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Thursday September 14, 2006 5:20 am Lethbridge Sunrise 7:05 Sunset 19:48 Hours of daylight: 12:43
A. Morning Musings
5:20 am It is + 6 C with a light drizzle. The forecast high is 9 C. This will be the first day where we fail to reach double digits. Summer appears to be over.
Yesterday was a day spent largely on the screen. I accomplished a fair amount but when the day was done the result was two more tasks, both substantial, that need to be carried out. I now need to complete my data base for bird watching for North America and I need to update my iPhoto library to include keywords for all the images. At the moment I have 3232 images with 560 keyworded.
Today Phyllis and I will toast Fran who will be beginning a new chapter in her life. Best wishes Fran!
B. Plan
Immediate Health Walk & exercise 1 hr Puzzles The Orange Puzzle Cube: puzzle #6 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
6:00 am I now have a cuppa in hand and have almost two hours before meeting Frank for a coffee. Puzzle 6 has turned into a delight. It is an exercise in detailed logic involving the integers 1 - 25. Each number must be placed in one cell such that a number of constraints are satisfied.
The trick is to come up with an appropriate representation to keep track of everything.
Here is a table where each row corresponds to a cell in the puzzle. An number in the cell (e.g. 2) signifies that step #2 implies that the number for that column cannot be a number in the cell. These cells are highlighted in mauve to emphasize that they are are eliminated.
A1 A2 A3 A4 A5 B1 B2 B3 B4 B5 C1 C2 C3 C4 C5 D1 D2 D3 D4 D5 E1 E2 E3 E4 E5 Part of the solution representation should describe each step. This allows one to review the steps if it becomes apparent that there is a logical error in the table.
Step 1 The 3 in cell C4 implies that a 3 cannot be in any other cell. Place a 1 in column 3 and in row C4 Step 2 The four corner numbers are all multiples of 3. Therefore A1, A5, E1 and E5 must be 3, 6, 9, 12, 15, 18, 21, or 24. Place an 2 in each of these four rows for all other numbers. Step 3 No other multiples of 3 are adjacent to them ... Therefore cells A2, A4, B1, B5, D1, D5, E2 & E4 cannot be 3, 6, 9, 12, 15, 18, 21 or 24. Put a 3 in these cells. Step 4 ... or each other. What does this imply? Cells B4, C3, C5 & D4 cannot be 3, 6, 9, 12, 15, 18, 21 or 24. Place a 4 in these cells. Furthermore, any time a multiple of 3 is identified for a cell it will imply that adjacent cells cannot be a multiple of 3. I must remember this as I proceed. Step 5 In column 1 the lowest number is 8. Therefore cells A1, B1, C1, D1 & E1 cannot be 1, 2, 3, 4, 5, 6, or 7. Place a 5 in these cells. Step 6 25 is somewhere in column 2. Therefore 25 is in one of cells A2, B2, C2, D2 or E2 and cannot be in any of the other cells. Place a 6 in all of these cells for column 25 7:40 am A promising beginning. Now to go for a coffee. With my system of representation of 3 separate tables I will be able to return to this later and pick up where I left off. Excellent.
10:30 am Coffee was great. On the way back I identified the following 4 birds:
- Blue Jay in Lafayette park
- Canada Goose overhead
- Ring-billed Gull in parking lot of apartment complex
- Black-billed Magpie on Nevada Road W
This may be bean-counting, but every bean counts.
Now to continue filling in the above tables for Puzzle #6.
Step 7 in row E the highest number is 17. Therefore cells E1, E2, E3, E4 & E5 cannot have numbers 18 - 25. Place a 7 in all of these cells. Step 8 The number in B2 is twice D2 ... Therefore cell B2 cannot be an odd number. Place an 8 in all the odd columns for row B2. Step 9 ... and half of E2. Therefore cell E2 is an even number. Place a 9 in all odd columns for E2. Also E2 is 16 or less. Therefore B2 is 8 or less. Place a 9 in all the cells of row B2 that are greater than 9 Step 10 The number in D4 is 4 higher than in D1. D1 is at least 8, therefore D4 is at least 12. Put a 10 in all the cells of row D4 that are less than 12. Step 11 D3 is half of B5. Therefore B5 is even. Place an 11 in all cells of row B5 that are odd. Step 12 D4 is D1 + 4, it is also B4 + D3 Step13 C1 is 5 higher than D1. D1 is at least 8. Therefore C1 is at least 13. Place a 13 in all columns of row C1 that are less than 13. Step 14 13 and 19 are in the row below. Therefore they are not in the first row. Place a 14 in all cells for rows A1 - A5 for columns 13 and 19 Step 15 A3 + A4 is 23. Let's see what pairs are still possible. (1, 22), (2, 21), (3, 20), (4, 19), (5, 18), (6, 17), (7, 16), (8, 15), (9, 14), (10, 13), (11, 12), (12, 11), (13, 10), (14, 9), (15, 8), (16, 7), (17, 6), (18, 5), (19, 4), (20, 3), (21, 2) & (22, 1). Of these, only (1, 22), (6, 17), (7, 16), 9, 14), (12,11), (15,8), (16,7), (18, 5), (21, 2) & (22, 1) are still possible). The other pairs are impossible because one of the two numbers is no longer valid be steps 1 - 14). In the case of the impossible pairs, place a 15 in the cell of the number that is paired with a number that is already impossible). Step 16 Each diagonal contains two consecutive numbers. One diagonal consists of the cells A1, B2, C3, D4 & E5. Here are the valid numbers for A1 (9, 12, 15, 18, 21, 24). There are still valid combinations possible for all numbers.
