Sturmian word: Difference between revisions
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In summary, <math>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>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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=={{header|BASIC}}== |
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==={{header|BASIC256}}=== |
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{{trans|FreeBASIC}} |
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<syntaxhighlight lang="vbnet">fib = fibWord(10) |
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sturmian = sturmian_word(13, 21) |
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if left(fib, length(sturmian)) = sturmian then print sturmian |
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end |
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function invert(cadena) |
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for i = 1 to length(cadena) |
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b = mid(cadena, i, 1) |
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if b = "0" then |
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inverted += "1" |
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else |
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if b = "1" then |
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inverted += "0" |
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end if |
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end if |
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next |
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return inverted |
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end function |
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function sturmian_word(m, n) |
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sturmian = "" |
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if m > n then return invert(sturmian_word(n, m)) |
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k = 1 |
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while True |
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current_floor = int(k * m / n) |
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previous_floor = int((k - 1) * m / n) |
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if k * m mod n = 0 then exit while |
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if previous_floor = current_floor then |
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sturmian += "0" |
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else |
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sturmian += "10" |
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end if |
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k += 1 |
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end while |
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return sturmian |
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end function |
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function fibWord(n) |
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Sn_1 = "0" |
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Sn = "01" |
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for i = 2 to n |
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tmp = Sn |
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Sn += Sn_1 |
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Sn_1 = tmp |
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next |
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return Sn |
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end function</syntaxhighlight> |
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{{out}} |
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<pre>01001010010010100101001001010010</pre> |
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==={{header|FreeBASIC}}=== |
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{{trans|Python}} |
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<syntaxhighlight lang="vbnet">Function invert(cadena As String) As String |
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Dim As String inverted, b |
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For i As Integer = 1 To Len(cadena) |
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b = Mid(cadena, i, 1) |
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inverted += Iif(b = "0", "1", "0") |
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Next |
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Return inverted |
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End Function |
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Function sturmian_word(m As Integer, n As Integer) As String |
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Dim sturmian As String |
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If m > n Then Return invert(sturmian_word(n, m)) |
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Dim As Integer k = 1 |
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Do |
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Dim As Integer current_floor = Int(k * m / n) |
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Dim As Integer previous_floor = Int((k - 1) * m / n) |
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If k * m Mod n = 0 Then Exit Do |
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sturmian += Iif(previous_floor = current_floor, "0", "10") |
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k += 1 |
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Loop |
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Return sturmian |
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End Function |
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Function fibWord(n As Integer) As String |
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Dim As String Sn_1 = "0" |
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Dim As String Sn = "01" |
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For i As Integer = 2 To n |
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Dim As String tmp = Sn |
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Sn += Sn_1 |
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Sn_1 = tmp |
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Next |
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Return Sn |
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End Function |
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Dim As String fib = fibWord(10) |
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Dim As String sturmian = sturmian_word(13, 21) |
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If Left(fib, Len(sturmian)) = sturmian Then Print sturmian |
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Sleep</syntaxhighlight> |
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{{out}} |
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<pre>01001010010010100101001001010010</pre> |
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==={{header|Yabasic}}=== |
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{{trans|FreeBASIC}} |
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<syntaxhighlight lang="vbnet">fib$ = fibword$(10) |
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sturmian$ = sturmian_word$(13,21) |
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if left$(fib$,len(sturmian$)) = sturmian$ print sturmian$ |
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end |
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sub invert$(cadena$) |
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for i = 1 to len(cadena$) |
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b$ = mid$(cadena$,i,1) |
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if b$ = "0" then |
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inverted$ = inverted$ +"1" |
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elsif b$ = "1" then |
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inverted$ = inverted$ +"0" |
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fi |
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next i |
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return inverted$ |
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end sub |
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sub sturmian_word$(m,n) |
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sturmian$ = "" |
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if m > n return invert$(sturmian_word(n,m)) |
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k = 1 |
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while true |
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current_floor = int(k*m/n) |
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previous_floor = int((k-1)*m/n) |
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if mod(k*m, n) = 0 break |
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if previous_floor = current_floor then |
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sturmian$ = sturmian$ +"0" |
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else |
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sturmian$ = sturmian$ +"10" |
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fi |
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k = k +1 |
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end while |
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return sturmian$ |
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end sub |
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sub fibword$(n) |
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sn_1$ = "0" |
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sn$ = "01" |
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for i = 2 to n |
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tmp$ = sn$ |
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sn$ = sn$ + sn_1$ |
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sn_1$ = tmp$ |
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next i |
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return sn$ |
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end sub</syntaxhighlight> |
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{{out}} |
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<pre>01001010010010100101001001010010</pre> |
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=={{header|Julia}}== |
=={{header|Julia}}== |
Revision as of 00:26, 22 February 2024
A Sturmian word is a binary sequence, finite or infinite, that makes up the cutting sequence for a positive real number x, as shown in the picture.
