Talk:Chess960 starting position

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Move to Wikibooks?[edit]

I saw a "move to Wikibooks" proposal, but this appears to be an encyclopedia article, not a Wikibooks article. It's not so much a "how to do something" as a compare-and-contrast of different systems. If there's a critical need to move it out, I'd like to hear why.

WikiProject class rating[edit]

This article was automatically assessed because at least one WikiProject had rated the article as start, and the rating on other projects was brought up to start class. BetacommandBot 07:56, 10 November 2007 (UTC)

Credit for single-die randomization method[edit]

Would like to call attention to this edit. In addition, here is some info on Ingo Althöfer. And here is some info about CSS magazine. The Chess960 article was started by User:Dwheeler with ref to Hans Bodlaender's 2002 article here. Ihardlythinkso (talk) 09:53, 4 January 2013 (UTC)

For User:Calmapalma to explain why he is trashing standard WP markup[edit]

Please justify here, Calmapalma, why you are creating your own markup instead of using standard markup at WP:LAYOUT. Ihardlythinkso (talk) 07:08, 27 January 2013 (UTC)

User:Calmapalma cards methods[edit]

User:Calmapalma, most of the cards-method explanations you have added appears to be your own WP:OR work and analysis. You cannot add your work here, you need to have WP:RS for it. (Can you supply?) Ihardlythinkso (talk) 09:30, 27 January 2013 (UTC)

Using a regular six-sided die, you don't have to re-roll...[edit]

It's possible to randomly generate a position with equal possibility with a regular six-sided die, one piece at a time, and never have to do a re-roll. The only thing is that your order in which you're determining which piece you're rolling for has to change depending on where the pieces have landed. It's complicated to write out, but it's quite simple visually. Basically you always choose to roll for a piece that has either 2, 3, or 6 squares to choose from, you never have to re-roll, since 6 is divisible by 2, 3, a six-sided die can act as a 3 sided die, and a coin toss all at the same time.

Example of one order you could do it in: Roll 1. The king can only be in one of six squares, so you start with placing him. B1 for rolling a 1, C1 for rolling a 2, etc Ok so now you have your king placed and 7 squares empty. Roll 2. Next, place a rook. If your king is on b1 or g1, then a rook must go in the corresponding corner square the king is near. If you need to roll for the rook then one side of the king will have either 2 or 3 squares empty, if there are two empty squares to the king's left then rolling a 1-3 will mean the rook goes on a1, rolling a 4-6 will mean a rook goes on b1. 3 empty squares you would attribute 1/2, 3/4, 5/6, to the corresponding first second and third square. You can do the same thing if there's two or three empty squares to the king's right.

Now there is a King and a rook and 6 empty squares. Roll 3, a queen: Place a queen on 1 through 6 of its corresponding roll. Now there is three pieces on the board.

Roll 4a: If there is 2 and 3 dark and light squares left respectively, roll for the dark or light squared bishop, whichever one has 2 possible squares. 1-3 for first open square from left, roll 4-6 for the second. Roll 4b: If there is only one dark or light square left, place a bishop on it. Roll 5: You'll either have either 2 or 3 possible squares for the bishop now, or in case you have 4, all same-colored squares left, you'll have 3 possible squares for the Rook. Roll for either piece which has three available squares. Roll 6: You'll have have either a rook with 2 or 3 possible squares or bishop with 2 or 3 possible squares. Roll for it in the same fashion as we have been. Place Knights in remaining squares. Like I said, it sounds complicated to give instructions for but when I set my board up for chess 960, I do it this way, and it's pretty intuitive and quick, you never have to roll twice for a piece, and it gives all possible positions equal opportunity. Note: It takes six rolls but there's a 1 in 3 chance roll 2 will not be necessary and a small chance that one of the bishops will not need a roll so the average number of rolls with my above described method is less than 6. Dancindazed (talk) 20:50, 23 April 2013 (UTC)

