Generative AI for Beginners - Full Videos Series Released!
With so many new technologies, tools and terms in the world of Generative AI, it can be hard to know where to start or what to learn next. "Generative AI for Beginners" is designed to help you on your learning journey no matter where you are now. We are happy announce that the "Generative AI for Beginners" course has received a major refresh - 18 new videos for each lesson.Building the Ultimate Nerdland Podcast Chatbot with RAG and LLM: Step-by-Step Guide
Large Language Models (LLMs) are popular in tech. In Belgium and the Netherlands, the podcast "Nerdland" is a favorite for tech and science fans. It covers topics like bioscience, space, robotics, and AI. With over 100 episodes, "Nerdland" is a goldmine of information. So, why not create a chatbot for "Nerdland" fans? This chatbot uses podcast content to engage and inform users. It allows the "Nerdland" community to interact with the content in new ways and makes the information accessible in many languages, thanks to LLMs' multi-language capabilities. This blog post explains the project's technical details, including the LLMs used, integration process, and deployment on Azure.How Reasoning Models are transforming Logical AI thinking
Introduction Reasoning models are a new category of specialized language models. They are designed to break down complex problems into smaller, manageable steps and solve them through explicit logical reasoning (This step is also called “thinking”). Unlike general-purpose LLMs which might generate direct answers, reasoning models are specifically trained to show their work and follow a more structured thought process. Some models don’t show their logical reasoning phase while others explicitly show their logical reasoning phase. The reasoning phase shows how the model can break down the problem stated into smaller problems (Decomposition), try different approaches (Ideation), choose the best approaches (validation), reject invalid approaches (possibly backtracking) and finally choose the best answer (execution/solving). Reasoning models like OpenAI’s o1, o1-mini, o3-mini, and DeepSeek-R1 are available in the AI Toolkit for VS Code model catalog, Azure AI Foundry model catalog as well as online for free usage (rate-limited) in the Github Marketplace for models. There are some key differences between reasoning models and general-purpose LLMs. These are summarized in the following table. Summary Table Feature Reasoning Models General Purpose LLMs Primary Purpose and Strengths Explicit step-by-step problem solving and logical reasoning General-purpose text generation and understanding Problem-Solving Approach Break down problems into smaller sub-steps and shows intermediate reasoning (“thinking” phase) Show the output directly without showing intermediate steps. Mostly pattern based next token prediction Output Structure Highly structured, with clear delineation between reasoning steps More flexible and conversational, may mix reasoning with other content Training Specifically trained on step-by-step reasoning tasks and formal logic Trained on a diverse corpus of text data with many different writing styles and purposes Usage of Chain-of-Thought Built into the core architecture and training process Can use chain-of-thought prompting but not inherently designed for it. Must specifically ask for this in the prompt. Interpretability and Error Detection Detecting errors is easier due to explicit reasoning steps Black-box nature makes reasoning for output harder to trace Computational Efficiency May require more computing resources due to step-by-step processing, especially at inference time Relatively lower computing requirements at inference time and usually more efficient for direct responses Latency for response May be slower for simple tasks that don't require explicit reasoning May struggle with complex logical problems requiring step-by-step analysis and interpretation. Examples OpenAI o1, o1-mini, o3-mini, DeepSeek-R1 GPT-4o, Llama3.3, Claude Use cases Scientific reasoning, legal analysis, complex problem-solving, AI agents with multi-step decision-making Chatbots, Text summarization, content generation, Q&A, code completion Since we know that reasoning models excel at logical step-by step problem solving, we will give them tasks. We will be using DeepSeek-R1 from the GitHub model marketplace and AI Toolkit for VSCode for this analysis. Task 1 – Solve a logical puzzle We will use a variation of the river crossing problem that has some interesting constraints and needs some level of backtracking which humans excel at. This is the prompt that we will use is Can you solve this logic puzzle and show it as a diagram "A farmer with a wolf, a goat, and a cabbage must cross a river by boat. The boat can carry only the farmer and a single item. If left unattended together, the wolf would eat the goat, or the goat would eat the cabbage. How can they cross the river without anything being eaten?" AI Toolkit Session follows below: User Can you solve this logic puzzle and show it as a diagram "A farmer with a wolf, a goat, and a cabbage must cross a river by boat. The boat can carry only the farmer and a single item. If left unattended together, the wolf would eat the goat, or the goat would eat the cabbage. How can they cross the river without anything being eaten?" DeepSeek-R1 <think> Okay, so the user wants me to solve this logic puzzle about the farmer crossing the river with a wolf, goat, and cabbage. Let me start by recalling the problem. The farmer can only take one item at a time in the boat. If left alone, the wolf eats the goat, and the goat eats the cabbage. The goal is to get everyone across safely. First, I need to break down the constraints. The boat can carry the farmer and one item. So each trip either the farmer goes alone or takes one item. The main issue is the forbidden pairs: wolf and goat can't be left alone, goat and cabbage can't be left alone. So whenever the farmer moves, the remaining items on either bank must not include these pairs. Let me think through the possible steps. Let's consider the initial state: all four (farmer, wolf, goat, cabbage) are on the left bank. The goal is to get all to the right bank. The first thought: if the farmer takes the goat first. Then left bank has wolf and cabbage. They are safe because wolf and cabbage don't interact. So after first trip, farmer and goat are on the right, wolf and cabbage on the left. Then the farmer goes back alone to the left. Now, the right bank has just the goat. Then the farmer can take either the wolf or the cabbage next. Suppose he takes the wolf. Then on the right bank would be wolf and goat. Oh no, that's a problem because the wolf would eat the goat. So that's forbidden. Alternatively, after