text:
Public Comment Period opens as Oregon Psilocybin Services (OPS) announces public hearings for proposed rules
The Oregon Health Authority and Department of Human Services sent this bulletin at 11/01/2022 09:39 AM PDT
Contact: Erica Heartquist, 503-871-8843, PHD.Communications@dhsoha.state.or.us
What: Today, the Oregon Psilocybin Services (OPS) Section within the Oregon Health Authority (OHA) released a new set of proposed rules.
All members of the public are invited to comment on the proposed rules during the public comment period scheduled for Nov. 1 – Nov. 21, 2022.
During this public comment period, people can provide their comments in writing and by e-mailing publichealth.rules@dhsoha.state.or.us or by participating in one of the three virtual public hearings scheduled for:
Tuesday, Nov. 15, 2022, 6:00 PM - 8:00 PM (Pacific Time) Wednesday, Nov. 16, 2022, 12:00 PM - 2:00 PM (Pacific Time) Thursday, Nov. 17, 2022, 10:00 AM - 12:00 PM (Pacific Time)
Spanish and American Sign Language interpretation, and CART captioning services, will be provided. If you have any questions about accommodations or need any assistance to participate please contact the Oregon Psilocybin Services team at 971-341-1713, 711 TTY or OHA.Psilocybin@odhsoha.oregon.gov and every effort will be made to respond to requests received in advance. Submitting your request as early as possible is greatly appreciated.
More information about the public comment period and how to attend the public hearings can be found here:
https://www.oregon.gov/oha/PH/PREVENTIONWELLNESS/Pages/Psilocybin-Administrative-Rules.aspx
Background: Ballot Measure 109 (M109), also known as the Oregon Psilocybin Services Act, was voted into law by Oregonians in November 2020. M109 directs the Oregon Health Authority (OHA) to license and regulate psilocybin products and the provision of psilocybin services.
OHA is currently in a two-year development period, from Jan. 1, 2021 to Dec. 31, 2022. During this time, OHA is working to build a new Oregon Psilocybin Services (OPS) Section and working with the Oregon Psilocybin Advisory Board and Rule Advisory Committees to establish rules for the production of psilocybin and provision of psilocybin services in the state of Oregon.
For the latest updates, subscribe to the distribution list at: oregon.gov/psilocybin
Everyone has a right to know about and use Oregon Health Authority (OHA) programs and services. OHA provides free help. Some examples of the free help OHA can provide are:
Sign language and spoken language interpreters.
Written materials in other languages.
Braille.
Large print.
Audio and other formats.
If you need help or have questions, please contact Meredith Rider at
971-341-1713, 711 TTY, or meredith.rider@dhsoha.state.gov.
You are subscribed to Oregon Health Authority News Releases. View all OHA news releases.
Contact Stay informed!


text:
Home/Tag: Sanofi
Tag Archives: Sanofi
FDA Approves Vaxelis, Sanofi and Merck's Pediatric Hexavalent Combination Vaccine
PARIS and KENILWORTH, N.J. – December 26, 2018 – The U.S. Food and Drug Administration has approved VAXELIS(TM) (Diphtheria and Tetanus Toxoids and Acellular Pertussis Adsorbed, Inactivated Poliovirus, Haemophilus b Conjugate [Meningococcal Protein Conjugate] and Hepatitis B [Recombinant] Vaccine) for use in children from 6 weeks through 4 years of age (prior to the 5th …
Denali Announces Positive Results with its Lead RIPK1 Inhibitor Molecule and Intention to Initiate Patient Studies in Multiple Indications in with Sanofi
SOUTH SAN FRANCISCO, Calif., Nov. 19, 2018 (GLOBE NEWSWIRE) — Denali Therapeutics Inc. (NASDAQ: DNLI), a biopharmaceutical company developing a broad portfolio of product candidates for neurodegenerative diseases, today announced positive results from its Phase 1 healthy volunteer study of DNL747, a brain penetrant small molecule inhibitor of receptor-interacting serine/threonine-protein …
FDA Grants Priority Review for Dupixent as Potential Treatment for Adolescents with Uncontrolled Moderate-to-Severe Atopic Dermatitis
Paris and Tarrytown, NY – November 6, 2018 – The U.S. Food and Drug Administration (FDA) has accepted for Priority Review the supplemental Biologics License Application (sBLA) for Dupixent® (dupilumab) in adolescent patients 12 to 17 years of age with moderate-to-severe atopic dermatitis, whose disease was inadequately controlled with topical therapies or …
Sanofi and Denali Therapeutics Partner to Develop Treatments for Neurological and Inflammatory Diseases
PARIS, France – November 1, 2018 – Sanofi plans to collaborate with Denali Therapeutics Inc. on the development of multiple molecules with the potential to treat a range of neurological and systemic inflammatory diseases. The two lead molecules (DNL747 and DNL758) target a critical signaling protein known as the receptor-interacting serine/threonine-protein …
FDA Approves Libtayo as First and Only Treatment for Advanced Cutaneous Squamous Cell Carcinoma
Regeneron Pharmaceuticals, Inc. (NASDAQ: REGN) and Sanofi today announced that the U.S. Food and Drug Administration (FDA) has approved Libtayo® (cemiplimab-rwlc) for the treatment of patients with metastatic cutaneous squamous cell carcinoma (CSCC) or locally advanced CSCC who are not candidates for curative surgery or curative radiation. Libtayo is a fully-human monoclonal antibody targeting the immune checkpoint …
Sanofi and REVOLUTION Medicines Partner to Develop Targeted Therapies for Multiple Cancers
CAMBRIDGE, Mass. & REDWOOD CITY, Calif.–(BUSINESS WIRE)–Sanofi and REVOLUTION Medicines, Inc. today announced an exclusive worldwide partnership to develop and commercialize targeted therapies, based on the biology of the cellular enzyme SHP2, for patients with non-small lung cancer and other types of cancer carrying certain mutations. This collaboration builds on precision oncology discoveries …
Adaptive and Sanofi Collaborate to Use Adaptive's clonoSEQ Assay in Active and Future Isatuximab Multiple Myeloma Trials
SEATTLE–(BUSINESS WIRE)–Adaptive Biotechnologies, the leader in combining Next Generation Sequencing (NGS) and expert bioinformatics to profile T- and B-cell receptors of the adaptive immune system, announced today that it entered into an agreement with Sanofi to utilize Adaptive's NGS-based clonoSEQ® Assay to assess minimal residual disease (MRD) status in response to …
FDA Grants Priority Review to Regeneron's Cemiplimab for Treatment of Advanced Cutaneous Squamous Cell Carcinoma
TARRYTOWN, N.Y. and PARIS, April 30, 2018 /PRNewswire/ — Regeneron Pharmaceuticals, Inc. (NASDAQ: REGN) and Sanofi today announced that the U.S. Food and Drug Administration (FDA) has accepted for priority review the Biologics License Application (BLA) for cemiplimab for the treatment of patients with metastatic cutaneous squamous cell carcinoma (CSCC) …
Sanofi to Acquire Hemophilia Company Bioverativ for $11.6 Billion
Paris (France) and Waltham, Mass. – January 22, 2018 – Sanofi and Bioverativ Inc., a biopharmaceutical company focused on therapies for hemophilia and other rare blood disorders, have entered into a definitive agreement under which Sanofi will acquire all of the outstanding shares of Bioverativ for $105 per share in cash, representing …
Alnylam and Sanofi Enter Strategic Restructuring of RNAi Therapeutics Rare Disease Alliance
CAMBRIDGE, Mass. & PARIS–(BUSINESS WIRE)–Alnylam Pharmaceuticals, Inc. (Nasdaq: ALNY), the leading RNAi therapeutics company, and Sanofi announced today a strategic restructuring of their RNAi therapeutics alliance to streamline and optimize development and commercialization of certain products for the treatment of rare genetic diseases. Specifically, Alnylam will obtain global development and commercialization rights …

redpajama_set_name: RedPajamaCommonCrawl


Find low prices on drum lifts, drum llifters, drum stackers and drum lifting equipment at Material Handling Solutions. Here you will find underdrum lifters. If you are looking for overhead lifts or cranes please see our cranes or our hoists section. Material Handling Solutions is a large distributor of material handling equipment, industrial supplies, office furniture, office supplies, and office electronics.


Conservative political commentator Tomi Lahren has been suspended from The Blaze, according to the Daily Caller.
The suspension is in effect for at least a week, the Daily Caller reported.
The fallout for Lahren, who has her own show on The Blaze, began when she said she was pro-choice on The View.
Among those criticizing her: The Blaze founder Glenn Beck.
Rule another label out. I am NOT a "constitutional". I believe in LIFE, liberty and property. Just an old fashioned "constitutionalist."
Lahren tried to defend herself over the weekend, tweeting that she would not apologize for "being an independent thinker."
Lahren's contract with The Blaze is up in September, the Daily Caller reported.


text:
#ifndef __MP_ADDITIONS_H__
#define __MP_ADDITIONS_H__

int
mp_singlecheck(const char *data, const char *end);

uint32_t
mp_sizeof_extl(uint32_t len);

size_t
mp_sizeof_ext(uint32_t len);

uint32_t
mp_decode_extl(const char **data);

uint8_t
mp_decode_ext(const char **data, const char **val, uint32_t *val_len);

char *
mp_encode_extl(char *data, uint32_t len);

char *
mp_encode_ext(char *data, uint8_t type, char *val, uint32_t val_len);

#endif /* __MP_ADDITIONS_H__ */

redpajama_set_name: RedPajamaGithub


text:
April 23, 2019 12:00AM EDT
European Union: Human Rights Questions for Candidates
Questionnaire Seeks Rights Positions from Major Parties
A general view shows the plenary room of the European Parliament during a voting session in Strasbourg, France, May 20, 2015.
© 2015 REUTERS/Vincent Kessler
(Brussels) – Candidates for the European Parliament should clarify their positions on key human rights issues at a time when respect for human rights and the rule of law are under attack from some European leaders, Human Rights Watch said today. Human Rights Watch has sent a questionnaire to leading candidates from European political groupings running for election and plans to make the responses public ahead of the 23-26 May, 2019 polls.
"The human rights stakes are high and European citizens should know how candidates view those issues ahead of the elections," said Hugh Williamson, Europe and Central Asia director at Human Rights Watch. "We hope to see candidates commit to protect the rights and dignity of everybody living in the EU."
The questionnaire, sent to lead candidates of six European groupings, urges them to state their positions on key human rights concerns. It includes questions on protecting the rule of law and democratic checks-and-balances in EU countries, addressing discrimination and inequalities, the rights of migrants and asylum seekers, counter-terrorism policies, and the place of human rights in EU's foreign policy.
The six European groupings are the Alliance of Conservatives and Reformists (ACRE), the Alliance of Liberals and Democrats in Europe (ALDE), the European Green Party (EGP), the European Left (EL), the European Peoples Party (EPP), and the Party of European Socialists (PES).
In May, Human Rights Watch will also release a series of videos in eight European languages highlighting the importance of human rights in the work of the European Parliament and urging European voters to think about human rights issues as they choose their future representatives in the European Parliament.
In recent years, several EU leaders have openly attacked democratic values, justifying policies that eroded the independence of court systems, and limited media diversity and the capacity of nongovernmental organizations to operate or receive funding. Populist political movements promoting agendas that weaken protection for women's rights or the rights of LGBTI people gained ground in several national elections and exerted a growing influence on the EU's political debates.
Despite manageable numbers of asylum seekers at EU's external borders, the EU focus has been on closing those borders, with little attention to the growing death toll at sea, the humanitarian impact at some EU borders, and abuses against migrants in places like Libya. Although the EU remains a leading force in promoting human rights globally, its resolve has been challenged by security and migration concerns, and by the withdrawal of the United States from international human rights forums.
On many issues the European Parliament has proved to be an important body in defending democratic institutions and human rights. For example, in September 2018, it initiated a political sanction process against the decline of the rule of law in Hungary. It has also helped to build pressure to respond to the attacks on the court system in Poland and has convened debates on the situation in Romania.
"We hope responses from the candidates help European citizens make informed choices about what their vote will mean for human rights," Williamson said. "We believe that given the challenges ahead for the new European Parliament, human rights are key to Europe's future."
Europe/Central Asia
EU: Leaders Stoke Fear, Ignore Rights
March 6, 2019 Dispatches
Time for European Conservatives to Take a Stand on Hungary
Russia: Escalating Persecution of Jehovah's Witnesses
Arrests, Prison, Harassment for Peaceful Religious Practice
Greece: Unaccompanied Children at Risk
Arriving Alone in Island Camp, They Face Insecurity, Neglect


