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goetzpartners CORPORATE FINANCE advised Ringier on its disposal of the Swiss and international programme press
Ringier AG is selling its paid programme press assets consisting of TV programme magazines in Switzerland, Romania, Slovakia and the Czech Republic. The Bauer Verlagsgruppe takes over the titles in Romania, Slovakia and the Czech Republic – Swiss titles are being sold to Axel Springer AG.
goetzpartners CORPORATE FINANCE acted as the exclusive financial advisors to Ringier in this transaction.
With 13 publications in four countries Ringier's paid programme press division consists of successful and popular weekly, bi-weekly and monthly TV guides. In the paid programme press segment Ringier is the market leader in Switzerland, Slovakia and Romania as well as one of the Top3 players in the Czech Republic.
The Swiss titles Tele, TV vier and TV2 will be published from July 1, 2007 onwards by Axel Springer through its recently acquired Swiss publishing house Jean Frey AG. The team of 27 Swiss employees will also be taken over by Axel Springer. The TVtäglich supplement will continue to appear in a joint venture of Ringier with Tamedia AG. The title TV8, in the western part of Switzerland, remains in the portfolio of Ringier Romandie.
Furthermore, Bauer Verlagsgruppe signed an agreement to take over the paid programme magazines Euro Televizia, Tele Magazin and TV Max in Slovakia; TV Max, TV Plus, TV Revue and Tydenik Televize in the Czech Republic, as well as TV Mania and TV Satelit in Romania. The TV supplements in the Ringier newspapers and magazines will not be sold along with the magazines.
The disposal of the paid programme press activities is in line with the Ringier's strategic refocusing.
goetzpartners advised the seller in the preparation and successful execution of the disposal process within seven months. During a controlled auction process, the paid programme press division of Ringier attracted strong interest from regional and international strategic buyers.
The transaction is subject to competition authorities' approval in the respective countries.
About Ringier AG
Ringier AG is the largest media company and publishing house in Switzerland. For more than 170 years, Ringier has been providing information and entertainment through diverse media. Ringier publishes more than 100 publications in 12 countries, operates eleven printing works, produces and markets more than 20 TV channels and maintains more than 50 websites. It has annual revenues of approximately CHF 1.3 billion with profits of around CHF 68 million and over 6,500 employees.
About Axel Springer AG
Axel Springer AG is Germany's largest newspaper publisher and the third biggest publishing house. With over 170 newspapers and magazines in 33 countries and approximately 9,700 employees, Axel Springer is one of the leading international media enterprises. The Group's principal activities are printing, publishing and sale of newspapers, magazines, books and periodicals carried out through four divisions: Newspapers, Magazines, Printing and Services/holding.
About Jean Frey AG:
Jean Frey AG is a Swiss publisher and issues the consumer and service magazine BEOBACHTER, the TV guide TV STAR and the business magazine BILANZ and also supplies services for trade journals. In 2006 the company had 240 employees generating revenues of approximately CHF 70 million. Jean Frey AG was acquired by Axel Springer Verlag in December 2006.
About Bauer Verlagsgruppe:
Bauer Verlagsgruppe was founded in 1875 and is one of the largest magazine publishing houses in Europe. Bauer Verlagsgruppe is the market leader in German TV Guides. Heinz Bauer is heading Bauer Verlagsgruppe in fourth generation of private ownership. The Company publishes 166 magazines in 14 countries on three continents and employees over 6,500 people worldwide.
About goetzpartners:
With a team of 150 investment bankers and consultants in Germany, France, the UK, Spain and the
Czech Republic, as well as co-operations in Hungary, Russia and India, goetzpartners is a leading independent European advisory firm offering M&A services through goetzpartners CORPORATE FINANCE, consulting services through goetzpartners MANAGEMENT CONSULTANTS, and interim management services through goetzpartners INTERIM MANAGERS.
Dr Gernot Wunderle, Managing Director, goetzpartners CORPORATE FINANCE GmbH
Tel: +49 (89) 290725 122 Email: wunderle@goetzpartners.com
Press_Release_-_Project_Turicum_-_210607.pdf 21 KB


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Griezmann eyes fight to the finish for Atletico
Written By Dom Farrell
Getty Images https://images.daznservices.com/di/library/omnisport/88/17/antoinegriezmannfernandotorres-cropped_e3fsq78mvgqvzymfz0b3uf36.jpg?t=-1785696380&w=500&quality=80
Antoine Griezmann issued a rallying call to his Atletico Madrid team-mates following Tuesday night's routine 2-0 La Liga win over Real Sociedad.
Atletico did all the damage early on, as Griezmann netted from close range against his former club to complete the scoring in the 10th minute, after Mikel Gonzalez had already headed Koke's corner into his own net.
The reigning champions are highly unlikely to retain their crown but a third consecutive Liga triumph moved them to within two points of Real Madrid in second place, while all eyes will be on the Spanish capital when its two clubs face off in the UEFA Champions League quarter-finals next week.
Griezmann was encouraged by Atletico's efforts at both ends of the field against Sociedad and urged his colleagues to fight until the end of the campaign.
"We know that it is normal to have some downs in a season, but we are up there, fighting for the Champions League, and we have to continue like this," the France international told his club's official website.
"Yes, we have to be strong. We know that if we keep a clean sheet, it is easier to win, we have to continue like this."

redpajama_set_name: RedPajamaCommonCrawl


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CIB to push ahead with expansion plans -Hisham Ezz El Arab
With ample liquidity, Egypt's leading private-sector bank looks to push ahead with expansion plans: CIB is committed to pushing ahead with its expansion strategy once it finalizes its recent acquisition of a 51% stake in Kenya's Mayfair Bank, Chairman Hisham Ezz El Arab told Bloomberg TV (watch, runtime: 8:08). The bank currently has ample capital and more local currency liquidity than required by the Central Bank of Egypt. "I don't see any use for the excess capital we have, unless we carry on with our expansion policy," the CIB chairman said. The bank is also "well-placed" to pay out dividends, but noted this isn't on the corporate agenda until 1Q2021; CIB has declared dividends once per year since 2011.
Africa expansion plans still on course: CIB sees plenty of possibilities for Egypt in Africa, and believes that the country can provide benefits to its neighboring countries through cooperation, Ezz El Arab said. Africa has been at the heart of CIB's expansion plans — in addition to its majority stake purchase in Mayfair Bank. CIB, Egypt's largest private-sector bank, opened a representative office in Ethiopia last April.


