text:
# Chemistry:Livermorium

Livermorium, 116Lv
Livermorium
Pronunciation/ˌlɪvərˈmɔːriəm/
Mass number[293]
Livermorium in the periodic table
 Hydrogen Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium Magnesium Aluminium Silicon Phosphorus Sulfur Chlorine Argon Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton Rubidium Strontium Yttrium Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon Caesium Barium Lanthanum Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutetium Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury (element) Thallium Lead Bismuth Polonium Astatine Radon Francium Radium Actinium Thorium Protactinium Uranium Neptunium Plutonium Americium Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium Rutherfordium Dubnium Seaborgium Bohrium Hassium Meitnerium Darmstadtium Roentgenium Copernicium Nihonium Flerovium Moscovium Livermorium Tennessine Oganesson
Po

Lv

(Usn)
moscoviumlivermoriumtennessine
116
Groupgroup 16 (chalcogens)
Periodperiod 7
Block  p-block
Element category  p-block, but probably a post-transition metal
Electron configuration[Rn] 5f14 6d10 7s2 7p4 (predicted)[1]
Electrons per shell2, 8, 18, 32, 32, 18, 6 (predicted)
Physical properties
Phase at STPsolid (predicted)[1][2]
Melting point637–780 K (364–507 °C, 687–944 °F) (extrapolated)[2]
Boiling point1035–1135 K (762–862 °C, 1403–1583 °F) (extrapolated)[2]
Density (near r.t.)12.9 g/cm3 (predicted)[1]
Heat of fusion7.61 kJ/mol (extrapolated)[2]
Heat of vaporization42 kJ/mol (predicted)[3]
Atomic properties
Oxidation states(−2),[4] (+2), (+4) (predicted)[1]
Ionization energies
• 1st: 663.9 kJ/mol (predicted)[5]
• 2nd: 1330 kJ/mol (predicted)[3]
• 3rd: 2850 kJ/mol (predicted)[3]
• (more)
Other properties
Natural occurrencesynthetic
CAS Number54100-71-9
History
Namingafter Lawrence Livermore National Laboratory,[6] itself named partly after Livermore, California
DiscoveryJoint Institute for Nuclear Research and Lawrence Livermore National Laboratory (2000)
Main isotopes of livermorium
Isotope Abundance Physics:Half-life (t1/2) Decay mode Product
290Lv syn 8.3 ms α 286Fl
291Lv syn 19 ms α 287Fl
292Lv syn 13 ms α 288Fl
293Lv syn 57 ms α 289Fl
294Lv syn 54 ms? α 290Fl
Category: Livermorium
| references
Lv
in calc from C diff report ref
C 364
K 637 637 0
F 687 687 0
max precision 0
WD

input C: 364–507, K: 637–780, F: 687–944
comment (extrapolated)[2]
Lv
in calc from C diff report ref
C 762
K 1035 1035 0
F 1403 1404 -1 delta
max precision 0
WD

input C: 762–862, K: 1035–1135, F: 1403–1583
comment (extrapolated)[2]

Livermorium is a synthetic chemical element with the symbol Lv and has an atomic number of 116. It is an extremely radioactive element that has only been created in the laboratory and has not been observed in nature. The element is named after the Lawrence Livermore National Laboratory in the United States, which collaborated with the Joint Institute for Nuclear Research (JINR) in Dubna, Russia to discover livermorium during experiments made between 2000 and 2006. The name of the laboratory refers to the city of Livermore, California where it is located, which in turn was named after the rancher and landowner Robert Livermore. The name was adopted by IUPAC on May 30, 2012.[6] Four isotopes of livermorium are known, with mass numbers between 290 and 293 inclusive; the longest-lived among them is livermorium-293 with a half-life of about 60 milliseconds. A fifth possible isotope with mass number 294 has been reported but not yet confirmed.

In the periodic table, it is a p-block transactinide element. It is a member of the 7th period and is placed in group 16 as the heaviest chalcogen, although it has not been confirmed to behave as the heavier homologue to the chalcogen polonium. Livermorium is calculated to have some similar properties to its lighter homologues (oxygen, sulfur, selenium, tellurium, and polonium), and be a post-transition metal, although it should also show several major differences from them.

## History

### Unsuccessful synthesis attempts

The first search for element 116, using the reaction between 248Cm and 48Ca, was performed in 1977 by Ken Hulet and his team at the Lawrence Livermore National Laboratory (LLNL). They were unable to detect any atoms of livermorium.[7] Yuri Oganessian and his team at the Flerov Laboratory of Nuclear Reactions (FLNR) in the Joint Institute for Nuclear Research (JINR) subsequently attempted the reaction in 1978 and met failure. In 1985, in a joint experiment between Berkeley and Peter Armbruster's team at GSI, the result was again negative, with a calculated cross section limit of 10–100 pb. Work on reactions with 48Ca, which had proved very useful in the synthesis of nobelium from the natPb+48Ca reaction, nevertheless continued at Dubna, with a superheavy element separator being developed in 1989, a search for target materials and starting of collaborations with LLNL being started in 1990, production of more intense 48Ca beams being started in 1996, and preparations for long-term experiments with 3 orders of magnitude higher sensitivity being performed in the early 1990s. This work led directly to the production of new isotopes of elements 112 to 118 in the reactions of 48Ca with actinide targets and the discovery of the 5 heaviest elements on the periodic table: flerovium, moscovium, livermorium, tennessine, and oganesson.[8]

In 1995, an international team led by Sigurd Hofmann at the Gesellschaft für Schwerionenforschung (GSI) in Darmstadt, Germany attempted to synthesise element 116 in a radiative capture reaction (in which the compound nucleus de-excites through pure gamma emission without evaporating neutrons) between a lead-208 target and selenium-82 projectiles. No atoms of element 116 were identified.[9]

### Unconfirmed discovery claims

In late 1998, Polish physicist Robert Smolańczuk published calculations on the fusion of atomic nuclei towards the synthesis of superheavy atoms, including elements 118 and 116.[10] His calculations suggested that it might be possible to make these two elements by fusing lead with krypton under carefully controlled conditions.[10]

In 1999, researchers at Lawrence Berkeley National Laboratory made use of these predictions and announced the discovery of elements 118 and 116, in a paper published in Physical Review Letters,[11] and very soon after the results were reported in Science.[12] The researchers reported to have performed the reaction

8636Kr + 20882Pb293118Og + Neutron → 289116Lv + α

The following year, they published a retraction after researchers at other laboratories were unable to duplicate the results and the Berkeley lab itself was unable to duplicate them as well.[13] In June 2002, the director of the lab announced that the original claim of the discovery of these two elements had been based on data fabricated by principal author Victor Ninov.[14][15]

### Discovery

Livermorium was first synthesized on July 19, 2000, when scientists at Dubna (JINR) bombarded a curium-248 target with accelerated calcium-48 ions. A single atom was detected, decaying by alpha emission with decay energy 10.54 MeV to an isotope of flerovium. The results were published in December 2000.[16]

24896Cm + 4820Ca296116Lv* → 293116Lv + 3 10n289114Fl + α

The daughter flerovium isotope had properties matching those of a flerovium isotope first synthesized in June 1999, which was originally assigned to 288Fl,[16] implying an assignment of the parent livermorium isotope to 292Lv. Later work in December 2002 indicated that the synthesized flerovium isotope was actually 289Fl, and hence the assignment of the synthesized livermorium atom was correspondingly altered to 293Lv.[17]

Two further atoms were reported by the institute during their second experiment during April–May 2001.[18] In the same experiment they also detected a decay chain which corresponded to the first observed decay of flerovium in December 1998, which had been assigned to 289Fl.[18] No flerovium isotope with the same properties as the one found in December 1998 has ever been observed again, even in repeats of the same reaction. Later it was found that 289Fl has different decay properties and that the first observed flerovium atom may have been its nuclear isomer 289mFl.[16][19] The observation of 289mFl in this series of experiments may indicate the formation of a parent isomer of livermorium, namely 293mLv, or a rare and previously unobserved decay branch of the already-discovered state 293Lv to 289mFl. Neither possibility is certain, and research is required to positively assign this activity. Another possibility suggested is the assignment of the original December 1998 atom to 290Fl, as the low beam energy used in that original experiment makes the 2n channel plausible; its parent could then conceivably be 294Lv, but this assignment would still need confirmation in the 248Cm(48Ca,2n)294Lv reaction.[16][19][20]

The team repeated the experiment in April–May 2005 and detected 8 atoms of livermorium. The measured decay data confirmed the assignment of the first-discovered isotope as 293Lv. In this run, the team also observed the isotope 292Lv for the first time.[17] In further experiments from 2004 to 2006, the team replaced the curium-248 target with the lighter curium isotope curium-245. Here evidence was found for the two isotopes 290Lv and 291Lv.[21]

In May 2009, the IUPAC/IUPAP Joint Working Party reported on the discovery of copernicium and acknowledged the discovery of the isotope 283Cn.[22] This implied the de facto discovery of the isotope 291Lv, from the acknowledgment of the data relating to its granddaughter 283Cn, although the livermorium data was not absolutely critical for the demonstration of copernicium's discovery. Also in 2009, confirmation from Berkeley and the Gesellschaft für Schwerionenforschung (GSI) in Germany came for the flerovium isotopes 286 to 289, immediate daughters of the four known livermorium isotopes. In 2011, IUPAC evaluated the Dubna team experiments of 2000–2006. Whereas they found the earliest data (not involving 291Lv and 283Cn) inconclusive, the results of 2004–2006 were accepted as identification of livermorium, and the element was officially recognized as having been discovered.[21]

