Hey friends today, we are hanging out at Disney's Hollywood Studios There are a bunch of new food items and menu changes that I wanted to come out and show you I'm finally gonna try the famous Cocoa puff top to black calf that they used to only sell in Disneyland now They sell it here and then also sci-fi has a whole bunch of new burgers So we've got a reservation a little bit later on and I am so excited Anywho's let's go do this Today is June 2nd and like I said there are a lot of new food items and most of the restaurants have new menus So we're gonna try to explore as many as we can, but I'm excited to try that Star Wars coffee probably the most For the longest time they were only selling this coffee over in Disneyland at Galaxy's Edge And whenever I went over to Disneyland, I never really spent time in there because I could do it here So everybody always says how amazing it is and we're finally gonna get to try it I think we're gonna head straight down to Galaxy's Edge to get the black calf coffee But like I said, we do have a reservation for sci-fi dine-in theater a little bit later on And I'm excited because there's a whole new menu here They've got some wings some spinach artichoke dip and a couple of new burgers and it's gonna be exciting on top of all the new treats and Menu updates. It's also the start of pride month So there's a lot of photo opportunities and specially pride snacks So we might try a couple of those I do hear about a new pop tart over at Woody's lunch box And that's kind of like a seasonal item. They always have a seasonal flavor. So we might check that one out as well On our way down to Galaxy's Edge I wanted to show you some of the cool pride photo opportunities that you can find right here Where streets of America used to be and Muppets Courtyard is they have these really awesome pride Mickey balloons And you just stand up there and hold them The cold brew black calf is located at Katzaka's here at Hollywood Studios, but in Disneyland I think it was over at docking Bay 7 so I'm excited to see what it's like like I said This is the first time it's ever been here And here it is they actually have a non-alcoholic version and a regular version and Here is the menu the cold brew black calf $5.49 Joffreys coffee cold brew topped with sweet cream cheese and chocolate puffs And then you get the spiked version and it comes with Godiva chocolate liquor for $14 So I'm pumped. I think I'm just gonna go with the regular one. I don't need the the Godiva chocolate liquor at the moment Look at that back there a whole bunch of cocoa puffs Oh, thank you. Wow, they look so good That's the way oh, yes All right, so here they are this is the spiked version it comes in a clear cup and then this is the non-spiked version and it's basically just the Joffreys cold brew coffee cocoa puffs and a cream cheese foam and look at how it's just going down like that And this one has Godiva chocolate liquor in it So we're gonna try them both But I wanted to get this one but I felt bad because this one's actually in a see-through cup But that's how you tell whether or not things have alcohol on it We're gonna try them both and they said it's best when you mix it around But you got to take your photos with them first so I'm gonna mix around the one that has the spike one There it is That's how you do we got to mix it around it looks really good though, right? It's like a cereal on top Oh, wait, it is a cereal on top Cheers to boozy coffee Oh, that's good. Oh, wow I don't taste the Godiva chocolate liquor or I might have gotten a little cream cheese foam in my beard Now we're gonna try the other one and this one I don't know I can't wait to see which one I like better You know what I mean? I'm sure the chocolate liquor has a little effect on the taste, but we're gonna we're gonna test it out We're gonna give it the official taste test Cheers I like this one better This one is really really good. There's a slight kickback on that one, but this one just it's it's very creamy very delicious Mmm, I also suggest that you maybe scoop out the cocoa puffs like cereal and eat it like cereal. Look at that It's breakfast and coffee in one Wait a second no joke this is amazing like I can just sit here at eating I wonder if you can get extra cocoa puffs because that would be so awesome, but I love this I think I'm gonna start having coffee in my cereal Honestly, this might be my favorite part right here All right now that the coffee business is taking care of and let me tell you something I love that coffee not the spike version, but the regular version I can see myself getting it a lot more whenever I come over here I think we're gonna make our way over to Toy Storyland and see about a Pop-Tart Maybe like everything all breakfast we have cereal coffee Pop-Tarts As we're making our way over to Toy Storyland It seems the stormtroopers are actually out and about so I want to go ask them if they got a chance to try the cold brew black calf Did you guys try the cocoa puff coffee? No, no coffee It looks like the stormtroopers are super cranky today. They definitely did not have their coffee this morning And here we are Woody's lunchbox. Oh by the way, I'm with Kristen right down Main Street And we're gonna get a Pop-Tart. We had coffee together. Did you like the coffee? I love the coffee. I thought it was great. I was so excited to try it I've been eyeing it since I saw it on the menu at Disneyland so I'm really happy it's here Yeah, oh good good now. Let's go get that Pop-Tart The lack of shade in Toy Storyland is so fierce It is extremely hot out right now and standing in line for Woody's lunchbox Because when you mobile order they give you a return time and the mobile order pickup times like an hour from now So we're gonna try to do the walk-up, but it is so hot And here it is the Pride Pop-Tart This is the seasonal Pop-Tart here at Woody's lunchbox and look at a little Mickey on there It looks so fancy. Normally I don't like the Pop-Tarts. I think they're a little bit dry But I'm intrigued to see what the filling is on this one It's a guava-filled pastry coated in purple vanilla fondant. That's it Guava-filled. Oh guava. Oh, and JoJo's here. Hey everyone. Guava Look at us. We're all like hovering around that gather around the Pop-Tart Yep All right, we're giving it a taste test here Still a little dry But you can taste the guava at least like you can really taste it I don't know. Maybe they should have had a little bit more filling to it What do you think JoJo? I honestly don't mind it. It's pretty good as for a Pop-Tart I like the little blue thing on it. Yeah, more flavor. Oh, what do you think?  Oh, I know. Oh She's processing It's good. It's not my favorite one that they have. I still like the lemon blueberries my favorite Raspberry is number two, but this one is good. I love the outside like dough part the pastry dough or whatever It reminds me of like an Italian cookie that they fill with jelly back home. Oh It's a pop of the pie It's a pop of the pie The best thing about coming to the parks and trying snacks with a bunch of friends means you don't need to buy all The snacks you can all share them. So that's really cool that JoJo and Kristen are here So that way we don't eat too much snacks and we don't have to spend a lot of money And we all get to try a little bit of them That's the way that is the way Now we're gonna move along and I noticed that we were here before 5 o'clock and last time I came to Hollywood Studios I got really excited to go meet Sully, but then I found out he only works 9 a.m. To 5 p.m. So today I think we're gonna do it We're gonna meet Sully and it's JoJo's first time meeting Sully JoJo! I love the queue to actually meet Sully because all the lights change and there's a bunch of different concept arts Look the lights are changing now. It's a very cool effect And then you have to go through a door and meet Sully like you actually have to go through a door And I guess it would be my door There it is that must be my door It looks like the meter is pretty full too Yes, please tell me the joke All right Well, there he is That's the way Very fancy, how are you good? Oh, you like the hat? I love it Well, I'm just so happy to see you today. I really am. Can we do a hug? Yeah, oh, yes, I'm all about it Oh Yo, the best hug right here the best hug Thanks, pal A lot of paperwork Always watching Thank you Yeah, tell boo I said hello too. Oh Man the best bye Sully. Thank you so much I Absolutely love meeting up with Sully. How cool was that? Jojo showed Sully And he jumped back. He was like, what is that a freak because they know what it was. Yeah And I said it was a night light. Yeah kind of is it kind of is a night light People use them as night lights.  I have a little monster fur in my mouth.  Yeah, so you're gonna like it Yeah, we're gonna watch some horror movies and just chill out and 1930s California now, I know what you think you think in charge. Did you just say California? That's so far away Well, don't worry for us movie magic. I'll get you there in seconds. All right Filming in progress do not enter when red light is flashing. I love it in here so much Looks like we're gonna be sitting in the back at the picnic tables We could sit down on the table, but it would have been a tight fit because there's three of us so I kind of like sitting back here anywhere because It's a little bit more well lit I Love how you have like your little speaker right here so you can hear the the movie Then you just get to watch Seriously, this is such a cool experience in here Here comes Walt Disney Here is a look at the menu and like I said some of the new items are the sci-fi smoked wings smoked chicken wings tossed in Porter barbecue sauce and a housemaid pineapple poppy seed coleslaw for $15. They also have a black garlic Caesar salad and a Spinach artichoke dip for $13 and then the featured film burger is a surf and turf burger Which is their classic beef patty with a shrimp and crab cake on top I always like to get the feature film burger, but I'm not too sure if I like crab cakes So I don't know we have to think about it and also they have tons of really amazing desserts So we're gonna try to get as much as we can We're definitely gonna be getting the smoked wings and I'm torn between the featured Burger, which is the surf and turf burger and then the drive-in barbecue burger I do love me a barbecue burger, but the featured burger the surf and turf is the special so like Yeah, kind of tossing a turn in there News of the future Don't they make that noise The news for the future Baby Last chance to visit the snack bar It used to be out front they actually used to give the free popcorn out Oh looks like here comes the wings on fans. Oh wow Those look fantastic Oh That's the spinach. Oh Wow The wings definitely look very very saucy very saucy. Is that Tom and Jerry look at that is Tom and Jerry up there Very saucy saucy wings, so I can't wait to dive into those really need lots of napkins And then the spin dip looks good, too. Got some chips there. I don't know. I guess we're gonna dive in We're gonna dive into the wings, but like I said, they're very saucy and I'm kind of shocked that it is $15 for four wings. They get four wings, but I'm going in for the first bite here prepare for a mess prepare for the wing mess They're better than I thought they would be but I need a wet nap Look at it.  I can't touch it It's the off button right there. No, that's it that way Yep, that's it $14. I mean $15 for four wings. I can't justify it.  