It's official! The Socratica Python Kickstarter was a success! Thank you to all of our supporters. Because of you, many more Python videos coming soon!! 💜🦉
As for the "even numbers get you closer" thing: Since the direction of the steps is evenly distributed, in the even case, we can expect the number of steps going up to be the same to those going down. But in the odd case, we can expect the number of steps in one direction to be one larger than the other, leading us further away from the origin. The same is true for the left/right dimension. This phenomenon should be less and less noticeable, as we increase the number of steps.
Yes, but this is only because it was considered the result "distance = abs(x) + abs(y)", removing the influence of the sign of x and y, which can influence your distance in each direction. Interesting.
Yes., Also this is dependent on the number of blocks chosen, take for instance 5 blocks away from home, We expect the even number to be further aways..
I think it is because I’m the second step you have a only a 25% chance to be 2 blocks away and 75% to be one block away or back to the starting point. For example if your first step is W only if the second step is W you will be 2 blocks away, if it is S or N you will be 1 block away and if it is E you will be back at the starting point. So this tendency will impact the subsequent results as more steps are added.
High class material, reminds me of Morpheus explaining the Matrix. Excellent, to the point and does not waste any time of the listener. Keep the good work.
The Best Ever Tutorials, entertaining ,CLASSY, always waiting for her deadpan "The city of Monte Carlo... you'll never find a more wretched hive of scum and villany." or the like. And I actually learn a lot. Thanks!
As a beginner in programming this took me a bit of time to grasp, but the way you explained was phenomenal. It really boost up my confidence. Thanks for this.!
I think the reason that random walks with even numbers of steps lies in the trivial cases wherein 0 steps lands you right at home, 1 step definitely won’t end up back home, 2 steps may or may not, and so forth.
if someone had taught me random walk in this way earlier, I would have been a prodigy by now. thanks for this awesome video. I have already subscribed and believe this video to be more useful for me than previous ones.
To answer @Benjamin Voll, my initial guess is that the odd distances (numbers) are not perfectly divisible by two like even distance(numbers). P.S. Great videos. I like the emphasis on computer science and not just "learning python." Also, great videos on mathematics, too.
This lady is not just a programmer, but she is actually a real scientist I have been searching for more than ten years for an efficient and logical way to implement Monte Carlo simulation
I appreciate all the efforts you put into these videos. And I'm impressed with how supportive the community was for your kickstarter campaign. I wonder if that suggests a new model for how educational content will be developed. I'm also thinking about the aesthetic you've established here and the pedagogical efficacy. I wonder if you've studied whether any increased learning efficiency emerges and it generates a return on your production costs. If so, you might have a scalable business model.
From what I have seen on the Socratica Channel, you have a "WINNER"... Hope you get funding to do great and wonderful things. Thank you for that "gamble" .
You are the only one who keeps the ability to make me understand all the difficult problems of the world .You are great ,awesome ..I dont know how to express my respect ,gratitude towards you . Thank you mam ...pls keep uploading more videos on python
I wish I watched this video like 5 years ago, the Monte Carlo method is a cornerstone of many important fields of research, but I had felt locked out of it for the longest time because classrooms simply dismissed it as an abstract idea from which we derive a single average value. Whelp, time to jump back into polymer science!
The difference in probability between an even/odd number of moves may be related to the fact that the origin is special: Every move from there increases the distance to the origin with 100% probability. The first move (odd) increases the distance, and if you happen to get back to the origin, this will happen again. You need an even number of moves to get back to the origin, therefore every move from the origin is an odd move. So, odd moves are slightly move "evil". The further you get away from the origin, the less relevant this becomes, therefore the probability difference is reduced with the number of moves.
Can you make a full course on Machine Learning and AI from scratch, to advanced with few project, so that a lot of other people will benefited from it. There are also some other video on the said above, but your video is stand alone. I really appreciate you for doing hard work for us. Thank You.
Hi Socratica, very impressive and that makes it all the more difficult to have a different understanding. 22 seems to be the highest number that you may be able to come back with no transportation. In other words, above that number you will have to pay for transportation. That is not the same as what the problem is asking - which is the highest number that with which you'll end up 4 blocks or less from home. This number seems to be 14. In between those two numbers the probability seems to oscillate up and down, in both runs, but that's a separate observation. I'd appreciate a reply for a sanity check. Thanks
Oh wow i enjoyed this video so much. Very educative and well though out. The female voice is catchy and the background noise actually is not disturbing but sort of stimulating. Awesome.
To see if I have an accurate understanding: To get an accurate estimate of the ratio of paths fewer than 4 blocks from home to paths more than 4 blocks from home for a given walk size, you perform Monte Carlo simulation. That ratio is basically a Bernoulli distribution which is the true underlying distribution that we're trying to estimate, unknown to us, for each walk size. What we could do is simply exhaustively generate every permutation of paths for each walk size, and divide by the total to get that distribution. But in higher dimension, the number of permutation blows up, and is intractable to solve. So we do Monte Carlo simulation which is just drawing a large number of samples to approximate the underlying distribution. The more trials we do, the more our estimation converges toward the true value according to the law of large numbers. The problem of longest walk size over 50% is irrelevant. The heart of Monte Carlo simulation is accurate estimation of a probability distribution via efficient random sampling to overcome intractability, right?
