## Kenken

For fans of Killer Sudoku, Samurai Sudoku and other variants
Those of you who are struggling with subtraction, multiplication, division or even addition, might find this "Square Wisdom" calculator useful.
djape

Posts: 34
Joined: 27 September 2005

KENKEN 1

Those of you who have done my KiMos will know I like to give minimal information, So in this kenken I have not bothered to tell you the operators - so as they say Enjoy!

HATMAN

Posts: 276
Joined: 25 February 2006
Location: Saudi Arabia

I've posted a much harder KenKen on DJApe's site:

http://www.djape.net/sudoku/forum/viewtopic.php?p=9097#p9097
HATMAN

Posts: 276
Joined: 25 February 2006
Location: Saudi Arabia

Triple click to see the solution I wrote:123456
234561
345612
456123
561234
612345

BTW the 6/4 (sum) cage is not needed.
udosuk

Posts: 2698
Joined: 17 July 2005

Done a few Times kenkens recently and I note the algebraic operator is not normally needed.

If you notice any interesting ones where it is needed - please post.
HATMAN

Posts: 276
Joined: 25 February 2006
Location: Saudi Arabia

Two points:

If anyone comes accross a KenKen that is not unique without the operators please post the fact.

I've done a couple of days this week and in both cases found the difficult easier to solve without operators than the simpler one. In fact on todays medium I'm stuck, and will start again in the morning sans C2H5OH.
HATMAN

Posts: 276
Joined: 25 February 2006
Location: Saudi Arabia

HATMAN

I find the C2H5OH approach often to be my most powerful technique, in many forms of puzzle. I recommend it, although there are some experts who equate this to T&E

Bigtone53

Posts: 413
Joined: 19 September 2005

HATMAN wrote:If anyone comes accross a KenKen that is not unique without the operators please post the fact.

JC's very first kenken seems pretty impossible without operators (the only operators I can determine "on the spot" are 1- (two), 18x, 20x, 11+ plus 9+ at the bottom row, all other cages are quite ambiguous). Haven't work out alternative solutions yet though.

ETA:

Just confirmed that JC's very first kenken has a unique solution even without operators, so the conjecture is still alive. To solve JC's puzzle without operators I had to apply some T&E though. Don't know if it's possible to solve it using pure logic.

I guess one might need to design a puzzle using many ambiguous cages to disprove the conjecture (that a unique kenken with operators must also be unique when all operators are removed).

Also solved Times 111-114 without operators (and no T&E). 111 is by far the most tricky but doesn't take any wings/fishes.
udosuk

Posts: 2698
Joined: 17 July 2005

RIP the operator-uniqueness conjecture:

Unique with operators, 5 solutions without.

But most interesting kenken puzzles I encountered had a unique solution either way. And the easier the operator-version is the harder it gets when the operators are taken off. Don't know if it's a general rule though.
udosuk

Posts: 2698
Joined: 17 July 2005

Very nicely constructed - I did not think the conjecture had much hope!
HATMAN

Posts: 276
Joined: 25 February 2006
Location: Saudi Arabia

Note I've solved the no-operator versions of JC's 3 kenken puzzles using "pure logic". I've written walkthroughs for the 1st & 3rd, and is currently writing one for the 2nd. I recommend every keen kenken players to try them. The satisfaction of solving them is enormous.

You should also solve his 2nd one with operators, without peeking at my hint first.
udosuk

Posts: 2698
Joined: 17 July 2005

Will do
HATMAN

Posts: 276
Joined: 25 February 2006
Location: Saudi Arabia

After solving JC's 3 puzzles without operators, I tried the 2 puzzles cited on the previous page (one from the Times, one by Pyrrhon) and both of them figure to be very challenging also.

In particular, the Times one with just 2 ambiguous cages is surprisingly tricky to crack (considering I've already found the way to solve the operator version). However, if you do it the right way it only takes a few lines of work to determine the operators of the 2 cages. It's almost like a nice brainteaser, and I highly recommend all serious kenken/killer players to try this challenge.

ETA:
Triple click to read the hint I wrote:139314069504000000

udosuk

Posts: 2698
Joined: 17 July 2005

udosuk wrote:Answers to be revealed later.

Okay, here is the complete walkthrough for Times Kenken #16 (April 10), no-op version.

