Jubilee tourney Tadashi Wakashima - 50 C 30.4.2003

h2# - Theme: Switchback
Problems: 85
Authors: 34

First of all, let me express my heartfelt thanks to all the participants. Their contributions are a real delight to me. And my special thanks goes to Kohey Yamada, who successfully directed this tourney.
I started to devote my time and energy to chess problems about ten years ago. At that time, I never dreamed that I could make friends with so many problemists all ower the world. Their everlasting friendship is what sustains me in my activities as a problemist.

The general quality of the entries is fairly good, and I ranked them as follows:
Section A Orthodox

Jorge Lois
Jorge Kapros
1st Prize
Michel Caillaud
2nd Prize
h2#                                   9 + 9
h2#                               6 + 9
B: Pe3 -> e4
1.Kd5 Sxb5 2.Kc5 Sd6#
1.Qxa7 Rxa4 2.Ra5 Rxa5#
1.Rxa5 Bb8 2.Qa7 Bxa7#
1.Kd4 Sb6 2.Kc5 Sc4#

A fine example of "the
Helpmate of the Future"
which is currently in
fashion, and these four
solutions give as a strong
sense of unity. The drawback
here is that black has not
much choice, but perhaps
that cannot be helped with
this tempo motivation.
The joint authors attached
the diagram below by Almiro
Zarur for comparison. and
altough the matrixes certainly
look similar, Iagree with their
judgement that H02 different
solutions and thus can be
considered as an original
B: a1=b1
1.Bh3 Bxd6 2.Rg4 Bf4#
1.Ra4 Sxe6 2.Bb4 Sc5#

A well-known hideaway
mechanism, but artistically
blended with white's line
opening and closing
switchbacks, it has an
optically pleasing efect. One
only wonders it this can be
done with fewer pieces ...
Almiro Zarur
Guaratingueta CC
8th TT 1965-66
2nd Prize
h2#                                   6 + 4
** 2 solutions
1... Bb8 2.Ba7 Bxa7#
1... Rxa4 2.Ra5 Rxa5#
1.Rxa5 Bb8 2.Rb5 Bd6#
1.Bxa7 Rxa4 2.Bb6 Rc4#
Michel Caillaud
Special Prize
Jorge Lois
Jorge Kapros
1st Honourable Mention
h2#                                   3 + 15
h2#                               8 + 9
1.Sa6+ Ka8 2.Bb8 Qb7#

"Where is switchback?" you
may ask, but first let us
consider how this position can
be attained. Black I's made 13
captures, which explains all
captures by black. The only
missing black piece is Q. And
now, what is white's last move?
It cannot be 1.Pb7xQc8=Q so is
must be 1.Ka8-b8. After that,
the retro play can be easily
deducted like this:
1.Sa6-c7+ 2.Pb7xQc8=Q
(1.Pc7-c8=Q ? Illegal !)
Bb8-d6+ 3.Bd6-b4.
So what have we got here?
At the first glance, it seems
that there is no switchback at
all in the solution, but after
taking back four half-moves
before this position, it turns
out that there are actually
four switchbaks by four
different pieces in the solution!!

Totally amazing conception,
and it stands out among all the
entries in its originality. The
only problem for the judge is
whether it is classified as an
orthodox or a fairy. Taking it as
an orthodox may not do justice
to other orthodox entries, and I
decided to give it Special Prize.
Anyway this problem is by far
the most original entry in this
tourney, and I thank the author
for this wonderful birthday
2 solutions
1.Rg2 Bxg3 2.Rh2+ Rh4#
1.Bd7 Re7 2.Be8+ Bf7#

Many participants tried this
unpin mechanism by black,
but this one is the best in its
graceful setting using only
RBPs. In addition white's B/R
replacing R/B is a nice
Comel Pacurar
2nd Honourable Mention
Chris J. Feather
3rd Honourable Mention
h2#                                  5 + 12
h2#                               7 + 8
2 solutions
1.Se4 Sg6 2.Sd6+ Se4#
1.Sg6 Se4 2.Se7+ Sg6#

