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Jan and 5 other children are in a classroom. The principal o

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Joined: 07 Jun 2014
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Jan and 5 other children are in a classroom. The principal o [#permalink]  30 Jul 2018, 11:02
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Question Stats:

62% (00:52) correct 37% (01:19) wrong based on 24 sessions
Jan and 5 other children are in a classroom. The principal of the school will choose two of the children at random. What is the probability that Jan will be chosen?

(A) $$\frac{4}{5}$$
(B) $$\frac{1}{3}$$
(C) $$\frac{2}{5}$$
(D) $$\frac{7}{15}$$
(E) $$\frac{1}{2}$$
[Reveal] Spoiler: OA

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Re: Jan and 5 other children are in a classroom. The principal o [#permalink]  31 Jul 2018, 07:10
There are altogether 6 children hence possible outcome = 6*5 = 30
If Jan is to be chosen.
Then Either Jan is chosen and someone else is chosen or Someone else is chosen first and then Jan is chosen
hence possible outcome = 1*5 + 5*1 = 10

probability = 10/30 =1/3
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Re: Jan and 5 other children are in a classroom. The principal o [#permalink]  21 Aug 2018, 18:29
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Explanation

The probability of any event equals the number of ways to get the desired outcome divided by the total number of outcomes.

Start with the denominator, which is the total number of ways that the principal can choose two children from the classroom. Use the fundamental counting principle. There are 6 possible options for the first choice and 5 for the second, giving (6)(5) = 30 possibilities.

However, this double-counts some cases; for example, choosing Jan and then Robert is the same as choosing Robert and then Jan.

Divide the total number of pairs by 2: $$\frac{6 \times 5}{2}= 1$$. Alternatively, use the formula for a set in which the order doesn’t matter: $$\frac{total!}{in! \times out!}$$.

In this case: $$\frac{6!}{2! \times 4!}=15$$.

Now compute the numerator, which is the number of pairs that include Jan. Since the pair only includes two children and one is already decided (Jan), there are exactly 5 options for the other child. Thus, there are 5 total pairs that include Jan: Jan with each of the other students.

The probability of choosing a pair with Jan is $$\frac{1}{3}$$.
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Re: Jan and 5 other children are in a classroom. The principal o [#permalink]  17 Sep 2018, 06:12
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Is this method right?

Probablility of Jan not being chosen: 5/6 *4/5=4/6=2/3
Probability of Jan being chosen = 1 - 2/3=1/3
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Re: Jan and 5 other children are in a classroom. The principal o [#permalink]  02 Jan 2019, 22:04
Expert's post
kruttikaaggarwal wrote:
Is this method right?

Probablility of Jan not being chosen: 5/6 *4/5=4/6=2/3
Probability of Jan being chosen = 1 - 2/3=1/3

Yes! much simpler too! Thanks !!
Re: Jan and 5 other children are in a classroom. The principal o   [#permalink] 02 Jan 2019, 22:04
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# Talent

### Sidebar

Nuclear Talent
Course 7
NT4A

baboons

This is an old revision of the document!

This is the project page for BaBooNs.

Kaitlin Cook: Parameters

Justin Michael Brown: Code

Rob Almus: Analysis

Aim: Investigate the effects of non-thermal baryon injection on BBN.

Introduction If dark matter particles (e.g. WIMPS) exist, they inevitably did so during the big bang. They also may decay during BBN into non-thermal particles - this may be in the form of photons, leptons (electron/positron pairs and/or neutrinos) or into hadronic channels (n/p, mesons, nuclei). These decays may have two influences into BBN:

1) Change the expansion rate due to the injection of relativistic species.

2) Effect the abundance of light nuclei produced in BBN via reactions with these decay particles.

In case (1), this effect is important only if the decay channels are relativistic, thus adding a lot of entropy into the universe. In this project, we will consider the injection of neutrons, so we will not consider this effect.

It is thus the goal of this project to investigate the effect of the injection of non-thermal neutrons on the light element abundances.

Key parameters:

1. The mean life of the decaying particle X → 2n (some magic decay. Not very physical, probably. But hey, we don't actually know what WIMPS are).
2. The energy of these neutrons (although if it is high enough, it may not matter too much?)
3. The abundance of X.
4. n interaction rates with every other nucleus.
5. Mass of the WIMP. Chosen to be 100 GeV, based on the wikipedia WIMP entry, the font of all wisdom.

This figure:

from CDMS in 2004 suggests that this value isn't too far off.

Reference Papers:

Cyburt et. al. 2009. Nucleosynthesis Constraints on a Massive Gravitino in Neutralino Dark Matter Scenario http://arxiv.org/pdf/0907.5003v1.pdf

Kamionkowski 1994. Diffuse Gamma Rays from WIMP Decay and Annihilation http://arxiv.org/pdf/astro-ph/9404079v1.pdf

Pierre Salati, 2014. Dark Matter Annihilation in the Universe http://arxiv.org/pdf/1403.4495.pdf

Logbook

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# How do you simplify \frac { x ^ { 2} } { y ^ { - 3} } ( \frac { x ^ { 2} } { y } - \frac { 3z ^ { - 2} y } { x } )?

Oct 21, 2017

${x}^{2} / {y}^{-} 3 \left({x}^{2} / y - \frac{3 {z}^{-} 2 y}{x}\right) = {x}^{4} - \frac{x {y}^{4}}{9 {z}^{2}}$

#### Explanation:

This may seem like a complicated problem due to all the variables and exponents. But, it's not.

${x}^{2} / {y}^{-} 3 \left({x}^{2} / y - \frac{3 {z}^{-} 2 y}{x}\right)$

Let's first use the Distributive Property to simplify things a little bit:

$\left({x}^{2} / {y}^{-} 3\right) \left({x}^{2} / y\right) - \left({x}^{2} / {y}^{-} 3\right) \left(3 {z}^{-} 2 \frac{y}{x}\right)$

We can then use some rules about multiplying together exponents with the same base as well as negative exponents to simplify a little bit more:

1) ${a}^{b} \cdot {a}^{c} = {a}^{b + c}$

2) ${a}^{b} / {c}^{b} = {\left(a - c\right)}^{b}$

3) ${a}^{-} b = \frac{1}{a} ^ b$

$\frac{{x}^{2} \cdot {x}^{2}}{{y}^{-} 3 \cdot {y}^{1}} - \left({x}^{2} / {y}^{-} 3\right) \left(3 {z}^{-} 2 \frac{y}{x}\right)$

Let's take care of the left side of the expression first:

${x}^{4} / {y}^{-} 2 = {x}^{4} / \left(\frac{1}{y} ^ 2\right) = {x}^{4} \cdot {y}^{2} = {x}^{4} {y}^{2}$

Now let's work on the right side:

$\left({x}^{2} / {y}^{-} 3\right) \left(3 {z}^{-} 2 \frac{y}{x}\right) = \left({x}^{2} / \left(\frac{1}{y} ^ 3\right)\right) \left(\frac{1}{{\left(3 z\right)}^{2}} \cdot \frac{y}{x}\right) = \left({x}^{2} {y}^{3}\right) \left(\frac{y}{9 x {z}^{2}}\right) = \left(x {y}^{3}\right) \left(\frac{y}{9 {z}^{2}}\right) = \frac{x {y}^{4}}{9 {z}^{2}}$

After all that, let's put them together, while not forgetting the $-$ sign in the middle:

${x}^{4} - \frac{x {y}^{4}}{9 {z}^{2}}$

And, you're done!


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# Making arrows not touch in TikZ

I'm trying to make a fairly simple diagram using TikZ, along the lines of the following examples:

\begin{tikzpicture}
\draw [->] (0,0) node[anchor=south]{A} -- (1,-2) node[anchor= north]{D};
\draw [->] (1,0) node[anchor=south]{B} -- (1,-2);
\draw [->] (2,0) node[anchor=south]{C} -- (1,-2);
\end{tikzpicture}

and

\begin{tikzpicture}
\draw [->] (1,0) node[anchor=south]{E} -- (0,-2) node[anchor= north]{F};
\draw [->] (1,0) -- (1,-2) node[anchor=north]{G};
\draw [->] (1,0) -- (2,-2) node[anchor=north]{H};
\end{tikzpicture}

What I don't like in the first picture is hopefully clear: I would much prefer that the arrows be spaced out above D, rather than all collapse to a single point.

The second picture doesn't bother me as much, since the clustering here is just of line segments and now arrows, but if there were a way to make the three edges spaced out as well, that would be quite nice. (See the top diagram in the chosen answer to this post for an illustration of something I would like.)

I can of course achieve all these results artificially, just by placing nodes separately, and then mucking about with the coordinates of the ends of the arrows. But it would be great if there were a simpler solution. Many thanks for any insight you may have.

-

I would use a \node to place D, and then draw the lines using the name of the node. This automatically solves the problem. Another option would be to use shorten=<dimension> to shorten the end of the arrows by the given dimension. You can also control the ending points of the arrows explicitly, as my third example suggests.

\documentclass{article}
\usepackage{tikz}

\begin{document}

\begin{tikzpicture}
\node[anchor=north] at (1,-2) (D) {D};
\draw [->] (0,0) node[anchor=south]{A} -- (D);
\draw [->] (1,0) node[anchor=south]{B} -- (D);
\draw [->] (2,0) node[anchor=south]{C} -- (D);
\end{tikzpicture}

\begin{tikzpicture}
\node[anchor=north] at (1,-2) (D) {D};
\draw [->,shorten >=3pt] (0,0) node[anchor=south]{A} -- (D);
\draw [->,shorten >=3pt] (1,0) node[anchor=south]{B} -- (D);
\draw [->,shorten >=3pt] (2,0) node[anchor=south]{C} -- (D);
\end{tikzpicture}

\begin{tikzpicture}
\node[anchor=north] at (1,-2) (D) {D};
\draw [->] (0,0) node[anchor=south]{A} -- (D.120);
\draw [->] (1,0) node[anchor=south]{B} -- (D.90);
\draw [->] (2,0) node[anchor=south]{C} -- (D.60);
\end{tikzpicture}

\end{document}

-
Superb. Thanks a lot! :) –  Skeptic Sep 23 '11 at 19:14
Any reason for having a space between anchor= and south? –  N.N. Sep 23 '11 at 19:22
@N.N. no. I just reused some of the OP's original code, where the space is already present. –  Gonzalo Medina Sep 23 '11 at 19:27
@N.N. I don't mind you editing my answer to suppress the spaces, but I'd prefer edits which are really necessary. –  Gonzalo Medina Sep 23 '11 at 19:35
@GonzaloMedina Sorry, I removed the spaces from your answer and the question because they confused me and I thought others might feel the same. –  N.N. Sep 23 '11 at 19:38

One possibility is to use the shorten >= parameter.

