This site now hosts a complete archive of PDF versions of The Problemist Supplement, from 1992 to the current year. This magazine is aimed mainly at newcomers to chess composition, so it is recommended for those who want to find out more about chess problems. You can find the archive here.
Helpmates are different from all other kinds of chess composition. In many countries they
are called something like “cooperative mates”, since Black and White cooperate
to reach a position in which Black is mated. This is the most popular type of chess
problem. It doesn't make the mistake of trying to imitate the game. It just does its own
thing. The solver gets a feeling of being in control of both sets of pieces, something
which never happens over the board!
If you think it's perverse to abandon the idea of struggle, you'll have to take issue with
some rather good players. Former British Champions Jonathan Penrose and Bill Hartston have
both composed helpmates, and Bob Wade has recently become a keen solver of them.
Grandmaster Pál Benkö is a helpmate composer of the first rank. Oh, and do you recognise
this?:– 1.e4 e5 2.Nf3 Nc6 3.Bc4 Nf6 4.d4 exd4 5.0-0 Bc5 6.e5 etc. That's right, the Max
Lange attack, named after ... the inventor of the helpmate! (From now on we'll change to
problemists' notation – rather than N for knight – however.)
So how does it work? BLACK plays first and helps White to mate the black king in the
stated number of moves. The fact that Black is to play means that the number of moves is
the same for each side, which gives a good balance. Consequently, for neatness of
notation, Black's moves are written first, which may at first seem confusing. Thus for
example a two-move helpmate solution reads: 1.black move, white move 2.black move, white
mate! Well, I did say helpmates were something completely different ... and unless you're
the sort of person who buys a Norwegian Blue Parrot I guess you'll be able to cope. Have a
look at no. 1.
(1) Jacob Mintz
1st Prize, The Problemist, 1982
Helpmate in 3 (see text)
The solution runs 1.a1Q+ Se1 2.Qb2 Sc2 2.Qb5 Ra3#. Get the idea? Black helps
by blocking b5, while White gets his pieces into position for an economical
mate. The solver has to visualise the mate (often much harder than this!) and
then see how the two sides can intertwine their manoeuvres to reach it. The
move order must be unique, with no transpositions or alternative routes. Did
you notice why 1...Se3? and 2.Qb1? were no good?
Ok, that was the easy part. Now take the same problem, change the knight's
position to f5 and try again. Remember, Black starts. A bit trickier, this.
The previous solution 1.a1Q? etc. fails because even though the sides are
cooperating, the rules of chess still apply, so White cannot ignore the check,
and if he moves the king or rook to answer it he will no longer have time for
the other necessary moves. Aha, but now the knight is a file nearer the black
king, and so can usefully guard squares with a single step, thus giving the
rook greater freedom: 1.a1R+ Rc1 2.Rb1 Sd4 3.Rb4 Ra1#.
Now the quiz question: how many more parts do you think there are? (Hint: how many things
can a BP on the second rank do?) Well, for the next part we start with the knight on g7.
That may seem a long way away, but notice that for the first time we can now attack c5
after one move. Somehow I think you might have guessed the first move: 1.a1B, of course!
But can you see the mate? That's right: 1...Rb3 2.Bc3 Se6 3.Ba5 Sc5#. After that you just
KNOW there must be one more part. You may be surprised at where the knight starts this
time, though: c4. Now it is too close to reach the squares it used before! The solution to
this part is surprisingly hard to see, and you may already call yourself a helpmate solver
if you can find it within a few minutes. If you haven't worked out the first move by now
(1.a1S) you had better go back to reviving your Norwegian Blue, but what about the rest?
The natural thing is to try to find a useful square for the black knight to block, but the
fact is that it has already found one! The remaining moves are 1...Rc1 2.Kb3 Rb1+ 3.Ka2
That problem had twelve moves for each side, and only five pieces! Helpmates are often
delightfully simple and clear. Here's another helpmate in three (no. 2, this time
with all of six pieces!).
