*greatest lower bound for *longest* game length, rather.
I agree. For me my explanation is just more "understandable" though maybe not that strict. Of course, it's a deal of taste for a reader.
I think, however, thanks to examples of game_spectator and schachfl, it is seen, that in total 4 pawns (as minimum) should be captured on 6-th row to make position playable. Therefore with ideal play we still come to these 192 - 4 = 188 half-moves regardless on the amount of units left (between 16 and 24), as was demonstrated practically.
I think, however, thanks to examples of game_spectator and schachfl, it is seen, that in total 4 pawns (as minimum) should be captured on 6-th row to make position playable. Therefore with ideal play we still come to these 192 - 4 = 188 half-moves regardless on the amount of units left (between 16 and 24), as was demonstrated practically.
#40
> Basically that section of the proof is explaining not all the uncaptured pieces can make their maximum moves; only 16 at most can, and therefore we lose 24-16 = 8 halfmoves, regardless of whether they are still on the board at the end or not.
Then the proof has a flaw. The proof state that there is at most 96 moves - 8 halfmoves = 92 moves. But see my example of 94 moves in this thread.
> Basically that section of the proof is explaining not all the uncaptured pieces can make their maximum moves; only 16 at most can, and therefore we lose 24-16 = 8 halfmoves, regardless of whether they are still on the board at the end or not.
Then the proof has a flaw. The proof state that there is at most 96 moves - 8 halfmoves = 92 moves. But see my example of 94 moves in this thread.
#42
> I think, however, thanks to examples of game_spectator and schachfl, it is seen, that in total 4 pawns (as minimum) should be captured on 6-th row to make position playable.
Your statement is very near to proof of "96-impossibility" (as I understand it). It is an answer to the key question (as I named it, see the thread). Here is my statement:
At least one unit should be captured on 6th row or earlier to make position playable.
Here:
"6th row" means the sixth row when you count from the side from which the unit starts to move;
"or earlier" means 1st, 2nd, 3rd, 4th, 5th rows
"unit" means piece or pawn; that is important --- not only pawn may be captured on 6th row to make position playable (in the thread I sometimes used word "piece" instead of "unit", that is misleading)
"make position playable" means ... basically, it means "to open lines for rooks"; however, as I see, here one needs additional arguments
> I think, however, thanks to examples of game_spectator and schachfl, it is seen, that in total 4 pawns (as minimum) should be captured on 6-th row to make position playable.
Your statement is very near to proof of "96-impossibility" (as I understand it). It is an answer to the key question (as I named it, see the thread). Here is my statement:
At least one unit should be captured on 6th row or earlier to make position playable.
Here:
"6th row" means the sixth row when you count from the side from which the unit starts to move;
"or earlier" means 1st, 2nd, 3rd, 4th, 5th rows
"unit" means piece or pawn; that is important --- not only pawn may be captured on 6th row to make position playable (in the thread I sometimes used word "piece" instead of "unit", that is misleading)
"make position playable" means ... basically, it means "to open lines for rooks"; however, as I see, here one needs additional arguments
#43:
Not 96 - 4 moves, but as I explained, 104*2 - (2+8+2+8) = 188 halfmoves = 94 moves. There is no problem here.
The 104 comes from assuming all units reach the last rank, whether or not there is actually room to fit them. (8 pieces * 7 + 8 pawns * 6) That is why we need to subtract the 8 halfmoves due to crowdedness. This is done after working out all necessary captures, to avoid subtracting too many moves at this stage. Whereas if you had began with assuming 16 units * 6 = 96 moves, this subtraction is not necessary but it becomes harder to figure out which captures lose moves and avoid double-counting.
The other 8 halfmoves are from the pawns (or other units, but pawns are most efficient) being captured and not making their final 2 moves.
Then 2 more halfmoves from the rooks, and 2 more from the edge pawns. Thus 104 full moves - 20 half moves = at most 94 moves, which is justified possible by any number of proof games in this thread (or mine on the other site.)