\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\begin{scope}[->,shorten >=9pt]
\draw (0,0) node[anchor= south]{A} -- (1,-2) node[anchor= north]{D};
\draw (1,0) node[anchor= south]{B} -- (1,-2);
\draw (2,0) node[anchor= south]{C} -- (1,-2);
\end{scope}
\begin{scope}[->,shorten <=9pt,xshift=3cm]
\draw (1,0) node[anchor= south]{E} -- (0,-2) node[anchor= north]{F};
\draw (1,0) -- (1,-2) node[anchor= north]{G};
\draw (1,0) -- (2,-2) node[anchor= north]{H};
\end{scope}
\end{tikzpicture}
\end{document}

-
Tip: If you're using scopes you may as well put the arrow tips in the scopes and remove them from \draw, e.g. \begin{scope}[->,shorten >=9pt]. –  N.N. Sep 23 '11 at 19:02
Many thanks, this is a useful trick to know. I'm going with Gonzalo's answer, since that applies more directly, but yours is a great tip. Thanks again. –  Skeptic Sep 23 '11 at 19:14
@N.N. You're right, of course. It crossed my mind, but for some reason I didn't change it ... I'll update the code. –  Torbjørn T. Sep 23 '11 at 19:14


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# On the Puzzle of Odd-Frequency Superconductivity - Condensed Matter > Superconductivity

On the Puzzle of Odd-Frequency Superconductivity - Condensed Matter > Superconductivity - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.

Abstract: Since the first theoretical proposal by Berezinskii, an odd-frequencysuperconductivity has encountered the fundamental problems on its thermodynamicstability and rigidity of a homogenous state accompanied by unphysical Meissnereffect. Recently, Solenov {\it et al}. Phys. Rev. B {\bf 79} 2009 132502.have asserted that the path-integral formulation gets rid of the difficultiesleading to a stable homogenous phase with an ordinary Meissner effect. Here, weshow that it is crucial to choose the appropriate saddle-point solution thatminimizes the effective free energy, which was assumed {\it implicitly} in thework by Solenov and co-workers. We exhibit the path-integral framework for theodd-frequency superconductivity with general type of pairings, including anargument on the retarded functions via the analytic continuation to the realaxis.

Autor: Hiroaki Kusunose, Yuki Fuseya, Kazumasa Miyake

Fuente: https://arxiv.org/


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## Algebra 2 (1st Edition)

$(1,0),(0,1)$
When squaring $(1-y^2)$ he got an incorrect result; the correct result is: $1-2y^2+y^4$, then our equation is: $1-2y^2+y^4+y^2-2+2y^2-2y=-1\\y^4+y^2-2y=0\\y(y^3+y-2)=0\\y(y-1)(y^2+y+2)=0$ The third term has a positive discriminant, so it cannot be $0$. Thus, $y=0$ or $y=1$. Thus the solutions are: $(1,0),(0,1)$


text:
GACODE User Agreement
Initial version (2013.07.02)
Updated version (2018.11.06)

GACODE development is funded by the US Department of Energy (DoE),
and distribution is subject to DoE requirements and restrictions.

agrees:

1. The above copyright notice, this list of conditions and the
following disclaimer shall be included in all copies or substantial
portions of the Software.

2. To compile/use the Software source code AS IS for publication
purposes; users however are welcome to request changes, to make
code modifications for testing purposes, or issue pull requests.

3. If the authorized user’s copy of Software is made available to third
parties, to ensure that the user agreement is followed by the third
parties.

4. To email a draft of any article/letter/note based on the Software
use to gacode@fusion.gat.com

5. To include in published results or presentations the code name(s)
and appropriate references (see below)

Appropriate References

Physics code

Topical Area

Required citation

CGYRO

gyrokinetic simulation

[CBB16]

GYRO

gyrokinetic simulation

[CW03b]

NEO

neoclassical simulation

TGLF

transport model

TGYRO

profile evolution

[CHW+09]

## Source code¶

Sources are found in the gacode repository on GitHub.

## Repository access¶

Get a (free) account at GitHub and send your GitHub username to gacode-AT-fusion.gat.com with a request for user or developer access.


text:
Article | Open

# Antarctic sea ice region as a source of biogenic organic nitrogen in aerosols

• Scientific Reportsvolume 7, Article number: 6047 (2017)
• doi:10.1038/s41598-017-06188-x
Accepted:
Published online:

## Abstract

Climate warming affects the development and distribution of sea ice, but at present the evidence of polar ecosystem feedbacks on climate through changes in the atmosphere is sparse. By means of synergistic atmospheric and oceanic measurements in the Southern Ocean near Antarctica, we present evidence that the microbiota of sea ice and sea ice-influenced ocean are a previously unknown significant source of atmospheric organic nitrogen, including low molecular weight alkyl-amines. Given the keystone role of nitrogen compounds in aerosol formation, growth and neutralization, our findings call for greater chemical and source diversity in the modelling efforts linking the marine ecosystem to aerosol-mediated climate effects in the Southern Ocean.

## Introduction

Antarctic sea ice covers between 1% (summer) and 5% (winter) of the global ocean, and influences an even larger area, the sea ice region. It is a vast biome composed of multiple habitats such as the upper and lower ice surfaces, snow cover, brine channels, melt ponds, ice openings and ice floes of all sizes, and the surrounding sea water, all hosting rich microbial communities1. Climate warming-derived changes in sea ice extent2, thickness and texture are expected to exert feedback effects on climate through changes in surface albedo and emission of climate-active substances, such as those forming cloud-seeding aerosols3, 4. An accurate projection of climate dynamics in the Southern Ocean, therefore, requires a sound assessment of aerosol precursor emissions from the sea ice region ecosystem. Here we report results of a study conducted during a research cruise across the Subantarctic and Antarctic waters of the Atlantic sector of the Southern Ocean in summer 2015, where low troposphere aerosols were characterised, source apportioned and compared with concurrent measurements in sea water and sea ice.

## Results and Discussion

Air mass trajectories were used to define “open water” (OW) and “sea-ice” (SI) source regions according to the characteristics of the area overflown (Supplementary Figure 1 and Table 1). A thorough examination of air mass trajectory heights showed that > 80% of the time during the 5 days prior to sampling, the collected air had travelled within the marine boundary layer, with very little intrusion from the free troposphere, and there was no significant differences between SI and OW air masses in this respect (Figures S2, S3 and Table S2). Filter-collected aerosols arriving from OW regions had ten-fold more sea salt and similar amounts of water soluble organic carbon (WSOC) than aerosols originated in the SI region (Fig. 1). By contrast, SI aerosols contained 5 times more water soluble organic nitrogen (WSON). Chemical speciation analyses of WSON revealed the presence of biogenic low molecular weight alkyl-amines, including mono-, di- and trimethylamine. Methylamines released by plankton incorporate into aerosols through reaction with sulphuric or methanesulphonic acids5, which are formed via oxidation of dimethylsulphide (DMS), another volatile of planktonic origin6. In fact, both methanesulphonic acid (MSA) and non-sea-salt sulphate (nss-SO4) were also enhanced in the ice-influenced air masses (Fig. 1).

In situ high time resolution monitoring of aerosol composition by High Resolution Time of Flight Aerosol Mass Spectrometry (HR-ToF-AMS) also revealed the ubiquitous occurrence of organic nitrogen (ON) compounds and their enhancement under sea ice influence (Fig. 1). To our knowledge, these are the first high time resolution aerosol composition measurements reported in the sea ice region of the Weddell Sea. As for the Atlantic Subantarctic region, a study conducted on Bird Island in the early summer reported aerosol compostion consisting of 47% sea salt, 21% nss-sulfate, 22% organics (including MSA and ON) and 8% ammonium7. These figures reflect a higher organic load than in our open-ocean aerosols, which average 85% sea salt, 5% nss-sulfate, 5% WSOC, 2% MSA, 2% ON and 1% ammonium (Fig. 1). In the previous study, however, the high content in organic nitrogen components - including amino acids and amines - was attributed to emissions from densely populated colonies of the island fauna7.

In our work, a geographical source apportionment for our high-resolution aerosol components was conducted by Probability Source Contribution Function analysis of air mass back trajectories. This analysis showed that the most probable, homogeneous source region for the measured aerosol ON was the ice-covered Weddell Sea and its marginal zone of ice-influenced ocean, along with some spots around the Antarctic Peninsula and the South Georgia phytoplankton bloom (Fig. 2).

Biogenic organic nitrogen and sulphur compounds have been shown to play synergistic roles in new particle formation8. We monitored the number of particles in the smallest size range (1–3 nm) in real time, and found an enhancement of almost 3 fold (189 cm−3 versus 69 cm−3) in SI air relative to OW air (Fig. 1). Although we cannot directly link any compound to the observed new particles, these data provide additional evidence of secondary aerosol production under sea ice influence.

