Thread: Logic riddle

lets look at 4 columns, if we can't find a winning strategy for that then more columns won't have a winning strategy either.

Label the columns 1,2,3,4

Shoot column 2 three times, if he started at 1 or 2 he's dead.

if he started at 3 he winds up on 4. shloot 4 he's hit

If he started at 4 he winds up at 3, shoot 3 he's hit

since you can't be sure where he is you can't gaurantee to hit him

if the number of shots you take at column 2 is odd, he winds up at the other available position, if even he winds up at the same position. Since you don't know where he started then you can't gaurantee to hit him.

I would be very interested to see a winning strategy for 4 columns

I take it back...this is not a winning strategy. I don't think there is one. You can determine whether he's on evens or odds at any given moment, but you can't pin him down absolutely. This is solvable by random chance over time, though.

for the case of 4 columns...suppose you shoot column 2 twice...then you either hit him or he's on 3 or 4...if you could dtermine whether he's on even or odd... your next shot would hit him.

I don't think there is a way to determine the even-odd parity of his column.

Yes you can. Also, I was wrong; there IS a strategy to kill him given four columns! I was overthinking things.

Shoot at 2 twice, and he cannot be on 1 or 2; he must be on 3 or 4. So he then moves to 2, 3, or 4. Shoot at 3, and he must have been on 2 or 4. You have now established parity. He then can move only to 1 or 3 on every other turn. Shoot at 3 a second time. Now he must be at 2. Shoot at 2 and you've killed him.

2, 2, 3, 3, 2 is a winning strategy for the shooter, guaranteed.

(1) 2 (2) 2 = dead
(2) 2 = dead
(3) 2 (2) 2 = dead
(3) 2 (4) 2 (3) 3 = dead
(4) 2 (3) 2 (2) 3 (3) 3 = dead
(4) 2 (3) 2 (2) 3 (1) 3 (2) 2 = dead
(4) 2 (3) 2 (4) 3 (3) 3 = dead

There's all the possibilities. It takes five shots maximum.

Originally Posted by Annatala

Yes you can. Also, I was wrong; there IS a strategy to kill him given four columns! I was overthinking things.

Shoot at 2 twice, and he cannot be on 1 or 2; he must be on 3 or 4. So he then moves to 2, 3, or 4. Shoot at 3, and he must have been on 2 or 4. You have now established parity. He then can move only to 1 or 3 on every other turn. Shoot at 3 a second time. Now he must be at 2. Shoot at 2 and you've killed him.

2, 2, 3, 3, 2 is a winning strategy for the shooter, guaranteed.

(1) 2 (2) 2 = dead
(2) 2 = dead
(3) 2 (2) 2 = dead
(3) 2 (4) 2 (3) 3 = dead
(4) 2 (3) 2 (2) 3 (3) 3 = dead
(4) 2 (3) 2 (2) 3 (1) 3 (2) 2 = dead
(4) 2 (3) 2 (4) 3 (3) 3 = dead

There's all the possibilities. It takes five shots maximum.

Well done!!! I have confirmed your strategy and am very impressed. Do you think your strategy can be extended to more columns?