That's an expensive chicken. That's an expensive chicken wing Now I'm gonna try to spin dip and I'm not the biggest fan of spinach artichoke dips So this is kind of adventurous for me, but I'm gonna give it a go Jojo will you hold that way I dip I got you now watch me dip dip It's very clever we're very fast very clever people here. So here we go So it's pretty good though, yeah, I don't think I would order it but that's because I don't like spinach it but I like it's not bad Looks like the burgers have arrived and I love the little plates that they get lucky. It has sci-fi on it Oh, it's cute. It's like little car plates. Yeah, everyone's got the good stuff. Yes, sir I can't believe that's an impossible burger. It looks so realistic doesn't it? Yeah, it's like Yeah, I was gonna try to make a joke about it being impossible. That's impossible And then you got the barbecue burger just like me. Yes, sir. I'm so excited. We're diving in All right, here we go. I'm going in for a bite on the barbecue a barbecue driving burger It's got both pork on there and everything Oh That's so good Wow Look at that Can I have an onion ring? Yeah And look at the inside of that burger there Vegan it's impossible. So impossible. I Do have to say I should have gotten the onion rings though. I think it's a two-dollar up charge But I think it's well worth it. They're unique. They're very very good The pulled pork on the burger is a nine-hour Smoked pulled pork nine hours and it comes with the steak fries which the steak fries are good It's just those onion rings are better. I Love the different horror movies that they actually show here. Look at this. It's like a double feature Space Mickey should be popping up any second now Oh There he is Sci-fi was actually really fun I just love sitting in there watching the movies the burgers were good the wings were good They were just expensive like they said I think was four dollars a wing four dollars for one wing I mean, I know the price of wings went up But that's that's crazy and now we didn't get dessert in there I think we're gonna head down and get another dessert down at scoops And just kind of share it because I'm kind of full. I mean, we're all full those burgers are very filling like I said, we're gonna make our way down sunset and Over to scoops because they have a special milkshake that they just brought out and I'm pretty sure it's a key lime Milkshake with the donut on top like the orange bird one that they had so I'm excited Also, I wanted to point out the Captain America shard be on the lookout the Captain America line with Roosevelt drops June 9 And it's probably gonna sell out fast. Like it's a popular thing. I'll probably post Instagram So follow me on Instagram and follow Roosevelt if you want to find them in it's your first time I have a promo code PMM 20 for your first order 20% off, but I love it. Look at this very very fancy Hollywood scoops ice cream is always super busy here It's always a line and it's actually pretty good ice cream. They always have unique flavors I'd like to see them working up in there. Look at them. I don't know. I see an orange ice cream with a cone on it. Oh That's not me I'm a hide over here As we're waiting for our ice cream from Hollywood scoops I wanted to point out the Tower of Terror is still the longest wait in the whole entire park still holding on strong to that top spot 130 minute waits 130 and it still looks like only half the ride is running because you see the elevator doors are only opening on this side I don't think you're opening on that side. So that means half the the ride is running at half capacity And here it is the key lime milkshake Wow, this actually looks phenomenal. I love how they add like a little donut on top of it They're like you that's what you need a little donut with your milkshake. How awesome is that? I like how they have little crumbs there on the side. I can't wait to try it. We're gonna have to dive in All right, we're going in for a sip here That's actually a really good milkshake key lime milkshake. I didn't think it would be good, but this is good Yeah, it's definitely not a heavy heavy milkshake it is very light But I do like the key lime taste to it and we're gonna dive into the donut Even though the donut's starting to fall in like it's starting to sink in there, right? But you got to get the donut in there you got to dunk the donut dunking donuts dunking don't know Sean's gonna dunk the donut I'm gonna do a little dunk of donuts too. There we go. Oh Yeah I think I like the donut better than the milkshake Yeah Let's take a look at some of the wait times smugglers run is 75 minutes rise of the resistance is closed Mickey and Minnie's run over where we're always only 30 minutes Rock and roller coaster was a hundred minutes Slinky dog dashes a hundred minutes Toy Story mania 45 minutes So it's kind of like if he I feel like if rise of the resistance was open a lot of these wait times would be a little bit lower You know what I mean because rise of the resistance takes a big chunk of that wait time and it being down makes everything else busy I always feel like Slinky dog dash and Tower of Terror always the busiest rides here other than rise All right now that we got all of the snack coverage out of the way I would love to go ride a ride. Let's go see what we can get on Toy Story midway mania or Mickey's Mickey and Minnie's runaway railway whatever has a low wait time Looks like Mickey and Minnie's runaway railway is like a 10 minute wait It says 40 minutes, but when there's no queue outside that basically means it's like a walk-on and look at this It's exactly what's happening right now Look at how crazy this is Seriously It's so funny when the wait times are posted and they're just so inaccurate 40 minutes is not here Look at that We're going right on in Well, it wasn't exactly a walk-on It was about maybe a 20 minute wait I would say because you had to go through the pre-show, but we're here. They keep me trying to run away away Oh Now that we got to ride Mickey and Minnie's runaway railway it looks like it's gonna start raining So we're gonna go seek some shelter probably at the best place to seek shelter during a thunderstorm at Hollywood Studios baseline oh No, it looks like the rain is coming and baseline is closed So we have to find somewhere else, but we're gonna get caught in the rain We really wanted to go to baseline and like I said the rain's coming down So we decided to run over to tune in lounge But as we were running we found a reservation for the canteen for oldest canteen and 905 So I guess we're just really lucky Now that we made it to the tune-in lounge We've got a little TV. We're sitting in the living room Sean's over here. Yep And then here is the family portrait is that you when you were younger Sean? Yes, that's me my mom That's actually you I believe oh, that's me. Yeah, I could see it came over the gather What is this the gathering storm? Wow? This is a good collection here Well now it looks like the rain has officially Gone away, so we are going to make our way back to Batu back to Batu All right, it's time to head on in Wow, I think we got one of the best seats in the house That sounds exciting We've got a really awesome mix here a little guy up super excited DJ Rex those lightsabers down He's confiscating the lightsabers he's giving them the kid Look at that wine like it nice. We're doing a group cheer with everybody Would you our own chairs here? Cheers with the fuzzy Tom Tom, and now I'm about to go numb Serena got the ogre's obsession, which is basically Jell-O Look at that, and then I got some Batu bits, and then we got the charcuterie. Look at this This is a fun table right here My favorite out of the Batu bits are these ones right here You guys should turn on your lightsabers Serena is trying the ogre's obsession. What do you think? Yeah, crunch berries on there. It looks good. I Want to try some Alright now that we're leaving the canteen We're gonna do my favorite part which is walk out of the park when it's empty And the funny thing is is we're not the last ones in there, and it's still not even the last Reservation the last reservation the park closes at nine o'clock last reservation is at like 955 So there's still people like going to show up to start their experience It's so cool to be able to exit out of here without anybody It's literally just us Got all of galaxies edge to ourselves. Isn't that so awesome, and it looks so different at night Don't you love it? No people till the spires and with that I think we are done here look at us It's so awesome. It's one of the best things to do, but overall I hope you guys enjoyed the video I enjoyed making it. We'll see you next time Bye


Hello and welcome to the special edition of the Fort Report where we are at the A.M.G. Carter exhibit hall at the 2011 Waterama event. I'm here with Hilda Zaniga with the Fort's water department and she has spearheaded the Waterama event for this year. Hilda, how many years has Waterama been going on? This is our 12th year. And what, tell me about it, what is it about? Waterama is a festival where we try to teach 4th graders the importance about water. We have 28 stations, meaning 28 agencies, all related to water that have come today and yesterday and tomorrow to showcase their information, all related to water. We have the Texas Department of Wildlife, they're talking about aquatic animals. We have the Corps of Engineers and they're talking about swimming safety, the importance of life jackets. We also have the Texas AgriLife and they're helping the kids learn about water and farming. They have, they're making a little grass there and so they teach them about the planting and all that. And then we also have the EPA and they're bringing their information. So we have just an array of agencies all related to water showing their kids how important water is in our lives and really sometimes we don't think about everything that goes behind opening a faucet and this is what we want them to take with them today. And also the importance of conserving water. Right now the state of Texas and other states are seeing drought and we might see that all through the summer so that's another message we want them to take. Be safe and also be aware of how you're using water. So these fourth graders when they leave here today what are some main things that you want them to take with them? That water is important and we need to take care of our water. That's really it. We all need to pay attention to how we use water especially now that summer is coming up and the kids are out. It's going to be hot and they will want to use the water so we want them to know how to use the water during summer. I'm with Ricky Cotto with the Neighborhoods Division of Fort Worth. Tell me what you got going on in your booth today. We are playing a game called Water Jeopardy. So pretty much we divide the kids up in the teams and they answer the questions. What's one of the questions you might ask the kids? We'll see if I've got my water shut down. When sliding down a slide. Head first always. Head first always. No. You always want to go feet first so that we don't hurt yourself. Okay. So what are two ways that floaties do not stop you from drowning? It's like I don't know center of gravity thing. All right well actually yeah I mean some people are too heavy for floaties.  So even if they keep your arms above water you know you can't breathe so there's no point to it. Okay I'm 0 for 2. Let me see if I can pull out of this. That's right. What should a baby never do by himself in the bathroom? Water skiing. Water skiing. That was very close. Take a bath. All right. You never let a little kid take a bath by himself in the bathroom. You need to go around and you need to take some notes at these booths so that way you'll learn a little bit more water safety. Okay I think you might be right. Awesome. He's got water in the right way. Watering the right way for 500. Of the following 30%, 10% or less than 5%. How much water can you save at home through conservation? 30% yell it out. Everyone yell it out if you have an answer. Okay very good. I never get picked. When they come to your booth what do you want them to leave with? What knowledge will help them outside of here? One of the things that we do in the National Weather Service is we try to teach people about the dangers of flooding. Flash flooding, river flooding, the creeks and streams in this area can all react very very fast to the heavy rain events we have in North Texas. So what we're trying to do is educate the kids on the dangers of flooding when they get old enough to drive. We want them to know if you come across a flooded roadway to turn around don't drown and be safe that way. What do we have right here in this bottle? Gravy. I don't remember where to eat before I go film these things. I'm here with Jesus Rodriguez with the 4th Water Department and his booth is all about grease abatement. We got this cool picture right here that looks a lot like my arteries but that's different story. What are you trying to get the 4th graders to learn about today? Well you mentioned that it's like your arteries that's not a bad analogy and we correlate that with how the city sewer works and we're just pretty much trying to protect the city sewer, the treatment plant and in the end the environment from grease. It has a significant impact on the environment. We try to practice or express practices such as throwing away grease instead of washing it down, promote non-use of disposals and that's all we learn. So when I fry up an entire slab of bacon which is you know pretty much that previous problem I mentioned. Breakfast? Yeah and I eat the bacon, all of it when sitting. Again, I've got this thing of grease I'm not pouring it down the drain. It goes down there so easily so thin and you just wash it down. Yeah that's the ease of it is what we're trying to get. Yeah what we would like is if you would wipe down the pan throw it in the trash or recycle it. Everything is recyclable. This is a miniature model of a grease trap. I'm gonna put some blue dye in here. It's gonna circulate over here and imagine that being greasy water. That's pretty much the whole premise of it. Okay the blue water's gonna come. You see? Imagine that's greasy water. It gets stalled there. The grease separates and goes to the top. Slightly less dirty water will go over and back into the sanitary sewer and then back to our treatment plant. And that is how we would like all grease traps to work and that's why we stress installation and maintenance of these for all restaurants or businesses that we deem grease generators. What are you gonna do whenever you get your Olympic medals? Okay when I tell it to you you're gonna make a quack monster. I'm gonna give you this. All right Matty you're first. You gotta do like this watch. You gotta go like this. Quack! Quack! Quack! Quack! Quack! Quack! Quack! Quack! Quack! Quack! Quack! Well I hope you learned something of 2011 Waterama. I know I did. I'm gonna quit watering it noon even though that's the best time to run through the sprinklers.  If you'd like to find out more just go to Fortwithgov.org and look up conservation tips. I'm Chris Connelly and thanks for watching.