Single particle analysis of the ambient aerosols with on-line Aerosol Time-Of-Flight Mass Spectrometry (ATOFMS) further showed an 8-fold enhancement of ON in ice-influenced air masses. Moreover, of all ON-containing particles analysed, 11% were internally mixed with sea salt. Whilst this confirms a major secondary aerosol pathway for aerosol ON formation, it also shows an unneglectable contribution of sea-spray-associated ON. The surface ocean is rich in colloidal and particulate organic matter including proteinaceous gels9, bacteria and viruses, which can act as sources of primary N-containing aerosol by bursts of entrained air bubbles10, 11. Whether sea ice could also be a source of primary ON aerosol was unknown. To test this, we collected sea-ice samples from the northernmost margin of the Weddell Sea. Sample chunks were selected amongst ubiquitous, rafted sea ice floes discoloured by ice algae colonization at freeboard level. They were melted in a bubbling tank designed to produce sea spray. Mass spectrometric analyses of generated particles were compared with those produced by bubbling sea water. AMS characterisation showed that sea spray from melted ice contained more than twice as much organic carbon as seawater spray (11% vs 5%) and as much as 0.3–1% of ON, which was almost undetectable from seawater (Fig. 3). ATOFMS single-particle information on the mixing state revealed that 21% of the generated particles from sea ice were an internal mixture of sea salt with organic nitrogen and carbon, versus only 5% in seawater spray (Fig. 3). Measurements of black carbon and metals in the same sea-spray aerosols ruled out contamination artifacts.

These results indicate that sea ice contained ON that, upon release by ice melt, was susceptible to incorporation into aerosol. Indeed, sprayed aerosol ON content was proportional to the ON content of melted sea ice (Fig. 4). We then explored the origin of ON compounds in the sea ice. Brine channels in ice harbour dense microbial communities1. Chlorophyll a concentrations in the sea ice samples (8–48 µg L−1) were higher than those found in surrounding seawater (0.3–2 µg L−1) and within the range previously found in similar rafted sea ice12. Ice algae were largely dominated by diatoms of the genera Fragillariopsis and Nitzschia, commonly found in polar sea ice and surrounding waters13. Brine channels are also known to harbour abundant organic matter1 that remains largely uncharacterized. Consistent with previous reports14, 15, we measured concentrations of carbon-rich exopolymeric particulate matter in the ice which were ten times higher than in seawater samples. However, neither these C-rich particles nor chlorophyll a concentrations in sea ice samples were positively correlated to endogenous and sprayed ON (Fig. 4). This suggests that algal biomass and production in the sea ice brines is not a good predictor for aerosol-forming ON, probably because the physiological state of microorganisms plays an important role. Indeed, the parallel patterns of ON and the ratio of large viruses to chlorophyll a, and the opposite pattern to nitrate concentrations, suggest that the maturity and decay of the algal assemblage would enhance sea ice ON concentrations. The fluorescence and light absorbance signals for protein-like dissolved organic matter were positively correlated to sprayed ON (Fig. 4). Fluorescence has been previously used to suggest peptide enrichments in Antarctic sea ice16. Moreover, ice brine phytoplankton are known to produce copious amounts of mycosporine-like amino acids (MAA) as chemical sunscreens for protection from UV radiation17. MAA have a distinct light absorption pattern that allows relative quantification18. MAA in our ice samples were proportional to the ON contents in the sprayed aerosols (Fig. 4). Altogether, these results clearly depict sea ice as an organic nitrogen-rich matrix due to the embedded microbiota.

When sea ice breaks and melts, a water of lower salinity and higher buoyancy tends to accumulate at the ocean’s surface among the ice floes, and similar behaviour is expected for the released biogenic matter1, 9. It is plausible, therefore, that the non-volatile forms of organic nitrogen contained in ice floes, like proteins, MAA and nitrogen osmolytes, are readily susceptible to air-sea exchange by sea spray generation upon bubble bursting. Our knowledge of sea spray production in sea ice regions is very limited, except for open leads in the Arctic19, 20. Interaction of ocean waves with sea ice edge and floes, including splashes, whirls and other turbulent structures, as well as bubbles generated underneath the sea ice surface, may provide significant pathways for the transfer of biological material and sea salt from the surface polar water into the atmosphere21.

Moreover, the most labile of the ice-released ON will be quickly degraded into volatile compounds, including alkyl-amines, by marine bacteria22, 23, thus favouring gaseous nitrogen emission fluxes from ice-pack openings and the marginal ice zone. This is similar to what occurs with sulphur: sea ice is an organic sulphur-rich environment because ice algae generally contain high intracellular levels of dimethylsulphoniopropionate (DMSP) for osmoregulation and cryoprotection12, 24. Release of DMSP from ice enhances the emission of its volatile breakdown product, DMS, through microbial action6, 24. Besides this degradation of organic material released from brine channels, sea ice promotes formation of nitrogen and sulphur volatiles in a more indirect way, by triggering oceanic phytoplankton blooms along the seasonally receding ice edge1. In the Weddell / Scotia Sea region of our study, the border of sea ice influence in ice-free waters was defined by the position of the Southern Boundary of the Antarctic Circumpolar Current (SBACC)25, around 60°S (Supplementary Figures 4 and 5). It is worth noting that, although waters south of the SBACC had lower chlorophyll a and organic carbon concentrations, and similar concentrations of exopolymers to those of the northern side, they had higher occurrence of MAA, higher proportions of the algal osmolytes glycine betaine, choline and DMSP, and higher concentrations of their degradation volatiles methylamines and DMS (Fig. 5). Bubbling tank experiments with ice-free waters showed that waters nearer the sea ice edge were more likely to supply ON in sprayed aerosols: according to ATOFMS analyses, 31% of total particles in the S-SBACC region contained ON, versus 15% in the N-SBACC region.

This study demonstrates that the microbiota throughout the sea ice region are an important contributor to aerosols by emission of aerosol-forming volatile and non-volatile ON, which adds to previously reported emissions of volatile sulphur24 and organic carbon microgels19. The relative importance of biogenic sources of primary versus secondary N in aerosols cannot be fully assessed at present, but our single-particle analyses of ambient aerosols discussed above (Fig. 3) indicates that, whilst secondary aerosols comprise the vast majority, primary ON aerosols should not be neglected. Amines are known to trigger formation of new aerosols4, 8 and allow their further growth into cloud-seeding particles. Whether primary ON-rich particles behave similarly remains unknown but, in the Arctic, biogenic C-rich primary aerosols from the sea ice region have already been identified as an efficient source of cloud condensation nuclei19, 26.

Past and on-going climatic changes are amplified in the polar regions2. As climate warms, energetic wave events are expected to occur more frequently27. Even though there is still much unknown complexity in the atmosphere–ocean–sea-ice system to enable accurate projections of Antarctic sea ice extent28, 29, increased storm frequency and reduced sea ice thickness will facilitate the breakup of sea ice by waves30 and strengthen sea ice-air interactions. Our results show that the sea ice region is a previously overlooked31 source of organic nitrogen in climate-relevant aerosols. This study calls for better representation of the sea ice-ocean-atmosphere biogeochemistry in Earth system models.

## Methods

### The cruise

The PEGASO (Plankton-derived Emissions of trace Gases and Aerosols in the Southern Ocean) cruise was conducted on board de RV Hesperides in the regions of Antarctic Peninsula, South Orkney and South Georgia Islands from 2 January to 11 February 2015.

### Aerosol counts and sizing

Particles in the range 1–3 nm were measured every 2 minutes with a Particle Size Magnifier (PSM, consisting of a Airmodus A10 operated with a TSI3775 condensation particle counter). Particle size distributions in the 10–500 nm size range were measured with a Scanning Mobility Particle Sizer (SMPS; DMA TSI 3080 and CPC TSI 3025) at 5 minute resolution.

### Aerosol sampling on filters

Ambient aerosol was sampled by means of a multistage impactor and a high volume sampler in parallel, from the ship upper deck. The Berner impactor (type LPI80, Hauke), mounting Tedlar foils as sampling substrates, collected particles on five stages with 50% particle upper cut off points at 0.14, 0.42, 1.2, 3.5 and 10 μm aerodynamic diameter (Dp), at a flow rate of 80 LPM. The high volume sampler (TECORA ECO-HIVOL, equipped with Digitel PM1 sampling inlet) collected ambient aerosol particles with Dp < 1 μm on pre-washed and pre-baked quartz-fibre filters, at a controlled flow of 500 LPM. Sampling was allowed only when the samplers were upwind the ship exhaust with a relative wind speed threshold of 5 m s-1. Due to the necessity of collecting sufficient amounts of samples for detailed chemical analyses, sampling time was of the order of ~50 h for each sample. Samples were stored at −20 °C until the chemical analyses.

### Chemical analyses of air filters

Aerosol samples were extracted in MilliQ water by sonication (30 min). Extracts were analysed by ion chromatography for the quantification of water soluble inorganic ions, organic acids (acetate, formate, methanesulphonate, oxalate)32 and low molecular weight alkyl-amines (methyl-, ethyl-, dimethyl-, diethyl- and trimethylamine)5. An IonPac CS16 3 × 250 mm Dionex separation column with gradient MSA elution and an IonPac AS11 2 × 250 mm Dionex separation column with gradient KOH elution were deployed for cations and anions, respectively. The water soluble organic carbon (WSOC) and water soluble total nitrogen (WSTN) content of the samples were quantified using a TOC-TN thermal combustion analyser (Multi N/C 2100 by Analytik Jena)33. The water soluble organic nitrogen (WSON) was calculated as the difference between the WSTN and the inorganic nitrogen deriving from ammonium, nitrate and nitrite ions, as detailed elsewhere5, 34. All the chemical species, with the exception of WSOC, were determined from the Berner impactor samples. WSOC concentration was determined from the high volume samples in order to improve accuracy. WSOC measurements are less sensitive than Ion Chromatography ones, therefore dividing the sample over the different stages of the impactor caused an enhancement of the measurement uncertainty, because of the presence of stages with WSOC concentration below or very close to the detection limit. This is not the case with the high volume sampler, which collects all the submicron aerosol mass in only one filter. Comparison of the results for ionic species measured on both substrates, demonstrate consistency between the two samplers.

