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Malcy's Blog
Malcy's Blog – Oil price, Union Jack Oil, Chariot, IOG, Rockhopper Exploration, Jadestone Energy, Cornerstone Resources & finally
WTI $109.56 -$8.03, Brent $113.12 -$6.69, Diff -$3.56 +$1.34.
Author @mgrahamwood
USNG $6.88 -58c, UKNG 205.0p -28.62p, TTF €127.5 +€9.762
After oil took a caning last week on Fed and sundry lousy interest rate news it is important to notice that Opec+ is still missing by a distance its own quotas, WTI fell $11.11 and Brent by $8.89 The. price fell sharply on Friday as the Fed said that it would be 'more aggressive' in getting inflation back to 2%.
The rig count is yet to help Sleepy Joe as last week it was up 7 overall at 740 units and rose only 4 in oil to 584.
Union Jack Oil
Union Jack has announced that material landmark net revenues of US$7,000,000 have been achieved from the Wressle hydrocarbon development, located within licences PEDL180 and PEDL182 in North Lincolnshire on the western margin of the Humber Basin.
Union Jack holds a 40% economic interest in this development.
· Landmark US$7,000,000 revenues generated to Union Jack since re-commencement of production on 19 August 2021
· Current daily production figures continue to be an average constrained flow-rate of 750 barrels of oil per day, well in-excess of the prognosed 500 bopd from the Ashover Grit reservoir
· Well continues to produce under natural flow with zero water cut
· Union Jack continues to be cash flow positive covering all G&A, OPEX and contracted or planned CAPEX costs, including any budgeted drilling activities for at least the next 12 months
· At 20 June 2022, cash balances and short term receivables stood at in-excess of £8,400,000
· Union Jack expects to report a maiden profit for the forthcoming unaudited half year results ending 30 June 2022
· Debt free
Executive Chairman of Union Jack, David Bramhill, commented:
"The ongoing excellent operational and financial performance at Wressle continues to bolster the Company's cash position, balance sheet and income statement.
"Net revenues from Wressle have now exceeded US$7,000,000 and, as a result of this exceptional performance, plus revenue contributions from the Keddington oilfield, the Fiskerton Airfield oilfield and North Sea Royalties
"The Board now expects to report a maiden profit in the forthcoming unaudited half year results ending 30 June 2022".
This is yet more good news from Union Jack where Wressle continues to be the gift that keeps on giving, now to the tune of over $7m. As Chairman David Bramhill says above, with other contributions the company now expects to report a maiden profit in the current period.
There is so much exciting news around for UJO at the moment as well as other UK domestic hydrocarbon players that I am quite surprised that the shares have not risen more. It is 6 months since it went into the updated Bucket List and the shares have almost tripled but I would expect plenty more given news announced since then.
Chariot has announced that it has signed a front-end engineering and design ("FEED") agreement with Schlumberger and Subsea 7, as part of a consortium, for the Anchois gas development project in Morocco. Chariot, Schlumberger, and Subsea 7 will continue to adopt a "one-team" integrated and collaborative approach to safely fast-track first gas to maximise the return on investment.
The scope of the Agreement covers:
• Front-end engineering support and design work for the Anchois gas development, incorporating:
o offshore components including well completions, subsea production systems, and subsea umbilicals, risers, and flowlines ("SURF") that will be delivered by Subsea Integration Alliance, and
o onshore components including a central processing facility ("CPF") and flowlines and controls from the CPF to the shore crossing that will be delivered by Schlumberger.
• Generation of deliverables, such as engineering, procurement, construction, installation, and commissioning ("EPCIC") costs and schedules, required prior to final investment decisions ("FID").
• Opportunity for Chariot to directly-source EPCIC services with Schlumberger and Subsea 7 for the field development and operations and maintenance ("O&M") of the facilities during commissioning and the early production phase.
• Commitment to environmental, social, and governance ("ESG"), minimising emissions and contributing to social development through the creation of direct and indirect jobs in Morocco.
Beyond this Agreement, Chariot is managing the additional FEED scopes required for the development, including well construction and onshore infrastructure, including fixed pipelines, to deliver gas to customers.
Olivier Blaringhem, CEO, Subsea Integration Alliance, commented:
"We are delighted to work with Chariot on this FEED project. We remain fully aligned with their objectives and will build upon the work undertaken to date to optimise the design and development of the Anchois Gas Project. We look forward to playing a key role in bringing this project to fruition."
Adonis Pouroulis, Acting CEO, Chariot Limited, commented:
"Signing this agreement with Schlumberger and Subsea 7 is further evidence that we have accelerated development plans for the Anchois Gas Project, a key tenet of our recent fundraising. It is a pleasure to work closely with both companies on this integrated design phase, to benefit from their experience in working on several similar projects globally, and to leverage their standardised technology which is directly applicable to the Anchois development.
We look forward to building on this relationship and to realise the benefits of an integrated project execution. Reduced interfaces, fewer contingencies, and strong leverage over procurement and the offshore construction schedule will help shorten time to first gas. This streamlined approach will benefit all stakeholders which is a key objective of fast-tracking the project towards cashflow."
Signing up a FEED agreement is a key part of the timeline of the Anchois gas development and the speed with which it is being done is very exciting for shareholders in Chariot. With definition of progress being made in the key areas of such pre-FID activity the signs are highly encouraging.
With these encouraging moves being made it must be likely that Chariot is either up with or ahead of the game in the process of developing Anchois and I expect more good news on this front. Despite the tripling of the shares since the December Bucket list I think that the shares have a long way to go up from here.
IOG has provided an update on current activities and operational guidance for the remainder of 2022.
Saturn Banks Phase 1 Production
Further to the 1 June 2022 operational update, in which production was confirmed to be restored to an initial gross 30 mmscf/d, temporary modification of the second recycle compressor in the Bacton terminal condensate stabilisation unit (CSU) was completed by its operator Perenco (UK) Limited. In line with the announced plan, this has alleviated liquids constraints and enabled production to be steadily increased, reaching 54 mmscf/d gross by 17 June.
In June to date, gross production has averaged 44 mmscf/d, with uptime at 84% for Blythe and 94% for Elgood (100% for both over the past week). Factoring in the May shutdown and other previous outages, uptime has been 45% and 65% for Blythe and Elgood respectively since First Gas, giving gross average production to date of 32 mmscf/d. In addition, IOG sold 1,341 tonnes of condensate (its net share) in April and May at an average price of $882/tonne.
Significant progress has been made so far in June in improving liquids management and onshore reliability at Bacton and the remaining restriction to gas flow is expected to be fully lifted over the coming weeks. To underpin uptime, work is underway to improve the Blythe platform's electrical stability and reduce the risk of emergency shutdowns. The Company has also enhanced its helicopter and vessel access to enable faster offshore restarts in the event of any further unplanned outages. Permanent chemical injection modifications are expected to reinstate full remote restart capabilities by early Q4.
The reservoir productivity observed to date at Blythe and Elgood is consistent with the updated management estimated recoverable reserve range in the 2021 Annual Report released on 17 March 2022. These ranges are gross 1P/2P/3P 25.4/42.5/55.8 billion cubic feet (BCF) for Blythe and 9.6/14.1/18.3 BCF for Elgood.
Based on reservoir analysis and ongoing facilities optimisation, gross average 2H22 Saturn Banks gas production is projected to be in the 45-60 mmscf/d range. Average condensate production is projected to be 250-350 bbl/d over 2H22 as condensate rates are expected to decline fairly quickly. These ranges make reasonable allowances for both planned (aggregate of 15 days to complete the above works, including a Bacton terminal shutdown) and unplanned outages over 2H22.
Total unit operating costs in 2022 are projected to be in the 10-15p/therm range (c.$8-12/boe). Expected total capex net to IOG for 2022 remains within the previously guided range of £70-85 million.
Gas Market
Pricing under IOG's gas sales agreement with BP Gas Marketing Limited is directly linked to the UK NBP Day-ahead price. This has been exceptionally volatile since First Gas in mid-March, with very strong LNG supply into the UK market in conjunction with limited export and storage capacity, alongside unexpected events such as the recent Freeport LNG outage and NordStream 1 capacity issues. The NBP Day-ahead price has traded in a 10-224p/therm range this month. IOG's volume weighted average monthly realised gas prices so far have been 232p/therm for March, 161p/therm for April, 83p/therm for May and 106p/therm for June to date (135p/therm combined since First Gas).
The UK NBP forward curve priced the Winter 2022 contract at over 300p/therm and Summer 2023 over 190p/therm on 17 June, indicating an expectation of a very supply constrained market over the coming year. These levels fluctuate significantly and are not a forecast of future day-ahead prices. Whilst in principle the Company remains inclined towards commodity hedging, with such exceptionally high volatility, pricing and execution for typical structures are currently prohibitive.
Subsea and pipelay work has gone to plan in Q2, ensuring progress continues towards Southwark First Gas in Q4. Notably, the 6km 24″ extension section from the Saturn Banks Pipeline System (SBPS) to the Southwark platform was successfully installed by the Seven Borealis S-lay vessel. The Seven Kestrel diving support vessel (DSV) has also installed the necessary mechanical connectors at both ends of SBPS outer section – another important element of the programme. The Kestrel is currently installing the spools to connect the 6km extension to the Southwark platform. The final closing spools will be installed later in the hook-up and commissioning programme, which is currently underway with risers being welded on the Southwark platform.
Drilling of the Southwark East production well has continued into the reservoir section. The well has reached Total Depth at 14,330ft MD and the production liner is being prepared to be run. The Noble Hans Deul rig will then drill the Southwark West well (at which the top-hole section had already been drilled earlier in the campaign). Both wells are due to be completed by Q4.
Appraisal campaign
After Southwark, the Noble Hans Deul is contracted to drill both the Goddard and Kelham North/Central appraisal wells in direct continuation at the same attractive day-rate as the Phase 1 wells. Both the geophysical and geotechnical site survey campaigns for the Goddard and Kelham wells are complete, while planning and procurement for both wells is also well underway.
The Goddard appraisal well is intended to define the south-east extent of the main Goddard discovery and de-risk its two flank structures, thereby helping to determine the optimal field development plan, including platform location and number of wells, to maximise investment returns. 3D seismic reprocessing over the structure in 2020-21 has improved subsurface understanding and enabled a choice of well location. At a 45p/therm price deck, the internal rate of return (IRR) from final investment decision (FID) on a Goddard development is provisionally estimated to be 19-40%, depending on whether the appraisal well proves up a larger structure. This rises to 50-80% at a 75p/therm price deck. The current forward curve average out to the end of 2027 is 158p/therm. Subject to further drilling, there are also other potential tie-back opportunities to a Goddard development.
The Kelham North/Central appraisal well is intended to prove up both accumulations as part of a potential three-field "Southern Hub" gas production cluster including the Abbeydale discovery on the same P2442 licence. Again, 3D seismic reprocessing work has opened up extensive, previously unidentified resource potential on this licence, which lies well within tie-back range of the SBPS. If successfully appraised, there would be potential to fast-track into production for a quick-payback high-return investment. Based on pre-drill mid-case resource estimates, this is provisionally estimated at 33% IRR at a 45p/therm price deck, rising to 75% IRR at a 75p/therm price deck. There are also several other potential future additions to such a Southern Hub on the licence, subject to further exploration and appraisal drilling.
Based on the current drilling schedule, the two appraisal wells are expected to be completed by Q1 2023.
Phase 2 and portfolio activities
Active regulatory engagement continues for the planned Nailsworth development, with the concept select report submitted to the North Sea Transition Authority. Meanwhile an expression of interest for Front-End Engineering and Design (FEED) work has been issued. A concept select decision remains expected by the end of Q2 based on a tie-back to the Southwark platform, with FID targeted by year end 2022 leading to first gas in 2H24.
The 3D seismic reprocessing of the Panther-Grafton area (licence P2589) is expected to be completed by around the end of Q3. This will be followed by an evaluation of the newly reprocessed data to inform a new assessment of the area's potential, expected by year end, seeking a similar uplift in resource potential as with the P2442 (Kelham/Abbeydale) licence. The Panther-Grafton area lies close to the Company's Southwark, Elland and Nailsworth assets, creating potential development synergies.
IOG also intends to be active in new licensing opportunities, with the NSTA recently confirming the 33rd Offshore Round for this year. The Company has identified a clear set of value-adding priority targets that can contribute to the UK energy resilience goals laid out in the British Energy Security Strategy announced in April. Further potentially synergistic nearby business development opportunities are also being actively evaluated in collaboration with our joint venture partner CalEnergy Resources (UK) Limited.
Andrew Hockey, CEO of IOG, commented:
"I am pleased to say that Saturn Banks production continues to be restored as planned, up to 54 mmscf/d gross currently from 30 mmscf/d at the start of June. We are taking the necessary actions both to improve facilities uptime and minimise restart times after any outages. Allowing for planned and unplanned downtime, we expect to produce 45-60 mmscf/d gross on average over 2H22.
UK Day-ahead gas pricing remains extremely volatile, trading in a 10p to 224p/therm range this month. Our average realised price to date is 135p/therm and forward prices for this winter are increasingly elevated, closing at over 300p/therm last Friday. We expect 2022 opex to be 10-15p/therm.
In parallel, the 24″ extension of the Saturn Banks Pipeline System to Southwark has been successfully laid, while drilling of the Southwark East well is at an advanced stage. With Southwark expected onstream in Q4, we aim to exit 2022 with a higher, more diversified production stream.
This is part of an exciting growth period ahead, with the Goddard and Kelham appraisal wells following straight after Southwark, plus a Nailsworth final investment decision and seismic re-evaluation results for the Panther/Grafton area both targeted by year end.
As the government encourages reinvestment in the North Sea to support energy security, we also have a clear view of value-adding new licence targets, alongside other business development opportunities under active evaluation."
This is a long and detailed update from across the IOG portfolio but what is most important is that production is picking up well after some early teething troubles. This means that 54 mmscf/d is up and running now and they are aiming to continue higher from here.
IOG is very busy and it is worth noting that in this update they are showing just how much else that there is in the portfolio, how diversified it is and how low its costs are. This is important as CEO Andrew Hockey says that low-carbon intensity domestic gas producers like IOG can help deliver the UK's energy security objectives.
Rockhopper Exploration
Rockhopper has announced that further to the announcements made on 15 June 2022 and 16 June 2022 concerning the proposed Capital Raising, the Circular containing further details of the Open Offer, together with an Open Offer Application Form for Qualifying Non-CREST Shareholders, will be posted to Shareholders later today and will also be made available on the Company's website at www.rockhopperexploration.co.uk shortly. The full details of the Open Offer timeline are included below.
The Open Offer
The Company considers it important that Shareholders who were not able to take part in the Placing and/or the Subscription have an opportunity to participate in the Capital Raising. The Company is therefore providing Qualifying Shareholders with the opportunity to subscribe for Units at the Issue Price pursuant to an Open Offer, to raise up to approximately US$5 million (approximately £4.0 million) if fully taken-up.
Subject to fulfilment of the conditions of the Open Offer, the Open Offer will provide Qualifying Shareholders with the opportunity to apply to acquire Open Offer Units at the Issue Price pro rata to their holdings of Existing Ordinary Shares against all Existing Ordinary Shares held by Qualifying Shareholders as at the Open Offer Record Date on the following basis:
1 Open Offer Unit for every 8 Existing Ordinary Shares held by Qualifying Shareholders
Each Open Offer Unit will comprise one Open Offer Share and, for every two Open Offer Shares subscribed for, one Warrant.
Importance of Vote at the 2022 Annual General Meeting
The Resolutions being proposed at the 2022 Annual General Meeting are, (i) in respect of the resolution relating to the granting of authority to Directors to allot shares (or rights to subscribe for or to convert any security into shares), an ordinary resolution which, to be passed, will require the support of a simple majority of the total voting rights of Shareholders who (being entitled to do so) vote on such resolution at the 2022 Annual General Meeting, and (ii) in respect of the resolution relating to the granting of authority to Directors to disapply pre-emption rights, a special resolution which, to be passed, will require the support of a majority of not less than 75 per cent. of the total voting rights of Shareholders who (being entitled to do so) vote on such resolution at the 2022 Annual General Meeting. The Open Offer is conditional, inter alia, on the passing of the Resolutions.
In the event that the Resolutions are not passed at the 2022 Annual General Meeting, the Company will not be able to proceed with the Open Offer and Shareholders will not get the opportunity to participate in the Open Offer, with the result that the anticipated net proceeds of the Open Offer will not become available to the Company.
The Directors consider that the scenario described above would not be in the best interests of the Company or its Shareholders as a whole as the Company considers it important that Shareholders who are not able to take part in the Placing and/or the Subscription have an opportunity to participate in the Capital Raising. Accordingly, the Directors believe that the passing of the Resolutions are in the best interests of Shareholders and recommend that Shareholders vote in favour of the Resolutions at the 2022 Annual General Meeting.
Nothing to add here, after the raise last week which was significantly oversubscribed and brought in new shareholders, it seems to be an opportunity for retail holders to take their rights.
Jadestone has provided the following update on operations at the Montara field offshore Australia.
On 17 June 2022 during routine production and crude oil cargo operations, crude oil was being transferred between two crude oil tanks onboard the Montara Venture FPSO. During the transfer, oil was observed on the surface of the sea adjacent to the FPSO. Transfer operations were ceased immediately and production from the Montara fields was shut in as a precautionary measure.
The release of oil to sea was quickly halted by pumping water into the tank, indicating a leak somewhere at the tank base. A subsequent inspection using an ROV confirmed a small 30mm diameter hole in the bottom of the tank. Having now controlled the release of oil, the next step will be to apply a temporary repair in order to remove the remaining oil from the tank. Following this, the tank will be accessed and cleaned in order to complete an inspection and permanent repair. This is the same work plan which has already been applied to all the other main crude oil storage tanks, as part of the five yearly inspection and repair programme which has been underway since Jadestone assumed operatorship of the Montara assets, and which is a normal and accepted process approved by regulators and Class inspectors globally.
The volume of released oil is estimated at three to five cubic metres, which was monitored and had fully dispersed by the morning of 19 June 2022. The National Offshore Petroleum Safety and Environmental Management Authority ("NOPSEMA") was notified of the incident immediately and will initiate an onsite inspection of the Montara facilities on 21 June 2022.
Jadestone's preliminary estimate is that it will take approximately four weeks to complete the tank inspection and effect repairs sufficient to re-commence production at Montara, with appropriate isolation of the crude oil tank. Thereafter, following consultation with key stakeholders, a full remediation programme will be conducted, similar to that which has been completed on all the other main crude oil tanks. At this early stage, Jadestone estimates the cost of securing the leak site and permanent repair of the crude oil tank at US$2-3 million. This is consistent with the overall expenditure to date on the tank inspection and repair work, which totals US$11 million over the past three years.
On the basis that full production from the Montara fields can be restarted as estimated, Jadestone expects that full-year 2022 production would be around the lower end of the previously announced guidance range of 15,500 – 18,500 boe/d.
"Whilst we deeply regret this event, immediate and decisive action by the Jadestone team limited the quantity of oil released and made the facility safe, and I commend the work of both the offshore crew and the onshore operational staff for their quick response in managing this incident.
The integrity of our facilities is at the heart of our strategy and this oil tank was due for inspection within the next few weeks as part of our five-yearly maintenance cycle, which would have been the final key step in the significant inspection and repair plan that has been underway on the Montara Venture since we assumed operatorship in 2019. Over this period, we have worked extensively and spent significant sums to restore the vessel and associated infrastructure to the condition required in order to deliver a long, safe and productive field life.
Safety and sustainability are core values for Jadestone, and we will now work closely with the Class inspector, NOPSEMA and other stakeholders to develop and execute a plan to safely and permanently repair the hole in the crude oil tank. There are many precedents for this type of activity, both within the Asia-Pacific region and globally, and we will deploy the necessary expertise to provide confidence in the solution we implement. It's particularly frustrating for this event to occur so close to the planned inspection and repair activity, but our investment in the facilities to date is bearing fruit, with facility uptime now over 90% and with further improvements in the planning stage."
This spill, as with any such unexpected events is extremely disappointing and I can imagine that given how tight a ship the management runs they will be furious. However the spill has shut-in production whilst repairs are underway and whilst guidance is not expected to be lowered it must be possible.
If there is no change in guidance then I expect little effect on revenues or profitability and accordingly the 16% fall in the share price may have been a bit overdone. However Jadestone will be extremely upset about this even though it is likely that no damage has been done to the environment, something the company has always had front and centre in its policies.
Cornerstone Resources Group
Cornerstone has announced that it has completed the acquisition of Block 47/3f (as well as certain other blocks) in offshore Licence P.2433, UK Southern North Sea from Painted Wolf Resources Limited.
Block 47/3f, which lies to the North of blocks already owned by Cornerstone, contains the low risk Baker prospect, a part of the low risk Carnaby prospect that Cornerstone did not already own and a small extension of the Company's flagship Abbey discovery.
The Abbey discovery, on which the Company is planning to drill three horizontal production wells has audited proven and probable reserves of 111 Bscf of sales gas and 0.6 MMbbl of condensate. A Concept Select Report has been submitted to the NSTA and engineering studies are underway for the development of Abbey, which would utilise existing infrastructure in the area.
The Baker prospect, which will be tested by an exploration well in due course, is intended to be developed with a single horizontal well drilled from the Abbey infrastructure. Baker has been ascribed 57 Bscf of Mean Prospective Resources and is anticipated to flow similarly to wells on surrounding fields such as Apollo, Minerva and York.
Peter Young, Chief Executive Officer of Cornerstone Resources Group commented:
"Block 47/3f forms a key part of our plans for a new gas development around the Abbey discovery and we are delighted to have signed this agreement with Painted Wolf Resources.
"This area of the North Sea is a location in which our team have long seen significant potential. Developments in both technology and nearby infrastructure have positioned our Abbey development as one of the most exciting opportunities in the Southern North Sea both as a potential deliverer of profitable returns as well as low-carbon gas; in line with the UK's planned energy transition strategy.
"We are extremely excited about the potential for this area and are looking forward to a busy year ahead as we progress our development."
Cornerstone Resources Group looks to be an unquoted company not to be confused with others of the same name as it appears to be Cornerstone Oil & Gas and the vehicle of serial energy investor Peter Young. Whilst I wouldn't normally cover unquoted's it is interesting to see what he is up to right now…
The Canadian GP went to Max which was no surprise with Sainz second but it was good to see Lewis on the podium and a shame that Alonso couldn't make his grid position count.
Better still was the sight of Matt Fitzpatrick from Sheffield winning the US Open.
The opinions expressed here are those of the author
Disclaimer: Malcy's Blog is provided for general information about the international oil and gas industry and the companies that operate within it. It does not constitute investment advice and Malcy does not buy or sell shares, warrants or bonds in any company written about within the blog. Information is taken from publicly available sources and any comment is that of the author who does not take any third party comment in the blog
Tags: Chariot, Cornerstone Resources Group, IOG, Jadestone Energy, Rockhopper Exploration, Union Jack Oil
Traders Cafe with Zak Mir: Bulletin Board Heroes, Monday 20th June 2022 via Vox Markets
China takes on Russian oil at a cheap price


