No Weekender Again??
If I continue like this pretty soon even the people who are interested in the weekenders will lose interest.
With Mom and Dad leaving this weekend, the weekend was quite busy. We did some shopping to keep ourselves occupied too. After working with Bluetooth for last 2 years, I finally have a Bluetooth phone. It's a motorola and I would have uploaded its picture but the only way for me to do that right now is to use the bathroom mirror and I will wait till I have some other camera handy.
I also played some "traditional" football this weekend. Axe and I bought this together but *never* played against each other. Although it is very difficult to score a goal in this release compared to other ones, that is what makes the game special. You have to earn every goal. The 10-2 scorelines are a thing of the past. And once you score a goal, you are sure to spend next 10 seconds pumping your fist. Well, at least I am sure about myself.
And to make up for the weekender here is a link to something weekender regulars would enjoy.
http://www.relpats-eht.com/
It gets pretty tough later on. I do not know anyone who has seen beyond the last page. So a good incentive and we can keep this going on in the background. This is NOT the next weekender.
With Mom and Dad leaving this weekend, the weekend was quite busy. We did some shopping to keep ourselves occupied too. After working with Bluetooth for last 2 years, I finally have a Bluetooth phone. It's a motorola and I would have uploaded its picture but the only way for me to do that right now is to use the bathroom mirror and I will wait till I have some other camera handy.
I also played some "traditional" football this weekend. Axe and I bought this together but *never* played against each other. Although it is very difficult to score a goal in this release compared to other ones, that is what makes the game special. You have to earn every goal. The 10-2 scorelines are a thing of the past. And once you score a goal, you are sure to spend next 10 seconds pumping your fist. Well, at least I am sure about myself.
And to make up for the weekender here is a link to something weekender regulars would enjoy.
http://www.relpats-eht.com/
It gets pretty tough later on. I do not know anyone who has seen beyond the last page. So a good incentive and we can keep this going on in the background. This is NOT the next weekender.
posted by Nikhil at 10:43 AM
8 Comments:
Check out the couple of interesting puzzles (not difficult) on this website. More interesting is how to get to the solution.
http://www.chriscoyne.com/puzzle/
They are quite interesting. I got both (haven't done the second solution of #2), but no, I do not have a formal proof yet. It was only some feedback thinking in my mind that got me the solutions.
Is there only one solution of #1?
I don't know if there is only one solution to 1. How did you get to it? I kept giving the last output as the input, and it converged to 1210.
Btw, similar to that relpats website, IIM-Indore guys have created some puzzles. They are a lot simpler though, at least the initial ones. Check out.
http://www.iimi-iris.com/iris/irising/klueLESS/
ok, then there are multiple solutions to 1. I figured out the function and then thought that I need to have one zero at least, then could not do it in a three digit number and then thought about 2020 somehow after giving a few choices.
What is the function?
The number of digits in f(x) is at least equal to number of digits in x.
the first digit(Most significant) in f(x) is the number of zeros in x, next is number of '1's in x, next is number of '2's and so on...
I believe
{1, 7, 3, 2, 1, 1, 1, 2, 1, 1}
is the only solution for #2.
It's easy to write a program to verify it. Instead of writing an O(9^10) algo, can you think of a property these numbers should satisfy to reduce the recursion depth?
Took me 16:19 but I found the other solution. I am not sure if there are more.
Hint: Think inside the Box.
About the property, first thing that comes to mind is the sum of these numbers should be 20 (at least 1 being already considered with 9^10).
For a value 'v' for number 'n', 20 -((v-1)*n) should be at least 7 if n != v. ( I am not sure myself about what I said, but you get the idea. Basically, apply pigeon-hole to this and start from the highest value of 9, that is sure to reduce the number of combinations.)
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