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Topic: A Complete Analysis of Normal Sprouts for n=7
Replies: 0

 Danny Purvis Posts: 176 Registered: 12/6/04
A Complete Analysis of Normal Sprouts for n=7
Posted: Jan 21, 1999 4:45 PM

S<p,q,...> - The dots p,q,... , all dots accessible to any of these,
and the associated space. S stands for "sphere". Pivot a dot
contained in 2 distinct spheres. H<p,q,...> - The sphere which
contains pivots p,q,... and which contains no nonpivots. S<p,q,...>=n
is the claim that S<p,q,...> will yield n survivors. T - In the
present context, a sphere consisting of 2 unattached dots or 2
unattached dots and a pivot. Let "a move to a given T" mean any move
affecting a given T except a move affecting only a pivot of that T.
It is easily shown that the second player to make a move to a given T
can force his choice of 1 or 2 survivors from that T. Countable - A
dot with 0 or 2 lines attached. N - A sphere containing r countables
and yielding s survivors such that r and s are either both even or
both odd. N+T - A total position consisting of an N and one or more
T's, where multiple T's do not share pivots. It is easily shown that
the player who leaves N+T wins. Player Z has TM - The claim that any
immediate resolution of the present position into N+T will be a case
of Z leaving N+T. X - The 1st player. Y - The 2nd player.

(L1) 7+ 1(8)1 1(9)8[2&3] II (S<4> has 5 countables and S<4>=3.
Thus, Y has left N+T.)

(L2) 7+ 1(8)1[2] 1(9)8[3&4] II (S<2,5> has 5 countables and
S<2,5>=3. Y has left N+T.)

(L3) 7+ 1(8)1[2&3] 1(9)8[4] II (S<4,5> has 5 countables and
S<4,5>=3. Y has left N+T.)

(L4) 7+ 1(8)1[2&3&4] 1(9)8[5] II (S<2,5> has 5 countables and
S<2,5>=3. Y has left N+T.)

(L5) 7+ 1(8)2 1(9)1[3&4]! (This forceful move makes it simple. Y
now intends S<2>=3 and S<3>=2. S<2> has 5 countables, so if Y is
correct in believing he can force S<2>=3, he has already left N+T. So
the question is, does S<2>=3?) 2(10)2? (This move does not put up
any fight at all. X fails to note that Y has TM. Similar blunders
would be 2(10)5, 2(10)8, 2(10)9, 5(10)5, 5(10)8, and 5(10)9)
9(11)10[5] II (S<5> has 3 countables and S<5>=1. Y has left N+T.)

(L6) 7+ 1(8)2 1(9)1[3&4]! 5(10)6 (Clearly X's only hope in this
position.) 5(11)6 (Now Y can (and of course must) force H<5>=1. So
the question now has become, does (the new) S<2>=2? I believe a
little time spent examining the position will convince the reader that
the answer is yes.) 3(12)9 (The battle has boiled down to whether
S<2>=1 or S<2>=2. Clearly S<2>=3 is out of the question. So X might
as well take dot 9 out of the fray "with tempo" (as they say in
chess).) 3(13)3[4] (Y defends properly. Now X needs to "dry up"
S<2>, and he has one "free shot" with which to do it.)
5(14@2)10 (or 5(14)2, 5(14)7, or 5(14)8) 6(15)11 II (S<2> is just
a little too fecund. S<2>=2.)

All of the foregoing analysis collapses nicely into:

7+ 1(8)2 (else Y leaves N+T) 1(9)1[3&4]! II (S<2>=3,S<3>=2 (Y has
TM) (5(10)6 5(11)6 3(12)9 3(13)3[4] and S<5>=3))