GATE CSE 2010 | Question: 37

Question

The program below uses six temporary variables $a, b, c, d, e, f$.

a = 1 
b = 10
c = 20 
d = a + b
e = c + d 
f = c + e 
b = c + e 
e = b + f 
d = 5 + e
return d + f 

Assuming that all operations take their operands from registers, what is the minimum number of registers needed to execute this program without spilling?

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Comments

@jatinmittal199510   can you explain how you write number on nodes abcdf….

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@jugnu1337 this is nothing but graph coloring.And whatever coloring he has done is wrong..You try to color this graph , you will get ans as 3. 

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A little Error correction, you were updating the R2 value @e=f+b , so we would loose the f/b value @R2, so we better bring R3 here itself, so that we can use the f/b from R2 at last step

Answer 1

Here in these types of compiler questions, idea is "map/assign multiple temporaries to one registers."

here $a, b,$ and $c$ all are having $3$ different values so i need atleast $3$ registers $r1$, $r2$ and $r3$.
$a$ is mapped to $r1$, $b$ to $r2$ and $c$ to $r3$.

$d = a + b$, after this line if u notice '$a$' is never present on right hand side, so I can map (register of $a$ which is $r1$ ) $d$ to $r1$.
$e = c + d$, after this line '$d$' is never present on rhs, so I can map (register of $d$ which is $r1$ ) $e$ to $r1$.

at this time mapping is
$r1 --- e$
$r2 --- b$
$r3 --- c$

(at this moment I have registers for $e, b$ and $c$. if I introduce new variable then I may need different register)
now at this point if u see
$f = c + e$
$b = c + e$
these two are essentially doing same thing, after these two line '$b$' and '$f$' are same so I can skip computing '$f$'. and whereever $f$ is present I will replace it with '$b$'. (because neither of '$f$' and '$b$' are changing after these two lines, so value of these will be '$c+e$' forever)

(seems like I introduced one more variable $f$, and register is needed for that, but actually I did not really introduce '$f$'. I am skipping computation of '$f$')
now at second last line "$d = 5 + e$"
here I introduced '$d$', I can map it to any of the register $r1$ or $r3$, because after this line neither of '$e$' or '$c$' is required. (value of '$b$' is required because I need to return '$d+f$', and '$f$' is essentially equal to '$b$')

finally code becomes

$r1 = 1$
$r2 = 10$
$r3 = 20$
$r1 = r1 + r2$
$r1 = r3 + r1$

(skipping '$f$' computation)
$r2 = r3 + r1$
$r2 = r3 + r1$
$r1 = r2 + r2$
$r3 = 5 + r1$
return $r3 + r2$

Therefore minimum $3$ registers needed.

Correct Answer: $B$

Comments

@sachin

So, removing $f = c + e$ or  $b = c + e$ looks sensible here, as both are same.

So, if same lines come, we can safely remove in the exam and replacing that $f$ by $b$ or $b$ by $f$ in other lines also, right ?

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Depends.
Here value of b and f are not changing anywhere after that assignment and through out the program f and b are same. So i can supply value of b whenever some one requires value of f.
If these two values are changing later in the program, then u need to take extra precaution for that. U can use same register upto whereever these values are same, and u may need different register later.
I am not expert in this :D But this makes sense :)

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You are right and same precaution we take while drawing DAG. Anyway, this method suits here best :-)

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I just have a small doubt ::

Why are you assigning 3 different registers to a,b,and c initially.

In order to compute d=a+b we need only 'a' and 'b' and later in the program nowhere we are using the initial values of 'a' and 'b'.

My point is that , what we can do is assign 'a' and 'b' to 2 different registers initially.And instead of immediately assigning third register to 'c' . We can overwrite one of the registers used for 'a' or 'b'.

Later we may introduce a third register when it is absolutely needed.

Main idea is to avoid using new registers and reuse the old ones as long as possible.

Although even if we compute the way I am suggesting the ans would remain the same.But going the way i suggest will avoid us ending with more number of registers than absolutely necessary.

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Are the instructions required to be executed in sequential order? Doesn't the code Optimizer re-shuffle instructions in order to use less no of registers? Bcz if we do a bit re-shuffle it can be done using two registers.

Should we assume that in this type of questions code optimizer be there?

I did this assuming code being optimized:

a=1

b=10

b=a+b=11

a=20

b=a+b=31

a=a+b=51

b=a=51

b=a+b=102

b=5+b

return b+a

Where am i wrong?Is it the assumptions?