(2) H Ternblad
1st Prize, feenschach, 1954
Helpmate in 3, with set play (see text)
It shows a similar idea to the first one, but with a twist. It's a pretty
obvious idea to promote the f-pawn and mate on a8, and you may quickly see
that a rook promotion will do better than a queen, so as not to check the
black king too soon. What piece can block b2? You've got it, a bishop. But you
haven't solved the problem! Why not? What's wrong with the sequence
f7-a1B-f8R-Bb2-Ra8#? Nothing, except that it's BLACK's move, remember? So
please try again.
How on earth does Black use up three moves rather than two? The hint was in
the previous problem, of course! Try the OTHER two promotions: 1.a1S f7 2.Sc2
f8Q+ 3.Ka4 Qb4#, with a surprising pin.
By the way, the line which failed because of a missing first move for Black is
known as a set play.
One of the most attractive features of helpmates is the intricacy of the interplay. White
and Black are trying to cooperate, but it isn't easy! (Who said helpmates weren't like
(3) Peter Kniest
4th Prize, Kennst Du die Bibel?, 1966
Helpmate in 4
All you have to do in the next problem (no. 3) is get the BK to e4 and the
Rf1 to f4 and that's mate. Go on, then! That really is what happens, so what
are you waiting for? Keep those moves legal, mind! Watch the checks. If you
can do it without getting your knickers (sorry, KNIGHTS) in a twist, you may
notice that the two knights have changed places: 1.Kf2 (unpinning) Sg7 2.Ke3+
Sg3-f5+ (selfpinning) 3.Ke4 Sh5 4.Rf4 Sh5-g3#. That really is the only way to
do it. By the way, all the problems quoted here have been computer-tested, so
if you think you have found alternative solutions to any of them, think again!
Many (but not all!) helpmates are easy to solve, for one thing because once you have seen
a fair number of them you can usually spot the composer's idea. The artistic aspect of
chess composition requires that a problem (of any type) should have a discernible theme
and some kind of unity. (For more on these ideas see the wonderful book
Secrets of Spectacular Chess, details of which are in the book list.) We have seen
this thematic unity in the four parts of no. 1, which are connected together by the
idea of all the four possible promotions of the black pawn, in the same idea with a
mixture of black and white promotions in no.2, and in the exchange of places in no.3. An
experienced solver would have seen the idea of that last problem immediately. If you can
guess the theme of a problem you have a head start towards solving it, but some composers
try to hide the idea as best they can. Maybe they just like to puzzle, or maybe they are
composing for a solving contest. International solving contests always contain a helpmate
round, and it is not always the easiest! Two of our strongest competitive solvers, John
Nunn and the editor of this website, Michael McDowell, are both helpmate composers, but
even they have occasionally been known to have trouble with helpmates in solving contests.
No. 4 was used in a solving competition in Finland where at least one Solving World
Champion was present. A Finnish chess problem magazine reported that it “caused
trouble for a room full of experts”. Nevertheless with a few hints you might just
manage it. Let's see.
The first point is that all the pieces must be necessary. Some of them may just be there
to stop unwanted solutions (usually called ‘cooks’) but where possible only black material
is used for this purpose, so as to achieve the final position with as little mating force
as possible. There's a good chance, then, that four or five white units (possibly not the
king) will be used for the mate. That should suggest the file where the mate takes place.
The pawn on f5 looks as though it might be there to block a square, and the bishop is well
placed to block e4, all of which points to e5 for the black king. Where would the rook
guard as many as possible of the squares around e5? So how does the c-pawn participate?
And why is f4 guarded by two white pawns? What must the mating move be?
(4) C. J. Feather
British Chess Magazine, 1999
Helpmate in 4
What a lot of questions! Time for a few answers: The rook is best placed on d6, and
the only unit which can guard it there is the c-pawn, moving up to c5. That
accounts for three white moves, so the mate must be by a pawn capturing on d4
or f4. Can it be a capture of the knight on f4? No, if the bishop is blocking
e4 as planned, that will cause an unwanted check to the white king. So the
mate must be by exd4, which explains the presence of the black queen, the only
possible unit for the pawn's capture. So there we are: WRd6, WPc5, BKe5, BQd4,
BBe4 and exd4#. We even seem to have a spare black move! Well, try putting it
all together into a legal sequence (Black to play!) and see what happens.
