Dominoes

[cinelli]

I have laid out all 28 pieces of a standard set of dominoes in an 8 by 7 array as follows (blanks are shown as zeros):

. 1   6   3   3   5   6   6   0

  2   6   2   5   4   4   3   3

  3   6   6   5   0   4   1   3

  2   4   5   0   0   1   1   4

  4   4   1   1   0   2   0   6

  0   4   3   2   0   2   2   6

  2   1   5   5   5   5   1   3

This puzzle is to indicate how all the pieces are oriented. For instance at the top left corner, is the 1-6 piece horizontal or is the 1-2 piece vertical? I have set a similar puzzle before but I think this one is quite hard.

Cinelli

[UncleEbenezer]

A reference to what a standard domino set looks like wouldn't go amiss. I understand a domino consists of two numbers from 1 to 6, but my first guess at a set would be one of each 1-6 combo, which would imply 21 tiles. 28 works with that formula if you have 1-7, or 0-6, but you indicated your zeros mean something else? Is that a red herring, so 0 is indeed its own number?

Some of the double-X tiles look like a good startingpoint, if I've got the right end of that stick.

[staffordian]

A reference to what a standard domino set looks like wouldn't go amiss. I understand a domino consists of two numbers from 1 to 6, but my first guess at a set would be one of each 1-6 combo, which would imply 21 tiles. 28 works with that formula if you have 1-7, or 0-6, but you indicated your zeros mean something else? Is that a red herring, so 0 is indeed its own number?

Some of the double-X tiles look like a good startingpoint, if I've got the right end of that stick.

A standard set of dominos has one of each combination of 0 to 6 together; each number represented by the number of dots on that half of the domino, so a zero has no dots and is therefore called a blank.

So there are:
6:6. 6:5. 6:4. 6:3. 6:2. 6:1. 6:0
5:5. 5:4. 5:3 and so on until...
2:2. 2:1. 2:0
1:1. 1:0
0:0

So total number of dominos is 7+6+5+4+3+2+1=28

The i newspaper has a similar puzzle every day, and as far as I can see, the only way to solve it is by brute force, looking at each possible domino combination in turn until you find one which occurs only once. Draw round this then rinse and repeat, though at certain points the drawn patterns will isolate others.

I'm now waiting for Gengulphus to show us a neat mathematical solution

[UncleEbenezer]

A reference to what a standard domino set looks like wouldn't go amiss. I understand a domino consists of two numbers from 1 to 6, but my first guess at a set would be one of each 1-6 combo, which would imply 21 tiles. 28 works with that formula if you have 1-7, or 0-6, but you indicated your zeros mean something else? Is that a red herring, so 0 is indeed its own number?

Some of the double-X tiles look like a good startingpoint, if I've got the right end of that stick.

A standard set of dominos has one of each combination of 0 to 6 together; each number represented by the number of dots on that half of the domino, so a zero has no dots and is therefore called a blank.

So there are:
6:6. 6:5. 6:4. 6:3. 6:2. 6:1. 6:0
5:5. 5:4. 5:3 and so on until...
2:2. 2:1. 2:0
1:1. 1:0
0:0

So total number of dominos is 7+6+5+4+3+2+1=28

Thanks. So the OP's making a special case of 0 was indeed a red herring.

The i newspaper has a similar puzzle every day, and as far as I can see, the only way to solve it is by brute force, looking at each possible domino combination in turn until you find one which occurs only once. Draw round this then rinse and repeat, though at certain points the drawn patterns will isolate others.

I'm now waiting for Gengulphus to show us a neat mathematical solution

Well, brute force can be seeded. In this case we can place a double-5 (and the 5 of 5-2) among those on the bottom row, which in turn tells us places it isn't. Double-zero is among the big patch of those, etc. And you can seed it with observations like the unique candidate for 3-0. But I haven't pursued it far: a few such observations still leave an awful lot of entropy to brute-force.

More interesting: how does a puzzle-setter ensure the solution is unique?

[staffordian]

More interesting: how does a puzzle-setter ensure the solution is unique?

This is a very good question, and one that has often puzzled me in connection with puzzles generally.

A related question is grading of puzzles such as sudokus, where it doesn't always seem to be directly proportional to the number of populated squares.

[UncleEbenezer]

More interesting: how does a puzzle-setter ensure the solution is unique?

This is a very good question, and one that has often puzzled me in connection with puzzles generally.

A related question is grading of puzzles such as sudokus, where it doesn't always seem to be directly proportional to the number of populated squares.

For what it's worth, my favourite sudoku site does grade its puzzles by number of populated squares. But there's still a lot of apparent pot-luck, with the hardest category ranging from very easy through to a real challenge depending largely on how far you can get with the simple moves.

[jfgw]

More interesting: how does a puzzle-setter ensure the solution is unique?

In this instance, I think the puzzle can be solved one step at a time. For example, there is only one possibility for 0|3. Once this is placed, there is only one possibility for 3|3, and so on. It gets a little more complex than that but I do not believe that brute force (other than, maybe, looking a few moves ahead) is necessary.

I have reached a paradox so I must have made a mistake somewhere. I will look again tonight.


Julian F. G. W.

[9873210]

More interesting: how does a puzzle-setter ensure the solution is unique?

They almost certainly have a computer with a brute force solver. These types of problems can be brute forced in far less than a second on a modern PC. This ensures the existence and uniqueness of the solution.

There should also be a direct digital path, with no keyboard data entry or fax machines, from checking the puzzle to typesetting it. Lists of numbers are notoriously difficult to proof read and editors do not want to deal with enraged puzzle fans.

[9873210]

A related question is grading of puzzles such as sudokus, where it doesn't always seem to be directly proportional to the number of populated squares.

A good ranking usually requires a program that solves the puzzle the same way a typical human would. If you have a sudoku solver that gives hints you may see how this can be done.

For example a sudoku solver might have a list of rules:
0) Use known values to eliminate possibilities from all vacant cells
1) Find any cell where there is only one possible value.
2) Search every row for missing values with a single location
3) Search every column for missing values with a single location
4) search for doubles,
...
Q) search for XYZ forced chains.
...

They would apply rules from a list step by step until a single additional cell is solved, then start over at step 0). The solver records the steps that are used. Once the answer is arrived at the particular rules that were used and the order they were used in is evaluated to generate a difficulty. This algorithm would be heavily weighted towards puzzles that need rules late in the list, particularly if those rules are used early in the solution.

Which rules are considered more difficult is subjective, but most people will be fairly similar. For example some people may apply rule 2) before rule 1) but almost all of us will apply rules 1) and 2) before Q).

