14 replies to this topic

[Yoruichi-san]

Err...desolee, mon capitaine...

Spoiler for Sorry...I sank your ship...

Actually, my method gives a max of 13, not 14...doh...you don't need to test the last pair, if it's not one of the other 50 squares, it has to be one of the remaining 2, so you use the 13th test to find which one.

I don't think there's a more efficient packing of the crosses, though (I think holes of 2 near eachother are the best you can do), and it makes sense in terms of the math, ie 5*10=50,+3 to figure out which one of the 5.

Edit: And I don't think you need the 13th test either? You don't need to get a 'hot' to figure out which it is, you can assume the last group is 'hot' by the process of elimination.

Edited by Yoruichi-san, 22 June 2012 - 09:35 PM.

[CaptainEd]

And so it remains a team effort. But your arrangement wins by one test, I believe.
Your 13th test distinguishes the last pair, so no 14th test, agreed.

My 13th test distinguishes my only pair, so no 14th test. Unfortunately, my 12th test touches 3, so it takes two more to find which one for sure.

This is why your solution ending in 3 pairs is better than mine, ending in a trio and a pair.

Good job, Y-san. Hontoo ni utsukushii desu!

[plainglazed]

right on - had a minor brainurism when first reading Y-san's solution. couldn't quite visualize her layout and quickly accepted her claim for fourteen. my solution for thirteen is similar in concept but slightly different in layout.

Spoiler for WIW and since i spent some time on the drawing

good work you two. thanks for playing

[CaptainEd]

Thanks for the pretty puzzle!

[Yoruichi-san]

And so it remains a team effort. But your arrangement wins by one test, I believe.
Your 13th test distinguishes the last pair, so no 14th test, agreed.

My 13th test distinguishes my only pair, so no 14th test. Unfortunately, my 12th test touches 3, so it takes two more to find which one for sure.

This is why your solution ending in 3 pairs is better than mine, ending in a trio and a pair.

Good job, Y-san. Hontoo ni utsukushii desu!

Spoiler for Alternative...

Actually, I redrew your solution, and I think if you zig-zag your crosses instead of spiraling (i.e. C(s,r), C(s+1,r+2). C(s, r+4), etc), you end with 3 pairs as well.

Edit: Lol...great minds think alike and reply at the same time, it appears...yeah, thanks for the puzzle, and the brainurism was mine . I actually liked both versions of the puzzle .

Edited by Yoruichi-san, 22 June 2012 - 10:46 PM.