great game. kind of addictive. got to 382 pretty easily, but afraid of what it is going to take to make the next couple bigger numbers... my question is what the symbols are all the way to the right on the home screen (if you scroll right of the puzzlejuice ad) - looks like a triangle, lemon, headphones, bird, cat ears, spider, etc
Not sure why but my games have been extremely glitchy today, joined cards are showing all white with no number on until I do another move and the tiles themselves seem non responsive when I move them, sometimes I think the swipe hasn't registered and swipe again and then two or three moves all happens at once. Husband has complained of similar on his device over the last few days, I've given up for now until the update is out as I fear my data will vanish again.
You are very much correct! Luckily, the poetic touch still holds. Spoiler In this case, "slightly" means "by a factor of 192".
Not really a fan of number/word puzzle genre, but whoa, this is truly brilliant; can't put it down. Best App Store purchase I've made in a while.
Sweet! Getting down to 7 tiles is pretty good. I think my best is 8 (at least that late in the game).
I'm certain it is impossible to make a board with fewer than 7 tiles. Reasoning: there will always be at least one tile in each row and column, because it is impossible to create an empty row or column. (To convince yourself; suppose the current board has at least one tile in each row and column, and you swipe down. Each column will stay non-empty, and the top three (non-empty) rows will shift down to make the bottom three rows non-empty, unless blocked (in which case the blocking row is already non-empty). Meanwhile a new tile appears in the top row, making it no longer empty. Similarly, it is impossible to make a "singleton" tile that is alone in both its row and column. Each incoming tile will share a row (if incoming horizontally) or column (if incoming vertically) with at least one other tile, and it will continue to share the row/column with that tile unless one or the other is blocked from sliding, in which case it now shares a row/column with the blocking tile. Can anyone help me complete the proof?
Sure. But first, we have to deal with the exceptions, namely the starting position. The game starts with 9 tiles on the board. If you were lucky and you could create 4 merges with one move, you would end up with 6 cards (9-4+1). Edit: For the following proofs, we will use the following definition of a legal starting condition: • Every row and column contains at least one card. • Every card has at least one neighbor (another card in the same row or column). • There are exactly 9 cards on the board.Thanks to @tommygents for pointing out a flaw that made this clarification necessary. Edit 2: Thinking about this some more actually makes the second point redundant. There is no way to place 9 cards on a 4 by 4 grid in such a way that a card has no neighbors. Edit 3: Looking at BenW's post below shows that my thought from Edit 2 is wrong. You can have 9 cards on the board with no empty row/column and still have a lonely card (one without neighbors). Lemma #1: Once a row/column is non-empty it will stay non-empty. Proof: Suppose a row/column were empty after it was non-empty before. The only way this could happen is when all cards in this row/column were moved vertically/horizontally during a move. However, a new card appears at the end of said move in this emptied row/column rendering it non-empty again. This proves the lemma by contradiction. ∎ Corollary #1: Once every row/column was non-empty, every card has at least one other card in its row or column. Proof: Suppose there was a card alone in its row and column. This could either be a card that was added after a move (new card) or a card that was already on the board (old card). If it was a new card it would have appeared after either a vertical or horizontal move but since all rows and columns were non-empty at some point and any move can only clear either the row (horizontal move) or column (vertical move) of this new card it must still have another card in its row or column (contradiction). If it was an old card it could only move to another row or column relative to its previous neighbors if said neighbors were blocked by other cards. These blocking cards would then be the new neighbors of the moving card (contradiction). This completes the proof. ∎ Theorem #1: The lowest number of cards on the board is 7 (apart from lucky starting conditions). Proof: By Lemma #1 and Corollary #1, every row/column must contain at least one card and every card must have at least one neighbor in its row or column. The minimum number of cards to accomplish this is 7 with a full row and a full column intersecting in one space. It is impossible to have an empty space at the intersection because this would violate the rules of the game: In order to produce such an empty space, an old card would have to move relative to its neighbors without blocking cards. ∎ Edit: I have an idea for a more elegant proof that I will share with you after I get some sleep.
Counterexample: Code: 12 _ _ 3 _ 2 3 _ 6 _ _ _ _ 3 _ _ (Twelve in the top left.) The trick is to show that such 6-tile boards cannot be constructed from a "valid" mid-game configuration (which I'll define as 9 tiles, no rows/cols empty, no singletons). In this example, the swipe that created this board must have come from the right or from the bottom. (Because if it came from e.g. the top, either the top left 12 or top right 3 would have moved down.) Suppose it came from the right. Then the preceding board may have been: Code: 6 6 _ _ _ _ 2 3 3 3 _ _ _ _ 3 _ And the board preceding this may have been: Code: 6 3 3 _ _ _ _ 2 1 2 3 _ _ _ _ 3 which could have come from: Code: 3 1 3 _ 3 2 _ _ _ _ _ 2 1 2 3 _ which has 9 tiles and no empty rows/cols, though it violates the no-singletons rule. (So close and yet so far!) In general though, it's not immediately obvious why a configuration like this should not be possible to construct from a "legal" starting point (9 starting tiles, all rows/cols filled, no singletons). So perhaps 6 tiles may be possible after all? FWIW, It's much easier to prove that 5 tiles is impossible: any board with 5 tiles covering all rows/columns will have exactly one row with 2 tiles, and exactly one column with 2 tiles. The 4 (or sometimes 3) tiles on these rows/columns will have "partners" and not be singletons. But the remaining tile(s) must be singletons, violating the no-singleton lemma.
