Could someone provide the reasoning behind the answer 60?

[nasi112]

In Chess960, the starting position must meet specific criteria, including placing bishops on opposite-colored squares. How many of the 960 legal starting positions feature the bishops specifically on squares c1 and f1?

I wrote all combinations.

01. RQBKRBNN
02. RQBKNBRN
03. RQBKNBNR
04. QRBKRBNN
05. QRBKNBRN
06. QRBKNBNR
07. RKBQRBNN
08. RKBQNBRN
09. RKBQNBNR
60. RKBNQBNR

My answer is overly complicated.

 

[Dr.Peterson]

Have you tried any other approach?

You could start with the complete rules, which you chose not to state:

Chess960 - Wikipedia

en.wikipedia.org

That shows how they get 960; just remove the [imath]4\times4[/imath] at the start of the calculation (because the bishops have already been placed), and you get 60.

 

[nasi112]

Did you read the page you sent me? I have a question about it if you don't mind.

 

[Dr.Peterson]

I didn't read all of it, but you can still try asking.

 

[JeffM]

I am not even sure what your question is. Dr. Peterson gave you a very clear-cut answer to the question asked in the title of this thread. If you are asking how they get 60 from 960, then the answer is that one bishop can go on any one of the four white squares and the other bishop can go on any one of the four black squares for a total of 16 configurations for the pair of bishops. And it is simple arithmetic to compute

[math]\dfrac{960}{16} = 60.[/math]
If you are asking about how 960 is derived, you never asked that question, but I might start with thinking about the king and the rook pair.

 

[Dr.Peterson]

I am not even sure what your question is.

As I understand it, the question is,

How many of the 960 legal starting positions feature the bishops specifically on squares c1 and f1?

and specifically how you can get the answer, 60, without listing.

I searched just to find the rules, and accidentally found that Wikipedia shows the calculation for the 960, so I figured I'd just give that as a hint. It was not intended as a primary answer.

But if I were working it out directly, I would start with the rules:

White's pieces (not pawns) are placed randomly on the first rank, following two rules:​

1. The bishops must be placed on opposite-color squares.
2. The king must be placed on a square between the rooks.

Since the bishops have already been placed, we just need to count the ways to place the rooks, king, queen, and knights. Placing the king and rooks amounts to choosing any 3 spaces of the remaining 6, which can be done in C(6,3) = 20 ways. Then the knights can be placed in any 2 of the remaining 2 places: C(3,2) = 3. The queen then has one place to go.

This gives us 20 * 3 = 60 ways to do it.

I was hoping to get an actual response to what I'd said.

But welcome back, @JeffM! I just visit this desert from time to time to see if there's anyone out here needing help ...

 

[JeffM]

I am not even sure what your question is. Dr. Peterson gave you a very clear-cut answer to the question asked in the title of this thread. If you are asking how they get 60 from 960, then the answer is that one bishop can go on any one of the four white squares and the other bishop can go on any one of the four black squares for a total of 16 configurations for the pair of bishops. And it is simple arithmetic to compute

[math]\dfrac{960}{16} = 60.[/math]

As I understand it, the question is,

and specifically how you can get the answer, 60, without listing.

I searched just to find the rules, and accidentally found that Wikipedia shows the calculation for the 960, so I figured I'd just give that as a hint. It was not intended as a primary answer.

But if I were working it out directly, I would start with the rules:

White's pieces (not pawns) are placed randomly on the first rank, following two rules:​

1. The bishops must be placed on opposite-color squares.
2. The king must be placed on a square between the rooks.

Since the bishops have already been placed, we just need to count the ways to place the rooks, king, queen, and knights. Placing the king and rooks amounts to choosing any 3 spaces of the remaining 6, which can be done in C(6,3) = 20 ways. Then the knights can be placed in any 2 of the remaining 2 places: C(3,2) = 3. The queen then has one place to go.

This gives us 20 * 3 = 60 ways to do it.

I was hoping to get an actual response to what I'd said.

But welcome back, @JeffM! I just visit this desert from time to time to see if there's anyone out here needing help ...

First, I hope that you did not view my response as being critical in any way of your response, which I thought on point and totally clear. I was merely objecting to the OP’s vagueness as to what his(her) question(s) might be.

Second, I remember your re-iteration of the point that combinatorics and probabilities do not seem to match up with native intuition and that therefore they require unusually disciplined thinking. My point about the rooks and the king led me down this path: once the bishops are located, there are six squares left, and the king and rooks take three of them. So, that is 6! / 3!, or 120 possible selection of squares. But we do not care whether the three squares are c1, b1, and d1 or, for example, d1, c1, and b1. So, there are 20 unique ordered sequences of squares. And the king must be in the middle of the sequence.

