New answers tagged optimization
1
A simple bound in the more general case:
3
Rawrdon Mamsay needs to stay at least
days to identify the head chef, and
days to identify the cooking order.
Reasoning:
Rawrdon Mamsay samples the following meals:
Rawrdon Mamsay cannot complete his task in fewer days:
4
[EDITED to add:] Seeing Magma's answer, posted about a minute after mine, I realise that the question asked two questions and I've only answered one of them. I shan't modify my question to answer the first as well because Magma's answer already does that perfectly well.
I am not certain that I've understood the explanation in the question quite correctly, ...
1
I believe this qualifies:
14
I managed to find the 8 in
So we are trying to find the 5th largest card in a bunch of 12, by measuring them in batches of four. Here's my strategy:
Now we have identified, for certain, a couple of cards we can exclude: A1 and A2 (both have at least 8 cards smaller than them), and B4, C2, C3 and C4 (all have at least 5 cards bigger than them). We also have ...
5
I can do it in
Note:
Continuing...
Assume the worst case. Any other results would obviously be easier to solve.
For any of these options:
Thus, we have completed the process and identified card 8 in at most
Note, that as a side-effect of identifying card 8, we also identified cards 12, 11, 10, & 9. There might be a more efficient option to ...
5
I can do it in:
2
Nobody has a offered a bishop construction yet. Here is one, though imperfect:
White to move and mate in 3
Solution:
It is not great because
The difficulty to build a problem with BB or RR promotions is that
1
Short and Concise solution,
Finally,
1
Promotion to rook, 5 pieces (note this may not qualify)
2
Currently on mobile, so I'll have to provide the full list of moves later, but I think I have it ending with:
The full list of moves:
3
Promotions to knights, 8 pieces
White to play and mate in 3.
Solution:
22
(Kind of) analytical solution that only requires small amount of calculations, (potentially) doable by hand.
First step: we can safely drop 2, 3 and 7 from the equation as those digits are used in 23 and 17. Now, we need to build a prime from: 5, 11, 13, 17, 19, 23, 29 and 31.
Second step: let's try to build the shortest number possible from these numbers. ...
6
Assuming "transportation cost" means sum of distances to each of the three roads, and the side of the equilateral triangle has length $1$:
22
The answer is
I wrote a Java program to find it:
14
Shuffling the sequence of 5 and the non- overlapping 19, 23, and 29 by trial and error produces:
15
So I can't yet prove this is the smallest, but it's at least an upper bound:
Reasoning:
4
For promotion to queen, a solution with four pieces:
Best play is
For promotion to knight, one with nine pieces:
The solution here is
2
Asnwer
Diagram explaination...
Therefore my answer is...
The general formula is that with n straight lines we can form as many
as if n is ODD.
if n is even
If there are m straight lines at right angles to n straight lines, m being less
than n, the
8
I got:
Counting:
As a picture
1
I assume we have to draw exactly 111 squares and the lines are finite.
My answer:
How I find that:
class Program
{
static void Main(string[] args)
{
Console.WriteLine("***********");
printGrids(111);
Console.WriteLine("***********");
printGrids(112);
Console.WriteLine("*******...
11
EDIT: Some have suggested that the question might be to find exactly 111 squares. In this case,
ORIGINAL ANSWER
Let's count
5
I think the best you can do is
As follows
Counting
Top 50 recent answers are included