If's not a counterexample if it does not contradict the proof. You could uye the Code-Tags to clearly format your board like so: Code: 12 _ _ 3 _ 2 3 _ 6 _ _ _ _ 3 _ _ When you quote this part of my post, you'll see the formatting. Now on to this. The general idea behind this simpler, more elegant proof is that we simplify the game rules and show that this simplification does not affect our intended goal. Let's define "Zeroes!" - a pointless puzzler that doesn't grow at all. Simplification #1: There is only one type of card with value 0. Merging two 0s gives another 0. This effectively eliminates blocking since all cards always merge. Proof: This change to the rules obviously only increases our chances of reducing the number of cards on the board. ∎ It is important to note that this is the only change to the rules of the game. Moving still works the same and there will also always appear another 0 card with every move in the same way as before. Also, to formally exclude any lucky starting position, I now define a legal starting position for the following proofs as follows: C1: Every row and column contains at least one card. C2: Every card has at least one neighbor (another card in the same row or column). C3: There are exactly 9 cards on the board. Lemma #1 still holds in Zeroes! since it did not rely on blocking. But we can now prove a stronger form of Corollary #1: Corollary #2: In Zeroes!, once a card has neighbors (cards in the same row or column) it cannot get rid of any one of them. Proof: Since there is no blocking (only merging) there is no way to move a card relative to its neighbors. ∎ Now we can prove Theorem #1 within Zeroes easily. Theorem #2: In Zeroes!, the lowest number of cards on the board is 7. Proof: Say we started with a completely full board of 0s. Based on Corollary #2 there must always be one full row and one full column on the board since no card can lose any of its neighbors but only merge with other cards which also merges parts of their neighborhoods. Therefore, the minimum number of cards on the board is 7 since the full row and column must intersect in one space. The above case is the easiest to understand but it represents the worst starting condition. It therefore does not complete the proof. We now construct a legal starting position that satisfies C1-C3 and show that the Theorem still holds. From Corollary #2 it follows that the neighborhood of any card cannot decrease in size. When two cards merge, a part of their neighborhood merges as well. The only way to reduce the overall number of cards on the board is therefore merging cards (along with parts of their neighborhoods). This will eventually lead to the same situation as before: 7 cards in a full row and a full column intersecting in one space. Once this 7 card configuration is reached no move will further reduce the number of cards on the board. This can be proven by exhaustive search*. ∎ *: This is left as an exercise to the inclined reader. Now since we know that Zeroes! is a relaxation of Threes! we know that the result of Theorem #2 also holds for Threes!.
This isn't true, counterexample: Code: 3 3 3 3 - - - - 1 1 1 1 2 2 2 2 Here row 3 is non-empty. If you move down, you'll get e.g.: Code: 3 - - - 3 3 3 3 - - - - 3 3 3 3 And row 3 is now empty. So I would state it like this: Theorem #1: Once all rows and columns are non-empty, they will stay non-empty. Proof: Suppose row A is non-empty. An horizontal move cannot leave it empty, because even after a merge it will always leave at least one card in the row. So assume that you make a vertical move, for example down. If A is the top row, a new card will be inserted from the top, so A will not be empty. If A is not the top row, there is a row B directly above it. B must contain at least one card C, because all rows and columns are not empty. During the move, C will either move down into A (possibly merging with a card in A), or will remain in B because it is blocked by a card D in A. Either way, A will not be empty. ∎ This isn't true, counterexample: Code: - - 3 3 - - 3 3 - - 3 3 1 2 - - Here every row and column is not empty, and every card has at least one other card in its row or column. But if you move left, you can get e.g. this: Code: - 3 3 3 - 3 3 - - 3 3 - 3 - - - where the 3 in the bottom left corner has no other card in both the row and column.
Both of your counterexamples fail the conditions for a legal starting state I had to add later because of these flaws. Your first counterexample violates C1 (Every row and column contains at least one card.) and your second counterexample violates C3 (There are exactly 9 cards on the board.).
Now? Really? After we dissected the game and beat the average human player with an AI? Now you're backing away from the thread because we're starting to formulate theorems and proofs? I'd take this as an opportunity to see that math / logic doesn't hurt. Stating theorems, proving them and poking holes in other peoples proofs is a different kind of puzzle (or mental challenge). But in stark contrast to Threes! or other puzzle games, developing the ability to state and understand proofs makes you win at the game of life by not being easily deceived.
Thanks for the formatting tip. The "counterexample" was specifically to the lemma that no-singletons entails at least 7 cards; this arrangement has no singletons with only 6 cards, and I show how it can be derived from earlier 7-card and 8-card configurations also with no singletons, though the 9-card precursor to that does have one singleton. It's not obvious to me that there couldn't be a no-singleton 9-card starting point that could yield a final 6-card configuration like this. Also, Nicola's observation that it is possible to create singletons by merging does throw a wrench in things. It means that the definition of "legal starting state" probably has to change. And if it has to change sufficiently to make the 9-card configuration I found "legal", then we've shown that attaining 6 cards is possible. On the other hand, if we abandon both the 'singleton' and empty-row requirements (neither of which I believe are enforced for actual starting boards for Threes!), then for certain 9-card starting configurations, it is trivially possible to shrink the board down to 5 cards, e.g.: Code: 1 1 1 _ 2 2 2 _ 3 3 3 _ _ _ _ _ becomes e.g. Code: 6 6 6 _ _ _ _ _ _ 1 _ _ _ 2 _ _ But if we start with 9 cards without empty rows, I don't see a way to do it, and suspect that 6 is the minimum.
