@kiran, your thinking is correct, because they are the same precedence, need to remove from stack (doesn't really matter with + on + or - on -) but is for + on - or - on +
I had the same question as the student in the video about the left bracket which goes on the stack, but I think I understand now; basically the left bracket 'has a precedence' which other operators compare their own precedence against, but the left bracket ignores its own precedence (i.e. never compares its own precedence against any of the operators (even though it could)) and just goes to sit on top of the stack. Thanks for your great example, with an assignment and other tricky bits! It's really helped me.
Trigonometric functions, or functions in general, go above exponents. For example: 2 + 3^4 * cos(0) Steps: 1. 2 is an operand; Add 2 to the output expression 2. + is an operator; Stack = empty; push + onto the stack 3. 3 is an operand; Add 3 to the output expression 4. ^ is an operator; Top of stack has lower precedence; push ^ onto stack 5. 4 is an operand; Add 4 to the output expression 6. * is an operator; Top of stack has higher precedence; pop stack until lower precedence is found; push * onto stack 7. cos is treated as an operator; top of stack has lower precedence; push cos onto stack 8. "(" = push onto stack regardless of lower or higher precedence or stack is empty 9. 0 is an operand; Add 0 to output expression 10. ")" = Pop and add to output expression until "(" = top of stack; pop top of stack 11. Since ")" is the end of the expression, pop the stack until stack = empty Result: 2 3 4 ^ 0 cos * +
WooHoooP Don't understand your example that would come from the infix (p - q) / (r - s) But to answer your question: if you have an expression of this form a - - b the minus in front of the b is actually unary minus (negation, applied to one operand) we would re-write that expression like this a - ~ b (using tilda to represent the unary minus) this would generate as unary minus is higher precedence that + or minus. a b ~ - which would negate the b before attempting to subtract it from a, which is logical in this example The precedence of unary minus is open to debate in some environments its placed higher than exponent (raising to the power ^) but others may place it below this but above multiply and divide. ~ b ^ a if unary minus was higher we would get b ~ a ^ which would negate b then raise to the power of a, but some would argue that we should have exponent higher than unary minus meaning this b a ^ ~ which would raise the b to the power of a and negate the result
XXxlightmarex just keep applying the rules. every left bracket goes on top of stack. when you get a right bracket remove contents of stack untill you find a left bracket, then discard that snd continue to parse the expression
work through remember operands just get written down as rpn (don't involve the stack at all) ((a + b * c) / (d + e * (f - g ^ h))) reaching first right bracket stack contains * + ( ( remove up to topmost left bracket and discard giving: abc*+ stack only contains ( carrying on until we reach 2nd right bracket, stack should contain: ^ - ( * + ( / ( dump stack upto topmost left bracket and discard giving us this much of the rpn: abc*+defgh^- stack now contains * + ( / ( parse 3rd right bracket so remove stack contents again upto topmost left bracket giving rpn of: abc*+defgh^-*+ stack contains / ( process final right bracket by removing upto topmost left bracket giving rpn of: abc*+defgh^-*+/ stack now empty hope this helps
+Jatin Rajput To be fair nothing makes up for bunk'd classes, it's amazing the nuances and added extra's you'll get when someone goes through a topic with you. I always find that I talk about things I never planned.
Another burgermate you can only place an operator on top if it is higher so remove the item on top of stack snd write it down in the rp expression, then see if the new operator can now go on the stack (like normal)
Left brackets are always going to override precedence, it goes on top of everything, it's also there as a marker for when we encounter a right bracket while parsing, so we know when to stop pulling operators off the stack
The diagram shows that all operators go on top of left bracket if it seen on top of the stack. Parsing a left bracket in the expression doesn't look at precedence as you automatically push it onto the stack
You can only push onto the stack if the precedence is higher, if it is the same you must pop the stack until the operator is higher than the contents of the top
you made a 3 hour lecture in only 9 minutes, thats talent
That was the intention
Ive seen several tutorials here but this one is crystal clear
thanks, that's the idea
It's much clearer than virgin's teardrop
Thank you, made it a lot easier than my teacher
hello
love your english accent too xD i can't understand those Indian tutorials
thanks for the "fack"ing help hehehe
@kiran, your thinking is correct, because they are the same precedence, need to remove from stack (doesn't really matter with + on + or - on -) but is for + on - or - on +
Thanks a lot! I'm taking part in CIE exams computer science tomorrow... you make it more easier
every year, when I have to teach RPN, I always end up watching this video. Cheers mate great vid :)
+Joseph Burke
No problems, pity it's come out of the new specifcations. Favourite topic
you"re such a charismatic, well explaining teatcher.
Finally an understandable explanation of this. Thanks so much.
that video made everything crystal clear, and loved the accent hehe ^^ cheers mate!
Really nice expiation, just implemented it in my semiconductor device model in C.
Nice
I had the same question as the student in the video about the left bracket which goes on the stack, but I think I understand now; basically the left bracket 'has a precedence' which other operators compare their own precedence against, but the left bracket ignores its own precedence (i.e. never compares its own precedence against any of the operators (even though it could)) and just goes to sit on top of the stack.
Thanks for your great example, with an assignment and other tricky bits! It's really helped me.
glad it helped
Yeah, the left bracket sits on the stack so further operators can override whatever precedence was already on top.
For the good of humanity, such lectures should records and presents.
amazing, Thank you for this great explanation
Ohh thank you now i understood this concept....really helpful for my exams...thanks a lot...
Well thank you :D Definitely saved me from failing an exam completely
glad to help
wow thank you so much! helped a lot!
this just saved my life, thanks you so much Mr Banana =)
No problem, my favourite topic of all time :)
I wish this was longer. Love the way you actually taught this.