Not 96 - 4 moves, but as I explained, 104*2 - (2+8+2+8) = 188 halfmoves = 94 moves. There is no problem here.
The 104 comes from assuming all units reach the last rank, whether or not there is actually room to fit them. (8 pieces * 7 + 8 pawns * 6) That is why we need to subtract the 8 halfmoves due to crowdedness. This is done after working out all necessary captures, to avoid subtracting too many moves at this stage. Whereas if you had began with assuming 16 units * 6 = 96 moves, this subtraction is not necessary but it becomes harder to figure out which captures lose moves and avoid double-counting.
The other 8 halfmoves are from the pawns (or other units, but pawns are most efficient) being captured and not making their final 2 moves.
Then 2 more halfmoves from the rooks, and 2 more from the edge pawns. Thus 104 full moves - 20 half moves = at most 94 moves, which is justified possible by any number of proof games in this thread (or mine on the other site.)
#45
I do not see proof in your text. I see a messy way to present a correct generic formula and an suspicious accounting of your specific personal game.
Your formula: M = ½ ( 8·7 + 8·6 + 8·7 + 8·6 + (8-d₁) + (8-d₂) + ... + (8-d₃₂) ) = 104 - ½( (8-d₁) + (8-d₂) + ... + (8-d₃₂) )
where d₁ ... d₃₂ are destination ranks, and M is number of moves. (Destination is the last place of a unit, regardless of unit liveness. Destination is either the place of unit capture or the place of unit at the end of the game).
My "96-impossibility" proof in your terms is the following:
A. No more than 16 of d₁ ... d₃₂ are equal to 8 (obviously).
B. At least one of d₁ ... d₃₂ is equal to 6 or less (need easy explanation).
C. Assume that all other d₁ ... d₃₂ are equal to 7 (they can not be equal to 8, see A)
Then use your formula and see that M ≤ 95 + ½
To present a "94-maximum" proof you need to present your inequalities for d₁ ... d₃₂ and, the most important, justify why they are so.
I agree with your idea that destination ranks of rooks may not be better than 8, 8, 7, 7. However, from this one can not conclude that M ≤ 95 + ½ - 1, because 7-th destination ranks are fairy legal even for 96 move game.
I do not see proof in your text. I see a messy way to present a correct generic formula and an suspicious accounting of your specific personal game.
Your formula: M = ½ ( 8·7 + 8·6 + 8·7 + 8·6 + (8-d₁) + (8-d₂) + ... + (8-d₃₂) ) = 104 - ½( (8-d₁) + (8-d₂) + ... + (8-d₃₂) )
where d₁ ... d₃₂ are destination ranks, and M is number of moves. (Destination is the last place of a unit, regardless of unit liveness. Destination is either the place of unit capture or the place of unit at the end of the game).
My "96-impossibility" proof in your terms is the following:
A. No more than 16 of d₁ ... d₃₂ are equal to 8 (obviously).
B. At least one of d₁ ... d₃₂ is equal to 6 or less (need easy explanation).
C. Assume that all other d₁ ... d₃₂ are equal to 7 (they can not be equal to 8, see A)
Then use your formula and see that M ≤ 95 + ½
To present a "94-maximum" proof you need to present your inequalities for d₁ ... d₃₂ and, the most important, justify why they are so.
I agree with your idea that destination ranks of rooks may not be better than 8, 8, 7, 7. However, from this one can not conclude that M ≤ 95 + ½ - 1, because 7-th destination ranks are fairy legal even for 96 move game.
It seems that 94 is really maximum. Why?
A. No more than 16 of d₁ ... d₃₂ are equal to 8 (obviously).
B. At least 4 of d₁ ... d₃₂ is equal to 6 or less (sketch of a proof below).
C. Assume that all other d₁ ... d₃₂ are equal to 7 (they can not be equal to 8, see A)
Then use formula that Illion had in mind, but did not write explicitly, and see that M ≤ 94.
