2
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A game of 2048 goes like this:

  • Every move you get another piece - either 2 90% of the time, or 4 10% of the time
  • for each 2 equal pieces combined you get a piece of the sum of the two and the sum gets added to your score

Your task

given a board of a 4x4 game, give the average score of what A game like that would result in.

To calculate it, go the other way around of the game. For example, if there's an 8 piece, it's made from 4s. The average score of a 4 piece is 4*90% (90% of the time it's made from combining two 2 pieces together and it the other 10% a 4 isn't made, but ) = 3.6. As to make an 8 you need two 4s, and then you need to combine them to an 8, the result is 3.6*2 + 8 = 15.2.

In pseudocode a calculation for one number could be this:

function value (x) {
  if (x > 4) return value(x / 2) * 2 + x
  else if (x == 4) return 4*90%
  else return 0
}

but you're free to choose any approach as long as it works the same.

Clarifications

  • The input board will always be from a 4x4 board.
  • You can take in the board as an array of the 16 values, as 16 different inputs or even a 2D array if that somehow helps you.
  • The input numbers will always be zero or 2 to the power of a positive integer.
  • You can discard any errors due to floating point operations (but theoretically the program must implement a correct algorithm).
  • This is , the shortest code wins!

Test-cases

[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0] -> 0
[2,4,2,4,4,2,4,2,2,4,2,4,4,2,4,2] -> 28.8
[4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4] -> 57.6
[0,0,0,4,0,0,0,8,0,0,0,16,2,2,0,2048] -> 20340.4
[2,16,2048,4096,4,32,1024,8192,8,64,512,16384,16,128,256,32768] -> 845465.2
[8,8,4,4,0,0,0,0,0,0,2,2,2048,65536] -> 996799.2
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  • \$\begingroup\$ @LuisMendo 90% of the time a given 4 gets made from combining two 2 pieces, while 10% of the time they appear as one and give no score \$\endgroup\$ – dzaima Aug 31 '17 at 16:28
  • \$\begingroup\$ @LuisMendo A 4 piece isn't always made from combining two 2's, as 10% of the time when a piece spawns in after a move, it starts as a 4 \$\endgroup\$ – dzaima Aug 31 '17 at 16:33
  • \$\begingroup\$ Out of 9 spawned 2's one can only create 4.5 4's, not 9. Shouldn't the average score of a 4 piece be 9/11 * 4 = 3.2727... ? \$\endgroup\$ – ovs Aug 31 '17 at 17:18
  • \$\begingroup\$ @ovs Even if my math is wrong, the challenge will stay this way. Trying to figure what you mean (and now even why I thought my method was correct) \$\endgroup\$ – dzaima Aug 31 '17 at 17:25
  • \$\begingroup\$ if 20 new pieces spawn, 2 4's and 18 2's. We can create additional 9 4's by merging the 2's. We now have 11 4's with a total score of 36 and an average score of 36/11 = 3.2727... \$\endgroup\$ – ovs Aug 31 '17 at 17:32
4
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Python 3, 51 48 bytes

-3 bytes thanks to Mr. Xcoder!

lambda b:sum(n*max(0,len(bin(n))-4.1)for n in b)

Try it online!

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  • 1
    \$\begingroup\$ 48 bytes. n.bit_length() means len(bin(n))-2, which we can compress into len(bin(n))-4.1 \$\endgroup\$ – Mr. Xcoder Aug 31 '17 at 16:43
  • \$\begingroup\$ @Mr.Xcoder Nice. I was sure there was some easier way to do number of bits but I blanked for some reason. \$\endgroup\$ – notjagan Aug 31 '17 at 16:44
1
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C# (.NET Core), 101 100 bytes

n=>{double k=0,l=0;for(int i=0,j;i<n.Length;l+=k,++i)for(j=4;j<=n[i];j*=2)k=j>4?k*2+j:3.6;return l;}

Try it online!

Lambda function. Takes in an array for the board, and returns a double for the score

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