45 posts
• Page **2** of **3** • 1, **2**, 3

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

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

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

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

- HATMAN
**Posts:**219**Joined:**25 February 2006

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.

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:**219**Joined:**25 February 2006

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

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.

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

- udosuk
**Posts:**2698**Joined:**17 July 2005

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.

Answers to be revealed later.

ETA:

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.

Answers to be revealed later.

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?

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.

Once upon a time I was a teenager who was active on here 2007-2011

ocean and eleven should have paired up to make a sudoku-solving duo called Ocean's Eleven

- 999_Springs
**Posts:**473**Joined:**27 January 2007**Location:**In the toilet, flushing down springs, one by one.

45 posts
• Page **2** of **3** • 1, **2**, 3