text:
Iran Tightens Security as Subsidy Cuts Loom
BEIRUT—Iranian authorities are taking extraordinary security measures ahead of cuts to energy and food subsidies this month, in an effort to prevent unrest by a public upset about rising expenses and inflation.
President Mahmoud Ahmadinejad cautioned the public on Wednesday against politicizing the economic reforms and threatened to severely punish businesses that raise prices of consumer goods in reaction to subsidy cuts.
In a nationally broadcast speech to a rally in the northeastern city of Bojnurd, he said government agents would "catch and fine anyone who abused the situation and make them regret it forever."
The government, in a five-year phaseout plan, seeks to eliminate up to $100 billion a year in food and energy subsidies that keep costs down for consumers. About 65 million Iranians, out of a population of 75 million, will receive cash payments of about $40 a month to ease the economic pain of lost subsidies. Cash payments in some provinces have already begun.
PrevPreviousMohammad Seifzadeh
NextKhamenei the world's 40 th most powerful people!Next


text:
Nearly 4,000 children have been evacuated from both sides of the Crimean border due to pollution-related health and safety concerns. The industrial pollution has long been a problem but the ecological crisis caused officials to start taking notice in August when metal items began rusting, fields suddenly changed colors, and the air became dense with mist.
According to doctors and ecologists, the disaster is the result of massive industrial pollution stemming from a Titanium Dioxide plant in Armyansk, Crimea. Titanium Dioxide is a powder used in the manufacturing of everyday items like paint and toothpaste. Due to the area's current state of political unrest, both countries have weighed in on possible causes and, while their explanations may differ, they do agree that the plant is the original source of the pollution. Ukrainian officials believe the pollution is originating from the current operations of the chemical factory, while the Russians claim that a dried up sulfur waste effluent pond is the source.
Determining the actual cause is critical at this point because of both safety and politics. If Russian officials are correct and the effluent pond is the source, that would mean that the pollution is the direct result of area conflicts. For many years, the effluent pond created by Titan was fed by River Dnieper of Ukraine during the dry summer seasons. Once Russia annexed Crimea, Ukraine cut off the water supply in an act demonstrating their refusal to accept the annex. The lack of water flowing to the pond enabled it to dry out, thereby increasing its acid concentration. Windy conditions easily blew the residue out of the pond bed.
The blowing residue has negatively impacted individuals on both sides of the Crimean border. Nearly 22,000 citizens living in the Russian-controlled city of Armyansk are forced to keep their doors and window shut tight for safety. There are frequent reports of shortness of breath, burning sensations in their nose and throat, and allergy development. Ukrainian physicians also fear that complaints of stomach ailments will rise as an after-effect of eating contaminated fruits and vegetables or other inadvertent ingestion. A state of emergency has been declared in Armyansk.
Citizens are concerned about their safety and their health, although not many of them have the capability to leave the area. Russian authorities in Crimea worked quickly to shut down the factory, which has been in operation for nearly 50 years and employs more than 4,000 people, in early September. It remains unclear, however, whether the shut-down will help if the source is indeed the effluent pond.


text:
Chesapeake Chamber Music Festival / Stamitz, Previn, Brahms
Artists: Mark Hill (Oboe); Marc Goldberg (Bassoon); Patrick Pridemore (French Horn); Ieva Jokubaviciute (Piano); David Bowlin (Violin); Daniel Phillips (Violin); Catherine Cho (Viola); Todd Phillips (Viola); Michael Kannen (Cello); Marcy Rosen (Cello)
Stamitz: Trio for Horn, Violin and Basso Continuo in E flat major N/A
Composer: Carl Stamitz
Artists: Patrick Pridemore (French Horn); David Bowlin (Violin); Michael Kannen (Cello)
Venue: Newberry Auditorium at MEBA Engineering School (Easton)
Recording Date: Sat 14 Jun 2008
II - Polonaise (1:32) N/A
III - Vivace (1:57) N/A
Previn: Trio for Piano, Oboe and Bassoon N/A
Composer: André Previn
Artists: Mark Hill (Oboe); Marc Goldberg (Bassoon); Ieva Jokubaviciute (Piano)
I - Lively (6:04) N/A
II - Slow (7:01) N/A
III - Jaunty (5:14) N/A
Brahms: Sextet for Strings no 2 in G major, Op. 36 N/A
Composer: Johannes Brahms
Artists: David Bowlin (Violin); Daniel Phillips (Violin); Catherine Cho (Viola); Todd Phillips (Viola); Michael Kannen (Cello); Marcy Rosen (Cello)
I - Allegro non troppo (15:02) N/A
II - Scherzo: Allegro non troppo (7:51) N/A
III - Poco adagio (9:31) N/A
IV - Poco allegro (8:43) N/A


text:
Q: ng datepicker css customization and change the selected date from sqaure to circle i want to custom the calendar from this[enter image description here][1]
to this[enter image description here][2]
Can anyone please help me out. as i am not able to understand how to change selected day cell from sqaure to circle and select the range in same way.
[1]: https://i.stack.imgur.com/JRprR.png
[2]: https://i.stack.imgur.com/mnQ9P.png

A: ngbCalender loads inner style in your page when page loads in browser. If you inspect and in Element tab search for .customer-day class and append border-radius : 50% ... See image below , and for Range background there will be class .range change it according to your desire color and background shape.