I'm not sure if your fundamental idea is valid or not, however, I do see some bugs in the instructions you describe. I'll list a couple cases ... Ihardlythinkso (talk) 21:38, 23 April 2013 (UTC)
  • You state in roll 5, in the case of 4 same-colored squares left, there will be 3 possible squares for the rook. But there are only 2 possible squares for the rook in this case: Rb1, Kd1, Qf1, Bh1.
  • In this position: Ra1, Kd1, Qf1, your roll 4a rule states to roll for the bishop with two possible squares. If the bishop goes to h1, your roll 5 states to roll for the piece which has three available squares. That would be the remaining bishop, let's say it goes to e1. But then your roll 6 states there will be left "either a rook with 2 or 3 possible squares or a bishop with 2 or 3 possible squares", but, both bishops are already placed, and the rook has only 1 square possible, so the roll 6 instructions that you've written are wrong in that case.
No, sorry, it's hard to explain but what I said was that you'll either have 2 or 3 possible squares for the bishop OR in case you have 4 possible squares for the bishop then you MUST have 2 or 3 possible places for the rook. In other words in the one case where you don't have 2 or 3 places for the bishop, you can revert to the rook which will have either, 2 or 3 possible squares, depending on where the king is. As for the second part, you are correct. If there is only one possible square for a piece during any of these instructions, then that is a given, and can be added. But my point does stand that 6 total rolls are the max possible with this method, and sometimes you can roll only 5 times. Dancindazed (talk) 23:28, 23 April 2013 (UTC)
So just to elaborate further, with your pointed things I left out, here's a better break down of the instructions once the method is understood.
1. King
2. If King on Left half of board, place left rook now, If king on right half of board, place right rook
3. then Queen
4. Place Bishop which has fewest possible squares (1 or 2)
5. Place final rook (it may have one possible square at this point)
6. Place final bishop and knights fall into place
Now that you point that out about step 5, I see it's actually possible to have it randomly selected in 4 rolls(!) all though it hasn't happened to me yet. Further edit: No I don't think it's possible with 4 because only one of the two rooks can ever be placed without a roll, either the first or the second... And the other thing is that basically for steps 5 and 6, in cases where the rook has 4 possible squares on step 5, reverse the order in which they're placed. Do the bishop first. That's the part that's confusing to write out. Dancindazed (talk) 23:42, 23 April 2013 (UTC)

Here's a problem: Kc1, Ra1, Qb1. And now according to 4 above, we place a dark-squared B. (Let's say it goes to e1.) Now step 5 says "Place final rook (it may have one possible square at this point)." First, I'm not sure what scenario would leave only one possible square left, but forget that for a minute ... there are four possible squares for the rook now. (How do you intend to place the rook with a single roll of the die?) Ihardlythinkso (talk) 01:13, 24 April 2013 (UTC)

I explained that. I said if the rook has four possible squares you go to the bishop. the last bishop and final rook can always be reversed. Dancindazed (talk) 02:03, 24 April 2013 (UTC)
When you have a chessboard handy, get a die and do it. You'll see it works quite briskly. As long as you remember the first King, Rook, Queen, Bishop set up, the rest will be obvious that you'll either do the bishop next or the rook next... and the Knights are just placed in the final two squares. Dancindazed (talk) 02:05, 24 April 2013 (UTC)
I don't see anywhere where you said "if the rook has four possible squares". I was following your latest list of instructions. (Perhaps make a complete instruction set, otherwise it is too confusing for me to examine/test.) Ihardlythinkso (talk) 02:16, 24 April 2013 (UTC)
Update: Ok, I see it. (I was working with your listed instructions only. You need to incorporate "exceptions" into the listed instructions, instead of as a "side-line", for clarity. I'll try and do that for you maybe ...) Ihardlythinkso (talk) 02:21, 24 April 2013 (UTC)
Just thinking about it further, I guess I should explain there are nearly equal scenarios in which the 5th piece placed will be the bishop or the second rook. It will be whichever one has the fewest available squares. It's almost a rule except the Queen, you're always placing next the piece in which has the fewest possible squares on which it's legally able to be placed. (The king has the fewest possibles squares and exactly six squares to start, the next placement of the rook has either 1, 2 or 3 possible squares, the Queen is the only one that breaks the rule it can go in all six squares, then from there on you're always placing the piece which has the least possible legal squares. It's the rule of 6 or fewest possible, you could say. Dancindazed (talk) 02:28, 24 April 2013 (UTC)
Question: Why in step 5 are you attempting to place the rook? (When conditionally that step will be switching to placing the 2nd bishop.) Why not place the second bishop in that step? (Is there a problem with doing that, according to your method? If no problem with placing the bishop in that step, then you get to skip the conditional complexity you have currently built-in, which was confusing me.) Ihardlythinkso (talk) 02:31, 24 April 2013 (UTC)
Update: Ok, I see that doesn't work. I see your point now. (What you need is a concise, complete instruction list including the conditionals. Do you want help writing that?) Ihardlythinkso (talk) 02:41, 24 April 2013 (UTC)
Argh, I wrote out a better, concise conclusion with the help seeing the confusion you've given, and I lost it in the update conflict (I thought I had it copied to my clipboard for backup and I didn't) Ok so working on writing it again. one sec..