The synthesis of livermorium has been separately confirmed at the GSI (2012) and RIKEN (2014 and 2016).[23][24] In the 2012 GSI experiment, one chain tentatively assigned to 293Lv was shown to be inconsistent with previous data; it is believed that this chain may instead originate from an isomeric state, 293mLv.[23] In the 2016 RIKEN experiment, one atom that may be assigned to 294Lv was seemingly detected, alpha decaying to 290Fl and 286Cn, which underwent spontaneous fission; however, the first alpha from the livermorium nuclide produced was missed, and the assignment to 294Lv is still uncertain though plausible.[25]

### Naming

Robert Livermore, the indirect namesake of livermorium

Using Mendeleev's nomenclature for unnamed and undiscovered elements, livermorium is sometimes called eka-polonium.[26] In 1979 IUPAC recommended that the placeholder systematic element name ununhexium (Uuh)[27] be used until the discovery of the element was confirmed and a name was decided. Although widely used in the chemical community on all levels, from chemistry classrooms to advanced textbooks, the recommendations were mostly ignored among scientists in the field,[28][29] who called it "element 116", with the symbol of E116, (116), or even simply 116.[1]

According to IUPAC recommendations, the discoverer or discoverers of a new element have the right to suggest a name.[30] The discovery of livermorium was recognized by the Joint Working Party (JWP) of IUPAC on 1 June 2011, along with that of flerovium.[21] According to the vice-director of JINR, the Dubna team originally wanted to name element 116 moscovium, after the Moscow Oblast in which Dubna is located,[31] but it was later decided to use this name for element 115 instead. The name livermorium and the symbol Lv were adopted on May 23,[32] 2012.[6][33] The name recognises the Lawrence Livermore National Laboratory, within the city of Livermore, California, USA, which collaborated with JINR on the discovery. The city in turn is named after the American rancher Robert Livermore, a naturalized Mexican citizen of English birth.[6] The naming ceremony for flerovium and livermorium was held in Moscow on October 24, 2012.[34]

## Predicted properties

Other than nuclear properties, no properties of livermorium or its compounds have been measured; this is due to its extremely limited and expensive production[35] and the fact that it decays very quickly. Properties of livermorium remain unknown and only predictions are available.

### Nuclear stability and isotopes

Main page: Physics:Isotopes of livermorium
The expected location of the island of stability is marked by the white circle. The dotted line is the line of beta stability.

Livermorium is expected to be near an island of stability centered on copernicium (element 112) and flerovium (element 114).[36][37] Due to the expected high fission barriers, any nucleus within this island of stability exclusively decays by alpha decay and perhaps some electron capture and beta decay.Template:Fricke1975 While the known isotopes of livermorium do not actually have enough neutrons to be on the island of stability, they can be seen to approach the island, as the heavier isotopes are generally the longer-lived ones.[16][21]

Superheavy elements are produced by nuclear fusion. These fusion reactions can be divided into "hot" and "cold" fusion,[lower-alpha 1] depending on the excitation energy of the compound nucleus produced. In hot fusion reactions, very light, high-energy projectiles are accelerated toward very heavy targets (actinides), giving rise to compound nuclei at high excitation energy (~40–50 MeV) that may either fission or evaporate several (3 to 5) neutrons.[39] In cold fusion reactions (which use heavier projectiles, typically from the fourth period, and lighter targets, usually lead and bismuth), the produced fused nuclei have a relatively low excitation energy (~10–20 MeV), which decreases the probability that these products will undergo fission reactions. As the fused nuclei cool to the ground state, they require emission of only one or two neutrons. Hot fusion reactions tend to produce more neutron-rich products because the actinides have the highest neutron-to-proton ratios of any elements that can presently be made in macroscopic quantities.[40]

Important information could be gained regarding the properties of superheavy nuclei by the synthesis of more livermorium isotopes, specifically those with a few neutrons more or less than the known ones – 286Lv, 287Lv, 288Lv, 289Lv, 294Lv, and 295Lv. This is possible because there are many reasonably long-lived isotopes of curium that can be used to make a target.[36] The light isotopes can be made by fusing curium-243 with calcium-48. They would undergo a chain of alpha decays, ending at transactinide isotopes that are too light to achieve by hot fusion and too heavy to be produced by cold fusion.[36]

The synthesis of the heavy isotopes 294Lv and 295Lv could be accomplished by fusing the heavy curium isotope curium-250 with calcium-48. The cross section of this nuclear reaction would be about 1 picobarn, though it is not yet possible to produce 250Cm in the quantities needed for target manufacture.[36] After a few alpha decays, these livermorium isotopes would reach nuclides at the line of beta stability. Additionally, electron capture may also become an important decay mode in this region, allowing affected nuclei to reach the middle of the island. For example, it is predicted that 295Lv would alpha decay to 291Fl, which would undergo successive electron capture to 291Nh and then 291Cn which is expected to be in the middle of the island of stability and have a half-life of about 1200 years, affording the most likely hope of reaching the middle of the island using current technology. A drawback is that the decay properties of superheavy nuclei this close to the line of beta stability are largely unexplored.[36]

Other possibilities to synthesize nuclei on the island of stability include quasifission (partial fusion followed by fission) of a massive nucleus.[41] Such nuclei tend to fission, expelling doubly magic or nearly doubly magic fragments such as calcium-40, tin-132, lead-208, or bismuth-209.[42] Recently it has been shown that the multi-nucleon transfer reactions in collisions of actinide nuclei (such as uranium and curium) might be used to synthesize the neutron-rich superheavy nuclei located at the island of stability,[41] although formation of the lighter elements nobelium or seaborgium is more favored.[36] One last possibility to synthesize isotopes near the island is to use controlled nuclear explosions to create a neutron flux high enough to bypass the gaps of instability at 258–260Fm and at mass number 275 (atomic numbers 104 to 108), mimicking the r-process in which the actinides were first produced in nature and the gap of instability around radon bypassed.[36] Some such isotopes (especially 291Cn and 293Cn) may even have been synthesized in nature, but would have decayed away far too quickly (with half-lives of only thousands of years) and be produced in far too small quantities (about 10−12 the abundance of lead) to be detectable as primordial nuclides today outside cosmic rays.[36]

### Physical and atomic

In the periodic table, livermorium is a member of group 16, the chalcogens. It appears below oxygen, sulfur, selenium, tellurium, and polonium. Every previous chalcogen has six electrons in its valence shell, forming a valence electron configuration of ns2np4. In livermorium's case, the trend should be continued and the valence electron configuration is predicted to be 7s27p4;[1] therefore, livermorium will have some similarities to its lighter congeners. Differences are likely to arise; a large contributing effect is the spin–orbit (SO) interaction—the mutual interaction between the electrons' motion and spin. It is especially strong for the superheavy elements, because their electrons move much faster than in lighter atoms, at velocities comparable to the speed of light.[43] In relation to livermorium atoms, it lowers the 7s and the 7p electron energy levels (stabilizing the corresponding electrons), but two of the 7p electron energy levels are stabilized more than the other four.[44] The stabilization of the 7s electrons is called the inert pair effect, and the effect "tearing" the 7p subshell into the more stabilized and the less stabilized parts is called subshell splitting. Computation chemists see the split as a change of the second (azimuthal) quantum number l from 1 to 12 and 32 for the more stabilized and less stabilized parts of the 7p subshell, respectively: the 7p1/2 subshell acts as a second inert pair, though not as inert as the 7s electrons, while the 7p3/2 subshell can easily participate in chemistry.[1][43][lower-alpha 2] For many theoretical purposes, the valence electron configuration may be represented to reflect the 7p subshell split as 7s27p21/27p23/2.[1]

The inert pair effects in livermorium should be even stronger than for polonium and hence the +2 oxidation state becomes more stable than the +4 state, which would be stabilized only by the most electronegative ligands; this is reflected in the expected ionization energies of livermorium, where there are large gaps between the second and third ionization energies (corresponding to the breaching of the unreactive 7p1/2 shell) and fourth and fifth ionization energies.Template:Fricke1975 Indeed, the 7s electrons are expected to be so inert that the +6 state will not be attainable.[1] The melting and boiling points of livermorium are expected to continue the trends down the chalcogens; thus livermorium should melt at a higher temperature than polonium, but boil at a lower temperature.[2] It should also be denser than polonium (α-Lv: 12.9 g/cm3; α-Po: 9.2 g/cm3); like polonium it should also form an α and a β allotrope.Template:Fricke1975[45] The electron of the hydrogen-like livermorium atom (oxidized so that it only has one electron, Lv115+) is expected to move so fast that it has a mass 1.86 times that of a stationary electron, due to relativistic effects. For comparison, the figures for hydrogen-like polonium and tellurium are expected to be 1.26 and 1.080 respectively.[43]

### Chemical

Livermorium is projected to be the fourth member of the 7p series of chemical elements and the heaviest member of group 16 in the periodic table, below polonium. While it is the least theoretically studied of the 7p elements, its chemistry is expected to be quite similar to that of polonium.Template:Fricke1975 The group oxidation state of +6 is known for all the chalcogens apart from oxygen which cannot expand its octet and is one of the strongest oxidizing agents among the chemical elements. Oxygen is thus limited to a maximum +2 state, exhibited in the fluoride OF2. The +4 state is known for sulfur, selenium, tellurium, and polonium, undergoing a shift in stability from reducing for sulfur(IV) and selenium(IV) through being the most stable state for tellurium(IV) to being oxidizing in polonium(IV). This suggests a decreasing stability for the higher oxidation states as the group is descended due to the increasing importance of relativistic effects, especially the inert pair effect.[43] The most stable oxidation state of livermorium should thus be +2, with a rather unstable +4 state. The +2 state should be about as easy to form as it is for beryllium and magnesium, and the +4 state should only be achieved with strongly electronegative ligands, such as in livermorium(IV) fluoride (LvF4).[1] The +6 state should not exist at all due to the very strong stabilization of the 7s electrons, making the valence core of livermorium only four electrons.Template:Fricke1975 The lighter chalcogens are also known to form a −2 state as oxide, sulfide, selenide, telluride, and polonide; due to the destabilization of livermorium's 7p3/2 subshell, the −2 state should be very unstable for livermorium, whose chemistry should be essentially purely cationic,[1] though the larger subshell and spinor energy splittings of livermorium as compared to polonium should make Lv2− slightly less unstable than expected.[43]

