I solved this puzzle in 25 moves.
Using r for the R in Martin and b for blank...
The letters before each move are
(locations of other coins/blocked locations)
(GDER)
(GDE/ATI) RbNr
(GEr/AIN) DT
(GrT/AND) EIRb
(Grb/ANR) TDIE
(GrE/ANI) bR
(GER/AIb) rN
(ERN/Ibr) GAM
(ERM/IbA) Nr
(EMr/IAN) Rb
(Mrb/ANR) EIDT
(MrT/AND) bRIE
(MTE/ADI) rNb
(MTb/ADNR) EI
(MTIb)
Joseph DeVincentis

I found it easier to rearrange the graph like this:
E

T  D  I  r
 /  
M  A  R  N  X

G
where r is the lowerrow R and R is the middlerow R, and X is the empty
circle.
solution:
DT
rXNR
EIrX
TDIE
Xr
RN
GAM
NR
rX
EIDT
XrIE
RNX
EI
Thanks,
___
X Trygve Flathen

1 R 0
2 0 N
3 N R
4 D T
5 E I
6 I R
7 R 0
8 T D
9 D I
10 I E
11 0 R
12 R N
13 G A
14 A M
15 N R
16 R 0
17 E I
18 I D
19 D T
20 0 R
21 R I
22 I E
23 R N
24 N 0
25 E I
Greetings from Barcelona, two days before our book's day.
Jordi Domθnech

Hi, I've never written in before, but I browsed by the site earlier and
saw this puzzle, looked interesting so I graphed it with less tangled
nodes and solved in a few minutes, figured I'd send in my solution. The
crux was figuring out what the board needed to look like to move M
through A to G, after that, it almost solves itself. Great puzzle. :)
Keep 'em coming!
Sincerely,
Michael Ferry
I'll call the left R, R1, and the right R, R2, and the blank spot I'll
call _.
Coin moves: (one row means you keep moving the same coin)
D > T
R2 > _ > N > R1
E > I > R2 > _
T > D > I > E
_ > R2
R1 > N
G > A > M
N > R1
R2 > _
E > I > D > T
_ > R2 > I > E
R1 > N > _
E > I
Less tangled nodes looked something like this:
G M
\ /
A  D
\ /
R T 
 
N  I  E
 
()  R

use * for the empty circle
although there are two R's,
it is clear which is used
DT
R*NR
EIRN
TDIE
*R
RN
GAM
NR
R*
EIDT
*RIE
RN*
EI
Rolan Christofferson

Debashish Majumdar sent a solution.

Ed
I think the shortest solution to R.A. Hearns MARTIN GARDNER puzzle is 14
steps as follows:
Let X represent the Blank cell; then the sequence is
R  X  N  R, D  T , E  I  R  X , T  D  I  E , X  R , R  N ,
G  A  M ,
N  R , R  X , E  I  D , R  T , X  R  I  E , R  N  X , E 
I
Although there are two 'R's in the puzzle, the above sequence should be
fairly easy to follow
Rgdz
Pete Kogel

Hi Ed,
I solved the Robert Hearn's puzzle in 27 moves. 'B' stands for "blank"
and although there are two R's, the moves should be unambiguous given
the edges.
R  B
B  N
D  T
N  R
E  I
I  R
T  D
R  B
D  I
I  E
B  R
R  N
G  A
A  M
N  R
R  B
E  I
I  D
D  T
B  R
R  I
I  E
T  D
R  N
D  T
N  B
E  I

Nathan Stohler

R1 is the R on the left of the puzzle, R2 is on the right.
R2Blank
DT
BlankNR1
EIR2
TD
R2Blank
DIE
BlankR2
R1N
GAM
NR1
R2Blank
EID
BlankR2
R1N
DT
NR1
R2IE
R1NBlank
EI
Warren Phillips

let empty space = S
R  S  N  R
D  T
E  I  R  S
T  D  I  E
S  R
R  N
G  A  M
N  R
R  S
E  I  D  T
S  R  I  E
R  N  S
E  I
Hope that works! :)
Yours,
Shahrul Mt.Isa

