Skip to content
  • Categories
  • Recent
  • Tags
  • Popular
  • World
  • Users
  • Groups
Skins
  • Light
  • Cerulean
  • Cosmo
  • Flatly
  • Journal
  • Litera
  • Lumen
  • Lux
  • Materia
  • Minty
  • Morph
  • Pulse
  • Sandstone
  • Simplex
  • Sketchy
  • Spacelab
  • United
  • Yeti
  • Zephyr
  • Dark
  • Cyborg
  • Darkly
  • Quartz
  • Slate
  • Solar
  • Superhero
  • Vapor

  • Default (No Skin)
  • No Skin
Collapse
Code Project
  1. Home
  2. The Lounge
  3. Mathematicians - Treat for the weekend / might drive you crazy

Mathematicians - Treat for the weekend / might drive you crazy

Scheduled Pinned Locked Moved The Lounge
game-devhelpquestion
40 Posts 8 Posters 0 Views 1 Watching
  • Oldest to Newest
  • Newest to Oldest
  • Most Votes
Reply
  • Reply as topic
Log in to reply
This topic has been deleted. Only users with topic management privileges can see it.
  • D David ONeil

    Is there more than one answer?

    R Offline
    R Offline
    Raj Lal
    wrote on last edited by
    #23

    only one i presume "Not everything that counts can be counted..." -Albert Einstein

    D 1 Reply Last reply
    0
    • R Raj Lal

      only one i presume "Not everything that counts can be counted..." -Albert Einstein

      D Offline
      D Offline
      David ONeil
      wrote on last edited by
      #24

      You presume? Please see my email to you, and then clarify your presumption. Thanks, David

      R 1 Reply Last reply
      0
      • D David ONeil

        You presume? Please see my email to you, and then clarify your presumption. Thanks, David

        R Offline
        R Offline
        Raj Lal
        wrote on last edited by
        #25

        thanks for the link actually i got what you are saying its the domain which is not mentioned there. If the Domain is 2 to 100 ( which is in this case ) then there is a unique solution If you expand the domain to 2 to 1000 then the there are two solutions :) the problem you sent the link it does not gives any MAX Range only thing it gives is the MIN Range

        The Link says:

        PUZZLE STATEMENT: A and B are positive integers greater than 1.B Professor Sum "S" knows only the sum of A and B. Professor Product "P" knows only the product of A and B. The following conversation occurs between "S" and "P" "S" says to "P": You don't know A and B. "P" says to "S": Now I do know A and B. "S" says to "P": Now I know A and B as well. What are the values of A and B?

        SO TWO SOLUTIONS THERE but in our problem there is only one :) So i think that removes the ambiguity About wraping it the cache page says the problem was

        The Link also have:

        Assigned: Wednesday, Sept. 25, 2002 Due: Thursday, Oct. 10, 2002

        So i think that makes it clear :) , by the way i liked to see that it was an assignment because to find a solution u need to do some exhaust search with some constraints. Cheers "Not everything that counts can be counted..." -Albert Einstein

        D 1 Reply Last reply
        0
        • R Raj Lal

          thanks for the link actually i got what you are saying its the domain which is not mentioned there. If the Domain is 2 to 100 ( which is in this case ) then there is a unique solution If you expand the domain to 2 to 1000 then the there are two solutions :) the problem you sent the link it does not gives any MAX Range only thing it gives is the MIN Range

          The Link says:

          PUZZLE STATEMENT: A and B are positive integers greater than 1.B Professor Sum "S" knows only the sum of A and B. Professor Product "P" knows only the product of A and B. The following conversation occurs between "S" and "P" "S" says to "P": You don't know A and B. "P" says to "S": Now I do know A and B. "S" says to "P": Now I know A and B as well. What are the values of A and B?

