Printable View
Well I may have not understood your problem because I only needed the first equation (3 equation for 3 unknow variable)..
Without calcul error solution may looks like...(and assuming cross product is vectorial product)
x1=(b1a1+b2a2+b3a3)/(a1a3-a2a3);
x2=(b1a1+b2a2+b3a3)/(a2a1-a3a1);
x3=(b1a1+b2a2+b3a3)/(a3a2-a1a2);
where 1 2 3 are indices for x y z axes...
A x X = B (x is cross-product)
A . X = b (. is vector dot product)
Solve for X in terms of A, B, b, and the magnitude of A.
Now
||A x X|| = ||A||*||X||*sina (alfa is angle formed between A and X)
||A . X|| = ||A||*||X||*cosa (alfa is angle formed between A and X)
Thus:
||A . X||² + ||A x X||² = ||A||²*||X||²*(cos²a + sin²a) = ||A||²*||X||²
or
sqrt(B . B + b*b)/(||A||) = ||X||
The direction of X is given by B x A. So we have the direction and the size, which should be sufficient.
False.Quote:
Originally posted by Simon666
The direction of X is given by B x A. So we have the direction and the size, which should be sufficient.
Let A=[1 0 0] and X=[1 1 0]; Then B=[0 0 1] and BxA is not in the direction of X.
Use the Vector Triple Product rule.
Ax(BxC)=B(A.C)-C(A.B)
Using AxX=B and A.X=b .
Ax(AxX)=A(A.X)-X(A.A)
Ax(B)=A(b )-X||A||^2
so X=(b *A-AxB)/||A||^2
I'm soooo sorry I didn't get back earlier! I didn't think anyone would respond so quickly (normally this thread lies dormant for quite some time) and I've been busy with work, and I didn't feel like it, and my chi was all bunched up, and... Anyway, Tom Frohman got the full answer, but I actually had solved it pretty much along the lines of Simon666 (but with the correct direction calculated by rotating the frame so that A and X were in the (x,y) plane -- thus B is along the z axis -- and then re-rotating on completion). But I really like Tom's answer, since its so much more direct and elegant (using the triple product rule seems to bring out the reasoning for the direction of X much better than the rotation).
Pierre31, notice that there are actually an entire family of solutions to each equation (the tranformation in each equation is not invertible).
Note that both Tom's and Simon's answers are the same for length. Since A and B are perpendicular, AxB has length ||A|| ||B|| along the direction perpendicular to both A and B (using the right-hand rule). b A is perpendicular to this last vector (since its in the direction of A), so the sum (or difference as this case is) is just that of two edges of a right-triangle and the theorem of Pythagoras applies. You can then factor out a length of ||A|| which cancels a term in the denominator.
Anyway, I think the point belongs to Tom, but either Tom Frohman or Simon666 can post the next question if they like.
Well I'm not much on constructing puzzles so I'll give you a checker problem. Sorry its either that or a chess problem.
This is a stroke problem. White is moving up the board and red is moving down.
White has a sequence of moves that takes all of the red pieces by force. White will make 7 moves all told(multiple jumps still count as 1 move in this tally).
White moves first.
Checker notation. If you look at the board and position in the enclosed file you will see the squares of the board are numbered 1 to 32 starting at the top left green square and moving left to right and then down.
If white moves the piece at the lower left out of the corner the move is recorded as 29-25 (not the right first move.) For jumps and multiple jumps just name the squares that the piece lands on. Omit the square jumped over.
I was able to remove all red pieces by force but I moved the red pieces in such a way that if red was actually trying to win then they would never do this. Here goes.
W: 6->2
R: 13->17
W: 21->14
R: 24->27
W: 14->10
R: 20->24
W: 10->3->12->19->28
R: 27->31
W: 22->18
R: 31->26
W: 30->23
That's only 6 moves by White. One assumption that I made was the White piece that made it to spot 3 it was kinged and allowed to keep going with the jumping. If not then it would be 7 moves. Just keep the white there and continue the jump on the next turn while red moves continue in the order listed. It’s been a while since I played checkers so I might be way off.
-Ben
Not the solution.
In the actual solution red can only make one choice at each move.
That is a huge hint.
On each turn, red will only have one choice as to what move he may legally make. For this to be true either there is only 1 move available to red as all the other pieces are gone or tied up or else red has to take a jump (if a jump is available you are required to take it.)
Tom I'm not really good familiar with the rules of this game so tell me:
If I can jump over to take a piece of my opponent am I obliged to by these rules, or can I make another move?
If a jump is available, you must take it.
If during a jump you land on the last rank and are "kinged", the move ends there.
A piece that is alreay a king keeps jumping if it can.
Forcing your opponent to jump is the most important principle of this combination.
I play by the rules of English Draughts. I think there are 14 million variants of checkers played around the world. I guess I really didn't think about this. In the US almost everybody has played checkers when they were young and almost everybody knows the rules.
Ok I got it. I'll write it out in columns, 1st white, 2nd red, with all jump locations for clarity.
I didn't know that one must jump if available. Are those tournament rules? Also, the big thing that kept sticking me was I really had this desire to go 06 - 02 instead of 06 - 01, and I have no idea why. It wasn't until I saw this that I was able to solve it today (I had the set up ready last night and it took me that long to see that step -- I think its the first step where white has a choice on how to force black). Anyway, it was a nice problem indeed!Code:22 - 17 13 - 22
29 - 25 22 - 29
30 - 25 29 - 22
21 - 17 22 - 13
06 - 01 13 - 06
01 - 10 - 03 - 12 - 19 - 28 20 - 24
28 - 19 win
Well done, we have a winner.
Having to take a jump is a rule in most checker variants. It is the basis of most checker combinations.
ok I have a riddle now. Just wait a sec for me to post it ;)
there are four men that need to pass a bridge through the night. They are soldiers and two of them are injured, one - seriously.
they have a flashlight and the bridge cannot be passed without a flashlight and it can carry no more than two men.
the first onw can pass the bridge for 1 minute, the second for 2 the third, slightly injured man, for 5 and the fourth for 10.
In 17 minutes the birdge will be blown to pieces so they need to pass it within that time.
How will they do it?
Let me name these 4 soldiers for A( 1minute needed to pass the bridge) , B(2 min), C(5 mins), D(10mins).
First, A and B pass the bridge with the flashlight.
Then A comeback with flashlight.
Time Now: 3 mins.
Second, C and D pass the bridge withe the flashlight carried back by A.
B carry the flashlight and pass the bridge to get A.
Time Now: 3+10+2 = 15 min.
Last, A and B pass the bridge in the left 2 min..
Time Now: 15+2 = 17 min..
The bridge is blown.. No enemy is coming..