Dancindazed (talk) 02:48, 24 April 2013 (UTC)

It's very interesting. I'm guessing your method is sound. (You came up with it on your own? What got you going on coming up with it? [This article? Or?] It falls under WP:OR, you know. [But hey!]) Ihardlythinkso (talk) 02:54, 24 April 2013 (UTC)
Roll 1: Place the King on one of its 6 legal squares based on its corresponding rolled number; 1 for B1, 2 for c1, 3 for d1, etc
Roll 2: Place the first rook. If the King is on the right half of the board, you're now placing the right-most Rook. If King is on the left, you're now placing the left-most Rook. The Rook will have 1, 2, or 3 possible squares. If 1, place the Rook. If 2 possible squares, use the die with the low 3 numbers (1-3) representing the left square, and the high 3 numbers (4-6) representing the right-most square. If there are three possible squares use the die split in thirds, (1,2 represent the left-most square, 3,4 represent the middle, 5,6 represent the rightmost) I will from here on be referring to this use of the die as "High/low" or "High/middle/Low"
Roll 3: Place the Queen. It has six possible squares, each number on the die has a corresponding square.
Roll 4: If only one light or dark square is left unused, place a bishop on it. If 2 dark squares are left for a bishop, roll high/low for its bishop. Do the same if 2 light squares are left available for a bishop.
Roll 5: Roll for or place the remaining Bishop or Rook, whichever has the fewest available squares. If either has 1 available square, place it. If either has two available squares, Roll high/low for it.
Roll 6: The Final Bishop or Rook may have 1, 2, or 3 available squares. If 1, Place it. If 2, Roll high/low for it. If 3, roll high/low/middle for it, and place the Knights in the empty squares.

Re; As to place it in the article, I think it's 100% original research so I don't think it can be put in the article. Maybe I can get a chess site to publish it and become referenced material. re; I did come up with it. It was just intuitively how to do it with a die to me. I know a six-sided die is very flexible as to how many numbers it can represent, and when I heard people were placing the pieces with dice, this is just how it was being done. Maybe a lucky guess, who knows! I've always had quite the analytic mind. That's why I like chess! Dancindazed (talk) 03:02, 24 April 2013 (UTC)

I want to quibble with your Roll 5. Take this case: Kd1, Ra1, Qc1. Now (Roll 4) we place dark-square bishop. (Let's say it goes to g1.) Now there are three possible squares for the remaining rook, and three possible squares for the remaining bishop. So your Roll 5 instructions aren't compatible for that case. (I suggest, you always place the remaining rook in Roll 5. Because there is never a problem in placing the remaining rook at that point, but there can be a problem placing the remaining bishop.) I could explain more but I think you understand what I mean. What do you think? (Have I overlooked any problem placing the rook always, in Roll 5 step? I don't see any issue.) Ihardlythinkso (talk) 03:48, 24 April 2013 (UTC)

No you're right, that was an oversight in the cleanup I did. Fact is a tie of 3 squares remaining, you can roll for either piece, the Bishop or the Rook, that's why I said that's the most confusing part, but, you can't screw it up because in the case of 3 and 3 rolling for either piece is fine. And if one has fewer, you can always roll for that one. Dancindazed (talk) 04:42, 24 April 2013 (UTC) p.s. re; "there is never a problem in placing the remaining rook at that point" Yes there is. Sometimes the rook has exactly 4 possible squares (Kc1 Ra1 Qb1 and Be1). In that case you're rolling for the bishop which has fewer legal squares (3) on roll 5. It's the rule of fewer.. Again tie goes to whatever you feel like on 5. It's only an issue with giving linear instructions. It's not an issue with the concept. The concept is this. Roll for the piece that has A: exactly 6 legal squares (that starts with the King and makes the Queen on roll 3) or B: fewest legal squares (That's what causes the nearest to the king's side rook to go second) or C: in case of a tie for fewest possible moves, you can pick either (only can occur when placing the 5th piece anyways) Dancindazed (talk) 04:50, 24 April 2013 (UTC)

