By Josef Kling. White to mate in two moves.
By Josef Kling. White to mate in two moves.
In 2005 Keele University computer scientist Gordon Rugg published two ciphers to the web.
The first is called the Penitentia Manuscript. The image above is only one panel; you can view and download the whole thing here. Rugg’s website provides one clue: “Most modern codes are based on a shared set of underlying assumptions. He wondered what would happen if you deliberately ignored those assumptions. What sorts of code might that produce?” There’s some more info here.
The second cipher, called the Ricardus Manuscript, was inspired by Rugg’s work on another famous puzzle: “When Gordon was working on the Voynich Manuscript, he started wondering what a real code based on the components of the Voynich Manuscript would look like. This code is the result.” Again, the image below is only a sample; you can find the whole thing here. More info here.
Both of these ciphers have been freely available on the web for more than 10 years, and both remain unsolved. Any takers?
Charles Trigg proposed this festive cryptarithm in the American Mathematical Monthly in 1956:
MERRY XMAS TO ALL
If each letter is a unique representation of a digit, and each word is a square integer, what are these four numbers?
In 1987, Chris Cole posted a message to the sci.crypt Usenet group:
When I was a graduate student at Caltech, Professor Feynman showed me three samples of code that he had been challenged with by a fellow scientist at Los Alamos and which he had not been able to crack. I also was unable to crack them. I now post them for the net to give it a try.
The first, marked “Easier,” was this:
MEOTAIHSIBRTEWDGLGKNLANEA INOEEPEYSTNPEUOOEHRONLTIR OSDHEOTNPHGAAETOHSZOTTENT KEPADLYPHEODOWCFORRRNLCUE EEEOPGMRLHNNDFTOENEALKEHH EATTHNMESCNSHIRAETDAHLHEM TETRFSWEDOEOENEGFHETAEDGH RLNNGOAAEOCMTURRSLTDIDORE HNHEHNAYVTIERHEENECTRNVIO UOEHOTRNWSAYIFSNSHOEMRTRR EUAUUHOHOOHCDCHTEEISEVRLS KLIHIIAPCHRHSIHPSNWTOIISI SHHNWEMTIEYAFELNRENLEERYI PHBEROTEVPHNTYATIERTIHEEA WTWVHTASETHHSDNGEIEAYNHHH NNHTW
Jack Morrison of NASA’s Jet Propulsion Laboratory had it solved the next day: “It’s a pretty standard transposition: split the text into 5-column pieces, then read from lower right upward.” This yields the opening of Chaucer’s Canterbury Tales:
WHANTHATAPRILLEWITHHISSHOURESSOOTET HEDROGHTEOFMARCHHATHPERCEDTOTHEROOT EANDBATHEDEVERYVEYNEINSWICHLICOUROF WHICHVERTUENGENDREDISTHEFLOURWHANZE PHIRUSEEKWITHHISSWEETEBREFTHINSPIRE DHATHINEVERYHOLTANDHEETHTHETENDRECR OPPESANDTHEYONGESONNEHATHINTHERAMHI SHALVECOURSYRONNEANDSMALEFOWELESMAK ENMELODYETHATSLEPENALTHENYGHTWITHOP ENYESOPRIKETHHEMNATUREINHIRCORAGEST HANNELONGENFOLKTOGOONONPILGRIM
But the other two ciphers have never been solved, despite 30 years of trying. Here they are:
XUKEXWSLZJUAXUNKIGWFSOZRAWURORKXAOS LHROBXBTKCMUWDVPTFBLMKEFVWMUXTVTWUI DDJVZKBRMCWOIWYDXMLUFPVSHAGSVWUFWOR CWUIDUJCNVTTBERTUNOJUZHVTWKORSVRZSV VFSQXOCMUWPYTRLGBMCYPOJCLRIYTVFCCMU WUFPOXCNMCIWMSKPXEDLYIQKDJWIWCJUMVR CJUMVRKXWURKPSEEIWZVXULEIOETOOFWKBI UXPXUGOWLFPWUSCH
#3 (“New Message”)
WURVFXGJYTHEIZXSQXOBGSVRUDOOJXATBKT ARVIXPYTMYABMVUFXPXKUJVPLSDVTGNGOSI GLWURPKFCVGELLRNNGLPYTFVTPXAJOSCWRO DORWNWSICLFKEMOTGJYCRRAOJVNTODVMNSQ IVICRBICRUDCSKXYPDMDROJUZICRVFWXIFP XIVVIEPYTDOIAVRBOOXWRAKPSZXTZKVROSW CRCFVEESOLWKTOBXAUXVB
Feynman, apparently, couldn’t break them either.
