Keeping Old Buildings?

On our recent trip to Montreal we were told that in Old Montreal they are not allowed to take down old buildings. It certainly makes for some great renovations...and also some 'interesting' ones:

Here's a restaurant behind a façade (after a fire I believe).

Edmonton could certainly have gained from not destroying the old courthouse in 1972, and the Tegler Building, and...and...and...  (see link here for some lost buildings in Edmonton).

Nice to see a 'bit' more saving of old buildings in Toronto, but it's a whole different (legislated) level in Montreal (or at least, in parts of Montreal).

Conservative Government embarrasses us and disgraces itself

Perhaps I don't watch enough Question Period, and maybe this kind of thing goes on all the time, but this seems to take it to a new level.

See: http://www.earljwoods.com/2014/09/inartful-dodging.html

Mulcair asks 3 questions about Iraq, and gets 3 responses from Paul Calandra to do with an NDP fundraiser and his views on Israel. Is Calandra even going to pretend to participate in the democratic process? Is he going to write in his memoirs that it was dictated to him from Harper?

Fine foods(?) and food pricing

I was naturally attracted to this food, and also found it interesting that:
a) the original price ends in the classic -,99 (so it's 'just a little more than $9')
b) the reduction amount is $1.21 (so it's 'much more than $1.00, and even more than $1.20')

Perspective

Found this photocopy from 27 years ago - as you can see from the scan. I really have to get all this useless stuff digitized so it can be saved forever.

'Drumming up Business'

We were going to the CAA at Lawrence and Don Mills and happened to walk by this:

Doesn't show up too well with my crappy cell phone camera, but there was one heck of a lot of noise by the person with the pneumatic drill, combined with the occasional noise from the 'Cat'.

(ps If you're actually looking for interesting comments on slightly bizarre things found in Toronto, I recommend Jim Arnott's blog. Features pictures taken with a cell phone camera - though a few zillion times better than mine)

Specific Solutions for the Broken Combination Lock problem

So how to come up with an actual solution for an n x n x n cube? (Assume n is odd, it's easier/symmetrical if even). At a high level:

1. Pick a corner (let's say, starting at (1,1,1)) of a cube of size (n+1)/2 that is in a corner of the overall large cube), and solve according to below, then
2. Pick the 'opposite' cube (the opposite corner of the overall large cube - say, starting at ((n+1)/2,(n+1)/2,(n+1)/2)) which will be a cube of size (n-1)/2), and solve it as below too.
3. The combination is the total solution. 

So let's do it. Starting at one corner (for simplicity, assume (1,1,1) then keeping 'z' constant, for each guess add 1 to each of x and y a total of (n+1)/2 times (or (n-1)/2 times for the second cube...let's just call it 'N' for now).
Then add 1 to z and start x,y at x+1, y+0 from the starting point of your first guess at z=1. Continue adding 1 to each of x and y, 'rolling over' if you go over N.
Continue for each z until z= N, starting each level with x at 1 more than the previous level.
So for n=7, a solution is:
(1,1,1), (2,2,1), (3,3,1), (4,4,1),
(2,1,2), (3,2,2), (4,3,2), (1,4,2),
(3,1,3), (4,2,3), (1,3,3), (2,4,3),
(4,1,4), (1,2,4), (2,3,4), (3,4,4),
- the above handles the cube of size (n+1)/2
(5,5,5), (6,6,5), (7,7,5),
(6,5,6), (7,6,6), (5,7,6),
(7,5,7), (5,6,7), (6,7,7)
- handles the rest

Two notes initially:
1. The selection of an (x,y,z) covers all guesses (x,y,?), (x,?,z), and (?,y,z) where ? represents any value from 1 to n
2. The guess can be seen to be within a 'sub-cube' of m x m x m (where 'm' = max value of x,y,z).

(sidebar) To help viusalize, imagine initially that n is even, and you divide the cube into 8 equal 'sub-cubes' of size n/2 x n/2 x n/2 by cutting the larger cube along the lines of symmetry. (See below...also the second diagram in my previous post)

a) The selection of an (x,y,z) within a 'sub-cube' covers guesses in 4 of the 8 'sub-cubes' (the one in which the selection is made, plus the three others that contain the same 'x and y', 'x and z' and 'y and z'...in other words, the volume shown within the red lines above), but will never cover any guesses in the other 4 'sub-cubes' (symmetrical).
b) The minimum number of guesses to cover the selected 'sub-cube' (and the 3 other sub-cubes affected) must be equal to the area of one side of the 'sub-cube' (n/2)**2, since each guess can only cover one square in this cross-sectional area.
c) To complete the overall cube, you must also cover the other 4 'sub-cubes' that are 'opposite' from your first guesses. As noted, this can be done in (n**2 / 4) guesses by selecting the 'sub-cube' that contains (n/2+1,n/2+1,n/2+1) through (n,n,n). The selection of guesses in any other 'sub-cube' will, by definition, not complete the cube.

