At the same time lit the second rope at one end. When rope 1 is done (burnt through) light the other end of the 2nd rope, so the remainder will only burn 15m

Hence 45m total.

Does that explain how it's done without going down to the atomic level of how burning and fire and heat works?

1. Nc3 Nf6 2. Nd5 Ne4 3. Nxe7 Nxf2 4. Nd5 Qh4 5. g4 Nxh1#.

Claim: The rook must be captured on h1, d3 or f3.

Claim: There's no solution with a pawn promotion.

Claim: The king must be mated on e1 or f2.

Claim: For the mates with the rook on d3 or f3, either Black's knight must capture the e-pawn, or Black must set up a double check with the knight and queen for Nxd3#.

Claim: If the knight delivers the mate to a king on f2, either White must develop the bishop to e3 and the queen to e1, or White must open the e-file and draw Black's queen to e7, and Black's knight must maneuver to h1.

Claim: If the knight delivers the mate to a king on e1 by capturing a rook on h1, it will be a variant of the discovered check solution with a queen on h4.

Type 2: White K on f2, Black Q on e7, N on h1
Example: 1. e4 Nf6 2. f3 Nxe4 3. Qe2 Ng3 4. Qxe7+ Qxe7+ 5. Kf2 Nxh1#

Type 3: White K on e1, Black N on f3 or d3 after capturing e-pawn Example: 1. g3 Nc6 2. a4 Nd4 3. Ra3 Nxe2 4. Rf3 Nd4 5. Ne2 Nxf3#

Type 4: White K on e1, Black N on d3 and Q on a5 with double check Example: 1. d4 c5 2. Nf3 Nc6 3. a4 Nb4 4. Ra3 Qa5 5. Rd3 Nxd3#

We've probably already found all the ways with capturing the rook on h1, d3 or f3, but can it be taken somewhere else? Here are some tries, but they take 6 moves.

Capture on g2: 1. g3 Nf6 2. e3 b6 3. h3 Nh5 4. Rh2 Nf4 5. Ba6 Bxa6 6. Rg2 Nxg2#

Capture on c2: 1. c4 Nf6 2. b3 c5 3. d3 Nc6 4. a3 Nb4 5. Ra2 Qa5 6. Rc2 Nxc2#

Capture on g4, h3 or e4: 1. h4 Nf6 2. g4 Nxg4 3. Rh3 e6 4. Rg3 Nxf2 5. Rg4 Qxh4 6. Nc3 Nxg4# (Note: similar method also works for rook on h3 or e4)

Capture on a2: 1. d3 c5 2. Kd2 d5 3. a3 Nc6 4. Kc3 Nb4 5. Bd2 Qb6 6. Ra2 Nxa2#

Similarly, can the king be mated on other squares besides the ones we've seen? Here are some tries.

Mate on g3: 1. f3 g5 2. Kf2 d5 3. Kg3 Nf6 4. b3 Ng4 5. Bb2 Nf2 6. Bxh8 Nxh1#

Mate on c2: 1. d3 Nc6 2. Kd2 Nd4 3. c3 Nb3+ 4. Kc2 e5 5. Nf3 Qg5 6. g3 Nxa1#

Mate on b3: 1. d3 c5 2. Kd2 Nc6 3. Kc3 Nb4 4. Kb3 Qa5 5. c4 Nc2 6. Nc3 Nxa1#

Mate on d2: 1. d3 Nf6 2. Nc3 Ne4 3. h4 Ng5 4. Rh3 g6 5. Rf3 Bh6 6. Kd2 Nxf3#

Mate on c3: See example above with rook capture on a2

Mate on e3: I couldn't even manage to find a way in 6 moves, but I'll post a couple of 7-move examples in case anyone wants to take on the daunting challenge of making it faster.

1. h4 Nf6 2. h5 b6 3. Rh4 Bb7 4. f4 c5 5. Kf2 c4 6. Ke3 e5 7. Rg4 Nxg4#

For completeness, I should mention I also looked for a mate on e2, but didn't find one shorter than 7 moves.

I'm ready to say our five known patterns are probably all the solutions, because we've done enough due diligence to look at other scenarios, we have good explanations why the other scenarios take more than 5 moves, and nobody has found any other solutions after some time. Besides, if anyone does somehow find new solutions, we can just add them to the count.

The next step is to calculate the number of combinations for the known patterns. I will post my calculations for each of the cases in a series of kibitzes probably over about a week, unless any other kibitzer beats me to it. I plan to cover Type 1 (see above) tomorrow.

Spoiler alert: I'm going to start showing how I count the solutions. If you want to work it out yourself, read no further.

Boredom alert: If you aren't interested in counting problems, read no further. It's a wall of math.

