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Saturday September 16, 2006 6:00 am Edmonton
A. Morning Musings
6:00 am It is dark outside and my computer is not picking up a wireless signal.
The drive up to Edmonton yesterday was fine until Calgary but we had rain the rest of the way, not hard but enough to cause limited visibility due to the spray from the other vehicles, particularly the trucks.
While getting our stuff ready for the trip I found a notebook that had some notes about birds we had seen while on other trips. I think we have these in our data base for North American birds, but I want to verify this this morning.
I also have an idea for the puzzle I am working on that might help me fill in a few more cells. This will involve setting up another grid to keep track of where the multiples of three are, or are not located.
The coffee is almost ready. I might create the web site for science notes this morning and then begin work on some notes for "Science and the Akashic Field".
B. Plan
Today's activities will depend largely on the weather. As near as I can tell at the moment, there appears to be a very light drizzle or mist outside. It might be a good day for indoor pursuits.
Immediate Health Walk & exercise 1 hr Puzzles The Orange Puzzle Cube: puzzle #6 6 hr Technology add keywords to iPhoto records 1 hr Birds Add June birds to North American data base 1 hr History Continue reading "Citizens" 1 hr Mathematics Read "The Computational Beauty of Nature" Chap 3 1 hr Literature Continue reading "All the Men Are Sleeping" 1 hr Read "In Praise of Folly" by Erasmus 1 hr Read "The Art of Living" by Epictetus 1 hr Read "The Song of Roland" 1 hr Science Make notes for "Science and the Akashic Field" by Ervin Laszlo 1 hr Later Chores Investigate water softeners for home 1 hr Technology Read manual for cell phone 1 hr Make notes for chap. 4 of "Switching to the Mac" 2 hr Begin reading "iPhoto" 1 hr digital photography - learn about using the various manual settings Mathematics Larson "Calculus" Gardner "The Colossal Book of Short Puzzles" History Watson "Ideas" Model Trains Build oil refinery diorama: add ground cover Add blue backdrop to layout Assemble second oil platform kit Redraw diagram for Lower Mainline control panel Wire Lower Mainline turnouts
C. Actual/Notes
6:35 am I compared the notes for bird watching in the notebook with the records in my database and was able to add 1 bird to my files. This was an Evening Grosbeak that we positively identified at Post Falls, Idaho while on a weekend trip to Spokane. It also was a new lifer, bringing our total to 81.
Now to play with my puzzle for a bit.
Step 34 The four corner numbers are all multiples of three and no other multiples of three are adjacent to them or each other. Let's try mapping this out and see if I notice anything. Let green represent cells with a multiple of 3 in them and red represent cells where a multiple of 3 is not allowed
Now I need to review what I know about each cell and see if I can identify any other numbers that can be identified as either being a multiple of three or as not being a multiple of three. If so, place the number 34 in those cells in the large table (which I have copied below.
The multiples of 3 are 3, 6, 9, 12, 15, 18, 21 & 24. There are 8 such numbers. I have identified 5 cells that contain these numbers.
Look at the white cells. Are any of them red or green based on what I already know? YES! D2 is red, so are B2 & D3. C1 is green. This implies that C2 cannot be red. Therefore place a 34 in all cells for C2 that are multiples of 3.
This also implies that E3 must be green. Place a 34 in all cells for E3 that are not multiples of 3. One of A3 or B3 is also a mutiple of 3, but I am not yet able to determine which one.
Let's try placing potential numbers in the green cells and see if anything becomes apparent.
This table is promising, but at the moment each number may appear in more than one cell
Step 35 The long diagonal from top left to bottom right totals 72. Each diagonal contains two consecutive numbers.
9, 12, 15, 18, 21, 24 2, 4, 8 1, 5, 7, 8, 10, 13, 14, 16, 17, 19, 20, 22, 23 14, 17, 20, 23 6, 9, 12, 15 72 Here are the possible consecutive numbers: (1, 2), (6, 7), (7, 8), (8, 9), (9, 10), (12, 13), (13, 14), (14, 15), ((15, 16), (16, 17), (17, 18), (18, 19), (19, 20), (20, 21), (21, 22), (22, 23), (23, 24). Almost anything is still possible.