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here instead of using the same value of f for b we can temporarily use the register of b for f as the old value of b is not used again and then b value is over written by c+e and then we can put b in place of register c and then the live variables will be b,f,e and then onewards we can continue .. i dont think the compiler will think intelligently to reuse the same value of b instead of f so this should be right approach

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@VS please look into the below snippet:-

It says only we require 2 registers and it is quite logical. But my question is why we require 3 registers at all

a = 1        ===>>   R1=a;
b = 10       ===>>   R2=b;
c = 20       ===>>   
d = a + b    ===>>   R1{D}=R1{A}+R2{B};
e = c + d    ===>>   R2=c;   R1{E}=R1{D}+R2{C};
f = c + e    ===>>   R1{F}=R2{C}+R1{E};
b = c + e    ===>>   R2{B}=R1{F};
e = b + f    ===>>   R2{E}=R2{B}+R1{F};
d = 5 + e    ===>>   R2{D}=5{CONST}+R2{E};
return d + f ===>>   RETURN R2{D}+R1{F};

Can you please look and elaborate it..

Thanks in advance ...!

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someone please tell what is wrong in above approach by @vinil .Am getting same

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Yes. I think we only need 2 registers.

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@Vinil-

But you have changed order of instructions.

The instruction that loads c from memory is at step 3, but now you have moved to step 5.

The questions says we have to execute the given program, but here we are executing the "modified" version of program.

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But that really matter because  question is about minimum number of registers right ??

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@Tuhin Dutta @Vinil @rahul sharma

Yes, Code motion is one of the compiler optimizations, we can use it !

Ref : https://gateoverflow.in/1556/gate2013-48

I think every body is doing the same mistake.

Read the question carefully,

Assuming that all operations take their operands from registers, what is the minimum number of registers needed to execute this program without spilling?

d = 5 + e    =>>   R2{D}=5{OPERAND}+R2{E}; We need Register R3 for Value 5 as it is an operand
return d + f ===>>   RETURN R2{D}+R1{F};

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Someone please show how to draw the  DAG for it

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So What is final answer 2 or 3?

Can someone please look at this

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how is it possible? when compiler will look  'c' and'e' for b=c+e ..then how we can tell to compiler to take b=f ?... instead allocating b's register to f and then same register to b may be an approach.  am i right?

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what does "program without spilling" means?

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When a register content is saved to memory and then reloaded again it constitutes a spill. Basically reducing spills enhances the running speed of an application as register is faster than memory (including cache)

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 We need Register R3 for Value 5 as it is an operand

Isn't 5 an immediate value? How can we say it is an operand? I thought operand is anything which is a variable. @VS

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When i traced back for "d+f," i got the return value as "6c+3a+3b+5"

can i use this expression to solve the problem..

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Why we are not storing 5 also in some register??

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Here we need to apply greedy method to avoid register spilling.

Live variable and live ranges concept comes handy here.

 

https://web.stanford.edu/class/archive/cs/cs143/cs143.1128/lectures/17/Slides17.pdf

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@vinil you need to first keep $5$ in a register than only you can add it to $e$ as the question states that ALL the operands are taken from register only. Once you fix this, yoh will get answer as $3$.

We can also solve this using Sethi-Ullman Algorithm and will get the same answer as $3$.

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/*~~~~~~~~~~~~~~
a = 1 
b = 10
c = 20 
d = a + b
e = c + d 
f = c + e 
b = c + e 
e = b + f 
d = 5 + e
return d + f
*/

R1 := 1         // a
R2 := 10        // b
R1 := R1+R2     // d
R2 := 20        // c
R1 := R1+R2     // e
R1 := R2+R1     // f
R2 := R1        // b
R2 := R1+R2     // e
R3 := 5         // Since CPU requires each operand to be in registers
R2 := R3+R2     // d
R2 := R2+R1     // d+f, the return value

/* 3 registers are required */

 

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Great explanation sir 👏

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@Deepak Poonia Sir , @Arjun Sir @Sachin Mittal 1 Sir is this correct?

I’m getting option A) $2$ registers.

If we break the expression from bottom to top, we will get:-

$d+f$

$=5+e+f$

$=5+b+2f$

$= 5+c+e+2f$

$= 5+c+e+2(c+e)$

$= 5+3(c+e)$

$= 5+3(2c+d)$

$=5+3(2c+a+b)$

$=5+6*c+3(a+b)$

So,

finally code becomes

$r1=1$ 
$r1=r1+10$                  // $r1 = a+b$ now
$r2=20$                       //$r2 = c$ now
$r1 = 3*r1$                 // $r1 = 3(a+b)$ now
$r2=6*r2$                  // $r2 = 6c$ now

$r1 = r1+r2$              // $r1 = 6c+3(a+b)$ now

$r1 = r1+5$              // $r1 = 5+6c+3(a+b)$ now

So we need only $2$ registers.