The other diagonal consists of the cells A5, B4, C3, D2 & E1. Here are the valid numbers for E1 ( 9, 12, 15). There are still lots of valid combinations.
Step 17 10 is further to the left than 11 although they are in the same row. Therefore 11 cannot be in the first column. That is, cells A1, B1, C1, D1 & E1 cannot be 11. Put a 17 in these cells. Step 18 24 is a chess knights move from 25. There are 3 possible cells for 25: A2, C2, D2. From A2 the possible knight moves are cells C1, C3 & B4. From C2 the possible knight moves are A1, A3, B4, D4, E1 & E3. From D2 the possible knight moves are cells B1, B3 & E4. There is one possibility for each of the 3 cells for 25. If 25 is in cell A2, then 24 is in cell C1; if 25 is in cell C2, then 24 is in cell A1; if 25 is in cell D2, then 24 is in cell B3. 24 is in one of 3 cells (C1, A1 or B3). Place an 18 in all the other rows for column 24. Step 19 The long diagonal from top left to bottom right totals 72. This is cells A1 + B2 + C3 + D4 + E5 Step 20 The long diagonal from top right to bottom left totals 61. This is cells A5 + B4 + C3 + D2 + E1 Step 21 Each diagonal contains 2 consecutive numbers. The long diagonal from top left is A1, B2, C3, D4, E5. There are lots of possiblities. The other diagonal is A5, B4, C3, D2, E1. There are lots of possibilities. Step 22 C1 is 5 higher than D1. D1 may not be 20, 22 or 23 as C1 cannot be 25, 27 or 28. Place 22 in columns 20, 22 and 23 for row D1. Step 23 No two consecutive numbers are adjacent in any direction. 3 is in C4. Therefore B4, C3, C5 & D4 cannot be 2 or 4. Step 24 B2 = 2D2. Therefore D2 must be less than 13. Place 24 in all columns greater than 12 for row D2 Step 25 B2 = 2D2 =E2/2. Therefore E2 = 4D2. Therefore D2 is less than 7. Place 25 in all columns greater than 6 for row D2. Step 26 24 is a chess knights move from 25. D2 cannot be 25. Then 24 cannot be in cell B3. Put a 26 in column 24 for row B3. Step 27 24 is either in cell A1 or cell C1. Suppose it is in cell C1. Then cells B1, C2 and D1 cannot contain a multiple of 3. Step 28 D4 = D1 + 4 ; D4 = B4 + D3; D3 = B5/2. The latter implies B5 = 2 D3. Therefore D3 must be less than 12. Place 28 in all cells in row D3 that are greater than 11.
D4 can only be 13, 14, 16, 17, 19, 20, 22 or 23. This implies values of D1 of 9, 10, 12, 13, 15, 16, 18, 19. But D1 cannot be 9, 12, 15 or 18. Therefore D4 cannot be 13, 16, 19 or 22. Place a 28 in each of these four cells.
D1 can only be 8, 10,13, 14, 16, 17, or 19. This implies D4 is 12, 14, 17, 18, 20, 21 or 23, But D4 cannot be 12, 18, or 21. Therefore D1 cannot be 8 14 or 17. Place a 28 in these three cells.
D4 = B4 + D3. D3 cannot be 12, 11, 10, 8 or 5 as the equation will not be satisfied by D4 and B4, Place a 28 in columns 12, 11, 10, 8 & 5 for row D3.
Step 29 10 is to the left of 11 although they are in the same row. Examine columns 10 and 11 and ensure that 10 is to the left of 11 and that they are not adjacent. Put a 29 in all other cells in columns 10 and 11. Step 30 13 and 19 are in the same row. Step 31 C1 is 5 higher than D1. D1 is 10, 13, 16 or 19. Therefore C1 is 15, 18, 21 or 24. Place a 31 in all other cells in row C1 Step 32 D3 =B5/2. This implies B5 = 2 D3. But D3 = 1, 2, 4, 6, 7 & 9. Therefore B5 = 2, 4, 8, 12, 14 & 18. But B5 cannot be 12 or 18. Therefore D3 cannot be 6 or 9. Place 32 in these two cells. Also B5 can only be 2, 4, 8 & 14. Step 33 B2 = 2D2; E2 = 4D2. D2 is 1 2 4 5 6. Therefore B2 is 2 4 8 10 12. and E2 is 4 8 16 20 24. But the only values that satify this are D2 = 1 2 4, B2 = 2 4 8 and E2 = 4 8 16. Place 33 in the remaining cells for B2, D2 & E2. 8:50 PM I have put more time and effort into this than any sane person. So far I have not hit a contradiction, but neither have I been able to determine where any of the numbers go. However once I get a few of them, I expect much of the remainder will simply cascade toward a solution. Cells A3 and A4 form a pair (if I know one, I know the other). Similarly numbers 24 and 25 form a pair. (if I know where one goes, I know where the other goes).
D. Reflection