redpajama_set_name: RedPajamaStackExchange


DIY and Food. Cooking tutorials and reviews on food. Pinterest ideas.
I love to create new ways to eat vegetables - from seasoning and herbs, to creative uses and ways to cook them, there seems to be endless options for me to try. However, when it comes to autumn and winter squashes and other seasonal veggies, I tend to go back to a veggie medley that is seasoned with an unexpected dash of yum: Jamaican jerked seasoning.
Whaaat? Yes, Jamaican jerked seasoning. With its subtle hints of cinnamon and spice, it pairs so well with the sweetness of the sweet potatoes and squash, but adds enough savory notes with its spicy pepper seasonings for other veggies, such as red-skinned potatoes and zucchini.
Preheat oven to 325 degrees F. Lay parchment paper down on baking sheet. Chop up your veggies. Mix oil and seasonings in a large bowl until well mixed. Toss veggies into oil mixture until well coated and place onto baking sheets evenly. Bake for 30-40 minutes or until fully cooked. Enjoy!
Let me know if you try and enjoy it!
Every year for the Christmas season, I bake up a storm. With all the powdered sugar I use up between all my sugar cookies and my chocolate crinkles, I am not surprised that my kitchen does actually look as though a snow storm hit!
My mom used to bake a lot when I was growing up, and she'd make chocolate crinkles. Every Christmas without fail, there were chocolate crinkles scattered on her giant platters, amongst all the other decadent candies, breads, cookies, and various assortments of other confections. Of all, these were my absolute favorite and still are.
I don't actually know the exact recipe she used though, but I did find a Betty Crocker version that works wonderfully, albeit a few modifications that I like to make (such as using coffee to intensify the bold cocoa flavor) and adding more vanilla to give a nice hue of dreamy chocolate flavor.
1. Mix oil, melted and cooled chocolate, sugar, cold coffee, and vanilla together.
2. Add in your eggs, one at a time to allow for even mixing.
3. Add flour, baking powder, and salt gradually. Mixture should form a ball in bowl. Cover and refrigerate for about 2 hours.
5. Line cookie sheet with parchment paper (or grease sheets evenly).
7. Place on cookie sheets about 2 inches apart and bake for 10-12 minutes.
8. Immediately transfer baked cookies to cooling racks.
9. Take caution - these are addicting little treats that are best served with hot cocoa or steamed milk.
These are so so good! You will not be disappointed.
If you want to try some different intensities of chocolate oomph, I recommend trying it with semi-sweet chocolate - 2 oz semi-sweet, and 2 oz unsweetened. Or, if you want to die from chocolate over-load, try milk chocolate instead (but reduce the amount of sugar used and add a little extra flour instead).
Let me know what you think and if there are any other types of yummy recipes you'd like to see, or any fun DIY projects.
Hello dear readers! It has been quite some time since I have posted any recipes or food ideas here on this blog, but I do have one for you now using a vegetable that a lot of people completely misunderstand or undervalue, which is a shame since these veggies are packed with high vitamins and other nutrients.
When I was little, I wasn't really picky about the vegetables I ate, except for in the case of beets and canned spinach. Some veggies, such as Brussels sprouts, were never really an item that made it onto the dinner table since my parents didn't have a huge love for them. Eventually, I grew up to love eating them and am always on Pinterest finding new ways to eat these (and other) delicious veggies.
For Friendsgiving Day, (Thanksgiving for all our friend group hosted just before or after Thanksgiving day) I decided to make a veggie side with Brussels sprouts. At first, I wanted only to make a balsamic-maple glaze, but decided on a more "southern" style instead - utilizing the double-b's of fat: butter (affectionately known as, 'buddah') and bacon.
After rinsing and straining nearly 2 lbs of B. Sprouts, I added them into a small pot and flash boiled them until the were almost fully cooked. I left them somewhat undercooked though to prevent them from getting mushy from overcooking. These were boiled for about two minutes.
During earlier prep work, adding 1 lb of bacon to a parchment paper-lined cookie sheet to bake in the oven on 350 degrees F. will create a crispy bacon that can later be chopped up and used in the dish later.
Out of all the steps, this is probably the most satisfying. It smells completely fantastic as you add half a stick of butter to a hot skillet and add chopped up onion to caramelizing leeks. This will made your stomach grumble with hunger after smelling this amazing olfactory pleasure.
Not that the buttery leeks is enough, I created this cheesy buttery sauce using the remaining half stick of butter with half a package of cream cheese. What's important here is that using a whisk will prevent the cheese from getting chunky from the heat, and will emulsify the butter into the creamed cheese.
Next, mix the leeks in to make sure that any residue of butter is absorbed by the Brussels sprouts. Then, afterwards, add the remaining ingredients and mix together well. It makes me hungry just to think about the bacon and leek scents combining with the browned butter. Nom nom nom.
Finally, add your butter-cheese sauce and coat everything evenly. Add your sprouts to an oven-friendly casserole dish, or else line a pan with foil (for easier cleanup) and add in the sprouts. If using foil, be sure to keep the sides opposite from each other long so that you can use it to cover the sprouts when you are baking them. Bake at 350 degrees for about 15-20 minutes.
As I said, my friends, even a very picky eater, complimented me on this dish and asked me for the recipe. Yay!
I hope you enjoy this post. Please let me know if you have any other requests!
Wow, it has been a long time since I've blogged here in this particular blog, sadly to say. As I've started to upload more videos to my All Things Emmy YouTube channel, those uploads will coincide with blog posts here.
I recently uploaded a new video to YouTube for a super easy DIY project in making a tiered stand. What's great about this project is that it will cost less than $10 to make, including the glue.
I bought the large plates and wine glasses at my local Dollar Tree, but the smaller plate at Value Village. The glue was something I purchased at WalMart, I believe.
I hope you enjoy this tutorial! Please let me know if there is anything else you'd like to see.
Now before I begin to give you the steps in preparing this delicious and somewhat healthy snack, I must warn you that are dangers to making these cookies. Not only are you going to be tempted, but those around you will be tempted too. They are nutritious, yes, but if you have too many, you'll be in trouble. They are so addicting that I have to call these my little purple cookie monster creations...And yes, they are a very dark purple hue.
At one point, I had an extremely bad sweet tooth--I mean, it was so insatiable that even a Yoplait yogurt cup couldn't help me. I sought out to make a delicious and quick confection with the ingredients I had available to me, which ultimately made my confection experiment a 'vegan' one. Even the chocolate used doesn't have dairy... I have never baked vegan before, nor did I have an example recipe for referral. I took the few skills and knowledge I possess in regards to the composition of a cookie and went from there.
Having staved off of eggs and butter recently meant I had to find another way to add in the moisture...I went to my freezer and found that I had some organic frozen wild blueberries. I also have agave syrup, vanilla, and water on hand. Almond butter also came to mind as it has an amazing creaminess that I wanted to replicate in the cookies' texture.
So the dry ingredients felt less intimidating since I did already have flour, baking powder, stevia, (which is easily measured cup-per-cup comparably to sugar) unsweetened cocoa powder, salt, and oats! Chocolate and oats are so delicious together and the last time I had an oatmeal chocolate chip cookie was so easily forgotten because it has been so very long! I knew where to draw inspiration.
Yes, you can drool now.
The best part about these cookies are their low sugar and high nutrient contents. Using stevia and agave syrup together helps to curb the amount of sugar released into your blood, reducing a sugar spike. The blueberries also have amazing antioxidants and fiber. I even had a couple for breakfast the day after making them. One of my roommates is eating a vegan diet and is very conscientious of what she is eating, and the other eats somewhat healthily too. I got asked for seconds. That is a sure sign of making some delectable treats.
The recipe I threw together makes approximately 36 cookies. This a drop-dough cookie and a small teaspoon is about all you need per cookie. Each cookie has approximately 70 calories (yes, I carefully calculated this) and a few grams of sugar (I haven't calculated the sugar content though).
Mix all dry ingredients together. Then add in the thawed blueberries. Mix.
Mix almond butter, vanilla, agave syrup, and blueberry puree together in separate bowl. Using a fork significantly helps if you do not have an electric mixer.
Cut wet ingredients into dry. Because the almond butter is a little tough, you may have to add a little water into the mix to help smooth out the consistency.
Use a tablespoon to measure out cookie dough and drop onto greased cookie sheet. Dough will not spread out much, so you can have as many as 15-18 cookies on a standard sheet.
Bake 10-12 minutes. They will be firm to the touch if cooked thoroughly. Remove from heat.
Optional topping: Melt semi-sweet squares on double boiler or in microwave on High for 1 1/2 minutes. Use teaspoon to apply. Optional garnishment: chopped nuts sprinkled on hardening chocolate.
This is a recipe that I had to carefully remember since I rarely like something enough to have to recreate it for the world in written form. I am so glad I did! Please let me know what you think if you do make them!
Good morning and happy almost-summer to you all!
Today's yummy concoction is inspired by trying to eat healthily without excess sugar, and incorporating more vegetarian eating habits into my diet. It is very nutritious and delicious, not to mention, there's a huge variety of things you can do to change it up!
Experts say that eating oatmeal is extremely good for your health, particularly if you have cholesterol issues. Well, I don't know about you, but oatmeal can get quite bland and boring with the same old milk and sugar routine--even if you add in brown sugar and maple. With all that sugar, however, it starts to deplete the health benefits a little.
One of my favorite ways to add some variety to my oatmeal is to add fruit. I find that cooking it down a little reduces the tartness and brings out the natural sugars in the fruit. No need to even add any artificial sweeteners! Yay!
Here are the instructions (given to you in the form of pictures). Enjoy!
With a little water in the pan, add in your oats. Cook on medium heat.
Add in your favorite serving of yogurt. It can be low-fat or even Greek. I used Yoplait's Strawberry Banana flavor. Continue to cook on medium heat.
Now add in some chopped fruit. I chopped up an apricot that was still a little tart. You can use berries or other varieties of fruit. Let this cook in the yogurt for a few minutes, especially if the fruit is a little tart and/or hard.
This part is optional, but you can chop up your favorite nut and add it into the mix as well for a little added protein and texture. These are hazelnuts. Almonds, walnuts, and even pecans work well.
Let everything cook together until the fruit is softened and has absorbed sugar from the yogurt. Reduce heat if necessary.
Now you're ready to eat! Have fun mixing flavors and fruits!
Oatmeal has 0 grams of sugar, so the only sugar you're eating is from the yogurt--the apricot or other fruits have natural sugars, which are better for you, nutritionally as they come paired with vitamins and minerals, not to mention, fiber. Greek yogurt would be great to use as well, but keep in mind that once you heat it up, the probiotics are depleted. It does have more protein than traditional yogurt, and if you buy it plain, you can add stevia extract for sweetness, and vanilla flavoring. Ta-da!
*A note for vegans: this recipe is easily interchanged with a vegan alternative to yogurt. Just find a coconut, almond, or soy-based yogurt and substitute that instead.
I know, I know. It has been forever ago since I have updated this blog--between school, Simply Bellezza, moving, and church, I've come to find myself busier than I could have imagined.
So, if you are a follower who is not familiar with my more personal blog or my more personal YouTube channel, Emalyce89, then you have missed out on what this year has meant for me. I started off the New Year by attempting the ever-so-popular resolution to lose weight--a desired total of 75 pounds lost this year. I am down approximately 30 or so pounds and did so by beginning the first 35 days or so of the year with a book called The 17 Day Diet, by Dr. Mike Moreno. I won't go much into details about the diet or specifically how it works, but I will say that one of the big changes is to NOT eat any wheat products for the first two cycles (a total of 34 days). In the first month, I dropped a significant amount of weight and have since felt amazingly well. However, upon reintroducing wheat into my diet, I felt nauseated and bloated, in addition to actually gaining weight for the first time in the 2012 weight loss endeavors. Not good.
I noticed that whenever I ate any wheat products, I had these types of digestive symptoms; however, when I didn't eat wheat, I didn't have any digestive issues whatsoever. I linked that to gluten. I haven't been formally diagnosed with gluten-sensitivity, but I do notice an enormous difference. My mom was diagnosed with gluten intolerance (not Celiac's though) nearly six months ago. She had gone through her regularly scheduled colonoscopy to be told that she had numerous amounts of polyps forming throughout her colon. She has had digestive problems for years, including nearly fatal bacterial infections that started from inside her colon. Her doctor attributes these issues to gluten-sensitivity.
Since she and I have similar symptoms regarding gluten, I have come to the conclusion that I need to back off the wheat products and opt for products that are gluten-free and won't upset my sensitive tummy. So how to do that and where to begin?
Well, let's first talk about what gluten is.
Gluten is a protein found in wheat and other grains (such as barley), giving it the elasticity and chewiness that you would notice in a slice of white bread. For example, gluten is what gives breads the ability to rise since it holds gases that expand when reacting with yeast. Just as an added tidbit, gluten is not soluble in water, that is, it doesn't dissolve in it. I like to think of this as an oil and water solution--they just don't mix. Gluten will float out and off a ball of dough consisting of wheat flour. Pretty interesting, right?
So I wanted to share with you my latest and yummiest gluten-free discovery. It's pretty popular now, but either way, I think I'll share it with you.
If you are a sandwich lover like me, giving up bread sounds like being cursed to embrace salads for lunch ALL the time! Not true. There are a few different brands that carry gluten-free bread; it becomes more difficult finding one that actually tastes good without being as hard as cardboard.
My solution: Udi's White Sandwich bread.
It doesn't gum up at the top of your mouth either, which is the best part! (Especially if you are an avid eater of PB&B (that's peanut butter and banana) like I am!) The serving size is two slices (which is 140 calories). Their slices are about 3/4 the size of regular sandwich bread, which is fine as a dieter, being the perfect size for me, but for anyone else who wants to experience a fuller, larger sandwich, this will be a slight problem for you. The rest of the pertinent dietary information: 3g of sugar, 22g carbs, 3g protein, 1 g fiber. It does contain eggs, so if you are vegan, this is not for you either.
Price: According to the Udi's website, it is $5/loaf; however, depending on where you purchase your groceries, it may be less. I bought mine at Fred Meyer for about $4.50.
I hope this helps you and if you have any questions or comments, feel free to leave them below! I'll get back to you when I can!
Looking to Buy? Shop Amazon!


text:
\section{1.~Introduction}

\noindent
The Sherrington-Kirkpatrick model for spin glasses ([\number\SK],
SK model in the following) was proposed in 1975 as a simple and exactly
solvable model of a disordered frustrated system.  This is a
model with very long range interaction,
for which the mean field theory is supposed to hold exactly, and
therefore it should be easy to handle. But the hopes of the authors were
soon disappointed: the model is exactly solvable, but its solution
turned to be much more complex than what initially thought. Only in
1980 the work of Parisi [\number\Parisi]
and the complicated and mysterious ``breaking of the replica symmetry''
solved the problem, and only a few years later
the so called {\sl cavity method}
[\number\Overlap,\number\Ultra,\number\RFE,\number\Cavity]
revealed the physics hidden behind the intricate mechanism of the
replica formalism.

Two features appear to be essential in the model, both linked to a highly
non-trivial decomposition of the Boltzmann measure in pure states:

\item{1)} the pure states are organized in an ultrametric scheme
[\number\Overlap,\number\Ultra];

\item{2)} the free energies at each level of this organization are
Poissonian random variables with exponential density, and the levels of
organization are connected through a probability cascade {\sl \`a la} Ruelle
[\number\RFE,\number\Ruelle].

It is not yet clear whether this picture can be applied to other frustrated
disordered models, such as 3-dimensional spin glasses with nearest
neighbours interaction [\number\CPPS,\number\FH].\footnote{${}^{(*)}$}
{Models like the REM and the GREM of Derrida [\number\REM,\number\GREM],
though very instructive, are rather artificial.}
For this reason, we have considered a model which preserves many features of
the SK model, but is much simpler to be analyzed numerically.
Our hope is
twofold: to clarify if and how one can apply the same ``ultrametric''
picture as the SK model, and to study the nearest neighbours
interaction case, taking
advantage of a better numerical tractability. This letter
represents a first step in this direction.

\medskip

The outline of the paper is as follows. In section~2 we define the model,
in section~3 we clarify the concept of frustration in the context of our
model, in section~4 we propose two approximate analytical solutions of the
model, in section~5 we show some numerical results, as a test for the
analytical ones.