film on kawakia and other... it's mind-reproducing I mean... mental Would you like some more? I'll try one more


So what I decided to do for this audience is to focus on one of the many controversial points in my recent reconstruction of ProtoBask or the oldest form of the Aeuskadian language. Sorry, I know it was I was here. Just for those of you who don't do much with Bask, some of the reasons why it's thought to be sort of a difficult language to work on are no longer probably that relevant. I mean the language is still spoken by very many people, under a million but still it's quite healthy. It's spoken as you probably know in the Bask country, there are at least seven major dialects. Some of the big dialects have disappeared in the last hundred or so years but there's still a lot of diversity. There's also the Aquitanian language which most people agree is a relation of modern Bask and this language is attested in Latin texts dating from about zero to 300 before the common era. There are also many place names from antiquity that can be used in the reconstruction. I think some of the medieval data will be discussed in this workshop as well. And yes, well the language is considered an isolate by many, including those who are producing the etymological dictionary that will appear this year. There's been a lot of work on Bask historical phonology, the best known being the work by Martinet and Michelina. There's been work on Aquitanian. There's been work on the root structure of proto-Bask on the accentual system starting with the different dialects and reconstructing back from there. And there's also been work on the phonetic basis of sound change in the history of the Bask language and again you'll hear more about that later in the workshop. And the reason I said that I think the situation has changed from earlier, from early in the 20th century is because there are now a great deal of lexical resources on the language that are searchable, including the 16-volume orotarico escaricegia.  And now we also have collections of Aquitanian data. So, and the last, actually the last thing on this slide is the collected digital version of the cartularies compiled around 1100, which contain place names and personal names, many of which maintain sounds that have been lost in modern Bask. So, for the past five or so years, I've been working on reconstructing proto-Bask, which would be meaning methodologically using both methods of internal reconstruction and the comparative method where the Aquitanian and dialect data is available. And much of this builds on the work of Michelina in particular, as you'll see, the debu-calization of initial P-T-K to H plays a central role in my reconstruction. It's something that Michelina thought was going on, but I think I found many more examples of that. So, what I'm presenting today is a very small piece of the big puzzle relating to the status of sibilance in the oldest form of the Bask language. So, I'm just reviewing here some of the problems with reconstructing isolates. I think most of us are aware of these here, so I'm just going to go quickly the comparative method. I didn't know whether we would all be neogrammarians in using the same method, so these are reviewed here. But there is a problem in working on Bask, and that is in identifying inherited words, because we know from the known history of Europe that Aeuskadian has been in contact with at least three branches of Indo-European. The Celtic languages, the Romance languages, beginning in about 200 BC and continuing, right? And then Germanic languages as well. So, a big part of the difficult work is separating out potential loans from inherited vocabulary. And of course, it really helps to have a good sense of what the proto-language looked like to start weeding these things out. But I think you have to try to be doing both things at the same time. In the etymologies that I offer at the end of the book that I've published, I've tried to stay away from any term which has what looks like a potential cognate in Celtic Romance or Germanic, unless I have a good argument why it could not be borrowed. So here, words that appear to be native may be loans, and seeming loans may be native. So here we have a word that you might see later today or tomorrow, Ibaii, that's medieval Bask for river. So it's attested with that H in the ultimate syllable. And this is considered by most Baskologists to be a native term. But there's a Celtic form or proto-Celtic form which looks very similar. We can't definitively rule out that it could have been loan from some Celtic language. We have other words, if you go skip down to the bottom here, like the word barra, it looks like the word for bar or barrier in vulgar Latin. But I think there are reasons to believe that actually bar is a native Bask root. It occurs in constructions like this term el bar, which has a semantic range that's sort of been extended outside of barring with an obstacle. But what's important about the word el bar is the initial part, the el, which I believe comes from del, which is a proto-Bask root. So this just shows you the difficulties inherent in this kind of exercise, especially if it turns out that proto-Bask is related to proto into European. That means there are going to be roots that look similar in the two languages. So as a background, and some of this is on the handout, the proto-Bask vowel system, it's a five vowel system. On this slide, the diphthongs are in parentheses because in many cases, things that look like diphthongs in modern Bask would have had an H or an N between the two vowels. But I'm following Michelina for the most part in this five vowel system. And the big differences that come up are in the consonant system. So here you see Michelina's 1977 system. There's no P, but there's TK, BDG, and then at least two series of affricates and fricatives. That's the focus here, the two fricative sounds. You see that these are also reconstructed in La Cara's work, and they also are in the forthcoming Bask etymological dictionary. There is a contrast in all of those systems between an apical S-like sound and a laminal S-like sound which still persists in some modern Bask dialects, and that is reconstructed back to the oldest form of the language. I am contesting that. This is my revised consonant inventory on this slide. As you see, there's only a single sibilant. Okay? So the problem here is how do we get from a one sibilant system to a two sibilant system? And that's what I'd like to talk about. Summarizing agreement with earlier proposals, everybody I think agrees that Proto Bask had a voice series BDG, sonarance NLR, and H, which was not reconstructed by for instance Trask. He did not have an H. That there was a single rhodic in Bask, not two is in the modern language, and that all consonants except for this rhodic could occur in the onset. Codas were limited to R, L, N, S. And in the coda you could have R, L, N followed by S. The differences here between my proposal and others are I'm arguing that Proto Bask had an M, that Proto Bask had a P, a voiceless bilabial stop, that Proto Bask stops were aspirated. I think you'll hear more about this when on the Agro-Tsegi speaks, and the focus today, that there was only a single Sibilant in Proto Bask, and that the origin of the contrast between the apical and the laminal Sibilant originates from Sibilant consonant clusters in the Proto language. So in reconstructing Proto Bask I have come up with sort of syntagmatic versus paradigmatic properties that aid reconstruction. Syntagmatically most of these are the same as those used by earlier linguists such as Michelina that you could have a bare root, a root in a compound as first or second element and reduplicated roots. What's different are the paradigmatic properties. I have identified what I think are two well-attested prefixes, one which is an nominalizer, often expressed as A in modern Bask without the H. A collective prefix, H-I, the verbal prefix H-E which usually again is surfacing without the H, is a verbalizer which has been well-recognized in Bask historical linguistics as has the participial Agitival suffix and then the two elements at the bottom, number four and number five on this slide which are new is an S which serves as a nominalizer suffix and an S prefix which can produce an SC cluster from a single consonant initial root. And here this is illustrated with the Proto Bask root BIL which I think is going to appear in the Bask etymological dictionary. It's a well-recognized root meaning turn around as in the modern verb E-B-D to go or go around. And you can see that this root occurs with the H-A prefix, the A prefix, it occurs as an Agitival or participial form. It occurs in the word, I believe, it occurs in the word beats which means like to meet or to come together, it can also be to round up. And perhaps even in the word Si, which is the word for belly button. Okay. Another small difference between this approach and others is what roots look like. So I've identified, yes, monosyllabic roots but also disyllabic roots. And you see in the first column the last two entries here are roots that are claimed to begin with SC clusters. Okay. So let us talk about the single-siblin hypothesis. All the previous reconstructions of Proto Bask that I'm aware of using the comparative method, I should say, posit a contrast between an apical S and a laminal S that's continued in common Bask. And the motivation for this contrast are minimal and near minimal pairs in the modern language where the two sibilants contrast. So we have words like shu for fire versus su. And as I said, in many modern varieties, the contrast is preserved. Okay. That between the apical and the laminal. This contrast also exists in medial position, in inter vocalic position. And in final position where the sibilant is africated. Okay. So this is a general property of Bask that a sibilant fricative becomes an africate in word final position. Okay. So from hutch, you get hutch. And so this is the big stumbling block to reconstructing a single sibilant. These two sounds contrast. Nevertheless, I believe there are five distinct types of arguments for a single sibilant in the proto language. Okay. The first argument here says, well, if you have a vowel, a sonar and a sibilant, you never get a contrast between the apical and the laminal in this position. Okay. So that's one position where the two sounds do not appear to contrast. Furthermore, there are synchronic alternations between roots or stems that show a sonarant followed by the laminal