INITIALIZATION Knowledgebase: ki-dev-large Base Query: hi what is are the contents? Model: gemini-1.5-pro-preview-0409 Use Curl?: None ================================================== **Elapsed Time: 0.00 seconds** ================================================== ROUTING Query type: list ================================================== **Elapsed Time: 1.88 seconds** ================================================== PRIMER Primer: You are Simon, a highly intelligent personal assistant in a system called KIOS. You are a chatbot that can read knowledgebases through the "CONTEXT" that is included in the user's chat message. Your role is to act as an expert at summarization and analysis. In your responses to enterprise users, prioritize clarity, trustworthiness, and appropriate formality. Be honest by admitting when a topic falls outside your scope of knowledge, and suggest alternative avenues for obtaining information when necessary. 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Use as much tokens as possible to provide a detailed response. ================================================== **Elapsed Time: 0.46 seconds** ================================================== FINAL QUERY Final Query: CONTEXT: ########## File: curseforge-com-minecraft-mc-mods-productivebees-48193.txt Page: 1 Context: Updated Aug 29, 2024 Project ID 377897 License [All Rights Reserved](/minecraft/mc-mods/productivebees#license) ### Game Versions[View all](/minecraft/mc-mods/productivebees/files/all?page=1&pageSize=20) * [1.21.1](/minecraft/mc-mods/productivebees/files/all?page=1&pageSize=20&version=1.21.1) * [1.21](/minecraft/mc-mods/productivebees/files/all?page=1&pageSize=20&version=1.21) * [1.20.1](/minecraft/mc-mods/productivebees/files/all?page=1&pageSize=20&version=1.20.1) * [1.20](/minecraft/mc-mods/productivebees/files/all?page=1&pageSize=20&version=1.20) * [1.19.4](/minecraft/mc-mods/productivebees/files/all?page=1&pageSize=20&version=1.19.4) * \+ 13 Versions ### Mod Loaders[View all](/minecraft/mc-mods/productivebees/files/all?page=1&pageSize=20) * [Forge](/minecraft/mc-mods/productivebees/files/all?page=1&pageSize=20&gameVersionTypeId=1) * [NeoForge](/minecraft/mc-mods/productivebees/files/all?page=1&pageSize=20&gameVersionTypeId=6) ### Categories * [Ores and Resources](/minecraft/search?page=1&pageSize=20&sortBy=relevancy&class=mc-mods&categories=world-ores-resources) * [Cosmetic](/minecraft/search?page=1&pageSize=20&sortBy=relevancy&class=mc-mods&categories=cosmetic) * [Mobs](/minecraft/search?page=1&pageSize=20&sortBy=relevancy&class=mc-mods&categories=world-mobs) * [Genetics](/minecraft/search?page=1&pageSize=20&sortBy=relevancy&class=mc-mods&categories=technology-genetics) ### Main File #### 1.21.1 #################### File: curseforge-com-minecraft-mc-mods-productivebees-48193.txt Page: 1 Context: ### Main File #### 1.21.1 [productivebees-1.21.1-13.4.2.jarLatest releaseReleaseR1.21.1NeoForgeAug 29, 2024](/minecraft/mc-mods/productivebees/files/5674301) ### Recent Files[View all](/minecraft/mc-mods/productivebees/files/all?page=1&pageSize=20&sortBy=dateCreated&sortOrder=desc) #### Minecraft 1.21 [productivebees-1.21.1-13.4.2.jarLatest releaseReleaseR1.21.1NeoForgeAug 29, 2024](/minecraft/mc-mods/productivebees/files/5674301)[productivebees-1.21.0-13.3.2.jarLatest releaseReleaseR1.21NeoForgeAug 4, 2024](/minecraft/mc-mods/productivebees/files/5594966) #### Minecraft 1.20 [productivebees-1.20.1-12.6.0.jarLatest releaseReleaseR1.20.1NeoForgeForge + 1Jul 25, 2024](/minecraft/mc-mods/productivebees/files/5566102)[productivebees-1.20.1-12.1.2.jarLatest releaseReleaseR1.20.1ForgeJul 5, 2023](/minecraft/mc-mods/productivebees/files/4626156)[productivebees-1.20.1-12.3.0.jarLatest releaseAlphaA1.20.1NeoForgeForge + 1Feb 17, 2024](/minecraft/mc-mods/productivebees/files/5115329) #### Minecraft 1.19 #################### File: curseforge-com-minecraft-mc-mods-productivebees-48193.txt Page: 1 Context: #### Minecraft 1.19 [productivebees-1.19.4-11.6.2.jarLatest releaseBetaB1.19.4ForgeMay 23, 2023](/minecraft/mc-mods/productivebees/files/4549743)[productivebees-1.19.3-0.11.1.0.jarLatest releaseReleaseR1.19.3ForgeFeb 19, 2023](/minecraft/mc-mods/productivebees/files/4403646)[productivebees-1.19.2-0.10.7.2.jarLatest releaseReleaseR1.19.2ForgeJun 4, 2023](/minecraft/mc-mods/productivebees/files/4566382) ### Members * [LobsterJonnOwner](/members/lobsterjonn) [Report](https://www.curseforge.com/api/v1/auth/login?returnUrl=https://www.curseforge.com/minecraft/mc-mods/productivebees) * [Description](/minecraft/mc-mods/productivebees) * [Comments (960)](/minecraft/mc-mods/productivebees/comments) * [Files](/minecraft/mc-mods/productivebees/files/all?page=1&pageSize=20) * [Images (18)](/minecraft/mc-mods/productivebees/screenshots) * [Relations](/minecraft/mc-mods/productivebees/relations/dependencies) * [Source](https://github.com/JDKDigital/productive-bees) * [Issues](https://github.com/JDKDigital/productive-bees/issues) ## Description Productive Bees builds on the content added by Minecraft 1.15 and adds additional bees and advanced beehives allowing you to put your bees to work creating resources in a fully automated setup. Productive bees are just as derpy as their vanilla cousins, but makes up for it in usefulness and beeuty. This mod adds bees for most vanilla resources as well as all your favorite modded resources like copper, lead, silver etc. There are also a lot of bees for mod specific resources like Mystical Agriculture essence, Silents Gems, Powah, Tinkers Construct, Thermal and many more. In total you can find more than 200 bees if the relevant mods are installed. #################### File: curseforge-com-minecraft-mc-mods-productivebees-48193.txt Page: 1 Context: We also add up to 32 variants of a new advanced beehives for the bees to live in, most of which are made from wood added by other mods. **The mod is fully customizable, you can add your own bees, disable existing bees and change the production resources and rates.** It is recommended to use JEI for looking up breeding recipes and Patchouli for ingame documentation. If you have questions check the FAQ below or hop on over to the [discord server](https://discord.gg/x822BKu624) FAQ **Q: Why do my bees not go into my hive?** \- Most of the bees found in the overworld are solitary bees and will not live in a hive. They need nests to live in, don't produce resources and are mostly used for breeding other bees. \- Each bee has specific flower requirements, make sure they are met. Flower requirements can bee seen in JEI by right clicking or pressing 'U' on the bee \- If your bees were living in a different hive, they might still remember it as their home. Put the bee in a cage and release it while sneaking next to the new hive. \- If all else fails, try breaking the hive and place it again. Bees stay in the hive when broken, no need for silk touch. \- If you have Radium installed, add this to the radium config file mixin.ai.poi=false  **Q: How to get X bee?** \- Most of the bees are obtained from breeding two other bees together, recipes for breeding can be found in JEI or the Big Book of Bees. \- Solitary and wild bees are obtained from nests found around the world. You can make a new bee move into a nest by placing the nest in the right biome and then use one or more Honey Treats on the nest. Each additional treat will lower the countdown by 10%. #################### File: curseforge-com-minecraft-mc-mods-productivebees-48193.txt Page: 1 Context: **Q: How to get a Neon Cuckoo or a Nomad Bee** \- Neon Cuckoo and Nomad bees are cuckoo bees and will only spawn from nests of other bees. Neon Cuckoo bee will spawn from the nest of a Blue Banded Bee and Nomad bees from the nest of an Ashy Mining Bee. There's a chance for a cuckoo bee to spawn each time the other bee brings back nectar, so make sure you have flowers around. **Q: How do I get Skeletal or Zombees** \- Skeletal and ZomBees spawn during the night in empty advanced hives left in the dark \- The spawn rate is rather low, so set up 6-8 hives in the dark and make sure the hives are chunkloaded \- Vanilla hives can't be used **Q: JEI is not showing all the bees** \- The bees are not always indexed in JEI the first time. If you are an op you can run the /reload command \- If you are not op, it usually works to disconnect and connect again **Q: I can't find vanilla bees!!!** from Minecraft wiki: "As of Java Edition 1.15.2, occupied bee nests additionally have a 5% chance to spawn on oak or birch trees in any biome that were grown from saplings within two blocks of any flower at the same y-level. With a bit of luck and a large enough supply of oak or birch saplings and bone meal, one can obtain bees relatively easily without having to travel, by repeatedly growing trees near flowers until one spawns a bee nest as it grows. This is also an ideal way to obtain bees in older worlds created prior to Java Edition 1.15, as it saves players from having to locate anything." **Q: How do I summon and kill bees** Most of the bees are NBT based and will need more detailed commands to summon and kill \- /kill @e\[type=productivebees:configurable\_bee, name="Diamond Bee"\] \- /summon productivebees:configurable\_bee \~ \~ \~ {"type":"productivebees:diamond"} #################### File: curseforge-com-minecraft-mc-mods-productivebees-48193.txt Page: 1 Context: **Q: How do I get bee genes for the Incubator** Put a piston above a bottler with bottles in it, squash the bee you want the genes of and you get a bottle of squashed bee. Put that in the centrifuge and you get the different bee attributes as genes as well as a bee type gene. Type genes can be combined with a honey treat and used in the Incubator to make spawn eggs of that type. The other genes can be put on honey treats and fed to other bees to transfer the attributes. You can also use the pokey pokey upgrade in a hive to peacefully and slowly extract genes or just use a command /give @p productivebees:gene{productivebees\_gene\_purity:100,productivebees\_gene\_attribute:"productivebees:crystalline",productivebees\_gene\_value:0} **Q: How to get Nocturnal and Metaturnal bees** Leash a Diurnal bee to something during night and it has a chance to turn Nocturnal Leash a Nocturnal bee during day and it has a chance to turn Metaturnal Be careful though as they can take damage with leashed in an environment they are not accustomed to **Q: How to improve weather tolerance** Leash a bee outside during rain or storm and the bee might become accustomed to the weather Be careful as it may take damage while leashed in a hostile environment **Here's a playlist with German tutorials: ** [**Unofficial getting started guide**](/linkout?remoteUrl=http%253a%252f%252ftoxicfrog.github.io%252fgetting-started-with-productive-bees%252f) **Mod integrations:** [JEI](https://www.curseforge.com/minecraft/mc-mods/jei): Absolutely needed to look up breeding recipes and bee requirements [Patchouli](https://www.curseforge.com/minecraft/mc-mods/patchouli): Install to get access to the in-game guide book #################### File: curseforge-com-minecraft-mc-mods-productivebees-48193.txt Page: 1 Context: [The Bumblezone](https://www.curseforge.com/minecraft/mc-mods/the-bumblezone-forge): Productive bees and their comb blocks has a chance to spawn in The Bumblezone dimension. [The One Probe](https://www.curseforge.com/minecraft/mc-mods/the-one-probe) or [Jade](https://www.curseforge.com/minecraft/mc-mods/jade): Displays extra information for nests and hives Special thanks to Bagel, author of Buzzier Bees, for permission to reuse beehive textures for advanced versions of the beehives introduced in BB. [](/linkout?remoteUrl=https%253a%252f%252fwww.akliz.net%252flobstermods) [](/) CurseForge - a world of endless gaming possibilities for modders and gamers alike. 