Fit digits 1 to 8 in such a way that no two consecutive numbers are next to each other.

An example is also attached.

Please count your attempts!

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- Mar 17th 2009, 01:38 AMu2_waBrainy 2
Fit digits 1 to 8 in such a way that no two consecutive numbers are next to each other.

**An example is also attached.**

Please count your attempts!

- Mar 17th 2009, 07:18 AMSoroban
Hello, u2_wa!

This is an excellent puzzle!

Quote:

Place the digits 1 to 8 so that no two consecutive numbers are adjacent.

("Adjacent" numbers are in cells that share a side**or**a vertex.)Code:`*---*`

| |

*---*---*---*

| | | |

*---*---*---*

| | | |

*---*---*---*

| |

*---*

I solved this while in college (back in the Jurassic Period)

. . and have a*logical*solution.

I'll post it later, after everyone has tried it.

- Mar 17th 2009, 07:57 AMmasters
Hi u2_wa,

I have no logical approach to this one, but through trial and many errors, I came up with this. I'll 'white' it out, in case others are still struggling and want to solve it.

Code:

| 2 |

--------------

| 6 | 8 | 5 |

--------------

| 4 | 1 | 3 |

--------------

| 7 |

- Mar 17th 2009, 09:37 AMADARSH
- Mar 17th 2009, 11:22 PMu2_wa
- Mar 18th 2009, 01:27 PMSoroban

Okay, here's my "logical" approach to this puzzle.

We have the eight cells labeled $\displaystyle a$ to $\displaystyle h$.

There are two "central cells", $\displaystyle c$ and $\displaystyle h.$Code:`*---*`

| a |

*---*---*---*

| b | c | d |

*---*---*---*

| e | f | g |

*---*---*---*

| h |

*---*

There are eight digits to install:

. . two "end" numbers and six "middle" numbers.

. . $\displaystyle \underbrace{1}_{\text{end}},\underbrace{2,3,4,5,6, 7}_{\text{midde}},\underbrace{8}_{\text{end}}$

Place a middle number in a central cell, say, 4 in cell $\displaystyle c$.Code:`*---*`

| x |

*---*---*---*

| x | 4 | x |

*---*---*---*

| x | x | x |

*---*---*---*

| h |

*---*

We see that its neighbors (3 and 5) cannot be placed

. . in cells $\displaystyle a,b,d,e,f,g.$

So both 3 and 5 cannot be installed.

Conclusion: the central cells mustcontain a middle number.__not__

Hence, 1 and 8 must go in cells $\displaystyle c$ and $\displaystyle f.$

Code:`*---*`

| a |

*---*---*---*

| b | 1 | d |

*---*---*---*

| e | 8 | g |

*---*---*---*

| h |

*---*

2 cannot be in cells $\displaystyle a,b,d,e,g \quad\Rightarrow\quad 2 \in h.$

7 cannot be in cells $\displaystyle b,d,e,g \quad\Rightarrow\quad 7 \in a.$Code:`*---*`

| 7 |

*---*---*---*

| b | 1 | d |

*---*---*---*

| e | 8 | g |

*---*---*---*

| 2 |

*---*

3 cannot be in $\displaystyle e$ or $\displaystyle g$ . . . 3 may be in $\displaystyle b$ or $\displaystyle d.$

Pick one: say, $\displaystyle 3 \in b$Code:`*---*`

| 7 |

*---*---*---*

| 3 | 1 | d |

*---*---*---*

| e | 8 | g |

*---*---*---*

| 2 |

*---*

6 cannot be in $\displaystyle d.$

Also, 6 cannot be in cell $\displaystyle e.$

Else 4 and 5 will be in cells $\displaystyle d$ and $\displaystyle g$ (and be adjacent).

. . Hence: .$\displaystyle 6 \in g.$Code:`*---*`

| 7 |

*---*---*---*

| 3 | 1 | d |

*---*---*---*

| e | 8 | 6 |

*---*---*---*

| 2 |

*---*

4 cannot be in cell $\displaystyle e$ . . . $\displaystyle 4 \in d$

And finally: .$\displaystyle 5 \in e$Code:

*---*

| 7 |

*---*---*---*

| 3 | 1 | 4 |

*---*---*---*

| 5 | 8 | 6 |

*---*---*---*

| 2 |

*---*

Solution

Disregarding rotations and reflections, there is**one**solution.