Black and White

kling chess problem

By Josef Kling. White to mate in two moves.

	Select Click for Answer
1. f8=B! Kxf6 2. Rf4# From The Chess Euclid, 1849.

December 17, 2017December 16, 2017 | Puzzles

Two Puzzles

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?

December 16, 2017December 15, 2017 | Puzzles

‘Tis the Season

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?

	Select Click for Answer
L.A. Ringenberg of Eastern Illinois State College showed that there are two solutions: A table of squares shows that ALL must be 100, 144, 400, or 900. If it’s either of the first two, then XMAS has to be 2916 or 9216, and TO can’t be a square. If ALL is 400, then XMAS has to be 1849, 3249, 6241, or 8649. Taking these four possibilities in turn: If ALL = 400 and XMAS = 1849, then M = 8, MERRY = 81225, and E and X now both represent 1, so that can’t be right. If ALL = 400 and XMAS = 3249, then M = 2, MERRY = 27556, and TO = 81, so produces a valid solution: 27556 3249 81 400. If ALL = 400 and XMAS = 6241, then TO isn’t a square. If ALL = 400 and XMAS = 8649, then M = 6 and L = 0 and there’s no solution for MERRY. Finally, going back to the possibility that ALL = 900, in this case XMAS = 1296 or 7396: If ALL = 900 and XMAS = 1296, then TO isn’t a square. If ALL = 900 and XMAS = 7396, then M = 3, MERRY = 34225, and TO = 81, giving a second solution: 34225 7396 81 900. (“Problems for Solution,” American Mathematical Monthly 63:10 [December 1956], 723.) 12/08/2017 UPDATE: A variant with a unique solution, from reader Jean-Noël Monette: MERRY XMAS TO YA ALL 27556 3249 81 64 400

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December 7, 2017December 8, 2017 | Puzzles

The Feynman Ciphers

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:

#2 (“Harder”)

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.

December 5, 2017 | Puzzles

Decisions

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?

	Select Click for Answer
Let a, b, and c be the respective speeds of Able, Baker, and Charlie, and let x and y be the times required for Able to catch the first and second person. If Able chases Baker first, then $\displaystyle x=\frac{bT}{a-b} \ \mathrm{and} \ y=\frac{aT(b+c)}{\left ( a-c \right )\left ( a-b \right )}$ If Able chases Charlie first, then $\displaystyle x=\frac{cT}{a-c} \ \mathrm{and} \ y=\frac{aT(b+c)}{\left ( a-c \right )\left ( a-b \right )}$ “Since both y’s are equal, the x’s show that the least time will elapse when Able chases Charlie first.” (From Pi Mu Epsilon Journal 6:5 [Fall 1976], 317.)

Click for Answer

November 30, 2017November 29, 2017 | Puzzles

Black and White

William Shinkman published this problem in the St. Louis Globe Democrat in 1887. White is to mate in 8 moves:

shinkman chess problem - 1

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#

shinkman chess problem - 2

Remarkably, though the problem position looks contrived, it’s reachable in a legal game (discovered by Bader Al-Hajiri):

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

(Thanks, Florian.)

November 26, 2017November 25, 2017 | Puzzles

Quickie

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.

	Select Click for Answer
Rotate the triangle until one of the two equal sides is horizontal, as above. Then rotate the other side through 180 degrees. The triangle’s area is half its base times its height, and this value is maximized when its height in this orientation is maximized, that is, when the rotating side is vertical, producing an isosceles right triangle whose hypotenuse is $\sqrt{2}$ . (From Wheels, Life and Other Mathematical Amusements, 1983.)

Click for Answer

November 22, 2017November 24, 2017 | Puzzles · Science & Math

Unsolved

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.”

November 22, 2017November 22, 2017 | Crime · Puzzles

Podcast Episode 178: Lateral Thinking Puzzles

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.

See full show notes …

November 20, 2017November 20, 2017 | Podcast · Puzzles

The Grapevine

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?

	Select Click for Answer
We know that A is at the start of the sequence, and E is at the end. Of the others, F and H are each speaking to only two other councilors. So F must pass the rumor between B and I (in one direction or the other, BFI or IFB), and H must pass it between B and C (BHC or CHB). Putting these fragments together, we must have either CHBFI or IFBHC. Also, this larger fragment can’t connect directly to either of the endpoints A or E, because neither C nor I is speaking to either of them. That means that the remaining councilors, D and G, must go on either end of our fragment before we can attach the endpoints. So the extended fragment is either GCHBFID or DIFBHCG, and with the endpoints A and E added we get the full sequence AGCHBFIDE or ADIFBHCGE. In either case, if A is counted as zero then the fourth person is B. (Crux Mathematicorum 27:3 [April 2001], 194.):

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November 16, 2017 | Puzzles