第10回(フィボナッチ数列)

お題:Fibonacci数列

fib(n) = fib(n-1)+fib(n-2)
0 1 1 2 3 5 8 13 21 ...

をrecursion(再帰)で求めなさい．

解説

fib(0) = 0

Fibonacci数列の初項は0

def fib(n)
  if n==0
    return 0
  end
end

前回作成したassert_equal.rbで出力が期待値と一致しているかを確認.

require './assert_equal'
puts assert_equal(0, fib(0))

結果

expected :: 0
result   :: 0
succeeded in assert_equal

fib(1) =1

次はfib(1)=1.まずはassertion

puts assert_equal(1, fib(1))

もちろん失敗するので書き換える.

def fib(n)
  if n==0
    return 0
  end
  if n==1
    return 1
  end
end

結果

expected :: 0
result   :: 0
succeeded in assert_equal
expected :: 1
result   :: 1
succeeded in assert_equal

これでOK.

テスト側の重複が気になってきたので，配列に

[[0,0],[1,1]].each do |pair|
  puts assert_equal(pair[0], fib(pair[1]))
end

結果は同じ

fib(2) = fib(1) + fib(0) = 1

まずはテスト

[[0,0],[1,1],[2,1]].each do |pair|

これではダメ.

> ruby fibonacci.rb
expeced:0
result :0
succeeded in assert_equal

expeced:1
result :1
succeeded in assert_equal 

expeced:2
result :1
failed in assert_equal

条件分岐をもう少しいじって，

def fib(n)
  if n==0
    return 0
  end
  if n<=2
    return 1
  end
end

これでいけると思うので実行.

> ruby fibonacci.rb
expeced :: 0
result  :: 0
succeeded in assert_equal

expeced :: 1
result  :: 1
succeeded in assert_equal 

expeced :: 2
result  :: 1
failed in assert_equal

あれ?

実は，assert_equalは(expect, result)とうけとっているため配列変数pairの示数indexが逆.

require './assert_equal'
[[0,0],[1,1],[2,1]].each do |pair|
  puts assert_equal(pair[1], fib(pair[0]))
end

が正解．

refactoring

もう少し配列の受け取りを明示的にすると，

index, expected = pair

と修正できて

require './assert_equal'
[[0,0],[1,1],[2,1]].each do |index, expected|
  puts assert_equal(expected, fib(index))
end

こっちのほうが分かりやすい.

refactoring

メソッドが長くなってきたので短く書き直す.

def fib(n)
  return 0 if n==0
  return 1 if n<=2
end

と2行に削減

fib(3) = 2 = fib(2) + fib(1)

テストは

[[0,0],[1,1],[2,1],[3,2]].each do |index, expected|

return 2になってほしいのでfib(2)+fib(1)にする.

def fib(n)
  return 0  if n==0
  return 1  if n<=2
  return fib(2) + fib(1)
end

結果

> ruby fibonacci.rb
expeced :: 0
result  :: 0
succeeded in assert_equal

expeced :: 1
result  :: 1
succeeded in assert_equal 

expeced :: 1
result  :: 1
succeeded in assert_equal 

expeced :: 2
result  :: 2
succeeded in assert_equal

fib(4) = fib(3) + fib(2) = 2 + 1 = 3

次は4. 期待値は3

[[0,0],[1,1],[2,1],[3,2],[4,3]].each do |index, expected|

テスト結果は当然fail.最後のreturnを定義通りに

fib(n) = fib(n-1) + fib(n-2)

と修正

結果

> ruby fibonacci.rb
expeced :: 0
result  :: 0
succeeded in assert_equal

expeced :: 1
result  :: 1
succeeded in assert_equal 

expeced :: 1
result  :: 1
succeeded in assert_equal 

expeced :: 2
result  :: 2
succeeded in assert_equal 

expeced :: 3
result  :: 3
succeeded in assert_equal

全てsucceed

そのまま続きもやってみる.

[[0,0],[1,1],[2,1],[3,2],[4,3],
[5,5],[6,8],[7,13],[8,21],[9,34]
].each do |index, expected|
  puts assert_equal(expected,fib(index))
end

結果

expected :: 0
result   :: 0
succeeded in assert_equal

expected :: 1
result   :: 1
succeeded in assert_equal

expected :: 1
result   :: 1
succeeded in assert_equal

expected :: 2
result   :: 2
succeeded in assert_equal

expected :: 3
result   :: 3
succeeded in assert_equal

expected :: 5
result   :: 5
succeeded in assert_equal

expected :: 8
result   :: 8
succeeded in assert_equal

expected :: 13
result   :: 13
succeeded in assert_equal

expected :: 21
result   :: 21
succeeded in assert_equal

expected :: 34
result   :: 34
succeeded in assert_equal

全部いけた.

ただ一つ抜けているのは，

fib(2) = fib(1) + fib(0) = 1 + 0

なんで，fibのなかの条件分岐はもう少し狭められて，

return 1  if n == 1

で十分

最終的に

require './assert_equal.rb'

def fib(n)
  return 0 if n == 0
  return 1 if n == 1
  return fib(n-1)+fib(n-2)
end

[[0,0],[1,1],[2,1],[3,2],[4,3],
[5,5],[6,8],[7,13],[8,21],[9,34]
].each do |index, expected|
  puts assert_equal(expected,fib(index))
end

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• source ~/grad_members_20f/members/yoshida/c5_fibonacci.org