### Real time mass spectrometry

The size resolved non-refractory chemical composition of submicron aerosol particles was measured with an Aerodyne High Resolution Time of Flight Aerosol Mass Spectrometer (HR-ToF AMS, Aerodyne, Billerica, MA). The instrument was described by DeCarlo et al.35. HR-ToF AMS was routinely calibrated according to the methods described by Jimenez et al.36. ON AMS is defined as total ON containing m/z fragments detected online by HR-ToF-AMS. The aerosol time-of-flight mass spectrometer (ATOFMS, TSI model 3800) collects bipolar mass spectra of individual aerosol particles. Ambient aerosol is focused into a narrow particle beam for sizes between 0.1 and 1.5 μm.

### Black carbon and metals

Black Carbon was measured by Single Particle Soot Photometer37 (SP-2) and metals by ATOFMS38.

### Air mass trajectories

Five-day back trajectories arriving at the ship’s position at 00:00, 06:00, 12:00 and 18:00 every day were calculated using the BADC Trajectory Service. In total, 117 air mass back trajectories were obtained. Following the Bergeron classification, two main air masses described the study area: continental Antarctic - cAA - (air masses that are extremely cold and dry due to their continental source region between 60°S and 90°S) and Maritime Polar - mP - (from open water regions). Simple calculations were made to assign each 5-days trajectory to the types of surface cover overflown. The Polar Stereographic map classified each of 1024 × 1024 24 km grid cells as land, sea, ice and, from this information, the percentage of time each trajectory spent over each surface type, and particularly over sea ice, could be calculated. A similar calculation was repeated but using daily maps of sea ice percentage concentration measured on a 12.5 km grid. The percentages assigned to each trajectory step allow a ‘spectrum’ of sea ice concentration from 0 to 100% (5% width) to be calculated, as shown in Supplementary Figure 1. Sea ice concentration was derived from satellite microwave data39 available at IFREMER. This same analysis allowed assigning air mass trajectories to the aerosol samples collected on filters. Of a total of 8 filter samples, three (PE24–28–06) had spent most of the time (79%) over open water, and three (PE09–13–18) had spent 73% of the time over sea ice and the sea ice marginal zone (Supplementary Table 1).

### Probability Source Contribution Function analysis

To identify the potential source areas of the observed ON-rich aerosol, potential source contribution function (PSCF) analysis was applied40. The procedure follows closely the one described by Chang et al.41. In the present work, five-day back trajectories (HYSPLIT4 with GDAS data)42 were calculated four times a day during the cruise period, with arrival position 100 m above the ship. Organic nitrogen AMS data, filtered for black carbon (BC) < 1 ng m−3 to avoid contaminations from the ship, were averaged over 6 hours periods (from 3 hours before the arrival time of each back trajectory to 3 hours afterwards). The southern hemisphere was divided into 2.5° latitude x 7.5° longitude grid cells. The conditional probability that the air passing through the ij-th cell had a high aerosol concentration when arriving to the ship position is given by the ratio nij/mij, where nij is the number of trajectories with segment endpoints in cell ij, and mij is the subclass (mij < nij) of trajectories connected to organic nitrogen concentrations above a threshold defining “high concentrations” (in this case, the 3d quartile of the ON database).

Weighting factors were applied to the grid cells, according to the following scheme:

$W = 1.0 when nij > N 2 = 0.8 when N 1 < nij < N 2 = 0 when nij < N 1.$

where nij is the number of trajectories passing for each cell in the study period and N1 = 60 * cos(latitude), and N2 = 300 * cos(latitude). The cosine factor is used to account for the changing grid cell size with varying latitude.

### Aerosol generation tank experiments

An airtight high grade stainless steel tank (200 L) was half filled with a constant flow of seawater pumped from a depth of 4 m by the flow-through pumping system of the ship. Water was dropped from the top of the tank as a plunging jet at a flow rate of 20 L min−1. The entrained air formed bubbles that, upon bursting, produced sea spray aerosol, as reported in O’Dowd et al.43. Particle-free compressed air was blown into the tank headspace (120 ml min−1), which had outlet ports leading to an Aerosol Chemical Speciation Monitor (ACSM) and a Scanning Mobility Particle Sizer (SMPS) for aerosol size distributions (10–500 nm). In parallel, bubble-bursting aerosol generation experiments with the same samples were performed using a square glass tank (10 L) that was filled with about 3 L of either seawater or melted sea ice, sealed with a stainless steel lid and continuously flushed with particle-free air (11 L min−1) as described elsewhere44. Sprayed aerosol size and composition was monitored with a HR-ToF-AMS and an ATOFMS.

### Hydrographic measurements and seawater sampling

Seawater salinity and temperature were recorded continuously via the flow-through thermosalinograph SBE 21 SeaCAT. The water current velocity was measured with the Shipboard Acoustic Doppler Current Profiler (SADCP) “Ocean Surveyor” at 75 khz. Seawater samples were collected from a depth of 4 m using either the uppermost Niskin bottle of the CTD rosette or the ship’s flow-through underway pumping system.

### Allocation of main surface ocean currents and fronts

Positions of the main hydrographic fronts and surface ocean water masses were determined on the basis of the seminal scheme of Orsi et al.25. Gradients in the continuous record of seawater, salinity (Supplementary Figure 3) and surface current direction and velocity along the cruise track were combined with synoptic modeling data obtained from the Global Real-Time Ocean Forecast System (Global RTOFS) to allocate the critical hydrographic boundaries (Extended Data Figure 4).

### Sea Ice samples

Three sea ice samples were collected on three different days at the northern edge of the Weddell Sea marginal ice zone, south west of South Orkney Islands. From a zodiac boat, ice chunks were selected for their brownish colour indicating colonisation by algae. Once in the on board laboratory (within one hour of collection), about 0.1 m3 of sea ice was introduced into the aerosol generation tank and was melted in the dark for 24 hours. Initial melt was facilitated by adding about 1/5 volume of sea water collected at the ship location near sea ice. Ice melt water was recirculated into plunging jet to generate bubbles and aerosols.

### Phytoplankton biomass and composition

Surface ocean and melted ice chlorophyll a concentrations were determined by filtration on glass fibre filters followed by extraction with 90% acetone at 4 °C in the dark for 24 hours. Fluorescence of extracts was measured with a calibrated Turner Designs fluorometer. No phaeophytin corrections were applied. Samples were also taken and fixed with formalin-hexamine 1% for quantitative phytoplankton determination by the inverted microscope method45.

### Viral abundances

Samples (2 ml) were fixed with glutaraldehyde (0.5% final concentration), refrigerated, quick frozen in liquid nitrogen and stored at −80 °C. In the lab, they were stained with SYBR Green I and counted with FACSCalibur flow cytometer with a blue laser (488 nm). Only high-fluorescence viral populations, considered to infect eukaryotes, were used in this study46.

### Sea-ice total organic nitrogen

Aliquots of melted sea ice were frozen at −20 °C until analysis of nitrate, nitrite and ammonium with an automated segmented flow analyser. Organic nitrogen was determined upon digestion (120 °C for 30 min) with alkaline persulfate and subtraction of inorganic nitrogen concentrations47.

### Chromophoric (CDOM) and fluorescent (FDOM) organic matter

CDOM absorption spectra were measured by spectrophotometry on board, and the peak intensity of absorption at 330 nm was extracted from the baseline absorption as a measure of mycosporine-like aminoacid (MAA) concentration. FDOM excitation/emission matrices were also analysed on board using a Perkin Elmer LS55 luminescence spectrometer equipped with a xenon discharge lamp equivalent to 20 kW for a duration of 8-µs. Excitation at 280 nm and emission at 350 nm (so-called peak T) was taken as indicator of protein-like substances48.

### Transparent exopolymer particles (TEP)

TEP were analysed following the colorimetric method49. Samples (160–450 mL) were filtered through 25 mm diameter 0.4 µm pore size polycarbonate filters (DHI). The filters were stained with 500 µL of pre-calibrated (with a xanthan gum solution) Alcian Blue (0.02%, pH 2.5) for 5 s and rinsed with MilliQ water. They were soaked in 80% sulphuric acid for 3 h and the absorbance of the extract was determined at 787 nm in a Varian Cary spectrophotometer. Duplicates were taken for each sample. Results were corrected for blanks consisting of unused filters stained with Alcian blue.

### Total (TOC) and particulate (POC) organic carbon concentrations

For TOC, 30 mL samples of sea water or melted sea ice were collected in acid-cleaned polycarbonate bottles, and stored in the dark at −20 °C until analysis. They were analysed in triplicate with a Shimadzu TOC-LCSV, with MilliQ water as a blank, potassium hydrogen phthalate as the calibration standard, and deep Sargasso Sea water as the reference. For POC, 250–2000 mL of seawater were filtered through pre-combusted (450 °C, 4 h) 25 mm glass fiber filters, frozen at −20 °C and analysed with a Perkin Elmer 2400 CHN analyzer.

### Aqueous MA and DMS concentrations

To determine methyl-amine concentrations, 850 mL seawater samples were gravity filtered through 47 mm GF/F filters into 1 L high density polyethylene (HDPE) bottles containing HCl (10 mL, 11.6 M), and stored at 4 °C until analysis. The seawater was saturated with NaCl and adjusted to pH 13.4 with sodium hydroxide. Methylamines were extracted from the water sample by solid phase microextraction approach50. A polydimethylsiloxane/divinylbenzene fibre was exposed in the headspace above the water sample, and the extraction performed for 2.5 h at 60 °C. The methylamines were then resolved and detected using a GC with NP detector. Calibrations were performed using external matrix-matched standard solutions containing mono-, di- and trimethylamine in the range 0.13–13.3 nM. DMS was determined, along with other volatiles, with a purge and trap GC with MS detector. Samples of 25 ml of seawater pre-screened through 200 µm were filtered on-line through GF-F, injected and sparged with He for 12 min. Calibration was performed with DMS solutions generated by dissolution and hydrolysis of solid DMSP in high purity water.