          SO TWO SOLUTIONS THERE but in our problem there is only one :) So i think that removes the ambiguity About wraping it the cache page says the problem was

          The Link also have:

          Assigned: Wednesday, Sept. 25, 2002 Due: Thursday, Oct. 10, 2002

          So i think that makes it clear :) , by the way i liked to see that it was an assignment because to find a solution u need to do some exhaust search with some constraints. Cheers "Not everything that counts can be counted..." -Albert Einstein

          D Offline
          D Offline
          David ONeil
          wrote on last edited by
          #26

          As I pointed out in my email, that page gives two solutions where the numbers are both under 100, therefore, as you worded it, it is an ambiguous problem, unless the page is incorrect? David

          R 1 Reply Last reply
          0
          • D David ONeil

            As I pointed out in my email, that page gives two solutions where the numbers are both under 100, therefore, as you worded it, it is an ambiguous problem, unless the page is incorrect? David

            R Offline
            R Offline
            Raj Lal
            wrote on last edited by
            #27

            Well , i try to clearify The Domain actually restricts the unique solution though i understand both the solutions is in the range and that makes it feel that both are correct. but the solution space is not so its not what it looks like , if the solution space is 2 - 100 you will never get the other solution unanimously. and ALSO it contradicts the problem it self becuse if there are two solutions then how come the the two persons are sure about the anwser they will still have to decide between the two solutions right ? I just wrote this without thninking -- modified at 19:18 Sunday 13th November, 2005

            D 1 Reply Last reply
            0
            • R Raj Lal

              Well , i try to clearify The Domain actually restricts the unique solution though i understand both the solutions is in the range and that makes it feel that both are correct. but the solution space is not so its not what it looks like , if the solution space is 2 - 100 you will never get the other solution unanimously. and ALSO it contradicts the problem it self becuse if there are two solutions then how come the the two persons are sure about the anwser they will still have to decide between the two solutions right ? I just wrote this without thninking -- modified at 19:18 Sunday 13th November, 2005

              D Offline
              D Offline
              David ONeil
              wrote on last edited by
              #28

              You are being unclear.

              Quartz... wrote:

              but the solution space is not so its not what it looks like , if the solution space is 2 - 100 you will never get the other solution unanimously.

              I think you meant "if the solution space is 2 - 100 you will never get the other solution by itself." Yet both the solutions occur 'by themselves' in the solutions given in that page, and I cannot see ANY difference in the solution spaces between your posting of the problem and the posting on that page.

              and ALSO it contradicts the problem it self becuse if there are two solutions then how come the the two persons are sure about the anwser they will still have to decide between the two solutions right ?

              Wrong. Each of them has a clue to the answer, and when A knows the answer, the answer becomes unambiguous to B based upon B's information.

              R 1 Reply Last reply
              0
              • D David ONeil

                You are being unclear.

                Quartz... wrote:

                but the solution space is not so its not what it looks like , if the solution space is 2 - 100 you will never get the other solution unanimously.

                I think you meant "if the solution space is 2 - 100 you will never get the other solution by itself." Yet both the solutions occur 'by themselves' in the solutions given in that page, and I cannot see ANY difference in the solution spaces between your posting of the problem and the posting on that page.

                and ALSO it contradicts the problem it self becuse if there are two solutions then how come the the two persons are sure about the anwser they will still have to decide between the two solutions right ?

                Wrong. Each of them has a clue to the answer, and when A knows the answer, the answer becomes unambiguous to B based upon B's information.

                R Offline
                R Offline
                Raj Lal
                wrote on last edited by
                #29

                There is a slight difference in the question itself The Problem given in that page is

                A and B are positive integers greater than 1 ( there is no upper limit )

                And the Problem here is

                2 >= a,b ** >= 100**

                I think thats the ony difference which filters out the second solution I am compilng the solution with a detailed explaination hope it will clearify it, sorry if i got you confused "Not everything that counts can be counted..." -Albert Einstein

                D 1 Reply Last reply
                0
                • R Raj Lal

                  There is a slight difference in the question itself The Problem given in that page is

                  A and B are positive integers greater than 1 ( there is no upper limit )

                  And the Problem here is

                  2 >= a,b ** >= 100**

                  I think thats the ony difference which filters out the second solution I am compilng the solution with a detailed explaination hope it will clearify it, sorry if i got you confused "Not everything that counts can be counted..." -Albert Einstein

