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Okay, I figured out why 14 spaces doesn't guarantee a P2 win.
Move 1:
P1 plays an O on either space 7 or 8, directly blocking P2 from using that side (1-7 or 8-14) to set up their S _ _ S trap (since 7 spaces are needed). We'll use space 7 for this example, but this and all other moves could be mirrored and it still holds true.
_ _ _ _ _ _ O _ _ _ _ _ _ _
Move 2: P2 plays an S four spaces from the other end (space 11) since that's the only way remaining for P2 to make S _ _ S without P1 directly interfering. If P2 puts it in 1-6, P1 can block with an O two or three spaces away toward the more empty area. If P2 plays S in 8-10 or 12-14, P1 can play an O using the same strategy as on the other side. If P2 plays an S in 11, then whichever way P1 goes, P2 can play another S in 8 or 14 to make S _ _ S... In theory, at least. In practice, not so much...
_ _ _ _ _ _ O _ _ _ s _ _ _
Move 3: P1 can now break the S _ _ S strategy by playing an O in 14.
_ _ _ _ _ _ O _ _ _ s _ _ O
Move 4: P2 can't use the S _ _ S strategy any more. If P2 puts an S in 8 like the strategy calls for, P1 can simply end the game with an SOS in 6-8. Putting S in any of the spots 1-6 is still just as blockable as in Move 2. A different strategy must now be used.
Any more than 14 spaces makes one half of the field unblockable for P1. P2 can always just put an S four spaces from the other end of where P1 initially played, and P1 can't block a successful S placement either in that very last spot, or in the spot three more inward, on P2's next move.
Basically, the strategy boils down to figuring out whether you always move with an odd or even number of spaces open to you, and then trying to either force S _ _ S to be created (odd) or keep it from being created (even). If there are more than 14 spaces on the board to start and you're even, you're screwed.
@TheFinalChapters Says:
Fun fact:
If you use a random number of blocks, player 1 is actually slightly more likely to win than lose.
While the minimum even number that player 2 can force a win is 16, the minimum odd number that player 1 can force a win is only 7.
@hericide1 Says:
i like how i knew the solution immediately
@Cryft Says:
i think it's worth mentioning that there is no setup of THREE blanks is a "move and lose" situation. that is, that there are no poisoned squares EXCEPT [S] [ ] [ ] [S] configurations that would allow player 1 to try and swap the polarity back to making player 2 forced to make a losing move. thanks for the video, had fun with the problem
@GummiSammi Says:
should have realized something was off when he didn't realize the winning move for player 2 on the 4 square case example. (No sure fire winning move, but if Player 1 plays S in first square, you can win for sure).
Thought this was a math channel. there's 13 min I won't get back.
@Ainogard Says:
Wait... for the 4 square example at 5:30: If player 1 put in a S in square 1, player 2 would be guaranteed to win by placing a S in square 4 (as shown in the next example). Of course if both player would play optimally, player 1 would never place a S in square 1.
@khoernchen3490 Says:
unless i am stupid player 1 wins on an uneven board starting at 7 fields and player 2 wins on an even board starting at 14 fields.
can someone explain what he meant with "greater then 15"?
@XoIoRouge Says:
At first, I only saw the "Zach Star Himself" channel and it wasn't my cup of tea. I saw this channel and was like "Ugh, Zach Star... but... Math game? I LOVE math games. Let's embrace it and see what happens."
It's awesome and I'm learning and having fun. Love this channel.
@oakley6889 Says:
Kinda an extension question , Now what if the row of squares gets connected at both ends, effectively like a loop? Is it still possible?
@temmiekssj Says:
In the 4 square, if player 1 starts with an S on the edge can’t player 2 always win? That is if they put an S on the other end because then no matter what p1 does p2 can always make sos
@damyankorena Says:
For the four block example player 2 could just win after player 1 played the first move
@Patricknotized Says:
Great video. I nearly just looked at the solution, and clicked off.I didn't feel like you presented the value of your own thought process well enough at the start. In retrospect I would have regretted just seeing the solution
@stephenalexander9558 Says:
Highly amused when you showed the answer at the end even SHORTER than the first time showing! Yea! Hahahahaha! O:-)
@Blockenheimer Says:
This man sounds too much like Stratzenblitz75
@samuskiii Says:
What if they just started trolling and made all of them the same letter
@DarionOshi Says:
The strategy of player 1 is wrong.
If He always lose in your optimal strategy.
Then he would look for a tie instead of win
@nimeshkansagara Says:
stupid explanation and stupid video
@chair547 Says:
Game tree is only 3 million states (probably fewer because of wins).
just brute force that shit.
