please empty your brain below |
I've got an answer, but I'm not going to put it here because it would just upset people. |
Oh come on, someone else have a go. You know you want to. |
Hmmm. Even with the help of my nice playing cards from Amsterdam I can only do this down to the last 2 cards then it all falls apart. I tried to do it logically by using the cards of each sort (K, Q, J, A) in diagonals, each time using them in the same order (H, D, C, S). I would probably have been more successful being random. Never mind. |
Yep, got it second time round (schoolboy error on the first). A tougher question, though, is how many possible answers are there? I may have a maths degree but even I'm not going to try to work that one out. |
I think there's at least 144 ways of doing it, and more than 2000 if the Ace of Diamonds isn't forced into that corner. |
Then again, I never was one to resist a challenge like that. Okay, assuming that there is one and only one solution for every permutation of the leading diagonal (I think that's correct haven't proven it) then the number of possible solutions is equal to the number of permutations for the leading diagonal. Which is 36. Simple really. *cowers in case someone discovers flaw in logic* |
All right, after a bit more playing around I can see that my assumption was false and therefore my answer was wrong. However, I think that there are only two possible answers for any permutation of the leading diagonal which would mean the answer is 72. Again, this is unproven but supported by experimental evidence. |
With the bottom right-hand card fixed, there are indeed 36 possible permutations of the leading diagonal. With that fixed, there are then exactly 4 possibilities for the card in the top right-hand corner, each of which leads to a unique solution. So that's 36 x 4 = 144 possible solutions. |
Not true. If you have KC and AD at the extremes of the leading diagonal then the top right card must be either a queen or a jack and either a spade or a heart. Four choices: QS, QH, JS & JH. However, two of those already make up the rest of the leading diagonal so there are in fact only two choices available. |
You're right, sorry, there are only two possibilities for the top right card (and the one that doesn't go top right has to go bottom left). Small comfort for anyone out there who hasn't solved the problem yet though, now knowing that there are 72 ways to do it and they can't even find one of them |
Ooh, you are awful. Better hope BW doesn't pop back anytime soon. |
I must have been distracted by the pictures on those cards. I seem to remember having shown them to DG at some stage in the dim and distant past, so he might understand |
You stopped us going to bed at a reasonable time last night. Mr BW couldn't do it either. It's like a bloody Rubik's cube it is. Frustrating and pointless. Cross Witch |
Here's a solution. For there are many. KC - JD - AS - QH QS - AH - JC - KD JH - KS - QD - AC AD - QC - KH - JS |
Except that you can't actually type that 'solution' into the boxes... |
Damn. Hadn't spotted that little complexity. Mean Geezer. Rotate the whole solution through 90 degrees to widdershins, then. |
Found some late diversion from work! Alternate solution to that posted above: KH QD AC JS AS JC KD QH JD AH QS KC QC KS JH AD Found by filling in from corners - bound to be more mathematical method though. |
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