一些逻辑训练的谜题：娜塔莎、十二点钟和茶
我的同事，娜塔莎喜欢小孩子，就给他们出了一些题。而我在十二点钟和娜塔莎闲谈时，总会为她补充一些，而她也觉得这还不错。到了下午，我们一起喝茶时，我们会把饼干放在鸟笼边，猫头鹰则会将其啄成碎片。我们的生活就是这样悠闲地度过的，和这所大学里的本科生的终日忙碌仿佛处在两个世界，他们没有心情欣赏校园里的建筑多么有艺术感，也看不见柳树和池塘蕴含着十四行诗——就算你不是莎士比亚也可以发现——只等着被抓出来和记录下来，但我们是刚好反过来的。
在这里，我会归纳一些我们设计的，助益于逻辑能力的谜题。
在这里，我会归纳一些我们设计的，助益于逻辑能力的谜题。
1179 个评论
>>Sorry, but I don't use any.
I see, had you one I would be able to create a puzzle about that name, just like what I did for okamiyu. But anyway, I'm very thankful you can participate in this quiz game.
https://www.youtube.com/watch?v=7dIZv5ec2gY
分享一个逻辑问题喜剧，虽然是日文的，希望你能喜欢。
分享一个逻辑问题喜剧，虽然是日文的，希望你能喜欢。
2024年6月（14-1）日
https://pincong.rocks/question/item_id-753691
https://pincong.rocks/question/item_id-753691
>>2024年6月（14-1）日
Q1: okamiyu
Check:
他的用户名由7个各不相同的小写英文字母（将a,e,i,u,o视为元音字母，其余是辅音字母）组成，且含有一个字母y。 TRUE
所有的元音字母前面必然是一个辅音子母，除非这里提到的元音字母是名字开头的字母，
o - ka - mi - yu TRUE
最终元音字母多于辅音字母。
4 vowels (o, a, i, and u) 3 consonants (k, m, and y). TRUE
"y" can also be a vowel depending on the language, but still TRUE
该用户名第5、2、4、1个字母在a-z的字母顺序上是用跳过一个字母的方式依次连续出现的，
i (j) k (l) m (n) o TRUE
且该用户名中比第4个字母在a-z顺序上靠前和靠后的元音字母数量相等，
a, i, k < m < o, u, y TRUE
该用户名第3个字母在a-z顺序上和其它6个字母中任意一个的距离（例如a与b的距离是1，e与h的距离是3）大于那6个字母之中任意两个之间的距离。
i-a = 8
k-i = 2
m-k = 2
o-m = 2
u-o = 6
y-u = 4
TRUE
Answer: okamiyu
QED
Q2: ismynewmail (RS = 535)
a b c
5 3 5
a+b+c = 13 (prime)
b = 3 (prime)
c = 5 (prime)
b+c = 8 (even)
c < 27a-26b+b*a^2-a^3
5 < 7
TRUE
b<27c-26a-c*a^2+a^3
3 < 5
TRUE
Answer: ismynewmail (RS = 535)
QED
>>Q1: okamiyuCheck:他的用户名由7个各不相同的小写英文字母（将a,e,i,u,o视为元...
You got both answers correct, but if you can provide proof, that will be even more magnificent.
>>You got both answers correct, but if you can provi...
That'd be too much for me this time, I got this one mainly through familiarity, so I worked backward. I had a hunch who they were, jotted down their name or RS, and then checked the answer against your terms. And voila. :p
>>That'd be too much for me this time, I got this on...
Amazing, it seems you know people here really well! And as for today's logic puzzle, it's just the very first question which is more about logic reasoning, because Q2 is more like a test on purely math skills.
>>Amazing, it seems you know people here really well...
For the first one, when I saw the "y" requirement and the distribution between vowels and consonants, I thought of "okamiyu" almost immediately. Since it is a 7-letter name that meets the most obvious terms, I went straight to the final checking. Lazy me.
>>2024年6月（14-1）日
a+b+c、b、c都为质数，b+c为偶数，c<27a-26b+b*a^2-a^3，且b<27c-26a-c*a^2+a^3。
b+c为偶数，10以内的偶数且质数只有2，不存在偶数+偶数，那么b和c都是奇数。
c-b<27(a-b)+b*a2-a31️⃣
27a-26b+b*a^2-a^3-c>0②
b-c<27(c-a)-c*a2+a3③
27c-26a-c*a^2+a^3-b>0④
1️⃣+③
27(c-a+a-b)-c*a2+a3+b*a2-a3>0
27（c-b）-a2（c-b）>0
（c-b）（27-a2）>0
c-b>0且（27-a2）>0
或者c-b<0且（27-a2)<0
10以内的质数且奇数3,5,7
a+b+c<22且>8
11,13,17,19。
a可能是1,3,5,7，9
他的声望目前大于400，小于600
那么a只会是5
27-25>0
那么c-b>0
c>b
>>a+b+c、b、c都为质数，b+c为偶数，c<27a-26b+b*a^2-a^3，且b<27c-26...
Good Job, 100% Correct!
Actually I designed this problem to test on knowledge about inequation theorems, but you see, with simple calculation it can also be solved.
>>a+b+c、b、c都为质数，b+c为偶数，c<27a-26b+b*a^2-a^3，且b<27c-26...
Do you mind if you check out an alternative question, when you're free?
https://pincong.rocks/article/70474
>>Good Job, 100% Correct!Actually I designed this pr...
是的，我觉得这还包括了因数分解的知识点，数字有诱导你去因数分解，
看的出来你是有降低难度的。www
Let me offer a puzzle, if I may.
I am looking for an English word that appeared in a Pincong article (replies included).
The word contains 14 letters, 6 of which are vowels.
There is 1 letter that appears three times in the word.
There are 4 letters that appear twice in the word.
You can see the word on Pincong following this clue:
It appeared under an article (replies included) with an article ID number of 5 digits.
The ID number is divisible by 83.
The ID number follows this pattern: c b a c b,
whereas c = b + 1, and b = a + 1.
The sum of all the digits is 16.
Question 1: What is the word?
Question 2: Is it possible to identify the word without going through the scavenger hunt for the post?
Question 3: Is there any redundant clue in this question?
I am looking for an English word that appeared in a Pincong article (replies included).
The word contains 14 letters, 6 of which are vowels.
There is 1 letter that appears three times in the word.
There are 4 letters that appear twice in the word.
You can see the word on Pincong following this clue:
It appeared under an article (replies included) with an article ID number of 5 digits.
The ID number is divisible by 83.
The ID number follows this pattern: c b a c b,
whereas c = b + 1, and b = a + 1.
The sum of all the digits is 16.
Question 1: What is the word?
Question 2: Is it possible to identify the word without going through the scavenger hunt for the post?
Question 3: Is there any redundant clue in this question?
且该用户名中比第4个字母在a-z顺序上靠前和靠后的元音字母数量相等
这句话是什么意思，a-z一共有5个元音字母是奇数除非第四个字母为元音，
不如前后的元音字母是不会相等的，但根据条件可以得知第4个字母是辅音，
是不是我的理解有偏差？
>>Let me offer a puzzle, if I may.I am looking for a...
c=b+1, b=a+1, and a+2b+2c=16, so a=2, b=3, c=4
according to the description and the artcle&replies, I think
Q1: Interpretation
Q2: No, because
Q3: Obviously the "divisible by 83" is redunant
>>这句话是什么意思，a-z一共有5个元音字母是奇数除非第四个字母为元音，不如前后的元音字母是不会相等的...
感谢你为谜题付出的时间。
“比第4个字母在a-z顺序上靠前和靠后的”是定语，你可以视作一个复杂的形容词，形容两类事物。它的修饰对象是“该用户名中的元音字母”，数量比较发生在用户名中而非字母表里。
>>感谢你为谜题付出的时间。“比第4个字母在a-z顺序上靠前和靠后的”是定语，你可以视作一个复杂的形容词...
如果是那样这个条件有什么意义呢，
从前面的条件就可以推出用户名的排列是
元辅元辅元辅元
第四个字母本身就前后两个元音，这个条件无法进一步缩小范围。
第五个和第一个字母可以是u和o也可以是o和i，
从最后的第三个字母的信息来推导吗
>>靠前的与靠后的数量相等，这样第四个字母就会在字母表中有一个参考的位置。
第5、2、4、1个字母在a-z的字母顺序上是用跳过一个字母的方式依次连续出现的
这个条件就可以得出这四个位置的信息了不是吗。
这句话的意思是在a-z的字母表里面，5,2,，4,1是连续的且相间一个字母。
那么只要能推出5和1就可以推出4，但追加的4的条件我不知道有什么意义。
就像你说代表4以前有2个元音，后面有2个元音。这无法作为参考。
>>如果5、2、4、1分别意味着O、Q、S、U，那么就为问题增加了一种可能性
所以我开始说可能是oqsu也可能是ikmo，需要进一步的筛选，
而我提问的那句话指得是s或m在用户名的中前后元音数目一样。
但无论怎么样都是元辅元辅元辅，这个条件没法筛选掉其中一个可能性。
>>第5、2、4、1个字母在a-z的字母顺序上是用跳过一个字母的方式依次连续出现的这个条件就可以得出这四...
举个例子，假如去掉这个条件，你的名字既可以叫okamiyu，也可以叫uqasoyi，都符合题目条件。
>>靠前和靠后的元音字母 数量 相等可是你的条件说的是元音的数目相等啊，这和顺序有什么关系？
这个名字之内的元音分成两组，一组是比名字的第4个字母在字母表中顺序靠前的，一组是靠后的，两组数目一样。
那么可知第4个字母所在的字母表位置不是O-U之间，因为这样的话，比它在字母表顺序中靠后的元音字母至多在名字中出现一个，那就是字母U。
>>这个名字之内的元音分成两组，一组是比名字的第4个字母在字母表中顺序靠前的，一组是靠后的，两组数目一样...
我大概懂你的意思了，如果u是第5个字母，第7个字母就没有可以满足条件的元音了，
第4字母前后的两个元音，在a-z顺序中分别在它的前后各两个。
我的语文水平果然不行了，www
谢谢你的夸奖，我也很期待可以与你相见的一天。
【我本来也是想写的后面又不好意思没发www】
最终元音字母多于辅音字母且用户名是由7个字母组成，所有的元音字母前面必然是一个辅音子母
可以得知第一个字母一定是元音，只有这样元音的数目才会多于辅音。
元音字母前面必然是一个辅音子母，所以第二个为辅音，同理后面的排序为
元辅元辅元辅元
1 2 3 4 5 6 7
辅音后是不可能会是辅音，因为元音总数大于辅音。
第5、2、4、1个字母在a-z的字母顺序上是用跳过一个字母的方式依次连续出现的
元，辅，辅，元
那么5和1之间相差6个字母。
a+6=g不可
e+6=k不可
i+6=o
o+6=u
第4个字母在a-z顺序上靠前和靠后的元音字母数量相等
如果第5个字母是u那么第7个字母将无法成为第4个字母之前的元音。
【用户名字母各不相同所以不会出现重复】
那么5241分别为ikmo第7个字母为u
已知有一个字母y。未知的序号为3和6，3为元音，所以6为y
3与其他6个字母的距离大于其中6个中任意两个
比i和y之间的距离还要大且是元音的字母为a
用户名是我的okamiyu
【我本来也是想写的后面又不好意思没发www】
最终元音字母多于辅音字母且用户名是由7个字母组成，所有的元音字母前面必然是一个辅音子母
可以得知第一个字母一定是元音，只有这样元音的数目才会多于辅音。
元音字母前面必然是一个辅音子母，所以第二个为辅音，同理后面的排序为
元辅元辅元辅元
1 2 3 4 5 6 7
辅音后是不可能会是辅音，因为元音总数大于辅音。
第5、2、4、1个字母在a-z的字母顺序上是用跳过一个字母的方式依次连续出现的
元，辅，辅，元
那么5和1之间相差6个字母。
a+6=g不可
e+6=k不可
i+6=o
o+6=u
第4个字母在a-z顺序上靠前和靠后的元音字母数量相等
如果第5个字母是u那么第7个字母将无法成为第4个字母之前的元音。
【用户名字母各不相同所以不会出现重复】
那么5241分别为ikmo第7个字母为u
已知有一个字母y。未知的序号为3和6，3为元音，所以6为y
3与其他6个字母的距离大于其中6个中任意两个
比i和y之间的距离还要大且是元音的字母为a
用户名是我的okamiyu
>>c=b+1, b=a+1, and a+2b+2c=16, so a=2, b=3, c=4acco...
I realized that (5a + 6) = 16 betrayed the article ID at elementary-level math as soon as I posted it. Thank you for indulging me with my amateurish tricks. It almost felt like playing physics with Sheldon of the Big Bang Theory, a.k.a. playing carpentry with Master Lu Ban.
>>I realized that (5a + 6) = 16 betrayed the article...
You're sooooo modest and polite, sir/mam, thank you for the puzzle.
2024年6月15日
这一账号@开心的苹果 开设的目的是恶作剧以及研究如何增加论坛上的误导性信息，从而增大网警压力，它要么由一位品葱上的美国公民运营，要么由一位品葱上的中国公民运营，不存在双国籍的情况，每个品葱账号也只被一个人运营。假如品葱上有某位美国公民没说过英语，ta必然知道南京现在的羊肉串价格，当且仅当某位中国公民说过英语，ta知道南京10年前的羊肉串价格。品葱上所有知道南京现在的羊肉串价格的人中不可能既有美国公民又有中国公民，所有知道南京10年前的羊肉串价格的人中不可能既有说过英语的人又有中国公民。
若本账号说过英语，则账号主人不知道10年前的羊肉串价格。
若本账号没说过英语，则账号主人知道南京10年前的羊肉串价格，而且某些中国公民知道南京10年前、现在之中至少一个时间的羊肉串价格。
请问账号主人到底知不知道10年前的羊肉串价格？
这一账号@开心的苹果 开设的目的是恶作剧以及研究如何增加论坛上的误导性信息，从而增大网警压力，它要么由一位品葱上的美国公民运营，要么由一位品葱上的中国公民运营，不存在双国籍的情况，每个品葱账号也只被一个人运营。假如品葱上有某位美国公民没说过英语，ta必然知道南京现在的羊肉串价格，当且仅当某位中国公民说过英语，ta知道南京10年前的羊肉串价格。品葱上所有知道南京现在的羊肉串价格的人中不可能既有美国公民又有中国公民，所有知道南京10年前的羊肉串价格的人中不可能既有说过英语的人又有中国公民。
若本账号说过英语，则账号主人不知道10年前的羊肉串价格。
若本账号没说过英语，则账号主人知道南京10年前的羊肉串价格，而且某些中国公民知道南京10年前、现在之中至少一个时间的羊肉串价格。
请问账号主人到底知不知道10年前的羊肉串价格？
if （美国公民没说英语）{必然知道南京现在羊肉串价格}
if （中国公民说过英语）{必然知道南京10年前羊肉串价格}
if（知道现在前南京羊肉串价格）{！美国人and中国人}
if（知道10年前南京羊肉串价格）{！说英语and中国人}
if（本人说英语）{本人不知道10年前南京羊肉串价格}
if（！本人说英语）{本人知道10年前南京羊肉串价格}
some中国人知道现在or10年前的羊肉串价格。
以上是为了我看的方便的整理 @新用户12555 可以不用看，看下面的黑体字就可以了
仅当中国公民说英语才能知道10年前烧烤价格，且知道10年前烧烤价格不会既有英语又有中国人
可以得知中国人不知道10年前烧烤价格，知道10年前烧烤价格的一定是美国人，
本人不会说英语则本人知道10年前烧烤价格，那么本人为美国人且知道现在烧烤价格，
中国人必定不知道现在烧烤价格，因为知道现在烧烤价格的只有中国人和美国人其中之一，
这使得中国人既不知道现在烧烤价格也不知道10年前烧烤价格，
与后面的至少知道一方的条件矛盾，则本人不会没说过英文，本人不知道10年前烧烤价格。
if （中国公民说过英语）{必然知道南京10年前羊肉串价格}
if（知道现在前南京羊肉串价格）{！美国人and中国人}
if（知道10年前南京羊肉串价格）{！说英语and中国人}
if（本人说英语）{本人不知道10年前南京羊肉串价格}
if（！本人说英语）{本人知道10年前南京羊肉串价格}
some中国人知道现在or10年前的羊肉串价格。
以上是为了我看的方便的整理 @新用户12555 可以不用看，看下面的黑体字就可以了
仅当中国公民说英语才能知道10年前烧烤价格，且知道10年前烧烤价格不会既有英语又有中国人
可以得知中国人不知道10年前烧烤价格，知道10年前烧烤价格的一定是美国人，
本人不会说英语则本人知道10年前烧烤价格，那么本人为美国人且知道现在烧烤价格，
中国人必定不知道现在烧烤价格，因为知道现在烧烤价格的只有中国人和美国人其中之一，
这使得中国人既不知道现在烧烤价格也不知道10年前烧烤价格，
与后面的至少知道一方的条件矛盾，则本人不会没说过英文，本人不知道10年前烧烤价格。
>>if （美国公民没说英语）{必然知道南京现在羊肉串价格}if （中国公民说过英语）{必然知道南京10...
很好的尝试，不过“一些中国人知道现在和10年前其中一个”是“若本账号没说过英语”的条件下才发生的。
>>当然，我们明天来谈论关于它吧
当你说较长的中文句子且表达比较细腻观点的时候，你的句子的定语会变得非常复杂，
这使得你所说的话的可读性变得很差，让中学生都可以看出你的对话风格。
不过直到你出题之前我都没有发现，反过来想，你不清楚中国白酒和烧烤非常经典的内脏这点
我就应该怀疑的。www
>>很好的尝试，不过“一些中国人知道现在和10年前其中一个”是“若本账号没说过英语”的条件下才发生的。
这有矛盾，
当且仅当某位中国公民说过英语，ta知道南京10年前的羊肉串价格
和
所有知道南京10年前的羊肉串价格的人中不可能既有说过英语的人又有中国公民。
可以推导出中国人不知道10年前，因为中国人只有会英语的前提下才能知道，
但英语和中国人在知道10年前这个条件下不能共存，即使中国人都知道，
那么他们一定不会英语，这就会与之前条件矛盾，所以必须中国人不知道10年前。
那么只有美国人知道10年前，本人如果说过英文则不知道10年前，反之知道，
那么非英语美国人才满足条件，但是非英语美国人如果存在，那么一定知道现在
而知道现在的只有美国人和中国人一种，这与本人不会英文条件下，
中国人知道现在和过去一种矛盾。陷入无解。
>>这有矛盾，当且仅当某位中国公民说过英语，ta知道南京10年前的羊肉串价格和所有知道南京10年前的羊肉...
“本人如果说过英文则不知道10年前”代表有些说过英语的美国人不知道10年前，而不是所有说过英语的美国人这样，所以不能推出：
非英语美国人才满足条件
另外，当你最后推出矛盾的时候，已经走入了“本人不会英文”的假设之中，并不代表本题有矛盾，或许这代表着这种假设被证伪了。
除此之外，“只有美国人知道10年前”应当作“若一个人知道10年前，则其为美国人”解（尽管我实际上没有写“总人群只包括中国人和美国人”，而是“本账号拥有者要么是中国人，要么是美国人”，但此处不妨当作前者是题意），不能说明“存在一些美国人知道10年前”，因为“知道10年前”的人群是否存在是未得到证明的。
假如品葱上有某位美国公民没说过英语，ta[1]必然知道南京现在的羊肉串价格，当且仅当某位中国公民说过英语，ta[2]知道南京10年前的羊肉串价格。
I am confused by the "ta" in the above quote.
Q1. There are three commas and only one period. Does that mean the entire line is one sentence, therefore the "ta" is the same person under difference conditions? Are ta[1] and ta[2] the same person?
Q2. Following that, if ta[1] and ta[2] are the same, do they refer to the post maker, i.e. 开心的苹果?
Q3. Or rather, ta[1] means the American citizen who never spoke English, and ta[2] means the Chinese citizen who spoke English?
Which one is true, Q2 or Q3?
所有知道南京10年前的羊肉串价格的人中不可能既有说过英语的人又有中国公民。
Depending on the interpretation of ta[2], the quoted line may or may not fit in the question. If we accept that the question is well-written and does not contradict its own terms, then it seems ta[2] is not the Chinese citizen who spoke English.
I'm not sure if I'm still on the right path.
>>除此之外，“只有美国人知道10年前”应当作“若一个人知道10年前，则其为美国人”解（尽管我实际上没有...
仅当中国公民说英语才能知道10年前，且知道10年前不会既有英语又有中国人
可以得知中国人不知道10年前，知道10年前的一定是美国人，
本人不会说英语则本人知道10年前，那么本人为美国人且知道现在，中国人必定不知道现在，
因为知道现在的只有中国人和美国人其中之一，这使得中国人既不知道现在也不知道10年前，
与后面的至少知道一方的条件矛盾，则本人不会没说过英文，本人不知道10年前。
>>I am confused by the "ta" in the above quote. Q1. ...
"ta"代表的是“若某人。。。”这种句式中的“某人”。
然后，条件语句的前提是不一定为真的，哪怕这个条件语句是真的——反而你可以通过某种方法来证明该语句的前提部分是假的。
比如当我说，“若牛奶是红色的，则Amy喜欢它，并且Amy喜欢苹果”是已知条件，
且已知Amy只喜欢苹果和牛奶之中的一个东西，那么我就可以推导出“牛奶不是红色的”
Wait. Is this actually a joke?
https://pincong.rocks/article/item_id-1205994#
若本账号说过英语，则账号主人不知道10年前的羊肉串价格。
https://pincong.rocks/article/item_id-1205994#
Given:
P(now) knower = either American or Chinese <*>
P(-10) knower = either English speaker or Chinese <**>
Examine If-clause:
If and ONLY if that a Chinese is also an English speaker, ta knows P(-10). Contradicts <**>
-> Chinese do not know P(-10). <***>
If there is an American who is NOT an English speaker, ta definitely knows P(now). <****>
According to <*>
-> If there is an American who is not an English speaker, no Chinese knows P(now).<*****>
Falsification:
<H2> "If 开心的苹果 does not speak English, 开心的苹果 knows P(-10) AND
Some Chinese citizens know P(now) and/or (P-10), at least one."
-> Someone who does NOT speak English knows P(-10).
-> According to <**> and <***>, 开心的苹果 is NOT Chinese.
-> 开心的苹果 is an American who does not speak English. <+>
-> According to <****>, 开心的苹果 knows P(now).
If <H2> is TRUE that "Some Chinese citizens know P(now) and/or (P-10), at least one."
According to <***> Chinese do not know P(-10),
-> There must be some Chinese who know P(now).
Then, both 开心的苹果, an American <+>, and some Chinese know P(now), which contradicts <*> given.
-> <H2> is FALSE.
-> only <H1> can be TRUE.
-> 开心的苹果 speaks English, and 开心的苹果 does not know P(-10). QED
<H1> Checking process omitted given that 开心的苹果 did post in English before. LOL
品葱上所有知道南京现在的羊肉串价格的人中不可能既有美国公民又有中国公民，
P(now) knower = either American or Chinese <*>
所有知道南京10年前的羊肉串价格的人中不可能既有说过英语的人又有中国公民。
P(-10) knower = either English speaker or Chinese <**>
Examine If-clause:
当且仅当某位中国公民说过英语，ta知道南京10年前的羊肉串价格。
If and ONLY if that a Chinese is also an English speaker, ta knows P(-10). Contradicts <**>
-> Chinese do not know P(-10). <***>
假如品葱上有某位美国公民没说过英语，ta必然知道南京现在的羊肉串价格，
If there is an American who is NOT an English speaker, ta definitely knows P(now). <****>
According to <*>
-> If there is an American who is not an English speaker, no Chinese knows P(now).<*****>
Falsification:
<H2> "If 开心的苹果 does not speak English, 开心的苹果 knows P(-10) AND
Some Chinese citizens know P(now) and/or (P-10), at least one."
-> Someone who does NOT speak English knows P(-10).
-> According to <**> and <***>, 开心的苹果 is NOT Chinese.
-> 开心的苹果 is an American who does not speak English. <+>
-> According to <****>, 开心的苹果 knows P(now).
If <H2> is TRUE that "Some Chinese citizens know P(now) and/or (P-10), at least one."
According to <***> Chinese do not know P(-10),
-> There must be some Chinese who know P(now).
Then, both 开心的苹果, an American <+>, and some Chinese know P(now), which contradicts <*> given.
-> <H2> is FALSE.
-> only <H1> can be TRUE.
-> 开心的苹果 speaks English, and 开心的苹果 does not know P(-10). QED
<H1> Checking process omitted given that 开心的苹果 did post in English before. LOL
>>Given: P(now) knower = either American or Chinese ...
Awesome, you did it clearly and rigorously, and the process is 100% correct.
In fact "ta" means "they" in English, it's pinyin and used when you don't know or don't specify someone's gender. Like, "If there is an American who is NOT an English speaker, they definitely know P(now)."
>>Awesome, you did it clearly and rigorously, and th...
Thank you. About the "ta," that was my customary joke. Due to the gender pronoun wars, especially the fact that the word "they" is ridiculed by some English-speaking comedians as a politically correct word that forces one to give up singular/plural forms, I sometimes use "ta" and include it as a singular non-gender pronoun that the English language should have. Well, but some use "ze" now. smh.
>>Thank you. About the "ta," that was my customary j...
Get it, this joke is so funny, I thought you didn't know "ta", because it's mostly appearing in Simplified Chinese community on the Internet.
In this rigidly singular sentence, given as an important premise to the puzzle, a "they" does make the grammar awkward though.
If and ONLY if that a Chinese is also an English speaker, ta knows P(-10). This is a consistent singular third-person sentence.
If and ONLY if that a Chinese is also an English speaker, they know P(-10). The verb "know" is no longer singular due to "they" but the sentence is still singular. In common English, it is just a sloppy custom. In comedian's language, this is a joke. In old academic papers, people say "he/she knows," which is no longer politically correct. Then, there's ze. Then, my "ta" becomes useful.
If and ONLY if that a Chinese is also an English speaker, ta knows P(-10). This is a consistent singular third-person sentence.
If and ONLY if that a Chinese is also an English speaker, they know P(-10). The verb "know" is no longer singular due to "they" but the sentence is still singular. In common English, it is just a sloppy custom. In comedian's language, this is a joke. In old academic papers, people say "he/she knows," which is no longer politically correct. Then, there's ze. Then, my "ta" becomes useful.
>>In this rigidly singular sentence, given as an imp...
I see, but my grasp on this is that "they" is not plural in people's mind when used, but a different declension form, just like "you" can mean both single and plural, they're just written and used the same, although in different places in the declension table.
>>I see, but my grasp on this is that "they" is not ...
I agree. It was like that, until the pronoun wars. Yet still, it can be sloppy in certain contexts when one seems to forget that a gender has already been identified previously. For instance, A referred B to C, and A mentioned that B is male. C would make a sloppy mistake if they write these words to A:
"Thank you for introducing him at the event. Get them to send a cv, and I'll take a look."
I see such mistakes quite often.
>>I agree. It was like that, until the pronoun wars....
OMG things really change, although I respect and speak according to others' gender identity, I just use one pronoun all the time once I know how people call themselves. And I think the pronoun war sucks, politeness and courtesy of any kind should be voluntary, as long as you don't intentionally insult people.
>>OMG things really change, although I respect and s...
This is a fine example of linguistic fingerprints that has been going on in the recent discussion on Pincong. When one takes "they" as the correct word subconsciously and uses it without thinking, "he" and "she" become anomalies that require extra attention to apply. In the above example, the C was clearly using "they" more often and more automatically than gender-specific singular pronouns. The "he" in the first sentence was an evidence that A's message was taken, and C was paying attention to A's words about B.
I would love some help with this:
Let it be defined that an inconsistent trinity is a triangle where one of the three corners wear a different sign than the other two.
Let it also be defined that a consistent trinity is a triangle where all the three corners wear the same sign.
As shown:
Triangles such as ABC, ABD, ABE, etc. are inconsistent trinities. Triangles such as ACD and BEF are consistent trinities.
The problem is to prove or falsify the following statements:
1. It is impossible to make all triangles into inconsistent trinities no matter how we distribute the signs.
2. As we connect, mark, and sign all the dots (created by crossed lines), we can create infinitely more crossing points and triangles inside the hexagon. At any given moment, with the best effort<*>, there are potentially more inconsistent trinities than consistent trinities. True or false?
<*> "the best effort" means to create as many inconsistent trinities as possible.
Is it possible to show? Q2 is truly blowing my mind. Thanks in advance!
Let it be defined that an inconsistent trinity is a triangle where one of the three corners wear a different sign than the other two.
Let it also be defined that a consistent trinity is a triangle where all the three corners wear the same sign.
As shown:
Triangles such as ABC, ABD, ABE, etc. are inconsistent trinities. Triangles such as ACD and BEF are consistent trinities.
The problem is to prove or falsify the following statements:
1. It is impossible to make all triangles into inconsistent trinities no matter how we distribute the signs.
2. As we connect, mark, and sign all the dots (created by crossed lines), we can create infinitely more crossing points and triangles inside the hexagon. At any given moment, with the best effort<*>, there are potentially more inconsistent trinities than consistent trinities. True or false?
<*> "the best effort" means to create as many inconsistent trinities as possible.
Is it possible to show? Q2 is truly blowing my mind. Thanks in advance!
>>I would love some help with this:Let it be defined...
Wow topological problem, I love it!
Thank you for providing this puzzle, and I'm just busy these couple of days, yet about to update this post today. And I'll solve this later on, may I ask, can I change the given six signs on the big hexagon?
>>Wow topological problem, I love it!Thank you for p...
Sure, please take your time, and there is no hurry for this problem. Thanks for helping.
And sure, you can change the given six signs on the big hexagon ABCDEF, as long as it is to the best effort to create the most inconsistent trinities. Actually I placed them there just to start, but I'm still not sure if I placed them to the best effort, which is why Q2 blows my mind from the get go.
A1: Oczywiście(Obviously), no matter how we place the signs, the six corners of the biggest hexagon will have at least 3 minuses or at least 3 pluses, and any given three corners among them can form a triangle, thus it's impossible to keep all triangles in the picture inconsistent trinities.
A2: Actually, if we place the signs this way(the central one is a minus sign),
all those signs of the same kind are well segregated, and through the only lines they're connected with each other via, we can easily see there're only 7 consistent trinities of the minus symbol, and 4 consistent trinities of the plus symbol.
And we sure don't need to count or calculate to know there're more than 22 triangles in this picture, so there will be more than 11 inconsistent trinities. That is to say, without further constructing new points and lines, we already have more inconsistent trinities than the consistent ones.
A2: Actually, if we place the signs this way(the central one is a minus sign),
all those signs of the same kind are well segregated, and through the only lines they're connected with each other via, we can easily see there're only 7 consistent trinities of the minus symbol, and 4 consistent trinities of the plus symbol.
And we sure don't need to count or calculate to know there're more than 22 triangles in this picture, so there will be more than 11 inconsistent trinities. That is to say, without further constructing new points and lines, we already have more inconsistent trinities than the consistent ones.
>>A1: Oczywiście(Obviously), no matter how we place ...
Thanks for this. In the case that I add a whole set of new lines that create a fully connected hexagon just like the outer hexagon ABCDEF, I think we can cite the result that you have shown, that there are more inconsistent trinities than consistent ones.
How about the "at any given moment" part in Q2? Can there be a moment when certain lines are added (while others are not) without forming a completely connected hexagon inside, there are more potential consistent trinities and inconsistent ones?
>>Thanks for this. In the case that I add a whole se...
I see, you mean ∀ by "any", I thought you meant ∃(by that time I perceived your question as "at some moments...").
KK, I'll work this out.
>>Thanks for this. In the case that I add a whole se...
so for your second question first:
Can there be a moment when certain lines are added (while others are not) without forming a completely connected hexagon inside, there are more potential consistent trinities and inconsistent ones?
There is definitely such moment that satisfies this, for example, based on the picture of my original answer, we can connect a minus point and a plus point without crossing a line. All new formed triangles will have this new formed line as their edges, otherwise they are not new formed, and they're fated to be inconsistent trinities. Considering at the beginning we had more inconsistent trinities, now we still have more of them than the consistent ones.
>>I see, you mean ∀ by "any", I thought you meant ∃(...
Indeed. In case I didn't make it clear enough, here's how it all started:
Edit: the above image was a mistake and not the best effort. See below:
In the case of a David Star without the outer hexagon, we have 8 inconsistent trinities vs 0 consistent trinities.
Every time we add a line to connect any unconnected crossing points, we create new triangles. The last part of Q2 is whether, in this process, at any given moment (with any 1, 2, 3, or n lines added), that we always have more potential inconsistent trinities than consistent ones. It is the uncertain dynamic that overwhelmed me. @-@
>>so for your second question first:Can there be a m...
we can connect a minus point and a plus point without crossing a line. All new formed triangles will have this new formed line as their edges, otherwise they are not new formed, and they're fated to be inconsistent trinities.
This is gold! I should have thought of that.
So it is perfectly possible to predict just by reading the signs of unconnected pairs, am I right?
>>Thanks for this. In the case that I add a whole se...
and may I ask,
1, the best effort means we add as more incomplete trinities as possible at every step, or I can make my own strategy to keep the inconsistent trinities more than consistent trinities?
2, does one new moment mean one new line? Or I can make several lines in one step?
>>This is gold! I should have thought of that. So it...
we can use this to analyze some certain steps(and I just talked about one more step, so definitely no new fully connected hexagons are formed inside), but if we crossed a line by connecting two points, things will change, we have another way to calculate this.
>>and may I ask,1, the best effort means we add as m...
1. The best effort means to create as many inconsistent trinities as possible at every step.
2. Every new line creates a new moment.
To add, I find that the best effort of a given "moment" is not only affected by the lines we add, but also the signs we give.
Not the best effort: 6 inconsistent vs 2 consistent
The best effort: 8 inconsistent vs 0 consistent
Not the best effort: 6 inconsistent vs 2 consistent
The best effort: 8 inconsistent vs 0 consistent
>>To add, I find that the best effort of a given "mo...
Here is a thing, even if the signs are determining the best decision each step, the fact that new lines can create new points with new signs makes it necessary to consider that lines we've added.
And I highly doubt whether we can finally create and connect all the points if we pursue the best solution at every single step, even if there is some way to finish the construction of all lines and dots by steps, I guess I need to prove it somehow.
But I just proved that this david star can create an infinity whose cardinality is larger than the natural number set, of the steps theoretically needed. So technically it's impossible to make sure all lines and dots are made in our process made of steps, which are also infinite though. Are you sure we are going to connect all points and lines by steps, with one line each step?
>>Here is a thing, even if the signs are determining...
Re
So technically it's impossible to make sure all lines and dots are made in our process made of steps, which are also infinite though.
I'm excited by this finding.
Re
Are you sure we are going to connect all points and lines by steps, with one line each step?
To be honest, no.
In fact, this whole problem stemmed from a metaphor in social science, not math or any hard science. I lay out the problem as a thought process in the making. The point was to see whether, at any given point, the analogy breaks in a way that undermines the original point in social science.
If, however, we find the David Star problem is already leading to infinity, I'd be more than happy to conclude it works, at least for now. And thanks again for all the effort and time you spared for me. Please don't let this distract you too much.
2024年6月18日
昨天和阿拉伯国家的爱国爱政府爱强制头巾的网民吵了一架，今天阿列克谢买了一个以色列国旗，并在六角星内部交叉了三笔用蓝笔，让图案看起来更饱满。在不重复涂抹相同线段（不含节点）的前提下，请问这个新的六角星有可能被一笔画完吗？每个节点已经用红色标出。
昨天和阿拉伯国家的爱国爱政府爱强制头巾的网民吵了一架，今天阿列克谢买了一个以色列国旗，并在六角星内部交叉了三笔用蓝笔，让图案看起来更饱满。在不重复涂抹相同线段（不含节点）的前提下，请问这个新的六角星有可能被一笔画完吗？每个节点已经用红色标出。
>>Here is a thing, even if the signs are determining...
If I may probe further, could you please help me understand what "cardinality" means in this context?
But I just proved that this david star can create an infinity whose cardinality is larger than the natural number set, of the steps theoretically needed.
Does it mean that, stemming from the David Star, the steps we need to test how to assign the signs per added line is approaching infinity?
>>2024年6月18日昨天和阿拉伯国家的爱国爱政府爱强制头巾的网民吵了一架，今天阿列克谢买了一个以色列...
The answer should be negative, given that there are 6 points (more than 2) where 5 lines (odd number) are stemmed from the point. I learned that for any shape with more than 2 points where an odd number of lines stem from it, the shape cannot be drawn in one go without repeating any lines. But to be honest, I don't know how to prove it.
>>If I may probe further, could you please help me u...
The infinities are different, you see, it's possible to assign every natural number with a real number, obviously. But we can't assign every real number with a natural number without repeating, otherwise we'll be able to give every real number bigger than 0 smaller than 1 a natural number to make the order.
Let's assume we can give all real numbers in (0,1) a name associated with a unique natural number, then we can write them like,
the 1st number 0.a(1,1)a(1,2)a(1,3)a(1,4)a(1,5)a(1,6).......
the 2nd number 0.a(2,1)a(2,2)a(2,3)a(2,4)a(2,5)a(2,6)......
.......
the a(x,y) above means a digit(ranged from 0 to 9) of such number, a(1,1) means the first digit after the decimal separator of the first number for example. All numbers in this list are supposed to cover all the real numbers in (0,1).
Then let's check out this number
W=0.b1b2b3b4b5....
where b1=a(1,1)+1(if a(1,1) is 9, then let b1 be 0, the same works below)
b2=a(2,2)+1
b3=a(3,3)+1
.....
W is well defined but different from any given kth number at the digit bk, so W won't be in the list, contradicting to our assumption.
just like the case above, we say the cardinality of the R set is bigger than the N set, because we can give each element in the latter one specific and unique element in the former, but we can't do the contrary. And obviously, when two lines are not resembling, we need two or more steps to draw them all. And you see, we need to draw a total number of both dots and lines(of different directions or locations) even bigger than the infinity of natural numbers, while our steps are always like, step 1, step 2, step 3 and so on, thus its cardinality is the same as natural numbers.
>>The infinities are different, you see, it's possib...
Thanks a lot! Now it makes perfect sense. I've learned so much today.
>>Thanks a lot! Now it makes perfect sense. I've lea...
You're the most welcome, I'm so happy to chat with you!
>>2024年6月18日昨天和阿拉伯国家的爱国爱政府爱强制头巾的网民吵了一架，今天阿列克谢买了一个以色列...
非常抱歉
我不擅长需要构图的几何题，
但我觉得这是不可能的，这是一个六芒星，
即使有中间的三个交叉也无法使其一笔完成。
>>非常好，这是正确答案
我的逻辑是，如果知道羊肉串的价格，那是谁啊？
想了一圈，都不符合逻辑和常识，那就不是！
（你编题也不会去反着常识瞎编，我就想了想，你晓得那么久远以前羊肉串价格就怪！我，我都不晓得呢！）
>>你说，一个真正在南京土生土长的人会记得这种细节吗？
如果是我的话，我会的。
我还记得小学的时候门口的辣蘑菇多少钱！（1元一串，当时感觉好贵）还有炸小鱼这些。
如果那个学校的视频你看了，你也大概晓得那是多久以前了（许多许多年了啊！）
这些碎片就是成长过程中比较愉快自然有人性的部分了吧（和小学头顶上慈父画像相比较的话）
但是我不在南京，而且我住的那个地方羊肉串很少有人卖。我现在回想起来都觉得尴尬，当时的羊肉串少见还很小，小得如半块Turkish delight, 说明那边羊肉产量和销量都不怎么样。
我现在闭上眼睛，能想起来，90年代末期，南京街头还有卖塑料拼装玩具的商店，汤包铺，还有卖鸭血粉丝的（但是我很讨厌吃粉丝，我吃掉鸭血和喝掉汤），那个时候南京的公交车还是美国设计的Flxible（车头车标就很古典），但是扣在长江客车的破底盘上面。 南京还有地铁等等。江苏其他的地方也是这样，没有空调很热
https://commons.wikimedia.org/wiki/File:Hangzhou_trolleybus_5708,_a_Changjiang_Flxible_CJWG110,_on_route_151_in_2013.jpg
90年代中期的南京街头还大概是这种味道，
后来要“引进外资”，和“国际”接轨，一下子把社会主义模样的公交车换成了资本主义的模样，社会主义时期的车身颜色，白底彩色条纹也改了，变成各种颜色的广告，毕竟要“市场经济”改革嘛，要学习“资本主义”！
你看车头车标，一看就很高档，身份不一般，比落后的东方集团的车好多了
下面这玩意儿也挂过上面的那个美国标，连开车的司机整个人的精神都不一样了，GM Allison变速器的标闪闪发亮，你看看不只是壳子是美国模样的了，里面也美国化了！
这种一看就是资本主义程度高的体面人！
大概二十年前就是这个态度吧
2024年6月18日谜题的参考解答：
对于图中标为黄色的点，我们可以发现一个关键的事实，假如我们开始画的起点并不在某黄色点延展出的五条线段之一或那个黄色点本身上（哪怕绘画起点在一个黄色点上，对于另外一些黄色点，终归是这样的），那么对于该黄色点，首先会被我们的笔到来，这会涂抹其中的一条线段；然后注定下一步是被我们的笔离开，这会涂抹五条中的另外一条线段。下一次我们的笔发生涂抹这五条线段之一的时候，也会重复相同的事。那么这意味着对于五条中的最后的那条线段，必然是伴随着笔的到来而非离开的，而在这一步发生后，我们将无路可走，所以最后那次涂抹将是整个绘图的最后一笔。但是黄色的点有整整6个，这会引发矛盾。
对于图中标为黄色的点，我们可以发现一个关键的事实，假如我们开始画的起点并不在某黄色点延展出的五条线段之一或那个黄色点本身上（哪怕绘画起点在一个黄色点上，对于另外一些黄色点，终归是这样的），那么对于该黄色点，首先会被我们的笔到来，这会涂抹其中的一条线段；然后注定下一步是被我们的笔离开，这会涂抹五条中的另外一条线段。下一次我们的笔发生涂抹这五条线段之一的时候，也会重复相同的事。那么这意味着对于五条中的最后的那条线段，必然是伴随着笔的到来而非离开的，而在这一步发生后，我们将无路可走，所以最后那次涂抹将是整个绘图的最后一笔。但是黄色的点有整整6个，这会引发矛盾。
如果只論那個店員的行為，其實我會把他視為中共教育讀壞腦子的對象。他並沒有意識到自己的所作所為，他自身並沒有打算對客人做不好的事。他在進行挑釁，但是他並不是打算進行挑釁。這是因為他認為他兢兢業業務實地工作，對自己仍然遭到指責感到不滿的風涼話。大概他把客人視作老闆。中國裡面經常會發生這種身份錯亂的現象。老闆你這是看不到我兢兢業業在手沖嗎，你為什麼指責我？提醒老闆你是不是連店裡的流程都忘記了。如果他是工廠裡面的工人，在完成老闆要求的任務，那麼他是正確的。我已經正確地回覆了8分多完成，我所做的完全符合要求。但，他現在並不是一個工人，客人也不是老闆。服務不是要完成指標，還要對客人進行服務和帶領。否則，這就不是服務了。對此，我只能對中國教育進行搖頭。