Chess Problem

Consider a 5×5 chess board. Is it possible to place five queens on the board such that three pawns can safely be placed on the board? In other words, by carefully choosing where to place the five queens, can you arrange it so that there are three squares on the board that none of the queens can attack?
chess_board

71 Comments

  1. Steven Miller on March 11, 2019 at 3:16 pm

    nope, can put 5 queens with 3 pawns safe. email me at [email protected] for a hint



  2. isaac on March 11, 2019 at 3:11 pm

    you can only put 4 queens down for the three pawns to be safe, so no.



  3. Steven Miller on September 19, 2016 at 1:24 am

    Email me at [email protected] — I don’t want to post the soln



  4. Steven Miller on August 10, 2015 at 9:00 am

    no, that will kill every square.



  5. andre123 on August 10, 2015 at 8:24 am

    place the five queens vertically at the side of the board… thats all!!1



  6. andre123 on August 10, 2015 at 8:23 am

    place the 5 queens on the side vertically…. alll



  7. Steven Miller on January 18, 2014 at 6:25 pm

    Sadly very busy and cannot update more than once a month or so; hope there is enough there for you to enjoy for awhile.



  8. Isidro on January 18, 2014 at 1:07 pm

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    daily. It contains pleasant data.



  9. Kelvin Kakos on October 8, 2013 at 12:58 am

    almost…HaHa!) Great job. I really enjoyed what you had to say



  10. Steven Miller on September 10, 2012 at 3:22 pm

    sent hint (sjm1 AT williams.edu)



  11. viren on September 10, 2012 at 1:52 pm

    after days of trying i realized that i need the soln. can you mail it.



  12. Steven Miller on August 15, 2012 at 1:30 pm

    there is a way — will send a hint first.



  13. mukhtar on August 15, 2012 at 12:05 pm

    there is no answer for this riddle. maximum safe point for this riddle is 2. if this is wrong can you email me the correct answer.
    thanks



  14. Steven Miller on July 31, 2012 at 1:20 am

    ok, hint/answer sent (sjm1 AT williams.edu)



  15. lusho on July 30, 2012 at 4:40 pm

    We tried a lot and we didnt find the answer.



  16. Steven Miller on July 24, 2012 at 3:11 am

    sure, one of my favorites — sent (sjm1 AT williams.edu)



  17. Shane on July 23, 2012 at 9:27 pm

    I also would like to know the calculations/algorithms behind such a problem. I could only figure it out through logical trial and error, as well.



  18. Steven Miller on May 3, 2012 at 6:02 pm

    there’s ways to cut down on the trial and error //s



  19. Truman P on May 3, 2012 at 5:42 pm

    Steve,

    Could you email me the solution? This one has been bothering me for a while. Also, is there any method to go about finding the solution besides simple trial and error?

    thanks



  20. Steven Miller on May 1, 2012 at 2:48 pm

    you didn’t include your email — email me at sjm1 AT williams.edu



  21. Anonymous on May 1, 2012 at 1:01 pm

    hi sir steve! Its almost two weeks i am trying to solve this problem. Usually i hate to give up but this time i lost. I am unable to find the answer. If u dont mind will u pls send the answer to my email.



  22. Steven Miller on April 9, 2012 at 4:49 pm

    sure. //s



  23. Shane Fisher on April 9, 2012 at 4:36 pm

    Steve,

    Please send me the answer. can only get two pawns or 4 queens!!!

    Shane



  24. Steven Miller on April 3, 2012 at 1:47 pm

    it’s a 5×5 chessboard. you can’t place anything in F, G, or H as they don’t exist.



  25. Luke on April 3, 2012 at 1:16 pm

    My solution can’t possibly be right because I found a way to place 6 pawns safely within seconds of reading the problem. My instinct tells me that the problem can’t possibly be that easy, or that I must have read it wrong, but for the life of me I can’t figure out what’s wrong with my solution.

