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This is an entry to the 17th fortnightly topic challenge.


Alice and Bob are playing connect four. Bob is tired of losing all the time, so he suggests changing the rules. He wants to drop 2 pieces every turn. Alice thinks this would make it too easy for him, so she wants to change the win conditions. They agree that she only has to get 3 in a row to win, while Bob needs 5 in a row. Alice goes first.

Assuming perfect play, who will win?


  • Standard connect four rules apply, unless something else is specified
  • The game is played on a 7 wide and 6 high grid
  • Bob will make 2 moves every time it's his turn
  • Alice will only make 1 move when it's her turn
  • Bob wins if he gets 5 in a row
  • Alice wins if she gets 3 in a row
  • If there is no moves left (the grid is full) without a winner, the game is a draw
  • Alice makes the first move
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  • $\begingroup$ Can Bob run out of pieces (standard set only has 21 of each color, and a full grid of Connect 4++ would have 14 of Alice's pieces and 28 of Bob's), or has he borrowed a few from another set? $\endgroup$ – Fillet Oct 13 '16 at 12:38
  • $\begingroup$ @Fillet They have unlimited pieces $\endgroup$ – Kruga Oct 13 '16 at 12:55
  • $\begingroup$ Just to confirm that Bob's two moves are disjoint and not necessarily adjacent, right? $\endgroup$ – Ian MacDonald Oct 13 '16 at 13:49
  • $\begingroup$ @IanMacDonald Yes, Bob's 2 moves are independant. He can put his second piece in any free column, regardless of his first move. $\endgroup$ – Kruga Oct 13 '16 at 13:55
  • $\begingroup$ Do you know the answer to this puzzle yourself ? $\endgroup$ – Evargalo Oct 23 '18 at 8:13
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This is just a quick guess but... I think that

Alice would win because she starts.
Bob might get 2 turns(pair), but needing 5 in row(odd) would still take him 3 turns to set a full chain same as Alice. Also, needing a chain of 5 will often make the size of the board a big problem blocking his chains. Of course having 2 turns can allow you to do more clever things so I might be wrong too.

EDIT

After reading the question again it would seem highly possible that a PERFECT play would result in a draw. Bob's long chains would quickly fill up the board and easily be blocked by Alice. While Bob having 2 turns could also easily block all of Alice's chains.

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I'm reasonably certain the result is

Bob wins

Below is an example game, in an obviously winning but not won position:

. . . . . . .
. . . . . . .
. B A A . A .
. B B B . B .
. B B A . B .
B B B A A B .

I'm not certain I have the mathematical chops to prove it definitively, but it feels as if

Bob can both answer Alice's threat on a turn, and develop one of his own, until Alice must make moves that don't result in direct threats, in which case Bob can work on setting up two threats at once.

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  • $\begingroup$ Hasn't Alice won in the 1st set? $\endgroup$ – bleh Oct 13 '16 at 2:05
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I'm not entirely sure what determines "perfect play" here but I'll do my best.

First off, a 7x6 board has 42 spaces, with 3 pieces being played per turn this means the game will end in 14 turns with 14 pieces played by Alice and 28 by Bob.

I claim...

Alice can't win. So can she force a draw?

The why...

Essentially, by only being able to place 1 piece per move and needing 3 in a row to win, for Alice to win she needs to already have 2 pieces in a row (duh Cpt. Obvious). With 2 pieces in a row she has 2 possible outs, one on either side. Because Bob can play 2 pieces per turn he can always cover those 2 outs. The only way Alice could possibly have more than 2 outs is to have something like...

...

- - - - -
- - - - -
- - - A A
B - a B B
A - B A B

.

where Alice has just dropped the "a" and has 3 outs. But as you can see this requires multiple Alice pieces in a scenario that would be easily spotted early and cut off by Bob.

So...

How can Alice force a draw? Essentially, she needs to cut Bob off from the center. Because she moves first her first move has to be the middle. This means Bob can't possibly win horizontally on the bottom row. Alice then needs to essentially build up pieces in the middle 3 columns (to shut off both rows and diagonals) only straying to shut off a column when Bob has 3 consecutive vertically (he can play 2 on top in a row and win). I'm fairly certain in this way Alice can force a draw.

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  • 2
    $\begingroup$ I feel like this is still a little too much speculation. It's hard to exhaustively prove that Alice can't make a triple threat. $\endgroup$ – greenturtle3141 Oct 13 '16 at 4:24

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