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PRIZE PUZZLE
Puzzle 1. Homebrew encryption
[IMAGE: The bytes below. Please space out as shown, in groups of 8 bits, and rows of four bytes – except the last row which has two bytes. If room for an additional image, perhaps a dystopian scene in a US in which all emails are routinely monitored. By the way Puzzles 1 to 3 form a little story about the Schmidt family and the President.]
In the months after President Burp signed the Unamerican Communications executive order, banning any criticism of himself, homebrew encryption schemes mushroomed, as dissenting citizens struggled to prevent their emails being read by the government.
The following encrypted message was discovered sent from the computer of the Schmidt family of Boston. What does it mean?
10111011100100001000100010010001
11011111100010001001011010001011
10010111110111111011101110010000
10010001100111101001001110011011
11011111101111011000101010001101
1000111111011110
[PRINT UPSIDE DOWN]
HINT
Use a standard.
Puzzle 2.Eye in the sky
[DESIGN NOTE: A reconnaissance satellite or spy satellite, or its large camera
]
The President wanted a new spy satellite. What is the minimum height it would have to be above the Earth’s surface to have line of sight to one-third of the surface of the Earth?
[PRINT UPSIDE DOWN]
HINT
Triangles help.
Puzzle 3.Waiting for a knock on the door
[IMAGE IDEA: An educated and respectable American family looking anxious, perhaps expecting an unwelcome knock on the door. By the way is the same family as in Puzzle 1 if you want to create a story.]
Following evidence gathered from the Schmidt family’s emails and satellite images of their daily activity, they were informed that they would receive an investigative visit at their home, one day next week, Monday through Friday. ‘We shall not tell you which day, and we shall arrive without notice; but be very sure that we shall visit on a day that surprises you,’ the official stated.
Mr Schmidt tried to reassure his family: ‘They cannot be visiting on Friday, because if they have not visited by the end of Thursday, we would know that Friday is the day, so it would not surprise us,’ he said. ‘The same logic applies to Thursday. If they have not visited by the end of Wednesday, we would know, having already ruled out Friday, that Thursday is the day; but in that case we would not be surprised by their visiting on Thursday, so they cannot be visiting on Thursday. The same logic appliesto all the other days, working backwards. Therefore they cannot be visiting on any day next week.’
What, if anything, is wrong with his argument?
[PRINT UPSIDE DOWN]
HINT
Logic or psychology?.
True or False
In seventeenth century Boston it was illegal to celebrate Christmas.
ANSWERS
PRIZE PUZZLE
Puzzle 1. Homebrew encryption
The ciphertext consists in groups of 8 binary digits, or bytes, and the plaintext is presumably characters (English letters) or perhaps alphanumeric (letters and numbers). This suggests that the ASCII standard may have been employed. However, letters in ASCII run from 01000001 (capital A) up to 01111010 (lower case z), and the bytes in the ciphertext all strat with 1 so seem too high.
The crucial step is to see that the bits are flipped, that is, each 0 is replaced by 1, and each 1 by 0. Equivalently, an XOR (exclusive or) operation has been carried out with a key of 11111111. Both procedures result in:
01000100011011110111011101101110
00100000011101110110100101110100
01101000001000000100010001101111
01101110011000010110110001100100
00100000010000100111010101110010
0111000000100001
Looking up an ASCII table reveals the plaintext:
Down
SPACEwit
hSPACEDo
nald
SPACEBur
p!
Puzzle 2. Eye in the sky
The distance is twice the Earth’s radius, or about 12,7426371 km.
Represent the Earth as a circle, and project a line from its centre C through the circumference at point D and out to the satellite S. Draw anequilateral triangle with corners C, D and another point of the circumference, E. Drop a line from the apex of the triangle to be perpendicular to DC, bisecting meeting it at B. The length of CE and the length of CD each side of the triangle isare both the radius of the Earth, r.
Compare the triangle CSE to the triangle CEB. They are similar, therefore:
CS/CE = CE/CB
We know that CE is the radius of the Earth r and CB is half the radius, so the above simplifies to:
CS / r = r / (r/2)CB
From the diagram it can be seen that CS = DS + r, and that CB = r – BD, so the above may be written:
(DS + r)/r = r/ (r – BD)
Multiply across:
r^2 = (DS+ r)(r – BD)
Expand:
r^2 = DSr –DSBD – rBD + r^2
Rearrange, and the two terms r^2 fall away:
DS(r-BD) = rBD
DS = rBD/(r-BD) (Result 1)
Now, the question states that the surface area of the cap (ie the area seen by the satellite) is a third of the surface area of the whole sphere (ie the Earth).
Area of the cap = 1/3 Area of the sphere
Applying the standard formulae for the area of a cap* and for the area of a sphere:
2 Pi r BD = 4/3 (Pi r^2)
Rearrange:
BD = (4/3 (Pi r^2)) / (2 Pi r)
BD = 4r/6
BD = 2r/3 (Result 2)
Substitute Result (2) into Result (1):
DS = r(2r/3) / (r-(2r/3))
DS = r(2r/3) / (r/3)
DS = (2r/3) x 3
DS = 6r/3
DS = 2r
r^2 = CS (r/2)
where ^2 means squared
CS = 2r
Or, the distance between the centre of the Earth and the satellite is 2 times the radius. Therefore So the distance from the surface of the Earth to the satellite is twice theoneEarth’s radius.
