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Solve any rows with 0 voltorbs and key those in as well. This is not perfect, it will try to calculate the best choice guess that has the least risk associated with it. It will.

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Best move calculator for Voltorb Flip: 212063.ruβVoltorb_Flip. 212063.ru import sys, math. class Board(object). def __init__(self, row.

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Best voltorb flip calculator? I made a topic, but it often left the main page. Anyway, I heard of a good one.

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Best move calculator for Voltorb Flip: 212063.ruβVoltorb_Flip. 212063.ru import sys, math. class Board(object). def __init__(self, row.

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Remember, the Voltorb counts listed in the Too High for Ones rule assume a total of five squares in the row - with four in our imaginary row, two Voltorb leave two number squares. In other words, you don't want to flip over the Voltorb. People really don't seem to point this out a lot, but you can rule out 3 for any square in a row whose total and Voltorb numbers add up to six. Note also that this means you can eliminate the possibility of 2s as well if the total number is just one higher than the minimums indicated above, since that means the row is all 3s and Voltorb. Simple and straightforward here: basically, the first thing you should do is always to look at all the Voltorb numbers and check if there are any rows with no Voltorb at all. Likewise, though all the examples display horizontal rows, they could just as well be vertical. But don't give up just yet. And from this, you need to work it out. Here's the thing. Note that this is also the Four Voltorb rule generalized to apply to reduced rows - for four squares in a row, it's essentially the "Three Voltorb" rule. But just to make the imagery a bit stronger, the zeroes are Voltorb that will violently explode and end your game immediately after all, there's no point in continuing when your score has a multiplier of zero in it. This happens if the number of number squares i. The most basic rule for guessing is to flip over a square where two rows with few Voltorb and a high total intersect - it's fairly likely to be a 2 or 3 and unlikely to be a Voltorb.{/INSERTKEYS}{/PARAGRAPH} Whenever you eliminate possibilities, you may be uncovering the mysteries of an entire row. You can of course do the same thing when you've flipped over more than one number in the same row - again, just subtract them all from the total and pretend the row is only however many squares remain when you've done that. It's always nasty to trip a Voltorb by accident on a square you thought was safe. Therefore, you should do just that if you have any zero-Voltorb rows. This is because five the total number of squares in the row minus the Voltorb number gives the number of squares containing some number, and if that number is the same as the total, they all inevitably have to be 1s. Basically, if you've already got some markers on a square and you find out something about that square's row or column, this should only result in the removal of markers, not new ones being added. The top number in each of the squares is the sum of all the numbers in that row or column, while the bottom number is the number of Voltorb in that row or column. That's where those numbers on the right and bottom sides come in. Don't you love how we've come this far in the guide with absolutely no mention of guesswork? There are two possibilities - , or Notice anything? How many ways can this happen? This is basically when the row is either all 3s or one 2 with the rest 3s; if there are at least two 2s, you can't immediately outrule that those two 2s could be a 1 and a 3 instead. So what do you do then? Therefore, the goal of the game is to flip over all the 2s and 3s on the table, and once you've done that, no matter how many 1s are left, you'll get your payout and advance to the next level. For instance, say you have a row like this:. You can compound what you know from the row and column this way, giving you still more information about a square. Bring up the memo pad X or the button at the top right and mark all the squares in that row with a Voltorb. You could be misled to think this means the row is irrelevant and you can ignore it, but don't; yet again, open the memo pad and mark all the remaining, if you've already flipped some over or marked them as definitely Voltorb squares in the row with both the Voltorb and 1 marks, to say that they're either Voltorb or 1s but not 2s or 3s. You can apply the same reasoning after you've flipped over a square by simply subtracting the number you've flipped from the total and then treating the row as if it were only four squares, with the appropriate tweaks to the rules. Therefore, we can safely flip it over. In a similar fashion to the Total Six rule, sometimes you can eliminate the possibility of 1s in the row. Sometimes, even though you can't tell exactly which number a square is hiding, you can tell that it must be some number - that is, it can't be a Voltorb, and thus it is safe to flip it over. So how do you know which squares to flip? Meanwhile, the 1s don't really matter; multiplying your score by one just keeps it the same. Be sure you're not making a mistake before you do that, though - double-check whether the row truly must have that many of that number still unflipped and whether the other squares truly have it eliminated. And you can do the same when you've confirmed for sure that a square is a Voltorb by mentally subtracting from the Voltorb count instead of the total number. If a row must have at least X 1s, 2s or 3s, then any time you have exactly X squares left with that number as a possibility, you can safely flip those squares over, because that number simply must be there. In general, every time you've flipped over a square or eliminated some possibilities, reevaluate both that square's row and column in light of the new changes, reducing them to fewer squares as necessary and applying all nine rules in order. This will simplify your rows and is often the key to solving the puzzle. If this row has a 1 and two 2s, there are only two possible places those 2s can go - they must be the fourth and fifth squares. This explains itself best with an example. Note that a square containing all four