Christopher2222

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15 years, 313 days

MaplePrimes Activity


These are replies submitted by Christopher2222

So it's been 10 years and it doesn't appear Maple has the ability to strikethrough a character.  2022 doesn't and I don't believe 2023 has that capability either.

I was generating a solution to a physics problem for some students, and although not entirely necessary, it would have been helpful to visualize what variables were being cancelled out.

I tried using Word 2007 with it's strikethrough option and tried copying and pasting it over into math or text but I just get the characters with no strike.  

This strikethrough option doesn't appear to be one of a concern to many or anyone else since there has been no upvotes and no replies on the topic. But this is mainly for document creation - it's so much easier to create a math document in Maple than it is in Word but I suppose people just resort to other methods or do without is my guess.

I've discovered Microsoft Edge works better than Firefox.  But the searches don't bring up all the applications by a given author.

Hopefully just a browser firefox issue.
Pick Samir Khan for example, he has a number of excellent applications.  I can scroll forever down but all I see is 18 of his applications and I go to select one and it throws me back to the top, quite frustrating to say the least.

@C_R "Probably because the readers of your original post immediately jumped on rendering planets."

No, but rendering world maps to a sphere is a logical and obvious choice of reasoning.  

Probably uncurl the picture or transform to a flat surface might make more sense.

A rough 2 dimensional mathematical formula of a vase

f(x) = (-16/45) x3 + (14/9)x2 - (8/5)x + 1

Measuring would get the formula to right proportions of the vase in the picture  - roughly - I'm not looking for exactness. 

@acer So imagining the picture in 3 dimensional space and mapping it to a flat 2d space. 

I was just starting to get used to unapply.  The problem with that now is that future applications by people who prefer to use, or start using, MakeFunction will fail in older versions of Maple. 

In the first example on the help page I can see the logic of the new function.

This is the way I interpret it anyways - f equals make a function out of p in terms of x.  But it can be confusing because literally it says MakeFunction(p,x) - ok so what happened to p, I would kind of expect the variable p to be in the output. 

Anyways, it's a play on words.

@Carl Love Thanks Carl.  Indeed I thought order was playing a factor, and of course always a consideration.

Of course, the simulations are the more nteresting parts.

@mmcdara So problem 2 is what I was getting at.  And I see you've arrived at the same probability I came up with 1.11E-06 Setting the containers to be analogous to colored balls.  Carl love and you had some great code for me to digest. 

Thanks!

@Carl Love So I did this while I was falling asleep, and had to calculate it in the morning.  I get a different result than the one you found at the party. 

I think the analogy is close to a container with colored balls each.  Just picking all of family B and all of family C the probabilty works out as 4 choose 16 times 3 choose 15 etc ...

4/16*3/15*2/14*1/13*4/12*3/11*2/10*1/9 = 1.11*10^(-6)  Not sure yet, exactly why my result is 6x less in probability than your result. 

@Carl Love Yes you are correct the permutations do not have to be symmetric.  Thanks for the different approach.

Also thanks Mmcdara for the code you created, symmetric or not - a second drawing of the same family is not impossible but still highly improbable and all the more reason to be skeptical of the draw. 

@Carl Love ok thanks for randperm, in the meantime I've fixed what I had and gave the variables some random names.

restart

hat := [a1, a2, a3, a4, b1, b2, b3, c1, c2, c3, c4, d1, d2, d3, d4, e1]

[a1, a2, a3, a4, b1, b2, b3, c1, c2, c3, c4, d1, d2, d3, d4, e1]

(1)

with(combinat)

with(StringTools)

 
c := []; hat1 := hat; for i in hat do b := [i, hat1[(rand(1 .. nops(hat1)))()]]; while Has(b[1], b[2]) = true do b := [i, hat1[(rand(1 .. nops(hat1)))()]] end do; c := [op(c), b]; hat1 := [op(`minus`({op(hat1)}, {op(b[2])}))] end do; c

[[a1, d2], [a2, b3], [a3, c4], [a4, e1], [b1, a2], [b2, d4], [b3, c1], [c1, b2], [c2, a4], [c3, a1], [c4, d1], [d1, b1], [d2, c3], [d3, c2], [d4, a3], [e1, d3]]

(2)

names := {a1 = Thomas, a2 = Christine, a3 = Nolan, a4 = Michael, b1 = Charlie, b2 = Karen, b3 = Sandra, c1 = Brad, c2 = Kristen, c3 = Laurie, c4 = Austin, d1 = Sam, d2 = Natalie, d3 = Tamara, d4 = Zoey, e1 = Barbra}

subs(names, c)

[[Thomas, Natalie], [Christine, Sandra], [Nolan, Austin], [Michael, Barbra], [Charlie, Christine], [Karen, Zoey], [Sandra, Brad], [Brad, Karen], [Kristen, Michael], [Laurie, Thomas], [Austin, Sam], [Sam, Charlie], [Natalie, Laurie], [Tamara, Kristen], [Zoey, Nolan], [Barbra, Tamara]]

(3)

``

Download Christmas_draw-4.mw

So Thomas will buy for Natalie, Christine will buy for Sandra, etc ...

So just developed a simple sample drawing from a hat.

Maybe not efficient but it's easy to follow.  So a little code for one random draw.

restart

hat := [a1, a2, a3, a4, b1, b2, b3, c1, c2, c3, c4, d1, d2, d3, d4, e1]

[a1, a2, a3, a4, b1, b2, b3, c1, c2, c3, c4, d1, d2, d3, d4, e1]

(1)

with(combinat)

with(StringTools)

c := []

for i to 8 do b := randcomb(hat, 2); while Has(b[1], b[2]) = true do b := randcomb(hat, 2) end do; c := [op(c), b]; hat := [op(`minus`({op(hat)}, {op(b)}))] end do; c

[[b3, d2], [c4, d4], [a2, b2], [b1, c2], [a4, c3], [a1, d3], [c1, d1], [a3, e1]]

(2)

NULL

Download Christmas_draw.mw

edit *** oops, sorry that's only half of the draws. 

@Carl Love yes forensic mathematics!

So the odds if it happening 2 years in a row is so small that the chance that the drawing was fixed is a very high probability.  So caught red handed by induction.

Thanks both to your contributions. 

The number of adults and children I guess doesn't really matter.  What matters is that families can't choose other family members for a trade. 

I didn't give historical results, but for just one year back lets just say all of Family B had members trading with all of Family C the previous Christmas, and the same thing happened this Christmas.

How likely was it that the same Family B just so happens to trade with all of Family C again two years in a row?

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