New answers tagged combinatorics
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The sequence is:
and we want the coefficient of $x^{10}$.
The sequence equals:
which is the;
We therefore want:
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The number of ways to achieve $10$ ounces is:
To make it general:
So:
Note that:
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Generalizing my comment on Gareth's solution, we can
arrange Pascal's triangle as a right triangular array and ignore the right half ($n < 2k$) to obtain something like this:
1
1
1 2
1 3
1 4 6
...
We then, for any $N$,
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Gareth has found the optimal solutions, but here is an R script if anyone wants to mess around with the upper bounds for n, just change the value of the variable"UpperBound".
require(pracma)
UpperBound<-500
n<-rep(1:UpperBound,each=UpperBound)
k<-rep(1:UpperBound,times=UpperBound)
data<-as.data.frame(cbind(n,k))
colnames(data)<-c("n","k")
...
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Choosing
gets to within about
of the desired answer. I think this is best possible with <= 100 cards.
Found with the help of a computer, but purely as an aid to calculation. My approach was to
[EDITED to add:]
Out of curiosity, I also ran a more automated search for the larger bound of n=500 mentioned in the OP. For this,
The automated search also ...
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An initial lower Bound
Finding a better lower bound
Doing a bit better with some brute force
The code used : https://pastebin.com/8nQ9UgBP
Notes about the code:
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Edit: as @DarkThunder pointed out, this is incorrect.
My most perfect soup contains
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since they are 1/64th increments, you can be left with a minimum of 1/64th of a cup. Add 1 cup, remove 1/2 cup, remove 1/4 cup, remove 1/8 cup, remove 1/16 cup, remove 1/32 cup, remove 1/64 cup, there is 1/64th of a cup left.
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An upper bound
This can surely be improved substantially because
Another (smaller) upper-bound
Applying a bit of brute force,
A smaller upper bound still
And smaller still
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0k I'll start with my intuition which probably is not the best one, but you can outdo me now ;) .
Idea:
Java code:
Result:
Note:
Other idea:
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Let’s start with notation:
I just tried to
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I can do it in
Basically, working from the front
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