\section{2.~Definition of the model}

\noindent
We consider a set of points $V$, connected through links of a
given length, defining the distance between each pair of points of $V$.
We start by stating the problem with the greatest generality,
with the points of $V$ not necessarily
distributed in the real space, and their relative
distances not necessarily being the euclidean ones.
To be precise,
let us recall a few definitions. A {\sl weighted graph} is a triple
$ G=(V,A,d) $ where $V$ is a finite set of {\sl nodes} or {\sl vertices},
$ A \subset V \times V $ is the set of the {\sl arcs} ({\sl i.e.} the links)
which connect the
nodes of $V$ and the map $ d : A \to {\bf R}^+ $ gives the length
(or {\sl weight}) of an arc, {\sl i.e.} the distance between
the nodes of an (ordered) pair.
A {\sl path} of $n$ steps in the graph
starting from $u$ and leading to $v$ is a sequence of $n+1$ nodes
$ \{ u_i \}_{i=0}^n $ such that
$$
\eqalign{
u_0 = u \; , \quad u_n & = v \; ,
\quad u_i \not= u_j \; \; \forall \; i,j = 0 , \dots , n \; \; i\not=j \cr
& (u_i,u_{i+1}) \in A \; \; \forall \; i = 0 , \dots , n-1 \cr
} \eqno (\number\defpath)
$$
Let now $ C_{u,v} $ be the set of the paths joining $u$ to $v$,
irrespective of the number of their steps. We suppose that the graph is
connected, {\sl i.e.} that there is at least one path joining $u$ to $v$.
The length of paths in $ C_{u,v} $ is represented by a map
$ l $ defined on $ C_{u,v} $ in a natural way, {\sl i.e.}
$ l : C_{u,v} \to {\bf R}^+ $ such that
$$
c \in C_{u,v} \; , \; c = \{ u_i \}_{i=0}^n \; ,
\quad l[c] \equiv \sum_{i=0}^{n-1} d(u_i,u_{i+1})
\eqno (\number\defl)
$$
Our system has phase space
$$
\Omega = C_{s,t}
$$
with $s$ and $t$ fixed in advance, and Hamiltonian
$$
{\cal H} = l
$$
The ground state of the system is the shortest
path in $G$ between $s$ and $t$. Using
a standard algorithm, the Dijkstra's algorithm
(see for instance [\number\Papa] and references therein), it can be
determined in a time which grows not too fast with the size of the
problem, for any realization of the distance matrix $ d(u,v) $.
On the contrary, to find the ground state of the SK model is a so
called $NP$-complete problem (see [\number\Papa] and also
[\number\Barahona]) and therefore it is very unlike
that a ``good'' algorithm can be found for it.

The distances $ d(u,v) $ play the same
role of the couplings $J_{ij}$ among
the spins in a generic model of interacting Ising spins.
In the case of our interest
they will be random variables and will represent
the ``disorder'' of the model. How frustration arises in this context will
be discussed in the next section.
As anticipated, our main interest in studying the problem is to
gain informations about the behaviour of disordered and frustrated systems,
in a context which has a rather rich structure, but nevertheless is
simpler to handle than spin systems. Apart from this rather abstract
motivation, the study of our problem could be useful in the framework of
high-temperature expansion for disordered systems of interacting Ising
spins. Moreover, also the problem of directed polymers in a random medium
(see, for some recent results, the work in [\number\dyrpol] and references
therein) could be stated as a particular case of our problem. In fact, a
directed polymer is a path in a particular graph which has a
privileged direction of motion: that is to say, with the
previously introduced notation, that if the
link $ (u,v) \in A $ points in the privileged direction then
$ (v,u) \not\in A $ (the reversed link is absent). With our notation
the random medium is represented by a distance matrix written in terms of a
single random function $ \phi $, the local disorder,
$$
d (u,v ) = { \phi(u) + \phi(v) \over 2 }
$$
so that equation~(\number\defl) for the length of a path $c$ of
$n$ steps becomes
$$
l[c] = \sum_{i=0}^n \phi(u_i) - { \phi(u_0) + \phi(u_n) \over 2 }
$$
which is the usual potential energy term in the hamiltonian of a polymer in
a local disorder field, apart from a contribute vanishing in the limit
$ n \gg 1 $.

The model we shall consider in the following is defined by:

\item{1)} $ (u,v) \in A $ for every $ u,v \in V $, with $ u\not=v $;

\item{2)} the $ d(u,v) $'s are independent identically
distributed random variables (with $ d(u,v) = d(v,u) $).

\noindent
Statements~1) and~2) are the counterparts
of the defining characterization of
the SK model, the ones by which mean field theory turns out to hold
exactly. This analogy makes us confident that one could find an exact
solution for our model, although the meaning of
``mean field theory'' in our context is not clear.
Moreover, the fact that from a numerical point of view the problem
is much simpler to deal with than the SK model, lead us to hope
that the same happens also in an analytical approach.
We did not succeed in finding such an
exact solution, nevertheless thanks to the~1) and~2) above we can obtain an
approximate solution (see section~4) which fits well enough the numerical
data (see section~5).

We choose as probability distribution for the $d$
$$
p_\alpha (x) = ( \alpha+1 ) \cdot x^\alpha \quad , \quad x \in [ 0,1 ]
\eqno (\number\defp)
$$
In the analytical computation of section~4, $\alpha$ remains a free
parameter, whereas in the numerical simulation of section~5 we fix, just
for concreteness, $\alpha=0$ and $\alpha=1$, {\sl i.e.} we consider,
respectively,
uniform and linear probability distribution of the distances.

\section{3.~Frustration}

\noindent
It is not immediate to understand how the usual concept of frustration,
developed in the framework of spin systems, applies to this context. What
naturally characterizes frustration, in a context-free way, is the
existence of a lot of local minima, hampering the selection of the global
one. Although in our model we can not precisely identify the local minima,
we propose a generic definition of frustration, which easily
implies that our model is frustrated.

It is now useful to introduce another model, structurally
similar to ours.
Let us recall that a {\sl tree} is a connected graph without cycles
and a {\sl spanning tree} of a graph is a tree which has the same set of
nodes as the graph. The phase space of this second model is the set
${\cal S}T$ of all the spanning trees of $G$, and the Hamiltonian is the map
{\sl weight} $ w : {\cal S}T \to {\bf R}^+ $
naturally defined by
$$
T \in {\cal S}T \; , \; T = (V,A_T) \; ,
\quad w[T] \mkern 5mu \equiv \mkern -15mu \sum_{(u,v) \in A_T} d(u,v)
\eqno (\number\defw)
$$
In the following we shall call {\sl minimal spanning tree} ($MST$) the ground
state of this system. As a matter of fact, the Dijkstra's algorithm,
which determines the shortest path from $s$ to $t$,
gives also the shortest path from $s$ to any other $ u \in V $.
These paths form a tree, the so called {\sl shortest path tree} ($SPT$).
Thus, to a certain extent, the $SPT$ problem and the $MST$ problem are similar,
as their solutions are both represented by trees on the graph (they do not
coincide, however, as it is easily shown by a counterexample).
On the other hand the algorithms which solve them are rather different.
The Dijkstra's algorithm is somewhat tricky.
On the contrary, the algorithm which solves the $MST$ problem is much simpler:
$$
\beginalgol
\+ $ MST := \emptyset $ \resword{;} \quad $ A' := A $ \resword{;} \cr
\+ \resword{repeat} \cr
\+ & \resword{let} $a$ the arc in $A'$ of minimal length \resword{;} \cr
\+ & $ A' := A' \setminus \{ a \} $ \resword{;} \cr
\+ & \resword{if} $ MST+a $ is a tree
  \resword{then} $ MST := MST + a $ \resword{;} \cr
\+ \resword{until} $ A' = \emptyset $ \resword{;} \cr
\egroup
\eqno (\number\greedy)
$$
This algorithm is called {\sl greedy} for obvious reasons.

Systems for which the greedy algorithm to find the ground state
works are called {\sl matroids}. Though this seems to be an informal
characterization, it turns to be equivalent to a purely algebraic one (see
[\number\Papa]).
For matroids there exists a natural way to give a notion of
neighbourhood between the points in the phase space, that for the spanning
trees of a graph results in:
$$
\mathinsert{If $ T \in {\cal S}T $ then the neighbours of $T$ are all the
$ T' \in {\cal S}T $ obtained from $T$ as follows: add an arc
to the tree $T$, producing a cycle; then delete one other arc on the cycle.}
\eqno (\number\neighMST)
$$
If we consider the local minima of the Hamiltonian $w$ with respect to this
``topology'', it is an amusing exercise
[\number\Papa, exercise~3 chap.~1]
to show that, apart from accidental degenerations, there exists a
unique local minimum, which therefore is a global one. Thus the following
{\sl steepest descent} algorithm is equivalent to the greedy one, in order
to determine the $MST$:
$$
\beginalgol
\+ \resword{let} $ T \in {\cal S}T $ \resword{;} \cr
\+ \resword{repeat} \cr
\+ & \resword{let} $T'$ the neighbour of $T$ of minimal weight \resword{;} \cr
\+ & \resword{if} $T'\not=T$ \resword{then} $ T := T' $ \resword{;} \cr
\+ \resword{until} $ T' = T $ \resword{;} \cr
\+ $ MST := T $ \resword{;} \cr
\egroup
\eqno (\number\steepest)
$$
Now, it seems natural to consider the matroids as the archetypical example
of a non-frustrated system, in contrast with the frustrated ones, where, as
it is well known, the huge number of local minima causes the steepest
descent method to fail. We remark that, to adapt the
greedy algorithm to a spin system, it is necessary to orient subsequently
the spins of the system, so that the couplings with the already
oriented ones are satisfied.
It is not difficult to make the above procedure formal, and
to verify that it works if and only if for every $n$-tuple of lattice sites
$ \{ i_1, \dots, i_n \} $ the following condition holds:
$$
J_{i_1 i_2} \cdot J_{i_2 i_3} \cdots J_{i_n i_1} \ge 0 \; ,
$$
where $ J_{ij} $ is the coupling constant between the spin $\sigma_i$ and
the spin $\sigma_j$. This is nothing but the usual definition of absence of
frustration. In conclusion, two facts lead us to conjecture that matroids
are the most general non-frustrated systems:

\item{-} there exists a
unique local minimum, which therefore is a global one;

\item{-} the spin systems which are matroids are the non-frustrated ones,
in the usual meaning.

\noindent
{}From this point of view our model is frustrated, and the $MST$ problem
becomes
its non-frustrated counterpart.
To prove that our system is frustrated it is sufficient to show
that it is not a matroid, and this can be done in two ways:
either by its algebraic characterization,
or by showing with a counterexample that to be greedy
does not work. Roughly speaking, in order
to find the shortest path between $s$ and
$t$, it is not sufficient in general to proceed na\"\i vely, {\sl i.e.} by
choosing always the shortest arc when going out from a node.

\medskip

In the following, we shall fix our attention on the shortest path between
two fixed nodes in a graph, which we denote with $s$ and $t$. The size
of the graph, {\sl i.e.} the number of its nodes, will be denoted by $N$, and
we
shall be interested in the asymptotic properties in the limit
$ N \to \infty $. Moreover, as it will be clear in the
following, we shall study, together with the shortest path, also the paths
in the graph between $s$ and $t$ which are next to the shortest one,
{\sl i.e.} we order the paths by increasing length, and we shall be interested
not only in the first one, but also in the second one, the third one,
and so on.
These will be called ``subminimal'' or ``suboptimal'' paths, whereas the
shortest path is the minimal, {\sl i.e.} the optimal, one. We do not attempt to
give a rigorous definition of this notion, {\sl i.e.} we do not
say how close a path should be to the shortest one, to be called a
subminimal one. However, if we order by increasing length all the paths
between $s$ and $t$ in the graph, it is clear that the $h$-th one, with $h$
fixed, is always subminimal in the limit $N\to \infty$, since it becomes
{\sl infinitely} close to the shortest one. In fact,
as far as we could say about our problem, by a partial analytical solution
and a numerical one, all the shortest $k$ paths in the graph, with $k$
fixed, in the limit $ N \gg 1 $ share the same leading behaviour
{\sl versus} $N$ (see (\number\lnresult) below, and section~5). Thus we
could take as an informal characterization of the subminimal paths the
fact that they all have the same asymptotic behaviour {\sl versus} $N$, when
$N$ goes to infinity. It is clear that the number of subminimal paths
must grow with $N$, but we do not know at all how.

