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Brainteaser thread
http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=17&t=32		 Page 11 of 14
Author:  udosuk [ Sun Jun 01, 2008 6:01 pm ]
Post subject:  Re: Brainteaser thread
Glyn wrote:
Edit. {1..21} is manually checkable and has no solution. Failure to place 7 removes the last option.

Really? How about the following configuration:

{1} {..,2,20,7,..} {..,3,21,5,..} {..,4,19,6,..} ...

I think you need to prove it more vigorously. Meanwhile Richard please don't stop your searches (both of them) until you have found at least 2-3 solutions or it comes to the end. :salute:

Working on your division equation. :study:
Author:  Glyn [ Sun Jun 01, 2008 9:10 pm ]
Post subject:  Re:Six Golden Rods
udosuk wrote:
Really? How about the following configuration:

{1} {..,2,20,7,..} {..,3,21,5,..} {..,4,19,6,..} ...


I think this configuration cannot be reached as shafts of increasing length are introduced. Following the procedure again several times to check it actually fails at a shaft length of 6. The status at each stage is :-

Hidden Text:
Only possibles for 1
{1}
{1,21...}
Only possibles for 1,2
{1}{2,20..}
{1}{2,21..}
{1,21,2}
{2}{1,21..}
Only possibles for 1,2,3
{1}{2,21..}{3,19..}
{1}{2,20,3}
{2}{1,21..}{3,20..}
{1}{2,20..}{3,21..}
{1}{2,21,3}
{2}{1,21,3}
Only possibles for 1,2,3,4
{1}{2,21,3}{4,18..}
{1}{2,20..}{3,21..}{4,19..}
{2}{1,21,3}{4,19..}
{1}{2,21..}{3,19..}{4,20..}
{2}{1,21..}{3,20,4}
{1}{2,21,4}{3,19..}
{2}{1,21,4}{3,20..}
{1}{2,20..}{3,21,4}
Only possibles for 1,2,3,4,5
{1}{2,20..}{3,21,4}{5,18..}
{2}{1,21,4}{3,20..}{5,19..}
{1}{2,21,3}{4,18..}{5,20..}
{1}{2,20,5}{3,21..}{4,19..}
{1}{2,21..}{3,19..}{4,20,5}
{1}{2,20..}{3,21,5}{4,19..}
{1}{2,21,5}{3,19..}{4,20..}
No possibles for 1,2,3,4,5,6


Each newly introduced shaft must generate a measurement greater than 21, that is not repeated. In some cases the elimination takes place because the only placement closes a second 3-shaft rod. Hope it makes sense.
Author:  udosuk [ Mon Jun 02, 2008 3:14 am ]
Post subject:  Re: Re:Six Golden Rods
Glyn wrote:
Hidden Text:
Only possibles for 1,2,3
{1}{2,21..}{3,19..}
{1}{2,20,3}
{2}{1,21..}{3,20..}
{1}{2,20..}{3,21..}
{1}{2,21,3}
{2}{1,21,3}

I still see many cases you haven't considered, for example:
Hidden Text:
{3}{1,21,2}

Perhaps at the moment I'm still too thick to "see past the obvious", but will appreciate much if you can make the proof tighter without gaps. Thanks! :thumbs:
Author:  Glyn [ Mon Jun 02, 2008 7:38 am ]
Post subject:  Re: SixGolden Rods
Matt, thanks to that example, I now see a whole class of arrangements I have missed. :oops:

So it's back to the drawing board, there appear to be a lot of them as well. :brickwall:

Edit: Far too many they are breeding too fast for pen and paper, and I'm missing some using Notepad, which defeats the purpose. I guess it could be automated. I'll wait for Richard to continue the complete search.
Author:  udosuk [ Mon Jun 02, 2008 7:54 am ]
Post subject:  Re: Brainteaser thread
Answer to Richard's very nice exercise: :king:
Hidden Text:
Code:
        ???????
    -----------
??? ) 5????????
      ???
      ---
        ????
         ???
        ----
           5??
           ???
           ---
           ?5??
           ????
           ----
              0

It can be trivially reduced to:
Code:
        DEFGHIJ
    -----------
ABC ) 5??00????
      5??
      ---
        100?
         99?
        ----
           5??
           ???
           ---
           ?5??
           ?5??
           ----
              0

ABC*D=5?? (can't be 499) => A<=5, D<=5
ABC*EFG=99? => E=F=0, ABC*G=99?, G>D
ABC*HI=2??|3??|4?? => H=0, ABC*I=2??|3??|4?? => A=1..4, I=1..4, I<D
ABC*J=?5?? => J>G

=> DEFGHIJ=D00G0IJ, 0<I<D<G<J, I=1..4, D=2..5, G=3..8, J=4..9
=> ABC*D=5?? => A=1..2

ABC*G=99?:
992=248*4=124*8
994=142*7
995=199*5
996=249*4=166*6

ABC=124,248,249,166 can't make ABC*D=5??
ABC=142 can't make ABC*J=?5??

