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Arithmetic Killer
http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=13&t=281		 Page 1 of 1
Author:  Jean-Christophe [ Sat Sep 27, 2008 1:01 pm ]
Post subject:  Arithmetic Killer
Arithmetic Killer, Sep 27 2008

Here is an Arithmetic Killer using the operations: addition (+), subtraction (-), multiplication (x) or division (รท).
Unlike regular killer, digits may possibly repeat in cages.
Unlike regular Kenken, there are 3x3 blocks which contain exactly one of each digit.

Image
Code for JSudoku:
SquareWisdomV1=13+0=15*2=20=16*5=21=2=16+9+9+4+4=3*5+7-8=16+18+9=28*21/14+7+7=15+18=15+28*21=25=13+32=16*26=15+36=8+31+31+31+32+34+34=15+36+38+38+31=13=13+51=16+45=15+55=6+50+50=22+53+53=4+55=15-57=24*67+60+60+60-63+55+65+65*67=1-77=1-79

Remember to add rectangular blocks

Solution:
673584129
125936487
498172563
369418275
254697318
781325946
847253691
932861754
516749832

Rating: not too easy, but not too hard either ;)
Enjoy :cheesey:
Author:  udosuk [ Mon Sep 29, 2008 1:44 am ]
Post subject:  Re: Arithmetic Killer
Nice puzzle! :pirate:

For past kenken puzzles I'd probably explore the possibilities of solving without operators, but most probably not for this one (too little time available these days). But even with operators this one is very enjoyable to solve (and nope I didn't use software help on this).

One thing though - the blue outlines of nonets gets quite confusing with the cage lines. :rambo: When I look at the pic the first time I made a few misjudgements (such as mistaking a cage having 2 cells when it in fact has 3 with one "crossing" a blue line). In future perhaps it'd be better if you give n2468 a shading of grey instead. Or use the old killer format where you have dotted lines around the cages. :geek:
Author:  udosuk [ Wed Oct 01, 2008 11:47 am ]
Post subject:  Re: Arithmetic Killer
Well, I've written a short but complete walkthrough for this. :ugeek:

My walkthrough:
15x(2) @ r1c3={35} (NP @ r1)
16x(3) @ r1c6={128|[242|414]}
28x(3) @ r3c4={147|[722]}
=> r3c45 from {1247} can't be {12}
=> 20+(3) @ r1c5 can't be {479}, can't have 4
=> 20+(3) @ r1c5={389|569|578} has 3|5
=> r1c4 & 20+(3) form KNP {35} @ n2
3/(2) @ r2c6 from {1246789}={26} (NP @ c6,n2)
=> 20+(3) @ r1c5={389|578} (8 @ n2 locked)
=> r1c6+r3c45={147} (NT @ n2)
=> 20+(3) @ r1c5={389} (NT @ n2)
=> 15x(3) @ r1c3=[35]

Outies @ n1: r4c1=3
=> 15x(2) @ r3c9=[35]
Innies @ n3: r12c79=21
But 2-(2) @ r12c9 must be even
=> r12c7 must be odd, can't be {28}
=> r1c6 can't be 1, must be 4
=> 16x(3) @ r1c6=[414]
=> 13+(2) @ r1c1 from {26789}={67} (NP @ r1,n1)
=> r2c79=21-1-4=16=[97]
=> r1c58=[82], r3c45={17}
=> 28x(3) @ r3c4: r4c4=28/1/7=4

Now r2c4 from {39}, r3c4 from {17}
=> 6-(2) @ r7c4 can't be {17|39}, must be {28} (NP @ c4,n8)
=> 24x(3) @ r8c5 from {1345679} must be {146}
=> r8c6=1, r89c5={46} (NP @ c5,n8)
=> min r7c56=3+5=8
13+(3) @ r6c6: max r6c6=13-8=5
Innie-outies @ n6: r4c6=r6c9+2
=> r4c6+r6c9=[86]
=> 13+(3) @ r4c6: r45c7=13-8=5 from {23789}=[23]
=> 13+(2) @ r6c7 from {14789}=[94]
Innies @ n5: r6c46=8=[35]
=> r29c4=[97], 8+(3) @ r5c3: r56c3=8-3=5=[41]
13+(3) @ r6c6: r7c56=13-5=8 from {359}=[53]
=> 1-(2) @ r9c6=[98]
Innie-outies @ n9: 1-/2 @ r9c8=r6c9-1=6-1=5=[32]
15+(3) @ r8c3: r89c3=15-7=8 from {256789}=[26]
Innies @ n4: 15+(2) @ r6c1=[78]
Innies @ n7: 4-(2) @ r8c1=14=[95]

All naked singles from here.
Author:  HATMAN [ Wed Oct 01, 2008 8:43 pm ]
Post subject:  Re: Arithmetic Killer
JC

Nice puzzle - unlike Matt I did not find it straightforward and admit lazily using "big clue" a couple of times.

I'm trying to create a Pro-kNight puzzle i.e. a set of cells must each hav a cell a knights move away with the same value. Clearly it cannot apply to the corners but actually the set is much smaller. I'm currently down to applying it to the four window shapes (but not with the windoku constraint of course) and keep almost getting there.

Maurice
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