A puzzle puzzle

Jigsaw puzzles that say they have 1,000 pieces have approximately 1,000 pieces, but probably not exactly 1,000. Jigsaw puzzle pieces are typically arranged in a grid, so the number of pieces along a side has to be a divisor of the total number of pieces. This means there aren’t very many ways to make a puzzle with exactly 1,000 pieces, and most have awkward aspect ratios.

Since jigsaw pieces are irregularly shaped, it may be surprising to learn that the pieces are actually arranged in a regular grid. At least they usually are. There are exceptions such as circular puzzles or puzzles that throw in a couple small pieces that throw off the grid regularity.

How many aspect ratios can you have with a rectangular grid of 1,000 points? Which ratio comes closest to the golden ratio? More generally, answer the same questions with 10^n points for positive integer n.

More puzzles:

A knight’s random walk
Peculiar property of 3909511
Roman numeral problem
A perspective problem

7 thoughts on “A puzzle puzzle

  1. A grid of 10^n points has (n + 1)^2 possible aspect ratios (Assuming A:B and B:A are each unique), and the closest ratio will be 1:1 (A square of sides 10^(n-1).

  2. public class _Main_ {

    private java.util.Vector _factors;
    private java.util.Iterator _factorsI;
    private double _diff;
    private int _num;
    private int _total;
    private int _theOne;
    public static final double _GoldenRatio = 1.618034;
    public _Main_(int total){
    _total = total;
    _diff=100;
    _factors = new java.util.Vector();
    for(int k =1; k<java.lang.Math.sqrt((double)_total);k++){
    if(_total%k == 0){
    _factors.add(k);
    }
    }
    _num=0;
    _factorsI = _factors.iterator();
    while(_factorsI.hasNext()){
    int temp = _factorsI.next();
    if(java.lang.Math.abs(_GoldenRatio-(double)(temp/(_total/temp)))<_diff ||java.lang.Math.abs(_GoldenRatio-(double)((_total/temp)/temp))<_diff){
    if(java.lang.Math.abs((double)(temp/(_total/temp)))<java.lang.Math.abs((double)((_total/temp)/temp))){
    _diff = java.lang.Math.abs(_GoldenRatio-(double)(temp/(_total/temp)));
    }else{
    _diff = java.lang.Math.abs(_GoldenRatio-(double)((_total/temp)/temp));
    }
    _theOne=temp;
    }
    _num+=1;
    }
    System.out.println("there are "+_num+" arrangements for "+_total+" pieces.");
    System.out.println(_theOne+"x"+_total/_theOne+" is the closest to the golden ratio");
    }

    public static void main(String[] argvs){
    new _Main_(1000);
    //There are 8 arrangements for 1000 pieces.
    //25×40 is the closest to the golden ratio
    }
    }

  3. For the generalization of the first question, it’s ceil((n+1)^2/2). The prime factorization of 10 is 2*5, so the prime factorization of 10^n is 2^n * 5^n. For a 2 dimensional jigsaw puzzle, what we want to do is find the number of ways to partition those 2’s and 5’s among the two different axes, so there are n+1 ways to divide up the 2’s, and n+1 ways to divide up the 5’s. However, a 25×40 jigsaw puzzle has the same aspect ratio as a 40×25 puzzle, so you divide by 2, except a 10×10 puzzle is the same as a 10×10 puzzle, so we take the ceiling, since for even n, we would otherwise end up with a fractional number of aspect ratios.

    Incidentally, I’m pretty sure most 1000 piece puzzles I’ve done are 25×40, but perhaps I’m misremembering the 1500 piece 30×50 puzzles, which is much closer to phi.

  4. For 10^3, 25 x 40 isn’t so bad. I will check the next 1000 piece puzzle I do and report back :)

  5. There are exceptions such as circular puzzles or puzzles that throw in a couple small pieces that throw off the grid regularity.
    Actually, quite a few commercial jigsaw puzzles have non-rectangular arrangements, not-fully-interlocking (or even non-interlocking) pieces, and other nefarious ways of making it more difficult to spot a piece’s correct orientation. Of the 2 dozen or so rectangular jigsaw puzzles in my closet, probably about 1/3 depart from a strictly rectangular grid.

  6. I prefer Robbie’s point of view that A:B and B:A are unique. I’ll label the following as conjecture for the ratio closest to the golden ratio for some given n.

    For odd n >= 3:

    40 x 10^(n-3) :: 25 x 10^(n-3)

    For even n >= 4

    125 x 10^(n-4) :: 80 x 10^(n-4)

    A proof of this might arise from examining a fraction composed of a bunch of 2’s and 5’s. It doesn’t appear possible to make such a fraction whose value is close to 1.011 or 1.036. That is what would be required to produce a different aspect ratio that is closer to phi than the 40×25 or 125×80 (respectively)

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