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It is depicted with a playful expression and feather details visible on its body. 3. **Scene and Activity Analysis:** - **Scene Description:** The scene is simple, with a solid yellow background and minimalistic design elements: a text title and a rubber duck illustration. - **Activity:** This image appears to be a slide from a presentation, serving as a title or cover page. 4. **Text Analysis:** - **Detected Text:** - "Topic title" - large, centered at the top. - Additional text content ". . ." - **Content Significance:** - "Topic title" indicates this is likely the title slide of a presentation. - The additional text at the bottom pertains to the software and version used to create the slide. 6. **Anomaly Detection:** - **Noticeable Elements:** - The text "..." could be considered an anomaly in the context of presentation slides, as it suggests this slide is from a software evaluation version rather than a finalized product. 8. **Color Analysis:** - **Color Composition:** The dominant color is a solid yellow background, which gives a bright and energetic feel to the slide. - **Impact on Perception:** The vibrant yellow dominates the visual experience, making the slide feel cheerful and attention-grabbing. 9. **Perspective and Composition:** - **Perspective:** The image is a flat, 2D presentation slide. - **Composition:** - The text "Topic title" is centrally aligned at the top, easily noticeable due to its large font size and black color against the yellow background. - The rubber duck illustration is positioned at the bottom center, adding a visual element that draws attention without overwhelming the slide. 10. **Contextual Significance:** - **Overall Document Context:** This slide appears to be the introductory or title slide of a presentation, setting the stage for the content to follow. - **Contribution to Theme:** The simplicity and bright color scheme suggest a light-hearted, easy-to-follow presentation. The rubber duck adds a playful touch, potentially indicating a friendly or informal presentation style. #################### File: PPT%2010%20duckietown-master.pptx Page: 2 Context: # Short topic title ## Explains - Concept 1 - Concept 2 ## Prerequisites - Concept 1 - Concept 2 ## Credits - Author Name (Affiliation) - first version - Author Name (Affiliation) - second version, another date These slides are part of the Duckietown project. For more information about Duckietown, see the website [http://duckietown.org](http://duckietown.org). Image Analysis: ### Localization and Attribution: - **Image 1:** - The entire page is one image, so it is referred to as Image 1. ### Object Detection and Classification: - **Image 1:** - **Objects Detected:** - Text blocks - Bullet points - Titles - Credits section ### Scene and Activity Analysis: - **Image 1:** - **Scene Description:** - The image is a slide from a presentation template. - **Activities:** - Providing an overview of a topic in sections labeled as "Explains" and "Prerequisites." - Lists under these sections contain concepts to be covered. - A "Credits" section is present, mentioning author affiliations. ### Text Analysis: - **Image 1:** - **Text Detected:** - "Short topic title" - "Explains" - "Concept 1" - "Concept 2" - "Prerequisites" - "Concept 1" - "Concept 2" - "Credits" - "Author name (affiliation) - first version" - "Author name (affiliation) - second version, another date" - "These slides are part of the Duckietown project. For more information about Duckietown, see the website http://duckietown.org" - **Significance of Text:** - The text outlines the structure of a presentation slide intended to cover various concepts and prerequisites related to a short topic. - The credits section attributes the content to specific authors and their affiliations. - The final line provides information about the Duckietown project, directing readers to a website for further details. ### Color Analysis: - **Image 1:** - **Color Composition:** - Dominant color is yellow, used as the background. - Text is primarily black, with some in red for emphasis. - **Impact on Perception:** - The yellow background creates a bright, engaging appearance. - Black text ensures readability against the bright background. - Red text for the project's additional information draws attention. ### Perspective and Composition: - **Image 1:** - **Perspective:** - Standard eye-level perspective typically used in slide presentations. - **Composition:** - The slide is divided into three main sections: - Top section contains the title. - Middle section is split into two columns for "Explains" and "Prerequisites." - Bottom section includes "Credits" and additional project information. - The layout is organized to provide clear structure and easy readability. ### Contextual Significance: - **Image 1:** - **Context:** - Likely a part of a larger presentation, possibly an academic or research-oriented series given its formal structure and inclusion of author credits and project information. - **Contribution to Overall Message:** - This slide sets the stage for the topics and concepts to be discussed in the presentation while crediting contributors and directing the audience to additional resources. #################### File: PPT%2010%20duckietown-master.pptx Page: 3 Context: # A regular slide - This is a regular slide - Another point - Yet another point - Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. - Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Image Analysis: ### Localization and Attribution - **Image Number**: This image is identified as **Image 1** as it is the only image provided. ### Object Detection and Classification - **Identified Objects**: - **Text Elements**: Various text components on the slide. - **Bullet Points**: Black bullet points indicating the start of each item on the list. ### Scene and Activity Analysis - **Scene Description**: The image represents a slide, presumably from a presentation, with a title and bulleted text points. - **Activities Taking Place**: The image appears to be a regular slide within a presentation, likely aimed at providing information or outlining points in a talk or lecture. ### Text Analysis - **Detected Text**: - **Title**: "A regular slide" - **Bullet Points**: - "This is a regular slide" - "Another point" - "Yet another point" - "Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua." - "Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat." - **Footer Text**: "Duckietown" - **Page Number**: "3" - **Text Significance**: The text provides an example of a standard slide format with multiple bullet points typically used in presentations. The "Lorem ipsum" text serves as placeholder text, commonly used in design to suggest textual layout. ### Color Analysis - **Color Composition**: - **Dominant Color**: Yellow border around the slide. - **Text Color**: Black, ensuring good readability against the white background. ### Perspective and Composition - **Perspective**: The image is presented from a straightforward, head-on view, typical of a slide presentation. - **Composition**: - **Title**: Positioned at the top of the slide, bold, and larger than the body text. - **Body**: Comprised of bullet points, aligned to the left for a structured and easy-to-follow format. - **Footer**: Contains additional text (Duckietown) and page number, aligned left and right at the bottom of the slide. ### Contextual Significance - **Overall Document**: As a regular slide within a presentation, this image contributes to the flow of a lecture or a talk, providing key points or main ideas in a clear bullet point format. - **Message or Theme**: The image emphasizes a standard format for professional or academic presentations, facilitating the delivery of information in a concise manner. ### Anomaly Detection - **Notable Element**: The presence of "Lorem ipsum" text, which is typically used as placeholder text. This might indicate that the slide is a template or draft rather than a final presentation slide. ### Conclusion - **Summary**: Image 1 is a typical slide from a presentation, featuring a title, several bullet points, and placeholder text. The primary color scheme includes a yellow border and black text, laid out in a clear and readable format, adhering to standard presentation practices. #################### File: PPT%2010%20duckietown-master.pptx Page: 4 Context: # A slide with a diagram - This type of slide has half of the space reserved for a diagram. - Avoid putting math inline in the text. This is not supported by Keynote, and if you hack it, it will get screwed up when the slide template is changed. - Example of what **not** to do: \( (d o y) \) Suppose \( \mathbb{R}^2 \) has a couple of coordinates \( (x, y) \) that belong to: | Better | Use table | |----------------|--------------------| | Space | \( \mathbb{R}^2 \) | | Coordinates | \( (x, y) \) | Image Analysis: ### Comprehensive Examination of the Image: **Image Number:** Image 1 #### 1. Localization and Attribution: - **Location on the Page:** This image occupies the entire slide. - **Numbering:** Image 1 #### 2. Object Detection and Classification: - **Objects Identified:** - Text elements - Mathematical notations - Diagram with labels - **Key Features:** - The text elements and mathematical notations are distinct and separated into various sections. - The diagram shows a coordinate system and a labeled diagram of a robot or an object moving in a plane. #### 3. Scene and Activity Analysis: - **Scene Description:** - The scene is an educational or instructional slide discussing how to properly format a slide that includes mathematical content and diagrams. - **Activities Taking Place:** - The slide provides guidelines on using diagrams and mathematical notations correctly. #### 4. Text Analysis: - **Detected Text:** - "A slide with a diagram" - "This type of slide has half of the space reserved for a diagram." - "Better: use table space coordinates" - "Avoid putting math inline in the text. This is not supported by Keynote, and if you hack it, it will get screwed up when the slide template is changed." - "Example of what not to do: Suppose we have a couple of coordinates that belong to \(\mathbb{R}^2\)" - **Content Analysis:** - The main text advises against integrating mathematical content inline with regular text, suggesting the use of tables instead. - Additional text explains the risks of inline mathematical expressions not rendering well in different software. - There is a specific example given on what not to do, illustrated with a diagram. #### 5. Diagram and Chart Analysis: - **Diagram:** - A diagram illustrating a coordinate system \((X_w, Y_w)\) and what appears to be a labeled robotic system. - It includes variables such as \(θ_t\) and movement vectors. #### 9. Perspective and Composition: - **Perspective:** - The image is taken from a straight-on perspective typical for slide presentations. - **Composition:** - The slide is divided into text on the left side and a diagram on the right side. - Key points are highlighted in different colors (e.g., "Better: use table" in red) to draw attention. #### 11. Metadata Analysis: - **Capture Date, Camera Settings:** Not applicable here, as the context is an educational or instructional slide. #### 13. Graph Numbers: - **No Numerical Graph Data:** The image does not contain numerical graph data. However, it includes illustrative examples and notations. #### Additional Aspects: - **Ablaufprozesse (Process Flows):** - The slide suggests a proper way to present information, highlighting a process of do's and don'ts for educational content creation. - **Prozessbeschreibungen (Process Descriptions):** - Describes the process of correctly integrating diagrams and mathematical content into educational slides. - **Typen Bezeichnung (Type Designations):** - Defines proper and improper ways (types) to present mathematical content in slides. - **Trend and Interpretation:** - The text encourages consistent and clear formatting for better understanding. - **Tables:** - Mentioned in the context as a preferred method to organize mathematical content. This examination highlights how the image serves as a guide for creating well-organized instructional content, emphasizing proper formatting practices and the use of diagrams. #################### File: PPT%2010%20duckietown-master.pptx Page: 5 Context: # How to use the theme - Double click on the file `themes/DuckietownColorfulTheme2.kth` - You can add it to your themes library. - To convert an existing presentation, use **File > Change Theme.