If you're like me you will never apply rule Q) since I almost always resort to trial and error (a.k.a. brute force) once I see that "guessing" a single cell will force a large number of the remaining unknowns. But the fact that I resort to trial and error still makes the rating based on use of rule Q) reasonable.

The differences in the way different people work accounts for the cases where you find a very hard puzzle to be easy, you may simply be applying different rules or in a different order from the rankers view of a typical human. You might also find that one rotation or reflection or reordering of a puzzle is easier to solve than another. Most raters will consider them equivalent, but most people start at, say, the top left, if the early steps are all in the top left you may find it easier than the rotation where all the early steps are in the bottom right.

[Gengulphus]

Well, brute force can be seeded. In this case we can place a double-5 (and the 5 of 5-2) among those on the bottom row, which in turn tells us places it isn't. Double-zero is among the big patch of those, etc. And you can seed it with observations like the unique candidate for 3-0. But I haven't pursued it far: a few such observations still leave an awful lot of entropy to brute-force.

More interesting: how does a puzzle-setter ensure the solution is unique?

For many puzzles, the way I'd do it is not by setting a puzzle, then trying to determine whether it has a unique solution, but by picking a solution, then trying to find a puzzle to which it is the unique solution.

For instance, suppose I decide to treat a sudoku puzzle as 'gentle' if it can be solved entirely by spotting an empty cell which must be a particular digit because it's the only cell in its row (or column, or block) that can contain that digit, or because all eight other digits already appear at least once in the cell's row, column and/or block and that digit itself doesn't, filling the empty cell with that digit, and repeating. I can easily write a program to solve 'gentle' sudoku puzzles - it just needs an inner loop that runs through the 81 cells, checking whether each of them can be filled in by that rule, and an outer loop that keeps invoking that inner loop until either the entire board is filled or it reports that it cannot fill any of the remaining cells. If that results in the entire board being filled, I know that the puzzle has a unique solution that can be found using 'gentle' steps only and so it is a 'gentle' puzzle; if it reports that there are remaining empty cells but it cannot fill any of them, I know that it's not a 'gentle' sudoku puzzle (though whether that's because it has no solutions, multiple solutions or a unique solution that requires non-'gentle' logical steps to get to it may not be obvious). And that program won't take long: the outer loop is invoked a number of times that is at most the number of empty cells in the puzzle, and each invocation of the inner loop applies a fairly simple test a number of times that is also at most the number of empty cells in the puzzle. So the program does less than 81^2 = 6561 such tests, and each such test just involves checking up on the contents or possible values of 20 cells. Well programmed and run on a decent computer, such a program could decide whether hundreds of thousands of proposed sudoku puzzles are 'gentle' per second, quite possibly even millions.

Still, the chances of a sudoku puzzle grid filled with a random collection of blanks, 1s, 2s, 3s, 4s, 5s, 6s, 7s, 8s and 9s being a solvable sudoku puzzle are almost certainly absolutely tiny, and the chances of it having a unique solution that can be arrived at by using only 'gentle' steps smaller still. So even being able to check millions of such grids per second for being 'gentle' would probably take a very long time to find a 'gentle' puzzle with a unique solution. Obviously a human puzzle setter could use their intelligence to increase those chances enormously, but using a human to generate just one possible puzzle is a great deal more expensive (in both monetary and elapsed-time terms) than using a computer to generate and check many millions, possibly even billions.

But turn that process of starting with a proposed puzzle and trying to check whether it has a unique solution that can be reached by a number of 'gentle' steps around: instead, start with a solved sudoku grid and try to find a sudoku puzzle from which it can be reached by a series of 'gentle' steps. One such 'puzzle' is the solved sudoku grid you start with - it is of course an extremely uninteresting one, requiring zero 'gentle' steps to solve. But we can make it more interesting (albeit very marginally so) by emptying one of the filled cells. That empty cell can of course always be filled by a 'gentle' step, so we've arrived at a (still excessively) 'gentle' sudoku puzzle with the desired solution. And we can repeat that step of emptying a filled cell to get a slightly more interesting 'gentle' sudoku puzzle (because it takes two 'gentle' steps to solve rather than one) with the desired solution, then again to get one that takes three 'gentle' steps to solve, etc.

In the early stages of that process of emptying filled squares, it is easy to see that the filled cell we've just emptied is certain to be refillable with a 'gentle' step. It becomes harder to see that as more and more cells are emptied, and at some stage, it actually ceases to be true. That stage can arrive as early as when the fourth cell is emptied - e.g. if my desired solution contains two blocks fitting the following pattern:

+-----------+-----------+
| x   x   x | x   x   x |
|           |           |
| x   x   1 | x   2   x |
|           |           |
| x   x   2 | x   1   x |
+-----------+-----------+

and I happen to choose to empty the four cells containing red 1s and 2s, none of the emptied cells can be filled with a 'gentle' step, so the resulting puzzle is not a 'gentle' one - and indeed it doesn't even have a unique solution, because as well as having the desired solution we started with, it has another, namely that solution with the two red 1s replaced by 2s and the two red 2s replaced by 1s.

That issue can however be easily dealt with: start with a desired solution, and repeatedly empty a cell chosen at random, then check the resulting sudoku puzzle for being 'gentle' using the program described above, until that check fails, at which point you output the puzzle you had before that last cell-emptying. That will probably go a long way, though there is a chance it will output a very unchallenging puzzle with just 3 empty cells. That process only needs to run the 'gentleness' checker a number of times equal to the number of empty cells in the puzzle you end up with for the successfully-emptied cells plus 1 for the unsuccessfully-emptied cell, so I'd be confident that it could generate thousands of 'gentle' puzzles per second, probably tens or even hundreds of thousands - though with no guarantees about their quality!

To improve the quality, don't just pick a random cell to empty at each stage. Instead, start by choosing a random order in which to try to empty the cells. At each stage, choose the first remaining cell in that order, empty it (keeping a record of the digit you removed from it) and test the resulting puzzle for being 'gentle'. If it is, leave the cell emptied; if not, replace the removed digit in it; either way, move on to the next cell in the random order. When you've finished dealing with all 81 cells in that way, you know that you've arrived at a 'gentle' puzzle with the desired solution that cannot be converted into any better such puzzle just by emptying one of its still-filled cells. There's a cost to doing that - you do 81 tests for 'gentleness' rather than the number of cells successfully emptied plus one, so I'd expect it to be a bit slower at producing 'gentle' sudoku puzzles - but I'd still expect it to be able to produce at least thousands of them per second...

