What configurations, with 10 or less buttons in each pile, could be classified as winning configurations, where you could be absolutely sure of winning?
Source: Puzzlegrams, Pentagram Design Limited, 1989, #90. Used as Ken's POTD 9/9/94.
To win the game as described above, take one button from the pile with 8. All analysts showed there are many winning positions and only a few losing positions. For 10 buttons or less in each pile, the losing positions are 1-2, 3-5, 4-7 and 6-10.
Denis Borris' exhaustive (and moderately humorous) analysis follows. (I liked the first line, so chose to share it here.)
Well, when I 1st saw it, I thought: what a dumb puzzle...
Then I went back: WHAT A GOOD PUZZLE!!
ANSWERS:
Part 1 : take one from the 8pile
Part 2 : ALL of them, except 1-2, 3-5, 4-7 and 6-10.
The Borris NIMematics
=====================
This takes a look at the first 20 "you lose if you go 1st" cases:
(top line is simply the differences from L to L...)
L = Low pile H = High pile N = Number
1 2 1 2 2 1 2 1 2 2 1 2 1 2 2 1 2 1 2 2
L 0 1 3 4 6 8 9 11 12 14 16 17 19 20 22 24 25 27 28 30 32
N 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
H 0 2 5 7 10 13 15 18 20 23 26 28 31 33 36 39 41 44 46 49 52
So we clearly see a repeating "1 2 1 2 2" making up the L's,
so an 8 accumulation at each 5th, and the H's are simply N + L.
From that, we get the following math-shattering formula:
L = N + 3A
H = L + N
if B = 2 or 3, then increase L and H by 1
if B = 4, then increase L and H by 2
where:
A = int(N / 5)
B = N - 5A
Example: find 17th combo;
A = int(17/5) = 3
B = 17 - 5*3 = 2
L = 17 + 3*3 = 26
H = 26 + 17 = 43
Since B = 2 then L = 27, H = 44 (see N=17 above)
AND: you can use this to figure out wether you should start
or not when someone says: wanna play a game of 2pile NIM
for the next round?
Here's what you do:
1- say: "OK, set up the piles"
2- assume he sets up 54 and 88 !
You now go figure in your head:
- ok, that's an N of 34 (88-54)
- so my A is 6 (int(34/5))
- gives me a B of 4 (34-30)
- the "Borris method" says N = L - 3A
- so my N = 36 (54-18)
- my B is 4, so I subtract 2
- so this gives me 34...that's a "loser"
And you say innocently: OK, YOU start !!
Naturally, if opponent sets up 2 piles where
your 2 N's are not equal, you say: OK: I'll start :>)
For those interested in a "table-like" show:
gives an "eye-view" of how the "losers" come about;
losers are asterisked; example: in row 6, 6-10 is a loser:
pick to
P1(L) ........................P2(H)...................... leave:
1 2* 3 4 5 6 7 8 9 10 11 12 ...
2 3 4 5 6 7 8 9 10 11 12 13 ... 1-2
3 4 5* 6 7 8 9 10 11 12 13 14 ...
4 5 6 7* 8 9 10 11 12 13 14 15 ...
5 6 7 8 9 10 11 12 13 14 15 16 ... 3-5
6 7 8 9 10* 11 12 13 14 15 16 17 ...
7 8 9 10 11 12 13 14 15 16 17 18 ... 4-7
8 9 10 11 12 13* 14 15 16 17 18 19 ...
9 10 11 12 13 14 15* 16 17 18 19 20 ...
10 11 12 13 14 15 16 17 18 19 20 21 ... 6-10
11 12 13 14 15 16 17 18* 19 20 21 22 ...
12 13 14 15 16 17 18 19 20* 21 22 23 ...
13 14 15 16 17 18 19 20 21 22 23 24 ... 8-13
14 15 16 17 18 19 20 21 22 23* 24 25 ...
15 16 17 18 19 20 21 22 23 24 25 26 ... 9-15
16 17 18 19 20 21 22 23 24 25 26* 27 ...
17 18 19 20 21 22 23 24 25 26 27 28*...
18 19 20 21 22 23 24 25 26 27 28 29 ... 11-18
The Borris 2pile Nim table
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