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The Bricks of Amenhotep

Few people are aware that the origin of high-card ranks is credited to Pharaoh Amenhotep IV. In his palace were 16 bricks of four different colors and sizes. Etched on each large brick was a circle depicting the sun god Aten; the second largest bricks had etchings of Amenhotep I to IV; the next largest, their four wives; and the smallest, their sons. Each year Amenhotep IV would waive taxation to any landowner who could arrange the bricks in a diagram (usually enhanced by pawns) to satisfy his cryptic whim.

S win 6 S 2
H 3
D
C
S
H
D 3
C 2
Table S
H 2
D
C 3
Tax Waiver
Form 1040

S 3
H
D 2
C

“With shovels in control, and southerly instigation, hands along the Nile must render oppositional forces brickless. Landlords, place your bricks!”

© 1339 BC Amenhotep IV

Distribute the 16 high cards (A-K-Q-J of each suit) to create a six-card ending in which North-South can win all the tricks against any defense with spades trump and South to lead. Many solutions exist, so the goal is to accomplish it with (1) fewest N-S HCP, (2) fewest South HCP, and (3) fewest West HCP, in that order of priority. To be legal at bridge, the ending must have at least three void suit holdings.

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Tim Broeken Wins!

In April-May 2011 this puzzle was presented as a contest, with 84 participants from 24 locations. Thanks to all who entered, and congratulations to the 18 who found the optimal solution, and the 13 others who came close. Solvers are ranked by the fewest HCP (N-S, South, and West in that priority) with remaining ties broeken, er, broken by date and time of entry.

Tim Broeken was the first in the brick yard, as well as the first double winner in this contest series. Do you suppose he bought his bricks at the Yarborough Fair?

RankNameLocationN-SSW
1Tim BroekenNetherlands14512
2Zla KhadgarOhio14512
3Charles BlairIllinois14512
4Edouard BonnetFrance14512
5Reint OstendorfNetherlands14512
6Gareth BirdsallEngland14512
7David BrooksAustralia14512
8Dan DangBritish Columbia14512
9Audrey KuehEngland14512
10Jonathan MestelEngland14512
11Hendrik NigulEstonia14512
12Jonathan BussOntario14512
13Simon CreaseyEngland14512
14Pavel StrizCzech Republic14512
15Ufuk CotukEngland14512
16Julian WightwickEngland14512
17James LawrenceEngland14512
18Nick JacobNew Zealand14512
19Alberto GioaItaly14513
20Dan BakerTexas14513
21Jeffrey TsangPennsylvania14513
22Jacco HopNetherlands14811
23Radu VasilescuPennsylvania14811
24John ReardonEngland14812
25Paul NelsonCalifornia14812
26Thomas HauklandNorway15510
27Jon GreimanIllinois15510
28Jonathan FergusonTexas15612
29Jim MundayCalifornia15710
30Colin SchlossPennsylvania15710
31Richard SteinCalifornia15813

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Solution

A change of pace, as this amenipotent challenge seems more like Sudoku than bridge; but then, bridge wasn’t invented yet. The Origin of Bridge came 22 years later thanks to Amenhotep IV’s son, better known as Tutankhamun. (Not all historians agree, but that’s my story and I’m stickin’ to it.) Indeed, some people were disappointed to find the solution based on a simple crossruff, rather than some amenruffensqueeze. To them, I would point out that we’re talkin’ about bricks here, not precious gems.

Jonathan Mestel: Ra, Ra, PavCo! I was hoping for a sacred crocodile coup.

The most obvious requirement is that N-S must have the S A, as even the High Priest of Amun couldn’t overcome that loser; and further, it should go to North to minimize South’s HCP. The next idea might be to give E-W the strongest capturable trump holding of S K-Q, each a mummy, er, stiff. Several respondents went this route on their first attempt (superseded by a better entry) so I’ll attribute this first example to the most feared mummy of all time, limping off into the night:

S win 6 S A 2
H K Q J 3
D
C
Trick
1. S
2. N
W 2 L 0
Lead
S 3
H K
2nd
Q
A
3rd
A
S J
4th
K
C 2
S Q
H
D K 3
C K Q 2
Table S K
H A 2
D A
C A 3
Kharis
Tomb KV13

S J 3
H
D Q J 2
C J

The above HCP counts of 15-5-10 (NS-S-W) total one point fewer than the winning solution of 14-5-12, but of course lose out in the stipulated priority.

Another way of achieving the same HCP counts is to give N-S all the trumps and minimal HCP elsewhere. Put it all together and you have Norwegian Wood, better known as a crossruff:

S win 6 S A K 2
H Q J 3
D
C
Trick
1. S
2. N
W 2 L 0
Lead
D 2
H 3
2nd
3
2
3rd
S 2
S 3
4th
K
K
S
H K
D Q 3
C K Q 2
Table S
H A 2
D A K
C A 3
Thomas Haukland
Norway

S Q J 3
H
D J 2
C J

Declarer could even succeed in the above layout if he led a trump to dummy.

Jim Munday: Even Fritz could bring this one home. If I find the correct solution can you waive taxation for me this year?

Sure, just open an account in any PavCo off-shore bank.

Bricklaying by the numbers

In theory, the minimum HCP for N-S is 14, since the S A must be held, and the weakest the remaining honors could be is four jacks and three queens. As for South, holding four jacks would mean no void suit, rendering the 14 HCP impossible, so his theoretical minimum is 5. The remaining 26 HCP (three aces, four kings and a queen) dictate West’s theoretical minimum as 11 (three kings and a queen). All but the last stage can be achieved (West must settle for 12) as the winning entry shows:

S win 6 S A Q 2
H Q J 3
D
C
Trick
1. S
2. N
3. S
4. N
W 4 L 0
Lead
D 2
H 3
D J
H J
2nd
3
2
C 2
C 3
3rd
S 2
S 3
S Q
S J
4th
K
K
A
A
S K
H A K
D 3
C Q 2
Table S
H 2
D A K
C A K 3
Tim Broeken
Netherlands

S J 3
H
D Q J 2
C J

West has no effective defense as declarer crossruffs. If he pitches a heart on the second diamond preparing to overruff, the North hand becomes high and the last trump is drawn. If he pitches a club (as shown), South safely ruffs a heart to establish the H Q. South next leads the good D Q, and West can only choose which red queen will win a trick.

Charles Blair: According to Geza Ottlik, West is squeezed in hearts on the second round of diamonds.

I have a theory about this too. The Great Pyramid is located in Giza, which translates to Geza in Hungarian. Trust me, when Ottlik penned Adventures in Card Play, he was high on tana leaves.

The optimal solution (14-5-12) was discovered by each of the top 18 solvers. The ending is unique, and testing by combinatorial exhaustion has proved it.

The condition of “at least three voids” has no bearing on the solution. Any construction with two or fewer voids, besides being illegal*, would require more HCP for North-South to win six tricks. For instance, with two voids the lowest possible N-S HCP total is 16.

*To be legal, an ending must be reachable from a full deal without a revoke or other irregularity. While this is not necessarily a restriction for puzzles, particularly for fantasies about bricks, I try to keep things legal around here — if only that I might pass for a bridge player in a dimly lit room.

Wayne Somerville: How did King Amenhotep manage to get a copyright in 1339 BC? How did they even know it was 1339 BC?

Kharis? We have a new mission; an infidel dares to question our veracity.

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© 2011 Richard Pavlicek