@tollspiller2043 Says:
The example you showed for 4 isn't played optimally, as 2 can easily create such a configuration.
@makkusaiko Says:
When Zach showed up on camera it kinda felt like one of his skits
@kennethdang4446 Says:
Nearly 9 years ago I played this game with my friend in middle school. I only know appreciate his smartness. He recently graduated Stanford.
Instead of doing SOS, we circled in dots. We can circle 2 dots or 1 dot. The last one to circle the dots losses or win or something, I forgot. He was an Olympiad winner.
@lesnt4449 Says:
technically you played sos in 1d
@ozargaman6148 Says:
What if player 1 manages to make "S--S--S"? The parity is ruined now and player 1 wins
@user-pr6ed3ri2k Says:
isn't this just 011001100010101
@everythingforeveryonegamer Says:
In 4 square player 2 could win
@WebSoak Says:
I’m glad you dedicated an entire video to this amazing song!
@godfreypigott Says:
I guess we can declare this channel dead.
@GregHuffman1987 Says:
Ima be honest... this doesn't sound like a fun game
@mightycannon1512 Says:
How did i notice THIS IN HALF A MIN WTF
@FURYBrenton Says:
There is a winning strategy for player 2 in 4 squares. Player 1 puts S in first spot, player 2 puts S in last spot. No matter what player 1 does next, player 2 wins
@kowshikmass5954 Says:
Which is best after btech mechatronics .ms in robotics ir mechatronics
@arontinkl8782 Says:
I am not sure whether i am dumb but my brain says that the first player can simply put only S or only O and then its drawn any way, or am i dumb?
@xxxcxcv8837 Says:
What is the use of Buffer in memory management discuss with example? pls help me 😢
@MagusAce Says:
I've been nearly shouting at my phone when you went from the 4 tiles to 5 tiles. All I could think of was S _ _ S, whilst wondering if there was maybe another way, which you were getting at.
@Nicoder6884 Says:
5:24 Correct me if I'm wrong, but I think player 2 could just put an S in the rightmost box and have a guaranteed win instead of a draw.
@kvsconstructionwork Says:
😘😘
@8-bitfox716 Says:
I rather have the O replaced with a U. Because there's people going crazy over sus EVERYWHERE
@pierrefraisse8610 Says:
...---...
@JohnPaulBuce Says:
amog
@Ryan-mw4zv Says:
Can you prove the same if opponents wanted to get SSS or OOO instead? What about any n-factor combination of SO?
@handanyldzhan9232 Says:
If a game with n cells is lost by P1, the latter can win the "n+1-game" by placing an O in the leftmost or rightmost cell, making P2 the first player of the remaining n-game.
If a game with n cells is won by P1 with the right first moves, and drawn or lost with the others, P2 can prevent their defeat by picking one of the losing or drawing options for P1. So there can't be any non-winning first move for P1. Even if P1 places an O in the leftmost or rightmost cell, they'll win, which means P2 wins the n-1-game.
We can figure out that an n-game and an n+1 game can't both be won or lost by P1. All odd numbers of cells lead to P1 winning, and all even numbers of cells lead to their defeat, because P1 wins a 7-cell game, so P1 loses a 2000-cell game.
@CarrotCakeMake Says:
You also have to show that you can always play a move that doesn't give the opponent a win on his next move. Placing an O next to an O never gives a win, and placing an S next to an S never gives a win unless you are in the SBBS case, which is always avoidable due to parity.
@icedragon5743 Says:
5:46 since player 1 played S on the first box you automatically win if you put s on the 4th box
@mga5534 Says:
in a 4 square game player one putting an s in the first square would result in player 2 placing an a in the last square forcing a win
@MultiDominic111 Says:
whyyy??
@Kyo_Him Says:
Aren't you that guy do those parodies?
@skarma9673 Says:
5:16 player 2 puts a S in the 4th square and it’s a guaranteed win...
@3RR0R415 Says:
in the case with 4 boxes, where p1 puts an s on either end, p2 places an s at the other end and forces a win...
@godfreypigott Says:
Should we be unsubscribing? Because it seem you've lost interest in this channel.
@Simqer Says:
You made a mistake at the 4 tiles at 5:15. If player 1 puts S in the first tile or last tile (S_ _ _), player 2 will always win if he puts another S at the other end (S_ _ S). It won't matter whether player 1 puts an O or S anywhere in the 2 remaining slots, player 2 will always be able to complete the SOS on his turn.
If player 1 makes any other move, player 2 will always be able to tie. So player 1 has no chance at winning, while player 2 does.
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