11
Suppose we start with a full diagonal (A1-H8), and then see which queens we can remove.
Removing any one queen is obviously fine. The other 7 queens cover all the other rows and columns, leaving only the removed queen's own square which is covered diagonally.
If we remove two queens,
If we remove more than two queens,
9
An obvious lower bound for 1 is
It is not hard to see that
Graphically for $k = 3$:
So the answers are
1a
1b
When all vertices can be multiply occupied, as @KyleParsons pointed out in the comments, no edge can be traversed by two people at a time (I had overlooked this in an earlier version).
2
9
TL;DR
First, and easiest, we can do it in:
Using the principle of the:
The method is as follows:
You then just need to rearrange your pieces.
Why does this work?
Now that we understand that method, how can we do it better?
We now need to prove that we cannot do better.
However, this is impossible. The following is a sketch of a proof:
8
The smallest number is going to be the one with the least digits. We need to make use of all the digits as efficiently as possible. Therefore, we should:
Therefore, the solution is
I’ll admit I used a computer (despite the no-computers tag) to generate the output, since a Bash command is much faster and more accurate than typing that up by hand.
I ...
5
The maximum is
1 2 2 3 3 . 4 4 5 5
1 1 2 3 6 6 . 4 5 7
8 8 . 9 6 10 10 . 7 7
8 11 12 9 9 10 13 14 15 15
11 11 12 12 . 13 13 14 14 15
16 16 17 17 18 19 20 20 21 21
22 16 17 18 18 19 19 20 21 23
22 22 . 24 24 25 25 . 23 23
26 26 27 28 24 25 29 30 31 31
26 27 27 28 28 29 29 30 30 31
I obtained this solution via ...
5
If the number of people per table is $p$, then to cover each pair we must have $$\frac{600}{p} \binom{p}{2} \ge \binom{30}{2},$$ which implies that $p \ge \lceil 49/20 \rceil = 3$. The following set of $200$ triples of years covers every pair and contains each year exactly $20$ times:
{{1,2,22},{1,3,5},{1,3,7},{1,4,20},{1,6,8},{1,7,15},{1,9,10},{1,11,12},{1,...
5
With the "Use as many and any black or white pieces (both kings included)" restriction, the theoretical maximum becomes
Obviously, there can't be any legal position with that many white pieces, so we throw legality out of the window and start from here:
And as long as we are careful to eat the bottom two pawn rows in good order, we are good to go ...
5
Lowest I was able to get it to (so far) was 5:
Initial logic to this solution:
4
Converting a Disconnect Four puzzle to a SAT instance is very easy.
Based on this
4
The maximum number of coins is
with a minimum total value of
achieved by
The integer linear programming solution approach I used might be of interest. Let nonnegative integer decision variable $x_c$ be the number of coins of type $c$. The first problem is to maximize $\sum_c x_c$ subject to $\sum_c c\cdot x_c \le 495$ and "compactness" ...
4
proof-without-words (almost...)
4
Variation 1. position must be legal:
As there are at most 15 capturable white pieces this is a hard upper bound on the chain length.
This bound can
be achieved.
Documentation:
Variation 2. position need not be legal:
I can do
4
Let's start by removing the lower bound - say you can go into stamina-debt as far as you need to, and you keep playing until you escape. Then:
Okay, but what does this say about the actual problem?
In other words:
4
The best strategy is to
This will take
steps, which is optimal: it breaks the least possible number of eggs, and if you were to use some other strategy to choose the floors,
so the expected number of steps won't be any less for that other strategy, either.
This happens because "infinity is really, really big", or more specifically:
And so is, ...
4
Given an unbounded number of trials, we could simply test one egg over and over and over again until it breaks, and find our floor.
"But that's too slow", you say, and you're right - so instead we
The only information we need to solidify our method is...
3
Here is one possible arrangement:
Proof of optimality:
In view of the answer by loopy walt, this also proves optimality for the case of decagon.
3
To kickstart,
But as we are given 7 empty squares,
Which gives a total of
3
I think the answer is
because the maximal increments at the four foldings are
Here the powers $2^k$ arise as the
The binomial coefficients count
This concludes proof (well, sketch of proof) that the number I give is an upper bound.
To construct a maximal example simply
3
Optimizing for number of throws t given a large amount of eggs, starting at throw n = 0 until success at throw t:
This finds the max floor for arbitrarily high floor F in
2
Congratz to @Anon for finding the intended solution. I'm posting this just to show off a to scale picture I happened to have lying around:
2
Here's another 40 moves sequence that I found using the simulator:
Note the brackets are [column, row]
2
Using the simulator, I was able to find 2 sequences requiring 41 moves each (see edit history for details, formatted T[y, x]). I was also able to find a single sequence requiring 40 moves (formatted t[x, y]):
2
My best shot:
And for 10:
2
2
One way to think about this is that the first drink the Cow has propels her 4 stalls, and each drink thereafter propels her 2 stalls. A one way trip of 10 stalls means she has to drink from 4 stalls, so for two trips of 10 stalls she has to drink from 8 stalls. One possible configuration of drinks is:
2
I believe the maximum, if the plane is not unfolded between consecutive folds, is
See picture below for the folds I made, and where the creases/intersections lie after unfolding the plane.
Note: It is reasonable to consider only a finite sub-region of the entire plane, so long as you ensure that all the crease-intersections formed lie within that finite ...
2
Trial answer with
1
[NOTE: After writing this, I checked with a computer. The following is wrong. I shall not say how it is wrong, of course. Nor shall I delete it, because I believe in not hiding my errors.]
It looks to me as if
Something like this:
It's not much better
If instead
First,
Next,
Or
1
I think the answer is
Trying to solve the puzzle from scratch doesn't give us much:
However, case analysis on
gives very interesting results.
Note that
Since we already forced a cell by case analysis, the total number of cells we need to force to get a unique solution is
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