Logarithmic run time for calculating prime numbers?

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Here's the function I'm currently analyzing. I know it's not the most optimal but I'm not understanding the $theta()$ of this algorithm. I've been told that it's not actually $theta(n)$ but instead a logarithmic run-time and it had something to do with the bits of the value passed in. I'm not understanding how the bits are causing the logarithmic so if someone can explain that it'd be great.



 function isPrime(n):
1 if n = 2 return true
2 for i from 2 to n
3 if n % i = 0 return false
4 return true





runtime-analysis primes







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    up vote
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    down vote

    favorite












    Here's the function I'm currently analyzing. I know it's not the most optimal but I'm not understanding the $theta()$ of this algorithm. I've been told that it's not actually $theta(n)$ but instead a logarithmic run-time and it had something to do with the bits of the value passed in. I'm not understanding how the bits are causing the logarithmic so if someone can explain that it'd be great.



     function isPrime(n):
    1 if n = 2 return true
    2 for i from 2 to n
    3 if n % i = 0 return false
    4 return true





    runtime-analysis primes







    share|cite|improve this question









    New contributor




    Justin Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Here's the function I'm currently analyzing. I know it's not the most optimal but I'm not understanding the $theta()$ of this algorithm. I've been told that it's not actually $theta(n)$ but instead a logarithmic run-time and it had something to do with the bits of the value passed in. I'm not understanding how the bits are causing the logarithmic so if someone can explain that it'd be great.



       function isPrime(n):
      1 if n = 2 return true
      2 for i from 2 to n
      3 if n % i = 0 return false
      4 return true





      runtime-analysis primes







      share|cite|improve this question









      New contributor




      Justin Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      Here's the function I'm currently analyzing. I know it's not the most optimal but I'm not understanding the $theta()$ of this algorithm. I've been told that it's not actually $theta(n)$ but instead a logarithmic run-time and it had something to do with the bits of the value passed in. I'm not understanding how the bits are causing the logarithmic so if someone can explain that it'd be great.



       function isPrime(n):
      1 if n = 2 return true
      2 for i from 2 to n
      3 if n % i = 0 return false
      4 return true





      runtime-analysis primes




      runtime-analysis primes






      share|cite|improve this question









      New contributor




      Justin Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Justin Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 7 hours ago









      David Richerby

      62.3k1595179




      62.3k1595179






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      asked 8 hours ago









      Justin Li

      61




      61




      New contributor




      Justin Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Justin Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Justin Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.




















          2 Answers
          2






          active

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          up vote
          2
          down vote













          Let's use a different parameter for your input, $m$. The number of instructions that your algorithm executes in the worst case is $O(m)$.



          We often measure the running time of algorithms in terms of the length of their input (measured either in bits or in words), often denoted by $n$. The length of your input is $n approx log_2 m$, since this is how long it takes to encode the integer $m$ in binary. Therefore your algorithm actually performs an exponential number of instructions $O(2^n)$.



          A different issue is the cost of arithmetic operations, such as the modulo operation on line 3. In most models of computation (and in real life), the cost of such an operation depends on the length of the numbers being operated on. Therefore your algorithm runs in time which is worse than $2^n$ by a polynomial factor $n^O(1)$ (the exact polynomial factor depends on the algorithm you use to perform the division). Such a running time is usually denoted $tildeO(2^n)$.






          share|cite|improve this answer




















          • So does O(m) have anything to do with the O(2^n) at all or are they just 2 ways of measuring the algorithm. And how does the encoding work exactly, I didn't think it was necessary to count from 0 all the way to n in binary
            – Justin Li
            3 hours ago











          • These are two ways to measure the running time. As for the encoding, it’s really an implementation detail.
            – Yuval Filmus
            18 mins ago

















          up vote
          1
          down vote













          Well, first, the algorithm is incorrect: it claims that every integer less than or equal to $2$ is prime, and every integer greater than–$2$ is composite.



