Total Invisible: The fairy piece Total Invisible is a piece which stands somewhere on the board, but whose color, identity and whereabouts are not known. The real identity of Total Invisible is any ordinary piece (including K if there is no K on the board). It is assumed that the initial position and the sequence of moves must be legal after the true identity of every Total Invisible is revealed. After all aspects of a Total Invisible are revealed, it becomes visible and turns into an ordinary piece. In an initial position, only the total number of Total Invisibles is given. (see Sake Tourney 2018 Announcement | Award)

White Ke1 TIf1 TIh8 TIg3 TIa1 TIc3
Black Kg4 Pf4h3
Neutral Qh1 Pg2

h#2.5 2 solutions (1+3+2)
4 Total Invisibles

{JV: On the diagram Total Invisibles remain invisible}
1...nQh1-g1 {means there is an Invisible on f1,
which is not bRf1/bQf1
} 2.nQg1-c5 0-0-0 {TId1=wR and there is an Invisible between wKc1 and nQc5,
but it is not bR or bQ, neither bTIc2, bSc3, bBc3.
} 3.f4*g3 {the fourth Invisible is captured on g3
} Rd1-d4# {none of the remaining Invisibles (TIf1 and TIc3/c4) can prevent
the mate by the white Rook
The Total Invisibles of the first solution:
wRa1, wTIg3, TIf1, TIc2/c3/c4
}
1...TIh8*h3 {the first two Invisibles are partially identified:
wTIh3 and one TIf1 or TIg1
} 2.nPg2*f1=nR+ nRf1-f3 {means TIf1 has been captured and there is
another Invisible on g1, but not bR/bQ
} 3.f4*g3 {the fourth Invisible was on g3 and has just been captured
} TIh3*g3=R {the only Invisible that can capture on g3 is TIh3 because TIg1
is pinned hence TIh3=wR!
The Total Invisibles of the second solution:
wR somewhere on the h-file capturing bPh3, wTIf1, wTIg3, TIg1}

The solutions are C+ by Popeye 4.83, which is not 100% trustworthy for Total Invisibles.
Hopefully they will offer some fun for solvers. (Author)

As there are neutral pieces on the board, the TIs may be neutral (why not?) and then second solution does not work, as the final TI may be a neutral rook (not a white one).

Perhaps it should be stipulated that as only neutral pawn and queen are there the TI can only be that, not a neutral rook !!

Thomas Maeder

December 17, 2019 20:08

> the TIs may be neutral (why not?)

Obviously because I won’t program them 🙂

Eric

December 17, 2019 21:56

I have published other originals with Total Invisibles in Problem Paradise 86, Apr-Jun. 2019 and it was then already mentioned that “none of the Total Invisibles can be neutral.”

My concise definition of a Total Invisible would be the following:

“A white or black orthodox piece that stands somewhere on the board, but whose color, identity and whereabouts are not known. The initial position and sequence of moves of the solution must be legal. After all aspects of a Total Invisible are revealed, it becomes visible and turns into an ordinary piece.”

If you allow the Invisibles to be also neutral, then you need more moves to identify the White Invisibles that can give mate and exclude the neutral Invisibles that can prevent mate.
The solutions would be more diluted and would lose most of their charm.

Longer solutions would also mean more cooks, all the more so as Popeye doesn’t check problems with neutral Invisibles and can’t help you.

Thomas Maeder

January 3, 2021 22:08

BAD NEWS!
Testing problem 1572 with Popeye (and vice versa!) revealed (pun intended) a number of bugs in the Popeye implementation of TotalInvisibles.
For this problem, Popeye now also finds: 1…0-0-0 2.nPg2*f1=nQ nQf1*f4 3.nQf4-e5 Rd1-d4 # other TIs: g1(not Q or R), e3 (b[SP] or w[QRBP]) 1…nQh1-g1 2.f4*g3 0-0-0 3.nQg1-c5 Rd1-d4 # alternative move order for 1st solution

Thomas Maeder

August 20, 2024 23:20

I am making fixes to Popeye’s implementation of TotalInvisibles. Popeye now finds another solution of this problem (with a dual):
1…[+wRa1]0-0-0[d1=wR] 2.[+wTIg3]f4*g3 nQh1-g1 3.nQg1-c5 Rd1-d4 #
1…[+wRa1]0-0-0[d1=wR] 2.nQh1-g1 nQg1-c5 3.[+wTIg3]f4*g3 Rd1-d4 #
The other 2 TIs are on f1 and c2-c4. They may be black, but if they were of a type that disturbs the mate, they would also have disturbed the castling.