Z where there may or may not be a sonar and present, suggesting a sound change where erotic was lost in erotic sibilant cluster. I'm going to go through the evidence for these. Third, there are root doublets where a Z in one form corresponds to an SC cluster in another. Fourth, there seems to be a distributional asymmetry between the S and the Z sound. The Z sound seems to be at least twice as common in basic vocabulary items. And finally, there appear to be related stop and laminal sibilant initial roots where, well, that capital T represents a historical PTK that's undergone in most cases, initial debucalization. Okay. I mentioned the initial debucalization at the start as something that Michelina had discussed briefly in his work on Bask historical phonology. And so I spent a lot of time trying to dig up roots that when they were not in initial position, had initial PTK in initial position. They have H, okay, to support this general sound change of PTK to H in initial position. If we see then T or PTK with Z initial roots, it suggests a possible spustusca to Z, the laminal sound change. Okay. So the first thing I mentioned were inherited monosyllables with vowel sonar and sibilant rhymes where there's no contrast between S and Z in this position. Here are some roots that I have reconstructed. We've already seen the second one, that root, BIL. Okay. And what you notice here is that after a sonarant, you find the laminal africate, not the apical, the TZ form. So the suggestion here is that at the bottom, these all derive from sonarant plus what that really is, is S, right, because the africation is a late rule. So to go through the steps, right, the distributional asymmetry suggests a single sibilant, S in the coda. There was a sound change then of coda cluster S-laminalization taking S to a laminal when it's preceded by a consonant in the coda. And then the common bass final africation, which results in now a contrast between two africates as opposed to the two fricates. Just to show how this works, right, we have a form like protobascus with S not in a cluster versus the word for alder, which has an S that is in a cluster, okay? And what we get on the surface is two africates that are distinct because one originates in a cluster and one does not. So then the obvious question, one poses is, wait a second, you said that the africates contrast post-vocalically. So what do we say about words like guts, right? This is a word for salt and huts or ahts, the word for finger. Well, in this case, I'm arguing that there was a sonorant there. It's been lost, but we have other arguments that it actually exists. In the case of the word for salt, ghar is a root, which means grain, okay, as in a grain of salt. And there's a lot of evidence supporting that root in bass. So I said at the start that there was other evidence for this coda cluster simplification, okay, namely actual stem variants within the language, okay? So you can see that whatever the simplification is, it's sort of continued to a point where we can recognize it. So we have the word, for instance, ahts, bear. These are modern bass terms. The word for badger, which seems to be derived from bear, right, has many forms actually, but one of them of interest to us has an R, and one does not have an R. And the one that does not have an R is showing this laminal fricative exactly where we would expect it, right, because historically the claim is it's laminal because of the preceding sonorant, okay? The same is true for these forms for aldergrove and the word for butter, okay? So these alternations motivate a prosodically conditioned coda cluster simplification. Notice the simplification is going on in disyllabic or longer forms, okay? So the actual exceptions, I would say, are words like ahts and ahts. We don't expect the R to be lost in the monosyllable, but it is. In disyllables, this seems like it was a fairly regular process, okay? So again, just showing how this would work in the case of the finger powered in the last column of this little table, you see the reconstruction with a root pa, which means like foot, actually. And then a stem, final R, the nominalizer s, okay? So from this we get things like modern pats ahts or a patsch in the form with the historical ha prefix, okay? The other two words you already saw. With the coda cluster s-laminalization process and coda cluster simplification, we're now in a situation where actually we can show there's no evidence for an s versus z contrast in the coda in protobasque, right? All of the instances of the laminal can be derived from clusters, laminalization in clusters, okay? So we're now left with the question of, well, what is going on in other positions of the word, okay? The, in medial position, my claim is that the medial position of these roots are cases which were historically final. They've had suffixes added. So the position we're left to deal with is initial position. The only other position where s and z could contrast is in the syllable onset. S is well attested in the onset, sometimes affricated to ts, and you have some examples here of what I think are true continued s from protoforms with s as in the word for fire or the word for moth, ts. But obviously you see there were words that begin with z in the modern language. So what do we say about these? Well, the third argument for a single sibilant is that there are things that look suspiciously like doublets in the language, okay? The pair in a on this slide, you have two words, ashtun, which is the common word for heavy. This would be a word that would be on a swadesh list, okay? It has an s-t cluster exactly in the position where I would expect a root initial segment or series of segments because a looks like the ha prefix that I mentioned. And then you also see there is a word, not very common in modern times, assun, which means full loaded pregnant. So there's an overlapping meaning here between the pregnancy of people and animals. There's a semantic field, which is not uncommon cross linguistically, right? Heavy meaning pregnant, weighted down and so on. It very much looks like a set of doublets. And the same is true for the word for skin shell bark, assun, okay? I'm comparing this to that s-t cluster in as-s-skal assuming that there's been civil and harmony in this word. Civil and harmony is a common process in VASC. Okay, there are more of these doublets in the languages, in the language. Here's just a few more sets. Istil is mud, filth, puddle, pond, effluent. Sindhu means to dirty or muddy. So the idea here is that s-t-i-l is the historical form of this root. Okay, I spent some time talking about these things that there are doublets. So if we have doublets, we propose then s-pustuska clusters. What happens to them? S becomes z in a cluster. And then we have simplification of the cluster where the output is just the laminal sibilant, okay? On the handout that you have, there's a paragraph or so about the phonetic basis of this kind of sibilant retraction. Cross-linguistically, there's been quite a bit of work on the phonetics of sibilant retraction in clusters. It is not an uncommon process. I mean, I know we have many people here working on Indo-European languages, so I'm sure you can think of many cases in Indo-European, including things like German s-k-tusha, okay? That's exactly the kind of sound change I'm talking about here, except that in Basque, all three clusters, s-tus-pus-ka, became something like s-a. I mentioned that there are asymmetries in the distribution of the sounds s and z in Basque, and it's hard to express that by just showing lists of words, but there are many words like s-a-r, s-a-r, s-a-r, s-a-n, and so on and so forth, which would be considered inherited Basque words. They begin with z, and I imagine this will be a big letter in the Basque etymological dictionary, okay? Because these are all considered native words. There are many more than, many more beginning with z than beginning with s. And finally, and this is sort of a very interesting property. There are roots that appear with z, which then can be related to things that occur on the surface with a p-t-k that has been debutalized in initial position, so it shows up as h. And I think I, yes, I've included here sort of my favorite word because if this etymology is correct, it's very interesting also from a comparative perspective. So I think the word for 10 in Basque, which is at the top of this list here, a-la-r, has puzzled people. There are many different etymologies of the word for 10. Basque has a base 20 counting system, so 10 is sort of only halfway, but under my analysis, this word for 10 actually means at the top, or top of the count. The question is, on what basis do I reconstruct a t for this form? Well, I do it on the basis of some other words that don't really look necessarily like they're related to the word 10, but the one that really caught my interest was the fourth entry down here where you see tamar in a word, shash-tamar, which means residue remains, waste, flotsam, the things that float on top of the water, the dirty things. Shash in Basque means dirty, and it shows up in lots of words. It shows up in the word for broom to clean up and so on and so forth. But the point here is literally this would be, means something like dirt at the top, okay? So it's an attestation of tamar, which I'm relating to amar, the word for 10, meaning top. There are other words like gardama, which seems to mean the milk fat, and it means milk fat in modern Basque, but literally the drops things at the top. And what's interesting here is that there's a very common Basque word, samarra, which is the Basque sheepskin coat. It has many other meanings in Basque, including skullcap, frontal piece put on an ox, blacksmith's apron, a smock. It is a piece of clothing or paraphernalia that goes on top of something, okay? That seems to be the general meaning of the term if we put everything together. So why does it have a Z? It has a Z, I'm saying because it is historically derived from an ST cluster. So this is the point at which there's some other forms like this that I've given you information on. This is the point at which I say, well, this is all very interesting, the internal reconstruction of ST clusters in Basque because with things like sklustaska, we have more material for a comparison between this language and this proto-Basque language and proto-Indo-European. And for this, because I'm out of time, I can refer you to the handout and to more information in the 2018 book. And thank you all very much for listening.


