Comments on: Peons http://yaniv.leviathanonline.com/blog/riddles/peons/ Strain your Brain Tue, 31 Aug 2010 17:39:33 -0700 hourly 1 http://wordpress.org/?v=3.0.1 By: srulix http://yaniv.leviathanonline.com/blog/riddles/peons/comment-page-1/#comment-8981 srulix Fri, 16 May 2008 11:27:43 +0000 http://yaniv.leviathanonline.com/blog/riddles/peons/#comment-8981 The core of the solution is this: Assume your target is (c,5) for some c. To each square assign "potential" function p(x,y). The potential of the board is the sum of p(x,y) for (x,y) which have a pawn on them. The potential function must be carefully-chosen, you need to show that the potential of the board stays constant (or decreases) for each move. Now, you're left with showing that p(c,5) is greater than the original potential (or exactly equal, since the number of steps must be finite). The initial thought is to set p(x,y) with something that resembles Fibonacci series, since p(x,y)>=p(x,y+1)+p(x,y+2). But this results with an alternating and unbounded potential :(. That thought must be a tad-bit refined, and you'll have the solution. Great riddle, I'm putting it on my page :) The core of the solution is this: Assume your target is (c,5) for some c. To each square assign “potential” function p(x,y). The potential of the board is the sum of p(x,y) for (x,y) which have a pawn on them. The potential function must be carefully-chosen, you need to show that the potential of the board stays constant (or decreases) for each move. Now, you’re left with showing that p(c,5) is greater than the original potential (or exactly equal, since the number of steps must be finite).

The initial thought is to set p(x,y) with something that resembles Fibonacci series, since p(x,y)>=p(x,y+1)+p(x,y+2). But this results with an alternating and unbounded potential :( . That thought must be a tad-bit refined, and you’ll have the solution.

Great riddle, I’m putting it on my page :)

]]> By: yaniv http://yaniv.leviathanonline.com/blog/riddles/peons/comment-page-1/#comment-386 yaniv Thu, 22 Nov 2007 00:18:28 +0000 http://yaniv.leviathanonline.com/blog/riddles/peons/#comment-386 Well, I think its time I posted a hint. <strong>WARNING - SEMI-SPOILER!</strong> It is actually impossible to achieve the required task! You cannot move a peon to 5 squares above the line. Now try to prove that... PS - Seb, you're numbers are pretty close. I do not really remember, but I think that the required number for getting to 4 squares above the line is 20. Anyways, proving 5 is impossible is much nicer than calculating the required number for 4. 8-) Well, I think its time I posted a hint.

WARNING – SEMI-SPOILER!

It is actually impossible to achieve the required task!
You cannot move a peon to 5 squares above the line.
Now try to prove that…

PS – Seb, you’re numbers are pretty close. I do not really remember, but I think that the required number for getting to 4 squares above the line is 20. Anyways, proving 5 is impossible is much nicer than calculating the required number for 4. 8-)

]]> By: Seb http://yaniv.leviathanonline.com/blog/riddles/peons/comment-page-1/#comment-297 Seb Wed, 07 Nov 2007 21:03:43 +0000 http://yaniv.leviathanonline.com/blog/riddles/peons/#comment-297 If n is the “height” of a single peon above the line (n=1 being the first square above the red line), then the first few terms are n=0 –> 0 (no peons are needed) n=1 –> 2 (the first example) n=2 –> 4 (the second example) n=3 –> 8 (since you first need a n=2 peon and need to reconstruct the “cannon” below the line) It would of course be esthetical to have n=4 –> 16 but this is clearly not the case. I have found a n=4 --> 21 solution, so it is clear that the solution for n=5 is at least higher than 29. If n is the “height” of a single peon above the line (n=1 being the first square above the red line), then the first few terms are
n=0 –> 0 (no peons are needed)
n=1 –> 2 (the first example)
n=2 –> 4 (the second example)
n=3 –> 8 (since you first need a n=2 peon and need to reconstruct the “cannon” below the line)

It would of course be esthetical to have n=4 –> 16 but this is clearly not the case. I have found a n=4 –> 21 solution, so it is clear that the solution for n=5 is at least higher than 29.

]]> By: Seb http://yaniv.leviathanonline.com/blog/riddles/peons/comment-page-1/#comment-296 Seb Wed, 07 Nov 2007 20:25:38 +0000 http://yaniv.leviathanonline.com/blog/riddles/peons/#comment-296 Sorry about that. I thought all comments went automatically to the second page. My mistake - but since it's far from a solution, I guess I won't have spoiled the fun. Sorry about that. I thought all comments went automatically to the second page. My mistake – but since it’s far from a solution, I guess I won’t have spoiled the fun.

]]> By: yaniv http://yaniv.leviathanonline.com/blog/riddles/peons/comment-page-1/#comment-294 yaniv Wed, 07 Nov 2007 19:32:36 +0000 http://yaniv.leviathanonline.com/blog/riddles/peons/#comment-294 Hi Seb, Your numbers are right ;-) Now, although your comment is not a solution, (it is at most a very slim hint), in the future, please post such comments on the second page of riddles, so as not to spoil the riddle for people that haven't solved it yet. That said, this type of comments is excellent, and please keep posting such insights. On article posts, you are welcome to comment anywhere (i.e. you cannot "spoil" an article). Thanks! Hi Seb,

Your numbers are right ;-)

Now, although your comment is not a solution, (it is at most a very slim hint), in the future, please post such comments on the second page of riddles, so as not to spoil the riddle for people that haven’t solved it yet. That said, this type of comments is excellent, and please keep posting such insights.

On article posts, you are welcome to comment anywhere (i.e. you cannot “spoil” an article).

Thanks!

]]> By: Seb http://yaniv.leviathanonline.com/blog/riddles/peons/comment-page-1/#comment-292 Seb Wed, 07 Nov 2007 16:25:51 +0000 http://yaniv.leviathanonline.com/blog/riddles/peons/#comment-292 If n is the "height" of a single peon above the line (n=1 being the first square above the red line), then the first few terms are n=0 --> 0 (no peons are needed) n=1 --> 2 (the first example) n=2 --> 4 (the second example) n=3 --> 8 (since you first need a n=2 peon and need to reconstruct the "cannon" below the line) It would of course be esthetical to have n=4 --> 16 but this is clearly not the case. If n is the “height” of a single peon above the line (n=1 being the first square above the red line), then the first few terms are
n=0 –> 0 (no peons are needed)
n=1 –> 2 (the first example)
n=2 –> 4 (the second example)
n=3 –> 8 (since you first need a n=2 peon and need to reconstruct the “cannon” below the line)

It would of course be esthetical to have n=4 –> 16 but this is clearly not the case.

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