MotoGoogle: Good, bad, or indifferent for Microsoft?
Debra Littlejohn Shinder analyzes how the deal between Google and Motorola Mobility will affect Microsoft and its efforts to regain a foothold in the smartphone business.
The tech press is divided over whether the Google/Motorola deal announced this week is a good thing or a bad thing for Android. They're also in disagreement over whether and how the deal will affect Microsoft and its efforts to regain a foothold in the smartphone business.
Given the nature of my column, the first thing I thought when I heard the news that Google was acquiring Motorola Mobility was "Is that a good thing or a bad thing for Microsoft?" It seems most of my fellow tech analysts and "opinionators" are thinking along the same lines; many of the early headlines regarding the Google/Motorola deal mention or even feature the potential effects on Microsoft.
Those opinions seem to be all over the board, even within the same publication. Zack Whittacker over at ZDnet suggests it might motivate Microsoft to buy Research in Motion (RIM), whereas Mary Jo Foley says "Nah." They both make good points, but I'll throw in with Mary Jo on this one.
Buying Nokia might have made sense, although I think the decision to form a strategic partnership instead was the better way to go. Nokia was ripe to switch to a new operating system; they already seemed headed in that direction when they started developing Maemo (a smartphone/tablet OS based on Debian Linux) and later MeeGo (which Maemo morphed into), despite Nokia's declaration of allegiance to Symbian.
Despite the fact that Symbian in 2010 had the largest market share of any mobile operating system (37.6%, compared to 22.7% for Android, 16% for RIM, 15.7% for iOS, and 4.2% for Windows phones), it has never seemed to have the kind of brand loyalty that other platforms elicit from their users.
Blackberry users are outdone only by iPhone fans when it comes to their strong feelings about their OS of choice. I've never heard anyone enthuse about the Symbian OS, even those who use it. On the other hand, many of my "Crackberry" friends will tell you that "you can take it when you pry it from my cold, dead hands."
Thus the announcement that Nokia was killing Symbian (at least in the United States and Canada) and would be running Windows on all their smartphones didn't raise much of a hue and cry from Symbian users. But can you imagine the outcry if Blackberry users discovered that their favorite OS was going to be replaced by Windows Phone? Yes, yes — we know RIM already has plans to pry the BB OS from users' hands and replace it with QNX. But I don't think that has the same psychological effect as replacing it with Windows.
Many BB loyalists are not at all fond of Microsoft; the rivalry hearkens back to when, not so long ago, RIM and Microsoft were considered the top players in the U.S. smartphone market. Of course, then the iPhone came along and turned the smartphone world on its head. In fact, for a while there it seemed as if the only people who didn't embrace the iPhone were Blackberry users.
While some see the Google/Motorola deal as a nail in RIM's coffin, others predict that both RIM and Microsoft will benefit from it. Interestingly, following the announcement, stock prices for both Nokia and Microsoft rose, suggesting that the market thinks they might be on the winning end of the deal. RIM's stock was up, too. More telling was the fact that Google's own stock took a fairly significant dive. Motorola Mobility's stock, on the other hand, was up slightly more than Google's was down. Of course, the stock market is volatile and initial reactions can easily reverse.
Nokia representatives were upbeat, with spokesman James Etheridge going so far as to say that this could end up being "a massive catalyst for the Windows Phone ecosystem."
Android's loss = Microsoft's gain?
The Android platform has been steadily gaining market share, it was on 48% of smartphones worldwide last quarter, according to research firm Canalys. But much of that success has been due to the popularity of "superphones" made by vendors HTC (Incredible/Incredible 2, Thunderbolt, EVO) and Samsung (Galaxy S, Droid Charge, Infuse). Motorola has had some hits, as well — particularly the Droid X — and hopes are high for the dual-core LTE Droid Bionic that's expected to go on sale at Verizon next month.
However, anecdotally speaking, I've heard far more complaints about the original Droid, Droid 2, and other Motorola handsets than about other manufacturers' phones. The MotoBlur UI is generally recognized as the least polished of the three big vendors' overlays, and based on my own hands-on testing of various Android handsets (thanks to Verizon) and the experiences of friends and family members who own the phones, performance often seems sluggish in comparison to HTC and Samsung models. There were high hopes for the Xoom tablet and the Atrix phone with its innovative laptop dock, but sales didn't reflect the enthusiasm of early reviewers.
Of course, there's a good chance that if Google takes over the Moto Droid line, they'll ditch MotoBlur and go with the standard Android interface, which would make Android purists very happy. But whatever they do, there's also the possibility that — despite Google's public statements that its other hardware partners are completely "on board" with the Motorola deal — those vendors may be a little less enthusiastic about developing fantastic Android phones.
That doesn't mean they're suddenly going to put their blood, sweat, and tears into making killer Windows phones (after all, given Microsoft's deal with Nokia, they have the same demotivation there). However, our own Jason Hiner has speculated that the Google/Motorola alliance may push at least one current Android vendor, HTC, in Microsoft's direction — and maybe even make that company amenable to a buyout offer from Microsoft.
If Microsoft did indeed buy both HTC and Nokia, as Jason suggests at the end of his article, it could potentially put them in a far stronger position in the smartphone market. However, to really compete with Android, I still think Microsoft needs to reexamine its platform philosophy and decide whether Windows Phone is going to compete with Android or with the iPhone, because, as I see it, the two target audiences are very different ones.
Those who like the iPhone are willing to sacrifice freedom and flexibility for a smooth, "cool" experience. Those who prefer Android abhor the idea of being locked into iTunes for updates, want to be able to mount the file system and directly transfer files between phone and computer, want to have the ability to swap out microSD cards to add storage space or to move files, and so forth. In other words, all those things we also used to be able to do with Windows Mobile.
Thus far, the new and improved (in many ways) Windows Phone has been more like an iPhone than a Droid. If Microsoft wants to go after Apple customers, that's probably the right path. But if they want to take advantage of a possible opportunity created by Google's deal to grab some of the Android market, Windows Phone 8 is going to need to have those features that are driving many to buy those HTC, Samsung, and Motorola Android-based phones in the first place.
Poll: Will Google's Motorola patent acquisition gambit pay off?