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Biblioteca Nazionale di Napoli
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Daddy Yankee Breaks Down His 'Barrio Fino' Album Track by Track, 15 Years Later: Exclusive
By Ann Ferguson On 15 July 2019
The King of Reggaeton revisits each track of his album that kickstarted a global Latin movement.
Fifteen years ago, Daddy Yankee became the first reggaeton act to debut at No. 1 on Billboard's Top Latin Albums chart with Barrio Fino (released July 13, 2004). It would become the top-selling Latin album of the entire decade. Even more importantly, Barrio and its lead single, "Gasolina," detonated a global reggaeton explosion that irrevocably altered the business, sound and aesthetic of Latin music. (And that was way before "Despacito.")
It was an eminently commercial take on what was then an underground, subversive genre shunned by major labels. Reggaeton, with its aggressive, danceable dembow beat, had been percolating for years in Puerto Rico's underground. But it took Daddy Yankee's "Gasolina" to make it explode in the Latin market and beyond. With its intro of rapid-fire rapping over the sound of gunning motors, "Gasolina" climbed both Latin and mainstream charts and traveled the world, opening the door for a reggaeton movement led by Yankee, Don Omar, Wisin & Yandel, Tito El Bambino, and others. And it's still going strong.
But it all started with Barrio Fino — or "Elegant Barrio" in English — an album full of hits that liberally fused street reggaeton with other tropical rhythms. It wasn't just "Gasolina." Tracks like the infectious "Lo Que Pasó, Pasó," the haunting "Corazones" and the uplifting "Salud y Vida" are classics of Latin urban music. "It was a movement," says Yankee. "Barrio Fino brought glamour to the barrio. And it gave kids the possibility to say, 'Man, if Yankee can, I can.'"
Below, Daddy Yankee gives a track-by-track breakdown of his reggaeton classic, in his own words.
"Intro": "It's a poem by Gavilán [Hawk], an ex-convict poet who talks about the humble face of the barrio."
"King Daddy": "I describe my career and I also predicted this album would change the genre and validate my name — Daddy Yankee. King Daddy is also the title of my [2013] mixtape, which includes the hit 'La Nueva y la Ex.'"
"Dale Caliente": "A dance track inspired by dancehall music. That's why you hear the Jamaican in the background together with me the entire song."
"No Me Dejes Solo (Don't Leave Me Alone)" ft. Wisin y Yandel: "A club anthem. This track helped Wisin & Yandel gain international recognition because it was part of the promo video for Barrio Fino, along with 'Gasolina.'"
"Gasolina": "Innovative with a very catchy chorus. It describes the woman who likes to go out and have a good time. The chorus — 'A mi me gusta la gasoline, dame más gasoline" (I like gasoline, give me more gasoline) — is a very popular Puerto Rican saying which inspired the track. 'Gasolina' became the most famous track in the history of the [Latin] urban movement. Thanks to this song, the world got to know reggaeton. The word gasoline — everyone in the world knew what it meant. And I think part of the success of the track was people looking for some hidden meaning: Was I talking about alcohol, about drugs? That track is completely literal. It's the most innocent song I've ever written."
"Like You": "This was my first Spanglish track. A mix of R&B and reggaeton. I made the decision to write a track with English in it so I could be understood by people who liked reggaeton but didn't speak Spanish."
"El Muro (The Wall)": "After the fusion of 'Like You,' I needed to quickly get back to basics with something hardcore. This is the classic sound of our genre."
"Lo Que Pasó, Pasó (What Happened, Happened)": "The album's second single. It became a Latin anthem. The fusion of merengue and reggaeton give the music a Caribbean tropical sound."
"Tu Príncipe (Your Prince)" ft Zion y Lennox: "A romantic track that describes how someone falls in love with his best friend but is afraid to tell her because he doesn't want to ruin the friendship and lose her. This song got into radio on its own, thanks to the fans."
"Cuéntame (Tell Me)": "Because 'Tu Príncipe' was a romantic track, I made this one similar, to create a romantic block within the album and balance the production."
"Santifica Tus Escapularios (Bless Your Icons)": "A rap track allows me to vent against all spiritual evil."
"Sabor a Melao (Taste of Honey)" ft Andy Montañez: "A mix of reggaeton with salsa. I had the opportunity to record it with one of my musical heroes and salsa pioneer. I used one of the choruses he sang with the Batacumbele orchestra to create the track."
"El Empuje (The Push)": "After a fusion track, I once again balance the album, bringing back a classic hardcore reggaeton track."
"Qué Vas a Hacer? (What Are You Going to Do?)": "Speaks about violence against women. You hear the dialogue between man and woman the entire song. The chorus was recorded by a female singer and I do the verses. It's a blend of R&B and reggaeton."
"Salud y Vida (Health & Life)": "Latin hip-hop with Mexican banda influence. It's a West Coast sound. It speaks about the greatest treasure we have — and don't value enough: our health and living our life to the fullest above all else. The title, 'Health & Life,' became a popular slogan among Latinos. If you want to wish someone well, when you say goodbye you add "salud y vida."
"Interlude": "Gavilán reappears with another poem, this time speaking about the roughness of the barrio. It paves the way for the next track."
"Corazones (Hearts)": "'Corazones' describes how every heart in the world is different and feels different things. So intentions, emotions and even the most similar of situations are always unique to each individual. The track touches on politics, crime, hope and Christian spirituality."
"Golpe de Estado": "It's all about how I took total control over the movement with Barrio Fino. "
"Dos Mujeres (Two Women)": "A humorous reggaeton that describes how a guy maintains a relationship with two women, his wife and his lover. It describes each scene and the difference between them. This track was the inspiration behind 'La Nueva y La Ex.'"
"Saber Su Nombre (Know Her Name)": "It's a dancehall song in Spanish. These are the kinds of tracks you do for yourself, for your own enjoyment. I've been a big dancehall fan since I was a teen."
"Historia (Outro)": "I wrote this story by putting a face to all the names of the neighborhoods in Puerto Rico. I wanted to bring them to life in a story that's very humble but full of pride. It closes the album and brings together the sound and concept of the production."
A version of this article originally ran on Billboard.com in 2014.
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One of the questions native enthusiasts are asking of everyone, "Will you plant milkweed"?
Nature left to it own devices without constant intervention.
If returning areas to native flora and fauna, what things should one consider? We looked at this a bit in Whose Got the Blues? Native or Non-Native.
First, what is a native plant?
As defined by the USDA, "Native species" means, with respect to a particular ecosystem, a species that, other than as a result of an introduction, historically occurred or currently occurs in that ecosystem.
Is it even possible in a nature that is all about change to have an exclusively native plant habitat?
What is considered introduction and as to when?
Can plants grown hundreds of years ago find conditions suitable with the seemingly changing effects of climate change? Can we restore native plants where they are no longer adapted?
Do birds and insects adapt when confronted with hardship or do they seek out more favorable conditions?
Are humans more the problem than any other factor of change, even by precipitating change?
Should non-native plants be studied for benefit to a habitat BEFORE removal? Most restoration projects seemingly focus entirely on the native plant with less concern to the animals that inhabitant the area of study. Plant and they will come. Plant and some will leave. Seems simplistic.
Why is it that natural, restored habitats touted to be sustainable often seem to have to be perpetually managed? If historical prairies were man-made initially, how is it even possible to restore those "original" habitats?
Why is it that working habitats need to be completely destroyed to make more room for native plants?
Is it worth the money sunk into many of these projects that do not produce empirical results?
Is it refutable at all that insects prefer native plants?
Each question would make an interesting post. The funny thing about wildlife, they seem to make do. But the question always is how do we want to see wildlife? Like this?
Or this image below? Some things will always be beyond what humans are willing to do to make places native habitat. The tower above happens to be in a nature preserve. We see this on many occasions.
Is habitat all about native plants? For instance, many gardens don't have the native soils of woodlands and their exposure may not favor many of the woodland natives found today. Is it not more prudent assessing species based on what they do rather than from where they come?
I know myself I think of quite a few non-native plants as natives. I have seen them in meadows since I was a kid. Queen Anne's Lace, White Clover and Dandelion are considered naturalized. What a nice welcoming term. If it hinders land management practices, does it become invasive instead? Seems so.
I think what all this study is really missing is that there is too many uncertain factors playing in on the results to either side's findings, climate change being one, and humans themselves being the other. There are just too many of us with too many detrimental actions, the majority of whom just do not care. We will always put ourselves first.
Animals and plants that are native can become excessive in a habitat too. They are termed "aggressive". This also alters the ability of other native creatures to live and grow. It can be overlooked by those restoring a habit, returning it to "like it was meant to be", especially when unwilling to return all species that once kept a habitat in check. What determines what it "is meant to be," is nature.
See if you can see above the "detrimental, yet native" plant and animal?
It seems that some native studies only happen with the objective to prove some plant objectionable, they just keep looking until something negative is found. It seems more prudent not to remove plants providing insects and birds food and shelter, no matter whether native or non-native. There will always be organisms that lose if a habitat is taken back to some arbitrary time in history.
I enjoy having a garden filled with wildlife and make no apologies for having named cultivars of many native species. I make no apologies for having non-natives either. If the insects and birds use them, that is all that matters, at least to me that is. There will always be those with a different view.
I make a point to watch, listen and learn in my garden.
What if every landowner had a portion of their property like shown below? That would be great for wildlife, but it is just not feasible for most.
Having a garden of any size is a luxury in many parts of the world. Having preserved land as parks and forests are not found everywhere. In some places, the forest might be cut to heat a home, a field plowed to grow food, and the wildlife might be sacrificed to feed the family. It is all relative to circumstance, culture, and location, and coincidentally, how it was hundreds of years ago. Can we go back ourselves?
No doubt restoring native landscapes is a difficult path to travel. I can support restoration efforts, they just need to carefully consider more than the one plant they seek to remove.
Yes, this is knapweed, invasive or honeybee plant?
Should it be eliminated? Depends on who you ask and what organism they want to protect.
Appears that it is not only bees that like it either.
Knapweed is a problem so what gets done? Well they bring in an insect and populations are under control.
Knapweed is not the only invasive plant to tempt beekeepers. Beekeepers at one time were even encouraged to plant purple loosestrife to provide honeybees with flowers to visit late in the season. It also opened up wetlands to foraging since most wet meadows are considered unproductive for foraging bees. But is it ripe for removal when a plant clogs wetland waterways and displaces native plants and animals? Such a tricky dilemma when one sees bees utilizing a plant late in the season.
Bottom line…Here's a novel idea, let's just not asphalt over any more natural habitats, that itself might make a big difference. Think bigger in that more habitats need to be created not always destroyed. It is not JUST the plants in them, they need to be there in the first place. Maybe we just make better decisions as we move forward.
This entry was posted in garden, Invasive plants and tagged beekeeping and knapweed, environment, garden, Invasive plants, knapweed, natural landscape, nature, non-native plants, photos, plants for wildlife, Purple Loosestrife, wildlife garden. Bookmark the permalink.
Yes, a good article. We need to stop labeling things in nature I think (I don't mean in "categorizing" like you do in your profession). What is non-native one place is native somewhere else, where it breaks down simply to non-native equals bad, native denotes good. Like you noted, the Australian plant is a savior in times of dry. We just have to stop determining that a certain plant is bad, because it will surely have a benefit to some organism. Kudzu is the plant in the US that runs amok and there seems to be little to control it. And how did it get here?
Yes, it is more an issue of good plant/bad plant sometimes! There is even an endangered bird species (southwestern willow flycatcher) dependent on the invasive tamarisk. Invaders have the advantage that they can sometimes persist where natives cannot! They can provide supplemental or replacement habitat. Kudzu was accidentally introduced, I think.
Fabulous, thought-provoking post! Lots of excellent questions! The black swallowtail butterfly is one insect that has adapted. It uses fennel, parsley, Queen Anne's Lace as host plants, non of which are native. When you look at what we have created it is truly amazing how the animals and insects have adapted to our invasion. But can they continue to do so at such a rapid pace as our environment changes faster and faster?
The last question you noted is the absolute ultimate question. Will it be too fast a change? With what I have been reading, it seems it might. It seems to be that two issues (each with many sub-reasons built in), us and climate change are "the straw that broke the camel's back". We gobble up land for development at unprecedented amounts at the very same time the world is drying in places and getting washed out in others. Man has helped that along in so many ways too, not just our use of fossil fuel. Looking at just these two factors makes it apparent, we are at a critical point. Those that don't care to recognize will not even notice the changes around them, and that is the majority of people. We can only make small improvements. There will be those that say many small improvements make a large effect, but going against a nature in flux, I can't see it much mattering in the long term. Many are optimistic, but certainly not the ones that matter. We have a world of uncaring individuals. There, my soapbox, not like it matters anyway.
I agree with your bottom line. But I would like it if you wrote a post about each of the questions. It is an interesting subject. Beautiful photographs, also the one with all the birds on the metal tower. I prefer birds in trees to birds in metal towers (I don't know the proper word). But a metal tower with birds is always nicer than a metal tower without birds.
I had to laugh. You are right. I would not have taken the photo if a flock of Cormorants were not on this huge structure. But the electrical tower mars the preserve. They are having issues now with Eco-friendly wind turbines and birds flying into them and getting killed, our Bald Eagle is one of the birds that drew attention to this. Funny it took the bird of our nation to bring this to light. Many smaller birds have died too. They install these wind mills many times on flyways. It seems everything we do to confront issues of fossil fuel use seems to backfire in some way. Now studies are looking to relocate the towers to places where the birds are not migrating. But the reason they are there is the wind currents. Both the structure and birds need it.
At my Master Gardener meeting last night we had a speaker from a lavender farm in the area. He also keeps honeybees. Around the edges of his property he grows wildflowers, native and non-native, that bloom in the fall to help feed the bees after the lavender is finished at the end of summer. His farm is buzzing all the time.
I can hear the frustration in what you say. Development worldwide does not care for our wildlife whether it is big beasts or little ones. Creating wildlife corridors may be an answer although I fear that is too late already as it takes enormous effort on government. As gardeners we, intentionally or not, are creating wildlife corridors even though they are not ideal and can not appeal to all. The world can never have too many gardeners!
i so agree. The world needs more gardeners. I think of that in my job as I create gardens, but it is ultimately the homeowners that maintain them (or their gardeners). I cannot think of even one of hundreds that gardens pesticide free. So being a gardener is not always a positive thing.
I worked on wildlife corridor creating in Costa Rica. It was an amazing project. Wildlife bridges were built over roads to keep the wildlife from being roadkill. I was not there long enough to see if the bridges were used. I would believe the arboreal animals would use them like monkeys, but I always had my doubts on animals that had established ground trails they follow everyday. We were working with biologists and also landscape architects. I was there as an architect for building. Animals don't go where we expect them too. The corridor was laid out in likely paths of travel. Nature reserves are in Africa and the US, but how many times do you read of an elephant wandering off and being killed. The same with our wolves. They are a prime target leaving the preserve. There is nothing to do about this either. Plant corridors make sense for insects, but again it matters on maintenance. It also matters on migratory patterns and if you read the comment above, I noted how our best intentions can have dire consequence.
Phrases like "restoring land to how it was meant to be" are sentimental and not very useful. Yes, humans are just one species in this intricate ecosystem. But humans, like the members of any other species, are going to use the environment to their best advantage. Other creatures might overforage and eat plants in a small area until there is nothing left. Unlike other creatures, though, we can see the wide ranging and long-term consequences of what we do. We need to strike a balance to maintain the environment for ourselves and for future generations.
Yes, humans are just one species in this intricate ecosystem. But humans, like the members of any other species, are going to use the environment to their best advantage. Other creatures might overforage and eat plants in a small area until there is nothing left. Unlike other creatures, though, we can see the wide ranging and long-term consequences of what we do. We need to strike a balance to maintain the environment for ourselves and for future generations.
We are one, but also the most pervasive ( well technically that is wrong, insects are more numerous than us, but you know what I mean). I agree all species use what is available to them, but most of nature has checks and balances to keep species from overpopulating any one area if not interfered. Like the deer image I showed, if wolves were here, there would not be so many deer. We would have less pet cats and dogs though since habit is cramped and wolves would not stay in small patches of nature. Since we like our cats and dogs and our kids too, wolves are not allowed. Nature is just too complex. The talk of having a balance and being part of the web… well that seems, well, not even a option anymore. Do you really think us as a whole are preparing for future generations? I think sometime in the future those generations will not have much to say complimentary to our actions on nature. We look at previous generations for the harm they caused. They probably did the same. It is human nature to create, to push science, to build, to grow and expand. It is human nature to solves problems too. I hope I am around long enough to see the way climate change is dealt with. Maybe nature will just deal with us. A really simple and quick solution. Who knows?
This is an excellent synthesis of all the questions that should be asked and answered about the crusade against non-native plants and the fruitless effort to eradicate them.
It took me years of research to identify all of those issues. You are a quick study. Surely the time you spend in nature has accelerated your learning curve. If you walk in nature with your eyes open and with an open mind, you can't help but notice that non-native plants and trees are performing many valuable functions for humans as well as for animals.
Thank you for bringing these issues to the attention of your gardener colleagues.
Oh, I have been harping for a long time because I have been involved with professional projects on wetlands and even a summer in Costs Rica on creating wildlife corridors in Monteverde. Building has a lot of responsibility associated with it beyond the property the building or structure occupies. When I was in school, this was what stuck with me the most. What we do affects much around us, and how do we do it with less environmental impact? I don't talk about my real profession much on this blog since the blog is about nature, photography and gardening, but I have on occasion. It is usually prompted by a post I read where some home gardener goes off on a rant about some issue and takes such a one-sided or ethereal view. Then I might cite projects I worked on to show I am not just a home gardener, just to give a little credibility to my argument.
My favorite is always when creating urban meadows. I have seen books on this and the plants they use are many times non-native plants. They show bulbs from abroad and call it a meadow. I have talked with a horticulturalist and a entomologist to get their views. I have talked with naturalists. It humors me to think that a 15 foot by 15 foot space can be called a meadow. I have designed large meadows for commercial projects and spoke of that. It is NOT the same by a long shot as an urban front yard. I myself have my city yard almost completely planted with many species cultivars, but I never call it a meadow. I have stirred the pot many times.
Thank you for the mention on your blog. It is much appreciated..
Totally agree with put a hold on asphalt/concrete. in little rock the downtown has been reviving and many developers are trying and have succeeded in building, developing next to the ar . nothing wrong with some natural nature!
That has always been an issue for me in a building profession. Finding ways to minimize what is necessary by codes and ordinances. Parking is a necessary evil. Permeable is a better choice, but also more expensive to install. Large properties we do minimize lawns where we can too. The building have many Eco-friendly considerations as well, some by code others by long term cost savings. Always about money…..
Excellent info with terrific photos!
Oh, I hope we do "make better decisions as we move forward".
Me too. But the greater "we" is the stumbling block.
OK, here are my answers.
1. I'm not trying to have an exclusively native garden, so I wouldn't know. I grow lots of exotics, but I guess you could still call me a native plant enthusiast. My guess is most such enthusiasts are more like me than not.
2.I would say plants introduced with or after the arrival of Europeans.
3. Not sure what you mean by "plants grown hundreds of years ago". The natives I grow have grown continuously in this or a nearby region.
4. I'm not an expert, but it's fair to say that some birds adapt if they are capable of doing so, others will move to other areas if they can, and others will go into decline or extinction. I think that pretty well describes what has happened. There has been a decline in species diversity with the most adaptable species becoming the most numerous.
5. What problem do you mean? Certainly, humans cause environmental degradation. But I think it makes no sense to say (if I understand you) that because human activity caused the degradation, there should be no human activity to mitigate it. Using that reasoning there is absolutely no hope at all of mitigating climate change or any other environmental problem.
6. I would say both non-native and native species should be studied for their impact on the environment before action is taken regarding that plant.
7. For reasons you explain so well, it isn't possible to maintain any woods, prairie, wetland, etc. without substantial maintenance. Some processes, like prairie fires, can no longer occur without human management. And invasive plants will pop up over and over. That doesn't mean it's not worth managing some acreage to preserve this type of environment. And the amount of such acreage, sadly, is pretty tiny in the grand scheme of things.
8. Not sure what you mean by "working habitats". I would support removal of plants such as buckthorn or kudzu that tend to create monocultures or otherwise severely reduce diversity. I've seen this happen near my home and the results have been very positive – more plant species, more animal life.
9. I guess it depends on the project. I was under the impression that there have been some studies that can be used as a guide for the likelihood of a positive impact.
10. I followed your link and it sounds like there are conflicting results. I'd have to read more to have a better sense of the relative validity of the work referenced in your link and the research done by Douglas Talamy.
As always, thanks for an extremely thoughtful post with fantastic photos!!!
Wow, I guess I was not expecting anyone would address each question because there really is little answer to each that does not require documentation for the most part. Some have no real answer either. It was more an invitation for bloggers to adopt a question and research both sides to come to some consensus, rather than taking sides to this delicate issue.
It is like Laura did in the very first comment. She has practical experience and can, from a scientific viewpoint, show examples of things addressed in this post. Death of a Million Trees has tons of research compiled on their site, they have a perfect example of question 8 and I believe 9 too. I think question 5 is the answer itself – people are both the cause and the solution. I saw a study on plants long gone from a habitat that researchers were trying to reintroduce without luck. I have to try and find it again. That would be a reference for question 3, but I bet others could supply more examples than the one I found. This also goes hand in hand with question 9, money spent on a failed initiative.
I am very middle of the road on my thoughts and actions on the native/non-native issue because I think the problems in habitats are much larger than the plants. They are affected by too many external forces there is no way to measure or control. I find many quoting Tallamy's book for instance, like no other authority exists, yet many do. He is probably the most quoted author on planting native plants that I have read. I don't have any issue other than some insects will feed on non-native plants, so there is nothing wrong with planting some.
Like Karin, I grow fennel for the butterflies and I learned today that invasive Tamarisk benefits a flycatcher. That is my point, there is always much to learn and it takes learning from many different sources, often from both sides of the discussion. We as informed gardeners, do our best generally, but I bet in your neighborhood, you might be the only one not using pesticides. I believe I am in my neighborhood. I am also the only one to my knowledge using some native plants, well again, not the weedy versions of the plants.
Like I said in other posts, the home gardener is too entrenched in the Scotts line of garden products (and Monsanto products like Roundup) because of quick results, and will NEVER change as long as these products are manufactured. It is not a battle that can be won. I have had hundreds and hundreds of clients through the years, and know of none that do not have lawn services or do not use pesticides and herbicides. Even properties that have native plants installed still spray. People see insects and out comes the sprays. A caterpillar chewing is a goner in most home gardens. Do they have to spray? No, but it is a quick convenience and having a tidy garden better than the neighbors.
I commend you for putting forth answers to the questions. Maybe you should a do posts addressing the issues.
Maybe I will during the long, cold winter! I appreciate your response. By the way, have you ever read the novel "When the Killing is Done" by TC Boyle? It addresses exactly these issues and is basically sympathetic to both sides. If you are a fiction reader and haven't already read this, I recommend it.
No I have not, but will look into it. I like reading articles and books that look at both sides. Things are too complex to have such narrow views. I finally got some time and just started to get back to reader's blogs. Ironically, your blog was next. See you in a minute.
Ironic that the first plant is milkweed which is the one that monarch butterflies like and their paths are changing. Good questions posed here, one that we all need to deal with as climates and vegetation change. Also liked your comment about what we perceive to be a native plant because it has been there as long as we remember doesn't make it true.I will try to keep my mind open and read various views.
I placed the milkweed first in the post because it signals such an important issue this year. The Monarchs are in serious trouble with reducing numbers and vanishing habitat. It is a subject that native enthusiasts of all levels from those that dabble to the zealots are talking about. But just planting milkweed in gardens is not the issue. What I have read is they need it in greater numbers where they found it before, in the farming field hedgerows for instance. But again it is much more that has been affecting them including the uncertain factor or weather and the uncontrollable loss of their wintering grounds. I keep reading that home gardens should be planting native plants and while I wholeheartedly agree, gardeners are not going to save the Monarch I believe with too many other negative obstacles facing the butterfly.
The other point you brought up, yes, it is how I bet most people see these naturalized plants. Would we see meadows the same without Queen Anne's Lace? I would miss it and so would the insects. Check out posts from Million Trees. They offer a very balanced view.
I think it was you that wrote about our killer cats. Since then I read how bird populations are dwindling by "tens of thousands" and the biggest reason other than man made habitats and poisons was cats!
Yes, I featured an article by Cornell Lab of Ornithology director, John Fitzpatrick. It is funny cats are being singled out lately since this report has been around for quite a while, but you are right, bird numbers are dwindling and studies are looking closer for the reasons. BTW, yesterday I saw another Monarch in my garden. It is really late for them to be here. Last year they were traveling through Pennsylvania at this time.
Thank you, Natalie. Blessings to you also.
Ahh your questions are thought provoking, indeed. And I'm hoping many more posts will come from them. As for milkweed, I didn't plant it in my gardens, but I certainly do nurture it so that it will thrive. Five monarch sightings this year up from zero the year before… and more than a dozen eggs left behind.
Thank you. I too hope others take on the task and find a way to be objective. There is a lot of emotion in this subject and often, that is all that drives some to write about it. It is like Connie said above, when others say a habitat is to be "planted as it was meant to be", it is not helpful at all. That means nothing since no one can produce that state but nature herself.
I saw a Monarch in the garden yesterday. I think it is late for his travels, but glad he was on his way South.
So informative and thoughts that need to be put forward to 'restorers'. Great photography as well. Landscape photography always works if I feel that I would like to go there.
Thank you Diana. Living in Niagara Falls and a few minutes walk from the Falls, I can take lots of photos to make people visit. But like I think you might mean as a photographer, getting them to want to be in a place through imagination and the sense of place. I may never visit some of those iconic places around our country or the world, but have no problem imagining myself there with some of the images that photographers put out there for our enjoyment.
I had no idea that dandelions aren't native! This is a very interesting post, Donna. It's given me a lot to think about.
Your last point is so important Donna…let's stop invading every wild place and leave some. I think it is one reason I started planting more natives. I was tired of all the grass and chemicals and lack of wildlife. Now with a mix of both native and non, I have more wildlife to enjoy. Feels better than just some grass and bushes.