### Phytoplankton DMSP, glycine betaine and choline content

For DMSPt (particulate + dissolved, largely particulate) analysis, two pellets of NaOH were added to 30 mL of sample for hydrolysis to DMS, for at least 24 h at room temperature in the dark. Aliquots of 0.1 to 1 mL were injected into a purge flask with high purity water, sparged for 4–6 min. and analysed for evolved DMS with a purge and trap GC system with FP detector51. DMSPt concentrations were calculated by subtraction of the endogenous DMS. Calibration as for DMS. For glycine betaine and choline, 10 mL of sample was gravity filtered onto 47 mm GF/Fs. Filters were snap frozen and stored at −80 °C until analysis. Glycine betaine and choline were extracted from filters into 15 mL 12:5:1 methanol:chloroform:water containing internal standard (d11-glycine betaine) by soaking (1 hr). Solutions were clarified by centrifugation and analysed by LC/MS52. To calculate the proportion of POC contributed by DMSP or the N-osmolytes, their concentration in nmol L−1 was multiplied by 5 to account for 5 C atoms per molecule, and divided by 1000 and by the POC concentration in µmolC L−1.

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## Acknowledgements

The cruise was funded by the Spanish Ministry of Economy through projects PEGASO (CTM2012-37615) and Bio-Nuc (CGL2013-49020-R), and by the EU though the FP7-PEOPLE-2013-IOF programme (Project number 624680, MANU – Marine Aerosol NUcleations). We wish to thank the Spanish Armada, and particularly the captains and crew of the BIO A-33 Hesperides, for their invaluable collaboration. We are also indebted to the UTM, and especially Miki Ojeda, for logistic and technical support. The Spanish Antarctic Programme and Polar Committee provided context and advice. We thank Martí Galí for providing daily satellite imagery, Pablo Rodríguez-Ros and Sarah-Jeanne Royer for help with analyses, and Encarna Borrull and Carolina Antequera for sampling assistance. Marina Serrano Cueva and Maximino Delgado helped with viral and phytoplankton counts, respectively. We would like to thank Dr. K. Sellegri (LaMP/CNRS, France) for lending a small sea spray tank for lab studies. The NUI Galway and ISAC-CNR Bologna groups acknowledge funding from the European Union’s Seventh Framework Programme (FP7/2007-2013) project BACCHUS under grant agreement n° 603445. The work was further supported by the CNR (Italy) under AirSEaLab: Progetto Laboratori Congiunti. The National Centre for Atmospheric Science NCAS Birmingham group is funded by the UK Natural Environment Research Council. Thanks to the British Atmospheric Data Centre, which is part of the NERC National Centre for Atmospheric Science (NCAS), for the calculation of trajectories and access to ECMWF data. CC, MFF and RA acknowledge funding from the Marine Institute, University of Plymouth to enable participation in PEGASO.

## Author information

### Affiliations

1. #### Institut de Ciències del Mar, CSIC, Pg. Marítim de la Barceloneta 37-49, 08003, Barcelona, Catalonia, Spain

• Manuel Dall’Osto
• , Pau Cortés
• , Marina Zamanillo
• , Sdena O. Nunes
• , Eva Ortega-Retuerta
• , Mikhail Emelianov
• , Dolors Vaqué
• , Cèlia Marrasé
• , M. Montserrat Sala
•  & Rafel Simó
2. #### School of Physics and Centre for Climate and Air Pollution Studies, Ryan Institute, National University of Ireland Galway, University Road, Galway, Ireland

• , Darius Ceburnis
•  & Colin O’Dowd
3. #### Institute of Atmospheric Sciences and Climate, National Research Council, Bologna, 40129, Italy

• Marco Paglione
• , Matteo Rinaldi
• , Stefano Decesari
•  & Maria Cristina Facchini
4. #### National Centre for Atmospheric Science, The School of Geography, Earth and Environmental Sciences, The University of Birmigham, Edgbaston, Birmingham, B15 2TT, United Kingdom

• David C. S. Beddows
•  & Roy M. Harrison
5. #### Biogeochemistry Research Centre, University of Plymouth, Drake Circus, Plymouth, PL4 8AA, UK

• Charlotte Cree
•  & Mark F. Fitzsimons
6. #### Instituto INIBIOMA, CRUB Comahue, CONICET, Quintral 1250, 8400S.C. de Bariloche, Rio Negro, Argentina

• Gonzalo L. Pérez
7. #### Department of Evolutionary Biology, Ecology and Environmental Sciences, Universitat de Barcelona, Av. Diagonal 643, 08028, Barcelona, Catalonia, Spain

• Montserrat Vidal
8. #### Plymouth Marine Laboratory, Prospect Place, Plymouth, PL1 3DH, UK

• Rachael Beale
•  & Ruth Airs
9. #### Department of Environmental Sciences / Center of Excellence in Environmental Studies, King Abdulaziz University, PO Box 80203, Jeddah, 21589, Saudi Arabia

• Roy M. Harrison

### Contributions

M.D. and R.S. designed and organised the study. R.S. led the cruise. J.O. and D.C. ran the AMS, ACSM and tank experiments. M.P. collected aerosol filter samples and ran the analyses with M.R. and S.D. S.D. did the PSCF analysis. D.B. and M.D. ran the ATOFMS and PSM, and carried out the air mass trajectory and source region analysis. C.C. analysed amines with supervision by M.F. and R.A. R.A., R.B. and C.C. analysed N-osmolytes. P.C. analysed DMS and DMSP. M.Z. and E.O.-R. analysed TEP. S.N. and M.Es. determined chla and phytoplankton. G.P., C.M. and E.O.-R. analysed and processed CDOM and FDOM. M.Em. processed the hydrographic data. D.V. determined viruses. M.M.S. and M.V. assisted with data interpretation. M.R., S.D., M.C.F., C.O., R.H., J.O., D.C., M.D. and R.S. discussed the results and outlined the paper. M.D. and R.S. wrote the paper, with contributions from all co-authors. M.D., J.O., M.P., D.C. S.B., D.C., C.C., P.C., M.Z., S.O.N., G.L.P., E.O.R., M.E., D.V., C.M., M.E., M., M.S., M.V., M.F.F., B.B., R.A., M.R., S.D., M.C.F., R.M.H., C.O’D., R.S. read and edited the paper.

### Competing Interests

The authors declare that they have no competing interests.

### Corresponding author

Correspondence to Manuel Dall’Osto.


text:
# Riddles that can be solved by meta-assumptions

The website of my university posted a riddle that goes something like this:

Riddle

There are three men named 1,2 and 3 and each one has two colored dots on his forehead. Possible colors are black and red. No color is used more than four times. The men can see the colors on the other mens head, but not the ones on their own. A game-master asks the men in the order 1,2,3,1,2,3,... whether they know the colors of their dots. If one says "no", he proceeds with the next man. If one says "yes", he has to state his guess. If he is correct, all men win the game, if he is wrong then all men lose. The men were not given any time to make up a strategy.

As it turnes out the men answer "no", "no", "no", "no", "yes". What is the color on the head of man 2?

I was going to solve this riddle systematically by trying to reason what color combinations make sense and what information each man can deduce from the others saying "no".

However, I came to a much more direct solution, that completely ignores most of the riddle and only works by the assumption that the riddle is solvable.

Solution: Man 2 must have a red and a black dot on his forehead.

Reason: I (from the meta perspective) only have the information about the answer-pattern and no actual colors. Since the game is symmetric w.r.t. the colors red and black, the game would have the same answer-pattern if we switch the colors. This means, any reasoning that explains that man 2 has e.g. two red dots, will also work to reason that he has two black dots. If any of these two would be the solution, I would not be able to conclusively find it. Hence the solution must be "one red and one black dot". Note that I cannot answer how man 2 came to his conclusion, but this was not part of the question.

For me, this kind of solving the riddle was very interesting and I wondered if there are more such riddles out there which (intentionally or unintentionally) can be solved by such meta-assumptions.

Question: Are there more such riddles which have a surprisingly easy solution by meta-assumptions like e.g. "the riddle has a solution" or "the riddle is solvable with reasonable effort" etc.

Update

Because it was asked in the comments, and maybe needs a clarification, here is what I am looking for in other words:

I ask for riddles or math-problems

• which are formulated in a way that suggests some specific meta-assumption (e.g. the riddle must have a solution which I as a reader can find).
• the meta-information is not obviously presented to the reader as something that should be used to solve the riddle (preferably might not even be needed).
• the meta-information turns out to be unexpectedly helpful.

I am not specifically asking about the assumptions of "existence of a unique solution", but also other meta-assumptions, like e.g. the context of the question, the number of answers (if it is multiple-choice), the expected time-frame of solution, maybe even the color of the paper the question was asked on, etc.

Some type of examples which came up a lot (and which I liked) are riddles/problems where the meta-information was the absence of some information. The absence implied that the solution will probably not depend on this missing information and hence we can choose a very simple instance of the problem.

One other example that came to my mind (but it does not fit so perfectly) is the graph-problem presented in this video of 3Blue1Brown. It can (and must) be solved by the meta-information that it is presented on a cup and not on a piece of paper. Unfortunatly the riddle can only be solved by recognizing that it is on a cup. I would prefer that the meta-way is not the intended solution.