Yup, you're right. (I overlooked the Kc1, Ra1, Qb1 example.) Agree with the validity now. Let me ask you this, do you agree with statement that your procedure requires a max of 6 die rolls, and can have a minimum of 5 die rolls (i.e., 4 die rolls aren't possible)? Ihardlythinkso (talk) 05:12, 24 April 2013 (UTC) p.s. Do you want to work here on the conciseness of the language of the instructions?
Just thinking, this is the only scenario in which the roller could have a choice, and if anyone would bring up the roller could somehow influence the final position from there, with his choice, it's not possible. Let's say you have the (RA1 QB1 KD1 AND BF1) set up, and you're looking at the 3 and 3 scenario. Let's say you would like it better if the bishop was on c1 for some reason now, between the queen and king. So you're thinking if you place the rook first, you have a 2 in 3 potential of decreasing the bishop's chances of going on g1 or e1. Well the odds come out the same. 2 in 3 that the Rook will be on e1 or g1, and then 1 in 2 if that succeeds that the bishop will be on one of the two remaining legal squares, the math comes out as 1 in 3 that the bishop will end up there if you place the rook first anyways. re; minimum/maximum rolls. Yes there's no way to do it in 4 rolls, now that I've looked over all the scenarios we've talked about. Whenever you skip the rolling for the first rook, you have to roll for the rest. And once you have to roll for the first rook, it's not possible to not need to roll twice after that, as far as I can see. However there are many places along the way in which no roll is a potential, and for each of the steps where its possible to place a piece on its "only legal square" and not roll, it brings the average rolls closer to 5. I'm okay at math, but I don't think I want to bother working on the average for that one. I'm sure it comes out somewhere between 5.5 and 5.75 rolls. Dancindazed (talk) 05:36, 24 April 2013 (UTC)
Ok to avoid another big wall of posting the steps again, I put something re-written on your talk page Ihardlythinkso, here -Dancindazed (talk) 05:52, 24 April 2013 (UTC)
Whelp, it pains me to say this, but our discussion has led me to realize that my entire method is flawed. And the conclusion came from trying to figure out why the bishop3 rook3 scenario occurs. Look at the first 2 placements. In my method, the first two pieces placed MUST be on the same half of the board. This gives a fixed chance that the queen will be on the same side of the King as 2 in 6 or 1 in 3 (there's always 2 squares empty on the same half of the board as the King and 6 total empty squares) after the first rook is placed. However the correct odds should be that the queen has a 3 in 7 chance of being on the same half of the board as the King. The "rule of the fewest" is fundamentally flawed. The Rule of the more needs to be followed to preserve complete randomness. And thus, 5 or 6 possible squares must be a possibility and re-rolls are probably needed with a six sided die. Oh well. I'm glad I figured out I was wrong. Dancindazed (talk) 16:33, 24 April 2013 (UTC)

Using polyhedrals[edit]

I don't understand the entry for this. If you were going to use all these odd shaped dice, why wouldn't you just use an 8-sided die for the first Bishop, a 4-sided for the next (just as it says) a six-sided for the queen (just as it says) but when they get to the Knight's dice they lose me. Just use a 5-sided die for the next Knight and a 4-sided die for the last Knight and place the King and rooks in the empty squares. Where's the need for a 12-sided and 20-sided die and counting and looping back like that? Dancindazed (talk) 17:25, 24 April 2013 (UTC)

The "five Platonic solids" are a polyhedral dice set consisting of one each of d4/d6/d8/d12/d20. (Check online to confirm. When you buy a polyhedral dice set they usually come as seven, with the five Platonics and two d10 decahedrons [one 0-9, one 00-90]. [1]) So using a d4 for the 2nd N would require a spare die of a different color (a convenience issue). As far as a "d5" -- is there such a thing? (Might be, I just dunno.) Does this answer? Ok, Ihardlythinkso (talk) 21:44, 24 April 2013 (UTC)
Yes, there is such a thing as a d5 (triangular prism of the right proportions that landing on the edges or faces are equally likely). You don't see it very often because small variations in the environment where it is rolled will make it unfair. That's a general problem with all traditional dice with an odd number of faces (except 1). You can get around this by using an n-gonal prism with the ends somehow modified so that they cannot be landed on, but why go to all that trouble when it's a lot easier to just use a 2n-sided die? Double sharp (talk) 10:13, 24 December 2013 (UTC)

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