** Image Analysis: ### Image Analysis #### Localization and Attribution: - **Image Number**: Image 1 #### Text Analysis: - **Text Detected**: - **Title**: "How to use the theme" - **Instructions**: 1. "Double click on the file `themes/DuckietownColorfulTheme2.kth`" 2. "You can add it to your themes library." 3. "To convert an existing presentation, use File > Change Theme." - **Content Significance**: - The text provides step-by-step instructions on how to use a specific theme file (`DuckietownColorfulTheme2.kth`). It guides users on how to add this theme to their library and how to convert an existing presentation to utilize the new theme by changing the theme. #### Scene and Activity Analysis: - **Scene Description**: - The scene is a simple instructional slide with a white background bordered by a yellow frame. - Main activity is providing instructions on the usage of a presentation theme. - No actors are present in the image. #### Color Analysis: - **Color Composition**: - The dominant color is white (background), along with noticeable use of black (text) and yellow (border). - The color combination gives a clean and professional look, emphasizing readability and the instructional nature of the content. #### Perspective and Composition: - **Perspective**: - The image is taken from a straight-on, flat perspective, typical for presentation slides. - **Composition**: - The title is centered at the top, making it immediately visible and setting context. - The instructions are listed as bullet points, ensuring clarity and ease of following steps. - The border frames the content, drawing attention to the central information. #### Contextual Significance: - **Overall Contribution**: - The image serves as an instructional guide within a larger document or presentation, focusing on helping users implement a specific presentation theme. - It likely fits into a series of instructional slides aimed at beginners or those unfamiliar with theme application in presentation software. #################### File: PPT%2010%20duckietown-master.pptx Page: 6 Context: # Organization of the lectures/repo - **0_drafts/**: anything goes here. Feel free to make subfolders. - **1_beta/**: put things here that must be reviewed by others. - **2_given/**: - when you give the lecture, save it here with the timestamp and the name of the location. - Also save a `.pdf` version. - These timestamped `.key` and `.pdf` files are the ones to be linked to from the class page. - These files are never modified after they are timestamped. - **3_ideal/**: later, here we will assemble the “ideal” version of the slides. Image Analysis: ## Image Analysis ### Localization and Attribution - **Image Number**: Image 1 - **Position**: Sole image on the page. ### Text Analysis - **Detected Text**: ``` Organization of the lectures/ repo. • 0_drafts/: anything goes here. Feel free to make subfolders. • 1_beta/: put things here that must be reviewed by others. • 2_given/: • when you give the lecture, save it here with the timestamp and the name of the location. • Also save a .pdf version. • These timestamped .key and .pdf files are the ones to be linked to from the class page • These files are never modified after they are timestamped. • 3_ideal/: later, here we will assemble the “ideal” version of the slides. ``` - **Content Analysis**: - The text outlines the organizational structure of lecture or repository files, specifying where different types of files or stages of development should be stored: - **0_drafts/**: Initial drafts and subfolders. - **1_beta/**: Files that require review. - **2_given/**: - Saving lectures with timestamp and location. - Saving .pdf versions. - Timestamped .key and .pdf files to be linked from the class page. - Files are not modified after being timestamped. - **3_ideal/**: Final assembly of the slides. - **Significance**: - This organization structure helps maintain a clear methodology for preparing, reviewing, presenting, and archiving lecture materials. It ensures orderly progression from drafts to the final version of the slides while keeping timestamps and original versions intact. ### Color Analysis - **Dominant Colors**: - **Background**: Yellow border with a white central area. - **Text**: Black. - **Impact on Perception**: - The yellow border frames the content well, making the white central area stand out and improving readability. The black text on the white background is highly legible. ### Perspective and Composition - **Perspective**: Eye-level view. - **Composition**: - The image features a centered block of text with a clear heading and bullet points, aiding readability. The yellow border encapsulates the content, providing focus and structure. ### Contextual Significance - **Overall Message**: - The image provides a clear guideline on how lectures or repository files should be managed, aiming for organized progression and seamless accessibility of resources. ### Anomaly Detection - No anomalies or unusual elements identified in the image. Everything appears to be correctly formatted and presented. ### Diagram and Chart Analysis - No diagrams or charts to analyze. ### Graph and Trend Analysis - No graphs or trends to analyze. #################### File: PPT%2010%20duckietown-master.pptx Page: 7 Context: # Getting the URL to share - Go to GitHub and click on the file. - On the file page, copy the link that is on the Download button. - They look like this: - [Download PDF](https://github.com/duckietown/lectures/raw/master/duckietown-master.pdf) - [Download KEY](https://github.com/duckietown/lectures/raw/master/duckietown-master.key) Image Analysis: ### Image Analysis #### 1. Localization and Attribution - **Image 1**: The analyzed content is a single image occupying the entire page. #### 2. Object Detection and Classification - **Image 1**: - **Objects Detected**: - Text - Hyperlinks #### 3. Scene and Activity Analysis - **Image 1**: - **Scene Description**: The image is a slide-style content with textual instructions and hyperlinks. - **Activities**: The activities described are centered around obtaining a URL to share from Github. #### 4. Text Analysis - **Image 1**: - **Extracted Text**: ``` Getting the URL to share • Go to Github, click on the file. • On the file page, copy the link that is on the Download button. • They look like this: • https://github.com/duckietown/lectures/raw/master/duckietown-master.pdf • https://github.com/duckietown/lectures/raw/master/duckietown-master.key ``` - **Significance**: The text provides instructions on how to obtain and share a URL from Github. It is straightforward and instructional, intended for guiding users through a specific process. #### 5. Diagram and Chart Analysis - **Not Applicable**: There are no diagrams or charts in this image. #### 6. Product Analysis - **Not Applicable**: There are no products depicted in this image. #### 7. Anomaly Detection - **Not Applicable**: No anomalies or unusual elements are noted in this image. #### 8. Color Analysis - **Image 1**: - **Dominant Colors**: - **Background**: White - **Borders**: Yellow - **Text**: Black and Blue (for hyperlinks) - **Impact on Perception**: The use of a white background with black text is standard for readability. The yellow border adds a touch of visual interest to differentiate it as an instructional slide. Blue links are typical for indicating hyperlinks. #### 9. Perspective and Composition - **Image 1**: - **Perspective**: The perspective is that of a standard presentation or instructional slide, viewed head-on. - **Composition**: The image is composed with a prominent title at the top, followed by bullet points providing instructions, and hyperlinks at the bottom. #### 10. Contextual Significance - **Image 1**: - **Overall Document Contribution**: The image's instructions are part of a larger instructional document or presentation, likely intended to inform users about accessing and sharing content from Github repositories. - **Theme**: The theme is educational, focusing on providing technical instructions. #### 11. Metadata Analysis - **Not Available**: There is no metadata available directly from the image itself. ### Additional Aspects #### Ablaufprozesse (Process Flows) - **Image 1**: The text describes a simple process flow for obtaining a URL from Github, involving navigation steps and actions to be performed. #### Prozessbeschreibungen (Process Descriptions) - **Image 1**: - **Step 1**: Go to Github and click on the file. - **Step 2**: On the file page, copy the link that is on the Download button. ### Conclusion This image serves as an instructional guide for obtaining a URL from Github. It clearly lays out the steps in a simple and visually clean format, ensuring the information is easily accessible and understandable. #################### File: DOCX%204%20Pb%20LCF%20data%20for%20Port%20Pirie%20soil%20Table.docx Page: 1 Context: Table 1: Elemental concentration, particle size and Pb speciation in PP-D Elemental concentration (mean ± SEM, n = 3)Trace elements (mg/kg)Major elements (g/kg)Arsenic (As)Cadmium (Cd)Copper (Cu)Lead (Pb)Zinc (Zn)Aluminium (Al)Iron (Fe)Phosphorus (P)586 ± 16231 ± 4.92740 ± 3132045 ± 39229305 ± 63014.0 ± 2.135.5 ± 1.41.86 ± 0.05Pb speciation (weighted %)Organic matter bound PbHydrocerussite Pb3(CO3)2(OH)2Galena PbSChloropyromorphite Pb5(PO4)3ClMineral sorbed Pb (Pb oxides and adsorbed onto clays)4421111013Particle size distribution (µm) (n = 3)Mean ± SEM Mode Median95% of the particles are 2.26 ± 0.060.570.93< 9.71 µm #################### File: PDF%205%20.pdf Page: 1 Context: # Linked List Problems By Nick Parlante Copyright ©1998-2002, Nick Parlante ## Abstract This document reviews basic linked list code techniques and then works through 18 linked list problems covering a wide range of difficulty. Most obviously, these problems are a way to learn about linked lists. More importantly, these problems are a way to develop your ability with complex pointer algorithms. Even though modern languages and tools have made linked lists pretty unimportant for day-to-day programming, the skills for complex pointer algorithms are very important, and linked lists are an excellent way to develop those skills. The problems use the C language syntax, so they require a basic understanding of C and its pointer syntax. The emphasis is on the important concepts of pointer manipulation and linked list algorithms rather than the features of the C language. For some of the problems we present multiple solutions, such as iteration vs. recursion, dummy node vs. local reference. The specific problems are, in rough order of difficulty: Count, GetNth, DeleteList, Pop, InsertNth, SortedInsert, InsertSort, Append, FrontBackSplit, RemoveDuplicates, MoveNode, AlternatingSplit, ShuffleMerge, SortedMerge, SortedIntersect, Reverse, and RecursiveReverse. ## Contents 1. **Review of basic linked list code techniques** 3 2. **18 list problems in increasing order of difficulty** 10 3. **Solutions to all the problems** 20 This is document #105, Linked List Problems, in the Stanford CS Education Library. This and other free educational materials are available at [cslibrary.stanford.edu](http://cslibrary.stanford.edu). This document is free to be used, reproduced, or sold so long as this notice is clearly reproduced at its beginning. ## Related CS Education Library Documents Related Stanford CS Education library documents: - [Linked List Basics](http://cslibrary.stanford.edu/103/) Explains all the basic issues and techniques for building linked lists. - [Pointers and Memory](http://cslibrary.stanford.edu/102/) Explains how pointers and memory work in C and other languages. Starts with the very basics, and extends through advanced topics such as reference parameters and heap management. - [Binary Trees](http://cslibrary.stanford.edu/110/) Introduction to binary trees. - [Essential C](http://cslibrary.stanford.edu/101/) Explains the basic features of the C programming language. #################### File: PDF%205%20.pdf Page: 2 Context: # The Great Tree List Problem [The Great Tree List Problem](http://cslibrary.stanford.edu/109/) presents the greatest recursive pointer problem