There seem to be so many ways of getting the BQ to d4, but there can be only
one that works, or else the problem is unsound. This is probably what confused
the room full of experts, since a multiplicity of routes to a particular
square is often an indication to the solver that he is on the wrong track! But
it was the intention of the composer (known to be a joker, and suspected of
being a Monty Python fan) to cause precisely that confusion ...
Consider the possibilities. If you play Qg7-d4 in a single move, you must play it before
the black king moves and blocks the path. But that means that you cannot play Rd1-d6, for
which the d-file must be clear. So try going to d4 in two moves. If you play the queen to
d4 via f6 (or e5) you will once again find that either the black king gets in the queen's
way or the queen gets in the rook's way. If you play the queen to d4 via a7 you will find
that the white pawn has got in the way on c5. If you try going via d7 you will find that
the queen and rook want to move in opposite directions on the d-file and are thwarting
each other. If you play Qg7-h8(check!)-d4, you will have to waste time moving the white
king. If you play Qg7-a1-d4 you will find that when you want to move the rook, it is
pinned. If you play Qg7-c3-d4 you will find that when you want to move the c-pawn, the
queen is in the way. So what is left? 1.Qb2! c4 2.Ke5 c5 3.Be4 Rd6 4.Qd4 exd4#.
After that we had better have something simpler. Short helpmates often have several
(related) solutions, which increases their thematic interest but usually makes them easy
to solve. An attractive way to relate the solutions is to have the mating positions echo
(5) Matti Myllyniemi
Satakunnan Kansa, 1968
Helpmate in 2: 4 solutions
No.5 shows exactly the same mate four times, three shifted diagonally across
the board, and one reflected and rotated through 90 degrees! Solutions:
1.Kd5 Qe7 2.Qc4 Be4# &
1.Bc3 Qd6 2.Qb3 Bd3# &
1.Bb2 Qc5+ 2.Kb3 Bc2# &
1.Qg2+ Kh4 2.Qd5 Bd3#.
Doesn't it look easy to do? It isn't. Ensuring that the problem is correct (no transpositions,
no unwanted solutions) can be a real challenge for the composer. But the result is very
Finally, three problems for you to try to solve on your own. Or nearly. First the
diagrams, then some hints (click the 'show hint' button), and finally the solutions
(click the 'show solution' button). If you want more, you can find plenty in both the
(6) Christopher J. A. Jones
British Chess Magazine, 1997
Helpmate in 2: 2 solutions
Get a white rook to the f-file.
1.Bc1 Rd1 2.Qb1 Rf1# &
1.Se6 Rd7 2.Qd5 Rf7#.
The composer is the helpmates editor of The Problemist and the leading British
composer of helpmates, having been placed 10th in the recent world
(7) Pal Benkö
Helpmate in 2 (b) wB from c1 to f1
A very paradoxical problem. Try starting with the least helpful-looking move in each
a) 1.Sxd6 Kd3 2.Ke5 Bb2# (b) 1.Qxd6+ Sd5+ 2.Ke6 Bh3#
Amazingly, there is no way to solve either part of the problem using the strongest white
Pieter ten Cate
2nd Prize, Diagramme und Figuren, 1965
Helpmate in 6: 2 solutions
In one solution you promote the pawn (but how do you get rid of the knight?) and in
the other you don't (so where will the BK have the least room for escape?)
1.Kd8 Ke6 2.Ke8 g4 3.Kf8 g5 4.Kxg8 g6 5.Kf8 g7+ 6.Ke8 g8Q/R# &
1.Kd7 Kf8 2.Ke6 Sh6 3.Kf6 g4 4.Kg6 g5 5.Kh7 g6+ 6.Kh8 Sf7#.
The kings have to back off to allow each other to manoeuvre.