ADN

Here's a riddle for the REAL genuises of the board...Not the photographic memory fakers like me, lol...

Let‚´s assume you got a chessboard and destroy it so that you have 32 white and 32 black parts.

Now you want to put it together again and every part must be put in the position where it was before

(rotation possibilities off the single stone not included).

How many combinations do you have to try in the worst case..?

to be honest, your riddle has no answer. there are too many process indicators that need to be collected./

for example:

WORST CASE implies too many variances.

A person of fit mind and ability has many less scenarios than, say, a blind quadrapalegic with schizophrenia working in an anti-grav field?

but even more practically, you did not mention if the squares are double sided, and what color the background or border is.

But if you simply want to know how many possible combinations of assigning a set of 32 black squares and 32 white squares into a 8x8 grid, the answer is 528+528, also written as 2(32!), or summed as 1056.

but that is not a riddle, that is just a factorial sum, specifially the factorials of two subsets of squares. Some people would say 2080, the factorial of 64, but that is not so. You see, the black squares and white squares have to be considered independently as two subsets.

This means there are 32! + 32! combinations, not 64! combinations.

So, at most, there are 2080 combinations of 64 squares into a specific grid, and if these are reduced to black and white subsets and "checkered", than you have 528 possibilities for the black and 528 possibilities for the white. This makes 1056 possiblities, assuming you know the simple rule that the king's starting square is opposite his own color, and you had the same orientation as when the board was disassembled.

beyond that, we go into innumerable process indicators... like was the board numbered?

hmm, see this is gonna bug me,.. to many details missing.

I presumed in worst case there would be an indicator that it was in the right square ( like a bing bing noise, or a approving nod of some kind.)

but if there is no indicator a piece is right, the person could work on this until their death, and still not solve it.

but Strictly speaking if it were that featureless and there was no indication that a specific piece was positioned as the original, if you take that the only criteria for attribution is that it have the white/black checker pattern, the person could say "by the authority vested in the rules of chess, this is a chess-board that I have fashioned from a chess board. It is accomplished." and it would be right on the basis it could function as the original, and contained no new parts, or lacked any original.

and see, then we go into process indicators, like "if there is an indication that a single row is correct" we get a much larger number than the "if there is an indication for each correct square"

raum, i think what ayhja meant to say is you got the 64 pieces and you have to try all the possible combinations of them so to reach the original state, this number is reduced by knowing that the order is white, black, white, black, ... etc. the worst case is the maximum number of different combination of the pieces (with proper alternating colours)

are we supposed to account for the possibility of rotating a square or are we to presume that each square has been maintained in it's original orientation and cannot be changed? In other words, even if I only have 1 square and one place to put it, I still have four possible ways of placing it (even if the result looks the same under each scenario)... If the squre is the same color on both sides, I've doubled my possibilites to 8 all with just one square

...being able to rotate each square with a different side forward greatly complicates matters as would having the squares color repeated on both sides....


*edit* sorry, just re-read the original where rotation was specifically addressed and not part of the equation. I guess reading is NOT over-rated... LOL

you end discovering an algorithm to do the process

you pick a random first piece (this number must be multiplied by 64), so the next should only be a white or a black (so you multiply by 32 since you have 32 pieces left of the other colour), then (by 31 since you retired the first piece of the other colour in the first place), then (31), then (30), then (30) ...

fuck it's kind of harsh for me to explain it in english but i think you can have an idea of what i meant.

it's simple factorials;

you take say, a specific black piece, and find the place where the white king would be...

the one that "lights up" is the one you put there,.. at most you have to try all 32. then, you go to the next place a black square would be, and try the 31 remaining.

so you would have AT MOST (assuming it lets you know you have the right one)

32 tries
31 tries
30 tries
29 tries
28 tries
27 tries...

and all the way down, to where you only have one square and one space. if you get one right, there will always be one less to try...

Therefore, you have the sum of the numbers 32 through 1... That is 32!, or 528 tries, for the black squares... and another 528 for the white.

that is the bold mathematics of it:

(32+31+30+29+28+27+26+25+24+23+22+21+20+19+18+17+16+15+14+13+12+11+10+9+8+7+6+5+

4+3+2+1)*2

Solution = 64*32*31*31*30*30*....*2*2*1*1

QUOTE(raum @ Oct 26 2006, 03:51 PM) it's simple factorials;

...

... That is 32!, or 528 tries ...

(32+31+30+29+28+27+26+25+24+23+22+21+20+19+18+17+16+15+14+13+12+11+10+9+8+7+6+5+

4+3+2+1)*2

CDR!!!!!! "Simple" factorials that's a new term? you better pray to your God to make that true "32!=528"

I hate to be the one to point this out to you, Raum....but.....
...
...
...
you forgot to add 0 to BOTH the white AND the black squares....