Here again we have black's
unpin move, but this time it
is interception instead of
withdrawl. The most
interesting point is black's
second move. Black wants
to move Qc7 away and
unguard the squares e5/g7,
but unfortunately, Q has no
decent hideaways. Therefore
he needs to shut off the
Q-lines by S switchbacks.
The symmetry certainly
detracts a little, but overall
this is a good composition.
Umnov effect also leaves a
nice aftertaste.
2 solutions
1.Bg4 Bxf7 2.Bf3+ Bd5#
1.Sg5 Sxf5 2.Sf3 Sg7#

Black's half-pinned pieces
move twice to get behind
white's pinning piece (Rf4
in this case). The idea looks
new to me, but compared to
this, switchbacks by white
are rather poorly presented
and their unity is imperfect.
Jorge Lois
Jorge Kapros
1st Commendation
René J. Millour
2nd Commendation
h2#                                   4 + 12
h2#                               9 + 6
2 solutions
B: Pe3 -> c5
1.Se6 Rxd6 2.Sc5 Rd4#
1.Se4 Rxf4 2.Sc5 Rf6#
1.Sd5 Rxd6 2.Se3 Rf6#
1.Sf5 Rxf4 2.Se3 Rd4#

Again black's half-pinned
pieces move twice to arrive
the destination squares, but
this time, theoretically
speaking, two half-moves are
not necessarily needed. So
there are still further spaces
to treat this idea. The
symmetry here is too strong
and obviously a minus.
2 solutions
1.e1R fxg8Q 2.Re2 Qf7#
1.e1B a8Q 2.Be2 Qa7#

Switchback by pawn with
promotion can be an interesting
idea (for good example, please
see Special Prize in this section).
Here in this problem, is author
exhausts the possibility of such
pawn switchbacks in orthodox
H#2 (namely, Q/R/B promotions).
The effort certainly deserves
honorable mention, but it is the
great pity that interest lies solely
in the realization and we can find
no move with attractive
motivations. One feels that further
investigation is needed in this area.
Jorma Pitkänen
3rd Commendation
Christer Jonsson
4th Commendation
h2#                                   3 + 8
h2#                               6 + 7
6 solutions
1.Rxd1 Ra1 2.Rh1 Rxh1#
1.Sbxd1 Ra1 2.Sb2 Rxh1#
1.Qxd1 Ra1 2.Qd2 Rxh1#
1.exd1B Ra1 2.Be2 Rxh1#
1.Bxd1 Ra1 2.Bb3 Rxh1#
1.Scxd1 Ra1 2.Sc3 Rxh1#

A task six switchbacks. Do
I need anything further to
add ?
2 solutions
1.Ba1 Bg8 2.Sxb4 Bh7#
1.Bxg5 Rg8 2.Bf6 Rg1#

This has absolutely no unity
between two solutions except
that one has black's switchbacks
and another white's. Amazing in a
sense. I believe this is a deliberate
joke on the part of its author. It
makes me wonder... and say under
my breath, "To hell with unity!"

Section B: Fairies

Harry Fougiaxis
1st Prize
Juraj Lörinc
2nd Prize
h2#                                   7 + 8
h2#                               9 + 7
B: a3 R -> B
Rook-Locust 1+2
Bishop-Locust 2+3

1.Ra6 Sg5 2.LRxc2-c1 Sh3#
(2... Sge6 ? 3.BLxe6-d5 !)
1.Bc1 Sg6 2.LBxb5-a6 Sf8#
(2... Sgf4 ? 3.RLxf4-e4 !)
Anticipatory line closings
with fairy specific blocks on
Locust capture squaresa6/c1.
White switchbacks mates with
fine dual avoidance. Unpins to
create white batteries,
excanging roles of white's RLb5
BLc2 and black's RLc5/BLd3.
Everything is just perfect.
A worthy first prize winner.
B: CGb6 - > b8
Grasshopper 4+0
Contra Grasshopper 2+0
Non-stop Equihopper 0+3
1.Sf5 Sd6 2.EQf7-b5 Sc4#
1.EQf1-f5 Sd5 2.EQf3-b7 Sc3#
This one deserves a good
comparison to the first prize
winner above. White's first
moves unpin NEf7/NEf3 and
provide hurdles for them. In
their turn. both of black's
second moves selfpin Sb4
along the lineCGb6(b8). A fine
geometry, and the construction
is also superb. In my opinion,
the only weakness is in black's
first moves. A certain unity is
acheived by arrival on the same
square f5, that is for sure, but
one wishes they have more
thematical motivations other
than mere unguard and indirect
line opening by CGg6.
Michal Dragoun
3rd Prize
Michal Dragoun
1st Honourable Mention
h2#                                   5 + 5
h2#                               8 + 12
2 solutions