Triple click to read the walkthrough I wrote:9/4 @ r1c1={{12}{15|24}|{13}{14|23}}+|[1313|3131]x
6/4 @ r5c4=[1212|2121]+|{{12}{13}}x

All other cages must be {....}x|{.....}x

720 = 2^4 * 3^2 * 5
400 = 2^4 _____ * 5^2
144 = 2^4 * 3^2
432 = 2^4 * 3^3
300 = 2^2 * 3^1 * 5^2
360 = 2^3 * 3^2 * 5

Tot. = 2^21 * 3^10 * 5^6
Grid = 2^24 * 3^12 * 5^6 = 139314069504000000

139314069504000000-rule:
Product of 9/4 @ r1c1 & 6/4 @ r5c4=2*2*2*3*3=72
=> 9/4 @ r1c1 can't be {1125|1224}+
(can't have a factor of 5 or 16)

Product of 6/4 @ r5c4={1122}+|{1123}x is 4 or 6
=> Product of 9/4 @ r1c1=18 or 12
=> 9/4 @ r1c1 can't be {1133}x
=> 9/4 @ r1c1={{13}{14|23}}+

Now 400/4 @ r1c6=[4455|5544]x

r12c6={45} (NP @ c6)
r23c5={45} (NP @ c5)
r2c56={45} (NP @ r2)
=> r2c23 can't be {14}

720-rule on r1 & 720/5 @ r1c3:
Product of r1c126 = Product of r23c4
r2c4 from {1236}
=> Product of r23c4 can't be 4*5=20
=> Product of r1c126 can't be 20
=> r1c126 can't be [{14}5]

r1c6 from {45}
=> r1c12 can't be {14}

Therefore 9/4 @ r1c1 can't be {{13}{14}}+
=> 9/4 @ r1c1={{13}{23}}+ has a product of 18
=> Product of 6/4 @ r5c4=4=[1212|2121]+

r5c45={12} (NP @ r5)
r6c56={12} (NP @ r6)
r56c5={12} (NP @ c5)

r1c12={13|23} (3 @ r1 locked)
r2c23={13|23} (3 @ r2 locked)
=> r14c5=[63]

720-rule on c6 & 432/5 @ r3c6:
Product of r123456c6 = 720
Product of r345c6 & r4c45 = 432
=> 720/Product of r126c6 = 432/Product of r4c45
=> Product of r4c45 = Product of r126c6 * 432/720
=> Product of r4c45 = 4*5*r6c6*3/5 = r6c6*12

r6c6=1 or 2
=> Product of r4c45 = 12 or 24

r4c5=3
=> Product of r4c45 = 12 = [43]
=> r6c6=1
=> 6/4 @ r5c4=[2121]+

720-rule on c4 & 720/5 @ r1c3:
Product of r456c4 = Product of r1c35
r456c4 = [423|425|426] has a product of 24|40|48
Product of r1c35 can't be 40|48, must be 24
=> r456c4 = [423]

r1c5=6
=> Product of r1c35 = 24 = [46]
=> r1c6=5
=> 400/4 @ r1c6=[5544]x
=> 720/5 @ r1c3=[41665]x

r1c12={23}
=> 9/4 @ r1c1=[{23}{13}]+
=> r2c23={13} (NP @ r2)
=> r2c1=2
=> r1c12=[32]

300/4 @ r4c2 from {13456}={3455}x
=> r4c2=5

r56c1 from {45}
=> r5c2=3

Completion:

9/4 @ r1c1=[3213]+
144/5 @ r2c1=[21626]x
432/5 @ r3c6=[34326]x
360/5 @ r4c3=[15463]x
300/4 @ r4c2=[5435]x

Only a reasonable Killer Sudoku player can appreciate the techniques I applied in this solving route.
udosuk

Posts: 2698
Joined: 17 July 2005

Recently those Times KenKens are being rather boring (and easy) in my opinion, so I have started trying to solve them for speed.

These are my best times:
Easy: 0:55 [edit: 0:54]
Medium: 1:38 [edit: 1:27 [edit: 1:20]]
Difficult: 2:03 [edit: 1:35 [edit: 1:29]]

Anyone else taking up the speed challenge?
Last edited by 999_Springs on Tue Sep 02, 2008 9:55 am, edited 3 times in total.
999_Springs

Posts: 585
Joined: 27 January 2007
Location: In the toilet, flushing down springs, one by one.

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