We believe that this characterization, though informal, is not trivial.
Roughly speaking, we do not know exactly which paths are the subminimal
ones, but we can argue which of them are surely not. In fact,
let us suppose we are
looking for the shortest path between $s$ and $t$ which visits also
{\sl any} other node in the graph. Such a path is the so called
{\sl Hamilton path}; to find it is a problem very close to the well
known {\sl travelling salesman problem} ($TSP$), which is a so called
{\sl $NP$-complete} problem (see [\number\Papa]), {\sl i.e.} so hard to solve
as to
find the ground state of the $SK$ model. On the other hand, our problem,
{\sl i.e.} the problem of finding the shortest path between two nodes in a
graph, is
one of the so called {\sl class-$P$} problems, and it is much easier to
solve numerically. It is commonly believed (see [\number\Papa]) that
$NP$-complete problems are drastically different from class-$P$ ones, at
least as far as the cardinality
of the continuum ({\sl i.e.} the number of points on the real line),
is drastically different from the cardinality of
natural numbers. Thus we argue that the class of
subminimal paths, {\sl i.e.} the paths close to the shortest
one, is drastically different from the class of the paths close to the
Hamilton path.

\section{4.~Two partial solutions}

\noindent
The following heuristic argument gives some insight on the ground state
of our model; but, in fact, it is easy to see that the argument should work
only in the non-frustrated case, {\sl i.e.} in the $MST$ problem.
Nevertheless we will see later that
the resulting relations fit the numerical data well enough
(see (\number\nftype) and next section). We try now
to estimate the length of the shortest path between $s$ and $t$.
Starting from $s$,
we select the outgoing arcs shorter than $\epsilon$ (to be determined),
and so on, until we reach $t$.
At each step the average number of selected outgoing arcs is
$$
N_\epsilon = N \; \int_0^\epsilon p_\alpha (x) \, dx
$$
where $ N = | V | $ is the number of nodes in $G$. After $n$ steps, the
number of reached nodes is about $ N_\epsilon^n $
(if $ N_\epsilon \gg 1 $). The probability of reaching $t$ in $n$ steps is
one if $ N_\epsilon^n = N $, {\sl i.e.}
$$
n = { \ln N \over \ln N_\epsilon }
$$
The length of the path thus constructed is approximately
$$
l = n \cdot { \int_0^\epsilon x p_\alpha (x) \, dx \over
       \int_0^\epsilon p_\alpha (x) \, dx }
$$
Taking the minimum as $ \epsilon $ varies, we get
$$
\eqalign{
n & = { \ln N \over \alpha + 1 } \cr
{ l \over n } & = { K \over N^{1/(\alpha+1)} } \qquad
\hbox{($ K = { \alpha + 1 \over \alpha + 2 } {\rm e} $)} \cr
}
\eqno (\number\nftype)
$$
We have performed a numerical analysis, measuring the average length of the
shortest path, and the average number of its arcs, for various values of $N$
and both uniform and linear probability distribution of the distances,
{\sl i.e.} for $\alpha=0$ and $\alpha=1$. The result of the analysis is
shown in next section: for the time being we stress that the numerical
data can be fitted with the following functional form, for $ N \gg 1 $
(see figures~1 and~2)
$$
\eqalign{
n & \approx c \cdot \ln N + C \cr
{ l \over n } & \approx { K \over N^s } \cr
}
\eqno (\number\nftype')
$$
This is in close agreement with (\number\nftype), and, fairly surprisingly,
also the numerical values of $c$ and $s$ agree with the ones in
(\number\nftype), {\sl i.e.} $ c = s \simeq 1/(\alpha+1) $. This fact will
be discussed in greater detail in the following.

\medskip

This approximate computation can be replaced with a more precise analysis
as follows. Consider the set $C_{s,t}$
of all the paths from $s$ to $t$, ordered by increasing length, and denote
with $ P^{(h)} ( n , \{ x_i \}_{i=1}^n ) $ the probability distribution
that the $h$-th path is composed of $n$ arcs of lengths
$ x_1, \dots, x_n $. We shall need the following notations: let $a_k$ be
the length of the arc $ k \in A $; given a path $ c \in C_{s,t} $,
let $l_c$ be its length, $n_c$ the number of its arcs, and
$ \{ k^i_c \}_{i=1}^{n_c} \subset A $ the sequence of arcs which it is made of
(so that $ l_c = \sum_{i=1}^{n_c} a_{k^i_c} $); finally we need the
characteristic function
$$
I_c^h \mkern 20mu = \sum_{c_1, \dots, c_{h-1}}
        \prod_{ c' \not= c_1, \dots, c_{h-1},\, c }
        \Theta(l_{c'} - l_c) \cdot
        \Theta(l_{c} - l_{c_{h-1}}) \cdots \Theta(l_{c_2} - l_{c_1})
\eqno (\number\Ihc)
$$
(which is 1 if $c$ is the $h$-th path, $0$ elsewhere). Then it is immediate
to see that:
$$
P^{(h)} ( n , \{ x_i \} ) = \int
   \bigl[
       \sum_c \,
       \delta_{n,n_c} \prod_{i=1}^n \delta(x_i - a_{k_c^i}) \, I^h_c
   \bigr] \prod_{k\in A} p_\alpha(a_k) \, d a_k
\eqno (\number\pexact)
$$
In order to make the computation of the integral in (\number\pexact)
feasible, we shall make the additional assumption
$$
\prod_{ k \in A } p_\alpha ( a_k ) =
\prod_{ c \in C_{s,t} } \prod_{i=1}^{n_c} p_\alpha ( a_{k^i_c} ) \; ,
\eqno (\number\approne)
$$
{\sl i.e.} two different paths do not have common arcs. Assumption
(\number\approne) should
be given a suitable sense ``in the average'' for $ N \gg 1 $ (since, as it
stands, it is evidently false).
However this is a minor problem, firstly
because we could not find better simplifying assumptions; secondly, because
we believe that to neglect the correlations between the paths (as in
(\number\approne)) corresponds to a kind of non-frustrated
approximation, which do not suitably fits our problem. We hope that
this point will become clearer in the following.
We omit here the cumbersome algebra, and give only the final results:
for $ N \gg 1 $ with $h$ fixed\footnote{${}^{(*)}$}
{$\gamma_E = 0.5772157 \dots $ is the Euler's gamma constant.}
$$
\eqalign{
\langle n \rangle^{(h)} & =
   { \ln N \over \alpha + 1 } + { 1 \over \alpha + 1 }
   \bigl[
      \ln ( \alpha + 1 ) + 1 - \gamma_E + \sum_{n=1}^{h-1} { 1\over n }
   \bigr] + O \bigl( { \ln N  \over N } \bigr) \cr
{ \langle l \rangle^{(h)} \over \langle n \rangle^{(h)} } & =
   { \alpha + 1 \over \chi \, N^{1/(\alpha+1)}} \;
   \bigl[
      1 - { 1 \over (\alpha+1) \langle n \rangle^{(h)} }
   \bigr]
   \qquad \chi = (\alpha + 1)!^{1/(\alpha+1)} \cr
}
\eqno (\number\lnresult)
$$
and
$$
{ \sigma^2_n \over \langle n \rangle^2 } = O \bigl( { 1 \over \ln N } \bigr)
\quad , \quad
{ \sigma^2_l \over \langle l \rangle^2 } = O \bigl( { 1 \over \ln^2 N } \bigr)
$$
where $ \langle n \rangle $ and $ \langle l \rangle $ denote the mean
value, respectively, of $n_c$ and $l_c$ in the distribution
(\number\pexact).

The relations (\number\lnresult) give to leading order the
non-frustrated behaviour of (\number\nftype). In other words, this more
refined computation with the simplifying assumption~(\number\approne) gives
again, as far as the leading behaviour is concerned, the result of the
previous heuristic argument, {\sl i.e.} a functional form of the
type~($\number\nftype'$) with $ c = s = 1/(\alpha+1) $, which, as we have
already pointed out, fits well enough the numerical data
(see next section and figures~1 and~2). An interesting feature of the
form~(\number\lnresult) is that the leading behaviour {\sl versus} $N$,
$ N \gg 1 $, of the number of arcs and of the length of the
$h$-th path, is not dependent on $h$: this is evident from
the fact that the dependence on $h$ in the right hand side of
the~(\number\lnresult) is only in the next-to-leading terms. This
feature is
very well confirmed by the numerical data (see again next section and
figures~1 and~2), so that, as we anticipated at the end of
previous section, we feel authorized to study all the subminimal paths
as a whole.
As far as the complete probability distribution of $l_c$ is concerned,
it results:
$$
\eqalign{
& P^{(h)} (l) = { H'_\alpha(l) \over (h-1)! } [ H_\alpha (l) ]^{h-1} \cdot
   \exp - H_\alpha (l) \cr
& H_\alpha (l) = { {\rm e}^{ {\cal N} l } \over N ( \alpha+1 ) } \qquad
   {\cal N} = (\alpha + 1)!^{1/(\alpha+1)} \, N^{1/(\alpha+1)} \cr
}
\eqno (\number\phl)
$$
The plots of the distributions $ P^{(h)} (l) $,
for $N=100$ and $ h = 1 $,~2 and~3, are shown in
figure~3, whereas in figure~5 a comparison with the experimental
results is exhibited. One can see that
the qualitative shapes of the distributions~(\number\phl) and the
experimental ones are rather
different, though their mean values are very close. For this reason we
believe that the assumption~(\number\approne), that we called
``non-frustrated'' and that was at the basis of
this computation, is not completely satisfactory to fit our problem.

\medskip

Another approach to the problem is
suggested by the cavity method of M\'ezard, Parisi and Virasoro
[\number\Cavity]. We can not pursue this approach beyond a kind of
non-frustrated approximation, to some extent equivalent to
the~(\number\approne) above, and very reminiscent of the
replica-symmetric approximation for the SK model.
We shall describe briefly
the procedure. Let $ G_N = ( V_N, A_N ) $ be a graph with $N$ nodes,
and fixed $s \in G_N$ let $\{ l_u \}_{u \in V_N \setminus \{ s \} }$ be the
lengths of the shortest paths in $G_N$ from $s$ to $u$. Let us denote with
$ {\cal P}_N ( \{ l_u \} ) $ their joint probability distribution. Let now
$ G_{N+1} = ( V_{N+1}, A_{N+1} ) $ be the graph with $N+1$ nodes obtained
by adding a node $u_0$ to $V_N$, and all the arcs between $u_0$ and the
nodes of $V_N$ to $A_N$; finally, let
$\{ {\cal L}_u \}_{u \in V_N}$ be the lengths of these new arcs, with
distribution $ P ( \{ {\cal L}_u \} ) = \prod p_\alpha ({\cal L}_u) $. The
shortest path from $s$ to $u_0$ in $G_{N+1}$ has length
$$
l_0 = \min [ {\cal L}_s , \min_{u \in V_N \setminus \{ s \} }
                    ( l_u + {\cal L}_u ) ]
\eqno (\number\lzero)
$$
As above, it is straightforward to write the exact form of the probability
distribution of $l_0$, but there is no point in using it
without suitable
simplifying assumptions. Namely, we shall assume the $ \{ l_u \} $ to be
uncorrelated, that is to say
$$
{\cal P}_N ( \{ l_u \} ) =
    \prod_{ u \in V_N \setminus \{ s \} } P_N ( l_u )
\eqno (\number\apprtwo)
$$
At first glance it is not clear which is the relation between this
approximation and the~(\number\approne) above: the latter
consists in neglecting the correlations among the paths between
the same pair of nodes, the first one
ignores the correlations between the shortest paths connecting a fixed
starting node with all the other ones of the graph.
However, as we shall see,
the resulting behaviour for the mean length of the shortest path is
exactly the same, though the two
complete probability distributions differ qualitatively.
Now, using the~(\number\apprtwo) one can translate the recursive
relation~(\number\lzero) in a recursive relation for the
probability distribution of the shortest path: writing
$$
F_N (l) = 1 - \int_0^l P_N (t) \, dt
$$
we obtain, in a more or less straightforward way,
$$
F_{N+1} (l) = \bigl[ 1 - \int_0^l p_\alpha \bigr] \cdot
   \exp - N \bigl(
               \int_0^l p_\alpha -
               \int_0^l p_\alpha ( l - x ) F_N ( x ) \, dx
            \bigr)
\eqno (\number\FNrel)
$$
If we now assume, apart from negligible terms,
$$
F^\infty_N (l) \simeq F_{N+1} (l) \simeq F_N (l) \qquad \hbox{for $N\gg1$,}
\eqno (\number\FNinfty)
$$
(\number\FNrel) becomes an integral (non-linear) equation in $F^\infty_N$.
It is possible to justify (\number\FNinfty) {\sl a posteriori}.