Solution Type 1: White K on f2, Q and B blocking e1 and e3, Black N on h1

I break it into cases depending on the route taken by Black's knight. The following routes are feasible: Nc6-d4-f5-g3-h1, Ne7-f5-g3-h1, Nf6-e4-g3-h1, Nf6-h5-g3-h1, Nh6-f5-g3-h1.

Nc6-d4-f5-g3-h1: For this route, Black has only one choice of moves. White has to play the sequence (d3 or d4), Be3 in that order, and the sequence f3, Kf2, Qe1 in that order, and these two sequences can be interleaved. For the number of interleavings I get 5!/(2!*3!), and multiply this by 2 to account for the choice of d3 or d4: 2*5!/(2!*3!) = 20.

Ne7-f5-g3-h1: For this route, Black must start with 1...e6 or 1...e5, and has only one choice for the rest of the moves. White has the same 2*5!/(2!*3!) choices as above, and multiply this by 2 to account for the choice of ...e6 or ...e5: 2*2*5!/(2!*3!) = 40.

Nf6-e4-g3-h1: For this route, White can play Kf2 only after Black has played ...Ng3, so the moves must be 1...Nf6, 2...Ne4, 3...Ng3 4. Kf2 5. Qe1 Nxh1#. Black's fourth move must be a filler with knight or pawn, and there are 1 knight and 8 pawns with 2 options each for 2*(1+8) options. For the first three moves White must play (d3 or d4), Be3 in that order and f3. There are 3 options when to play f3, and 2 options for the d-pawn so 3*2 options for White. Put them together: 2*(1+8)*3*2 = 108.

Nf6-h5-g3-h1: For this route, there's no interaction between Black's and White's moves, so White has 2*5!/(2!*3!) or 20 options again. For Black's filler move, we must distinguish the general fillers from the moves ...f6 or ...h5 that intersect the knight's route. For the general case, the knight and 6 of the pawns have 2 choices each, and the f- and h-pawns have 1 choice each, and the filler can be played on any of the first 4 moves, so there are 4*(2*(1+6)+2) choices for those. If the filler move is ...f6, it must be played after 2...Nh5 and before 5...Nxh1# so there are 2 choices. If the filler moves is ...h5, it must be played after 3...Ng3 and before 5...Nxh1# so it must be the 4th move. Combining all these with White's choices, we get 2*5!/(2!*3!)*[4*(2*(1+6)+2)+2+1] = 1340.

Nh6-f5-g3-h1: For this route, as before White has 2*5!/(2!*3!) or 20 options again, and the logic with the fillers is same as above with the 2*(1+6)+2 for those that don't intersect the knight's path, and 2 options when ...h6 could be played and 1 for ...f5, so we again get 2*5!/(2!*3!)*[4*(2*(1+6)+2)+2+1] = 1340.

Putting all these together, I get the total number of options for Solution Type 1 as 20+40+108+1340+1340 = 2848.

I will now explain my mistake above. Spoiler alerts apply.

For the Type 1 solutions in the Nc6-d4-f5-g3-h1 route, I argued that White has a 2-move sequence (d-pawn, Be3) and a 3-move sequence (f3, Kf2, Qe1) and there are 5!/(2!*3!) = 10 interleavings of them. This part, I think, is valid. In fact the number is small enough one can even list out the 10 variations by hand.

But then I said that White can play either d3 or d4, so we can double that number and get 20 variations. Here's the flaw.

The game may begin 1. f3 Nc6 2. Kf2 Nd4.

click for larger view

As we can see, White can't play 3. d4 here because the knight already occupies the square. This means the formula with doubling will count some spurious solutions, variations that can't actually be played.

To correct for this, we need to subtract the cases that start with 1. f3 Nc6 2. Kf2 Nd4 3. d4. In those, we find White plays Be3 either on move 4 or 5, so there are 2 of them.

After some time, nobody has found any more mistakes in the Type 1 count above, so I'll move on now to Type 2. Spoiler alert and boredom alert apply.

Solution Type 2: White K on f2, Black Q on e7, N on h1 with the e-pawns gone.

Black's moves are fixed: 1...Nf6, 2...Nxe4, 3...Ng3, 4...Qxe7, 5...Nxh1#. By move 3, White has to get pawns on e4 and f3 and queen on e2 and reach the following position:

click for larger view

So in the first 3 moves White has to play e4 and Qe2 in that order, and has to play f3. One could note that f3 can be on any of the moves, so the answer is 3 (similar to the way I showed the count is 2 for the spurious solutions above).

Or one could see it as interleaving a 2-move sequence and a 1-move sequence and use our now-familiar formula for that: 3!/(1!*2!) = 3. I could have done that also above (interleaving a pair of 1-move sequences or 2!/(1!*1!) = 2) but felt ridiculous applying such overkill on something all of us can solve in our heads.

It was just a curious sidenote I found researching a player who was a mathematician - but you showed it was a lot more challenging than I first thought.

It would be nice to collect your posts as a coherent whole (maybe slighted edited down?).