          • Any $nleq1$ fails the test at line $1$, does zero iterations of the loope at lines $2$–$3$ and returns true at line $4$.


          • $n=2$ triggers the return true on line $1$.

          • Any $n>2$ fails the test at line $1$. If it is composite, it returns false at line $3$ for its smallest prime factor; if it is prime, then the loop will get as far as $i=n$, at which point the n%i=0 is true, so we return false at line $3$.

          So, suppose we correct line $2$ to be for i=2 to n-1 and assume the input is greater than $1$. Then, for any prime input $n>2$, the loop at lines $2$–$3$ will run for each value $2, dots, n-1$, which is $n-2 = Theta(n)$ iterations. So your suspicion that the running time is $Theta(n)$ is correct: this algorithm is not logarithmic.



          In general, as Yuval explains, we measure the running time of algorithms in terms of the length of the input as written in binary. Since the number $n$ takes $Theta(log n)$ bits to write down, and the running time of the algorithm is $Theta(n) = Theta(2^textinput length)$, this algorithm would normally be described as having exponential running time.






          share|cite|improve this answer




















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote













            Let's use a different parameter for your input, $m$. The number of instructions that your algorithm executes in the worst case is $O(m)$.



            We often measure the running time of algorithms in terms of the length of their input (measured either in bits or in words), often denoted by $n$. The length of your input is $n approx log_2 m$, since this is how long it takes to encode the integer $m$ in binary. Therefore your algorithm actually performs an exponential number of instructions $O(2^n)$.



            A different issue is the cost of arithmetic operations, such as the modulo operation on line 3. In most models of computation (and in real life), the cost of such an operation depends on the length of the numbers being operated on. Therefore your algorithm runs in time which is worse than $2^n$ by a polynomial factor $n^O(1)$ (the exact polynomial factor depends on the algorithm you use to perform the division). Such a running time is usually denoted $tildeO(2^n)$.






            share|cite|improve this answer




















            • So does O(m) have anything to do with the O(2^n) at all or are they just 2 ways of measuring the algorithm. And how does the encoding work exactly, I didn't think it was necessary to count from 0 all the way to n in binary
              – Justin Li
              3 hours ago











            • These are two ways to measure the running time. As for the encoding, it’s really an implementation detail.
              – Yuval Filmus
              18 mins ago














            up vote
            2
            down vote













            Let's use a different parameter for your input, $m$. The number of instructions that your algorithm executes in the worst case is $O(m)$.



            We often measure the running time of algorithms in terms of the length of their input (measured either in bits or in words), often denoted by $n$. The length of your input is $n approx log_2 m$, since this is how long it takes to encode the integer $m$ in binary. Therefore your algorithm actually performs an exponential number of instructions $O(2^n)$.



            A different issue is the cost of arithmetic operations, such as the modulo operation on line 3. In most models of computation (and in real life), the cost of such an operation depends on the length of the numbers being operated on. Therefore your algorithm runs in time which is worse than $2^n$ by a polynomial factor $n^O(1)$ (the exact polynomial factor depends on the algorithm you use to perform the division). Such a running time is usually denoted $tildeO(2^n)$.






            share|cite|improve this answer




















            • So does O(m) have anything to do with the O(2^n) at all or are they just 2 ways of measuring the algorithm. And how does the encoding work exactly, I didn't think it was necessary to count from 0 all the way to n in binary
              – Justin Li
              3 hours ago











            • These are two ways to measure the running time. As for the encoding, it’s really an implementation detail.
              – Yuval Filmus
              18 mins ago












            up vote
            2
            down vote










            up vote
            2
            down vote









            Let's use a different parameter for your input, $m$. The number of instructions that your algorithm executes in the worst case is $O(m)$.



            We often measure the running time of algorithms in terms of the length of their input (measured either in bits or in words), often denoted by $n$. The length of your input is $n approx log_2 m$, since this is how long it takes to encode the integer $m$ in binary. Therefore your algorithm actually performs an exponential number of instructions $O(2^n)$.