Livermorane (LvH2) would be the heaviest chalcogen hydride and the heaviest homolog of water (the lighter ones being H2S, H2Se, H2Te, and PoH2). Polane (polonium hydride) is a more covalent compound than most metal hydrides because polonium straddles the border between metals and metalloids and has some nonmetallic properties: it is intermediate between a hydrogen halide like hydrogen chloride (HCl) and a metal hydride like stannane (SnH4). Livermorane should continue this trend: it should be a hydride rather than a livermoride, but would still be a covalent molecular compound.[46] Spin-orbit interactions are expected to make the Lv–H bond longer than expected simply from periodic trends alone, and make the H–Lv–H bond angle larger than expected: this is theorized to be because the unoccupied 8s orbitals are relatively low in energy and can hybridize with the valence 7p orbitals of livermorium.[46] This phenomenon, dubbed "supervalent hybridization",[46] is not particularly uncommon in non-relativistic regions in the periodic table; for example, molecular calcium difluoride has 4s and 3d involvement from the calcium atom.[47] The heavier livermorium dihalides are predicted to be linear, but the lighter ones are predicted to be bent.[48]

## Experimental chemistry

Unambiguous determination of the chemical characteristics of livermorium has not yet been established.[49][50] In 2011, experiments were conducted to create nihonium, flerovium, and moscovium isotopes in the reactions between calcium-48 projectiles and targets of americium-243 and plutonium-244. The targets included lead and bismuth impurities and hence some isotopes of bismuth and polonium were generated in nucleon transfer reactions. This, while an unforeseen complication, could give information that would help in the future chemical investigation of the heavier homologs of bismuth and polonium, which are respectively moscovium and livermorium.[50] The produced nuclides bismuth-213 and polonium-212m were transported as the hydrides 213BiH3 and 212mPoH2 at 850 °C through a quartz wool filter unit held with tantalum, showing that these hydrides were surprisingly thermally stable, although their heavier congeners McH3 and LvH2 would be expected to be less thermally stable from simple extrapolation of periodic trends in the p-block.[50] Further calculations on the stability and electronic structure of BiH3, McH3, PoH2, and LvH2 are needed before chemical investigations take place. Moscovium and livermorium are expected to be volatile enough as pure elements for them to be chemically investigated in the near future, a property livermorium would then share with its lighter congener polonium, though the short half-lives of all presently known livermorium isotopes means that the element is still inaccessible to experimental chemistry.[50][51]

## Notes

1. Despite the name, "cold fusion" in the context of superheavy element synthesis is a distinct concept from the idea that nuclear fusion can be achieved in room temperature conditions (see cold fusion).[38]
2. The quantum number corresponds to the letter in the electron orbital name: 0 to s, 1 to p, 2 to d, etc. See azimuthal quantum number for more information.

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21. Barber, R. C.; Gaeggeler, H. W.; Karol, P. J.; Nakahara, H.; Verdaci, E.; Vogt, E. (2009). "Discovery of the element with atomic number 112" (IUPAC Technical Report). Pure Appl. Chem. 81 (7): 1331. doi:10.1351/PAC-REP-08-03-05.
22. Hofmann, S.; Heinz, S.; Mann, R.; Maurer, J.; Khuyagbaatar, J.; Ackermann, D.; Antalic, S.; Barth, W. et al. (2012). "The reaction 48Ca + 248Cm → 296116* studied at the GSI-SHIP". The European Physical Journal A 48 (5): 62. doi:10.1140/epja/i2012-12062-1. Bibcode2012EPJA...48...62H.
23. Morita, K. (2014). "Measurement of the 248Cm + 48Ca fusion reaction products at RIKEN GARIS". RIKEN Accel. Prog. Rep. 47: 11.
24. Kaji, Daiya; Morita, Kosuke; Morimoto, Kouji; Haba, Hiromitsu; Asai, Masato; Fujita, Kunihiro; Gan, Zaiguo; Geissel, Hans et al. (2017). "Study of the Reaction 48Ca + 248Cm → 296Lv* at RIKEN-GARIS". Journal of the Physical Society of Japan 86 (3): 034201–1–7. doi:10.7566/JPSJ.86.034201. Bibcode2017JPSJ...86c4201K.
25. Seaborg, Glenn T. (1974). "The Search for New Elements: The Projects of Today in a Larger Perspective". Physica Scripta 10: 5–12. doi:10.1088/0031-8949/10/A/001. Bibcode1974PhyS...10S...5S.
26. Chatt, J. (1979). "Recommendations for the Naming of Elements of Atomic Numbers Greater than 100". Pure Appl. Chem. 51 (2): 381–384. doi:10.1351/pac197951020381.
27. Folden, Cody (31 January 2009). "The Heaviest Elements in the Universe". Saturday Morning Physics at Texas A&M.  "
28. Hoffman, Darleane C.. "Darmstadtium and Beyond". Chemical & Engineering News.
29. Koppenol, W. H. (2002). "Naming of new elements(IUPAC Recommendations 2002)". Pure and Applied Chemistry 74 (5): 787. doi:10.1351/pac200274050787.
30.  : Mikhail Itkis, the vice-director of JINR stated: "We would like to name element 114 after Georgy Flerov – flerovium, and another one [element 116] – moscovium, not after Moscow, but after Moscow Oblast".
31. Loss, Robert D.; Corish, John. "Names and symbols of the elements with atomic numbers 114 and 116 (IUPAC Recommendations 2012)". IUPAC.
32. Cite error: Invalid <ref> tag; no text was provided for refs named Bloomberg
33. Zagrebaev, Valeriy; Karpov, Alexander; Greiner, Walter (2013). "Future of superheavy element research: Which nuclei could be synthesized within the next few years?". 420. IOP Science. pp. 1–15. Retrieved 20 August 2013.
34. Considine, Glenn D.; Kulik, Peter H. (2002). Van Nostrand's scientific encyclopedia (9th ed.). Wiley-Interscience. ISBN 978-0-471-33230-5. OCLC 223349096.
35. Fleischmann, Martin; Pons, Stanley (1989). "Electrochemically induced nuclear fusion of deuterium". Journal of Electroanalytical Chemistry and Interfacial Electrochemistry 261 (2): 301–308. doi:10.1016/0022-0728(89)80006-3.
36. Barber, Robert C.; Gäggeler, Heinz W.; Karol, Paul J.; Nakahara, Hiromichi; Vardaci, Emanuele; Vogt, Erich (2009). "Discovery of the element with atomic number 112 (IUPAC Technical Report)". Pure and Applied Chemistry 81 (7): 1331. doi:10.1351/PAC-REP-08-03-05.
37. Armbruster, Peter; Munzenberg, Gottfried (1989). "Creating superheavy elements". Scientific American 34: 36–42.
38. Zagrebaev, V.; Greiner, W. (2008). "Synthesis of superheavy nuclei: A search for new production reactions". Physical Review C 78 (3): 034610. doi:10.1103/PhysRevC.78.034610. Bibcode2008PhRvC..78c4610Z.
39. Thayer, John S. (2010). "Relativistic Effects and the Chemistry of the Heavier Main Group Elements". Challenges and Advances in Computational Chemistry and Physics. 10. 83. doi:10.1007/978-1-4020-9975-5_2. ISBN 978-1-4020-9974-8.
40. Faegri, K.; Saue, T. (2001). "Diatomic molecules between very heavy elements of group 13 and group 17: A study of relativistic effects on bonding". Journal of Chemical Physics 115 (6): 2456. doi:10.1063/1.1385366. Bibcode2001JChPh.115.2456F.
41. Eichler, Robert (2015). "Gas phase chemistry with SHE – Experiments". Texas A & M University.
42. Nash, Clinton S.; Crockett, Wesley W. (2006). "An Anomalous Bond Angle in (116)H2. Theoretical Evidence for Supervalent Hybridization.". The Journal of Physical Chemistry A 110 (14): 4619–4621. doi:10.1021/jp060888z. PMID 16599427. Bibcode2006JPCA..110.4619N.
43. Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.). Butterworth-Heinemann. p. 117. ISBN 978-0-08-037941-8.
44. Van WüLlen, C.; Langermann, N. (2007). "Gradients for two-component quasirelativistic methods. Application to dihalogenides of element 116". The Journal of Chemical Physics 126 (11): 114106. doi:10.1063/1.2711197. PMID 17381195. Bibcode2007JChPh.126k4106V.
45. Düllmann, Christoph E. (2012). "Superheavy elements at GSI: a broad research program with element 114 in the focus of physics and chemistry". Radiochimica Acta 100 (2): 67–74. doi:10.1524/ract.2011.1842.
46. Eichler, Robert (2013). "First foot prints of chemistry on the shore of the Island of Superheavy Elements". Journal of Physics: Conference Series 420 (1): 012003. doi:10.1088/1742-6596/420/1/012003. Bibcode2013JPhCS.420a2003E.
47. Moody, Ken (2013-11-30). "Synthesis of Superheavy Elements". in Schädel, Matthias; Shaughnessy, Dawn. The Chemistry of Superheavy Elements (2nd ed.). Springer Science & Business Media. pp. 24–8. ISBN 9783642374661.


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# How do you graph r=3-2costheta?

Aug 1, 2018

See the graph of this dimpled limacon.