Suppose that you instead want 'medium' sudoku puzzles, which you decide mean those that cannot be solved by 'gentle' steps alone, but that can be if you also do 'medium' steps - those might be steps of the form "if all the cells within a block that can hold digit N are in the same row (or column), then the six cells in that row (or column) that are outside that block cannot contain the digit N" and the similar "if all the cells within a row (or column) that can hold digit N are in the same block, then the six cells in that block that are outside that row (or column) cannot contain the digit N". 'Medium' sudoku puzzles with a desired solution can be generated by writing a similar 'medium' puzzle checker that tries to solve a sudoku puzzle using both 'medium' and 'gentle' steps rather than just using 'gentle' steps, and then following the same procedure as above: start with the desired solution, choose a random order in which to try to empty the cells, and try emptying each cell in turn in that order, using the 'medium' puzzle checker after each emptying to determine whether one still has a 'medium' puzzle with a unique solution - if not, back off from that emptying; otherwise, leave the cell empty; and either way, move on to the next cell in the order. When one gets to the end of that, one has a puzzle that can be solved with a sequence of 'gentle' and/or 'medium' steps and that cannot be reduced to a more challenging such puzzle by emptying any of the remaining filled cells. That means that it's a 'gentle' or 'medium' puzzle, so as a final check, run the 'gentle' puzzle checker on it and either separate out the 'gentle' puzzles if you want 'gentle' as well as 'medium' puzzles, or reject them if you only want 'medium' puzzles.

One can clearly extend that idea up to 'challenging', 'hard', etc, levels of puzzle where the solution can contain further types of step. And indeed, one can produce a 'full backtracking' puzzle checker quite easily, that simply picks an empty cell to try all possible digits for (preferably choosing one with the smallest number of possible digits for efficiency), and for each of those possible digits, fills it in, removes it as a possible digit for all other cells in the same row, column or block and then invokes itself recursively on the now more-constrained puzzle. One has to take a little bit of care about writing such a puzzle checker, basically to stop it exploring further possibilities once it's found enough solutions: if one just wants to know whether a puzzle has a unique solution, stop it with a "multiple solutions" result as soon as it finds a second solution rather than letting it proceed with its search to find all solutions (the reason why can be seen if one considers the completely unconstrained 'sudoku puzzle' which has all 81 cells blank: https://en.wikipedia.org/wiki/Mathematics_of_Sudoku and/or https://oeis.org/A107739 tell me that it has 6,670,903,752,021,072,936,960 solutions, so letting it find all of them might take inordinately long...). That allows one to produce 'ultimate' sudoku puzzles: ones that have a unique solution, but not one that's findable by using all of the types of solution steps that your lower levels of difficulty explore.

I'll also mention that this way of generating sudoku puzzles extends easily to the job of generating sudoku puzzles with a symmetrical pattern of empty and filled squares: rather than emptying cells one at a time, empty them a complete set of symmetrically placed cells at a time.

I'm not saying that every type of puzzle can be generated in that way, i.e. by starting with a solution as a "no work needed at all" puzzle, successively removing constraints from the puzzle until it no longer has just one solution, and finally backing off from the last constraint removal. But a lot can, and for those that can, it's generally much easier to start with a valid solution and use it to come up with a puzzle to which it's the unique solution, than it is to start with a puzzle, hope it turns out to have a unique solution, and repeat if that doesn't happen until it does.

Gengulphus

[modellingman]

Spoiler....




The initial grid is GRID1

1   2   3   4   5   6   7   8
  +---+---+---+---+---+---+---+---+
1 | 1 | 6 | 3 | 3 | 5 | 6 | 6 | 0 |  
  +---+---+---+---+---+---+---+---+
2 | 2 | 6 | 2 | 5 | 4 | 4 | 3 | 3 |
  +---+---+---+---+---+---+---+---+
3 | 3 | 6 | 6 | 5 | 0 | 4 | 1 | 3 |
  +---+---+---+---+---+---+---+---+
4 | 2 | 4 | 5 | 0 | 0 | 1 | 1 | 4 |
  +---+---+---+---+---+---+---+---+
5 | 4 | 4 | 1 | 1 | 0 | 2 | 0 | 6 |
  +---+---+---+---+---+---+---+---+
6 | 0 | 4 | 3 | 2 | 0 | 2 | 2 | 6 | 
  +---+---+---+---+---+---+---+---+
7 | 2 | 1 | 5 | 5 | 5 | 5 | 1 | 3 |
  +---+---+---+---+---+---+---+---+

A domino is represented as [a,b] where a>=b and the dominoes are ordered as:

1    [0,0]
2    [1,0]
3    [1,1]
4    [2,0]
5    [2,1]
6    [2,2]
...
27   [6,5]
28   [6,6]

For a domino lying within row a of the grid and occupying columns b and c (where c=b+1), the position is represented by (a,b&c). For a domino lying with column a of the grid and occupying rows b and c (where c=b+1), the position is represented by (b&c,a)

Take each domino in turn in the order listed above and systematically search the grid by rows and columns to determine the number of possible positions for that domino. Doing this reveals that [3,0] is the first domino with only one possible position in the grid. This position is (1&2,8). The grid can now updated to become GRID2

1   2   3   4   5   6   7   8
  +---+---+---+---+---+---+---+---+
1 | 1 | 6 | 3 | 3 | 5 | 6 | 6 | 0 |  
  +---+---+---+---+---+---+---+   +
2 | 2 | 6 | 2 | 5 | 4 | 4 | 3 | 3 |
  +---+---+---+---+---+---+---+---+
3 | 3 | 6 | 6 | 5 | 0 | 4 | 1 | 3 |
  +---+---+---+---+---+---+---+---+
4 | 2 | 4 | 5 | 0 | 0 | 1 | 1 | 4 |
  +---+---+---+---+---+---+---+---+
5 | 4 | 4 | 1 | 1 | 0 | 2 | 0 | 6 |
  +---+---+---+---+---+---+---+---+
6 | 0 | 4 | 3 | 2 | 0 | 2 | 2 | 6 | 
  +---+---+---+---+---+---+---+---+
7 | 2 | 1 | 5 | 5 | 5 | 5 | 1 | 3 |
  +---+---+---+---+---+---+---+---+

As well as showing [3,0] in its position, GRID2 also eliminates 3 possible domino positions present in GRID1: these eliminated positions are

(1) domino [6,0] at (1,7&8)
(2) domino [3,3] at (2,7&8)
(3) domino [3,3] at (2&3,8)

The second and third of these are helpful because continuing the search from domino [3,1] onwards reveals that GRID2 now has only one possible position for [3,3], this being position (1,3&4). The result is GRID3.