Step 36 B2 = 2D2 Since D2 is 1, 2 or 4 then B2 is 2, 4 or 8
B2 = E2/2 Since B2 is 2, 4 or 8 then E2 is 4, 8 or 16. Nothing new here.
Step 37 A3 + A4 totals 23. All combinations are still possible. Step 38 10 is further left than 11 although they are in the same row, 13 and 19 in the row below
There are no 13s in row E. Therefore there cannot be any 10s or 11s in row D. Place a 38 in these cells.
Step 39 C1 is 5 higher than D1. D1 cannot be 10, therefore C1 cannot be 15. Place a 39 in this cell. Step 40 D4 = D1 +4. D1 cannot be 10, therefore D4 cannot be 14. Place a 40 in this cell. Step 41 D4 = B4 + D3.
B4 1, 5, 7, 8, 11, 13, 14, 16, 17, 19, 20, 22, 23 D3 1, 2, 4, 7 D4 17, 20, 23 Therefore B4 cannot be 1, 5, 7, 8, 11, 17, 20 or 23. Place a 41 in these cells.
Since B4 cannot be 11, then in step 38, cell B4 should be simple red
Step 42 D3 = B5/2. Therefore D3 is 1, 2, 4 or 7. Nothing new here. Step 43 The long diagonal from top right to bottom left totals 61. Each diagonal contains two consecutive numbers.
A5 6, 9, 12, 15, 18, 21 B4 13, 14, 16, 19, 22 C3 1, 5, 7, 8, 10, 13, 14, 16, 17, 19, 20, 22, 23 D2 1, 2, 4 E1 9, 12, 15 61 Here are the possible consecutive numbers: (1, 2), (4, 5), (5, 6), (6, 7), (8, 9), (9, 10), (12, 13), (13, 14), (14, 15), (15, 16), (16, 17), (17, 18), (18, 19), ( 19, 20), (20, 21), (21, 22), (22, 23)
Let's try looking at the result of trying these pairs:
(1, 2) = 3 [C3, D2] 6, 9, 12, 15, 18, 21 13, 14, 16, 19, 22 9, 12, 15 3 61 okay Let's try another pair:
(4, 5) = 9 [C3, D2] 6, 9, 12, 15, 18, 21 13, 14, 16, 19, 22 9, 12, 15 9 61 Nothing obvious here at the moment.
Step 44 No two consecutive numbers are adjacent in any direction.
Now consider 23. Cell A2 is not possible under either of the above two possibilities. Therefore 23 cannot be in cell A2. Place the number 44 in that cell. Cell B1 is also not possible under either possibility. Place the number 44 in that cell. Cell B3 is possible for both possibilities. Cell C2 is not possible under either possibility. Place the number 44 in that cell. Cells C3, D4 & D5 are possible for both possibilities.
Step 45 In column 1 the lowest number is 8. Therefore 8 is in column 1. Therefore 8 is not in columns 2 - 5. Put a 45 in these cells. Breakthrough!! This means that 8 is in cell B1. Therefore no other number is in cell B1. Put the number 45 in these cells. Step 46 Since B1 is an 8, this means it cannot be a 10 or 13 or 19 (see step 38). Therefore since 10 cannot be in B1, 11 cannot be in B3 or B4. Step 47 Since 10 and 11 cannot be in row B, 13 and 19 cannot be in row C. Place a 47 in these cells Step 48 No two consecutive numbers are adjacent in any direction. Since B1 is an 8, cells A1, B2 and C1 cannot be either a 7 or a 9. Place the number 48 in these cells. Step 49 The long diagonal from top left to bottom right totals 72. Each diagonal contains two consecutive numbers. A1 cannot be a 9, B2 cannot be an 8, C3 cannot be an 8
12, 15, 18, 21, 24 2, 4 1, 5, 7, 10, 13, 14, 16, 17, 19, 20, 22, 23 14, 17, 20, 23 6, 9, 12, 15 72 Here are the possible consecutive numbers: (1, 2), (6, 7), (9, 10), (12, 13), (13, 14), (14, 15), ((15, 16), (16, 17), (17, 18), (18, 19), (19, 20), (20, 21), (21, 22), (22, 23), (23, 24). Almost anything is still possible.