Now, back to the general proof. The 'sub-cube' selected (for your x,y,z guess) does not have to be of size n/2. In fact, the cube can be solved by solving any two 'sub-cubes' where one is of size 'i' (i<=n) (starting at 1,1,1 for simplicity) and the other of size n-i which is 'opposite' to the first 'sub-cube' within the large cube. In these cases, the same '3-dimensional L' visualization applies (although the big cube is not divided into 8 sub-cubes now, just 2 sub-cubes of different size plus the other volumes that get 'filled' as you solve each sub-cube). The solution is
S = solution for first sub-cube + solution for second sub-cube
 = i**2 + (n-i)**2
 = n**2 -2ni +2(i**2)
==> if n is even, this is optimal (minimum) at i=n/2

If n is odd (actually, even is just a special case of odd)
S = solution for first sub-cube + solution for second sub-cube
 = i**2 + (n-i)**2 (where i <> n-i)
 = n**2 -2ni +2(i**2)  (as before)
==> this is optimal (minimum) at |n-2i|=min, or in other words, when the two sub-cubes are closest in size.

This comes down to the familiar equations of
S = n**2 / 2   (for n=even)
S = (n**2+1) / 2  (for n=odd)

The Broken Combination Lock problem

(well...*one* of the broken bicycle lock problems)
(AKA rooks on a 3-D chessboard problem)

You have a combination bike lock of the type which has 3 rings, each with numbers 1 through 8. Normally with this kind of lock you have to line up 3 numbers in the right order to open it. The number of possible combinations is 8x8x8.

But, this lock is broken in such a way that you only have to match any 2 of the 3 numbers in order to open it. Obviously you could do it in 64 tries (8x8) by just using any 2 of the rings, but surely you can do better than that.

1. What is the minimum number of combinations you have to try to ensure that you can open the lock?

2. What is the general solution for a 3-ring lock 'broken' as above, but with 'n' numbers on each ring
 a) if 'n' is even?
 b) if 'n' is odd?

(This is the same type of problem as the 'rooks on a 3-dimensional chessboard'. It is known that N rooks may be placed on a 2-dimensional NxN chessboard so that every square is controled by at least one rook. What's the minimal number of rooks on 3-D chessboard NxNxN to meet this condition?)

One thing I found interesting was how to visualize this problem. Basically, you start with an 8 x 8 x 8 cube. Each guess at a combination 'cuts out' a line of 1 x 1 x 1 cubes in each of the 'x', 'y' and 'z' directions from the specific (x, y, z) cube selected. That is, if you select a cube - say (1,1,1) - then you are selecting all cubes that have the same x,y but a different z (the vertical bar in the diagram below), all cubes that have the same x,z but different y (the bar going into the page) and all cubes that have the same y,z but different x (the bar going left to right). Together, these make a '3-dimensional L' shape.

Here's a diagram (it's 4 x 4 x 4, but you get the idea).

In this case, you can see that all guesses where (say) x, y, z <= 2 will carve out a larger '3-dimensional L' shape.

So you only need to make guesses in the x, y, z <= 2, plus (from symmetry) guesses in the x, y, z >= 3 quadrant to complete the solution.

To help visualize, think of a 2 x 2 x 2 cube first. If you pick (1,1,1) and then (2,2,2) you make two '3-dimensional L' shapes that cover the entire cube.

Now looking at the 4 x 4 x 4 cube...the number of guesses you need to cover the x, y, z <= 2 quadrant cannot be less than 4 (the area of any 2 x 2 'end' to the 3-dimensional L). It's easy to see there are many solutions of 4 (more on this later), so the full solution is (2x2)x2 = 8. And for an 8 x 8 x 8 cube the answer is (4x4)x2 = 32.

For an n x n x n cube where n is even, this gives (n/2)(n/2)x2 = (n**2)/2.