ever devised. ## Why Linked Lists Are Great To Study Linked lists hold a special place in the hearts of many programmers. Linked lists are great to study because... - **Nice Domain** The linked list structure itself is simple. Many linked list operations such as "reverse a list" or "delete a list" are easy to describe and understand since they build on the simple purpose and structure of the linked list itself. - **Complex Algorithm** Even though linked lists are simple, the algorithms that operate on them can be as complex and beautiful as you want (See problem #18). It’s easy to find linked list algorithms that are complex and pointer intensive. - **Pointer Intensive** Linked list problems are really about pointers. The linked list structure itself is obviously pointer intensive. Furthermore, linked list algorithms often break and re-weave the pointers in a linked list as they go. Linked lists really test your understanding of pointers. - **Visualization** Visualization is an important skill in programming and design. Ideally, a programmer can visualize the state of memory to help think through the solution. Even the most abstract languages such as Java and Perl have layered, reference-based data structures that require visualization. Linked lists have a natural visual structure for practicing this sort of thinking. It’s easy to draw the state of a linked list and use that drawing to think through the code. Not to appeal to your mercenary side, but for all of the above reasons, linked list problems are often used as interview and exam questions. They are short to state, and have complex, point-intensive solutions. No one really cares if you can build linked lists, but they do want to see if you have programming agility for complex algorithms and pointer manipulation. Linked lists are the perfect source of such problems. ## How To Use This Document Try not to use these problems passively. Take some time to try to solve each problem. Even if you do not succeed, you will think through the right issues in the attempt, and looking at the given solution will make more sense. Use drawings to think about the problems and work through the solutions. Linked lists are well-suited for memory designs, so these problems are an excellent opportunity to develop your visualization skill. The problems in this document use regular linked lists, without simplifications like dummy headers. ## Dedication This Jan-2002 revision includes many small edits. The first major release was Jan 17, 1999. Thanks to Negar Shamma for her many corrections. This document is distributed for the benefit and education of all. Thanks to the support of Eric Roberts and Stanford University. That someone seeking educational should have the opportunity to find it. May you learn from it in the spirit of goodwill in which it is given. Best Regards, Nick Parlante nick parlante@cs.stanford.edu #################### File: PDF%205%20.pdf Page: 3 Context: # Section 1 — Linked List Review This section is a quick review of the concepts used in these linked list problems. For more detailed coverage, see [Link List Basics](http://cslibrary.stanford.edu/103) where all of this material is explained in much more detail. ## Linked List Ground Rules All of the linked list code in this document uses the "classic" singly linked list structure: A single head pointer points to the first node in the list. Each node contains a single `.next` pointer to the next node. The `.next` pointer of the last node is `NULL`. The empty list is represented by a `NULL` head pointer. All of the nodes are allocated in the heap. For a few of the problems, the solutions present the temporary "dummy node" variation (see below), but most of the code deals with linked lists in their plain form. In the text, brackets `{}` are used to describe lists — the list containing the numbers 1, 2, and 3 is written as `{1, 2, 3}`. The node type used is... ```c struct node { int data; struct node* next; }; ``` To keep things simple, we will not introduce any intermediate typedefs. All pointers to nodes are declared simply as `struct node*`. Pointers to pointers to nodes are declared as `struct node**`. Such pointers to pointers are often called "reference pointers". ## Basic Utility Functions In a few places, the text assumes the existence of the following basic utility functions: - `int Length(struct node* head);` Returns the number of nodes in the list. - `struct node* BuildOneTwoThree();` Allocates and returns the list `{1, 2, 3}`. Used by some of the example code to build lists to work on. - `void Push(struct node** headRef, int newData);` Given an int and a reference to the head pointer (i.e. a `struct node**` pointer to the head pointer), add a new node at the head of the list with the standard 3-step-in: create the new node, set its `.next` to point to the current head, and finally change the head to point to the new node. (If you are not sure of how this function works, the first few problems may be helpful warm-ups.) #################### File: PDF%205%20.pdf Page: 4 Context: # Use of the Basic Utility Functions This sample code demonstrates the basic utility functions being used. Their implementations are also given in the appendix at the end of the document. ```c void BasicsCaller() { struct node* head; int len; head = BuildOneTwoThree(); // Start with [1, 2, 3] Push(&head, 13); // Push 13 on the front, yielding [13, 1, 2, 3] // (The 'g' is because head is passed // as a reference pointer.) Push(&(head->next), 42); // Push 42 into the second position // yielding [13, 42, 1, 2, 3] // Demonstrates a use of '&' on // the .next field of a node. len = Length(head); // Computes the length as 5. } ``` If these basic functions do not make sense to you, you can (a) see [Linked List Basics](http://cslibrary.stanford.edu/103/) which explains the basics of linked lists in detail, or (b) do the first few problems, but avoid the intermediate and advanced ones. # Linked List Code Techniques The following list presents the most common techniques you may want to use in solving the linked list problems. The first few are basic. The last few are only necessary for the more advanced problems. ## 1. Iterate Down a List A very frequent technique in linked list code is to iterate a pointer over all the nodes in a list. Traditionally, this is written as a `while` loop. The head pointer is copied into a local variable `current` which then iterates down the list. Test for the end of the list with `current != NULL`. Advance the pointer with `current = current->next`. ```c // Return the number of nodes in a list (while-loop version) int Length(struct node* head) { int count = 0; struct node* current = head; while (current != NULL) { count++; current = current->next; } return(count); } ``` Alternatively, some people prefer to write the loop as a `for`, which makes the initialization, test, and pointer advance more centralized, and so harder to omit... ```c for (current = head; current != NULL; current = current->next) { // Loop body here } ``` #################### File: PDF%205%20.pdf Page: 5 Context: # 2. Changing a Pointer Using a Reference Pointer Many list functions need to change the caller's head pointer. In C++, you can just declare the pointer parameter as an `&` argument, and the compiler takes care of the details. To do this in the C language, pass a pointer to the head pointer. Such a pointer to a pointer is sometimes called a "reference pointer." The main steps for this technique are: - **Design the function to take a pointer to the head pointer.** This is the standard technique in C — pass a pointer to the "value of interest" that needs to be changed. To change a `struct node*`, pass a `struct node**`. - **Use `&` in the caller** to compute and pass a pointer to the value of interest. - **Use `*` on the parameter in the callee function** to access and change the value of interest. The following simple function sets a head pointer to NULL using a reference parameter: ```c // Change the passed in head pointer to be NULL // Uses a reference pointer to access the caller's memory void ChangeToNull(struct node** headRef) { // Takes a pointer to the value of interest *headRef = NULL; // use '*' to access the value of interest } ``` ```c void ChangeCaller() { struct node* head1; struct node* head2; ChangeToNull(&head1); // use '&' to compute and pass a pointer to ChangeToNull(&head2); // the value of interest // head1 and head2 are NULL at this point } ``` Here is a drawing showing how the `headRef` pointer in `ChangeToNull()` points back to the variable in the caller: ``` Stack ChangeCaller() head1 | ChangeToNull(&head1) headRef ``` #################### File: PDF%205%20.pdf Page: 6 Context: Many of the functions in this document use reference pointer parameters. See the use of `Push()` above and its implementation in the appendix for another example of reference pointers. See problem #8 and its solution for a complete example with drawings. For more detailed explanations, see the resources listed on page 1. ## 3. Build — At Head With `Push()` The easiest way to build up a list is by adding nodes at its "head end" with `Push()`. The code is short and it runs fast — lists naturally support operations at their head. The disadvantage is that the elements will appear in the list in the reverse order that they are added. If you don't care about order, then the head end is the best. ```c struct node* AddAtHead() { struct node* Head = NULL; int i; for (i=1; i<6; i++) { Push(&Head, i); } // head == [5, 4, 3, 2, 1]; return Head; } ``` ## 4. Build — With Tail Pointer What about adding nodes at the "tail end" of the list? Adding a node at the tail of a list most often involves locating the last node in the list and then changing its `.next` field from `NULL` to point to the new node, such as the `tail` variable in the following example of adding a "3" node to the end of the list (`1, 2...`). ### Stack ``` head → tail → newNode ``` ### Heap ``` 1 → 2 → 3 ``` This is just a special case of the general rule: to insert or delete a node inside a list, you need a pointer to the node just before that position, so you can change its `.next` field. Many list problems include the sub-problem of advancing a pointer to the node before the point of insertion or deletion. The one exception is if the operation falls on the first node in the list — in that case the head pointer itself must be changed. The following examples show the various ways code can handle the single head case and all the interior cases... #################### File: PDF%205%20.pdf Page: 7 Context: # 5. Build — Special Case + Tail Pointer Consider the problem of building up the list `{1, 2, 3, 4, 5}` by appending the nodes to the tail end. The difficulty is that the very first node must be added at the head pointer, but all the other nodes are inserted after the last node using a tail pointer. The simplest way to deal with both cases is to just have two separate cases in the code. Special case code first adds the head node `{1}`. Then there is a separate loop that uses a tail pointer to add all the other nodes. The tail pointer is kept pointing at the last node, and each new node is added at `tail->next`. The only "problem" with this solution is that writing separate special case code for the first node is a little unsatisfying. Nonetheless, this approach is a solid one for production code—it is simple and runs fast. ```c struct node* BuildWithSpecialCase() { struct node* head = NULL; struct node* tail; int i; // Deal with the head node here, and set the tail pointer Push(&head, 1); tail = head; // Do all the other nodes using 'tail' for (i = 2; i < 6; i++) { Push(&(tail->next), i); // add node at tail->next tail = tail->next; // advance tail to point to last node } return(head); // head == {1, 2, 3, 4, 5} } ``` # 6. Build — Temporary Dummy Node This is a slightly unusual technique that can be used to shorten the code: Use a temporary dummy node at the head of the list during the computation. The trick is that with the dummy, every node appears to be added after the `.next` field of some other node. That way the code for the first node is the same as for the other nodes. The