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Wednesday September 20, 2006 6:00 am Lethbridge Sunrise 7:14 Sunset 19:35 Hours of daylight: 12:21
A. Morning Musings
6:00 am It is + 6 C at the moment. The forecast is for periods of rain with a high of + 14 C.
B. Plan
I have another dental appointment this morning and do not expect to be doing much more for the rest of the day.
Immediate Health Walk & exercise 1 hr Puzzles The Orange Puzzle Cube: puzzle #7 2 hr Technology add keywords to iPhoto records 2 hr Birds Add August birds to North American data base 2 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
1:10 PM Here are my notes for "Science and the Akashic Field":
1:35 PM Now to have a look at the puzzles. Puzzle #8 only took about 10 seconds. It was a picture matching puzzle with the added clue that the letters of the matching images would spell the name of a famous university. Since the name was 6-letters, I guessed it might be Oxford and this checked with the images.
Puzzle #7 is another logic puzzle. As with puzzle #6, the trick is to have a good representation system for keeping track of the options.
A |
B |
C |
D |
Josef |
Karl |
Maria |
William |
swords |
dragon |
shield |
wreath | |
Josef |
8 |
8 |
8 |
7 |
||||||||
Karl |
13 |
42 |
43 |
8 |
||||||||
Maria |
14 |
15 |
15 |
8 |
||||||||
William |
6 |
41 |
42 |
8 |
||||||||
swords |
5 |
42 |
43 |
21 |
3 |
2 |
3 |
3 |
||||
dragon |
4 |
5 |
5 |
5 |
17 |
3 |
16 |
6 |
||||
shield |
5 |
41 |
42 |
11 |
19 |
3 |
17 |
18 |
||||
wreath |
5 |
21 |
21 |
20 |
18 |
3 |
12 |
19 |
||||
1745 |
23 |
40 |
39 |
25 |
27 |
36 |
1 |
37 |
33 |
31 |
32 |
10 |
1785 |
23 |
25 |
25 |
24 |
26 |
27 |
1 |
27 |
10 |
10 |
10 |
9 |
1825 |
22 |
23 |
23 |
23 |
27 |
29 |
28 |
29 |
31 |
30 |
31 |
10 |
1865 |
11 |
39 |
38 |
25 |
27 |
35 |
29 |
36 |
34 |
31 |
11 |
10 |
| Step 1 | No female monarch reigned during the 18th century. Therefore a 1 in cells (1745, Maria) and (1785, Maria). |
| Step 2 | Karl has swords. Therefore a 2 in cell (swords, Karl) |
| Step 3 | A 2 in cell (swords, Karl) means an x in cells ((swords, Joseph), (swords Maria), (swords, William) as well as in cells ((dragon, Karl), (shield, Karl), (wreath, Karl) |
| Step 4 | Dragon on coin A. Therefore a 4 in cell (dragon, A). |
| Step 5 | A 4 in cell (dragon, A) means an 5 in cells (dragon, B), (dragon, C), (dragon, D) as well as in cells ( swords, A), (shield, A), (wreath, A) |
| Step 6 | Coin A not William, dragon not William. Therefore an 6 in cells (William, A), (dragon, William) |
| Step 7 | Coin D is Josef. Therefore a 7 in cell (Josef, D). |
| Step 8 | A 7 in cell (Josef, D) means an 8 in cells ((Josef, A), (Josef, B), (Josef, C) as well as ((Karl, D), (Maria, D), (William, D). |
| Step 9 | 1785 coin is a wreath. Therefore a 9 in cell (1785, wreath) |
| Step 10 | A 9 in cell (1785, wreath) means a 10 in cells (1785, swords), (1785, dragon), (1785, shield) as well as (1745, wreath), (1825, wreath), ((1865, wreath) |
| Step 11 | 1865 is to the right of the shield. Therefore 1865 cannot be coin A. Place an 11 in cell (1865, A). And the shield cannot be coin D. Place an 11 in cell (shield, D). And 1865 cannot have a shield. Place an 11 in cell (1865, shield). |
| Step 12 | Maria cannot be 1785 but 1785 is the wreath. Therefore Maria cannot be the wreath. Place a 12 in cell (wreath, Maria) |
| Step 13 | Swords cannot be A, but swords are Karl. Therefore Karl cannot be A. Place a 13 in cell (Karl, A). |
| Step 14 | A cannot be Josef, Karl or William. Therefore A is Maria. Place a 14 in cell (Maria, A). |
| Step 15 | A 14 in cell (Maria, A) means a 15 in cells (Maria, B), (Maria, C) |
| Step 16 | Maria is a dragon. Therefore place a 16 in cell (dragon, Maria) |
| Step 17 | A 16 in cell (dragon, Maria) means a 17 in cells (dragon, Joseph), (shield, Maria) |
| Step 18 | shield & wreath are either Josef or William. But Josef is D and shield cannot be D. Therefore shield is William and Josef is wreath. Place an 18 in cell (shield, William) and (wreath, Josef) |
| Step 19 | An 18 in cell (shield, William) means a 19 in cell (shield, Josef), and an 19 in cell (wreath, William) |
| Step 20 | Josef is in D and is a wreath. Therefore a wreath is in D. Place a 20 in cell (wreath, D). |
| Step 21 | Place a 21 in appropriate cells for the symbols. |
| Step 22 | Maria cannot be 1745, 1785 or 1865. Therefore Maria is 1785. Place a 22 in cell (1825, A) |
| Step 23 | Place a 23 in appropriate cells for the dates. |
| Step 24 | The wreath is in 1785. Place a 24 in this cell. |
| Step 25 | Place a 25 in the appropriate cells for the dates. |
| Step 26 | Josef is in D and D is 1785. Therefore Josef is 1785. Place a 26 in this cell. |
| Step 27 | Place a 27 in the appropriate cells. |
| Step 28 | Maria is in A and A is 1825. Place a 28 in this cell. |
| Step 29 | Place a 29 in the appropriate cells. |
| Step 30 | The dragon is in 1825. Place a 30 in this cell. |
| Step 31 | Place a 31 in the appropriate cells. |
| Step 32 | The shield is in cell 1745. Place a 32 in this cell. |
| Step 33 | Place a 33 in cell (1745, sword) |
| Step 34 | Place a 34 in cell (1865, sword) |
| Step 35 | Karl has the sword and the sword is in 1865. Therefore Karl is in 1865. Place a 35 in this cell. |
| Step 36 | Place a 36 in the appropriate cells. |
| Step 37 | Therefore William is in 1745. Place a 37 in this cell. |
| Step 38 | The coin dated 1865 is to the right of the shield. The shield is not in A. Therefore 1865 must be in C. Place a 38 in this cell. |
| Step 39 | Place 39 in the appropriate cells. |
| Step 40 | Therefore 1745 is in B. Place a 40 in this cell. |
| Step 41 | The shield is in 1745. William has the shield. They are both in B. Place a 41 in these cells. |
| Step 42 | Place a 42 in the appropriate cells |
| Step 43 | Place 43 in the remaining two cells. Done. |
7:15 PM Phyllis bought a bird feeder this afternoon which we assembled and filled with oiled sunflower seed. We put it out in the yard at 5 PM and already have seen a Chickadee eating from it. I do not recall ever seeing a Chickadee in our yard before. A quick check of the database confirms this. This will be a new siting for 2006.
8:00 PM I completed puzzle #7 this evening. I put a little over 2 hours on this. These puzzles are fun and often challenging but it is a time sink. Once I know that I have an approach that works, I will pass on other problems that are of the same type.
D. Reflection
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