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Пусти ме унутра () је америчко-британски љубавни хорор филм из 2010. године, римејк шведског филма Отвори врата правом из 2008. Редитељ и сценариста филма је Мет Ривс, а у главним улогама су Коди Смит-Макфи и Клои Грејс Морец. Радња је смештена у Нови Мексико почетком 1980-их и прати 12-годишњег дечака по имену Овен, који је жртва вршњачког насиља. Он се спријатељује са девојчицом свог узраста, Еби, за коју се испоставља да има натприродне моћи јер је вампир.

Иако није успео да понови успех оригинала, филм је добио позитивне критике. На сајту Ротен томејтоуз оцењен је са 88%, а Метакритику са 85%. Добио је Награду Сатурн за најбољи хорор филм, а Морец је на истој додели награда добила признање у категорији најбољих младих глумаца. У истој категорији, Морец је тријумфовала и на додели Награда Емпајер. Поред два освојена Сатурна, филм је био номинован у још 5 категорија и то: за најбољу режију, за најбољег младог глумца (Коди Смит-Макфи), најбољи сценарио, најбољу музику и за најбољу шминку.

Филм је премијерно приказан 13. септембра 2010, на Филмском фестивалу у Торонту у продукцији куће Хамер и дистрибуцији Парамаунт пикчерса. Са буџетом од 20 милиона долара, филм је остварио зараду од 24,1 милиона долара.

Радња

Радња је смештена у Лос Аламос, Нови Мексико, 1983. године. Овен је 12-годишњак, чији су се родитељи развели и он је остао да живи са мајком. Пошто га други дечаци из разреда непрестано злостављају, Овен проводи вечери сам, напољу, замишљајући како им се свети. Једне ноћи упознаје се са бледом девојчицом његових година, по имену Еби. Она му говори да се недавно доселила у његово суседство и да живи са својим оцем, Томасом. Њих двоје се брзо спријатеље и проводе ноћи размењујући поруке Морзеовом азбуком.

Овен убрзо сазнаје да је Еби вампир, али упркос томе, њихов однос остаје непромењен. Једног дана, детектив који испитује убиства Ебиних жртава умало је спали сунчевом светлошћу и она остаје приморана да га убије пред Овеном. Након тога, она одлучује да се одсели из Лос Аламоса и Овен поново остаје сам.

Када група насилника покуша да удави Овена у базену, Еби се враћа и све их убија, спасавши Овена. На крају филма њих двоје путују возом у непознатом правцу.

Улоге

|-
| Коди Смит-Макфи || Овен
|-
| Клои Грејс Морец || Еби
|-
| Елајас Котис || детектив
|-
| Кара Буоно || Овенова мајка
|-
| Саша Барезе || Вирџинија
|-
| Ричард Џенкинс || Томас
|-
| Дилан Минет || Кени
|-
| Ричи Костер || господин Зорић
|-
| Џими Пинчак || Марк
|-
| Николас Доријан || Доналд
|-
| Колин Морец || касир
|-
| Дилан Кенин || Лари
|-
| Брет дел Буоно || Џими
|-
| Крис Броунинг || Џек
|}

Референце

Спољашње везе

Филмови 2010.
Амерички филмови
Британски филмови
Филмови на енглеском језику
Амерички хорор филмови
Британски хорор филмови
Готички хорор филмови
Филмови продукцијске куће Хамер
Римејкови хорор филмова
Филмови Мета Ривса
Хорор филмови 2010-их
Филмови за које је музику компоновао Мајкл Ђакино

redpajama_set_name: RedPajamaWikipedia


text:
\section{Introduction}

We will consider plane connected trees with an additional
structure: edges and vertices have positive integer weights,
moreover, for each vertex the sum of weights of outgoing edges is
equal to the weight of this vertex. Such tree will be called a
\emph{w-tree}.

Each plane connected tree has a binary structure: its vertices can
be colored in black and white so, that adjacent vertices have
different colors. We will study the following problem: enumerate
pairwise non-isotopic w-trees with a given list of weights of
black vertices and a given list of weights of white vertices. The
set of such w-trees will be called a type. A type will be denoted
$\langle k_1,\ldots,k_s\,|\,l_1,\ldots,l_t\rangle$, where
$k_1,\ldots,k_s$ are weights of white vertices, written in non
decreasing order, and $l_1,\ldots,l_t$ are weights of black
vertices (also in non decreasing order). Obviously,
$k_1+\ldots+k_s=l_1+\ldots+l_t$. The number of vertices of the
type $\Xi$ will be denoted $v(\Xi)$. The cardinality of a type
$\Xi$ will be denoted $|\Xi|$.
\begin{ex} There are 6 w-trees in the type $\langle 5,12\,|\,
1,7,9\rangle$:
\[\begin{picture}(330,70) \put(0,35){\circle*{4}} \put(30,35){\circle{4}}
\put(30,5){\circle*{4}} \put(60,35){\circle*{4}} \put(90,35){\circle{4}}
\put(0,35){\line(1,0){28}} \put(30,5){\line(0,1){28}}
\put(32,35){\line(1,0){56}} \put(-2,39){\small 1} \put(26,39){\small 12}
\put(58,39){\small 9} \put(89,39){\small 5} \put(34,3){\small 7}
\put(14,28){\scriptsize 1} \put(44,28){\scriptsize 4} \put(74,28){\scriptsize
5} \put(24,16){\scriptsize 7}

\put(120,35){\circle*{4}} \put(150,35){\circle{4}} \put(150,65){\circle*{4}}
\put(180,35){\circle*{4}} \put(210,35){\circle{4}} \put(120,35){\line(1,0){28}}
\put(150,65){\line(0,-1){28}} \put(152,35){\line(1,0){56}} \put(118,25){\small
1} \put(146,25){\small 12} \put(178,25){\small 9} \put(209,25){\small 5}
\put(154,63){\small 7} \put(134,37){\scriptsize 1} \put(164,37){\scriptsize 4}
\put(194,37){\scriptsize 5} \put(144,48){\scriptsize 7}

\put(240,35){\circle*{4}} \put(270,35){\circle{4}} \put(270,5){\circle*{4}}
\put(300,35){\circle*{4}} \put(330,35){\circle{4}} \put(240,35){\line(1,0){28}}
\put(270,5){\line(0,1){28}} \put(272,35){\line(1,0){56}} \put(238,39){\small 1}
\put(266,39){\small 12} \put(298,39){\small 7} \put(329,39){\small 5}
\put(274,3){\small 9} \put(254,28){\scriptsize 1} \put(284,28){\scriptsize 2}
\put(314,28){\scriptsize 5} \put(264,16){\scriptsize 9} \end{picture}\]

\[\begin{picture}(350,50) \put(0,15){\circle*{4}} \put(30,15){\circle{4}}
\put(30,45){\circle*{4}} \put(60,15){\circle*{4}} \put(90,15){\circle{4}}
\put(0,15){\line(1,0){28}} \put(30,45){\line(0,-1){28}}
\put(32,15){\line(1,0){56}} \put(-2,5){\small 1} \put(26,5){\small 12}
\put(58,5){\small 7} \put(89,5){\small 5} \put(35,42){\small 9}
\put(14,18){\scriptsize 1} \put(44,18){\scriptsize 2} \put(74,18){\scriptsize
5} \put(24,27){\scriptsize 9}

