Javascript while loop count higher than expected











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0
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I'm writing a function which takes in a positive number and returns the number of times you must multiply the digits in num until you reach a single digit.



example:



test(39) === 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
// and 4 has only one digit//if single digit return num


The code:



function test(num) {
if (num > 10) { return num} else {
var j = num;var a;var count = 0;
while ( j > 10){
a = num.toString().split('').map( e=> parseInt(e)).reduce((a,c)=> a*c);
num = a;
count++;
j--;
}
return count;
}}

test(39) //outputs 29 expected 3;


I have fixed the above by adding an array and filtering for unique values but would still like to know why the code is giving me a much higher count than expected.










share|improve this question




















  • 1




    If num is already below 10, shouldn't it return 0 and not the number itself? I mean you describe your function as returning the number of iterations, not a result from a calculation.
    – Lennholm
    Nov 20 at 10:21






  • 2




    The j-- part is the bit that's making the result far too high. You can add console.log(a,num,j) inside the while to see what it's doing (or step through with the debugger). Also return num at the start makes no sense, should be return 0 as num<10 means it's already single digit so doesn't need to run your algorithm.
    – freedomn-m
    Nov 20 at 10:24












  • You're right,dunno why i even put that j part and forgot about it.Works without it.
    – dan brown
    Nov 20 at 11:44















up vote
0
down vote

favorite












I'm writing a function which takes in a positive number and returns the number of times you must multiply the digits in num until you reach a single digit.



example:



test(39) === 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
// and 4 has only one digit//if single digit return num


The code:



function test(num) {
if (num > 10) { return num} else {
var j = num;var a;var count = 0;
while ( j > 10){
a = num.toString().split('').map( e=> parseInt(e)).reduce((a,c)=> a*c);
num = a;
count++;
j--;
}
return count;
}}

test(39) //outputs 29 expected 3;


I have fixed the above by adding an array and filtering for unique values but would still like to know why the code is giving me a much higher count than expected.










share|improve this question




















  • 1




    If num is already below 10, shouldn't it return 0 and not the number itself? I mean you describe your function as returning the number of iterations, not a result from a calculation.
    – Lennholm
    Nov 20 at 10:21






  • 2




    The j-- part is the bit that's making the result far too high. You can add console.log(a,num,j) inside the while to see what it's doing (or step through with the debugger). Also return num at the start makes no sense, should be return 0 as num<10 means it's already single digit so doesn't need to run your algorithm.
    – freedomn-m
    Nov 20 at 10:24












  • You're right,dunno why i even put that j part and forgot about it.Works without it.
    – dan brown
    Nov 20 at 11:44













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I'm writing a function which takes in a positive number and returns the number of times you must multiply the digits in num until you reach a single digit.



example:



test(39) === 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
// and 4 has only one digit//if single digit return num


The code:



function test(num) {
if (num > 10) { return num} else {
var j = num;var a;var count = 0;
while ( j > 10){
a = num.toString().split('').map( e=> parseInt(e)).reduce((a,c)=> a*c);
num = a;
count++;
j--;
}
return count;
}}

test(39) //outputs 29 expected 3;


I have fixed the above by adding an array and filtering for unique values but would still like to know why the code is giving me a much higher count than expected.










share|improve this question















I'm writing a function which takes in a positive number and returns the number of times you must multiply the digits in num until you reach a single digit.



example:



test(39) === 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
// and 4 has only one digit//if single digit return num


The code:



function test(num) {
if (num > 10) { return num} else {
var j = num;var a;var count = 0;
while ( j > 10){
a = num.toString().split('').map( e=> parseInt(e)).reduce((a,c)=> a*c);
num = a;
count++;
j--;
}
return count;
}}

test(39) //outputs 29 expected 3;


I have fixed the above by adding an array and filtering for unique values but would still like to know why the code is giving me a much higher count than expected.







javascript






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 20 at 11:41

























asked Nov 20 at 10:12









dan brown

5616




5616








  • 1




    If num is already below 10, shouldn't it return 0 and not the number itself? I mean you describe your function as returning the number of iterations, not a result from a calculation.
    – Lennholm
    Nov 20 at 10:21