    Queens: A1, B2, C3, D4, E5
    Pawns: F7, F8, G6, G8, H6, H7



  26. Steven Miller on April 2, 2012 at 1:43 pm

    The problem is there are so many possibilities, if you don’t have a good way to explore all you could miss one. I’ll email you. //s



  27. Tai Nishiuchi on April 2, 2012 at 1:40 pm

    I have no idea how to solve this problem using math (or otherwise) please E-mail math solution. From a chess perspective I’ve been placing pieces one at a time to try to cover the same area with different pieces. ex. Q (A-1) and Q (A-2) both cover (A-3,4,5) and (B-1,2) but that alone isn’t math no matter how much I will it so.



  28. Steven Miller on March 27, 2012 at 6:26 pm

    well done! enjoy the test! I’ll send comments on my soln. //s



  29. Steven Miller on March 27, 2012 at 6:26 pm

    ok

    send it as a board, with places for Q and P marked. For example:

    Q – – – –
    Q Q – – –
    Q Q – – –
    – – – – P
    – – – – –



  30. sauravshakya on March 27, 2012 at 5:06 am

    I WAS NOT ABLE TO CONCENTRATE ON MY PHYSICS TEST UNTIL I SEND THE ANSWER. SO I DECIDED TO SEND THE ANSWER AND GO BACK TO STUDY.
    BYE…
    SEE YOUR PROBLEMS AFTER 2 WEEKS



  31. sauravshakya on March 27, 2012 at 4:20 am

    i have the solution for this question but i dont know how to send it.
    Now i will be missing for next two weeks as my exam is starting from tomorrow. But i promise that i can solve your all problems and i will do it when i will return.



  32. Steven Miller on March 13, 2012 at 4:01 am

    sure — drop me an email at sjm1 AT williams.edu



  33. prashant kumar on March 13, 2012 at 1:40 am

    Hi Steven Miller, I have figured it out. I would like to know the mathematical proof of this problem.



  34. Steven Miller on February 21, 2012 at 8:57 pm

    Andrew: correct: email me at [email protected] if you want my comments on the soln. //s



  35. Steven Miller on February 6, 2012 at 3:41 am

    sure, sending //s



  36. fathi on February 6, 2012 at 12:27 am

    almost a week i am trying to solve it. it was a hard one. i gave up i want to know the solution pleeeese.



  37. Steven Miller on February 4, 2012 at 7:30 pm

    It’s a hard riddle, but the solution is connected with some of the most important and applicable math there is. Email me at [email protected] if you want a hint.



  38. mukuro on February 4, 2012 at 4:06 pm

    oh, your other puzzle is hard, you know



  39. Steven Miller on January 10, 2012 at 3:08 pm

    That’s true (like tic-tac-toe), but there’s something even BETTER than symmetry which reduces the number of trials even further! Email me at [email protected] if you want to chat more.



  40. Ike on January 10, 2012 at 11:51 am

    Great riddle! Keeping symmetry in mind really lets you reduce the number of trials you need.



  41. Steven Miller on December 31, 2011 at 3:06 pm

    glad you’re enjoying it — it’s related to some of the most important applied math in use!



  42. mukuro on December 31, 2011 at 11:29 am

    ahahaha, this puzzle is soooooooooo good!



  43. Steven Miller on December 12, 2011 at 2:25 am

    check again — that won’t leave 3 squares safe; it will leave at most 2.



  44. Kareem on December 11, 2011 at 10:27 pm

    put 4 queens in the top left corner. one queen bottom left corner. you have exactly 3 space that are untouchable.



  45. Steven Miller on December 7, 2011 at 3:56 pm

    haha, cute.



  46. Jim on December 7, 2011 at 3:43 pm

    If the Queens are all white and the pawns are too, then your all set.



  47. Steven Miller on December 1, 2011 at 2:07 am

    sure, I’ll email. this is one of my favorite problems. you need to do some brute force computations, but if you’re clever you can cut down a lot. this leads to a lot of great math. I’ll eventually write a lot about this for the teacher’s corner (which I hope to do this spring).



  48. Jake on November 30, 2011 at 10:07 pm

    After struggling for a couple days with this problem, I figured it out, but only through logical trial and error. I’d love to know a combinatorical/algorithmic/logical solution to this problem. Also, I really enjoy many of your other riddles as well.