\put(120,15){\circle*{4}} \put(145,15){\circle{4}} \put(170,15){\circle*{4}}
\put(195,15){\circle{4}} \put(220,15){\circle*{4}} \put(120,15){\line(1,0){23}}
\put(147,15){\line(1,0){46}} \put(197,15){\line(1,0){23}} \put(118,19){\small
7} \put(141,19){\small 12} \put(168,19){\small 9} \put(193,19){\small 5}
\put(218,19){\small 1} \put(132,8){\scriptsize 7} \put(156,8){\scriptsize 5}
\put(182,8){\scriptsize 4} \put(207,8){\scriptsize 1}

\put(250,15){\circle*{4}} \put(275,15){\circle{4}} \put(300,15){\circle*{4}}
\put(325,15){\circle{4}} \put(350,15){\circle*{4}} \put(250,15){\line(1,0){23}}
\put(277,15){\line(1,0){46}} \put(327,15){\line(1,0){23}} \put(248,19){\small
1} \put(273,19){\small 5} \put(298,19){\small 7} \put(321,19){\small 12}
\put(348,19){\small 9} \put(262,8){\scriptsize 1} \put(288,8){\scriptsize 4}
\put(312,8){\scriptsize 3} \put(338,8){\scriptsize 9} \end{picture}\] \end{ex}

\section{Plane graphs, Belyi functions and anti-Vandermonde systems}

To a w-tree we can correspond a tree-like plane graph with multiple edges :
\[\begin{picture}(300,50) \put(0,15){\circle*{4}} \put(30,15){\circle{4}}
\put(60,15){\circle*{4}} \put(90,15){\circle{4}} \put(60,45){\circle{4}}
\put(0,15){\line(1,0){28}} \put(32,15){\line(1,0){56}}
\put(60,15){\line(0,1){28}} \put(-1,5){\small 2} \put(29,5){\small 4}
\put(59,5){\small 5} \put(89,5){\small 2} \put(64,43){\small 1}
\put(14,18){\scriptsize 2} \put(44,18){\scriptsize 2} \put(74,18){\scriptsize
2} \put(55,30){\scriptsize 1}

\put(120,13){$\Rightarrow$}

\put(160,20){\circle*{4}} \put(190,20){\circle{4}} \put(220,20){\circle*{4}}
\put(250,20){\circle{4}} \put(220,45){\circle{4}} \put(220,20){\line(0,1){23}}
\qbezier(160,20)(175,35)(189,21) \qbezier(160,20)(175,5)(189,19)
\qbezier(191,21)(205,35)(220,20) \qbezier(191,19)(205,5)(220,20)
\qbezier(220,20)(235,35)(249,21) \qbezier(220,20)(235,5)(249,19)
\put(172,17){\scriptsize $O_1$} \put(202,17){\scriptsize $O_2$}
\put(232,17){\scriptsize $O_3$} \end{picture}\] Here the weight of a vertex
becomes its valency and the weight of an edge becomes its multiplicity. In our
example the complement to the graph has 4 components: 3 are bounded (components
$O_1,O_2$ and $O_3$) and one unbounded.

Belyi function of this graph is a rational function $\varphi$ with critical
values $0,1,\infty$ such, that our graph and the graph $G=\varphi^{-1}[0,1]$
are isotopic. In our case $\varphi$ has 3 simple poles in points $c_1\in O_1$,
$c_2\in O_2$ and $c_3\in O_3$. White vertices of $G$ are inverse images of $0$
and black --- of $1$. Without loss of generality we can assume that point $0$
is the white vertex of valency 4 and point $1$ is the black vertex of valency
5. Let white vertex of valency 2 be at point $a_1$, white vertex of valency 1
--- at point $a_2$ and black vertex of valency 2 --- at point $b$. Thus,
$$\varphi(z)=\frac{\alpha z^4(z-a_1)^2(z-a_2)}{(z-c_1)(z-c_2)(z-c_3)},\quad
\varphi(z)-1=\frac{\alpha (z-1)^5(z-b)^2}{(z-c_1)(z-c_2)(z-c_3)}.$$ Hence,
$$\alpha z^4(z-a_1)^2(z-a_2)-(z-c_1)(z-c_2)(z-c_3)=\alpha (z-1)^5(z-b)^2.$$
Equaling coefficients at sixth, fifth and forth powers of $z$ we obtain a
system:
$$\left\{\begin{array}{l}2a_1+a_2=2b+5\\ a_1^2+2a_1a_2=b^2+10b+10\\
a_1^2a_2=5b^2+20b+10\end{array}\right.\eqno(1)$$ As
$$(2a_1+a_2)^2-2(a_1^2+2a_1a_2)=2a_1^2+a_2^2\text{ and } (2b+5)^2-2(b^2+10b+10)=
2b^2+5,$$ then the second equation can be written in the form
$2a_1^2+a_2^2=2b^2+5$. Analogously, as
$$3a_1^2a_2-3(2a_1+a_2)(a_1^2+2a_1a_2)+(2a_1+a_2)^3=2a_1^3+a_2^3$$ and
$$3(5b^2+20b+10)-3(2b+5)(b^2+10b+10)+(2b+5)^3=2b^3+5,$$  then the third equation
can be written in the form $2a_1^3+a_2^3=2b^3+5$. Thus, we obtain so called
\emph{anti-Vandermonde} \cite{Ko} system:
$$\left\{\begin{array}{l}2a_1+a_2-2b-5=0\\ 2a_1^2+a_2^2-2b^2-5=0\\
2a_1^3+a_2^3-2b^3-5=0\end{array}\right.$$ The number of finite solutions of
this system is $\leqslant 6$ (in our case it is 4).

In general case to a w-tree from the given type $\langle
k_1,k_2,\ldots,k_s\,|\, l_1,l_2,\ldots,l_t\rangle$ we correspond a plane
tree-like graph with $s$ white vertices of valences $k_1,k_2,\ldots,k_s$ and
$t$ black vertices of valences $l_1,l_2,\ldots,l_t$. Such graph has
$n=k_1+\ldots+k_s=l_1+\ldots+l_t$ edges and its complement is a union of
$m=n+1-s-t$ bounded domains and one unbounded domain. Belyi function of such
graph is a rational function $\varphi$ with simple poles at points
$c_1,\ldots,c_m$
--- one pole in each bounded domain. Let white vertices of the graph are at
points $a_1,\ldots,a_s$ and black --- at points $b_1,\ldots,b_t$.
So
$$\varphi(z)=\frac{\alpha (z-a_1)^{k_1}\cdots\ldots\cdot(z-a_s)^{k_s}}
{(z-c_1)\cdot\ldots\cdot(z-c_m)}= \frac{\alpha
(z-b_1)^{l_1}\cdots\ldots\cdot(z-b_t)^{l_t}}
{(z-c_1)\cdot\ldots\cdot(z-c_m)}+1.$$ Hence,
$$\alpha (z-a_1)^{k_1}\cdots\ldots\cdot(z-a_s)^{k_s}= \alpha
(z-b_1)^{l_1}\cdots\ldots\cdot(z-b_t)^{l_t}+ (z-c_1)\cdot\ldots\cdot(z-c_m).$$
Equaling coefficients at powers $z^{m+1},\ldots,z^{n-1}$ at both sides of
equation, we obtain a system with $n-1-m=s+t-2$ equations and $s+t$ unknowns
$a_1,\ldots,a_s,b_1,\ldots,b_t$. Without loss of generality, we can assume that
$a_s=0$ and $b_t=1$ (for example) and obtain a polynomial system of $s+t-2$
equations with $s+t-2$ unknowns that has $\leqslant (s+t-2)!$ solutions.

In next section we'll demonstrate that such system can always be transformed
into anti-Vandermonde system.

\section{Reduction to an anti-Vandermonde system}

Let
$$s=\prod_{i=1}^m (x-a_i)^{k_i}=x^n-s_1x^{n-1}+s_2x^{n-2}+\ldots,\, \text{ where }
n=\sum_{i=1}^m k_i.$$

\begin{theor} There exist polynomials $q_i$ with rational coefficients: $q_1$ from
one variable, $q_2$ from two variables and so on, such, that
$q_i(s_1,\ldots,s_i)=k_1a_1^i+\ldots+k_ma_m^i$. Moreover, coefficients of
polynomial $q_i$ depend only on $i$. \end{theor}

\begin{proof} Let us consider the product
$$s'=\prod_{i=1}^m (1-xa_i)^{k_i}=1-s_1x+s_2x^2+\ldots.$$ Then
\begin{align*}\ln(s')=\sum_{i=1}^m&k_i\ln(1-xa_i)=\\=&-\left(\sum_{i=1}^m
k_ia_i\right)x- \frac 12 \left(\sum_{i=1}^m
k_ia_i^2\right)x^2-\frac 13 \left(\sum_{i=1}^m k_ia_i^3\right)x^3-
\ldots.\hspace{10mm}(2)\end{align*} On the other hand,
$$\ln(s')=-(s_1x-s_2x^2+\ldots)-\frac 12 (s_1x-s_2x^2+\ldots)^2- \frac 13
(s_1x-s_2x^2+\ldots)^3-\ldots.\eqno(3)$$ Equaling coefficients at powers of $x$
in (2) and (3), we obtain the required polynomials. Thus,
$$\begin{array}{l} q_1=x_1,\,q_2=-2x_2+x_1^2,\,q_3=3x_3-3x_1x_2+x_1^3,\\
q_4=-4x_4+4x_1x_3+2x_2^2-4x_1^2x_2+x_1^4,\\
q_5:=5x_5-5x_1x_4-5x_2x_a3+5x_1^2x_3+5x_1x_2^2-5x_1^3x_2+x_1^5,\\
q_6=-6x_6+6x_1x_5+6x_2x_4+3x_3^2-6x_1^2x_4-\\\hspace{5cm}
-12x_1x_2x_3-2x_2^3+6x_1^3x_3 +9x_1^2x_2^2-6x_1^4x_2+x_1^6.
\end{array}$$ \end{proof}
\par\noindent
Let
$$\prod_{i=1}^s (x-a_i)^{k_i}=x^n-s_1x^{n-1}+s_2x^{n-2}+\ldots$$ and
$$\prod_{i=1}^t (x-b_i)^{l_i}=x^n-t_1x^{n-1}+t_2x^{n-2}+\ldots.$$ Then
$s_i=t_i$ for $i=1,\ldots,s+t-2$, and the system
$$s_1=t_1,\quad q_2(s_1,s_2)=q_2(t_1,t_2),\quad q_3(s_1,s_2,s_3)=
q_3(t_1,t_2,t_3),\quad\ldots$$ is the required ant-Vandermonde system.