I don’t see anything in the definition above the diagram that forbids this.The link to the original definition mentioned in the JF definition no longer works, so perhaps it’s in there?
But if capturing invisibles if you don’t know for certain that there’s an invisible there is forbidden, then why is fxg3 a legal move?

2.nQh1xg1 is impossible because we don’t know there is a TI on g1: there may as well be a TI (of either colour) on f1.
fxg3 is legal because Pf4 is visible, so we can prove the P moved to g3 just by nplaying that move. Which in turn proves that there was a TI on g3. (This is a very normal TI tactic.)
A related question is whether a TI can capture another TI that we know is situated on a certain square. There have been some such problems published, but they violate the original TI rules so they must be called “TI Type 2″ or similar (I haven’t seen a full definition of this type – what is legal and what is not?).
For the original definition I can quote the award of the Sake tourney in Ohrid 2018 by the inventer Tadashi Wakashima:
” First, we must apologize for the ambiguity in the definition which caused confusion to some composers. In the Total Invisible system as well as the Invisible system, the easiest way to grasp the notion intuitively is to imagine what happens on the board where some pieces are invisible. Nothing happens there when a Total Invisible cantures an opposite Total Invisible or simply moves to somewhere. Thus, both cases should be written as TI–, not TIxa1 even when we know there is a Total Invisible on a1 (in other words, we cannot claim the capture of a Total Invisible without proof).”
So for example, if a visible white piece on f5 was just captured (by a black TI), so we know at this moment that there is a TI on f5, we still cannot play TIxf5. For we cannot prove that the TI moved to f5 rather than some other square when nothing is visible on the chessboard!

The most convenient way to imagine invisible pieces is (for me) as follows. Imagine that two players play the game (and see all the pieces), while a kibitzer is standing nearby and some pieces are invisible to him. Something happens (revealing the piece, declaring the mate etc.) only if the kibitzer can deduce for sure that this is what happened on the board.
Therefore, fxg3 is legal, as this move can be legally played on the board, and once the kibitzer sees the pawn moving diagonally, he concludes that this has to be a capture. But if he sees the queen moving from h1 to g1, he cannot possibly know whether this is a capture or not, and this is why a solution/cook cannot be based on the move nQh1*g1.

I agree with Bojan below, but let me also answer my own way:
No, we don’t prove that the TI is on g1 by playing nQh1-g1. It is still possible that there was a TI on g1 (which we just captured) OR that there is a TI on f1. In fact, we don’t prove anything by playing nQh1-g1.
The idea of Total Invisibles is to prove what really happened on the board just by what happens visibly. (We don’t always have to prove something exactly; for example, it is enough if there are still N possibilities but in all those cases there is a mate.)

As there are neutral pieces on the board, the TIs may be neutral (why not?) and then second solution does not work, as the final TI may be a neutral rook (not a white one).

Perhaps it should be stipulated that as only neutral pawn and queen are there the TI can only be that, not a neutral rook !!

> the TIs may be neutral (why not?)

Obviously because I won’t program them 🙂

I have published other originals with Total Invisibles in Problem Paradise 86, Apr-Jun. 2019 and it was then already mentioned that “none of the Total Invisibles can be neutral.”

My concise definition of a Total Invisible would be the following:

“A white or black orthodox piece that stands somewhere on the board, but whose color, identity and whereabouts are not known. The initial position and sequence of moves of the solution must be legal. After all aspects of a Total Invisible are revealed, it becomes visible and turns into an ordinary piece.”

If you allow the Invisibles to be also neutral, then you need more moves to identify the White Invisibles that can give mate and exclude the neutral Invisibles that can prevent mate.

The solutions would be more diluted and would lose most of their charm.

Longer solutions would also mean more cooks, all the more so as Popeye doesn’t check problems with neutral Invisibles and can’t help you.