Thereupon, the queen can be ion any of the three still-unused squares. Once a square is selected for the queen, where the knights go is fully determined, and, because the two knights are indistinguishable in terms of chess, their placement can be ignored. That leads to:

[math]16 \times 20 \times 3 = 320 \times 3 = 960.[/math]
I suspect that you can analyze this in different ways.

P.S. The completely pointless hack to this site (and the failure to repair it quickly) created the desert. One hopes that such a once rich site can be restored.

 

[Dr.Peterson]

First, I hope that you did not view my response as being critical in any way of your response, which I thought on point and totally clear. I was merely objecting to the OP’s vagueness as to what his(her) question(s) might be.

Not at all; I was merely clarifying my own understanding of the question, and providing the background to my answer and the somewhat odd direction from which I approached it. And my original answer was in part my own way of commenting on the vagueness of the question, which didn't even state the rules.

 

[nasi112]

I didn't read all of it, but you can still try asking.

To get the 960 positions, the page started with bishop one, bishop two, queen, knights and lastly king between rooks. Then it started again with bishop one, bishop two, knights, queen and lastly king between rooks. The same answer was obtained. It always chose bishops first and king between rooks last. What happen if I change the order completely? Is it possible to get the 960 positions with choosing king between rooks first?

 

[Dr.Peterson]

To get the 960 positions, the page started with bishop one, bishop two, queen, knights and lastly king between rooks. Then it started again with bishop one, bishop two, knights, queen and lastly king between rooks. The same answer was obtained. It always chose bishops first and king between rooks last. What happen if I change the order completely? Is it possible to get the 960 positions with choosing king between rooks first?

Probably. Give it a try.

Usually some orders will be easier to work out than others, but there isn't one correct way. In this case, the difficulty will be that you don't know how many of each color square you have left when you get to the bishops, so you'll have to consider separate cases.

And when someone asks, "is it possible to do it this other way?", I always say, try it! You learn a lot more that way than by asking others to do the thinking for you. You'll either learn why they didn't do it that way, or discover how much freedom you have in solving that kind of problem.

 

[nasi112]

Probably. Give it a try.

Usually some orders will be easier to work out than others, but there isn't one correct way. In this case, the difficulty will be that you don't know how many of each color square you have left when you get to the bishops, so you'll have to consider separate cases.

And when someone asks, "is it possible to do it this other way?", I always say, try it! You learn a lot more that way than by asking others to do the thinking for you. You'll either learn why they didn't do it that way, or discover how much freedom you have in solving that kind of problem.

I asked because I don't know how to calculate how many ways to place the king between rooks. Apparently the picture in the op is one way if we fix the placement of the rooks. The pattern I will follow is rooks, king, dark bishop, light bishop, queen, knights.

[imath]1 \times 1 \times 1 \times 4 \times 3 \times 1 + 1 \times 2 \times 2 \times 3 \times 3 \times 1 = 12 + 36 = 48[/imath]

There are 48 different positions if the two rooks are fixed on the squares a1 and e1.

 

[Dr.Peterson]

The pattern I will follow is rooks, king, dark bishop, light bishop, queen, knights.

[imath]1 \times 1 \times 1 \times 4 \times 3 \times 1 + 1 \times 2 \times 2 \times 3 \times 3 \times 1 = 12 + 36 = 48[/imath]

There are 48 different positions if the two rooks are fixed on the squares a1 and e1.

As I understand it, you are not actually placing the king and rooks first (that is counting ways to do so), but just picking the placement of the rooks in the picture.

But I can't tell what you are thinking, just seeing your calculation. Why do you need to add two terms? What does each represent? where does each number come from?

I'd expect you to be saying there are two ways to place the king using a light square, and one using a dark square, and so on. Maybe that's what you did.

Anyway, if this is part of a larger attempt to count the 960, then you need to repeat this for each possible placement of the rooks ...

 

[nasi112]

Yes I chose the placement of the rooks in the picture. With this set up, there are only three squares for the king between the two rooks but I can't just multiply by 3 because I am restricted by the squares left for the two bishops in each placement of the king. I need two terms because a king placed on a dark square will give different ways to place the two bishops than a king placed on a light square. Each number comes from the fact that after each placement the next number represents the squares that can be occupied by the next piece. I will explain [imath]1 \times 1 \times 1 \times 4 \times 3 \times 1[/imath]. The first 1 is because the rooks are fixed. The second 1 is the king can occupy only one dark square. When this happens, the dark bishop can only occupy one square which is the third 1 and the 4 because the light square can occupy four squares. The 3 is there because the queen now can occupy 3 squares and when the queen is placed, there are two squares left for the two knights which they can occupy them only one time and that's why there is 1 at the last. If you can't understand my explanation, I will try to explain it differently next time. Let's assume there are 50 valid ways to place the two rooks. Then my choice to place them in the squares a1 and e1 is just one way of the 50. If this one way gives me 48 positions, I assume the 49 remaining ways will give me the 960 - 48 other positions. It's cumbersome to calculate it in this way and I thought this is what was meant by placing king and rooks first. If this wasn't placing king and rooks first, how to do it?