Don't fool yourself because I'll be honest with you the probably one of the worst mistakes that a parent can make in today's time is say not mine because this stuff can come to any home. I have doctors, lawyers, nurses, police officers. We have been incarcerated today from drug abuse you know so don't think that it can't come to your home and one of the best ways to prevent it is do things with your kids get out here be active in your community introduce your kids to good stuff.


Learning requires a healthy degree of vulnerability, a willingness to change our minds, to re-examine something we think we've got all figured out and there could be nothing else to learn or say and we can't learn from those people and we can't learn from those people who are not our tribe or not throughout from our group or it's a voice that we're not familiar or comfortable with. I think without a healthy sense of vulnerability we exclude and we remove ourselves from so much learning, so much new information, so much more wisdom than we have.


There's been an increase in the number of retractions. How can peer review play a role in this to arrest this trend? So I think the rate of retractions is not necessarily increasing but we are seeing more retractions for two reasons. One because the literature output is growing and two we're using different technologies or new technologies to be able to identify cases of largely where publication ethics have been flouted. I'm not sure that it's a burden reviewers can stand to police everything that they look at. I'm not sure that's the function of peer review to police it in that way and they have an obligation to look at the quality and the content of a manuscript for sure but as a couple of people have said to me on different occasions often peer review in the whole editorial process is really about trust so we have to be as publishers and as managing editors we have to be as watertight as we can be in our processes we have to we have to be careful about what we're looking at and we have to make sure we have tight processes in place. We have to make it easy for reviewers to be able to look at for example images to make sure that there's no clear evidence of image manipulation. We have to make sure that reviewers if they've got doubts about whether a paper has been page-rised and we give them the tools to be able to check that or we check it ourselves. Peer review is one thing and editorial governance is another thing and editorial control is another thing.


Okay, so Lecture 5, Stochastic Hamiltonians, I'll define them, we'll motivate them, we'll talk about some of the complexity. And like last yesterday's lecture, this is a special case that has some classical features to it and maybe is a little bit easier than the full blown quantum version of the local Hamiltonian problem. Okay, so let's start with a definition. So we have a Hamiltonian and we're going to specify it in the standard basis. It doesn't really have to be the standard basis, any basis we'll do, but it's by default we typically use the standard basis. And the definition of stochastic is that the Hamiltonian is stochastic if all the off-diagonal entries are less than or equal to zero. What's that? There's a late bus. Oh, okay, okay, so made a second, okay. And now this mic is flashing red, which is a bad sign, so maybe I should just change it out. All right, that works. Should we proceed? Okay, so definition of stochastic is that all the off-diagonals are greater than or equal to zero. We're going to actually consider a stronger version of it in which each individual local Hamiltonian term has the stochastic property. And as I understand it, there are examples where the global is, but the terms are not necessarily. Is that right, Barbara? Okay. But we will assume the stronger definition, and there was a very good question about why stochastic. This is on the later slide, but I think it's an important thing to say right away, is that the thing about, and we will prove a version of this, is that the ground states have non-negative, the ground states are non-negative, meaning that the amplitudes are all real and non-negative. So they look more like probability distributions, hence the stochastic nature, but their quantum, so stochastic. Just some motivating examples, we've already seen some stochastic Hamiltonians, the H propagation term from the circuit to Hamiltonian construction is a stochastic Hamiltonian, and the Laplacid of matrixes in general random walks are stochastic in nature. We've talked about adiabatic computation, and if my end, my final Hamiltonian, if I'm using adiabatic computation to solve a classical optimization problem, the final Hamiltonian is just diagonal. And if I start with a Hamiltonian, which starts in the all plus states, and I evolve over time, for all points of time in that adiabatic evolution, the Hamiltonian is stochastic. So that's another sort of important, and we'll actually talk about adiabatic evolution in connection with stochastic Hamiltonians as we go. So this is another situation, and then many Hamiltonians from physics have this property. It comes with a small local change of basis, and one of the exercises you have is to actually do that local change of basis and take a Hamiltonian that appears not to be stochastic and represent it in a basis in which it is stochastic. So it is well motivated from the physics point of view. Okay, so the general setup here is we have general quantum Hamiltonians, we have the classical on the inside and stochastic sits somewhere in between, much like the community case which is sort of in between classical and quantum. And we'll show a version of this, is that the ground states of stochastic Hamiltonians are what we call, and I will call these states non-negative if the amplitudes are real and greater than or equal to zero. Just a little bit now, I'm just going to do a little bit of survey, give you the lay of the land, and we'll take a little deeper dive into a couple of these results as we go. Just to do a side-by-side comparison of general quantum Hamiltonians and the stochastic version, if we have the local Hamiltonian, and we phrase it as a decision problem to determine if the ground energy is at most A or at least B, and typically we want that gap to be one over a poly, as we've seen already in great detail, that problem for general Hamiltonians is QMA complete, and for stochastic, it's complete for this funny class stochastic MA, which I'll define on the next slide. And then the frustration-free case is particularly interesting here, because there seems to be sort of a difference between the frustration-free. So frustration-free for the general Hamiltonian problem is what we call QMA 1 complete, and all the 1 means is that the verifier accepts yes instances with probably 1. So if you know the ground energy is supposed to be 0, and you're handed a state, you'll measure that the energy is indeed 0, and there's no probably of error there. And this is the result we'll sort of talk about more in depth, is that on the stochastic side, if we're looking at frustration-free Hamiltonians, this is complete for MA, okay? And this was the first MA complete problem, which is sort of significant, that sort of this comes from the quantum world, that sort of a natural complexity class. The first example of a complete problem comes from the quantum complexity. Okay, so we've seen Merlin Arthur. Merlin Arthur is our sort of classical verifier, so if an instance is a yes instance, there's a string that causes a classical probabilistic verifier to accept with probably at least two-thirds, and if it's no for every witness, the probability of acceptance is at most one-third. So MA contains NP because a probabilistic verifier is a special case of a deterministic one. And then stochastic MA is the same except that the verifier can have this sort of funny quantum-like format to it. So this is what the, and it's sort of a generalization of a probabilistic verifier. So the verifier computation is a classical reversible circuit.  Did I get this right? The witness has to be quantum. The witness has to be quantum, wait, wait, sorry. And then we execute a classical reversible circuit. Right, because it's gonna be, in general, entangled. And then we can measure in the head of mind basis. And so the general stochastic Hamiltonian is complete for the special class. Now, I think it looks to be all tied up. I think there's some question, maybe whether these two are perhaps equal. I don't think any big complexity assumptions would be violated if that were to be shown. So there's not some big complexity barrier to show that there's equality here. But this is generally open, okay? But we do have sort of characterizations of each version and completeness for it. Okay, yes. How much quantumness in the witness helps us? I think there's also quantumness in this verifier. That's an important, it's not just the fact that the input is. The measurement is sort of the key part. Yeah, right, okay. So it's good we have the experts in the audience this morning. Yeah, it is very funny. It would be nice to understand it better and to sort of have a better sort of more natural characterization for it. So on the numerical side, physicists have been working with stochastic Hamiltonians quite a while for general quantum Hamiltonians. Sort of the best classical, there are lots of tensor network techniques and everything but for in terms of provable algorithms, if brute force the diagonalization can handle 20 to 30 qubits. On the other hand, there's a very nice quantum Monte Carlo algorithm that can be applied to stochastic Hamiltonians. And this has been used for a long time in numerical condensed matter physics. And the nice thing is, this is what they call it sign free. So the ground state that they're looking for doesn't have signs in it. So you can implement this with a random walk. So I'm not gonna cover this method, but just numerically, there's a lot more that you can do with the stochastic ones that have been very successful. There's not a lot in terms of proofs of convergence. But heuristically, these have been handled quite well for larger systems. And then adiabatic, I will discuss this a little bit more. We talked about, I guess it was Tuesday, Wednesday. The fact that the adiabatic model for quantum computing is equal to the circuit model, okay? On the stochastic side, and we'll discuss this today, for the special case of frustration free stochastic Hamiltonians, these can be simulated classically. So if there is an adiabatic path, you don't need a quantum computer. You can just use your classical computer to simulate that evolution, okay? We don't always understand when the adiabatic path exists. But if it does exist, then it can be simulated classically, yes? Yeah, if you're using a numerical method, say quantum autocollow, how are you sure that you're actually in the ground state? Well, frustration free, I think you can just empirically measure. Now, I don't think you ever really know numerically whether you've hit the ground state, you just get sort of papers come out. We got a lower energy on this Hamiltonian model. So I think there's no guarantees necessarily. Maybe there, I think there is sort of an interplay though, between the numerics and the experimental physicists. So some of these things can be sort of experimentally verified in the lab. So, okay? So let's look at sort of a simpler version of this claim. So suppose I have a stochastic H, then we're just gonna prove a simple version and I'm gonna quote sort of the more general version. There exists some state that's non-negative, that's the ground state of my Hamiltonian, okay? So just stuffing through proof, we'll start with any ground state. So I'm expressing it here in the standard basis. And now I'm gonna define this new state, which is the same state, but I'm just now taking the magnitudes of the complex amplitudes instead. And what we're gonna show is that the energy of my new state is at most the energy of the first state I started with, okay? So if you just kind of multiply this out, you get the diagonal terms, you get the off diagonal terms. And the