Anyway, the problem is: come up with an algorithm, better than plain brute force, to find two single digit 2x2 matrices whose product is obtained by concatenating the base 10 digits of each element.
We want to find A = [a b / c d] and B = [e f / g h] such that AB = [10a+e 10b+f / 10c+g 10d+h], with a,b,c,d from 1 to 9 and e,f,g,h from 0 to 9. Note that AB-10A-B = [0 0 / 0 0], and so by adding 10I to each side (where I is the identity matrix [1 0 / 0 1]) and factoring, we have:
(A-I)(B-10I)=10I
Letting A'=A-I and B'=B-10I we have A'B'=10I. It follows that det(A') must divide 100. Furthermore, let A'= [a' b / c d'], where a'=a-1 and d'=d-1. Then B' is given by:
B' = A'-1 10I = (10/det(A')) [d' -b / -c a']
Note that the signs within B' must be [- + / + -] by restriction (it is impossible for them to be 0 and satisfy the restrictions on e,f,g,h), and so det(A')=a'd'-bc must be negative. Let k = -det(A'). Then we can solve for everything as follows:
a=a'+1 , b=b , c=c , d=d'+1 , e = 10 - 10d'/k , f = 10b/k , g = 10c/k , h = 10 - 10a'/k
Note that since det(A')det(B')=det(10I)=100, we have that k=bc-a'd' must be a positive divisor of 100. This restricts the possibilities greatly and now it is possible to enumerate everything by hand.
It turns out that k can only be 5, 10 or 20, and bc must be greater than k. Furthermore, if k=5, all of b,c,a',d' are between 1 and 4. If k=20, all of b,c,a',d' must be even. Finally, none of b,c,a'd' can be 0, and a' and d' cannot be 9. Because b and c are interchangeable, as well as a' and d', we classify the following according to b>=c and a'>=d', with up to four possibilities for each:
k bc b c a' d'
5 ==========
6 3 2 1 1 [2 3 / 2 2][8 6 / 4 8] [2 2 / 3 2][8 4 / 6 8]
8 4 2 3 1 [4 4 / 2 2][8 8 / 4 4] [4 2 / 4 2][8 4 / 8 4] [2 4 / 2 4][4 8 / 4 8] [2 2 / 4 4][4 4 / 8 8]
9 3 3 4 1 [5 3 / 3 2][8 6 / 6 2] [2 3 / 3 5][2 6 / 6 8]
3 3 2 2 [3 3 / 3 3][6 6 / 6 6]
10 ==========
12 6 2 2 1 [3 6 / 2 2][9 6 / 2 8] [3 2 / 6 2][9 2 / 6 8] [2 6 / 2 3][8 6 / 2 9] [2 2 / 6 3][8 2 / 6 9]
4 3 2 1 [3 4 / 3 2][9 4 / 3 8] [3 3 / 4 2][9 3 / 4 8] [2 4 / 3 3][8 4 / 3 9] [2 3 / 4 3][8 3 / 4 9]
14 7 2 4 1 [5 7 / 2 2][9 7 / 2 6] [5 2 / 7 2][9 2 / 7 6] [2 7 / 2 5][6 7 / 2 9] [2 2 / 7 5][6 2 / 7 9]
7 2 2 2 [3 7 / 2 3][8 7 / 2 8] [3 2 / 7 3][8 2 / 7 8]
15 5 3 5 1 [6 5 / 3 2][9 5 / 3 5] [6 3 / 5 2][9 3 / 5 5] [2 5 / 3 6][5 5 / 3 9] [2 3 / 5 6][5 3 / 5 9]
16 8 2 6 1 [7 8 / 2 2][9 8 / 2 4] [7 2 / 8 2][9 2 / 8 4] [2 8 / 2 7][4 8 / 2 9] [2 2 / 8 7][4 2 / 8 9]
