New answers tagged

3

Let's consider the triangular grid by column. Each column in the left half has one more left-pointing triangle than right-pointing triangles. In the right half there is an excess of one right-pointing triangle. Diagonal lozenges contribute to exactly one left-pointing and one right-pointing triangle in a column. Let's ignore these. You are left with left ...


6

UPDATE: Found optimal solution Simple upper bound: Better upper bound: Optimal solution: Structure and function: Optimal because:


-2

In order to prove this principle, through Pascal programming to generate different diamond layouts, through different colors, you will find that this 2D paving problem has become a 3D model generation problem, and these models are very similar to urban planning or architecture. A trial calculation of the layout of the tower and the podium. Another feature is ...


0

It's easy to show that the minimal number of problems is at least because


11

First, We then In fact, Alright, Because the question is modified,


3

With the normalization that the first color occurring (starting from top left) should be R and the second G there are $358,108,246$ positions. This is brute-forceable. I wrote a program that first finds all $342,074$ end positions, then those $914,980$ one step away from an end, then those $3,747,392$ two steps away and so on. Note that I did not enforce ...


2

This puzzle is a nice variation of the Hailstone Problem. Not quite the same, since every timer value can be increased and decreased. I solved it by C program, not by logic, assuming that "halve the time" does not mean "halve the number". I think it would be rather hard to work them out by hand. Here is the (recursive) C code. #include &...


1

I had trouble understanding the problem statement. It looks like it was translated at some point. How I understand it, you must follow a path that reads the number, moving in any direction, but not reusing any digit, and moving only to contiguous digits. The solution is as follows:


2

I suspect this question can only be definitively answered with some clever programming, but here's a slight improvement using just manual placement of flags:


1

...would you mind if I gave a score of...


3

Two pretty solutions with mirror and rotation symmetric outcomes: Note on optimality:


2

Since the question does not request that all of the balls are part of a component, I will go with 3 moves. YGBR GBGR RYGY GBRR to YGBR GYGR RBGY GBRR to YYBR GGGR RBGY GBRR and YYBR GGGR RBGR GBRY Compenents are YY B R GGG R R B G R G B RY If more than 4 components are allowed, the last step is not necessary, and the total count is 2.


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3

This can be done in which I believe is pretty close to optimal, if not already. Denoting the four colors as R, G, B, Y respectively, the initial state is Y G B R G B G R R Y G Y G B R R Now, Then, Finally,


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6

Number of fives: One such grid:


11

One possible answer is:


1

We can obtain four triangles, specifically two equilaterals ABG and ECG, one isosceles triangle EFD and one right angle triangle ABC. If we draw the other four missing chords and the one missing radius, we obtain too many triangles to count (I stopped at thirty).


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Question 1 Question 2 Jean Hominal makes an important point in the comments


4

I think the answer is Reasoning:


1

You can solve the problem via integer linear programming as follows. For state $s$, let $v_s$ and $r_s$ be the numbers of votes and rallies, respectively. Let binary decision variable $x_s$ indicate whether I win state $s$. The problem is to minimize $\sum_s (r_s+1) x_s$ subject to $$\sum_s v_s x_s \ge 1 + \sum_s v_s (1-x_s)$$ The unique optimal solution ...


3

I think the answer is Reasons:


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0

I say the minimum weight is 4. Since we have five different weights and we take two at a time, then if we apply the formula of combinatorics with no repetitions, the number of combinations is ten. Since the question assigns the weights the letters A,B,C,D,E we obtain the following ten combinations. AB BC CD DE AC BD CE AD BE AE Let AB be the ...


1

I feel like I misunderstood something, but I think this works (?)


2

Using the corrected weights:


0

Here is my brute-force method for those who are interested in brute-force solutions (only takes two seconds to return the output)! Output:


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