Fibonacci word: Difference between revisions
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=={{header|Delphi}}==
See [[#Pascal]]
=={{header|EasyLang}}==
{{trans|Nim}}
<syntaxhighlight>
func log2 x .
return log10 x / log10 2
.
func entropy s$ .
l = len s$
if l <= 1
return 0
.
for v$ in strchars s$
cnt0 += if v$ = "0"
.
cnt1 = l - cnt0
return -(cnt0 / l * log2 (cnt0 / l) + cnt1 / l * log2 (cnt1 / l))
.
a$ = ""
b$ = ""
func$ fibword .
if a$ = ""
a$ = "1"
return a$
.
if b$ = ""
b$ = "0"
return b$
.
a$ = b$ & a$
swap a$ b$
return b$
.
numfmt 6 8
print " n length entropy"
print " ——————————————————————"
for n to 37
s$ = fibword
print n & " " & len s$ & " " & entropy s$
.
</syntaxhighlight>
=={{header|EchoLisp}}==
Line 4,852 ⟶ 4,893:
36 14930352 0.959418728222743 <too long>
37 24157817 0.959418728222745 <too long>
</pre>
=={{header|Uiua}}==
Memoisation saves the day :-)
<syntaxhighlight lang="Uiua">
# Build the string recursively.
F ← |1 memo⟨⟨⊂∩F-1.-1|"0"◌⟩=2.|"1"◌⟩=1.
# General entropy formula - quite slow for this task.
Egen ← /+(¯×ₙ2.)÷/+.≡(⧻⊚=)⊃◴¤
# Specific entropy formula for a binary string.
E ← ⍥(0◌)=NaN.+∩(¯×ₙ2.)⟜(¯-1)÷⊃⧻(⧻⊚="1")
# Much faster approach -- don't even build the string, just count
# how many "0"s and "1"s the string will have.
Fx ← |1 memo⟨⟨+∩Fx-1.-1|[1 0]◌⟩=2.|[0 1]◌⟩=1.
Ex ← ⍥(0◌)=NaN./+(¯×ₙ2.)÷/+.
# Print and time it
⍜now(≡(⇌[⊃/+ (⍜(×1e8)⁅Ex)Fx.])+1⇡37)
</syntaxhighlight>
{{out}}
<pre>
╭─
╷ 1 0 1
2 0 1
3 1 2
4 0.91829583 3
5 0.97095059 5
6 0.954434 8
7 0.9612366 13
8 0.95871188 21
9 0.95968689 34
10 0.95931603 55
...etc...
35 0.95941873 9227465
36 0.95941873 14930352
37 0.95941873 24157817
╯
0.0029999999678694-ε
</pre>
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