tail pointer plays the same role as in the previous example. The difference is that it now also handles the first node as well. ```c struct node* BuildWithDummyNode() { struct node dummy; // Dummy node is temporarily the first node struct node* tail = &dummy; // Start the tail at the dummy int i; dummy.next = NULL; for (i = 1; i < 6; i++) { Push(&(tail->next), i); tail = tail->next; } // The real result list is now in dummy.next return(dummy.next); // dummy.next == {1, 2, 3, 4, 5} } ``` #################### File: PDF%205%20.pdf Page: 8 Context: Some linked list implementations keep the dummy node as a permanent part of the list. For this "permanent dummy" strategy, the empty list is not represented by a NULL pointer. Instead, every list has a heap allocated dummy node at its head. Algorithms skip over the dummy node for all operations. That way, the dummy node is always present to provide the above sort of convenience in the code. I prefer the temporary strategy shown here, but it is a little peculiar since the temporary dummy node is allocated in the stack, while all other nodes are allocated in the heap. For production code, I do not use either type of dummy node. The code should just cope with the head node boundary cases. ## 7. Build — Local References Finally, here is a tricky way to unify all the node cases without using a dummy node at all. For this technique, we use a local "reference pointer" which always points to the last pointer in the list instead of to the last node. All additions to the list are made by following the reference pointer. The reference pointer starts off pointing to the head pointer. Later, it points to the `.next` field inside the last node in the list. (A detailed explanation follows.) ```c struct node* BuildWithLocalRef() { struct node* head = NULL; struct node** lastPtrRef = &head; // Start out pointing to the head pointer int i; for (i = 1; i <= 6; i++) { Push(lastPtrRef, i); // Add node at the last pointer in the list lastPtrRef = &((*lastPtrRef)->next); // Advance to point to the new last pointer } // head == {1, 2, 3, 4, 5}; return (head); } ``` This technique is short, but the inside of the loop is scary. This technique is rarely used, but it’s a good way to see if you really understand pointers. Here’s how it works... 1. At the top of the loop, `lastPtrRef` points to the last pointer in the list. Initially, it points to the head pointer itself. Later it points to the `.next` field inside the last node in the list. 2. `Push(lastPtrRef, i);` adds a new node at the last pointer. The new node becomes the last node in the list. 3. `lastPtrRef = &((*lastPtrRef)->next);` Advances the `lastPtrRef` to now point to the `.next` field inside the new last node — that `.next` field is now the last pointer in the list. Here is a drawing showing the state of memory for the above code just before the third node is added. The previous values of `lastPtrRef` are shown in gray... #################### File: PDF%205%20.pdf Page: 9 Context: # Stack and Heap ```plaintext LocalRef() head lastPtrRef ``` This technique is never required to solve a linked list problem, but it will be one of the alternative solutions presented for some of the advanced problems. The code is shorter this way, but the performance is probably not any better. ## Unusual Techniques Both the temporary-stack-dummy and the local-reference-pointer techniques are a little unusual. They are cute, and they let us play around with yet another variation in pointer-intensive code. They use memory in unusual ways, so they are a nice way to see if you really understand what's going on. However, I probably would not use them in production code. #################### File: PDF%205%20.pdf Page: 10 Context: # Section 2 — Linked List Problems Here are 18 linked list problems arranged in order of difficulty. The first few are quite basic and the last few are quite advanced. Each problem starts with a basic definition of what needs to be accomplished. Many of the problems also include hints or drawings to get you started. The solutions to all the problems are in the next section. It's easy to just passively sweep your eyes over the solution — verifying its existence without letting its details touch your brain. To get the most benefit from these problems, you need to make an effort to think them through. Whether or not you solve the problem, you will be thinking through the right issues, and the given solution will make more sense. Great programmers can visualize data structures to see how the code and memory will interact. Linked lists are well suited to that sort of visual thinking. Use these problems to develop your visualization skill. Make memory drawings to trace through the execution of code. Use drawings of the pre- and post-conditions of a problem to start thinking about a solution. "The will to win means nothing without the will to prepare." - Juma Ikan gaa, marathoner (also attributed to Bobby Knight) ## 1 — Count() Write a `Count()` function that counts the number of times a given int occurs in a list. The code for this has the classic list traversal structure as demonstrated in `Length()`. ```c void CountTest() { List myList = BuildOneTwoThree(); // build {1, 2, 3} int count = Count(myList, 2); // returns 1 since there's one '2' in the list } /* Given a list and an int, return the number of times that int occurs in the list. */ int Count(struct node* head, int searchFor) { // Your code } ``` #################### File: PDF%205%20.pdf Page: 11 Context: ## 2 — GetNth() Write a `GetNth()` function that takes a linked list and an integer index and returns the data value stored in the node at that index position. `GetNth()` uses the C numbering convention that the first node is index `0`, the second is index `1`, and so on. So for the list `{42, 13, 666}`, `GetNth(0)` should return `42`. The index should be in the range `[0..length-1]`. If it is not, `GetNth()` should assert(fail) (or you could implement some other error case strategy). ```c void GetNthTest() { struct node* myList = BuildOneTwoThree(); // build {1, 2, 3} int lastNode = GetNth(myList, 2); // returns the value 3 } ``` Essentially, `GetNth()` is similar to an `array[i]` operation — the client can ask for elements by index number. However, `GetNth()` no list is much slower than `1[i]` on an array. The advantage of the linked list is its much more flexible memory management — we can `Push()` at any time to add more elements and the memory is allocated as needed. // Given a list and an index, return the data // in the nth node of the list. The nodes are numbered from 0. // Assert fails if the index is invalid (outside `[0..length-1]`). ```c int GetNth(struct node* head, int index) { // Your code } ``` ## 3 — DeleteList() Write a function `DeleteList()` that takes a list, deallocates all of its memory and sets its head pointer to `NULL` (the empty list). ```c void DeleteListTest() { struct node* myList = BuildOneTwoThree(); // build {1, 2, 3} DeleteList(myList); // deletes the three nodes and sets myList to NULL } ``` ### Post DeleteList() Memory Drawing The following drawing shows the state of memory after `DeleteList()` executes in the above sample. Overwritten pointers are shown in gray and deallocated heap memory has an 'X' through it. Essentially, `DeleteList()` just needs to call `free()` once for each node and set the head pointer to `NULL`. #################### File: PDF%205%20.pdf Page: 12 Context: # Stack ```c DeleteListTest() { myList // myList is overwritten with the value NULL. } ``` # Heap ``` 1 -> 2 -> 3 ``` *The three heap blocks are deallocated by calls to free(). Their memory will appear to be intact for a while, but the memory should not be accessed.* ## DeleteList() The `DeleteList()` implementation will need to use a reference parameter just like `Push()` so that it can change the caller's memory (`myList` in the above sample). The implementation also needs to be careful not to access the `.next` field in each node after the node has been deallocated. ```c void DeleteList(struct node** headRef) { // Your code } ``` ## 4 — Pop() Write a `Pop()` function that is the inverse of `Push()`. `Pop()` takes a non-empty list, deletes the head node, and returns the head node's data. If all you ever used were `Push()` and `Pop()`, then our linked list would really look like a stack. However, we provide more general functions like `GetNth()` which make what our linked list more than just a stack. `Pop()` should assert(`fail`) if there is not a node to pop. Here's some sample code which calls `Pop()`: ```c void PopTest() { struct node* head = BuildOneTwoThree(); // build {1, 2, 3} int a = Pop(&head); // deletes "1" node and returns 1 int b = Pop(&head); // deletes "2" node and returns 2 int c = Pop(&head); // deletes "3" node and returns 3 int len = Length(head); // the list is now empty, so len == 0 } ``` ## Pop() Unlink `Pop()` is a bit tricky. `Pop()` needs to unlink the front node from the list and deallocate it with a call to `free()`. `Pop()` needs to use a reference parameter like `Push()` so that it can change the caller's head pointer. A good first step to writing `Pop()` properly is making the memory drawing for what `Pop()` should do. Below is a drawing showing a `Pop()` of the first node of a list. The process is basically the reverse of the 3-Step-Link used by `Push()` (would that be "Ni Ni Petz-?"). The overwritten pointer value is shown in gray, and the deallocated heap memory has a big 'X' drawn on it... #################### File: PDF%205%20.pdf Page: 13 Context: # Stack and Heap ## PopTest() ``` head ``` The head pointer advances to refer to the node after the unlinked one. ## Pop() /* The opposite of Push(). Takes a non-empty list and removes the front node, and returns the data which was in that node. */ ```c int Pop(struct node** headRef) { // your code... } ``` ## 5 — InsertNth() A more difficult problem is to write a function `InsertNth()` which can insert a new node at any index within a list. `Push()` is similar, but can only insert a node at the head end of the list (index 0). The caller may specify any index in the range [0..length], and the new node should be inserted so as to be at that index. ```c void InsertNthTest() { struct node* head = NULL; // start with the empty list InsertNth(&head, 0, 13); // build (13) InsertNth(&head, 1, 42); // build (13, 42) InsertNth(&head, 1, 5); // build (13, 5, 42) DeleteList(&head); // clean up after ourselves } ``` #################### File: PDF%205%20.pdf Page: 14 Context: InsertNth() is complex — you will want to make some drawings to think about your solution and afterwards, to check its correctness. /* A more general version of Push(). Given a list, an index 'i' in the range 0..length, and a data element, add a new node to the list so that it has the given index. */ void InsertNth(struct node** headRef, int index, int data) { // your code... } ## 6 — SortedInsert() Write a `SortedInsert()` function which, given a list that is sorted in increasing order and a single node, inserts the node into the correct sorted position in the list. While `Push()` allocates a new node to add to the list, `SortedInsert()` takes an existing node and just rearranges pointers to insert it into the list. There are many possible solutions to this problem. ```c void SortedInsert(struct node** headRef, struct node* newNode) { // your code... } ``` ## 7 — InsertSort() Write an `InsertSort()` function which, given a list, rearranges its nodes so they are sorted in increasing order. It should use `SortedInsert()`. // Given a list, change it to be in sorted order (using `SortedInsert()`). ```c void InsertSort(struct node** headRef) { // your code... } ``` ## 8 — Append() Write an `Append()` function that takes two lists, 'a' and 'b', appends 'b' onto the end of 'a', and then sets 'b' to NULL (since it is now trailing off the end of 'a'). Here is a drawing of a sample call to `Append(a, b)` with the start state in gray and the end state in black. At the end of the call, the 'a' list is {1, 2, 3, 4}, and 'b' list is empty. ### Stack ``` a ``` ### Heap ``` 1 → 2 3 → 4 ``` #################### File: PDF%205%20.pdf Page: 15 Context: It