taking the goat over and returning alone, the farmer could take the cabbage. If he takes the cabbage to the right bank, then left bank has wolf. Right bank has goat and cabbage. But when he takes the cabbage over, the goat is now on the right with the cabbage. If the farmer leaves them to go back, the goat would eat the cabbage. So that's not allowed either. Wait, this seems confusing. Let's track step by step. State 0: All on left (F, W, G, C). Right is empty. Step 1: Farmer takes goat to right. Now left has W, C; right has F, G. That's safe. Then Step 2: Farmer goes back alone to left. Now left has F, W, C; right has G. Step 3: Farmer takes either W or C. Suppose he takes W. Now right has F, W, G. Left has C. But when he takes W, he leaves G and W on the right? Wait, no. When farmer takes W to the right, after step 3, left has C, right has F, W, G. But if he leaves W and G on the right, that's a problem (wolf and goat). Wait, no, the farmer is there too. So actually, if the farmer is present, they can't eat. The problem occurs only when unattended. So when the farmer takes W to the right, he's with W and G on the right. Then he can leave W there and take G back? Wait, but that would complicate things. Wait, let's step through again. After step 2: left is F, W, C; right is G. Step 3: Farmer takes W to right. Left has C. Right has F, W, G. Since farmer is on the right, wolf and goat are not left alone. Then the farmer leaves W on the right, goes back with G. Wait, but if he leaves W on the right, then the right bank has W and G, but farmer is taking G back. Wait, no. If the farmer takes W to the right, then right bank now has W and G. But the farmer is there. So he can take W to the right bank, and then he stays there, but then he can't take G back. Wait, no. Wait, no. Let me correct this. If the farmer is taking the wolf from left to right, starting at left (F, W, C) and right (G). After moving W: Left has C (and no farmer), right has F, W, G. Wait, but left would have C alone, which is safe. Then the farmer is on the right with W and G. Then he has to leave either W or G to go back. But if he leaves W and G on the right, unattended, that's a problem. So he can't leave them. Thus, the farmer needs to take something with him. Wait, so step 3 alternatives: after step 2 (left has F, W, C; right has G), farmer takes cabbage to the right: Left becomes W (and no farmer), right has F, C, G. But then when he takes C to the right, the right bank would have C and G, which would be a problem if left unattended. Because farmer is on the right, so they are attended. Then he leaves C there, takes G back to left. But that seems steps going back and forth. This is getting a bit tangled. Let's try outlining possible valid moves. The key insight here is that the goat is the only item that can't be left with either wolf or cabbage. So the farmer must ensure that whenever the goat is on a bank without the farmer, there are no other dangerous items there. So, the strategy that typically works is: Take goat over first. Return alone. Take either wolf or cabbage over. Bring goat back. Take the other item (cabbage or wolf) over. Return alone. Take goat over again. Let me check this. Step-by-Step: Start: All on left (F, W, G, C) Take G to right. State: Left: W, C; Right: F, G. Safe, because Left has W and C (no conflict). Farmer returns to left alone. State: F, W, C | G. Farmer takes W to right. State: Left: C; Right: F, W, G. But on the right, when the farmer leaves W there, does he take G back? No. Wait, after bringing W to right, the farmer would have to leave something. If he takes W to the right and leaves it there while he goes back, then right has W and G. The farmer is on the way back, so they are left alone. That's a problem because wolf and goat would be together. Therefore, this is invalid. Hence, step 3 taking W leads to a problem. Ah, so to avoid that, after step 2, instead of taking W, the farmer should take C. Alternative Step 3: After returning, take C to right. Take C to right. State: Left: F, W; Right: G, C. But when he takes C to right, the farmer leaves left bank as W. Which is safe, just wolf alone. Right bank has F, C, G. But when he leaves C and G on the right (since farmer is there, they are safe). Wait, the farmer is present, so they don't eat. So he can leave them, then go back. So step 4: return alone. State: Left: F, W; Right: C, G. Wait, no. Let me track correctly: After Step 1: left: W, C; right: F, G. Step 2: return to left alone. Now left: F, W, C; right: G. Step 3: take C to right. After moving, left: W; farmer is taking C to right. So left becomes W and farmer is moving to right, so left has W alone (which is safe). Right now has F, G, C. But upon arrival, farmer is with G and C. So they are safe. Step 4: farmer returns to left alone. Now left: F, W; right: G, C. But on the left, F and W are there, which is okay. On right, G and C are unattended – that's a problem! Goat would eat cabbage. Wait, even though the farmer has left, he's on left now. So the right bank has G and C without the farmer. So that violates the conditions. So this path would result in the goat eating the cabbage. Hence invalid. Therefore, taking the cabbage on step 3 leads to a problem when returning. Therefore, neither taking W nor taking C works here. Hmm. So alternative approach: after step 2, when the farmer is on the left with W and C, instead of taking C or W, he needs to take the goat back. Wait, but that seems to reverse step 1. That can't help. Wait, maybe the key is that after taking the goat over first, then the farmer must bring it back later to avoid leaving dangerous pairs. Let's try a different sequence: Take goat to right. Left: W, C; Right: F, G. Return alone to left. Left: F, W, C; Right: G. Take wolf to right. Left: C; Right: F, W, G. But as before, after this, farmer is on right with W and G. If he leaves W there and takes G back, then right has W alone, which is okay. So: Take goat back to left. Left: F, C, G; Right: W. Now left has C, G, F; right has W. Then take cabbage to right. Left: G; Right: F, W, C. Then return alone to left. Take goat to right. Now all are on right. Done. This seems to work. Let's verify each step: Take G over. Left: W, C (safe); Right: F, G. Return alone. Left: F, W, C; Right: G. Take W over. Left: C (safe). Right: F, W, G. Farmer is present, so W and G safe. Take G back. Left: F, G, C; Right: W (safe). Wait, on left, when farmer takes G back, left now has G and C with farmer. So when he arrives, left has F, G, C; right has W. But then when he leaves again, left would have G and C, which would be a problem (goat eats cabbage). Wait, no. After step 4, after taking G back, the state is: Left: F, G, C (farmer is there); Right: W. Then the farmer leaves to take C over: Step 5: Take C