            A different issue is the cost of arithmetic operations, such as the modulo operation on line 3. In most models of computation (and in real life), the cost of such an operation depends on the length of the numbers being operated on. Therefore your algorithm runs in time which is worse than $2^n$ by a polynomial factor $n^O(1)$ (the exact polynomial factor depends on the algorithm you use to perform the division). Such a running time is usually denoted $tildeO(2^n)$.






            share|cite|improve this answer












            Let's use a different parameter for your input, $m$. The number of instructions that your algorithm executes in the worst case is $O(m)$.



            We often measure the running time of algorithms in terms of the length of their input (measured either in bits or in words), often denoted by $n$. The length of your input is $n approx log_2 m$, since this is how long it takes to encode the integer $m$ in binary. Therefore your algorithm actually performs an exponential number of instructions $O(2^n)$.



            A different issue is the cost of arithmetic operations, such as the modulo operation on line 3. In most models of computation (and in real life), the cost of such an operation depends on the length of the numbers being operated on. Therefore your algorithm runs in time which is worse than $2^n$ by a polynomial factor $n^O(1)$ (the exact polynomial factor depends on the algorithm you use to perform the division). Such a running time is usually denoted $tildeO(2^n)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 7 hours ago









            Yuval Filmus

            183k12171332




            183k12171332











            • So does O(m) have anything to do with the O(2^n) at all or are they just 2 ways of measuring the algorithm. And how does the encoding work exactly, I didn't think it was necessary to count from 0 all the way to n in binary
              – Justin Li
              3 hours ago











            • These are two ways to measure the running time. As for the encoding, it’s really an implementation detail.
              – Yuval Filmus
              18 mins ago
















            • So does O(m) have anything to do with the O(2^n) at all or are they just 2 ways of measuring the algorithm. And how does the encoding work exactly, I didn't think it was necessary to count from 0 all the way to n in binary
              – Justin Li
              3 hours ago











            • These are two ways to measure the running time. As for the encoding, it’s really an implementation detail.
              – Yuval Filmus
              18 mins ago















            So does O(m) have anything to do with the O(2^n) at all or are they just 2 ways of measuring the algorithm. And how does the encoding work exactly, I didn't think it was necessary to count from 0 all the way to n in binary
            – Justin Li
            3 hours ago





            So does O(m) have anything to do with the O(2^n) at all or are they just 2 ways of measuring the algorithm. And how does the encoding work exactly, I didn't think it was necessary to count from 0 all the way to n in binary
            – Justin Li
            3 hours ago













            These are two ways to measure the running time. As for the encoding, it’s really an implementation detail.
            – Yuval Filmus
            18 mins ago




            These are two ways to measure the running time. As for the encoding, it’s really an implementation detail.
            – Yuval Filmus
            18 mins ago










            up vote
            1
            down vote













            Well, first, the algorithm is incorrect: it claims that every integer less than or equal to $2$ is prime, and every integer greater than–$2$ is composite.



            • Any $nleq1$ fails the test at line $1$, does zero iterations of the loope at lines $2$–$3$ and returns true at line $4$.


            • $n=2$ triggers the return true on line $1$.

            • Any $n>2$ fails the test at line $1$. If it is composite, it returns false at line $3$ for its smallest prime factor; if it is prime, then the loop will get as far as $i=n$, at which point the n%i=0 is true, so we return false at line $3$.

            So, suppose we correct line $2$ to be for i=2 to n-1 and assume the input is greater than $1$. Then, for any prime input $n>2$, the loop at lines $2$–$3$ will run for each value $2, dots, n-1$, which is $n-2 = Theta(n)$ iterations. So your suspicion that the running time is $Theta(n)$ is correct: this algorithm is not logarithmic.