Step 50 In row E the highest number is 17. Therefore 17 is not in rows A - D. Place a 50 in these cells. This means 17 is in cell E4. Step 51 A3 + A4 = 23. A4 cannot be 17. Therefore A3 cannot be 16. Place a 51 in that cell Step 52 D4 = B4 + D3.
B4 13, 14x, 16, 19, 22 D3 1, 2x, 4, 7 D4 17, 20, 23 Therefore B4 cannot be 14 and D3 cannot be 2. Place a 52 in these two cells.
Step 53 D3 = B5/2. Therefore B5 can be 2, 4 or 14. And D3 can be 1, 2 or 7. D3 cannot be 4. Place 53 in that cell. Also B5 cannot be a 4. Therefore D3 cannot be 2. Place 53 in that cell. Step 54 The long diagonal from top left to bottom right totals 72. Each diagonal contains two consecutive numbers. A1 cannot be a 9, B2 cannot be an 8, C3 cannot be an 8
12, 15, 18, 21, 24 2, 4 1, 5, 7, 10, 14, 16, 20, 22, 23 20, 23 6, 9, 12, 15 72 Here are the possible consecutive numbers: (1, 2), (6, 7), (9, 10), (14, 15), ((15, 16), (20, 21), (21, 22), (22, 23), (23, 24). Almost anything is still possible.
Step 55 The long diagonal from top right to bottom left totals 61. Each diagonal contains two consecutive numbers.
A5 6, 9, 12, 15, 18, 21 B4 13, 16, 19, 22 C3 1, 5, 7, 10, 14, 16, 20, 22, 23 D2 1, 2, 4 E1 9, 12, 15 61 Here are the possible consecutive numbers: (1, 2), (4, 5), (5, 6), (6, 7), (9, 10), (12, 13), (13, 14), (14, 15), (15, 16), (18, 19), ( 19, 20), (20, 21), (21, 22), (22, 23).
Only one of the above pairs can appear in this diagonal. All of the rest are impossible. Suppose it is (1, 2). Then the largest 3 remaining numbers would be 21, 19 & 15. Therefore (1, 2) cannot be the consecutive numbers.
Let's try the next largest pair (4, 5). The three largest remaining numbers would be 21, 19 & 15. This doesn't seem to be leading anywhere quickly.
Step 56 B2 = 2D2 D2 is 1, 2 or 4. Therefore B2 is 2, 4 or 8.
B2 = E2/2 E2 is 4 or 16. Therefore B2 is 2 or 8. Therefore B2 is not 4. Put a 56 in that cell. Therefore B2 is 2. And B5 is 14.
Step 57 No two consecutive numbers are adjacent in any direction. Since B2 is 2, A2, B1, B3 & C2 cannot be 1 or 3. Place 57 in these cells. Since B5 is 14, A5, B4 & C5 cannot be 13 or 15. Place 57 in these cells. Step 58 The long diagonal from top left to bottom right totals 72. Each diagonal contains two consecutive numbers. Since B2 is 2 the table can be simplified to
12, 15, 18, 21, 24 2 1, 5, 7, 10, 14, 16, 20, 22, 23 20, 23 6, 9, 12, 15
72 I have just realized that the phrase "each diagonal contains two consecutive numbers" is potentially ambiguous. Does it mean that the two numbers must be next to each other or can they be anywhere among the five? If they must be next to each other, the only possibilities are (2, 1) and (22, 23). (2, 1) is impossible as the total will be less than 72. (22, 23) is also impossible as I cannot make a total of 72 with the remaining numbers. Therefore the two consecutive numbers do not have to be next to each other.
Thus the consecutive numbers can be any of (1, 2), (5, 6), (6,7), (9, 10), (14, 15), (15, 16), (20, 21), (21, 22), (23, 24).
If the consecutive numbers are (1, 2) then the largest three remaining numbers would be 24, 23 and 15 which is less than 72. Therefore C3 cannot be 1.