turns out that both of the head pointers passed to `Append(a, b)` need to be reference parameters since they both may need to be changed. The second `b` parameter is always set to `NULL`. When `a` is changed? That case occurs when the `a` list starts out empty. In that case, the `a` head must be changed from `NULL` to point to the `b` list. Before the call, `b` is `{3, 4}`. After the call, `a` is `{3, 4}`. ``` Stack Heap a ├───┐ 3 4 └───┘ b // Append 'b' onto the end of 'a', and then set 'b' to NULL. void Append(struct node** aRef, struct node** bRef) { // Your code... } ## 9 — FrontBackSplit() Given a list, split it into two sublists — one for the front half, and one for the back half. If the number of elements is odd, the extra element should go in the front list. So `FrontBackSplit()` on the list `{2, 3, 5, 7, 11}` should yield the two lists `{2, 3, 5}` and `{7, 11}`. Getting this right for all the cases is harder than it looks. You should check your solution against a few cases (length=2, length=3, length=4) to make sure that the list gets split correctly near the short-list boundary conditions. If it works right for length=4, it probably works right for length=1000. You will probably need special case code to deal with the (length < 2) cases. Hint: Probably the simplest strategy is to compute the length of the list, then use a for loop to hop over the right number of nodes to find the last node of the front half, and then cut the list at that point. There is a trick technique that uses two pointers to traverse the list. A "slow" pointer advances one node at a time, while the "fast" pointer goes two nodes at a time. When the fast pointer reaches the end, the slow pointer will be about half way. For either strategy, care is required to split the list at the right point. /* Split the nodes of the given list into front and back halves, and return the two lists using the reference parameters. If the length is odd, the extra node should go in the front list. */ void FrontBackSplit(struct node* source, struct node** frontRef, struct node** backRef) { // Your code... } ``` #################### File: PDF%205%20.pdf Page: 16 Context: # 10 RemoveDuplicates() Write a `RemoveDuplicates()` function which takes a list sorted in increasing order and deletes any duplicate nodes from the list. Ideally, the list should only be traversed once. ```c /* Remove duplicates from a sorted list. */ void RemoveDuplicates(struct node* head) { // Your code... } ``` # 11 — MoveNode() This is a variant on `Push()`. Instead of creating a new node and pushing it onto the given list, `MoveNode()` takes two lists, removes the front node from the second list and pushes it onto the front of the first. This turns out to be a handy utility function to have for several later problems. Both `Push()` and `MoveNode()` are designed around the feature that list operations work most naturally at the head of the list. Here’s a simple example of what `MoveNode()` should do... ```c void MoveNodeTest() { struct node* a = BuildOneTwoThree(); // the list [1, 2, 3] struct node* b = BuildOneTwoThree(); // b = {2, 3} MoveNode(&a, &b); // a = {1, 2, 3} // b = {} } /* Take the node from the front of the source, and move it to the front of the dest. It is an error to call this with the source list empty. */ void MoveNode(struct node** destRef, struct node** sourceRef) { // Your code } ``` # 12 — AlternatingSplit() Write a function `AlternatingSplit()` that takes one list and divides up its nodes to make two smaller lists. The sublists should be made from alternating elements in the original list. So if the original list is {a, b, a, b, a}, then one sublist should be {a, a}, and the other should be {b, b}. You may want to use `MoveNode()` as a helper. The elements in the new lists may be in any order (for some implementations, it turns out to be convenient if they are in the reverse order from the original list.) ```c /* Given the source list, split its nodes into two shorter lists. If we number the elements 0, 1, 2, ..., then all the even elements should go in the first list, and all the odd elements in the second. The elements in the new lists may be in any order. */ void AlternatingSplit(struct node* source, struct node** aRef, struct node** bRef) { // Your code } ``` #################### File: PDF%205%20.pdf Page: 17 Context: # 13 — ShuffleMerge() Given two lists, merge their nodes together to make one list, taking nodes alternately between the two lists. So `ShuffleMerge()` with `{1, 2, 3}` and `{7, 13, 1}` should yield `{1, 7, 2, 13, 3, 1}`. If either list runs out, all the nodes should be taken from the other list. The solution depends on being able to move nodes to the end of a list as discussed in the Section 1 review. You may want to use `MoveNode()` as a helper. Overall, many techniques are possible: dummy node, local reference, or recursion. Using this function and `FrontBackSplit()`, you could simulate the shuffling of cards. ```c /* Merge the nodes of the two lists into a single list taking a node alternately from each list, and return the new list. */ struct node* ShuffleMerge(struct node* a, struct node* b) { // Your code... } ``` # 14 — SortedMerge() Write a `SortedMerge()` function that takes two lists, each of which is sorted in increasing order, and merges the two together into one list which is in increasing order. `SortedMerge()` should return the new list. The new list should be made by splicing together the nodes of the first two lists (use `MoveNode()`). Ideally, `Merge()` should only make one pass through each list. `Merge()` is tricky to get right—it may be solved iteratively or recursively. There are many cases to deal with: either `a` or `b` may be empty, during processing either `a` or `b` may run out first, and finally there’s the problem of starting the result list empty, and building it up while going through `a` and `b`. ```c /* Takes two lists sorted in increasing order, and splices their nodes together to make one big sorted list which is returned. */ struct node* SortedMerge(struct node* a, struct node* b) { // Your code... } ``` # 15 — MergeSort() (This problem requires recursion.) Given `FrontBackSplit()` and `SortedMerge()`, it's pretty easy to write a classic recursive `MergeSort()`: split the list into two smaller lists, recursively sort those lists, and finally merge the two sorted lists together into a single sorted list. Ironically, this problem is easier than either `FrontBackSplit()` or `SortedMerge()`. ```c void MergeSort(struct node** headRef) { // Your code... } ``` #################### File: PDF%205%20.pdf Page: 18 Context: # 16 — SortedIntersect() Given two lists sorted in increasing order, create and return a new list representing the intersection of the two lists. The new list should be made with its own memory — the original lists should not be changed. In other words, this should be `Push()` list building, not `MoveNode()`. Ideally, each list should only be traversed once. This problem along with `Union()` and `Difference()` form a family of clever algorithms that exploit the constraint that the lists are sorted to find common nodes efficiently. ```c /* Compute a new sorted list that represents the intersection of the two given sorted lists. */ struct node* SortedIntersect(struct node* a, struct node* b) { // Your code } ``` # 17 — Reverse() Write an iterative `Reverse()` function that reverses a list by rearranging all the `.next` pointers and the head pointer. Ideally, `Reverse()` should only need to make one pass of the list. The iterative solution is moderately complex. It's not so difficult that it needs to be this late in the document, but it goes here so it can be next to #18 Recursive Reverse which is quite tricky. The efficient recursive solution is quite complex (see next problem). (A memory drawing and some hints for `Reverse()` are below.) ```c void ReverseTest() { struct node* head; head = BuildOneTwoThree(); Reverse(&head); // head now points to the list [3, 2, 1] DeleteList(&head); // clean up after ourselves } ``` ## Stack ``` ReverseTest() head ``` ## Heap - List reverse before and after. - Before (in gray), the list is [1, 2, 3]. - After (in black), the pointers have been rearranged so the list is [3, 2, 1]. ## "Push" Reverse Hint Iterate through the main list. Move each node to the front of the result list as you go. It's like doing a `Push()` operation with each node, except you use pointer re-arrangement on. #################### File: PDF%205%20.pdf Page: 19 Context: # 3 Pointers' Hint This strategy is not as good as the "Push" strategy, but it's the first one I thought of (thanks to Owen Astrachan for pointing out the better solution). Instead of running a single "current" pointer down the list, run three pointers (front, middle, back) down the list in order: front is on one node, middle is just behind it, and back in turn is one behind middle. Once the code to run the three pointers down the list is clear and tested in a drawing, add code to reverse the `.next` pointer of the middle node during the iteration. Add code to take care of the empty list and to adjust the head pointer itself. ```c /* Reverse the given linked list by changing its .next pointers and its head pointer. Takes a pointer (reference) to the head pointer. */ void Reverse(struct node** headRef) { // your code... } ``` ## 18 — RecursiveReverse() (This problem is difficult and is only possible if you are familiar with recursion.) There is a short and efficient recursive solution to this problem. As before, the code should only make a single pass over the list. Doing it with multiple passes is easier but very slow, so here we insist on doing it in one pass. Solving this problem requires a real understanding of pointer code and recursion. ```c /* Recursively reverses the given linked list by changing its .next pointers and its head pointer in one pass of the list. */ void RecursiveReverse(struct node** headRef) { // your code... } ``` ## The Tree-List Recursion Problem Once you are done with these problems, see the best and most complex list recursion problem of all time: the great Tree-List-Recursion problem at [http://cslibrary.stanford.edu/109/](http://cslibrary.stanford.edu/109/) #################### File: PDF%205%20.pdf Page: 20 Context: # Section 3 — Solutions ## 1 — Count() Solution A straightforward iteration down the list — just like Length(). ```c int Count(struct node* head, int searchFor) { struct node* current = head; int count = 0; while (current != NULL) { if (current->data == searchFor) count++; current = current->next; } return count; } ``` Alternatively, the iteration may be coded with a `for` loop instead of a `while`... ```c int Count2(struct node* head, int searchFor) { struct node* current; int count = 0; for (current = head; current != NULL; current = current->next) { if (current->data == searchFor) count++; } return count; } ``` ## 2 — GetNth() Solution Combine standard list iteration with the additional problem of counting over to find the right node. Off-by-one errors are common in this sort of code. Check it carefully against a simple case. If it's right for n=0, n=1, and n=2, it will probably be right for n=1000. ```c int GetNth(struct node* head, int index) { struct node* current = head; int count = 0; // the index of the node we're currently looking at while (current != NULL) { if (count == index) return(current->data); count++; current = current->next; } assert(0); // if we get to this line, the caller was asking for a non-existent element so we assert fail. } ``` #################### File: PDF%205%20.pdf Page: 21 Context: # 3 — DeleteList() Solution Delete the whole list and set the head pointer to NULL. There is a slight complication inside the loop, since we need extract the `.next` pointer before we delete the node, since after the delete it will be technically unavailable. ```c void DeleteList(struct node** headRef) { struct node* current = *headRef; // deref headRef to get the real head struct node* next; while (current != NULL) { next = current->next; // note the next pointer free(current); // delete the node current = next; // advance to the next node } *headRef = NULL; // Again, deref