 

[Dr.Peterson]

It's cumbersome to calculate it in this way and I thought this is what was meant by placing king and rooks first. If this wasn't placing king and rooks first, how to do it?

If I wanted to do it by placing rooks and king first (which I would not, because it will clearly be too complicated), I would want to add up as many cases as necessary, counting the number of ways to place these three pieces that would result in each possible condition for the bishops.

As soon as I thought about how many such cases there would be, I would change my plans! Having realized that the bishops are the hardest to handle, I would deal with them first!

 

[nasi112]

If I wanted to do it by placing rooks and king first (which I would not, because it will clearly be too complicated), I would want to add up as many cases as necessary, counting the number of ways to place these three pieces that would result in each possible condition for the bishops.

As soon as I thought about how many such cases there would be, I would change my plans! Having realized that the bishops are the hardest to handle, I would deal with them first!

Can you please show me one case of placing rooks and king first so I can see and tell why the method I used in my last post is not such one? I would like to see your calculations because I'm confused and I still think what I did was placing rooks and king first.

 

[Dr.Peterson]

Can you please show me one case of placing rooks and king first so I can see and tell why the method I used in my last post is not such one? I would like to see your calculations because I'm confused and I still think what I did was placing rooks and king first.

If the goal is to count all ways of placing all the pieces, then you have to include all ways of starting, not just one. You did show one case; but you'd need to list all cases, and work out the number of ways to do each of them. What you did could be a start, but before you work out one such case, you really need to talk about how many cases there are. And you might find you can reduce the number of distinct cases.

I have no desire to even start it, because I'd have to think about all distinct ways to start. In particular, there are C(8,3) = 56 ways to place the rooks and king, so you might list all 56 of those and make a term for each (rather than the two terms you wrote); or I might realize I could classify those according to how many light squares they use, and break it up into just 4 cases (those that use 0, 1, 2, or 3 light squares) and work from there. Maybe it's not really so bad, but it will be much more work than starting with the bishops.

Okay, since I've made this much progress, let's start there. My first step is to break it into those 4 cases. Then, for the first case, there are C(4,3) = 4 ways to place the rooks and king all on light squares; that leaves one light square for a bishop, and 4 dark squares for the other bishop, which can therefore be placed in 1*4 = 4 ways. Then we can count ways to place the 3 other pieces.

Then go on to the next case. And if you think enough, you may realize that in a sense there are only two cases ...

The point is, counting ways to place everything else for one given placement of the rooks does not count as "placing rooks and kings first", in the sense of counting ways to to that.

 

[nasi112]

I calculated manually the number of ways to place the king between the rooks and I found out it was 56. The same number you got with a simple calculation. Interesting! My method implemented symmetry. It made the calculations a little bit faster. It is like I calculated from the left side to the right side and I know the same will happen from the right side to the left side and I just multiplied the result by 2. This is actually allowing me to work on only 28 cases, and then double my result because of symmetry. In these 28 cases, I need only to focus on one colored-square at one time, dark, or light occupy by the king, and then double the result because of symmetry. With careful but somehow cumbersome tries, this method can get the 960 ways. Another easier way to get the 960 ways is to follow your other suggestions. Thanks, I think you explained your point well. We wasted a lot of time going far from the core topic of the op but gained new ideas. However, the 960 ways is also a core idea to understand the 60 ways of the op. The main point of all of this I knew why they chose to work with the bishops first. These two bishops become problematic if we postpone their order. All I need now is to go back to what @JeffM gave and try to understand it.

 

[Dr.Peterson]

The main point of all of this I knew why they chose to work with the bishops first. These two bishops become problematic if we postpone their order.

Yes, a key idea in learning to do this sort of thing is to be willing to see the same problem in multiple ways, and to recognize which are likely to be easiest. Sometimes none are easy, and you just have to work through a long series of cases!

 

[nasi112]

I am not even sure what your question is. Dr. Peterson gave you a very clear-cut answer to the question asked in the title of this thread. If you are asking how they get 60 from 960, then the answer is that one bishop can go on any one of the four white squares and the other bishop can go on any one of the four black squares for a total of 16 configurations for the pair of bishops. And it is simple arithmetic to compute

[math]\dfrac{960}{16} = 60.[/math]
If you are asking about how 960 is derived, you never asked that question, but I might start with thinking about the king and the rook pair.

I agree I didn't state my question clearly in the op as I already gave the answer. As the Doctor said I wanted to get the 60 ways without listing. I can see now how easy to get the 60 when you divide 960 by 16. Thanks for this answer. One can do that only when she knows the rules of the bishops in the Chess960. Did you know them in advance to reading the op or after?