over. Left: G (alone on left? Wait, no. Wait, if farmer takes C to right, he leaves G on left. Wait, no. Let's see: After step 4: farmer is on left with G, having brought G back from the right. So left has F, C, G; right has W. Then, step 5: farmer takes C to right. So left becomes F, G on left? Wait no: starting from: Left: F, C, G; Right: W. When farmer takes C to right, farmer and C move to right. Left becomes G alone, but without the farmer. That's bad because goat alone on left but nothing else. So it's safe. Wait, goat alone is okay. Goat on left alone is harmless. So after step 5, right has F, C, W. Left has G. But the farmer is on the right. Then step 6: farmer returns alone to left. Left: F, G; Right: W, C. Step 7: Take G to right. Now all on right: F, G, W, C. Done. But let's check at each step: Step 1: left (W, C), right (F, G) - safe. Step 2: left (F, W, C), right (G). Farmer is on left. Step 3: take W to right. Left has C (safe). Right has F, W, G. Farmer there. Step 4: take G back. Left: F, G, C (farmer there). Right: W. So now right has W alone, safe. Step 5: take C to right. Left: G alone (safe). Right: F, W, C. Farmer there. Step 6: return alone. Left: G (farmer on right). So G alone on left is safe. Wait no, after step 6: farmer returns to left, now left has F, G; right has W, C. Farmer is on left with G, right has W and C (separately, so no problem). Safe. Step 7: take G to right. Now all on right. Done. Yes, that works. So the steps are: Farmer and Goat → right. Farmer ← left. Farmer and Wolf → right. Farmer and Goat ← left. Farmer and Cabbage → right. Farmer ← left. Farmer and Goat → right. Each step maintains safety on both banks. Hence the solution is this sequence. Then, to represent it as a diagram, perhaps using a flowchart showing each state with left and right banks, connected by arrows indicating the steps. </think> To solve the puzzle, follow these steps systematically, ensuring that no dangerous pairs are left unattended: Cross with Goat: Farmer takes the Goat to the right bank. Left: Wolf, Cabbage | Right: Farmer, Goat Return Alone: Farmer returns to the left bank alone. Left: Farmer, Wolf, Cabbage | Right: Goat Cross with Wolf: Farmer takes the Wolf to the right bank. Left: Cabbage | Right: Farmer, Wolf, Goat Return with Goat: Farmer brings the Goat back to the left bank. Left: Farmer, Goat, Cabbage | Right: Wolf Cross with Cabbage: Farmer takes the Cabbage to the right bank. Left: Goat | Right: Farmer, Wolf, Cabbage Return Alone: Farmer returns to the left bank alone. Left: Farmer, Goat | Right: Wolf, Cabbage Final Trip with Goat: Farmer takes the Goat to the right bank. Left: (empty) | Right: Farmer, Wolf, Goat, Cabbage Diagram (simplified as a flowchart): <>Plain text Start: F, W, G, C | [Left] → [Right] (empty) Take G ➔ [W, C] | [F, G] Return ↵ [F, W, C] | [G] Take W ➔ [C] | [F, W, G] Take G ↵ [F, G, C] | [W] Take C ➔ [G] | [F, W, C] Return ↵ [F, G] | [W, C] Take G ➔ [ ] | [F, W, G, C] All cross safely! 🚣✨ Token used: 3278 (11693 characters) Conclusion As you can see it has arrived at the right answer, and you can see it taking the following steps Reformulating the problem as a set of constraints Adding symbols for each of the interacting entities. Then it takes a series of steps to see if they satisfy the constraints and discards ones that do not. Then it layers each of these steps to build an efficient solution based on backtracking and trial-and-error. Finally, once it has a probable answer, it crosschecks again on the constraints and draws the diagram as we instructed it (with symbols for each). A nice touch is that it has also shown the answer in human readable text before to make it easy to follow. Task 2 – Explain obfuscated code Another scenario that we can often face in our lives as programmers is stumbling across a piece of code in our codebase that does something clever but alas, it has no documentation. Sounds familiar? Let's see how DeepSeek-R1 can tackle this task. This is the prompt we will use What does this program do? What language is it in? And what are such programs called? s='s=\47%s\47;printf "$s" "$s"';printf "$s" "$s" AI toolkit Session transcript follows: User What does this program do? What language is it in? and what are such programs called? s='s=\47%s\47;printf "$s" "$s"';printf "$s" "$s" DeepSeek-R1 <think> Okay, so I need to figure out what this program does, what language it's written in, and what such programs are called. Let me start by looking closely at the code. First, the code starts with s='s=\47%s\47;printf "$s" "$s"'; followed by printf "$s" "$s". Hmm, the initial part is assigning a string to variable s. The string seems to have some escape characters. The \47 is probably an octal escape in some programming language. Since 47 in octal is 39 in decimal, which is the ASCII code for a single quote ('). So s='s=\47%s\47;printf "$s" "$s"' would actually translate to s='s='%s';printf "$s" "$s"'. Wait, let me check that. So the single quote in the string is escaped using \47, which would allow the string to include a single quote without closing the string prematurely. So the value of s is a string that starts with 's=' followed by a single quote (from \47), then %s, then another single quote (again \47), followed by ;printf "$s" "$s". Then, the next part is printf "$s" "$s". So they're using printf with the format string stored in s and the argument is the same s variable. Let's think step by step. Let's substitute s into the printf command. The s variable is: 's=\47%s\47;printf "$s" "$s"'. When printf is executed, the format string is s, which includes a %s placeholder. The arguments passed to printf are "$s", so the first %s in the format string is replaced by the value of s. Wait, but in the first line, the code is written as s='s=\47%s\47;printf "$s" "$s"'; which when interpreted would need to handle the single quotes correctly. Let me parse this. The assignment is s='...' where the content inside the single quotes includes \47, which as I thought is a single quote. So the actual value of s when assigned is: s='s='%s';printf "$s" "$s"'. Wait, maybe not. Let me think again. The original assignment is s='s=\47%s\47;printf "$s" "$s"'; So inside the single-quoted string, \47 is an escape sequence. But in some shells like Bash, inside single quotes, escapes are not processed unless the single quote is closed. Wait, but the code is using printf, which is a shell command, suggesting maybe this is a shell script? But in the shell, single quotes don't allow escaping, except for the single quote itself by using '''. Wait, but maybe the code is in a different language where \47 represents a single quote. For example, in C, octal escapes are allowed in strings. Wait, but the code uses printf with "$s" as the format and "$s" as the argument. Let