(i was just thinking that where could i post this )
This is for those who are newcomers and are exhausted trying that cube. Rubics Cube is not just a Theoritical Math Problem. So you can do physical torture with it , if it makes you angry
You've to move any layer(only one) by 45 degree. Then, take a screw-driver(not star screw-driver!) or anything like that; and pull the edge cubie! Don't worry it'll not break.
Walla!!! yo have your seperate cubie and you can press it any where you like. MAKE SURE you DO NOT pull the centre piece as it is still. Pulling that will break your cube.
Another important thing is that you have to remember from where have you pulled out the cubie. Or else your cube will turn into a disaster. Your cube will not be matched ever if you arrange
the cubies incorrectly. And if you do forget from where you have taken any cubie, then just match the whole cube; and you're done!
Though it's cheating, it is NOT so easy as it seems .But atlest doing this wont make you cry
Disclaimer: I am totally NOT RESPONSIBLE for any kind of disorder/breaking of your cube.
Have Fun ▓░▒╬▒░▓
Easiest Way of Solving RC
r@k€€/|/
- nafistiham
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Re: Easiest Way of Solving RC
i also had done this technique before i learned the algorithms.but after learning that i would say it was lot easier than the breaking-making game.though,people not wanting to learn the algorithms should buy first class rubic's cube.unless.... it is bound to break.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
- nafistiham
- Posts:829
- Joined:Mon Oct 17, 2011 3:56 pm
- Location:24.758613,90.400161
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Re: Easiest Way of Solving RC
same here.i am really curious to see how a $4^{3}$,$5^{3}$ or even bigger cubes function like.wanna solve them ,too
বাংলাদেশে কি এগুলো কোথাও পাওয়া যায় বলে আপনার জানা আছে?
[একটা হাস্যকর পোস্ট দিয়ে আমার $century$ হল..... ]
বাংলাদেশে কি এগুলো কোথাও পাওয়া যায় বলে আপনার জানা আছে?
[একটা হাস্যকর পোস্ট দিয়ে আমার $century$ হল..... ]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.