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#!/usr/bin/env ruby | |
# coding: utf-8 | |
N=100 # infty | |
def gen | |
ar = N.times.map{ %w(U D)[rand(2)]}.join '' # infinite seq | |
memo = {} | |
for i in 0..(N-3) | |
sub = ar[i..i+2] | |
memo[sub] = i + 1 if memo[sub] == nil | |
break if memo.size == 8 | |
end | |
memo | |
end | |
M=20000 # trial times | |
count={} | |
M.times { | |
m = gen | |
for k in m.keys | |
count[k] = 0 if count[k] == nil | |
count[k] += m[k] | |
end | |
} | |
for k in count.keys | |
puts "#{k} #{count[k] / M.to_f}" | |
end |
考察.
部分文字列 w について、 w を部分文字列として含まない長さmの文字列の数をC(m; w) とする.
C(m; UUU) と C(m; DDD) は1,2,4 で始まるtri-bonacci
そうでないときは、微妙にアレンジされたtri-bonacci よくわからん?!
w=DUU
for i in {U,D}; do echo $i; done | grep -v $w | wc -l
for i in {U,D}{U,D}; do echo $i; done | grep -v $w | wc -l
for i in {U,D}{U,D}{U,D}; do echo $i; done | grep -v $w | wc -l
for i in {U,D}{U,D}{U,D}{U,D}; do echo $i; done | grep -v $w | wc -l
for i in {U,D}{U,D}{U,D}{U,D}{U,D}; do echo $i; done | grep -v $w | wc -l
for i in {U,D}{U,D}{U,D}{U,D}{U,D}{U,D}; do echo $i; done | grep -v $w | wc -l
for i in {U,D}{U,D}{U,D}{U,D}{U,D}{U,D}{U,D}; do echo $i; done | grep -v $w | wc -l
アルファベット {U,D} 上の文字列を考える.
無限長の文字列 s (=s[1],s[2],...) 中に有限長文字列 w が m番目に現れるとは、
- (s[m], s[m+1] .., s[m+|w|-1]) = w であってかつ、
- (s[1], s[2], .., s[m+|w|-2]) は部分文字列として w を持たないこと
例. s=DUUUDDDUDUDUDU 中に DUD は7番目に出現する (たぶん漫画とは数え方が微妙に異なる)
問題. 長さ3のwが何番目に出現するか、sが一様分布で与えられる下でその期待値を求めよ.
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Simulated Expected Time (exclusive)