 

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here \(a\), \(b\), and \(c\) all are having \(3\) different values so i need atleast \(3\) registers \(r1\), \(r2\) and \(r3\).

@Sachin Mittal 1 Sir, would this always be true? For example – suppose that the code was such that after all the assignment statements to the variables \(a\), \(b\) and \(c\), only the variable \(b\) was never again used later in the code. Then, during the compiler's code optimization phase, the variable \(b\)’s assignment statement is removed from the code (dead code elimination), and in which case we would require only two registers, though three variables are getting initialized with different values, right?

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@Sachin Mittal 1 @GO Classes  Why we can’t just store value of f in register containing original b,as original b value will not be used again?

So without ignoring any step we can do it with 3 registers.

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@Abhrajyoti00 ig if we change the initial values of a,b,c then your new code wont work as you have hardwired the initial value of b to 10 by writing r1 = r1 +10 . Therefore only 2 registers wont do .

Answer 2

All of the given expressions use at-most 3 variables, so we never nee more than 3 registers.

See  http://en.wikipedia.org/wiki/Register_allocation

It requires minimum 3 registers.

Principle of Register Allocation : If a variable needs to be allocated to a register, the system checks for any free register available, if it finds one, it allocates. If there is no free register, then it checks for a register that contains a dead variable ( a variable whose value is not going to be used in future ), and if it finds one then it allocates. Otherwise it goes for Spilling ( it checks for a register whose value is needed after the longest time, saves its value into the memory, and then use that register for current allocation, later when the old value of the register is needed, the system gets it from the memory where it was saved and allocate it in any register which is available ).

But here we should not apply spilling as directed in the question.

Let’s allocate the registers for the variables.

a = 1 ( let’s say register R1 is allocated for variable ‘a’ )

b = 10 ( R2 for ‘b’ , because value of ‘a’ is going to be used in the future, hence can not replace variable of ‘a’ by that of ‘b’ in R1)

c = 20 ( R3 for ‘c’, because values of ‘a’ and ‘b’ are going to be used in the future, hence can not replace variable ‘a’ or ‘b’ by ‘c’ in R1 or R2 respectively)

d = a+b ( now, ‘d’ can be assigned to R1 because R1 contains dead variable which is ‘a’ and it is so called because it is not going to be used in future, i.e. no subsequent expression uses the value of variable ‘a’)

e = c+d ( ‘e’ can be assigned to R1, because currently R1 contains value of varibale ‘d’ which is not going to be used in the subsequent expression.)

Note: an already calculated value of a variable is used only by READ operation ( not WRITE), hence we have to see only on the RHS side of the subsequent expressions that whether the variable is going to be used or not.

f = c+e ( ‘ f ‘ can be assigned to R2, because vaule of ‘b’ in register R2 is not going to be used in subsequent expressions, hence R2 can be used to allocate for ‘ f ‘ replacing ‘b’ )

b = c+e ( ‘ b ‘ can be assigned to R3, because value of ‘c’ in R3 is not being used later )

e = b+f ( here ‘e’ is already in R1, so no allocation here, direct assignment )

d = 5+e ( ‘d’ can be assigned to either R1 or R3, because values in both are not used further, let’s assign in R1 )

return d+f ( no allocation here, simply contents of registers R1 and R2 are added and returned)

hence we need only 3 registers, R1 R2 and R3.

ref-http://quiz.geeksforgeeks.org/gate-gate-cs-2010-question-37/

Comments

simple and clear,great job....

Answer 3

After making the interference graph it can be colored with 3 different colors. Therefore minimum number of color needed to execute the program without spilling would be 3 (B).

Comments

with interfacing(DAG) diagram also it will take only 2 registers

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See this:  https://web.stanford.edu/class/archive/cs/cs143/cs143.1128/lectures/17/Slides17.pdf

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Thanks Nirmal that was helpful.

Answer 4

3 registers.

here,

d = a + b
e = c + (a+b) 
f = c + (c+ (a+b)) 
b = c + (c+ (a+b) = f 
e = b + (c+(c+(a+b))) 
d = 5 + (b+(c+(c+(a+b))))
return d + f 

first evaluate b using two registers and then store its value in f because f and b are equal.

now evaluate d with two registers. then return d+f

hence 3 registers.

Comments

thanx @ jayendra

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In the step d=5+e 5 should be in a register as it is given in the question that all operations take their operands from registers

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@ Aishwarya Gujrathi 1
Operands should be in Memory not constant values its like immediate adressing mode right?