S2T2

Jim Randell 7:09 am on 16 August 2026 Permalink Reply
Tags: by: Victor Bryant ( 151 )   

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  Jim Randell 7:13 am on 16 August 2026 Permalink | Reply

  Here is a straighforward constructive solution, that just counts the number of squares in each candidate grid.

  The following Python program runs in 71ms. (Internal runtime is 1ms).

  from enigma import (irange, printf)
  
  # calculate the number of squares on an <a> by <b> grid
  def nsquares(a, b):
    t = 0
    for k in irange(min(a, b)):
      t += (a - k) * (b - k)
    return t
  
  # check the given arrangement
  assert (nsquares(3, 5) == 26)
  
  # target number of squares
  T = 10000
  
  # consider number of cols
  for nc in irange(1, 49):
    # and number of rows
    for nr in irange(1, nc - 1):
      t = nsquares(nr, nc)
      if t < T: continue
      if t > T: break
      # output solution
      printf("rows = {nr}, cols = {nc} -> nsquares = {t}")
  

  Solution: [To Be Revealed]

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  • 

    Jim Randell 7:38 am on 16 August 2026 Permalink | Reply

    See also: Teaser 1995, Enigma 1086.

    Analytically, we can show that for a grid with a rows and b columns (where a ≤ b), then the number of squares in the grid is:

    S(a, b) = a (a + 1) (3b − a + 1) / 6

    (See: OEIS A082652).

    So we can look for solutions to the equation:

    S(a, b) = 10000
    or:
    a (a + 1) (3b − a + 1) = 60000

    The following Python program implements a generic solver to find candidate grid dimensions given the total number of squares required. And then solves this particular puzzle by looking for grids that give exactly 10000 squares, and selecting those that meet the remaining requirements of the puzzle.

    It has an internal runtime of just 76µs.

    from enigma import (divisor, tuples, printf)
    
    # find (a, b) grids with a total of <t> squares
    def solve(t):
      n = 6 * t
      # look for consecutive divisors of 6t
      for (a, a1) in tuples(divisor(n)):
        if a1 - a != 1: continue
        (b, r) = divmod(a - 1 + n // (a * a1), 3)
        if b < a: break
        if r == 0:
          yield (a, b)
    
    # solve the puzzle
    for (a, b) in solve(10000):
      if a < b < 50:
        printf("rows = {a}, cols = {b}")
    

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    • 

      Frits 11:00 pm on 16 August 2026 Permalink | Reply

      @Jim, it doesn’t seem to happen but a check for b > a wouldn’t hurt.

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      • 

        Jim Randell 7:41 am on 17 August 2026 Permalink | Reply

        @Frits: Good point. I’ve added in a check for that.

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  Ruud 7:46 am on 16 August 2026 Permalink | Reply

  Brute force:

  for ncols in range(1, 50):
      for nrows in range(1, ncols):
          if sum((nrows - size + 1) * (ncols - size + 1) for size in range(1, nrows + 1)) == 10000:
              print(f'{nrows=} {ncols=}')
  

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