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pnm
Posted on: Wed Dec 04, 2013 9:38 pm
Posts: 2281 Joined: Thu May 12, 2011 11:58 pm

an interesting 4x4
I made a note a few months ago to put this 4x4 up on the forum, because it was interesting (I don't remember why (?!)). Because it was set for publication in a Dutch paper (de Volkskrant), I had to wait until it was actually published before I could put it online..: The challenge here is to come up with a short & elegant & logical solution (I have one here but I'm sure there are better ones )
Last edited by pnm on Thu Dec 05, 2013 10:30 am, edited 1 time in total.




bram
Posted on: Thu Dec 05, 2013 12:26 am
Posts: 193 Joined: Tue May 24, 2011 4:55 pm

Re: an interesting 4x4
OK, here goes (using clm's notation, in which columns from left to right are called a, b, c and d, and rows from top to bottom are numbered 1, 2, 3 and 4, meaning that the upper left cell is called a1, etc): In the left column, 1 can't be in cage "1" since a3a4 = [1,2] >>> a1a2 = [3,4], which would mean that the sum of the three numbers in cage "7+" would be greater than 7. Therefore a1 = 1. The two remaining 1s in the lower right part of the puzzle can't be in c4 and d3 since that would mean that the numbers in b4 and d4 would both be 2. Therefore c3 = 1, d4 = 1 and d3 = 2. The sum of the numbers in cages "7+", "1" (the singlecell cage) and "17+" is 25. Subtract from that the sum of all numbers in rows 1 and 2, which is 2 x (1 + 2 + 3 + 4) = 20, and it emerges that the sum of the "outies" in cells b3c3 is 5. Given that c3 = 1, it follows that b3 = 4 and (to finish row 3) a3 = 3. In column b, 3 can't be in the top cell because b1 = 3 >>> a2 = 3, which is impossible because a3 =3, so 3 must be in the bottom cell of column b: b4 = 3. The rest is trivial.




pnm
Posted on: Thu Dec 05, 2013 10:37 am
Posts: 2281 Joined: Thu May 12, 2011 11:58 pm

Re: an interesting 4x4
bram wrote: OK, here goes (using clm's notation, in which columns from left to right are called a, b, c and d, and rows from top to bottom are numbered 1, 2, 3 and 4, meaning that the upper left cell is called a1, etc): thanks, I updated the image to include the coordinates. bram wrote: In the left column, 1 can't be in cage "1" since a3a4 = [1,2] >>> a1a2 = [3,4], which would mean that the sum of the three numbers in cage "7+" would be greater than 7. Therefore a1 = 1. A bit better than what I had: the "7+" cage is either 2,3,2, 3,1,3 or 1,2,4. The first two options are impossible because of the a3a4 "1" cage, so 7+ is 1,2,4, so a1 = 1. bram wrote: The two remaining 1s in the lower right part of the puzzle can't be in c4 and d3 since that would mean that the numbers in b4 and d4 would both be 2. Therefore c3 = 1, d4 = 1 and d3 = 2. I had exactly the same step :) bram wrote: The sum of the numbers in cages "7+", "1" (the singlecell cage) and "17+" is 25. Subtract from that the sum of all numbers in rows 1 and 2, which is 2 x (1 + 2 + 3 + 4) = 20, and it emerges that the sum of the "outies" in cells b3c3 is 5. Given that c3 = 1, it follows that b3 = 4 and (to finish row 3) a3 = 3. Nice one. I did something similar with columns c and d, concluding that b3 = c4 (both 3 or both 4). Next a step that if b3 = c4 = 3, then there would have two be two 4's in the block c1c2d1d2, which is not possible because the 7+ has 1,2,4 ... etc.




clm
Posted on: Thu Dec 05, 2013 11:00 pm
Posts: 710 Joined: Fri May 13, 2011 6:51 pm

