BMAD Posted March 8, 2013 Report Share Posted March 8, 2013 Take from a pack of cards all the Aces, Kings, Queens and Jacks. Arrange them in a 4 × 4 square so that every row, column and diagonal contains one card of each value (A,J,Q,K) and one card of each suit (Heart, Spade, Diamond, Club). Quote Link to comment Share on other sites More sharing options...
0 BMAD Posted March 8, 2013 Author Report Share Posted March 8, 2013 nice. now how many different 'correct' answers are there? HA SK DJ CQ CJ DQ SA HK SQ HJ CK DA DK CA HQ SJ AS QC JH KD JD KS AC QH KC JS QD AH QH AD KS JC IN SECOND ROW SECOND COLUMN IT SHOULD BE 'KH'. I TRIED TO EDIT THE POST BUT EDITING FAILED. As Kh Qd Jc Jd Qc Ks Ah Kc Ad Jh Qs Qh Js Ac Kd Quote Link to comment Share on other sites More sharing options...
0 vistaptb Posted March 8, 2013 Report Share Posted March 8, 2013 SA HK DQ CJ CK DA HJ SQ DJ CQ SK HA HQ SJ CA DK (H=hearts, S=spades, D=diamonds, C=clubs, A=ace, Q=queen, K=king, J=jack) Quote Link to comment Share on other sites More sharing options...
0 BMAD Posted March 8, 2013 Author Report Share Posted March 8, 2013 (edited) SA HK DQ CJ CK DA HJ SQ DJ CQ SK HA HQ SJ CA DK (H=hearts, S=spades, D=diamonds, C=clubs, A=ace, Q=queen, K=king, J=jack) your diagonals don't meet the requirement of the op Edited March 8, 2013 by BMAD Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 8, 2013 Report Share Posted March 8, 2013 As Kh Qd Jc Jd Qc Ks Ah Kc Ad Jh Qs Qh Js Ac Kd Quote Link to comment Share on other sites More sharing options...
0 bhramarraj Posted March 8, 2013 Report Share Posted March 8, 2013 (edited) AS QC JH KD JD KS AC QH KC JS QD AH QH AD KS JC Edited March 8, 2013 by bhramarraj Quote Link to comment Share on other sites More sharing options...
0 bhramarraj Posted March 8, 2013 Report Share Posted March 8, 2013 AS QC JH KD JD KS AC QH KC JS QD AH QH AD KS JC IN SECOND ROW SECOND COLUMN IT SHOULD BE 'KH'. I TRIED TO EDIT THE POST BUT EDITING FAILED. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 8, 2013 Report Share Posted March 8, 2013 AS QC JH KD JD KS AC QH KC JS QD AH QH AD KS JC IN SECOND ROW SECOND COLUMN IT SHOULD BE 'KH'. I TRIED TO EDIT THE POST BUT EDITING FAILED. Hint: To edit after time elapses, copy / edit / paste onto a new post. 1 Quote Link to comment Share on other sites More sharing options...
0 dark_magician_92 Posted March 8, 2013 Report Share Posted March 8, 2013 HA SK DJ CQ CJ DQ SA HK SQ HJ CK DA DK CA HQ SJ Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 8, 2013 Report Share Posted March 8, 2013 nice. now how many different 'correct' answers are there? Excluding rotations and mirror images, There is only one solution. Without loss of generality, we can fill in the top row with choices. Eliminating duplicates in rows, columns and diagonals we have We see that the center and corner squares have only two possibilities. Choosing each leads to two solutions: ........ Note the patterns of the a cells are diagonal mirror images. Call the patterns CW and CCW respectively. In the CCW case the c cells include the opposite corner cell. They have a CCW pattern as well. Together they look like this. . Now it is clear that the remaining cells can be filled in with CW patterns for b and d. So this is the basic pattern for isolating four entities in a 4x4 grid. It would apply to card rank or card suit equally well. To isolate two sets of four entities, one would simply overlay one of the above choices for Rank with the other choice for Suit. That is the construction for the single solution to this puzzle. It has only reflection and rotation variants. Quote Link to comment Share on other sites More sharing options...
0 18 IXOHOXI 81 Posted April 19 Report Share Posted April 19 There are 144 distinct solutions times 8 rotations and reflections for a total of 1152 solutions for the 4x4 Magic Court Card Puzzle. Solution # 1: AS KH QD JC QC JD AH KS JH QS KC AD KD AC JS QH. Solutions # 2: AS KH QD JC JD QC KS AH KC AD JH QS QH JS AC KD. Plus there's 1150 more solutions that satisfies the conditions that each of the four card ranks and four card suits appears only once in each of the four rows, four columns, and the two diagonals. This also works if the cards came from a four color deck. Quote Link to comment Share on other sites More sharing options...
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BMAD
Arrange them in a 4 × 4 square so that every row, column and diagonal contains one card of each value (A,J,Q,K) and one card of each suit (Heart, Spade, Diamond, Club).
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