Thanks to Bill Brading for his comments on an earlier version of this answer.
*See
Puzzle 3.Waiting for a knock on the door
Surprise is a psychological state and we ought to be suspicious of an argument that appears logical yet relies so heavily on predictions about psychology. However let us allow this: if Thursday evening is reached with no visit, and there must be a visit, then we accept Schmidt’s claim that a Friday visit would, in those circumstances, no longer be a surprise. (However, we do not accept that means the visit would not happen – see below).
Schmidt then claims that at the end of Wednesday(with no visit yet), two days – both Thursday and Friday – must be ruled out. This is less plausible. Friday was only ruled out in the different situation where Thursday ended with no visit. Although the two hypothetical situations are consistent, they are not the same. Schmidt’s rational reaction at the end of Wednesday (with no visit yet) is merely that the visit will occur either on Thursday or on Friday. As at the end of Wednesday the probabilities are already high (0.5 for Thursday and 0.5 for Friday), the surprise will not be as great as a day earlier, when the probabilities were 0.333 for Wednesday, 0.333 for Thursday and 0.333 for Friday. At the end of Monday the probabilities were 0.25 for Tuesday, 0.25 for Wednesday, 0.25 for Thursday and 0.25 for Friday, so the uncertainty, and hence the surprise, would be greater. Before the week starts the probabilities are just 0.20 for each day, and so the surprise would be greatest. Surprise is not an absolute, but has degrees.
And what if, after all, the visit occurred on Friday? Then perhaps the official lied about its being a surprise. Or perhaps the official merely meant that before the week started it would be a surprise, allowing that by the end of Thursday it would no longer be. Or if the argument is defined so that on Thursday evening, no visit having occurred, a Friday visit is both certain and a surprise, then the officials’ statements are found, in those circumstances, to be self-contradictory. Their self-contradictory nature would not have been exposed if the visit had been earlier in the week.
This puzzle is a variant on the ‘unexpected hanging paradox’. Opinions about it differ, some maintaining that the argument by Schmidt is fallacious, others holding that there is a deep philosophical paradox.
True or False
True. A law of 1659 punished the celebration of Christmas with a five shilling fine. The Puritans passed a similar law in England in 1647.
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Regulars,Columnist
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Thinking Cap
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Thinking Cap
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Teaser/Standfirst
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David Sandham
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Body copy
One reader who sends the correct answer to the prize puzzle to ith ‘MONTH Thinking Cap’ by CLOSING DATE will win copies of four books from Princeton University Press – ‘Across the Board: the Mathematics of Chessboard Problems’ by John J Watkins, ‘Why Cats Land on Their Feet, and 76 Other Physical Paradoxes and Puzzles’ by Mark Levi, ‘Magical Mathematics: The Mathematical Ideas That Animate Great Magic Tricks’ by Persi Diaconis and Ron Graham, and ‘Wizards, Aliens and Starships: Physics and Math in Fantasy and Science Fiction’ by Charles L Adler. The winner will appear in the May 2017 issue of E&T. For solutions, email: after the closing date. Full terms and conditions at bit.ly/eandt-competitions. MONTH issue winner was WINNER.
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Puzzle 1Homebrew encryption
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In the months after President Burp signed the Unamerican Communications executive order, banning any criticism of himself, homebrew encryption schemes mushroomed, as dissenting citizens struggled to prevent their emails being read by the government.
The following encrypted message was discovered sent from the computer of the Schmidt family of Boston. What does it mean?
10111011100100001000100010010001
11011111100010001001011010001011
10010111110111111011101110010000
10010001100111101001001110011011
11011111101111011000101010001101
1000111111011110
[PRINT UPSIDE DOWN]
HINT
Use a standard.
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Puzzle 2Eye in the sky
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The President wanted a new spy satellite. What is the minimum height it would have to be above the Earth’s surface to have line of sight to one-third of the surface of the Earth?
[PRINT UPSIDE DOWN]
HINT
Triangles help.
@@@BoxoutHeadline
Puzzle 3Waiting for a knock on the door
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Following evidence gathered from the Schmidt family’s emails and satellite images of their daily activity, they were informed that they would receive an investigative visit at their home, one day next week, Monday through Friday. ‘We shall not tell you which day, and we shall arrive without notice; but be very sure that we shall visit on a day that surprises you,’ the official stated.
Mr Schmidt tried to reassure his family: ‘They cannot be visiting on Friday, because if they have not visited by the end of Thursday, we would know that Friday is the day, so it would not surprise us,’ he said. ‘The same logic applies to Thursday. If they have not visited by the end of Wednesday, we would know, having already ruled out Friday, that Thursday is the day; but in that case we would not be surprised by their visiting on Thursday, so they cannot be visiting on Thursday. The same logic applies to all the other days, working backwards. Therefore they cannot be visiting on any day next week.’
What, if anything, is wrong with his argument?
[PRINT UPSIDE DOWN]
HINT
Logic or psychology?.
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Puzzle 4True or False
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In seventeenth century Boston it was illegal to celebrate Christmas.
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