Announcement

Collapse
No announcement yet.

How many puzzles?

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • How many puzzles?

    I'm guessing that the Numbergrids are created by algorithm. There are approximately 39 shaded blocks in each (fiendish) puzzle but this is not always the case. Does anyone know the actual algorithm or if they are created manually? These 39 blocks are not randomly placed. I've never seen a 6 with any other number and I've only once come across a row/column with 7 shaded (2-2-2-1) and I've never seen a 7, 8, 9 or 10.

    If there were always 39 shaded blocks and if they could be anywhere (allowing 8 in a row for example) then there would be around 1,84 x 10 to the 74 possible puzzles.

    I occasionally see my own comments and I see others do as well so the actual number must be very much lower.

    What do you think? I would love to hear from any mathematicians or administrators.

    Also for the mathematicians / statisticians, how many of my own comments do I need to see in a row to be 80 / 90 / 95 / 99% certain that I have tried every puzzle?
    Last edited by Robin; 02-13-2015, 04:35 AM.

  • #2
    First approximation...

    I imagine they're created and tested by algorithm, but weeding out the un-solvable puzzles must be an important step.

    Ok, so I did 100 puzzles and kept track of whether or not I'd posted my score before, or posted it this time around. I've been doing 10x10 challenging puzzles, and have been posting scores from pretty much the start.

    So, of the 100 puzzles, I'd posted scores on 18 before. I also posted scores on another 69 (happy with that!). Half way through the 100 puzzles, I'd done a total of 606 puzzles.

    So if I post scores on 69% of the puzzles I do, that would be 606 x 0.69 = 418 puzzles I've posted scores on so far.

    I'd posted scores on 18% of the puzzles I saw, so the total number of 10x10 challenging puzzles must be somewhere near 418 / 0.18 = 2323 puzzles in total?

    Now to go and find my life again, I'm sure I left it around here somewhere...

    All the best,

    RMK.

    Comment


    • #3
      ...Another Approximation

      Now I've done more puzzles I thought I'd give it another go! Of the 100 puzzles I tried this time, I'd posted scores on 19 before and posted scores on 61 this time around. Half way through I'd done 2604 puzzles.

      So I've probably posed scores on about 2604 x 0.61 = 1588 puzzles, and there are probably about 1588 / 0.19 = 8360 puzzles in total. Quite a few more than the last estimate!

      Maybe I'll get bored and have another go in a few months time...

      Comment


      • #4
        An Upper Bound on the Possible Number

        Hello!

        I've been thinking about the number of possible puzzles. It seems to me that we should first determine the number of nonograms of a particular size. I'm only an undergraduate in math, so I can only submit a fairly basic calculation---in this case the highest upper bound for the possible number. Here's my logic:

        Each square of a puzzle has two possible "states": white or black. If we have a one-square puzzle, a 1x1, then there are obviously only two possible puzzles, since that square can be either black or white. Considering the next square, 2x2, there are 4 squares. Each of these can be in either state. Multiplication gives us all possible puzzles: 2*2*2*2 = 16. It's easy to check this with some graph paper. The multiplication can be expressed as exponentiation, where the base is 2 (the number of states) to the power of the number of squares, itself a square of puzzle's dimension.

        This means that for a puzzle of "n by n" dimensions, there are 2^(n^2) possible puzzles. Now this number does not take into account that many of these puzzles will be ambiguous (will result in multiple "solutions") or would be impossible to solve by logic (brute force testing). But it does give us a sense of the enormity of the number of puzzles. Here are some results:

        1x1: 2^(1^2) = 2^1 = 2
        2x2: 2^(2^2) = 2^4 = 16
        3x3: 2^(3^2) = 2^9 = 512
        4x4: 2^(4^2) = 2^16 = 65,536

        Puzzlebaron Sizes:
        5x5: 2^(5^2) = 2^25 = 33,554,432
        10x10: 2^(10^2) = 2^100 = 1,267,650,600,228,229,401,496,703,205,376
        15x15: 2^225 = 5.391989 * 10^67
        20x20: 2^400 = 2.582249 * 10^120
        25x25: 2^625 = 1.392346 * 10^188

        I doubt even the most avid numbergrids player would be able to exhaust the 10x10s, and a planet of players working simultaneously would probably not scratch the 25x25s...

        NL

        Comment

        Working...
        X