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Interview follow up

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  • L leppie

    Was given a problem that was suppose to come from the 'easy' bin. Two hours later I was still not finished... On the plus side, the interviewer attempted the problem at the same time, and also got stuck :) He said the problem must have been miscategorized. He told me to finish at home and to send the solution when done ;p Anyways, after a good nights sleep (with some wack apocalyptic dream I had), I started from scratch and coded a solution in about an hour this morning. The given problem: Binary period (dont bother googling for answers as you will likely just end up getting astronomy results or if you have google foo, a few very mathematical papers mostly related to cryptography) Given a number N larger than 0 (range was given, but does not matter), find the minimum period of repeating bit sequences or -1 if none is found. Examples

    110110110 => 3 (as 110 repeats)
    11011011 => 3 (still 3 and valid as the remaining subset, 11, is in 110)
    110110111 => -1 (no period)
    101 => -1 (no repeats, repeat implies at least 2 sequences)
    10101 => 2 (repeat at least over 2 sequences)
    11 => 1

    10101010101 => 2
    111011011101101110 => 7
    111111111111111110 => -1

    My answer in Scheme can be viewed and run here[^]. Edit: Web host seems to be down now... Edit 2: Seems to be back up again, please only look one at a time ;p Damn you shared hosting. If above link is down, view here[^].

    IronScheme
    ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

    M Offline
    M Offline
    Member 2053006
    wrote on last edited by
    #16

    Is the simplistic solution of starting with length 1 and checking each character, then increasing to length 2 etc too processor intensive? e.g. for 11011011 1 1 0 O O - Not 1, so period is not 1. K K 11 01 10 11 OK - Not 11, so period is not 2 110 110 11 All OK - Period is 3. Failing that would a FFT work on this information?

    L 1 Reply Last reply
    0
    • M Member 2053006

      Is the simplistic solution of starting with length 1 and checking each character, then increasing to length 2 etc too processor intensive? e.g. for 11011011 1 1 0 O O - Not 1, so period is not 1. K K 11 01 10 11 OK - Not 11, so period is not 2 110 110 11 All OK - Period is 3. Failing that would a FFT work on this information?

      L Offline
      L Offline
      leppie
      wrote on last edited by
      #17

      Member 2053006 wrote:

      Failing that would a FFT work on this information?

      That was my first thought, but coding one from scratch with no external help is a bit tough I think :) That is probably the most efficient way to do it.

      IronScheme
      ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

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      • J Jonathan Nethercott

        This looks like correlation (autocorrelation?) to me. I would look for a solution along the lines of: num xor (num shift left 1) num xor (num shift left 2) etc. repeat until (number of bits / 2) Not a very good bit of pseudo code, but hopefully you get the idea ;) I'm not sure how you'd deal with leading and trailing parts of a sequence, but I'm sure there must be an elegant way of incorporating that...

        Jon CodeWrite

        L Offline
        L Offline
        leppie
        wrote on last edited by
        #18

        Jon Nethercott wrote:

        I'm not sure how you'd deal with leading and trailing parts of a sequence, but I'm sure there must be an elegant way of incorporating that...

        There are no leading parts (thank FSM!). As for the trailing bit, my elegant solution was to do a negative right shift (see line 11, when rs is negative) that fits into the rest (test on line 17). Brain too tired to remember exactly what I did there now ;p

        IronScheme
        ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

        J 1 Reply Last reply
        0
        • S Septimus Hedgehog

          When will you hear if you've got the job? Best of luck.

          "I do not have to forgive my enemies, I have had them all shot." — Ramón Maria Narváez (1800-68). "I don't need to shoot my enemies, I don't have any." - Me (2012).

          L Offline
          L Offline
          leppie
          wrote on last edited by
          #19

          PHS241 wrote:

          When will you hear if you've got the job?

          He wants to have another telephonic discussion next week and I guess an offer will be made at that stage, if they so desire.

          IronScheme
          ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

          1 Reply Last reply
          0
          • L leppie

            Jon Nethercott wrote:

            I'm not sure how you'd deal with leading and trailing parts of a sequence, but I'm sure there must be an elegant way of incorporating that...

            There are no leading parts (thank FSM!). As for the trailing bit, my elegant solution was to do a negative right shift (see line 11, when rs is negative) that fits into the rest (test on line 17). Brain too tired to remember exactly what I did there now ;p

            IronScheme
            ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

            J Offline
            J Offline
            Jonathan Nethercott
            wrote on last edited by
            #20

            I think this should do it:

            private int BinaryPeriod(int num)
            {
            int result = -1;
            int numBits = (int)((Math.Log(num) / Math.Log(2))) + 1;

            for (int i = 1; i <= numBits / 2 && result == -1; i++)
            {
                if (((num & ((1 << (numBits - i)) - 1)) ^ (num >> i)) == 0)
                    result = i;
            }
            return result;
            

            }

            I've done it in C#, but hopefully it should be fairly generic code. The important line is:

            if (((num & ((1 << (numBits - i)) - 1)) ^ (num >> i)) == 0)

            which strips the top i bits and XORs that with num right shifted. If the result is 0 then we have found an autocorrelation.

            Jon CodeWrite

            L 2 Replies Last reply
            0
            • J Jonathan Nethercott

              I think this should do it:

              private int BinaryPeriod(int num)
              {
              int result = -1;
              int numBits = (int)((Math.Log(num) / Math.Log(2))) + 1;

              for (int i = 1; i <= numBits / 2 && result == -1; i++)
              {
                  if (((num & ((1 << (numBits - i)) - 1)) ^ (num >> i)) == 0)
                      result = i;
              }
              return result;
              

              }

              I've done it in C#, but hopefully it should be fairly generic code. The important line is:

              if (((num & ((1 << (numBits - i)) - 1)) ^ (num >> i)) == 0)

              which strips the top i bits and XORs that with num right shifted. If the result is 0 then we have found an autocorrelation.