=> ABC=199 => ABC*D=199*3=597, ABC*G=199*5=995, ABC*I=199*2=398, ABC*J=199*8=1592
Code:
        3005028
    -----------
199 ) 598000572
      597
      ---
        1000
         995
        ----
           557
           398
           ---
           1592
           1592
           ----
              0

PS: Glyn by all means carry on your manual approach. Meanwhile I'm still looking forward to Richard's brute-force confirmation. :pray:
Author:  udosuk [ Tue Jun 03, 2008 3:25 am ]
Post subject:  Re: Brainteaser thread
Since Richard has confirmed through PM that I solved his riddle, here is another one for you (and probably the last one from me for a while):

Dicey Calendar

I have a set of 4 cubic dice working as a calendar for me on my office desktop. 2 of them have letters on their faces and 2 of them have numbers. I call the letter-dice the "month dice" and the number-dice the "date dice". So by turning the 6 faces of each die and rearranging I can form the dates 01..31 using the 2 "date dice" and a set of 2-letter abbreviations of the 12 months (in English) using the 2 "month dice".

My scheme to abbreviate the months is very specific, with the following rules:

  1. Only letters of the name of the month can be included in its abbreviation.
  2. The first letter of the abbreviation must always be the first letter of the name of the month.
  3. There must be no ambiguity whatsoever about which month the abbreviation is representing.
    (e.g. JU is not allowed as you don't know if it's representing January, June or July)
  4. Each die must be used the same number of times in each position overall.
    (i.e. each die must be used 6 times as the first letter and 6 times as the second)
  5. Of each die, 2 faces must be used exactly thrice, 2 faces twice and 2 faces once overall.

Can you work out what letters/numbers each of my dice have? :ugeek:
Author:  rcbroughton [ Tue Jun 03, 2008 8:02 pm ]
Post subject:  Re: Brainteaser thread
Not sure if this is the solution you had in mind - I think it works . . .

Hidden Text:
Two number dice are 0 1 2 3 6 7 and 0 1 2 4 5 8 (I turn the 6 upside down when I need a 9)

Two month dice are J R S Y O D and A N L M R F

Die 1: Jx3 Rx3 Sx2 Yx1 Ox2 Dx1
Die 2: Ax3 Nx2 Lx1 Mx2 Rx3 Fx1

JA=January
FR=February
MR=March
AR=April
MY=May
JN=June
JL=July
AS=August
SR=September
OR=October
NO=November
DR=December
Author:  udosuk [ Wed Jun 04, 2008 4:57 am ]
Post subject:  Re: Brainteaser thread
rcbroughton wrote:
Hidden Text:
JN=June

It doesn't work because:
JN can be January or June. I don't allow ambiguity on individual abbreviations even if you've already assigned JA as January. Consider I see JN04 on my desktop without any knowledge of other months. How could I tell if it's January 4 or June 4?

:ugeek:

PS: Still no news about the rods? Perhaps the solution you found is the unique one after all. :geek:
Author:  udosuk [ Wed Jun 04, 2008 6:42 am ]
Post subject:  Re: Brainteaser thread
Dearie dearie me... :santa:

With some enlightenment I tried to google with the series of the numbers of rods' solutions as cited by Richard "1,2,5,47,136". Bingo! Finally retrieved the lost webpage of that result. :king:

Prepared to be shocked. :alien:

But Richard by all means carry on the searches till the end to verify that result (if it won't cause too much inconvenience). :ugeek:
Author:  rcbroughton [ Wed Jun 04, 2008 8:02 am ]
Post subject:  Re: Brainteaser thread
udosuk wrote:
Consider I see JN04 on my desktop without any knowledge of other months. How could I tell if it's January 4 or June 4?
Curses!

However, if you live in England, just look out of the window: If it is cold and raining, it's June.
:lol:
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