RE: Stuck on a Very Hard PicAPix

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RE: Stuck on a Very Hard PicAPix  1/6/2015 9:19:08 PM



arik_iis
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row 18 column 7 should be black. This is because the sequnce of row 18 is 2 4 2 3 2 summing up to 13 + 4 spaces for a total of 17 out of the 20 squares... so it means 3 freedom squares, meanig the numer 4 in that row has one square you can black... If what I wrote made no sense, please ask more... Once you mark this black, the whole column 7 is resolved. See if it gives you enough boost. Good luck! ... and you should try to solve such puzzles on this web site  it is more fun on the computer than on paper... Arik



RE: Stuck on a Very Hard PicAPix  1/6/2015 9:53:46 PM



debragordon18
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Thank you, Arik, but I am not sure I understand. Are you saying that the 4th box in row 18 should be shaded? I understand that there will be 3 freedom squares but I am not understanding the logic as to which single box will absolutely be shaded? Appreciate your response!



RE: Stuck on a Very Hard PicAPix  1/6/2015 10:02:32 PM



debragordon18
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I misread your response, I am sorry. You are saying that the box at the intersection of row 18 and column 7 should be shaded but I do not understand how you knew that should be the one we could shade?



RE: Stuck on a Very Hard PicAPix  1/7/2015 2:54:25 AM



debragordon18
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Thank you for your help, Charlene! Taking the screen shot worked!



RE: Stuck on a Very Hard PicAPix  1/7/2015 4:27:15 AM



CharleneTX
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I'm glad it finally worked; although I was assuming you were doing these on the computer rather than paper. :) Arik, you could please explain your logic process. I agree that 18:7 should be shaded, but I got the answer differently. I don't quite understand your method. Debra, here's how I concluded that 18:7 should be shaded. I hope this makes sense; it's easy to do, but difficult to describe. I count the clues from both the left and right hand sides and see where they overlap. In this case, from the lefthand side count the 2 and a space, then the 4 could possibly go in the 4th, 5th, 6th, and 7th squares. Doing the same from the righthand side count 2, a space, 3, a space, 2, a space, then the 4 could possibly go in the 10th, 9th, 8th, and 7th squares. Since the 7th square overlaps in both directions it must be black. Using this same method, you should be able to fill in some of the squares for the 8 in the 5th column. Charlene



RE: Stuck on a Very Hard PicAPix  1/7/2015 4:49:05 PM



arik_iis
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Yes, I was assuming it is on the computer too... So, Charlene, both of us are based on the same 'overlap' theory which is the basis of PicAPix / interpreting the hints. Did I understand you correclty that for each, say, row, you will count twice  once lefttoright and once righttoleft for EACH of the hints? (i.e. if there are 4 hints for the row, you will repeate this doublecounting 4 times? ) or do you have a method to do all 4 hints in one counting? The arithmetic I am using mimics what you do, but in one pass: I calculate the 'freedom' / the 'uncertanty' of the placement (which is the same number in this row or column for each hint), and then, in one pass, I shade all cells for each hint above the freedom threshold. As a sanity check, I then assure freedom was kept at the end too. Let me illustrate it with a drawing of 4 simple cases  we have one row with 10 cells (columns): So, case 1: 1 hint of "7" I am guessing your method  counting 1 through 7 from left, and then again from right, identifying the 4 overlaps (#4 through 7), so you shade these 4 cells. I am doing a calculation: 10 cells, hint of 7 ==> 10 minus 7 is 3 freedom (uncertenty) cells; so, counting from left, I leave the 3 freedom cells empty, and shade 4 through 7. As a sanity check, after the 7th cell is shaded, I count the remaining cells to the right, and it is 3 = same as the freedom count; case 2: 2 hints, 4, 4 again, I am guessing the left & right counting; are you counting it twice from left and from right, each time for a different hint, and mark the overlap? Again, I calculate: I have 4 +1space +4 = 9, ==>10  9 = 1 freedom cell... So, I count from left  1 freedom, then 2,3,4 shaded, 1 space, 1 freedom, then 2,3,4 shaded, and sanity check of 1 cell left = freedom count. case 3: 3 hints, 1, 4, 1 are you counting three times from left and right, to cover all hints? Again, I calculate: I have 1 +1space +4 + 1space, +1 = 8, ==> 10  8 = 2 freedom cell... So, I count from left  1 freedom, (hint1 is under the thershold / counting for it is done) then space, 2 freedom and 3,4 shaded from the 2nd hint (which is above the thershold), then space, 1 freedom (and the 3rd hint is below thershold / counting for it is done)  sanity check  2 cells left to the right = freedom... case 4: 3 hints, 2, 2, 3 I assume, again, you are counting three times from left and right, to cover all hints? My calculation: I have 1 +1space +4 + 1space, +1 = 8, (actually I sum 2,2,3 and add 2 spaces to it) ==> 10  9 = 12 freedom cell... meaning all three hints are above the freedom threshold So, I count from left  1 freedom, and shade number 2 (which is above the thershold), then space, 1 freedom and 1 shaded from the 2nd hint (which is above the thershold), then space, 1 freedom and 2 shaded from hint 3 (which is above the thershold)  sanity check  1 cells left to the right = freedom... {Since the first and last hints are above the thershold / left shaded cells, we see a symmetry = freedom cells left on the left and the right side) I truely hope I managed to explain myself, and not just confused you... Good luck! Arik
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RE: Stuck on a Very Hard PicAPix  1/8/2015 1:18:04 AM



