Part 2？区块节点为假？
接下来我们来看一种新结构，在此之前，我们必须要引入一个新的节点类型：区块。
Next, we look at a new structure, before we have to introduce a new node type: blocks.
来看这一则示例。
Here's an example.
如图所示，我们可以观察到链头是r3c4(9)。按照链的逻辑，我们写出文本写法。
As the figure shows, we can see the chain head is r3c4 (9). In the logic of the chain, we write the text.
当推理到r7c9(9)的时候，按照顺序，这里应该是为真的，但发现b9里含有三个候选数9，按照之前的链的逻辑，我们只可以让其中一个9为假，然后继续推理。但实际上，我们不论指定链从r8c7继续延伸还是r9c7继续，都无法继续推理，因为我们此时仅能让r8c7或r9c7一个单元格的候选数9为假，但c7有三处9，我们不可能因为一处9为假就否定了剩下两处的9，所以我们不得不采用一种逻辑：既然单独看r8c7或r9c7都不行，干脆我们就直接把它们看作一起，当成一个区块。
When it comes to r7c9 (9), it should be true, in order of order, but it was found that there were three candidates in b9 and, according to the logic of the previous chain, we could only let one of them be faked, and then we could continue the reasoning. But in fact, we could not continue the reasoning, regardless of whether we specified the chain to continue from r8c7 or from r9c7, because at this point we would be able to make 9 out of only R8c7 or r9c7 cells, but there were 9 out of c7, and we could not have rejected the remaining 9 out of 9 because one is fake, so we would have to adopt a logic: since it is not possible to look at r8c7 or r9c7 alone, we would simply see them together as a block.
假设到r7c9(9)为真时，便同时可以使得r89c7(9)均为假。此时，当它们都为假后，我们才可以继续向下推理，c7只有三处9，其中的两处都为假了，自然只能让r2c7(9)为真。那么链得以继续持续到链尾，并且确实拥有删数，且删数是正确的。这条链就此分析完成，并且我们就引入了一种新的节点类型：区块。
If the r7c9 (9) is true, then it is possible to make r89c7 (9) false at the same time. At this point, when they are all fakes, we can continue to deduce that there are only three nine c7, two of which are fakes, and naturally only the r2c7 (9) is real. Then the chain can continue to the end of the chain, and indeed has the number of deletions, and the number is correct. This chain is done, and we have introduced a new node type: blocks.
不过区块不是整体分析的吗？如此分析已经拆解区块结构了，或者换句话说，实际上前面的逻辑是分了两个“支线”，让r7c9(9)同时分别用虚线箭头（弱关系）指向r8c7(9)和r9c7(9)，这并不能说就算真正的区块结构了。那么，既然是节点，就有真假性，而按照之前的逻辑，既然看作整体，那么区块这种类型的节点为真或为假都分别表示哪一种情况呢？下面我们就来看看下面这一则示例里，我们会得到什么新鲜的东西。
But the blocks are not analysed as a whole. Such an analysis has broken down the structure of the blocks, or, in other words, the logic ahead is to divide the two “sublines” so that r7c9 (9) points to r8c7 (9) and r9c7 (9), respectively, which does not mean that even if the structure of the blocks is real. So, since it is a node, there is a real falseness, and according to the previous logic, if it is taken as a whole, the node of the blocks is true or false.
如图所示，这条链和之前的例子不一样的地方是，它从区块节点开始推导的。
As the figure shows, this chain, unlike the previous example, is derived from block nodes.
那么，初始状态要求区块节点为假，那么区块节点为假是什么？我们回顾一下之前的理解：区块为真是什么，区块为假又是什么。区块是一种结构，我们人为定义的候选数节点为“真”就表示这个节点就填这个数，这个数是对的；那么相反，为“假”就表示不填，删除这个数。可以看到，它们是互斥的。那么区块节点为真，我们人为规定为真即表示“这个区块成立”，即区块所属的单元格里必须有一个单元格就填入这个数。那么为假就取这个说法的相反情况，即“没有单元格填这个数”或“有至少两个单元格填入这个数”
So, the initial state requires block nodes to be fake, so what is block nodes? Let's look back at the previous understanding: what is a block really and what is a block fake. Blocks are a structure, and our artificially defined candidate nodes to be “real” means this number, which is correct; instead, “false” means not to fill it out and delete this number. As you can see, they are mutually repulsive. Then block nodes are real, and we're artificially set to mean “the block is formed”, which means that the cell to which the block belongs must have a cell to fill this number. So the opposite of the phrase is “no cell fills this number” or “there are at least two cells to fill that number”.
当然，你也可以这么去想：区块为真表示区块成立，那区块为假则表示区块不成立，那么就思考哪些情况属于区块不成立。
And, of course, you can think of that: blocks are created to be true, blocks are created to be false, blocks are not, so think about what is not.
而显然，我们这里指代的区块结构是不可能出现“至少两个单元格填入这个数”的情况的，所以只可能是“没有单元格填这个数”。那代入到图里，就表示r3c79都不能填5。那这么一来，链就成立了：r3c2=5，即得到了“区块节点为假”得到“候选数节点为真”的结果，即从假过渡到真，此时这个形式为强关系。那么，后面的逻辑就不难了，并最终得到了r2c789(5)全部为假。
And it is clear that the block structure that we refer to here is unlikely to be “at least two cells to fill this number”, so it may simply be “no cells to fill this number”. And then, when it comes to the figure, it means that r3c79 cannot be filled in 5. So the chain is set up: r3c2=5, i.e. the result that the “block nodes” are faked with the result that the “cancell nodes are authentic” is obtained, that is, from the false to the true, and this form is strong at this point. So the logic behind it is not difficult, and eventually it is all faked by r2c789(5).
从例子里，我们得到了一些结论，下面来总结一下。当区块作为节点时：
From the example, we have some conclusions, which are summarized below. When blocks serve as nodes:
节点为真表示区块成立，即区块涉及的单元格里恰好有一个单元格填入该数字，即只有一处候选数为真；
Node is the true indication that the block is formed, i.e. that there is just one cell in the cell involved in the block
节点为假表示区块涉及的单元格里，没有一个单元格能放该数字，即每一个候选数全部为假。
Node is a false reference to the cell involved in block, and none of the cells can place the number, i.e., every number of candidates is false .
那么此时区块为假的逻辑就可以套用到最开始的那一则示例里理解了：r89c7(9)区块为假，指的是这两个候选数全部为假。这完全符合我们之前推理的要求。
So the logic of block sizes can then be applied to the first example: r89c7 (9) blocks are fakes, which means that both candidates are fakes. This is exactly what we've been arguing about.
所以，区块节点的基本逻辑就只有上述两种情况了。
Therefore, the basic logic of block nodes is the only two cases mentioned above.
下面我们来看一则区块节点放入不连续环的逻辑。
Here's the logic of putting a block node in an incoherent ring.
如图所示，链的表述如下：
As shown in the figure, the chain is presented as follows:
区块为假，指的是无法使得结构构成区块。无法使得区块结构成立，即r45c6同时都没有候选数7。这个时候，由于都没有7的缘故，在c6之中，7只能填入到r2c6之中。所以此时r45c6(7)区块不成立，即r45c6(7)区块为假时，r2c6(7)为真。这样一来，r45c6(7)就和r2c6(7)构成了特别的强关系。于是接着向下推导。
Blocks are false, meaning that it is not possible to make the structure form blocks. It is not possible to make the structure of blocks, i.e., r45c6 without a candidate number 7. At this point, because there is no 7 at all, 7 can only be filled into r2c6. The r45c6(7) block is not valid at this time, i.e., r45c6(7) is authentic when the r45c6(7) block is false. Thus, r45c6(7) constitutes a special strong relationship with r2c6(7).
那么，这种结构怎么删数呢？我们可以知道的是，AIC的逻辑，导致的是链头和链尾至少一个为真，这里应该指的是r45c6(9)这个区块和r6c5(5)至少一个为真。既不同数，要找的又是一个区块和一个候选数的交集，这怎么找？
So, how can this structure be deleted? What we can see is that the AIC logic leads to at least one true head and end of the chain, and this should refer to the r45c6 (9) block and r6c5(5) at least one true. How can we find this if there is a difference in numbers and a combination of blocks and a candidate number?
区块内有且仅有一处填数位置是为真的，也就是说，r45c6只有一格是7，但是我们无法确定哪个是7。那么链头的确切位置就无法确定，到底是r4c6还是r5c6；但是链尾处，是唯一确定的候选数r6c5(5)。当区块成立时，b5内一定不可以填7了，这包括了r6c5(7)；而r6c5(5)为真时，r6c5当然也就不可以填7。所以r6c5 <> 7。
It is true that there is only one filled position in the block, i.e. only 7 in r45c6, but we can't determine which is 7. The exact position of the chain head is uncertain, i.e. r4c6 or r5c6; but the end of the chain is the only identified number of r6c5(5). When the block is formed, it must not be filled in 7, which includes r6c5(7); and r6c5(5) when it is true, r6c5 is certainly not. R6c5 & lt; > 7. So r6c5 & lt; & gt; 7.
这个结构是不连续环的拓展版本，因为带了一个区块，所以这个结构也被直接称为区块不连续环（Grouped Discontinuous Nice Loop）。
This structure is an extended version of a non-continuous ring, and because it has a block, it is also known directly as (Grouped Discontinous Nice Loop).
特别需要引起注意的是，区块节点如果动用首尾异数链的逻辑时，区块并不是所涉及的单元格，每个单元格都能删除候选数，例如此例，删数并不会产生在区块节点上。
Particular attention is drawn to the fact that when block nodes are drawn to the logic of the end-of-the-synchronous chain, blocks are not cells involved and each cell is able to delete the number of candidates, for example, in this case, the number of deletions does not result in the number of blocks nodes.
4-1?空矩形
"font-size-20" <4-1? Empty rectangle
我们试想一下，如果一条链带若干个区块，而且区块之间有重叠的单元格被共同使用了，这将如何理解？
Let us imagine how this would be understood if a chain of several blocks, with overlapping cells between blocks, were to be used together.
如图所示，如果r8c5(9)为假，则r8c7(9)真，于是将r123c7(9)看作区块，这个区块节点为假，接着得到b3里还剩下r1c89(9)，它们组成区块结构，于是可以保证两处有一处为真，故组成的r1c89(9)区块节点可以表示为节点为真。最终该结构的删数就是链头r8c5(9)和链尾r1c89(9)的交集。显然，这个区块和候选数的交集只有r1c5(9)，所以它便是删数。
As shown in the figure, if r8c5(9) is false, then r8c7(9) is true, so that r123c7(9) is treated as a block, this block node is false, and then it is obtained that r1c89(9) is left behind in b3 and that they form a block structure that can be assured that there is one in both, so that the node of r1c89(9) may be expressed as a node. Ultimately, the number of deletions of the structure is the intersection of chain head r8c5(9) and chain end r1c89(9). Obviously, the sum of blocks and the number of candidates is only r1c5(9) and thus it is.
这个技巧叫做空矩形（Empty Rectangle），虽然和“矩形”沾边，但跟唯一矩形、拓展矩形甚至是可规避矩形里的“矩形”没有任何关系，它仅仅指的是一个结构，即宫内L形状的五个单元格，即此处的{r1c789, r23c7}五个单元格。另外，空矩形一词还可以用来指代这个链技巧。不过……
This technique, known as empty rectangle, which, although tied to a rectangle, has nothing to do with a single rectangle, an extended rectangle, or even a evadable rectangular rectangle, refers only to a structure, i.e. five cells in the intra-uterine L shape, i.e. {r1c789, r23c7} five cells here. Furthermore, the word empty rectangle can also be used to denote this chain technique.