• @dEmigOd I do not take this into account. You are right that by seeing four dots of the same color he would have concluded that he had two of the same color. But we as an "external solver of the riddle" had no way to conclude whether he saw four red or four black dots because the riddle has no further information for us which is asymmetric w.r.t. the colors. A real situation could be staged in such a way that the second man concludes e.g. red-red, but in the riddle there is no fixed distribution of colors. Feb 12, 2018 at 12:55
• Plenty of standardized, quantitative multiple choice questions (SAT, GRE general and subject, etc) can be attacked with meta-reasoning... "I know the answer is one of the listed choices, so ...," for example. This is one reason why such tests aren't great: you can get the right answer for the "wrong" reason (wrong in the sense that no such meta-reasoning is available if you had to answer an open-ended question with no answer choices). Feb 12, 2018 at 14:07
• Relevant and possibly interesting: An exam question that asks about property of a group of a certain order, but without saying what the group is, On an exam one can assume it doesn't matter which group it is, take the group to be whichever one happens to make the answer easiest to find, and then write down the answer for that particular group. I said: "Reasoning from the premise that the question is well-posed is fair game on a written exam, but is not legitimate mathematics."
– MJD
Feb 12, 2018 at 16:06
• In contrast, here's a puzzle that many people get wrong, because they assume (incorrectly) that the puzzle is solvable. The puzzle has a unique solution, but only under the (possibly false) assumption that it has a unique solution.
– MJD
Feb 12, 2018 at 16:15
• Do let us know whether this belongs to Puzzling.SE Feb 13, 2018 at 5:44

Having solved my share of logic puzzles, I can report that logic puzzles can be solved using the meta-assumption that the logic puzzle is solvable surprisingly often (in fact, I had the occasion of 'solving' the very puzzle you mention just a few months ago in exactly the way you did). I use 'solving' since of course you are not really solving the riddle at all, but merely showing that 'if the riddle has a solution, then it has to be ...'. Here is another example:

In the Sudoku-solving (and other grid-filling puzzles) community the meta-assumption that a Sudoku puzzle has a unique solution is called the Uniqueness assumption and can be used to great effect. For example, suppose you know that you have a row where there are two open cells left (say in columns $2$ and $3$) which must contain the numbers $1$ and $4$. Suppose you have another row with three open cells left (say columns $2$, $3$, and $5$), which must contain the numbers $1$, $4$, and $7$. Then using the Uniqueness assumption we can immediately say that the cell in that second row in column $5$ does not contain a $7$, since if it did, you could place the remaining $1$'s and $4$'s in two different ways, and clearly this has no bearing on the solvability of the rest of the puzzle.

• Thanks for adding the piece about Sudoku's. I considered posting it as an isolated example, but felt that I wouldn't have too much to say about it. Feb 15, 2018 at 17:53
• Do you have a reference? I've solved some Sudoku previously, but I've not heard of this, and it would have made some seemingly impossible puzzles solvable if I'm not misunderstanding you. Feb 16, 2018 at 6:50
• I have actually seen Sudoku's before that have more than 1 solution. I don't think this would work on them, would it? Feb 16, 2018 at 11:23
• @Belle Correct! Applying the Uniqueness Assumption to a puzzle that does not have a unique solution can quickly lead to trouble. Take the situation as described in my post, for example. Suppose that particular Sudoku actually has exactly two solutions, with one solution having the one row having a $1$ in column $2$, and a $4$ in column $3$, and the other row having a $4$ in column $2$, and a $1$ in column $3$, while the second solution has those $1$'s and $4$'s swapped. Then both solutions do have a $7$ in column $5$ of that second row, but the Uniqueness Assumption said there is not a 7! Feb 16, 2018 at 14:32

Here's a riddle that fits right into this category. Hope you like it:

Begin with a sphere. Drill a circularly-cylindrical hole directly through the center of the sphere [i.e the result is still a 'solid of revolution'], and remove the drilled material. Measure the height of the hole from one circular edge through the body of the sphere to the other circular edge; the measured height is 10 centimeters.

What is the volume of the remaining solid?

• I know this from a VSauce video :D. Despite from concluding that the the volume seems to be independent of the exact sphere-radius and drill-hole-radius (as long as the height is 10cm), can we somehow trivially conclude the exact volume? Feb 12, 2018 at 14:58
• @M.Winter - Consider the case when the drill-hole radius is $0$. If there is a solution, it has to work for this case (or at least for cases approximating this one to any resolution). Feb 12, 2018 at 15:24
• This was the puzzle I was thinking of. I got it from one of Martin Gardner's books of mathematical puzzles from the 1960-80s (don't know exact year, book may have been old when I read it. Not sure if its original to him, or came from elsewhere before him. Feb 13, 2018 at 16:28
• @Shufflepants It definitely is. The assumption tells us that the answer must be a number. If the hole radius is 0, the sphere radius is 5 and the volume is 4/3*PI*5^3. As the "question must be solvable" assumption tells us it does not depend on the hole/sphere radius, the value must be 4/3*PI*5^3 for any hole/sphere radius values. If the question were not phrased the way it was. I would probably resort to calculus. Feb 13, 2018 at 17:33
• A similar problem: suppose you draw two concentric circles such that a chord of the larger circle is tangent to the smaller circle and has length 1. What is the area of the region between the two circles? Feb 13, 2018 at 21:11

Here's one example.

The Riddle

Assume that the Earth is a perfectly round sphere and that you are constrained to the surface of the Earth. Is there a point P on Earth such that if you start at P, walk one mile North, then walk one mile East, and then walk one mile South, you will end up exactly where you started, at point P?

The Solution

The South Pole is one solution. Any direction away from the South Pole is North if you are standing at the South Pole. Just walk along a longitude, then a latitude, then a longitude and you will be back at the South Pole.

It turns out that there are infinitely many other solutions. Any point on a certain circle near the North Pole will work. But besides that, the number of circles which will work is also infinite. All of them have the North Pole as their center.

Solving the Riddle Just by Assuming that a Solution Exists

I asked this question to a fellow mathematician. She immediately assumed that a solution exists and (being a good mathematician) considered an extreme case. She was talking out loud, rapid-fire comments and questions so I know what she was thinking. She said it must be one of the poles and figured out in two seconds that it must be the South Pole.

Then I told her, "what if I told you that there is another entirely different solution". She thought about it for a few seconds and asked me if the number of solutions is finite or infinite. I thought about it and replied, "I can't tell you. Because even that would give away too much information." She then assumed that it is probably an infinite number of solutions because of the symmetry of the sphere. If (besides the poles) one point will work then there should be another that works just as well, and another, and another, and then maybe the locus of points forms a circle on the sphere and so on.

This all took about sixty seconds.

It's cool to realize that here, not just some meta-information helped her solve the riddle, but rather some meta-meta-information helped her as well. The fact that me telling her the number of solutions would give her too much information, gave her some information too.

Explaining the Complete Solution

Consider the following picture.

In this picture, start on the black dot, go North for one mile along the red arc. Then go East for one mile and you will have traversed the entire red circle. Then go one mile South, which is along the red arc again, and you will be at the black dot again. Therefore the black dot is another solution to the riddle (as the riddle is stated). In fact, any point on the blue circle is a valid solution.

Besides these solutions, there is another red circle, centered at the North Pole, such that its circumference is 1/2 mile so that going East on that circle will make you go around twice. Therefore you simply start a mile below it.

Similarly there is another smaller red circle with circumference 1/3 mile and you can start a mile below it. And so on.

There is a blue circle for every natural integer. So there are countably many (blue) circles centered around the North Pole, converging on the circle one mile South of the North Pole, which are all valid solutions for this riddle. Some of them are shown here.

• That's a cool example, because I know this riddle but never thought about othern solutions than the south pole! :D Shame on me. Feb 12, 2018 at 20:00
• @CiaPan I think you misunderstand. You walk one mile north, till you're far enough from the northpole that walking one mile east is a circle with circumference one mile. (1/2pi mile) Then you walk one mile south. You never go cross the northpole. (basically you start at 1 mile+ 1/2pi mile from the northpole.) Feb 13, 2018 at 14:09
• @PieterB Actually, you could start at a point that leaves you on a circle of 1/2 mile (walking East, you'll go round twice) circumference or 1/3 mile circumference, or 1/4 mile etc etc etc. Feb 13, 2018 at 16:33
• @CiaPan PieterB and JeremyP are both correct. I have added a picture in the post to show what the solutions are. Feb 13, 2018 at 19:37
• This one was also mentioned by Martin Gardner, IIRC. Feb 14, 2018 at 14:05

In the spirit of providing more examples, I see this come up a lot in the puzzle Slant from Simon Tatham's Portable Puzzle Collection. The rules of this puzzle are simple:

Fill in a diagonal line in every grid square so that there are no loops in the grid, and so that every numbered point has that many lines meeting at it.

The meta-information is that each grid square can be filled in unambiguously.

You can use this information to deduce that the upper left corner must be the NW-SE ("\") diagonal, because otherwise it could form a loop somewhere (in fact, it must otherwise form a loop somewhere, otherwise the corner would be ambiguous.)

Without this meta-information, you would have to wait until later in the game to place this piece.

I was taking (decades ago) an advanced topics math course in high school. We got a test with several problems, one of which was "Prove that ...". I forget what we were expected to prove, but at the time I had no clue how to approach it. I went on to the other problems, solved them all, and came back to this one. I still had no clue. When the hour was just about up, I wrote "If it weren't true, you wouldn't ask us to prove it." I don't remember what score I got, but all future questions were "Prove or disprove that ..."

