Four die but six live

    Keywords: Life & Death, Proverb

Table of contents Table of diagrams
Four stones on the third line die
Special case
Six stones on the third line: Segoe's sequences
Six stones on the third line another try
White's option
[Sente seki]
Dubious ko
Five stones on the third line : B to play and live

On the third line or in the corner on the second line a string of four stones is dead, and a string of six stones is alive. A string of five stones lives or dies according to sente. Similar proverb is Six Die But Eight Live (on second line)



Four in row

[Diagram]

Four stones on the third line die

Black tries to expand his eyespace with B1 and B3, but White plays W2 and W4 and Black dies. B3 at a or B5 is met with W6.

[Diagram]

Special case

If double hane on one side is sente, a ko fight for life is possible. In the diagram, B1 and B3 take advantage of the board edge (more specifically, the 1-2 point) to start a ko. If White protects with W6, B7 makes two eyes.


Six in row

[Diagram]

Six stones on the third line: Segoe's sequences

Bill: This appears to be White's best play, leaving a possible seki. See [ext] Segoe's sequences.

[Diagram]

Six stones on the third line another try

Another attempt. Black is still alive. A is correct, B seems inferior. For continuations see Analysis page.

[Diagram]

White's option

Robert Pauli:

The second attempt is better (sides exchanged to be compareable). The descent W1 in the first diagram is aji keshi, it destroys the option shown here (W5 instead of a). If Black now can't escape or make a second eye by jumping to b with B8, he dies (pushing at a instead dies after b).

So if Black can't hope for something at the left side, he has to play the hane B4, which leads to seki (missing in the analysis, by the way). The kaketsugi B6 is better than the descent left to it since it leaves the option a (and keeps the dubious ko below indirect).

That's one point better for White than Segoe's sequence:

  • after Black's hanetsugi there (left side), White lost 2 and Black counts 5: 2 + 5 = 7
  • compared to White already having lost 2 here (W5 and a) and losing another 4 after Black's hane at a: 2 + 4 = 6

White risks nothing. If B4 at B12 is a threat, she still can switch to the initial sequence by answering at B4. White can also choose the side.

[Diagram]

Dubious ko

Black won't risk this ko: Black starts and White owes an approach at a, but nevertheless ko.



Five in row

Obviously black needs sente to save her stones. Black can unconditionaly live. Diagram shows the simplest solution.

[Diagram]

Five stones on the third line : B to play and live

See The comb formation is alive and /Analysis


This is a copy of the living page "Four die but six live" at Sensei's Library.
(OC) 2024 the Authors, published under the OpenContent License V1.0.
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