  49. Steven Miller on November 28, 2011 at 2:14 am

    I do not believe that works. Try putting the 5 queens on a board as you say — I believe you’ll find that you cannot place 3 pawns



  50. Ben on November 28, 2011 at 12:22 am

    Bunch the queens up on the black spaces around the main white diagonal or vice versa as close to on corner as possible. The pawns go on the diagonal as close to the opposite corner as possible.



  51. Steven Miller on November 23, 2011 at 2:37 am

    If you fill up an entire row or column then all squares are killed and you cannot place any pawns safely



  52. Josh H on November 23, 2011 at 1:11 am

    Put queens in A1, A2, A3, A4, A5, which leaves B7, B8 and C8 open for the pawns



  53. Steven Miller on November 22, 2011 at 11:58 am

    I don’t think this leads to 3 pawns safely positioned — it isn’t my soln and I thought the soln was unique — can you double check?



  54. Jeremy Lichtman on November 21, 2011 at 11:07 pm

    A1.A2. B1,B2,D1

    in other words 1st Column 2 spaces are queens, same for Column 2 then Column 4 1st space is a queen.



  55. Steven Miller on November 7, 2011 at 1:51 am

    no — this only leaves two squares open — E2 isn’t open



  56. Tommy Le on November 6, 2011 at 6:42 pm

    Easy. Again. Put the queens in an X on the bottom left of the chess board (A1 A3 B2 C1 C3) This will leave a D5, E4, E2 open for the pawns.



  57. Steven Miller on October 25, 2011 at 7:01 pm

    mukuro: nice job, glad you enjoyed it



  58. Steven Miller on October 25, 2011 at 1:57 am

    this covers all spaces — it doesn’t leave 3 spaces free for pawns.



  59. Yorik on October 24, 2011 at 7:05 pm

    first row: 2
    2nd row: 4
    3rd: 1
    4th: 3
    5th: 5

    right? and i have no idea how to do this by math, i just play chess sometimes and the solution just jumped into my head when i read the riddle.



  60. Steven Miller on October 19, 2011 at 2:09 am

    Sure. This is a great example of duality — solve a related problem. Try instead to place 3 queens so that 5 pawns are safe, and then flip the pawns and queens!



  61. Jon on October 19, 2011 at 12:37 am

    I would love some logic as to how to solve this problem other than guess and check. I have solved the riddle, but am unsatisfied not knowing the logic used.



  62. Steven Miller on August 30, 2011 at 3:01 am

    OK. I can give a hint that doesn’t give away too much, just some advice on how to help with the trial and error. The idea behind the hint is one of THE most important observations in mathematics!



  63. Charity on August 30, 2011 at 2:44 am

    Okay, I can only get it so there are 2 open spaces. I put queens in all 4 corners and then you can put the remaining queen in the middle of any side. I took combinatorics in college (math ed BA and MA) but I would have to pull out my book to use anything other than trial and error and logic which is what I’m relying on right now and not getting very far. I don’t know if I want a hint yet. I just wanted to leave a comment I guess.



  64. Steven Miller on August 18, 2011 at 7:39 pm

    Not following — please send more details to me at sjm1 AT williams.edu



  65. Anonymous on August 18, 2011 at 6:07 pm

    a1 a2 a3 a4 a5 = queens



  66. Steven Miller on June 28, 2011 at 12:50 pm

    Not sure what you mean — can you email me a board with the pieces placed?



  67. jim on June 28, 2011 at 12:37 pm

    A1,2,3,4,5?



  68. Steven Miller on June 21, 2011 at 8:53 pm

    Are the solutions distinct or are they just rotated / reflected versions of each other? There are 9 possible first moves in tic-tac-toe, but really there are only 3 by symmetry.



  69. Korbin on June 21, 2011 at 7:12 pm

    if you number them A-E and 1-5 C1,D1,A2,D2,A3 an interesting question would be how many posible solutions are there. from my solution there are at least 8 i believe.



  70. Steven Miller on May 3, 2011 at 1:49 am

    I’ll send you the answer. This is one of my favorite problems in terms of applications.



  71. Petur on May 2, 2011 at 9:01 pm

    I rly want to know the answer 🙂



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