数独で数えて

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カラーブラインドバージョン:

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Here’s a sudoku puzzle with a twist!

四角で示された数字は、実際にはその地域の数字の合計です。例:18 = 4つの黄色い正方形の数字の合計、または12 =
2つの緑色の正方形の数字の合計など。

ルールは以下のとおりです:

  1. 行または列に何度も数字を入力する必要はありません。
  2. 各3×3ボックスには、1〜9の各桁が正確に1回必要です。
  3. 地域内の数字は、指定された数字になるはずです。
  4. 地域内で繰り返される番号はありません。例:10 = 6 + 4、5 + 5ではなく。

私はルールが明確だと信じています。
それでは、それを解決してください。受け入れる答えは、ほとんどのステップをはっきりと示しているはずです(少なくとも最初のいくつかのステップは、少し難しいためです)。

ベストアンサー

ソリューション:

273|964|581
189|357|264
564|821|793
---+---+---
418|632|957
952|718|436
736|495|128
---+---+---
891|273|645
647|589|312
325|146|879

より読みやすくするための@Ianありがとうございます。私は急いで静かだった…

パズルで始めるように。このパズルを開始するには、いくつかのルールに従う必要があります。

– 1行、3列、3行3列の各ボックスは1、2、.. 9 = 45ポイントの価値があります。

これは、すでにいくつかの数字を記入できることを意味します。 ボックスの左下のパズルを開始すると、次のことが分かります。

The 17 value consists of 9 and 8. The bottom 5 and 6 pairs must
be 1, 2, 3 and 5 and the 16 above the 17 consists of a 9 and 7.
When following the rule I just described, we know there is an
amount of points missing in the bottom left box: 45 – 17 – 17 – 5 =
6. With the group of 5 having a 2 and 3, we know the remaining 6
points needs to be 1 and 5. Therefore, we put the 1 and 5 in both
the group of 8 and the group of 6. We can now see the
following:
First step

ステップ2。

Let’s continue with the 9 above the 16. As said, each box has 45
points, and we do know one number is left. 45 – 19 – 8 – 8 = 4. So,
we can fill in both 4 and 5. We now know the 12 in the same box as
well. 45 – 18 – 19 – 5 = 3.
Top center box contains a 3, that has to contain of a 1 and 2. We
can now already assume what numbers are in the different boxes.
Step 2

ステップ3。

Continuing on the top right box. The 23 has to be a 6, 8 and 9.
We can also fill in the 5 and 2, as the 5 can only be in one place.
Now the 2 is taken, we know the group of 5 consists of a 1 and 4.
The 3 and 7 are left and placed below the 2 and 4. With this
information we can fill in most of the top left box as well.
Starting with the 7 (middle top) and 2 (same row). We can now fill
in the box of 18. 18 – 2 – 7 leaves 9. The only possibility of this
9 is a 1 and 8.
Step 3

ステップ4。

We continue with the middle left box. The 8 can only be placed
within the group consisting of 9. We fill in both 8 and 1 here. Now
the 3 and 5 has to go to the left of this group of 8, while the 2
and 6 are placed right. This allows us to complete the top left box
as well.
I checked the middle box as well and I found out I can fill in one
more free number. 45 – 16 – 11 – 13 = 5. This places the 5 on three
new locations.
Step 4

ステップ5.それはまだ掛かっていますか? ;)

Let’s take a look at the 21 of the box right top. If I would
fill in a 3, the puzzle cannot be solved, as I cannot fill in a 13.
I have to add the 7 there, completing the top right box with a 3 in
the 10 group. I placed a 7 below. In this row I now have left a 2,
3 and 6 which I am able to fill in the middle box. I tried putting
a 2 in the group of 16 in the box, but I can’t. 14 consists of 6
and 8, which is impossible to do. So the 2 goes in the box of 11. 3
and 6 makes 9, so we can add the 7 in the group of 16 as well.
Step 5

ステップ6。

We are able to finish the middle box now.
Step 6

ステップ7。

The 13 in the middle right box can only be a 3, 4 and 6. And,
since the 3 and 4 are used in the last column, we know the exact
location of the 6. I did exactly the same on the 19 in the bottom
middle box.
Step 7

ステップ8。

As said before, a 23 can only exist of a 6, 8 and 9. We are able
to fill in the 23 in the bottom box now as the 8 must be placed in
the box bottom right. With this information, we are able to add
more numbers in the bottom box and in the middle right box.
Step 8

ステップ9。

I assume the last part is the easiest one: bottom right box. The
16 of the middle box already has a 5, 1 and 3. 6 is missing there.
The group of 8 can only have a 1, 3 and 4 combination. Let’s fill
that in, as well as the remaining box of 18. Now we are almost done
– do not forget the 3 and 4 in the middle right box!
Step 9

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