August 2020
Beginner to intermediate
332 pages
8h 16m
Japanese
book
動かして学ぶ量子コンピュータプログラミング ―シミュレータとサンプルコードで理解する基本アルゴリズム
by Eric R. Johnston, Nic Harrigan, Mercedes Gimeno-Segovia, 北野 章, 丸山 耕司
Content preview from 動かして学ぶ量子コンピュータプログラミング ―シミュレータとサンプルコードで理解する基本アルゴリズム
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6.5
振幅増幅を使う
119
レジスタの値の半分が有標値である場合、振幅増幅の演算を何回行っても構いませんが、乱数を得
るようなことしかできないということになります。
対称性のために、レジスタ値の
50 %
以上が位相反転した場合に何が起こるかを考慮する必
要はありません。例えば、
16
個の値のうち
12
個が有標値である場合、成功しているかどう
かを読み取ろうとするのは、そのうちの
4
個の値を有標値として観察することと同じです。
単に成功と失敗を入れ替えて考えればよいということです。
興味深いことに、これまでの例から、成功確率の振動周波数は反転した値の数にのみ依存し、反転
される値には依存しません。式6-1を拡張すると、有標値が複数個ある場合にも成立するようにした式
6-2が得られます。ここで、
n
はキュビットの数、
m
は対応する有標値の数です。
式6-2
N
AA
=
π
2
n
4
m
m
がわかれば、式 6-2を使用して、高い成功確率を示す振幅増幅の回数が求められます。これは興
味深い問題を提起します。いくつの状態が反転されたかわからない場合、成功確率が最大になる振幅
増幅の回数をどのように求められるか、という問題です。
10
章の応用に振幅増幅を用いる際に、この
問題を再び提起し、他のプリミティブがそれにどのように役立つかを検討することにします。
6.5
振幅増幅を使う
これまでの説明から、「振幅増幅で何ができそうか？」がわかってきたことでしょう。読み出すことの
できない位相を読み出すことのできる振幅
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I wanted to learn C and C++, but it didn't click for me until I picked up an O'Reilly book. When I went on the O’Reilly platform, I was astonished to find all the books there, plus live events and sandboxes so you could play around with the technology.Addison B.
I’ve been on the O’Reilly platform for more than eight years. I use a couple of learning platforms, but I'm on O'Reilly more than anybody else. When you're there, you start learning. I'm never disappointed.Amir M.
I'm always learning. So when I got on to O'Reilly, I was like a kid in a candy store. There are playlists. There are answers. There's on-demand training. It's worth its weight in gold, in terms of what it allows me to do.