            In general, as Yuval explains, we measure the running time of algorithms in terms of the length of the input as written in binary. Since the number $n$ takes $Theta(log n)$ bits to write down, and the running time of the algorithm is $Theta(n) = Theta(2^textinput length)$, this algorithm would normally be described as having exponential running time.






            share|cite|improve this answer
























              up vote
              1
              down vote













              Well, first, the algorithm is incorrect: it claims that every integer less than or equal to $2$ is prime, and every integer greater than–$2$ is composite.



              • Any $nleq1$ fails the test at line $1$, does zero iterations of the loope at lines $2$–$3$ and returns true at line $4$.


              • $n=2$ triggers the return true on line $1$.

              • Any $n>2$ fails the test at line $1$. If it is composite, it returns false at line $3$ for its smallest prime factor; if it is prime, then the loop will get as far as $i=n$, at which point the n%i=0 is true, so we return false at line $3$.

              So, suppose we correct line $2$ to be for i=2 to n-1 and assume the input is greater than $1$. Then, for any prime input $n>2$, the loop at lines $2$–$3$ will run for each value $2, dots, n-1$, which is $n-2 = Theta(n)$ iterations. So your suspicion that the running time is $Theta(n)$ is correct: this algorithm is not logarithmic.



              In general, as Yuval explains, we measure the running time of algorithms in terms of the length of the input as written in binary. Since the number $n$ takes $Theta(log n)$ bits to write down, and the running time of the algorithm is $Theta(n) = Theta(2^textinput length)$, this algorithm would normally be described as having exponential running time.






              share|cite|improve this answer






















                up vote
                1
                down vote










                up vote
                1
                down vote









                Well, first, the algorithm is incorrect: it claims that every integer less than or equal to $2$ is prime, and every integer greater than–$2$ is composite.



                • Any $nleq1$ fails the test at line $1$, does zero iterations of the loope at lines $2$–$3$ and returns true at line $4$.


                • $n=2$ triggers the return true on line $1$.

                • Any $n>2$ fails the test at line $1$. If it is composite, it returns false at line $3$ for its smallest prime factor; if it is prime, then the loop will get as far as $i=n$, at which point the n%i=0 is true, so we return false at line $3$.

                So, suppose we correct line $2$ to be for i=2 to n-1 and assume the input is greater than $1$. Then, for any prime input $n>2$, the loop at lines $2$–$3$ will run for each value $2, dots, n-1$, which is $n-2 = Theta(n)$ iterations. So your suspicion that the running time is $Theta(n)$ is correct: this algorithm is not logarithmic.



                In general, as Yuval explains, we measure the running time of algorithms in terms of the length of the input as written in binary. Since the number $n$ takes $Theta(log n)$ bits to write down, and the running time of the algorithm is $Theta(n) = Theta(2^textinput length)$, this algorithm would normally be described as having exponential running time.






                share|cite|improve this answer












                Well, first, the algorithm is incorrect: it claims that every integer less than or equal to $2$ is prime, and every integer greater than–$2$ is composite.



                • Any $nleq1$ fails the test at line $1$, does zero iterations of the loope at lines $2$–$3$ and returns true at line $4$.


                • $n=2$ triggers the return true on line $1$.

                • Any $n>2$ fails the test at line $1$. If it is composite, it returns false at line $3$ for its smallest prime factor; if it is prime, then the loop will get as far as $i=n$, at which point the n%i=0 is true, so we return false at line $3$.

                So, suppose we correct line $2$ to be for i=2 to n-1 and assume the input is greater than $1$. Then, for any prime input $n>2$, the loop at lines $2$–$3$ will run for each value $2, dots, n-1$, which is $n-2 = Theta(n)$ iterations. So your suspicion that the running time is $Theta(n)$ is correct: this algorithm is not logarithmic.



                In general, as Yuval explains, we measure the running time of algorithms in terms of the length of the input as written in binary. Since the number $n$ takes $Theta(log n)$ bits to write down, and the running time of the algorithm is $Theta(n) = Theta(2^textinput length)$, this algorithm would normally be described as having exponential running time.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 7 hours ago









                David Richerby

                62.3k1595179




                62.3k1595179




















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