\section{Simple, non-decomposable types}

\begin{defin} A type $\Xi=\langle k_1,k_2,\ldots,k_s\,|\, l_1,l_2,\ldots,l_t\rangle$ will
be called \emph{simple}, if numbers $k_1,k_2,\ldots,k_s$ are pairwise different
and numbers $l_1,l_2,\ldots,l_t$ also are pairwise different. $\Xi$ is called
\emph{decomposable}, if there exists a proper subset $I\subset\{1,\ldots,s\}$
and a proper subset $J\subset\{1,\ldots,t\}$ such, that
$$\sum_{i\in I} k_i=\sum_{j\in J} l_j.$$ In this case
$\langle\{k_i\}_{i\in I}\,|\,\{l_j\}_{j\in J}\rangle$ is also a type, which
will be called a \emph{subtype} of the type $\Xi$. \end{defin}

\begin{theor} Let $\langle k_1,k_2,\ldots,k_s\,|\,
l_1,l_2,\ldots,l_t\rangle$ be a simple non-decomposable type. Than
its cardinality is $(s+t-2)!$.
\end{theor}

\begin{rem} This theorem was proved by V.Dremov \cite{Dr} in his thesis in assumption that the
corresponding anti-Vandermonde system doesn't have any multiple solutions. \end{rem}

\begin{proof} Let us assume that the white vertex of the biggest weight $k_s$ is at
point $0$ and the black vertex of the biggest weight $l_t$ --- at $1$. The
corresponding anti-Vandermonde system is of the form:
$$\left\{\begin{array}{l} k_1x_1+\ldots+k_{s-1}x_{s-1}-l_1y_1-\ldots-
l_{t-1}y_{t-1}-l_t=0\\ \\ \qquad \cdots \qquad \cdots \qquad
\cdots \qquad\cdots\qquad\cdots\qquad\cdots\\ \\k_1x_1^{s+t-2}+
\ldots+k_{s-1}x_{s-1}^{s+t-2}-l_1y^{s+t-2}-\ldots-
l_{t-1}y_{s-1}^{l+s-2}-l_t=0\end{array}\right.$$ At first we'll
prove that the system has no solutions at infinity.
\par\medskip\noindent
A solution at infinity satisfies the system
$$\left\{\begin{array}{l} k_1x_1+\ldots+k_{s-1}x_{s-1}-l_1y_1-\ldots-
l_{t-1}y_{t-1}=0\\ \\ \qquad \cdots \qquad \cdots \qquad \cdots \qquad\cdots
\qquad\cdots\qquad\cdots\\
\\k_1x_1^{s+t-2}+ \ldots+k_{s-1}x_{s-1}^{s+t-2}-l_1y^{s+t-2}-\ldots-
l_{t-1}y_{s-1}^{l+s-2}=0\end{array}\right.$$ We can consider it as a
homogeneous system with unknowns $k_i,l_j$. As this system has a nonzero
solution, then the determinant of the system is zero, i.e.
$$\prod_{i=1}^{s-1}x_i\prod_{j=1}^{t-1}y_j\prod_{1\leqslant i<j\leqslant
t-1}(x_i-x_j)\prod_{1\leqslant i<j\leqslant
t-1}(y_i-y_j)\prod_{\scriptsize\begin{array}{l}1\leqslant i\leqslant s-1\\
1\leqslant j\leqslant t-1\end{array}}(x_i-y_j)=0.$$
\begin{itemize}
    \item If some $x_i=0$ (or $y_j=0$), then we eliminate terms $k_ix_i^r$ (or
    $l_jy_j^r$), $r=1,\ldots,s+t-2$, from the system.
    \item If $x_i=x_j$ (or $y_i=y_j$, or $x_i=y_j$), then we eliminate terms
    $k_jx_j^r$ (or $l_jy_j^r$) and coefficient $k_i$ ($l_i$) we replace by
    $k_i+k_j$ (by $l_i+l_j$, or by $k_i-l_j$).
\end{itemize} Also we eliminate the last equation.

We continue in the same way. As all variables $x_i$ and $y_j$ cannot be zeroes,
then at some step the coefficient at some variable $x_i$ or $y_j$ will become
zero. But this means that the type is decomposable.

To prove that our system has no multiple solutions let us consider a
anti-Vandermonde system
$$\begin{array}{l}k_1x_1+\ldots+k_nx_n+1=0\\ \qquad
\cdots \qquad \cdots \qquad \cdots \qquad\\
k_1x_1^n+\ldots+k_nx_n^n+1=0\end{array}$$ It defines a multivalued map from
$n$-dimensional space $K$ with coordinates $k_1,\ldots,k_n$ to $n$-dimensional
space $X$ with coordinates $x_1,\ldots,x_n$. The Jacobian of this map is
$$\prod_{i=1}^n x_i \prod_{1\leqslant i<j\leqslant n} (x_i-x_j)/n!
\prod_{i=1}^n k_i \prod_{1\leqslant i<j\leqslant n} (x_i-x_j).$$ We see that a
multiple solution can occur only when one of unknowns is zero,  or when values
of two unknowns coincide. In both cases we can diminish the number of unknowns
by one.

In the end we will obtain a system of the form
$$ly+1=ly^2+1=\ldots=ly^n+1=0,$$ that has a solution only if $l=-1$, but then our
type is decomposable.

We see that our system does not have solutions on infinity and does not have
multiple solutions. Hence, it has $(s+t-2)!$ finite pairwise different
solutions. \end{proof}

\section{Simple decomposable types}

If a type $\Xi$ is simple, but decomposable, then the the number
of its w-trees is less then $(v(\Xi)-2)!$.
\begin{ex} We have:
$$\begin{array}{ll}|\langle 1,5,7\,|\,2,4,7\rangle|=18, & |\langle
1,3,11\,|\,4,5,6\rangle|=20,\\
|\langle 1,2,4\,|\,1,2,4\rangle|=11, & |\langle 1,2,4,5\,|\,1,2,9\rangle|=72,\\
|\langle 1,2,3\,|\,1,2,3\rangle|=7, & |\langle 1,2,6,10\,|\,4,5,10\rangle|=96.
\end{array}$$
\end{ex}

\begin{defin} We say that a type $\Xi=\langle k_1,k_2,\ldots,k_s\,|\,
l_1,l_2,\ldots,l_t\rangle$ admits a $n$-partition if there exit
disjoint subsets $I_1,\ldots,I_n$ of the set $\{1,\ldots,s\}$, and
disjoint subsets $J_1,\ldots,J_n$ of the set $\{1,\ldots,t\}$ such
that
$$\bigcup_{p=1}^n I_p=\{1,\ldots,s\}, \quad \bigcup_{p=1}^n J_p=
\{1,\ldots,t\}$$ and
$$\sum_{i\in I_1}k_i=\sum_{j\in J_1}l_j,\quad\ldots,\quad \sum_{i\in I_n} k_i=
\sum_{j\in J_n}l_j.$$ We will say that the type $\Xi$ is a union of subtypes
$\Xi_1,\ldots,\Xi_n$: $\Xi=\Xi_1\cup\ldots\cup\Xi_n$, where
$$\Xi_1=\langle \{k_i\,|\,i\in I_1\}\,|\,\{l_j\,|\,j\in J_1\}\rangle,\ldots, \Xi_n=
\langle \{k_i\,|\,i\in I_n\}\,|\,\{l_j\,|\,j\in J_n\}\rangle.$$
\end{defin}

\begin{rem} The type itself can be considered as its 1-partition.\end{rem}

\begin{theor} The number of w-trees in a type $\Xi$ is a sum by
partitions of $\Xi$, where to each partition $\Xi=\Xi_1\cup\ldots\cup\Xi_n$,
$n=1,2,\ldots$, corresponds a summand
$$(-1)^{n-1}(v(\Xi)-1)^{n-2}\prod_{m=1}^n (v(\Xi_i)-1)!$$ \end{theor}

\begin{ex}
\par\noindent
\begin{itemize}
    \item The type $\langle 7,5,1\,|\,7,4,2\rangle$ has only one nontrivial partition
    (2-partition): $\langle 7,5,1\,|\,7,4,2\rangle
    =\langle 7\,|\,7\rangle\cup \langle 5,1\,|\,4,2\rangle$. Hence,
    $|\langle 7,5,1\,|\,7,4,2\rangle=4!-3!=18$.
    \item The type $\langle 4,2,1\,|\,4,2,1\rangle$ has following nontrivial
    partitions:
    \begin{enumerate}
        \item 2-partition $\langle 4\,|\,4\rangle\cup\langle
        2,1\,|\,2,1\rangle$;
        \item 2-partition $\langle 2\,|\,2\rangle\cup\langle
        4,1\,|\,4,1\rangle$;
        \item 2-partition $\langle 1\,|\,1\rangle\cup\langle
        4,2\,|\,4,2\rangle$;
        \item 3-partition $\langle 4\,|\,4\rangle\cup \langle
        2\,|\,2\rangle\cup \langle 1\,|\,1\rangle$.
    \end{enumerate} Hence, $|\langle 4,2,1\,|\,4,2,1\rangle|=
    4!-3!-3!-3!+5=11$.
    \item The type $\langle 3,2,1\,|\,3,2,1\rangle$ has following nontrivial
    partitions:
    \begin{enumerate}
        \item 2-partition $\langle 3\,|\,3\rangle\cup\langle
        2,1\,|\,2,1\rangle$;
        \item 2-partition $\langle 2\,|\,2\rangle\cup\langle
        3,1\,|\,3,1\rangle$;
        \item 2-partition $\langle 1\,|\,1\rangle\cup\langle
        3,2\,|\,3,2\rangle$;
        \item 2-partition $\langle 3\,|\,2,1\rangle\cup\langle
        2,1\,|\,3\rangle$;
        \item 3-partition $\langle 3\,|\,3\rangle\cup \langle
        2\,|\,2\rangle\cup \langle 1\,|\,1\rangle$.
    \end{enumerate} Hence, $|\langle 3,2,1\,|\,3,2,1\rangle|=
    4!-3!-3!-3!-2!\cdot 2!+5=7$.
\end{itemize} \end{ex}

\begin{rem} We can consider w-trees with non integral weights: the sum of
white weights must be equal to the sum of black. A solution of the
corresponding anti-Vandermonde system gives us positions of vertices.
\end{rem}

\begin{proof} \underline{The first step.} Let our type admits only one
partition, namely 2-partition $\Xi=\Xi'\cup\Xi''$, $v(\Xi')=k$, $v(\Xi'')=l$.

Let us define a "deformation" $\widetilde{\Xi}$ of our type $\Xi$: we increase
the weight of some white vertex $v$ from $\Xi'$ by small $\varepsilon$ and
increase the weight of some black vertex $u$ from $\Xi''$ by the same
$\varepsilon$. Thus obtained type is non-decomposable and contains
$(v(\Xi)-2)!$ w-trees. If $T$ is a w-tree from $\widetilde{\Xi}$, then weights
of its edges are positive integers, except edges of the path from $v$ to $u$.
The weight of such edges differs from integer on $\varepsilon$. If some edge of
this path has weight $\varepsilon$ (such edge will be called
$\varepsilon$-bridge), then such tree doesn't exist in the type $\Xi$
(connectivity is lost, when $\varepsilon$ becomes zero).

Thus, we need to find the number of w-trees in the type $\widetilde{\Xi}$ that
have $\varepsilon$-bridge. Let us note that $\varepsilon$-bridge connects a
white vertex $x$ from the subtype $\Xi'$ and a black vertex $y$ from the
subtype $\Xi''$: to a white vertex $x$ a path from $v$ comes with weight
$p-\varepsilon$ (where $p$ is an integer) and to a black vertex $y$ a path from
$u$ comes with weight $q-\varepsilon$. Hence, we can construct a tree with
$\varepsilon$-bridge in the following way: we choose a tree $T'$ from $\Xi'$
and a tree $T''$ from $\Xi''$ ($(k-2)!(l-2)!$ variants of such double choice);
we choose a white vertex of $T'$ and a direction of outgoing
$\varepsilon$-bridge (the number of such choices is equal to the number of $T'$
edges, i.e. $k-1$); we choose a black vertex from $T''$ and the direction of
ingoing $\varepsilon$-bridge ($l-1$ variants). Thus, $|\Xi|=(s+t-2)!-(k-1)!
(l-1)!$.
\par\medskip\noindent
\underline{Second step.} Let our type admits only 2-partitions and
$\Xi'_1\cup\Xi''_1,\ldots,\Xi'_n\cup\Xi''_n$ are all 2-partitions of the type
$\Xi$. Let $k_i$ be the number of vertices of subtype $\Xi'_i$ and $l_i$ be the
number of vertices of subtype $\Xi''_i$.

Let us increase the weight of a white vertex $x$ from $\Xi'_n$ by $\varepsilon$
and the weight of a black vertex $y$ from $\Xi''_n$ by the same $\varepsilon$.
Thus obtained deformed type $\widetilde{\Xi}$ can have only 2-partitions (the
number of such 2-partitions is strictly less, then $n$). Let us assume that
exactly first $k$ partitions of the type $\Xi$ are partitions of
$\widetilde{\Xi}$, i.e. vertices $x$ and $y$ both belong to $\Xi'_i$ or to
$\Xi''_i$, if $i\leqslant k$, and if $i>k$ then one of these two vertices
belongs to $\Xi'_i$ and another --- to $\Xi''_i$.

