You are working with a 4x4 grid as in this image:
Place one of these four characters: @ # $ % in each of the sixteen squares so that no character appears more than once in any row, any column, or any of the two main diagonals.
Use Excel or any suitable app to complete the puzzle.
Then take a screen capture/image of your solution and paste it into this thread (between spoiler tags) to supply your answer.
There may be more than one solution based on different combinations of the four characters, but any valid solution must comply with the parameters above.
Unique Symbol Combination
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- gamma jay
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Unique Symbol Combination
Regards,
Rudi
If your absence does not affect them, your presence didn't matter.
Rudi
If your absence does not affect them, your presence didn't matter.
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Re: Unique Symbol Combination
Spoiler
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StuartR
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- gamma jay
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Re: Unique Symbol Combination
Excellent Stuart, but you didn't have to name the solution with some swear word!!
Regards,
Rudi
If your absence does not affect them, your presence didn't matter.
Rudi
If your absence does not affect them, your presence didn't matter.
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- Administrator
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- gamma jay
- Posts: 25455
- Joined: 17 Mar 2010, 17:33
- Location: Cape Town
Re: Unique Symbol Combination
Regards,
Rudi
If your absence does not affect them, your presence didn't matter.
Rudi
If your absence does not affect them, your presence didn't matter.
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Re: Unique Symbol Combination
Spoiler
If we take @ # $ % as the first row, as Stuart has done, there are only 2 solutions: the one given by Stuart, and
i.e. a cyclic shift of rows 2, 3 and 4.
Since there are 4*3*2*1 = 24 possible permutations of the characters in row 1, the total number of solutions is 2*24 = 48.
i.e. a cyclic shift of rows 2, 3 and 4.
Since there are 4*3*2*1 = 24 possible permutations of the characters in row 1, the total number of solutions is 2*24 = 48.
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Best wishes,
Hans
Hans
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- gamma jay
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Re: Unique Symbol Combination
Thanks for those mathematical insights. I'll take your word for it.
My attempt uses a different combination from Stuart's (and yours), but nevertheless, a valid solution to...
My attempt uses a different combination from Stuart's (and yours), but nevertheless, a valid solution to...
Spoiler
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Regards,
Rudi
If your absence does not affect them, your presence didn't matter.
Rudi
If your absence does not affect them, your presence didn't matter.
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- Administrator
- Posts: 78595
- Joined: 16 Jan 2010, 00:14
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Re: Unique Symbol Combination
That's a valid solution indeed. It can be obtained from the one given by me by changing # to $, $ to % and % to #, so it's one of the 24 permutations of that one.
You can get any solution by taking either of the two solutions given earlier and shuffling the characters.
You can get any solution by taking either of the two solutions given earlier and shuffling the characters.
Best wishes,
Hans
Hans