brian bovril

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16 years, 88 days

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These are replies submitted by brian bovril

@acer thank

@acer Ok, I *thought* the way to get rid of the spaces was to convert the output to a string, because the DeleteSpace command was only available under the Strings package. 

The csv export works, I was just wanting an easier way to copy the maple output to my software.

Anyway, this is Tom's code from a previous post to illustrate.

restart:
  with(combinat):
  abil:=[1407,1408,1409,1411]:
    DBL:=choose(abil,2):
  score:=[2,0]:
    ans:=seq( [DBL[j][],score[1],DBL[numelems(DBL)-j+1][],score[2]], j=1..numelems(DBL)/2),seq( [DBL[numelems(DBL)-j+1][],score[1],DBL[j][],score[2]], j=1..numelems(DBL)/2);

#then to get rid of the square brackets and the spaces (because the software I am trying to copy-paste to doesn't like them)

#I do this

for i to 6 do
seq(ans[i][j],j=1..6);
next i
od

#Ideally I would like to copy this maple output and paste into the software, but it won't accept it, because its not plain text

@Carl Love @Kitonum thanks for the help solving my problem.

Edit: I have a supplementary question.

The end of the first 3 singles block ends with a doubles, and the second block ends with another doubles. If the first doubles involve XY and AB, is there a combination that minimizes the occurrence of XYAB in the preceding and post singles. Same thing for the 2nd doubles; this will involve XZ and AC.

I'm just trying to find a combination that will maximize rest times between opponents playing singles pre and post-doubles. 

@Kitonum I forgot to mention, each block of 3 singles matches is separated by a doubles match. So each 3 match batch is independent of the previous 3 match batch. So you could have one block end with [C,Z] and the next block start with [A,Z]. Basically you need A,B,C, X,Y,Z featured in each 3 singles batch. Apologies for not giving you all the info at the beginning..PS I'm glad I do have the iterator package, because it's a lot quicker..

@Carl Love thanks for your input. Is there any way to refine the list further, so no player from either team features twice in each block of 3. 

eg

[[A,X],[B,Y],[C,Z],[B,X],[A,Z],[C,Y],[B,Z],[C,X],[A,Y]]

@mmcdara and @Kitonum 

But what I was really after was a listing of 'practical' permutations, w.r.t the convention: A or B or C on the left, X or Y or Z on the right.

This is an example of an order I don't want:

 [[A, X], [A, Y], [A, Z], [B, X], [B, Y], [B, Z], [C, X], [C, Y],[C, Z]]

player A isn't going to play 3 times in a row ....(nor is B or C)

Here are two valid ones, i'm sure theres more

 [[A,X],[B,Y],[C,Z],[A,Y],[B,Z],[C,X],[A,Z],[B,X],[C,Y]]
 [[A,X],[B,Y],[C,Z],[B,X],[A,Z],[C,Y],[B,Z],[C,X],[A,Y]]


what I want is a method to display them all.

@mmcdara Thank you sir.

@tomleslie the error message was not so cryptic (in retrospect). Thanks

@tomleslie I did overcomplicate it. After I posted I thought, there must be a way to dispense with the x's, but I thought it needed to be a function to use in Explore.

In fact, you can dispense with variable x altogether. In your code set x=l*sin(theta), then it will be only in terms of theta, l and w.

Thanks to Scot too.

@mmcdara The other way to get a is more direct, albeit taking longer. I initially did this but thought I could get 'a' from the physical properties of the material.

restart:
DIV := diff(y(x), x, x) = a*sqrt(1 + diff(y(x), x)^2):
RV  := y(0)=2.6, y(3)=2.1:
sol := rhs(dsolve({DIV, RV}, y(x)));
L   := Int(sqrt(1+diff(sol, x)^2), x=0..3);
a   := select(is, [solve(value(L)=3.6)], positive)[];

EDIT

Just used this method for the triangular sail. For L=4.33m, got a= -1.015016, but with your method +1.015016

Anyway, the sag was ~1.4m for this case

@brian bovril C is a scaling factor

@mmcdara Whats big c? "for instance, let C > 0, and denote..."

@vv Thats got it!  thanks vv.......

@Carl Love I could do ST[2] and get the result I wanted. I did give an upvote if it's any consolation.

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