markers is equivalent to an unmarked square here; I personally like to fill out the whole board with markers on every remaining square after I've applied the five most basic rules, but you don't have to do that if you're satisfied with knowing an empty square means you've eliminated nothing. The nine rules above don't just apply to the rows as we have them at the beginning. You'll just have some identically marked squares and absolutely no way to tell which of them hides the Voltorb. So when must a row have at least X 1s, 2s or 3s? And yes, this means it is useful to mark those squares. It can be a little time-consuming to work this out, so I recommend only doing it if none of the other rules are giving you anything - hence my placing this as the last rule. It may seem like this will take absolute ages, but you start to see very quickly whether most of the rules apply - the first five need only a glance at the numbers or mental numbers , while the latter four are something you get used to applying after a while. We're getting to slightly less obvious now, but nonetheless still straightforward. Remember that when you place markers, you are eliminating the numbers you don't mark. They are functionally interchangeable, so all the same rules apply vertically. No, it's not total five, but it's total four, and now there are only four squares in this imaginary row, so the same principle applies. You've got a memo pad to assist you, which allows you to give any square markers corresponding to Voltorb, 1, 2 or 3. We have three number squares and must get a total of five from them. But, well, in most games not all - I've gotten through many rounds on logic alone without guessing at all , eventually none of the nine rules will lead you anywhere with any row or column. At any point in the game, your score is equal to all the numbers you've flipped over multiplied together. Note that there can be more 1s, 2s and 3s than indicated here; these are just minimums that must be present in the row when these rules apply. Either one of them will give you the same result, since this is an overlapping case. That also means you're not going to add any of them back in unless you realize you made a mistake in eliminating them in the first place. Even better, whether this number happens to be a 2 or a 3, we find out more about the row - if it's a 3, the other numbers in the row are both 1s and so we can eliminate the 2 from the fifth square, while if it's a 2, we know there's another 2 and that it must be the fifth square, allowing us to flip it over as well. So rule six tells us the remaining squares in this row can only be Voltorb, 2s or 3s Is that the Total Five rule I see? Using just the above rules, we'd be at a loss here. This inevitably means that the total number is equal to that one number to be found in the row, and therefore, for every square in the row, you can eliminate all possibilities other than Voltorb and the total number. Likewise, the Total Six rule applies to a total of five when you have four squares and, just like with five squares, gives you the ability to eliminate 3s from all the remaining squares where it applies. Example: becomes. If you don't want to have to work out the possible combinations every time, here's a quick cheat sheet of the stuff you need this rule to solve - in the rest, there is only one possible combination and rule eight can tell you everything you need:. This happens when, out of all the possible combinations of numbers the row could have to reach the given total, they all need this square to be included, on the basis of what has been eliminated in the other squares. There are twenty-five green squares, and these squares each contain either a Voltorb equivalent to zero or the numbers 1, 2 or 3. And thus, it cannot possibly be a Voltorb. This will tell you that you know these squares are Voltorb. When you tap a square, you will flip it over to see what it contains. {PARAGRAPH}{INSERTKEYS}Unlike the slots, this is actually a logic puzzle that requires some thinking, and thus I thought it would be appropriate to write a guide about it. Note that even if this doesn't apply, it is still possible there are no 1s in the row; it's just that you can't rule out that there might be a one. There are several rules you can follow to narrow it down for you. And people complain this game is just luck like the slots. Another example: becomes. Logic, my friend. And hey, what's that? We have two number squares and a total of at least five! Say you have a row like this:. The reason for this is that if you take away the Voltorb squares, this leaves you with one less square than the total number, and the only way this can happen is if one of those squares is a 2 and the others are 1s. None of our current rules tell us how to deal with this. However, because we've flipped over that 1 already, we can mentally pretend this row looks like this:. Use the following rules, given that we call the number of number squares not-Voltorb N and the total number for the row T:. We can eliminate all those 2s, giving the whole row Voltorb-one markers. It's total seven so that doesn't help us; there are Voltorb but not four or five of them; 1s, 2s and 3s are all possible; and while there must be at least one 1, that doesn't help us here since we haven't eliminated 1s from any of the squares, while we have no guarantees about the presence of either 2s or 3s. If any row's total and Voltorb numbers add up to five , then it's clear that that row contains only Voltorb and ones. If you flip over a Voltorb at any point, however, you'll lose all your winnings for this round, and if the number of squares you flipped before this happened is less than the level you're currently on, you'll drop back to the level corresponding to the number of flipped squares though of course, since there is no level zero, if the first thing you flip over is a Voltorb you'll be sent to level one instead. If a row has no Voltorb, it is obvious it is safe to just flip over the whole row. This inevitably means that once you flip over a zero, you've lost all your points and can never get them back. If any row has four Voltorb, you know that only one of its squares is a number. Meanwhile, if this row has two 1s and a 3, there is only one place that 3 can go - the fourth square. So either way, the fourth square must be either a 2 or a 3! But no, that does not mean you just leave it alone and think about something else.