1   2   3   4   5   6   7   8
  +---+---+---+---+---+---+---+---+
1 | 1 | 6 | 3   3 | 5 | 6 | 6 | 0 |  
  +---+---+---+---+---+---+---+   +
2 | 2 | 6 | 2 | 5 | 4 | 4 | 3 | 3 |
  +---+---+---+---+---+---+---+---+
3 | 3 | 6 | 6 | 5 | 0 | 4 | 1 | 3 |
  +---+---+---+---+---+---+---+---+
4 | 2 | 4 | 5 | 0 | 0 | 1 | 1 | 4 |
  +---+---+---+---+---+---+---+---+
5 | 4 | 4 | 1 | 1 | 0 | 2 | 0 | 6 |
  +---+---+---+---+---+---+---+---+
6 | 0 | 4 | 3 | 2 | 0 | 2 | 2 | 6 | 
  +---+---+---+---+---+---+---+---+
7 | 2 | 1 | 5 | 5 | 5 | 5 | 1 | 3 |
  +---+---+---+---+---+---+---+---+

Amongst the possibilities that the placement of [3,3] eliminates in GRID3 (when compared to GRID2) are two possibilities for [5,3]. This leaves only one possibility for [5,3] at position (6&7,3). This is the third domino to be positioned in the grid.

Continuing the approach the fourth placement is [6,0] at position (5,7&8), whilst the fifth is [6,1] in position (1,1&2).

After these 5 placements, the grid is now GRID4

1   2   3   4   5   6   7   8
  +---+---+---+---+---+---+---+---+
1 | 1   6 | 3   3 | 5 | 6 | 6 | 0 |  
  +---+---+---+---+---+---+---+   +
2 | 2 | 6 | 2 | 5 | 4 | 4 | 3 | 3 |
  +---+---+---+---+---+---+---+---+
3 | 3 | 6 | 6 | 5 | 0 | 4 | 1 | 3 |
  +---+---+---+---+---+---+---+---+
4 | 2 | 4 | 5 | 0 | 0 | 1 | 1 | 4 |
  +---+---+---+---+---+---+---+---+
5 | 4 | 4 | 1 | 1 | 0 | 2 | 0   6 |
  +---+---+---+---+---+---+---+---+
6 | 0 | 4 | 3 | 2 | 0 | 2 | 2 | 6 | 
  +---+---+   +---+---+---+---+---+
7 | 2 | 1 | 5 | 5 | 5 | 5 | 1 | 3 |
  +---+---+---+---+---+---+---+---+

This exhausts the process of searching the grid for sole possible positions for unplaced dominoes. Starting the search again starting with domino [0,0] does not identify any further sole position possibilities.

There is, though, something interesting going on with domino [3,2]. There are only 2 possible positions for this: either (2&3,1) or (3&4,1). The second of these possibilities can be eliminated by noting that there are also only two possible positions for [4,2]: either (4,1&2) or (4&5,1). The 2 at (row,col) position (4,1) is the other half of a domino covering either (4,1&2) or (4&5,1). This means the 2 at (4,1) cannot form domino [3,2] at (3&4,1). Therefore, the second possible position for domino [3,2] is eliminated. The sixth domino placed on the grid is therefore [3,2] at position (2&3,1).

After this placement the grid is GRID5

1   2   3   4   5   6   7   8
  +---+---+---+---+---+---+---+---+
1 | 1   6 | 3   3 | 5 | 6 | 6 | 0 |  
  +---+---+---+---+---+---+---+   +
2 | 2 | 6 | 2 | 5 | 4 | 4 | 3 | 3 |
  +   +---+---+---+---+---+---+---+
3 | 3 | 6 | 6 | 5 | 0 | 4 | 1 | 3 |
  +---+---+---+---+---+---+---+---+
4 | 2 | 4 | 5 | 0 | 0 | 1 | 1 | 4 |
  +---+---+---+---+---+---+---+---+
5 | 4 | 4 | 1 | 1 | 0 | 2 | 0   6 |
  +---+---+---+---+---+---+---+---+
6 | 0 | 4 | 3 | 2 | 0 | 2 | 2 | 6 | 
  +---+---+   +---+---+---+---+---+
7 | 2 | 1 | 5 | 5 | 5 | 5 | 1 | 3 |
  +---+---+---+---+---+---+---+---+

In this grid, none of the unplaced 22 dominoes have a sole possible position within the grid.

However, there are a number of dominoes which in GRID5 have just a pair of possible positions. These include domino [4,2] in which cell (4,1) is common to both its possible positions. Similarly, cell (6,6) is common to the two possible positions for domino [2,2]. Dominoes with a pair of possible positions but without a common cell include [6,3] (at top of col 7 or bottom of col 8) and [4,3] (in row 2 or col 8).

Although the existing of dominoes with pairs of possible positions might suggest searching for a solution using some sort of branching algorithm, this does not seem much within the spirit of a puzzle.

The solution approach below was instead based on an observation about domino [4,4].

In GRID5 there are two regions of the grid where domino [4,4] might be positioned. The first involves cell (5,2) whilst the second involves cell (2,6). After some stumbling around, I found that the assumption that [4,4] is NOT located in the first of these regions leads to a single, feasible solution. I have not considered whether the opposite assumption leads to infeasibility. Instead reliance is placed on the usual norm that puzzles have only one solution.

The logic showing the positioning of all 28 dominoes in the grid is listed. The logic for the first 6 placements is that set out previously and which results in GRID5.