headRef to affect the real head back // in the caller } ``` # 4 — Pop() Solution Extract the data from the head node, delete the node, advance the head pointer to point at the next node in line. Uses a reference parameter since it changes the head pointer. ```c int Pop(struct node** headRef) { struct node* head; int result; head = *headRef; assert(head != NULL); result = head->data; // pull out the data before the node is deleted *headRef = head->next; // unlink the head node for the caller // Note the `*` uses a reference-pointer // just like Push() and DeleteList(). free(head); // free the head node return(result); // don't forget to return the data from the link } ``` # 5 — InsertNth() Solution This code handles inserting at the very front as a special case. Otherwise, it works by running a current pointer to the node before where the new node should go. Uses a for loop to march the pointer forward. The exact bounds of the loop (the use of `x <= n` vs. `n - 1`) are always tricky—the best approach is to get the general structure of the iteration correct first, and then make a careful drawing of a couple test cases to adjust the `n` vs. `1` cases to be correct. (The so-called "OBOB"—Off By One Boundary cases.) The OBOB cases are always tricky and not that interesting. Write the correct basic structure and then use a test case to get the OBOB cases correct. Once the insertion point has been determined, this solution uses `Push()` to do the link in. Alternately, the 3-Step Link In code could be pasted here directly. #################### File: PDF%205%20.pdf Page: 22 Context: ```markdown void InsertNth(struct node** headRef, int index, int data) { // position 0 is a special case... if (index == 0) Push(headRef, data); else { struct node* current = *headRef; int i; for (i = 0; i < index - 1; i++) { assert(current != NULL); // if this fails, index was too big current = current->next; } assert(current != NULL); // tricky: you have to check one last time Push(&(current->next), data); // Tricky use of Push() // The pointer being pushed on is not // in the stack. But actually this works // fine -- Push() works for any node pointer. } } 6 — SortedInsert() Solution The basic strategy is to iterate down the list looking for the place to insert the new node. That could be the end of the list, or a point just before a node which is larger than the new node. The three solutions presented handle the "head end" case in different ways... // Uses special case code for the head end void SortedInsert(struct node** headRef, struct node* newNode) { // Special case for the head end if (*headRef == NULL || (*headRef)->data >= newNode->data) { newNode->next = *headRef; *headRef = newNode; } else { // Locate the node before the point of insertion struct node* current = *headRef; while (current->next != NULL && current->next->data < newNode->data) { current = current->next; } newNode->next = current->next; current->next = newNode; } } // Dummy node strategy for the head end void SortedInsert(struct node** headRef, struct node* newNode) { struct node dummy; struct node* current = &dummy; dummy.next = *headRef; while (current->next != NULL && current->next->data < newNode->data) { current = current->next; } newNode->next = current->next; current->next = newNode; } ``` #################### File: PDF%205%20.pdf Page: 23 Context: # HeadRef - dummy.next; ```c // Local references strategy for the head end void SortedInsert3(struct node** headRef, struct node* newNode) { struct node** currentRef = headRef; while (*currentRef != NULL && (*currentRef)->data < newNode->data) { currentRef = &((*currentRef)->next); } newNode->next = *currentRef; // Bug: this line used to have // an incorrect (*currentRef)->next *currentRef = newNode; } # 7 — InsertSort() Solution // Start with an empty result list. Iterate through the source list and SortedInsert() each of its nodes into the result list. Be careful to note the .next field in each node before moving it into the result list. // Given a list, change it to be in sorted order (using SortedInsert()). void InsertSort(struct node** headRef) { struct node* result = NULL; // build the answer here struct node* current = *headRef; // iterate over the original list struct node* next; while (current != NULL) { next = current->next; // tricky - note the next pointer before we change it SortedInsert(&result, current); current = next; } *headRef = result; } # 8 — Append() Solution // The case where the 'a' list is empty is a special case handled first — in that case the 'a' head pointer needs to be changed directly. Otherwise we iterate down the 'a' list until we find its last node with the test (current->next != NULL), and then tack on the 'b' list there. Finally, the original 'b' head is set to NULL. This code demonstrates extensive use of pointer reference parameters, and the common problem of needing to locate the last node in a list. (There is also a drawing of how Append() uses memory below.) void Append(struct node** aRef, struct node** bRef) { struct node* current; if (*aRef == NULL) { // Special case if a is empty *aRef = *bRef; } else { // Otherwise, find the end of a, and append b there current = *aRef; while (current->next != NULL) { // find the last node current = current->next; } current->next = *bRef; } } ``` #################### File: PDF%205%20.pdf Page: 24 Context: ``` current->next = *bRef; // hang the b list off the last node } *bRef = NULL; // NULL the original b, since it has been appended above } # Append() Test and Drawing The following `AppendTest()` code calls `Append()` to join two lists. What does memory look like just before the call to `Append()` exits? ```c void AppendTest() { struct node* a; struct node* b; // set a to {1, 2} // set b to {3, 4} Append(&a, &b); } ``` As an example of how reference parameters work, note how reference parameters in `Append()` point back to the head pointers in `AppendTest()`... ## Stack ``` AppendTest() a b Append(&a, &b) aRef bRef current ``` ## Heap ``` 1 2 | | 3---4 ``` # 9 — FrontBackSplit() Solution Two solutions are presented... ```c // Uses the "count the nodes" strategy void FrontBackSplit(struct node** source, struct node** frontRef, struct node** backRef) { int len = Length(source); int i; struct node* current = source; ``` ``` #################### File: PDF%205%20.pdf Page: 25 Context: # 10 — RemoveDuplicates() Solution Since the list is sorted, we can proceed down the list and compare adjacent nodes. When adjacent nodes are the same, remove the second one. There's a tricky case where the node after the next node needs to be noted before the deletion. ## Function Definitions ### FrontBackSplit ```c void FrontBackSplit(struct node* source, struct node** frontRef, struct node** backRef) { struct node* fast; struct node* slow; if (source == NULL || source->next == NULL) { // length < 2 cases *frontRef = source; *backRef = NULL; } else { slow = source; fast = source->next; // advance 'fast' two nodes, and advance 'slow' one node while (fast != NULL) { fast = fast->next; if (fast != NULL) { slow = slow->next; fast = fast->next; } } // 'slow' is before the midpoint in the list, so split it in two *frontRef = source; *backRef = slow->next; slow->next = NULL; } } ``` ### RemoveDuplicates ```c void RemoveDuplicates(struct node* head) { struct node* current = head; while (current != NULL && current->next != NULL) { if (current->data == current->next->data) { struct node* next_next = current->next->next; // store the next node free(current->next); // free the duplicate current->next = next_next; // link to the next distinct node } else { current = current->next; // move to the next node } } } ``` #################### File: PDF%205%20.pdf Page: 26 Context: ```markdown if (current == NULL) return; // do nothing if the list is empty // Compare current node with next node while (current->next != NULL) { if (current->data == current->next->data) { struct node* nextNext = current->next->next; free(current->next); current->next = nextNext; } else { current = current->next; // only advance if no deletion } } ## 11 — MoveNode() Solution The `MoveNode()` code is most similar to the code for `Push()`. It’s short — just changing a couple pointers — but it’s complex. Make a drawing. ```c void MoveNode(struct node** destRef, struct node** sourceRef) { struct node* newNode = *sourceRef; // the front source node assert(newNode != NULL); *sourceRef = newNode->next; // Advance the source pointer newNode->next = *destRef; // Link the old dest off the new node *destRef = newNode; // Move dest to point to the new node } ``` ## 12 — AlternatingSplit() Solution The simplest approach iterates over the source list and uses `MoveNode()` to pull nodes off the source and alternatively put them on `a` and `b`. The only strange part is that the nodes will be in the reverse order that they occurred in the source list. ```c void AlternatingSplit(struct node* source, struct node** aRef, struct node** bRef) { struct node* a = NULL; // Split the nodes to these 'a' and 'b' lists struct node* b = NULL; struct node* current = source; while (current != NULL) { MoveNode(&a, ¤t); // Move a node to 'a' if (current != NULL) { MoveNode(&b, ¤t); // Move a node to 'b' } } *aRef = a; *bRef = b; } ``` ``` #################### File: PDF%205%20.pdf Page: 27 Context: # AlternatingSplit() Using Dummy Nodes Here is an alternative approach which builds the sub-lists in the same order as the source list. The code uses a temporary dummy header nodes for the 'a' and 'b' lists as they are being built. Each sublist has a "tail" pointer which points to its current last node — that way new nodes can be appended to the end of each list easily. The dummy nodes give the tail pointers something to point to initially. The dummy nodes are efficient in this case because they are temporary and allocated in the stack. Alternately, the "local references" technique could be used to get rid of the dummy nodes (see Section 1 for more details). ```c void AlternatingSplit2(struct node* source, struct node** aRef, struct node** bRef) { struct node* aDummy; struct node* aTail = aDummy; // points to the last node in 'a' struct node* bDummy; struct node* bTail = bDummy; // points to the last node in 'b' struct node* current = source; aDummy.next = NULL; bDummy.next = NULL; while (current != NULL) { MoveNode(&(aTail->next), ¤t); // add at 'a' tail aTail = aTail->next; // advance the 'a' tail if (current != NULL) { MoveNode(&(bTail->next), ¤t); bTail = bTail->next; } } *aRef = aDummy.next; *bRef = bDummy.next; } ``` # 13 SuffleMerge() Solution There are four separate solutions included. See Section 1 for information on the various dummy node and reference techniques. ## SuffleMerge() — Dummy Node Not Using MoveNode() ```c struct node* SuffleMerge(struct node* a, struct node* b) { struct node* dummy; struct node* tail = &dummy; dummy.next = NULL; while (1) { if (a == NULL) { // empty list cases tail->next = b; break; } else if (b == NULL) { tail->next = a; break; } else { // common case: move two nodes to tail tail->next = a; tail = a; a = a->next; } } } ``` #################### File: PDF%205%20.pdf Page: 28 Context: ```markdown # SuffleMerge() — Dummy Node Using MoveNode() Basically the same as above, but use `MoveNode()`. ```c struct node* SuffleMerge(struct node* a, struct node* b) { struct node dummy; struct node* tail = &dummy; dummy.next = NULL; while (1) { if (a == NULL) { tail->next = b; break; } else if (b == NULL) { tail->next = a; break; } else { MoveNode(&(tail->next), &a); tail = tail->next; MoveNode(&(tail->next), &b); tail = tail->next; } } return (dummy.next); } ``` # SuffleMerge() — Local References Uses a local reference to get rid of the dummy nodes entirely. ```c struct node* SuffleMerge(struct node* a, struct node* b) { struct node* result = NULL; struct node** lastPtrRef = &result; while (1) { if (a == NULL) { *lastPtrRef = b; break; } else if (b == NULL) { *lastPtrRef = a; break; } else { MoveNode(lastPtrRef, &a); lastPtrRef = &((*lastPtrRef)->next); MoveNode(lastPtrRef, &b); lastPtrRef = &((*lastPtrRef)->next); } } return result; } ``` #################### File: PDF%205%20.pdf Page: 29 Context: # SuffleMerge() — Recursive