Project Euler: Problem 8

Problem 8 in Project Euler is this

Find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

The brute-force approach to this problem would be to calculate the product of all the sets of five consecutive digits in the number. The first set would be from the first digit to the fifth, the next set would be from the second digit to the sixth, the final set being the digits from position 996 to position 1000. We could ignore any sets that had one or more zeros, because the product would then be zero. The answer would be the highest product.

I coded this algorithm and it did produce the correct result, but I found that it was quicker to simply calculate all the products, probably because of the overhead of checking each set for zero. That is the algorithm presented here:

  1. Set the result to 0
  2. Set the counter to 1
  3. Get the 5 digits from counter to counter + 4
  4. Calculate the product of these digits and if is greater than result replace result with this product
  5. Increment the counter
  6. If the counter is less then 997 go to step 3
  7. Display the result

There is another optimisations that we could apply to this algorithm. If we keep a record of the current 5 digits together with the next digit, we can calculate the new product by dividing the old product by the first digit and then multiplying by the newest digit. The drawback to this is that we would have to check for zeros, to avoid a ‘divide by zero’ error.

Since the algorithm above produces the result in 920 milliseconds (less than 1 second), I have decided to stick with this. This is the Delphi code:

procedure TfrmMain.Problem8;
var
  x, y: integer;
  maxResult, thisResult: int64;
  s, t : AnsiString;
begin
  //initial string (code truncated for clarity)
  s :=  '731671765313306...';
  for x  := 1 to 996 do begin
    t := copy(s, x, 5);
    thisResult := StrToInt(copy(t, 1, 1));
    for y := 2 to 5 do
      thisResult := thisResult * StrToInt(copy(t, y, 1));
    if thisResult > maxResult then maxResult := thisResult;
  end;
  ShowMessage(IntToStr(maxResult));
end;

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