13
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You may only use pieces in the original set, and all your pieces must be the same color. The king is not allowed. Pawns do not get promoted. Pieces do not control the square they occupy.

Accepted Answer goes to the person that has the least score.

Piece cost:

  • Pawn - 1
  • Knight - 3
  • Bishop - 3
  • Rook - 5
  • Queen - 9
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  • 1
    $\begingroup$ That's... Unreasonable. They are still puzzles -_- $\endgroup$ – warspyking Oct 14 '14 at 0:53
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    $\begingroup$ Yep, and they are tricky logic-based, constraint satisfaction puzzles - they aren't riddles however. Maybe you could try to pitch it as a riddle. Good night! $\endgroup$ – d'alar'cop Oct 14 '14 at 0:54
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    $\begingroup$ e.g. give the knight a character and personality... the are bounding across the kingdom as spies. some of the spies are double-agents so you don't want to encounter any of your own team - now it sounds less like a dry chess position where knight can't attack eachother $\endgroup$ – d'alar'cop Oct 14 '14 at 1:07
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    $\begingroup$ @justhalf: But if the original set cannot control the whole board, you pay for each piece you add. This restriction forces people out of uniform solutions based on lots of bishops. I don't have an answer, but think it is a good puzzle for that reason. $\endgroup$ – Ross Millikan Oct 14 '14 at 3:04
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    $\begingroup$ What do you mean by "You may not use any less than the original set."? $\endgroup$ – IQAndreas Oct 14 '14 at 6:13
14
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I think I've found a cost-33 solution:

Cost-33

As far as I can tell it's valid, but I've been staring at it for so long that I kind of don't trust myself anymore... so if you see a mistake, please tell me.

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  • $\begingroup$ New Accepted Answer! Great job! I wonder if less than 33 is possible. What do you think @d'alar'cop $\endgroup$ – warspyking Oct 17 '14 at 20:16
  • $\begingroup$ Do you think it's possible to beat 33? $\endgroup$ – d'alar'cop Oct 18 '14 at 4:45
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    $\begingroup$ Thanks guys. I don't think less than 33 is possible... this configuration has several pieces (Queen, a Bishop and a Rook) covering its maximum nr of squares, plus several paws that cover 2 squares that wouldn't otherwise be covered. But then again, it only requires one less pawn for a better solution :) $\endgroup$ – Astrotrain Oct 18 '14 at 7:48
11
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The best I've got at the moment is a cost-34 solution:

enter image description here


Due to clarifications, I have found a cost-35 solution that uses the pieces from one side (no king):

enter image description here

It was surprisingly hard, and I'd really like to see better.

Old answer:

I have a solution (no restriction on pieces - only goal is to minimise cost) that costs 30. It is actually from wikipedia:

enter image description here

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  • 1
    $\begingroup$ This does not use the full original set of pieces as asked. $\endgroup$ – Ross Millikan Oct 14 '14 at 2:50
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    $\begingroup$ The problem is OP also said "Accepted Answer goes to the person that has the least score", which implies that not every piece should be used. $\endgroup$ – justhalf Oct 14 '14 at 3:00
  • $\begingroup$ You have the best answer so far! Is it possible to get a score of 34? $\endgroup$ – warspyking Oct 14 '14 at 22:43
  • $\begingroup$ @warspyking I think it is possible to beat this solution. so I think 34 is possible.. but I cannot find it just now $\endgroup$ – d'alar'cop Oct 14 '14 at 22:50
  • $\begingroup$ @d'alar'cop: Btw, where do you create your image? It's good if we can have the same tool to generate the images :) $\endgroup$ – justhalf Oct 15 '14 at 2:28
6
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I believe I've found a cost-34 solution:

Chess puzzle 34


That was challenging. Just from what I've tried my (uneducated) guess would be that a score < 33 is not possible. In case it might give people ideas to try, here's another cost-35 solution:

Chess puzzle 35

And another cost-35 solution, which has the advantage of being a completely legal chess position (no pawns on first and last rows, bishops on opposite colours):

Legal 35 point solution

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  • $\begingroup$ Great job! But @d'alar'cop is trying to beat you. $\endgroup$ – warspyking Oct 16 '14 at 14:46
  • $\begingroup$ There is a lower. (Astrotrain right now) $\endgroup$ – kaine Oct 17 '14 at 14:43
  • $\begingroup$ Your 34-peice solution has only one flaw (not addressed by the question) and it's this: How do you get the pawn into that corner? $\endgroup$ – Zibbobz Oct 17 '14 at 19:46
  • $\begingroup$ @kaine I noticed, which is cool 'cause I was always just short of managing 33. I don't think it's beatable, but when I have time (life got busy), I'm going to attempt to find another 33 solution. $\endgroup$ – DPenner1 Oct 22 '14 at 1:41
1
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Invalid answer, need to be updated to conform with latest edits in the question