  • 2




    The j-- part is the bit that's making the result far too high. You can add console.log(a,num,j) inside the while to see what it's doing (or step through with the debugger). Also return num at the start makes no sense, should be return 0 as num<10 means it's already single digit so doesn't need to run your algorithm.
    – freedomn-m
    Nov 20 at 10:24












  • You're right,dunno why i even put that j part and forgot about it.Works without it.
    – dan brown
    Nov 20 at 11:44














  • 1




    If num is already below 10, shouldn't it return 0 and not the number itself? I mean you describe your function as returning the number of iterations, not a result from a calculation.
    – Lennholm
    Nov 20 at 10:21






  • 2




    The j-- part is the bit that's making the result far too high. You can add console.log(a,num,j) inside the while to see what it's doing (or step through with the debugger). Also return num at the start makes no sense, should be return 0 as num<10 means it's already single digit so doesn't need to run your algorithm.
    – freedomn-m
    Nov 20 at 10:24












  • You're right,dunno why i even put that j part and forgot about it.Works without it.
    – dan brown
    Nov 20 at 11:44








1




1




If num is already below 10, shouldn't it return 0 and not the number itself? I mean you describe your function as returning the number of iterations, not a result from a calculation.
– Lennholm
Nov 20 at 10:21




If num is already below 10, shouldn't it return 0 and not the number itself? I mean you describe your function as returning the number of iterations, not a result from a calculation.
– Lennholm
Nov 20 at 10:21




2




2




The j-- part is the bit that's making the result far too high. You can add console.log(a,num,j) inside the while to see what it's doing (or step through with the debugger). Also return num at the start makes no sense, should be return 0 as num<10 means it's already single digit so doesn't need to run your algorithm.
– freedomn-m
Nov 20 at 10:24






The j-- part is the bit that's making the result far too high. You can add console.log(a,num,j) inside the while to see what it's doing (or step through with the debugger). Also return num at the start makes no sense, should be return 0 as num<10 means it's already single digit so doesn't need to run your algorithm.
– freedomn-m
Nov 20 at 10:24














You're right,dunno why i even put that j part and forgot about it.Works without it.
– dan brown
Nov 20 at 11:44




You're right,dunno why i even put that j part and forgot about it.Works without it.
– dan brown
Nov 20 at 11:44












4 Answers
4






active

oldest

votes

















up vote
2
down vote



accepted










You might be looking to use recursion here - if the num is smaller than 10, return 0, otherwise, perform the needed multiplication, and then return 1 + the result of calling test on the product:






function test(num) {
return num < 10
? 0
: 1 + test(
num.toString().split('').reduce((a, c) => a * c)
)
}

console.log(test(39));





(note that * will coerce strings to numbers already, no need for parseInt)






share|improve this answer





















  • What happened to if (num < 10) { return num }?
    – Peter B
    Nov 20 at 10:22












  • return num doesn't make sense if OP is trying to figure out the number of iterations required to produce a single digit number
    – CertainPerformance
    Nov 20 at 10:23












  • @PeterB that's handled with the ?: = if (num<10) return 0;
    – freedomn-m
    Nov 20 at 10:26










  • True, description by OP does not seem to match what they tried.
    – Peter B
    Nov 20 at 10:28










  • Thanks guys.This works.Also the > sign in the first line was a type,now corrected
    – dan brown
    Nov 20 at 11:42


















up vote
1
down vote













Correction in your code, you can compare



    function test(num) {
if (num < 10) { return num} else {
var a;var count = 0;
while ( num > 10){
a = num.toString().split('').map( e=> parseInt(e)).reduce((a,c)=> a*c);
num = a;
count++;

}
return count;
}}

test(39) //outputs 3;





share|improve this answer




























    up vote
    0
    down vote













    You're decrementing j, which is initially 39. It will reach 10, then it will return the number of times it was decremented (29 times, which coincides with your result). You need to do the following, assuming the split part is correct:



    function test(num) {
    if (num < 10) { return num} else {
    var j = num;var a = 11; // just to enter the while loop at least once
    var count = 0;
    while ( a > 10){
    a = num.toString().split('').map( e=> parseInt(e)).reduce((a,c)=> a*c);
    num = a;
    count++;
    j--;
    }
    return count;
    }}





    share|improve this answer




























      up vote
      0
      down vote













      You decremented your num by 1 each iteration. It gives you 29 because 39 - 10 is 29. You didn't update the actual num which you need to compare if it's still greater than 10.