text:
Exodus: Movement of the People
Tag: Ernest
Immigrants Made Good – Ernest Cassel
Alec Tritton
Immigrants Made Good 1 Comment
As one of the richest men of his day, Sir Ernest Cassel left a lasting mark on Britain's history. He was a generous philanthropist, a wildly successful businessman, a close friend of royalty, and recipient of numerous honors and distinctions. Few would have expected such a legacy from a young, penniless German immigrant.
A portrait of banker and capitalist Sir Ernest Cassel. (Photo credit: Wikipedia)
Ernest was the youngest of three children born to Amalia and Jacob Cassel. He was born on March 3, 1852, in Cologne, Germany. His father owned a small bank, and Ernest received a standard education up until the age of 14. At 14, he quit school and started an apprenticeship under J.W. Eltzbacher, a local banker who specialized in financing foreign business and large industrial ventures. Ernest had a natural knack for business, and with his outstanding capacity for hard work he quickly learned the ropes in the world of finance.
Shortly before his seventeenth birthday, Ernest set out for England. He arrived in Liverpool penniless, but with plenty of ambition. He quickly found a job working for a firm of grain merchants where he was paid £2 a week.
His true gifts lay in the world of banking, and he was soon working in Paris for a bank. His stint in Paris was short lived though, as the Franco-Prussian War soon broke out. Since he was born in Prussia, Paris was no longer safe for him, and he was forced to return to England where he soon found work in a London bank.
He began working as a confidential clerk for Louis Bischoffsheim in the financial house of Bischoffsheim and Gildschmidt. He became a fast friend of the Bischoffsheim family, and this led to rapid promotion through the ranks of the financial firm. By the time he was 22, Ernest was managing the bank at a salary of £5000 plus commission.
In 1878, Ernest was married to Annette Mary Maud Maxwell in a ceremony at Westminster. They had one child – a daughter, who they named Amalia Mary Maud Cassel. Tragedy struck before long, when Annette contracted tuberculosis and died within three years of their marriage. Ernest's widowed sister and her children soon came to live in London, where she helped to run the household and look after little Amalia.
Ernest Cassel funerary monument, Kensal Green Cemetery, London (Photo credit: Wikipedia)
Despite Ernest's personal losses, he prospered in the offices of Louis Bischoffsheim and in 1884 he began putting together his own financial deals. He developed business ventures in Turkey at first, and soon had substantial ventures in Sweden, South America, Egypt, South Africa, and the United States. By 1898, his independent business was so successful that left Bischoffsheim and Gildschmidt to open his own offices.
Ernest developed a friendship with Lord Willoughby de Broke and they jointly started a stud farm for breeding race horses. They began racing their bred horses, and it was at the tracks that Ernest met the Prince of Wales (soon to become King Edward VII), and a fast friendship formed between them.
Ernest Cassel received a number of public recognitions as well as foreign decorations. The first was the K.C.M.G. awarded by Queen Victoria for his importance in the financial world. King Edward awarded him the further distinctions of K.C.V.O., G.C.M.G., G.C.V.O., G.C.B., and a privy councillorship.
Throughout his lifetime, Sir Ernest Cassel gave away nearly £2 million to public works and charities.
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text:
\section{Introduction}

The study of the Generalized Hamming Weights (GHWs) has been motivated by several applications in cryptography \cite{W:91} and they  characterize the performance of a linear code when it is used for a given channel. There are few families of codes for which it is known the complete generalized weight hierarchy as for example: first-order Reed-Muller codes, binary Reed-Muller codes, Hamming code and its dual, Extended Hamming codes, Golay code, see \cite{W:91}. On the other side,  there has been an extensive research on GHWs and the second distance in particular for some  classes of codes, see for example \cite{GHW1,GHW5,GHW9,GHW8,GHW2,GHW6,GHW10,GHW7,GHW4,GHW3} and the references therein. However, for the general case of a  linear code, few properties are known.