me try to simulate this. Let's suppose s has the value: s='s=\47%s\47;printf "$s" "$s"'. Then, when the printf is executed: printf "$s" "$s". So substituting variables, the format string is the value of s, which is 's=\47%s\47;printf "$s" "$s"'. Then, the first argument is also s. So the %s in the format will be replaced by the value of the first argument, which is s. Wait, but the substitution happens inside the format string. Let me try writing it out. Let's say s is the string 's=\47%s\47;printf "$s" "$s"'. Then, when we do printf "$s" "$s", the %s in the format string will be replaced by the argument "$s". So the output would be the format string with %s replaced by s's value. Let's compute that. Replacing the %s with the value of s gives: s='\47 followed by the value of s (which includes \47%s\47) followed by \47'; then ;printf "$s" "$s". Hmm, that's getting a bit tangled. Maybe this program is a quine? A quine is a program that outputs its own source code. So perhaps this code, when run, generates its own source as output. Let's test this idea. Let's see what the output would be. The initial value of s is s='s=\47%s\47;printf "$s" "$s"'. Then printf is called with the format string being s and the argument being s. So substituting, the format string is 's=\47%s\47;printf "$s" "$s"', and the %s will be replaced with the same string. Let's do that: replacing %s with the value of s (i.e., the string 's=\47%s\47;printf "$s" "$s"') into the format string. So the result would be s='\47's=\47%s\47;printf "$s" "$s"'' followed by ;printf "$s" "$s". Wait, no. Let me actually plug in the substitution. The original s is: s=\47%s\47;printf "$s" "$s" So when we replace the %s in the format with the s itself, the line becomes s='s=\47%s\47;printf "$s" "$s"';printf "$s" "$s" Wait, but wait. Let me process this step by step. Let me consider what happens when the printf command runs. The first printf is given the format string in variable s, which is 's=\47%s\47;printf "$s" "$s"' (the outer single quotes in the assignment are part of the shell syntax, but the actual content is s=\47%s\47;printf "$s" "$s"). Wait, no. Let's clarify: in shell scripting, when you assign s='some string', the variable s holds 'some string' without the single quotes. Wait, no. Wait in shell scripting, variable assignments like s='value' assign s the literal 'value', with the quotes used to delimit the string but not part of the value. So in this code, s is assigned the string s=\47%s\47;printf "$s" "$s" where \47 is a literal \ followed by 47, but when interpreted inside a shell, what's the actual value? Wait, but shell does not interpret octal escapes in normal strings unless using certain commands like echo with flags. Oh, but here they're using printf, which does interpret backslash escapes. Wait, but variable assignment in shell doesn't process escapes unless in certain contexts. So if s is assigned as s='s=\47%s\47;printf ...', then \47 is literally a backslash followed by 4 and 7, not as an octal escape. Hmm, that complicates things. So in that case, when the printf is executed, the format string would have \47 which printf would interpret as octal 47 (if the format string is processed with printf's escape handling). Wait, but the shell's printf command processes the escapes in the format string. Wait, the shell's printf uses the format string and replaces escape sequences in it if they are present. But in this code, the format string is stored in $s, which is being passed as the first argument to printf, and the second argument is $s. Let's take an example. Let's say s is the string 's=\47%s\47;printf "$s" "$s"'. Then, when you run printf "$s" "$s", the first argument (the format) is "s=\47%s\47;printf "$s" "$s"" and the second argument is the same string. So the %s in the format will be replaced by the second argument. Wait, but the format has \47 which, when interpreted by printf, is octal 47, which is a single quote. So in the format string, s=\47%s\47 becomes s='%s' after replacing \47 with '. Then the %s is replaced by the argument, which is the original s value. Wait, this is getting a bit confusing. Maybe I should write down the actual evaluation step by step. Let me try to reproduce the program and see what the output is. The program is: s='s=\47%s\47;printf "$s" "$s"';printf "$s" "$s" Assuming this is a shell script (Bash, perhaps), let's parse this. The first part assigns the variable s to the string 's=\47%s\47;printf "$s" "$s"'. Then, the printf command is executed with the format string "$s" and the argument "$s". So, expanding variables, the printf command becomes: printf 's=\47%s\47;printf "$s" "$s"' "$s" So the format string is 's=\47%s\47;printf "$s" "$s"' with %s, and the argument is the value of s (same string). Now, when printf processes the format string, it replaces the %s with the argument. But the format string includes \47, which in the printf command is interpreted as an octal escape for the character with octal value 47, which is 39 in decimal, which is ASCII for '. Therefore, \47 becomes ', so the format string would be 's='%s';printf "$s" "$s"'. So substituting the argument (which is s='s=\47%s\47;printf "$s" "$s"') into the %s: s='s=\47%s\47;printf "$s" "$s"';printf "$s" "$s" Wait, but substituting the value of s into the format, which is now 's=%s' with the %s being replaced by s, which includes the same string. Wait, this seems like a quine, where the program generates its own source code. When you run the printf command, it replaces %s with the original s variable, but within a structure that reconstructs the original program. Let me try creating the output step by step. The original code is: s='s=\47%s\47;printf "$s" "$s"';printf "$s" "$s" The printf command is: printf "s=\47%s\47;printf "$s" "$s"" "$s" Because the $s in the format is replaced by the variable s. Wait, but variables inside single quotes are not expanded. Wait, no, in the assignment, the string is in single quotes, so variables inside aren't expanded. Then in the printf "$s" "$s", the $s are in double quotes, so they get expanded. Wait, but the first line is s='s=\47%s\47;printf "$s" "$s"'; Then, the second command is printf "$s" "$s". Because $s is in double quotes, the shell will replace $s with its value, which is 's=\47%s\47;printf "$s" "$s"'. Wait, but not exactly, because in the assignment, the variable s is assigned the literal string between the single quotes. Let's clarify with an example: suppose s is assigned to the literal characters s=\47%s\47;printf "$s" "$s". Because the assignment is s='...', so the single quotes around the value when assigning mean that everything inside is literal, so \47 is just three characters: backslash, 4, 7. But when you use printf "$s" "$s", the shell first substitutes the $s variables. So