                  D Offline
                  D Offline
                  David ONeil
                  wrote on last edited by
                  #30

                  Now you are making no sense, and even contradicting your earlier posts. In the first post you said:

                  2 <= a,b <= 100

                  Now you say

                  2 >= a,b >= 100

                  I believe you meant the first one. And a and b in BOTH of the solutions on the page I showed are between 2 and 100. David

                  R 2 Replies Last reply
                  0
                  • D David ONeil

                    Now you are making no sense, and even contradicting your earlier posts. In the first post you said:

                    2 <= a,b <= 100

                    Now you say

                    2 >= a,b >= 100

                    I believe you meant the first one. And a and b in BOTH of the solutions on the page I showed are between 2 and 100. David

                    R Offline
                    R Offline
                    Raj Lal
                    wrote on last edited by
                    #31

                    Sorry it was the latter

                    David O`Neil wrote:

                    2 <= a,b <= 100

                    and you were right these makes no sense, i wrote that without thinking and ALSO it contradicts the problem it self becuse if there are two solutions then how come the the two persons are sure about the anwser they will still have to decide between the two solutions right ? Midnight jitters :) -- modified at 19:22 Sunday 13th November, 2005

                    1 Reply Last reply
                    0
                    • R Raj Lal

                      (Actually, This is an old problem but just to stop smartsy CPians from using the word "nanosecond" forever let's blast them with it! ). There are 2 integers a and b (Though programmers like i and j more as integers) And two CPian A and B. (The result will decide who are they) Rules of the game ------------------------------- I. 2 <= a,b <= 100 II. A is told the product of a and b III. B is told the sum of a and b. IV. Neither is told the values The cPians conversation ::: ------------------------------- 1=> B: I cannot determine a, b. 2=> A: I cannot determine a, b 3=> B: I already knew that. 4=> A: In that case, I now know them 5=> B: In that case, I too now know a, b. What are the numbers ??? ------------------------------- Is'nt it simple and beautiful ? I can hear you saying " huh! " cheers. "Not everything that counts can be counted..." -Albert Einstein

                      R Offline
                      R Offline
                      Raj Lal
                      wrote on last edited by
                      #32

                      check the Solution here Check[^] "Not everything that counts can be counted..." -Albert Einstein -- modified at 8:09 Monday 14th November, 2005

                      1 Reply Last reply
                      0
                      • R Rui A Rebelo

                        Quartz... wrote:

                        where r u guys !

                        Thinking! Obsessivelly!:mad: "Thanks" for wrecking my weekend!:mad: X| Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer)

                        R Offline
                        R Offline
                        Raj Lal
                        wrote on last edited by
                        #33

                        check the Solution here Check[^] "Not everything that counts can be counted..." -Albert Einstein

                        1 Reply Last reply
                        0
                        • D David Stone

                          Quartz... wrote:

                          2 <= a,b <= 100

                          Do you mean 2 <= a <= 100 and 2 <= b <= 100 or 2 <= a and b <= 100 ?


                          Picture a huge catholic cathedral. In it there's many people, including a gregorian monk choir. You know, those who sing beautifully. Then they start singing, in latin, as they always do: "Ad hominem..." -Jörgen Sigvardsson

                          R Offline
                          R Offline
                          Raj Lal
                          wrote on last edited by
                          #34

                          check the Solution here Check[^] "Not everything that counts can be counted..." -Albert Einstein

                          1 Reply Last reply
                          0
                          • G Gary R Wheeler

                            OK, your post has been up for several hours now. How about posting your solution? Personally, I think the problem is indeterminate, at least without additional constraints.