• I'd hope that you got a zero score for that proof. It is only probably true that he wouldn't ask you to prove it if it weren't true. For one, the Teacher could have been wrong or he could just be devious. So, not only is your proof wrong, it's not even a proof in a mathematical sense. It's only circumstantial evidence pointing toward a likely conclusion. Feb 13, 2018 at 16:41
• @Shufflepants and yet, if Ross had disproven it thoroughly and rigorously, he would not have followed the instructions on the exam paper, would he? The response is brilliant and shows thinking outside the box. Feb 13, 2018 at 23:09
• Even "Prove or disprove" still leaves you as an implicit given that the proposition is provable or disprovable. If you can then somehow show that it isn't disprovable, you've proven it, and vice versa. Feb 14, 2018 at 21:04
• @Shufflepants: You missed MontyHarder's point there. He's suggesting that one could write on the answer sheet a proof that the proposition is not disprovable, and then follow it with the sentence "Therefore the proposition must be provable, or else you wouldn't have asked us to either prove or disprove it. Q.E.D." It's just a one-level-more-meta version of Ross's original snark. Feb 16, 2018 at 5:41
• Monty added a (non-trivial) proof that X was not disprovable, to deal with the test question "Prove or disprove X." Note that if the prof had phrased it as "Prove that X is either true or false," or (more contrived) "Prove that X is either provable or disprovable," Ross's original answer would have kept working — via the same mechanism that lets us answer "Yes" to "Would you like coffee or tea?" The prof's pivot from "Do X" to "Do either X or Y" is important! Shades of ZKP here. Feb 16, 2018 at 18:31

Bertrand's paradox (not to be confused with Russel's paradox) is a paradox in probability based on the following problem:

Given a circle, choose a chord at random. What is the probability that the chord is longer than the side of an equilateral triangle inscribed in the circle?

where, depending on how you choose the chord, you may end up with wildly varying answers (the wikipedia article mentions three examples, all of which may be justified as uniform, giving $\frac12, \frac13$ and $\frac14$ respectively).

One possible resolution, also detailed in that Wikipedia article, uses a technique similar to yours, arguing that this problem is symmetric in the choice of circle, so the distribution of possible chords as lines in the plane should look the same regardless of which circle we end up with. Here it is called the "maximum ignorance principle".

Since this is a Math forum, let me give a mathematical answer, instead of just another example of a 'riddle'. My example conveys the heart of the issue you are interested in, and stresses an unrelated lesson that math majors often struggle with.

Induction: In order to prove that the sum of the first $n$ integers is $n(n+1)/2$, you assume the formula is correct for a given $n$ and use this information to prove the same formula for $n+1$. In essence you say "IF there is a formula, it has to be this one..."

Extremal problems: Which 2D figure of perimeter 1 covers the most area? Which line between 2 points is the shortest? You usually get a proposed solution and prove that it does not improve upon infinitesimal deformations. In essence, you prove that "IF there is an optimal shape, it has to be the one you propose..."

There are many situations like those above. A complete proof requires proving that there IS a solution (induction base, or a compactness argument). This is the lesson that math majors should internalize.

But most often, a proof assumes implicitly that there is a solution. In fact, one uses that assumption without even noticing it.

• I realise this is mostly a matter of phrasing, but I just want to make something absolutely clear: a correct proof never actually assumes there is a solution. At some point, you may assume that a solution exists and deduce what it must then be; however, you then must check that this is actually a solution. This is equivalent to showing that there is only one solution, and then showing a solution, thus finding all the solutions. (I'm not trying to contradict your answer, just I feel the wording of the final paragraph may lead someone to think that proofs assume a solution) Feb 12, 2018 at 22:38
• This induction example seems not to fit - you have to somehow come up with the idea for what formula you're going to prove before you prove it. The direct non-inductive argument (doubling a right triangle to make an $n$ by $n + 1$ rectangle) actually produces the formula. Probably many examples of induction are like this. Feb 13, 2018 at 6:10
• But induction still involves proving that your formula is correct for a specific case, before extrapolating to the general by showing that if it's true for n it's also true for n+1. I don't really think it that counts as a meta assumption Feb 13, 2018 at 17:28
• Insightful answer. And outside the realm of mathematics, for instance in the humanities, the basic assumption most often used is that there is no true solution. (Examples: war, murder, insanity.) Of course, when it's assumed there is no solution, it's hardly surprising that no solution is found. Feb 14, 2018 at 4:28
• This is certainly not how I think of induction. Induction is "prove a property for everything by means of first proving the property for whatever is the most primitive kind of thing, then proving that all possible ways of constructing a less primitive thing from a more primitive thing must preserve the property". No assumption; just construction. Feb 14, 2018 at 22:18

Since it doesn't seem to have been mentioned anywhere yet, if anyone is curious here's how to solve OP's puzzle without the meta assumption (though note how the black-black vs. red-red ambiguity naturally pops up at the end, and that one might naturally be lead to OP's argument as a refinement at that point):

If any participant sees 4 blacks or 4 reds combined on the other two (ie, they each have 2 blacks, or they each have 2 reds), then he knows that he must have two of the other color. So a first round of all "no" answers says that no two participants have two of the same color. The colors on each of the three players head must then combine to form one of the following four rows (potentially after reordering):

$$\begin{array}{ccc} (b,b)&(b,r)&(b,r)\\ (r,r)&(b,r)&(b,r)\\ (b,b)&(b,r)&(r,r)\\ (b,r)&(b,r)&(b,r)\\ \end{array}.$$

So #1, on his second turn, looks at the other two participants heads. By picking any two of the entries in each of the above rows, we see that the possibilities that #1 can see on #2 and #3 form the first two entries of one of the following rows, with the third entry telling us which of the possible total state spaces we could be in: $$\begin{array}{cc} A & B & \text{appears in row...}\\ (b,b) & (b,r)&1,3\\ (b,r) & (b,r)&1,2,4\\ (b,r) & (r,r)&2,3\\ (b,b) & (r,r)&3\\ \end{array}.$$ Since #1 says "no" the second time around, this means he does not see (b,b)+(r,r) on the other two participants.

So when #2 gets his second chance, he knows combined the three of them have at least two (b,r)'s at most one of (b,b) or (r,r). If he saw that #1 and #3 had (b,r) each, he would not know if he had two blacks or two reds, or even (b,r), and would have to answer "no". However, since he answered "yes" then he must have only seen one (b,r) among the other two, and so could deduce that he must have the second mandatory (b,r) pair on his own head.

• This approach extends to $n \geq 3$ men under the assumption that there are at most $n+1$ dots of each color. It's always the second in line to get the answer on the second turn. Feb 15, 2018 at 8:35

While the other very entertaining answers have been providing more examples of problems/riddles where similar techniques work, I think that it's just as important to recognize that each one works (or not) on a case-by-case basis, and there is no universal magic that works in all cases.

Consider for example the simple problem of "determine the maximum area of a rectangle with integer sides and a perimeter of $20$". A solution must obviously exist, since there are only a finite number of such rectangles. Then, applying the same line of reasoning as here...

Since the game is symmetric w.r.t. the colors red and black, the game would have the same answer-pattern if we switch the colors.

...it follows that neither side is distinguished in any way, so the problem is entirely symmetric in the two sides, and therefore the solution must be the one where the sides are equal so the rectangle is a square with side length $\,5\,$ and area $\,25\,$.

It turns out that that is indeed the correct answer, right? Yes, but had the problem asked for the rectangle with "the minimum area", instead, the same line of reasoning would have lead to the same square as the answer, which is wrong in that case - correct answer is the rectangle $\,1$x$9\,$.

Point of the above is that symmetry in the conditions does not always imply a corresponding symmetry in the solutions. For a more involved math example borrowed from another MSE question, both extrema of $\,(a^2-ab+b^2)(b^2-bc+c^2)(c^2-ca+a^2)\,$ for non-negative $\,a,b,c\,$ that satisfy $\,a+b+c=3\,$ do in fact exist, but neither is attained at $\,a=b=c=1\,$.

• The faulty assumption here is that there is only one solution. The question's reasoning would not work if the correct riddle solution was "either two red or two black", but by the form of the riddle it can be inferred that is not a valid solution. The minimum area rectangle is "1x9 or 9x1" Feb 13, 2018 at 1:10
• @immibis That's a good math point, but it's not always obvious in more casual speak. Even in my (very simplistic) rectangle example, the question supposedly asks for the "minimum area", not a specific rectangle, and even if it were to ask for a specific rectangle, 1x9 and 9x1 could easily be taken to be the "same" rectangle depending on how you tilt your head. My intention was not to indiscriminately put down symmetry-based arguments at large, but to show why certain care is needed when using them.
– dxiv
Feb 13, 2018 at 1:24
• Is this supposed to be a perimeter of 20? Feb 13, 2018 at 4:27
• @PeterKagey Or a semiperimeter of 10 ;-) Thanks for catching that, corrected now.
– dxiv
Feb 13, 2018 at 4:29
• The symmetric assumption only works for the dots because there is no geometric structure and the two types of dots are distinguishable. In the case of rectangles, there are geometric considerations which invalidate the assumption. It's not that the symmetric assumption doesn't hold in the dots case, it's just that you'd need to go through a little bit more work to rigorously prove that assumption. Feb 13, 2018 at 16:53

Maybe it is not exactly what you're after, but here is one meta-argument that I like:

Q: Are there two irrational numbers $x$ and $y$ such that $x^y$ is a rational?

A: Either $x=\sqrt{2}^\sqrt{2}\in \mathbb{Q}$, in which case the answer is yes, or $x=\sqrt{2}^\sqrt{2}\not\in \mathbb{Q}$, in which case $x^\sqrt{2}=2$ implies that the answer is also yes.

• @ToddSewell It is transcendental by Gelfond-Schneider theorem :) en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem Feb 12, 2018 at 16:33
• How is that a meta-argument? Feb 12, 2018 at 18:26
• But this is just the law of excluded middle, probably THE classical argument of propositional logic. Feb 12, 2018 at 20:08
• This doesn't quite seem appropriate. What is the meta-assumption here? As far as I can tell, this is a complete and correct non-constructive proof of the above theorem. You do not need to know anything about the existence of such a proof, or the solvability of the problem (as being asked this question might indicate) Feb 13, 2018 at 13:30
• @Hyperplane And the one which spectacularly gives you the wrong solution to this 'puzzle' previously commented under the question. Feb 14, 2018 at 4:55

This is why I think all puzzle/riddle compilations should include ones impossible to solve, with no solution or simply "not interesting" at all.

