
I’ll leave puzzle 37 open for now.
Puzzle 38:
What comes fourth in the sequence, taking the place of the question mark?


2004, yes. Stonehenge, no. More likely the Parthenon, or something else in Athens, Greece.


I mean, obviously. Samsters love the Olympics. I heard they’re going to host it in London in 2012.


Originally posted by LouWeed:
2004, yes. Stonehenge, no. More likely the Parthenon, or something else in Athens, Greece.
That is the correct answer!
Remember, puzzle 37 is still open.


We’re going to need a hint or at least a 1 number jumping point.


Puzzle 39:
Complete both these word puzzles for a point.
1) Each of the following sets of letters can be made into a real word by adding three letters to the beginning, and the same three letters in the same order to the end. For example, ANGLEM can have ENT added at the start and the end to become ENT + ANGLEM + ENT = ENTANGLEMENT.
ERTAINM
ISCOVE
ROW
IOXID
ROCOS
2) These words can all have a letter added and then be rearranged to make a new 5letter word. The letters added spell a 9letter word. What are the new words?
HERB
BANG
KEEN
SUIT
QUAD
WHEY
CELL
TOIL
WILD


1)
Entertainment
Rediscovered
Ingrowing
Antioxidant
Microcosmic
For the second 1, is the 9 letter word in order descending down the list or do I have to rearrange it too.


2)
BER T H
B E GAN
KNEE L
SUIT E
S QUAD
C HEWY
CELL O
P ILOT
WI E LD
TELESCOPE


MYGODITHINKIDIDIT
I don’t see any mistakes but there’s always a chance
I don’t know what happened to that one square so just pretend it’s normal.
Also thanks to LouWeed for providing the template


Originally posted by funiax:
MYGODITHINKIDIDIT
I don’t see any mistakes but there’s always a chance
In addition, once completed correctly, no row nor column will contain a duplicate number within a white square.
Column B has two 1s
Column D has two 5s
Row 4 has two 6s
Row 6 has two 5s
I’m unsure if this matters.


Hmmm. Maist might have a point. I’d assumed the columns & rows referred to were the rows & columns that didn’t have black squares, but if not the whole thing becomes a bit more solvable.


Originally posted by funiax:
MYGODITHINKIDIDIT
I don’t see any mistakes but there’s always a chance
Sorry to bash your hopes, funi, but…
Does not add up to 39.


Originally posted by MaistlinRajere:
Originally posted by funiax:
MYGODITHINKIDIDIT
I don’t see any mistakes but there’s always a chance
In addition, once completed correctly, no row nor column will contain a duplicate number within a white square.
Column B has two 1s
Column D has two 5s
Row 4 has two 6s
Row 6 has two 5s
I’m unsure if this matters.
If this is the case I give up on the puzzle.
Originally posted by MaistlinRajere:
Sorry to bash your hopes, funi, but…
Does not add up to 39.
Well shit. I blame LouWeeds spreadsheet. If it’s been doing that this whole time it’s been sabotage!


You could also give some insight on how this could be solved logically as opposed to just entering in random numbers…


I guess there’s a way to solve this via brute force more easily.
Write a program that finds the partitions of all the numbers, that consist of 8 terms where none of the terms exceed 7. Use another program to find similar partitions in adjacent cells, thereby cutting down the amount of combinations to about a thousandth of what it is now (theoretically, I think it is 7^33, which is stupidly high if you are stupidly bad at maths).
Point by Face, though. This puzzle is better suited for solving by the computer, not logical enough to be suited by human beings (by human beings, I mean ordinary human beings who have 24 hours in 1 day and dedicate 1/3rd to sleeping and eating) unless I am missing something obvious here. Brute forcing this without the help of a program will take approximately a week. Maybe a few days less.
EDIT; Ran a minor test.
Meeting the criteria…
Partitions of 26 = 46
Partitions of 35 = 94
Partitions of 37 = 90
Partitions of 31 = 81
Partitions of 36 = 94
Partitions of 39 = 81
Partitions of 38 = 88
Partitions of 26 = 46
Partitions of 24 = 31
Currently, number of possible combinations: 7730993719707444524137094407
Utilizing only the info given above: 30118149433278720
I was making an understatement above.


Originally posted by AdeebNafees:
30118149433278720
Thats what i came up with but i`m too lazy to post a spreadsheet.


Originally posted by racefan12:
You could also give some insight on how this could be solved logically as opposed to just entering in random numbers…
I’m gonna be honest, it’s pretty much trial and error. I’ll post the solution tomorrow.
Puzzle 40:
Complete this mixed sudoku – it works exactly the same as a normal sudoku but one overlaps the other.