If $ \alpha = 0 $ the equation reduces to
$$
F^\infty_N (l) = (1-l) \cdot \exp - N ( l - \int_0^l F_N^\infty (x) \, dx )
$$
which is equivalent to
$$
\cases{
\displaystyle
{ d \over d l } F^\infty_N (l) =
   - { F^\infty_N \over 1-l } + F^\infty_N \cdot N [ F^\infty_N -1 ] & \cr
F^\infty_N (0) = 1 & \cr
}
\eqno (\number\FNzero)
$$
and can be linearized by putting $ T=1/F $. At the end we get
$$
F^\infty_N (l) = { 1 \over 1 +
  { \displaystyle \exp (Nl) - 1 \over \displaystyle N(1-l) }
  } \quad , \quad P^\infty_N (l) = - { d \over d l } F^\infty_N (l)
\eqno (\number\plzero)
$$
If $ \alpha \not = 0 $ the equation
(\number\FNrel), again under the hypothesis (\number\FNinfty), must be
solved numerically, thus giving
$ F^\infty_N (l) $ for each fixed value of $N$ and $\alpha$.
The resulting distributions $P^\infty_N(l)$ differs qualitatively from
the $P^{(h)}_N(l)$'s of equation~(\number\phl) above, even for $h=1$.
Nevertheless, as we shall see soon, their mean values coincide. A
comparison between these two distributions and the experimental one is shown
in figure~5, for $N=100$ and
both uniform and linear probability distribution of the distances,
{\sl i.e.} for $\alpha=0$ and $\alpha=1$.

As far as the mean value of $l$ is concerned, if $ \alpha = 0 $
the~(\number\plzero) can be easily integrated and gives
$$
\langle l \rangle = \int l P^\infty_N (l) \, d l \simeq { \ln N \over N }
\eqno (\number\lmean)
$$
in perfect
agreement with what is predicted by equation~(\number\lnresult).
If $ \alpha \not = 0 $ we must solve numerically the integral
equation~(\number\FNrel) for some selected values of $N$. Then we
compute $ \langle l \rangle $ and we
fit the resulting data with a functional form of the type
$$
\langle l \rangle = K' { \ln N \over N^s } + \hbox {subleading terms}
$$
We have performed
such an analysis for $\alpha = 0$ and $ \alpha=1$, computing
$ \langle l \rangle $ for the same values of $N$ of the experimental data,
thus giving to the finite-size effects the same relevance. The
details are given in next section. For $ \alpha = 0 $ we obtain
$$
K' = 1 \qquad s = 1
$$
with negligible corrections, in perfect agreement with~(\number\lmean),
thus showing that the selected values of $N$ are not too small. For
$ \alpha=1$, fairly
surprisingly, we have found again a very good
agreement with what is predicted by equation~(\number\lnresult), {\sl i.e.}
$$
K' = \sqrt{2}/2 \qquad s = 1/2
$$
Thus this
approach (suggested by the cavity method of [\number\Cavity] with the
assumption~(\number\apprtwo)), as to the leading behaviour
of the length of the shortest path,
is equivalent at all to the one which leads to
equation~(\number\lnresult), {\sl i.e.}
the direct one with the assumption~(\number\approne).

\medskip

As anticipated,
in next section we shall see that the non-frustrated behaviour of
equation~(\number\lnresult), shared by both our analytical approaches,
fits well enough the
experimental data for $ \langle l \rangle $ and $ \langle n \rangle $.
As we shall see, however,
the agreement is not beyond any doubts, so we believe that
it would be essential to improve the rough approximation (\number\apprtwo),
thus obtaining a more refined prediction than equation~(\number\lnresult).
An idea for a future work could be the following.

Let us assume that the Dijkstra's algorithm has given the $SPT$ for the
graph $G_N$
with $N$ nodes; then we dispose of the exact numerical values of the
lengths $l_u$ of the shortest paths from $s$ to any other node
$ u \in V_N $. If now we add the node $u_0$, obtaining $G_{N+1}$, the
length $l_0$ of the shortest path from $s$ to $u_0$ is given by
(\number\lzero). What
can be said now about the other lengths $ {\tilde l}_u $ of the shortest
paths from $s$ to the nodes of $V_N$? Of course, they will be affected by
the addition of $u_0$, in analogy to what happens in the SK model when
applying the cavity method. In fact the addition of one spin to an $N$-spin
system produces a rearrangement of the relative weights of the
configurations in a pure state, thence a minor rearrangement of the relative
weights of the pure states in a cluster and so on. The cavity method allows
to keep track of this cascade of rearrangements (see [\number\Cavity]
for details). The crucial observation was the discovery of
the cluster structure of the pure states in the SK model, the so called
{\sl ultrametric} organization (see [\number\Ultra]).
We do not yet know how to implement an
analogous procedure in our model, though it seems to be natural that an
improvement of the rough approximations
(\number\apprtwo) or (\number\approne) should be required.
In some sense the approximations we used here correspond to a
``replica-symmetric'' solution, which does not take into account the
rearrangement of the pure states. This would be exact if the system was
not frustrated ({\sl i.e.} with only one pure state, apart from accidental
degenerations).

The first condition, in order to start the program sketched
above, is to find out what plays the role of the SK pure states in our
model. Let us therefore consider the following argument. After adding the
node $u_0$ to the graph $G_N$ obtaining $G_{N+1}$, one can apply the
Dijkstra's
algorithm to find the $SPT$ in $G_{N+1}$. Take now a node $u$ in $V_N$. If
the shortest path in $G_{N+1}$ from $s$ to $u$ does not pass trough $u_0$,
then it will coincide with the old shortest path in $G_N$.
If, on the contrary, it
passes through $u_0$, it is interesting to see how much ``close'' (in a
suitable sense) it is to the old shortest path in $G_N$. In general, there
is no reason to expect these two paths to be close, because of the
frustration of the model, but one can reasonably suppose that the new path
runs close to an old subminimal path in $G_N$ from $s$ to $u$.
Therefore in our numerical analysis we have pointed out our interest on the
suboptimal paths a little longer than the minimal one: can we interpret the
suboptimal paths as the pure states (in a suitable sense) of our model?
Let us stress that to leading order for $N\gg1$ the lengths
$ \langle l \rangle^{(h)} $ of the suboptimal paths coincide, as it
results from
(\number\lnresult) and is confirmed by numerical analysis
(see next section). Otherwise
the identification between subminimal paths and pure states will fail:
in the SK model the free energies are equal in the thermodynamical limit.

\section{5.~Numerical results}

\noindent
As pointed out above, we have performed some
numerical computations to test the validity of the
scaling laws (\number\nftype), and to measure the probability distribution
for the lengths of the subminimal paths. The algorithm used is a clever
generalization of the Dijkstra's algorithm [\number\PDG].
It determines, for a given realization of the
distance matrix $d(u,v)$, the first $k$ paths of $C_{s,t}$ ordered by
increasing
length. We have considered only $k=3$, because the CPU-time
spent in the computation grows rather fast with $k$.
We made several runs for $\alpha = 0 $ and $ \alpha = 1 $, {\sl i.e.}
with the
probability distribution in (\number\defp) respectively uniform and linear,
and for different values of $N$, the size of the
graph, ranging from $N=10$ to $N=500$. In each run we measure the
length and the number of arcs of the paths for a large number of
independent
realizations of the distance matrix. More precisely we made
$200.000$ iterations for each value of $\alpha $
for $ N = 10, 20, 50, 100 $; $100.000$ iterations for $ N = 200 $;
$ 40.000 $ iterations for $ N = 500 $.
The runs required about 150 hours of CPU on a DEC VAX 6000-520,
running VAX Pascal. The runs to measure the probability distribution, being
purely qualitative and not requiring a high precision, were made only for
$N=100$ and low statistics:
$30.000$ iterations for each value of $\alpha$.

\medskip

In figure~1 and~2 we plot respectively
$ \langle n \rangle^{(h)} $ and
$ \langle l \rangle^{(h)} / \langle n \rangle^{(h)} $ {\sl versus} the size
of the graph $N$, for $h=1,2,3$ and both probability distributions for the
distances ($\alpha=0$ and $\alpha=1$). The scaling laws that qualitatively
appear in these plots are the ones anticipated in
equation~($\number\nftype'$), {\sl i.e.} for $N \gg 1 $
$$
\eqalign{
\langle n \rangle^{(h)} & \approx c \cdot \ln N + C^{(h)} \cr
{ \langle l \rangle^{(h)} \over \langle n \rangle^{(h)} }
                        & \approx { K \over N^s }
                        \quad \hbox{independent on $h$} \cr
}
$$
We want to stress once more that to
leading order in $N$ these scaling laws do not depend on $h$.

To be sure that the leading behaviour for the ``physical'' quantities
$\langle n \rangle$ and $\langle l \rangle$ is the non-frustrated one of
equation~(\number\nftype) and~(\number\lnresult), we
should be more quantitative, and show that
$$
c = s = {1 \over \alpha + 1 }
\eqno (\number\csrel)
$$
We cannot simply perform a na\"\i ve least-squares fit, as our
data could not have reached their asymptotic behaviour, because of the low
involved values of $N$. Therefore our data have been fitted with curves of the
form
$$
\eqalign{
\langle n \rangle & = c \cdot \ln N + C + B { \ln N \over N } \cr
{ \langle l \rangle \over \langle n \rangle } & = { K \over N^s }
                  \bigl( 1 + { B \over \ln N } \bigr) \cr
}
\eqno (\number\fit)
$$
for various {\sl fixed} values of $B$, applying the {\sl flatness criterion}
advocated in [\number\BS, sections~4.2 and~5.3]. The $B$-dependent term
should take into account the corrections to the leading behaviour,
and its form was suggested by equation~(\number\lnresult).
Although we believe that this technique has been now generally accepted,
in the appendix we shall spend some words of explanation.
The result of the analysis for
$\langle n \rangle$ and $\langle l \rangle / \langle n \rangle$
is given in table~1 and~2. From these data we see that
relation~(\number\csrel) is verified well enough. Sufficiently well verified
is also the fact that $K$ is independent on $h$, at least for
$\alpha=1$. As to the value of $C^{(h)}$ the agreement with
equation~(\number\lnresult) is only qualitative.