One example is chess puzzles. I am yet to find a chess puzzle library that both includes extremely normal game conditions, where e.g. indeed taking that pawn is the right thing to do, and other variants (where there's actually a winning queen sacrifice). This I feel would make me a better chess player abecause it reflects what you experience across the game.

I also feel the same with my job. I'm a researcher and also a riddle lover. The real-life "riddles" I come across everyday at work might not have a solution at all! I might be a better researcher if I train my critical thinking with a collection of riddles that may have an ingenious solution, a dumb solution, or no solution at all


text:
### Verizon’s net neutrality chickens come home to roost

The FCC's emerging net neutrality plan appears to be closely modeled on the partial reclassification plan proposed by Mozilla.…

### Why do U.S. ISPs want to charge for peering? Peering makes the internet cheaper. Here’s how

The practice of network peering is gaining ground, which is good news for everyone on the web except for those companies providing transit. Will that continue?…

### Peering, under the hood

The FCC's findings regarding "path specific," congestion problems comports quite closely with Verizon's version of what was going on in its dispute with Netflix.…


text:
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# TOC-grand test Q4

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+1 vote

What is the smallest possible number of states needed in constructing DFA for the language $\small \Sigma^*1(\Sigma\Sigma^*)1\Sigma^*$ , where $\small \Sigma = \{0,1\}$. ?

reshown Sep 8, 2019

The following DFA with four states recognizes L=Σ*1(ΣΣ*)1Σ*

By the Myhill–Nerode theorem, no smaller DFA exists because each of the four strings "epsilon,1,10,11" is in a different equivalence class of L. Their distinguishing suffixes are as follows:

answered Sep 11, 2019 by (91,540 points)

ANS . 4

Expression will become -  (0+1)* 1 (0+1)^+  1  (0+1)*

Langauges = { 101, 111............}

Min. DFA will contain 4 states.

answered Sep 8, 2019 by (14,660 points)
may be 3 states


text:
# What is 8- 6( 7- 3)?

Sep 28, 2016

$- 16$

#### Explanation:

When working with expressions which have different operations, always count the number of terms first.

Each term simplifies to a single answer and these are only added or subtracted in the last step.
Within each term the order of operations is:
Brackets first, then powers and roots, then multiply and divide.

$8 - 6 \left(7 - 3\right) \text{ } \leftarrow$ there are only 2 terms, keep them separate

$\textcolor{b l u e}{8} \textcolor{red}{- 6 \left(7 - 3\right)}$

$= \textcolor{b l u e}{8} \textcolor{red}{- 6 \left(4\right)}$

$= \textcolor{b l u e}{8} \textcolor{red}{- 24}$

$= - 16$


text:
# How does the Rocket Equation work when you use boosters?

For simple rockets, using the rocket equation is simple enough.

$$\Delta v = v_e ln\left(\frac{m_i}{m_f}\right) = I_{sp} \ g \ ln\left(\frac{m_i}{m_f}\right)$$

plug in full and empty weights and Isp, and you're done.

But what to do when a rocket uses boosters? You get 2 Isp values. I expect you can't just add the delta-V of the main stages to the delta-V of the individual boosters.

Is there a way to use the Rocket equation on rockets that use boosters?

• I'd imagine you'd do a weighted average based on the fuel mass. Just a guess on my part, though. Feb 9 '18 at 15:52
• Not at all an answer, but there's some deriving going on here.
– uhoh
Feb 9 '18 at 16:04

For each phase of flight (stage or throttle variation or what have you) you weight the specific impulse (= exhaust velocity) by the propellant mass flow rates (i.e. consumption rates) of the different engines operating.

So if, for example the boosters are consuming 5000kg/s at 240s specific impulse and the core is consuming 1000kg/s at 310s, it works out to

$$\frac {5000 \frac {kg} {s} * 240 {s} + 1000 \frac {kg} {s} * 310{s} } {5000 \frac {kg} {s} + 1000 \frac {kg} {s} } = 251.66 {s}$$

For 251.66s specific impulse.

• What does that equation mean? What are the units? Feb 9 '18 at 20:32
• I reckon the units are (kg/s * s) / (kg/s) = s, yielding 251.66s combined Isp. Just a guess. Feb 9 '18 at 21:28
• @Brian Good guess. Feb 10 '18 at 1:05

You have to consider each "phase" of flight.

For each phase of flight you need to work out the initial mass, the final mass and the overall specific impulse (total thrust/total propellant flow rate). You can then apply the rocket equation to each phase of the flight and add up the results.

• This is also true, but the problem is that the boosters can have different Isp and dm/dt also.
– uhoh
Feb 9 '18 at 16:34
• If you install the Flight Engineer mod for Kerbal Space Program, it will tell you the ISP, the dV, the TWR, and burn time for every stage. That's how I get it to work ;) Feb 10 '18 at 0:03

The derivation is based on the rate of change of total momentum

$$\frac{dp_{tot}}{dt}$$.

For each engine,

$$\frac{dp}{dt}=v_{ex}\frac{dm}{dt},$$

the exhaust velocity times the rate that the propellants' mass is ejected (kg/sec). You calculate that for each engine, then add them up.

So where you would normally have

$$m \frac{dV}{dt} = -v_{ex}\frac{dm}{dt},$$

$$m \frac{dV}{dt} = -\Sigma \left(v_{ex, i}\frac{dm_i}{dt}\right),$$

where $$v_{ex,i}$$ and $$\frac{dm_i}{dt}$$ are the exhaust velocity and mass rate of the ith engine. You can then introduce the weighted exhaust velocity:

$$v_{ex,m} = \frac{\sum v_{ex,i}\frac{dm_i}{dt}}{\sum \frac{dm_i}{dt}} = \frac{\sum v_{ex,i}\frac{dm_i}{dt}}{\frac{dm}{dt}}$$

Hence you have

$$m\frac{dV}{dt} = -v_{ex,m}\frac{dm}{dt}$$

which is the same equation as in the single engine case. Integrating this equation yields the following rocket equation:

$$\Delta V = v_{ex,m} ln\big(\frac{m_i}{m_f}\big)$$

• Oh, that's much nicer, as well as pretty obvious in retrospect. Thank you Community♦
– uhoh
Feb 10 '18 at 0:19
• "exhaust" - such a depressing term. Feb 10 '18 at 2:14
• @OrganicMarble lol, indeed.
– uhoh
Feb 10 '18 at 3:34


text:
Plots quantiles obtained from a quantile regression forest. Additionally insets the continuous rank probability score (crps), a useful diagnostic of accuracy.

plot.quantreg(x, prbL = .25, prbU = .75,
m.target = NULL, crps = TRUE, subset = NULL, ...)

## Arguments

x

A quantile regression object obtained from calling quantreg.

prbL

Lower quantile (preferably < .5).

prbU

Upper quantile (preferably > .5).

m.target

Character value for multivariate families specifying the target outcome to be used. If left unspecified, the algorithm will choose a default target.

crps

Calculate crps and inset it?

subset

Restricts plotted values to a subset of the data. Default is to use the entire data.

...

Further arguments passed to or from other methods.

## Author

Hemant Ishwaran and Udaya B. Kogalur

quantreg.rfsrc


text:
LLVM  14.0.0git

An LinkGraph pass configuration, consisting of a list of pre-prune, post-prune, and post-fixup passes. More...

#include "llvm/ExecutionEngine/JITLink/JITLink.h"

Collaboration diagram for llvm::jitlink::PassConfiguration:
[legend]

Public Attributes

Pre-prune passes. More...

Post-prune passes. More...

Post-allocation passes. More...

Pre-fixup passes. More...

Post-fixup passes. More...

Detailed Description

An LinkGraph pass configuration, consisting of a list of pre-prune, post-prune, and post-fixup passes.

Definition at line 1554 of file JITLink.h.

◆ PostAllocationPasses

Post-allocation passes.

These passes are called on the graph after memory has been allocated and defined nodes have been assigned their final addresses, but before the context has been notified of these addresses. At this point externals have not been resolved, and symbol content has not yet been copied into working memory.

Notable use cases: Setting up data structures associated with addresses of defined symbols (e.g. a mapping of __dso_handle to JITDylib* for the JIT runtime) – using a PostAllocationPass for this ensures that the data structures are in-place before any query for resolved symbols can complete.

Definition at line 1585 of file JITLink.h.

◆ PostFixupPasses

Post-fixup passes.

These passes are called on the graph after block contents has been copied to working memory, and fixups applied. Blocks have been updated to point to their fixed up content.

Notable use cases: Testing and validation.

Definition at line 1605 of file JITLink.h.

Referenced by llvm::orc::EHFrameRegistrationPlugin::modifyPassConfig().

◆ PostPrunePasses

Post-prune passes.

These passes are called on the graph after dead stripping, but before memory is allocated or nodes assigned their final addresses.

Notable use cases: Building GOT, stub, and TLV symbols.

Definition at line 1570 of file JITLink.h.

◆ PreFixupPasses

Pre-fixup passes.

These passes are called on the graph after memory has been allocated, content copied into working memory, and all nodes (including externals) have been assigned their final addresses, but before any fixups have been applied.

Notable use cases: Late link-time optimizations like GOT and stub elimination.

Definition at line 1596 of file JITLink.h.


text:
# 25th July: Linear regression

## Recovering data points: using $(\overline{x}, \overline{y})$

By now, we should be familiar with substituting numbers into our regression lines to estimate values.

However, to recover an unknown data point, we cannot substitute values into our regression line because our regression line generally do not pass through the data points. This is where the theory that $(\overline{x}, \overline{y})$ lies on our regression lines come in.

### Example

Given $y=22.51355 – 4.908387x$, it is very tempting to sub in $x=2$ into the equation. We cannot do that. Instead, let’s calculate the mean $\overline{x}$. Now we can substitute this into the regression line to get $\overline{y}$. Finally we can work backwards to get $k$: give it a try.