Yesterday's 7x7 difficult (that's Feb 11, 2015)
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tomas
Posted on: Thu Feb 12, 2015 11:56 am
Posts: 20 Joined: Tue Nov 22, 2011 4:13 pm

Yesterday's 7x7 difficult (that's Feb 11, 2015)
Hello everyone, I've never posted in this forum before, because I've never been as stuck as I am now. So let me give a little bit of introduction: Of course I'm a great fan of these puzzles, and many other types of logic puzzles as well. I teach Computer Science at the TU Delft, and I sometimes show these puzzles to my students, to give them another take on the topic of Natural Deduction in my course on formal logic. (They consider this a very difficult and esoteric topic, but reasoning is really very "natural", as the puzzles allow me to illustrate.) My history on the site tells me I solved my first puzzle here in August 2010 (!)... So I'm stuck trying to solve the 7x7 difficult puzzle from February 11, 2015. It has lots of subtractions, which I always find very challenging (read: horrible)I never seem to be able to reduce the number of possibilities enough. I think TAE will not get me anywhere (yet) because there are just too many possibilities. Still, this puzzle has already been solved 180 times, so others are apparently not having as many difficulties, and I hope I can learn some subtractionsolving strategies from any tips and hints you may have. This is the status so far... (I hope this embedding works, the picture is at https://www.dropbox.com/s/gjhe53n78glqncd/20150211difficult.png?dl=0.) So I've gone through all possibilities for the numbers that can be filled in in cages that are limited somehow, some "1"'s don't appear to have any limitations whatsoever, so I haven't filled anything in there. Can anybody shed some light on this? Thanks a lot! Bye, Tomas




eclipsegirl
Posted on: Thu Feb 12, 2015 9:31 pm
Posts: 240 Joined: Wed Apr 16, 2014 9:20 pm

Re: Yesterday's 7x7 difficult (that's Feb 11, 2015)
my response was in error. I apologize.
Sometimes, I find myself reduced to using brute force to find the solution. I end up choosing one cage where I test the possibilities and see it it solves the puzzle.




tomas
Posted on: Thu Feb 12, 2015 10:11 pm
Posts: 20 Joined: Tue Nov 22, 2011 4:13 pm

Re: Yesterday's 7x7 difficult (that's Feb 11, 2015)
eclipsegirl wrote: my response was in error. I apologize. Which response is that? eclipsegirl wrote: Sometimes, I find myself reduced to using brute force to find the solution. I end up choosing one cage where I test the possibilites. Yeah, that's probably what I'll end up doing... Thanks for the reply.




bram
Posted on: Thu Feb 12, 2015 10:42 pm
Posts: 192 Joined: Tue May 24, 2011 4:55 pm

Re: Yesterday's 7x7 difficult (that's Feb 11, 2015)
tomas wrote: Which response is that? eclipsegirl overwrote her initial (incorrect) response so as not to lead you astray Hints coming up …




firefly
Posted on: Thu Feb 12, 2015 10:44 pm
Posts: 37 Joined: Mon May 05, 2014 4:59 am

Re: Yesterday's 7x7 difficult (that's Feb 11, 2015)
Combined parity is a very helpful tool on these puzzles. Working on a writeup as well...




bram
Posted on: Thu Feb 12, 2015 10:47 pm
Posts: 192 Joined: Tue May 24, 2011 4:55 pm

Re: Yesterday's 7x7 difficult (that's Feb 11, 2015)
The sum of the numbers in a subtraction cage is even if the number in front of the operator is even and uneven if the number in front of the operator is uneven. For this reason the parity rule is often useful when you're dealing with puzzles featuring many subtraction (and/or addition) cages. Hint no. 1 for "your" puzzle: The sum of all the numbers in the second and third columns from the left is even according to the parity rule. Does this help us deduce something about the number at the bottom of the third column? Answer: Since the number of uneven subtraction cages within the two columns is itself even (there are four of them), the sum of the sums of the numbers in those cages must be even – they even each other out And seeing that the sum of the numbers in the cages that cover all cells but one in the two columns is even, the number at the bottom of the third column must be even (in this case, 4) for the parity rule to hold. Hint no. 2: Looking at the second and third rows from the bottom in the same way, can it be determined whether the number in the first cell of the third row from the bottom is even or uneven?




tomas
Posted on: Thu Feb 12, 2015 11:25 pm
Posts: 20 Joined: Tue Nov 22, 2011 4:13 pm

