Confident Measurements - Artificial Intelligence for Robotics Udacity English Japanese

Confident Measurements - Artificial Intelligence for RoboticsYouTube↗ IA↗(long shot)
Education 5.3K views 7 likes 0 dislikes 31 May 2012

Udacity 651K subscribers United States YouTube↗

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I'm going to give you a glimpse as to why it works.

これがうまくいく理由を簡単に説明しましょう

Suppose we have two robot positions, x0 and x1, and we know they're 10 apart

ロボットの2つの位置X₀、X₁があり間の距離が10とします

with some Gaussian noise, and we know the Gaussian noise in exportation

ガウス分布に従うノイズがありその伝播が右のロボットの位置を

moves the right robot position 10 off the left robot position, but there is some uncertainty.

左のロボットより10だけ先に動かすのには若干の不確実性があります

When we talked about Kalman filters, we talked about Gaussians,

カルマンフィルタについてお話しした時にガウス分布についても触れました

and this uncertainty might look at follows: There is a constant exponential, and the expression that x1 minus x0

不確実性は次のような式で表せます定数項、指数項、 X₁からX₀を引いて10へ緩和されますが

should relax to 10 but might deviate from it.

逸脱してしまうこともあります

This Gaussian constraint over here characterizes the constraint between x1 and x0

このガウス分布の制約は X₁とX₀の間の制約を特徴づけて

and wishes them to be exactly 10 apart.

正確に10離れるようにしています

The Gaussian is maximum where this equation is fulfilled,

この等式が満たされればガウス分布は最大になりますが

but if the residual is not equal to 0, there is still a probability associated with it.

残差がゼロではなければ確率が関係してきます

Let's now model a second motion. Say x2 is 5 apart.

2つ目の動きのモデルを作りましょう X₂の位置で5離れます

We now get an even bigger Gaussian relative to the very first one,

最初のガウス分布に従うノイズよりも大きくなりますが

but the local constraint over here reads just like the constraint over there.

X₁とX₂の間の制約はX₀とX₁の間のものと同じです

Let me just write it down.

式を書きます

X2 minus x1 minus 5 squared over sigma-squared.

（X₂－X₁－5）²／σ²です

Now, the total probability of this entire thing over here is the product of these two things.

この全確率は2つの式の積で求めることができます

If we want to maximize the product, we can play a number of interesting tricks.

積を最大にしたければいくつかの面白い技が使えます

First, the constant has no bearing on the maximum, just on the absolute value.

まず定数は最大値には何の影響もなく絶対的な値です

If we want to find the best values for x0 and x1 and so on, we can drop the constant.

X₀やX₁について最適な値を求めるときは定数を取ってしまいます

Secondly, we can drop the exponential

積を和に変えたい場合には

if we're willing to turn the product into an addition.

今度は指数を取ってしまいます

Remember, we added things in omega and in sigma. That's why.

Ωやσを加算できたのはこういった理由からです

Finally, we can actually drop the -1/2.

最後に－1／2を取ります

It turns out that also plays no role in the maximization of this expression.

この式の最大値には何の影響も与えません

It turns out what you added where constraints just like these,

これがこのような制約における加算です

and you even added them at a certain strength of 1 over sigma-squared.

1／σ²の強度でこれらを加算しました

In particular, if you really believe that a constraint is true,

特に制約が真だと信じるなら

you should add a larger value in this matrix over here,

行列の中により大きな値を足します

and on the right side you should multiply the right constraint with an even larger value.

そして右側の行列により大きな値の正しい制約を掛けます

Put differently, take an expression like this and multiply in the sigma-squared

つまり上の式から2乗を取るとこのような式が得られ

you get something of this nature over here where 1 over sigma regulates how confident you are.

1／σが信頼度を規定します

For a small sigma, 1 over sigma becomes large.

σが小さければ1／σは大きな値になるので

So 5 is much larger than 1.

5は1よりもかなり大きくなります

That means you have much more confidence.

これは信頼度が大きいということです

Let's go back to the code and modify the code

ではプログラムに戻って修正を加えましょう

so the last measurement has a really high confidence.

最後の観測の信頼度は高いものです

I want you to multiply the last measurement between x2 and our landmark with a factor of 5 in your code. Hard code it.

X₂とランドマークの間の最後の観測を5で掛けてプログラムの中に組み込んでみてください

That is, go in somewhere over here where the last measurement is being applied

最後の観測が適用されるのはこの辺りです

and do the trick that I just showed you and see what the outcome is.

ここで教えた技を使って結果を確認してください

When I do this, I get -3, 2.1, 5.714, and 6.821 as the answers.

私が実行してみると結果は－3、2．1、5．714、6．821です

You'll see in this final result

この最終結果から

the final robot position of 5.714 and the landmark position of 8.821

ロボットの最終的な位置は5．714でランドマークは6．821です

are really close to 1 in difference, which was the measurement,

観測の結果差が1に近くなったのは

because you know believe this measurement over-proportionally over other measurements and motions.

他の観測と動作に比べてこの観測値の重みが飛び抜けて大きいからです

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