By induction
$$|\widetilde{\Xi}|=(s+t-2)!-\sum_{i=1}^k(k_i-1)!(l_i-1)!.$$ Now it remains to
find the number of trees in the type $\widetilde{\Xi}$ with
$\varepsilon$-bridge.

Let $T$ be such tree. If $\varepsilon=0$, then $T$ loses connectedness and is a
union of (exactly) two subtrees $T_1$ and $T_2$, where $T_1$ belongs to subtype
$\Xi'_i$ and $T_2$ --- to subtype $\Xi''_i$ for some $i>k$. As above we have
that the number of such trees $T$ is $(k_i-1)!(l_i-1)!$. And summation by $i$,
$k<i\leqslant n$, gives us the required formula.
\par\medskip\noindent
\underline{Third step.} Let now the type $\Xi$ has exactly one 3-partition
$\Xi=\Xi_1\cup\Xi_2\cup\Xi_3$ and the only 2-partitions of $\Xi$ are
$$\Xi_1\oplus(\Xi_2\cup\Xi_3),\,\,\Xi_2\oplus(\Xi_1\cup\Xi_3),\, \text{ and }
\Xi_3\oplus(\Xi_1\cup\Xi_2),$$ where by $\Xi'\oplus\Xi''$  we denote the type
whose set of white (black) weights is a union of set of white (black) weights
of type $\Xi'$ and set of white (black) weights of type $\Xi'$. Let subtypes
$\Xi_1$, $\Xi_2$ and $\Xi_3$ have $k$, $l$ and $m$ vertices respectively.

Let us increase weight of some white vertex from $\Xi_1$ by $\varepsilon$ and
weight of some black vertex from $\Xi_2$ by the same $\varepsilon$. Thus
obtained type $\widetilde{\Xi}$ has only partition --- 2-partition
$(\widetilde{\Xi_1}\oplus\widetilde{\Xi_2})\cup\Xi_3$. Hence,
$|\widetilde{\Xi}|=(k+l+m-2)!-(m-1)!(k+l-1)!$.

Now we must find the number of trees with $\varepsilon$-bridge. There can be
one bridge or two. In the first case either a tree from $\Xi_1$ is connected by
$\varepsilon$-bridge with a tree from $\Xi_2\oplus\Xi_3$, or a tree from
$\Xi_2$ is connected by $\varepsilon$-bridge with a tree from
$\Xi_1\oplus\Xi_3$. Hence, the number of trees with one $\varepsilon$-bridge is
\begin{multline*}(k-1)!(l+m-1)((l+m-2)!-(l-1)!(m-1)!)+\\
+(l-1)!(k+m-1)((k+m-2)!-(k-1)!(m-1)!).\end{multline*}

A tree with two $\varepsilon$-bridges can be constructed in the following way:
we choose a tree $T_1$ from $\Xi_1$, a tree $T_2$ from $\Xi_2$ and a tree $T_3$
from $\Xi_3$. A white vertex from $T_1$ we connect by $\varepsilon$-bridge with
a black vertex from $T_3$ and a white vertex from $T_3$ --- with a black vertex
from $T_2$. Hence the number of trees with two $\varepsilon$-bridges is
$(k-1)!(l-1)!(m-1)!(m-1)$. Now it is easy to check that we obtain the required
formula.

\par\medskip\noindent
\underline{Forth step.} Let the type has only 2- and 3-partitions. Then we
increase weight of some white vertex by $\varepsilon$ and weight some black
vertex by $\varepsilon$ with the aim to decrease the number of 3-partitions.
After that we use induction as in the second step.
\par\medskip\noindent
\underline{Fifth step.} Let type $\Xi$ has one 4-partition
$\Xi=\Xi_1\cup\Xi_2\cup\Xi_3\cup\Xi_4$ and all its 2- and 3-partitions are
constructed from subtypes $\Xi_i$. Let  the number of vertices in $\Xi_1$ be
$k$, in $\Xi_2$ --- $l$, in $\Xi_3$ --- $m$ and in $\Xi_4$ --- $n$.

As above let us define a deformation $\widetilde{\Xi}$ of the type $\Xi$, by
increasing the weight of some white vertex in $\Xi_1$ by $\varepsilon$ and some
black vertex in $\Xi_2$ by the same $\varepsilon$. After that we must find the
number of trees in $\widetilde{\Xi}$ with $\varepsilon$-bridges and subtract
this number from $|\widetilde{\Xi}|$. It is enough to check that thus obtained
expression contains the term $(k+l+m+n-1)^2(k-1)!(l-1)!(m-1)!(n-1)!$ (other
terms are related to 2- and 3-partitions). We will enumerate classes of trees
with bridges and for each class we will study with what coefficient the product
$(k-1)!(l-1)!(m-1)!(n-1)!$ appears in the formula for number of trees in this
class.
\begin{itemize}
    \item Class: $(\Xi_1\oplus\Xi_3\oplus\Xi_4)\cup\Xi_2$. One bridge.
    Coefficient: $(k+m+n-1)^2$.
    \item Class: $(\Xi_1\oplus\Xi_3)\cup(\Xi_2\oplus\Xi_4)$. One bridge.
    Coefficient: $(k+m-1)(l+n-1)$.
    \item Class: $(\Xi_1\oplus\Xi_4)\cup(\Xi_2\oplus\Xi_3)$. One bridge.
    Coefficient: $(k+n-1)(l+m-1)$.
    \item Class: $\Xi_1\cup(\Xi_2\oplus\Xi_3\oplus\Xi_4)$. One bridge.
    Coefficient: $(l+m+n-1)^2$.
    \item Class: $(\Xi_1\oplus\Xi_3)\cup\Xi_4\cup\Xi_2$. Two bridges.
    Coefficient: $-(k+m-1)(n-1)$.
    \item Class: $(\Xi_1\oplus\Xi_4)\cup\Xi_3\cup\Xi_2$. Two bridges.
    Coefficient: $-(k+n-1)(m-1)$.
    \item Class: $\Xi_1\cup(\Xi_3\oplus\Xi_4)\cup\Xi_2$. Two bridges.
    Coefficient: $-(m+n-1)^2$.
    \item Class: $\Xi_1\cup\Xi_3\cup(\Xi_2\oplus\Xi_4)$. Two bridges.
    Coefficient: $-(l+n-1)(m-1)$.
    \item Class: $\Xi_1\cup\Xi_4\cup(\Xi_2\oplus\Xi_3)$. Two bridges.
    Coefficient: $-(l+m-1)(n-1)$.
    \item Class: $\Xi_1\cup\Xi_3\cup\Xi_4\cup\Xi_2$. Three bridges.
    Coefficient: $2(m-1)(n-1)$.
\end{itemize}
It is easy to find the sum of coefficients
\begin{multline*} (k+m+n-1)^2+(k+m-1)(l+n-1)+(k+n-1)(l+m-1)+(l+m+n-1)^2-\\
-(k+m-1)(n-1)-(k+n-1)(m-1)-(m+n-1)^2-(l+m-1)(n-1)-\\-(l+n-1)(m-1)+2(m-1)(n-1)
=(k+l+m+n-1)^2.\end{multline*} There is a geometrical interpretation of this
combinatorial identity.

Let us consider a square $A=[0,k+l+m+n-1]\times
[0,k+l+m+n-1]\subset\mathbb{R}^2$ and four rectangles in it:
$$\begin{array}{l}A_1=[0,k+m+n-1]\times [0,k+m+n-1],\\
A_2=[k,k+l+m+n-1]\times[k,k+l+m+n-1],\\
A_3=[0,k+m-1]\times[k+m,k+l+m+n-1],\\ A_4=[k+n,k+l+m+n-1]\times[0,k+n-1].
\end{array}$$  It is easy to see that $A=A_1\cup A_2\cup A_3\cup A_4$ and
$$\begin{array}{l} A_1\cap\ A_2=[k,k+m+n-1]\times[k,k+m+n-1],\\ A_1\cap
A_3=[0,k+m-1]\times[k+m,k+m+n-1],\\ A_1\cap A_4=[k+n,k+m+n-1]\times
[0,k+n-1],\\ A_2\cap A_3=[k,k+m-1]\times[k+m,k+l+m+n-1],\\ A_2\cap A_4=
[k+n,k+l+m+n-1]\times[k,k+n-1],\\ A_3\cap A_4=\varnothing,\\ A_1\cap A_2\cap
A_3=[k,k+m-1]\times[k+m,k+m+n-1],\\ A_1\cap A_2\cap A_4=[k+n,k+m+n-1]\times
[k,k+n-1].\end{array}$$ Thus, our identity is simply the inclusion-exclusion
formula.
\par\medskip\noindent
\underline{Sixth step.} In the general case we consider a partition of the type
$\Xi$, analogous to the partition in the fifth step, but (n+2)-partition.
Analogously we define deformation $\widetilde{\Xi}$. We must find the
coefficient at product $\prod_{i=1}^{n+2}(m_i-1)!$ in the formula, that
describe the number of trees with bridges (here $m_i=v(\Xi_i)$,
$i=1,\ldots,n+2$). Let $M=m_1+\ldots+m_{n+2}-1$.

As above we will enumerate classes of trees with bridges and for
each class we will find the corresponding coefficient. A class is
defined by such partition (admissible partition) of the set
$\{1,\ldots,n+2\}$ into disjoint subsets, that numbers 1 and 2
belong to \emph{different} subsets. To a given admissible
partition we correspond a product by subsets of this partition and
then we compute a sum by all admissible partitions.

To a subset $\{s_1,\ldots,s_p\}\subset\{1,\ldots,n+2\}$ of given partition we
correspond the factor
$$(m_{s_1}+\ldots+m_{s_p}-1)^p,\text{ if }\{1,2\}\cap\{s_1,\ldots,s_p\}=
\varnothing,$$ and the factor
$$(m_{s_1}+\ldots+m_{s_p}-1)^{p-1},\text{ if
} s_1=1 \text{ or } s_1=2.$$ Thus obtained product by subsets of the partition
we multiply by $(-1)^{q-2}(q-2)!$, where $q$ is the number of subsets in the
partition.

We must prove, that the sum by all admissible partitions is equal
to $M^n$. The proof will use the inclusion-exclusion formula.

Let us consider an $n$ dimensional cube $C$ and right parallelopes in it. The
cube $C$ is  the Cartesian product: $C=[0,M]^n$. A right parallelotope is a
Cartesian product of $n$ segments. Parallelotopes under consideration are
enumerated by subsets (proper and non-proper) of the set $N=\{3,\ldots,n+2\}$.
Given the set $A=\{i_1,\ldots,i_s\}\subset\{3,\ldots,n+2\}$ we define segments
$[a_i,b_i]$, $i=1,\ldots,n$,
$$a_i=\begin{cases}\begin{array}{ll}0,&\text{ if } i+2\in A,\\ m_1+S_A,&\text{ if }
i+2\notin A;\end{array}
\end{cases}\quad b_i=\begin{cases}\begin{array}{ll}m_1+S_A-1,&\text{ if }
i+2\in A,\\ M,&\text{ if } i+2\notin A.\end{array}\end{cases}$$ and
parallelotope $P_A=[a_1,b_1]\times\ldots\times[a_n,b_n]\subset C$ (here
$S_A=\sum_{i\in A}m_i$) with volume
$$\text{volume}(P_A)=\left(m_1+S_A-1\right)^{|A|}\left(M-m_1-S_A\right)^{n-|A|}.$$
We must prove that $\bigcup_A P_A=C$. Let $X=(x_1,\ldots,x_n)$ be a point in
$C$. If $x_i\leqslant m_1+m_{i+2}-1$, $i=1,\ldots,n$, then $X\in P_N$. If
$x_i\geqslant m_1+m_{i+2}$, $i=1,\ldots,n$, then $X\in P_\varnothing$. Let now
$$x_i\leqslant m_1+m_{i+2}-1,\,i=1,\ldots,s\text{ and } x_i\geqslant m_1+m_{i+2},\,
i=s+1,\ldots,n$$ and let us denote by $F$ the set $\{3,\ldots,s+2\}$. If
$x_i\leqslant m_1+S_F-1, i=s+1,\ldots,n$, then $X\in P_N$. If $x_i\geqslant
m_1+S_F, i=s+1,\ldots,n$, then $X\in P_F$. If $m_1+S_F-1\leqslant
x_{i-2}\leqslant m_1+S_F$, for $i\in G\subset \{s+3,\ldots,n+2\}$, then instead
of $F$ we will consider the set $F\cup G$. And so on.