I wrote a tcl script to solve it for me:
variable grid
set grid(1,1) 2
set grid(1,2) 0
set grid(1,3) 0
set grid(1,4) 0
set grid(1,5) 0
set grid(1,6) 0
set grid(1,7) 7
set grid(2,1) 0
set grid(2,2) 26
set grid(2,3) 0
set grid(2,4) 35
set grid(2,5) 0
set grid(2,6) 37
set grid(2,7) 0
set grid(3,1) 0
set grid(3,2) 0
set grid(3,3) 0
set grid(3,4) 0
set grid(3,5) 0
set grid(3,6) 0
set grid(3,7) 0
set grid(4,1) 0
set grid(4,2) 31
set grid(4,3) 0
set grid(4,4) 36
set grid(4,5) 0
set grid(4,6) 39
set grid(4,7) 0
set grid(5,1) 0
set grid(5,2) 0
set grid(5,3) 0
set grid(5,4) 0
set grid(5,5) 0
set grid(5,6) 0
set grid(5,7) 0
set grid(6,1) 0
set grid(6,2) 38
set grid(6,3) 0
set grid(6,4) 26
set grid(6,5) 0
set grid(6,6) 24
set grid(6,7) 0
set grid(6,7) 0
set grid(7,1) 7
set grid(7,2) 0
set grid(7,3) 0
set grid(7,4) 0
set grid(7,5) 0
set grid(7,6) 0
set grid(7,7) 3
# check if the number is already used in this column/row
proc check { x y new_value } {
variable grid
foreach location [concat [array names grid $x,*] [array names grid *,$y]] {
if { $grid($location) == $new_value } {
return 0
}
}
return 1
}
proc print_grid {} {
variable grid
puts "$grid(1,1) $grid(1,2) $grid(1,3) $grid(1,4) $grid(1,5) $grid(1,6) $grid(1,7)"
puts "$grid(2,1) $grid(2,2) $grid(2,3) $grid(2,4) $grid(2,5) $grid(2,6) $grid(2,7)"
puts "$grid(3,1) $grid(3,2) $grid(3,3) $grid(3,4) $grid(3,5) $grid(3,6) $grid(3,7)"
puts "$grid(4,1) $grid(4,2) $grid(4,3) $grid(4,4) $grid(4,5) $grid(4,6) $grid(4,7)"
puts "$grid(5,1) $grid(5,2) $grid(5,3) $grid(5,4) $grid(5,5) $grid(5,6) $grid(5,7)"
puts "$grid(6,1) $grid(6,2) $grid(6,3) $grid(6,4) $grid(6,5) $grid(6,6) $grid(6,7)"
puts "$grid(7,1) $grid(7,2) $grid(7,3) $grid(7,4) $grid(7,5) $grid(7,6) $grid(7,7)"
}
proc check_limits { square } {
}
proc solve { squares } {
variable grid
if { $squares == {} } { return 1 }
check_limits x
set this_square [lindex $squares 0]
set other_squares [lrange $squares 1 end]
set thisX [lindex $this_square 0]
set thisY [lindex $this_square 1]
# which squares can we change?
set sum 0
set unused_squares {}
for {set x [expr {$thisX  1}]} { $x <= $thisX + 1 } { incr x } {
for {set y [expr {$thisY  1}]} { $y <= $thisY + 1 } { incr y } {
if { $grid($x,$y) == 0 } {
lappend unused_squares [list $x $y]
} elseif { $x != $thisX  $y != $thisY } {
incr sum $grid($x,$y)
}
}
}
#puts $unused_squares
#puts $sum
#print_grid
set count 0
set index 0
set value 1
set end [expr {[llength $unused_squares]  1}]
while { 1 } {
#puts "count [incr count]"
#puts "index $index"
#puts "value $value"
#puts "sum $sum"
#print_grid
set location [lindex $unused_squares $index]
set x [lindex $location 0]
set y [lindex $location 1]
while { $value <= 7 && ![check $x $y $value] } {
incr value
}
if { $value > 7 } {
if { $index == 0 } {
break
}
set grid($x,$y) 0
incr index 1
set location [lindex $unused_squares $index]
set x [lindex $location 0]
set y [lindex $location 1]
set value $grid($x,$y)
incr sum $value
incr value
continue
}
set grid($x,$y) $value
incr sum $value
if { $index == $end } {
if { $sum == $grid($thisX,$thisY)} {
set result [solve $other_squares]
if { $result != 0 } {
return $result
}
}
incr sum $value
incr value
continue
}
incr index
set value 1
}
foreach square $unused_squares {
set x [lindex $square 0]
set y [lindex $square 1]
set grid($x,$y) 0
}
puts "unsolved $this_square for:"
print_grid
return 0
}
if { [solve { {2 2} {2 4} {2 6} {4 2} {4 4} {4 6} {6 2} {6 4} {6 6}}] } { puts "solved" }
print_grid
Must get back to PE some time.
EDIT: I swapped around the evaluation order of the input squares to see if it would run quicker (it takes about a minute), it turns out the puzzle doesn’t even have a single unique solution:
EDIT #2: Runs about twice as fast if i remove the clue that Samster gave us:
Not sure how many unique solutions there are, I’m guessing a lot.


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