In their seminal paper \cite{JV:2013}, Johnsen and Verdure showed how the GHWs of a linear code could be computed from a free resolution of a monomial ideal associated to the set of codewords of minimal support of the code provided we know this last set. That paper has produced a great avenue of research, see for example \cite{JVrel6,GS:2020,JV:2014,JVrel7,JVrel5,JVrel4,JVrel3,JVrel2,JVrel1}.

In the present work,  we explore if one can find a set (smaller than  the set of all codewords of minimal support) that provides us some information on the GHWs in the case of binary codes. The selected set of codewords is the so called Gr\"obner test set related to the binomial ideal associated to a code defined in \cite{BBFM:2008}. In that paper it was proven that one can decode using that set and that the minimal distance of the code (thus the first GHW) was reflected on it. Thus, somehow some of the relevant information of the code lies in it. Moreover, in \cite{BBFM:2008} it was also shown how to compute the  Gr\"obner test set avoiding some of the most common disadvantages when one uses Gr\"obner basis. In this paper we will show how one can compute also the  second GHW of a binary linear code   from that binomial ideal associated with  the code without the need of computing the complete set of codewords of minimal support of the code as in \cite{JV:2013}.  Moreover, in Theorem~\ref{teo:gordo} we bound the remaining GHWs with the resolution of a monomial ideal associated to this new set.

The structure  of the contribution is simple, in Section~\ref{sec:intro} we revise some results on GHW, free resolutions and the ideal associated with a code. Section~\ref{sec:conjfalse} shows briefly the false conjecture and experiments that drive us to this study.
In Section~\ref{sec:main} we show our main result and finally in Section~\ref{sec:last} we show some future lines of research and some conjectures related to the topic that we hope will be helpful for future research.

\section{GHW and minimal supports}\label{sec:intro}

Let $\mathbb F_q$ be a finite field with $q$ elements. Given two vectors $\mathbf x = (x_1, \ldots, x_n)$ and $\mathbf y = (y_1, \ldots, y_n)\in \mathbb F_q^n$, the \emph{Hamming distance} between $\mathbf x$ and $\mathbf y$ is defined as
$$d_H(\mathbf x, \mathbf y) = |\{ i \mid x_i \neq y_i\}|,$$
where $|\cdot|$ denotes the cardinality of the set. The \emph{Hamming weight} of $\mathbf x$ is given by $\mathrm{w}_H(\mathbf x) =d_H (\mathbf x, \mathbf 0)$, where $\mathbf 0$ denotes the zero vector in $\mathbb F_q^n$. The \emph{support} of $\mathbf x$ is the set $\mathrm{supp}(\mathbf x) = \{ i \mid x_i \neq 0\}$.
A linear subspace $\mathcal C$ in $\mathbb F_q^n$ is called a \emph{linear code}. The elements of $\mathcal C$ are called \emph{codewords}. The \emph{basic parameters} of $\mathcal C$ are its length, its  dimension and its minimum distance, which are denoted by $n(\mathcal C)$, $k(\mathcal C)$ and $\delta(\mathcal C)$, respectively. In this case, we call $\mathcal C$ an $[n(\mathcal C), k(\mathcal C), \delta(\mathcal C)]_q$ linear code. We define a \emph{generator matrix} of $\mathcal C$ to be a  matrix $G$ over $\mathbb F_q$ of size $k(\mathcal C) \times n(\mathcal C)$ whose row vectors span $\mathcal C$, while a \emph{parity check matrix} of $\mathcal C$ is a matrix $H$ over $\mathbb F_q$ of size $(n(\mathcal C) - k(\mathcal C)) \times n(\mathcal C)$ whose null space is $\mathcal C$.

{
\begin{definition}
Let $\mathcal C$ be a linear code, we will say that a codeword $\mathbf m$ has minimal support if it is non-zero and $\mathrm{supp}(\mathbf m)$ is not contained in the support of any other codewords. We will denote by $\mathcal M_{\mathcal C}$ the set of codewords of minimal support of $\mathcal C$.
\end{definition}
}

{Note that computing a set of codewords of minimal support is a hard problem for a general linear code, in fact it implies finding the minimum distance of the code and, thus, it is harder that the problem of maximum-likelihood complete decoding, see \cite{Hard,B98}.}

\begin{definition}
	Let $\mathcal C$ be a linear code and $D$ be a subcode of $\mathcal C$, we define \emph{the support of $D$}, denoted $\mathrm{supp}(D)$ as the set of not-always-zero bit positions of $D$, i.e.,
	$$\mathrm{supp}(D) = \{ i \mid \exists \mathbf c \in D \hbox{ with } c_i \neq 0\}.$$
\end{definition}

It is clear that  if $D$ is a one-dimensional subcode then, the support of $D$ is equal to the Hamming weight of any of its   nonzero codewords, i.e.  $d_1(\mathcal C) = \delta (\mathcal C)$. Based on this idea, the \emph{$h$-th generalized Hamming weight}  of $\mathcal C$, denoted $d_h(\mathcal C)$, is the size of the smallest support of an $h$-dimensional subcode of $\mathcal C$ with $h=1, 2, \ldots, k(\mathcal C)$. That is, if $D_h$ is the set of all linear subspaces of the linear code $\mathcal C$ of dimension $h$, then
$$d_h (\mathcal C) = \min \{ |{\rm supp} (E)| \mid E \in D_h\}.$$
Some basic facts on the Generalized Hamming Weights (GHW) are provided in the following proposition.
\begin{proposition}[\cite{W:91}]
	Let $\mathcal C$ be a linear code. Then:
	\begin{enumerate}
		\item $1\leq d_1(\mathcal C) < d_2(\mathcal C) < \ldots < d_{k(\mathcal C)}(\mathcal C) \leq n(\mathcal C)$
		\item  {(Generalized Singleton Bound)} $d_h(\mathcal C) \leq n(\mathcal C) - k(\mathcal C) + h$
	\end{enumerate}
\end{proposition}
From now on we assume that   $\mathcal C$ is a nondegenerate code, that is,   $d_{k(\mathcal C)}(\mathcal C)~=~n(\mathcal C)$.

The GHWs $d_1(\mathcal C)$, \ldots, $d_{k(\mathcal C)}(\mathcal C)$ are completely determined by the underlying linear matroid structure of the code in a non-trivial manner, and the way to obtain them is not efficient as it was already pointed in \cite{JV:2013}. Of course, as pointed before, calculating $d_1(\mathcal C)$  is equivalent to the problem of complete decoding linear codes~\cite{B98}, hence  one can not expect a computational efficient approach.
From now on, given a positive integer $\ell,$ we define $\left[\ell\right]=\left\{1,\ldots,\ell\right\}$ and $\left[\ell\right]_0=\left\{0,\ldots,\ell\right\}$.

\begin{definition}
Let $\mathcal C$ be a $[n,k]_q$ code and let $H$ be a parity check matrix of $\mathcal C$. Let $H_i$ denote the $i$-th column of $H$ and define the simplicial complex
$$\Delta= \left\{ \sigma \in 2^{\left[n\right]} \mid \{ H_i \mid i \in \sigma\} \hbox{ is linearly independent over } \mathbb F_q\right\}.$$
Then, the pair $\mathcal M= (\left[n\right], \Delta)$ is the linear \emph{matroid associated with the code} $\mathcal C$. The collection $\Delta$ of subsets of $\left[n\right]$ are called \emph{independent sets} of this matroid. A subset of $\left[n\right]$ that does not belong to $\Delta$ is called \emph{dependent set}. Minimal dependent subsets of $\left[n\right]$ are known as \emph{circuits} of $\mathcal M$. A set is said to be a \emph{cycle} if it is a disjoint union of circuits.
\end{definition}
We refer to \cite{ES:2005} for a brief introduction on the theory of simplicial complexes, and to \cite{OX} for a thorough study of matroids.

\begin{definition}
Let $\mathbb K$ be any field. We denote by $I_{\Delta}$ the ideal in  the polynomial ring $R = \mathbb K [X_1, \ldots, X_n]$ over a field $\mathbb K$ generated by all square-free monomials supported on elements that are not in $\Delta$, i.e.
$$\prod_{i \in \tau} X_i \hbox{ with } \tau \in 2^{\left[n\right]} \setminus \Delta$$
That is, $I_{\Delta}$ is the ideal generated by monomials supported on the circuits of $\mathcal M$, or equivalently in a matroid associated with a code, supported on codewords of minimal support of $\mathcal C$.
\begin{equation}
I_{\Delta} = \left\langle \prod_{i \in \mathrm{supp}(\mathbf c)} X_i \mid \mathbf c \in \mathcal M_{\mathcal C}\right\rangle.
\label{ideal-matroid}
\end{equation}
\end{definition}
The quotient $R_{\Delta} = R / I_{\Delta}$ is called the \emph{Stanley-Reisner} ring associated with $\Delta$. $R_{\Delta}$ is a finitely generated standard graded $\mathbb K$-algebra of dimension $n(\mathcal C)- k(\mathcal C)$. Thus, it has a minimal graded free resolution. Moreover, since the generators of $I_\Delta$ are supported in the set of circuits of a matroid, by \cite{B:1992}, one has that $\Delta$ is shellable and this implies that $R_{\Delta}$ is Cohen-Macaulay. So, by the Auslander-Buchsbaum formula, the projective dimension of $R_{\Delta}$ (i.e., the length of any minimal graded free resolution of $R_{\Delta}$) is $k(\mathcal C)$ and it looks like
\begin{equation}
\label{MFR-eq}
\begin{array}{ccccccccccccccc}
0 & \longrightarrow & F_{k(\mathcal C)} & \longrightarrow & F_{k(\mathcal C)-1} & \longrightarrow & \cdots & \longrightarrow & F_1 & \longrightarrow & F_0 & \longrightarrow & R_{\Delta} & \longrightarrow 0
\end{array}
\end{equation}
where $F_0 = R$ and each $F_i$ is a graded free $R$-module of the form
$$F_i = \bigoplus_{j \in \mathbb N} R(-j)^{\beta_{i,j}} \hbox{ for } i \in \left[k(\mathcal C)\right]_0.$$
We will refer to Equation~\eqref{MFR-eq} as a graded minimal free resolution of $\mathcal C$.
The nonnegative integers $\beta_{i,j}$ are called \emph{Betti numbers} of $\mathcal C$ and they depend only on $\mathcal C$ and not on the choice of the parity check matrix $H$, or the minimal free resolution of $R_{\Delta}$ or  even the chosen  field $\mathbb K$ (this is because $I_\Delta$ comes from the set of circuits of a matroid, see \cite{B:1992}).

