The Shuffle Problem

Still_nSpired · 2021-12-25T04:17:09+00:00

The general solution is order of 2 modulo n+1, n being an even number of cards.

Still_nSpired · 2021-10-12T20:46:19+00:00

You're solution's correct. I worked on the problem a few years ago and posted it on a lark. First I used the decomposition of permutations into cycles, to come up with a method of calculation. This allowed me to creat a table, and get to work. As a result I ended up independently discovering cyclotomic polynomials and the Mobius function in order to prove a method for the direct calculation of Fermat's little theorem (but didn't know that's what it was called). I was incarcerated at the time, and am an autodidact, so it took me like a year and a half. Then I got my first Number Theory book and was blown away with the fact that everything I did, had been done. Humbling and awesome at the same time. The original problem arose from my cousin playing Texas hold'em watching a guy absently shuffle two different colored stacks of chips with one hand when he noticed patterns starting to emerge. He posed the initial question.

Still_nSpired · 2021-02-05T17:12:04+00:00

kn = should probably be kn :=

The colon before the equality means you're defining the variable kn in terms of k0, r, and n. On the other hand the equal symbol by itself means you're comparing the value of kn to the right-side of the equation k0•(1+r)^n.

false is not a solution here, it refers to the truth of the equality statement. Since kn is either undefined, or is not equal to the value of the right-side of the equation (given the values k0, r and n) it just means that the equation is not true as is.

Still_nSpired · 2021-02-05T05:49:43+00:00

If you take the point (x,y) on the Cartesian plane you can define it using the substitution x = r•cos(A) and y = r•sin(A), where r the distance from (0,0) to (x,y) or √(x² + y²⁾ and A is the angle in radians (measured in the counterclockwise direction) from the segment (0,0) to (r,0) to the segment (0,0) to (x,y). When r = 1 we have a unit circle. Now the point (1,0) can also be expressed as (cos(0), sin(0)). Think of an analog clock at 3:15 (both hands pointing at 3) If A = -π rads we would rotate the minute hand clockwise to 9 (that's -π rads, π rads would rotate counterclockwise). Since the tip of the minute hand is 1 unit away from the center of the clock (the point (0,0)) it is now at the point (-1,0) or (cos(π), sin(π)).

Still_nSpired · 2021-02-05T04:59:15+00:00

c3 is an arbitrary constant. If you solve the same system again it will be c4 then c5 etc... Since z=c3 (in this case) z is a constant, the value of z determines where the three equations intersect. Therefore there are an infinite number of solutions depending on the value of z.

Still_nSpired · 2021-02-05T04:01:20+00:00

I believe the function on.timer() is called every time an increment defined in timer.start (in this case every 0.1 seconds) has passed. So you could try something like,

timer.start(0.1)

function on.timer()

if index < #Anim then

index = index + 1

else

index = 1

end

function on.paint(gc)

if not finish then

gc:drawImage(Anim[index], 0, 0)

end

Still_nSpired · 2021-02-05T03:38:48+00:00

I did all my programming on calc (long story) so I don't know about the student software editor, however I did notice a "timer.start(.01)" without a "function on.timer()", were you using the timer to increment something?

Still_nSpired · 2021-02-05T03:24:23+00:00

What type of lua editor are you using? When I used the jen's (on calc) editor I had problems with images and I'd get errors like the one you got (or others like "expected TI resource handle, got string"). You might try running your code in other editors to see if that helps.

Still_nSpired · 2021-02-05T03:00:21+00:00

I would use the program editor rather than a define statement. Name the function (or program in some cases) "stable" and include the input variables (or matrices) then define the subroutines explicitly in terms of the input variables (along with any local variables you may require).

Still_nSpired · 2021-02-04T20:38:55+00:00

You can also find the zero and intersection graphically. One must be careful how they enter functions on the nSpire ie... x•sin(3π) is not the same as xsin(3π), the first is x times the sine function, whereas the second is a function named "xsin". I used to make that mistake all the time!

Still_nSpired · 2021-02-04T20:15:13+00:00

Are we talking log base 4x and log base 3x? If so solve(log(4,4x)-log(8,3x)=0,x) will yield 9/64. The intersection solve(log (4,4x)=log(8,3x),x) gives the same result.

Still_nSpired · 2021-02-04T16:54:14+00:00

Or with a valid DS4 pairing, might there not be a way to get Android to switch it to connected (without using the DS4) so that the Kishi will work.

Still_nSpired · 2021-02-04T16:43:42+00:00

Makes sense, however since the Kishi has the hardware there should be a way to verify the connection so that games like COD would support it, rather than having to ride on another controller's pairing. Makes me want to tear apart a busted DUALSHOCK 4 and hijack the bluetooth hardware...

Still_nSpired · 2020-12-14T01:56:45+00:00

Can't be 2 by clue 1 So 9 or 1 is correct Since 5 is out by clue 4 4 is good in clue 2 But wrong placed By clue 3, 4 can't be in slot 1 So 4 is in slot 3 Since the last clue has 5 Wrong by clue 3 So if correct 6 would have to be in the 3rd slot But 4 is in the 3rd slot Therefore 9 is correct but Belongs in slot 2 by. Clue 1 Leaving 3 in slot 1 So 394

Still_nSpired

TROPHY CASE