8 2 3 2 [4 8 / 2 3][8 8 / 2 7] [4 2 / 8 3][8 2 / 8 7] [3 8 / 2 4][7 8 / 2 8] [3 2 / 8 4][7 2 / 8 8]
4 4 6 1 [7 4 / 4 2][9 4 / 4 4] [2 4 / 4 7][4 4 / 4 9]
4 4 3 2 [4 4 / 4 3][8 4 / 4 7] [3 4 / 4 4][7 4 / 4 8]
18 9 2 8 1 [9 9 / 2 2][9 9 / 2 2] [9 2 / 9 2][9 2 / 9 2] [2 9 / 2 9][2 9 / 2 9] [2 2 / 9 9][2 2 / 9 9]
9 2 4 2 [5 9 / 2 3][8 9 / 2 6] [5 2 / 9 3][8 2 / 9 6] [3 9 / 2 5][6 9 / 2 8] [3 2 / 9 5][6 2 / 9 8]
6 3 8 1 [9 6 / 3 2][9 6 / 3 2] [9 3 / 6 2][9 3 / 6 2] [2 6 / 3 9][2 6 / 3 9] [2 3 / 6 9][2 3 / 6 9]
6 3 4 2 [5 6 / 3 3][8 6 / 3 6] [5 3 / 6 3][8 3 / 6 6] [3 6 / 3 5][6 6 / 3 8] [3 3 / 6 5][6 3 / 6 8]
20 5 4 5 2 [6 5 / 4 3][8 5 / 4 5] [6 4 / 5 3][8 4 / 5 5] [3 5 / 4 6][5 5 / 4 8] [3 4 / 5 6][5 4 / 5 8]
24 8 3 7 2 [8 8 / 3 3][8 8 / 3 3] [8 3 / 8 3][8 3 / 8 3] [3 8 / 3 8][3 8 / 3 8] [3 3 / 8 8][3 3 / 8 8]
6 4 7 2 [8 6 / 4 3][8 6 / 4 3] [8 4 / 6 3][8 4 / 6 3] [3 6 / 4 8][3 6 / 4 8] [3 4 / 6 8][3 4 / 6 8]
25 5 5 5 3 [6 5 / 5 4][7 5 / 5 5] [4 5 / 5 6][5 5 / 5 7]
28 7 4 6 3 [7 7 / 4 4][7 7 / 4 4] [7 4 / 7 4][7 4 / 7 4] [4 7 / 4 7][4 7 / 4 7] [4 4 / 7 7][4 4 / 7 7]
30 6 5 5 4 [6 6 / 5 5][6 6 / 5 5] [6 5 / 6 5][6 5 / 6 5] [5 6 / 5 6][5 6 / 5 6] [5 5 / 6 6][5 5 / 6 6]
35 7 5 5 5 [6 7 / 5 6][5 7 / 5 5] [6 5 / 7 6][5 5 / 7 5]
40 8 5 6 5 [7 8 / 5 6][5 8 / 5 4] [7 5 / 8 6][5 5 / 8 4] [6 8 / 5 7][4 8 / 5 5] [6 5 / 8 7][4 5 / 8 5]
42 7 6 8 4 [9 7 / 6 5][6 7 / 6 2] [9 6 / 7 5][6 6 / 7 2] [5 7 / 6 9][2 7 / 6 6] [5 6 / 7 9][2 6 / 7 6]
45 9 5 7 5 [8 9 / 5 6][5 9 / 5 3] [8 5 / 9 6][5 5 / 9 3] [6 9 / 5 8][3 9 / 5 5] [6 5 / 9 8][3 5 / 9 5]
20 ==========
24 6 4 2 2 [3 6 / 4 3][9 3 / 2 9] [3 4 / 6 3][9 2 / 3 9]
32 8 4 6 2 [7 8 / 4 3][9 4 / 2 7] [7 4 / 8 3][9 2 / 4 7] [3 8 / 4 7][7 4 / 2 9] [3 4 / 8 7][7 2 / 4 9]
36 6 6 4 4 [5 6 / 6 5][8 3 / 3 8]
6 6 8 2 [9 6 / 6 3][9 3 / 3 6] [9 6 / 6 3][9 3 / 3 6]
We only need to consider when x>0, so the curves we need to intersect are:
y=x3
x2+(y-r)2=r2
The intersection point between the two functions is given as a polynomial equation: x6-2rx3+x2=0
Furthermore, there is a multiple zero at that point. So the x-derivative must also be 0: 6x5-6rx2+2x=0
After factoring x2 and 2x, respectively, and removing them (since x>0), we must solve the system of equations:
x4-2rx+1=0
3x4-3rx+1=0
Multiplying the first equation by 3 and subtracting the two equations results in the equation 3rx=2. Plugging that into the second equation and solving gives x=3-1/4 (about 0.759836), and r=2*3-3/4 (about 0.877383). This gives an area of the circle as pi*4*sqrt(3)/9 (about 2.418399).