Nevertheless the fitted values for $c$ and $s$ in table~1 and~2 are
different
from what is predicted by equation~(\number\csrel) by more than their
error, though this should be a 95\% confidence limit. Thus we conclude
that, as we anticipated at the end of the previous section,
the use of the non-frustrated leading behaviour~(\number\lnresult) to
fit our experimental data is not completely satisfactory.
We strongly believe that the finite size errors are not
underestimated, because the same procedure of fitting
performed on the theoretical data leads to confirm the validity
of~(\number\csrel) exactly, {\sl i.e.} at 95\% confidence level.
More precisely we have proceeded as follows. We have generated a set of
values for $ \langle l \rangle $ integrating the theoretical
distributions of equation~(\number\phl) and~(\number\FNrel), for the same
values of $N$ of the real data,
and with fictitious statistical errors equal to the real ones. Then we have
performed in parallel the same fit, with the same procedure of the previous
analysis, on
the experimental data and the theoretical ones,
which so work as a {\sl placebo}. The result is
given in table~3: the data have been fitted with the functional form
$$
\langle l \rangle = K' { \ln N \over N^s }
                  \bigl( 1 + { B \over \ln N } \bigr)
\eqno (\number\fit')
$$
where again
the corrections are suggested by equation~(\number\lnresult).
We see that the theoretical data give the expected values
well within their error, with some troubles for $h=1$, not only for
the exponent
$ s $ ({\sl i.e.} $ 1/(\alpha+1) $), but also for the constant
$ K' $ ({\sl i.e.} $1$ for $ \alpha = 0 $ and
$ \sqrt{2}/2 $ for $ \alpha = 1 $, see~(\number\lnresult)).
As to the experimental data, though
the flatness region in $B$ is larger, thus producing a systematic error
larger, the fitted value for
$ s $ differs from $ 1/ (\alpha+1) $ by more than its error, and
the discrepancy becomes stronger when $ h $ increases.

\medskip

In figure~3, 4~and~5 we show the various
probability distributions for the length of the subminimal paths for
$ N = 100 $ and both $\alpha=0$ and $\alpha=1$.
In figure~3 we plotted the distributions
$ P^{(h)} (l) $ of equation~(\number\phl) for $ h = 1$, 2~and~3.
Figure~4 presents the experimental distributions for $ h = 1$, 2~and~3,
obtained by smoothing the histogram of the experimental data. More
precisely we filled a histogram with the experimental data for
$ l $, for different realizations of the distance matrix (200 channels
for $30.000$ data, no one of the data falls out of the considered range),
then we averaged the counting of each channel with
the $H$ preceeding and the $H$ following it
($ H = 5 $ for $ \alpha = 0 $ and $ H = 2 $ for $ \alpha = 1 $), thus
obtaining a smoother plot.
Lastly, as far as the length of the shortest path is concerned,
in figure~5 we compare
the distribution $ P^{(h)} (l) $ with $ h = 1 $ of figure~3,
the experimental distribution of figure~4,
and the distribution of the ``cavity method'', {\sl i.e.} the one
obtained by solving equation~(\number\FNrel), exactly for $ \alpha = 0 $
(see~(\number\plzero)), and numerically for $ \alpha = 1 $.

As a comment to figure~5 we remark again (this fact has been yet noted in
the previous section) that the two theoretical distributions and the
experimental one differ qualitatively from each others,
at $N$ fixed, though the first two distributions
share the same leading behaviour {\sl versus} $N$ as to their mean
value, whereas the third one gives a
significatively different behaviour.

\section{Appendix}

\noindent
In this appendix we sketch briefly the contents of the analysis of the
corrections to the scaling {\sl \`a la} Berretti-Sokal~[\number\BS].
In our case we
measure $ \langle l \rangle $ and $ \langle n \rangle $ for some values of
$N$, the size of the graph, and we would extrapolate from the data
the scaling behaviour {\sl versus} $N$. Because of the involved relatively
small values of $N$, we can not be sure that the data have reached their
asymptotic behaviour, so we must take into account possible corrections to
the scaling. Firstly we make an Ansatz for
the corrections to the leading behaviour, as in~(\number\fit).
The coefficient $B$ of the corrections in~(\number\fit) is {\sl not}
subject to a fit, although
this would be easily made by well known non-linear methods of fitting,
because this would not solve our problem, {\sl i.e.} to understand
to what extent the fitted parameters are reliable as true asymptotic
values, and not merely as effective
values, changing as the range in $N$ increases. In fact we do not
know exactly which the subleading corrections are, and also, if we knew them,
we could not include in a fit other but the first relevant ones, as our
data are only for a few values of $N$ (6, in our case).

Thus we fix the value of $B$, we fix also a value $N_{cut}$ for $N$, and we
fit the data with the curves in~(\number\fit) only for $ N > N_{cut} $.
Varying $N_{cut}$ at $B$ fixed, if we observe a systematic dependence on
$N_{cut}$ of the fitted parameters ({\sl e.g.} a systematic decrease
of the fitted exponent $s$), we say that the chosen value of $B$ can not
take into account the corrections to the scaling. Now, by varying $B$,
we select the range in $B$ for which the fitted parameters are flat
({\sl i.e.} do not vary in a systematic way), with respect to
$N_{cut}$. We say, following~[\number\BS], that for $B$ in this range
equation~(\number\fit) takes into account the corrections to the scaling as
an effective correction, even if the exact form for the corrections is
different from that we imposed, and even if we rule out some other (more
irrelevant) terms. In this range of $B$, which we call the
``flatness region'', for each fitted parameter we select the maximum and
the minimum value. The best fit for the parameter will be simply the
arithmetic mean of these values, whereas their difference will be
considered as the systematic error due to unconsidered corrections to
the scaling, or to imperfect knowledge of the form of the corrections
(95\% subjective confidence limit as defined in
[\number\BS, footnote 17]). In addition we quote the usual statistical
error for the fit, at 95\% confidence level (2$\sigma$).

\section{Acknowledgements}

\noindent
I am greatly indebted to Giorgio Parisi
for having originally proposed this work, and
for many stimulating discussions
and helpful suggestions throughout all its development.
I am also grateful to Sergio Caracciolo and Andrea Pelissetto for their
kind collaboration in some aspects of this work.
All numerical computations were performed at Scuola Normale Superiore in
Pisa.

\section{References}

\parskip=5pt

\item{[\number\SK]} \author {D.Sherrington and S.Kirkpatrick} \hfill\break
      \revtit {Solvable model of a spin glass}
      {Phys.Rev.Lett.} {35} {1792} {1975}

\item{[\number\Parisi]} \author {G.Parisi}
      \revtit {A sequence of approximated solutions to the SK model for
               spin glasses}
      {J.Phys.A} {13} {L115} {1980} \hfill\break
      \revtit {The order parameter for spin glasses: a function on the
               interval $[0,1]$}
      {J.Phys.A} {13} {1101} {1980} \hfill\break
      \revtit {Magnetic properties of spin glasses in a new mean field
               theory}
      {J.Phys.A} {13} {1887} {1980}

\item{[\number\Overlap]} \author {G.Parisi}
      \revtit {Order parameter for spin glasses}
      {Phys.Rev.Lett.} {50} {1946} {1983}

\item{[\number\Ultra]} \author {M.M\'ezard, G.Parisi, N.Sourlas, G.Toulouse
                                and M.A.Virasoro}
      \revtit {Replica symmetry breaking and the nature of the spin glass
               phase}
      {J.Physique} {45} {843} {1984}

\item{[\number\RFE]} \author {M.M\'ezard, G.Parisi and M.A.Virasoro}
      \revtit {Random free energies in spin glasses}
      {J.Physique Lettres} {46} {L217} {1985}

\item{[\number\Cavity]} \author {M.M\'ezard, G.Parisi and M.A.Virasoro}
      \revtit {SK model: the replica solution without replicas}
      {Europhys.Lett.} {1} {77} {1986}

\item{[\number\Ruelle]} \author {D.Ruelle}
      \revtit {A mathematical reformulation of Derrida's REM and GREM}
      {Commun.Math.Phys.} {108} {225} {1987}

\item{[\number\CPPS]} \author {S.Caracciolo, G.Parisi, S.Patarnello and
                               N.Sourlas}
      \revtit {Low temperature behaviour of 3-D spin glasses in a magnetic
               field}
      {J.Physique} {51} {1877} {1990}

\item{[\number\FH]} \author {D.S.Fisher and D.A.Huse} \hfill\break
      \revtit {Ordered phase of short-range Ising spin glasses}
      {Phys.Rev.Lett.} {56} {1601} {1986}

\item{[\number\REM]} \author {B.Derrida}
      \revtit {Random-energy model: an exactly solvable model of disordered
               systems}
      {Phys.Rev.B} {24} {2613} {1981}

\item{[\number\GREM]} \author {B.Derrida}
      \revtit {A generalization of the random energy model which includes
               correlations between energies}
      {J.Physique Lettres} {46} {L401} {1985}

\item{[\number\Papa]} \author {C.H.Papadimitriou and K.Steiglitz}
      \revbook {Combinatorial optimization: algorithms and complexity}
      {Prentice-Hall, Inc.} {1982}

\item{[\number\Barahona]} \author {F.Barahona}
      \revtit {On the computational complexity of Ising spin glass models}
      {J.Phys.A} {15} {3241} {1982}

\item{[\number\dyrpol]} \author {M.M\'ezard and G.Parisi}
      \revtit {Interfaces in a random medium and replica symmetry breaking}
      {J.Phys.A} {23} {L1229} {1990}

\item{[\number\PDG]} \author {P.Digianantonio} private communication.

\item{[\number\BS]} \author {A.Berretti and A.D.Sokal}
      \revtit {New Montecarlo method for the self-avoiding walk}
      {J.Stat.Phys.} {40} {483} {1985}

\vfill \eject

\nopagenumbers

\section{Captions for the figures and tables}

\noindent
{\bf Figure 1.} $ \langle n \rangle^{(h)} $ {\sl versus} $N$,
with uniform ($a$) and linear ($b$)
probability distribution for the distances.

\noindent
{\bf Figure 2.} The same as in figure~1 for
$ \langle l \rangle^{(h)} / \langle n \rangle^{(h)} $.

\noindent
{\bf Figure 3.} Distributions $ P^{(h)} (l) $ for the length of the $h$-th
path as in equation~(\number\phl), for $N=100$ and
uniform ($a$) and linear ($b$)
probability distribution for the distances.

\noindent
{\bf Figure 4.} The same as in figure~3 for the
measured distributions $ P_{exp}^{(h)} (l) $,
obtained by smoothing the histogram of the experimental data.

\noindent
{\bf Figure 5.} Three probability distributions for the length of the
shortest path, with $N=100$ and
uniform ($a$) and linear ($b$)
probability distribution for the distances.
$ P^{(1)} (l) $ is the distribution quoted in (\number\phl), as in
figure~3, for $h=1$;
{\tt exp} denotes the experimental distribution, as in figure~4, for $h=1$;
{\tt cav} denotes the distribution
of the ``cavity method'' (see equation~(\number\FNrel)).

\noindent
{\bf Table 1.} Best estimates for $c$ and $C$ in the fit
$ \langle n \rangle^{(h)} = c \cdot \ln N + C $, with corrections to the
leading behaviour inserted (see~(\number\fit)).
The first error is the systematic error due to unconsidered corrections
to the scaling (95\% subjective confidence limit as defined in
[\number\BS, footnote 17]) and the second error is the usual
statistical error (95\% confidence interval). It is also quoted the range
in $B$ which obeys to the flatness criterion.

\noindent
{\bf Table 2.} Best estimates as in table~1 for $s$ and $K$ in the fit
$ \langle l \rangle^{(h)} / \langle n \rangle^{(h)} = K / N^s $,
with corrections to the leading behaviour inserted (see~(\number\fit)).

