New answers tagged optimization
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There is no unique way to label the vertices. So does the question mean “minimum” for a specific consistent cube? Or “minimum” over all consistent cubes? I choose the former because it’s more interesting.
The labels 1...5 are fixed, up to rotation and reflection. But the other 3 vertices can be labelled 6...8 in any order, without risk of any triple being ...
5
Old school proof of optimality:
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Another solution, with smallest possible maximum entry subject to minimizing the sum:
If you ignore the sum, the smallest possible maximum entry is smaller by $1$:
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I decided to let my computer solve it, and it found the following solution:
This is the same as hexomino's latest solution, who updated his post with this shortly before I posted this answer.
Assuming my program is correct, this is optimal.
To show this is indeed a solution:
The second-best solution has box sizes that are coprime:
Using 4 or 5 display ...
-1
If the store is a stationery or office supply shop, then the following possibilities are valid for the number of boxes and the number of pencils they contain. (The $b$s represent the boxes and the numbers represent the corresponding numbers of pencils.)
$b_1=5 \times 2^0$
$b_1‹b_2 \le5\times2^1$
$b_2‹b_3\le5\times2^2$
$b_3‹b_4\le5\times2^3$
$b_4‹b_5\le5\...
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Minimum amount of boxes on display
I haven't found a proven minimum for question 2 yet but the lowest I've gotten so far is
With display box sizes of
3
It's
Note that
and then just
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The permutations of X linear tile placements in Y symmetrical array grid depends on the qty of Y-X combinations >= X to determine the minimum qty of X tiles in Y array.
4x X=3 on a Y=5 2D array will never block insertion of another 1x3 tile without straddling.
Many solutions exist. The proof requires permutation simplification.
Here’s another solution.
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I'd like to give an alternative answer in the style of Hagen von Eitzen.
If there is a horizontal blocking tile on row 3, then there are 2 horizontal tiles aligned above and 2 aligned below that can only be blocked by horizontal tiles. Covering all these requires 5 blocking tiles.
If there is a horizontal blocking tile on line 2, the same tile just above ...
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Assume you manage to solve the problem with four tiles only.
If there is no horizontal tiles, you cover at most four columns, hence one column is still free
If there is exactly one horizontal tile, there are four potential positions for parallel tiles in the same three columns, and it takes at least two of the vertical tiles to block these. Then the last ...
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Assuming you mean that the tiles must consist of grid squares, this is the independent domination number in a graph with one node per tromino and an edge between each pair of conflicting trominoes. You can solve the problem via integer linear programming, as shown in my answer for $1 \times 4$ in an $n \times n$ grid.
Unfortunately, the linear programming ...
22
UPDATE 2: To put OP out of their misery find now at the very bottom of this post an answer to what they probably mean.
UPDATE: OP has changed the rules, so this is no longer valid, but see bottom of this post for an answer to the modified question which I assume is still not what OP has in mind.
I say it is
"Proof":
With the new rules the answer ...
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As an addition to the existing answer, I would say that
However, in the real-life situation with the same traffic flow it would be better
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I say
which is
Generally,
For example:
For any algorithm,
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Here is one neat way of organising the matches, which is optimal.
First I'll describe a general method that works for any number 2n that is
Method:
This works because:
By leaving out some of the players in the matches you can use same team arrangement with any smaller number of players.
So for $2n$ players this method uses k matches where
Here is the n=...
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Notation: let $p_i$ be the player number i and n the size of a team.
First approach, very inefficient:
Second approach:
This approach has been proposed by Jaap Scherphuis (see comment bellow):
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I believe this is a perfect score:
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I believe this one
scores
Intended proof:
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I'm afraid I have to answer your question with a question of my own: how would you score this impossible cube:
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You put two pieces of type 1 together to make a 2x3x2 brick.
So the answer is
Can it be done with less?
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My inner little sister can do it in
Replay
Not sure this is optimal, though. Bonus answer is given as a variation.
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