#### Explanation:

$0 \le r = 3 - 2 \cos \theta \in \left[1 , 5\right]$

Period = period of $\cos \theta = 2 \pi$.

Using $r = \sqrt{{x}^{2} + {t}^{2}} \mathmr{and} \cos \theta = \frac{x}{r}$,

the Cartesian form is obtained as

${x}^{2} + {y}^{2} - 3 \sqrt{{x}^{2} + {y}^{2}} = 2 x = 0$

The graph of this dimpled limacon is immediate.

graph{ x^2 + y^2 -3sqrt ( x^2 + y^2 )+ 2x = 0[-20 20 -10 10]}

format: plain


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# Recent Progress on the Dynamic Optimality Conjecture

## Place: CUNY Graduate Center, Rm 4419

The dynamic optimality conjecture is the most fundamental open question about binary search trees (BST). It postulates the existence of an asymptotically optimal online BST, i.e., one that is constant factor competitive with any BST on any input sequence, even with one that knows the entire sequence in advance. The two main candidates for dynamic optimality are splay trees [Sleator and Tarjan, 1985] and GREEDY [Lucas, 1988; Munro, 2000; Demaine et al. 2009]. Despite BSTs being among the simplest data structures in computer science and despite extensive effort over the past three decades, the conjecture remains elusive. Dynamic optimality is achieved for almost all sequences, as the majority of them are hard and have optimum cost $\Theta(n \log n)$, achievable by any balanced BST. Thus the missing step towards the conjecture is understanding “easy” access sequences.

One prominent example is the traversal conjecture [Sleator and Tarjan, 1985] which states that preorder sequences are accessed in linear time by splay trees; no online BST was known to satisfy this conjecture until now.

In this talk I will survey the literature on dynamic optimality and present a proof of the traversal conjecture for GREEDY. I will also present a proof that Splay trees are not black magic, and end with some promising open problems. The talk will be accessible to anyone who knows what a binary tree is.

This talk is based on joint work (appearing in WADS, ESA and FOCS 2015, ISAAC 2018) with P. Charlemsook, K. Mehlhorn, L. Kozma and T. Saranurak.


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Definitions
Nearby Words

# Jounce

[jouns]
In physics, jounce or snap is the fourth derivative of the displacement vector with respect to time, with the first, second, and third derivatives being velocity, acceleration, and jerk, respectively; in other words, the jounce is the rate of change of the jerk with respect to time.

$vec s=frac \left\{d vec j\right\} \left\{dt\right\}=frac \left\{d^2 vec a\right\} \left\{dt^2\right\}=frac \left\{d^3 vec v\right\} \left\{dt^3\right\}=frac \left\{d^4 vec r\right\} \left\{dt^4\right\}$

Currently, there are no well accepted designations for the derivatives of jounce. The fifth and sixth derivatives of displacement as a function of time are "sometimes somewhat facetiously" referred to as "Crackle" and "Pop"; however, these terms have not gained widespread acceptance.


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# Math Help - L'Hospital's Rule Square Root Problem

1. ## Re: L'Hospital's Rule Square Root Problem

Ok sorry, missed that. Anyhow, what about the original problem dealing with L' Hospital's rule. So how did the "x over the square root fraction" come about?

As for the original problem, it does come out to be indeterminate before factoring out. But after factoring, it's not.

2. ## Re: L'Hospital's Rule Square Root Problem

Look back to my very first post (post #3). What is the derivative of the expression $\sqrt{x^2 + 3} - 2$? (in terms of x)

3. ## Re: L'Hospital's Rule Square Root Problem

1/(b/a) = a/b

would be the reason for the fraction under x.

4. ## Re: L'Hospital's Rule Square Root Problem

Yes...and you know how to compute the derivative of the denominator, right?

Page 2 of 2 First 12


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×

# Microeconomic Theory - Production Function

Consider a simple production function with only one factor of production L, where L is Labor and Q is the total output.

Q=18L$$^{2}$$ -L$$^{3}$$

If L =0, Q= 0. If L=1, Q=17. If L=2, Q=64. If L=3, Q= 135 and so forth....

The Marginal Product of Labor, MPL, is defined as the change in output that results from employing an added unit of labor. If that is the case, we can easily calculate the MPL given the production function Q.

When L goes from 0 to 1, Q went from 0 to 17. When L goes from 1 to 2, Q went up by 64 as shown previously. Therefore, the MPL when Labor=2 is (64-17/1-0)= 47. You can also do the same when you add the third labor, or when L=3. The corresponding MPL is (135-64)/(3-2)= 71.

We can also compute MPL as the first derivative of the Production Function Q where Q is Q=18L$$^{2}$$ -L$$^{3}$$ Therefore, dQ/dL= 36L - 3L$$^{2}$$

So, when L = 1, plug this into dQ/dL and we get 33. When L=2, dQ/dL= 36(2)-3(2)$$^{2}$$= 60.

See associated production function table here.

Question: Why do the results differ ( 17 against 33 and 47 against 60 and so forth) when we calculated MPL manually versus using the first derive function of Q?

Note by Venture Hi
2 years, 11 months ago


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# How do you evaluate 2x - 4, if x = 3?

May 31, 2016

You have to replace $x$ with $3$.

#### Explanation:

where do you read $x$ you have to substitute the value $3$ remembering that the number in front of the $x$ is multiplying the $x$.

$2 x - 4$

I substitute where I read $x$ the $3$

$2 \cdot 3 - 4 = 6 - 4 = 2$.


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# Probability: tree diagram

• Dec 8th 2009, 10:32 PM
thcbender
Probability: tree diagram
A bag contains six white marbles, and four blue marbles. Three marbles are drawn in succession. At each draw, if the marble is white, it is replaced, and if its blue it is not replaced.
.
a) draw a tree diagram with this information with all probabiltiies shown (2)
b) find the probability of drawing:
1) no mblue marbles (1)
ii) two blue marbles (1)

another question:

a comittee of 4 people is to be chosen at random from among 10 people.
what is the probability that both James and Sarah will be chosen.
• Dec 9th 2009, 06:38 AM
Hello thcbender
Quote:

Originally Posted by thcbender
A bag contains six white marbles, and four blue marbles. Three marbles are drawn in succession. At each draw, if the marble is white, it is replaced, and if its blue it is not replaced.
.
a) draw a tree diagram with this information with all probabiltiies shown (2)
b) find the probability of drawing:
1) no mblue marbles (1)
ii) two blue marbles (1)

another question:

a comittee of 4 people is to be chosen at random from among 10 people.
what is the probability that both James and Sarah will be chosen.

I attach a partly completed tree diagram. I hope that you can complete it.

(b) (i) The probability of drawing no blue marbles is, as I have shown: $\frac{6}{10}\times\frac{6}{10}\times\frac{6}{10} =$ ... ?

(ii) There are three routes through the probability tree that correspond to two blue marbles: BWW, WBW and WWB. Work out the probabilities of each of them and add the resulting fractions together. I'll start you off with BBB:
$\frac{4}{10}\times\frac{3}{9}\times\frac{2}{8}= \frac{1}{10}$
Can you complete it?

Question 2. (On an administrative note, you're asked not to post two different questions in the same thread. See Rule 14.)

There are ${^{10}C_{4}}$ ways of choosing a committee of $4$ people from $10$.

If James and Sarah have been chosen, we need to choose $2$ from the remaining $8$ to complete the committee. In how many ways can this be done?

Divide the second number by the first to get the probability that James and Sarah are on a committee chosen at random.

I make the answer $\frac{2}{15}$. Do you agree?


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# Expected value of maximum of two random variables from uniform distribution

If I have two variables $X$ and $Y$ which randomly take on values uniformly from the range $[a,b]$ (all values equally probable), what is the expected value for $\max(X,Y)$?

-

Here are some useful tools:

1. For every nonnegative random variable $Z$, $$\mathrm E(Z)=\int_0^{+\infty}\mathrm P(Z\geqslant z)\,\mathrm dz.$$
2. As soon as $X$ and $Y$ are independent, $$\mathrm P(\max(X,Y)\leqslant z)=\mathrm P(X\leqslant z)\,\mathrm P(Y\leqslant z).$$
3. If $U$ is uniform on $(0,1)$, then $a+(b-a)U$ is uniform on $(a,b)$.

If $(a,b)=(0,1)$, items 1. and 2. together yield $$\mathrm E(\max(X,Y))=\int_0^1(1-z^2)\,\mathrm dz=\frac23.$$ Then item 3. yields the general case, that is, $$\mathrm E(\max(X,Y))=a+\frac23(b-a)=\frac13(2b+a).$$

-

I very muck liked Martin's approach but there's an error with his integration. The key is on line three. The intution here should be that when y is the maximum, then x can vary from 0 to y whereas y can be anything and vice-versa for when x is the maximum. So the order of integration should be flipped:

-
Thanks for pointing that out. As your answer corrects the mistake, I'll simply delete mine. –  Martin Van der Linden Nov 28 '14 at 16:00
E(max(x,y)) should read E(max(X,Y)). –  Did Aug 22 at 22:24

did's excellent answer proves the result. The picture here

may help your intuition. This is the "average" configuration of two random points on a interval and, as you see, the maximum value is two-thirds of the way from the left endpoint.

-
-1 Sorry for the down vote. I know what you mean, but I hate such examples, as it confuses the situation for those who do not fully understand what an "average" configuration means. –  Calvin Lin Jul 1 '13 at 2:57
I don't follow. In what sense is this an average configuration? –  Jonah Jul 9 '13 at 18:02


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Change of electron spin in Carbon after Hybridization?

During bond formation in Carbon, one of the electrons from the 2s orbital jumps to one the empty p orbitals creating a hybrid orbital as seen in the figure below.