1	[3,0] at (1&2,8) See discussion leading to GRID5 
2	[3,3] at (1,3&4) See discussion leading to GRID5
3	[5,3] at (6&7,3) See discussion leading to GRID5
4	[6,0] at (5,7&8) See discussion leading to GRID5
5	[6,1] at (1,1&2) See discussion leading to GRID5
6	[3,2] at (2&3,1) See discussion leading to GRID5
7	[4,1] at (5,2&3) Avoid possibility of [4,4] in region centred on cell (5,2)
8	[4,0] at (6,1&2) To avoid 2nd [4,1] at (6&7,2)
9	[2,1] at (7,1&2) Domino must occupy (7,1&2)
10	[4,2] at (4&5,1) To avoid stranding cell (5,1)
11	[5,1] at (7,6&7) Sole position available for [5,1]
12	[6,3] at (6&7,8) To avoid stranding cell (7,8)
13	[2,2] at (6,6&7) To avoid stranding cell (6,7)
14	[6,6] at (1,6&7) To avoid 2nd [6,3] at (1&2,7)
15	[5,4] at (1&2,5) To avoid stranding cell (1,5)
16	[4,4] at (2&3,6) Sole position available for [4,4]
17	[3,1] at (2&3,7) To avoid stranding cell (2,7)
18	[4,3] at (3&4,8) To avoid stranding cell (3,8)
19	[1,1] at (4,6&7) To avoid stranding cell (4,7)
20	[2,0] at (5,5&6) To avoid stranding cell (5,6)
21	[5,0] at (6&7,5) To avoid 2nd [2,0] at (6,4&5)
22	[5,2] at (6&7,4) To avoid stranding cell (7,4)
23	[1,0] at (4&5,4) To avoid stranding cell (5,4)
24	[0,0] at (3&4,5) To avoid stranding cell (4,5)
25	[6,5] at (3&4,3) To avoid 2nd [5,4] at (4,2&3)
26	[6,4] at (3&4,2) To avoid stranding cell (4,2)
27	[6,2] at (2,2&3) To avoid stranding cell (2,2)
28	[5,5] at (2&3,4) Final unfilled position in grid

The final solution grid is GRID6

1   2   3   4   5   6   7   8
  +---+---+---+---+---+---+---+---+
1 | 1   6 | 3   3 | 5 | 6   6 | 0 |  
  +---+---+---+---+   +---+---+   +
2 | 2 | 6   2 | 5 | 4 | 4 | 3 | 3 |
  +   +---+---+   +---+   +   +---+
3 | 3 | 6 | 6 | 5 | 0 | 4 | 1 | 3 |
  +---+   +   +---+   +---+---+   +
4 | 2 | 4 | 5 | 0 | 0 | 1   1 | 4 |
  +   +---+---+   +---+---+---+---+
5 | 4 | 4   1 | 1 | 0   2 | 0   6 |
  +---+---+---+---+---+---+---+---+
6 | 0   4 | 3 | 2 | 0 | 2   2 | 6 | 
  +---+---+   +   +   +---+---+   +
7 | 2   1 | 5 | 5 | 5 | 5   1 | 3 |
  +---+---+---+---+---+---+---+---+

[Gengulphus]

I have laid out all 28 pieces of a standard set of dominoes in an 8 by 7 array as follows (blanks are shown as zeros):

. 1   6   3   3   5   6   6   0

  2   6   2   5   4   4   3   3

  3   6   6   5   0   4   1   3

  2   4   5   0   0   1   1   4

  4   4   1   1   0   2   0   6

  0   4   3   2   0   2   2   6

  2   1   5   5   5   5   1   3

This puzzle is to indicate how all the pieces are oriented. For instance at the top left corner, is the 1-6 piece horizontal or is the 1-2 piece vertical? I have set a similar puzzle before but I think this one is quite hard.

Spoiler...

A careful count shows that the numbers of times that pairs of digits appear next to each other are:

  | 6 5 4 3 2 1 0
--+--------------
0 | 2 4 4 1 6 4 4
1 | 1 3 5 4 5 3
2 | 4 3 2 4 2
3 | 4 3 3 3
4 | 3 3 5
5 | 3 4
6 | 5

These are the counts of possible locations for the corresponding dominoes, so the 1s in that table indicate that there is only one place for each of the 0-3 and 1-6 dominoes, in the top right and top left corners respectively. Placing them in those locations produces 6 'domino boundaries', each of which removes a possible digit location for a domino. So the 0-3 domino being placed reduces the counts of possible locations for the 0-6 domino by 1 and for the 3-3 domino by 2, and the 1-6 domino being placed reduces the counts of possible locations for the 1-2, 3-6 and 6-6 dominoes by 1 each. So a diagram showing the known domino boundaries and an updated table of counts of possible domino locations (with asterisks indicating dominoes that have already been placed) are:

+---+---+---+---+---+---+---+---+       | 6 5 4 3 2 1 0
| 1   6 | 3   3   5   6   6 | 0 |     --+--------------
+---+---+   +   +   +   +   +   +     0 | 1 4 4 * 6 4 4
| 2   6   2   5   4   4   3 | 3 |     1 | * 3 5 4 4 3
+   +   +   +   +   +   +   +---+     2 | 4 3 2 4 2
| 3   6   6   5   0   4   1   3 |     3 | 3 3 3 1
+   +   +   +   +   +   +   +   +     4 | 3 3 5
| 2   4   5   0   0   1   1   4 |     5 | 3 4
+   +   +   +   +   +   +   +   +     6 | 4
| 4   4   1   1   0   2   0   6 |
+   +   +   +   +   +   +   +   +
| 0   4   3   2   0   2   2   6 |
+   +   +   +   +   +   +   +   +
| 2   1   5   5   5   5   1   3 |
+---+---+---+---+---+---+---+---+

The counts of possible locations for the 0-6 and 3-3 dominoes have been reduced to 1, so they can be placed, and placing them fixes 5 and 3 more domino boundaries respectively, so that a total of 8 more count reductions by 1 happen:

+---+---+---+---+---+---+---+---+       | 6 5 4 3 2 1 0
| 1   6 | 3   3 | 5   6   6 | 0 |     --+--------------
+---+---+---+---+   +   +   +   +     0 | * 4 4 * 4 3 4
| 2   6   2   5   4   4   3 | 3 |     1 | * 3 5 4 4 3
+   +   +   +   +   +   +   +---+     2 | 4 3 2 3 2
| 3   6   6   5   0   4   1   3 |     3 | 3 1 3 *
+   +   +   +   +   +   +   +   +     4 | 2 3 5
| 2   4   5   0   0   1   1   4 |     5 | 3 4
+   +   +   +   +   +   +---+---+     6 | 3
| 4   4   1   1   0   2 | 0   6 |
+   +   +   +   +   +   +---+---+
| 0   4   3   2   0   2   2   6 |
+   +   +   +   +   +   +   +   +
| 2   1   5   5   5   5   1   3 |
+---+---+---+---+---+---+---+---+

And now the count of possible locations for the 3-5 domino has been reduced to 1, so it can be placed, and placing it fixes 5 more domino boundaries and causes 5 more count reductions by 1:

+---+---+---+---+---+---+---+---+       | 6 5 4 3 2 1 0
| 1   6 | 3   3 | 5   6   6 | 0 |     --+--------------
+---+---+---+---+   +   +   +   +     0 | * 4 4 * 4 3 4
| 2   6   2   5   4   4   3 | 3 |     1 | * 2 5 3 4 3
+   +   +   +   +   +   +   +---+     2 | 4 3 2 2 2
| 3   6   6   5   0   4   1   3 |     3 | 3 * 2 *
+   +   +   +   +   +   +   +   +     4 | 2 3 5
| 2   4   5   0   0   1   1   4 |     5 | 3 3
+   +   +   +   +   +   +---+---+     6 | 3
| 4   4   1   1   0   2 | 0   6 |
+   +   +---+   +   +   +---+---+
| 0   4 | 3 | 2   0   2   2   6 |
+   +   +   +   +   +   +   +   +
| 2   1 | 5 | 5   5   5   1   3 |
+---+---+---+---+---+---+---+---+

At this point, we've run out of dominoes with only one possible location that haven't already been placed, but a look through the dominoes with two possible locations reveals that:

* The 2-2 domino must consist of the second 2 on the seventh row and either the 2 above it or the 2 to its right, so there must be domino boundaries to its left and below it.
* The 2-3 domino must consist of the 3 at the left end of the third row and either the 2 above it or the 2 below it, so there must be a domino boundary to its right.
* The 2-4 domino must consist of the 2 on the fourth row and either the 4 to its right or the 4 below it, so there must be a domino boundary above it.
* Those last two domino boundaries force the location of a domino, which is the 2-3 domino, and placing it fixes another domino boundary.

After doing the reductions in possible location counts for those five newly-fixed domino boundaries, we've got to:

+---+---+---+---+---+---+---+---+       | 6 5 4 3 2 1 0
| 1   6 | 3   3 | 5   6   6 | 0 |     --+--------------
+---+---+---+---+   +   +   +   +     0 | * 4 4 * 3 3 4
| 2 | 6   2   5   4   4   3 | 3 |     1 | * 2 5 3 4 3
+   +   +   +   +   +   +   +---+     2 | 3 2 2 * 2
| 3 | 6   6   5   0   4   1   3 |     3 | 2 * 2 *
+---+   +   +   +   +   +   +   +     4 | 2 3 5
| 2   4   5   0   0   1   1   4 |     5 | 3 3
+   +   +   +   +   +   +---+---+     6 | 3
| 4   4   1   1   0   2 | 0   6 |
+   +   +---+   +   +   +---+---+
| 0   4 | 3 | 2   0 | 2   2   6 |
+   +   +   +   +   +---+   +   +
| 2   1 | 5 | 5   5   5   1   3 |
+---+---+---+---+---+---+---+---+

The techniques I've used so far, of placing dominoes in their only possible location and finding numbers that have to make up half of a particular domino, followed by identifying the domino boundaries they fix and tracking the restrictions those domino boundaries place on possible locations of dominoes, have run out of steam for the moment, so time to look for something new. The bottom left and bottom right corners of the position look quite constrained, so they're where I looked next, and I found a reasonably straightforward way of making progress in the bottom right corner, as follows:

Suppose the 3 in the bottom right corner combines with the 1 to its left to form the 1-3 domino. Going further leftward, that forces the next two 5s to their left to form the 5-5 domino, then the 5 even further left to combine with the 2 above it to form the 2-5 domino. And going upwards, it forces the 6 above it to combine with the 2 to its left to form the 2-6 domino. But that makes it impossible for the second 2 in the second row to combine with any of its neighbours.

So the 3 in the bottom right corner must instead combine with the 6 above it to form the 3-6 domino. Placing that domino fixes two more domino boundaries; furthermore, it means that the other previously-plausible location for the 3-6 domino must be blocked from forming a second 3-6 domino, allowing us to fix another domino boundary between the rightmost 6 on the top row and the 3 below it. That forces that 6 to combine with the 6 to its left to form the 6-6 domino, which in turn forces the 5 to their left to combine with the 4 below it to form the 4-5 domino. Preventing further 6-6 or 4-5 dominoes from being formed allows us to fix another 6 domino boundaries, and further forced domino locations cascade quite a way from there, ending up with (dropping the possible location counts because I'm not going to need them again):

+---+---+---+---+---+---+---+---+
| 1   6 | 3   3 | 5 | 6   6 | 0 |
+---+---+---+---+   +---+---+   +
| 2 | 6   2 | 5 | 4 | 4   3 | 3 |
+   +---+---+   +---+   +   +---+
| 3 | 6 | 6 | 5 | 0   4   1   3 |
+---+   +   +---+   +   +   +   +
| 2 | 4 | 5 | 0   0   1   1   4 |
+   +---+---+   +   +   +---+---+
| 4 | 4   1   1   0   2 | 0   6 |
+---+   +---+   +   +   +---+---+
| 0   4 | 3 | 2   0 | 2   2 | 6 |
+   +   +   +   +   +---+   +   +
| 2   1 | 5 | 5   5   5   1 | 3 |
+---+---+---+---+---+---+---+---+

Whichever of its neighbours the two in the bottom left corner combines with to form a domino, its other neighbour must combine with its diagonal neighbour to form a second domino, and forces the 1-4 domino to be formed by the next two unpaired digits upwards, and that in turn forces the two dominoes formed in the corner to be 1-2 and 0-4, not 0-2 and 1-4. Also, it means that the two 4s in that corner don't combine to form the 4-4 domino, so the only possibility left to get that domino is from the two unpaired 4s in the top right corner, and that results in another cascade of forced domino locations, ending at:

+---+---+---+---+---+---+---+---+
| 1   6 | 3   3 | 5 | 6   6 | 0 |
+---+---+---+---+   +---+---+   +
| 2 | 6   2 | 5 | 4 | 4 | 3 | 3 |
+   +---+---+   +---+   +   +---+
| 3 | 6 | 6 | 5 | 0 | 4 | 1 | 3 |
+---+   +   +---+   +---+---+   +
| 2 | 4 | 5 | 0 | 0 | 1   1 | 4 |
+   +---+---+   +---+---+---+---+
| 4 | 4   1 | 1 | 0   2 | 0   6 |
+---+---+---+---+   +   +---+---+
| 0   4 | 3 | 2   0 | 2   2 | 6 |
+---+---+   +   +   +---+   +   +
| 2   1 | 5 | 5   5   5   1 | 3 |
+---+---+---+---+---+---+---+---+

Not much remains to be done - the easiest way of completing the solution that strikes me is to observe that we've already placed the 5-5 domino, so there must be domino boundaries separating the unpaired 5s on the bottom row, to develop the final cascade of domino placements that results in, and to check that that has resulted in all 28 dominoes being placed, with no duplications (which it has):