In \cite{JV:2013}, Johnsen and Verdure described the GHWs of a linear code $\mathcal C$ in terms of the shifts of the minimal graded free resolution of a Stanley-Reisner ideal $I_\Delta$ associated to $\mathcal C$. One of the main disadvantages of this method is that the generators of $I_\Delta$ correspond to the supports of all minimal support of $\mathcal C$. In general, the whole set of codewords of minimal support can be huge and computationally expensive to obtain and, in many cases, computing a minimal graded free resolution of an ideal with that many generators is unaffordable. More precisely, their result was:

\begin{theorem} \cite[Theorem 2]{JV:2013}
\label{Verdure-Teo}
Let $\mathcal C$ be a $q$-ary linear code. Then,
$$d_i(\mathcal C) = \min \{ j \mid \beta_{i,j} \neq 0\} \hbox{ for } j \in \left[k(\mathcal C)\right]$$
\end{theorem}

\begin{example}[Toy example]
\label{toy-example1}
Let $\mathcal C$ be the binary non-degenerate $[6,3]$ code with generator matrix
$$G=\left( \begin{array}{cccccc}
1 & 0 & 0 & 0 & 0 & 1\\
0 & 1 & 1 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 1 & 1
\end{array}\right) \in \mathbb F_2^{3\times 6}$$
One can check that the set of codewords of minimal support is
$$\mathcal M_{\mathcal C} = \left\{ \begin{array}{ccc}
w_1= 100001, & w_2=100110, & w_3=011010, \\
w_4= 000111, & w_5 = 111100, & w_6= 011101
\end{array}
\right\}$$
Its Stanley-Reisner ring is $R_{\Delta} = R / I_{\Delta}$ where $R=\mathbb F_2[x_1, \ldots, x_6]$ and the ideal $I_{\Delta}$ is generated by the monomials associated to $\mathcal M_{\mathcal C}$. If we compute a graded minimal free resolution of $R_{\Delta}$ we get the following Betti diagram:
\[
\begin{array}{c|cccc}
& 0 & 1 & 2 & 3  \\ \hline
0 &  1 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 \\
2 & 0 & 3 & 2 & 0\\
3 & 0 & 2 & 7 & 4 \\
\end{array}
\]
where the entry of the row indexed by $i$ and column indexed by $j$ indicates the value $\beta_{j,i+j}$ (for example $\beta_{3,5} = 7$).
Hence, by Theorem \ref{Verdure-Teo}, we have $d_1(\mathcal C) = 2$, $d_2(\mathcal C) = 4$ and $d_3(\mathcal C) = 6$.
\end{example}
}

This result shows that the determination of the Betti numbers of the monomial ideal  $I_{\Delta}$  related to a code completely determine the weight hierarchy. However, as mentioned before, this is usually a hard problem \cite{B98} except in some special cases.
For example, Johnsen and Verdure in \cite{JV:2014} explicitly determine the Betti Numbers for MDS codes, since the minimal free resolution of these codes is linear.
Moreover, in \cite{JV:2014} the authors prove that the resolution of the first order Reed-Muller code is pure. And a similar result can be deduced for constant weight codes \cite{GS:2020}. Thus, simplex codes or dual Hamming codes, which are constant weight codes also have pure resolution, although not necessarily linear.

\begin{remark}The resolution \eqref{MFR-eq} is said to be \emph{pure} of type $(d_0, \ldots, d_{k(\mathcal C)})$ if for each $i\in \left[k\right]_0$, the Betti number $\beta_{i,j}$ is nonzero if and only if $j= d_i$. If, in addition $d_1, \ldots, d_k$ are consecutive, then the resolution is said to be \emph{linear}.
\end{remark}

The rest of this work is devoted to compute a simpler and smaller structure than the whole set of codewords of minimal support that allows, at least partially, know something about the GHW of the code.

\subsection{Test-Sets of a binary code}

From now on, we will restrict our study to binary codes.
Let $\mathcal C$ be a binary linear code. Let $X$ be a vector with $n = n(\mathcal C)$  variables $x_1, \ldots, x_{n}$. A monomial in $X$ is a product of the form $X^{\mathbf a} := \prod_{i = 1}^n x_i^{\,a_i},$ where $\mathbf a = (a_1, \ldots, a_{n}) \in \mathbb N^{n}$. The total degree of $X^{\mathbf a}$ is $\deg(X^{\mathbf a}) = \sum_{i=1}^{n} a_i$. When $\mathbf a = \mathbf 0$, note that $X^{\mathbf a} = \mathbf 1$. The polynomial ring $\mathbb K[X]$ is the set of all polynomials in $X$ with coefficients in $\mathbb K$, where $\mathbb K$ denotes an arbitrary finite field.

\begin{remark}
For abuse of notation we will write $X^{\mathbf a}$ with $\mathbf a \in {\mathbb F_2^{n}}$. In this case, we understand that the classes of $0,1$ are replaced by the same symbols regarded as integers. Moreover,  we will use the notation $X^I$ for the square free monomial with support $I \subset [n]$, that is,
$$X^I = \prod_{i \in I} x_i \hbox{ with } I \subseteq \left[ n \right].$$
Let $g=X^A - X^B$ be a binomial with $A,B\subseteq \left[ n\right]$, we define $\mathrm{supp}(g) = A \cup B$. We say that $g$ is in \emph{standard form} if $A\cap B = \emptyset$ or, equivalently, if $\mathrm{supp}(g) = A \sqcup B,$ where $\sqcup$ denotes the disjoint union of $A$ and $B$.
\end{remark}

\begin{definition}
Let $\mathbb K$ be any field, we define the \emph{ideal associated with $\mathcal C$ over $\mathbb K$} as the binomial ideal:
\begin{equation}
I(\mathcal C) = \langle  X^{\mathbf a} - X^{\mathbf b} \mid \mathbf a, \mathbf b \in \mathbb F_2^n,\, \mathbf a + \mathbf b \in \mathcal C  \rangle + \langle x_i^2 - 1 \mid i \in [ n ] \rangle \subseteq \mathbb K[X].
\label{ideal-code}
\end{equation}
\end{definition}
Note that $I(\mathcal C)$ is a zero-dimensional ideal since the quotient ring $R= \mathbb K[X] / I(\mathcal C)$ is a finite-dimensional vector space (i.e. $\dim_{\mathbb K}(R)< \infty$). Moreover, its dimension is equal to the number of \emph{cosets} in $\mathbb F_2^{n} / \mathcal C$.
For $\mathbf a, \mathbf b \in \mathbb F_2^n$ one has that $X^{\mathbf a}  - X^{\mathbf b} \in I(\mathcal C)$ if and only if $\mathbf a - \mathbf b \in \mathcal C$.
The following result shows how to obtain a set of generators of the ideal $I(\mathcal C)$ from a generator matrix of $\mathcal C$.

\begin{proposition}\cite[Theorem 1]{BBFM:2008}
Let $\{\mathbf w_1, \ldots, \mathbf w_k\}$ be the row vectors of a generator matrix for $\mathcal C$. Then
$$I(\mathcal C) = \left\langle \left\{ X^{\mathbf w_i}-1\right\}_{i \in \left[ k \right]} \cup \left\{ x_i^2 -1\right\}_{i \in \left[ n \right]} \right\rangle$$
\end{proposition}

If we fix a term order $\prec$, then the \emph{leading term} of a polynomial $f$ with respect to $\prec$, denoted by $\mathrm{LT}_{\prec}(f)$, is the largest monomial among all monomials which occur with non-zero coefficient in the expansion of $f$. Let $I$ be an ideal in $\mathbb K[X]$, then the \emph{initial ideal} $\mathrm{in}_{\prec}(I)$ is the monomial ideal generated by the leading term of all the polynomials in $I$, i.e., $\mathrm{in}_{\prec}(I) = \langle \left\{ \mathrm{LT}_{\prec}(f) \mid f\in I\right\} \rangle$. By definition, $\mathrm{in}_{\prec}(I)$ is a monomial ideal and, thus, it has a unique minimal generating set formed by monomials. These monomials will be called {\it minimal generators of} $\mathrm{in}_{\prec}(I)$.

\begin{definition}
A finite set of nonzero polynomials $\mathcal G=\{ g_1, \ldots, g_m\}$ of the ideal $I$ is a Gr\"obner basis of $I$ with respect to the term order $\prec$ if the leading terms of the elements of $\mathcal G$ generate the initial ideal $\mathrm{in}_{\prec}(I)$. Moreover $\mathcal G$ is reduced if
\begin{enumerate}
\item $g_i$ is monic (i.e., its leading coefficient is $1$) for all $i \in \left[m\right]$, and
\item none of the monomials appearing in the expansion of $g_j$ is divisible by $\mathrm{LT}_{\prec}(g_i)$ for all $i \neq j$.
\end{enumerate}
\end{definition}

For a given monomial order $\prec$, every ideal has a unique reduced Gr\"obner basis (see, e.g., \cite{AL}).
Since $I(\mathcal C)$ is generated by binomials (differences of monomials), then all its reduced Gr\"obner bases consist of binomials (see \cite{ES:96}). In \cite{BBFM:2008} it is shown that, if  $\mathcal{C}$ is a binary code and we fix a degree compatible term order $\prec$ on $\mathbb K[X]$, then the  reduced Gröbner basis $\mathcal G_\prec$ for the code ideal $I(\mathcal C)$ can be computed by a linear algebra (an FGLM-like) algorithm. Moreover, the reduction provided by $\mathcal G_\prec$ gives a decoding procedure. Along the way, they also prove that the support of every binomial in $\mathcal G_\prec$ different from $x_i^2 - 1$ for $i = 1,\ldots,n$ provides a codeword of minimal support of $\mathcal C$, and that there is a word of Hamming weight $d_1(\mathcal C)$ that can be obtained in this way. More precisely:

\begin{proposition}\cite{BBFM:2008}\label{pr:BBFM} Let $\mathcal G_\prec$ be the reduced Gr\"obner basis of $I(\mathcal C)$ with respect to a degree compatible term order $\prec$. For every binomial $X^{\mathbf a} - X^{\mathbf b} \in \mathcal G_\prec - \{x_i^2 - 1\, \vert \, i \in [n]\}$, then $\mathbf a + \mathbf b \in \mathbb F_2^n$ is a codeword of minimal support of $\mathcal C$. Moreover, there exists $X^{\mathbf a} - X^{\mathbf b} \in \mathcal G_\prec$ such that $\mathrm{w}_H(\mathbf a + \mathbf b ) = d_1(\mathcal C)$.
\end{proposition}
This result motivates the definition of a $\mathcal G_{\prec}$-test, which by the above proposition is a subset of the set of codewords of minimal support and contains a word of minimum weight.

\begin{definition}
Given a binary code $\mathcal C$ and a degree compatible term order $\prec$, we will call $\mathcal G_\prec$-test set of $\mathcal C$ to the subset of codewords of minimal support of $\mathcal C$ whose supports are given by those binomials in the reduced Gr\"obner basis of $I(\mathcal C)$ different from $x_i^2-1$
for all $i \in [n]$.
\end{definition}

\begin{example}
\label{toy-example2}
We continue the toy Example  \ref{toy-example1}. Then, the ideal associated with $\mathcal C$ over $R$ is defined as
$$I(\mathcal C) = \left\langle \left\{\begin{array}{ccc}
x_1x_6-1, &x_2x_3x_5-1, &x_4x_5x_6-1
\end{array}\right\} \cup \left\{ x_i^2-1 \right\}_{i \in \left[ 6 \right]}\right\rangle \subseteq R$$
Now we consider the degree reverse lexicographic order $\prec$ with $x_6\prec\ldots \prec x_1$. Then, the reduced Gr\"obner basis $\mathcal G_{\prec}$ of $I(\mathcal C)$ with respect to $\prec$ has $14$ elements. An a $\mathcal G_{\prec}$-test-set of $\mathcal C$ is given by codewords of $\mathcal M_{\mathcal C}$ whose supports are given by those binomials of $\mathcal G_{\prec}$ different from $x_i^2-1$, i.e. the set
$$ \left\{ \begin{array}{cccc}
w_1=100001, & w_3 = 011010, & w_4 = 000111, & w_6=011101
\end{array}\right\}\subseteq \mathcal M_{\mathcal C}.$$

\end{example}
}

\section{On a false conjecture}\label{sec:conjfalse}
 Note that the second and the third author of this work have conjectured for a long time the following: \\[0.025em]