      Do this instead:






      function test(num) {
      if (num < 10) {
      return num
      } else {
      var count = 0;
      while (num > 10) {
      num = num.toString().split('').map(e => parseInt(e)).reduce((a, c) => a * c);
      count++;
      }
      return count;
      }
      }

      console.log(test(39))








      share|improve this answer





















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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        2
        down vote



        accepted










        You might be looking to use recursion here - if the num is smaller than 10, return 0, otherwise, perform the needed multiplication, and then return 1 + the result of calling test on the product:






        function test(num) {
        return num < 10
        ? 0
        : 1 + test(
        num.toString().split('').reduce((a, c) => a * c)
        )
        }

        console.log(test(39));





        (note that * will coerce strings to numbers already, no need for parseInt)






        share|improve this answer





















        • What happened to if (num < 10) { return num }?
          – Peter B
          Nov 20 at 10:22












        • return num doesn't make sense if OP is trying to figure out the number of iterations required to produce a single digit number
          – CertainPerformance
          Nov 20 at 10:23












        • @PeterB that's handled with the ?: = if (num<10) return 0;
          – freedomn-m
          Nov 20 at 10:26










        • True, description by OP does not seem to match what they tried.
          – Peter B
          Nov 20 at 10:28










        • Thanks guys.This works.Also the > sign in the first line was a type,now corrected
          – dan brown
          Nov 20 at 11:42















        up vote
        2
        down vote



        accepted










        You might be looking to use recursion here - if the num is smaller than 10, return 0, otherwise, perform the needed multiplication, and then return 1 + the result of calling test on the product:






        function test(num) {
        return num < 10
        ? 0
        : 1 + test(
        num.toString().split('').reduce((a, c) => a * c)
        )
        }

        console.log(test(39));





        (note that * will coerce strings to numbers already, no need for parseInt)






        share|improve this answer





















        • What happened to if (num < 10) { return num }?
          – Peter B
          Nov 20 at 10:22












        • return num doesn't make sense if OP is trying to figure out the number of iterations required to produce a single digit number
          – CertainPerformance
          Nov 20 at 10:23












        • @PeterB that's handled with the ?: = if (num<10) return 0;
          – freedomn-m
          Nov 20 at 10:26










        • True, description by OP does not seem to match what they tried.
          – Peter B
          Nov 20 at 10:28










        • Thanks guys.This works.Also the > sign in the first line was a type,now corrected
          – dan brown
          Nov 20 at 11:42













        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        You might be looking to use recursion here - if the num is smaller than 10, return 0, otherwise, perform the needed multiplication, and then return 1 + the result of calling test on the product:






        function test(num) {
        return num < 10
        ? 0
        : 1 + test(
        num.toString().split('').reduce((a, c) => a * c)
        )
        }

        console.log(test(39));





        (note that * will coerce strings to numbers already, no need for parseInt)






        share|improve this answer












        You might be looking to use recursion here - if the num is smaller than 10, return 0, otherwise, perform the needed multiplication, and then return 1 + the result of calling test on the product:






        function test(num) {
        return num < 10
        ? 0
        : 1 + test(
        num.toString().split('').reduce((a, c) => a * c)
        )
        }

        console.log(test(39));





        (note that * will coerce strings to numbers already, no need for parseInt)






        function test(num) {
        return num < 10
        ? 0
        : 1 + test(
        num.toString().split('').reduce((a, c) => a * c)
        )
        }

        console.log(test(39));





        function test(num) {
        return num < 10
        ? 0
        : 1 + test(
        num.toString().split('').reduce((a, c) => a * c)
        )
        }

        console.log(test(39));






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 20 at 10:15









        CertainPerformance

        71.3k143453




        71.3k143453












        • What happened to if (num < 10) { return num }?
          – Peter B
          Nov 20 at 10:22












        • return num doesn't make sense if OP is trying to figure out the number of iterations required to produce a single digit number
          – CertainPerformance
          Nov 20 at 10:23












        • @PeterB that's handled with the ?: = if (num<10) return 0;
          – freedomn-m
          Nov 20 at 10:26










        • True, description by OP does not seem to match what they tried.
          – Peter B
          Nov 20 at 10:28










        • Thanks guys.This works.Also the > sign in the first line was a type,now corrected
          – dan brown
          Nov 20 at 11:42


