Re: Yesterday's 7x7 difficult (that's Feb 11, 2015)
bram wrote: The sum of the numbers in a subtraction cage is even if the number in front of the operator is even and uneven if the number in front of the operator is uneven. For this reason the parity rule is often useful when you're dealing with puzzles featuring many subtraction (and/or addition) cages. This is certainly a useful tool, thanks a lot! Also for the link to that other thread, I'll dive into this after I'm done grading exams... bram wrote: Hint no. 1 for "your" puzzle: The sum of all the numbers in the second and third columns from the left is even according to the parity rule. Does this help us deduce something about the number at the bottom of the third column? Yes, it has to be even. Because there are four cages with odd parity (summing to an even number), and there's the 4cage and two even paritycages, so that cell has to contain an even number to make the sum in the two columns add up to 72. bram wrote: Answer: Since the number of uneven subtraction cages within the two columns is itself even (there are four of them), the sum of the sums of the numbers in those cages must be even – they even each other out And seeing that the sum of the numbers in the cages that cover all cells but one in the two columns is even, the number at the bottom of the third column must be even (in this case, 4) for the parity rule to hold. Hint no. 2: Looking at the second a third rows from the bottom in the same way, can it be determined if the number in the first cell of the third row from the bottom is even or uneven? I'm not sure about just the second and third row from the bottom (yet), but looking at the bottom three rows, there are six odd paritycages (summing to an even number), and otherwise only even paritycages, so the first cell in the third row from the bottom must contain an even number to make the total of these three rows add up to 108 (so 2 or 6). This is indeed quite a useful technique, just what I needed. Thanks again!




firefly
Posted on: Fri Feb 13, 2015 12:01 am
Posts: 37 Joined: Mon May 05, 2014 4:59 am

Re: Yesterday's 7x7 difficult (that's Feb 11, 2015)
Bram had the same first two steps that I was going to post. Combined parity for rows 56 lead to a5=[2,6], which combined with the 4 in b5 and the 1 in ef5 will block the evens in row 5, which makes g5=1. The combined parity of columns bc lead to c7=4. The combined parity of rows 34 lead to d3=[2,6]. At this point, the puzzle looks like: A few more incidental tricks: ef6 can't have a 7 due to the 6 in g6; de4 can't have a 7 because that would leave both a4 and c4 as 1. This means that the 7's in ac are locked in rows 4, 6, and 7, eliminating the 7s in b2, c2, and 5c, also meaning c6 can't be 3. Since c6=[1,7] now, and since c34 has to be either [1,6] or [2,7], that blocks the 1's from c1 and c2. This leaves the puzzle looking like: I think honestly I might have brute forced the 4's in row 4 at this point, since it was such an obviously divergent binary option, but let me see if I can find a more elegant (ie logical) solution.




clm
Posted on: Fri Feb 13, 2015 12:06 am
Posts: 690 Joined: Fri May 13, 2011 6:51 pm

Re: Yesterday's 7x7 difficult (that's Feb 11, 2015)
The parity rule is absolutely useful for this type of puzzles (the perfect reasoning of bram to decide that c7 = 4). But, in addition of this, many times it's interesting to see if a determined number can be placed or not in a determined line. For instance, another hint:
Suposse that you make c3 =2, c4 = 7 then c5c6 = [1,5]. Since g5g6 = [1,6] and due to the three ones present in the three bottom rows, the 1 of column "b" would only be possible in cage "1" located in b3b4, that is, b3b4 = [1,2]. And then, where do we place a 6 in column "b"?, there is not a place for it, so cage "5" (c3c4) must necessarily be [1,6] and cage "4" (c5c6) = [3,7].
And from this point see that a 7 cann't be in b1, b2 or in cage "1" (b3b4) (that is, [6,7] there) since then a 5 would not be possible in either c1 or c2 and consequently b6b7 = [5,7].
Next: a5 must be even (using the parity rule and as explained in the other posts) but it cann't be a 6 (g5 = 1, g6 = 6) >>> a6a7 = [3,7] >>> c6 = 3 (c5 = 7) >>> a6 = 7 >>> b6 = 5 and now there is not a place for a 4 in row 6 so a5 = 2, a4 = 7
Last edited by clm on Fri Feb 13, 2015 12:32 am, edited 1 time in total.




tomas
Posted on: Fri Feb 13, 2015 12:26 am
Posts: 20 Joined: Tue Nov 22, 2011 4:13 pm

Re: Yesterday's 7x7 difficult (that's Feb 11, 2015)
clm wrote: The parity rule is absolutely useful for this type of puzzles (the perfect reasoning of bram to decide that c7 = 4). But, in addition of this, many times it's interesting to see if a determined number can be placed or not in a determined line. For instance, another hint:
Suposse that you make c3 =2, c4 = 7 then c5c6 = [1,5]. Since g5g6 = [1,6] and due to the three ones present in the three bottom rows, the 1 of column "b" would only be possible in cage "1" located in b3b4, that is, b3b4 = [1,2]. Why can the 1 of column b not appear in rows 1 or 2? clm wrote: And then, where do we place a 6 in column "b"?, there is not a place for it, I must confess I don't see that either. clm wrote: so cage "5" (c3c4) must necessarily be [1,6] and cage "4" (c5c6) = [3,7].






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