Train Carriage Game Solutions

I'm looking to program a function that can check if you can "make 10" from some 4 digit number (0000 to 9999).

The function should use these rules when performing the check:
https://www.facebook.com/pages/Train-Carriage-Game/108348062559174?sk=info

Does /prog/ know if there's a simple way to perform such a check short of checking every combination individually? I obviously don't want to be writing the following:


boolean makesTen (int a, int b, int c, int d) {
    if (a + b + c + d == 10) {
        return true;
    } else if (-a + b + c + d == 10) {
        return true;
    } else if (a - b + c + d == 10) {
        return true;
    }
    ...
    ...
    } else {
        return false;
    }
}

0⁰ isn't undefined. It's not 0 either. It's 1.

>>41

let p(t) = (0,t), with x(t) = 0, y(t) = t

lim t->0 x^y
= lim t->0 (0)^(t)
= lim t->0 0
= 0

let p(t) = (t,0), with x(t) = t, y(t) = 0

lim t->0 x^y
= lim t->0 (t)^(0)
= lim t->0 1
= 1

ЕСЛИ ТЫ НЕ ГОЛУБОЙ
НАРИСУЙ ВАГОН ДРУГОЙ


   ___________~~.____~~._____
  |  ___   ___    ___   ___  |
 ш| | | | |---|  |---| | | | |шш
 Н| | | | |___|  |___| | | | |НН
 м| |_|_| москва-адлер |_|_| |мм
  |_______|__________________|
Э==/ \ц/ \Т__Г       / \ц/ \==Е
   \_/ \_/           \_/ \_/

>>39 here
I realised I was making a mistake in evaluating expressions of the form:

-a^b

I was evaluating it as

(-a)^b
instead of

-(a^b)

The latter method allows me to take advantage of the way a carriage number (and it's sign -- +ve or -ve) is permuted, because

(b mod 2)==2
will always be positive.

I now think 9161 of 10000 carrage numbers from 0000 to 9999 make 10. Although (this time) I'm highly confident in my answer, I'd like someone else to confirm. The reason I'm confident in my answer is because the answers in the "cheat sheet" (http://tinyurl.com/everyfuckingsolution) seem reasonable. I also checked that the same 9161 carriage numbers make -10, though this could just mean I'm making the same mistake twice. (If a carriage number can make n, it can also make -n, right?)

Another thing. I was able to check (and show answers to) all 10000 carriage numbers in 4 minutes and a half on an stock i7 950. Can anybody check more than 37 solutions per second?

>>44
Are you just going to ignore my post entirely?

(Also, if you ended up defining 0^0 = 1 your results are still wrong. Either way... you should post a list of "unsolved" numbers.)

>>45
I don't understand.

I ended up changing it so that 0^0 was undefined. The cheat sheet does does contain unsolved games with the solution written "No solution." Unless you wanted me to post a list of numbers separately...

Am I being trolled? This is extremely easy.


#include <stdio.h>
#include <stdlib.h>

int main(void) {
  char buff[5];
  int n[4], i, num;
 
  int a, b, c, d;
 
  while(1) {
    fgets(buff, 5, stdin);
    buff[4] = '\0';
    if(!(num = atoi(buff))) {
      puts("NaN");
      continue;
    }
   
    num*=10;
    for(i = 0; i < 4; i++)
      n[i] = ((num/=10) % 10);
   
    for(a = -1; a <= 1; a += 2)
      for(b = -1; b <= 1; b += 2)
        for(c = -1; c <= 1; c += 2)
          for(d = -1; d <= 1; d += 2)
            if(a*n[0] + b*n[1] + c*n[2] + d*n[3] == 10) {
              puts("True");
              return(0);
            }
           
    puts("Nope");
  }
 
  return(1);
}

Also, to the person who suggested using recursion. Learn how to program.

>>47
Great! Now figure out how to include the other operators allowed in the train game (*, / and ^).

>>48
Would only be slightly more difficult. I didn't check out the rules because who doesn't have facebook blocked?

>>49
the forced partitioning into sums and products. thread made into an abstract syntax tree.

>>47
0 is not a number?

>>49
You don't need an account to view OP's link.

If you blocked Facebook in you hosts file or something, well then that's a pretty tin foil hat you have there.

>>52
fuck you faggot shit

>>46
Well, rule 6 on the Facebook page says you can use multiple digit base numbers, i.e., 0054 has (50-40) as a solution. (It appears I was looking at an old cached version of your answer sheet, so I hadn't noticed you're not using 0^0 anymore; my bad.)

If you blocked Facebook in you hosts file
Everyone needs to do this, in all honesty, if people weren't stupid then they would

>>54
rule 6
I don't like that rule. I suppose it shouldn't be too hard to solve though.

>>52
Enjoy your anal rape

>>57
Enjoy your anal. Only queers go to the trouble of blocking Facebook.

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