                            Software Zen: delete this;

                            R Offline
                            R Offline
                            Raj Lal
                            wrote on last edited by
                            #35

                            check the Solution here Check[^] "Not everything that counts can be counted..." -Albert Einstein -- modified at 8:09 Monday 14th November, 2005

                            1 Reply Last reply
                            0
                            • L leppie

                              2 & 3 or vica versa My reasoning the lowest 2 numbers that are not 2 & 2. 1. B: (a x b) = 6 2. A: (a + b) = 5 3. B: a != b and a < 4 and b < 4 4. A: Can only be 2 & 3 5: B: Can only be 2 & 3 :confused: Probably very incorrect... :p [update] Similar to Rui A. Rebelo's solution [update] xacc.ide-0.1-rc3 released! Download and screenshots -- modified at 12:23 Saturday 12th November, 2005

                              R Offline
                              R Offline
                              Raj Lal
                              wrote on last edited by
                              #36

                              check the Solution here Check[^] "Not everything that counts can be counted..." -Albert Einstein

                              1 Reply Last reply
                              0
                              • D David ONeil

                                Now you are making no sense, and even contradicting your earlier posts. In the first post you said:

                                2 <= a,b <= 100

                                Now you say

                                2 >= a,b >= 100

                                I believe you meant the first one. And a and b in BOTH of the solutions on the page I showed are between 2 and 100. David

                                R Offline
                                R Offline
                                Raj Lal
                                wrote on last edited by
                                #37

                                check the Solution here Check[^] "Not everything that counts can be counted..." -Albert Einstein

                                1 Reply Last reply
                                0
                                • L leppie

                                  Is A and B necesary CPians? Or can they be other kinds of people too? If the former, my original hunch was perhaps the user id of 2 specific users under 10000 that can be broken down to the requirements... (long shot :) ) xacc.ide-0.1-rc3 released! Download and screenshots

                                  R Offline
                                  R Offline
                                  Raj Lal
                                  wrote on last edited by
                                  #38

                                  Check the Solution here [^] "Not everything that counts can be counted..." -Albert Einstein

                                  1 Reply Last reply
                                  0
                                  • R Rui A Rebelo

                                    The numbers are 3 and 4. Explanation: B is told the sum: 7. He knows that only 2 different pairs of numbers greater than 2 can give a sum of 7 : 3+4 and 2+5. Therefore:

                                    Quartz... wrote:

                                    1=> B: I cannot determine a, b.

                                    A is told the product: 12. He knows that there are just 2 different pairs of numbers greater than 2 which give this result: 2*6 and 3*4. So:

                                    Quartz... wrote:

                                    2=> A: I cannot determine a, b

                                    Now, B knows 2 things. He knows A's ambiguity if the numbers are 3 and 4. And he knows that if they were 2 and 5 (as B considered) then A wouldn't have an ambiguity; these 2 numbers would generate a unique product of 10. Facing A's ambiguity:

                                    Quartz... wrote:

                                    3=> B: I already knew that.

                                    A perceives that B solved it's own ambiguity and the only possibility of this happening is that the numbers are 3 and 4. If they were 2 and 6 as he considered the sum would be 8, wich would give 3 diferent possibilities to B (4+4, 2+6, 3+5). So:

                                    Quartz... wrote:

                                    4=> A: In that case, I now know them

                                    And B already knew the answer. Makes sense? Did I preserve my sanity? Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer)

                                    R Offline
                                    R Offline
                                    Raj Lal
                                    wrote on last edited by
                                    #39

                                    Check the Solution here [^] "Not everything that counts can be counted..." -Albert Einstein

                                    1 Reply Last reply
                                    0
                                    • C ColinDavies

                                      Interpreting your qu differently gives 4, 13. I think this used to be called "the impossible problem" Regardz Colin J Davies The most LinkedIn CPian (that I know of anyhow) :-)

                                      R Offline
                                      R Offline
                                      Raj Lal
                                      wrote on last edited by
                                      #40

                                      You were right :) Check the Solution here[^] "Not everything that counts can be counted..." -Albert Einstein

                                      1 Reply Last reply
                                      0
                                      Reply
                                      • Reply as topic
                                      Log in to reply
                                      • Oldest to Newest
                                      • Newest to Oldest
                                      • Most Votes


                                      • Login

                                      • Don't have an account? Register

                                      • Login or register to search.
                                      • First post
                                        Last post
                                      0
                                      • Categories
                                      • Recent
                                      • Tags
                                      • Popular
                                      • World
                                      • Users
                                      • Groups