Public Access

# Solution to The Hound of the Vast-Cur Villes

by Katie Hamill and Kevin Wald
Art by Joe Cabrera

For each of the 9 parts, we deduce the sizes of the parties, and use D'Hondt allocation to allocate lanterns to them. (D'Hondt is older Dutch for "The Hound")

Each lantern amount is a single digit; we thus get 9 5-digit sequences.

## Afghan Hound

Under [1] and [2], the possibilities for lowest and highest panels are: 20 and 90 or 22 and 99. If it were 22 and 99, that plus 55 [6] would equal 176; the remaining 57 panels must consist of a perfect square [4] and a prime number [7], which is impossible without violating condition [1]. Therefore, the lowest number of panels is 20, and the highest is 90.

One knitter made 55 panels [6]; the remaining two knitters made a perfect square [4] and a prime number [7]. These have to be 25 and 43 to bring the total to 233 without violating condition [1].

Ember didn't make the striped or argyle panels [7] or the polka-dot or checkerboard panels [3], so they made the harlequin, and there are 20 of them, as deduced above.

By condition [5], then, Casey made 25, Drew made 43, and Alex made 55. Alex therefore made the polka-dot panels, and Drew made the checkerboard ones [3 and 6].

By condition [4], Casey did not make the striped panels; they made the argyle panels, and Blake made the striped ones.

Knitter:ABCDE
Pattern: polka-dot stripe argyle checkerboard harlequin
Number of panels knitted: 55 90 25 43 20
 Allocation of 19 lanterns: 5 8 2 3 1

## Dachshund

Determine how many bags of buns they have to buy, and how many buns are left over after each week's sausage fest.

Each week, they get p packs of s sausages and b bags of 16 buns, so they have 16b - ps buns remaining.

(You can also think of this as 16 minus the remainder of ps/16.)

Andouille: 3 packs of 19 sausages = 57 sausages. 57/16 = 3.5625, so they have to buy 4 bags of buns. 4*16 = 64 total buns. 64 - 57 = 7 leftover buns.

Bratwurst: 7*11 = 77. 77/16 = 4.8125. (5*16) - 77 = 3.

Chorizo: 7*9 = 63. 63/16 = 3.9375. (4*16) - 63 = 1.

Italian sausage: 6*10 = 60. 60/16 = 3.75. (4*16) - 60 = 4.

Kielbasa: 5*13 = 65. 65/16 = 4.0625. (5*16) - 65 = 15.

Type of Sausage:AndBratChorItalianKielbasa
Leftover buns: 7 3 1 4 15
 Allocation of 12 lanterns: 3 1 0 1 7

## Greyhound

All the shades of gray shown are achromatic (i.e., R=G=B); the number to input is that common value:

 RGB value: 85 54 240 160 130
 Allocation of 19 lanterns: 2 1 7 5 4

## Irish Wolfhound

Take the mean radius of the largest moon of each planet listed, multiply by the factor shown, and round:

Triton 1 1353.4 1353.4 1353
Titania 4 788.4 3153.6 3154
Titan 2 2574.73 5149.46 5149
Ganymede 1 2634.1 2634.1 2634
Phobos 80 11.267 901.36 901
 Allocation of 16 lanterns: 1 4 7 3 1

## Otter Hound

All the statistics referred to are for the noted Giants slugger Mel Ott. (Source)

StatYearNumber
At Bats 1926 60
Runs Batted In 1934 135
Walks (Bases on Balls) 1942 109
Home Runs 1929 42
Runs 1941 89
 Allocation of 20 lanterns: 3 6 5 2 4

## Pharaoh Hound

Solving the cryptic gives the party sizes in the unclued shaded entries:

### Across

1. A[narchi]C + DC
2. A + CE + D
3. A[d]A + A[v]A
4. P[ursue] + R[outine] + I[nvestigation] + OR
5. hidden in bED I THink
6. LEEWARD INN anagram
7. &lit: W in (DO + SE[arching])

### Down

1. reversed WE + FA
2. CO + DER
3. A NUDE anagram
4. C + RAT
5. [c]ACTI
6. reversed I + YAM
7. A + GREET + O
8. AN(TEN)NA
9. O(PEN)ED
11. OIL SO anagram
12. reversed E + V + LED
13. E[njoy] + WWW
14. ER + SE
15. VEE[p] + R
16. reversed SEAN
 1A 2C 3D 4C 5A F O U R 6M 7A C E D 8A A A 9A 10O G T W E N T Y N 11S 12P R I 13O R 14E 15D I T H E E 16E L 17E 18V E 19N E A 20N E W I R E L A N D E T W O S E V E N O 21D O W S E 22R E S A W
 Allocation of 13 lanterns: 1 7 3 0 2

## Plott Hound

Plotting each of these on the interval [0, 1] yields a figure which becomes a digit when you flip the axes:

 (4t + ln (1 - t^100 + e^-4)) / 8 7 (1/2) * (1 - cos(4pi * t))^(1/4) 3 sqrt(|4t^3 - 6t^2 + 2t|) 3 25(1 - 98t/99 - |1 - 100t/99|) 7 cos^2 t - (cos 2t) / 2 1
 Allocation of 12 lanterns: 4 2 2 4 0

## Shikoku

Solving the puzzle as a Sudoku using all numbers and a Shikaku using only the highlighted numbers gives:

 1 8 5 6 3 2 4 7 9 7 9 3 1 4 8 2 6 5 2 6 4 7 5 9 1 3 8 4 2 6 5 8 7 3 9 1 9 7 8 3 1 6 5 2 4 5 3 1 9 2 4 6 8 7 6 1 7 2 9 5 8 4 3 8 5 9 4 6 3 7 1 2 3 4 2 8 7 1 9 5 6

The party sizes are given by the 5-digit numbers in the size-5 areas

 Allocation of 29 lanterns: 2 7 9 3 8

## Trigg Hound

Taking the tan of each number (in radians) gives (to the precision possible) an integer:

Value Tangent
1.53235 26
1.53082 25
1.54699 42
1.47113 10
1.54858 45
 Allocation of 25 lanterns: 4 4 8 1 8

## The Grand Finale

Each 5-digit sequence is the zip code for a US location ending in -ville. These start with the letters A through I, giving an ordering.

Allocating 99 lanterns gives them numbers of lanterns as shown:

ZIP Code-VilleLanternsLetter
17302Airville, PA6F
44818Bloomville, OH15O
36524Coffeeville, AL12L
31017Danville, GA11K
14731Ellicottville, NY5E
58231Fordville, ND20T
27938Gatesville, NC9I
42240Hopkinsville, KY14N
21754Ijamsville, MD7G

Translating the numbers into letters gives the answer, FOLKETING (the Danish Parliament, which uses D'Hondt allocation in part to allocate seats to parties).