# [SOLUTION] Grid Folding (#63)

I wonder for 4x4 the total solution is 16 * 4! = 384

My solution about check_fold is try to find all combinations possible.

Below is modified version of my check_fold, it will print out all
possible combinations and total count number.
However I found for 4x4 there is only 96 possible not 384.
Can someone tell me that I am wrong ( missing some combination ) or
the previous math formula is not correct ?

Thanks,

def check_fold(row, col, result)

# find all combinations with binary 0 for row and 1 for column

operation
def all_orders(r, c) #
return [2c - 1] if (r <= 0) # c bits of 1 is 2c-1
return [0] if (c <= 0) # r bits of 0 is 0
table = []
all_orders(r-1,c).each { |t| table << ((t << 1) + 0) }
all_orders(r,c-1).each { |t| table << ((t << 1) + 1) }
table
end
=begin
if row <= 0 ||
col <= 0 ||
row * col != result.size ||
2 ** (Math.log(row)/Math.log(2)).to_i != row ||
2 ** (Math.log(col)/Math.log(2)).to_i != col
raise “Error: Parameters are not correct.”
end
=end
r = Integer(Math.log(row) / Math.log(2))
c = Integer(Math.log(col) / Math.log(2))
all_rc_orders = all_orders(r,c)
count = 0

row.times do |tb_operation|
col.times do |lr_operation|
all_rc_orders.each do |order|
operations = ‘’
tb_op = tb_operation
lr_op = lr_operation
(r+c).times do
if (order & 1 == 0)
operations += (tb_op & 1 == 0) ? ‘T’ : ‘B’
tb_op >>= 1
else
operations += (lr_op & 1 == 0) ? ‘L’ : ‘R’
lr_op >>= 1
end
order >>= 1
end
puts operations
count += 1
# return operations if fold(row, col, operations) == result
end
end
end
p count
end

check_fold(4,4, nil)

#==================================#
The output:

TTLL
TLTL
TLLT
LTTL
LTLT
LLTT
TTRL
TRTL
TRLT
RTTL
RTLT
RLTT
TTLR
TLTR
TLRT
LTTR
LTRT
LRTT
TTRR
TRTR
TRRT
RTTR
RTRT
RRTT
BTLL
BLTL
BLLT
LBTL
LBLT
LLBT
BTRL
BRTL
BRLT
RBTL
RBLT
RLBT
BTLR
BLTR
BLRT
LBTR
LBRT
LRBT
BTRR
BRTR
BRRT
RBTR
RBRT
RRBT
TBLL
TLBL
TLLB
LTBL
LTLB
LLTB
TBRL
TRBL
TRLB
RTBL
RTLB
RLTB
TBLR
TLBR
TLRB
LTBR
LTRB
LRTB
TBRR
TRBR
TRRB
RTBR
RTRB
RRTB
BBLL
BLBL
BLLB
LBBL
LBLB
LLBB
BBRL
BRBL
BRLB
RBBL
RBLB
RLBB
BBLR
BLBR
BLRB
LBBR
LBRB
LRBB
BBRR
BRBR
BRRB
RBBR
RBRB
RRBB
96

Actually, you are right, that my math was wrong. (All this talk about
loving math and here I am messing it up…)

Some of my 4! permutations overlapped with possible folds… That is,
“RR” backwards is still “RR”, etc.

So really, rather than 4! permutations, I have to not permute the
change the relative order of [LR] and [TB] folds. So a 4x4 would be
16 * 6 == 96 possible complete folds.

You are correct, sir.

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