Classes have started! And so does the boredom. I decided to doodle in my notebook and started to draw some dots in (IMO) an aesthetically pleasing way. I came up with these numbers: Dot Numbers

based on these conditions:
Given (n, m)
1) There must be n dots
2) All dots must lie on an m by m lattice
3) The position of the dots must minimize the spread* from the top left corner
4) The configuration of the dots must be diagonally symmetric as pictured:
Symmetry Example

*Spread (in this context) is defined as the sum of the distances from the top left dot position (whether or not there is a dot there). For example, the spread of the example above is 0 + 1 + 2 + 1 + √2 + 2 + 2√2 = 6 + 3√2

The Challenge

Come up with an algorithm using that uses natural numbers n and m (where it will always be that n <= m^2) to generate a configuration of dots that follow all the rules above.

Input can be received via STDIN, command-line argument, or function parameter.

Output the pattern to STDOUT or return a string with newlines. Any two different characters may be used in the output. (e.g.


is the same as


Shortest Code in Bytes Wins


STDIN: 6 3


The Challenge has Ended!!!

After seven days I decided to declare the winner, but first some special awards!
Fastest code goes to... orlp with his submission in Pyth !
Slowest code goes to... orlp with his submission in Pyth !
Most straightforward code goes to... orlp with his submission in Pyth !
Most confusing code goes to... orlp with his submission in Pyth !
Longest Code goes to... orlp with his submission in Pyth !

And last but not least

Shortest code goes to......... orlp with his submission in Pyth !


  • \$\begingroup\$ I updated the conditions so that 10 3 will never be an input. Also 0 is not a natural number so 0 0 will also never be an input \$\endgroup\$ Sep 15, 2015 at 0:32
  • 2
    \$\begingroup\$ natural numbers is ambiguous. The set-theoretic definition of natural number, e.g., includes 0. To avoid confusion, it's preferable to use (strictly) positive integers. \$\endgroup\$
    – Dennis
    Sep 15, 2015 at 3:26
  • \$\begingroup\$ @Dennis Also, if one specifically wants to include zero, nonnegative integers are a good choice. \$\endgroup\$ Sep 15, 2015 at 4:47
  • \$\begingroup\$ What if there there are several valid solutions? E.g. 24 6. \$\endgroup\$ Sep 15, 2015 at 8:37
  • \$\begingroup\$ @steveverrill I think they're sqrt(1^2+2^2)=sqrt(5) \$\endgroup\$
    – Cabbie407
    Sep 15, 2015 at 16:13

1 Answer 1


Pyth, 37 bytes


Slow solution. Generates all possible pairs of [0, m), generate all possible n-combinations of those pairs (without repetition). Filter each combination of pairs such that if P is in the combination, so is reverse(P). This guarantees diagonal symmetry. Sort by the sum of distances to the origin. Decode the pairs of numbers as base-m numbers, giving a list of indices. Create a string of spaces except for those indices, putting quotation marks instead. Finally chop into m pieces and join by newlines. Creates outputs like such:


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