Parallelotops $P_A$ and $P_B$ have a non-empty intersection only
if $A\subset B$ (or $B\subset A$). In this case
$$V(P_A\cap
P_B)=(k+S_B-1)^{|S_B|}(S_A-S_B-1)^{|S_A)-|S_B|}(M-k-S_A)^{n-|S_A|}.$$ Triple
intersections $P_{A\cup B\cup C}\cap P_{A\cup B}\cap P_A$ and $P_{A\cup B\cup
C}\cap P_{A\cup C}\cap P_A$, where $A,B,C\subset\{3,\ldots,n+2\}$ are pairwise
disjoint subsets, have the same volume. Analogously, for each non-empty
intersection $P_A\cap P_B\cap P_C\cap P_D$ there are five more with the same
volume.
\end{proof}

\section{Non-simple types}

If a type $\Xi$ is not simple, then a numeration of vertices (or a
labeling of vertices) makes the type simple. Thus obtained type
will be called a "derivative" type and will be denoted $\Xi'$.

\begin{ex} If $\Xi=\langle 6,1,1,1\,|\,3,3,3\rangle$, then
$\Xi'=\langle 6,1,1',1''\,|\,3,3',3''\rangle$. This type is simple, but
decomposable ---  it admits three 2-partitions:
\begin{enumerate}
\item $\langle 6\,|\,3,3'\rangle\cup\langle 1,1',1''\,|\,3''\rangle$; \item
$\langle 6\,|\,3,3''\rangle\cup\langle 1,1',1''\,|\,3'\rangle$; \item $\langle
6\,|\,3',3''\rangle\cup\langle 1,1',1''\,|\,3\rangle$.
\end{enumerate} Hence, there are 84
w-trees in this type. 72 of them are of the form
\[\begin{picture}(100,50) \put(0,15){\circle{4}} \put(20,15){\circle*{4}}
\put(50,15){\circle{4}} \put(50,45){\circle*{4}} \put(80,15){\circle*{4}}
\put(80,35){\circle{4}} \put(100,15){\circle{4}} \put(2,15){\line(1,0){46}}
\put(52,15){\line(1,0){46}} \put(50,17){\line(0,1){28}}
\put(80,15){\line(0,1){18}} \put(-1,3){1} \put(18,3){3} \put(48,3){6}
\put(78,3){3} \put(99,3){1} \put(55,41){3} \put(85,31){1}
\end{picture}\] and 12 are symmetric with the third order symmetry
\[\begin{picture}(90,50) \put(0,25){\circle{4}} \put(40,25){\circle{4}} \put(80,5){\circle{4}}
\put(80,45){\circle{4}} \put(20,25){\circle*{4}} \put(60,15){\circle*{4}}
\put(60,35){\circle*{4}} \put(2,25){\line(1,0){36}} \put(41,26){\line(2,1){38}}
\put(41,24){\line(2,-1){38}} \put(-1,12){1} \put(37,12){6} \put(18,12){3}
\put(53,5){3} \put(53,37){3} \put(85,2){1} \put(85,42){1}
\end{picture}\] Thus, the number of w-trees in the type $\langle
6,1,1,1\,|\,3,3,3\rangle$ is
$$\frac{72}{3!\,3!}+\frac{12}{\frac{3!\,3!}{3}}=3$$ --- the correct answer.
\end{ex}
Now we can formulate the statement.

\begin{theor} Let $\Xi$ be a non simple type and let $\{k_1,\ldots,k_s\}$ be the
set of pairwise different white weights and $\{l_1,\ldots,l_t\}$ be the set of
pairwise different black weights. Let there be $m_i$ vertices with weight
$k_i$, $i=1,\ldots,s$ and $n_j$ vertices with weight $l_j$, $j=1,\ldots,t$. Put
$p=\prod_{i=1}^s (m_i)!\prod_{i=1}^t(n_i)!$. If the derivative type $\Xi'$
contains $N$ nonsymmetric w-trees and $M_i$, $i=2,3,\ldots$, w-trees with
$i$-order symmetry, then
$$|\Xi|=\frac Np+\sum_i \frac{iM_i}{p}.$$ \end{theor}

\begin{rem} A type $\Xi$ contains a w-tree with $i$-order symmetry, if
\begin{enumerate}
    \item there exists a vertex $v$ with weight divisible by $i$;
    \item in the set of all other vertices for each color and each weight the
    number of vertices of these weight and color is divisible by $i$.
    and
\end{enumerate}

Such type generates the type $\Xi/i$, where the weight of the vertex $v$ is in
$i$ times smaller and the number of vertices of given color and weight is in
$i$ times smaller, then in the type $\Xi$. Thus, for the type $\Xi=\langle
6,1,1,1\,|\,3,3,3\rangle$ the type $\Xi/3=\langle 2,1\,|\,3\rangle$. Each
w-tree of type $\Xi/i$ generates the unique w-tree of the type $\Xi$. \end{rem}

\begin{ex} Let us consider the type $\Xi=\langle 5,2,1,1,\,|\,3,3,3\rangle$ and
its derivative type $\Xi'=\langle 5,2,1,1',\,|\,3,3',3''\rangle$. The
derivative type admits six 2-partitions, hence, $|\Xi'|=120-6\cdot 12=48$. As
there are no symmetric w-trees, then $|\Xi|=4$. Four w-trees from the type
$\Xi$ are presented below:
\[\begin{picture}(270,50) \put(0,15){\circle{4}} \put(30,15){\circle*{4}}
\put(60,15){\circle{4}} \put(90,15){\circle*{4}} \put(120,15){\circle{4}}
\put(60,45){\circle*{4}} \put(90,45){\circle{4}} \put(2,15){\line(1,0){56}}
\put(62,15){\line(1,0){56}} \put(60,17){\line(0,1){28}} \put(90,15){\line
(0,1){28}} \put(-2,5){\small 2} \put(28,5){\small 3} \put(58,5){\small 5}
\put(88,5){\small 3} \put(119,5){\small 1} \put(64,43){\small 3}
\put(94,43){\small 1}

\put(150,15){\circle{4}} \put(180,15){\circle*{4}} \put(210,15){\circle{4}}
\put(240,15){\circle*{4}} \put(270,15){\circle{4}} \put(180,45){\circle{4}}
\put(210,45){\circle*{4}} \put(152,15){\line(1,0){56}}
\put(212,15){\line(1,0){56}} \put(210,17){\line(0,1){28}} \put(180,15){\line
(0,1){28}} \put(149,5){\small 1} \put(178,5){\small 3} \put(208,5){\small 5}
\put(238,5){\small 3} \put(268,5){\small 2} \put(184,43){\small 1}
\put(214,43){\small 3} \end{picture}\]

\[\begin{picture}(280,50) \put(0,25){\circle{4}} \put(30,25){\circle*{4}}
\put(60,25){\circle{4}} \put(80,35){\circle*{4}} \put(80,15){\circle*{4}}
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redpajama_set_name: RedPajamaArXiv


I reached for my Model 29, but the cold press of a Glock in my side stopped me. I slowly put my hands back on the table, letting out a sigh. I knew when I was beat.
The dame reached into my pockets and inside my jacket, putting my Model 29 and 1911A on the table along with the tranq gun the doctor had given me. Taking the clips out, she tossed them to her sister, leaving the empty weapons on the table. Sophia strode over and retrieved the doctor's tranq gun. She studied it carefully.
"Recognize the craftsmanship, Jerry?" she asked, and he nodded.
"Then why does he think you're after him?" I asked. She looked at me in surprise, and her tone was one of utter incredulity.
"Constanzia?" she asked. I blinked and nodded, put off by how much she knew about me. I hadn't spoken with Constanzia since this whole mess had started.
I sized her up. Same dimensions as her sister, obviously, but with a different posture and demeanor — more confident and self-assured, carrying her body in a way that told me she knew how to use it. She would be far deadlier than her twin in combat, I presumed, and much more resilient under mental and emotional duress. There was almost an air of being the eldest about her, and I made a mental note to find out if she had come out first.
"And how are you sure he's planning anything?" I asked, feeling as though I was still in the dark.
"Is that wise?" Sophia asked.
I watched as Maria got up and walked into the second room, drawing a clip-on earpiece from her belt. I had no idea where she'd gotten it, but then again, I'd never searched her. She'd pulled a fricking Glock on me, for Pete's sake.
I shook my head and returned my attention to her sister, who had cleared my weapons from the table and was now sitting on it, facing me. She rested her Berretta on her lap. I studied her, and she let me – odd, although she did the same, for in her position she could easily have avoided my scrutiny.
Like her sister, she was tall, about five eight, and a nearly perfect 34-24-34, C-cup. Her face had a beauty to it that was more Russian than British, although it was clear from her accent that she was no newcomer to the island. High cheekbones, smooth curves and lines, nothing planar or angular about it, as there would be with a Brit, a Franc, or a German, but with a slight prominence to the chin and a narrowness to the jaw which made the overall impression distinct from the roundness of an American. The skin tone simply wasn't right for a Spaniard or an Italian, and the posture wasn't there for any of the above. Definitely Ukrainian. I could see it in the way she held her head and arms. Ukrainian, perhaps moving to the UK at the age of ten or twelve, learning a new language just as proficiency with the old was starting to set in. Coming from a place where freedom was not the norm and expectation, deciding to defend it with her life. I felt myself beginning to gain a modicum of respect for her. Then I realized I was getting a bit ahead of myself and decided to ask her outright.
"Thirteen," she replied. "I was taken to the States against my will and separated from my sister. I was rescued by a German businessman who was living here on the island, and he brought me home with him as his daughter. When I was old enough, I joined MI6 and started looking for my sister. I found her when I got word that someone was looking into me by my old name, the one I left behind me in the Ukraine. Sure enough, it was Maria." She paused, lost in reflection.
She cocked an eyebrow and looked me over for a second time. I let her do so, keeping my poker face on and meeting her eyes when she had finished. She looked me up and down a third time, and then she leaned closer, speaking softly enough that I had to learn in to hear.
I swallowed and composed myself before I spoke.
"Nothing she doesn't want," I promised. She chuckled in amusement.
"I like you, Mr. Naissance. You play a dangerous game." She draped her arms around me and planted a kiss on my cheek. Before I could recover, she detached herself and strode across the room to meet Maria, who apparently had finished with her call. I stared dumbfounded at her, then shook my head and snapped out of it, refocusing on her face, which I soon discovered wasn't any less distracting. I closed my eyes and took a few deep breaths. It seemed to help.
"What?" I asked, aware that I had missed something. Maria repeated herself.


Grant provides construction operations leadership to the company. As a Florida State Certified General Contractor and business manager he brings a wealth of operations and management experience to the table.
As a 10-year veteran carpenter, mason, electrician's apprentice and then as a supervisor for a General Contractor, Grant's prior experience includes roles of increasing responsibility in operations management, new store development and construction with high-end privately owned and corporate restaurant chains.
He is responsible for the management and operations of MGB Built projects including the development of short- and long-range strategic planning initiatives and procedures. He oversees pre-construction and project management services to ensure a smooth process and timely success.
The company's seasoned team of project managers report to Grant, as does a highly efficient small office staff that allows MGB Built to provide unparalleled service delivery, quality and open accountability throughout the building process.