debragordon18
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Thank you, Charlene. I believe I follow your logic. I have followed your suggestion and now have 11, 12, and 13 filled in column 5 to represent some part of the 8. Would you agree with that?



RE: Stuck on a Very Hard PicAPix  1/8/2015 1:31:03 AM



debragordon18
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Thank you, Arik. I believe I understand case 1. In case 2 I understand how you obtained 1 freedom cell. I do not understand how you knew to go 1 space and then one freedom then 2,3,4 shaded and sanity check of 1 cell left. Can you please define the terms you are using i.e. freedom cell and threshold. I appreciate your patience as I really want to understand your thinking.



RE: Stuck on a Very Hard PicAPix  1/8/2015 5:48:42 AM



CharleneTX
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I believe I mostly understand your method although to me there seems to be lots of room for mathematical errors. I count twice  once from the left and once from the right  regardless of how many clues are in the row or column. This is really so much easier to do than it is to describe. Using your examples of 10 boxes in one row: Case #1  7 : I would count from the left and in the 7th box I would shade the square. I would then count from the right. The previously shaded box is the 4th square from the right so I would fill in while counting 4, 5, 6, 7 and end up with the 4 shaded boxes. This describes the technique I use, but in reality with so simple an example I would mentally think 107 = 3 empty squares on each side similar to your method. Case #2  4 4 : I would count from the left and shade the 4th box. Then I would skip 1 blank square and start counting again and shade the next 4th box (which is the 9th box). I would then start counting from the right. The 1st box is blank. The next block is already shaded and is the first overlap box so I keep shading for a 4 count (boxes 3, & 4 from the right). Then just like the left, I skip a blank and start counting the next 4. The previously shaded box is an overlap so I keep filling in for the 4 count (boxes 8 & 9 from the right). Case #3  1 4 1 : I would count from the left: 1, skip, 1234 and shade the last box of the 4 (which is the 6th box from the left). Then I would count from the right: 1, skip, 1234. The 3 is already shaded so I only need to shade the 4 (which is the 6th box from the right). Case #4  2 2 3 : I would count from the left: 12 (shade the 2), skip, 12 (shade the 2), skip, 123 (shade the 3). Therefore, these boxes are shaded from the left: 2nd, 5th, and 9th. Then I would count from the right: 123 (the 2 is already shaded and I shade the 3), skip, 12 (the 2 is already shaded), skip, 12 (the 2 is already shaded). Basically, I when counting the clues from one end I shade the endpoint of each clue then coming from the other direction I know where the overlaps start. In reality, I don't often bother with the 2s and 3s except on the smallest puzzles because they rarely overlap. Also, I don't always shade every clue one the first pass. If only one or two boxes of a 10 overlap, then I know any smaller clue on that row (7, 6, etc.) won't overlap. Whenever someone posts asking for help, the first thing I do is recount each clue by row and column. Usually the problem is a miscount or mathematical error somewhere. I hope this makes sense. Charlene