The Sturmian word can be computed thus as an algorithm:
- If , then it is the inverse of the Sturmian word for . So we have reduced to the case of .
- Iterate over
- If is an integer, then the program terminates. Else, if , then the program outputs 0, else, it outputs 10.
The problem:
- Given a positive rational number , specified by two positive integers , output its entire Sturmian word.
- Given a quadratic real number , specified by integers , where is not a perfect square, output the first letters of Sturmian words when given a positive number .
(If the programming language can represent infinite data structures, then that works too.)
A simple check is to do this for the inverse golden ratio , that is, , which would just output the Fibonacci word.
Stretch goal: calculate the Sturmian word for other kinds of definable real numbers, such as cubic roots.
The key difficulty is accurately calculating for large . Floating point arithmetic would lose precision. One can either do this simply by directly searching for some integer such that , or by more trickly methods, such as the continued fraction approach.
First calculate the continued fraction convergents to . Let be a convergent to , such that , 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 , the sequence would stride right across the gap . Thus, we can take the largest such that , and we would know for sure that .
In summary, where is the first continued fraction approximant to with a denominator
BASIC
BASIC256
fib = fibWord(10)
sturmian = sturmian_word(13, 21)
if left(fib, length(sturmian)) = sturmian then print sturmian
end
function invert(cadena)
for i = 1 to length(cadena)
b = mid(cadena, i, 1)
if b = "0" then
inverted += "1"
else
if b = "1" then
inverted += "0"
end if
end if
next
return inverted
end function
function sturmian_word(m, n)
sturmian = ""
if m > n then return invert(sturmian_word(n, m))
k = 1
while True
current_floor = int(k * m / n)
previous_floor = int((k - 1) * m / n)
if k * m mod n = 0 then exit while
if previous_floor = current_floor then
sturmian += "0"
else
sturmian += "10"
end if
k += 1
end while
return sturmian
end function
function fibWord(n)
Sn_1 = "0"
Sn = "01"
for i = 2 to n
tmp = Sn
Sn += Sn_1
Sn_1 = tmp
next
return Sn
end function
- Output:
01001010010010100101001001010010
FreeBASIC
Function invert(cadena As String) As String
Dim As String inverted, b
For i As Integer = 1 To Len(cadena)
b = Mid(cadena, i, 1)
inverted += Iif(b = "0", "1", "0")
Next
Return inverted
End Function
Function sturmian_word(m As Integer, n As Integer) As String
Dim sturmian As String
If m > n Then Return invert(sturmian_word(n, m))
Dim As Integer k = 1
Do
Dim As Integer current_floor = Int(k * m / n)
Dim As Integer previous_floor = Int((k - 1) * m / n)
If k * m Mod n = 0 Then Exit Do
sturmian += Iif(previous_floor = current_floor, "0", "10")
k += 1
Loop
Return sturmian
End Function
Function fibWord(n As Integer) As String
Dim As String Sn_1 = "0"
Dim As String Sn = "01"
For i As Integer = 2 To n
Dim As String tmp = Sn
Sn += Sn_1
Sn_1 = tmp
Next
Return Sn
End Function
Dim As String fib = fibWord(10)
Dim As String sturmian = sturmian_word(13, 21)
If Left(fib, Len(sturmian)) = sturmian Then Print sturmian
