrence and floor functions agree for first 64 terms" |
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. |
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for i to 32 |
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pal$[] &= padolsys |
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. |
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print "\nFirst 10 strings produced from the L-system:" |
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for i to 10 |
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write pal$[i] & " " |
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. |
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print "" |
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for i to 32 |
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if len pal$[i] <> par[i] |
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break 1 |
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. |
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. |
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if i > 32 |
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print "\nLength of first 32 strings produced from the L-system = Padovan sequence" |
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. |
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</syntaxhighlight> |
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{{out}} |
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<pre> |
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First 20 terms of the Padovan sequence: |
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1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 |
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Recurrence and floor functions agree for first 64 terms |
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First 10 strings produced from the L-system: |
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A B C AB BC CAB ABBC BCCAB CABABBC ABBCBCCAB |
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Length of first 32 strings produced from the L-system = Padovan sequence |
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</pre> |
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=={{header|Factor}}== |
=={{header|Factor}}== |