@skullpatrol No, the term you mentioned is deliberately vague. Essentially they cover all the undergrad and first year grad algebra, analysis and geometry/topology.
@JasperLoy If I can told you a non requested advice, it should better if you divide the book (if they don't) by "difficulty level": starting from the first that one have to read to understand the succesive and go ahead.. I don't know if I explained :P
@OldJohn Aces : ) Here goes: Assume we put 5 times 5 coins on a table. A fly lands on one. It can hop from one coin to adjacent coins (next row or column but not diagonally). Can it visit all coins exactly once? (I'll give you some time to think before I tell you what I've worked out so far)
I don't know that problem but I thought it's the same as asking whether we can show that there is no Hamiltonian path starting at (1,2) in a graph where coins are vertices.
@OldJohn I thought you were saying that if we can cover the board using tiles of 2 then the answer is yes. But this would not take into account the starting point.
Is white = start of hop and black = landing square after hop of fly?
so - after we remove the initail landing square, there must be an equal number of black and white squares left - which will only happen if we remove a black square initially (the landing square) - otherwise we end up with 2 more black than white
@OldJohn I cannot prove that a solution always exists, but I think it does. It is always possible for a fly to go over all the coins exactly once, wherever it lands.
@OldJohn but, but this does not relate to jumps. If you land on a white tile, you do not take off from the adjacent black. :-/ you take off from that very white tile and jump onto the next black.
coins 2 and 3 can be covered with a domino - so can coins 4 and 5 - it always jumps from 2n to 2n+1 ... 2n is om one colour and 2n+1 is on the other colour
@OldJohn but then this solution should apply irrespective of where the fly landed the first time, and if the fly lands on the corner, a solution is possible and hence, that is a contradiction.
@Jayesh Hi jay!Today by the morning i thought:here comes the sun!The day is good to walk in strawberry fields,look up and see lucy in the sky with diamands.I'm as happy as i told you yesterday,Jay!
So far, I have proved that approx half of the starting points (the white squares) are impossible starting points - I have not yet proved it will always be possible if we start on a black square
@OldJohn No. I guess I can produce a solution. I am writing down a counterexample to your assertion. I have drawn a 5x5 chessboard with black at the top-left corner. Suggest any white starting point.
Try it by colouring a 5x5 grid and start at (1,2) - you will definitely end up with a black square you can't cover at the end - because it always hops from black to white etc
@OldJohn My point is not about the coloring problem of the grid, it is about the fact that the coloring is not equivalent to path we desire (or to our problem)
@OldJohn I'm trying the 3x3 case. But I still don't understand it fully. We start with 5 black and 4 white. Then one could argue that if we cover it we end up with one black. Hence we cannot cover it.
@Charlie Yup, graph theory is combinatorics many times. This is some sort of hamiltonian path, but even the edges are not specified, so it is not a hamiltonian path.
@MattN. If you have one more black than white, and start on a white, you cannot possibly visit them all, as each 2 consecutive moves cover one black and one white
e.g. on the 5x5, there are 25 squares, of which 13 are black and 12 are white, if we start on a white (so we now have 13 black and 11 white left to visit) and then jump (to a black of necessity) we now have 12 black and 10 white left to visit after the next 2 moves we then have 11 black and 9 white to visit etc etc
at the end, when we have visited all the whites, there will be 2 blacks left, and we can only visit one of them as each hop takes us from one square to a different colour
Also, I have some ideas as follows: Rudimentary and not complete, but I am thinking, for nxn. I am thinking like this, the diagonal moves do not change the parity of the co-ordinates, while the adjacent moves do. And, we have a solution when we start on a corner. So, now what we can do is, we can prove that the discrete can always be mapped equivalently to new co-ordinates such that we do not have to change the parity of any of the points.
I am assuming that $(1,1)$ is the bottom left and $(5,5)$ is the top right. So that I can make my moves in up, down, left , right to make it short to describe.
and after 2 more moves we have 11 black and 9 white left to visit ...
then after an even number of moves we end up with 2 black and 0 white left to visit - which is impossible as 2 consecutive squares visited have to have different colours
In theory, I could automate the process, but I'm not entirely sure it would take me less time. I can at least filter the nodes so that each vertex has only about 20 candidate pairs, but still.
@EdGorcenski Can you not get the intersections and the streets in the CSV format and then text process it (If you are planning to use more cities in future?)
@Charlie Nice! I was thinking of generating a random map and then hosting an online scotland yard board game for people to play. It would be nice. :-) What do you think?
Hmm. I cannot believe there is no CSV/VTK some other text-based representation of data. I was expecting in the least to have the intersections labeled numerically, and streets represented as the edges of the intersections. (graph style)
So there's an academic research component; we're working with a major university, but we're also doing it with considerations to practical applications
For instance, it would be really easy to do in a lab setting, but the people who have these disorders can't afford $15,000 worth of equipment, or 3 visits to a doctor per week
So it has to be cost-effective for the target demographic.
@JohnSmith I think you might have to try posting that one on the main site - I can't see a quick way to do it, unless the quadratic factorises when you put it all on one side
I did my degree in 1970-1973, and then in 1992-ish, I took some Open University courses with the aim of convincing people that I was capable of doing a PhD
The courses I did were the toughest ones I could find - mostly final year undergrad ones - and then I did one year of a taught MSc in functional analysis before I convinced them I could cope with a PhD
@OldJohn No, still not. If we hop we cover one new coin per hop! So how do we get your argument from there? (very sorry for being slow, I had a terrible day, the 21st or so in a row)
so- starting from a white, and odd by odd grid is impossible
No problem! - I am a patient man! (used to be a teacher!)
If we start with a grid with (at least ) one side even, I think it is easy to prove it is possible - the only case I haven't sorted in my mind is proving it is possible in an odd by odd if we start on a black
If one side is even, then I think there must be a solution starting in top left corner, which ends back where it started, and then we can change starting point to any random coin and get a solution that way
@PeterTamaroff what was the non-believers comment about ? :)
@OldJohn You said "I did my degree in 1970-1973, and then in 1992-ish, I took some Open University courses with the aim of convincing people that I was capable of doing a PhD"