If the consecutive numbers are (5,6) then the largest three remaining numbers would be 24, 2, 23 which is less than 72. Therefore C3 cannot be 5.
If the consecutive numbers are (6, 7) then the largest three remaining numbers would be 24, 2, 23 which is less than 72. Therefore E5 cannot be 6.
If the consecutive numbers are (9, 10) then the largest three remaining numbers would be 24, 2 23 which is less than 72. Therefore E5 cannot be 9.
If the consecutive numbers are (14, 15) then the largest three remaining numbers would be 24, 2, 23 which is more than 72.
I may come back to this approach, but at the moment it is still looking to be indeterminate.
Step 59 Suppose A1 is 24. (24 is in either A1 or C1). Then the diagonal numbers are 24, 2, C3, D4 (which is either 20 or 23) and E5. This total is 72. Then C3 + D4 + E5 = 46.
1, 5, 7, 10, 16, 20, 22, 23 20, 23 6, 9, 12, 15 46 If E5 is 6 then C3 + D4 = 40 which is impossible.
If E5 is 9 then C3 +D4 = 37 which is impossible.
If E5 is 12 then C3 + D4 = 34 which is impossible.
If E5 is 15 then C3 + D4 = 31 which is impossible. Therefore A1 is not 24. Place 59 in this cell.
Step 60 Step 59 implies that 24 is in cell C1 and 25 is in cell A2. Step 61 The number in B2 is twice D2. D2 is 1 or 4. B2 is 2. Therefore D2 is 1 and D3 is 7. Step 62 A3 + A4 = 23.
6, 9, 12, 15, 18, 21, 22 5, 11, 16, 22 23 Therefore A3 is not 22, 21, 15, 9, 6. If A3 = 12, then A4 = 11 which violates adjacent cells being consecutive.
If A4 = 5, then A3 = 18 and A4 = 5.
Step 63 (refer to step 38) 10 cannot be in cell B2. Therefore 10 must be in either C2 or C3. 11 must be in C5. Step 64 (refer to step 38) therefore 13 and 19 cannot be in cells B3 or B4. Therefore B4 is 16. Step 65 Since 16 is C5, E2 is 4. Step 66 D4 = D1 + 4. D1 is either 12 or 19. Therefore D4 is either 16 or 23. D4 is not 16. Therefore D4 is 23. Therefore D1 is 19. Step 67 Therefore D5 is 13. Step 68 The long diagonal from top left to bottom right totals 72. Each diagonal contains two consecutive numbers. Since B2 is 2 the table can be simplified to
12, 15, 21 2 10, 20, 22 23 6, 9, 12, 15
72 This simplifies to :
12, 15, 21 10, 20, 22 6, 9, 12, 15
47 There are three combinations that satisfy this: (12, 20,15), (15, 20, 12), ( 21, 20, 6). Therefore C3 is 20.
12, 15, 21 10, 20, 22 6, 9, 12, 15
47
Step 69 There must be two consecutive numbers in the diagonal. Therefore A1 is 21. Step 70 Therefore E5 must be 6. Step 71 Therefore C2 is 10. Step 72 Therefore B3 is 22. Step 73 The long diagonal from top right to bottom left totals 61. Each diagonal contains two consecutive numbers.
A5 9, 12 B4 16 C3 20 D2 1 E1 9, 12, 15 61 This simplifies to
A5 9, 12 E1 9, 12, 15 24 Therefore A5 is 9 and E1 is 15.
Step 74 Therefore E3 is 12.
A1 A2 A3 A4 A5 B1 B2 B3 B4 B5 C1 C2 C3 C4 C5 D1 D2 D3 D4 D5 E1 E2 E3 E4 50 E5 Here is the final solution:
9:55 PM I have finally (!) completed puzzle #6 from The Orange Puzzle Cube. This occupied a considerable amount of my time today (at least 6 hours). I was like a bull terrier on this one. I knew I had a representation system that would keep track of the steps and I knew that all that was required was a very basic knowledge of the numbers from 1 to 25. But the effort required and the amount of time it took to eliminate all of the alternatives took far longer than I would have guessed. Still it is a good feeling to have suceeded. The rain today helped keep me indoors.
D. Reflection