              Jon CodeWrite

              L Offline
              L Offline
              leppie
              wrote on last edited by
              #21

              Very good! Wish I knew of it :(

              IronScheme
              ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

              J 1 Reply Last reply
              0
              • J Jonathan Nethercott

                I think this should do it:

                private int BinaryPeriod(int num)
                {
                int result = -1;
                int numBits = (int)((Math.Log(num) / Math.Log(2))) + 1;

                for (int i = 1; i <= numBits / 2 && result == -1; i++)
                {
                    if (((num & ((1 << (numBits - i)) - 1)) ^ (num >> i)) == 0)
                        result = i;
                }
                return result;
                

                }

                I've done it in C#, but hopefully it should be fairly generic code. The important line is:

                if (((num & ((1 << (numBits - i)) - 1)) ^ (num >> i)) == 0)

                which strips the top i bits and XORs that with num right shifted. If the result is 0 then we have found an autocorrelation.

                Jon CodeWrite

                L Offline
                L Offline
                leppie
                wrote on last edited by
                #22

                And here we have it in Scheme :) http://eval.ironscheme.net/?id=59[^] Thanks again!

                IronScheme
                ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

                1 Reply Last reply
                0
                • L leppie

                  Very good! Wish I knew of it :(

                  IronScheme
                  ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

                  J Offline
                  J Offline
                  Jonathan Nethercott
                  wrote on last edited by
                  #23

                  Thank you :) I love solving problems like this - although that has distracted me a bit from what I'm supposed to be doing ;P Good luck with the job!

                  Jon CodeWrite

                  1 Reply Last reply
                  0
                  • L leppie

                    Was given a problem that was suppose to come from the 'easy' bin. Two hours later I was still not finished... On the plus side, the interviewer attempted the problem at the same time, and also got stuck :) He said the problem must have been miscategorized. He told me to finish at home and to send the solution when done ;p Anyways, after a good nights sleep (with some wack apocalyptic dream I had), I started from scratch and coded a solution in about an hour this morning. The given problem: Binary period (dont bother googling for answers as you will likely just end up getting astronomy results or if you have google foo, a few very mathematical papers mostly related to cryptography) Given a number N larger than 0 (range was given, but does not matter), find the minimum period of repeating bit sequences or -1 if none is found. Examples

                    110110110 => 3 (as 110 repeats)
                    11011011 => 3 (still 3 and valid as the remaining subset, 11, is in 110)
                    110110111 => -1 (no period)
                    101 => -1 (no repeats, repeat implies at least 2 sequences)
                    10101 => 2 (repeat at least over 2 sequences)
                    11 => 1

                    10101010101 => 2
                    111011011101101110 => 7
                    111111111111111110 => -1

                    My answer in Scheme can be viewed and run here[^]. Edit: Web host seems to be down now... Edit 2: Seems to be back up again, please only look one at a time ;p Damn you shared hosting. If above link is down, view here[^].

                    IronScheme
                    ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

                    E Offline
                    E Offline
                    Ennis Ray Lynch Jr
                    wrote on last edited by
                    #24

                    Heh, I had to do that in University.

                    Need custom software developed? I do custom programming based primarily on MS tools with an emphasis on C# development and consulting. I also do Android Programming as I find it a refreshing break from the MS. "And they, since they Were not the one dead, turned to their affairs" -- Robert Frost

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                    • E Ennis Ray Lynch Jr

                      Heh, I had to do that in University.

                      Need custom software developed? I do custom programming based primarily on MS tools with an emphasis on C# development and consulting. I also do Android Programming as I find it a refreshing break from the MS. "And they, since they Were not the one dead, turned to their affairs" -- Robert Frost

                      L Offline
                      L Offline
                      leppie
                      wrote on last edited by
                      #25

                      Was that some kind of homework or other task? Perhaps thesis? ;p

                      IronScheme
                      ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

                      E 1 Reply Last reply
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                      • L leppie

                        Was that some kind of homework or other task? Perhaps thesis? ;p

                        IronScheme
                        ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

                        E Offline
                        E Offline
                        Ennis Ray Lynch Jr
                        wrote on last edited by
                        #26

                        Well my degree was computer science. All we did was learn stuff like this. Wouldn't say I knew much about "programming" until after school. But solving problems with code, easy.

                        Need custom software developed? I do custom programming based primarily on MS tools with an emphasis on C# development and consulting. I also do Android Programming as I find it a refreshing break from the MS. "And they, since they Were not the one dead, turned to their affairs" -- Robert Frost

                        L 1 Reply Last reply
                        0
                        • E Ennis Ray Lynch Jr

                          Well my degree was computer science. All we did was learn stuff like this. Wouldn't say I knew much about "programming" until after school. But solving problems with code, easy.

                          Need custom software developed? I do custom programming based primarily on MS tools with an emphasis on C# development and consulting. I also do Android Programming as I find it a refreshing break from the MS. "And they, since they Were not the one dead, turned to their affairs" -- Robert Frost

                          L Offline
                          L Offline
                          leppie
                          wrote on last edited by
                          #27

                          Not even in my 2nd year course (just before I decided to go back to working class) we did stuff like this... I left when when we were busy with Data Structures and Algorithms.

                          IronScheme
                          ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

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