key observation here is that this X, H, Y is at most zero, okay? So if we can lower bound the product of these magnitudes by the complex conjugate times the number, you pick up a sign here of the angle between the two complex numbers when you multiply it out. This is gonna be, this value is gonna be a lower bound for this value, but since this multiplier is non-negative, the resulting energy is gonna be at least as large for Psi as it is for, at least as large for Phi as it is for Psi. And then this is exactly the energy for Phi. So any questions about this little, and so the key step here is this substitution. And I didn't multiply out the, but if you pick up basically a cosine term here, and so this makes this value less than or equal to that value, and since this is a non-negative multiplier, the resulting energy will be at least as large, okay? Questions? It's a little early in the morning to think that through. Sorry, the last day. Yes, yeah, but it's symmetric. So it cancels out, the imaginary part cancels out. So if I'm assuming that it's gonna be symmetric, I have a yx term, and so then the imaginary part cancels out, all right? And that's where you pick up, so then what's left is just the cosine between the two, so. All right, the Peron-Frobinius theorem sort of gives sort of a sharper and that picture of what's actually going on is that if H is reducible, meaning that so it's, there's a single, it's block. I want more than block diagonal. I want a single block here for reducible. I want it all to be connected. I just realized maybe this is not quite, yeah. So reducible means that my, it's all sort of connected to each other, and the resulting walk is connected. Oh, I want reducible to be a single block. No, irreducible to be a single block, it's early in the morning for me too, and reducible to be block diagonal. So if H is irreducible, meaning that everything's connected, then there's a single ground state, and that's not negative. If it's reducible, what I get is an orthonormal basis of these things. So each block corresponds to a different ground state. Each of these ground states is non-negative, okay? And we'll use this quite frequently throughout, okay? So the, in 2008, Sergey and Barbara came up with this very beautiful Markov chain that can be applied to, that converges to the ground state, okay? So remember, our ground states are really kind of probability distributions. So we're going to be looking at Markov chains that eventually sort of converge to this ground state distribution, okay? So I'm going to start with my Hamiltonian. And I'm just going to adjust it a little bit so that it's sort of more convenient to work with. We've talked about the fact that we can scale the terms so that they're norm at most one, then they're non-negative. And this gives me that the norm of my entire Hamiltonian is bounded by M. And I'm also going to offset it so that the energy is equal to zero. Now, you may wonder already, isn't energy sort of hard to find? This is going to be the least of our problems, right? As I will define it, the Markov chain is not easy to implement, but we'll show that in the special case of frustration free, that it can be implemented. So, and in this case of frustration free, E will be zero, okay? But for now, for convenience, I'm going to assume that I've sort of shifted my Hamiltonian so that it's now the ground state is zero. So the norm of H sits in between zero and M, and the ground energy is zero. So here's the definition of the random walk. I'm taking this G to be I minus H over 2M, and this guarantees me that the norm of G sits in between one and one-half. Because M is between zero and M, and I'm subtracting off from the identity. And so the random walk is defined as follows. I have some standard basis string X, and the probability of moving to Y is the following expression, where psi is a ground state of the Hamiltonian, okay? So it's the ratio of the amplitude of Y to the amplitude of X, multiplied by this cross diagonal term of G. And this is guaranteed to be non-negative, okay? So I don't think I put up here that I'm assuming that psi is a ground state. And again, I don't necessarily know how to implement this thing because I don't have a ground state in my hand, but we're just sort of abstractly defining a Markov chain at this point, okay? So we're gonna just step through the proof that this is indeed a well-defined Markov chain, and then we'll talk about why it's easier to implement in the frustration-free case. You have a question I can tell on your face. Oh, yeah, that's my handwriting, that's my, okay. So this is psi, a ground state. And I think those are the only, this is a psi too, but this is a string. This is the amplitude of Y in psi. This is the amplitude of X in psi. And this is just the off diagonal term of G between Y and X, okay? Thanks. Note to self, I have to make that clear or so. Okay, so the probability of transitioning from X to Y, we're gonna be defined to be this value, and this defines a Markov chain on the support of psi, okay? So first, let's establish that it's non-negative. If Y and X are both in the support of psi, then they're greater than zero. And we're assuming, we're using the fact here that this is a non-negative state. And this is now an off diagonal term, the probability of transitioning from X, Y to X. So actually, we'll cover both cases where Y is equal to X or they're unequal. So I have a delta X, Y from the identity, and then I have this off diagonal term divided by 2 over M. If X is not equal to Y, my off diagonal term is less than or equal to zero. If X is equal to Y, then the off diagonal term is less than or equal to M. And in either case, this is gonna be a non-negative number, okay? Between 0 and 1. So it's a well-defined probability. Now let's talk about the fact that it's a well-defined Markov chain. So given, if I'm at X, the sum of the probabilities of the strings that I might move to have to sum to 1, okay, for all X. So the probability, I'm summing up over the probabilities of Y that I would move to. And so now, I'm just gonna insert my definition of the probability of transition. And I've just moved the summation around a little bit. I've moved the Ys to the inside, and this becomes the identity then. Cuz I'm summing up over all Y. And so what I have in the numerator is just psi G inner product with X. And using the fact that the ground energy of H is 0. So if I have a ground state psi and I'm applying G, I just get it's an eigenvector with eigenvalue 1. And so I can replace this with the amplitude of X. And I just get 1 at the end, okay? No, I don't think so, because G applied to psi is just psi, yeah, here? So this is just the, if I bring the summation on the inside, I just get the identity, okay? All right, okay, so it turns out that the unique limiting distribution of this Markov chain, now we have the squared, or the amplitude squared, okay? And this just follows from the fact, we started assuming that H is irreducible, and it satisfies, it's not too hard to check that it actually satisfies detailed balance in this. So the probability of moving in one direction along an edge is the same as going backwards. And if it's not irreducible, then we can take our non-negative basis, and the walk is the support on each of these individual sets. So if I have a non-negative basis, and it's orthonormal, the supports are disjoint. And this random walk now, or this Markov chain, is defined on each separate connected component. And so each of these states defines a connected component, and it's a random walk on that, all right? So how does this Markov chain converge? We're just gonna pull out some sort of standard results from sort of Markov chain mixing theory. So if I'm interested in starting at string Z, and I'm interested in walking until I'm within epsilon of the stationary distribution, that's gonna be the smallest number, such that when I apply my random walk T times, I'm within epsilon of the true stationary distribution. And a standard mixing time bound is that this mixing time is bounded by the following expression. So pi of Z is the amplitude squared of my starting state. And lambda 2 is the second largest eigenvalue of the Markov chain. So what are these two? So what is lambda 2? Let's talk about that first. So the claim is that lambda 2 is just the gap of H over 2m. So if I have a 1 over poly gap in my Hamiltonian, I'm gonna get 1 over poly downstairs here, okay? And this just comes from sort of plugging in the definition of the probability distribution and observing that this probably distribution, I'm taking g and I'm multiplying on either side by the diagonal, the diagonal inverse, where this is just the amplitude of x in the state psi. And since g and p are similar, they have the same eigenvalues, and the eigenvalue of g is just gonna be 1 minus lambda over 2m, okay? So the upshot is that the mixing time is gonna be 2 over the gap of H times the log of this term. So what do I need for this random walk to converge in polynomial time? Well, I need a good starting place. So I need a starting place whose overlap with the two ground state is at least 2 to the minus poly, so exponentially small is okay. And I also need the gap to be 1 over poly. And in that case, this mixing time, this Markov chain, which I can't quite implement anyway, is at least converges in polynomial time, okay? Okay, caveat, a mixing time, this Markov chain requires knowing the probability distributions, obviously, if you're gonna implement it. And it requires knowing these overlaps and we don't even know the ground state, okay? And the beautiful thing that Sergei and Barbara showed is that in the case of frustration-free stochastic Hamiltonians, this ratio can be efficiently computed, okay? And I'm gonna sort of walk you through why that's the case. So in general, we don't know how to implement this Markov chain in general. It's a well-defined Markov chain, but we can use the frustration-free aspect to implement it. Okay, so I'm gonna start with some ground state of my Hamiltonian. And I'm gonna start with some x, which has non-zero overlap with that ground state, okay? And let's say my random walk is sitting at string x. And I wanna compute now these probability distributions, the transition probabilities moving out of x. How do I do that? Well, the initial definition of the Markov chain requires knowing this ratio, which we, in general, we might not know. But we're gonna make a couple of observations. So if the probability of transition is greater than zero, then it has to be true that this off-diagonal term is greater than zero. Because we have that as a multiplier in there. And just by definition of g, that means that the off-diagonal terms has to be less than zero, strictly less than zero. So what that means is that these two things are sort of connected in the random walk, okay? And therefore, for at least one of the terms, one of the individual terms in my Hamiltonian, the off-diagonal term has to be less than zero, okay? And this is now starting to look something that I can locally check, because it has to do with the individual terms and not the global Hamiltonian, okay? So the claim is that if h is stochastic and frustration free, and I have an x is connected to y by this little local term, then this ratio that looked difficult to compute can actually be computed by something about the local term. So I don't need to understand my global Hamiltonian in order to be able to do this. I just need to identify a single term where the off-diagonal is strictly negative. And then for that particular term, I can look at the diagonal entries of y and x and compute this term. So this is now looking like something that is feasible to compute. I don't need to know the global properties of the ground state. And the nice thing is that for a given x, there's at most polynomial many of these. I can look through all of the terms and each term is going to give me a constant number of strings that I can then walk to. So this is sort of now modulo proving this claim. This is now a random walk that I could potentially implement. I need a term where, so I need a term that has non-negative. And