the first argument to printf is the value of s: s=\47%s\47;printf "$s" "$s". Then the second argument is the same. The printf command now has a format string where %s is present. Also, in the format string, there are \47 sequences. But when the shell's printf processes the format string, it will interpret backslash escapes. So the \47 in the format string, as parsed by printf, represents an octal escape code for the single quote. So let me reconstruct the formatted output: The format string after substitution is "s=\47%s\47;printf "$s" "$s"" (wait, not exactly, because variables inside double quotes are expanded, but inside the format string here it's the literal s value). Maybe I need to consider that when the line is parsed, the printf command receives the expanded variables. Wait, step by step: Original code: s='s=\47%s\47;printf "$s" "$s"';printf "$s" "$s" The first part assigns s to the string "s=\47%s\47;printf "$s" "$s"" (but with the $s inside the string being literal because of the single quotes). So after assignment, the variable s contains the literal characters: s=\47%s\47;printf "$s" "$s" Now, the printf command is executed as printf "$s" "$s". The shell replaces $s with its value, so the command becomes: printf 's=\47%s\47;printf "$s" "$s"' 's=\47%s\47;printf "$s" "$s"' Notice that the first argument is the format string: 's=\47%s\47;printf "$s" "$s"' But when printf processes this format string, it replaces the \47 with the octal value 47, which is a single quote. So the format string becomes: s='%s';printf "$s" "$s" Then, the %s is replaced with the second argument: 's=\47%s\47;printf "$s" "$s"' So the resulting output is: s='s=\47%s\47;printf "$s" "$s"';printf "$s" "$s" Which is exactly the original code. Therefore, this program outputs its own source code. Therefore, it's a quine. Now, regarding the language: this uses printf, which is a shell command. The syntax of variable assignment and the use of quotes indicates that it's a shell script. The initial assignment uses single quotes to define the variable s, then uses printf in a way that constructs the output by embedding the variable's content. Therefore, this code is written in a shell scripting language, probably Bash or similar. Such programs that produce their own source code as output are called quines. So putting this all together: the program is a quine written in a shell scripting language (like Bash) that prints its own source code when executed. </think> The program is a quine, which is a program that outputs its own source code when executed. It is written in shell script (specifically compatible with Bourne shell or Bash). Here's how it works: Variable Assignment: The string s='s=\47%s\47;printf "$s" "$s"' assigns the variable s with a format string containing \47 (an octal escape for the single quote '). The %s acts as a placeholder for string substitution. Print Command: The printf "$s" "$s" uses the value of s as both the format string and the argument. The shell expands $s, replacing %s with the original value of s and interpreting \47 as '. Self-Referential Output: When executed, the program constructs its identical source code by substituting %s with the escaped version of s, reproducing the original code. Language: Shell script (Bash/Bourne-like shell). Program Type: Quine. Output: <>Plain text s='s=\47%s\47;printf "$s" "$s"';printf "$s" "$s" This matches the original code, confirming it's a quine. Token used: 3376 (12922 characters) Conclusion So, as you can see from the example above, the model followed a series of logical steps to arrive at the answer It examines the code, simulates some parts of the code to figure out what it does At the same time, it also sees certain functions such as print and single quotes and guesses that the language is bash scripting It then simulates the entire program and realizes that code output is same as the generating code. It knows from its world knowledge and training that such programs are called quines. Armed with all this information, it validates the entire chain of thought to be sure of its answer and then finally formulates and prints the answer. Task 3 – Solve an optimization problem with constraints In everyday life, we often must solve problems when we have budgets and cost constraints or want to maximize revenue or profit or any other such numerical target. We might have metrics to measure them and might guess some relationships between different variables. For example, when we have a set marketing budget and different channels such as online ads, Linkedin/X posts and emails to send to prospective attendees, how do we distribute this budget between different channels with slightly different audiences and efficacy of reach and signs ups (say for an online seminar or a conference). Let us look at a similar problem but in the manufacturing domain. This is the prompt that we will give. I want to sell a certain number of items (children's toys) in order to maximize my profit. Market research tells me that if I set the price at $1.50, I will be able to sell 5000 items, and for every 10 cents I lower the price below $1.50 I will be able to sell another 1000 items. Suppose that my manufacturing setup costs ( “start-up costs” ) total $2000, and the manufacturing cost per item ( “marginal cost” ) is $0.50. Find the price to set per item and the number of items that I need to sell in order to maximize my profit and also determine the maximum profit I can get. AI Toolkit Session follows: User I want to sell a certain number of items (children's toys) in order to maximize my profit. Market research tells me that if I set the price at $1.50, I will be able to sell 5000 items, and for every 10 cents I lower the price below $1.50 I will be able to sell another 1000 items. Suppose that my manufacturing setup costs ( “start-up costs” ) total $2000, and the manufacturing cost per item ( “marginal cost” ) is $0.50. Find the price to set per item and the number of items that I need to sell in order to maximize my profit and also determine the maximum profit I can get. Deepseek-R1 <think> Okay, let's try to figure out how to maximize the profit for selling these children's toys. Hmm, so the problem gives me some information about price, quantity sold, start-up costs, and marginal cost. I need to model this mathematically and then find the optimal price and quantity. Let's break it down step by step. First, the initial price is $1.50, and at that price, they can sell 5000 items. Then, for every 10 cents they lower the price below $1.50, they can sell an additional 1000 items. So, the price decrease affects the quantity sold. I need to model the relationship between price and quantity sold. Let me denote the number of times they lower the price by 10 cents as x. So, if x = 0, the price is $1.50, and quantity sold is 5000. If x = 1, the price is $1.40, and quantity sold is 5000 + 1000*1 = 6000. If x = 2, price is $1.30, quantity is 7000, and so on. So, in general, if they lower the price x times, the price becomes: Price = 1.50 - 0.10x dollars. And the quantity sold becomes: Quantity = 5000 + 1000x items. But x has to be a non-negative integer, right? Wait, or can x be any non-negative real number? The problem says "for every 10 cents I lower the price below $1.50 I will be able to sell another 1000 items." So, is the 10 cent decrease a discrete step or can we consider a continuous decrease? Hmm, the problem doesn't specify that we have to keep the price in multiples of 10 cents. Maybe it's continuous. But the phrasing is "for every 10 cents," which suggests that each 10 cent decrease gives an additional 1000 items. But maybe we can model it as a linear relationship. Let me check. If the price drops by 10 cents, quantity increases by 1000. So, per 1 cent decrease, quantity increases by 100 items? Wait, 10 cents correspond to 1000 items, so 1 cent decrease would correspond to 1000/10 = 100 items per cent. So, we can model the quantity sold as a linear function of the price decrease. Let me define the price p as a variable, then express the quantity sold in terms of p. Alternatively, if we consider p in dollars, the relationship is: when p decreases by $0.10, quantity increases by 1000. So, if we let p = 1.50 - 0.10x, then quantity q = 5000 + 1000x. Solving for x in terms of p: x = (1.50 - p)/0.10. Then plug into q: q = 5000 + 1000*(1.50 - p)/0.10. Let me compute that. q = 5000 + 1000*(1.50 - p)/0.10 = 5000 + 1000*( (1.50 - p) / 0.10 ). Let's compute the coefficient: 1000 / 0.10 = 10,000. So, q = 5000 + 10,000*(1.50 - p) = 5000 + 15,000 - 10,000p = 20,000 - 10,000p. Wait, that seems like a linear demand function where q = 20,000 - 10,000p. Let me check that. When p = 1.50, q = 20,000 - 10,000*(1.50) = 20,000 - 15,000 = 5000, which matches. If p decreases by $0.10 to 1.40, then q = 20,000 - 10,000*(1.40) = 20,000 - 14,000 = 6,000, which is 5000 + 1000. So that works. Therefore, the quantity sold is a linear function of price: q = 20,000 - 10,000p. So, we can model this as a linear demand curve. Alternatively, maybe they want us to model x as a continuous variable, so that x can take any non-negative real value, meaning that the price can be any value below $1.50, not necessarily in increments of 10 cents, and the quantity sold would adjust accordingly. So, if you lower the price by 5 cents, then you can sell 500 more items. But does the problem allow that? The market research says "for every 10 cents I lower the price below $1.50 I will be able to sell another 1000 items." So, it's a rate: 10 cents decrease => 1000 items more. So, that is $0.10 decrease corresponds to 1000 items increase. Therefore, the slope of the demand curve is dq/dp = -1000 / 0.10 = -10,000. So, the demand function is linear with slope -10,000. So, starting from p = 1.50, q = 5000. So, the equation is q = -10,000(p - 1.50) + 5000 = -10,000p + 15,000 + 5000 = 20,000 - 10,000p, which is the same as before. So, yes, continuous. Therefore, the quantity sold as a function of price is q(p) = 20,000 - 10,000p. Alternatively, we can write p as a function of q. Let's solve for p: q = 20,000 - 10,000p => 10,000p = 20,000 - q => p = (20,000 - q)/10,000 => p = 2 - q/10,000. So, the inverse demand function is p = 2 - q/10,000. Okay, so total revenue R is pq. So, R = pq = (2 - q/10,000)*q = 2q - q²/10,000. Total cost C consists of fixed costs (start-up costs) of $2000 and variable costs of $0.50 per item. So, total cost is C = 2000 + 0.50q. Therefore, profit π is total revenue minus total cost: π = R - C = (2q - q²/10,000) - (2000 + 0.50q) = 2q - q²/10,000 - 2000 - 0.50q Simplify: π = (2q - 0.50q) - q²/10,000 - 2000 = 1.50q - q²/10,000 - 2000. So, profit as a function of quantity q is π(q) = - (q²)/10,000 + 1.50q - 2000. To maximize profit, we need to find the value of q that maximizes π(q). Since this is a quadratic function with a negative coefficient on q², it opens downward, so the maximum is at the vertex. The vertex of a quadratic function ax² + bx + c is at x = -b/(2a). Here, a = -1/10,000 and b = 1.50. Therefore, the quantity that maximizes profit is: q = -b/(2a) = -1.50 / (2*(-1/10,000)) = -1.50 / (-2/10,000) = (1.50 * 10,000)/2 = (15,000)/2 = 7,500. So, the profit-maximizing quantity is 7,500 items. Then, the corresponding price can be found from the inverse demand function p = 2 - q/10,000 = 2 - 7,500/10,000 = 2 - 0.75 = $1.25 per item. Let me verify that. So, at q = 7500, p = $1.25. Then, total revenue R = p*q = 1.25 * 7500 = $9,375. Total cost C = 2000 + 0.50*7500 = 2000 + 3750 = $5,750. Profit π = 9,375 - 5,750 = $3,625. Is this the maximum profit? Let's check another point. Suppose q = 7000. Then, p = 2 - 7000/10,000 = 2 - 0.7 = $1.30. Revenue = 1.30 * 7000 = $9,100. Cost = 2000 + 0.50*7000 = 2000 + 3500 = $5,500. Profit = 9,100 - 5,500 = $3,600, which is less than $3,625. Similarly, q = 8000, which would be p = 2 - 8000/10,000 = 2 - 0.8 = $1.20. Revenue = 1.20 * 8000 = $9,600. Cost = 2000 + 0.50*8000 = 2000 + 4000 = $6,000. Profit = 9,600 - 6,000 = $3,600, again less than $3,625. So, 7500 seems to be the maximum. Alternatively, since we derived this using calculus (vertex of parabola), it's correct. But just to check. Alternatively, if we take derivative of π(q) with respect to q: dπ/dq = 1.50 - 2q/10,000. Set derivative equal to zero: 1.50 - (2q)/10,000 = 0 => (2q)/10,000 = 1.50 => q = (1.50 * 10,000)/2 = 7,500. So, same result. Wait, but in the original problem, they mention that for every 10 cents decrease, you can sell another 1000 items. So, starting at $1.50 with 5000 items. Then, decreasing by 10 cents increments gives 1000 more each time. So, how does it work at 7500 items? That would be an increase of 2500 items from the original 5000. Since each 10 cents decrease gives 1000 more, 2500 would be 2.5 times 10 cents. So, decrease price by 25 cents? Wait, but in the model we had, q = 20,000 - 10,000p. So, if p is $1.25, which is a $0.25 decrease from $1.50, then yes, (1.50 - 1.25)/0.10 = 2.5, so x=2.5. Then, quantity sold is 5000 + 1000*2.5 = 7500. So, even though x is 2.5, which is not an integer, the problem didn't specify that the price has to be in 10-cent increments, so allowing x to be a real number. So, in reality, you could set the price at $1.25, even though it's a 25-cent decrease, which is 2.5 times 10 cents. So, the model allows for continuous variables, which is fine. So, the conclusion is that the optimal price is $1.25 per item, selling 7500 items, resulting in a maximum profit of $3,625. Wait, let me just confirm profit calculation once again: Revenue: 7500 * 1.25 = 7500 * 1 + 7500 * 0.25 = 7500 + 1875 = 9375. Cost: 2000 + 7500 * 0.50 = 2000 + 3750 = 5750. Profit: 9375 - 5750 = 3625. Yes, that's $3,625. So, all checks out. So, the answer is price of $1.25, quantity of 7500, profit of $3,625. Final Answer The optimal price is \boxed{1.25} dollars per item, the number of items to sell is \boxed{7500}, and the maximum profit is \boxed{3625} dollars. </think> To determine the optimal price, quantity, and maximum profit for selling children's toys, we start by modeling the relationship between price and quantity sold. The market research indicates that for every $0.10 decrease in price below $1.50, the quantity sold increases by 1000 items. This results in a linear demand function. Demand Function: The price ( p ) and quantity ( q ) are related by the equation: [ q = 20,000 - 10,000p ] Solving for ( p ) gives the inverse demand function: [ p = 2 - \frac{q}{10,000} ] Profit Function: Total revenue ( R ) is given by ( R = p \times q ): [ R = \left(2 - \frac{q}{10,000}\right) q = 2q - \frac{q^2}{10,000} ] Total cost ( C ) includes fixed costs of $2000 and variable costs of $0.50 per item: [ C = 2000 + 0.50q ] Profit ( \pi ) is total revenue minus total cost: [ \pi = \left(2q - \frac{q^2}{10,000}\right) - (2000 + 0.50q) = 1.50q - \frac{q^2}{10,000} - 2000 ] Maximizing Profit: The profit function is a quadratic equation, and the maximum profit occurs at the vertex. For the quadratic equation ( \pi(q) = -\frac{q^2}{10,000} + 1.50q - 2000 ), the vertex is found at: [ q = -\frac{b}{2a} = -\frac{1.50}{2 \left(-\frac{1}{10,000}\right)} = \frac{1.50 \times 10,000}{2} = 7,500 ] Substituting ( q = 7,500 ) into the inverse demand function gives the optimal price: [ p = 2 - \frac{7,500}{10,000} = 2 - 0.75 = 1.25 ] Calculating Maximum Profit: Total revenue at ( q = 7,500 ) and ( p = 1.25 ): [ R = 1.25 \times 7,500 = 9,375 ] Total cost: [ C = 2000 + 0.50 \times 7,500 = 2000 + 3750 = 5,750 ] Maximum profit: [ \pi = 9,375 - 5,750 = 3,625 ] Final Answer The optimal price is \boxed{1.25} dollars per item, the number of items to sell is \boxed{7500}, and the maximum profit is \boxed{3625} dollars. Token used: 3399 (10017 characters) Conclusion As you can see the various steps logically laid out where the model uses its knowledge of the worlds (profits = sales - cost), generates an algebraic function to link the behavior of various variables and then uses calculus to maximize the profit function to arrive at the right price and quantity for maximum profit. Resources Reasoning models on OpenAI - https://platform.openai.com/docs/guides/reasoning Bring your own models on AI Toolkit - using Ollama and API keys Building Retrieval Augmented Generation on VSCode & AI ToolkitBring your own models on AI Toolkit - using Ollama and API keys
As we have seen in past blog posts, AI toolkit supports a range of models using Github Marketplace of models. However, you might require support for external models hosted by Google, Anthropic and Open AI which are either not available in the Github catalog of models or might want to use the models served by Ollama. We will cover both of these scenarios in this blog post. OpenAI, Anthropic and Google Hosted models Once you click on the Model catalog window and selected the models hosted by Google, Anthropic and OpenAI, you should see the following models selected. Also, you can add your APIs keys to the model in the following way On the above models, you should click on the "Try in playground", just below the model name model card and you should be able to see the following dialog box on the top search bar of the VS Code window. Here I have clicked on the "Try it in playground" link for Anthropic Claude 3.5 Sonnet model Enter your API Key and you are good to go. Also, as you can see in the dropdown text, you can also edit or change the value later. Similarly, you can perform the same action for the Google and OpenAI hosted models. Once you have done this you are free to use these models in the playground and for using the other features of AI toolkit extension. Using models served by Ollama Several developers are also using Ollama to experiment and play with models using the command line. Ollama is an open-source AI tool that allows users to run large language models (LLMs) on their local systems. It's a valuable tool for industries that require data privacy, such as healthcare, finance, and government which might need locally hosted models. So, AI toolkit already supports some locally downloadable models such as those in the Phi-series by Microsoft or those by Mistral. Ollama supports a wider variety of models especially those from Meta's Llama series of LLMs and SLMs. The complete list of models currently supported by Ollama can be found at Ollama library. We will run ollama on windows and when you run ollama and see help command you get the following output. Once you have selected the model from the library, you can use the ollama pull or ollama run to download the model. The run command will download the model and then run it if it's not already downloaded. The pull command will just download the model from the repository. Since I want you show you a multimodal model that can be run locally, I will be go to the command line and download and run llama3.2-vision model. See the commands for the same below We can also list the models already downloaded. As you can see, I have downloaded and tried a bunch of models. It might take a bit of time to download the models based on the speed of your internet connection. As you can see some of the models are quite large downloads. Also, you might need to make sure you have enough RAM on your laptop or desktop to run the models. So now we have downloaded the models, and we have run the models. Now, let's see how we can access them in AI Toolkit for VSCode. Go to the My models window as seen below. Click on the '+' symbol as seen below in the screenshot. Once you click on this you will see a dropdown in search. Click on add an Ollama model You will now see the choice to either select a model from ollama library or a custom ollama endpoint. For the purposes of this tutorial, we will select models. Let's select Multimodal modal we downloaded earlier. We should see the models we had seen in ollama list command earlier. See below. Now we can select the checkbox alongside llama3.2-vision:latest and select okay. You should now see the model appear in the My Models window like below. You can now right click on it and start using the model by loading in the playground. Since this is a multimodal model, you can use it to generate text as below. The screenshot below shows the model loaded via ollama and at the same time you can see that it is has the clip symbol activated in the window. Since this is a multi-modal model, we can give it an image and ask it questions. Which