tag:blogger.com,1999:blog-5573456362841621035.comments2013-10-04T12:42:39.375-07:00Puzzles for Puzzled Minds...Shashank Nnoreply@blogger.comBlogger111125tag:blogger.com,1999:blog-5573456362841621035.post-58929227944188681632013-10-04T12:42:39.375-07:002013-10-04T12:42:39.375-07:00how ?how ?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-10538404787463268612013-10-04T02:03:03.107-07:002013-10-04T02:03:03.107-07:00If (n-1)/6 is an integer, the first player loses.If (n-1)/6 is an integer, the first player loses.Bennoreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-873725705563349792013-09-22T20:42:20.302-07:002013-09-22T20:42:20.302-07:00I think Mai has it right, except I was leaning tow...I think Mai has it right, except I was leaning toward using intonation to indicate "same" or "different". Dwarf 10 (first one) says "Black" to indicate that the dwarf in front has a black hat. Dwarf 9 then says "Black?" to simultaneously note that his hat is black (correct) and the hat in front of him is not. If the hat in front of him was black he would say "Black" without the question mark. Dwarf 9 then says "White" if the hat in front of him is white, and "White?" if the hat in front of him is black. etc., this should reasonably save 9 dwarfs, but the first dwarf will have a 50/50 chance of dying. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-57320028208556167552013-09-16T04:21:52.982-07:002013-09-16T04:21:52.982-07:00You can save at least 9 dwarfs - even without assu...You can save at least 9 dwarfs - even without assuming there is an equal number of hats. The only dwarf that has a 50/50 chance is the tallest one. The key is telling the dwarf in front of you the color of their hat while naming the color of your hat at the same time. One way to do this may be to whisper for white and scream for black. You tell the dwarf in front of you that no matter which color I say, my pitch determines the color of your hat. The actual color that is said will be based on whether or not the dwarf behind me whispered or screamed. So for example - if the dwarf behind me screams white, I know I have a black hat on (scream = black). At the same time I see the dwarf in front of me has a white hat on. I would then whisper black.Maihttps://www.blogger.com/profile/07644960425125056327noreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-14315050674803813682013-09-11T00:36:10.988-07:002013-09-11T00:36:10.988-07:00You can save a minimum of 9 as follows:
Dwarf 10 ...You can save a minimum of 9 as follows:<br /><br />Dwarf 10 counts up the number of white hats in front of him. If there is an even number be says "black" otherwise if an odd number he says "white". He has a 50% chance of being right about his own hat, but passes crucial information to the remaining 9 dwarves.<br /><br />Dwarf 9 counts whether there are an even or odd number of white hats in front of him. If this differs from what dwarf 10 declared (encoded as black=even and white=odd), then he knows he has a white hat, otherwise a black hat. He gives the correct answer.<br /><br />All subsequence dwarves 8 down to 1 can calculate the number of white hats in positions 1 through 9, including their own, by counting both the hats they can see and the ones they have heard behind them (ignoring dwarf 10). If there is a disparity between the even/odd-ness of this number, and what dwarf 10 originally declared, then they know they have a white hat. If not, they have a black hat.<br /><br />(In boolean logic terminology they can use the XOR "exclusive or" to calculate the answer).Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-13552407024365435362013-09-11T00:19:16.239-07:002013-09-11T00:19:16.239-07:00One strategy that will save at least 7 is the foll...One strategy that will save at least 7 is the following:<br /><br />Dwarves 8, 9 and 10 count the number of white hats in positions 1 through 7. There can be anything from 0 to 7 hats. They then encode this number in binary, requiring 3 bits, and then answer "black" for a zero and "white" for a one.<br /><br />Dwarves 1 through 7 now know how may white hats (call it W) they have between them. Dwarf 7 counts the number of white hats in front of him. If the number is less than W, he knows his hat is white, otherwise it is black.<br /><br />Dwarf 6 follows the same procedure, remembering to subtract one from W each time a taller dwarf declares a white hat. And so on down to dwarf 1.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-41254919145484575692013-08-14T18:23:28.810-07:002013-08-14T18:23:28.810-07:00Although you can easily save 5, that's not the...Although you can easily save 5, that's not the optimum.<br />You can save at least 6 Dwarfs! (Splitting the Problem in 'save 3 out of 5'). Perhaps more.