if I know that the probability of moving to y is non-negative, if I have a positive probability of moving to y, that means that there has to be at least one term where the off-diagonal is negative. Yes. I'm not going to go through the proof in excruciating detail. I'm just going to wave my hands about it a little bit. It's fairly intuitive to see. And so here's the intuition. So pi is the projector onto the ground space of the big Hamiltonian, not a single term. And remember, I have some non-negative orthonormal basis, and I can express pi as sort of some of 1D projectors. Now let's take a look at sort of what that matrix looks like. So since they're non-negative and orthonormal, that means the support is disjoint. So if I look at this projector in sort of matrix form, it's block diagonal. I have a block for each one of these states in my orthonormal basis. The support is disjoint, so one of these terms is going to be a 1D projector. And it's defined on the support of that state. So far, so good. Now the ground space of my Hamiltonian can be any superposition of these psi a's. But if it's in the ground space somewhere, if I have two strings that sit inside the support of a single state, that means the ratio of their amplitudes is fixed. So I can certainly within this ground space vary the weight on this block versus that block versus that block. But for two given strings inside the same state, the ratio of those two strings is fixed once and for all by virtue of the fact that it's sitting inside here, and that's dictated by psi a. OK? This is sort of an important point. So let's say if I look at just psi a by itself, what's the ratio of the amplitude of x to the amplitude of y? So this is just the ratio of the diagonal terms of x and y. Now I can certainly have a ground state that's different superpositions of these different states, but the ratio of x and y doesn't change. So if I have some state that's sitting inside the ground space, and I know that the, this is just another way of saying exactly the same thing, and I know that the amplitude of x is non-zero in the state, and I know also that x and y sit inside the same block. That's what this says here. Then the ratio of their amplitudes in this psi is equal to the projection now of x using the big projector. So this is fixed and independent of which psi I happen to pick. OK? Are we good? Does anyone have any questions about that? OK. Good. Then since it's frustration free, I can make that exact same observation about every single individual term as well. So because psi has to sit inside the ground state of each individual term, if I look at the projector for an individual term, and I apply it to psi, I get the same state back. And this is the frustration free property. And if in this individual term, x and y are connected, meaning that the off diagonal term is strictly less than 0, then the projector, x and y, lie in the same connected component for that individual term. And for that individual term, the ratio of x and y is fixed. OK? So now I have all these terms that are dictating the ratio of x and y. And they have to be consistent with the global ratio. OK? Yes. I said, OK, so maybe I misspoke. The ratio of x and y within the same block is fixed. Within different blocks, yeah, you can vary depending on which state you happen to pick. But if they are inside the same block, then the ratio is fixed. And this is saying that they're inside the same block. OK? OK, so this is just kind of hand-wavy argument. The beauty of the frustration free case is that the individual terms are sort of fixing the ratio of these amplitudes. And so I only have to look at these sort of constant size terms to determine this ratio. OK? Questions? Yeah. Actually, if you think about it, if I'm taking a projector with these blocks, they're going to have no zeros inside them. Yeah, good point. You're awake? That's good to know. All right, great. So this is something that can be computed by a classical computer. So we have now that this random walk that converges to the stationary distribution corresponding to the ground state can be implemented by a classical computer. It converges if I have a good starting point and my Hamiltonian has a one over poly gap. That's where we are. OK? And they use this to show that if there's an adiabatic path, then it can be classically simulated. I'm not going to go through all the details. I'm just going to kind of state the result. So if I start at, if it's easy to find the ground state of the initial Hamiltonian, that's always the condition of adiabatic computation. And as I'm going from the initial Hamiltonian to the final Hamiltonian, if I remain stochastic and frustration free the entire time, that's this condition here. And the gap is always one over poly n. And the path has to be relatively smooth. That's usually a condition in adiabatic. Then a classical algorithm can approximately sample from this distribution. This is the final Hamiltonian as it goes from 0 to 1, or the ground state of the final Hamiltonian. And we can approximately sample. And the idea is that you can discretize your path. And at each point in your path, you can use your current ground state as sort of a warm start for the next one. So there's some work to do in showing that how finally you have to discretize this path in order for the ground state of your current one to be a decent start for the next one. But just noticing what this means is that without the stochastic frustration fee condition, this was universal for quantum computing. So this we could execute. If we could do this for general Hamiltonians, then we have a quantum computer. So this is sort of a stark difference between the two complexity. And an interesting question you might think about is that adiabatic evolution with general local H is equivalent to the quantum circuit model. And is there sort of a natural circuit model corresponding to adiabatic evolution for stochastic, not necessarily frustration free? So if I look at adiabatic evolution, it corresponds very cleanly to the quantum circuit model. The frustration free case is classically simulatable. But what about this sort of in-between where I have stochastic Hamiltonians that aren't necessarily frustration free? Is there some kind of natural circuit model that's equivalent to that adiabatic model? Something to think about. OK, so complexity of the local Hamiltonian for stochastic free. So given, let me just define the problem, I'm given a stochastic local Hamiltonian. And I want to determine, is the ground energy zero or is at least A for some A's, at least one over poly? So this is the local Hamiltonian problem now in the stochastic frustration free case. So it's going to be at least NP hard because boolean satisfiability is a special case. So we're not going to escape that. But a verifier could send you a nice starting position. So that's something a verifier could do. And you could potentially simulate this now random walk. We know how to simulate it. And the question is, how would I verify in that case? So let's imagine instead of an MA protocol in which the verifier sends you a good place to start and you execute this random walk. So one idea that doesn't work is start with an X. So it's sent by the prover. Implement the random walk long enough until you converge. Since it's gapped, we're going to assume we can do that. Measure the energy, and then repeat for accuracy. You could do this potentially to verify yes instances, although it occurred to me now that the promise gap is one over poly, but I don't necessarily know that the spectral gap is one over poly. Yeah, it doesn't necessarily, yeah. Okay, so that was another, and also there's an issue with verifying no instances too. Because we're using these local terms to define this random walk. If it's frustration free, that corresponds to the sensible random walk that we know that converges. But if it's not frustration free, it's not clear what it does. So we need a better verification procedure. So here's the idea is that we can define the set of good strings that have sort of a positive, for each individual term, they have a positive overlap with the projector. And this is certainly a polynomial time checkable condition. This is just term by term, I can check this. And any good string, I mean anything in the ground state is certainly gonna have to satisfy this property. Because it's in the projector for every term. And we're gonna find the bad strings as the opposite of good. So this is our picture here of all strings. We have the bad strings out here. We have a subset of strings that are good. And inside here is the actual support of the ground space. And one of the exercises to do, I've given you a particular Hamiltonian, you'll define the good strings, the support, and so the ground, as ground are strings that are in the support of the true ground state. And this superset is some local checkable condition that I can do, okay? So one observation is that this random walk is actually closed on the ground states. And we sort of saw that before.  And in the no instances, we wanna show that if you execute this random walk, after a certain amount of time with high probability you're gonna hit a bad string. And you're gonna be able to check that. So in the yes instances, the prover sends you something in here, you walk for a while that's closed, you never hit a bad string. You accept, okay? In the no instances, there's actually, this ground is zero. You're sitting here in the good space and you wanna show that if you execute that random walk for a polynomial number of steps, with high probability you end up in this bad space which you can then check, okay? Again, this is gonna be approved by hand waving in the end. So here's the verification procedure. Prover sends a starting string X to the verifier supposedly in the ground state. Verifier runs the random walk for T steps. And in each step we're gonna check did we hit the bad strings or not. If I hit a bad string I reject outright. And if after T steps I haven't hit a bad string I'm gonna accept. There's a little technical caveat which I'll sort of touch on in the last slide. There's an additional check that we need. And the good thing is that it's always satisfied for yes instances. So I'll hint at what that is. It's not kind of the crux of the matter here. And related to that technical point is the verifier doesn't send just any X. It sends the X that maximizes the overlap with the ground projector. Which supposedly the prover can do, okay? So how big is this T have to be? So remember, I think I used A for this before but now we're using epsilon as our promise gap. So in a no instance, the ground energy of my Hamiltonian is at least epsilon. And so what does that translate to in terms of G? Here's G, it's I minus H over two M. It means the maximum eigenvalue of G is at most one minus epsilon over two M, okay? The upshot is that it essentially means that this Markov chain is rapidly mixing. So I'm gonna hit all of the states with reasonably high probability within a polynomial number of steps. And it's gonna use this upper bound on the maximum eigenvalue, okay? So it's essentially it's an expander graph. I walk for a polynomial steps, I hit everything including a bad string if I'm connected to one, okay? So that's the short version of the argument. But if I start with something in X good, what's the likelihood that I stay within that set the entire time? I can imagine all possible sequences of good strings and I'm summing up over the probability that I follow that sequence of strings. Now if we unfold this probability, I get this ratio. This is exactly sort of how we've defined the random walk, okay? Now the glitch that required the technicality is that I'm gonna drop this term here in the next step, okay? Now let's just let me just sort of justify a little bit why I can do this. Notice I'm taking this product, okay? So if it's a yes instance, the product of all of these terms I get this sort of telescoping product and at the end I get the overlap of the final state with the overlap of the initial state, okay? If the adversary sent me the string that maximizes this