A minor Fibonacci issue

[pilun]

8 is a Fibonacci number. If you have a group of 8 hands, you can resolve them all within 5 steps. (3+2+1+1+1)

The question is. Should you?

 

Take the example of 3-suiters with both minors. There are eight of those.

This is a logical Fibonacci way to group them:

 

4-0-4-5, 4-0-5-4, 5-0-4-4 (so 3 hands with a heart void)

4-1-4-4, 1-4-4-4 (2 of those)

0-4-4-5

0-4-5-4

0-5-4-4

 

If 2♥ is the bid to show 3-suited with both minors, you can get them all out by 3♠, zooming with three shapes.

All good and easy enough to remember. Is it efficient?

 

4441 hands are more common. If we look at hands short in hearts, 4-1-4-4 is twice as common as the three heart void hands put together.

With the scheme above, 4-1-4-4 comes out at 3♥, 1-4-4-4 comes out 3♠ with zooming.

 

One asymmetric way to group them is like this

 

2N = 4 spades (4 hands)

3♣ = 1-4-4-4

3♦ = 0-4-4-5

3♥ = 0-4-5-4

3♠+ = 0-5-4-4

 

then

3♦ = 4-1-4-4

3♥ = 4-0-4-5

3♠ = 4-0-5-4

3NT = 5-0-4-4

 

So one hand finishes high. In fact all the heart void hands are a step higher. (There are 2 zooming shapes, not 3)

To compensate, the 4441s come out earlier, which has to be good.

 

1-4-4-4 at 3♣

4-1-4-4 at 3♦

 

That's a gain of two steps for both the 4441s. Seems a better deal; getting the more common hands out earlier.

 

For us, there is another reason for an asymmetric split. Two of our major openings deny four cards in the other major.

As a consequence, when describer shows a 3-suiter, there are only four shapes to show.

Four is not a Fibonacci number so there is no gain in splitting. Just show them in a line, naturally starting with the 4441.

[DinDIP]

KK Relay (Woolsey and McCallum) decided to show shapes symmetrically. To achieve that they

(a) use the cheapest step (2♠) after teller shows a three-suiter with both majors (1♣-1♥-1♠-1N-2♣-2♦) to cancel the message of a three-suiter and replace it with a 6+card semi-solid major; and

(b) use 3♣ to show a 4441 with a singleton A, K or Q (so that other 4441s are always shown by 3♦)

 

I think symmetry is a big plus but one step can matter so I'm not keen on losing space.