BAD NEWS!

Testing problem 1572 with Popeye (and vice versa!) revealed (pun intended) a number of bugs in the Popeye implementation of TotalInvisibles.

For this problem, Popeye now also finds:

1…0-0-0 2.nPg2*f1=nQ nQf1*f4 3.nQf4-e5 Rd1-d4 #

other TIs: g1(not Q or R), e3 (b[SP] or w[QRBP])

1…nQh1-g1 2.f4*g3 0-0-0 3.nQg1-c5 Rd1-d4 #

alternative move order for 1st solution

I am making fixes to Popeye’s implementation of TotalInvisibles. Popeye now finds another solution of this problem (with a dual):

1…[+wRa1]0-0-0[d1=wR] 2.[+wTIg3]f4*g3 nQh1-g1 3.nQg1-c5 Rd1-d4 #

1…[+wRa1]0-0-0[d1=wR] 2.nQh1-g1 nQg1-c5 3.[+wTIg3]f4*g3 Rd1-d4 #

The other 2 TIs are on f1 and c2-c4. They may be black, but if they were of a type that disturbs the mate, they would also have disturbed the castling.

Shouldn’t it also find the cook 1…[+wRa1]0-0-0[d1=wR] 2.(+wTIg1)nQh1*g1 ?

No. 2.(+wTIg1)nQh1*g1 is not a legal move according to the TI rules.

I don’t see anything in the definition above the diagram that forbids this.The link to the original definition mentioned in the JF definition no longer works, so perhaps it’s in there?

But if capturing invisibles if you don’t know for certain that there’s an invisible there is forbidden, then why is fxg3 a legal move?

2.nQh1xg1 is impossible because we don’t know there is a TI on g1: there may as well be a TI (of either colour) on f1.

fxg3 is legal because Pf4 is visible, so we can prove the P moved to g3 just by nplaying that move. Which in turn proves that there was a TI on g3. (This is a very normal TI tactic.)

A related question is whether a TI can capture another TI that we know is situated on a certain square. There have been some such problems published, but they violate the original TI rules so they must be called “TI Type 2″ or similar (I haven’t seen a full definition of this type – what is legal and what is not?).

For the original definition I can quote the award of the Sake tourney in Ohrid 2018 by the inventer Tadashi Wakashima:

” First, we must apologize for the ambiguity in the definition which caused confusion to some composers. In the Total Invisible system as well as the Invisible system, the easiest way to grasp the notion intuitively is to imagine what happens on the board where some pieces are invisible. Nothing happens there when a Total Invisible cantures an opposite Total Invisible or simply moves to somewhere. Thus, both cases should be written as TI–, not TIxa1 even when we know there is a Total Invisible on a1 (in other words, we cannot claim the capture of a Total Invisible without proof).”

So for example, if a visible white piece on f5 was just captured (by a black TI), so we know at this moment that there is a TI on f5, we still cannot play TIxf5. For we cannot prove that the TI moved to f5 rather than some other square when nothing is visible on the chessboard!

So by moving Qh1-g1, you prove that the TI is on f1 and not on g1? That seems ….. weird.

The most convenient way to imagine invisible pieces is (for me) as follows. Imagine that two players play the game (and see all the pieces), while a kibitzer is standing nearby and some pieces are invisible to him. Something happens (revealing the piece, declaring the mate etc.) only if the kibitzer can deduce

for surethat this is what happened on the board.Therefore, fxg3 is legal, as this move can be legally played on the board, and once the kibitzer sees the pawn moving diagonally, he concludes that this has to be a capture. But if he sees the queen moving from h1 to g1, he cannot possibly know whether this is a capture or not, and this is why a solution/cook cannot be based on the move nQh1*g1.

I agree with Bojan below, but let me also answer my own way:

No, we don’t prove that the TI is on g1 by playing nQh1-g1. It is still possible that there was a TI on g1 (which we just captured) OR that there is a TI on f1. In fact, we don’t prove anything by playing nQh1-g1.

The idea of Total Invisibles is to prove what really happened on the board just by what happens visibly. (We don’t always have to prove something exactly; for example, it is enough if there are still N possibilities but in all those cases there is a mate.)