+---+---+---+---+---+---+---+---+
| 1   6 | 3   3 | 5 | 6   6 | 0 |
+---+---+---+---+   +---+---+   +
| 2 | 6   2 | 5 | 4 | 4 | 3 | 3 |
+   +---+---+   +---+   +   +---+
| 3 | 6 | 6 | 5 | 0 | 4 | 1 | 3 |
+---+   +   +---+   +---+---+   +
| 2 | 4 | 5 | 0 | 0 | 1   1 | 4 |
+   +---+---+   +---+---+---+---+
| 4 | 4   1 | 1 | 0   2 | 0   6 |
+---+---+---+---+---+---+---+---+
| 0   4 | 3 | 2 | 0 | 2   2 | 6 |
+---+---+   +   +   +---+---+   +
| 2   1 | 5 | 5 | 5 | 5   1 | 3 |
+---+---+---+---+---+---+---+---+

Not the "neat mathematical solution" staffordian was hoping for, I'm afraid - like the solutions of many puzzles, it's just a matter of whittling away at the edges of one's ignorance about the solution until it's all gone, trying to spot the softest bits to remove at each stage...

Gengulphus

[modellingman]

For completeness, I thought it worth demonstrating that the solution I previously derived is unique. This is a slightly academic exercise in that the solution I posted agrees with the one Gengulphus posted but the logic underlying Gengulpus's solution leads to the conclusion that his solution is unique.

Here goes ...




The solution in GRID6 does appear to be unique. The solution was based on the observation that domino [4,4] must located in either the region around cell (5,2) or the region around cell (2,6) and then adopting the assumption that it is NOT in the first of these regions. The opposite assumption is that it is NOT in the second of these regions.

Under this opposite assumption there are two possibilities for cell (2,6): either it forms a domino with the 6 above it or with the 3 to its right.

Following the notation previously used, the first case leads to:

1	[3,0] at (1&2,8) See discussion leading to GRID6 
2	[3,3] at (1,3&4) See discussion leading to GRID6
3	[5,3] at (6&7,3) See discussion leading to GRID6
4	[6,0] at (5,7&8) See discussion leading to GRID6
5	[6,1] at (1,1&2) See discussion leading to GRID6
6	[3,2] at (2&3,1) See discussion leading to GRID6
7	[6,4] at (1&2,6) Defines the first case
8	[6,3] at (1&2,7) To avoid stranding cell (1,7)
9	[5,4] at (1&2,5) To avoid stranding cell (1,5)
10	[6,2] at (6,7&8) To avoid 2nd [6,3] at (6&7,8)
11	[3,1] at (7,7&8) To avoid stranding cell (7,8)
12	[2,2] at (5&6,6) Sole position available for [2,2]
13	[5,5] at (7,5&6) To avoid stranding cell (7,6)
14	[5,2] at (6&7,4) To avoid stranding cell (7,4)

At this point, the solution runs into the same impossibility that Gengulphus usefully spotted when he started from his bottom right position. This is that cell (2,3) must form either the [6,2] or the [5,2] domino. However, these have already been placed elsewhere in moves 10 and 14, respectively, so this rules out the first case.

For the second case we have

1	[3,0] at (1&2,8) See discussion leading to GRID6 
2	[3,3] at (1,3&4) See discussion leading to GRID6
3	[5,3] at (6&7,3) See discussion leading to GRID6
4	[6,0] at (5,7&8) See discussion leading to GRID6
5	[6,1] at (1,1&2) See discussion leading to GRID6
6	[3,2] at (2&3,1) See discussion leading to GRID6
7	[4,3] at (2,6&7) Defines the second case
8	[6,6] at (1,&&7) To avoid stranding cell (1,7)
9	[5,4] at (1&2,5) To avoid stranding cell (1,5)
10	[3,1] at (3,7&8) To avoid 2nd [4,3] at (3&4,8)
11	[4,1] at (4,7&8) To avoid stranding cell (4,8)
12	[6,4] at (3&4,2) Sole position available for [6,4]
13	[4,2] at (4&5,1) To avoid stranding cell (4,1)
14	[4,4] at (5&6,2) To avoid 2nd [4,1] at (5,2&3)

This then leaves 3 cells trapped in the bottom left corner of the grid, with no possibility of allocating dominoes correctly to these. This rules out the second case and in doing so shows that the solution in GRID6 is unique.

Not the "neat mathematical solution" staffordian was hoping for, I'm afraid - like the solutions of many puzzles, it's just a matter of whittling away at the edges of one's ignorance about the solution until it's all gone, trying to spot the softest bits to remove at each stage...

I wholeheartedly agree. I stumbled on the solution almost by accident. Once I had made my assumption about where a particular domino wasn't located (and it was after quite a few earlier assumptions I had abandoned) the solution then followed straightforwardly - much to my amazement.

There are sledgehammer approaches available to problems of this nature which will be familiar to those knowledgeable with the techniques of operational research, computer science and similar. Solutions? Yes. Mathematical? Possibly. Neat? Hardly.

[jfgw]

Spoiler,
https://imgur.com/Jp57N5K

Once I had done the easy ones, I noticed that there were two possible positions for 4|6 and that this limited the position of 6|6.
I placed 6|4 in the second column, which limited 6|6 to the top row.
I examined this for impossibilities and didn't find any. Solving the puzzle from here was easy.

I tried 4|6 in the alternative position (sixth column vertically, top two rows) hoping to quickly find an impossibility but I placed quite a few bones before I found that I had no-where to put 0|5.

The success was that there were only two branches with no sub-branches so, while there was some trial and error, only one guess had to be made. It is possible that the second branch could have been shorter than I made it.


Julian F. G. W.

[9873210]

Not the "neat mathematical solution" staffordian was hoping for, I'm afraid - like the solutions of many puzzles, it's just a matter of whittling away at the edges of one's ignorance about the solution until it's all gone, trying to spot the softest bits to remove at each stage...

Not all mathematics is neat. There is a lot of dirty grubbing around looking at all possible cases. One example is the classification of finite groups, the so called "Monster Theorem". The complete theorem is beyond the scope of this board, but classification of all groups up to order 6 is within the capabilities of many puzzlers, and the proof is exactly the sort of "if A then B then C but B and C means not A so not A" used to place dominoes.

There are sledgehammer approaches available to problems of this nature which will be familiar to those knowledgeable with the techniques of operational research, computer science and similar. Solutions? Yes. Mathematical? Possibly. Neat? Hardly.

