Simple LabVIEW Puzzle Challenge

[ShotSimon]

OK here is one solution attached below.  You need to find the center of your screen by checking your monitor resolution and dividing by 2 and adjusting to catch the pixels changing.  You need to run the "GetPixel Color to Morse Code.vi" then the ".-.vi".  You still need to translate the final product into Old English.

Or if you prefer to solve it yourself you can use the following key:

Dot = Red, 300ms

Dash = Green, 100ms

Space Between Letters = White, 300ms 

Space Between Words = Black, 700ms

Regards,

-SS

[ShotSimon]

For some reason I always come back to pi.  I was sent the following tidbit about the Feynman point.

So I was going through various math sites wondering what a simple algorithm was for generating the 1st 1000 digits of pi and came across this little chunk of c code from here:

long k=4e3,p,a[337],q,t=1e3;
  main(j){for(;a[j=q=0]+=2,--k;)
  for(p=1+2*k;j<337;q=a[j]*k+q%p*t,a[j++]=q/p)
  k!=j>2?:printf("%.3d",a[j-2]%t+q/p/t);}

Well I haven't been able to change it into LabVIEW (it does work in C) but thought it would make for a nice simple puzzle challege for the folks that know both C and LabVIEW.

So the challege is to convert the above C into LabVIEW.

As usual the prize is to pick the next puzzle if you choose.  Oh and before I forget... the password to the last puzzle is "MorseCode2012" and you will have to abort the vi with "ctrl+." to use it.

Regards,

-SS

[crossrulz]

---

@ShotSimon wrote:

For some reason I always come back to pi.  I was sent the following tidbit about the Feynman point.

So I was going through various math sites wondering what a simple algorithm was for generating the 1st 1000 digits of pi and came across this little chunk of c code from here:

long k=4e3,p,a[337],q,t=1e3;
  main(j){for(;a[j=q=0]+=2,--k;)
  for(p=1+2*k;j<337;q=a[j]*k+q%p*t,a[j++]=q/p)
  k!=j>2?:printf("%.3d",a[j-2]%t+q/p/t);}

Well I haven't been able to change it into LabVIEW (it does work in C) but thought it would make for a nice simple puzzle challege for the folks that know both C and LabVIEW.

So the challege is to convert the above C into LabVIEW.

As usual the prize is to pick the next puzzle if you choose.  Oh and before I forget... the password to the last puzzle is "MorseCode2012" and you will have to abort the vi with "ctrl+." to use it.

Regards,

-SS

---

Wow, there's some messed up syntax in that.  That's what could make it a challenge.

[Darin.K]

I win, that looks like a variation of the Pi spigot I posted here:

http://forums.ni.com/t5/BreakPoint/Happy-Pi-Day/m-p/1090569#M11227

Not nearly as concise, almost as unreadable!

[ShotSimon]

Darrin.K,

That is very cool that you did that already.  Somehow I missed it on the forums search...oh well.  I saw a Monte Carlo example but really like this one better.  Very nice code and allows you to find the Feynman point quickly.  I would love to see if Altenbach could speed it up. 

I have another puzzle based on a game I played over the weekend called "Fill or Bust".  Using LabVIEW show the odds of throwing at least one, (1 OR a 5) with either 1, 2, 3, 4, or 5 dice.

Regards,

-SS

[Darin.K]

In case anyone wants to see a G implementation of the Pi spigot C code:

long k=4e3,p,a[337],q,t=1e3;
  main(j){for(;a[j=q=0]+=2,--k;)
  for(p=1+2*k;j<337;q=a[j]*k+q%p*t,a[j++]=q/p)
  k!=j>2?:printf("%.3d",a[j-2]%t+q/p/t);}

Mine is faster (spitting out 4 digits at a time) and actually includes the 3 in the beginning (notice how I had to tack it on).

Some annoying abuse of the ternary operator there, along with the comma operator, and I rarely put more than my increment/decrement in the third slot of a for expression....

[Darin.K]

---

@ShotSimon wrote:

I have another puzzle based on a game I played over the weekend called "Fill or Bust".  Using LabVIEW show the odds of throwing at least one, (1 OR a 5) with either 1, 2, 3, 4, or 5 dice. 

---

[ShotSimon]

Darin,

Wow, Can you explain that solution?  It is correct but I don't understand it:)

Thanks,

-SS

[Darin.K]

---

@ShotSimon wrote:

Darin,

Wow, Can you explain that solution?  It is correct but I don't understand it:)

Thanks,

-SS

---

The first step in probability is to ask the correct question, in particular consider whether the inverse question is easier.  For example, let's say there are three dice and let P(#1,#5) be the probability of #1 1s and #5 5s.  The question you asked is what is P(1,0)+P(2,0)+P(3,0)+P(0,1)+P(0,2)+P(0,3)+P(1,1)+P(1,2)+P(1,3)+P(2,1)+P(2,2)+P(2,3)+P(3,1)+P(3,2)+P(3,3).  Yuk, especially with more than three dice.  Let's instead ask the question: What is the probability of rolling no 1's and no 5's, P(0,0)?  This is pretty easy, you have N dice, for each die there are 4 out of 6 results which are neither 1 nor 5.  P(0,0) is then (4/6)^N = (2/3)^N.  Then the probability you are looking for is simply 1-P(0,0), essentially saying the probability of more than 1 1 or 5 is 1 minus the probability of no 1s or 5s.

(I was actually playing a small joke as I figured the intent was to use LV to simulate the results and I used LV to implement the math).

[ShotSimon]

Darin,

I solved this problem with LabVIEW using simulation (i.e throwing the dice a couple thousand times) and with C with nested for loops counting all possible combinations with 1 or 5 divided total combinations.  Your solution is by far the slickest example of problem solving I have seen in a while..Throwing pure math at a problem is no joke and shows it's power is amazing.

Thanks,

-SS

P.S. Got any simple puzzles that have stumped you?