I found some other solutions which cost 30. I'm posting this in the hope someone else can improve any of these =)

Cost-30-attack-all-squares Cost-30-attack-all-squares-2 Cost-30-attack-all-squares-3

I found a near-solution which costs 28 (A6 and H6 are not covered):

Cost-28-attack-nearly-all-squares

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    $\begingroup$ nice. I reckon we can beat 30 $\endgroup$ – d'alar'cop Oct 14 '14 at 2:41
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    $\begingroup$ This does not use the full original set of pieces as asked. A full set is 39, so we can't beat that. $\endgroup$ – Ross Millikan Oct 14 '14 at 2:52
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    $\begingroup$ This is a great answer, and I really like it, however there are some clarifications you should look at, thanks for understanding. $\endgroup$ – warspyking Oct 14 '14 at 22:49
  • $\begingroup$ Ok, will edit later. $\endgroup$ – justhalf Oct 15 '14 at 2:28
1
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I found a nice solution, which definitely isn't perfect, but could evolve in a game (only pieces of one colour):

Ra8 Rh1 Qc3 Bd4 Bd5 Ne4 Ng3 Kf3 and pawns on c4, d6, e6, f5

 +--+--+--+--+--+--+--+--+
8|R*|  |  |  |  |  |  |  |
 +--+--+--+--+--+--+--+--+
7|  |  |  |  |  |  |  |  |
 +--+--+--+--+--+--+--+--+
6|  |  |  |p*|p*|  |  |  |
 +--+--+--+--+--+--+--+--+
5|  |  |  |B*|  |p*|  |  |
 +--+--+--+--+--+--+--+--+
4|  |  |p*|B*|N*|  |  |  |
 +--+--+--+--+--+--+--+--+
3|  |  |Q*|  |  |K*|N*|  |
 +--+--+--+--+--+--+--+--+
2|  |  |  |  |  |  |  |  |
 +--+--+--+--+--+--+--+--+
1|  |  |  |  |  |  |  |R*|
 +--+--+--+--+--+--+--+--+
  a  b  c  d  e  f  g  h

As I said, it isn't perfect, the score is 35 (39 - 4) and the King plays an active role in protecting g2 and g4. Still, assuming white pieces, the black king has nowhere to go.

After making the rules better understandable, and shooting the king, I have another solution that fits. Since the King is definitely more powerful than a pawn, albeit more vulnerable, I had to add a few pawns but could eliminate a knight.

Pieces are:

Ra8 Rh1 Qc3 Nc7 Be3 Bf3 and pawns on c5, d2, d6, e2, e6, f2, f5, g4

 +--+--+--+--+--+--+--+--+
8|R*|  |  |  |  |  |  |  |
 +--+--+--+--+--+--+--+--+
7|  |  |N*|  |  |  |  |  |
 +--+--+--+--+--+--+--+--+
6|  |  |  |p*|p*|  |  |  |
 +--+--+--+--+--+--+--+--+
5|  |  |p*|  |  |p*|  |  |
 +--+--+--+--+--+--+--+--+
4|  |  |  |  |  |  |p*|  |
 +--+--+--+--+--+--+--+--+
3|  |  |Q*|  |B*|B*|  |  |
 +--+--+--+--+--+--+--+--+
2|  |  |  |p*|p*|p*|  |  |
 +--+--+--+--+--+--+--+--+
1|  |  |  |  |  |  |  |R*|
 +--+--+--+--+--+--+--+--+
  a  b  c  d  e  f  g  h

This makes a score of 36 (39 - 3). Not a dream score but better than nothing.

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  • $\begingroup$ Nice, but the king didn't have a score, it cannot be used. $\endgroup$ – warspyking Oct 14 '14 at 18:16
  • $\begingroup$ @warspyking well, the entire challenge is a bit ambigious. At least it was as I searched for this solution. Every time I play chess, I have a King in my standard set and it is very well known that the King can play an important role in mating the opponent. (Try to mate an opponent with Q without K, just as a challenge). You changed the written rules (probably not the rules in your head), so I'll search for another solution and add it to this one. I still think it's nice. :-) $\endgroup$ – Ronald Oct 14 '14 at 18:28
  • $\begingroup$ Alright then :D Sorry about the confusion, the king had no score so I thought it was sorta implied. $\endgroup$ – warspyking Oct 14 '14 at 18:37
  • $\begingroup$ So far you've got the best answer, well done! $\endgroup$ – warspyking Oct 14 '14 at 18:59
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    $\begingroup$ well, actually it is (strictly speaking) the only answer (at the moment). The question is: should I feel flattered? ;-) $\endgroup$ – Ronald Oct 14 '14 at 19:01

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