Thanks, it's such a cool thing reverse polish
Thank you! Interesting.
Cheers, helped a lot to figure out an algorithm
holy crap this was so helpful
Thank you so much you explained it really well Sir
Well now I can do reverse polish. Top notch!
+CaptainClueless Everyone should be able to do it, it's cool
awesome!! thanks :)
purfect !
great explaination :) thank you!
No problem, happy it was of some use
Lindy approved video!
Please be my professor. Also the FACK part made me lol
What precedence would factorial "!" have. Does it go above exponents?
God bless You
As easier as it gets :) thnx
Great job! Thank you very much for help! ;)
No I understand it :D
is always better to invest time for finding a good tutorial :) ...and Yours is Awesome :) ...search is over :)
Thanks
plus cannot be placed on plus
minus cannot be placed on minus
right?????
"Facking" awesome !!
Any idea what order of precedence the trig functions sin, cos and tan fall under?
Trigonometric functions, or functions in general, go above exponents.
For example: 2 + 3^4 * cos(0)
Steps:
1. 2 is an operand; Add 2 to the output expression
2. + is an operator; Stack = empty; push + onto the stack
3. 3 is an operand; Add 3 to the output expression
4. ^ is an operator; Top of stack has lower precedence; push ^ onto stack
5. 4 is an operand; Add 4 to the output expression
6. * is an operator; Top of stack has higher precedence; pop stack until lower precedence is found; push * onto stack
7. cos is treated as an operator; top of stack has lower precedence; push cos onto stack
8. "(" = push onto stack regardless of lower or higher precedence or stack is empty
9. 0 is an operand; Add 0 to output expression
10. ")" = Pop and add to output expression until "(" = top of stack; pop top of stack
11. Since ")" is the end of the expression, pop the stack until stack = empty
Result: 2 3 4 ^ 0 cos * +
But how do you deal with expressions with the unary operator like:
-A + B?
Higher priority than multiply and divide. Only Pop one item from stack like store does
did he just give an example over "fack up"?
would like to know what if there is a negative sign meet with a negative sign ? or any other sign meeting the same ones
for example pq-rs-/
WooHoooP
Don't understand your example that would come from the infix
(p - q) / (r - s)
But to answer your question:
if you have an expression of this form
a - - b
the minus in front of the b is actually unary minus (negation, applied to one operand)
we would re-write that expression like this
a - ~ b (using tilda to represent the unary minus)
this would generate as unary minus is higher precedence that + or minus.
a b ~ -
which would negate the b before attempting to subtract it from a, which is logical in this example
The precedence of unary minus is open to debate in some environments its placed higher than exponent (raising to the power ^) but others may place it below this but above multiply and divide.
~ b ^ a
if unary minus was higher we would get
b ~ a ^
which would negate b then raise to the power of a, but some would argue that we should have exponent higher than unary minus meaning this
b a ^ ~
which would raise the b to the power of a and negate the result
What happens when there is a modulus sign present?
Depends what the precedence for modulus is in that language
What do i do when i have multiple open brackets e.g. cos((a + b*c)/(d + e*(f-g^h))) ?
XXxlightmarex just keep applying the rules. every left bracket goes on top of stack. when you get a right bracket remove contents of stack untill you find a left bracket, then discard that snd continue to parse the expression
work through remember operands just get written down as rpn (don't involve the stack at all)
((a + b * c) / (d + e * (f - g ^ h)))
reaching first right bracket stack contains
*
+
(
(
remove up to topmost left bracket and discard giving:
abc*+
stack only contains
(
carrying on until we reach 2nd right bracket, stack should contain:
^
-
(
*
+
(
/
(
dump stack upto topmost left bracket and discard giving us this much of the rpn:
abc*+defgh^-
stack now contains
*
+
(
/
(
parse 3rd right bracket so remove stack contents again upto topmost left bracket giving rpn of:
abc*+defgh^-*+
stack contains
/
(
process final right bracket by removing upto topmost left bracket giving rpn of:
abc*+defgh^-*+/
stack now empty
hope this helps
Thank you very much, you've been a great help
great video sir, make up for bunk classes.
+Jatin Rajput To be fair nothing makes up for bunk'd classes, it's amazing the nuances and added extra's you'll get when someone goes through a topic with you. I always find that I talk about things I never planned.
Saludos a la BUAP xd
How to implement unaries using this?
Higher precedence than multiply and divide but lower than exponent
I can't take it that the K is really an R
What happens whens 2 symbols hold the same line of precedence?
Another burgermate you can only place an operator on top if it is higher so remove the item on top of stack snd write it down in the rp expression, then see if the new operator can now go on the stack (like normal)
I see, thank you very much for this. Helped me out heaps. Was just doing past papers, came across this scenario again.
6:32 why is the left bracket of higher precedence than the power of?
Left brackets are always going to override precedence, it goes on top of everything, it's also there as a marker for when we encounter a right bracket while parsing, so we know when to stop pulling operators off the stack
@@HurrayBanana okay, the drawing in the top right corner misled me. thank you
The diagram shows that all operators go on top of left bracket if it seen on top of the stack. Parsing a left bracket in the expression doesn't look at precedence as you automatically push it onto the stack
somebody add please subtitles
why did you pop "*" before you push "/" if they are the same priority??? i dont really understand this!!
You can only push onto the stack if the precedence is higher, if it is the same you must pop the stack until the operator is higher than the contents of the top
HurrayBanana Thank you !
keke do you love me
Where is code?
saransh dhyani this process described here is reasonably easy to code up if you parse the expression first to tokenise each component