Puzzles with numbers and things

[snookersfun]

thought so
I guess, 'I tried 20 times' doesn't count then...

[davis_greatest]

snookersfun! If it is possible, I want to see how. And if it's not possible, I want to know why not.

[snookersfun]

The question is.... is it possible?

not possible, but somebody else give the explanation...

[Semih_Sayginer]

That's almost worth a point too! But cutting, tearing, ripping etc is not allowed.

i just want to say that youve got a funny shaped toilet base, if this is from experience.

[Semih_Sayginer]

perhaps you meant it was almost worth two points?

need to look out my ikea grid, and work it out.

on second thought, ill just keep winding cessy up

[davis_greatest]

That's almost worth a point too! But cutting, tearing, ripping etc is not allowed.

[snookersfun]

cut the last piece until it fits

[davis_greatest]

Hehe semih That's almost worth a point!

Edit: I've no idea why this has appeared twice. Is it possible to delete a post completely?

[davis_greatest]

Hehe semih That's almost worth a point!

[Semih_Sayginer]

drop these guys an e-mail:-

http://www.carpetfitter.co.uk/

theyll know best

[davis_greatest]

Question 7 - Snooker Tricolore

Mr Dotty's bathroom is an interesting affair. It's rectangular, 10 feet by 8 feet, with a bath 2 feet by 8 feet along one side, leaving a square 8 feet x 8 feet exposed floor. Actually, that's not entirely true - in the corner is a toilet 1 foot x 1 foot area, so there are 63 square feet of exposed floor. The door is painted red, the ceiling is white and the walls are blue.

One day, Mr Dotty decides to cover the floor in some green baize. He finds a snooker table, and cuts out some green baize rectangles from its cloth - 21 of them in fact. Each of them is 3 feet by 1 foot.

He then decides to lay the baize rectangles on his bathroom floor to cover it with no gaps. Sounds easy, but after trying for a couple of minutes, for some reason he hasn't managed to do it yet. He seems to be having some trouble with finding the right position.

The question is.... is it possible?

[davis_greatest]

Question 6 - Well done abextra!

abextra, excellent answer! I knew you would get there!

It took me a few moments to rearrange your slightly complicated expressions but they are bang on, for n dots - and they work for all n, not just n odd. Congratulations!!

I am sure you must have solved it with reasoning to come up with these formulae. Here is an answer to count the number of regions when there are n dots around the circle. This also shows why the pattern begins 1, 2, 4, 8, 16 and the "powers of 2" breaks down as it continues 31, 57, 99, 163, 256 (which interestingly is a power of 2), 386,…

First, define C(k,r) ("k choose r") to be the number of ways of choosing r objects from k objects (r<=k), where the order is irrelevant.

So C(k,r) = k!/r!(k-r)!
and, for instance C(k, 2) = k(k-1)/2 ...............[1]

Let C(k,r) = 0 if r>k : there are no ways of choosing r objects from k in this case.

A) Let's count the number of vertices v:

(i) There are n dots around the edge of the circle.

(ii) Within the circle, there is a vertex wherever two of the chords (joining the dots) meet. This happens whenever we choose 4 dots and draw two intersecting chords. So there are C(n, 4) vertices within the circle.

So, v = C(n, 4) + n ...........................[2]


B) Now let's count the number of vertices again, but this time counting each one EVERY time it appears on an edge - so we count each one more than once:

(i) The C(n,4) vertices within the circle each touch 4 edges, so this gives: 4C(n,4)

(ii) The n dots around the edge each connect to (n-1) lines joining the other dots, plus 2 curved edges around the circumference of the circle, i.e. each of the n dots connects to (n-1)+2 = n+1 edges. So this gives n(n+1).

Thus the number of vertices , counting each one EVERY time it appears on an edge, is
4C(n,4) + n(n+1) .............[3]

Now, each edge touches 2 vertices, one on each end, so this expression must equal 2e.

i.e. 2e = 4C(n,4) + n(n+1) ............ (using [3])

So e = 2C(n,4) + n(n+1)/2 ...................... [4]

C) Now, from Euler's formula, the number of regions f is
f = 1 + e - v
= 1 + 2C(n,4) + n(n+1)/2 - C(n, 4) - n ...............(using [2] and [4])
= 1 + n(n-1)/2 + C(n, 4)
= 1 + C(n,2) + C(n, 4) ................(using [1]) ............[5]


These expressions can be expanded to the same formulae that you gave, abextra.

We can also note something interesting with C(k,r).

C(k,r)

= the number of ways of choosing r objects from k objects

= [the number of ways of choosing r objects from k objects where object k is chosen] + [the number of ways of choosing r objects from k objects where object k is not chosen]

= [the number of ways of choosing object k and then r-1 objects from the remaining k-1 objects] + [the number of ways of not choosing object k and then choosing r objects from the remaining k-1 objects]

= C(k-1, r-1) + C(k-1, r) .................[6]

So, using (5) and (6), we can write

f = 1 + C(n-1,1)+ C(n-1,2)+ C(n-1,3)+ C(n-1,4)

= C(n-1,0) + C(n-1,1)+ C(n-1,2)+ C(n-1,3)+ C(n-1,4)

This is the number of ways of choosing anywhere between 0 and 4 objects from n-1 objects.

If n-1 <=4, i.e. if n<=5, this is the number of ways of either choosing or not choosing each object, so we get:

f = 2^(n-1) (where n<=5)

Hence, the number of regions starts with the powers of 2, for up to 5 dots, but breaks down for 6 or more dots.


The original question was:

How many regions can you make when the number of dots is equal to:

a) the lowest ever number of frames won in a Crucible semi-final by the loser;

b) the number of UK Championships won by Stephen Hendry;

c) the number of UK Championships won by Steve Davis;

d) the number of World Championships won by Stephen Hendry?


to which the answers are:

a) 4 dots -> 8 regions
b) 5 dots -> 16 regions
c) 6 dots -> 31 regions
d) 7 dots -> 57 regions


Next question will come soon!

I am also going to prepare a scoreboard….

[abextra]

I've found these formulas for v, e and f, but I'm not able to type them! The formulas are easier in case the number of dots n is odd and bigger than 1. So you have to be contented with them

v = [( n^4 - 6 n^3 + 11 n^2 - 6 n ) / 24 ] + n

e = [( n^4 - 6 n^3 + 17 n^2 - 24 n + 12 ) /12 ] + 2 n - 1

f = [( n^4 - 6 n^3 + 23 n^2 - 42 n + 24 ) / 24 ] + n

What do you think?

[davis_greatest]

Yes, there is certainly a formula. The sequence in fact begins 1, 2, 4, 8, 6, 31, 57,...

Here is a clue - use Euler's formula:

Let:
f = the number of regions ("faces")
v = the number of "vertices", i.e. the dots and the points of intersection where lines meet within the circle
e = the number of "edges", i.e. lines (not necessarily straight) joining 2 vertices, including the curved lines of the circle

Then Euler's formula says v-e+f=1 (usually written v-e+f=2 but we are ignoring here the region OUTSIDE the circle)

As examples:
- when there are 3 dots, v=3, e=6 (3 straight lines + 3 curved lines of the circle), f=4
- when there are 4 dots, v=5 (four dots and one point of intersection inside the circle), e=12 (8 straight lines + 4 curved lines of the circle), f=8

So you need to see if you can come up with a general formula for v and a formula for e. Then you will have a formula for f...!

[abextra]

But this is still a pattern - 1, 2, 4, 8, 16, x, 57, y, z, 256, ,,,( x > 30 ) ? I mean, there must be some kind of regularity?