Now clearly one electron from the 2s orbital jumped to the 2pz orbital.
But what happened to the spin?
Before hybridization, one electron had an opposite spin, but after hybridization all electrons have the same spin.
Almost everywhere the same thing is shown.
Did the spin of the promoted electron get reversed?

• That picture is a very crude oversimplification of many things. First of all, hybridisation is a function of geometry and will never happen in an atom by itself. The spin state is dependent on the excitation, it is possible, to excite from a triplet to a quintet state. May 20, 2015 at 4:50
• And that is why I want anyone who teaches chemistry to always emphasize that orbitals, hybridization, etc. are not real things and processes. They are just parts of modeling (and often oversimplified) description of the reality. May 20, 2015 at 7:21

Short answer: No it didn’t. But hybridisation also never happened in the way it is drawn.

The typical hybridisation picture from textbooks that you have shown us, is not ‘the truth’, it’s not ‘an approximation of the truth’ and not even ‘somewhere near to the truth’. It is no more and no less than a set of drawings trying to use one concept to explain an experiment.

The experiment in this case is the structure of methane, which is found to be tetrahedral, while the concept is that of orbitals as standing waves and solutions of the Schrödinger equation. The latter says that in a carbon atom floating around in sheer nothingness, there should be five standing waves produced by the electrons, that were given the name 1s, 2s, 2px, 2py and 2pz. But this is an entirely hypothetical construct from our understanding — not many people have been to outer space and nowhere have people observed and experimented with single, unbound carbon atoms.

Carbon is only found in compounds (on Earth and in the near space, that is). But once it’s in a compound, you need to write down a whole new Schrödinger equation, that not only includes the carbon but also all the other atoms in the molecule. If you solve that equation, you will again get a set of standing waves, but waves, that are spread out across the entire molecule. You need to perfom quite a few mathematical tricks to reduce those waves back to the orbitals you calculated for a single carbon atom — but by doing so, you have lost the predictive power of where the electrons are to go. There is no longer a way to say if any electron is spin $\alpha$ or spin $\beta$.

For methane, calculations that there are in fact four bonding orbitals at two (!) different energy levels. If hybridisation occurred, one would expect only one energy level, because all carbon orbitals would have been made equal. The orbitals generally represent a carbon ground-state orbital added to the corresponding hydrogen orbitals of the same phase (carbon’s p-orbitals oriented so that the stab in between two hydrogens). The lower energy state of course corresponds to carbon’s 2s plus the four hydrogens. (See pictures in this German lecture script, part A, section 4.8)

Take a different molecule, e. g. formaldehyde. Drawing hybridisation schemes, you would expect the oxygen to be something like $\mathrm{sp}^2$ and have a ‘rabbit ear’-like pair of free electrons. But solving the Schrödinger equation does not give you any free electrons. Instead you find a multitude of orbitals, none of which exclusively surrounds the oxygen — excepting the one that represents oxygen’s 1s in simplified diagrams. Again, reducing this to any type of structure that would tell you anything about the ‘shape’ of the free oxygen electrons means that you would lose predictive power. (Luckily in that case, though, all electrons are spin-paired, so they will remain spin-paired.)

• just a side note, Phil Skell studied the reactions of atomic carbon back in the 60's. It reacts with ethylene to yield spiropentane. The stereochemistry of the reactions with cis- and trans-butene is very interesting.
– ron
May 22, 2015 at 0:17
• @ron Oh wow, didn’t know that. Got a doi/source for me?
– Jan
May 22, 2015 at 0:18
• Just google "p. skell, atomic carbon" and you'll find some.
– ron
May 22, 2015 at 0:21


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# arithmetic mean and standard deviation formula

What is Arithmetic Mean? In a physical sense, the arithmetic mean can be thought of as a centre of gravity. Place the cursor where you wish to have the standard deviation appear and click the mouse button.Select Insert Function (f x) from the FORMULAS tab. The formula for the Standard Deviation is square root of the Variance. Population A higher standard deviation means the data points are spread out over a greater range. Work out the complete Standard Deviation, then work out a Sample Standard Deviation from just some of the 8 numbers.. 1. For a mean of 100 and standard deviation of 10, this probability is 0.84. In this article we will discuss about the calculation of simple and weighted arithmetic mean with the help of formulas. A dialog box will appear. The geometric standard deviation (GSD) is the same transformation, applied to the regular standard deviation. Solution : Standard deviation = 6.5. From the mean of a data set, we can think of the average distance the data points are from the mean as standard deviation. It divides the series into two halves by first arranging the items in ascending or descending order of magnitude and then locating the middle value and is denoted by the symbol $\tilde{X}$ or M. The arithmetic mean can also inform or model concepts outside of statistics. This number is relatively close to the true standard deviation and good for a rough estimate. If the data represents the entire population, you can use the STDEV.P function. What Does It "Mean" ? If the data are normally distributed, then about 68% of the data are within one standard deviation of the mean, which is the interval [m-s, m+s]. Statistics - Mean Deviation of Continuous Data Series - When data is given based on ranges alongwith their frequencies. Relative standard deviation is one of the measures of deviation of a set of numbers dispersed from the mean and is computed as the ratio of stand deviation to the mean for a set of numbers. Formula to find arithmetic mean x̄ is. Meaning of Standard Deviation: The best and most important measure of dispersion is standard deviation which was first worked out by Karl Pearson (1833). $σ=\sqrt{∑[(x – μ)2 ∙ P(x)]}\nonumber$ When all outcomes in the probability distribution are equally likely, these formulas coincide with the mean and standard deviation of the set of possible outcomes. The standard deviation, Σ, of the PDF is the square root of the variance. Press the "Submit Data" button to perform the computation. By putting one, two, or three standard deviations above and below the mean we can estimate the ranges that would be expected to include about 68%, 95%, and 99.7% of the observations. As for the arithmetic mean, you need to start by thinking about the location of the geometric mean (20.2). Mean / Median /Mode/ Variance /Standard Deviation are all very basic but very important concept of statistics used in data science. The formula of arithmetic mean is calculated by adding all the available periodic returns and divide the result by the number of periods. It is the positive square root of mean of deviations of individual values of a data series from the arithmetic mean of the series. Example 1 : The standard deviation and mean of a data are 6.5 and 12.5 respectively. Mean is an important measure of central tendency in statistics. Mean = 12.5. Note that the standard deviation in this example is for a sample, not a population. There are also other means of calculating averages such as Median and Mode. X i = each value of dataset. M D = The main features of the mean deviation are: All. Lower the deviation, the close the numbers are dispersed from the mean. The arithmetic mean is the most simple and commonly used measure of central tendency. Check out Median in the Bootstrap example. It is also called a coefficient of variation. Example 8.5 The amount of rainfall in a particular season for 6 days are given as 17.8 cm, 19.2 cm, 16.3 cm, 12.5 cm, 12.8 cm and 11.4 cm. We generally refer to it as Average. Arithmetic mean can be a simple arithmetic mean or weighted arithmetic mean. What is the probability of a student passing the test? Consequently, if we know the mean and standard deviation of a set of observations, we can obtain some useful information by simple arithmetic. Any suitable average among the mean, median or mode can be used in its calculation, but the value of the mean deviation is the minimum if the deviations are taken from the median. Here is a free online arithmetic standard deviation calculator to help you solve your statistical questions. Standard deviation The standard deviation is defined as the square root of the arithmetic mean of the squares of Arithmetic mean is a commonly used average to represent a data. It is not unduly influenced by large or small values. This means that most men (about 68%, assuming a normal distribution) have a height within 3 inches (7.62 cm) of the mean (67–73 inches (170.18–185.42 cm)) – one standard deviation – and almost all men (about 95%) have a height within 6 inches (15.24 cm) of the mean (64–76 inches (162.56–193.04 cm)) – two standard deviations. Standard Deviation Formulas. View math666.doc from MATH MAT130 at Azerbaijan State Oil and Industrial University. X - X n Mean Deviation 3- 26. Standard Deviation Formula. Note: the mean deviation is sometimes called the Mean Absolute Deviation (MAD) because it is the mean of the absolute deviations. Standard Deviation Formula. Following is an example of continous series: x̄ ( = the arithmetic mean of the data (This symbol will be indicated as the mean from now) N = the total number of data points ∑ (X i - x̄) 2 = The sum of (X i … Calculate Standard Deviation, Mean Median, Mode, & Variance. The mean will be displayed if the calculation is successful. To clear the calculator and enter new data, press "Reset". Find its standard deviation. The symbol for Standard Deviation is σ (the Greek letter sigma). This can also be used as a measure of variability or volatility for the given set of data. The standard deviation is a measure of the dispersion (i.e., the degree to which data are spread out) relative to the mean (i.e., arithmetic mean). values are used in the calculation. Standard Deviation. The square of standard deviation (i.e. The mean (average) for the list will appear in the cell you selected. Standard deviation, denoted by the symbol σ, describes the square root of the mean of the squares of all the values of a series derived from the arithmetic mean which is also called as the root-mean-square deviation. The sub procedure reads in the numbers from the array (A1..A10), calls up the function procedures and returns the mean and standard deviation value. + r n) / n. where Ri = return in the i th year and n = Number of periods. The absolute values are difficult to manipulate. It is obtained by simply adding all the values and dividing them by the number of items. But here we explain the formulas.. Also, suppose the test has a mean of m and a standard deviation of s. You have set the cut off as 90 marks for the test. To calculate standard deviation in Excel, you can use one of two primary functions, depending on the data set. Variance and standard deviation are measures of dispersion in statistics and various measures of concentration including quartiles, quintiles, deciles, and percentiles. Let us start learning more about the arithmetic mean formula in detail. Larger the deviation, further the numbers are dispersed away from the mean. If instead we first calculate the range of our data as 25 – 12 = 13 and then divide this number by four we have our estimate of the standard deviation as 13/4 = 3.25. However, What is the formula … Select STDEV.S (for a sample) from the the Statistical category. It overstates the true return and is only appropriate for shorter time periods. What does the geometric standard deviation mean? Find the coefficient of variation. $\text{GSD}[x] = e^{\text{SD}[\log x]}$ This is going to be useful if and only it was a good idea to use a geometric mean on your data, and particularly if your data is positively skewed.Make sure you realize what this is saying. It is given by: σ = standard deviation. It is also very easy to calculate. Determine mean and standard deviation of first n terms of an A.P. (In statistics and probability theory, the standard deviation measures the amount of variation or dispersion from the average.) Standard deviation is a measure of how much variance there is in a set of numbers compared to the average (mean) of the numbers. Mean Deviation The arithmetic mean of the absolute values of the deviations from the arithmetic mean. Standard deviation (by mean method) σ = If d i = x i – are the deviations, then . Standard deviation in Excel. Standard Deviation compared with Sample Standard Deviation For a sample standard deviation example, we'll look at a random list of 8 numbers. A lower standard deviation means the data points are distributed close to the mean. Almost all the machine learning algorithm uses these concepts in… October 11, 2019 by self Leave a Comment. Standard deviation is a measure of the dispersion of data points from the mean of a data set. Mean Deviation tells us how far, on average, all values are from the middle. 0 is the smallest value of standard deviation since it cannot be negative. How To Find (Calculate) The Mean for Ungrouped Data & Mean Deviation . Variables. What is the arithmetic mean. Related formulas. The formula of the mean deviation gives a mathematical impression that is a better way of measuring the variations in the data. The standard deviation of any arithmetic progression can be calculated by the nth term of the sequence. whose first term is a and common difference is d. asked Feb 26, 2018 in Class XI Maths by nikita74 ( -1,017 points) statistics Statistics is a way more beyond the topics mentioned, but here we stop for the “Mean” by Step Deviation method. Arithmetic mean = (r 1 + r 2 + …. Finding the Standard Deviation. E.g. Mean Formula. Arithmetic Median is a positional average and refers to the middle value in a distribution. The Standard Deviation is a measure of how spread out numbers are.. You might like to read this simpler page on Standard Deviation first.. There are various kinds of mean in various branches of statistics, especially statistics. The mean and standand deviation formulas are shown below. When the elements in a series are more isolated from the mean, then the standard deviation is also large. Another convenient way of finding standard deviation is to use the following formula. The arithmetic average return is always higher than the other average return measure called the geometric average return. These values have a mean of 17 and a standard deviation of about 4.1. The standard deviation formula is similar to the variance formula. Arithmetic average return is the return on investment calculated by simply adding the returns for all sub-periods and then dividing it by total number of periods. 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Open Access Publications from the University of California