A puzzle by David Silverman:
Able, Baker, and Charlie are playing tag. Able is faster than Baker, who’s faster than Charlie. All three of them start at point P, and Able is “it.” At time -T, Baker runs north and Charlie runs south. After a count that takes time T, Able starts chasing one of the two quarries. The game ends when Able has tagged both Baker and Charlie. If Baker and Charlie maintain their speeds and directions, who should Able chase first to minimize the time required to make the second tag?
William Shinkman published this problem in the St. Louis Globe Democrat in 1887. White is to mate in 8 moves:
It’s easier than it sounds — with the right approach, all Black’s moves are forced:
1. O-O-O Kxa7 2. Rd8 Kxa6 3. Rd7 Kxa5 4. Rd6 Kxa4 5. Rd5 Kxa3 6. Rd4 Kxa2 7. Rd3 Ka1 8. Ra3#
1. g4 e5 2. Nh3 Ba3 3. bxa3 h5 4. Bb2 hxg4 5. Bc3 Rh4 6. Bd4 exd4 7. Nc3 dxc3 8. dxc3 g3 9. Qd3 Rb4 10. Nf4 g5 11. h4 f5 12. h5 d5 13. h6 Bd7 14. h7 g2 15. h8B g1R!! 16. Bd4 Ba4 17. Rh4 Rg3 18. Bg2 gxf4 19. Be3 fxe3 20. Be4 fxe4 21. fxe3 exd3 22. exd3 c5 23. Rc4 dxc4 24. dxc4 b5!! 25. cxb4 Qa5 26. cxb5 Na6 27. bxa5 O-O-O!! 28. bxa6 Rd4 29. exd4 Rb3 30. cxb3 Ne7 31. bxa4 Nd5 32. dxc5 Nb6 33. cxb6 Kb8 34. bxa7+ Ka8
From Martin Gardner: Each of the two equal sides of an isosceles triangle is one unit long. How long must the third side be to maximize the triangle’s area? There’s an intuitive solution that doesn’t require calculus.
On June 30, 1999, the body of 41-year-old Ricky McCormick was discovered near a cornfield in West Alton, Missouri. He’d last been seen alive five days earlier; now he was 15 miles from home though he owned no car. In his pockets were two handwritten notes (click to enlarge):
In the ensuing 18 years both the FBI’s Cryptanalysis and Racketeering Records Unit and the American Cryptogram Association have failed to find any meaning in these messages. In 2011 the FBI appealed to the public for their insights: If you have any you can contact them via this page.
“We are really good at what we do,” said CRRU chief Dan Olson, “but we could use some help with this one.”
Here are six new lateral thinking puzzles to test your wits and stump your friends — play along with us as we try to untangle some perplexing situations using yes-or-no questions.
A problem from the British Columbia Colleges Senior High School Mathematics Contest, 2000:
Not all of the nine members on the student council are on speaking terms. This table shows their relationships — 1 means two members are speaking to each other, and 0 means they’re not:
A B C D E F G H I A - 0 0 1 0 0 1 0 0 B 0 - 1 1 1 1 1 1 1 C 0 1 - 0 0 0 1 1 0 D 1 1 0 - 1 0 1 0 1 E 0 1 0 1 - 0 1 0 0 F 0 1 0 0 0 - 0 0 1 G 1 1 1 1 1 0 - 0 0 H 0 1 1 0 0 0 0 - 0 I 0 1 0 1 0 1 0 0 -
Recently councilor A started a rumor, and it was heard by each councilor once and only once. Each councilor heard it from, and passed it to, another councilor with whom she was on speaking terms. If we count councilor A as zero, then councilor E was the eighth and last councilor to hear the rumor. Who was the fourth?