=begin
 
132 : 2023/10/16(月) 08:51:26.93 : デフォルトの名無しさん : ID:kgcCjrnK
    1)
    BY
    RG
 
    2)
    RG
    BY
 
    3)
    GR
    YB
 
    1-3を回転対称で同じとみなせるグループと線対称で同じとみなせるグループに分類せよ
 
    4色を2x2の升にランダムに一つづつ配置して4x3x2通りのパターンを作成し
    それぞれを上の基準でグループ分けせよ
 
=end
 
#> 4色を2x2の升にランダムに一つづつ配置して4x3x2通りのパターンを作成し
# この文は曖昧で分かりづらいと思う
# 二通りの解釈で回答してみる
 
 
def solution( m )
	m.each_with_index{|n,i|
		print "%3d) " % [i+1], n[0..1], "\n     ", n[2..3], "\n"
	}
	sym = Array.new(3){ [] }
	(0...m.size).to_a.combination(2){|i,j|
		mi = m[i].chars
		mj = m[j].chars
		s = case
		when mi == mj		# 同一
			0
		when mi[0] == mj[2] && mi[1] == mj[0] && mi[2] == mj[3] && mi[3] == mj[1] ||		# 回転対称
		     mi[0] == mj[3] && mi[1] == mj[2] && mi[2] == mj[1] && mi[3] == mj[0] ||
		     mi[0] == mj[1] && mi[1] == mj[3] && mi[2] == mj[0] && mi[3] == mj[2]
			1
		when mi[0] == mj[2] && mi[1] == mj[3] && mi[2] == mj[0] && mi[3] == mj[1] ||		# 線対称
		     mi[0] == mj[1] && mi[1] == mj[0] && mi[2] == mj[3] && mi[3] == mj[2] ||
		     mi[0] == mj[0] && mi[1] == mj[2] && mi[2] == mj[1] && mi[3] == mj[3] ||
		     mi[0] == mj[3] && mi[1] == mj[1] && mi[2] == mj[2] && mi[3] == mj[0]
			2
		else
			next
		end
		sym[s] << [ i, j ]
	}
	sym.each_with_index{|n,i|
		print "同回線"[i], ": ", n.map{|x| "#{x[0]+1}-#{x[1]+1}" }.join(', '), "\n"
	}
	puts
end
 
##############################################################################
 
# 1-3
	solution( %w{ BYRG RGBY GRYB } )
 
# ランダムに一つづつ配置して4x3x2通り
	solution( Array.new(4*3*2){ 'RGBY'.chars.shuffle.join } )
 
# 順列 (4P4通り)をシャッフルして配置
	solution( 'RGBY'.chars.permutation.map{|a| a.join }.shuffle )
 

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