There are tons of little ailments that, while not life-threatening, are annoying and just plain insufferable. Athlete's foot fits the bill for many.
More likely than not, athlete's foot won't lead to your death, but the infection can itch and burn to the point that you wish you didn't have feet. With that said, wouldn't it be great if you could get rid of athlete's foot as soon as possible? First, we should answer the question, what is athlete's foot, exactly?
From the signs of athlete's foot, to athlete's foot treatment, we'll cover everything you need to know to recognize and quickly treat athlete's foot and get on with your life.
Is Athlete's Foot Contagious? What Causes Athlete's Foot?
One of the first questions people have about athlete's foot is, "How do you get it?" Or, if you've already got it, "Is it contagious?" Unfortunately, the answer isn't a simple one. While there is just one cause of athlete's foot, a fungal infection, there are a number of ways you may have contracted it—with the added bonus of athlete's foot being contagious.
The fungus itself tends to be Trichophyton rubrum, but a few other fungi can cause athlete's foot as well, including Trichophyton mentagrophytes and Epidermophyton floccosum. The infection breeds on your skin, which can cause damage that leads to dry, cracked, itchy, and even blistered feet.
So, how did you get "fungus feet"? Well, a multitude of factors may be to blame. The fungus tends to find you if your feet are moist, so wearing shoes and/or socks that make your feet sweat and block proper ventilation can help foster the fungal growth.
Walking around barefoot in warm, moist places where the fungus likes to breed or near where it likes to breed can also cause an outbreak. Think of places like your gym's locker room and shower. If you have diabetes or a weakened immune system due to illness, you are more prone to athlete's foot.
Athlete's foot is indeed contagious, so you can acquire the fungal infection through direct contact. Sharing towels, shoes, or socks with someone already suffering from athlete's foot can transmit the fungus to your feet.
That's why the showers and locker room at your gym can be dangerous—they create a perfect storm for potential causes in one place.
The classic athlete's foot symptoms are pretty easy to recognize. Essentially, you'll want to be on the lookout for dry, cracked, itchy skin on your feet. The skin can appear dry, red, scaly and flaky, but it may also appear white and soggy in its beginning stages.
Moreover, the infected area can be sore and itchy, and it isn't uncommon to see small blisters appear in the region. This infection can spread around the foot, especially near the nails (often causing a fungal infection in the nail). Overall, the symptoms of athlete's foot can be pretty unpleasant, so you will want to take care of your feet as soon as possible. Luckily, there are various easy-to-make-or-purchase remedies that can get your feet back in prime shape.
The good thing about athlete's foot is that it's very treatable. There are several over-the-counter medications and treatments that can help clear up the fungal infection as well as a vast array of treatments that can be prescribed by your doctor.
If you've caught your athlete's foot early and you'd like to try a natural foot fungus treatment, the home remedies below may be able to relieve your discomfort.
A home remedy standby, tea tree oil could help eliminate many different types of infections, including those of a fungal nature. Place approximately 40 drops of tea tree oil in a water-filled bath or basin where you can soak your feet.
Soak your feet for 10 minutes and bring them out to dry. Rub a few drops of tea tree oil onto the infected area as it will not only help clean up the fungal infection, but it could also repair the damage done to the skin around that area.
Apple cider vinegar could be used as a pre-treatment athlete's foot cure. Take a half-cup of apple cider vinegar and mix it with two cups of water in a container large enough to soak your feet in. Soak your feet for 30 minutes and pat dry once they are out.
Garlic has many antibacterial properties that may help remove your athlete's foot, and the herb can be used easily. Put some crushed or freshly sliced garlic in between your toes for a few days (refreshing the garlic every day). In addition, if you take some garlic supplements or add garlic to the foods you are eating, this could also help with the athlete's foot.
A little bit of corn starch or baking soda sprinkled onto your feet before you put your socks on may help dry out the athlete's foot fungus, and soak up any excess moisture around your feet within your footwear. This can also be used as a preventative measure.
Athlete's foot doesn't have to be a cause for serious concern. It can be itchy and even painful, but as you can see by the remedies listed above, it's easily treated. However, there are a few things you can do to help prevent your feet from getting this fungal infection.
Simple good hygiene can help prevent athlete's foot. Soap and water. Make sure to clean your feet after strenuous exercise or whenever you're at the gym. After they are clean, make sure you dry them off well as moisture can breed fungus.
The fungus that leads to athlete's foot is easily transferable. To that end, don't share your shoes or socks with other people, especially if it's workout gear.
If you use a towel to dry your feet after showering, sharing one (especially at the gym) is an easy way to spread athlete's foot. The fungus left behind on the towel can also be transferred to others through direct contact.
These tips should help prevent you from getting athlete's foot, but if not, we have given you all the tools you need to tackle it head on.
"Athlete's Foot," NHS; http://www.nhs.uk/Conditions/Athletes-foot/Pages/Introduction.aspx, last accessed June 12, 2017.
"Is Athletes Foot Contagious?," Skin Care Guide; http://www.skincareguide.com/article/skin-conditions/athletes-foot/is-athletes-foot-contagious, last accessed June 12, 2017.
Katie, "5 Effective Natural Ways to Remedy Athlete's Foot for Good," Wellness Mama, June 9, 2017; https://wellnessmama.com/156131/athletes-foot/, last accessed June 12, 2017.
Group, E., "10 Natural Remedies for Athlete's Foot," Global Healing Center, first published September 1, 2009, last updated March 15, 2017; https://www.globalhealingcenter.com/natural-health/athletes-foot-home-remedy/, last accessed June 12, 2017.


text:
Two AEW Stars Injured At Full Gear, Miss Event As a Result
BY Thomas Hall – ON November 14, 2021 IN AEW, News
IMG Credit: All Elite Wrestling
That's always a possibility. Injuries can happen at anytime to any wrestler and that is never a good thing to see. Sometimes these things happen over time, but on occasion they are the sort of injury that happens all at once. Either way, they can lead to a wrestler being out of action for some time, which seems to be the case again after a pair of physical matches over the weekend.
According to PWInsider.com, both Malakai Black and Eddie Kingston were injured during this weekend's AEW Full Gear pay per view. The two of them were pulled from The Big Event convention in New York City. There is no word on what kind of injuries they have suffered or how long either of them will be out of action. During the show, Black teamed with Andrade El Idolo against Cody Rhodes and Pac in a lowing effort, which CM Punk defeated Kingston.
These two are rather talented. Check out what they can do when they are healthy:
Opinion: That's never great to hear and hopefully both of them are going to be able to be back as soon as possible. The good thing is that they might be able to miss some time to let things heal up without causing any major problems to their stories, both of which had some important moments at Full Gear. These two are very talented and it would be a shame for either of them to miss any kind of time from AEW.
What do you think of the two of them? What should they be doing next? Let us know in the comments below.
Thomas Hall has been a wrestling fan for over thirty years and has seen over 60,000 wrestling matches. He has also been a wrestling reviewer since 2009 with over 6,000 full shows covered. You can find his work at kbwrestlingreviews.com, or check out his- Amazon author page with 30 wrestling books.
Keep up with the LATEST WRESTLING RUMORS! Click here to sign up for the exclusive Wrestling Rumors daily newsletter, delivered right to your inbox.
Emotional Moment Takes Place After AEW Full Gear Wraps Up
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AEW Full Gear 2021 Preview, Predictions And Thoughts
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Tags: AEW, Eddie Kingston, Full Gear, Injuries, Malakai Black, News

redpajama_set_name: RedPajamaCommonCrawl


text:
Arts & CultureGuidesNew YorkNews
21 Art Installations and Exhibits Not to Miss in NYC December 2016
AFineLyne
2016 is going out with a bang, filled with not only our annual holiday favorites like Luminaries in the Winter Garden, and the New York Botanical Garden Holiday Train Show, but some installations that will bring us closer to nature, from Brooklyn and the Bronx to Staten Island. The art world continues to explore and enhance our urban environment in the ways we grow and eat our food, and how we design for play.
This month, you can explore the art of being a woman, as a kick-off to a permanent home to be named The Women's Building. Pay homage to the men, women and children who succumbed to AIDS at a permanent Memorial site on World AIDS Day, and the doors will open to a permanent home for the GrowNYC's Project Farmhouse. Here are 21 art installations, exhibits and projects not to miss in December, and on into the New Year.
21. AIDS Memorial Park Unveiled for World AIDS Day
The AIDS Memorial Park is located near the former St. Vincent's Hospital, at the intersection of West 12th Street and Greenwich Avenue, a block from the LGBT Community Center. It will be completed in time for an unveiling coinciding with World AIDS Day, December 1st, with a Dedication from 11 am-1 pm.
Created by a coalition of artists, health care providers, and historians, and inspired by two young urban planners, New York City will finally have an AIDS Memorial. The AIDS Memorial Park will honor the more than 100,000 New York City men, women and children who succumbed to the disease. The project began with a design competition in November, 2011, launched by the NYC AIDS Memorial and chaired by Michael Arad, the designer of the National September 11 Memorial. Nearly 500 architects from around the world submitted. The winner was Studio a+, based in Brooklyn, with the Memorial's granite pavers to be designed by visual artist Jenny Holzer, adding selections from Walt Whitman's Song of Myself.
Next: #20 New York Made Stanton Street Courts on the Lower East Side
Tags : AIDS Memorial ParkAnnie LeibovitzArts BrookfieldBayview Correctional FacilityBronx Botanical GardenBronx ZooBrookfield PlaceCarl Schurz ParkConference HouseCooper Hewitt Smithsonian Design MuseumDag Hammarskjold PlazaDUMBOFlatironGreenwich VillageGrowNYCJackson AvenueLower East SideManhattan Bridgemuseum of the city of new yorkNew York At Its CorephotographyProject Farmhouseriverside parkslideshowpageStanton Street Courtsstaten islandThe Women's Building
AFineLyne November 28, 2016


text:
Variety Plus Icon Read Next: Apple Taps Zhang Meng for Chinese New Year Film 'The Comeback'
Ventana Sur: HBO Latino Picks Up U.S. on 'My Love or My Passion' (EXCLUSIVE)
After a batch of successful sales at AFM, Argentina's Film Sharks announced the deal for one Latin America's biggest successes of 2017
By Jamie Lang
Jamie Lang
@idonthaveaband
'Feria: The Darkest Light' Creators Discuss Netflix's New Spanish Supernatural Thriller 11 hours ago
Prime Video Spain Executives María José Rodríguez, Koro Castellano Outline Streamers' Original Drama Production Plans 12 hours ago
Javier Fesser to Direct 'Shared Custody' for The Immigrant as Company Appoints Claudia Valdez Director of Physical Production (EXCLUSIVE) 12 hours ago
Courtesy: FilmSharks Intl.
As the world builds towards what some would regard as the near pathology sparked by a soccer World Cup, Buenos Aires-based FilmSharks Int. has sold a movie which captures soccer madness, "My Love or My Passion," to the U.S. to HBO Latino.
A blockbuster hit, in its native Argentina, the newest deal on "My Love or My Passion," (the Spanish title of the film translates to "Football or Me") adds to a bullish first batch of sales unveiled at the American Film Market, and which includes France (Koba Films), Spain (Festival Films) and China (Royal Seal). Greece (Tanweer), Central America (Weisner), and Chile, Peru and Colombia (Cinecolor).
At the time, Rud said that talks with a "major label" in the U.S. were "very advanced." The now-finished deal grants exclusive U.S. pay TV and VOD rights to HBO Latino.
Film Sharks CEO Guido Rud told Variety: "This film is the perfect parody portrait of the Latin American or European housewife suffering a husband's football addiction. It's perfect timing for a worldwide release as the 2018 FIFA World Cup in Russia approaches."
The film stars Adrián Suar as a recovering soccer-holic who has lost his job, his wife, and distanced himself from his daughters. "My Love or Passion" is already being distributed by Buena Vista in Latin America. Its Argentina premiere was in mid-August where it went on to become the second-highest grossing national movie of 2017, selling 1.1 million tickets (about $6 million in box office gross).
The continued international sales success of "My Love or My Passion," is positive proof that, when handled correctly, the right comedy can travel.
Filmsharks Intl.,
Hbo Latin America,
Ventana Sur