1.Qf3 gxf3 (Qf1) 2.Qxf3 (Pf7)
fxe8S (Bf1)#
1.Rh3 gxh3 (Rh1) 2.Rxh3 (Ph7)
Fairy condition can do wonders.
You can achieve what you cannot
do in orthodox problems. I think
that is what fairies are all about.
Here is a good example: blacks
switchbacks twice in three half-
You might say this is too
easy, and I must admit that this
has no depth whatever compared
to other two prize winners, but the
gist of the matter is to dream
an impossible dream. Certainly
Circe families are a perfect fit for
B: Ph3 - > h5
B: Pg5 - > h5
Anticirce Type Calvet
1.Qxb5 (Qd8) exd8Q (Qd1)
2.Rxf5 (Ra8) Qxd4 (Qd1)#
1.Rxf5 (Ra8) bxa8B (Bf1)
2.Bxe5 (Bf8) Bxc4 (Bf1)#
1.Bxe5 (Bf8) exf8 (Bc1)
2.Qxb5 (Qd8) Bxf4 (Bc1)#
Three pieces QRB along the pin
line constitute the black cycle.
Although the position is too
heavy, this idea is wonderfully
realized. My only contention is
that switchbacks are too
obvious by definition and have
no thematic contents. That is
exactly why I placed this lower
than the third prize winner by
the same author in spite of
being a better problem.
René J. Millour
2nd Honourable Mention
Michal Dragoun
Dieter Müller
1st Commendation
h2#                                   9 + 4
h2#                               7 + 5
2 solutions
Imitator c7
1.a1R (Ic6) hxg8Q (Ib7)
2.R1a2 (Ib8) Qh7 (Ic7)#
1.axb1B (Id6) e8Q (Id7)
2.B1a2 (Ic8) Qe7 (Ic7)#

1.Rb3 (Id7) ? e8Q (Id8)
2.Ra3 (Ic8) Qe7 (Ic7)+
3.a1Q (Ic6) !
1.Rb3 (Id7) ? hxg8Q (Ic8)
2.Ra3 (Ib8) Qh7 (Ic7)+
3.axb1Q (Id6) !
1.a1Q (Ic6) ? hxg8Q /Ib7)
2.Qa2 (Ib8) Qh7 (Ic7)+
3.Qxb1 (Id6) !
1.axb1Q ? (Id6) e8Q (Id7)
2.Qa2 (Ic8) Qe7 (Ic7)+
3.Qa1 (Ic6) !
Strategic switchbacks with
multiple promotions by three
pawns. The Imitator on c7 also
performs an additional
switchback if you consider it as
a fairy piece rather than a fairy
condition. I appreciate this
tremendous effort, but its overall
impression is as mechanical as
that of construction problems.
B: Pc6 - > f5
Castling Chess
Grasshopper 2+2
1 0-0 (Kc4-Pd5) dxc6
2.0-0 (Ke6-Gd5) Sd4#
1.0-0 (Ke4-Pe5) exf6
2.0-0 (Ke6-Gd5) Sc7#
Using Castling Chess, you can
have King's hopping switchback.
That is about all this problem
has in its content. It surely
attracts solvers' fancy, but one
wants much more.
János Buglos
2nd Commendation
György Bakcsi
3rd Commendation
h2#                                   9 + 7
h2#                               3 + 4
B: Bc8 - > c4
1.Kxg6 Sh6+ 2.Kf5 Rxg8#
1.Kxe6 Se7+ 2.Kf5 Bxg8#
Black king clears the paths for
white's RB. Discovered checks
secure the place for king's
switchbacks by self-paralization.
A lovely little affair.
B: a1=b1
1.Rc6 bxc6 (bRa8) 2.Rc8 Se6#
1.Bd6 cxd6 (bBf8) 2.f6+
Sxf6 (bPf7)#
Why not? I love the humor of
this, and there is no harm in