If n is odd, a similar approach gives the solution (noting that the two quadrants you are selecting in should be as near to the median as possible, so for a 9 x 9 x 9 cube you could select 16 guesses in the x, y, z <= 4 quadrant, and 25 guesses in the x, y, z >4 quadrant.

For an n x n x n cube where n is odd, this gives ((n-1)/2)**2+((n+1)/2)**2 = (n**2+1) / 2

And finally, the reason that the two quadrants should be the same size (if n is even) and as close as possible to the same size (if n is odd) is simply because the (sum of the) products of the sides of the quadrant (i.e. the sum of the area of one side of each quadrant) is a minimum in that case. 'An exercise left to the reader' - though I can send you the math if you want. See also my next post.

Next post will look at how to generate specific solutions, and more on why you need to select these 3-dimensional L's as close in size as possible to each other to minimize the number of guesses needed in the solution.

And it is nowhere near to clear (to me) for n x n x n x n. More on that even later.

Trees and Construction Damage

(Something I had put together when I saw our next door neighbour's house construction put a huge Norway Maple under a lot of stress)

Trees (especially in the urban environment) are a huge asset. Large trees are typically valued at $10K-$20K or more, but does that really reflect the enjoyment of the tree? Plus its air-conditioning effect on the street and houses? To say nothing of its climbability :-) Construction can obviously damage these trees and most municipalities have regulations as to the minimum allowed distance from the tree before allowing any construction, excavating, trenching, scraping, dumping or disturbance of any kind. One example of regulations are those for Toronto (see http://www.toronto.ca/trees/pdfs/TreeProtSpecs.pdf) which specify the 'tree protection zone' which is the zone within which there must be:

• no construction;
• no altering of grade by adding fill, excavating, trenching, scraping, dumping or disturbance of any kind.
• no storage of construction materials, equipment, soil, construction waste or debris.
• no disposal of any liquids e.g. concrete sleuth, gas, oil, paint.
• no movement of vehicles, equipment or pedestrians.
• no parking of vehicles or machinery.
• directional micro-tunnelling and boring may be permitted within the limits of a TPZ subject to approval by Urban Forestry.
• open face cuts outside a TPZ that are consistent with an approved plan and that require root pruning, require the services of a qualified arborist or approved tree professional. An exploratory dig, either by hand or using a low water pressure hydro vac method, must be completed prior to commencing with open face cuts outside the TPZ.

Certain types of trees are also more or less sensitive to root damage (see this link for info on tolerance of tree roots to damage) and may require additional distance and care. A tree's essential root system will be out to roughly the ‘dripline’ (the radius of the canopy of the tree) and ideally there is some protection out to this distance. Hope this helps those who have been wondering just how to quantify what protection needs to be given to our trees!

1973/74 - Lots of Snow

Continuing with some old posts, the winter of 1973/74 had the most snow that I can recall in Fort Saskatchewan. I took a couple of photos - some carefully positioned so it appeared that the snow banks were up to the top of the first story of our house - and showed them to various English/Dutch (see also my Ft Sask School Band tour) to show how tough we Albertans were. I’d throw in an occasional reference to Edmonton’s record cold spell if they didn’t seem impressed...does anyone else still have the “I survived Edmonton’s Record Cold Spell” certificate from the Edmonton Journal?

Fort Band Tours England and Holland

(Second clarinets draw raves)

In 1974 the Fort Saskatchewan Senior High School Band (with some members from the Junior High, and a few ringers from elsewhere) took the trip of the Erv Miller (conductor) years – to England and Holland.
We started in London and experienced some pretty ‘basic’ accommodation the first night (and that’s putting it mildly - it was some kind of youth hostel place that used to be army barracks in the Kensington area…near where Mum used to live anyway) until my uncle managed to find a university residence that we stayed in and that was much better! We visited Chelmsford and Cambridge (and possibly other places) from our base in London, then went up to the north-east of England to Teesside (near Middlesbrough) and played in a festival there. Stayed with wonderful people in a beautiful house. Managed to forget some stuff there which they sent back to Canada for us!
Next up was Holland, and we went to both Soest and Purmerend (playing in the International Youth Music Festival there) and also made a record at a studio in (I think) Amsterdam. The part of the trip I remember vividly is visiting some Canadian war cemeteries and seeing how well kept they were. All the people we met were extremely friendly, open, and still thankful for all that Canada had done for Holland during the Second World War! I have had the best impression of the Dutch ever since – and have never been disappointed.