Call the bottom right circle " R ", to distinguish it from the r on the
left. Also, the blank space is "  "
The first character in each line is the starting position of the move,
the subsequent characters are destinations.
Moves are as follows:
dt
Rnr
eir
die
R
rn
ga
(We now have coins at a,n,e, and R)
am
nr
R
eidt
Rie
rn
ei
 Matt Elder

I couldn't work this out in my head, so I did the next best thing and
wrote a program to solve it ;)
I replaced the second "R" (the one on the far right) with "Z", and used
"O" for the blank node, so that each node would have a unique letter to
identify it. At each step, I list the positions of the four coins (with
the order preserved), followed by the next move.
1. GDEZ D>T
2. GTEZ Z>O
3. GTEO O>N
4. GTEN N>R
5. GTER E>I
6. GTIR I>Z
7. GTZR Z>O
8. GTOR T>D
9. GDOR D>I
10. GIOR I>E
11. GEOR O>Z
12. GEZR R>N
13. GEZN G>A
14. AEZN A>M
15. MEZN N>R
16. MEZR Z>O
17. MEOR E>I
18. MIOR I>D
19. MDOR D>T
20. MTOR O>Z
21. MTZR Z>I
22. MTIR I>E
23. MTER R>N
24. MTEN N>O
25. MTEO E>I
MTIO
Interestingly, this is almost palindromic. The 3rd through 12th
moves are the mirror image of the 15th through 24th moves.
Bill Daly

R2ONR1
DT
EIR2O
TDIE
OR2
R1N
GAM
NR1
R2O
EIDT
OR2IE
R1NO
EI
Jim Hawkins

Note: "RL" is "R" on the left, "RR" is "R" on the right, "SP" is space.
Solution: RR  SP, SP  N, N  RL, D  T, E  I, I  RR,RR  SP,T  D,D  I,I  E, SP  RR, RL  N, G  A, A  M,N  RL, RR  SP, E  I, I  D, D  T, SP  RR, RR  I, I  E, RL  N, N  SP, E  I
Regards J.HAYES.

My students love it when I find little puzzles (usually suggested by you)
for them to try. Well, some of them do  the rest think I'm just a weird
math teacher.
A few of us solved it.
I asked one of my students to write out his solution and this is what he
gave me:
DT
R20NR1
EIR20
TDIE
0R2
R1N
GAM
NR1
R20
EIDT
0R2IE
R1N0
EI
R1 is the lefthand R
R2 is the righthand R
0 is the blank space
This is either 13 moves or 25 moves, depending on how you count them. I'm
not sure if it is the shortest solution.
Jeremy Galvagni
ps. How much extra credit should this be ;)

R2 : R on bottom
R1 : R in middle
R2 > Empty
Empty > N
N > R1
D > T
E > I
I > R2
T > D
R2 > Empty
D > I
I > E
Empty > R2
R1 > N
G > A
A > M
N > R1
R2 > Empty
E > I
I > D
D > T
Empty > R2
R2 > I
I > E
R1 > N
N > Empty
E > I
Clark Cooper