is what we will do next. Next, now let's attach an image and ask it questions about this image. This might take a bit more time as it will need to analyze the image and answer any questions. Since it is a generative AI model it can give slightly different inputs when given the same image. I can also ask the models questions about the image, and it will use the information from the image - the objects shown, the relationship between the objects and its world knowledge from its training to answer the questions. For example, see the session below. So, as you can the AI toolkit can be a fantastic place to try out different models from various sources. Ollama is also a great tool to try pre-built models locally securely without sending your data to the cloud, which can make it suitable for air-gapped environments and data privacy sensitive industries like fintech, healthcare and government. You have greater control over the models and the environment and the data they run on. It is also possible to customize the models and then serve them via ollama. This helps you choose the best model for your AI application. Resources AI toolkit for VSCode - https://marketplace.visualstudio.com/items?itemName=ms-windows-ai-studio.windows-ai-studio AI toolkit for VSCode on Github - https://github.com/microsoft/vscode-ai-toolkit Ollama - https://github.com/ollama/ollama Ollama library - https://ollama.com/library Azure AI Discord - https://aka.ms/AzureAI/Discord
Getting Started with the AI Dev Gallery
The AI Dev Gallery is a new open-source project designed to inspire and support developers in integrating on-device AI functionality into their Windows apps. It offers an intuitive UX for exploring and testing interactive AI samples powered by local models. Key features include: Quickly explore and download models from well-known sources on GitHub and HuggingFace. Test different models with interactive samples over 25 different scenarios, including text, image, audio, and video use cases. See all relevant code and library references for every sample. Switch between models that run on CPU and GPU depending on your device capabilities. Quickly get started with your own projects by exporting any sample to a fresh Visual Studio project that references the same model cache, preventing duplicate downloads. Part of the motivation behind the Gallery was exposing developers to the host of benefits that come with on-device AI. Some of these benefits include improved data security and privacy, increased control and parameterization, and no dependence on an internet connection or third-party cloud provider. Requirements Device Requirements Minimum OS Version: Windows 10, version 1809 (10.0; Build 17763) Architecture: x64, ARM64 Memory: At least 16 GB is recommended Disk Space: At least 20GB free space is recommended GPU: 8GB of VRAM is recommended for running samples on the GPU Visual Studio 2022 You will need Visual Studio 2022 with the Windows Application Development workload. Running the Gallery To run the gallery: Clone the repository: git clone https://github.com/microsoft/AI-Dev-Gallery.git Run the solution: .\AIDevGallery.sln Hit F5 to build and run the Gallery Using the Gallery The AI Dev Gallery has can be navigated in two ways: The Samples View The Models View Navigating Samples In this view, samples are broken up into categories (Text, Code, Image, etc.) and then into more specific samples, like in the Translate Text pictured below: On clicking a sample, you will be prompted to choose a model to download if you haven’t run this sample before: Next to the model you can see the size of the model, whether it will run on CPU or GPU, and the associated license. Pick the model that makes the most sense for your machine. You can also download new models and change the model for a sample later from the sample view. Just click the model drop down at the top of the sample: The last thing you can do from the Sample pane is view the sample code and export the project to Visual Studio. Both buttons are found in the top right corner of the sample, and the code view will look like this: Navigating Models If you would rather navigate by models instead of samples, the Gallery also provides the model view: The model view contains a similar navigation menu on the right to navigate between models based on category. Clicking on a model will allow you to see a description of the model, the versions of it that are available to download, and the samples that use the model. Clicking on a sample will take back over to the samples view where you can see the model in action. Deleting and Managing Models If you need to clear up space or see download details for the models you are using, you can head over the Settings page to manage your downloads: From here, you can easily see every model you have downloaded and how much space on your drive they are taking up. You can clear your entire cache for a fresh start or delete individual models that you are no longer using. Any deleted model can be redownload through either the models or samples view. Next Steps for the Gallery The AI Dev Gallery is still a work in progress, and we plan on adding more samples, models, APIs, and features, and we are evaluating adding support for NPUs to take the experience even further If you have feedback, noticed a bug, or any ideas for features or samples, head over to the issue board and submit an issue. We also have a discussion board for any other topics relevant to the Gallery. The Gallery is an open-source project, and we would love contribution, feedback, and ideation! Happy modeling!AI Toolkit for Visual Studio Code: October 2024 Update Highlights
The AI Toolkit’s October 2024 update revolutionizes Visual Studio Code with game-changing features for developers, researchers, and enthusiasts. Explore multi-model integration, including GitHub Models, ONNX, and Google Gemini, alongside custom model support. Dive into multi-modal capabilities for richer AI testing and seamless multi-platform compatibility across Windows, macOS, and Linux. Tailored for productivity, the enhanced Model Catalog simplifies choosing the best tools for your projects. Try it now and share feedback to shape the future of AI in VS Code!Building your own copilot – yes, but how? (Part 1 of 2)
Today, there’s a wide range of built-in services and features designed to enable organizations and developers to build their own copilots, able to answer questions based on their own knowledge bases and data sources. But how to choose the most suitable one for each scenario? This blog post wants to provide an overview of some of the main choices you have in the Microsoft technology ecosystem. Part 1 will look into low-code tools and out-of-the-box features, while part 2 will focus on code-heavy and extensible options.