RE: Stuck on a Very Hard PicAPix  1/8/2015 8:55:45 PM



arik_iis
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I understand most of what you are doing, Charlene... we are both doing the same, only I do it via arithmetic calculations, while you do more of a 'trial and error'; do I understand correctly that when you mark, from left to right, the last cell of each hint (well, assuming it is bigger than 2 or 3) that you might be ERASING these altogether when you count on the way back, from right to left, if you get no overlap? or do you do some calculation to decide overall if this row/column will have ANY overlap? (and if you are erasing, how do you avoid erasing a shading that came from a column hint into the row you are now working on, and find no overlap?) I think that especially in the large puzzles, and the hard ones, there are many rows/columns at the beginning where you get no shading from the hints we have, so it sounds like a lot of marking and erasing  unless I misunderstood you... If I understood you correctly, the first step you do is EXACTLY the calculation I do, but you achieve it by drawing and not calculating  namely, you count it from left, with spaces between hints, and when you've reach the end of the last hint, the number of cells which are left right of that last shaded cell is what I callled the "uncertanty" or "freedom" or "thershold' (three names for the same thing... or should I cal it "the minimum'?)  it tells you the minimum size of a hint that will have an overlap and thus also how many cells of overlap (shaded) there are; example: in case 1, if you counted from left, after counting 7 (and you've shaded cell #7) you are left with 3 cells until the end of line. This is the 'thershold' = it is 3: meaning, hints up to (including) 3 will NOT have ANY overlap / any shaded cells, and hints above it will have ovelap, which is calculated by reducing that thershold from the hint size; in this case, 7 will have overlap, and it's size will be 73 = 4 cells are shaded. Thus, after doing this math (in your head, without drawing), when you count from left to right, you count for the hint of 7 the 'thershold' 3 cells without marking, and then the other 4 cells of the 7 will be marked / shaded. you can check also for the other examples I did and see that after coutning all hints from left (with spaces between them  so for case 2 it will be 4 + space + 4 = 9, so 109 is 1...), you have found the 'thershold' So, Charlene  if for no other reason, you might want to try this calculation for large puzzles / long lines... like the example below from a puzzle I am now working on; if you look at the rows from the top, and want to decide swhich row to enter/shade cells, you calculate them: 1st row: 2 7 5 3 3 9 4 10 3 (9 hints)  summing up to 46 + 8 spaces (in between the 9 hints)  54; since the row is 60 cells, it means the freedom is 60  54 = 6, and this means the hints 7, 9 and 10 will have overlaps (of 1, 3 and 4 cells accordingly): so now you can go in with your method and draw, since you are sure to gain some shaded cells... Same for the second row  14 8 8 4 9 3 (6 hints)  summing up to 46 + 5 spaces (in between the 6 hints) = 51; so, for a row of 60 we get 609 freedom, thus only the hint 14 will have overlap (of 5 cells)  you can go in and draw your way for the last 2 rows, there will be no overlaps: 1+10+2+3+1+1+1 = 19+6spaces = 25 ==> 35 freedom; too many... 4+3+2+5+4+3+7+1+1 = 30+8 = 38 ==> 22 freedom; too many... ...But for as long as you've perfected your method, and are happy with it  I am sure you'll continue to use it and be happy with it. Different people think differently, and eventually you do find points I miss, so who am I to try and change your methods ;) Good luck to all of us, Arik P.S. if you are doing the math, you'll be amazed at how quickly you'll improve your ability to add this relatively small numbers in your head / a good exercise for the brain... ;)
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RE: Stuck on a Very Hard PicAPix  2/13/2020 6:38:01 PM



Rasheed
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try to take a screenshot and post it here



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