When watching the single screen mode (as is with the temp encode I made), it's watchable for the most part. However when I watched this in emulator with 3D mode turned on, I quickly realized it ISNT watchable because you can't make out what you see in the hallways. I know that youtube has a feature that lets you turn off 3D mode, but unfortunately that isnt a feature in youtube mobile last time I checked. So I have to give this a meh in entertainment because of this.
I don't know how to edit a 60fps video, so it is 30fps.
You'd think someone who has submitted on April Fools in the past would submit this also on April Fools (2021)... especially when the user has been part of the site for so long... April Fools man. April Fools is the time you submit things like these.
I kinda wish some more time was wasted before starting the loop. Time it so that the counter wraparound gets exactly to 1 when the time runs out, so that the one death in the "Maximum Lives" run actually leads to a game over
If e^z = ln(z), because e^z and ln(z) are inverse functions, then it is the case that also e^z = z. Then solving using the Lambert W function is rather trivial: ze-z = 1 -ze-z = -1 -z = W(-1) z = -W(-1)
The answer to the problem posed in the question is known, and is a large irreducible fraction
That got me thinking: Are there ellipses (other than the circle) where the major radius, the minor radius and the perimeter are all expressible with closed-form expressions? (In this context "closed-form expression" excludes integrals.)
Wikipedia's page on parabolas gives the expression for the vertex and focus of the parabola obtained by transforming the unit parabola using a given affine transformation. The vertex and focus of the unit parabola are (0,0) and (0,1/4) respectively, so all that is required now is to solve for the similarity transformation for those two points, which is relatively straightforward.
https://www.youtube.com/watch?v=R_VWFaw0_Es new glitch allows you to take zero to boss battles, i'd like to test it on PSX too
For instance y=x^2 and y = 1/2 x^2 are not technically similar, but they are related by an uneven scaling.
When I had written my last post I thought that It may be the case that "parabolaness" is preserved under all (non-degenerate) affine transformations, but when I attempted to prove it ...
A similarity transform is a little bit too restrictive. It doesn't allow for stretching in either the x or y direction.
All parabolas are similar. This means ...
... scaling it equally in all directions ...
(They are not affine transformations because those allow uneven scaling.)
All parabolas are similar. This means that you can take any parabola and by simply rotating it, flipping it, translating it and scaling it equally in all directions, you can map it exactly on any other parabola. I was wondering if there's a name for those allowed transformations. (They are not affine transformations because those allow uneven scaling.)
What is the smallest prime number p for which tan(p) > p? (p is interpreted as radians.)
1 1.5574077246549022305069748074584
260515 383610.70774372728510957205890168
37362253 37754853.361772908592175824730313
122925461 326900723.47983520354309169506516
534483448 1914547468.5368285042235547779774
3083975227 13356993783.764391329460871831004Evaluate:
So my thinking is: If we assume that the list is countable, then it shouldn't make a difference if we replace every element with a natural number (eg. its position in the list). It shouldn't make any difference what we replace every digit of every element in the list with, as long as the resulting remapping is unique. From the point of view of the proof, it shouldn't make a difference. So if we do that, we are replacing the original list of infinite strings with a list of finite strings, and the only thing that the diagonalization does is to tell us "there are no infinite strings in the list", which is self-evident to the point of being tautological. Of course there aren't, because we replaced them with finite lists in the first place. The only thing that the "proof" is now saying is, in essence, "if you replace all infinite strings with finite strings, there will be no infinite strings."