        • What happened to if (num < 10) { return num }?
          – Peter B
          Nov 20 at 10:22












        • return num doesn't make sense if OP is trying to figure out the number of iterations required to produce a single digit number
          – CertainPerformance
          Nov 20 at 10:23












        • @PeterB that's handled with the ?: = if (num<10) return 0;
          – freedomn-m
          Nov 20 at 10:26










        • True, description by OP does not seem to match what they tried.
          – Peter B
          Nov 20 at 10:28










        • Thanks guys.This works.Also the > sign in the first line was a type,now corrected
          – dan brown
          Nov 20 at 11:42
















        What happened to if (num < 10) { return num }?
        – Peter B
        Nov 20 at 10:22






        What happened to if (num < 10) { return num }?
        – Peter B
        Nov 20 at 10:22














        return num doesn't make sense if OP is trying to figure out the number of iterations required to produce a single digit number
        – CertainPerformance
        Nov 20 at 10:23






        return num doesn't make sense if OP is trying to figure out the number of iterations required to produce a single digit number
        – CertainPerformance
        Nov 20 at 10:23














        @PeterB that's handled with the ?: = if (num<10) return 0;
        – freedomn-m
        Nov 20 at 10:26




        @PeterB that's handled with the ?: = if (num<10) return 0;
        – freedomn-m
        Nov 20 at 10:26












        True, description by OP does not seem to match what they tried.
        – Peter B
        Nov 20 at 10:28




        True, description by OP does not seem to match what they tried.
        – Peter B
        Nov 20 at 10:28












        Thanks guys.This works.Also the > sign in the first line was a type,now corrected
        – dan brown
        Nov 20 at 11:42




        Thanks guys.This works.Also the > sign in the first line was a type,now corrected
        – dan brown
        Nov 20 at 11:42












        up vote
        1
        down vote













        Correction in your code, you can compare



            function test(num) {
        if (num < 10) { return num} else {
        var a;var count = 0;
        while ( num > 10){
        a = num.toString().split('').map( e=> parseInt(e)).reduce((a,c)=> a*c);
        num = a;
        count++;

        }
        return count;
        }}

        test(39) //outputs 3;





        share|improve this answer

























          up vote
          1
          down vote













          Correction in your code, you can compare



              function test(num) {
          if (num < 10) { return num} else {
          var a;var count = 0;
          while ( num > 10){
          a = num.toString().split('').map( e=> parseInt(e)).reduce((a,c)=> a*c);
          num = a;
          count++;

          }
          return count;
          }}

          test(39) //outputs 3;





          share|improve this answer























            up vote
            1
            down vote










            up vote
            1
            down vote









            Correction in your code, you can compare



                function test(num) {
            if (num < 10) { return num} else {
            var a;var count = 0;
            while ( num > 10){
            a = num.toString().split('').map( e=> parseInt(e)).reduce((a,c)=> a*c);
            num = a;
            count++;

            }
            return count;
            }}

            test(39) //outputs 3;





            share|improve this answer












            Correction in your code, you can compare



                function test(num) {
            if (num < 10) { return num} else {
            var a;var count = 0;
            while ( num > 10){
            a = num.toString().split('').map( e=> parseInt(e)).reduce((a,c)=> a*c);
            num = a;
            count++;

            }
            return count;
            }}

            test(39) //outputs 3;






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 20 at 10:21









            Anubrij Chandra

            7311719




            7311719






















                up vote
                0
                down vote













                You're decrementing j, which is initially 39. It will reach 10, then it will return the number of times it was decremented (29 times, which coincides with your result). You need to do the following, assuming the split part is correct:



                function test(num) {
                if (num < 10) { return num} else {
                var j = num;var a = 11; // just to enter the while loop at least once
                var count = 0;
                while ( a > 10){
                a = num.toString().split('').map( e=> parseInt(e)).reduce((a,c)=> a*c);
                num = a;
                count++;
                j--;
                }
                return count;
                }}





                share|improve this answer

























                  up vote
                  0
                  down vote













                  You're decrementing j, which is initially 39. It will reach 10, then it will return the number of times it was decremented (29 times, which coincides with your result). You need to do the following, assuming the split part is correct:



                  function test(num) {
                  if (num < 10) { return num} else {
                  var j = num;var a = 11; // just to enter the while loop at least once
                  var count = 0;
                  while ( a > 10){
                  a = num.toString().split('').map( e=> parseInt(e)).reduce((a,c)=> a*c);
                  num = a;
                  count++;
                  j--;
                  }
                  return count;
                  }}





                  share|improve this answer























                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    You're decrementing j, which is initially 39. It will reach 10, then it will return the number of times it was decremented (29 times, which coincides with your result). You need to do the following, assuming the split part is correct:



                    function test(num) {
                    if (num < 10) { return num} else {
                    var j = num;var a = 11; // just to enter the while loop at least once
                    var count = 0;
                    while ( a > 10){
                    a = num.toString().split('').map( e=> parseInt(e)).reduce((a,c)=> a*c);
                    num = a;
                    count++;
                    j--;
                    }
                    return count;
                    }}





                    share|improve this answer












                    You're decrementing j, which is initially 39. It will reach 10, then it will return the number of times it was decremented (29 times, which coincides with your result). You need to do the following, assuming the split part is correct:



                    function test(num) {
                    if (num < 10) { return num} else {
                    var j = num;var a = 11; // just to enter the while loop at least once
                    var count = 0;
                    while ( a > 10){
                    a = num.toString().split('').map( e=> parseInt(e)).reduce((a,c)=> a*c);
                    num = a;
                    count++;
                    j--;
                    }
                    return count;
                    }}






                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Nov 20 at 10:24









                    Bogdan

                    225




                    225






















                        up vote
                        0
                        down vote













                        You decremented your num by 1 each iteration. It gives you 29 because 39 - 10 is 29. You didn't update the actual num which you need to compare if it's still greater than 10.



                        Do this instead:






                        function test(num) {
                        if (num < 10) {
                        return num
                        } else {
                        var count = 0;
                        while (num > 10) {
                        num = num.toString().split('').map(e => parseInt(e)).reduce((a, c) => a * c);
                        count++;
                        }
                        return count;
                        }
                        }

                        console.log(test(39))








                        share|improve this answer

























                          up vote
                          0
                          down vote













                          You decremented your num by 1 each iteration. It gives you 29 because 39 - 10 is 29. You didn't update the actual num which you need to compare if it's still greater than 10.



                          Do this instead:






                          function test(num) {
                          if (num < 10) {
                          return num
                          } else {
                          var count = 0;
                          while (num > 10) {
                          num = num.toString().split('').map(e => parseInt(e)).reduce((a, c) => a * c);
                          count++;
                          }
                          return count;
                          }
                          }

                          console.log(test(39))








                          share|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            You decremented your num by 1 each iteration. It gives you 29 because 39 - 10 is 29. You didn't update the actual num which you need to compare if it's still greater than 10.



                            Do this instead:






                            function test(num) {
                            if (num < 10) {
                            return num
                            } else {
                            var count = 0;
                            while (num > 10) {
                            num = num.toString().split('').map(e => parseInt(e)).reduce((a, c) => a * c);
                            count++;
                            }
                            return count;
                            }
                            }

                            console.log(test(39))








                            share|improve this answer












                            You decremented your num by 1 each iteration. It gives you 29 because 39 - 10 is 29. You didn't update the actual num which you need to compare if it's still greater than 10.



                            Do this instead:






                            function test(num) {
                            if (num < 10) {
                            return num
                            } else {
                            var count = 0;
                            while (num > 10) {
                            num = num.toString().split('').map(e => parseInt(e)).reduce((a, c) => a * c);
                            count++;
                            }
                            return count;
                            }
                            }

                            console.log(test(39))








                            function test(num) {
                            if (num < 10) {
                            return num
                            } else {
                            var count = 0;
                            while (num > 10) {
                            num = num.toString().split('').map(e => parseInt(e)).reduce((a, c) => a * c);
                            count++;
                            }
                            return count;
                            }
                            }

                            console.log(test(39))





                            function test(num) {
                            if (num < 10) {
                            return num
                            } else {
                            var count = 0;
                            while (num > 10) {
                            num = num.toString().split('').map(e => parseInt(e)).reduce((a, c) => a * c);
                            count++;
                            }
                            return count;
                            }
                            }

                            console.log(test(39))






                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered Nov 20 at 10:39









                            ACD

                            714111




                            714111






























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