## Geometric Integrators for Stiff Systems, Lie Groups and Control Systems

• Author(s): Shen, Xuefeng
• Advisor(s): Leok, Melvin
• et al.
Abstract

The main idea of a geometric integrator is to adopt a geometric viewpoint of the problem and to construct integrators that preserve the geometric properties of the continuous dynamical system. For classical mechanics, both the Lagrangian and the Hamiltonian formulations can be described using the language of geometry. Due to the rich conservation properties of mechanics, it is natural to study the construction of numerical integrators that preserve some geometric properties, such as the symplectic structure, energy, and momentum maps. Such geometric structure-preserving numerical integrators exhibit nice properties compared to traditional numerical methods. This is especially true in galaxy simulations and molecular dynamics, where long time simulations are required to answer the corresponding scientific questions. Variational integrators have attracted interest in the geometric integration community as it discretizes Hamilton's principle, as opposed to the corresponding differential equation, to obtain a numerical integrator that is automatically symplectic, and which exhibits a discrete Noether's theorem. Besides classical mechanics, such an approach has also been applied to other fields, such as optimal control~\cite{junge2005discrete,leyendecker2010discrete}, partial differential equations~\cite{marsden1998multisymplectic}, stochastic differential equations~\cite{bou2009stochastic}, and so on.

In this thesis, we consider generalizations of geometric integrators that are adapted to three special settings. One is the case of stiff systems of the form, $\dot{q} = Aq + f(q)$, where the coefficient matrix $A$ has a large spectral radius that is responsible for the stiffness of the system, while the nonlinear term $f(q)$ is relatively smooth. Traditionally, exponential integrators have been used to address the issue of stiffness. In Chapter~\ref{exp}, we consider a special semilinear problem with $A=JD$, $f(q)=J\nabla V(q)$, where $J^T = -J, D^T=D$, and $JD=DJ$. Then, the system is described by $\dot{q} = J(Dq+\nabla V(q))$, which naturally arises from the discretization of Hamiltonian partial differential equations. It is a constant Poisson system with Poisson structure $J_{ij}\frac{\partial}{\partial x_i}\otimes \frac{\partial}{\partial x_j}$, and Hamiltonian $H(q) = \frac{1}{2}q^TDq + V(q)$. Two types of exponential integrators are constructed, one preserves the Poisson structure, and the other preserves energy. Numerical experiments for semilinear Possion systems obtained by semi-discretizing Hamiltonian PDEs are presented. These geometric exponential integrators exhibit better long time stability properties as compared to non-geometric integrators, and are computationally more efficient than traditional symplectic integrators and energy-preserving methods based on the discrete gradient method.

The other generalization is to Lie groups. When configuration manifold is a Lie group, we would like to utilize the group structure rather than simply regard it as embedded submanifold. This is particularly useful when codimension of the embedding is large. For the rigid body problem, the configuration space is $\mathbb{R}^3\rtimes SO(3)$, which is a Lie group. \citet{LeMcLe2005} were the first to directly use the Lie group structure of the rotation group to construct a Lie group variational integrator. In contrast, most prior approaches used the unit quaternion representation of the rotation group and applied symplectic integrators for constrained systems with the unit length constraint. In Chapter~\ref{quater}, we adopt the approach used in constructing Lie group variational integrators for rigid body dynamics on the rotation group and applied it to the unit quaternion representation. A Lie group variational integrator in the unit quaternion representation is derived, and it can be shown that our method is related to the RATTLE method applied to the rotation representation by the projection from unit quaternions to rotation matrices. The numerical results for our Lie group quaternion variational integrator are presented. The integrators constructed in Chapter~\ref{quater} are only second-order, and in Chapter~\ref{polar}, variational integrators of arbitrarily high-order on special orthogonal group $SO(n)$ are constructed by using the polar decomposition. It avoids the second-order derivative of the exponential map that arises in the traditional Lie group variational integrator method. Also, a reduced Lie--Poisson integrator is constructed. The resulting algorithms can naturally be implemented using fixed-point iteration. Numerical results are given for the case of $SO(3)$.

The last generalization is to control systems. We studied the problem of uncertainty propagation and measurement update for systems that are partially unobservable. We construct a method that satisfies the chain property that the unobservable subspace remains perpendicular to the measurement $dh$ during propagation. We characterize the unobservable subspace in terms of the group-invariance of the control system, and obtain a reduced control system on the observable variables. By decomposing the system explicitly into unobservable and observable parts $(x_N, x_O)$, the chain property can be naturally satisfied. Also, we propose a reduced Bayesian framework, where the update from the measurement is only applied to the observable variables $x_O$. In Chapter~\ref{geometric_reduce}, we consider a planar robot model, which has one odometry sensor and one camera. Odometry is used for propagation and the camera is used for measurement. In this model, the two-dimensional position as well as the orientation are all unobservable. We applied our technique to this model and performed numerical simulations. We tested this on straight line, circle, and general trajectories and found that the reduced Kalman filter that we proposed outperforms the classical Kalman filter and modifications that were proposed in the literature. In particular, it estimates the angle quite well, and as a result, yields a better estimate of the position as well.


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+0

# some math

0
91
10
+141

1. In the figure shown, $$ABCD$$ is a square and $$\triangle CDE$$ is equilateral. What is the degree measure of $$\angle CBE$$?

2. A triangle whose side lengths are whole numbers has one side which measures 25 inches and a perimeter of 80 inches. What is the fewest number of inches that can be the length of one of the remaining sides?

3.

In the diagram below, $$AM=BM=CM$$ and $$\angle BMC+\angle A = 201^\circ$$ Find  $$\angle B$$ in degrees.

Feb 20, 2019

#1
+4221
+3

1. Angle C measures 60+90=150 degrees. Since it forms a triangle, the answer is just 180-150=30/2=15 degrees.

Feb 20, 2019
#3
+100570
+2

Excellent, tertre!!!

CPhill  Feb 20, 2019
#4
+4221
+2

Thank you, CPhill!

tertre  Feb 20, 2019
#2
+22188
+5

3.

In the diagram below,  $$AM=BM=CM ={\color{red}\mathbf{r} }$$ and  $$\angle BMC+\angle A = 201^\circ$$

Find  $$\angle B$$  in degrees.

$$\text{\triangle ABC is a right-angled triangle, so \angle A = 90^\circ - \angle B } \\ \text{\angle CMA = 2\angle B }\\ \text{\angle BMC = 180^\circ - \angle CMA }$$

$$\begin{array}{|rcll|} \hline \angle BMC+\angle A &=& 201^\circ \quad | \quad BMC = 180^\circ - \angle CMA \\ 180^\circ - \angle CMA+\angle A &=& 201^\circ \quad | \quad CMA = 2\angle B \\ 180^\circ - 2\angle B+\angle A &=& 201^\circ \quad | \quad A = 90^\circ - \angle B \\ 180^\circ - 2\angle B+90^\circ - \angle B &=& 201^\circ \\ 270^\circ - 3\angle B &=& 201^\circ \\ 3\angle B &=& 270^\circ- 201^\circ \\ 3\angle B &=& 69^\circ \\ \mathbf{\angle B} & \mathbf{=} & \mathbf{23^\circ} \\ \hline \end{array}$$

$$\angle B$$  in degrees is $$\mathbf{23^\circ}$$

Feb 20, 2019
#5
+100570
+3

Very nice, Heureka!!!!