It's perhaps worth pointing out that the question is ambiguous. As stated, the question is: "What is the longest random walk you can take so that, on average, you will end up 4 blocks or fewer from home?" When I read this, I thought it concerned the expected (i.e. "average") net distance travelled given random walks of a particular length. That is, I interpreted the question as "What is the longest random walk you can take such that you end up 4 blocks or fewer from home in expectation?" On this interpretation, the answer does not seem to be 22 -- it is (I think!) only 12! Here are my full results: Number of steps = 1 / Expected blocks to home = 1.0 Number of steps = 2 / Expected blocks to home = 1.504775 Number of steps = 3 / Expected blocks to home = 1.87875 Number of steps = 4 / Expected blocks to home = 2.18895 Number of steps = 5 / Expected blocks to home = 2.460925 Number of steps = 6 / Expected blocks to home = 2.7177 Number of steps = 7 / Expected blocks to home = 2.93355 Number of steps = 8 / Expected blocks to home = 3.142575 Number of steps = 9 / Expected blocks to home = 3.341525 Number of steps = 10 / Expected blocks to home = 3.519475 Number of steps = 11 / Expected blocks to home = 3.70665 Number of steps = 12 / Expected blocks to home = 3.87785 Number of steps = 13 / Expected blocks to home = 4.02005 Number of steps = 14 / Expected blocks to home = 4.17785
I think it is because I’m the second step you have a only a 25% chance to be 2 blocks away and 75% to be one block away or back to the starting point. For example if your first step is W only if the second step is W you will be 2 blocks away, if it is S or N you will be 1 block away and if it is E you will be back at the starting point. So this tendency will impact the subsequent results as more steps are added.
Thank you for another great tutorial! For the challenge at the end, I get 31 as the answer for the largest random walk which will on average leave me 5 blocks or less from home. That is a slightly different challenge than stated - you said "less than 5 blocks" which i believe is the same as "4 blocks or less"...hence the same as the original question and same answer (22). What did others come up with?
The stunning beautiful woman initially caught my eye. The straight forward, easy to understand tutorials caused me to subscribe. The humor is a nice added bonus. Oh why we're YOU not my Pascal teacher in the 80's?
I think that the problem statement is not clear. The average length of a 22 block random walk is about 5.28. This is higher than 4. On the other hand, for 12 blocks, the average length is about 3.86. And for 13 blocks, the average is about 4.33. So the longest random walk you can take so that , on average, you end up at at most 4 blocks from the origin ist 12. The soulution of 22 is for a different problem: "What is the longest random walk you can take, so that more than half of the time you end up at at most 4 blocks away from the origin?"
this channel is the best among others. we need more videos on python.
Год назад
About the even/odd probability, I believe that the even number of steps has a higher chance of getting closer to the origin because even numbers can perfectly cancel the total displacement; It's easy to see if you reduce the dimension of the problem and put it on a line where you can go a positive or negative direction; Let's begin analyzing the 1,2 scenarios with one step you will in the best scenario be at at least one step away of the origin, with 2 steps you can be at 2 or zero steps away; Again with the 3,5,7... you cannot get zero displacements.
Good Demonstration of the technique, but I have a small restraint on the code! There should be no for loop for the number of blocks n in the second part of the program, and for n=30, the % of the no_transport is ~ 41 for 10,000 experiments.
Great video!! I love this series it is both entertaining with the great sense of humor and explains everything in a way that I can understand. Did anybody else get 31 as the answer to the problem at the end of the video?
More accurately: "you will seldom find a /less/ wretched hive of scum and villany..." It's the land of the super rich, also they have sun, and very large boats.
Nice intro video, but your code would never pass a linter. ;) For people new to Python (or coding, in general), don't write your methods like this. These are fine for examples, but if you are actually writing a project, you should be more explicit with your variable names. PEP8 is a fantastic guideline for coding style. It's targeting Python, but most of it can be applied generally. I would also recommend following the Zen of Python. You can read it by entering a Python interpreter and run "import this". Run the following in a terminal. python -c "import this" Always run a linter against your code before finishing. There are linters for all languages. A linter simply runs a check against style rules. For instance, if your variable names are correct, or lines are too long, or your inheritance structure isn't correct, etc. It won't check for runtime errors. It only checks for style and clarity. Both of which are important - especially if you work in a team. For Python, I recommend pylint.
meanwhile my code: from random import choice def random_walk3(n): """Return coordinates after 'n' block random walk""" return [sum(x) for x in zip(*(choice([(0, 1), (0, -1), (1, 0), (-1, 0)]) for _ in range(n)))]
What an awesome sci-fi format, love it! So the MC method as you present it is just a double for loop to run the simulation yes? I had imagined there would be a lot more to it.