Sleep
- Output:
01001010010010100101001001010010
Yabasic
fib$ = fibword$(10)
sturmian$ = sturmian_word$(13,21)
if left$(fib$,len(sturmian$)) = sturmian$ print sturmian$
end
sub invert$(cadena$)
for i = 1 to len(cadena$)
b$ = mid$(cadena$,i,1)
if b$ = "0" then
inverted$ = inverted$ +"1"
elsif b$ = "1" then
inverted$ = inverted$ +"0"
fi
next i
return inverted$
end sub
sub sturmian_word$(m,n)
sturmian$ = ""
if m > n return invert$(sturmian_word(n,m))
k = 1
while true
current_floor = int(k*m/n)
previous_floor = int((k-1)*m/n)
if mod(k*m, n) = 0 break
if previous_floor = current_floor then
sturmian$ = sturmian$ +"0"
else
sturmian$ = sturmian$ +"10"
fi
k = k +1
end while
return sturmian$
end sub
sub fibword$(n)
sn_1$ = "0"
sn$ = "01"
for i = 2 to n
tmp$ = sn$
sn$ = sn$ + sn_1$
sn_1$ = tmp$
next i
return sn$
end sub
- Output:
01001010010010100101001001010010
Julia
function sturmian_word(m, n)
m > n && return replace(sturmian_word(n, m), '0' => '1', '1' => '0')
res = ""
k, prev = 1, 0
while rem(k * m, n) > 0
curr = (k * m) ÷ n
res *= prev == curr ? "0" : "10"
prev = curr
k += 1
end
return res
end
function fibWord(n)
Sn_1, Sn, tmp = "0", "01", ""
for _ in 2:n
tmp = Sn
Sn *= Sn_1
Sn_1 = tmp
end
return Sn
end
const fib = fibWord(7)
const sturmian = sturmian_word(13, 21)
@assert fib[begin:length(sturmian)] == sturmian
println(" $sturmian <== 13/21")
""" return the kth convergent """
function cfck(a, b, m, n, k)
p = [0, 1]
q = [1, 0]
r = (sqrt(a) * b + m) / n
for _ in 1:k
whole = Int(trunc(r))
pn = whole * p[end] + p[end-1]
qn = whole * q[end] + q[end-1]
push!(p, pn)
push!(q, qn)
r = 1/(r - whole)
end
return [p[end], q[end]]
end
println(" $(sturmian_word(cfck(5, 1, -1, 2, 8)...)) <== 1/phi (8th convergent golden ratio)")
- Output:
01001010010010100101001001010010 <== 13/21 01001010010010100101001001010010 <== 1/phi (8th convergent golden ratio)
Phix
Rational part translated from Python, quadratic part a modified copy of the Continued fraction convergents task.
with javascript_semantics
function sturmian_word(integer m, n)
if m > n then
return sq_sub('0'+'1',sturmian_word(n,m))
end if
string res = ""
integer k = 1, prev = 0
while remainder(k*m,n) do
integer curr = floor(k*m/n)
res &= iff(prev=curr?"0":"10")
prev = curr
k += 1
end while
return res
end function
function fibWord(integer n)
string Sn_1 = "0",
Sn = "01",
tmp = ""
for i=2 to n do
tmp = Sn
Sn &= Sn_1
Sn_1 = tmp
end for
return Sn
end function
string fib = fibWord(7),
sturmian = sturmian_word(13, 21)
assert(fib[1..length(sturmian)] == sturmian)
printf(1," %s <== 13/21\n",sturmian)
function cfck(integer a, b, m, n, k)
-- return the kth convergent
sequence p = {0, 1},
q = {1, 0}
atom rem = (sqrt(a)*b + m) / n
for i=1 to k do
integer whole = trunc(rem),
pn = whole * p[-1] + p[-2],
qn = whole * q[-1] + q[-2]
p &= pn
q &= qn
rem = 1/(rem-whole)
end for
return {p[$],q[$]}
end function
integer {m,n} = cfck(5, 1, -1, 2, 8) -- (inverse golden ratio)
printf(1," %s <== 1/phi (8th convergent)\n",sturmian_word(m,n))
- Output:
01001010010010100101001001010010 <== 13/21 01001010010010100101001001010010 <== 1/phi (8th convergent)
Python
For rational numbers:
def sturmian_word(m, n):
sturmian = ""
def invert(string):
