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The side length of the largest square inscribed in a rectangle of width $a$ and height $b$ is surely $min(a,b)$. How is this proved? (Or if it be wrong, what is the correct result?)
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As an area maximization problem:
reg = Polygon[{{0, 1}, {0, 6}, {4, 10}, {8, 10}, {11, 7}, {11, 4}, {7, 0}, {1, 0}, {0, 1}}];
rnf = RegionNearest[RegionBoundary[reg]];
gendisk[{x_, y_}] := Disk[{x, y}, EuclideanDistance[{x, y}, rnf[{x, y}]]]
cost[{x_?NumericQ, y_?NumericQ}] := Area[gendisk[{x, y}]]
{err, sol} = NMaximize[cost[{x, y}], {x, y} [Element] reg,
Method -> "RandomSearch"];
Graphics[{FaceForm[None], EdgeForm[Black], reg, Yellow,
FaceForm[Yellow], gendisk[Values@sol], Red, Point[Values[sol]]}]
There’s a ridge at the centre of the distance transform, so I don’t think there is a unique solution but a family of disks with maximal area.
ImageAdjust@DistanceTransform@ColorNegate@Graphics[reg]
I don’t know if there is a readily available conclusion for this. Given a diagonal matrix $Sigma = operatorname{diag}(sigma_1, sigma_2, …, sigma_n)$ where the diagonal entries are in descending order, I’m trying to find a rank $k$ ($k le n$) matrix A to maximize
$$
lVert Sigma – A rVert_2,
$$
but I’m not sure whether I should try to maximize the largest singular value of $Sigma – A$, or try to use the definition of the spectral norm:
$$
lVert Sigma – A rVert_2 = max_{xne 0} frac{lVert (Sigma – A)x rVert_2}{lVert x rVert_2}.
$$
Any suggestion is appreciated.
For example, these are expected outputs:
3: 2, 1
4: 4
5: 4, 1
6: 4, 2
7: 4, 2, 1
8: 8
9: 8, 1
...
20: 16, 4
...
25: 16, 8, 1
...
36: 32, 4
...
50: 32, 16, 2
Up to the max of 32 being the largest subunit. So then we get larger:
100: 32, 32, 32, 4
...
201: 32, 32, 32, 32, 32, 32, 8, 1
...
What is the equation / algorithm to implement this most optimally in JavaScript? By optimal I mean the fastest performance, or fewest primitive steps for example, with the least amount of temporary variables, etc. I feel like my solution below is a “brute force” approach which lacks elegance and it seems like it could be optimized somehow. Ideally there would be no Math.floor or division as well, if possible to use some sort of bit magic.
log(20)
log(25)
log(36)
log(50)
log(100)
log(200)
function log(n) {
console.log(generate_numbers(n).join(', '))
}
function generate_numbers(n) {
const chunks = count_chunks(n)
const sum = chunks.reduce((m, i) => m + i, 0)
const result = new Array(sum)
const values = ( 1, 2, 4, 8, 16, 32 )
let i = chunks.length
let j = 0
while (i--) {
let x = chunks(i)
while (x--) {
result(j++) = values(i)
}
}
return result
}
function count_chunks(n) {
let chunks = (0, 0, 0, 0, 0, 0)
if (n >= 32) {
let i = Math.floor(n / 32)
chunks(5) = i
n = n - (i * 32)
}
if (n >= 16) {
chunks(4) = 1
n = n - 16
}
if (n >= 8) {
chunks(3) = 1
n = n - 8
}
if (n >= 4) {
chunks(2) = 1
n = n - 4
}
if (n >= 2) {
chunks(1) = 1
n = n - 2
}
if (n >= 1) {
chunks(0) = 1
}
return chunks
}Thanks for contributing an answer to Computer Science Stack Exchange!
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There are $n$ sets of distinct positive integers, $S_1,ldots,S_n$.
There is a set of recipes that allows us to construct tuples of integers from these sets. For example, the recipe {1,2} allows us to construct tuples of two integers: one from $S_1$ and one from $S_2$. Each integer can be used in at most one tuple. What is an algorithm that, given the sets and the recipes, constructs tuples such that the sum of integers in all tuples together is maximum? Here are some simple cases:
(a) There is a single recipe, e.g. {1,2}. Then the algorithm is simple: sort the integers in $S_1$ and in $S_2$ in descending order; pair the largest integers in each set; pair the second-largest integers in each set; keep pairing integers as long as the sum of the next pair is positive.
(b) There are two recipes {1,2} and {1,3}. Then, $S_2$ and $S_3$ are substitutes: combine them into a single set $S_4 = S_2cup S_3$, and proceed as in (a) with the recipe {1, 4}.
(c) There are two recipes {1,2} and {1,3,4}. Then, $S_3$ and $S_4$ are complements: sort each one in descending order, and construct a new set $S_5$ in which the largest element is the sum of the largest elements in $S_3,S_4$, the second-largest is the sum of the two second-largest elements, and so on. Proceed as in (b) with the recipes {1,2} and {1,5}.
(d) The above two operations can be generalized to the case in which the set of recipes has a tree structure: there is a tree in which each node is a set, and each recipe is a path from root to leaf in the tree. The tree can be “contracted” as follows: a leaf which is the unique child of its parent can be combined with its parent as in (c); two leafs of the same parent can be combined as in (b).
Is there an efficient algorithm that works for any set of recipes?
I tried to solve this problem but could not do it better than O(n^2).
My Algorithm:
1.calculate prefixsum
2.for i 1...n
for j 1...i
if(presum(i)-presum(j-1)>k)
ans=max(ans,i-j);
However, this is inefficient for large values of n.Can someone help me with optimized algorithm along with code preferably in c++.
I am currently designing a web app to cater for the following screen resolutions:
I’m wondering: