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November 29

Escape room balance puzzle

Escape room balance puzzle

The four colours represent integer weights, not necessarily distinct, from 0 to 9 inclusive. What's a simple way to deduce the weight of each colour without trial-and-error or backtracking? Thanks, cmɢʟeeτaʟκ 15:23, 29 November 2021 (UTC)

If I'm interpreting the diagram correctly, we get the following relations: b < g, y < g, 2r < y, b+g = y+2r, y+g = 3b. Looking at the second equation, since y and g are distinct (by the second inequality), their sum is at most 17. So b is at most 5.
Combining the first equation with the second inequality gives b < 2r.
Subtracting the two equations and simplifying gives y+r=2b. Combining with the third inequality gives 3r < 2b.
So our possibilities for (b, 2b) are (0, 0), (1, 2), (2, 4), (3, 6), (4, 8) or (5, 10). Only (5, 10) contains both a multiple of 2 and a multiple of 3, so b = 5, and r = 3. Plugging this into the earlier equations gives y = 7 and g = 8.--108.36.85.111 (talk) 16:41, 29 November 2021 (UTC)
You can approach such problems more systematically using integer programming techniques. Combining the two equations gives g = b + r. Also, because the values are integers, b + 1 ≤ g, y + 1 ≤ g, 2r + 1 ≤ y. This gives
0=−b+g−r
0=−3b+g+y
1=−b+g−s1
1=g−y-s2
1=y−2r-s3
where all variables are ≥0. Solving for b, g, y, r, s1 in terms of s2 and s3 gives
b=5+3s2+2s3
g=8+5s2+3s3
y=7+4s2+3s3
r=3+2s2+s3
s1=2+2s2+s3
This gives all non-negative integers solutions as s2 and s3 range over the non-negative integers. (The magic is deciding which variables to solve for. There are 7 choose 2 = 21 possibilities but the simplex algorithm can be used to avoid trial and error.) From the second equation it's clear that the only solution with g<10 has s2 = s3 = 0, making the only solution b=5, g=8, y=7, r=3. --RDBury (talk) 22:23, 29 November 2021 (UTC)
Thank you very much, RDBury and user at 108.36.85.111. Good insights, in particular that blue <= 5. Cheers, cmɢʟeeτaʟκ 00:27, 30 November 2021 (UTC)

December 4

What's the way to know what country is located 180 degrees from some specific city of another country?

What's the way to know what country is located 180 degrees from some specific city of another country? --ThePupil (talk) 00:05, 4 December 2021 (UTC)

@ThePupil: By 180 degrees, I assume you mean "directly opposite" (as in, if you theoretically were able to go through the core of the earth). Those places are called antipodes and there are online tools for finding them, though mathematically you could negate the latitude and subtract/add 180 from the longitude (depending on which hemisphere you're in). For example, Perth is at roughly -32,116, and its antipode is at roughly 32,-64. eviolite (talk) 00:28, 4 December 2021 (UTC)
Linguistic peeve — the alleged word antipode is a barbarism. The word antipodes is both singular and plural.
The singular back-formation antipode has become sanctioned by being so commonly used. If you are etymologically strict, singular antipodes is equally a solecism; Ancient Greek ἀντίποδες (antípodes) is strictly a plural noun, the plural of ἀντίπους (antípous). I have no qualms with pistachio and its pronunciation /pɪsˈtæʃioʊ/ either, although this is spelled pistacchio in Italian and pronounced /piˈstak.kjo/ (pi-STAAH-kyoh).  --Lambiam 11:20, 4 December 2021 (UTC)
Antipodes means "opposite feet". If you're standing at the antipodes from me, your feet (of which you have two) are opposite to mine, so I don't see a problem. I think antipode is clearly substandard; it's on a par with matrice or ephemeride or bicep — note that the last one is also very commonly used, but our biceps article correctly avoids it. The biceps is a single muscle with two heads, which is what the word means. --Trovatore (talk) 20:46, 4 December 2021 (UTC)
But I wouldn't write, "My esteemed opponent's feet (of which they have two, last time I counted them) is opposite mine." My beef is with the use of antipodes as a singular noun. In Ancient Greek, your ἀντίπους is someone having their feet opposite of you. Ancient Greek ἀντίποδες refers to a plurality of such counterfooters.  --Lambiam 23:41, 4 December 2021 (UTC)
Antipodes in English is a singular (and also plural) noun, just as is biceps. I hope you wouldn't write bicep in an encyclopedic context? I agree it could have happened that we had adopted antipus as the singular, but we didn't. I don't agree at all that "antipode has become sanctioned by being so commonly used"; that's just not true. --Trovatore (talk) 18:36, 5 December 2021 (UTC)
In Latin, biceps is singular, a nominalized adjective: "the two-headed one (viz. muscle)". The Latin plural is bicipites. Many muscles have medical names that are nominalized Latin adjectives, such as the longissimus.  --Lambiam 22:01, 5 December 2021 (UTC)
It remains the case that antipodes has long standing as the singular and plural form in English. On the other hand antipode looks terrible, as plural-to-singular back-formations generally do. Maybe a hundred years from now it will be unremarkable (like pea from pease), but for now it still comes across as informal or substandard usage. --Trovatore (talk) 22:05, 5 December 2021 (UTC)
Singular antipode was not invented yesterday either; it's almost as old as pea: [1], [2], [3] .  --Lambiam 22:57, 5 December 2021 (UTC)
It so happens that if you match up the globe with the reflected globe, you'll see that land mostly matches up with water. The antipodes of Los Angeles, for example, is in the Indian Ocean, hundreds of miles from land, and the closest land is some tiny island in the French Southern and Antarctic Lands. You can easily find the exceptions in the map to the right. Some of Argentina and Chile matching up with China/Mongolia, some of the rest of South America with Southeast Asia, New Zealand and Spain, some Pacific islands with Africa. --Trovatore (talk) 01:02, 4 December 2021 (UTC)
Using Eviolite's recipe, we find that the antipode of the Kaʿbah at 21.4225°N, 39.8262°E is at 21.4225°S, 140.1738°W (140.1738° = 180° − 39.8262°) in the southern Pacific Ocean, about 50 km (30 mi) from from the Tematagi atol; see also Tematagi § Antipode of Mecca. I suppose a Muslim performing salah right there could use any direction as the direction of prayer.  --Lambiam 11:39, 4 December 2021 (UTC)

Geometric shape for four (or more) cylinders around a central cylinder

I am sure it has a specific name that was linked somewhere on discussions of the Energia (rocket) but I can't find it again. JoJo Eumerus mobile (main talk) 20:56, 4 December 2021 (UTC)

Quincunx? Rojomoke (talk) 06:27, 5 December 2021 (UTC)
The term "quincunx pattern" is used for the arrangement of the rocket engines in the S-IC and S-II, two stages of the Saturn V.  --Lambiam 07:07, 5 December 2021 (UTC)
Yes, exactly that! JoJo Eumerus mobile (main talk) 21:31, 5 December 2021 (UTC)

December 5

Improving secrecy

This question is inspired by quantum cryptography, but you don't need to be familiar with that.

Alice and Bob have a method for generating a sequence of random bits ${\displaystyle b_{0},\dots ,b_{n-1}}$ that they both know, such that Eve only knows 50% of them. More formally, each ${\displaystyle b_{i}}$ is independently uniformly distributed, and Eve has probability 1/2 of knowing the value of ${\displaystyle b_{i}}$ (again, independent of anything else). For each bit, Eve knows whether or not she knows it.

If Alice and Bob want Eve to know fewer, they can split the sequence into two and perform bitwise addition mod 2. So ${\displaystyle a_{i}=b_{2i}+b_{2i+1}\mod 2}$, for ${\displaystyle i. Then Eve has probability 1/4 of knowing any given ${\displaystyle a_{i}}$, but the sequence is only half as long.

In general, Alice and Bob can create a sequence of length ${\displaystyle n/k}$ where Eve has probability ${\displaystyle 2^{-k}}$ of knowing any given bit. Note that if ${\displaystyle l}$ is the length of the sequence and ${\displaystyle p}$ is Eve's probability of knowing a bit, this maintains ${\displaystyle l(-\log _{2}p)=n}$.

Now the question: Can Alice and Bob do better? Is there a process such that ${\displaystyle l(-\log _{2}p)>n}$? I suspect the answer is no, but I don't know how to show this.--72.80.81.137 (talk) 21:02, 5 December 2021 (UTC)

I should add: it's important that Eve's probability of knowing a given bit remains independent of other bits. Otherwise you could do much better.72.80.81.137 (talk) 21:46, 5 December 2021 (UTC)
I assume that Eve knows the recipe. In the most general formulation, each output bit is produced by a different function, each of which is a function of some subsequence (not necessarily a contiguous segment) from the original sequence, containing the bits that are used in the computation. So taking the output to be ${\displaystyle c_{0},...,c_{m-1},}$ we could have ${\displaystyle c_{4}=f_{4}(S_{4})=f_{4}(b_{3},b_{7},b_{8}),}$ in which each input bit is "live", meaning it may influence the outcome. Assume that for some pair ${\displaystyle i,j,i\neq j,}$ ${\displaystyle S_{i}}$ and ${\displaystyle S_{j}}$ share some bit ${\displaystyle b_{k},}$ so ${\displaystyle c_{i}=f_{i}(...,b_{k},...)}$ and ${\displaystyle c_{j}=f_{j}(...,b_{k},...).}$ If we know that Eve knows ${\displaystyle c_{i},}$ this increases the likelihood she knows ${\displaystyle c_{j}}$ as well. This will not do for a solid proof, but the requirement of knowledge independence seems to imply that input bits cannot be "reused", but that each can contribute to one output bit only. This would means that you cannot do better than ${\displaystyle \textstyle {\sum _{i}-\log _{2}p_{i}=n}.}$  --Lambiam 22:38, 5 December 2021 (UTC)
Thinking about how to formalize your argument, I came up with the following: Consider the probability space ${\displaystyle \{0,1\}^{n}}$, where an element indicates which bits of the original sequence are known to Eve. The measure of a singleton is ${\displaystyle 2^{-n}}$. The probability of Eve knowing every bit of the new sequence is ${\displaystyle p^{l}>0}$, so there must be at least one point from the space giving this outcome, meaning ${\displaystyle p^{l}\geq 2^{-n}}$, which gives ${\displaystyle p(-\log l)\leq n}$, as desired. However, this is assuming that which bits Eve knows of the new sequence depends only on which she knew of the old sequence, and not on the values of the bits in the old sequence. I can't find a way to incorporate the values of the original bits without breaking independence, but I'm not seeing an argument that it's impossible.72.80.81.137 (talk) 05:35, 6 December 2021 (UTC)

Density of logical valid statements

Let say I can encode each statement in Propositional calculus, for example by letters and ignore syntax invalid statements (for example a->->b). I ask what is the density of the valid statements (in strings with size not larger than n)? --Exx8 (talk) 23:04, 5 December 2021 (UTC)

I don't think you're going to be able to come up with a satisfying expression in terms of n for a fixed n; it depends too much on the details of the coding.
You might have more luck with the asymptotic behavior — I would be strongly tempted to conjecture that logically valid statements have asymptotic density zero given any reasonable coding. That's basically because a valid statement has to be true in every structure of a given signature, even without imposing any non-logical axioms whatsoever on the structure. That seems like an unusual condition that's going to get more and more unusual as statements get more complex.
However I don't have either a proof or a source; maybe someone else does. --Trovatore (talk) 23:51, 5 December 2021 (UTC)

Oh, I just noticed that you specified the propositional calculus, whereas I was thinking of first-order predicate calculus. My conjecture is still the same, though. A valid statement has to be true no matter what values you assign to the propositional variables. Again that seems like something that's not going to happen very often. --Trovatore (talk) 23:53, 5 December 2021 (UTC)
Among the formulae of length 3, in which case it does not matter whether the notation is infix or Polish, when there are ${\displaystyle v}$ propositional variables, the density of tautologies is ${\displaystyle 1}$ over ${\displaystyle v{+}2.}$  --Lambiam 00:34, 6 December 2021 (UTC)
At least, this is the case using only logical constants ${\displaystyle \top }$ (true), ${\displaystyle \bot }$ (false), ${\displaystyle \neg }$, ${\displaystyle \wedge }$ and ${\displaystyle \vee }$. The density ${\displaystyle {\tfrac {1}{v{+}2}}}$ then also applies for formulae of length 1 and 2, making it plausible that this may remain constant for longer formulae. Adding ${\displaystyle \rightarrow }$ may make a difference, but if the ${\displaystyle {\tfrac {1}{v{+}2}}}$ conjecture holds for formulae with the original set of five constants, this too will have a non-vanishing asymptotic density.  --Lambiam 04:02, 6 December 2021 (UTC)