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Binary

cinelli
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Posts: 550
Joined: November 9th, 2016, 11:33 am
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Binary

#428019

Postby cinelli » July 16th, 2021, 10:19 am

.--- --- --- --- --- --- --- --- --- ---
| | 1 | | | | 0 | 0 | | | 0 |
--- --- --- --- --- --- --- --- --- ---
| | 1 | | | | | | 1 | | |
--- --- --- --- --- --- --- --- --- ---
| 0 | | | | | 1 | | | | |
--- --- --- --- --- --- --- --- --- ---
| | | | | | | | | 0 | |
--- --- --- --- --- --- --- --- --- ---
| | 0 | 0 | | | | | | 0 | |
--- --- --- --- --- --- --- --- --- ---
| | 1 | | 1 | 1 | | | | | |
--- --- --- --- --- --- --- --- --- ---
| 0 | | | | | 0 | | | | |
--- --- --- --- --- --- --- --- --- ---
| | | | | | | | | | |
--- --- --- --- --- --- --- --- --- ---
| 0 | | 0 | 0 | | | | 0 | | |
--- --- --- --- --- --- --- --- --- ---
| | | | | | 1 | | | | |
--- --- --- --- --- --- --- --- --- ---

This is a puzzle which appears in the i newspaper. Complete the grid so that each row and each column contains five 0s and five 1s. The same number cannot appear in more than two consecutive squares in any row or column.

Cinelli

Gengulphus
Lemon Quarter
Posts: 4255
Joined: November 4th, 2016, 1:17 am
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Re: Binary

#428076

Postby Gengulphus » July 16th, 2021, 1:12 pm

cinelli wrote:
.--- --- --- --- --- --- --- --- --- ---
| | 1 | | | | 0 | 0 | | | 0 |
--- --- --- --- --- --- --- --- --- ---
| | 1 | | | | | | 1 | | |
--- --- --- --- --- --- --- --- --- ---
| 0 | | | | | 1 | | | | |
--- --- --- --- --- --- --- --- --- ---
| | | | | | | | | 0 | |
--- --- --- --- --- --- --- --- --- ---
| | 0 | 0 | | | | | | 0 | |
--- --- --- --- --- --- --- --- --- ---
| | 1 | | 1 | 1 | | | | | |
--- --- --- --- --- --- --- --- --- ---
| 0 | | | | | 0 | | | | |
--- --- --- --- --- --- --- --- --- ---
| | | | | | | | | | |
--- --- --- --- --- --- --- --- --- ---
| 0 | | 0 | 0 | | | | 0 | | |
--- --- --- --- --- --- --- --- --- ---
| | | | | | 1 | | | | |
--- --- --- --- --- --- --- --- --- ---

This is a puzzle which appears in the i newspaper. Complete the grid so that each row and each column contains five 0s and five 1s. The same number cannot appear in more than two consecutive squares in any row or column.

Spoiler...

The rules allow us to take the following simple types of solution step:

1) Anywhere where three consecutive cells in a row or column contain two 0s, with the remaining cell blank, fill that blank cell with a 1, and similarly with the roles of 0s and 1s reversed.

2) If any row or column contains five 0s, fill any remaining blank cells in that row or column with 1s, and similarly with the roles of 0s and 1s reversed.

Apply solution steps of type 1 as far as possible:

.--- --- --- --- --- --- --- --- --- ---
| | 1 | | | 1 | 0 | 0 | 1 | | 0 |
--- --- --- --- --- --- --- --- --- ---
| 1 | 1 | 0 | | | | | 1 | | |
--- --- --- --- --- --- --- --- --- ---
| 0 | 0 | 1 | | | 1 | | 0 | 1 | |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 1 | 0 | 1 | 0 | | | 0 | |
--- --- --- --- --- --- --- --- --- ---
| 1 | 0 | 0 | 1 | 0 | 1 | | | 0 | |
--- --- --- --- --- --- --- --- --- ---
| | 1 | 0 | 1 | 1 | 0 | | | 1 | |
--- --- --- --- --- --- --- --- --- ---
| 0 | | 1 | 0 | 1 | 0 | | | | |
--- --- --- --- --- --- --- --- --- ---
| 1 | | | 1 | 0 | 1 | | | | |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | | |
--- --- --- --- --- --- --- --- --- ---
| | | | | | 1 | | | | |
--- --- --- --- --- --- --- --- --- ---