\noindent \textit{ \textbf{Conjecture (false)} If one considers the monomial ideal $M$ associated with the supports of the binomials in the $\mathcal G_\prec$-test set, then $d_i = {\rm min}\{j \, \vert \, \beta_{i,j}(R/M) \neq 0\}$ for $i \in \{1,\ldots,k(\mathcal C)\}$, that is, the $\mathcal G_\prec$-test set determines the GHWs of the codes.}\\[0.025em]

  \noindent This conjecture was supported by  computational evidence (see Examples \ref{toy-example3} and \ref{ex:ej1} for some example satisfying this conjecture),  but unfortunately this is not true as shown in the counterexample~\ref{ex:contraej}. Theorem \ref{teo:gordo} in Section~\ref{sec:main} in this paper  will prove this fact for $i = 2$ (and it was known for $i = 1$ \cite{BBFM:2008}). Note also that in \cite{Irene}, the authors show that from the Graver basis associated to $I(\mathcal C)$ one can purge the set of codewords of minimal support of $\mathcal C$, i.e. $\mathcal M_{\mathcal C}$.

	\begin{example}
		\label{toy-example3}
		We continue the Example \ref{toy-example1} where we have computed the Betti diagram associated to $R_{\Delta} = R / I_{\Delta}$ where $I_{\Delta}$ is generated by the monomials associated to $\mathcal M_{\mathcal C}$. Then, in Example \ref{toy-example2} we compute a $\mathcal G_{\prec}$-test-set $\mathcal T_{\mathcal C}$ of $\mathcal C$ with respect to the degree reverse lexicographic order $\prec$. Note that $\mathcal T_{\mathcal C}\subset \mathcal M_{\mathcal C}$ with just $4$ elements. If we consider $M$ the corresponding monomial ideal related to $\mathcal T_{\mathcal C}$ and compute a graded minimal free resolution of $R / M$ we get the following Betti diagram:
		\[
		\begin{array}{c|cccc}
		& 0 & 1 & 2 & 3  \\ \hline
		0 &  1 & 0 & 0 & 0 \\
		1 & 0 & 1 & 0 & 0 \\
		2 & 0 & 2 & 1 & 0\\
		3 & 0 & 4 & 4 & 2 \\
		\end{array}
		\]
		The Betti numbers of $R/M$ are smaller than those of $R_{\Delta}$ and in this example the sequence
		\[ ({\rm min}\{ j \, \vert \, \beta_{i,j}\neq 0 \}; 1 \leq i \leq 3) = (2,4,6), \]
		coincides with the GHW of $\mathcal C$
	\end{example}

	The following example is a less trivial case where the difference between our structure $\mathcal T_{\mathcal C}$ and the set of codewords of minimal support of $\mathcal C$, $\mathcal M_{\mathcal C}$, is larger.

\begin{example}\label{ex:ej1} Let $\mathcal C$ be the binary non-degenerate $[14,9]$-code with generator matrix
	\[G = \left(
	\begin{array}{cccccccccccccccccccccc}
	1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\
	1 & 1 & 1 & 1 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 1\\
	0 & 0 & 1 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 0\\
	1 & 1 & 0 & 1 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 1\\
	0 & 1 & 1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1\\
	0 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 1\\
	0 & 0 & 1 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 & 0 & 1 & 0\\
	0 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0\\
	1 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0\\
	\end{array} \right) \in \mathbb F_2^{\,9 \times 14}\]

	Its Stanley-Reisner ring is $R_{\Delta} = R/I_{\Delta}$ where $R=\mathbb F_2[x_1, \ldots, x_{14}]$ and $I_\Delta$ is minimally generated by $147$ monomials. If we compute a graded minimal free resolution of $R_{\Delta}$ we get the following Betti diagram:

	\[
	\begin{array}{c|ccccccccccc}
	& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline
	0 &     1   &  0 &    0  &   0  &   0  &   0   &  0  &   0     & 0    & 0 \\
	1 &     0   &  2 &    0  &   0  &   0  &   0   &  0  &   0     & 0    & 0 \\
	2 &     0   &  8 &    5  &   0  &   0  &   0   &  0  &   0     & 0    & 0 \\
	3 &     0   & 34 &   82  &  48  &   8  &   0   &  0  &   0     & 0    & 0 \\
	4 &    0   & 52 &  441  & 897  & 753  & 289   & 42  &   0     & 0    & 0 \\
	5 &     0   & 51 & 1345  & 7410 &18309 & 25248 & 21008 & 10579 & 2990 &  366

	\end{array}
	\]
	Hence, by Theorem \ref{Verdure-Teo}, we have that the GHW are $$(d_1(\mathcal C),\ldots,d_9(\mathcal C)) = (2,4,6,7,9,10,12,13,14).$$
	Moreover, if we consider $\prec$ the graded degree lexicographic order with $x_1 \succ \cdots \succ x_{14}$, we get that the $\mathcal G_\prec$-test set has 24 elements. If we consider $M$ the corresponding monomial ideal and compute a graded minimal free resolution of $R / M$ we get the following Betti diagram:

	\[
	\begin{array}{c|ccccccccccc}
	& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline
	0 &    1   &  0  &   0  &   0 &    0   &  0   &  0   &  0  &   0    & 0 \\
	1 &    0   &  2  &   0  &   0 &    0   &  0   &  0   &  0  &   0    & 0 \\
	2 &    0   &  6  &   3  &   0 &    0   &  0   &  0   &  0  &   0    & 0 \\
	3 &    0   & 13  &  38  &  17 &    2   &  0   &  0   &  0  &   0    & 0  \\
	4 &    0   &  3  &  92  & 194 &  130   &  35  &   3  &   0 &    0    & 0 \\
	5 &    0   &  0  &  83  & 599 & 1410   & 1621 & 1040 &  378 &    71  &   5 \\
	6 &    0   &  0  &   0  &   0 &    2   &   5  &   4  &   1  &    0   &  0 \\

	\end{array}
	\]
	As one can observe, the Betti numbers of $R/M$ are smaller than those of $R_\Delta$. Moreover, in this example the sequence
$ ({\rm min}\{ j \, \vert \, \beta_{i,j}\neq 0 \}; 1 \leq i \leq 9)$ is given by $ (2,4,6,7,9,10,12,13,14)$ and it
	coincides with the GHW of $\mathcal C$.
\end{example}

\begin{example}\label{ex:contraej}
	Let $\mathcal C$ be the binary non-degenerate $[10,7]$-code with generator matrix
	$$G = \left(\begin{array}{cccccccccc}
	1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 1 \\
	1 & 1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1 \\
	1 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 0 \\
	0 & 0 & 1 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \\
	1 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 \\
	0 & 0 & 1 & 1 & 0 & 1 & 1 & 1 & 0 & 0 \\
	0 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 \\
	\end{array}\right) \in \mathbb F_2^{\,7\times 10}$$

	Its Stanley-Reisner ring is $R_{\Delta} = R/I_{\Delta}$ where $R=\mathbb F_2[x_1, \ldots, x_{10}]$ and $I_\Delta$ is minimally generated by $42$ monomials. If we compute a graded minimal free resolution of $R_{\Delta}$ we get the following Betti diagram:
	$$
	\begin{array}{c|cccccccc}
	& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline
	0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
	1 & 0 & 4 & 0 & 0 & 0 & 0 & 0 & 0\\
	2 & 0 & 18 & 48 & 32 & 7 & 0 & 0 & 0\\
	3 & 0 & 20 & 214 & 637 & 874 & 637 & 242 & 38\\

	\end{array}
	$$-/
	Hence, by Theorem \ref{Verdure-Teo}, we have $d_1(\mathcal C) = 2$, $d_2(\mathcal C) = 4$, $d_3(\mathcal C) = 5$, $d_4(\mathcal C) = 6$, $d_5(\mathcal C) = 8$, $d_6(\mathcal C) = 9$, $d_7(\mathcal C) = 10$.

	If we compute a $\mathcal G_\prec$-test set $T$   with respect to the degree reverse lexicographical order, we get that $T$ has only 10 elements. Computing a minimal graded free resolution of $R/M$ being the ideal $M := \langle \{ X^{\mathbf c} \, \vert \, \mathbf c \in T \} \rangle$,   we get the following Betti diagram.
	$$
	\begin{array}{c|ccccccc}
	& 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
	0 & 1 & 0 &  0 & 0 & 0 & 0 & 0 \\
	1 & 0 & 4 & 0 & 0 & 0 & 0 & 0 \\
	2 & 0 & 4 & 14 & 5 & 0 & 0 & 0 \\
	3 & 0 & 2 & 23 & 56 & 48 & 17& 2 \\
	\end{array}
	$$

	Thus, we can recover the correct values of $d_1(\mathcal C), d_2(\mathcal C), d_3(\mathcal C)$ from this resolution but not $d_i(\mathcal C)$ for $i =4,5,6,7$.
\end{example}
Anyway, the experiments we dealt trying to prove that conjecture drove us the results in Section~\ref{sec:main} and to propose some further conjectures and future work based on experimental  evidences showed in Section~\ref{sec:last}.

\section{Second GHW obtained from a $\mathcal G_\prec$-test set}\label{sec:main}

In this section we will explain how to compute the second generalized Hamming weight from a $\mathcal G_\prec$-test set. Throughout this section, $\mathcal C$ will denote   a binary linear code and let $\mathcal G_{\prec}$ be the reduced Gr\"obner basis of the ideal $I(\mathcal C)$ with respect to $\prec$, where we take $\prec$ to be any degree compatible ordering on $\mathbb K[X]$.
The following result is a technical lemma whose proof is just an easy exercise in set theory. We will denote by $A\triangle B$ the \emph{symmetric difference} of the subsets $A, B \subseteq [n]$, which is the set of elements which are in the union of the two sets $A\cup B$, minus their intersection $A\cap B$. Moreover, for a set $A$, its cardinality is denoted by $|A|$.

\begin{lemma}
\label{SymetricDifLemma}
Let $A,B \subseteq X$ with $|A\cap B|> \frac{|A|}{2}$. Then $C=A\triangle B$ satisfies the following statements:
\begin{enumerate}
\begin{multicols}{3}
    \item $A\cup B=A\cup C$
    \item $|A\cap C| < \frac{|A|}{2}$
    \item $|C|< |B|$.
\end{multicols}
\end{enumerate}
\end{lemma}

Consider the set $\mathbf M$ of codewords in $\mathcal C$ belonging to a linear subspace of dimension two of minimal support, more precisely,
\begin{equation}
 \mathbf M = \{\mathbf m \in \mathcal C \, \vert \, \exists \mathbf m' \in \mathcal C \text{ such that }
d_2(\mathcal C) = \mathrm{supp}(\left \langle \mathbf m, \mathbf m' \right\rangle)\},
\label{eq:minimal} \end{equation}
and define $\mathbf m_1, \mathbf m_2 \in \mathbf M$ as follows:
\begin{itemize}
\item[(a)] $\mathbf m_1 := {\rm min}_{\prec}(\mathbf M)$, i.e., $\mathbf m_1$ is the smallest codeword with respect to $\prec$ in $\mathbf M$,
\item[(b)] $\mathbf m_2 := {\rm min}_{\prec}\{\mathbf m \in \mathbf M \, \vert \, d_2(\mathcal C) = \mathrm{supp}(\left \langle \mathbf m_1, \mathbf m \right)\rangle\}$, i.e, $\mathbf m_2$ is the smallest codeword with respect to $\prec$ such that
$d_2(\mathcal C) = \mathrm{supp}(\left \langle \mathbf m_1, \mathbf m_2 \right\rangle)$.
\end{itemize}
With these conditions we define:
$$\begin{array}{ccc}
I = \mathrm{supp}(\mathbf m_1) & \hbox{ and } &
J= \mathrm{supp}(\mathbf m_2).
\end{array}$$

\begin{remark}\label{remark1}
Since $\prec$ is degree compatible and, by Lemma \ref{SymetricDifLemma}, we have that $$|I\cap J|\leq \frac{|I|}{2}\leq \frac{|J|}{2}.$$
\end{remark}

\begin{proposition}\label{Prop1}
There exists a binomial $f\in \mathcal G_{\prec}\subseteq \mathbb K[X]$ such that $\mathrm{supp}(f) = I$.
\end{proposition}

\begin{proof}
Consider $f=X^{I_1}-X^{I_2} \in I(\mathcal C)$ any binomial with $I=I_1 \sqcup I_2$, $|I_1| - 1 \leq |I_2| \leq |I_1|$ and $X^{I_1}\succ X^{I_2}$, we will show that $f\in \mathcal G_{\prec}$. For proving this, it suffices to check that:
\begin{itemize}
\item[(a)] $X^{I_1} = \mathrm{LT}_{\prec}(f)$ is a minimal generator of $\mathrm{in}_{\prec}(I(\mathcal C))$, and
\item[(b)] $X^{I_2} \notin \mathrm{in}_{\prec}(I(\mathcal C))$.
\end{itemize}

Proof of (a). By contradiction suppose that there exists a binomial $h = X^{K_1}- X^{K_2}\in \mathcal G_{\prec}$ with
$K=K_1 \sqcup K_2$,
$\mathrm{LT}_{\prec}(h) = X^{K_1}$ (and, in particular, $|K_1| \geq |K_2|$) such that $K_1 \neq I_1$ and $X^{I_1}$ is divisible by $X^{K_1}$, i.e. $K_1 \subsetneq I_1$.