Sketch of a proof of B:
B1. one need to open "a" file for rooks, so one unit have to die at a3 or b3 or a6 or b6
B2. one need to open "h" file for rooks, so one unit have to die at h3 or g3 or h6 or g6
B3. after that "c", "d", "e", "f" files will have 1 white pawn and 1 black pawn each; pawns can not pass thru each other, so again at least 2 units have to die at 6-th rank
A. No more than 16 of d₁ ... d₃₂ are equal to 8 (obviously).
B. At least 4 of d₁ ... d₃₂ is equal to 6 or less (sketch of a proof below).
C. Assume that all other d₁ ... d₃₂ are equal to 7 (they can not be equal to 8, see A)
Then use formula that Illion had in mind, but did not write explicitly, and see that M ≤ 94.
Sketch of a proof of B:
B1. one need to open "a" file for rooks, so one unit have to die at a3 or b3 or a6 or b6
B2. one need to open "h" file for rooks, so one unit have to die at h3 or g3 or h6 or g6
B3. after that "c", "d", "e", "f" files will have 1 white pawn and 1 black pawn each; pawns can not pass thru each other, so again at least 2 units have to die at 6-th rank
The "suspicious accounting" of the proof game I had is easily explained. I wrote the text first, then constructed the game based on what I had written (as in maths, proving first there is a supremum, then showing by construction that it can be attained).
As for what I presented not being a proof... I'm in 2 minds there. The part of me that does mathematics says, "Yes, nowhere near rigorous enough", seeing that the weakest part of the argument is that part where I said "apparently optimal" - it is handwavy. Then the part of me that does retroanalysis problems says "It's obvious that 4 pawn captures like that is optimal, it's tedious and unnecessary to explain further." Which is often the case in solving retros, where the solutions are often given in abbreviated form that could take pages if written in full.
The sketch of the proof of B would indeed cover the hole. It explains precisely why the 4 pawn captures are required, something I took on intuition rather than rigour. And your formula is much simplet and more generalisable.
As for what I presented not being a proof... I'm in 2 minds there. The part of me that does mathematics says, "Yes, nowhere near rigorous enough", seeing that the weakest part of the argument is that part where I said "apparently optimal" - it is handwavy. Then the part of me that does retroanalysis problems says "It's obvious that 4 pawn captures like that is optimal, it's tedious and unnecessary to explain further." Which is often the case in solving retros, where the solutions are often given in abbreviated form that could take pages if written in full.
The sketch of the proof of B would indeed cover the hole. It explains precisely why the 4 pawn captures are required, something I took on intuition rather than rigour. And your formula is much simplet and more generalisable.
#44 "At least one unit should be captured on 6th row or earlier to make position playable. "
4 pawns as minimum to be precise. "Pawns", not "units". The rooks are not the only problem. All pawns should be able to pass at least until 7-th row to make maximal amount of moves. This is allowed when all lines are half-open, 4 pawns capture is an optimal optimal way to do it (at least as I see it).
See end position:
http://ru.lichess.org/1AiXx7pl
Note that the same pattern is present in all examples of 94-moves games presented in the thread.
4 pawns as minimum to be precise. "Pawns", not "units". The rooks are not the only problem. All pawns should be able to pass at least until 7-th row to make maximal amount of moves. This is allowed when all lines are half-open, 4 pawns capture is an optimal optimal way to do it (at least as I see it).
See end position:
http://ru.lichess.org/1AiXx7pl
Note that the same pattern is present in all examples of 94-moves games presented in the thread.
#49
> "Pawns", not "units".
You have to *prove* your proposition. This is a pain without gain (it does not simplify rest of proof, does not give better maximum value and so on)
BTW, it is possible to sacrifice bishop instead of pawn on b6. However, it seems to be worse than my way.
> "Pawns", not "units".
You have to *prove* your proposition. This is a pain without gain (it does not simplify rest of proof, does not give better maximum value and so on)
BTW, it is possible to sacrifice bishop instead of pawn on b6. However, it seems to be worse than my way.
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