Problem of the Day
A new programming or logic puzzle every Mon-Fri

Comments:

• Anonymous - 10 years, 4 months ago

  foldl1 haskell does some great work there

  run :: Int -> (Int -> Int -> Int) -> Int
  run x f = foldl1 f [1..x]
  

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• Johnathan - 10 years, 4 months ago

  It got overly complex because my for loop was messing up, but this works.

  num = int(raw_input('Please enter a number and I will find the summation: '))
  x = 0
  i = 1
  
  while i <= num:
      x = x + int(i)
      i = i + 1
  
  print x
  raw_input('')
  

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• Pufe - 10 years, 4 months ago

  Your example does not conform to your definition, if the range is exclusive (does not contain endpoints) then the sum should be 4950. My solution on emacs lisp uses an inclusive range to get the same answer as your example. It also evaluates the function at both endpoints (0 and i)

      (defun summation (i f)
        (if (< i 0)
            0
          (+ (funcall f i) (summation (- i 1) f))))
      (summation 100 'identity)
      ; >> 5050
  

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• Nick Krichevsky - 10 years, 4 months ago

  Quick and dirty python solution

  i=100
  print sum(range(i))
  

  As noted by Pufe, your example does not show counter-1, but rather just counter. If you were to do just counter, the code would be

  i=100
  print sum(range(i+1))
  

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• David - 10 years, 4 months ago

  Function Pointers are cool. I>o<I

  #include <stdlib.h>
  #include <stdio.h>
  
  int specialSumation (int counter, int (*mathExpr)(int) ) {
    int retVal = 0;
    for(int i = 0; i < counter; i++) {
      retVal += mathExpr(i);
    }
    return retVal;
  }
  
  int identity(int input) {
    return input;
  }
  
  int square(int input) {
    return input * input;
  }
  
  int history(int input) {
    static int past = 0;
    int retVal = input + (past % 3);
    past = input;
    return retVal;
  }
  
  //-------------------------------------------------------------------
  //--------------------------main loop--------------------------------
  //-------------------------------------------------------------------
  int main (int argc, char* argv[]) {
  
    int prnt = specialSumation(5, identity);
    printf("output: %i\n", prnt);
    prnt = specialSumation(5, square);
    printf("output: %i\n", prnt);
    prnt = specialSumation(5, history);
    printf("output: %i\n", prnt);
  
      return 0;
  }   //---------------------------------------------------------------
  
  
  

  I have included a couple use examples. The fact that you could pass in a mathematical expression seems to have been overlooked in the other answers. I opted to pass in a function, since I am not motivated to write a parser for math expressions.

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• Anonymous - 10 years, 4 months ago

  I guess you should have used a more complicated mathematical expression in your example ...

  function printSummation(counter, expr) { var sum = 0; for (i=0; i<counter; i++) { var curExpr = expr.replace('i', i); sum += eval(curExpr); }

  return sum;
  

  }

  printSummation(2, 'i*2') 2 printSummation(3, 'i*2') 6 printSummation(3, 'i/2') 1.5

  I'm wondering, is this even possible in Java?

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• Anonymous - 10 years, 4 months ago

  I think I'm the first person to post a O(1) solution - the rest have been O(n).

  def summation(to):
      n = to-1
      return n*(n+1) / 2
  
  >>> summation(100)
  4950.0
  

  This uses Gauss' method from the 1700s, using the observation that numbers can be paired (e.g. in summation(100) = 100+0 + 99+1 + 98+2...)

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