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Category: Community
Forum: Chat Room
Thread: Weekend Puzzles |
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Thread Status: Active Total posts in this thread: 206
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adriverhoef
Master Cruncher The Netherlands Joined: Apr 3, 2009 Post Count: 1978 Status: Offline Project Badges: 
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Hello and welcome to a new puzzle in which you will have to find the answer by using this(!) sentence -and- the following very necessary key:
03-1,01-2,02-1,16-6,03-2,05-1,01-3, 01-4,12-3,01-5,03-3,07-2,02-2,04-1, 03-7,06-2,24-3,16-3,06-3,08-1,13-1, 01-1,03-5,10-3,25-8,02-3,07-6,24-1, 11-2,09-4,15-3,18-2,03-4,10-2,03-6, 07-1,18-1,15-1,09-3,06-1,18-5,19-1, 12-2,19-4,26-1,20-1,26-4,21-1,19-7. That's all. Solution coming Thursday. Good luck! Adri |
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alanb1951
Veteran Cruncher Joined: Jan 20, 2006 Post Count: 729 Status: Recently Active Project Badges: 
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Adri,
Thanks for the link to Andrew Stuart's sudowiki.org site -- that's a rabbit hole I had to tear myself away from after reading the very interesting .pdf file you linked, and I'll have to go back to it when I've finished getting my WCG stats scripts and data up to date :-) I found it interesting that he implies that a proper Killer Sudoku has NO cell values given, just cages -- I have to believe that's quite a lot harder to organize without software help :-) As for the new puzzle -- nice one... Cheers - Al. |
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adriverhoef
Master Cruncher The Netherlands Joined: Apr 3, 2009 Post Count: 1978 Status: Offline Project Badges:
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Dear all,
----------------------------------------Thank you for your patience. Here is the solution to my puzzle. You had to use one provided sentence ("Hello and welcome …") and a 7-line key to decipher the code. First, number all words in the sentence, so 'Hello' is 01, 'and' is 02, 'welcome' is 03, etc., '-and-' is 21, 'very' is 24, 'key:' is 26. The key that I provided ("03-1,01-2,02-1,16-6" and so on) denotes 7 pairs of characters on 7 separate lines: 03-1 denotes the 3rd word, 1st character (of the word 'welcome'), that's a W. 01-2 denotes the 1st word, 2nd character (of the word 'Hello'), that's an E. And so on. (…) After you've found 46 characters, there are still three characters left. These three last ones are the somewhat more difficult ones, but not impossible. :-) 26-4 denotes the 26th word, 4th character (of the word 'key:'), that's a colon. 21-1 denotes the 21st word, 1st character (of the word '-and-'), that's a dash. 19-7 denotes the 19th word, 7th character (of the word 'this(!)'), that's a closing parenthesis. Now we have 49 characters without any spaces. Wouldn't it be nice if I'd just add them now? So I'm presenting this solution: WE ARE ALL VOLUNTEERS WITH OUR DEVICES COMPUTING TASKS :-) Peculiarities: - All 49 pairs each denoting one character are used only once. - The number of words in the sentence is 26, equal to the number of letters in the English alphabet. Thank you all for participating! Adri PS In my conviction that's correct English, meaning "devices that are computing tasks". [Edit 1 times, last edit by adriverhoef at May 16, 2024 10:40:06 AM] |
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alanb1951
Veteran Cruncher Joined: Jan 20, 2006 Post Count: 729 Status: Recently Active Project Badges:
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Adri,
Thanks for confirming the answer :-) I hadn't noticed that there were 26 words (after all, it didn't influence the solution)... Once again, thanks for setting a puzzle. Cheers - Al. P.S. Yes, that is correct/valid English usage :-) -- curious language, isn't it? |
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Sgt.Joe
Ace Cruncher USA Joined: Jul 4, 2006 Post Count: 7219 Status: Offline Project Badges: 
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Adri,
----------------------------------------Thanks for the puzzle. I did not devote a lot of time to solving it as this is the time of the year when I am doing much outside work. It is an interesting puzzle with an interesting solution. Your English is better than many native speakers. Cheers
Sgt. Joe
*Minnesota Crunchers* |
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adriverhoef
Master Cruncher The Netherlands Joined: Apr 3, 2009 Post Count: 1978 Status: Offline Project Badges:
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Thanks Sgt.Joe,
Indeed I have been missing your input here and hope you enjoyed the solution. :-) At this moment I don't have a new puzzle available. Maybe you remember you wrote in another post: "I did not take the time to see how many were waiting to be sent." This is like a puzzle. How would I do it? How would I count the number of needed wingmen? Apart from the "503 Service unavailable" errors, wcgstats seems like a good start to me. So this is how I counted my needed wingmen ('Waiting to be sent'): by issuing the command "wcgstats -w -sPQ -a0 -m0". (The option -w will show workunits that you select.) (The option -sPQ selects all results that are Pending Validation (P) and Pending Verification (Q).) (The option -a0 will select all current sciences, MCM1, OPN1, SCC1, etc.) (The option -m0 will select all your predefined devices.) (Let's do it.) (Of course, if you're not interested, you may skip the rest of this post.)  $ wcgstats -w -sPQ -a0 -m0(Now the program waits for input. The newest workunits are on page 1, the oldest on the last page. There is a high probability that the oldest results are our goal. So we will type 'l' (ell, for last) in response.): l (Now the program will show the last page.) * Showing page 32/32 of all tasks with status ’P/Q’ on all of your devices:(This page contains 2 workunits.) - Did this show the desired workunit? [Y (yes)/n (no, next)/p (previous)/l (last)/q (quit)/c (change)/* (match)/PAGENUMBER](There were only two workunits on the last page; however, luckily they are showing exactly what we are looking for.) (The program waits for input. Since we are on the last page, we want the program to show the previous page, so we type 'p' in response.): p * Showing page 31/32 of all tasks with status ’P/Q’ on all of your devices: (There are 15 workunits on a page; in this post I will show workunits <1> and <15> (from page 31) only.) <15> * MCM1_0216714_5535_0 Linux Fedora Pending Validation 2024-05-07T09:27:00 2024-05-08T22:43:48 (There are 15 workunits on a page; in this post I will show workunits <1> and <15> (from page 30) only.) <15> * MCM1_0216768_0575_0 Linux Fedora Pending Validation 2024-05-08T04:28:19 2024-05-09T16:45:38 (There are 15 workunits on a page; in this post I will show workunits <1> and <15> (from page 29) only.) <15> * MCM1_0216819_5764_0 Fedora Linux Pending Validation 2024-05-09T02:24:01 2024-05-11T07:13:33(All 15 workunits on this page are still In Progress; in this post I will show workunit <15> (from page 28) only.) <15> MCM1_0216858_9388_0 Linux Ubuntu In Progress 2024-05-10T08:27:58 2024-05-16T08:27:58(Typing 'q' to quit the program.) (Now the math is simple: all workunits on pages 29, 30 and 31 (15 workunits each) and page 32 (2 workunits) need a wingman. Total: 47 workunits.) (See post 696636: "47 tasks needing a wingman".) (Obviously, this search didn't take me more than a few minutes.)  (If you made it all the way up to here, congrats!)  Adri |
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