\noindent
{\bf Table 3.} Best estimates as in table~1 for $s$ and $K'$ in the fit
$ \langle l \rangle^{(h)} = K' \cdot \ln N / N^s $,
with corrections to the leading behaviour inserted (see~($\number\fit'$)).
The fit has been performed for both the experimental
data (denoted by {\tt exp}) and the theoretical ones. These last were
produced by integrating the two theoretical probability distributions for $l$,
the one of equation~(\number\phl) and the one of the cavity
equation~(\number\FNrel).

\vfill \eject

\line

  \hfil {\bf Figure~1.a}}

\setbox\BodyBox = \vbox{
   \epsfysize = 0.8\hsize
   \setbox\RotateBox=\hbox{\epsfbox[25 10 550 720] {fig1a.ps}}
   \rotr\RotateBox}

\centerline{\box\BodyBox}

\vfill

\line

  \hfil {\bf Figure~1.b}}

\setbox\BodyBox = \vbox{
   \epsfysize = 0.8\hsize
   \setbox\RotateBox=\hbox{\epsfbox[25 10 550 720] {fig1b.ps}}
   \rotr\RotateBox}

\centerline{\box\BodyBox}

\vfill \eject

\line
  \hfil {\bf Figure~2.a}}

\setbox\BodyBox = \vbox{
   \epsfysize = 0.8\hsize
   \setbox\RotateBox=\hbox{\epsfbox[25 10 550 720] {fig2a.ps}}
   \rotr\RotateBox}

\centerline{\box\BodyBox}

\vfill

\line
  \hfil {\bf Figure~2.b}}

\setbox\BodyBox = \vbox{
   \epsfysize = 0.8\hsize
   \setbox\RotateBox=\hbox{\epsfbox[25 10 550 720] {fig2b.ps}}
   \rotr\RotateBox}

\centerline{\box\BodyBox}

\vfill \eject

\line

  \hfil {\bf Figure~3.a}}

\setbox\BodyBox = \vbox{
   \epsfysize = 0.8\hsize
   \setbox\RotateBox=\hbox{\epsfbox[25 10 550 720] {fig3a.ps}}
   \rotr\RotateBox}

\centerline{\box\BodyBox}

\vfill

\line

  \hfil {\bf Figure~3.b}}

\setbox\BodyBox = \vbox{
   \epsfysize = 0.8\hsize
   \setbox\RotateBox=\hbox{\epsfbox[25 10 550 720] {fig3b.ps}}
   \rotr\RotateBox}

\centerline{\box\BodyBox}

\vfill \eject

\line
  \hfil {\bf Figure~4.a}}

\setbox\BodyBox = \vbox{
   \epsfysize = 0.8\hsize
   \setbox\RotateBox=\hbox{\epsfbox[25 10 550 720] {fig4a.ps}}
   \rotr\RotateBox}

\centerline{\box\BodyBox}

\vfill

\line
  \hfil {\bf Figure~4.b}}

\setbox\BodyBox = \vbox{
   \epsfysize = 0.8\hsize
   \setbox\RotateBox=\hbox{\epsfbox[25 10 550 720] {fig4b.ps}}
   \rotr\RotateBox}

\centerline{\box\BodyBox}

\vfill \eject

\line
  \hfil {\bf Figure~5.a}}

\setbox\BodyBox = \vbox{
   \epsfysize = 0.8\hsize
   \setbox\RotateBox=\hbox{\epsfbox[25 10 550 720] {fig5a.ps}}
   \rotr\RotateBox}

\centerline{\box\BodyBox}

\vfill

\line
  \hfil {\bf Figure~5.b}}

\setbox\BodyBox = \vbox{
   \epsfysize = 0.8\hsize
   \setbox\RotateBox=\hbox{\epsfbox[25 10 550 720] {fig5b.ps}}
   \rotr\RotateBox}

\centerline{\box\BodyBox}

\vfill \eject

\newbox\mystrutbox
\setbox\mystrutbox=\hbox{\vrule height15pt width0pt depth5pt}
\def\relax\ifmmode\copy\mystrutbox\else\unhcopy\mystrutbox\fi{\relax\ifmmode\copy\mystrutbox\else\unhcopy\mystrutbox\fi}

\line{}
\vfill

\line
  \hfil {\bf Table~1.}}
$$ \vbox { \offinterlineskip \tabskip=0pt
\halign {%
\relax\ifmmode\copy\mystrutbox\else\unhcopy\mystrutbox\fi \vrule # &
$ # $ \hss & \quad \vrule # &
\quad \hss $ # $ \hss &
\quad \hss $ # $ \hss &
\quad \hss $ # $ \hss \vrule \cr
\noalign {\hrule}
&            & &  c                          & C                  & B \cr
\noalign {\hrule}
& \; \alpha = 0 & &                          &                    &   \cr
& \mkern20mu h = 1 & & 1.0070 \pm 0.0079 \pm 0.0023 &
                       -0.475 \pm 0.041  \pm 0.008  &  0.87 \div  0.99 \cr
& \mkern20mu h = 2 & & 1.0363 \pm 0.0098 \pm 0.0024 &
                        0.316 \pm 0.051  \pm 0.009  & -0.20 \div -0.05 \cr
& \mkern20mu h = 3 & & 1.0627 \pm 0.0179 \pm 0.0125 &
                        0.650 \pm 0.095  \pm 0.060  & -0.60 \div -0.40 \cr
\noalign {\hrule}
& \; \alpha = 1 & &                          &                    &   \cr
& \mkern20mu h = 1 & & 0.5051 \pm 0.0108 \pm 0.0025 &
                        0.283 \pm 0.061  \pm 0.010  & -0.55 \div -0.30 \cr
& \mkern20mu h = 2 & & 0.5051 \pm 0.0087 \pm 0.0074 &
                        0.781 \pm 0.051  \pm 0.036  & -0.50 \div -0.30 \cr
& \mkern20mu h = 3 & & 0.5138 \pm 0.0129 \pm 0.0026 &
                        0.971 \pm 0.073  \pm 0.011  & -0.30 \div  0.00 \cr
\noalign {\hrule}
}}$$

\vfill

\line
 \hfil {\bf Table~2.}}
$$ \vbox { \offinterlineskip \tabskip=0pt
\halign {%
\relax\ifmmode\copy\mystrutbox\else\unhcopy\mystrutbox\fi \vrule # &
$ # $ \hss & \quad \vrule #
& \quad \hss $ # $ \hss
& \quad \hss $ # $ \hss
& \quad \hss $ # $ \hss \vrule \cr
\noalign {\hrule}
&            & &  s                          & K                  & B \cr
\noalign {\hrule}
& \; \alpha = 0 & &                          &                    &   \cr
& \mkern20mu h = 1 & & 1.0191 \pm 0.0118 \pm 0.0021 &
                        1.246 \pm 0.131  \pm 0.012  & 0.25 \div 0.50 \cr
& \mkern20mu h = 2 & & 0.9914 \pm 0.0106 \pm 0.0016 &
                        0.844 \pm 0.094  \pm 0.006  & 1.70 \div 2.10 \cr
& \mkern20mu h = 3 & & 0.9780 \pm 0.0070 \pm 0.0032 &
                        0.675 \pm 0.049  \pm 0.010  & 2.90 \div 3.20 \cr
\noalign {\hrule}
& \; \alpha = 1 & &                          &                    &   \cr
& \mkern20mu h = 1 & & 0.4981 \pm 0.0102 \pm 0.0018 &
                        1.361 \pm 0.119  \pm 0.011  & -0.08 \div  0.10 \cr
& \mkern20mu h = 2 & & 0.4920 \pm 0.0076 \pm 0.0013 &
                        1.265 \pm 0.084  \pm 0.008  &  0.20 \div  0.35 \cr
& \mkern20mu h = 3 & & 0.4948 \pm 0.0076 \pm 0.0012 &
                        1.290 \pm 0.086  \pm 0.007  &  0.20 \div  0.35 \cr
\noalign {\hrule}
}}$$

\vfill\eject

\line
 \hfil {\bf Table~3.}}
$$ \vbox { \offinterlineskip \tabskip=0pt
\halign {%
\relax\ifmmode\copy\mystrutbox\else\unhcopy\mystrutbox\fi \vrule # &
$ # \mkern10 mu $ \hss & \vrule #
& \quad \hss $ # $ \hss
& \quad \hss $ # $ \hss
& \quad \hss $ # $ \hss \vrule \cr
\noalign {\hrule}
&            & &  s                          & K'                 & B \cr
\noalign {\hrule}
& \; \alpha = 0 & &                          &                    &   \cr
& \mkern10mu {\tt exp} \; h = 1 & & 1.0077 \pm 0.0077 \pm 0.0013 &
                        1.098 \pm 0.074  \pm 0.007  & 0.20 \div 0.35 \cr
& \mkern10mu {\tt exp} \; h = 2 & & 0.9874 \pm 0.0081 \pm 0.0008 &
                        0.814 \pm 0.075  \pm 0.003  & 2.50 \div 2.90 \cr
& \mkern10mu {\tt exp} \; h = 3 & & 0.9667 \pm 0.0044 \pm 0.0005 &
                        0.571 \pm 0.034  \pm 0.001  & 5.50 \div 5.90 \cr
& \mkern10mu eq.~(\number\FNrel) & & 1 &  1 &       0        \cr
& \mkern10mu eq.~(\number\phl) \; h = 1 & & 0.9935 \pm 0.0038 \pm 0.0007 &
                        0.932 \pm 0.030  \pm 0.003  & -0.36 \div -0.43 \cr
& \mkern10mu eq.~(\number\phl) \; h = 2 & & 0.9998 \pm 0.0011 \pm 0.0003 &
                        0.997 \pm 0.009  \pm 0.001  & 0.415 \div 0.445 \cr
& \mkern10mu eq.~(\number\phl) \; h = 3 & & 0.9999 \pm 0.0013 \pm 0.0002 &
                        0.999 \pm 0.012  \pm 0.001  & 0.910 \div 0.935 \cr
\noalign {\hrule}
& \; \alpha = 1 & &                          &                    &   \cr
& \mkern10mu {\tt exp} \; h = 1 & & 0.5063 \pm 0.0041 \pm 0.0007 &
                        0.763 \pm 0.027  \pm 0.002  & 0.16 \div 0.24 \cr
& \mkern10mu {\tt exp} \; h = 2 & & 0.4990 \pm 0.0035 \pm 0.0004 &
                        0.692 \pm 0.025  \pm 0.001  & 1.51 \div 1.63 \cr
& \mkern10mu {\tt exp} \; h = 3 & & 0.4931 \pm 0.0033 \pm 0.0004 &
                        0.638 \pm 0.023  \pm 0.001  & 2.40 \div 2.55 \cr
& \mkern10mu eq.~(\number\FNrel) & & 0.4994 \pm 0.0025 \pm 0.0006 &
                        0.702 \pm 0.014  \pm 0.002  & 0.29 \div 0.33 \cr
& \mkern10mu eq.~(\number\phl) \; h = 1 & & 0.4951 \pm 0.0025 \pm 0.0006 &
                        0.670 \pm 0.013  \pm 0.002  & 0.25 \div 0.29 \cr
& \mkern10mu eq.~(\number\phl) \; h = 2 & & 0.4999 \pm 0.0016 \pm 0.0002 &
                        0.706 \pm 0.010  \pm 0.001  & 1.10 \div 1.14 \cr
& \mkern10mu eq.~(\number\phl) \; h = 3 & & 0.5000 \pm 0.0010 \pm 0.0001 &
                        0.707 \pm 0.007  \pm 0.001  & 1.60 \div 1.63 \cr
\noalign {\hrule}
}}$$

\bye

redpajama_set_name: RedPajamaArXiv