<br />The idea is, that you have more information by let a dwarf choose, if he uses his color, or if he scarifies himself. So you get more information!<br />The Theory of 9 saved Dwarfs can't be correct, according to you have ONE information and 2^9 possibilities.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-82596660036285291472013-08-08T12:39:58.667-07:002013-08-08T12:39:58.667-07:00Regardless of how many hats there are in each colo...Regardless of how many hats there are in each colour, there is actually a strategy that will save a _minimum_ of 9 dwarves.Perrynoreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-5603340858940548782013-08-03T12:23:37.026-07:002013-08-03T12:23:37.026-07:00Lieke.. considering the 5 black & 5 white assu...Lieke.. considering the 5 black & 5 white assumption not made and if tallest (10) guy speaks the color of 9th guy and if that doesn't matches with him. he is dead and 9th guy cannot hear that answer. so no one will survive. <br />The better way is the tallest(10) guy will know what is the majority no of hats in either or black white color. the two extreme cases are out of 9(1b, 8w) or (8b,1w)other way it could be 5w, 4b or reversal. If every one adopts the same strategy a minimum of 1 person and maximum of 9 persons can be savedbrave hearthttps://www.blogger.com/profile/00377460185862122058noreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-90518699693408048272013-08-01T05:19:39.868-07:002013-08-01T05:19:39.868-07:00That is not correct, because the giant starts aski...That is not correct, because the giant starts asking the tallest dwarf, who is able to see all the other dwarfs. So if there are 5 black hats and 5 white hats, the tallest counts the other nine and calls the one with only four in front of him. And so on. All dwarfs can be saved.<br /><br />If the assumption that there are 5 black hats and 5 white hats cannot be made, the following is the best strategy I think:<br />The tallest (nr 10) calls the color of the hat of the dwarf in front of him. That dwarf (nr9) calls the same color (his own hat) thereby saving himself. Nr 8 calls the color of nr 7. Nr 7 calls his own color and so on. All uneven numbered dwarfs are saved (so that makes 5) and the other ones have a 50% change of surviving.Liekenoreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-34571551678841451792013-07-26T11:23:39.029-07:002013-07-26T11:23:39.029-07:00There is one big assumption that has to be made he...There is one big assumption that has to be made here; namely that the giant has 5 black hats and 5 white hats that he puts on randomly. without that assumption, there is no strategy that works as there will be a 50-50 chance of a white or black hat on EVERY dwarf regardless of any or all prior hat colors.<br /><br />Assuming 5 black and 5 white, the next dwarf listens to all prior answers and calculates the odds remaining of what his hat color will be. At least 1 dwarf will be saved this way as the final dwarf will know all 9 hat colors and know if the black or white one remains. Previous dwarfs just play the odds. If for example, dwarf 6 heard 4 blacks and one white, he should guess white as there is only one black left and four whites left. Ditto for all other elves but the tallest -- who will have to make a 50/50 call.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-13603907172881391462013-07-25T13:34:12.226-07:002013-07-25T13:34:12.226-07:00Go back to sleep! :)Go back to sleep! :)Mehttps://www.blogger.com/profile/13253445338598917698noreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-86497069413185622672013-06-29T05:08:52.964-07:002013-06-29T05:08:52.964-07:00because question is asked for 10 min. right ?because question is asked for 10 min. right ?Shashank Narayanhttps://www.blogger.com/profile/08388805306629452557noreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-47163264674667878722013-06-29T04:37:46.952-07:002013-06-29T04:37:46.952-07:00Why intervals of 10 min and not 15 min?Why intervals of 10 min and not 15 min?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-76468261595485683332013-06-19T09:26:06.724-07:002013-06-19T09:26:06.724-07:00what is the answer to the problemwhat is the answer to the problemRahul Kumarhttps://www.blogger.com/profile/04523179481491831748noreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-84349671482045477372013-03-18T21:47:05.937-07:002013-03-18T21:47:05.937-07:00problem of variation:-