CPhill  Feb 20, 2019
#6
+22188
+3

Thank you, CPhill !

heureka  Feb 21, 2019
#7
0

Heureka, where was it stipulated that ABC is a right angle?  I'm going crazy trying to find that.

Feb 21, 2019
#8
+100570
+1

In Heureka's diagram, angle ACB intercepts  diameter  AB

So....the arc that it intercepts = 180°

But...since angle ACB is an inscribed angle interceptng this arc, it has 1/2 of its measure = 90°

So...remember....any inscribed angle intercepting a diameter = 90°

CPhill  Feb 21, 2019
edited by CPhill  Feb 21, 2019
#9
0

If we say that straight lines AC and BC in the diagram are elastic, and I move point C anywhere around the circle, angle ACB would always be 90o?  Have I grasped it?  I can't remember being taught that, but of course school was a long time ago for me.  Thanks.  Ron

Feb 21, 2019
edited by Guest  Feb 21, 2019
#10
+100570
0

Yep...C can be located  anywhere on the circle [ except at A or B ]  and angle ACB will be a right angle !!!

CPhill  Feb 21, 2019


text:
## Homework 1.69 [ENDORSED]

$E=hv$

Jimmy Zhou 3D
Posts: 30
Joined: Sat Jul 09, 2016 3:00 am

### Homework 1.69

Question is as follows:

In a recent suspense fi lm, two secret agents must penetrate
a criminal’s stronghold monitored by a lithium photomultiplier
cell that is continually bathed in light from a laser. If the beam
of light is broken, an alarm sounds. The agents want to use a
handheld laser to illuminate the cell while they pass in front
of it. They have two lasers, a high-intensity red ruby laser
(694 nm) and a low-intensity violet GaN laser (405 nm), but
they disagree on which one would be better. Determine
(a) which laser they should use and (b) the kinetic energy of
the electrons emitted. The work function of lithium is 2.93 eV.

I'm not sure how exactly to approach this problem as I'm not sure what qualities would deem a laser as "better"

Chem_Mod
Posts: 18400
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times

### Re: Homework 1.69  [ENDORSED]

Maggie Bui 1H
Posts: 35
Joined: Fri Jul 22, 2016 3:00 am
Been upvoted: 1 time

### Re: Homework 1.69

A general approach I typically use is writing down given information and thinking about what I know about the quantities given.

The problem gives us the wavelengths of light and mentions a lithium photomultiplier is used; I singled out these tidbits of the problem because they deal directly with the problem at hand: getting into the stronghold without breaking the beam of light. A photomultiplier detects light by absorbing photons and then emitting electrons, which calls to mind the photoelectric effect. To connect the photomultiplier and wavelength, think about absorption spectra: all elements have unique absorption spectra and and only absorb light of certain wavelengths. The fact that photomultiplier uses lithium is particularly telling. If the secret agents don't want to break the beam of light, they need to use light of a wavelength that will be absorbed by the lithium photomultiplier.

Jimmy Zhou 3D
Posts: 30
Joined: Sat Jul 09, 2016 3:00 am

### Re: Homework 1.69

Thanks for the Responses,

Maggie, your explanation makes sense however, the absorption spectra of the lithium photomultiplier is not known. I assumed from Chem Mod's work that as long as the energy emitted by the laser is higher than that of the work function, it should be fine.

Jimmy Z.

Jimmy Zhou 3D
Posts: 30
Joined: Sat Jul 09, 2016 3:00 am

### Re: Homework 1.69

Hopefully we will also be able to know how many joules per eV to make the conversion


text:
# Do all convergent sub-sequences of a sequence in a compact set $\mathcal K$ converge to the same element in $\mathcal K$?

I came upon this when trying to understand the proof to "Theorem on Existence of Best Approximations in a Metric Space" as given by Cheney (1981).

Let $$\mathcal K$$ denote a compact set in a metric space. To each point $$p$$ of the space there corresponds a point in $$\mathcal K$$ of minimum distance from $$p$$.

The proof begins with assuming a sequence $$x_1, x_2, \cdots, x_n$$ in $$\mathcal K$$ and continues as below.

"By the compactness of $$\mathcal K$$, we may assume that the sequence converges to a point $$x^*$$ of $$\mathcal K$$; for if necessary we may extract from the given sequence a subsequence with this property."

Now this last sentence comes quite problematic to me; as I imagine we may extract two (or arbitrarily many) different convergent sub-sequences from the original $$\{ x_n\}$$ each converging to a different $$x^*.$$ (Am I right about this?)

Thus I believe the answer to the question in the title is No. The proof then proceeds: "We will show that $$x^*$$ is a point of $$\mathcal K$$ of minimum distance from $$p$$." But if I am right in what I described earlier, then we may have two (or arbitrarily many) $$x^*$$s. Am I missing something here?

You are right that any two different convergent subsequences you extract may have different limits. This doesn't make a difference for the proof because the proof only needs a subsequence converging to some limit $$x^*$$ and then it shows $$x^*$$ has the properties you need. It doesn't matter at all that you could have chosen a different subsequence with a different limit $$y^*$$.

Notice that you shouldn't be worried by the possibility of arbitrarily many points $$x^*$$ at minimum distance from $$p$$ since such points aren't unique. For example, consider the unit circle $$K = \{x \in \mathbb{R}^2: |x| = 1\}$$ and $$p = 0$$. Then every point of $$K$$ minimises the distance to $$p$$.

• @neutrino (you've deleted the comment I replied to, but I'll leave this here in case it's useful) Ah, I think I understand your confusion now. You won't start by picking an arbitrary sequence in $K$ but rather one that minimises the distance to $p$. That is, by basic properties of $\inf$, you can find a sequence $(x_n)$ in $K$ such that $d(x_n,p) \to \inf \{d(x,p) : x \in K\} = \delta$. Then you can show that if $x^*$ is the limit of any convergent subsequence of $(x_n)$, you have $d(x^*,p) = \delta$ – Rhys Steele Mar 26 at 13:44
• Oops, sorry. So this was the original comment: Thanks. Yes, I understand that this is an "existence proof" and does not concern uniqueness. But then again, if the number of the convergent sub-sequences that we may construct and their corresponding $x^*$s is arbitrary, what guarantees that their distance from $p$ are the same? – Neutrino Mar 26 at 13:50

The (sequential) compactness tells us that a sequence has at least one convergent subsequence with limit in $$\mathcal{K}$$. The proof just picks one of those and goes on with the corresponding subsequence (as if that were the original sequence; an often recurring idea in proofs), and I suppose for the proof it does not matter which one it chose, but we know there is at least one. The "property" of the chosen sequence that the proof talks about, must then be one that is preserved by going to a subsequence, of course.

The sequence $$x_n = q_n$$ where $$q_n, n \in \mathbb{N}$$ is an enumeration of the rationals in $$[0,1]$$, say, has the property that all $$x \in [0,1]$$ can be limits of subsequences of $$(x_n)$$, so the answer to the question in the title is indeed no, there can be many subsequential limits.

And in the application (nearest point), many such nearest points are not unique. E.g. $$\mathcal{K}=[0,1]\cup[2,3]$$ and $$x=\frac32$$ is a simple example, both $$1$$ and $$2$$ are nearest.

• @Neutrino that depends in how the $x_n$ are chosen. You haven’t given the full construction. – Henno Brandsma Mar 26 at 13:46


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# Is it possible to design an efficient approximation algorithm for one NP-complete problem based on Shor's algorithm?

Is it possible to design an efficient approximation algorithm for an $\sf{NP\text{-}complete}$ problem based on reductions from Shor's algorithm?

Are known any (classical) approximation algorithms for an $\sf{NP\text{-}complete}$ problem that uses Factoring as oracle?

• While it might be possible, a deterrent to doing so would be that the true running time of the resulting approximation algorithm would be unknown (and it might not even be polynomial!) – Suresh Venkat Mar 30 '12 at 16:22
• I think it is not known if Factoring is $\sf{P\text{-}hard}$, so the likely answer is none at the moment. – Kaveh Mar 30 '12 at 17:15
• @Kaveh why do you say that? P-hardness is wrt logspace reductions. How are logspace reductions related to this question? – Sasho Nikolov May 17 '15 at 7:26
• @Sasho, this is from 3 years ago so I don't recall exactly, but I think I meant that we don't know how to do it for P so it is unlikely we know how to do it for NP. :) – Kaveh May 17 '15 at 7:31
• @Kaveh Fair enough. Greg is digging out old questions :) – Sasho Nikolov May 17 '15 at 8:28

Meanwhile Shor's algorithm does something very specific: It finds the period of a periodic function on the integers or on $\mathbb{Z}^n$. This problem is also in the complexity class SZK, for example. Maybe you could cook up an approximation problem to an NP-hard problem that lands you in SZK or period-finding, but I suspect that there aren't any known, natural examples of that.