Shortest possible answer seems to be 25 moves, and there are 16 shortest
paths.
To ease possible complications, we refer to the R on the middle row by lower
case, that on the bottom row by upper case. The blank is referred to by an
underscore.
Let us consider a sequence S of moves from GDER to MTI_ which takes the
shortest possible time. We will use (without proving) the obvious fact that
it is not possible that at two different stages in S the coins are on the
same spots.
Theorem 1: At some point in S, the coins are on GNER and the G coin moves
through A immediately to M.
Proof:
There is some last stage when there is a coin on G. Immediately after this
stage there must be a coin on A.
If there is a coin on A, there cannot be a coin on M,T,r,G or D. Thus the
other 3 coins must be on I,N,E,R or _.
I is next to three of the others, and thus cannot contain a coin. _ is next
to 2 of the others, and can also not contain a coin.
Therefore the other 3 coins are on NER. Therefore the move goes from
GNER>ANER.
Similarly at the first stage when there is a coin on M, the move has gone
ANER>MNER.
Since S only goes through 1 ANER stage, S has a sequence GNER>ANER>MNER.
QED
Corollary 2: S consists of the shortest possible sequence to GNER, then the
transfer GNER>ANER>MNER, then the shortest possible sequence from MNER.
Before GNER, the G coin does not move. After MNER, the M coin
(the same coin, as it happens) does not move.
Proof:
QED
So we split into finding the shortest sequence to GNER (S_1) and the
shortest sequence from MNER (S_2), noting that the Gcoin or Mcoin does not
move, and hence the spots G,A,M are out of bounds for moves in either
sequence. (A as there will be something in the [ahem] Gspot or Mspot and
G,M as the only moves go to A).
Theorem 3: S_1 contains the sequence GTrE, GTrI, GTrR.
Proof:
When the E coin first moves, it moves to I. Therefore the spots N,D,R are
empty, and the other 2 coins (ignoring the G coin for this theorem) must be
in
2 of T,R,_. Note that none of these spots is adjacent to any spot (other
than the forbidden A) which is not adjacent to I. Therefore the other coins
cannot
be next to move. Therefore the Icoin must move. It won't move back to E,
so it must move to either D,N or R.
The coin that started off at D MUST be at T: Without going through the
forbidden spots A (because of G) and I (because of E), it cannot move
anywhere
else, and we have already seen that it cannot still be at T.
Therefore the Icoin cannot move to D. It also cannot move to N, as this is
adjacent to both of r and _, one of which must have the original Rcoin in
it.
Therefore it must move to R, and the original Rcoin must be at r.
QED
Corollary 4: S_1 (and hence S) starts in an equivalent way to
GDER,GTER,GTE_,GTEN,GTEr.
(Equivalent means that the shifting of the Dcoin to T can happen at any
stage during the Rcoin's path to r  giving 4 solutions this far)
Proof:
We've seen in the proof of theorem 3, that the Dcoin moves to T and the
Rcoin moves to R before either of the other coins move. This takes a
minimum of
4 moves.
QED
Theorem 5: S_1 ends with the sequence GrI_,GrE_,GrER,GNER
Proof:
Similar to Proof of Theorem 3. Within S_1, consider when there is last no
coin on the Espot. It must be on the Ispot. The other 2 must therefore
be on 2 of T,r,_. If one is on T, it can never get to N or R without moving
G or E, both of which are forbidden. The other 2 must therefore be on
r and _. Therefore within S_1, it will go GrI_ > GEI_. Then we must still
get to GNER as soon as possible. Clearly it will take at least 2 moves,
as 2 coins are on spots they shouldn't be on. Furthermore the only way to
do it in 2 moves is to go via GrER.
QED
