Comments on: Euler Problem 206

By: fzz

fzz — Mon, 01 Jun 2009 02:23:00 +0000

No squared integer can end with 0 without ending with 00. So the max is actually 1929394959697989900. The answer is 10 N with N^2 satisfying 1_2_3_4_5_6_7_8_9, and if N^2 ends in 9, then N must end in either 3 or 7. So N = 10 M + {3|7} for some other integer M, so N^2 = 100 M^2 + {60|140} M + {9|49}. No need to check for the final 9, so M^2 + {.6|1.4} M + {0|.4} must be of the form 1_2_3_4_5_6_7_8.#, so it's sufficient to find such M.

The reason for the speed of your second macro is searching backwards from max to min. The answer is close to max.

Here's an alternative.

Sub foo()
Const p As String = "1?2?3?4?5?6?7?8"

Dim m As Double, dt As Double

dt = Timer

m = CLng(Sqr(CDec("192939495969798"))) 'start at the max

Do
If Int(m * (m + 0.6)) Like p Then
m = 100 * m + 30
Exit Do
End If

If Int(m * (m + 1.4) + 0.4) Like p Then
m = 100 * m + 70
Exit Do
End If

m = m - 1# 'and decrement

Loop

Debug.Print m, CDec(m) * CDec(m), Timer - dt
End Sub

By: Michael

Michael — Sun, 31 May 2009 16:10:00 +0000

Hi Dan -

Exactly so...that's why the step can go from -1 to -10. I didn't see it, but Mike W pointed the same out above. Thinking about it, it goes the other way too...if the square root ends in zero, the square must end in 00.

...mrt

By: Dan

Dan — Sun, 31 May 2009 13:54:00 +0000

Not sure if you've factored this in (I'm not a VB expert by any stretch), but the square root must end in a zero, given that its square ends in a zero.

Does this reduce your possible answers by a factor of ten?

By: Michael

Michael — Sun, 31 May 2009 00:50:00 +0000

Hi Mike -

I added one line of code below the MAX = Int(Sqr(Cdbl(...))) line:

Max = Max + 10 - (Max Mod 10)

And then I changed "Step -1? to "Step -10?

And now Timer - t is 0 seconds! Maximum cool! Thanks.

...mrt

By: Michael

Michael — Sat, 30 May 2009 19:00:00 +0000

Hi Mike -

Please optimize away. I asked Dick if I could do this just as much to learn as maybe to teach. I’m a systems engineer. Even with a CS grad degree, all my code is for my personal use. This makes me not too much more than a hobbyist conversing with pro’s.

Reading the Euler fine print, I saw that you could sort the problems by increasing difficulty, which in practice I think is a descending sort on number of solvers. Problem 206 was right near where my checkmarks ran out.

I added to the Problem 73 thread how I saw the recursive function working. Didn’t help doing problem 72. I’m stalled too. No more snowy weekends for a while to stay inside Chores aplenty await.

…mrt

By: Mike Woodhouse

Mike Woodhouse — Sat, 30 May 2009 10:07:00 +0000

I hadn't noticed this one; I've been busy getting stuck on some around the 100 point. Nice one.

Ever the optimiser of other peoples' code, however, I did notice the following: the answer has to be a multiple of 10 to give a square that ends in zero, so your step size can be -10 providing you adjust start and end values accordingly. That also means you don't need the i=19/j=0 test, which will always be true. Also, I wonder if the CStr(j) is slower than creating a string constant and applying Mid() to that. It would reduce the size of the code somewhat:

SAT = True
For i = 1 To 19 Step 2
SAT = ( Mid(Answer_Sqrd, i, 1) = Mid(TEST_STRING, i, 1) )
If Not SAT Then Exit For
Next i

...but it's a bit nit-picky for a 0.04 second runtime!

I think there's probably some more mathemtical work that can be done: my solution took 0.015 sec counting down, but 444 sec counting up.