New answers tagged combinatorics
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Here is a slightly easier proof than Rand al'Thor's.
Let's look at a simpler problem consisting of
The number of solutions to this simpler problem is
The original problem consists of
This easily generalizes to
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The answer is
the proof being as follows. (Thanks to @El-Guest for finding the error in my previous reasoning!)
This assumes that rotations and reflections of the same path count as different from each other.
For a general word of $n$ letters, laid out in a diamond configuration like this, the answer will be at least
but it may be more if doubling back ...
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No need for computers, hand solving will do (though I did use code in the end to avoid making any calculation mess).
It is trivial to see that you should start from the end. Then preferentially remove ones and decrement when you are out of them and have more than 3 beads left.
Thus, the observation leading to the optimal solution is pretty trivial:
This ...
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The minimum cost is 125 (credits to Ben Barden and Zizy Archer), with steps as follows:
Input: 18 1 16 19 6 22 14 15 2 12 27 18 11 1 14 9 23 1
Minimum cost: 125
Step 1: 1 2 5 10 1 6 Buy 1 2 5 10 1 6 (cost: 25, total: 25)
Step 2: 3 4 7 12 3 8 Increment 2 times (cost: 4, ...
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My C-coded solution improves on OP but is perhaps not optimal
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I got 128 through a fairly overkill approach suggested by the fact that
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Thinking about this a little further. First, @Chronocidal's approach is really the right one...instead of building up to a bracelet, we should start with the result and deconstruct it back. In this construct, swapping stays the same, but incrementing becomes decrementing, and buying becomes selling. If you think about it this way, there are a couple of ...
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This is an example of the
Using that,
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The grid looks like this:
Making the missing word
And listing the words:
Explanation of how I got the grid.
I know this logic is flawed as the missing word could fill in, but that was my thought process that got me the answer so I guess it works here :)
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(Not an answer but too long for a comment - computer based)
I checked all the possible games for $1 \leq k \leq 10000$ by computer. I discovered the following:
I have no idea why this is, but in case it is of some use, the following spoiler contains a table of the winning and losing positions for even $k$ between $2$ and $20$. Note that "Winning" means "If ...
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Partial answer: odd $k$ (no-computers)
has a winning strategy, namely
This wins because
If computers are required for even $k$, I'll leave that for someone else. I'm no good at programming.
General notes:
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Complete first answer:
Yes there exists one for order 5. Consider the combination $1,3,10,2,5$.
Partial second answer:
A perfect circle of order 98 :
$$1, 2, 34, 15, 139, 117, 24, 101, 481, 5, 65, 109, 62, 76, 7, 362, 78, 45, 9, 23, 18, 53, 104, 8, 161, 17, 25, 316, 255, 147, 199, 129, 279, 58, 131, 20, 73, 391, 6, 38, 114, 4, 217, 10, 72, 120, 57, 187, ...
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I could be wrong, but is the answer:
Because:
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First, a divisibility condition for $n$.
We now derive a bound.
We can now plug it into our first equality:
Now it gets somewhat messy.
Now that we have this condition:
Ok, let's do $n=4$ now.
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Partial solution
We have n+2 sums all adding up to the same sum, suppose this is s.
Add all the sums together to get (n+2)s. This means each edge has been counted exactly twice.
(n+2)s = (1+2+..3n)*2 = 3n(n+1)
s = 3n(n+1)/(n+2) < 3n
Now we need two sums of n distinct numbers that both add up to s. Add both sums. Let's take the first 2n numbers, this ...
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Let's say you turn this prism to $n$-angled prism, one $n$-gon face has edges filled by $a_1, a_2, \cdots, a_n$; the other $n$-gon face has edges filled by $c_1, c_2, \cdots, c_n$; then the rectangle face consists of edges $a_i$ and $c_i$ also has edges $b_i$ and $b_{i+1}$ (with $b_{n+1}$ is defined as $b_1$).
Let's say each face should have a same sum ...
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