return ''.join(list(map(lambda b: {"0":"1", "1":"0"}[b], string)))
if m > n:
return invert(sturmian_word(n, m))
k = 1
while True:
current_floor = int(k * m / n)
previous_floor = int((k - 1) * m / n)
if k * m % n == 0:
break
if previous_floor == current_floor:
sturmian += "0"
else:
sturmian += "10"
k += 1
return sturmian
Checking that it works on the finite Fibonacci word:
def fibWord(n):
Sn_1 = "0"
Sn = "01"
tmp = ""
for i in range(2, n + 1):
tmp = Sn
Sn += Sn_1
Sn_1 = tmp
return Sn
fib = fibWord(10)
sturmian = sturmian_word(13, 21)
assert fib[:len(sturmian)] == sturmian
print(sturmian)
# Output:
# 01001010010010100101001001010010
Raku
# 20240215 Raku programming solution
sub chenfoxlyndonfactorization(Str $s) {
sub sturmian-word($m, $n) {
return sturmian-word($n, $m).trans('0' => '1', '1' => '0') if $m > $n;
my ($res, $k, $prev) = '', 1, 0;
while ($k * $m) % $n > 0 {
my $curr = ($k * $m) div $n;
$res ~= $prev == $curr ?? '0' !! '10';
$prev = $curr;
$k++;
}
return $res;
}
sub fib-word($n) {
my ($Sn_1, $Sn) = '0', '01';
for 2..$n { ($Sn, $Sn_1) = ($Sn~$Sn_1, $Sn) }
return $Sn;
}
my $fib = fib-word(7);
my $sturmian = sturmian-word(13, 21);
say "{$sturmian} <== 13/21" if $fib.substr(0, $sturmian.chars) eq $sturmian;
sub cfck($a, $b, $m, $n, $k) {
my (@p, @q) := [0, 1], [1, 0];
my $r = (sqrt($a) * $b + $m) / $n;
for ^$k {
my $whole = $r.Int;
my ($pn, $qn) = $whole * @p[*-1] + @p[*-2], $whole * @q[*-1] + @q[*-2];
@p.push($pn);
@q.push($qn);
$r = 1/($r - $whole);
}
return [@p[*-1], @q[*-1]];
}
my $cfck-result = cfck(5, 1, -1, 2, 8);
say "{sturmian-word($cfck-result[0], $cfck-result[1])} <== 1/phi (8th convergent golden ratio)";
- Output:
Same as Julia example.
You may Attempt This Online!
Wren
The 'rational number' function is a translation of the Python entry.
The 'quadratic number' function is a modified version of the one used in the Continued fraction convergents task.
import "./assert" for Assert
var SturmianWordRat = Fn.new { |m, n|
if (m > n) return SturmianWordRat.call(n, m).reduce("") { |acc, c|
return acc + (c == "0" ? "1" : "0")
}
var sturmian = ""
var k = 1
while (k * m % n != 0) {
var currFloor = (k * m / n).floor
var prevFloor = ((k - 1) * m / n).floor
sturmian = sturmian + (prevFloor == currFloor ? "0" : "10")
k = k + 1
}
return sturmian
}
var SturmianWordQuad = Fn.new { |a, b, m, n, k|
var p = [0, 1]
var q = [1, 0]
var rem = (a.sqrt * b + m) / n
for (i in 1..k) {
var whole = rem.truncate
var frac = rem.fraction
var pn = whole * p[-1] + p[-2]
var qn = whole * q[-1] + q[-2]
p.add(pn)
q.add(qn)
rem = 1 / frac
}
return SturmianWordRat.call(p[-1], q[-1])
}
var fibWord = Fn.new { |n|
var sn1 = "0"
var sn = "01"
for (i in 2..n) {
var tmp = sn
sn = sn + sn1
sn1 = tmp
}
return sn
}
var fib = fibWord.call(10)
var sturmian = SturmianWordRat.call(13, 21)
Assert.equal(fib[0...sturmian.count], sturmian)
System.print("%(sturmian) from rational number 13/21")
var sturmian2 = SturmianWordQuad.call(5, 1, -1, 2, 8)
Assert.equal(sturmian, sturmian2)
System.print("%(sturmian2) from quadratic number (√5 - 1)/2 (k = 8)")
- Output:
01001010010010100101001001010010 from rational number 13/21 01001010010010100101001001010010 from quadratic number (√5 - 1)/2 (k = 8)