Re: an interesting 4x4
pnm wrote: I made a note a few months ago to put this 4x4 up on the forum, because it was interesting (I don't remember why (?!)). ... The challenge here is to come up with a short & elegant & logical solution (I have one here but I'm sure there are better ones ) I think the puzzle is interesting because it contains a big (6 cells) cage ("17+") and three "1" cages; in a 4x4, initially, "1" allows the maximum number of possibilities (3) for a subtraction cage. ****** The elegant and logical solution has been given by bram showing that the 1's must go to the main diagonal a1b2c3d4, further more I think this is the more logical way to start (certainly a3a4 <> [1,2], otherwise b2 [outie] = 10  10 = 0 [impossible] so a1 = 1). But in this process, once you have c3 = 1, d4 = 1 >>> d3 = 2 and, since b3 + c3 is odd (parity rule to the two lower rows), it follows b3 = 4 (unique being d3 = 2) >>> a3 = 3 ( without calculating the sum of the outies b3c3 of the cage "17+" which is not in fact necessary) and row 3 is fully defined. It's clear that the rest is trivial (as also exposed by bram), i.e., solving now "7+" with b1 = 2 and a2 = 4, etc. (or directly finding c4 = 4 via the addition rule to the two righmost columns, etc.). ****** However I would like to observe something interesting what could be a different line of thinking to solve this puzzle: If we consider the full puzzle (which total sum is 40) we derive that the three "1" cages sum 15 = 40  7  1  17 so b3 + c3 = 5 (20  15, addition rule to the two bottom rows). This pair cann't be [2,3], no matter now the order, because the 4 of row 4 would go to b4 or c4 (along with a 3) for this horizontal cage "1" and then no 1 could be placed in row 4 (a4 or d4) due to the 2 present in b3c3 so necessarily b3c3 = [1,4] with c3 = 1, b3 = 4 and the rest is trivial again since d4 = 1 (considering in this moment that a4 cann't be 1 due to b1 = 0 or, alternately, the Patrick's line since the other two 1's in the corners a4c1 would make "7+" unsolvable), etc..




clm
Posted on: Sat Dec 07, 2013 1:23 pm
Posts: 710 Joined: Fri May 13, 2011 6:51 pm

Re: an interesting 4x4
Something more on this puzzle. In the previous posts we saw that for the proposed puzzle: b3 + c3 = 5, either by calculating first the sum of the three “1“ cages which is 15 (40  7  1  17) or simply by considering these two cells as “outies”, that is, 7 + 1 + 17  20 (the sum of rows 1 and 2); later we demonstrated that this pair must be [1,4] while [2,3] is invalid. Now let’s consider the puzzle below (in which we have “1” instead of “7+” in the 3cell Lshape cage and in which we have left blank the 6cell cage originally “17+”), which is also interesting and “similar” to the one proposed. Can we solve this other puzzle being the solution unique?. The answer is yes, pse try it and you will find that surprisingly the solution is exactly the same obtained for the puzzle proposed by Patrick. Why?. Probably because the “essential info” has already been provided in the drawing. Anyway, all this suggests the following, more “abstract”, discussion: If a permutation is possible then we would have 4 different “solutions” due to the permutation in the other two parallel cages, i.e., [1,2] in the vertical cages and [3,4] in the horizontal cages. To analyze by “brute force”, in the thread previously followed for the pair [2,3], it’s easy to see that also [1,2] is invalid in the blue area: a 4 would go to a3 or d3 producing a 3 either in a4 or d4 and then the cage “1” in b4c4 cann’t contain a 3 so it would be [1,2] and the solution would not be unique (permutations in the two horizontal cages [1,2]); also [3,4] is not valid in the blue area: a 1 would go to a3 or d3 producing a 2 in a4 or d4 and then the cage “1” in b4c4 cann’t contain a 2 so it must be [3,4] and we have again the permutations. In summary, for the Patrick’s proposed puzzle (and for the one I propose above), just combining the required pair [1,4] with the 1 in b2 we are driven to the solution.






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