Change History

Hello Sumbloke,
Now I understand your way of thinking.....
That's not what I had in mind for this challenge. With just 1 combination the challenge is not fulfilled, because you need the whole column (and row and 3x3 square).
All 9 possible combinations for T1 must have a value (where 0 is allowed), the challenge is about these 9 combinations have different foundnumbers. And that for every column, row and 3x3 square (the complete matrix).

Let's stick to the example.....
If someone has 1 found on the 5/1 cache, like you say.... the challenge is not fulfilled if no other T1 combinations are found by this person. The matrix would show a zero on the other 8 combinations for T1. Since all 9 values should differ to meet the challenge, (and 8 times a zero doesn't differ), this person still needs different foundnumbers on the other 8 combinations for T1, so it actually encourages someone to find more caches, not prevent/discourage them.

The person could choose to leave one combination at zero (there is only 1 zero in a column, row or 3x3 square allowed for this challenge) then this person has 2 different foundnumbers, the 1 found on 5/1 and a zero on anothers combination. For the next combination the person needs 2 foundnumbers, since 2 differs from 0 and 1, still leaving 6 other T1 combinations to set with different foundnumbers than the ones before. So the foundnumber must go up to get all 9 combinations to a different number. Something like this perhaps:
D5 / T1 result of matrix = 1 found
D4,5 / T1 result of matrix = 0 found
D4 / T1 result of matrix = 2 or more founds (because it's different than 0 and 1)
D3,5 / T1 result of matrix = 3 ore more founds, but have to be different foundnumber as above
D3 / T1 result of matrix = and so on, and so on...

And the challenge is to get it right for all the other columns, rows and 3x3 squares of the foundmatrix. Not an easy task and not fulfilled by finding one cache in a row, column or 3x3 square.......

Hoping this makes sense.... awaiting your feedback, Rob.

Hello Sumbloke,
Now I understand your way of thinking.....
That's not what I had in mind for this challenge. With just 1 combination the challenge is not fulfilled, because you need the whole column.
column (and row and 3x3 square).
All 9 possible combinations for T1 must have a value (where 0 is allowed), the challenge is about these 9 combinations have different foundnumbers. And that for every column, row and 3x3 square (the complete matrix).

Let's stick to the example.....
If someone has 1 found on the 5/1 cache, like you say.... the challenge is not fulfilled if no other T1 combinations are found by this person. The matrix would show a zero on the other 8 combinations for T1. Since all 9 values should differ to meet the challenge, this person still needs different foundnumbers on the other 8 combinations for T1, so it actually encourages someone to find more caches, not prevent/discourage them.

The person could choose to leave one combination at zero (there is only 1 zero in a column, row or 3x3 square allowed for this challenge) then this person has 2 different foundnumbers, the 1 found on 5/1 and a zero on anothers combination. For the next combination the person needs 2 foundnumbers, since 2 differs from 0 and 1, still leaving 6 other T1 combinations to set with different foundnumbers than the ones before. So the foundnumber must go up to get all 9 combinations to a different number. Something like this perhaps:
D5 / T1 result of matrix = 1 found
D4,5 / T1 result of matrix = 0 found
D4 / T1 result of matrix = 2 or more founds (because it's different than 0 and 1)
D3,5 / T1 result of matrix = 3 ore more founds, but have to be different foundnumber as above
D3 / T1 result of matrix = and so on, and so on...

And the challenge is to get it right for all the other columns, rows and 3x3 squares of the foundmatrix. Not an easy task and not fulfilled by finding one cache in a row, column or 3x3 square.......

Hoping this makes sense.... awaiting your feedback, Rob.

Hello Sumbloke,
Now I understand your way of thinking.....
That's not what I had in mind for this challenge. With just 1 combination the challenge is not fulfilled, because you need the whole column.
All 9 possible combinations for T1 must have a value (where 0 is allowed), the challenge is about these 9 combinations have different foundnumbers. And that for every column, row and 3x3 square (the complete matrix).

Let's stick to the example.....
If someone has 1 found on the 5/1 cache, like you say.... the challenge is not fulfilled if no other T1 combinations are found by this person. The matrix would show a zero on the other 8 combinations for T1. Since all 9 values should differ to meet the challenge, this person still needs different foundnumbers on the other 8 combinations for T1, so it actually encourages someone to find more caches, not prevent/discourage them.

The person could choose to leave one combination at zero (there is only 1 zero in a column, row or 3x3 square allowed for this challenge) then this person has 2 different foundnumbers, the 1 found on 5/1 and a zero on anothers combination. For the next combination the person needs 2 foundnumbers, since 2 differs from 0 and 1, still leaving 6 other T1 combinations to set with different foundnumbers than the ones before. So the foundnumber must go up to get all 9 combinations to a different number. Something like this perhaps:
D5 / T1 result of matrix = 1 found
D4,5 / T1 result of matrix = 0 found
D4 / T1 result of matrix = 2 or more founds (because it's different than 0 and 1)
D3,5 / T1 result of matrix = 3 ore more founds, but have to be different foundnumber as above
D3 / T1 result of matrix = and so on, and so on...

And the challenge is to get it right for all the other columns, rows and 3x3 squares of the foundmatrix. Not an easy task and not fulfilled by finding one cache in a row, column or 3x3 square.......

Hoping this makes sense.... awaiting your feedback, Rob.