arithmetic – How Many Squares on the Peg Solitaire

For reference, let the pegs be numbered as proven under.

      00 01 02
      03 04 05
06 07 08 09 10 11 12
13 14 15    16 17 18
19 20 21 22 23 24 25
      26 27 28
      29 30 31

Initially, I neglected the presence of the {3 11 20 28} sq. in M Oehm’s reply, so assuming some squares had been neglected, determined to put in writing a program to seek out all of them. To my shock, I acquired precisely the identical variety of squares as in that reply; however upon studying it once more, noticed the {3 11 20 28} sq. listed pretty clearly within the (3,1) class. Anyhow, right here is a few brute-force python code that takes about 1/16 second to listing all of the squares.

#!/usr/bin/env python
# Re: https://puzzling.stackexchange.com/questions/51773/how-many-squares-on-the-peg-solitaire
# In this program, given a structure of pegs, we depend the variety of
# squares fashioned through the use of any 4 of these pegs as corners.
import time

# List the x, y values of all 32 pegs, treating one of many corners as
# 0,0 and one other nook as (6,6).
pegX = [2,3,4, 2,3,4, 0,1,2,3,4,5,6, 0,1,2,4,5,6, 0,1,2,3,4,5,6, 2,3,4, 2,3,4]
pegY = [0,0,0, 1,1,1, 2,2,2,2,2,2,2, 3,3,3,3,3,3, 4,4,4,4,4,4,4, 5,5,5, 6,6,6]
nPeg = len(pegX)

def dist2(m,n):
    return (pegX[m]-pegX[n])**2 + (pegY[m]-pegY[n])**2

# Test if factors p,q,r,s make one nook of a sq.
def stest(p,q,r,s):
    dq = dist2(p,q)
    dr = dist2(p,r)
    ds = dist2(p,s)
    dhi = max(dq, dr, ds)
    dlo = min(dq, dr, ds)
    if (dhi != 2*dlo): return False
    nlo = (dq==dlo) + (dr==dlo) + (ds==dlo)
    if nlo != 2: return False
    return True

# Cycle by all p,q,r,s combos, sustaining p<q<r<s
p, q, r, s = 0, 1, 2, 2        # Initial p,q,r,s tuple, -1
nSqu = 0
t0 = time.time()
whereas 1:
    s += 1
    if s >= nPeg:
        r += 1
        if r >= nPeg-1:
            q += 1
            if q >= nPeg-2:
                p += 1
                if p >= nPeg-3:
                    break
                q = p+1
            r = q+1
        s = r+1
    # If three corners work, the fourth is compelled
    if stest(p,q,r,s) and stest(q,p,r,s) and stest(r,p,q,s):
        nSqu += 1
        print 'S#{:3}  {:3}{:3}{:3}{:3}'.format(nSqu, p,q,r,s)

t1 = time.time()
print 'Elapsed seconds: ', t1-t0