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# Talk:Union of Inverses of Mappings is Inverse of Union of Mappings

## Merger with Preimage of Union under Mapping/Family of Sets

Unless I'm mistaken, Preimage of Union under Mapping/Family of Sets is at most a direct corollary to this theorem. Is this what is intended by the merge recommendation? --Keith.U (talk) 12:18, 2 July 2016 (UTC)

Difficult to tell. Links are needed to the concepts invoked. In particular, clarification as to exactly what is meant by the "inverse of a union of a family of mappings" -- inverse of mapping needed, union of family of mappings needed.
Also worth noting that the structuring of this area is suboptimal: the parent page concerns the inverses of two relations, this one is about the inverses of any arbitrary family of mappings. --prime mover (talk) 12:30, 2 July 2016 (UTC)
Would it be wise to create a definition page for the union of functions? When functions are considered as a set of ordered pairs, the definition is clear. However, many readers might be unfamiliar with this concept.
Should already exist. Should be able to find it in Category:Definitions/Set Union if I remember correctly. --prime mover (talk) 10:12, 3 July 2016 (UTC)
I can't seem to find it. Closest I've found is Definition:Union Mapping, though this is not the same. I have some texts that speak of unions of mappings, though none take the time to define it. --Keith.U (talk) 10:38, 3 July 2016 (UTC)
Definition:Union Mapping is indeed what it is. Same will apply to a family of sets, but unless all the mappings agree on all the sets in the family, the concept is of limited use. You will of course find that the union of the sets is a relation,, but then you might as well just rewrite this in the context of relations. And once you have established the need for the functions to agree, you're near enough in the realm of Union of Functions Theorem, which is probably as far as you can go. --prime mover (talk) 10:53, 3 July 2016 (UTC)
So would it be best for me to just reference Definition:Union of Family? The assumption is not made that the functions agree on the intersections of their domains. My reference for formatting is Union of Functions Theorem, though no definition of "union of family of mappings" is given here. I'll let your word be the last here: this has been a lot of discussion over a tiny detail :) --Keith.U (talk) 11:14, 3 July 2016 (UTC)
The fact that no mention of the fact that the mappings agree on the intersections of their domains is not a "tiny detail", it is an important detail -- in that if they do not agree, then their union is not a mapping. Let $A = \{1, 2\}, B = \{a, b\}$. Then let $f_1 = \{(1, a), (2, a)\}, f_2 = \{(1, b), (2, b)\}$. Then $f_1 \cup f_2 = \{(1, a), (2, a), (1, b), (2, b)\}$ and so is not a mapping. So this turns into nothing more interesting than Preimage of Union under Relation/Family of Sets applied to the complete domain and range.
And another noob instruction: please keep to the indentation convention in a discussion -- otherwise it becomes difficult to follow the thread of a conversation. --prime mover (talk) 14:12, 3 July 2016 (UTC)
My phrasing was just an attempt to defuse what I felt was me making too much fuss. In my experience, few (if any) details in mathematics are "tiny" :)
I still feel that this result is an equality of relations, while the theorem linked is only an equality of domains (from my understanding, one result is the equality of a set of ordered pairs, the latter is the equality of a set of singletons). I suppose I am not seeing an important detail; I confess this is not a field in which I am particularly well-versed. You won't hear any more protest from me. --Keith.U (talk) 17:40, 3 July 2016 (UTC)
Yes you're right, there does exist that distinction, I had mentally glossed over it in my mind. Yes, one result concerns the domain, one result concerns the mapping / relation itself. I confess to not getting excited over it because I confess that it's not very exciting. --prime mover (talk) 20:55, 3 July 2016 (UTC)
Thanks for the note on the structuring. I'll restructure shortly. --Keith.U (talk) 09:51, 3 July 2016 (UTC)


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# Exercise 7-6

Reimplement the gen_sentence program using two vectors: One will hold the fully unwound, generated sentence, and the other will hold the rules and will be used as a stack. Do not use any recursive calls.

# Solution

After spending hours scratching my head and still getting stuck on this question, I decided to turn to Google for help. Eventually I discovered this Github Solution Code (uploaded by the Author himself) to this exercise. For the sake of learning, I simply picked out the “bits and pieces” from the Author’s code (such as the core concepts and snippets) and integrated these to my own solution. So let me disclose this upfront – the solution that you are seeing in this page is a result of the great help and pointer from the Author himself. Nevertheless, I believe it is the right thing to do to at least try and understand how the core concept work. The mission of this post is to add some additional documentation to demonstrate that we actually understand the concepts and techniques presented by the Author.

In the followings, I shall:

1. Disclose the project (i.e. the C++ source and header files) that I use in my learning.
2. Test run the job to confirm that it works.
3. Demonstrate that we understand the new concept by manually draw out the step by step logic.

## The Project

First, open the baseline project (i.e. the C++ source and header files) as per Solution to Exercise 7-0 (Part 3 / 3). Once the baseline project is opened up in (say) Code::Block, replace the following files with the newer version.

#### gen_aux.h

#ifndef GUARD_GEN_AUX_H
#define GUARD_GEN_AUX_H

// gen_aux.h

#include <string>             // std::string
#include <vector>             // std::vector

void gen_aux(const Grammar&, const std::string&, std::vector<std::string>&,
std::vector<std::string>&);

#endif // GUARD_GEN_AUX_H

#### gen_aux.cpp

#include <string>            // std::string
#include <vector>            // std::vector
#include <stdexcept>         // logic_error

#include "read_grammar.h"    // Grammar, Rule, Rule_collection
#include "bracketed.h"       // bracketed
#include "nrand.h"           // nrand

using std::string;
using std::vector;
using std::logic_error;

// Look up the input Grammar, and expand
// (modified version of S7.4.3/132)
void gen_aux(const Grammar& g, const string& token,
vector<string>& sentence, vector<string>& tokens) {
if (!bracketed(token)) {
sentence.push_back(token);
} else {
// locate the rule that corresponds to token'
Grammar::const_iterator it = g.find(token);

if (it == g.end())
throw logic_error("empty rule");

// fetch the set of possible rules
const Rule_collection& c = it->second;

// from which we select one at random
const Rule& r = c[nrand(c.size())];

// push rule's tokens onto stack of tokens
// (in reverse order, because we're pushing and popping from the back)
for (Rule::const_reverse_iterator i = r.rbegin(); i != r.rend(); ++i)
tokens.push_back(*i);
}
}

#### gen_sentence.cpp

#include <vector>             // std::vector
#include <string>             // std::string
#include "gen_aux.h"          // gen_aux

using std::vector;
using std::string;

// Generate a sentence based on a Grammar object
// (modified version of S7.4.3/132)
vector<string> gen_sentence(const Grammar& g) {
vector<string> sentence;
vector<string> tokens;
tokens.push_back("<sentence>");

while (!tokens.empty()) {
string token = tokens.back();
tokens.pop_back();
gen_aux(g, token, sentence, tokens);
}

return sentence;
}

## Test Result

Note: the result might differ when you run the job due to the nrand(). Core concept remains true though.

<noun>          cat
<noun>          dog
<noun>          table
<noun-phrase>   <noun>
<verb>          jumps
<verb>          sits
<location>      on the stairs
<location>      under the sky
<location>      wherever it wants
<sentence>      the <noun-phrase> <verb> <location>
^Z
the large brown dog sits wherever it wants
the table sits under the sky
the cat jumps on the stairs
the brown cat jumps under the sky
the brown large brown cat sits wherever it wants

## Understanding the concept

Let’s use the first output sentence to illustrate our understanding of the core concept used by the gen_sentence and gen_aux functions.

the large brown dog sits wherever it wants

The billion dollar question, how did the program generate this line?

To demonstrate my understanding, I’ve manually put together a couple of simple Excel tables to walk through the logic.

Recall that this is the input grammar table:

The gen_sentence and gen_aux can give the following output values:

### Some Notes To This Example

For some extra clarity, I include here a brief explanation of what each column means.

#### tokens non-empty?

If non-empty, the while loop inside the gen_sentence step would keep expanding tokens and construct the string sentence. Otherwise, the while loop stop and we have our sentence! (Note that we start by expanding the top level “<sentence>”)

vector<string> sentence;
vector<string> tokens;
tokens.push_back("<sentence>");
while (!tokens.empty()) {
// some codes to expand <sentence> and construct the string sentence.
}
return sentence;

#### const string& token

This is the last element of tokens, implied by this statement within the gen_sentence step:

string token = tokens.back();

#### Is token bracketed?

If the token is bracketed, it implies the token corresponds to a Grammar category and need expanding. If it is not bracketed, we  simply append the text the the string sentence.

if (!bracketed(token)) {
// some codes to construct the string sentence.
} else {
// some codes to expand the token.
}
`

#### const Rule& r

If the token is bracketed (i.e. a Grammar category), the const Rule& r is essentially the Grammar rule. I highlight this in orange.

#### vector<string>& sentence

If the token is not bracketed, we simply append the text to the string sentence. This column shows what the constructed string sentence look like. I highlight the newly appended element in blue.

#### (Updated) vector<string>& tokens

This is the interesting part and is updated by a combination of the gen_sentence and gen_aux steps.

• The tokens.pop_back() statement remove the last element of tokens (hence the strike-through in the table).
• If token (i.e. the token column) is bracketed, the for loop within the gen_aux function appends the entire rule (highlighted in orange) in reverse order. This is a very clever technique to get the tokens expand and string sentence constructed in the expected order. (it’s easier to visualise this via the Excel table above instead of explaning in writing!).

## Summary

Walking through the logic as illustrated in the Excel table, the program generates and return the complete string sentence. The actual sentence may get constructed differently due to the nrand().

# Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000
The Github code solution uploaded by the author (A.Koening).

## 2 thoughts on “Accelerated C++ Solution to Exercise 7-6”

1. Mustapha says:

Hey man, when you gonna do the rest of the book? I’m nearly finished with chapter 7 I first do the question myself and then look at yours, it’s interesting to see different solutions and makes the learning process more complete.

1. Johnny says:

Hey thanks for the lovely comment :-) unfortunately last time when I touched C++ was nearly 2 years ago and I haven’t touched it since (have been playing with other languages and technologies instead – so many things to try and so little time!). But don’t despair! I do recall Koeing put together the full solutions on GitHub – use that as a backup!
Thanks again and best of luck!!! :-)