\smallskip

\noindent \emph{Claim:} $d_2(\mathcal C) \leq |I\cup K| -1$.

\noindent \emph{Proof of the claim:} We have that $|K_2| \leq |K_1|  \leq |I_1| - 1 \leq |I_2|$. Then $|K| < |I|$ and, in particular, $X^K \prec X^I \prec X^J$. Hence, by the choice of $\mathbf m_2$, we have that $|I \cup  J| < |I \cup K|$ and the \emph{Claim} follows.

By the previous \emph{Claim} we have that
\begin{eqnarray*}
|I \cup J| & = & d_2(\mathcal C) \leq |I\cup K| -1 ~~~~ \\
&=& |I\cup K_2|-1 ~~~~ \hbox{(since } K_1 \subsetneq I \hbox{)} \\
&=& |I| + |K_2| - |I\cap K_2|-1\\
&\leq &|I| + |K_1|- |I\cap K_2|-1 ~~~~ \hbox{(since } X^{K_1} \succ X^{K_2} \hbox{)}.
\end{eqnarray*}
Thus,
$$|I| + |J| - |I \cap J| = |I \cup J| \leq |I| + |K_1|- |I\cap K_2|-1$$
or equivalently $|J| - |I \cap J| \leq |K_1|- |I\cap K_2|-1$. Thus,
\begin{eqnarray*}
|I_1| - \frac{1}{2}  & \leq & \frac{|I|}{2} \leq  \frac{|J|}{2} \leq |J| - |I\cap J| ~~~~ \hbox{ (by Remark \ref{remark1})}\\
& \leq & |K_1|- |I\cap K_2|-1 \leq |K_1|-1.
\end{eqnarray*}
Therefore, $|I_1| \leq |K_1| - \frac{1}{2}$ and $K_1\subsetneq I_1$, a contradiction.

Proof of (b). By contradiction, we assume that there exists $h= X^{K_1} - X^{K_2}\in \mathcal G_{\prec}$ such that $\mathrm{LT}_{\prec}(h)=X^{K_1}$ divides $X^{I_2}$ or, equivalently, $K_1 \subseteq I_2$. Then $K_2 \prec K_1 \prec I_2 \prec I_1$ and, in particular, $X^K \prec X^I \prec X^J$.  Hence, by the choice of $\mathbf m_2$, we have that $|I \cup  J| < |I \cup K|.$
\begin{eqnarray*}
d_2(\mathcal C) &=& |I \cup J| \leq |I\cup K|-1 = |I \cup K_2|-1 = |I| + |K_2|-|I\cap K_2|-1 \\
&\leq& |I| + |K_2|-1 \leq |I|+ |K_1|-1.
\end{eqnarray*}
Therefore, $|J|-|I\cap J|\leq |K_1|-1$. And, by Remark \ref{remark1} we deduce that
$$|I_2|\leq \frac{|I|}{2}\leq \frac{|J|}{2}\leq |J|-|I\cap J| \leq |K_1| - 1.$$
Therefore, $|I_2|<|K_1|$ and $K_1\subseteq I_2$, which is a contradiction. \qed
\end{proof}

One can check that the same result (and the same proof) holds for all $I'$ such that $I' = {\rm supp}(\mathbf m)$ with $\mathbf m \in \mathbf M$ and $|I'| = |I|$. As a consequence of this observation we have that:
\begin{corollary}
If $|I| = |J|$. Then, we can always find binomials $f,g\in \mathcal G_{\prec}\subseteq \mathbb K[X]$ such that $\mathrm{supp}(f) = I$ and $\mathrm{supp}(g) = J$.
\end{corollary}
 We just found that $I$ is always involved in the supports associated with $\mathcal G_{\prec}$ and that if $J$ has the same cardinal as them, also the second GHW can be derived from the Gr\"obner basis. Let us prove now the general case.

 \begin{proposition}\label{Prop2}
There exists a binomial $g\in \mathcal G_{\prec}\subseteq \mathbb K[X]$ such that  $\mathrm{supp}(g) = J$.
 \end{proposition}

 \begin{proof} By Remark \ref{remark1} we know that $|I \cap J| \leq |J|/2$, then one may consider $$g=X^{I\cap J}X^{J_1}-X^{J_2} \in I(\mathcal C ), $$
 such that $J_1\cup J_2=J-I$, $J_1\cap J_2=\emptyset$ and $|J_2| + 1 \geq |I \cap J| + |J_1| \geq |J_2|$. Our goal is to prove that $g$ (or $-g$) is in $\mathcal G_{\prec}$. We split the proof in two:
 \begin{description}
\item[Case ${\rm LT}_{\prec}(g)= X^{I\cap J}X^{J_1}$.] We are going to see that $g \in \mathcal G_{\prec}$.
First, we show that $X^{I\cap J}X^{J_1}$ is a minimal generator of ${\rm in}_\prec (I(\mathcal{C}))$. By contradiction, suppose that there is a binomial $h=X^{K_1}-X^{K_2}\in \mathcal G_\prec$ with
$\mathrm{LT}_{\prec}(h) = X^{K_1}$ such that $K_1 \subsetneq (I\cap J)\cup J_1$, and denote $K=K_1 \sqcup K_2$.
We have that $|J_2| \geq |I \cap J| + |J_1| - 1 \geq |K_1| \geq |K_2|$ and, in particular, $|J| > |K|$.

However,
\[ \begin{array}{lcl}
d_2(\mathcal C) & = &  |I\cup J|=|I|+|J|-|I\cap J|\\
 & = & |I|+ |J_1|+|J_2| \\
 & \geq & |I|+|J_1|+|K_2|\\
 & \geq & |I\cup K|, \end{array}\]
which cannot happen by the choice of $J = {\rm supp}(\mathbf m_2)$. Therefore, $X^{I \cap J} X^{J_1}$ is a minimal generator of ${\rm in}_\prec (I(\mathcal{C}))$.

 Now, we will show that $X^{J_2}\notin {\rm in}_{\prec}(I(\mathcal{C}))$.
 Since $X^{I\cap J}X^{J_1}$ is a minimal generator of ${\rm in}_{\prec}(I(\mathcal{C}))$, then there exists $h=X^{I\cap J}X^{J_1}-X^{L}\in \mathcal G_{\prec}$. We have that  $X^L \notin {\rm in}_{\prec}(I(\mathcal{C}))$ and let us see that $L = J_2$. Suppose that $L \neq J_2$, then  $0 \neq h - g = x^{J_2} - x^L \in I(\mathcal C)$ and $X^{J_2} \succ X^L$, because $X^L \notin {\rm in}_{\prec}(I(\mathcal{C})).$ However, \[ |I \cup {\rm supp}(h)| \leq |I| + |J_1| + |L| \leq |I| + |J_1| + |J_2| = |I \cup J| = d_2(\mathcal C),\] which is a contradiction.

\item[Case ${\rm LT}_{\prec}(g)=  X^{J_2}$.] We are going to see that $-g \in \mathcal G_{\prec}$.  First, we show that $X^{J_2}$ is a minimal generator of ${\rm in}_\prec (I(\mathcal{C}))$. Suppose that there exists a binomial $h=X^{K_1}-X^{K_2} \in\mathcal G_\prec$ with $X^{K_1} \succ X^{K_2}$ and $K_1 \subsetneq J_2$. Consider $L := J\triangle K$, we have that $L \subseteq (J - K_1) \cup K_2$ and, hence, $X^L \prec X^{J - K_1} X^{K_2} \prec X^J$. However, \[ |I \cup L| \leq |I \cup (J - K_1) \cup K_2| \leq |I \cup J| - |K_1| + |K_2| \leq |I \cup J| = d_2(\mathcal C),\] and again it is
a contradiction.
\end{description}

 We will show now that $X^{I\cap J} X^{J_1}\notin \mathrm{in}_\prec (I(\mathcal{C}))$, by contradiction. Suppose that $X^{I\cap J} X^{J_1}\in \mathrm{in}_\prec (I(\mathcal{C}))$, then there exists a binomial $X^{I\cap J} X^{J_1}- X^{L} $ in $I(\mathcal{C})$. Consider now $K=(I\cap J)\cup J_1\cup L$, again one can compute that $|I\cup J|\geq |I\cup K|$ and that $K\prec J$, which again contradicts the choice of $\mathbf m_2$.

 \qed
 \end{proof}
 From the above results the main theorems of this paper will follow.

 \begin{theorem}\label{Thmdeglex}
Let $\prec$ a degree compatible order
in $\mathbb K[X]$, then there are $f, g$ in the reduced Gr\"obner basis $\mathcal G_\prec$ of $I(\mathcal C)$ with respect to $\prec$, such that $d_2=|{\rm supp}(f) \cup {\rm supp}(g)|$.
\end{theorem}

\begin{example}\label{ex:ejth2} Let $\mathcal C$ be the binary non-degenerate $[6,3]$-code with generator matrix
\[G = \left(
\begin{array}{cccccccccccccccccccccc}
1 & 0 & 0 & 1 & 1 & 0 \\
0 & 1 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 & 1
\end{array} \right) \in \mathbb F_2^{\,3 \times 6}\]

One can check that every codeword different from $0$ is a codeword of minimal support and, hence, there are $7$ codewords of minimal support, namely $w_1 = 111000, w_2 = 110011, w_3 = 101101, w_4 = 100110, w_5 = 101110, w_6 = 010101, w_7 = 001011$. Moreover, we have that
$d_1(\mathcal C) = 3$ and $d_2(\mathcal C) = 5$ and $d_3(\mathcal C) = 6$ and the set $\mathbf M$ described in (\ref{eq:minimal}) coincides with $\mathcal C - \{0\}$.

Consider the degree reverse lexicographic order $\prec_1$ with $x_6 \prec_1 \cdots \prec_1 x_1$, then \begin{itemize} \item $\mathbf m_1 := \min_{\prec_1}(\mathbf M) = 001011 = w_7$, and  \item $\mathbf m_2 := {\rm min}_{\prec_1}\{\mathbf m \in \mathbf M \, \vert \, d_2(\mathcal C) = \mathrm{supp}(\left \langle \mathbf m_1, \mathbf m \right)\rangle\} = 010101 = w_6.$ \end{itemize} The reduced Gr\"obner basis $\mathcal G_1$ of $I(\mathcal C)$ with respect to $\prec_1$ has 20 elements. As proved in Proposition \ref{Prop1}, the binomial $f = x_5 x_6 - x_3$ belongs to $G_1$ and has ${\rm supp}(f) = {\rm supp}(w_7) = \{3,5,6\}$ and, by Proposition \ref{Prop2}, the binomial $g = x_4 x_6 - x_2$ belongs to $\mathcal G_1$ and has  ${\rm supp}(g) = {\rm supp}(w_6) = \{2,4,6\}$. Moreover, $d_2 = |{\rm supp}(f) \cup {\rm supp}(g)| = |{\rm supp}(\langle w_6,w_7\rangle)| = |\{2,3,4,5,6\}| = 5$.