4-2 空矩形的疑问
"font-size-20" 4-2 empty rectangular question
可以看到，实际上结构还是很简单和清晰的，不过很典型，因为这个例子里连续用到了区块和区块之间的强弱关系。
As can be seen, the structure is, in fact, simple and clear, although it is typical, because this example is consistently used in strong and weak relationships between blocks and blocks.
那么，对于空矩形还有一个疑问。如果我把空矩形技巧画成这样下图，请问这个链是否能正常使用，并得到一样的删数呢？
So, there's a question about the empty rectangle. If I paint the empty rectangular technique like this, can this chain be used properly and get the same number of deletions?
为了表达清楚，这里也写出文本形式：
For the sake of clarity, the text is also written here:
实际上是可行的。虽说这里重叠了一点位置，不过现在要详细阐述此点逻辑。
It is actually possible. Although there is a little overlap here, it is time to elaborate on this logic.
首先，我们通过之前一样的推理可以到r123c7(9)处。此时得到的结论是节点为假，即r123c7没有一个单元格填9。接着，我们写了r1c789(9)，虽然其中r1c7是两个区块节点共用的部分，而且此时r1c7(9)已经为假了。但是这并不妨碍r1c789(9)这个节点的其它两个单元格为真。因为之前说过，区块为真表示区块结构成立，所谓的成立就指的是区块里有一个单元格填这个数就可以了。显然，b3里现在只剩下r1c89可以放下9了，也就直接意味着r1c89必然有一个9的出现，所以总的来说，r1c789里确实有一个9了，所以区块是成立的，也就意味着该区块结构为真。
First, we can go to r123c7 (9) through the same reasoning that we had before. The conclusion that was reached was that the node was false, i.e., that there was no single cell in r123c7. Then we wrote r1c789 (9), although r1c7 was a shared section of the two blocks, and at this point r1c7 (9) was false. But that did not prevent the other two cells in the node r1c789 (9). Because, as was said earlier, the node was actually formed, so that the so-called formation meant that there was a cell in the block. Obviously, only r1c89 is left in the b3 now, which means that there must be a 9 in the r1c89, so that, in general, there is a 9 in the r1c789, so that the block is established and so that the structure of the block is real.
正因为如此，空矩形才得以存在。实际上，在一些资料和文献之中（包括发现人对于空矩形的介绍）都是采用的重叠的形式介绍的。
That is why an empty rectangular exists. In fact, it is presented in an overlapping form in some information and literature, including the finder's presentation of an empty rectangular.
4-3?节点重叠
接下来我们就来介绍一个节点重叠的普通版示例。
Let us now turn to an example of a generic version of the overlapping nodes.
如图所示，链的表达如下。
As shown in the figure, the chain is expressed as follows.
可以从示例里得到，链头和链尾都是区块节点，但区块有所重叠：共用了r1c4(4)。那么，删数怎么看呢？实际上，删数就是这两个区块的交集，而这两个区块的交集正是b2的其余单元格，所以r2c5和r3c6自然不能够填入4。
From the examples, the chain head and the end of the chain are block nodes, but the blocks overlap: they share r1c4(4). So, how do you see the deletions? In fact, the deletions are the intersection of the two blocks, which are the rest of the b2 cells, so that r2c5 and r3c6 cannot naturally be filled in 4.
下面来看看这个鱼图测试题，和之前一样，“/”表示不能放候选数a的地方。这个测试题，你能找到合适的删数a的确切位置吗？
Now, look at this fish map test question, which, like before, "/" means you can't put a candidate number. Can you find the exact location of the right missing a?
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