overlap, that's gonna be less than at most one, okay? And that's a checkable condition, okay? And we're gonna always check that this ratio, the product of this ratio is at most one, which allows me then to sort of just drop these terms altogether. So I get this big product of terms and I'm gonna sort of drop at the end all of these terms, but I'm gonna verify that the product of all of these things at the very end is at most one. And that's a checkable condition. It's calculatable because I know my probability and I can calculate this so I can calculate that ratio. It's kind of a technical point, but basically modulo that check, the sequence, the probability of executing this sequence is upper bounded by this probability of all these inner products. And now I just use the fact that the maximum eigenvalue of G is at most one minus epsilon over two M. I'm multiplying times all the strings, but that's okay. As long as T is polynomial and as long as the gap is one over a poly, there's gonna be some polynomial T, which I can drive this thing down exponentially small. Okay? All right. So that basically says in the no instances with high probability I'm gonna hit a bad string. So yes, if epsilon is one over a poly, this will be exponentially small with T that's polynomial in the size of the, okay. Just mentioned some follow on work, some very nice paper by Dorit and Alex. So what we've shown here is that the hardness of the stochastic local Hamiltonian by the, oh, as an aside, we've showed containment. I didn't show the hardness, the hardness was, and it's actually achieved for situations where the ground state is a uniform superposition over a set of strings, okay. And Dorit and Alex showed that they can actually de-randomize this process when the Hamiltonian has a constant size gap in the case where the ground state is a uniform superposition, okay. So for this special case of Hamiltonians where the ground state is a uniform superposition, if I have a big gap, then I can completely de-randomize this process. And the upshot is that the path from the good string to the bad strings is not that long. So you can basically exhaustively search the entire neighborhood classically instead of following a random walk, okay. And it has a nice interesting complexity sort of implication which means that if you have gap amplification for this special class of uniform, frustration-free stochastic local Hamiltonians, then that basically collapses MA into NP because that process then can be de-randomized. So if I can take a generic uniform, frustration-free stochastic local Hamiltonian that's a mouthful, and if I can boost the gap, then I can all of a sudden have a classical verification procedure which will put MA into NP. All right, that's it. Any other questions? Feel free to chime in if you've got interesting open problems related to stochastic, the stochastic world. Anything, yes, questions? Promise gap, promise gap, yeah. No, it's the spectral gap. That's, yeah, I think promise gap, sorry, yes. Promise gap, sorry, yeah. So it's not a gap to Hamiltonian. It's if you have an instance which is hard and gapped, where the promise gap is constant, say. Yes, yeah, I think you don't have to because we know that if I look at the special case of uniform ground states, that's still hard for MA. So that's why they're not necessarily equivalent but in terms of the complexity, it is hard still for that special case, yes. Question? Uh-huh. Whether it can be highly entangled? I think that the question has to do with whether these ground states of stochastic Hamiltonians can be highly entangled or not. Do you know? Yeah, yeah. The tour code ground state isn't maximally entangled. I mean, yeah, you get global entanglement, is it? Yeah, yeah. Is it, yeah. All right, you had a question? Like the area law stuff or fancier ground state projectors? Would you get there faster? It's, so the, observe that the G operator looks like an approximate ground state projector, essentially. And the question is if you used a more sophisticated ground state projector, would it say speed up the convergence in some way of the random walk? Yeah. I mean, it's not gonna help you compute the Markov chain in the frustrated, you still have this ratio term that's problematic, but maybe there's potential that different terms there could speed up that convergence in some way, because the convergence was related to that term. So, yeah, you might be able to get sort of a better eigenvalue there, so yeah. Yeah, yeah. I mean, you could approve the polynomial in some way, so yeah. Question? Like preparing Gibbs states or in general, or, which is related to preparing Gibbs states, in AM, yeah. For stochastic? And for general stochastic? Okay, okay. Okay. Yeah, I didn't actually include that in my, yeah. So, acoustic MA is contained in AM, which is contained in QMA, yeah. Yeah. Uh-huh. Right, that was actually the goal. Yeah. Yeah, and that was the D-wave model, right? Was there, initially, that they were finding ground states of stochastic? Yeah, but not frustration, great. Yeah, with the frustrated stochastic, yeah. Any other questions? All right, great, thank you.


Let's jump in. Ben and Jerry. Ben and Jerry Ice Cream. You know Ben and Jerry Ice Cream, they're owned by Unilever, the European company that owns a lot of different brands. You probably don't know Unilever, but you know a lot of the brands that they own, food and other household items and so on. They're massive, a little bit like Proctin Gamble is in the sense of the diverse products that they sell. Well Ben and Jerry is the ice cream out of Vermont and I don't know how many of you know, but Ben and Jerry have been committed hippies, leftists for way back, way back. We're talking about, we're talking about the 1990s I remember Ben and Jerry. Ben and Jerry were way leftists back then. They're still way leftists and on the 4th of July they did what Ben and Jerry do. Nothing here surprises. They wrote, the 4th of July, this 4th of July, it's high time we recognize that the U.S. exists on stolen indigenous land and commit to returning it. Learn more and take action now and they give you a link to Ben and Jerry website that basically articulates the story of the different indigenous people and they particularly highlight the Mount Rushmore where we've got the statues of four presidents and they say, you know, this belongs to the, to the, you know, Tunka Sully, Scuppy, whatever, the six grandfathers, Sioux Nation and we should be returning it to them. So they are, they are generally, and then of course, you know, I saw some posts saying, hey, Ben and Jerry, are you, are you returning some of your stores that are on, on so-called native land or the factory in Vermont, which is on so-called native land?  And of course, Wangui media went up apoplectic and they've announced the boycott of Ben and Jerry and Unilever stock as of this morning is down significantly, which I don't, you know, we'll talk, we'll talk about it, but it's down significantly because this is, as, as the Wangui media has defined it, this is Ben and Jerry's Bud Light moment. This is Bud Light and we're going to see now Ben and Jerry sales tank and we're going to see, we're going to see Unilever stock tank and so on. And I actually don't think so. I think this is way overplayed in terms of the stock price and even the boycott of Ben and Jerry. The reality is, and, and it tells you a lot about the right, I think, the reality is I've been boycotting Ben and Jerry literally.  I've been boycotting Ben and Jerry since the mid 1990s.  They would regularly be interviewed on TV. They were CEOs and therefore they got some prestige from that. And therefore people took them seriously. And I said, I'm not giving these bastards money because, because they're advocating for philosophy and ideology that I oppose. And this is the brand that the brand is affiliated, associated with the left, the worst elements within the left. And then if you, if you, and then when they sold to Unilever, I said, ooh, maybe now I can buy Ben and Jerry ice cream. But then Unilever basically established a special board for Ben and Jerry to maintain their kind of corporate culture and to maintain their advocacy and their continued, you know, leftist agenda and policy. So I basically have not eaten Ben and Jerry since since the mid 1990s. And yes, their ice cream, I think, does taste good, although I don't eat ice cream anymore. Anyway, it's probably way too sweet for me. But, but yeah, I'm sure at the time, it was definitely a good tasting ice cream. But you know what, there's lots of good tasting ice creams. There's no shortage in great ice cream in America. You don't have to buy Ben and Jerry. And everybody's known they are crazy leftists from way back from, from, you know, decades ago, 90s, right, 30 years ago. So what did they say here? They set a standard leftist line about indigenous people and about they land and we stole their land. All right. Is this going to, is this changed my assessment of Ben and Jerry? No. I mean, this is no different than anything else they've said in the past. They hate capitalism. They hate Israel. They hate everything. They are very woke, they're pro-woke. They've always been politically correct when politically correct was the term. So yeah, they said in 4th of July, which is par for the course for them. I mean, they used to come on television in the 90s a lot, just slamming markets and capitalism and, and, and CEO pay and everything about corporations. So my view is they've always been left to somebody. Suddenly people have woken up because it was the 4th of July. Give me a break. So that's why I don't think, I don't think this is going to be the issue. If Ben and Jerry had sent something pro-trans or pro-LGBTQ plus, which I'm sure they hold, they have those views and made a big deal out of it, then I think the boycott would hold because I think the boycott is primarily driven not by these crazy leftist, not even by woke stuff. I think it's really driven by LGBTQ. That's why the successful ones are being Bud Light and Target because of their LBGTQ stuff, which I thought was pretty lame. Here's a company. Here's, here's how lame the right is. Here's a company that has dedicated all of its energy towards radical leftist, crazy ideas for 30 years. And the right's like, blah, who cares? Oh, now, now they suddenly wake up, but I don't think it's going to have any impact. Yeah, they have had rainbow flavors, they're huge pride and everything, but the whole corporate stick is crazy left from the beginning. And the right has ignored them, basically. Nobody's boycotted them. I'm sure all these TV hosts, they all eat Ben and Jerry ice cream because it tastes good.  And it's a leverage point and it is their thing. And that's why I don't think this'll have an impact because this is too generic left. This is too, they've always been like this. And again, I've been boycotting because they do believe in boycotting companies that are, that commit their profits to a social ideology.  And I don't boycott Target because the corporation is not dedicated to ideology dedicated to an ideology. I'm opposed to Ben and Jerry's whole all their profits or significant portion of the profits are dedicated to promoting ideology that I opposed from beginning to end. And I boycotted them. And I think you should too. And I think everybody should. I don't think this is going to have as much oomph on the right as Targeted and Bud Light because it doesn't have that trigger. The trigger is what LGBTQ issues? It's the same. If you watch my talk at Ocon, it's what Putin knows, right? Putin knows that the right in America is obsessed by this issue. So in every single speech, he talks about the decadent right. How is the decadent because of gays and because of LGBTQ stuff. And to a large extent, the American right supports Putin because, hey, he's on our side on the most important issue of the day, most important issue in the world today, which is LGBTQ. And because of that, because I should have ordered the topics differently.  Let's let's put DeSantis. Let's get DeSantis. We'll skip in my head a little bit. So anyway, that's Ben and Joey.