Theorem 6: The shortest routes from GTrR to GrI_ are 3 moves long, and there
are 2 options: GTrR,GTr_,GDr_,GIr_ or GTrR,GDrR,GDr_,GIr_.
Proof:
If either G or r move, then the solution will take at least 4 moves (1 to
initially move each coin that moves, and 1 to push a coin into the
G or r spot).
Therefore G and r are fixed, and the only valid squares are TI_DER. T is 3
squares from _ and therefore T must go to I (2 moves away), and R must
go to _ (1 move away). R must move to _ before T finally arrives at I.
QED.
Corollary 7: S_1 is 12 moves long, and there are 8 versions. 1 such version
is GDER,GTER,GTE_,GTEN,GTEr,GTrI,GTrR,GTr_,GDr_,GIr_,GRe_,GrER,GNER
Proof:
Follows from Corollary 4, Theorem 3, Theorem 6 and Theorem 5 (in that order
along the route).
QED
Now we look at S_2 (from MNER to MTI_ not moving the M and not using M,A or
G)
Theorem 8: S_2 starts MNER, MrER, MrE_, MrI_, MrD_.
Proof:
As in theorem 5, before E moves, spaces T,D cannot be used, and therefore
the other 2 coins must get to positions r and _. This will
take at least 2 moves and can be done so in precisely 1 way, as N must be
cleared before _ is filled. This gets you to MrE_. This was
in order for E to move, bringing you to MrI_. Now r and _ are not adjacent
to anything that is not adjacent to either I or M, and so I
must be the next coin to move, and it must move to D.
QED
Theorem 9: S_2 ends MTIr, MTEr, MTEn, MTE_, MTI_
Proof:
In order for the rcoin to leave the rspot, it must go to N. Therefore
I,R,_ must be empty. At most 1 coin can be in the T or D spots,
and therefore the other coin must be in the Espot. It must have moved
there from the I spot, and the other coin cannot have been in the D spot,
it must have been in the Dspot. Therefore, the situation must have been
MTIr,MTEr. The r coin is then 2 squares away from any finishing square, and
the E coin is 1 square away. Therefore at least 3 moves are needed to
complete the sequence. E's move to I must be the final one, as the N and I
squares
cannot be simultaneously full.
QED
Theorem 10: There are 2 paths of shortest length from MD_r to MTIr:
MD_r,MT_r,MTRr,MTIr or MD_r,MDRr,MTRr,MTIr.
Proof:
D is 1 move away from a destination and _ is 2 moves away from its nearest
destination. Therefore 3 moves are needed, and you
can only move _ or D. Furthermore the only destination within 2 moves of _
is I. Therefore it must go to I. Also it must go via R, as N is
a forbidden neighbour of r. D must go to T, and it must do so before the
later part of _'s path to I.
QED
Corollary 11: There are 2 possible S_2 paths:
MNER,MrER,MrE_,MrI_,MrD_,MrT_,MrTR,MrTI,MrTE,MnTE,M_TE,M_TI
and
MNER,MrER,MrE_,MrI_,MrD_,MrDR,MrTR,MrTI,MrTE,MnTE,M_TE,M_TI.
Proof:
Theorem 8, Theorem 10 and Theorem 9 in that order.
QED
Corollary 12: There are 16 possible S paths, and one such is:
GDER,GTER,GTE_,GTEN,GTEr,GTIr,GTRr,GT_r,GD_r,GI_r,GE_r,GERr,GERN,AERN,MERN,M
ERr,ME_r,MI_r,MD_r,MT_r,MTRr,MTIr,MTEr,MTEn,MTE_,MTI_
Proof:
Corollary 7 and 11
QED
Luke Pebody

I think I have a solution. In the following description I won't
distinguish the two R's since their moves are different. Moves on the same
line belong to a single coin.
4 : R > _, _ > N, N > R,
2 : D > T,
3 : E > I, I > R, R > _,
2 : T > D, D > I, I > E,
3 : _ > R,
4 : R > N,
1 : G > A, A > M,
4 : N > R,
3 : R > _,
2 : E > I, I > D, D > T,
3 : _ > R, R > I, I > E,
4 : R > N, N > _,
3 : E > I.
Wang Zirui

Phew! 25 moves! Excellent puzzle.
Solution:
R0NR
DT
EIR0
TDIE
0R
RN
GAM
NR
R0
EIDT
0RIE
RN0
EI
Thanks and keep up the good work!
ttfn,
Mark Ingram

Since the two Rs do not have any edges to a common vertex, the following is
unambiguous:
D>T
R> >N>R
E>I>R>
T>D>I>E
>R
R>N
G>A>M
N>R
R>
E>I>D>T
>R>I>E
R>N>
E>I

Trevor Green