* dogs increased- pounds wi...problem of variation:-<br />* dogs increased- pounds will increase<br />=> dogs proportional to pound consumed......(1)<br /><br />* dogs increase- days decrease<br />=> dogs inversly proportional to days..........(2)<br />from (1) & (2)<br /><br />dogs proportional to pound per day<br />this will solve the problemAMGhttps://www.blogger.com/profile/10997699345331882330noreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-12692782866947762142013-03-05T23:40:21.099-08:002013-03-05T23:40:21.099-08:00please paste the method .please paste the method .Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-29827103694823764702013-02-12T16:31:17.707-08:002013-02-12T16:31:17.707-08:00Take the sheep across. Then go back and take the f...Take the sheep across. Then go back and take the fox across. Take the sheep with you, and then exchange him for the hay and take the hay over. Then take the sheep over.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-58156482950897059432012-12-12T20:07:58.092-08:002012-12-12T20:07:58.092-08:00No idea what's going on in your step 4. I don&...No idea what's going on in your step 4. I don't think it's possible without some working up to the start of the timing:<br /><br />Start 4 and 7.<br />Start 4 again when it runs out.<br />7 runs out, 1m left on 4. Start timing the 9 minutes now.<br />Just let 4 run out, then run it twice more.<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-70610136874633637002012-11-26T22:11:15.582-08:002012-11-26T22:11:15.582-08:00Epoch electronic Aircraft weighing kit is designed...Epoch electronic Aircraft weighing kit is designed primarily to cater the accurate weighing needs of aircraft and aerospace vehicles, but can be used for other precision weighing application as well as for the calibration of force generating machines.<br /><br /><a href="http://www.epochloadcell.com/s_beam_load_cell.html" rel="nofollow"> Aircraft weighing</a><br />EPOCHLOADCELLhttps://www.blogger.com/profile/05664390405859969300noreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-40898134299267441482012-11-24T18:40:45.522-08:002012-11-24T18:40:45.522-08:00Wikipedia has good explantion for solution to this...Wikipedia has good explantion for solution to this puzzle. Find it here:<br />http://en.wikipedia.org/wiki/Monty_hall_problemPraveenhttp://dealsindiadeals.comnoreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-66054109302058123922012-11-07T05:45:27.891-08:002012-11-07T05:45:27.891-08:00In the first case we had only two options for each...In the first case we had only <i>two</i> options for each of the weight,<br />1)Place the weight on the left side<br />2)Don't place the weight<br />So the amount of weights required were in the terms of 2^0,2^1,...<br />But In the second case we had <b>three</b> options for each of the weight,<br />1)Place the weight on the left side<br />2)Place the weight on the right side<br />3)Don't place the weight anywhere<br />So the amount of weights required were in the terms of 3^0,3^1,...Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-33405893182788766192012-10-28T04:04:28.761-07:002012-10-28T04:04:28.761-07:00Number each bottle 1 to 1000. Now convert this num...Number each bottle 1 to 1000. Now convert this number of the bottle to its binary equivalent. You will have a 10 digit number comprising of 1s and 0s.<br /><br />Now, take 10 prisoners. Assign each of them a digit from 1 to 10. The prisoner n, will sip wine from every bottle whose number's binary equivalent has a 1 at nth digit. For instance, prisoner no. 3 will sip from each and every bottle whose number's binary equivalent has a 1 at third digit.<br /><br />Now, after a month, the prisoners that die are those who sipped the poisoned wine. The king can now derive the exact number of the bottle. For instance, if prisoner number 3, 5, 7, 9 died, the number of bottle of poisoned wine is:<br />0010101010, which is binary equivalent for 170. So bottle number 170 is poisoned.<br /><br />Happy birthday - long live the evil King!!<br /><br />PoloAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-26979988930307531072012-10-26T03:14:17.621-07:002012-10-26T03:14:17.621-07:00Thanks for the great post! I was asked this quest...Thanks for the great post! I was asked this question in an interview as well.<br /><br />This page also helped my understanding:<br /><br /><a href="http://www.programmerinterview.com/index.php/puzzles/2-eggs-100-floors-puzzle/" rel="nofollow">2 Eggs 100 Floors</a>Joehttps://www.blogger.com/profile/12568980537278420520noreply@blogger.comtag:blogger.com,1999:blog-5573456362841621035.post-106949527478654372012-10-04T17:24:40.405-07:002012-10-04T17:24:40.405-07:00Just a thought:
May be use EGG2 to identify block ...Just a thought:<br />May be use EGG2 to identify block of 10. That is test EGG2 to floor # 10, 20...100.<br />In worst case, it will require 10 attempts. Once it breaks at floor #100. Now we know the safe floor is between 91 and 99. <br /><br />Using EGG1 do linear search from 91 on wards. In worst case 9 more attempts. <br /><br /><b>Total attempts 19.</b>mandeep singhnoreply@blogger.com