The recursive solution is the most compact of all, but is probably not appropriate for production code since it uses stack space proportionate to the lengths of the lists. ```c struct node* ShuffleMerge(struct node* a, struct node* b) { struct node* result; struct node* recur; if (a == NULL) return(b); // see if either list is empty else if (b == NULL) return(a); else { // It turns out to be convenient to do the recursive call first -- // otherwise a->next and b->next need temporary storage. recur = ShuffleMerge(a->next, b->next); result = a; // one node from a a->next = b; // one from b b->next = recur; // then the rest return(result); } } ``` # 14 — SortedMerge() Solution ## SortedMerge() Using Dummy Nodes The strategy here uses a temporary dummy node as the start of the result list. The pointer `tail` always points to the last node in the result list, so appending new nodes is easy. The dummy node gives `tail` something to point to initially when the result list is empty. This dummy node is efficient, since it is only temporary, and it is allocated in the stack. The loop proceeds, removing one node from either `'a'` or `'b'`, and adding it to `tail`. When we are done, the result is in `dummy.next`. ```c struct node* SortedMerge(struct node* a, struct node* b) { struct node dummy; // a dummy first node to hang the result on struct node* tail = &dummy; // Points to the last result node -- dummy.next = NULL; while (1) { if (a == NULL) { // if either list runs out, use the other list tail->next = b; break; } else if (b == NULL) { tail->next = a; break; } } } ``` #################### File: PDF%205%20.pdf Page: 30 Context: ```markdown if (a->data <= b->data) { MoveNode(&(tail->next), &a); } else { MoveNode(&(tail->next), &b); } tail = tail->next; return(dummy.next); } # SortedMerge() Using Local References This solution is structurally very similar to the above, but it avoids using a dummy node. Instead, it maintains a `struct node**` pointer, `lastPtrRef`, that always points to the last pointer of the result list. This solves the same case that the dummy node did—dealing with the result list when it is empty. If you are trying to build up a list at its tail, either the dummy node or the `struct node**` "reference" strategy can be used (see Section I for details). ```c struct node* SortedMerge2(struct node* a, struct node* b) { struct node* result = NULL; struct node** lastPtrRef = &result; // point to the last result pointer while (1) { if (a == NULL) { *lastPtrRef = b; break; } else if (b == NULL) { *lastPtrRef = a; break; } if (a->data <= b->data) { MoveNode(lastPtrRef, &a); } else { MoveNode(lastPtrRef, &b); } lastPtrRef = &((*lastPtrRef)->next); // tricky: advance to point to the next ".next" field } return(result); } ``` # SortedMerge() Using Recursion Merge() is one of those recursive problems where the recursive solution code is much cleaner than the iterative code. You probably wouldn't want to use the recursive version for production code, however, because it will use stack space which is proportional to the length of the lists. ```c struct node* SortedMerge3(struct node* a, struct node* b) { struct node* result = NULL; } ``` ``` #################### File: PDF%205%20.pdf Page: 31 Context: ``` # MergeSort() Solution The MergeSort strategy is to split into sublists, sort the sublists recursively, and merge the two sorted lists together to form the answer. ```c void MergeSort(struct node** headRef) { struct node* head = *headRef; struct node* a; struct node* b; // Base case -- length 0 or 1 if ((head == NULL) || (head->next == NULL)) { return; } FrontBackSplit(head, &a, &b); // Split head into 'a' and 'b' sublists // We could just as well use AlternatingSplit() MergeSort(&a); // Recursively sort the sublists MergeSort(&b); *headRef = SortedMerge(a, b); // answer = merge the two sorted lists together } ``` (Extra for experts) Using recursive stack space proportional to the length of a list is not recommended. However, the recursion in this case is ok — it uses stack space which is proportional to the log of the length of the list. For a 1000 node list, the recursion will only go about 10 deep. For a 2000 node list, it will go 11 deep. If you think about it, you can see that doubling the size of the list only increases the depth by 1. # SortedIntersect() Solution The strategy is to advance up both lists and build the result list as we go. When the current point in both lists are the same, add a node to the result. Otherwise, advance whichever list is smaller. By exploiting the fact that both lists are sorted, we only traverse each list once. To build up the result list, both the dummy node and local reference strategy solutions are shown... ```c struct node* SortedIntersect(struct node* a, struct node* b) { struct node dummy; } ``` ``` #################### File: PDF%205%20.pdf Page: 32 Context: ```markdown struct node* tail = dummy; dummy.next = NULL; // Once one of the other list runs out -- we're done while (a != NULL && b != NULL) { if (a->data < b->data) { Push(&(tail->next), a->data); tail = tail->next; a = a->next; } else { b = b->next; } } return (dummy.next); // This solution uses the local reference struct node* SortedIntersect2(struct node* a, struct node* b) { struct node* result = NULL; struct node** lastPtrRef = &result; // Advance comparing the first nodes in both lists. // When one or the other list runs out, we're done. while (a != NULL && b != NULL) { if (a->data < b->data) { // found a node for the intersection Push(lastPtrRef, a->data); lastPtrRef = &((*lastPtrRef)->next); a = a->next; } else { b = b->next; } } return result; } # 17 — Reverse() Solution This first solution uses the "Push" strategy with the pointer re-arrangement hand coded inside the loop. There's a slight trickiness in that it needs to save the value of the "current->next" pointer at the top of the loop since the body of the loop overwrites that pointer. /* Iterative list reverse. Iterate through the list left-right. */ ``` #################### File: PDF%205%20.pdf Page: 33 Context: ```markdown Move/insert each node to the front of the result list -- like a Push of the node. ```c static void Reverse(struct node** headRef) { struct node* result = NULL; struct node* current = *headRef; struct node* next; while (current != NULL) { next = current->next; // tricky: note the next node current->next = result; // move the node onto the result result = current; current = next; } *headRef = result; } ``` Here's the variation on the above that uses MoveNode() to do the work... ```c static void Reverse2(struct node** headRef) { struct node* result = NULL; struct node* current = *headRef; while (current != NULL) { MoveNode(&result, ¤t); } *headRef = result; } ``` Finally, here's the back-middle-front strategy... ```c // Reverses the given linked list by changing its .next pointers and // its head pointer. Takes a pointer (reference) to the head pointer. void Reverse(struct node** headRef) { if (*headRef != NULL) { // special case: skip the empty list struct node* middle = *headRef; // the main pointer struct node* front = middle->next; // the two other pointers (NULL ok) struct node* back = NULL; while (1) { middle->next = back; // fix the middle node if (front == NULL) break; // test if done back = middle; middle = front; front = front->next; } *headRef = middle; // update headRef to new head } } ``` ``` #################### File: PDF%205%20.pdf Page: 34 Context: # 18 — RecursiveReverse() Solution Probably the hardest part is accepting the concept that the `RecursiveReverse(&rest)` does in fact reverse the rest. Then there's a trick to getting the one front node all the way to the end of the list. Make a drawing to see how the trick works. ```c void RecursiveReverse(struct node** headRef) { struct node* first; struct node* rest; if (*headRef == NULL) return; // empty list base case first = *headRef; // suppose first = {1, 2, 3} rest = first->next; // rest = {2, 3} if (rest == NULL) return; // empty rest base case RecursiveReverse(&rest); // Recursively reverse the smaller {2, 3} case // after: rest = {3, 2} first->next->next = first; // put the first elem on the end of the list first->next = NULL; // (tricky step – make a drawing) *headRef = rest; // fix the head pointer } ``` The inefficient solution is to reverse the last n-1 elements of the list, and then iterate all the way down to the new tail and put the old head node there. That solution is very slow compared to the above which gets the head node in the right place without extra iteration. #################### File: PDF%205%20.pdf Page: 35 Context: # Appendix ## Basic Utility Function Implementations Here is the source code for the basic utility functions. ### Length() ```c // Return the number of nodes in a list int Length(struct node* head) { int count = 0; struct node* current = head; while (current != NULL) { count++; current = current->next; } return(count); } ``` ### Push() ```c // Given a reference (pointer to pointer) to the head // of a list and an int, push a new node on the front of the list. // Creates a new node with the int, links the list off the next of the // new node, and finally changes the head to point to the new node. void Push(struct node** headRef, int newData) { struct node* newNode = (struct node*) malloc(sizeof(struct node)); // allocate node newNode->data = newData; // put in the data newNode->next = (*headRef); // link the old list off the new node (*headRef) = newNode; // move the head to point to the new node } ``` ### BuildOneTwoThree() ```c // Build and return the list (1, 2, 3) struct node* BuildOneTwoThree() { struct node* head = NULL; // Start with the empty list Push(&head, 3); // Use Push() to add all the data Push(&head, 2); Push(&head, 1); return(head); } ``` #################### File: XLSX%201%20Oranges-2022.xlsx Page: 1 Context: | Oranges—Average retail price per pound or pint and per cup equivalent, 2022 | Unnamed: 1 | Unnamed: 2 | Unnamed: 3 | Unnamed: 4 | Unnamed: 5 | Unnamed: 6 | |:-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|:---------------------|:-------------|:-------------------------|:-------------------------|:-------------|:---------------------------------| | Form | Average retail price | | Preparation yield factor | Size of a cup equivalent | | Average price per cup equivalent | | Fresh1 | 1.46236008399391 | per pound | 0.6799999999999999 | 0.4078551850420235 | Pounds | 0.8771046215520666 | | Juice | | | | | | | | Ready to drink2 | 1.05716147808459 | per pint | 1 | 8 | Fluid ounces | 0.528580739042295 | | Frozen3 | 0.733780736670315 | per pint | 1 | 8 | Fluid ounces | 0.3668903683351575 | | 1Includes navel oranges including cara cara navel and other navel orange varieties. Excludes other varieties like Valencia oranges. USDA, Agricultural Research Service’s (ARS) National Nutrient Database for Standard Reference (SR) reports that inedible peel and seeds account for 27 percent of the retail weight, implying a preparation yield of 73 percent, when oranges are eaten raw. | | | | | | | | 2Includes refrigerated and unrefrigerated juice. | | | | | | | | 3Includes juice sold as frozen concentrate. The consumer reconstitutes this juice after purchase by adding three containers of water per container of concentrate. Retail price is dollars per pint after reconstitution. | | | | | | | | Source: USDA, Economic Research Service calculations from 2022 Circana (formerly Information Resources, Inc. (IRI)) OmniMarket Core Outlets (formerly InfoScan) data; and USDA, ARS, National Nutrient Database for Standard Reference (SR), Legacy Release, Food Patterns Equivalents Database (FPED) 2017–18, and the FPED's accompanying Methodology and User Guide. | | | | | | | ########## """QUERY: hi what is are the contents?""" Consider the chat history for relevant information. Use all information included. Use as much tokens as needed. Important: If you find information separated by a | in the context, it is a table formatted in Markdown.Important: Take a look at the QUERY and only the QUERY. If this is vague or unclear, please ignore everything and ask a follow-up question instead! 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