Good point. Here is one step further:
3. Consider the center 4 numbers(regardless of order): {1,8,a,b},
containing 2 even and 2 odd.
and {a,b} can be chosen from {3,4,5,6}. Then there is only one case:
{1,8,3,6}.
The rest is trivial.
On 2010-11-12 13:26, Amod wrote:
Let me try to formalize it
1. Divide even and odd in halfs
Since any pair of consecutive number consists of an even and an odd
number hence these two cannot be in the same half.
2 Place 1 and 8 on the separation line between even and odd.
Since 1 and 8 have affinity for one number only hence solution is
possible.
Comments and inputs are welcome
On Nov 12, 8:18 am, Amod<gam...@gmail.com> wrote:
Gr8, guys :)
If any body can formalize the solution, then it would be great
On Nov 12, 12:17 am, Shiv Shankar Prajapati
<mca.shivshan...@gmail.com> wrote:
Several solutions are possible for it
5 1 8 2
7 3 6 4
Here we can swap the position of 5<->7, 1<->3 etc.
On Thu, Nov 11, 2010 at 11:48 PM, jagannath prasad das
<jpdasi...@gmail.com>wrote:
4 8 3 7
2 6 1 5
On Thu, Nov 11, 2010 at 5:57 PM, sunny agrawal<sunny816.i...@gmail.com>wrote:
@rohit
4 5 are diagonally adjacent .....
On Thu, Nov 11, 2010 at 5:09 PM, Rohit Singhal<rsinghal.it...@gmail.com>wrote:
1 5 2 6
- - - - - -
3 7 4 8
On Thu, Nov 11, 2010 at 3:16 PM, Abhilasha jain<
mail2abhila...@gmail.com> wrote:
solution is
5 1 6 2
_ _ _ _
7 3 8 4
On Thu, Nov 11, 2010 at 1:26 PM, Amod<gam...@gmail.com> wrote:
We have a rectangle
It is divided in eight parts by three vertical and one horizontal line
so that there are 8 chambers.
Now we have numbers from 1-8 to be filled in these chambers.
Rule : No two consecutive numbers must be present either side to side
or diagonal
Invalid situation example
Given 5 at position 2 then 4 cannot occur at any of the give position.
4 5 4
_ _ _ _
4 4 4
_ _ _ _
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