1/20/2014 7:53:33 PM
larrydevlin1770 Posts: 3

Hi, everyone!
I have a place cards puzzle that is stumping me. I don't even know how to begin. This puzzle can be found in the January '14 edition of Variety Puzzles, p. 130. I have figured out a few details from the clues given: row E contains only A5 and 7, and column J contains only 37, with two 5's. I made a list of card sums for clue six, and made notes on the grid to indicate the placement of the straights. I also figured out that Column I contains exactly 3 diamonds, and one each of the other suits. I can't figure out where to go from here! Help!!! If anyone knows any tutorials for this kind of puzzle, please let me know!
Thanks again.
Larry D

1/21/2014 4:22:19 PM
Amy Lowenstein Posts: 1599

Those of us that don't have the particular magazine, but might know a technique for help, won't be able to help unless given the puzzle. If nobody that happens to have that particular magazine comes forth with suggestions, you might try typing the intro and clues, so that others of us might be able to help.
 Amy

1/21/2014 8:15:32 PM
larrydevlin1770 Posts: 3

Dear Amy,
Thank you.
Here is the puzzle:
Place Cards ace through nine of the four suits in the 36 boxes on the diagram. No two adjacent squares, vertically or horizontally, should contain cards of the same number or suit. (Each row and each column contain all four suits.) Aces are low, the equivalent of one. [My note: The grid is 6 x 6 with rows lettered AF and columns lettered GL. There is also a list of all the cards to cross off as you use them.]
CLUES
1. Four fourcard straights (consecutive numbers listed from low card to high card), each containing one card of each suit can be found at these rowandcolumn coordinates: EK BJ DH FL BL FG DJ FK DI BG CL AL CJ DL FJ EG
2. One row and one column each contain six numbers in order, backward or forward.
3. Each diagonal from AG to FL and AL to FG alternates red and black cards. In one diagonal the only [club] is the only oddnumbered card, while in the other diagonal, the only [spade] is the only evennumbered card.
4. In row A the two [diamond]s are its only even numbers, while in row F the A[diamond] is its only odd number.
5. Row E's six numbers total 30 and are five consecutive numbers, while Column J's six numbers total 22 and are five consecutive numbers.
6. In row B the two [club]s total the [diamond]. In row C the two [heart]s total the diamond. In column I two of the [diamond]s total the third [diamond].
Hope that helps.
Thanks!

1/21/2014 10:00:30 PM
Amy Lowenstein Posts: 1599

I'm not sure about this, Larry, but I'd take an actual deck of cards and lay them out on a table so that I could experiment with different placements. I'm always intrigued with clues about diagonals, but of course you don't know which way the diagonals go until you experiment (which one is, from sortof left to sortright, Red Black Red Black Red Black, and which one is Black Red Black Red Black Red). But you might try arbitrarily setting up one of the diagonals as RedBlackRed etc, then the other as BlackRedBlack etc, and see if one of the other clues makes whatever you tried impossible, in which case you'd know to change the blacks & reds on the diagonals.
Not sure I will have time to try to do this puzzle myself, but the above is my first thought.
 Amy

1/21/2014 11:10:33 PM
Frances Posts: 703

I have done a few Place Cards and these are hard puzzles! I'm not sure I want to tackle another, but here's a bit that might help you. Let's look at clue #5 in relation to clue #1.
1. Row E's numbers=30 with 5 consecutive. I get: 3,4,5,5,6,7. They total 30, and have 3 to 7 consecutive.
2. Column J's numbers =22 with 5 consecutive. I get 2 possible: 1,2,3,4,5,7 or 2,2,3,4,5,6. In all three cases, there are no 8's or 9's in either row E or column J.
3. CJDLFJEG is 4 consecutive, and has 2 numbers in the J column, which we know has no 8 or 9. Therefore, it's 1234, 2345, 3456, or 4567. So: CJ=1,2,3 or 4; DL=2,3,4 or 5; FJ=3,4,5 or 6; and EG=4,5,6 or 7. I would lightly mark these possibilities in those squares.
Also: Since column I has 3 diamonds, and like suits can't touch, they are in either AI, CI and EI or BI, DI and FI.
I don't know of any tips or shortcuts with this puzzle, but I just keep picking along, trying to eliminate possibilities in both numbers and suits as I go.
P.S. From clue #4, the ace (1) of diamonds is the ONLY odd number in row F. What can you then determine about the possibilities I have listed above for CJDLFJEG? edited by Frances on 1/21/2014