Solution steps of type 2 now allow us to fill:

* the remaining empty cells on the 9th row with 1s;
* the remaining empty cells in the 2nd column with 0s;
* the remaining empty cells in the 5th column with 0s;
* the remaining empty cell in the 6th column with a 1;

resulting in:

.--- --- --- --- --- --- --- --- --- ---
| | 1 | | | 1 | 0 | 0 | 1 | | 0 |
--- --- --- --- --- --- --- --- --- ---
| 1 | 1 | 0 | | 0 | 1 | | 1 | | |
--- --- --- --- --- --- --- --- --- ---
| 0 | 0 | 1 | | 0 | 1 | | 0 | 1 | |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 1 | 0 | 1 | 0 | | | 0 | |
--- --- --- --- --- --- --- --- --- ---
| 1 | 0 | 0 | 1 | 0 | 1 | | | 0 | |
--- --- --- --- --- --- --- --- --- ---
| | 1 | 0 | 1 | 1 | 0 | | | 1 | |
--- --- --- --- --- --- --- --- --- ---
| 0 | 0 | 1 | 0 | 1 | 0 | | | | |
--- --- --- --- --- --- --- --- --- ---
| 1 | 0 | | 1 | 0 | 1 | | | | |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 |
--- --- --- --- --- --- --- --- --- ---
| | 0 | | | 0 | 1 | | | | |
--- --- --- --- --- --- --- --- --- ---

That allows us to do some more solution steps of type 1. Doing them as far as possible gives us:

.--- --- --- --- --- --- --- --- --- ---
| | 1 | | | 1 | 0 | 0 | 1 | | 0 |
--- --- --- --- --- --- --- --- --- ---
| 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | | |
--- --- --- --- --- --- --- --- --- ---
| 0 | 0 | 1 | | 0 | 1 | 1 | 0 | 1 | |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 1 | 0 | 1 | 0 | | | 0 | |
--- --- --- --- --- --- --- --- --- ---
| 1 | 0 | 0 | 1 | 0 | 1 | | | 0 | |
--- --- --- --- --- --- --- --- --- ---
| | 1 | 0 | 1 | 1 | 0 | | | 1 | |
--- --- --- --- --- --- --- --- --- ---
| 0 | 0 | 1 | 0 | 1 | 0 | | | | |
--- --- --- --- --- --- --- --- --- ---
| 1 | 0 | | 1 | 0 | 1 | | | | |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 |
--- --- --- --- --- --- --- --- --- ---
| | 0 | | | 0 | 1 | | | | |
--- --- --- --- --- --- --- --- --- ---

The 2nd row now contains five 1s, so a type 2 step completes it with 0s in its last two cells. A type 1 step then allows us to put a 1 in the last cell of the 3rd row, followed by a type 2 step filling its remaining blank cell with a 0:

.--- --- --- --- --- --- --- --- --- ---
| | 1 | | | 1 | 0 | 0 | 1 | | 0 |
--- --- --- --- --- --- --- --- --- ---
| 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 |
--- --- --- --- --- --- --- --- --- ---
| 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 1 | 0 | 1 | 0 | | | 0 | |
--- --- --- --- --- --- --- --- --- ---
| 1 | 0 | 0 | 1 | 0 | 1 | | | 0 | |
--- --- --- --- --- --- --- --- --- ---
| | 1 | 0 | 1 | 1 | 0 | | | 1 | |
--- --- --- --- --- --- --- --- --- ---
| 0 | 0 | 1 | 0 | 1 | 0 | | | | |
--- --- --- --- --- --- --- --- --- ---
| 1 | 0 | | 1 | 0 | 1 | | | | |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 |
--- --- --- --- --- --- --- --- --- ---
| | 0 | | | 0 | 1 | | | | |
--- --- --- --- --- --- --- --- --- ---

At this point, there are no more simple solution steps available of either type 1 or type 2, so we need to look for something not quite so simple. After some looking, I found a third, not-quite-so-simple type of solution step:

3) If a set of four consecutive cells in a row or column contains 0--1 or 1--0, the two middle blank squares in it cannot contain 00 or 11 without violating the 'no three consecutive identical digits' rule. So one of those digits must be a 0 and the other a 1 - and if there are four known 0s (or 1s) plus any other blank cells in the row or column, we can deduce that those other blank cells are all 1s (or 0s respectively.

