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Calculate the running time of the code snippet below

0

$begingroup$

for (i=2*n; i>=1; i=i-1)
 for (j=1; j<=i; j=j+1)
 for (k=1; k<=j; k=k*3)
 print(“hello”)

I sopused that its N^5 but Im not sure

recursive-algorithms

share|cite|improve this question

edited Mar 28 at 15:25

Julián Aguirre

69.5k24297

asked Mar 27 at 19:37

Harlinton Palacios MosqueraHarlinton Palacios Mosquera

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Example:stackoverflow.com/questions/18486543/…
  $endgroup$
  – NoChance
  Mar 27 at 23:12
  

  
  
  
  

add a comment | 

0

$begingroup$

for (i=2*n; i>=1; i=i-1)
 for (j=1; j<=i; j=j+1)
 for (k=1; k<=j; k=k*3)
 print(“hello”)

I sopused that its N^5 but Im not sure

recursive-algorithms

share|cite|improve this question

edited Mar 28 at 15:25

Julián Aguirre

69.5k24297

asked Mar 27 at 19:37

Harlinton Palacios MosqueraHarlinton Palacios Mosquera

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Example:stackoverflow.com/questions/18486543/…
  $endgroup$
  – NoChance
  Mar 27 at 23:12
  

  
  
  
  

add a comment | 

$begingroup$

for (i=2*n; i>=1; i=i-1)
 for (j=1; j<=i; j=j+1)
 for (k=1; k<=j; k=k*3)
 print(“hello”)

I sopused that its N^5 but Im not sure

recursive-algorithms

share|cite|improve this question

edited Mar 28 at 15:25

Julián Aguirre

69.5k24297

asked Mar 27 at 19:37

Harlinton Palacios MosqueraHarlinton Palacios Mosquera

1

$endgroup$

for (i=2*n; i>=1; i=i-1)
 for (j=1; j<=i; j=j+1)
 for (k=1; k<=j; k=k*3)
 print(“hello”)

share|cite|improve this question

edited Mar 28 at 15:25

Julián Aguirre

69.5k24297

asked Mar 27 at 19:37

Harlinton Palacios MosqueraHarlinton Palacios Mosquera

1

share|cite|improve this question

edited Mar 28 at 15:25

Julián Aguirre

69.5k24297

asked Mar 27 at 19:37

Harlinton Palacios MosqueraHarlinton Palacios Mosquera

1

edited Mar 28 at 15:25

Julián Aguirre

69.5k24297

edited Mar 28 at 15:25

Julián Aguirre

69.5k24297

asked Mar 27 at 19:37

Harlinton Palacios MosqueraHarlinton Palacios Mosquera

1

asked Mar 27 at 19:37

Harlinton Palacios MosqueraHarlinton Palacios Mosquera

1

1 Answer
1

active

oldest

votes

0

$begingroup$

Let's try it. The total number of iterations is given by:

$$I=sum_i=1^2nsum_j=1^isum_k=1^lfloor log_3(j) rfloor 1approxsum_i=1^2nsum_j=1^i log(j) approx
sum_i=1^2nlog(i!)underbraceapprox_textStirling sum_i=1^2nilog(i)underbraceapprox_textSum approx textIntegral n^2log(n) $$

So $O(n^2 log(n))$, but i'm not so keen on this type of exercise :

share|cite|improve this answer

edited Mar 27 at 22:40

answered Mar 27 at 21:49

EurekaEureka

907115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd say more O(n^2 log(n)) because the k=k*3 in the last loop
  $endgroup$
  – Robin Nicole
  Mar 27 at 22:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right, what a blunder! I thought for strange reason that it took all of the 3 multiples. Thank you :)
  $endgroup$
  – Eureka
  Mar 27 at 22:29
  

  
  
  
  

add a comment | 

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0

$begingroup$

Let's try it. The total number of iterations is given by:

$$I=sum_i=1^2nsum_j=1^isum_k=1^lfloor log_3(j) rfloor 1approxsum_i=1^2nsum_j=1^i log(j) approx
sum_i=1^2nlog(i!)underbraceapprox_textStirling sum_i=1^2nilog(i)underbraceapprox_textSum approx textIntegral n^2log(n) $$

So $O(n^2 log(n))$, but i'm not so keen on this type of exercise :

share|cite|improve this answer

edited Mar 27 at 22:40

answered Mar 27 at 21:49

EurekaEureka

907115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd say more O(n^2 log(n)) because the k=k*3 in the last loop
  $endgroup$
  – Robin Nicole
  Mar 27 at 22:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right, what a blunder! I thought for strange reason that it took all of the 3 multiples. Thank you :)
  $endgroup$
  – Eureka
  Mar 27 at 22:29
  

  
  
  
  

add a comment | 

0

$begingroup$

Let's try it. The total number of iterations is given by:

$$I=sum_i=1^2nsum_j=1^isum_k=1^lfloor log_3(j) rfloor 1approxsum_i=1^2nsum_j=1^i log(j) approx
sum_i=1^2nlog(i!)underbraceapprox_textStirling sum_i=1^2nilog(i)underbraceapprox_textSum approx textIntegral n^2log(n) $$

So $O(n^2 log(n))$, but i'm not so keen on this type of exercise :

share|cite|improve this answer

edited Mar 27 at 22:40

answered Mar 27 at 21:49

EurekaEureka

907115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd say more O(n^2 log(n)) because the k=k*3 in the last loop
  $endgroup$
  – Robin Nicole
  Mar 27 at 22:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right, what a blunder! I thought for strange reason that it took all of the 3 multiples. Thank you :)
  $endgroup$
  – Eureka
  Mar 27 at 22:29
  

  
  
  
  

add a comment | 

$begingroup$

Let's try it. The total number of iterations is given by:

$$I=sum_i=1^2nsum_j=1^isum_k=1^lfloor log_3(j) rfloor 1approxsum_i=1^2nsum_j=1^i log(j) approx
sum_i=1^2nlog(i!)underbraceapprox_textStirling sum_i=1^2nilog(i)underbraceapprox_textSum approx textIntegral n^2log(n) $$

So $O(n^2 log(n))$, but i'm not so keen on this type of exercise :

share|cite|improve this answer

edited Mar 27 at 22:40

answered Mar 27 at 21:49

EurekaEureka

907115

$endgroup$

share|cite|improve this answer

edited Mar 27 at 22:40

answered Mar 27 at 21:49

EurekaEureka

907115

answered Mar 27 at 21:49

EurekaEureka

907115

answered Mar 27 at 21:49

EurekaEureka

907115