One difference between the sledgehammer and neat solutions is that the sledgehammer includes the search as part of the solution. When you neatly place the first 3-1 domino you are ignoring the work used to build Gengulphus' supplementary table and then to search the table for zeros.

[Gengulphus]

Despoilered, because this extract doesn't IMHO spoil the puzzle at all:

The success was that there were only two branches with no sub-branches so, while there was some trial and error, only one guess had to be made. It is possible that the second branch could have been shorter than I made it.

It's a good question exactly what constitutes a "guess". Anything that splits into two or more possible cases and only pursues one of them is clearly a guess, and it results in an incomplete analysis of the puzzle. But what about something that splits into two or more possible cases, and analyses all of them properly? There's a tendency to think of them as making a "guess" if significant analysis is required for both cases and just being a deduction if the analysis required for one case is insignificant, and especially if that analysis is hidden in the proof of a theorem or lemma you're quoting. For example, if in a numerical puzzle I say "N must be even because N^2 is even", that's phrased as a deduction. If I instead say "N must be even or odd. If N were odd, N would equal 2M+1 for some integer M, so N^2 would be (2M+1)^2 = 4M^2 + 4M +1 = 2(2M^2+2M) + 1, which is odd because 2M^2+2M is an integer. Since we know that N^2 is even, we can dismiss the possibility that N is odd, and so only analyse the case that N is even any further.", that's phrased as a "guess" between two possibilities followed by analysis that eliminates one of them, to be followed by analysis of the other. But the two are essentially the same argument, just going into more detail of that argument for the latter than the former - so how can one be a "guess" and the other a deduction?

My inclination is to regard any argument that splits into two or more cases and analyses them properly, either explicitly or implicitly because they lie within a well-known result that it quotes and used, as not involving a "guess". Which implies that when you found there were two possible positions for one domino and found that the first one you thought of led to a solution, your solution involved a guess - but when you eventually found that the other one couldn't lead to a solution, I regard that as you having removed the guess from the solution, so that it no longer contains a guess. I can see though that you and/or others might still regard it as having involved a guess, and I'm not insisting that my preferred interpretation is correct. I'm just sharing my thoughts about the question of what guesses are - thoughts that arose from wondering whether people would regard my solution as involving a guess. I don't - but it certainly does contain a point at which I posit one positioning of the domino that covers a particular cell, find an argument that that positioning makes a solution impossible, and conclude that a solution must use the only other possible positioning of a domino covering that square.

Gengulphus

[Allitnil]

btw, if you like this sort of puzzle, I can recommend the 'Beyond Sudoku' magazine (https://www.puzzler.com/magazines/sudoku/beyond-sudoku). It's fairly expensive but each issue should keep most people going for a couple of weeks so cost per hour is low. Can sometimes be found in larger WH Smiths stores if you want to try a single issue.

Each issue usually has a couple of regular 6x6 domino puzzles and a larger 9x9 one. Lots of other less common puzzles too - out of the 100 or so puzzles only a couple are actual Sudokus and those are genuinely tough ones. The magazine is very much aimed at the 'hard' end of the market.

[no connection other than as a subscriber for several years]

[9873210]

IMHO the difference between a "guess" and a "deduction" is if you can hold the good branch in your head while rejecting the other branch. So this is really a discussion about the size of a human's working memory.

[Gengulphus]

Not the "neat mathematical solution" staffordian was hoping for, I'm afraid - like the solutions of many puzzles, it's just a matter of whittling away at the edges of one's ignorance about the solution until it's all gone, trying to spot the softest bits to remove at each stage...

Not all mathematics is neat. There is a lot of dirty grubbing around looking at all possible cases. One example is the classification of finite groups, the so called "Monster Theorem". The complete theorem is beyond the scope of this board, ...

An example that is closer to the scope of this board is the 4-colour theorem. Its statement is basically the answer to a puzzle about how many colours you need to be able to guarantee to be able to colour the countries on a map so that different colours are used for neighbouring countries. That puzzle needs a few extra conditions stated, such as that countries whose borders only meet at one or more isolated points don't count as "neighbouring", and that countries don't have more than one connected part (e.g. back in the days when Bangladesh and Pakistan were the two parts of a single country called Pakistan, that single country would have violated that condition), but it is essentially a simple enough question that it can be expressed reasonably succinctly and in a way that most people are capable of understanding what it's asking - so I feel that it can very reasonably be regarded as a "puzzle". And the theorem is that the answer to that puzzle is "4", also brief and easily understood.

The original proof of the four-colour theorem basically proved it by showing that it could be split into 1834 major cases, at least one of which must occur in any minimal counterexample to the theorem ("minimal" meaning "containing the smallest possible number of countries of any counterexample") and then proving for each of those major cases that the map can in fact be four-coloured - by arguments that involve large numbers of subcases for almost all of those major cases. So the proof is basically simple: "Suppose there is a map that is a counterexample to the 4-colour theorem, i.e. a map that cannot be coloured with just four colours. Then there must be such a map using a minimal number of countries, and it must involve at least one of the 1834 major cases - and so we've shown that one of the maps which we've supposed cannot be coloured with just four colours can actually be coloured with just four colours. That's an impossible conclusion, and it all follows from our supposition that there is a map that cannot be coloured with just four colours. So that supposition must be wrong - i.e. we've concluded that all maps can be coloured with just four colours".

But while it's basically simple, that argument does involve looking at a large number of major cases, almost all of them with large numbers of subcases - so a huge amount of "dirty grubbing around" is required. So much that AFAIAA, little of that "dirty grubbing around" has ever been done by a human - the vast majority of it has been done only by computers. Some progress has been made towards simplifying it - the number of major cases had been reduced to 633 by 1996 according to Wikipedia and may well have been further reduced since - but (barring any major breakthrough on simplifying the proof, which I'd expect to have heard about if it had happened) the amount of "dirty grubbing around" required is still enormously beyond the capability of any sane human being...

Gengulphus

[Gengulphus]

IMHO the difference between a "guess" and a "deduction" is if you can hold the good branch in your head while rejecting the other branch. So this is really a discussion about the size of a human's working memory.

Yes, that's one that occurred to me, and it's got a good deal to be said for it. It does however imply that one person's "deduction" can be another's "guess" even if they're using the same definition - which is fine as long as people recognise that that can be the case. Indeed, recognising that may be the best way of dealing with the fact that different people use the words differently because of using different definitions...

But I would advocate taking care about using the word "guess". I doubt it will cause any problems when people use it about their own solutions, as jfgw did in the quote that triggered my thoughts on the matter, but I'd beware of using it about other people's solutions,

Gengulphus