The 5th and 6th rows meet the conditions for type 3 solution steps, allowing us to fill the last cell of the 5th row with a 1 and the two end cells of the 6th row with 0s. Also, some somewhat similar reasoning can be applied to the four blank cells at the end of the 7th row: they need to contain three 1s and a 0, and cannot be 0111 or 1110, so must be 1011 or 1101, allowing us to deduce that the first and last of them contain 1s:

.--- --- --- --- --- --- --- --- --- ---
| | 1 | | | 1 | 0 | 0 | 1 | | 0 |
--- --- --- --- --- --- --- --- --- ---
| 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 |
--- --- --- --- --- --- --- --- --- ---
| 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 1 | 0 | 1 | 0 | | | 0 | |
--- --- --- --- --- --- --- --- --- ---
| 1 | 0 | 0 | 1 | 0 | 1 | | | 0 | 1 |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 0 | 1 | 1 | 0 | | | 1 | 0 |
--- --- --- --- --- --- --- --- --- ---
| 0 | 0 | 1 | 0 | 1 | 0 | 1 | | | 1 |
--- --- --- --- --- --- --- --- --- ---
| 1 | 0 | | 1 | 0 | 1 | | | | |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 |
--- --- --- --- --- --- --- --- --- ---
| | 0 | | | 0 | 1 | | | | |
--- --- --- --- --- --- --- --- --- ---

This enables us to do more type 1 and type 2 steps, and when we've done them as far as possible, we get:

.--- --- --- --- --- --- --- --- --- ---
| 1 | 1 | 0 | | 1 | 0 | 0 | 1 | | 0 |
--- --- --- --- --- --- --- --- --- ---
| 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 |
--- --- --- --- --- --- --- --- --- ---
| 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
--- --- --- --- --- --- --- --- --- ---
| 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 |
--- --- --- --- --- --- --- --- --- ---
| 0 | 0 | 1 | 0 | 1 | 0 | 1 | | | 1 |
--- --- --- --- --- --- --- --- --- ---
| 1 | 0 | 1 | 1 | 0 | 1 | 0 | | | 0 |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 |
--- --- --- --- --- --- --- --- --- ---
| 1 | 0 | 1 | | 0 | 1 | 0 | | | 1 |
--- --- --- --- --- --- --- --- --- ---

Finally, look at the 8th column. Its three blank cell must contain a 1 and two 0s, and only putting the 1 in the middle of them and the 0s in the other two avoids having three consecutive 0s in the column. A further few type 1 and 2 steps completes the solution:

.--- --- --- --- --- --- --- --- --- ---
| 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 |
--- --- --- --- --- --- --- --- --- ---
| 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 |
--- --- --- --- --- --- --- --- --- ---
| 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
--- --- --- --- --- --- --- --- --- ---
| 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 |
--- --- --- --- --- --- --- --- --- ---
| 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 |
--- --- --- --- --- --- --- --- --- ---
| 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 |
--- --- --- --- --- --- --- --- --- ---
| 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 |
--- --- --- --- --- --- --- --- --- ---
| 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 |
--- --- --- --- --- --- --- --- --- ---

Gengulphus

cinelli
Lemon Slice
Posts: 550
Joined: November 9th, 2016, 11:33 am
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Re: Binary

#428285

Postby cinelli » July 17th, 2021, 1:17 pm

This is a beautifully logical solution method from Gengulphus. Incidentally the i newspaper has four whole pages of puzzles in its weekday editions (I'm not sure about the weekend versions). I think the Minesweeper and Bridges puzzles are quite challenging and are the type of puzzle where it is difficult to recover from a mistake.

Cinelli


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