If we consider the degree reverse lexicographic order $\prec_2$ with $x_1 \prec_2 \cdots \prec_2 x_6$, then \begin{itemize} \item $\mathbf m_1' := \min_{\prec_2}(\mathbf M) = 111000 = w_1$, and  \item $\mathbf m_2' := {\rm min}_{\prec_2}\{\mathbf m \in \mathbf M \, \vert \, d_2(\mathcal C) = \mathrm{supp}(\left \langle \mathbf m_1', \mathbf m \right)\rangle\} = 100110 = w_4.$ \end{itemize} The reduced Gr\"obner basis $\mathcal G_2$ of $I(\mathcal C)$ with respect to $\prec_2$ has 20 elements. As proved in Proposition \ref{Prop1}, the binomial $f' = x_1 x_2 - x_3$ belongs to $ \mathcal G_2$ and has ${\rm supp}(f) = {\rm supp}(w_1) = \{1,2,3\}$ and, as proved in Proposition \ref{Prop2}, the binomial $g' = x_1 x_4 - x_5$ belongs to $\mathcal G_2$ and has  ${\rm supp}(g) = {\rm supp}(w_4) = \{1,4,5\}$. Moreover, $d_2 = |{\rm supp}(f') \cup {\rm supp}(g')| = |{\rm supp}(\langle w_1,w_4\rangle)| = |\{1,2,3,4,5\}| = 5$.

\end{example}
As a consequence, we have that the two first GHWs $d_1(\mathcal C)$ and $d_2(\mathcal C)$ can be obtained from the  minimal graded free resolution  associated with the supports in the $\mathcal G_\prec$-test set.
Moreover, from this resolution one  can also obtain upper bounds for all the  $d_i(\mathcal C)$ where  $i \in \left[ k(\mathcal C) \right]$. More precisely, we have the following.

\begin{theorem}\label{teo:gordo} Let $\mathcal C \subseteq \mathbb F_2^n$ be a binary code, $\prec$ a degree compatible monomial order in $R$. Let $T$ denote the $\mathcal G_\prec$-test, define the square-free monomial ideal \[ M := \langle \{ X^{\mathbf c} \, \vert \, \mathbf c \in T \} \rangle  \subseteq R,\] and consider
 a minimal graded free resolution of $R/M$:
\[ 0 \longrightarrow F_p \longrightarrow \cdots \longrightarrow  F_2  {\longrightarrow}  F_1 \longrightarrow R \longrightarrow R/M \longrightarrow 0,  \]
where each $F_i$ is a graded free $R$-module of the form
$$F_i = \bigoplus_{j \in \mathbb N} R(-j)^{\beta_{i,j}(R/M)} \hbox{ for } i \in \left[p\right]_0.$$
Then,
\begin{itemize}
\item[{\rm (a)}] $p \leq k(\mathcal C)$
\item[{\rm (b)}] $d_i(\mathcal C) \leq {\rm min}\{j \, \vert \, \beta_{i,j}(R/M) \neq 0\}$ for all $j \in \{3,\ldots,p\}$, and
\item[{\rm (c)}] $d_i(\mathcal C) = {\rm min}\{j \, \vert \, \beta_{i,j}(R/M) \neq 0\},$ for $i = 1, 2$.
\end{itemize}
\end{theorem}
\begin{proof} By Proposition \ref{pr:BBFM}, every $\mathbf c \in T$ is a codeword of minimal support and, hence, $\{X^{\mathbf c} \, \vert \, \mathbf c \in T\}$ is a subset of the generators of the monomial ideal supported on all the codewords of minimal support. Thus, (a) and (b) follow from  Theorem \ref{Verdure-Teo}.

The rest of the proof concerns (c). Again by Proposition \ref{pr:BBFM}, $\{X^{\mathbf c} \, \vert \, \mathbf c \in T\}$ is the minimal monomial generating set of $M$. Since $\beta_{1,i}(R/M)$ equals the number of minimal generators of $M$ of degree $i$, then,
$\beta_{1,i}(R/M) = | \{\mathbf{c} \in T \, \vert \, \mathrm{w}_H(c) = i\}. |$
By Proposition \ref{pr:BBFM}, there is a $\mathbf c \in T$ such that $\mathrm{w}_H(\mathbf c) = d_1$ and, thus, \[d_1(\mathcal C) = {\rm min}\{ \mathrm{w}_H(\mathbf c) \, \vert \, \mathbf c \in T\} = {\rm min}\{i \, \vert \, \beta_{1,i}(R/M) \neq 0\}.\]

Assume now that $T = \{\mathbf c_1,\ldots,\mathbf c_r\},$ and consider $\mathcal T$ the Taylor resolution of $M = \langle X^{\mathbf c_1},\ldots,X^{\mathbf c_r} \rangle$. The first steps of this resolution are given by
$$\mathcal T: \cdots \longrightarrow  F_2'  \overset{\varphi_2}{\longrightarrow}  F_1' \longrightarrow R \longrightarrow R/M \longrightarrow 0,$$
where $F_i' := \oplus_{|I| = i \atop I \subset [r]}  R (-|{\rm supp}\langle \mathbf c_j\, \vert \, j \in I \rangle|)$ and $F_1' := \oplus_{1 \leq i \leq r}  R (-|{\rm supp}(\mathbf c_i)|).$
Hence, the shifts in the second step of $\mathcal T$ are given by $|{\rm supp}\langle \mathbf c_i, \mathbf c_j \rangle|$ for $1 \leq i < j \leq r$ and, as a consequence, $d_2(\mathcal C) \leq {\rm min}\{|{\rm supp}\langle \mathbf c_i, \mathbf c_j \rangle|\, \vert \, 1  \leq i < j \leq r\}.$ Moreover, by Theorem \ref{Thmdeglex}, this is indeed an inequality.

In general, the Taylor resolution is not minimal (it is usually very far from minimal). However, it can be pruned to get a minimal one. Consider now $$\mathcal F: \cdots \longrightarrow  F_2  \longrightarrow  F_1 \longrightarrow R \longrightarrow R/M \longrightarrow 0,$$
a minimal graded free resolution of $R/M$ obtained after pruning the Taylor one. As we said before, $\{X^{\mathbf c_1},\ldots,X^{\mathbf c_r}\}$ is the minimal monomial generating set of $M$ and, hence, $F_1' = F_1$. As a consequence, ${\rm min}\{i \, \vert \, \beta_{2,i}(R/M) \neq 0\} = {\rm min}\{|{\rm supp}\langle \mathbf c_i, \mathbf c_j \rangle| \, \vert \, 1 \leq i < j \leq r\} = d_2(\mathcal C).$
\qed
\end{proof}

\section{Final remarks, future work \& new conjectures}\label{sec:last}

 In this work, we build on the results of Borges-Quintana et al. \cite{BBFM:2008} and propose $\mathcal G_\prec$-test sets as a smaller structure from where one can obtain the values of $d_1(\mathcal C)$ and $d_2(\mathcal C)$ for binary codes.
Several experiments with SageMath \cite{sagemath} suggest that Theorem \ref{teo:gordo} can also be extended for $i = 3$. More precisely:

\begin{question}
Let $\mathcal C \subseteq \mathbb F_2^n$ be a binary code, $\prec$ a degree compatible monomial order in $R$. Consider the $\mathcal G_\prec$-test set $T$ and define the square-free monomial ideal \[ M := \langle \{ X^{\mathbf c} \, \vert \, \mathbf c \in T \} \rangle  \subseteq R.\] Is
$d_3(\mathcal C) = {\rm min}\{i \, \vert \, \beta_{3,i}(R/M) \neq 0\}$?
\end{question}

In Example \ref{ex:contraej} one has that the projective dimension (i.e., the number of steps of the resolution) of $R/M$ is ${\rm pd}(R/M) = 6$, while the dimension of $\mathcal C$ is $k(\mathcal C) = 7$. In all the counterexamples to the original conjecture that we have found it turns out that ${\rm pd}(R/M) < k(\mathcal C)$. This motivates us to ask if the conjecture holds provided that ${\rm pd}(R/M) = k(\mathcal C)$. More precisely:

\begin{question}
Whenever ${\rm pd}(R/M) = k(\mathcal C)$, is it true that \[ d_i(\mathcal C) = {\rm min}\{j \, \vert \, \beta_{i,j}(R/M) \neq 0\} \] for all $i \in \{1,\ldots,k(\mathcal C)\}$?
\end{question}

Also, the following natural questions arise.
\begin{question} What is in between the test set and the complete set of codewords of minimal support? i.e. Can we characterize a mid-way structure that provides the complete set of GHWs?
\end{question}

A possible candidate fot that intermediate set could be the union of all $\mathcal G_\prec$-test sets for all $\prec$ degree compatible orderings. In general, this set can be smaller than the whole set of codewords of minimal support { and can be computed by the algorithm proposed in \cite{Natalia}}. For example, for the $[7,4]$ binary Hamming code, i.e., the code with  generator matrix

\[ \left( \begin{matrix}
1 & 0 & 0 & 0 & 0 & 1 & 1 \\
0 & 1 & 0 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 & 1 & 1 & 1 \\
\end{matrix} \right) \in \mathbb F_2^{\,4 \times 7},
\]
has $14$ codewords of minimal support, half of them with  weight $3$, and the rest with   weight $4$. Moreover, the sequence of GHW is $(d_1(\mathcal C),\ldots,d_4(\mathcal C)) = (3,5,6,7)$. One has that all the $\mathcal G_\prec$-test sets when $\prec$ ranges over all degree compatible orderings consists of the $7$ codewords of Hamming weight $3$. If one computes the Betti diagram of the monomial ideal $M \subset \mathbb F_2[x_1,\ldots,x_7]$ corresponding to this set, one gets the following

$$
\begin{array}{c|ccccccc}
& 0 & 1 & 2 & 3 & 4  \\ \hline
0 & 1 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0\\
2 & 0 & 7 & 0 & 0 & 0 \\
3 & 0 & 0 & 21 & 21 & 6 \\
\end{array}
$$
Hence, the sequence
$ ({\rm min}\{ j \, \vert \, \beta_{i,j}\neq 0 \}; 1 \leq i \leq 4) = (3,5,6,7)$
coincides with the sequence of GHWs of the code $\mathcal C$.

\begin{question} Can we say something in the non-binary case?  \end{question}
	In order to answer this question, one could try to apply the techniques in \cite{Irene}, where a generalization of the ideal $I(\mathcal C)$ for non-binary codes is studied.

\begin{question} Recently,  E. Gorla and A. Ravagnani in \cite{GR22} extend and generalize the results of Johnsen and Verdure \cite{JV:2013} to compute the generalized weights of a code with respect to a different notions of weight. It would be interesting to see if these generalized weights can be computed from a $\mathcal G_\prec$-test set of $\mathcal C$.  \end{question}

\section*{Acknowledgments}

We would like to thank E. Gorla (University of Neuchatel, Switzerland) for her valuable comments and suggestions.

\bibliographystyle{splncs04}

redpajama_set_name: RedPajamaArXiv


It's Sunday evening and I am absolutely exhausted! I feel a little out of sorts, considering all of our furniture is gone (except the bed)!
Trying to make Bean-bag chairs out of plastic bags and linens isn't exactly the most comfortable way to live, but it works for about 5 minutes... then I have to sit on my bed as my back kills me if I'm sitting anywhere else.


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